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\begin{document}

\section{Totally characteristic operators on the half
space}\label{totally-characteristic-operators-on-the-half-space}

\emph{Written by Claude Opus 5.5 (Anthropic), September 2026. Self-checked by the writing AI. Public domain (CC0).}

\emph{Edited and supplemented by Codex, September 2026. The additions
and editorial corrections are also public domain (CC0).}

This lesson builds the local calculus of \emph{totally characteristic}
pseudodifferential operators on the closed half space
\(\overline{\mathbb R}{}^n_+=\{x_n\geq0\}\). The differential operators
of this kind are generated by the vector fields tangent to the boundary,
which are the combinations of \(\partial_1,\ldots,\partial_{n-1}\) and
\(x_n\partial_n\) with smooth coefficients. They respect the boundary:
the boundary values of \(Pu\) depend only on the boundary values of
\(u\), and the normal derivatives of \(Pu\) of order \(k\) at the
boundary depend only on those of \(u\) of order at most \(k\). The
pseudodifferential operators of the class come from symbols \(a(x,\xi)\)
in which the normal frequency \(\xi_n\) is replaced by \(x_n\xi_n\)
before quantization. A condition on the Fourier transform of the symbol
in \(\xi_n\), called lacunarity, makes the output in the open half space
depend only on the input there.

We prove that these operators act on functions and on distributions on
the half space, and we compute their commutators and the boundary jets
of their outputs. We describe the kernels of the operators of order
\(-\infty\) near the corner \(x_n=y_n=0\) by blowing up the corner, and
we show that these kernels are conormal there. The class is closed under
adjoints and compositions, with asymptotic formulas for the symbols.
Operators of order 0 are bounded on \(L^2\), on Sobolev spaces of every
real order and on dyadic Besov spaces, and operators of every order
preserve the distributions that are conormal to the boundary. In one
respect the class differs from the ordinary calculus: operators of order
\(-\infty\) need not improve regularity at all, because their kernels
are singular at the corner.

These operators matter because they are the pseudodifferential
counterpart of the differential operators tangent to the boundary. Since
they respect the boundary, they can be combined with operators on the
boundary, such as taking boundary values. On manifolds with boundary
they form the b-calculus, which is used for instance in index theory.

The lesson assumes the symbol calculus on \(\mathbb R^n\) from the
lesson \href{euclidean-symbol-calculus.pdf}{From symbol estimates to
operators on every Sobolev scale}; conormal distributions and smooth
extension across a boundary from
\href{conormal-transmission.pdf}{Singularities along a submanifold and
smooth boundary passage}; and local Besov spaces from
\href{geometric-microlocal-calculus.pdf}{Detecting regularity without
choosing coordinates}. It also uses tempered distributions, the Fourier
transform and two facts from functional analysis. Section1 states the
precise facts used from the linked lessons and identifies their proof
sections.

The freely available author versions {[}Melrose{]} and {[}Loya{]}
provide further reading. The mathematical arguments used here are proved
in this lesson and its linked prerequisites.

\subsection{1. Conventions and
background}\label{conventions-and-background}

\subsubsection{Notation}\label{notation}

Throughout \(n\geq1\), \(x=(x',x_n)\in\mathbb R^{n-1}\times\mathbb R\),
and \[
\mathbb R^n_\pm=\{x:\pm x_n>0\},\qquad \overline{\mathbb R}{}^n_+=\{x:x_n\geq0\}.
\] We write \(D=-i\partial\),
\(\widehat u(\xi)=\int e^{-ix\cdot\xi}u(x)\,dx\) (inverse factor
\((2\pi)^{-n}\)), \(\langle\xi\rangle=(1+|\xi|^2)^{1/2}\), and
\((u,v)=\int u\overline v\), linear in the first argument. For a
function of \((x,\xi)\) we write
\(a^{(\alpha)}_{(\beta)}=\partial_\xi^\alpha\partial_x^\beta a\).

\(C^\infty(\overline{\mathbb R}{}^n_+)\) denotes the functions on
\(\overline{\mathbb R}{}^n_+\) that are smooth in \(\mathbb R^n_+\) and
whose derivatives all extend continuously to
\(\overline{\mathbb R}{}^n_+\). Every such function is the restriction
of a smooth function on \(\mathbb R^n\); this is the extension fact in
the list below. \(C^\infty_b\) means smooth with all derivatives
bounded.

\textbf{Sobolev and Besov spaces.} \(H_{(s)}\) is the space of tempered
distributions \(u\) whose Fourier transform is locally square integrable
and for which the norm
\(\|u\|_{(s)}^2=(2\pi)^{-n}\int\langle\xi\rangle^{2s}|\widehat u|^2d\xi\)
is finite. With the sharp annuli \(A_0=\{|\xi|<1\}\),
\(A_j=\{2^{j-1}\leq|\xi|<2^j\}\) and the Fourier projections \(\Pi_j\)
onto them, the dyadic Besov norm is \[
\|u\|_{B^s_{2,p}}=\big\|\big(2^{js}\|\Pi_ju\|_{L^2}\big)_{j\geq0}\big\|_{\ell^p},\qquad 1\leq p\leq\infty .
\tag{1.1}
\] \(B^s_{2,p}\) is the space of tempered distributions with locally
square-integrable Fourier transform for which this norm is finite. Since
\(\langle\xi\rangle\) is comparable to \(2^j\) on \(A_j\),
\(B^s_{2,2}=H_{(s)}\) with equivalent norms. A distribution \(u\) on an
open set \(\Omega\subset\mathbb R^n\) lies in the local space
\(B^s_{2,p,\mathrm{loc}}(\Omega)\) if \(\chi u\), extended by zero, lies
in \(B^s_{2,p}\) for every \(\chi\in C_0^\infty(\Omega)\).

\textbf{Values of symbols.} All symbols may take values in
\(L(\mathbb C^p,\mathbb C^q)\) for fixed finite \(p,q\). Then
\(|\cdot|\) is the operator norm, products keep their order, and complex
conjugation of a symbol is replaced by the conjugate transpose \(a^*\).
Every statement below holds in this generality with the same proof,
except the square-root step in the proof of Theorem 10.1, where we say
what changes. The reader may keep \(p=q=1\) in mind.

We also use Peetre's inequality
\((1+|\xi+\zeta|)^s\leq(1+|\xi|)^s(1+|\zeta|)^{|s|}\) for real \(s\),
which follows from \(1+|\xi|\leq(1+|\xi+\zeta|)(1+|\zeta|)\).

\subsubsection{Proofs used from earlier
lessons}\label{proofs-used-from-earlier-lessons}

Fourier inversion and Plancherel on \(\mathcal S\), \(\mathcal S'\) and
\(L^2\) are proved in \href{prerequisite-bridges.pdf\#section*.2}{the
Fourier lesson, Sections1--2 and7--8}. The Schwartz kernel theorem for
continuous linear maps
\(\mathcal S(\mathbb R^n)\to\mathcal S'(\mathbb R^n)\) is proved in
\href{weyl-metric-products.pdf\#section*.27}{the Weyl-product
lesson, Sections4.1--4.5}. Dominated convergence and Fubini's theorem
are proved in
\href{banach-foundation-bridges.pdf\#section*.118}{the measure
chapter, Sections16.1--16.5}, and Taylor's formula in
\href{metric-foundation-bridges.pdf\#section*.106}{the calculus
chapter, Section13.6}. The following statements specify the other exact
interfaces used below.

\emph{Symbols and operators on \(\mathbb R^n\).} These facts are proved
in the lesson \href{euclidean-symbol-calculus.pdf}{From symbol estimates
to operators on every Sobolev scale}.

\textbf{Symbol classes.} For \(m\in\mathbb R\),
\(S^m(\mathbb R^N\times\mathbb R^k)\) is the space of smooth functions
\(a(x,\xi)\) with
\(|\partial_\xi^\alpha\partial_x^\beta a(x,\xi)|\leq C_{\alpha\beta}(1+|\xi|)^{m-|\alpha|}\)
for all \(x\) and \(\xi\). The best constants are seminorms, and they
make \(S^m\) a Fréchet space. We put \(S^{-\infty}=\bigcap_mS^m\). If
\(a_j\in S^{m_j}\) and \(m_j\) decreases to \(-\infty\), there is an
\(a\in S^{m_0}\), unique modulo \(S^{-\infty}\), with
\(a-\sum_{j<l}a_j\in S^{m_l}\) for every \(l\); one writes
\(a\sim\sum_ja_j\). A symbol is \emph{polyhomogeneous of degree \(\mu\)
with step one} if \(a\sim\sum_ja_j\) with \(a_j\) homogeneous of degree
\(\mu-j\) in \(\xi\) for \(|\xi|\geq1\).

{\small\sloppy
\textbf{Quantization and kernels.} For
\(a\in S^m(\mathbb R^n\times\mathbb R^n)\), the operator
\(\operatorname{Op}(a)u(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}a(x,\xi)\widehat u(\xi)\,d\xi\)
maps \(\mathcal S\) into \(\mathcal S\), and
\((a,u)\mapsto\operatorname{Op}(a)u\) is continuous. More generally, let
\(a\) be any tempered distribution on \(\mathbb R^n\times\mathbb R^n\).
The formulas \[
K_a(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}a(x,\xi)\,d\xi ,\qquad a(x,\xi)=\int e^{-iz\cdot\xi}K_a(x,x-z)\,dz ,
\] in which the integrals are partial Fourier transforms of tempered
distributions, are inverse to each other. So every continuous linear map
\(\mathcal S\to\mathcal S'\) is the map \(\operatorname{Op}(a)\) with
kernel \(K_a\) for exactly one tempered distribution \(a\); in
particular the operator determines its symbol. If \(a\) is a measurable
function with polynomial bounds, then \(\operatorname{Op}(a)u\) is
given, for \(u\in\mathcal S\), by the absolutely convergent integral
above; one checks this by pairing with a Schwartz function and using
Fubini's theorem and Fourier inversion.
\par}

\textbf{The Gauss transform.} \(e^{i\langle D_y,D_\eta\rangle}\) is the
Fourier multiplier with symbol \(e^{i\langle p,q\rangle}\), where
\((p,q)\) are the variables dual to
\((y,\eta)\in\mathbb R^n\times\mathbb R^n\). It maps
\(S^m(\mathbb R^n_y\times\mathbb R^n_\eta)\) continuously into itself,
and for every \(N\) \[
e^{i\langle D_y,D_\eta\rangle}c-\sum_{|\alpha|<N}\frac1{\alpha!}\partial_\eta^\alpha D_y^\alpha c\in S^{m-N},
\] each seminorm of the remainder being bounded by finitely many
seminorms of \(c\). If a sequence is bounded in \(S^m\) and converges
locally uniformly with all derivatives, then so do its transforms.

\textbf{The pre-diagonal product estimate.} Let
\(a\in S^{m_1}(\mathbb R^n\times\mathbb R^n)\),
\(b\in S^{m_2}(\mathbb R^n\times\mathbb R^n)\), and let
\(B(x,\xi,y,\eta)\) be the transform \(e^{i\langle D_y,D_\eta\rangle}\)
of \(a(x,\eta)b(y,\xi)\) in the variables \((y,\eta)\). Then for all
multi-indices and all \(N\geq0\), at every point \((x,\xi,y,\eta)\), \[
\Big|\partial_\xi^\alpha\partial_x^\beta\partial_\eta^{\alpha'}\partial_y^{\beta'}\Big(B-\sum_{|\gamma|<N}\frac1{\gamma!}\partial_\eta^\gamma a(x,\eta)\,D_y^\gamma b(y,\xi)\Big)\Big|\leq C\langle\eta\rangle^{m_1-N-|\alpha'|}\langle\xi\rangle^{m_2-|\alpha|},
\] with \(C\) bounded by finitely many seminorms of \(a\) and \(b\). At
\(y=x\), \(\eta=\xi\), \(B\) is the symbol of
\(\operatorname{Op}(a)\operatorname{Op}(b)\).

\textbf{The Schur test.} If \(K\) is a measurable function on
\(X\times Y\), for two measure spaces \(X\) and \(Y\), with
\(\sup_x\int|K(x,y)|\,dy\leq A\) and \(\sup_y\int|K(x,y)|\,dx\leq B\),
then \(u\mapsto\int K(\cdot,y)u(y)\,dy\) is bounded from \(L^2(Y)\) to
\(L^2(X)\) with norm at most \(\sqrt{AB}\).

\emph{The preceding unrestricted raw-integral statement is retained for
the visible correction and complete proof in
\hyperref[AN03-TC-SCHUR-001]{Section1.1}. Its applications below use
Lebesgue output measure.}

\textbf{Sobolev and Besov continuity.} For
\(a\in S^m(\mathbb R^n\times\mathbb R^n)\), \(\operatorname{Op}(a)\)
maps \(H_{(s)}\) continuously into \(H_{(s-m)}\) for every real \(s\).
The spaces \(B^s_{2,p}\) are Banach spaces, and multiplication by a
\(C^\infty_b\) function is bounded on each of them. If
\((c_l)_{l\in\mathbb Z}\) is summable, then convolution with it is
bounded on \(\ell^p(\mathbb Z)\), \(1\leq p\leq\infty\), with norm at
most \(\sum_l|c_l|\), by the triangle inequality for translates.

\emph{Local spaces.} The local spaces \(B^s_{2,p,\mathrm{loc}}\) do not
depend on the choice of cutoffs, and they are preserved by
multiplication with smooth functions and by changes of coordinates. This
is proved in the lesson
\href{geometric-microlocal-calculus.pdf}{Detecting regularity without
choosing coordinates}.

\emph{Conormal distributions and extension.} These facts are proved in
the lesson \href{conormal-transmission.pdf}{Singularities along a
submanifold and smooth boundary passage}.

\textbf{Conormal distributions.} Let \(Y\) be a closed submanifold of
\(\mathbb R^N\) of codimension \(k\), and \(\mu\in\mathbb R\). The space
\(I^\mu(\mathbb R^N,Y)\) consists of the
\(u\in\mathcal D'(\mathbb R^N)\) such that
\(L_1\cdots L_Mu\in B^{-\mu-N/4}_{2,\infty,\mathrm{loc}}(\mathbb R^N)\)
for every \(M\geq0\) and all first-order differential operators
\(L_1,\ldots,L_M\) with smooth coefficients whose principal symbols
vanish on the conormal bundle \(N^*Y\). Such a \(u\) is smooth off
\(Y\), and its wave front set lies in \(N^*Y\). Suppose \(Y=\{t=0\}\) in
coordinates \(x=(t,z)\in\mathbb R^k\times\mathbb R^{N-k}\). Then every
\(u\in I^\mu(\mathbb R^N,Y)\) with compact support has the form
\(u(t,z)=\int e^{i\langle t,\tau\rangle}b(z,\tau)\,d\tau\) with
\(b\in S^{\mu+(N-2k)/4}(\mathbb R^{N-k}\times\mathbb R^k)\) of compact
support in \(z\), and \(b\) is \((2\pi)^{-k}\) times the Fourier
transform of \(u\) in \(t\). Conversely every such integral lies in
\(I^\mu(\mathbb R^N,Y)\).

\textbf{Hadamard's lemma.} A smooth function \(f(t,z)\) that vanishes on
\(t=0\) equals \(\sum_it_i\int_0^1(\partial_{t_i}f)(st,z)\,ds\). Hence
every smooth vector field tangent to \(Y=\{t=0\}\) is a combination,
with smooth coefficients, of the fields \(t_i\partial_{t_j}\) and
\(\partial_{z_l}\).

\textbf{Smooth extension.} A function on \(\overline{\mathbb R}{}^n_+\)
that is smooth in \(\mathbb R^n_+\), and all of whose derivatives extend
continuously to \(\overline{\mathbb R}{}^n_+\), is the restriction of a
smooth function on \(\mathbb R^n\).

\emph{Functional analysis.}

\textbf{The Hahn--Banach theorem, seminorm form.} Let \(p\) be a
seminorm on a complex vector space \(X\), \(M\subset X\) a subspace, and
\(f\) a linear functional on \(M\) with \(|f(x)|\leq p(x)\) for
\(x\in M\). Then \(f\) extends to a linear functional \(\Lambda\) on
\(X\) with \(|\Lambda(x)|\leq p(x)\) for all \(x\in X\). Its full
seminorm proof is
\href{banach-foundation-bridges.pdf\#section*.63}{Section14.7
of the Banach foundations}.

\textbf{Quotients of Fréchet spaces.} If \(N\) is a closed subspace of a
Fréchet space \(X\), then \(X/N\) with the quotient topology is a
Fréchet space. Its full quotient-topology and completeness proof is
\href{banach-foundation-bridges.pdf\#section*.67}{Section14.8
of the Banach foundations}.

\subsubsection{1.1. Schur's bound on the original measure
spaces}\label{schurs-bound-on-the-original-measure-spaces}
\hypertarget{AN03-TC-SCHUR-001}{}\label{AN03-TC-SCHUR-001}

\textbf{Editorial correction to the background statement.} The preceding
Schur statement, as originally written, places no restriction on either
measure space. Its assertion about the raw integral is false at that
generality. We retain that statement above so the correction is
identifiable. Here we prove the counterexample, the bounded operator
that exists on the original arbitrary spaces, and its exact relationship
with the raw integral. In particular, the raw-integral assertion holds
when the output measure is semifinite; the input measure can remain
arbitrary. Every application later in this lesson has Lebesgue output
measure and therefore satisfies this condition.

Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be arbitrary
measure spaces. In this section measurable kernels mean
\(\mathcal A\otimes\mathcal B\)-measurable, finite complex-valued
kernels. Retain the original two constants and both pointwise bounds \[
 \int_Y|K(x,y)|\,d\nu(y)\leq A\quad(x\in X),\qquad
 \int_X|K(x,y)|\,d\mu(x)\leq B\quad(y\in Y),
 \qquad 0\leq A,B<\infty .
 \tag{SC1}
\] All \(L^2\) spaces use the original measures and their
almost-everywhere equivalence classes; inner products are linear in
their first argument. We do not assume completeness, sigma-finiteness or
semifiniteness of either measure.

\paragraph{A measurable counterexample with both bounds at every
point}\label{a-measurable-counterexample-with-both-bounds-at-every-point}

Put \(I=[0,1]\). On its Borel sets define \[
 \mu(E)=
 \begin{cases}0,&E\text{ is meager in }I,\\
 \infty,&E\text{ is not meager in }I,\end{cases}
 \qquad \nu=\text{Lebesgue measure on the Borel sets of }I.
 \tag{SC2}
\] Here meager means a countable union of sets whose closures have empty
relative interior. This is a measure: a countable union of meager sets
is meager; if one member of a countable disjoint family is nonmeager,
its union is nonmeager and both the sum and union measure are infinite.
These two cases also prove countable additivity. The complete-metric
Baire theorem, proved in
\href{banach-foundation-bridges.pdf\#section*.14}{the Banach chapter,
Section6}, shows that \(I\) is not meager in itself. Thus
\(\mu(I)=\infty\), while every set of finite \(\mu\)-measure has measure
zero.

Enumerate the rationals as \((q_k)_{k\geq1}\), and set \[
 U_j=\bigcup_{k\geq1}
 (q_k-2^{-j-k-3},q_k+2^{-j-k-3}),\qquad
 G=\bigcap_{j\geq1}U_j,\qquad N=\mathbb R\setminus G .
 \tag{SC3}
\] Every \(U_j\) is open and dense, and its Lebesgue measure is at most
\[
 \sum_{k\geq1}2\cdot2^{-j-k-3}=2^{-j-2}.
 \tag{SC4}
\] Consequently \(G\) is a Borel null set. Baire's theorem on
\(\mathbb R\) also makes it dense. Each \(\mathbb R\setminus U_j\) is
closed with empty interior, so \(N\) is meager and its complement is
Lebesgue null. No finite approximation of these sets is being
substituted for them.

Take \(X=Y=I\) with the measures in (SC2), and define \[
 K(x,y)=1_N(x+y),\qquad u(y)=1 .
 \tag{SC5}
\] Addition is continuous and \(N\) is Borel, so \(K\) is jointly Borel
measurable. For every \(x\), the omitted set of \(y\)'s is
\((G-x)\cap I\), a Lebesgue null set; hence its row integral is one. For
every \(y\), the set of \(x\)'s where \(K\ne0\) is \((N-y)\cap I\).
Translation preserves closed sets with empty interior, and restriction
of any such set to the nondegenerate interval \(I\) has empty relative
interior. Thus this section is meager in \(I\) and its column integral
is zero. We have the exact values \[
 A=1,\quad B=0,\quad \|u\|_{L^2(\nu)}=1,\qquad
 \int_I K(x,y)u(y)\,d\nu(y)=1\quad\text{for every }x,\qquad
 \int_I1^2\,d\mu=\infty .
 \tag{SC6}
\] The original raw-integral assertion therefore fails. This is not a
failure of a matrix estimate or of the square-root constant: the two
measures detect different sections of the same measurable set.

\paragraph{\texorpdfstring{The exact \(L^2\) operator without measure
restrictions}{The exact L\^{}2 operator without measure restrictions}}\label{the-exact-l2-operator-without-measure-restrictions}

Each \(L^2(\nu)\) class has a finite-valued measurable representative
\(u\). An infinite value, if allowed in a representative, occurs on a
null set and can be changed to zero. Its actual nonzero support is
sigma-finite: \[
 S_u=\{u\ne0\}=\bigcup_{k\geq1}\{|u|>1/k\},\qquad
 \nu\{|u|>1/k\}\leq k^2\|u\|_2^2 .
 \tag{SC7}
\] The same construction applies to \(v\in L^2(\mu)\). On the original
restricted measures on \(S_v\times S_u\), both factors are sigma-finite.
Thus the product and Tonelli/Fubini proofs in
\href{banach-foundation-bridges.pdf\#section*.118}{the Banach
chapter, Sections16.2--16.4} apply there, with no assertion of Tonelli
on all of \(X\times Y\). Define \[
 b_K(u,v)=
 \int_{S_v}\int_{S_u}K(x,y)u(y)\overline{v(x)}\,d\nu(y)\,d\mu(x).
 \tag{SC8}
\] The two nonnegative integrals needed for product Cauchy--Schwarz obey
\[
 \begin{split}
 \int_{S_v\times S_u}|K(x,y)|\,|u(y)|^2\,d(\mu\otimes\nu)
 &\leq B\|u\|_{L^2(\nu)}^2,\\
 \int_{S_v\times S_u}|K(x,y)|\,|v(x)|^2\,d(\mu\otimes\nu)
 &\leq A\|v\|_{L^2(\mu)}^2,\\
 \int_{S_v\times S_u}|K(x,y)|\,|u(y)|\,|v(x)|\,d(\mu\otimes\nu)
 &\leq\sqrt{AB}\,\|u\|_{L^2(\nu)}\|v\|_{L^2(\mu)} .
 \end{split}
 \tag{SC9}
\] Here the first line integrates the column bound and the second
integrates the row bound. Cauchy--Schwarz uses the two displayed
functions \(|K|^{1/2}|u|\) and \(|K|^{1/2}|v|\), retaining both factors.
All integrals on the right are finite, including when \(A=0\) or
\(B=0\); in either zero case the third integral vanishes. These bounds
prove absolute convergence of (SC8). Changing \(u\) or \(v\) on its
original null set changes no pairing: on the sigma-finite union of the
old and new supports, those changes are product-null by the proved
product theorem. The same observation on finite unions of supports
proves sesquilinearity.

We include the Hilbert-space facts needed here at this generality.
Cauchy--Schwarz for any measure follows by integrating \(|f-tg|^2\geq0\)
with its minimizing complex \(t\), treating \(g=0\) separately; it gives
Minkowski by expanding \(\|f+g\|_2^2\). If a sequence is Cauchy in
\(L^2(\mu)\), choose a subsequence with successive differences \(d_j\)
of norms at most \(2^{-j}\). For \(s_m=\sum_{j\leq m}|d_j|\), Minkowski
and monotone convergence give \[
 \int\left(\sum_{j\geq1}|d_j|\right)^2d\mu
 =\lim_m\int s_m^2d\mu
 \leq\left(\sum_{j\geq1}2^{-j}\right)^2 .
 \tag{SC10}
\] The sum is finite off a measurable null set. There the subsequence
converges pointwise; define the limit to be zero on that null set.
Applying the same bound to each tail proves convergence in \(L^2\), and
the Cauchy property gives convergence of the whole sequence. This proves
completeness on arbitrary \(\mu\), and the proof for \(\nu\) is
identical.

For completeness, the representing-vector argument requires no
separability. For fixed \(u\), put \(\ell(v)=\overline{b_K(u,v)}\), a
bounded linear functional on this Hilbert space. If \(\ell=0\), choose
its representing vector to be zero. Otherwise the closed affine set
\(\ell(v)=1\) is nonempty and its distance \(d\) from zero is positive,
since \(1\leq\|\ell\|\|v\|\). A sequence in that set whose norms
decrease to \(d\) is Cauchy: the parallelogram identity gives \[
 \|z_j-z_k\|^2
 =2\|z_j\|^2+2\|z_k\|^2-4\|(z_j+z_k)/2\|^2
 \leq2\|z_j\|^2+2\|z_k\|^2-4d^2 .
 \tag{SC11}
\] Completeness gives a minimizing vector \(z\) in the set. Varying
\(z\) by \(tw\) and \(itw\), for real \(t\) and \(w\in\ker\ell\), shows
\(\langle z,w\rangle=0\). Since \(v-\ell(v)z\in\ker\ell\), \[
 \ell(v)=\left\langle v,\frac{z}{\|z\|^2}\right\rangle,\qquad
 b_K(u,v)=\left\langle\frac{z}{\|z\|^2},v\right\rangle .
 \tag{SC12}
\] The vector in the second pairing is unique, as its difference from
another would pair to zero with itself. Define it to be \(T_Ku\).
Sesquilinearity and uniqueness prove that \(T_K\) is linear. Taking
\(v=T_Ku\) in (SC9), and treating its zero norm separately, proves \[
 T_K:L^2(\nu)\longrightarrow L^2(\mu),\qquad
 \langle T_Ku,v\rangle=b_K(u,v),\qquad
 \|T_K\|\leq\sqrt{AB}.
 \tag{SC13}
\] This operator is defined on the original spaces. Its adjoint retains
the literal swapped kernel \(K^*(y,x)=\overline{K(x,y)}\): absolute
Fubini on the two supports gives \(b_K(u,v)=\overline{b_{K^*}(v,u)}\),
so \(T_K^*=T_{K^*}\). The swapped constants are \(B,A\).

\paragraph{The raw integral on every finite-measure
piece}\label{the-raw-integral-on-every-finite-measure-piece}

Fix \(u\) and its support in (SC7). For any nonnegative
product-measurable function \(h\) on \(X\times S_u\), its section
integral is a measurable function of \(x\), without a condition on
\(\mu\). Indeed, on a finite-\(\nu\) piece the section-measure argument
in Section16.3 of the linked chapter uses rectangles, finite subtraction
for complements, and countable sums for disjoint unions; its
generating-class proof in Section16.2 requires no outer measure
hypothesis. Increasing finite-measure exhaustions of \(S_u\), followed
by increasing simple approximations to \(h\), give the assertion. Real
and imaginary parts then give measurability of absolutely convergent
complex section integrals.

Consequently the following sets and functions are measurable: {\small
\[
 J_u(x)=\int_{S_u}|K(x,y)|\,|u(y)|\,d\nu(y),\quad
 D_u=\{J_u=\infty\},\qquad
 r_0(u)(x)=
 \begin{cases}
 \displaystyle\int_YK(x,y)u(y)\,d\nu(y),&x\notin D_u,\\
 0,&x\in D_u .
 \end{cases}
 \tag{SC14}
\]
} The value zero on \(D_u\) is an explicit convention; it is not an
assertion that the divergent raw integral exists there. For a measurable
\(E\subset X\) with \(\mu(E)<\infty\), Tonelli on \(E\times S_u\) and
the original column bound give \[
 \int_E\int_{S_u}|K(x,y)|\,|u(y)|^2d\nu(y)d\mu(x)
 \leq B\|u\|_2^2 .
 \tag{SC15}
\] The inner quantity is finite for almost every \(x\) in \(E\).
Weighted Cauchy--Schwarz in \(y\) gives, at those points, \[
 J_u(x)^2
 \leq\left(\int_Y|K(x,y)|d\nu(y)\right)
       \left(\int_{S_u}|K(x,y)|\,|u(y)|^2d\nu(y)\right).
 \tag{SC16}
\] If \(A=0\), every row is zero almost everywhere in \(\nu\), and
\(J_u=0\) at every \(x\); this also covers the zero-row case before
multiplying any infinite value. Otherwise (SC15)--(SC16) prove \[
 \mu(E\cap D_u)=0,\qquad
 \int_E|r_0(u)|^2d\mu\leq AB\|u\|_2^2
 \quad\text{for every }E\text{ of finite measure}.
 \tag{SC17}
\] For \(v\) supported in such \(E\), absolute Fubini and (SC9) identify
\(\langle r_0(u),v\rangle=b_K(u,v)\) on \(E\). Both \(r_0(u)1_E\) and
\((T_Ku)1_E\) lie in \(L^2(\mu)\); testing their difference proves \[
 \int_E|r_0(u)-T_Ku|^2d\mu=0
 \quad(\mu(E)<\infty).
 \tag{SC18}
\] Equality on all finite-measure pieces is the exact comparison proved
so far. In the counterexample it does not imply equality almost
everywhere for the original \(\mu\).

\paragraph{The space defined by this defect and its complete quotient
map}\label{the-space-defined-by-this-defect-and-its-complete-quotient-map}

Let \(F_\mu\) be the vector space of original almost-everywhere classes
of finite-valued measurable functions \(h\) for which \[
 p_\mu(h)=\sup_{\substack{E\in\mathcal A\\\mu(E)<\infty}}
              \left(\int_E|h|^2d\mu\right)^{1/2}<\infty,\qquad
 N_\mu=\{h\in F_\mu:p_\mu(h)=0\}.
 \tag{SC19}
\] Minkowski on each \(E\) proves that \(p_\mu\) is a seminorm. Its
kernel consists exactly of the functions invisible to every
finite-measure test. This definition retains the original global null
classes, so \(N_\mu\) can contain a class that is nonzero globally.

There is a canonical onto isometry \[
 i_\mu:L^2(\mu)\longrightarrow F_\mu/N_\mu,\qquad
 i_\mu(h)=h+N_\mu .
 \tag{SC20}
\] First, for \(h\in L^2(\mu)\), the finite-measure level sets in (SC7),
made increasing by finite unions, exhaust its support. Monotone
convergence gives \(p_\mu(h)=\|h\|_2\). Thus this map is injective and
isometric.

To prove surjectivity, take \(h\in F_\mu\), and put \(c=p_\mu(h)^2\). If
\(c=0\), its class in the quotient is zero. If \(c>0\), choose
finite-measure sets \(E_k\) such that \(\int_{E_k}|h|^2d\mu>c-2^{-k}\).
Replace these sets by the finite unions \(F_k=\bigcup_{j\leq k}E_j\),
and put \(S=\bigcup_kF_k\). Then \[
 \int_S|h|^2d\mu=\lim_k\int_{F_k}|h|^2d\mu=c,\qquad
 h_s=h1_S\in L^2(\mu).
 \tag{SC21}
\] For any finite-measure \(E\), the disjoint union
\(F_k\cup(E\setminus S)\) still has finite measure, so \[
 \int_{F_k}|h|^2d\mu+\int_{E\setminus S}|h|^2d\mu\leq c .
 \tag{SC22}
\] Letting \(k\) increase proves that the second integral is zero.
Therefore \(h-h_s\in N_\mu\), establishing surjectivity. Two choices of
\(h_s\) differ by an \(L^2\) element of \(N_\mu\), hence are equal in
the original \(L^2\) space. The inverse \(P_\mu=i_\mu^{-1}\) is
consequently canonical and linear, and the quotient is a complete
Hilbert space through this isometry.

Equations (SC17)--(SC18) now give the complete connecting maps \[
 r_0(u)\in F_\mu,\qquad
 r_0(u)+N_\mu=i_\mu(T_Ku),\qquad
 P_\mu(r_0(u)+N_\mu)=T_Ku .
 \tag{SC23}
\] In particular the raw assignment followed by the quotient is linear,
even though the conventions on divergent sets were only pointwise
choices. Any other finite measurable convention on \(D_u\) has the same
quotient class, because every finite-measure intersection of \(D_u\) is
null. In (SC2), \(p_\mu\) is zero on every finite-valued measurable
function, \(F_\mu=N_\mu\), and \(L^2(\mu)=\{0\}\). The raw constant one
in (SC6) is a nonzero original class in \(N_\mu\); (SC13) is the zero
operator. Thus (SC23) describes exactly what the counterexample loses
under finite-measure testing.

\paragraph{When the raw-integral assertion is
recovered}\label{when-the-raw-integral-assertion-is-recovered}

A measure is semifinite if each measurable set of positive measure
contains a measurable subset of finite positive measure. Under this
condition, a measurable set whose intersection with every finite-measure
set is null is itself null: if it were positive, the defining subset
would contradict the intersection property. Applying this first to
\(D_u\) and then to the sets \(\{|h|>1/k\}\) for \(h\in N_\mu\) gives {\footnotesize
\[
 \mu\text{ semifinite}\quad\Longrightarrow\quad
 \mu(D_u)=0,\quad N_\mu=\{0\},\quad
 r_0(u)=T_Ku\ \mu\text{-almost everywhere},\quad
 \|r_0(u)\|_2\leq\sqrt{AB}\|u\|_2 .
 \tag{SC24}
\]
} There is no restriction on the input measure \(\nu\) in this
implication.

The defect criterion is exact: \(N_\mu=\{0\}\) if and only if \(\mu\) is
semifinite. Indeed, if semifiniteness fails, its definition supplies a
positive measurable set \(H\) with no finite positive-measure subset.
Every finite-measure intersection \(E\cap H\) has measure zero. The
finite-valued function \(1_H\) therefore belongs to \(N_\mu\) and is a
nonzero original almost-everywhere class. This proves the converse.
Equivalently, semifiniteness gives \[
 \mu(H)=\sup_{\substack{E\subset H\\\mu(E)<\infty}}\mu(E).
 \tag{SC25}
\] For finite \(\mu(H)\) this is immediate. If \(\mu(H)=\infty\) and the
supremum \(c\) were finite, choose finite-measure \(E_k\subset H\)
approaching \(c\), take their increasing finite unions, and let their
union be \(S\). Measure continuity gives \(\mu(S)=c\). Any finite
positive-measure subset of \(H\setminus S\) would raise the supremum by
its disjoint union with \(S\), while \(H\setminus S\) still has infinite
measure. This contradicts semifiniteness and proves (SC25). We have
proved the exact vanishing criterion for the defect space; we do not
infer that every kernel on a nonsemifinite space must fail its
raw-integral bound.

Finally, the same construction works for the finite vector dimensions in
this lesson, retaining their order. If
\(K(x,y)\in L(\mathbb C^p,\mathbb C^q)\), replace \(|K|\) in (SC1) and
(SC9) by its original operator norm and use \[
 |\langle K(x,y)u(y),v(x)\rangle_{\mathbb C^q}|
 \leq\|K(x,y)\|\,|u(y)|_{\mathbb C^p}|v(x)|_{\mathbb C^q}.
 \tag{SC26}
\] This gives exactly (SC9), with no dimension factor. Vector-valued
\(L^2\) completeness follows by the same norm-sum proof (SC10); the
Hilbert representing argument (SC11)--(SC12) applies unchanged. The raw
convergence proof bounds the norm of its vector integral by \(J_u\).
Define \(F_\mu,N_\mu\) with the original Euclidean vector norm;
(SC21)--(SC22) prove the same onto isometry and comparison. The adjoint
kernel is exactly \(K(y,x)^*\), reversing the two vector dimensions, as
absolute Fubini in (SC8) shows. All subsequent Schur applications thus
retain the full scalar or matrix constants and the original Lebesgue
integral. No novelty claim is made for these measure-theoretic results.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Figure SC-F1. The exact kernel K(x,y)=1\_N(x+y) on I={[}0,1{]} has row bound A=1 and column bound B=0. Its raw integral sends the input constant one of Lebesgue L2 norm one to the constant one with infinite global output L2 integral. Every finite-measure output test is zero. The canonical Schur operator is zero, and the onto isometry F\_mu/N\_mu=L2(mu) identifies the raw output with zero through SC20--SC23. This is a diagram of the exact maps, not a sampled picture of the meager set N. Definitions and full proofs: SC2--SC6 and SC19--SC24. Reproducible figure source: figures/schur\_arbitrary\_measure.py.}]{../figures/schur_arbitrary_measure.png}}
\caption{Figure SC-F1. The exact kernel K(x,y)=1\_N(x+y) on I={[}0,1{]}
has row bound A=1 and column bound B=0. Its raw integral sends the input
constant one of Lebesgue L2 norm one to the constant one with infinite
global output L2 integral. Every finite-measure output test is zero. The
canonical Schur operator is zero, and the onto isometry
F\_mu/N\_mu=L2(mu) identifies the raw output with zero through
SC20--SC23. This is a diagram of the exact maps, not a sampled picture
of the meager set N. Definitions and full proofs: SC2--SC6 and
SC19--SC24. Reproducible figure source:
figures/schur\_arbitrary\_measure.py.}
\end{figure}

\subsection{2. Totally characteristic differential
operators}\label{totally-characteristic-differential-operators}

Let \(\mathcal V_b\) be the smooth vector fields
\(V=\sum_jv_j\partial_j\),
\(v_j\in C^\infty(\overline{\mathbb R}{}^n_+)\), that are tangent to the
boundary, that is, \(v_n(x',0)=0\). Let
\(\operatorname{Diff}_b(\overline{\mathbb R}{}^n_+)\) be the algebra of
operators on \(C^\infty(\overline{\mathbb R}{}^n_+)\) generated by
\(\mathcal V_b\) and by multiplication with functions in
\(C^\infty(\overline{\mathbb R}{}^n_+)\), and
\(\operatorname{Diff}^m_b\) the span of products containing at most
\(m\) vector fields. Its elements are the \emph{totally characteristic}
differential operators.

\textbf{Proposition 2.1} (Structure of totally characteristic
differential operators).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  \(\mathcal V_b\) is the
  \(C^\infty(\overline{\mathbb R}{}^n_+)\)-module generated by
  \(\partial_1,\ldots,\partial_{n-1}\) and \(x_n\partial_n\).
\item
  For every integer \(k\geq0\), \[
x_n^kD_n^k=\prod_{j=0}^{k-1}\big(x_nD_n+ij\big)=:q_k(x_nD_n).
\tag{2.1}
\] Hence \(\{x_n^jD_n^j:j\leq k\}\) and \(\{(x_nD_n)^j:j\leq k\}\)
  span the same space, with constant coefficients.
\item
  \(\operatorname{Diff}^m_b\) consists exactly of the finite sums \[
P=\sum_{|\alpha|\leq m}c_\alpha(x)\,x_n^{\alpha_n}D^\alpha,\qquad c_\alpha\in C^\infty(\overline{\mathbb R}{}^n_+),
\tag{2.2}
\] equivalently of the sums
  \(\sum_{|\alpha|\leq m}c'_\alpha(x)D'^{\alpha'}(x_nD_n)^{\alpha_n}\).
\item
  For \(P\) as in (2.2) and
  \(u\in C^\infty(\overline{\mathbb R}{}^n_+)\),
  \((Pu)(x',0)=\sum_{\alpha_n=0}c_\alpha(x',0)D'^{\alpha'}u(x',0)\): the
  boundary value of \(Pu\) depends only on the boundary value of \(u\).
\end{enumerate}

\textbf{Proof.} (a) If \(v_n(x',0)=0\), then \(v_n(x)=x_nw(x)\) with
\(w(x)=\int_0^1(\partial_nv_n)(x',\theta x_n)\,d\theta\in C^\infty(\overline{\mathbb R}{}^n_+)\).
Thus \(V=\sum_{j<n}v_j\partial_j+w\,x_n\partial_n\). Conversely each
generator is tangent.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  For \(k\geq0\) and \(u\) smooth,
  \(x_nD_n(x_n^kD_n^ku)=x_n^{k+1}D_n^{k+1}u+x_n(D_nx_n^k)D_n^ku=x_n^{k+1}D_n^{k+1}u-ik\,x_n^kD_n^ku\),
  because \(D_nx_n^k=-ikx_n^{k-1}\). So
  \(x_n^{k+1}D_n^{k+1}=(x_nD_n+ik)\,x_n^kD_n^k\), and induction gives
  (2.1). The polynomial \(q_k\) is monic of degree \(k\), so the
  triangular system can be inverted.
\item
  Moving a function to the left across a generator produces only
  multiplication operators: \(\partial_jc=c\partial_j+(\partial_jc)\)
  and \(x_n\partial_nc=c\,x_n\partial_n+x_n(\partial_nc)\). So a product
  of at most \(m\) vector fields and functions is a sum of terms
  \(c(x)M_1\cdots M_l\), \(l\leq m\), with each \(M_i\) one of the
  generators in (a). These generators commute pairwise, because
  \([x_n\partial_n,\partial_j]=0\) for \(j<n\). So each word is a
  constant times \(D'^{\beta'}(x_nD_n)^{\beta_n}\) with
  \(|\beta|\leq m\), and (b) rewrites it in the form (2.2). Conversely
  \(x_n^{\alpha_n}D^\alpha=D'^{\alpha'}q_{\alpha_n}(x_nD_n)\) is a
  product of \(|\alpha|\) generators.
\item
  At \(x_n=0\) every term with \(\alpha_n>0\) carries the factor
  \(x_n^{\alpha_n}\), which vanishes. \(\square\)
\end{enumerate}

Part (d) is the motivation for the whole lesson. An operator that
respects the boundary in this way can be followed by boundary operators.
Theorem 5.1(c) below extends (d) to the pseudodifferential operators of
this lesson and to normal derivatives of every order.

\subsection{3. Function spaces on the half
space}\label{function-spaces-on-the-half-space}

\subsubsection{Restrictions and
supports}\label{restrictions-and-supports}

\textbf{Two ways to attach a space to the half space.} Let \(F\) be a
space of distributions on \(\mathbb R^n\).

\begin{itemize}
\tightlist
\item
  \(\overline F(\mathbb R^n_+)\) is the space of restrictions
  \(U|_{\mathbb R^n_+}\), \(U\in F\), with the quotient topology of
  \(F/\{U\in F:U=0\text{ in }\mathbb R^n_+\}\).
\item
  \(\dot F(\overline{\mathbb R}{}^n_+)\) is the space of \(U\in F\) with
  \(\operatorname{supp}U\subset\overline{\mathbb R}{}^n_+\), with the
  topology of \(F\).
\end{itemize}

These are different objects and must be kept apart. A restriction of a
Schwartz function may have any boundary values. A Schwartz function
supported in \(\overline{\mathbb R}{}^n_+\) vanishes to infinite order
on \(x_n=0\), since all its derivatives are continuous and vanish for
\(x_n<0\). The zero extension of an element of
\(\overline{\mathcal S}(\mathbb R^n_+)\) is an integrable function in
\(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\); it lies in
\(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\) only when all its
normal derivatives vanish at the boundary. In this notation,
\(C^\infty(\overline{\mathbb R}{}^n_+)=\overline{C^\infty}(\mathbb R^n_+)\),
by the smooth extension fact of Section 1.

\textbf{Lemma 3.1} (Supports in the closed half space). Let
\(U\in\mathcal S'(\mathbb R^n)\) with
\(\operatorname{supp}U\subset\overline{\mathbb R}{}^n_+\), and let
\(\varphi\in\mathcal S(\mathbb R^n)\) vanish in \(\mathbb R^n_+\). Then
\(U(\varphi)=0\).

\textbf{Proof.} First, \(U(\psi)=0\) whenever \(\psi\in\mathcal S\)
vanishes on a neighbourhood \(W\) of \(\operatorname{supp}U\): for
\(\psi\in C_0^\infty\) this is the definition of the support, and in
general \(\psi\,\theta(\cdot/R)\to\psi\) in \(\mathcal S\) for a cutoff
\(\theta\) equal to 1 near 0, while each \(\psi\theta(\cdot/R)\)
vanishes on \(W\). Now put \(\varphi_\delta(x)=\varphi(x',x_n+\delta)\).
It vanishes on \(\{x_n>-\delta\}\), a neighbourhood of
\(\overline{\mathbb R}{}^n_+\), so \(U(\varphi_\delta)=0\); and
\(\varphi_\delta\to\varphi\) in \(\mathcal S\) as \(\delta\to0\).
\(\square\)

\subsubsection{Restricted Schwartz
functions}\label{restricted-schwartz-functions}

For \(v\in\overline{\mathcal S}(\mathbb R^n_+)\) and multi-indices
\(\alpha,\beta\) put \[
q_{\alpha,\beta}(v)=\sup_{x\in\mathbb R^n_+}|x^\alpha D^\beta v(x)| .
\]

\textbf{Lemma 3.2} (Restricted Schwartz functions).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Each \(q_{\alpha,\beta}\) is finite and continuous on
  \(\overline{\mathcal S}(\mathbb R^n_+)\). For every continuous
  seminorm \(q\) on \(\overline{\mathcal S}(\mathbb R^n_+)\) there are
  \(k\) and \(C\) with \[
q(v)\leq C\sum_{|\alpha|+|\beta|\leq2k}q_{\alpha,\beta}(v) .
\tag{3.1}
\] So the \(q_{\alpha,\beta}\) define the quotient topology.
\item
  \(\overline{\mathcal S}(\mathbb R^n_+)\) is a Fréchet space.
\item
  A function \(w\in C^\infty(\overline{\mathbb R}{}^n_+)\) is the
  restriction of a Schwartz function if and only if every
  \(q_{\alpha,\beta}(w)\) is finite.
\item
  If \(g\in\overline{\mathcal S}(\mathbb R^n_+)\), \(k\geq1\), and
  \(\partial_n^jg(x',0)=0\) for \(j<k\), then \(g=x_n^kh\) with
  \(h\in\overline{\mathcal S}(\mathbb R^n_+)\).
\end{enumerate}

\textbf{Proof.} (a) For every extension \(V\) of \(v\),
\(q_{\alpha,\beta}(v)\leq\sup_{\mathbb R^n}|x^\alpha D^\beta V|\); so
\(q_{\alpha,\beta}\) is bounded by a quotient seminorm, hence finite and
continuous. Conversely let \(q\) be continuous, and let \(\pi\) be the
restriction map. Then \(q\circ\pi\) is a continuous seminorm on
\(\mathcal S(\mathbb R^n)\), so there are \(k,C\) with
\(q(\pi V)\leq C\sum_{|\alpha|+|\beta|\leq k}\sup_{\mathbb R^n}|x^\alpha D^\beta V|\).
Fix \(V\in\mathcal S\). Put \(\tilde V=V\) on \(x_n\geq0\) and \[
\tilde V(x)=\theta(x_n)\sum_{j\leq k}\partial_n^jV(x',0)\frac{x_n^j}{j!}\qquad(x_n<0),
\] with \(\theta\in C_0^\infty(\mathbb R)\) equal to 1 on \((-1,1)\).
The two pieces have the same derivatives of order \(\leq k\) on
\(x_n=0\), so \(\tilde V\in C^k(\mathbb R^n)\). For
\(|\alpha|+|\beta|\leq k\),
\(\sup_{\mathbb R^n}(1+|x|)^{|\alpha|}|D^\beta\tilde V|\) is bounded by
a constant times
\(\sum_{|\alpha'|+|\gamma|\leq2k}q_{\alpha',\gamma}(\pi V)\): on
\(x_n\geq0\) this is clear, and on \(x_n<0\) the derivatives are
combinations of derivatives of \(\theta(x_n)x_n^j\) (bounded, with
support in a fixed interval) and of \(D'^{\beta'}\partial_n^jV(x',0)\),
\(|\beta'|+j\leq2k\), whose weighted suprema are limits from
\(\mathbb R^n_+\). Now take \(\phi\in C_0^\infty(\mathbb R^n_-)\) with
\(\int\phi=1\),
\(\phi_\varepsilon(x)=\varepsilon^{-n}\phi(x/\varepsilon)\),
\(0<\varepsilon\leq1\). Then
\(\tilde V*\phi_\varepsilon\in\mathcal S(\mathbb R^n)\). For
\(x\in\mathbb R^n_+\) the convolution only uses values at \(x-y\) with
\((x-y)_n>x_n>0\), so
\(\pi(\tilde V*\phi_\varepsilon)=\pi(V*\phi_\varepsilon)\). For
\(|\beta|\leq k\),
\(D^\beta(\tilde V*\phi_\varepsilon)=(D^\beta\tilde V)*\phi_\varepsilon\),
and \(1+|x|\leq(1+R)(1+|x-y|)\) for
\(y\in\operatorname{supp}\phi_\varepsilon\subset\{|y|\leq R\}\). Hence
\(q(\pi(V*\phi_\varepsilon))\leq C'\sum_{|\alpha|+|\beta|\leq2k}q_{\alpha,\beta}(\pi V)\),
uniformly in \(\varepsilon\). Since \(V*\phi_\varepsilon\to V\) in
\(\mathcal S\), (3.1) follows.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  The subspace \(\{V\in\mathcal S:V=0\text{ in }\mathbb R^n_+\}\) is
  closed, since point evaluations are continuous. The quotient of a
  Fréchet space by a closed subspace is again a Fréchet space (see the
  background list in Section 1).
\item
  Necessity is clear. Conversely let every \(q_{\alpha,\beta}(w)\) be
  finite, let \(w_0\) be the zero extension of \(w\), and let
  \(\phi_\varepsilon\) be as in (a). Then
  \(W_\varepsilon=w_0*\phi_\varepsilon\in\mathcal S(\mathbb R^n)\),
  because \(w_0\) is bounded and rapidly decreasing and all derivatives
  fall on \(\phi_\varepsilon\). If
  \(\operatorname{supp}\phi\subset\{y_n\leq-c\}\), then for \(x\) near a
  point of \(\mathbb R^n_+\) the integral
  \(\int w_0(x-y)\phi_\varepsilon(y)dy\) only involves points with
  \((x-y)_n\geq x_n+c\varepsilon\), so we may differentiate under it:
  \(D^\beta W_\varepsilon(x)=\int(D^\beta w)(x-y)\phi_\varepsilon(y)\,dy\).
  By the mean value theorem along segments, which stay in
  \(\mathbb R^n_+\), \[
|x^\alpha(D^\beta W_\varepsilon-D^\beta w)(x)|\leq C\varepsilon\sum_{|\alpha'|\leq|\alpha|,\,|\gamma|=|\beta|+1}q_{\alpha',\gamma}(w)\qquad(x\in\mathbb R^n_+).
\] So \(\pi W_\varepsilon\to w\) in every \(q_{\alpha,\beta}\). By (a)
  the family is Cauchy in \(\overline{\mathcal S}(\mathbb R^n_+)\), by
  (b) it converges to some \(\pi W\), and since \(q_{0,0}\) is
  continuous the limit agrees with \(w\) on \(\mathbb R^n_+\).
\item
  On \(x_n>1\) put \(h=g/x_n^k\). On \(0\leq x_n<2\) Taylor's formula
  with integral remainder and the vanishing jets give \(g=x_n^kh\) with
  \(h(x)=\frac1{(k-1)!}\int_0^1(1-\theta)^{k-1}(\partial_n^kg)(x',\theta x_n)\,d\theta\).
  The two definitions agree for \(1<x_n<2\). The second one is smooth up
  to \(x_n=0\), and both have finite weighted suprema of all derivatives
  (on \(x_n\leq2\) the weights are controlled by \(1+|x'|\)). So
  \(h\in\overline{\mathcal S}(\mathbb R^n_+)\) by (c). \(\square\)
\end{enumerate}

\subsection{4. Symbols, compressed quantization and
lacunarity}\label{symbols-compressed-quantization-and-lacunarity}

\subsubsection{The symbol class and its
quantization}\label{the-symbol-class-and-its-quantization}

\textbf{Definition 4.1} (The class \(S^m_+\)). For \(m\in\mathbb R\),
\(S^m_+\) is the set of
\(a\in C^\infty(\overline{\mathbb R}{}^n_+\times\mathbb R^n)\) such that
for all multi-indices \(\alpha,\beta\) and all integers \(\nu\geq0\) \[
p^m_{\alpha,\beta,\nu}(a)=\sup_{x\in\overline{\mathbb R}{}^n_+,\ \xi\in\mathbb R^n}(1+|\xi|)^{|\alpha|-m}(1+x_n)^{\nu}\,|a^{(\alpha)}_{(\beta)}(x,\xi)|<\infty .
\tag{4.1}
\] These seminorms make \(S^m_+\) a Fréchet space: a sequence that is
Cauchy for all of them converges locally uniformly with all derivatives,
and the weighted bounds pass to the limit. We put
\(S^{-\infty}_+=\bigcap_mS^m_+\). The estimates are uniform in \(x'\),
with no decay in \(x'\), and require rapid decay in \(x_n\).

\textbf{Compression and quantization.} For \(a\in S^m_+\) put \[
a^\flat(x,\xi)=a(x,\xi',x_n\xi_n)\quad(x_n\geq0),\qquad a^\flat(x,\xi)=0\quad(x_n<0),
\tag{4.2}
\] and, for \(u\in\mathcal S(\mathbb R^n)\), \[
T_au(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}a^\flat(x,\xi)\,\widehat u(\xi)\,d\xi .
\tag{4.3}
\] Since \(1+|(\xi',x_n\xi_n)|\leq(1+x_n)(1+|\xi|)\), the compressed
symbol grows at most polynomially and the integral converges absolutely.
We always write the last variable of \(a\) as \(\xi_n\); thus
\(\partial_{\xi_n}a\) is the derivative of \(a\) in its last slot,
\((\partial_{\xi_n}a)^\flat\) its compression, and
\(\partial_{\xi_n}(a^\flat)=x_n(\partial_{\xi_n}a)^\flat\). We also
write \(\xi_na\) for the symbol \((x,\xi)\mapsto\xi_na(x,\xi)\); its
compression is \(x_n\xi_na^\flat\).

Two remarks explain the choice of class. First, away from the boundary
\(T_a\) is an ordinary pseudodifferential operator. Indeed, for
\(x_n\geq1\) the compressed symbol obeys the ordinary estimates of
\(S^m\) uniformly. Each \(\partial_{\xi_n}\) of \(a^\flat\) brings a
factor \(x_n\), and each \(\partial_{x_n}\) brings
\((\partial_{x_n}a)^\flat\) or \(\xi_n(\partial_{\xi_n}a)^\flat\), where
\(|\xi_n|\leq(1+|(\xi',x_n\xi_n)|)/x_n\). Moreover
\((1+|\xi|)\leq1+|(\xi',x_n\xi_n)|\leq x_n(1+|\xi|)\). So every
derivative obeys the estimate of \(S^m\) up to powers of \(x_n\), and
the rapid decay in \(x_n\) absorbs every power of \(x_n\). Second, near
\(x_n=0\) the compressed symbol is not a classical symbol:
\(\partial_{\xi_n}a^\flat=x_n(\partial_{\xi_n}a)^\flat\) gains no power
of \(\langle\xi\rangle\).

If \(P=\sum_{|\alpha|\leq m}c_\alpha(x)x_n^{\alpha_n}D^\alpha\) with
\(c_\alpha\in C^\infty_b(\overline{\mathbb R}{}^n_+)\) vanishing for
\(x_n\geq R\), then \(P=T_p\) with
\(p(x,\xi)=\sum c_\alpha(x)\xi^\alpha\in S^{m}_+\): indeed
\(p^\flat=\sum c_\alpha(x)\xi'^{\alpha'}(x_n\xi_n)^{\alpha_n}\), and
left quantization places functions of \(x\) on the left. So Proposition
2.1 suggests the definition.

\textbf{The kernel.} The compressed symbol is a polynomially bounded
measurable function. So, by the facts on quantization and kernels in
Section 1, the operator \(T_a:\mathcal S\to\mathcal S'\) has the
tempered kernel
\(K_a(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}a^\flat(x,\xi)\,d\xi\). For
\(x_n>0\) the substitution \(\eta_n=x_n\xi_n\) suggests \[
K_a(x,y)=x_n^{-1}A\Big(x,\,x'-y',\,\frac{x_n-y_n}{x_n}\Big),\qquad A(x,z)=(2\pi)^{-n}\int e^{iz\cdot\xi}a(x,\xi)\,d\xi .
\tag{4.4}
\] For residual symbols this is an identity of functions (Theorem
6.2(b)). We want \(T_au\) to depend only on \(u|_{\mathbb R^n_+}\), that
is, \(K_a(x,y)=0\) for \(y_n<0\). With \(z_n=(x_n-y_n)/x_n\), the
condition \(y_n<0\) means \(z_n>1\). So \(A(x,\cdot)\) should vanish on
\(z_n>1\); this is a condition on the Fourier transform of \(a\) in its
last variable.

\subsubsection{Lacunary symbols}\label{lacunary-symbols}

\textbf{Definition 4.2} (Lacunary symbols). For \(a\in S^m_+\) and fixed
\((x,\xi')\), the function \(\xi_n\mapsto a(x,\xi',\xi_n)\) is tempered;
let \(\mathcal F_na(x,\xi',\cdot)\) be its Fourier transform, a tempered
distribution in the dual variable \(t\) (formally
\(\int e^{-it\xi_n}a\,d\xi_n\)). We call \(a\) \emph{lacunary} if \[
\operatorname{supp}\mathcal F_na(x,\xi',\cdot)\subset[-1,\infty)\qquad\text{for all }(x,\xi')\in\overline{\mathbb R}{}^n_+\times\mathbb R^{n-1},
\tag{4.5}
\] that is, \(\int a(x,\xi',\xi_n)\widehat\varphi(\xi_n)\,d\xi_n=0\) for
every \(\varphi\in C_0^\infty((-\infty,-1))\). We call \(a\)
\emph{strongly lacunary} if these supports lie in \([-\tfrac12,1]\).
\(S^m_{\mathrm{la}}\) denotes the lacunary elements of \(S^m_+\) and
\(S^{-\infty}_{\mathrm{la}}=\bigcap_mS^m_{\mathrm{la}}\).

Each defining condition is a continuous linear functional on \(S^m_+\),
since
\(|\int a\widehat\varphi|\leq p(a)\int(1+|\xi'|+|\xi_n|)^{|m|}|\widehat\varphi(\xi_n)|d\xi_n\).
So \(S^m_{\mathrm{la}}\) is a closed subspace and a Fréchet space.

The following closure properties are used constantly. If \(a\) is
lacunary (strongly lacunary), then so are \(\partial_x^\beta a\),
\(\partial_{\xi'}^\gamma a\), \(\partial_{\xi_n}a\), \(\xi^\gamma a\),
and \(c(x)a\) for \(c\in C^\infty_b(\overline{\mathbb R}{}^n_+)\).
Indeed \(\mathcal F_n\) commutes with operations in \(x\) and \(\xi'\),
turns \(\partial_{\xi_n}\) into multiplication by \(it\), and turns
multiplication by \(\xi_n\) into \(i\partial_t\); none of these enlarges
the support.

\textbf{Proposition 4.3} (Lacunarity is exactly the support condition).
For \(a\in S^m_+\) the following are equivalent.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(a\) is lacunary.
\item
  \(T_av=0\) in \(\mathbb R^n_+\) for every
  \(v\in\mathcal S(\mathbb R^n)\) that vanishes in \(\mathbb R^n_+\).
\item
  \(T_av=0\) in \(\mathbb R^n_+\) for every
  \(v\in C_0^\infty(\mathbb R^n_-)\).
\end{enumerate}

\textbf{Proof.} Fix \(x\in\mathbb R^n_+\) and \(v\in\mathcal S\). Let
\(w(\xi',y_n)=\int e^{-iy'\cdot\xi'}v(y',y_n)\,dy'\), so that
\(\widehat v(\xi',\xi_n)=\int e^{-iy_n\xi_n}w(\xi',y_n)\,dy_n\). By
Fubini, \[
T_av(x)=(2\pi)^{-n}\int e^{ix'\cdot\xi'}J(x,\xi')\,d\xi',\qquad
J(x,\xi')=\int a(x,\xi',x_n\xi_n)\,e^{ix_n\xi_n}\,\widehat v(\xi',\xi_n)\,d\xi_n .
\] Substitute \(\eta_n=x_n\xi_n\), and write
\(e^{ix_n\xi_n}\widehat v(\xi',\xi_n)=\int e^{-i(y_n-x_n)\xi_n}w(\xi',y_n)\,dy_n\).
With \(s=(y_n-x_n)/x_n\) one finds \[
J(x,\xi')=x_n^{-1}\int a(x,\xi',\eta_n)\,\widehat{\phi_{x,\xi'}}(\eta_n)\,d\eta_n,\qquad \phi_{x,\xi'}(s)=x_n\,w\big(\xi',x_n(1+s)\big).
\tag{4.6}
\] (1 \(\Rightarrow\) 2) If \(v=0\) in \(\mathbb R^n_+\), then
\(w(\xi',y_n)=0\) for \(y_n>0\), so \(\phi=\phi_{x,\xi'}\) vanishes for
\(s>-1\). The translates \(\phi_\delta(s)=\phi(s+\delta)\) vanish on
\((-1-\delta,\infty)\), a neighbourhood of \([-1,\infty)\), and
\(\phi_\delta\to\phi\) in \(\mathcal S(\mathbb R)\). Hence
\(\int a\,\widehat\phi\,d\eta_n=\langle\mathcal F_na,\phi\rangle=\lim_{\delta\to0}\langle\mathcal F_na,\phi_\delta\rangle=0\).
So \(J=0\) and \(T_av(x)=0\).

(2 \(\Rightarrow\) 3) is trivial.

(3 \(\Rightarrow\) 1) Take \(v(y)=v_1(y')v_2(y_n)\) with
\(v_1\in C_0^\infty(\mathbb R^{n-1})\) and
\(v_2\in C_0^\infty((-\infty,0))\). Then
\(w=\widehat{v_1}(\xi')v_2(y_n)\) and
\(J(x,\xi')=\widehat{v_1}(\xi')\,j(x,\xi')\) with
\(j(x,\xi')=x_n^{-1}\int a(x,\xi',\eta_n)\widehat{\psi_x}(\eta_n)\,d\eta_n\),
\(\psi_x(s)=x_nv_2(x_n(1+s))\). For fixed \(x\), the continuous
polynomially bounded function \(\xi'\mapsto e^{ix'\cdot\xi'}j(x,\xi')\)
annihilates every \(\widehat{v_1}\), and these are dense in
\(\mathcal S(\mathbb R^{n-1})\); so \(j(x,\cdot)=0\). As \(v_2\) runs
through \(C_0^\infty((-\infty,0))\), \(\psi_x\) runs through all of
\(C_0^\infty((-\infty,-1))\). This proves (4.5) for \(x_n>0\), and
continuity in \(x\) gives it at \(x_n=0\). \(\square\)

\subsubsection{Every symbol is lacunary up to a residual
symbol}\label{every-symbol-is-lacunary-up-to-a-residual-symbol}

\textbf{Lemma 4.4} (Lacunary modification of a symbol). Let
\(\rho\in\mathcal S(\mathbb R)\) with
\(\widehat\rho\in C_0^\infty((-\tfrac12,1))\) and \(\widehat\rho=1\)
near 0. For \(a\in S^m_+\) put \[
a_\rho(x,\xi)=\int a(x,\xi',\xi_n-t)\,\rho(t)\,dt .
\tag{4.7}
\] Then:

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  \(a\mapsto a_\rho\) is continuous \(S^m_+\to S^m_+\), and \(a_\rho\)
  is strongly lacunary, with
  \(\operatorname{supp}\mathcal F_na_\rho(x,\xi',\cdot)\subset\operatorname{supp}\widehat\rho\).
\item
  \(a-a_\rho\in S^{-\infty}_+\), and \(a\mapsto a-a_\rho\) is continuous
  from \(S^m_+\) into every \(S^{m'}_+\).
\item
  If \(a\) is lacunary (strongly lacunary), so is \(a-a_\rho\).
\item
  The kernel of \(T_{a_\rho}\) vanishes on the open set
  \(\{x_n>0,\ y_n/x_n\notin[\tfrac12,2]\}\).
\item
  The natural map
  \(S^m_{\mathrm{la}}/S^{-\infty}_{\mathrm{la}}\to S^m_+/S^{-\infty}_+\)
  is bijective.
\end{enumerate}

\textbf{Proof.} (a) By Peetre's inequality, \[
|(a_\rho)^{(\alpha)}_{(\beta)}(x,\xi)|\leq\int|a^{(\alpha)}_{(\beta)}(x,\xi',\xi_n-t)||\rho(t)|\,dt\leq p(a)(1+x_n)^{-\nu}(1+|\xi|)^{m-|\alpha|}\int(1+|t|)^{|m-|\alpha||}|\rho(t)|\,dt .
\] By the convolution theorem
\(\mathcal F_na_\rho=\widehat\rho\,\mathcal F_na\), whose support lies
in \(\operatorname{supp}\widehat\rho\subset(-\tfrac12,1)\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Since \(\widehat\rho(\tau)=\int e^{-i\tau t}\rho(t)dt\) equals 1 near
  0, \(\int\rho=1\) and \(\int t^j\rho(t)\,dt=0\) for \(j\geq1\). Hence,
  for every \(N\), \[
a_\rho(x,\xi)-a(x,\xi)=\int\Big(a(x,\xi',\xi_n-t)-\sum_{j<N}\partial_{\xi_n}^ja(x,\xi)\frac{(-t)^j}{j!}\Big)\rho(t)\,dt .
\tag{4.8}
\] Where \(|t|<(1+|\xi|)/2\), Taylor's formula bounds the bracket by
  \(|t|^N/N!\) times the supremum of \(|\partial^N_{\xi_n}a|\) on the
  segment from \(\xi\) to \(\xi-te_n\); there \(1+|\xi|\) and the norm
  of the point differ by a factor at most 2, so the bracket is at most
  \(C_Np(a)|t|^N(1+|\xi|)^{m-N}(1+x_n)^{-\nu}\). Where
  \(|t|\geq(1+|\xi|)/2\), each term of the bracket is at most
  \(Cp(a)(1+|t|)^{|m|+N}(1+x_n)^{-\nu}\), and \(1+|\xi|\leq2(1+|t|)\)
  gives
  \((1+|t|)^{|m|+N}\leq2^{N+|m|}(1+|\xi|)^{m-N}(1+|t|)^{2N+2|m|}\).
  Integrating against the rapidly decreasing \(|\rho|\) gives
  \(|a_\rho-a|\leq C_Np(a)(1+|\xi|)^{m-N}(1+x_n)^{-\nu}\) for all
  \(N,\nu\). Derivatives commute with the convolution, so the same
  argument applied to \(a^{(\alpha)}_{(\beta)}\in S^{m-|\alpha|}_+\)
  proves (b), with every seminorm controlled by finitely many seminorms
  of \(a\).
\item
  \(\mathcal F_n(a-a_\rho)=(1-\widehat\rho)\mathcal F_na\) has support
  inside that of \(\mathcal F_na\).
\item
  Fix \(x\) with \(x_n>0\) and let \(v\in\mathcal S\) vanish on the
  closed slab \(\{y:x_n/2\leq y_n\leq2x_n\}\). In (4.6) the function
  \(\phi_{x,\xi'}\) then vanishes on \([-\tfrac12,1]\), which is a
  neighbourhood of the compact set \(\operatorname{supp}\widehat\rho\).
  Hence \(J=0\) and \(T_{a_\rho}v(x)=0\). If \(\psi\in C_0^\infty\) and
  \(v\in C_0^\infty\) have
  \(\operatorname{supp}\psi\times\operatorname{supp}v\) inside the open
  set of (d), this gives \((T_{a_\rho}v,\psi)=0\); such products span a
  dense set of test functions, which proves (d).
\item
  The kernel of the map is
  \(S^m_{\mathrm{la}}\cap S^{-\infty}_+=S^{-\infty}_{\mathrm{la}}\);
  surjectivity is (a)--(b). \(\square\)
\end{enumerate}

By (e), the lacunary condition only restricts the residual part of a
symbol. It has no effect on principal symbols or asymptotic expansions.

\subsection{5. Action on restricted Schwartz
functions}\label{action-on-restricted-schwartz-functions}

\subsubsection{Continuity, commutators and boundary
jets}\label{continuity-commutators-and-boundary-jets}

\textbf{Theorem 5.1} (Action, commutators and boundary jets). Let
\(a\in S^m_{\mathrm{la}}\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  For \(u\in\overline{\mathcal S}(\mathbb R^n_+)\) and any
  \(U\in\mathcal S(\mathbb R^n)\) equal to \(u\) in \(\mathbb R^n_+\),
  the restriction \((T_aU)|_{\mathbb R^n_+}\) depends only on \(u\); we
  call it \(T_au\). It lies in \(\overline{\mathcal S}(\mathbb R^n_+)\),
  and \((a,u)\mapsto T_au\) is a continuous bilinear map
  \(S^m_{\mathrm{la}}\times\overline{\mathcal S}(\mathbb R^n_+)\to\overline{\mathcal S}(\mathbb R^n_+)\).
  More precisely, for every \((\alpha,\beta)\) there are a seminorm
  \(p\) of \(S^m_+\) and a continuous seminorm \(\bar p\) of
  \(\overline{\mathcal S}(\mathbb R^n_+)\), depending only on
  \(\alpha,\beta,m,n\), with
  \(q_{\alpha,\beta}(T_au)\leq p(a)\,\bar p(u)\).
\item
  As operators on \(\overline{\mathcal S}(\mathbb R^n_+)\), for \(j<n\),
  \[
[T_a,D_j]=iT_{\partial_{x_j}a},\qquad [T_a,x_j]=-iT_{\partial_{\xi_j}a},
\] and for the normal direction \[
[T_a,D_n]=iT_{\partial_{x_n}a}+iT_{\partial_{\xi_n}a}D_n,\qquad [T_a,x_n]=-i\,x_n\,T_{\partial_{\xi_n}a}.
\tag{5.1}
\]
\item
  For every integer \(k\geq0\) and
  \(u\in\overline{\mathcal S}(\mathbb R^n_+)\), {\footnotesize
\[
D_n^k(T_au)(x',0)=\sum_{j=0}^k\binom kj\,a_{kj}(x',D')\big(D_n^ju(\cdot,0)\big)(x'),\qquad
a_{kj}(x',\xi')=\sum_{i=0}^j\binom ji\big(D_{x_n}^{k-j}D_{\xi_n}^ia\big)(x',0,\xi',0).
\tag{5.2}
\]
} Here \(a_{kj}\in S^m(\mathbb R^{n-1}\times\mathbb R^{n-1})\), its
  \(i\)-th summand has order \(m-i\), and \(a_{kj}(x',D')\) is the left
  quantization on \(\mathbb R^{n-1}\).
\item
  If \(D_n^ju(\cdot,0)=0\) for \(j<k\), then \(D_n^j(T_au)(\cdot,0)=0\)
  for \(j<k\). In particular \(T_a\) maps
  \(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\) into itself.
\end{enumerate}

\emph{Correction:} The expression \(-i(1+x_n)T_{\partial_{\xi_n}a}\) is
not the commutator \([T_a,x_j]\) with \(j=n\); the correct term is
\(-i\,x_nT_{\partial_{\xi_n}a}\), as in (5.1) and Example 5.2.

\textbf{Proof.} (a) Let \(U\in\mathcal S\) and, for
\(x\in\overline{\mathbb R}{}^n_+\), put
\(W(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}a^\flat(x,\xi)\widehat U(\xi)\,d\xi\),
so \(W=T_aU\) in \(\mathbb R^n_+\). From (4.2), for \(x_n\geq0\), \[
D_{x_j}\big(e^{ix\cdot\xi}a^\flat\big)=e^{ix\cdot\xi}\big(\xi_ja^\flat+(D_{x_j}a)^\flat\big)\ (j<n),\qquad
D_{x_n}\big(e^{ix\cdot\xi}a^\flat\big)=e^{ix\cdot\xi}\big(\xi_na^\flat+(D_{x_n}a)^\flat+\xi_n(D_{\xi_n}a)^\flat\big).
\tag{5.3}
\] By induction,
\(D_x^\beta(e^{ix\cdot\xi}a^\flat)=e^{ix\cdot\xi}\sum_\gamma\xi^\gamma c_\gamma^\flat\),
a finite sum with \(|\gamma|\leq|\beta|\), where each \(c_\gamma\) is a
constant times some \(\partial_x^\mu\partial_{\xi_n}^ia\in S^{m-i}_+\).
Since \(1+|(\xi',x_n\xi_n)|\leq(1+x_n)(1+|\xi|)\), each term is at most
\(|\xi|^{|\gamma|}p(c_\gamma)(1+x_n)^{m_+-\nu}(1+|\xi|)^{m_+}\),
\(m_+=\max(m,0)\), for any \(\nu\). For the weight \(x'^{\alpha'}\) we
integrate by parts in \(\xi'\), using
\(x'^{\alpha'}e^{ix\cdot\xi}=D_{\xi'}^{\alpha'}e^{ix\cdot\xi}\);
\(\xi'\)-derivatives of \(c_\gamma^\flat\) are compressions of
\(\xi'\)-derivatives and obey the same bounds. For the weight
\(x_n^{\alpha_n}\) we take \(\nu\geq\alpha_n+m_+\). Thus \[
\sup_{x\in\overline{\mathbb R}{}^n_+}|x^\alpha D^\beta W(x)|\leq p(a)\,p'(U)
\] with \(p\) a seminorm of \(S^m_+\) and \(p'\) a Schwartz seminorm.
The same bounds justify differentiation under the integral, and the
integrands are continuous up to \(x_n=0\); so
\(W\in C^\infty(\overline{\mathbb R}{}^n_+)\), and
\(W|_{\mathbb R^n_+}\in\overline{\mathcal S}(\mathbb R^n_+)\) by Lemma
3.2(c). By Proposition 4.3, \(W|_{\mathbb R^n_+}\) depends only on
\(U|_{\mathbb R^n_+}\). Taking the infimum over all extensions gives
\(q_{\alpha,\beta}(T_au)\leq p(a)\bar p'(u)\) with the quotient seminorm
\(\bar p'\), and Lemma 3.2(a) turns this into joint continuity.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  For \(U\in\mathcal S\) and \(x_n>0\), differentiation under the
  integral gives
  \(D_jT_aU=\operatorname{Op}(\xi_ja^\flat+D_{x_j}(a^\flat))U\), while
  \(T_aD_jU=\operatorname{Op}(a^\flat\xi_j)U\). Hence
  \([T_a,D_j]=-\operatorname{Op}(D_{x_j}(a^\flat))=i\operatorname{Op}(\partial_{x_j}(a^\flat))\).
  For \(j<n\), \(\partial_{x_j}(a^\flat)=(\partial_{x_j}a)^\flat\). For
  \(j=n\),
  \(\partial_{x_n}(a^\flat)=(\partial_{x_n}a)^\flat+\xi_n(\partial_{\xi_n}a)^\flat\),
  and \(\operatorname{Op}(c^\flat\xi_n)=T_cD_n\). Next,
  \(\widehat{x_jU}=-D_{\xi_j}\widehat U\); integrating by parts in
  \(\xi_j\) gives
  \(T_a(x_jU)=(2\pi)^{-n}\int D_{\xi_j}(e^{ix\cdot\xi}a^\flat)\widehat U\,d\xi=x_jT_aU+\operatorname{Op}(D_{\xi_j}(a^\flat))U\),
  so \([T_a,x_j]=-i\operatorname{Op}(\partial_{\xi_j}(a^\flat))\). For
  \(j<n\) this is \(-iT_{\partial_{\xi_j}a}\). For \(j=n\),
  \(\partial_{\xi_n}(a^\flat)=x_n(\partial_{\xi_n}a)^\flat\), and the
  factor \(x_n\) stands on the left. The symbols on the right are
  lacunary, so the identities pass to
  \(\overline{\mathcal S}(\mathbb R^n_+)\).
\item
  By (5.3),
  \(D^k_{x_n}(e^{ix\cdot\xi}a^\flat)=e^{ix\cdot\xi}(\xi_n+D_{x_n})^ka^\flat\)
  for \(x_n\geq0\). Regard \(a\) as a function of \((x,\xi',\zeta)\),
  with \(\zeta=x_n\xi_n\) in the last slot and \(\xi_n\) as a parameter.
  Then \(D_{x_n}(a^\flat)=[(D_{x_n}+\xi_nD_\zeta)a]^\flat\), and
  \(D_{x_n}\), \(\xi_nD_\zeta\) commute; so
  \(D^\ell_{x_n}(a^\flat)=\sum_i\binom\ell i\xi_n^i(D^{\ell-i}_{x_n}D^i_\zeta a)^\flat\).
  At \(x_n=0\) (where \(\zeta=0\)), \[
(\xi_n+D_{x_n})^ka^\flat\big|_{x_n=0}=\sum_{\ell}\binom k\ell\xi_n^{k-\ell}\sum_{i\leq\ell}\binom\ell i\xi_n^i\big(D^{\ell-i}_{x_n}D^i_{\xi_n}a\big)(x',0,\xi',0).
\] The power \(\xi_n^j\) occurs for \(j=k-\ell+i\), and
  \(\binom k{k-j+i}\binom{k-j+i}i=\binom kj\binom ji\). So
  \(D^k_n(T_aU)(x',0)=\sum_j\binom kj(2\pi)^{-n}\int e^{ix'\cdot\xi'}a_{kj}(x',\xi')\xi_n^j\widehat U(\xi)\,d\xi\),
  and \((2\pi)^{-1}\int\xi_n^j\widehat U(\xi',\xi_n)d\xi_n\) is the
  Fourier transform in \(x'\) of \(D_n^jU(\cdot,0)\). This is (5.2). The
  symbol \(D^i_{\xi_n}D^{k-j}_{x_n}a\) lies in \(S^{m-i}_+\), and its
  restriction to \(x_n=0,\xi_n=0\) lies in
  \(S^{m-i}(\mathbb R^{n-1}\times\mathbb R^{n-1})\).
\item
  This is read off from (5.2). If all jets of \(u\) vanish, those of
  \(T_au\) vanish too; the zero extension of \(T_au\) is then smooth,
  with all weighted derivatives bounded, hence in
  \(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\). \(\square\)
\end{enumerate}

Formula (5.2) is the purpose of the construction: the normal derivatives
of the output at the boundary are obtained by letting pseudodifferential
operators on the boundary act on normal derivatives of the input of the
same or lower order. Proposition 2.1(d) is the case \(k=0\) for
differential operators.

\textbf{Example 5.2} (The factor \(x_n\) in the normal commutator). Take
\(\theta\in C_0^\infty(\mathbb R)\) with \(\theta=1\) near 0, and
\(a(x,\xi)=\theta(x_n)\xi_n\). This symbol lies in
\(S^1_{\mathrm{la}}\), because its normal Fourier transform is supported
at \(t=0\). Here \(T_a=\theta(x_n)x_nD_n\), and
\([T_a,x_n]u=\theta(x_n)x_nD_n(x_nu)-x_n\theta(x_n)x_nD_nu=-i\theta(x_n)x_nu\),
while \(T_{\partial_{\xi_n}a}=\theta(x_n)\). So
\([T_a,x_n]=-i\,x_nT_{\partial_{\xi_n}a}\), as (5.1) says, and there is
no term \(-iT_{\partial_{\xi_n}a}=-i\theta(x_n)\) without the factor
\(x_n\).

\subsubsection{Composition with totally characteristic
derivatives}\label{composition-with-totally-characteristic-derivatives}

Composing \(T_c\) on the right with a totally characteristic
differential operator gives again an operator of the class, with an
exact formula for its symbol.

\textbf{Lemma 5.3} (Composition with totally characteristic
derivatives). For \(c\in S^\mu_{\mathrm{la}}\), on
\(\overline{\mathcal S}(\mathbb R^n_+)\), \[
T_cD_j=T_{\xi_jc}\ (j<n),\qquad T_c\,x_nD_n=T_{\xi_n(1-i\partial_{\xi_n})c},\qquad T_c\,D_nx_n=T_{(\xi_n-i-i\xi_n\partial_{\xi_n})c},
\tag{5.4}
\] and more generally \[
T_c\,x_n^kD_n^kD'^{\beta'}=T_{\xi'^{\beta'}\xi_n^k(1-i\partial_{\xi_n})^kc}.
\tag{5.5}
\] \textbf{Proof.} For \(j<n\),
\(T_cD_j=\operatorname{Op}(c^\flat\xi_j)\) and
\(c^\flat\xi_j=(\xi_jc)^\flat\). By (5.1),
\(T_cx_n=x_nT_{(1-i\partial_{\xi_n})c}\), so
\(T_cx_n^k=x_n^kT_{(1-i\partial_{\xi_n})^kc}\). Moreover
\(x_n^kT_gD_n^k=\operatorname{Op}(x_n^k\xi_n^kg^\flat)=T_{\xi_n^kg}\),
because \(x_n^k\xi_n^kg^\flat=(\xi_n^kg)^\flat\). Together these give
(5.5) and the first two identities in (5.4). The third identity in (5.4)
follows from \(D_nx_n=x_nD_n-i\). \(\square\)

\subsection{6. Kernels near the corner}\label{kernels-near-the-corner}

\subsubsection{Coordinates at the
corner}\label{coordinates-at-the-corner}

Kernels of totally characteristic operators live on \[
Q=\{(x,y)\in\mathbb R^{2n}:x_n\geq0,\ y_n\geq0\},\qquad \partial_2Q=\{(x,y):x_n=y_n=0\},
\] the quarter space and its distinguished boundary. Near
\(\partial_2Q\) we use \[
t=\frac{x_n+y_n}2,\qquad r=\frac{x_n-y_n}t\quad(t>0);\qquad x_n=t\Big(1+\frac r2\Big),\quad y_n=t\Big(1-\frac r2\Big).
\tag{6.1}
\]

\textbf{Proposition 6.1} (Blow-up coordinates). Write
\(\Phi(t,r)=(t(1+r/2),t(1-r/2))\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\Phi\) maps \((0,\infty)\times\mathbb R\) diffeomorphically onto
  \(\{x_n+y_n>0\}\), with \(|\det\Phi'|=t\); hence
  \(dx_n\,dy_n=t\,dt\,dr\).
\item
  \(Q\setminus\{x_n=y_n=0\}\) corresponds to \(t>0\), \(|r|\leq2\). The
  face \(\{x_n=0<y_n\}\) is \(r=-2\), the face \(\{y_n=0<x_n\}\) is
  \(r=2\), the diagonal \(x_n=y_n\) is \(r=0\), and
  \(y_n/x_n=(2-r)/(2+r)\).
\item
  \(\Phi\) extends smoothly to \([0,\infty)\times\mathbb R\) and maps
  the whole line \(t=0\) to the corner: the corner is blown up into the
  \emph{front face} \(t=0\), of which the segment \(|r|\leq2\) lies over
  \(Q\).
\item
  Normal dilations \((x_n,y_n)\mapsto\lambda(x_n,y_n)\) are
  \((t,r)\mapsto(\lambda t,r)\), and the radial field is
  \(x_n\partial_{x_n}+y_n\partial_{y_n}=t\partial_t\).
\item
  On \(Q\), \(|w|/2\leq t\leq|w|\) for \(w=(x_n,y_n)\); \(r\) is
  homogeneous of degree 0, so
  \(|\partial_w^\beta r|\leq C_\beta|w|^{-|\beta|}\) on
  \(\{x_n+y_n>0\}\cap\{|r|\leq3\}\).
\item
  The rescaled normal variable in (4.4) is a function of \(r\) alone:
  \((x_n-y_n)/x_n=2r/(2+r)\).
\end{enumerate}

\textbf{Proof.} The Jacobian matrix of \(\Phi\) has rows
\((1+\tfrac r2,\tfrac t2)\) and \((1-\tfrac r2,-\tfrac t2)\), with
determinant \(-t\); the inverse is (6.1). Parts 2--4 and 6 are direct
substitutions. For part 5, \(x_n+y_n\geq|w|\) when both are nonnegative,
and \(x_n+y_n\leq\sqrt2|w|\). The set
\(\{|r|\leq3\}=\{|x_n-y_n|\leq\tfrac32(x_n+y_n)\}\) is a closed cone
that meets the line \(x_n+y_n=0\) only at the origin; its intersection
with the unit circle is a compact subset of the open set where \(r\) is
smooth. The derivatives of order \(k\) of \(r\) are homogeneous of
degree \(-k\), so they are bounded by \(C_k|w|^{-k}\) on that cone.
\(\square\)

The point of these coordinates is part 6. The kernel formula (4.4)
involves \(x_n^{-1}\) and a function of \((x_n-y_n)/x_n\), and neither
is smooth at the corner. But \(tK\) becomes a smooth function of
\((t,r)\), as Theorem 6.2 shows.

\subsubsection{Residual kernels}\label{residual-kernels}

\textbf{Theorem 6.2} (Residual kernels).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Let \(a\in S^{-\infty}_{\mathrm{la}}\) and
  \(A(x,z)=(2\pi)^{-n}\int e^{iz\cdot\xi}a(x,\xi)\,d\xi\). Then
  \(A\in C^\infty(\overline{\mathbb R}{}^n_+\times\mathbb R^n)\),
  \(A=0\) for \(z_n\geq1\), and for all \(\alpha,\beta,N\) \[
|\partial_x^\alpha\partial_z^\beta A(x,z)|\leq C_{\alpha\beta N}(1+|z|)^{-N}(1+x_n)^{-N},
\tag{6.2}
\] with \(C_{\alpha\beta N}\) bounded by seminorms of \(a\).
  Conversely every
  \(A\in C^\infty(\overline{\mathbb R}{}^n_+\times\mathbb R^n)\)
  satisfying (6.2) and vanishing for \(z_n>1\) comes in this way from
  exactly one \(a\in S^{-\infty}_{\mathrm{la}}\), namely
  \(a(x,\xi)=\int e^{-iz\cdot\xi}A(x,z)\,dz\).
\item
  For \(a\in S^{-\infty}_{\mathrm{la}}\) the kernel of \(T_a\) is the
  locally integrable function \[
K(x,y)=x_n^{-1}A\Big(x,\,x'-y',\,\frac{x_n-y_n}{x_n}\Big)\quad(x_n>0),\qquad K(x,y)=0\quad(x_n<0).
\tag{6.3}
\] For \(x_n>0\) and \(u\in\mathcal S\),
  \(T_au(x)=\int K(x,y)u(y)\,dy\), and
  \(\int|K(x,y)|\,dy=\int|A(x,z)|\,dz\leq C(1+x_n)^{-N}\). Moreover
  \(\operatorname{supp}K\subset Q\) and
  \(K\in C^\infty(\mathbb R^{2n}\setminus\partial_2Q)\).
\item
  The function
  \(F(x',y',t,r)=t\,K(x',t(1+\tfrac r2),y',t(1-\tfrac r2))\), \(t>0\),
  extends to a \(C^\infty\) function on
  \(\{t\geq0\}\times\mathbb R^{n-1}_{x'}\times\mathbb R^{n-1}_{y'}\times\mathbb R_r\),
  which vanishes for \(|r|\geq2\). For all
  \(\alpha,\beta,\tau,\rho,\nu\) and \(r>-2\), \[
|D^\alpha_{x'}D^\beta_{y'}D^\tau_tD^\rho_rF|\leq C(1+|x'-y'|+t)^{-\nu}(2+r)^{\nu},
\tag{6.4}
\] and in particular \[
|D^\alpha_{x'}D^\beta_{y'}D^\tau_tD^\rho_rF|\leq C'(1+|x'-y'|+t)^{-\nu}.
\tag{6.5}
\] On the front face, \[
F(x',y',0,r)=\frac{2}{2+r}\,A\Big(x',0,\,x'-y',\,\frac{2r}{2+r}\Big)\quad(r>-2).
\tag{6.6}
\]
\item
  Conversely, let \(K\in L^1_{\mathrm{loc}}(\mathbb R^{2n})\) with
  \(\operatorname{supp}K\subset Q\), and suppose that the function \(F\)
  of (c) agrees almost everywhere on \(t>0\) with a function in
  \(C^\infty(\{t\geq0\})\) that vanishes for \(|r|\geq2\) and satisfies
  (6.5). Then \(K\) is, almost everywhere, the kernel of \(T_a\) for
  exactly one \(a\in S^{-\infty}_{\mathrm{la}}\).
\end{enumerate}

\textbf{Proof.} (a) For \(a\in S^{-\infty}_+\), integration by parts
gives
\(z^\gamma\partial^\beta_z\partial^\alpha_xA=(2\pi)^{-n}\int e^{iz\cdot\xi}(-D_\xi)^\gamma[(i\xi)^\beta\partial^\alpha_xa]\,d\xi\),
with an integrand bounded by \(C(1+|\xi|)^{-n-1}(1+x_n)^{-N}\). This
proves smoothness and (6.2). For fixed \((x,\xi')\),
\(a(x,\xi',\cdot)\in\mathcal S(\mathbb R)\), so \(\mathcal F_na\) is a
continuous function; by (4.5) it vanishes for \(t<-1\). Since
\(A(x,z)=(2\pi)^{-n}\int e^{iz'\cdot\xi'}(\mathcal F_na)(x,\xi',-z_n)\,d\xi'\),
\(A=0\) for \(z_n>1\), and by continuity for \(z_n\geq1\). Conversely,
if \(A\) satisfies (6.2), then
\(\xi^\gamma\partial_\xi^\beta\partial_x^\alpha a=\int e^{-iz\cdot\xi}D_z^\gamma[(-iz)^\beta\partial_x^\alpha A]\,dz\)
is bounded by \(C(1+x_n)^{-N}\), so \(a\in S^{-\infty}_+\); Fourier
inversion recovers \(A\) from \(a\); and
\(\mathcal F_na(x,\xi',t)=2\pi\int e^{-iz'\cdot\xi'}A(x,z',-t)\,dz'\)
vanishes for \(t<-1\). Uniqueness is Fourier inversion.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\tightlist
\item
  For \(x_n>0\) fixed, \(a^\flat(x,\cdot)\in\mathcal S(\mathbb R^n)\),
  because \(a\) decreases rapidly in \((\xi',x_n\xi_n)\). So
  \(K(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}a^\flat(x,\xi)d\xi\)
  converges absolutely, and the substitution \(\eta_n=x_n\xi_n\) gives
  (6.3). Fubini gives \(T_au(x)=\int K(x,y)u(y)dy\). The change of
  variables \(z=(x'-y',(x_n-y_n)/x_n)\), \(dy=x_n\,dz\), gives
  \(\int|K(x,y)|dy=\int|A(x,z)|dz\). Hence
  \((T_au,v)=\iint K(x,y)u(y)\overline{v(x)}\,dy\,dx\) for
  \(u,v\in\mathcal S\), with absolute convergence; so \(K\), which is
  locally integrable, is the Schwartz kernel. If \(x_n<0\), \(K=0\). If
  \(x_n>0>y_n\), then \((x_n-y_n)/x_n>1\) and \(A=0\). So \(K\) vanishes
  outside \(Q\), up to the null set \(x_n=0\).
\end{enumerate}

Smoothness off \(\partial_2Q\). Near a point with \(x_n>0\), (6.3) is
smooth. Near a point with \(x_n<0\), or with \(x_n=0\) and \(y_n<0\),
\(K=0\). Let \(x^0_n=0<y^0_n\). For \(x_n>0\) small and \(y_n\) near
\(y^0_n\), the normal argument \(z_n=(x_n-y_n)/x_n\) satisfies
\(|z_n|\geq y^0_n/(2x_n)\). By the chain rule, a derivative of order
\(|\gamma|\) of \(K\) is a finite sum of terms
\(x_n^{-k}y_n^l(\partial A)(x,z)\) with \(k\leq1+2|\gamma|\),
\(l\leq|\gamma|\), and
\(|\partial A|\leq C_M(1+|z_n|)^{-M}\leq C_M(2x_n/y^0_n)^M\). So \(K\)
and all its derivatives tend to 0 as \(x_n\to0+\), uniformly near the
point. Since \(K=0\) for \(x_n\leq0\), \(K\) is smooth there.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  For \(t>0\) and \(r>-2\) we have \(x_n=t(1+r/2)>0\),
  \((x_n-y_n)/x_n=2r/(2+r)\) and \(t/x_n=2/(2+r)\). So by (6.3) \[
F=\frac2{2+r}\,A\Big(x',\,t\big(1+\tfrac r2\big),\,x'-y',\,\frac{2r}{2+r}\Big)=:G .
\tag{6.7}
\] The right side is smooth on \(\{t\geq0,\ r>-2\}\), because \(A\) is
  smooth up to \(x_n=0\). It vanishes for \(r\geq2\), where
  \(2r/(2+r)\geq1\). For \(t>0\), \(r<-2\) we have \(x_n<0\) and
  \(F=0\).
\end{enumerate}

Now let \(-2<r<2\) and write \(z_n=2r/(2+r)\), \(z'=x'-y'\),
\(x_n=t(2+r)/2\). Then \[
\frac1{2+r}\leq\frac{1+|z_n|}2,\qquad t=\frac{2x_n}{2+r}\leq x_n(1+|z_n|).
\tag{6.8}
\] The first inequality holds because \(1+|z_n|\geq1\geq2/(2+r)\) when
\(r\geq0\), and \(1+|z_n|=(2-r)/(2+r)\geq2/(2+r)\) when \(r<0\). By the
chain rule, \(D^\alpha_{x'}D^\beta_{y'}D^\tau_tD^\rho_rG\) is a finite
sum of terms
\(c\,t^k(2+r)^{-l}(\partial_x^\mu\partial_z^\kappa A)(x',x_n,z',z_n)\)
with \(0\leq k\leq\rho\) and \(l\leq1+2\rho\): an \(r\)-derivative may
hit \((2+r)^{-l}\), the argument \(t(1+r/2)\) (factor \(t/2\)) or the
argument \(2r/(2+r)\) (factor \(4/(2+r)^2\)); a \(t\)-derivative brings
the factor \((2+r)/2\). By (6.8) and (6.2), each term is at most \[
C\,x_n^k(1+|z_n|)^{k+l_+}(1+|z|)^{-M}(1+x_n)^{-M}.
\] On the other hand
\(1+|x'-y'|+t\leq(1+|z'|)(1+x_n)(1+|z_n|)\leq(1+|z|)^2(1+x_n)\) and
\((2+r)^{-\nu}\leq(1+|z|)^\nu\). Choosing \(M\) large gives (6.4) for
\(-2<r<2\); for \(r\geq2\) the left side vanishes. In particular every
derivative of \(G\) tends to 0 as \(r\downarrow-2\), locally uniformly
(also at \(t=0\)); so \(G\), extended by 0 to \(r\leq-2\), is smooth,
and it equals \(F\). Since \(2+r\leq4\) on the support, (6.4) implies
(6.5). Formula (6.6) is (6.7) at \(t=0\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{3}
\tightlist
\item
  Taylor's formula at \(r=-2\), where \(F\) vanishes to infinite order,
  turns (6.5) into (6.4):
  \(|\partial_r^\rho F(r)|\leq\sup_{[-2,r]}|\partial_r^{\rho+\nu}F|\,(r+2)^\nu/\nu!\).
  Define, for \(x\in\overline{\mathbb R}{}^n_+\), \[
A(x,z)=\frac2{2-z_n}\,F\Big(x',\,x'-z',\,\frac{x_n(2-z_n)}2,\,\frac{2z_n}{2-z_n}\Big)\quad(z_n<2),\qquad A(x,z)=0\quad(z_n>1).
\tag{6.9}
\] On \(1<z_n<2\) both definitions give 0, because then
  \(2z_n/(2-z_n)>2\); so \(A\) is smooth. For \(z_n\leq1\) put
  \(r=2z_n/(2-z_n)\in(-2,2]\) and \(t=x_n(2-z_n)/2\). Then
  \(r+2=4/(2-z_n)\leq8/(1+|z_n|)\) and \(x_n/2\leq t\leq x_n(1+|z_n|)\).
  Every derivative of \(A\) is a finite sum of terms (polynomial in
  \(x_n\)) \(\times\) (smooth function of \(z_n\) growing at most
  polynomially on \(z_n\leq1\)) \(\times\) (a derivative of \(F\) at the
  displayed point). By (6.4) such a term is at most
  \(C(1+x_n)^{k}(1+|z_n|)^{k}(1+|z'|+x_n/2)^{-\nu}(1+|z_n|)^{-\nu}\),
  and choosing \(\nu\) large gives (6.2). By (a), \(A\) comes from a
  unique \(a\in S^{-\infty}_{\mathrm{la}}\). Finally, for \(x_n>0<y_n\),
  inserting \(z=(x'-y',(x_n-y_n)/x_n)\) into (6.9) gives
  \(2/(2-z_n)=x_n/t\), \(x_n(2-z_n)/2=t\) and \(2z_n/(2-z_n)=r\), so the
  kernel (6.3) of \(T_a\) equals \(F/t=K\) almost everywhere; for
  \(y_n<0\) both vanish. \(\square\)
\end{enumerate}

\textbf{Remark 6.3} (Singular kernels of order \(-\infty\)). For an
ordinary pseudodifferential operator of order \(-\infty\) the kernel is
smooth. Here, by (c), \(K=F/t\) near the corner, and the leading part
\(F(x',y',0,r)/t\) is homogeneous of degree \(-1\) in \((x_n,y_n)\). It
is not smooth, and not even bounded, unless \(F\) vanishes on the front
face. This singularity is what makes residual operators fail to improve
regularity in Section 12.

\subsubsection{A kernel bound at finite negative
order}\label{a-kernel-bound-at-finite-negative-order}

\textbf{Proposition 6.4} (A kernel bound). Let
\(a\in S^{-n-2}_{\mathrm{la}}\). Then the kernel of \(T_a\) is a
function, and \[
|K_a(x,y)|\leq C\,(1+|x'-y'|)^{-n}\,\frac{x_ny_n}{(x_n+y_n)^3}\quad(x_n,y_n>0),\qquad K_a=0\ \text{elsewhere},
\tag{6.10}
\] with \(C\) bounded by a seminorm of \(a\). Consequently
\(\sup_x\int|K_a(x,y)|dy\) and \(\sup_y\int|K_a(x,y)|dx\) are at most
\(\tfrac12C\int_{\mathbb R^{n-1}}(1+|z'|)^{-n}dz'\).

\textbf{Proof.} For \(x_n>0\),
\(|a^\flat(x,\xi)|\leq C(1+|(\xi',x_n\xi_n)|)^{-n-2}\) is integrable in
\(\xi\), so \(K_a(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}a^\flat d\xi\)
converges absolutely, \(T_au(x)=\int K_a(x,y)u(y)dy\), and (6.3) holds
with the bounded continuous function
\(A(x,z)=(2\pi)^{-n}\int e^{iz\cdot\xi}a(x,\xi)d\xi\). As in Theorem
6.2(a), \(\mathcal F_na\) is now a continuous function, so \(A=0\) for
\(z_n\geq1\). Integration by parts gives \(|z^\gamma A|\leq C\) for
\(|\gamma|\leq n+2\), because
\(|\partial_\xi^\gamma a|\leq C(1+|\xi|)^{-n-2-|\gamma|}\) is
integrable; so \((1+|z'|)^n(1+|z_n|)^2|A|\leq C\). Likewise
\(\partial_{z_n}A=(2\pi)^{-n}\int e^{iz\cdot\xi}i\xi_na\,d\xi\) and
\(|z'^\gamma\partial_{z_n}A|\leq C\) for \(|\gamma|\leq n\). Since
\(A(x,z',1)=0\), the mean value theorem gives
\(|A(x,z)|\leq C(1+|z'|)^{-n}|1-z_n|\) for \(z_n\leq1\).

If \(0<x_n\leq y_n\), then \(z_n=(x_n-y_n)/x_n\leq0\),
\(1+|z_n|=y_n/x_n\), and
\(|K_a|\leq x_n^{-1}C(1+|z'|)^{-n}(x_n/y_n)^2=C(1+|z'|)^{-n}x_n/y_n^2\leq8C(1+|z'|)^{-n}x_ny_n/(x_n+y_n)^3\),
because \(x_n+y_n\leq2y_n\). If \(0<y_n<x_n\), then \(|1-z_n|=y_n/x_n\)
and
\(|K_a|\leq C(1+|z'|)^{-n}y_n/x_n^2\leq8C(1+|z'|)^{-n}x_ny_n/(x_n+y_n)^3\).
For the marginals, \(\int_{\mathbb R^{n-1}}(1+|z'|)^{-n}dz'<\infty\) and
\(\int_0^\infty x_ny_n(x_n+y_n)^{-3}dy_n=\int_0^\infty s(1+s)^{-3}ds=\tfrac12\);
the bound is symmetric in \(x_n,y_n\). \(\square\)

The two cases correspond to the two sides of the diagonal: for
\(y_n\geq x_n\) the decay of \(A\) in \(z_n\) is used, for \(y_n<x_n\)
the vanishing of \(A\) at \(z_n=1\), that is, lacunarity.

\subsubsection{Examples of residual
kernels}\label{examples-of-residual-kernels}

\textbf{Example 6.5} (Without lacunarity the operator sees below the
boundary). Let \(0\leq h\in C_0^\infty(\mathbb R^n)\) be supported near
\((0,\tfrac32)\), with \(h(0,\tfrac32)>0\), and
\(a=e^{-x_n}\widehat h(\xi)\in S^{-\infty}_+\). Then \(A=e^{-x_n}h\)
does not vanish on \(z_n>1\), so \(a\) is not lacunary. For
\(0\leq u\in C_0^\infty(\mathbb R^n_-)\) supported near
\((0,-\tfrac12)\), with \(u(0,-\tfrac12)>0\), formula (6.3), whose
derivation for \(x_n>0\) in Theorem 6.2(b) does not use lacunarity,
gives \(T_au(0,1)=e^{-1}\int h(-y',1-y_n)u(y)\,dy>0\), since \(1-y_n\)
is near \(\tfrac32\) there. So \(T_au\neq0\) in \(\mathbb R^n_+\)
although \(u=0\) in \(\mathbb R^n_+\): lacunarity cannot be dropped from
Theorem 5.1(a), in accordance with Proposition 4.3.

\textbf{Example 6.6} (A residual kernel in one dimension). Let \(n=1\),
\(0\neq h\in C_0^\infty((-\tfrac12,\tfrac12))\) and
\(a(x,\xi)=e^{-x}\widehat h(\xi)\). Then
\(K(x,y)=e^{-x}x^{-1}h((x-y)/x)\) for \(x>0\), and \[
T_au(x)=e^{-x}\int h(1-s)\,u(xs)\,ds,\qquad F(t,r)=\frac{2e^{-t(1+r/2)}}{2+r}\,h\Big(\frac{2r}{2+r}\Big).
\] The kernel vanishes unless \(\tfrac12<y/x<\tfrac32\), and \(F\)
vanishes unless \(-\tfrac25<r<\tfrac23\). Along the diagonal
\(K(x,x)=e^{-x}h(0)/x\), which is unbounded if \(h(0)\neq0\): a symbol
of order \(-\infty\) with an unbounded kernel. \(T_a\) is bounded on
\(L^2(0,\infty)\) by Proposition 6.4 and gains no derivative by Theorem
12.1. The model \(T_0u(x)=\int h(1-s)u(xs)ds\) on the front face (so
\(T_a=e^{-x}T_0\)) is not in the class, since its symbol does not decay
in \(x\). It commutes with the unitary dilations
\(u\mapsto\lambda^{1/2}u(\lambda\,\cdot)\) of \(L^2(0,\infty)\), and
Minkowski's inequality gives
\(\|T_0u\|\leq\int|h(1-s)|s^{-1/2}ds\,\|u\|\), since
\(\|u(\cdot\,s)\|=s^{-1/2}\|u\|\).

\textbf{Example 6.7} (The resolved kernel near \(r=-2\)). In Example
6.6, \(F\) vanishes identically near \(r=-2\) because \(h\) has compact
support. For a lacunary but not strongly lacunary symbol, take \(n=1\)
and \(A(x,z)=e^{-x}g(z)\) with \(g\in\mathcal S(\mathbb R)\) vanishing
for \(z\geq1\) but not near \(-\infty\), for instance
\(g(z)=e^{-1/(1-z)}e^{-z^2}\) for \(z<1\) and \(g(z)=0\) for \(z\geq1\).
Then \(F(t,r)=\tfrac2{2+r}e^{-t(1+r/2)}g(\tfrac{2r}{2+r})\), and as
\(r\downarrow-2\) the argument \(2r/(2+r)\to-\infty\), where \(g\)
decreases rapidly; this is the flatness at \(r=-2\) used in (6.4). At
\(r=2\) the argument tends to 1, where \(g\) vanishes to infinite order.

\subsubsection{Residual kernels are polyhomogeneous conormal
distributions}\label{residual-kernels-are-polyhomogeneous-conormal-distributions}

Smoothness of the resolved kernel \(F\) has an invariant meaning: it
says precisely that \(K\) is a polyhomogeneous conormal distribution of
order \(-n/2\) with respect to \(\partial_2Q\). We prove this now.

\textbf{The class.} Conormal distributions \(I^\mu(X,Y)\) are defined by
tangential regularity. A compactly supported \(u\in I^\mu\) has, in
coordinates, the normal form
\(u=\int e^{i\langle t,\tau\rangle}b(z,\tau)\,d\tau\) with
\(b\in S^{\mu+(N-2k)/4}\), where \(N=\dim X\), \(k\) is the codimension,
and \(b\) is \((2\pi)^{-k}\) times the Fourier transform of \(u\) in the
normal variables; conversely every such \(u\) is conormal. (All of this
is in the background list of Section 1.) The polyhomogeneous class
\(I^\mu_{\mathrm{phg}}\) requires in addition that these amplitudes be
polyhomogeneous with step one. Step one is needed here, because a term
of degree \(-\tfrac32\) in \(\tau\) would put a factor \(t^{1/2}\) into
\(F\). For \(X=\mathbb R^{2n}\), \(Y=\partial_2Q\) we have \(N=2n\),
\(k=2\), normal variables \(w=(x_n,y_n)\), tangential variables
\(z=(x',y')\), and \(\mu=-n/2\) gives amplitude degree \(-1\). So
\(K\in I^{-n/2}_{\mathrm{phg}}(\mathbb R^{2n},\partial_2Q)\) means:
\(K\) is smooth off \(\partial_2Q\), and for all
\(\phi\in C_0^\infty(\mathbb R^{2n-2}_z)\),
\(\psi\in C_0^\infty(\mathbb R^2_w)\), \[
(2\pi)^{-2}\widehat{\phi\psi K}(z,\tau)\sim\sum_{j\geq0}b_j(z,\tau),\qquad b_j\ \text{homogeneous of degree }-1-j\text{ in }\tau\text{ for }|\tau|\geq1,
\tag{6.11}
\] in the sense of asymptotic sums of symbols (Section 1); the Fourier
transform is taken in \(w\).

\textbf{Proposition 6.8} (Residual kernels are conormal). Let
\(K\in L^1_{\mathrm{loc}}(\mathbb R^{2n})\) with
\(\operatorname{supp}K\subset Q\), and let
\(F(z,t,r)=tK(x',t(1+r/2),y',t(1-r/2))\) for \(t>0\). Then
\(K\in I^{-n/2}_{\mathrm{phg}}(\mathbb R^{2n},\partial_2Q)\) if and only
if \(F\) agrees almost everywhere with a function in
\(C^\infty(\{t\geq0\}\times\mathbb R_r\times\mathbb R^{2n-2}_z)\); such
a function vanishes for \(|r|\geq2\). In particular
\(K_a\in I^{-n/2}_{\mathrm{phg}}(\mathbb R^{2n},\partial_2Q)\) for every
\(a\in S^{-\infty}_{\mathrm{la}}\).

The global decay (6.5) is not a conormal property. So Theorem 6.2 and
Proposition 6.8 together say: residual kernels are exactly the kernels
supported in \(Q\), polyhomogeneous conormal of order \(-n/2\) at
\(\partial_2Q\), with the uniform decay (6.5).

We need four lemmas. In them \(z\) ranges over \(\mathbb R^{m}\), all
functions have compact \(z\)-support, and every estimate holds with
\(z\)-derivatives, uniformly in \(z\).

\textbf{Lemma 6.9} (Homogeneous pieces). Let
\(h\in C^\infty(\mathbb R^m\times\mathbb R_r)\) vanish for \(|r|\geq2\),
let \(d>-2\), and let \(k(z,w)=t^dh(z,r)\) on \(Q\setminus\{0\}\),
\(k=0\) off \(Q\). Then \(k\) is smooth off \(w=0\), homogeneous of
degree \(d\) in \(w\), and locally integrable. If
\(\psi\in C_0^\infty(\mathbb R^2)\) equals 1 near 0, then
\(\widehat{\psi k}=g+e\), where \(g\) is smooth on
\(\mathbb R^m\times(\mathbb R^2\setminus0)\) and homogeneous of degree
\(-2-d\) in \(\tau\), and \(e\) is smooth with all derivatives
\(O(|\tau|^{-N})\) for \(|\tau|\geq1\). In particular
\(\widehat{\psi k}\in S^{-2-d}\).

\textbf{Proof.} \(h\) vanishes to infinite order at \(r=\pm2\), so \(k\)
is smooth across the faces of \(Q\); it is smooth elsewhere off \(w=0\)
by Proposition 6.1. Also \(|k|\leq C|w|^d\), so \(k\) is locally
integrable and tempered, and its Fourier transform \(\widehat k\) is
homogeneous of degree \(-2-d\) (compare \(k(\lambda\cdot)=\lambda^dk\)
with
\(\widehat{k(\lambda\cdot)}=\lambda^{-2}\widehat k(\cdot/\lambda)\)).
Write \(\widehat k=\widehat{\psi k}+\widehat f\), \(f=(1-\psi)k\). The
first term is smooth. The function \(f\) is smooth, with
\(|\partial^\beta_wf|\leq C_\beta|w|^{d-|\beta|}\) for \(|w|\geq1\). If
\(|\beta|>d+2+|\gamma|\), then \(D^\beta_w(w^\gamma f)\) is integrable,
so \(\tau^\beta\partial_\tau^\gamma\widehat f\) is a bounded continuous
function. Hence \(\widehat f\) is smooth on \(\tau\neq0\) and all its
derivatives are \(O(|\tau|^{-N})\) for \(|\tau|\geq1\). So
\(g=\widehat k|_{\tau\neq0}\) is smooth and homogeneous, and
\(e=-\widehat f\) on \(\tau\neq0\) (with \(\widehat{\psi k}=g+e\)
there). The symbol estimates follow from homogeneity on \(|\tau|\geq1\)
and smoothness on \(|\tau|\leq1\). \(\square\)

\textbf{Lemma 6.10} (Remainders). Let
\(R\in C^\infty(\{t\geq0\}\times\mathbb R_r\times\mathbb R^m)\) vanish
for \(|r|\geq2\), let \(J\geq1\), and let \(E=t^{J-1}R(z,t,r)\) on
\(Q\setminus0\), \(E=0\) off \(Q\). Then
\(\widehat{\psi E}\in S^{-J-1}(\mathbb R^m\times\mathbb R^2)\).

\textbf{Proof.} By Proposition 6.1(5), each \(w\)-derivative of \(t\) or
\(r\) costs at most \(C|w|^{-1}\), and \(t\) is comparable to \(|w|\) on
\(Q\). So \(f=\psi E\) satisfies
\(|\partial_w^\beta f|\leq C_\beta|w|^{J-1-|\beta|}\), and
\(g_\gamma=w^\gamma f\) satisfies
\(|\partial^\beta g_\gamma|\leq C|w|^{\nu-|\beta|}\) with
\(\nu=J-1+|\gamma|\geq0\). Fix \(|\tau|\geq1\) and
\(\chi_0\in C_0^\infty(\{|w|<2\})\) equal to 1 on \(|w|\leq1\). The
Fourier transform of \(\chi_0(|\tau|w)g_\gamma\) is at most
\(\int_{|w|\leq2/|\tau|}C|w|^\nu dw\leq C'|\tau|^{-\nu-2}\). For the
rest,
\(e^{-iw\cdot\tau}=(i|\tau|^{-2}\tau\cdot\nabla_w)e^{-iw\cdot\tau}\);
integrating by parts \(L>\nu+2\) times, and noting that derivatives of
\(\chi_0(|\tau|w)\) are \(O(|w|^{-k})\) where they do not vanish, gives
the bound
\(C|\tau|^{-L}\int_{1/|\tau|\leq|w|\leq R}|w|^{\nu-L}dw\leq C'|\tau|^{-\nu-2}\).
Since \(\partial_\tau^\gamma\widehat f=\widehat{(-iw)^\gamma f}\), we
get
\(|\partial^\gamma_\tau\widehat f(\tau)|\leq C|\tau|^{-J-1-|\gamma|}\).
\(\square\)

\textbf{Lemma 6.11} (Inverse transforms of homogeneous terms). Let
\(j\geq0\), let \(b^{\mathrm{hom}}\) be smooth on
\(\mathbb R^m\times(\mathbb R^2\setminus0)\) and homogeneous of degree
\(-1-j\) in \(\tau\), let \(\chi\in C_0^\infty(\mathbb R^2)\) equal 1
near 0, and put \(b=(1-\chi)b^{\mathrm{hom}}\) and
\(k(z,w)=\int e^{i\langle w,\tau\rangle}b(z,\tau)\,d\tau\). Then \(k\)
is smooth off \(w=0\), rapidly decreasing with all derivatives as
\(|w|\to\infty\), locally integrable, and on \(0<|w|<1\) \[
k=h-P\log|w|+s,
\tag{6.12}
\] where \(h\) is smooth off \(w=0\) and homogeneous of degree \(j-1\),
\(P\) is a homogeneous polynomial of degree \(j-1\) in \(w\) with
coefficients smooth in \(z\) (\(P=0\) when \(j=0\)), and \(s\) is smooth
on \(\{|w|<1\}\).

\textbf{Proof.} \(b\in S^{-1-j}\). For \(|\gamma|\) large,
\(w^\gamma\partial^\beta_wk\) is the absolutely convergent integral of
\(e^{i\langle w,\tau\rangle}\) against a constant times
\(D^\gamma_\tau(\tau^\beta b)\); this gives smoothness off 0 and rapid
decay. Put
\(\theta=(\tau\cdot\partial_\tau+1+j)b=-(\tau\cdot\partial_\tau\chi)\,b^{\mathrm{hom}}\),
which is smooth with compact support in \(\mathbb R^2\setminus0\)
(Euler's relation kills \(b^{\mathrm{hom}}\)), and
\(\Theta=\int e^{i\langle w,\tau\rangle}\theta\,d\tau\in\mathcal S(\mathbb R^2)\).
Since
\(\int e^{i\langle w,\tau\rangle}\tau\cdot\partial_\tau f\,d\tau=-(2+w\cdot\partial_w)\int e^{i\langle w,\tau\rangle}f\,d\tau\)
for tempered \(f\), \[
\big(w\cdot\partial_w-(j-1)\big)k=-\Theta .
\tag{6.13}
\] On a ray \(w=\sigma\omega\), \(|\omega|=1\), this says
\(\frac d{d\sigma}[\sigma^{1-j}k(\sigma\omega)]=-\sigma^{-j}\Theta(\sigma\omega)\).
Since \(\sigma^{1-j}k(\sigma\omega)\to0\) as \(\sigma\to\infty\),
\(k(\sigma\omega)=\sigma^{j-1}\int_\sigma^\infty s^{-j}\Theta(s\omega)\,ds\).
Write \(\Theta=T+\sum_{|\alpha|=j}w^\alpha r_\alpha\), where \(T\) is
the Taylor polynomial of \(\Theta\) of degree \(j-1\) at 0 and the
\(r_\alpha\) are smooth. For \(\sigma<1\) split
\(\int_\sigma^\infty=\int_1^\infty+\int_\sigma^1\):

\begin{itemize}
\tightlist
\item
  \(\sigma^{j-1}\int_1^\infty s^{-j}\Theta(s\omega)ds\) is homogeneous
  of degree \(j-1\) and smooth off 0.
\item
  For the degree-\(\ell\) part \(T_\ell\) of \(T\), \(\ell<j-1\):
  \(\sigma^{j-1}\int_\sigma^1s^{\ell-j}T_\ell(\omega)ds=(T_\ell(w)-\sigma^{j-1}T_\ell(\omega))/(j-1-\ell)\),
  a polynomial minus a homogeneous function of degree \(j-1\). For
  \(\ell=j-1\):
  \(\sigma^{j-1}T_{j-1}(\omega)\int_\sigma^1s^{-1}ds=-T_{j-1}(w)\log|w|\).
  So \(P=T_{j-1}\).
\item
  \(\sigma^{j-1}\int_\sigma^1s^{-j}\sum_\alpha(s\omega)^\alpha r_\alpha(s\omega)ds=\sum_\alpha\sigma^{j-1}\omega^\alpha\big(\int_0^1-\int_0^\sigma\big)r_\alpha(s\omega)ds\).
  The first part is homogeneous of degree \(j-1\). In the second,
  \(\sigma^{j-1}\omega^\alpha=\sigma^{-1}w^\alpha\) and
  \(\sigma^{-1}\int_0^\sigma r_\alpha(s\omega)ds=\int_0^1r_\alpha(uw)du\),
  which is smooth in \(w\).
\end{itemize}

Collecting terms gives (6.12). Local integrability follows from (6.12),
since \(j-1>-2\). \(\square\)

\textbf{Lemma 6.12} (Uniqueness of expansions). If
\(\sum_{i=-1}^L(\alpha_i+\beta_i\log\lambda)\lambda^i=o(\lambda^L)\) as
\(\lambda\to0+\), then all \(\alpha_i\) and \(\beta_i\) vanish.

\textbf{Proof.} Multiply by \(\lambda\):
\(\alpha_{-1}+\beta_{-1}\log\lambda\) tends to a finite limit (namely
0), so \(\beta_{-1}=0\) and then \(\alpha_{-1}=0\). Repeat with the next
power. \(\square\)

\textbf{Proof of Proposition 6.8.} (\(\Leftarrow\)) Off \(\partial_2Q\),
\(K=F/t\) is smooth in the interior of \(Q\), smooth across its faces
because \(F\) is flat at \(r=\pm2\), and zero outside \(Q\). Near
\(\partial_2Q\), fix cutoffs \(\phi(z)\), \(\psi(w)\). Taylor's formula
in \(t\) gives \(F=\sum_{j<J}t^jF_j(z,r)+t^JR_J(z,t,r)\), with \(F_j\)
and \(R_J\) smooth and vanishing for \(|r|\geq2\). Hence
\(\phi\psi K=\sum_{j<J}\phi\psi\,t^{j-1}F_j+\phi\psi\,t^{J-1}R_J\). By
Lemma 6.9, \((2\pi)^{-2}\widehat{\phi\psi t^{j-1}F_j}\) equals a
function homogeneous of degree \(-1-j\) for \(|\tau|\geq1\), up to
\(S^{-\infty}\); by Lemma 6.10 the last term has transform in
\(S^{-J-1}\). As \(J\) is arbitrary, (6.11) holds.

(\(\Rightarrow\)) Off \(\partial_2Q\), \(K\) is smooth, so \(F\) is
smooth on \(t>0\), and \(F=0\) for \(|r|>2\) because
\(\operatorname{supp}K\subset Q\). Fix \(z_0\) and choose \(\phi=1\)
near \(z_0\) and \(\psi=1\) on \(|w|\leq2\delta\). Let
\(b=(2\pi)^{-2}\widehat{\phi\psi K}\sim\sum b_j\), with
\(b_j=(1-\chi)b_j^{\mathrm{hom}}\). Let \(k_j\) be the inverse
transforms of Lemma 6.11 and
\(\rho_J=\int e^{i\langle w,\tau\rangle}(b-\sum_{j<J}b_j)d\tau\). Since
\(b-\sum_{j<J}b_j\in S^{-1-J}\) in two variables, \(\rho_J\in C^{J-2}\).
So for \(z\) near \(z_0\) and \(0<|w|<\delta\), \[
K=\sum_{j<J}\big(h_j-P_j\log|w|\big)+S_J,\qquad S_J=\rho_J+\sum_{j<J}s_j\in C^{J-2}.
\] Let \(U=\mathbb R^2\setminus Q\), an open cone on which \(K=0\). For
\(w\in U\), \(|w|<\delta\), and \(0<\lambda\leq1\) we have
\(K(z,\lambda w)=0\). Insert homogeneity,
\(\log|\lambda w|=\log\lambda+\log|w|\), and Taylor's formula
\(S_J(\lambda w)=\sum_{i\leq J-3}\lambda^iS_{J,i}(w)+o(\lambda^{J-3})\)
with homogeneous polynomials \(S_{J,i}\) of degree \(i\). The term
\(j=J-1\) is \(O(\lambda^{J-2}|\log\lambda|)=o(\lambda^{J-3})\). Lemma
6.12 gives, for \(j\leq J-2\): \(P_j=0\) on \(U\), hence \(P_j\equiv0\);
and \(h_j=-S_{J,j-1}\) on \(U\) (with \(S_{J,-1}=0\)). Thus
\(\tilde h_j=h_j+S_{J,j-1}\) is homogeneous of degree \(j-1\), smooth
off 0, and supported in \(Q\), and \[
K=\sum_{j\leq J-2}\tilde h_j+R_J,\qquad R_J=\Big(S_J-\sum_{i\leq J-3}S_{J,i}\Big)+\big(h_{J-1}-P_{J-1}\log|w|\big).
\] All derivatives of \(R_J\) of order \(\leq J-3\) are continuous and
tend to 0 at \(w=0\), so \(R_J\in C^{J-3}\). Now
\(t\tilde h_j(\Phi(t,r))=t^j\tilde F_j(z,r)\) with
\(\tilde F_j(z,r)=\tilde h_j(z,1+r/2,1-r/2)\), which is smooth and
vanishes for \(|r|\geq2\). Hence, for small \(t\) and \(|r|\leq3\),
\(F=\sum_{j\leq J-2}t^j\tilde F_j+t\,R_J\circ\Phi\) is \(C^{J-3}\), and
\(F=0\) for \(|r|\geq2\). Since \(J\) is arbitrary, \(F\) is smooth. The
last assertion of the proposition follows from Theorem 6.2(c).
\(\square\)

\subsection{7. Adjoints}\label{adjoints}

The adjoint of \(T_a\) with respect to \((u,v)=\int u\overline v\) is
again an operator of the class. We first compute it for strongly
lacunary residual symbols, where the transposed kernel can be read off
from Theorem 6.2. Then we extend the formula by an adjoint transform
defined on all of \(S^m_+\).

\subsubsection{The adjoint of a strongly lacunary residual
operator}\label{the-adjoint-of-a-strongly-lacunary-residual-operator}

\textbf{Proposition 7.1} (A residual adjoint formula). Let
\(a\in S^{-\infty}_{\mathrm{la}}\) be strongly lacunary, and let
\(\chi\in C_0^\infty((0,\infty))\) equal 1 on \((\tfrac12,2)\). There is
exactly one \(b\in S^{-\infty}_{\mathrm{la}}\) with \[
(T_au,v)=(u,T_bv)\qquad(u,v\in\mathcal S(\mathbb R^n)),
\tag{7.1}
\] and for \(x_n>0\), with the inner integral taken first, \[
b(x,\xi)=(2\pi)^{-n}\int\!\Big(\int e^{-i\langle y,\eta\rangle}\,\overline a\big(x'-y',\,x_n(1-y_n),\,\xi'-\eta',\,(1-y_n)(\xi_n-\eta_n)\big)\chi(1-y_n)\,d\eta\Big)dy
\tag{7.2}
\] \[
=\Big[e^{i\langle D_y,D_\eta\rangle}\big(\overline a(y',x_ny_n,\eta',y_n\eta_n)\chi(y_n)\big)\Big]_{y=(x',1),\ \eta=\xi}.
\tag{7.3}
\]

\textbf{Proof.} \emph{Existence and uniqueness.} The transposed kernel
\(K^*(x,y)=\overline{K_a(y,x)}\) is locally integrable and supported in
\(Q\), and its resolved form is
\(F^*(x',y',t,r)=\overline{F(y',x',t,-r)}\), which has all the
properties in Theorem 6.2(c). By Theorem 6.2(d), \(K^*=K_b\) for a
unique \(b\in S^{-\infty}_{\mathrm{la}}\), and Fubini's theorem,
justified by the bounds of Theorem 6.2(b), gives (7.1). An operator
determines its symbol (see quantization and kernels in Section 1), so
\(b\) is unique.

\emph{The formula.} By the inversion formula for kernels in Section 1,
for \(x_n>0\),
\(b^\flat(x,\xi)=\int e^{-iz\cdot\xi}\,\overline{K_a(x-z,x)}\,dz\), an
absolutely convergent integral. Strong lacunarity says that \(A\)
vanishes unless \(z_n\in[-1,\tfrac12]\); in \(K_a(y,x)\) the normal
argument is \((y_n-x_n)/y_n\), so \(K_a(y,x)=0\) unless
\(x_n/y_n\in[\tfrac12,2]\). So we may insert \(\chi((x_n-z_n)/x_n)\),
since it equals 1 almost everywhere on the support. Writing
\(\overline{K_a(x-z,x)}=(2\pi)^{-n}\int e^{i\langle z,\eta\rangle}\overline{a^\flat(x-z,\eta)}\,d\eta\)
and substituting \(\eta\mapsto\xi-\eta\), we get \[
b^\flat(x,\xi)=(2\pi)^{-n}\int\!\Big(\int e^{-i\langle z,\eta\rangle}\overline a\big(x-z,\xi'-\eta',(x_n-z_n)(\xi_n-\eta_n)\big)\chi\Big(\frac{x_n-z_n}{x_n}\Big)d\eta\Big)dz .
\] Now \(b(x,\xi)=b^\flat(x,\xi',\xi_n/x_n)\). Substitute
\(z_n=x_ny_n\), \(\eta_n\mapsto\eta_n/x_n\), \(z'=y'\): then
\((x_n-z_n)(\xi_n/x_n-\eta_n/x_n)=(1-y_n)(\xi_n-\eta_n)\), \(z_n\eta_n\)
becomes \(y_n\eta_n\), and \(dz_n\,d\eta_n=dy_n\,d\eta_n\). This is
(7.2). Finally, for a function \(c(y,\eta)\) that is a residual symbol,
\(e^{i\langle D_y,D_\eta\rangle}c(y,\eta)=(2\pi)^{-n}\iint e^{-i\langle w,\theta\rangle}c(y-w,\eta-\theta)\,d\theta\,dw\)
(test with \(c=e^{i(p\cdot y+q\cdot\eta)}\), for which both sides equal
\(e^{ip\cdot q}c\)). With
\(c(y,\eta)=\overline a(y',x_ny_n,\eta',y_n\eta_n)\chi(y_n)\), a
residual symbol for fixed \(x_n>0\), and \((y,\eta)=((x',1),\xi)\), this
is (7.2). \(\square\)

\subsubsection{The adjoint transform}\label{the-adjoint-transform}

\textbf{Lemma 7.2} (The adjoint transform). Let
\(\chi\in C_0^\infty((0,\infty))\) equal 1 near 1, with
\(\operatorname{supp}\chi\subset(M^{-1},M)\), \(M>1\). For
\(a\in S^m_+\) put \[
L_\chi a(x,\xi)=\Big[e^{i\langle D_y,D_\eta\rangle}c_{x_n}\Big]\big((x',1),\xi\big),\qquad c_{x_n}(y,\eta)=\overline a(y',x_ny_n,\eta',y_n\eta_n)\chi(y_n).
\tag{7.4}
\]

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  \(L_\chi\) is a continuous conjugate-linear map
  \(S^m_+\to S^m_{\mathrm{la}}\).
\item
  In \(S^m_+\), \[
L_\chi a\sim\sum_{j\geq0}\frac1{j!}\langle D_y,iD_\eta\rangle^j\,\overline a(y',x_ny_n,\eta',y_n\eta_n)\Big|_{y=(x',1),\,\eta=\xi},
\tag{7.5}
\] the \(j\)-th term lying in \(S^{m-j}_+\), with the remainder after
  \(N\) terms in \(S^{m-N}_+\) and controlled by finitely many seminorms
  of \(a\). For \(x_n>0\) the \(j\)-th term equals
  \(\frac1{j!}\langle D_y,iD_\eta\rangle^j\overline a(y,\eta',y_n\eta_n)\)
  at \(y=x\), \(\eta'=\xi'\), \(\eta_n=\xi_n/x_n\).
\item
  If \(a\in S^{-\infty}_+\), then
  \(\operatorname{supp}\mathcal F_n(L_\chi a)(x,\xi',\cdot)\subset[M^{-1}-1,\,M-1]\),
  and \(T_{L_\chi a}\) has the kernel
  \(\chi(y_n/x_n)\overline{K_a(y,x)}\) for \(x_n,y_n>0\) (and 0
  elsewhere).
\end{enumerate}

\textbf{Proof.} (a), (b) On \(\operatorname{supp}\chi\) we have
\(M^{-1}\leq y_n\leq M\), hence
\((1+|\eta|)/M\leq1+|(\eta',y_n\eta_n)|\leq M(1+|\eta|)\) and
\(1+x_ny_n\geq(1+x_n)/M\). A \(y_n\)-derivative of
\(\overline a(y',x_ny_n,\eta',y_n\eta_n)\) produces
\(x_n\partial_{x_n}\overline a\) (the factor \(x_n\) is absorbed by the
decay in \(x_ny_n\)) or \(\eta_n\partial_{\xi_n}\overline a\) (the
factor \(\eta_n\) is paid for by the lower order); an
\(\eta_n\)-derivative produces \(y_n\partial_{\xi_n}\overline a\). Hence
\((1+x_n)^\nu c_{x_n}\) is bounded in the classical class
\(S^m(\mathbb R^n_y\times\mathbb R^n_\eta)\), uniformly in \(x_n\geq0\),
for every \(\nu\), and so is every \(\partial_{x_n}^kc_{x_n}\) (it has
the same form, with \(y_n^k\partial_{x_n}^k\overline a\)).

By the facts on the Gauss transform in Section 1,
\(e^{i\langle D_y,D_\eta\rangle}\) is continuous on \(S^m\), with the
expansion
\(\sum_{|\alpha|<N}\frac1{\alpha!}\partial_\eta^\alpha D_y^\alpha c\)
and remainder in \(S^{m-N}\). The map \(x_n\mapsto c_{x_n}\) is
\(C^\infty\) into \(S^m\) (difference quotients converge, by the mean
value theorem and the bounds on the next derivative), so
\(C(y,\eta;x_n)=e^{i\langle D_y,D_\eta\rangle}c_{x_n}\) is smooth in all
variables, with
\(|\partial^k_{x_n}\partial^\alpha_\eta\partial^\beta_yC|\leq C(1+|\eta|)^{m-|\alpha|}(1+x_n)^{-\nu}\).
Evaluating at \(y=(x',1)\), \(\eta=\xi\) (so \(x'\)-derivatives are
\(y'\)-derivatives) gives \(L_\chi a\in S^m_+\), continuously in \(a\).

Since \(\chi=1\) near \(y_n=1\), the expansion terms at \(y_n=1\) are
those of (7.5). The \(j\)-th term is in \(S^{m-j}_+\): the operators
\(x_n\partial_{x_n}\) and \(\eta_n\partial_{\xi_n}\) produced by
\(D_{y_n}\) preserve \(S^m_+\), and each \(\partial_\eta\) lowers the
order by one (also when it hits a factor \(\eta_n\), since
\([\partial_{\eta_n},\eta_n\partial_{\xi_n}]=\partial_{\xi_n}\)). The
second form of the terms follows from
\(\overline a(y',x_ny_n,\eta',y_n\eta_n)=\overline a(Y,H',Y_nH_n)\) with
\(Y=(y',x_ny_n)\), \(H=(\eta',\eta_n/x_n)\), under which
\(D_{y_n}D_{\eta_n}=D_{Y_n}D_{H_n}\).

Lacunarity. First let \(a\in S^{-\infty}_+\). Then \(c_{x_n}\) is a
residual symbol and \(L_\chi a\) is given by the integral (7.2) with
this \(\chi\). Substitute \(\theta=\xi_n-\eta_n\) in the inner integral:
\(L_\chi a(x,\xi',\cdot)\) is the Fourier transform, in \(y_n\), of \[
G(y_n)=(2\pi)^{-n}\iiint e^{-i\langle y',\eta'\rangle+iy_n\theta}\,\overline a\big(x'-y',x_n(1-y_n),\xi'-\eta',(1-y_n)\theta\big)\chi(1-y_n)\,d\theta\,d\eta'\,dy' .
\] So \(\mathcal F_n(L_\chi a)(x,\xi',t)=2\pi G(-t)\), which vanishes
unless \(1+t\in\operatorname{supp}\chi\), that is
\(t\in[M^{-1}-1,M-1]\subset(-1,\infty)\). For general \(a\in S^m_+\),
take \(a_k=a\,\psi(\xi/k)\in S^{-\infty}_+\) with \(\psi\in C_0^\infty\)
equal to 1 near 0; then \(a_k\to a\) in \(S^{m+1}_+\), so
\(L_\chi a_k\to L_\chi a\) in \(S^{m+1}_+\) by (a), and \(L_\chi a\) is
lacunary because \(S^{m+1}_{\mathrm{la}}\) is closed.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{2}
\tightlist
\item
  The support statement was just proved. For the kernel, the computation
  in the proof of Proposition 7.1 applies to any \(a\in S^{-\infty}_+\)
  and shows that \((L_\chi a)^\flat(x,\cdot)\) is the transform
  \(\int e^{-iz\cdot\xi}k(x,x-z)dz\) of
  \(k(x,y)=\chi(y_n/x_n)\overline{K_a(y,x)}\). \(\square\)
\end{enumerate}

The kernel statement in (c) explains the construction: the cutoff
multiplies the transposed kernel by a function of the ratio of the
normal variables, and that makes the result lacunary.

\subsubsection{Adjoints of lacunary
operators}\label{adjoints-of-lacunary-operators}

\textbf{Theorem 7.3} (Adjoints).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  For every \(a\in S^m_{\mathrm{la}}\) there is exactly one
  \(a^\dagger\in S^m_{\mathrm{la}}\) with \[
(T_au,v)=(u,T_{a^\dagger}v)\qquad(u,v\in\mathcal S(\mathbb R^n)).
\tag{7.6}
\] The map \(a\mapsto a^\dagger\) is conjugate-linear and continuous
  \(S^m_{\mathrm{la}}\to S^m_{\mathrm{la}}\), \((a^\dagger)^\dagger=a\),
  and \(a^\dagger-\overline a\in S^{m-1}_+\).
\item
  If \(a\) is strongly lacunary and \(\chi\in C_0^\infty((0,\infty))\)
  equals 1 on \((\tfrac12,2)\), then \(a^\dagger=L_\chi a\); in
  particular \(a^\dagger\) has the expansion (7.5).
\item
  Since \(T_au=0\) on \(\mathbb R^n_-\), (7.6) says
  \((T_au,v)_{L^2(\mathbb R^n_+)}=(u,T_{a^\dagger}v)_{L^2(\mathbb R^n_+)}\)
  for \(u,v\in\overline{\mathcal S}(\mathbb R^n_+)\).
\end{enumerate}

\textbf{Proof.} (b) For residual \(a\) this is Proposition 7.1. Let
\(a\in S^m_{\mathrm{la}}\) be strongly lacunary, and let \(\rho\) be as
in Lemma 4.4. Take \(a_k=a\,\psi(\xi/k)\in S^{-\infty}_+\), so that
\(a_k\to a\) in \(S^{m+1}_+\) (the error \((1-\psi(\xi/k))a\) has
\(S^{m+1}\) seminorms \(O(k^{-1})\)). Then
\((a_k)_\rho\in S^{-\infty}_+\) is strongly lacunary, and
\((a_k)_\rho\to a_\rho\) in \(S^{m+1}_+\). By the residual case,
\((T_{(a_k)_\rho}u,v)=(u,T_{L_\chi(a_k)_\rho}v)\). Let \(k\to\infty\):
\(L_\chi(a_k)_\rho\to L_\chi a_\rho\) in \(S^{m+1}_{\mathrm{la}}\) by
Lemma 7.2, and Theorem 5.1(a) lets us pass to the limit on both sides.
So the formula holds for \(a_\rho\). The difference \(a-a_\rho\) is
residual and strongly lacunary (Lemma 4.4(b),(c)), so the formula holds
for it too, and \(L_\chi\) is additive and conjugate-linear.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Write \(a=a_\rho+(a-a_\rho)\). The first term is strongly lacunary, so
  it has the adjoint symbol \(L_\chi a_\rho\) by (b). The second is in
  \(S^{-\infty}_{\mathrm{la}}\); by Theorem 6.2 its transposed kernel is
  the kernel of \(T_{b'}\) for some \(b'\in S^{-\infty}_{\mathrm{la}}\),
  as in the proof of Proposition 7.1. Put
  \(a^\dagger=L_\chi a_\rho+b'\). Uniqueness follows because \(T_c\)
  determines \(c^\flat\) (an operator determines its symbol), hence
  \(c\) on \(x_n>0\), hence \(c\) by continuity. Additivity,
  conjugate-linearity and \((a^\dagger)^\dagger=a\) follow from
  uniqueness. Indeed \(T_{\lambda a+\mu b}=\lambda T_a+\mu T_b\) and the
  inner product is linear in its first argument, so the unique adjoint
  symbol is \(\overline\lambda a^\dagger+\overline\mu b^\dagger\). For
  continuity, each step is continuous: \(a\mapsto a_\rho\) and
  \(a\mapsto a-a_\rho\) by Lemma 4.4, \(L_\chi\) by Lemma 7.2, and the
  residual adjoint by the explicit formulas of Theorem 6.2
  (\(a\mapsto A\mapsto F\mapsto F^*\mapsto A^*\mapsto b'\), each with
  seminorm bounds). Finally
  \(L_\chi a_\rho=\overline{a_\rho}+S^{m-1}_+\) by (7.5), and
  \(\overline{a_\rho}-\overline a\in S^{-\infty}_+\).
\item
  is immediate. \(\square\)
\end{enumerate}

\subsection{8. Composition}\label{composition}

The composition of two operators of the class is again in the class. Its
symbol is the sum of a near part, given by a Gauss transform as in the
ordinary calculus, and a residual far part.

\textbf{Theorem 8.1} (Composition). Let
\(a_j\in S^{m_j}_{\mathrm{la}}\), \(j=1,2\), and let
\(\chi\in C_0^\infty((0,\infty))\) equal 1 near 1, with
\(\operatorname{supp}\chi\subset(M^{-1},M)\). Put \[
b_1(x,\xi)=\Big[e^{i\langle D_y,D_\eta\rangle}\big(a_1(x,\eta)\,a_2(y',x_ny_n,\xi',\xi_ny_n)\,\chi(y_n)\big)\Big]_{y=(x',1),\ \eta=\xi},
\tag{8.1}
\] where the Gauss transform acts in \((y,\eta)\) with \((x,\xi)\) as
parameters, and {\tiny
\[
b_2(x,\xi)=\int e^{-i\langle x'-y',\xi'\rangle-i(1-y_n)\xi_n}A_1(x,x'-y',1-y_n)\,a_2(y',x_ny_n,\xi',\xi_ny_n)\,dy,\quad
A_1(x,z)=\big(1-\chi(1-z_n)\big)(2\pi)^{-n}\!\int e^{iz\cdot\xi}a_1(x,\xi)d\xi,
\tag{8.2}
\]
} with the integrand taken to be 0 for \(y_n\leq0\). Then
\(b=b_1+b_2\in S^{m_1+m_2}_{\mathrm{la}}\); \(b_2\in S^{-\infty}_+\);
the map \((a_1,a_2)\mapsto b\) is continuous and bilinear; \(b\) does
not depend on \(\chi\); and \[
T_{a_1}T_{a_2}=T_b\quad\text{on }\overline{\mathcal S}(\mathbb R^n_+),\qquad
b\sim\sum_\alpha\frac1{\alpha!}\,\partial_\xi^\alpha a_1(x,\xi)\,D_{x'}^{\alpha'}D_s^{\alpha_n}\big[a_2(x',sx_n,\xi',s\xi_n)\big]_{s=1},
\tag{8.3}
\] the \(\alpha\)-term having order \(m_1+m_2-|\alpha|\). If
\(a_1\in S^{-\infty}_{\mathrm{la}}\) and \(a_2\) vanishes for large
\(|x|\), then for \(x_n>0\) \[
b(x,\xi)=(2\pi)^{-n}\iint_{y_n>0}e^{-i\langle x'-y',\xi'-\eta'\rangle-i(1-y_n)(\xi_n-\eta_n)}a_1(x,\eta)\,a_2(y',x_ny_n,\xi',\xi_ny_n)\,dy\,d\eta,
\tag{8.4}
\] an absolutely convergent integral. Formally,
\(b=e^{i\langle D_y,D_\eta\rangle}a_1(x,\eta)a_2(y',x_ny_n,\xi',\xi_ny_n)\)
at \(y=(x',1)\), \(\eta=\xi\); the sum (8.1)+(8.2) is the precise
meaning of this formula.

The truncation used below needs boundedness and pointwise convergence;
its lack of convergence in the full symbol topology is proved in Remark
8.2.

\textbf{Proof.} \emph{Step 1: the near part.} Put
\(g(y,\xi)=a_2(y',x_ny_n,\xi',\xi_ny_n)\chi(y_n)\), with \(x_n\geq0\) a
parameter. On \(\operatorname{supp}\chi\), \(M^{-1}\leq y_n\leq M\), so
\(1+|(\xi',y_n\xi_n)|\) is comparable to \(1+|\xi|\). A
\(y_n\)-derivative produces \(x_n\partial_{x_n}a_2\) (the factor \(x_n\)
is absorbed by the decay of \(a_2\) in \(x_ny_n\)) or
\(\xi_n\partial_{\xi_n}a_2\), and
\(|\xi_n|(1+|(\xi',y_n\xi_n)|)^{m_2-1}\leq M(1+|(\xi',y_n\xi_n)|)^{m_2}\).
A \(\xi\)-derivative lowers the order by one. So \(g\) is a classical
symbol of order \(m_2\) in \((y,\xi)\), uniformly in \(x_n\), and so are
its \(x_n\)-derivatives. Likewise \((1+x_n)^\nu a_1(x,\eta)\) is a
symbol of order \(m_1\) in \((x,\eta)\) for every \(\nu\). We apply the
pre-diagonal product estimate of Section 1 to the product
\((1+x_n)^\nu a_1(x,\eta)\,g(y,\xi)\), with the multiplier acting in
\((y,\eta)\) and \(x_n\) a passive parameter. That estimate holds at
every \((y,\eta)\); at \(y=(x',1)\), \(\eta=\xi\) it gives \[
\Big|\partial^\alpha_\xi\partial^\beta_x\Big(b_1-\sum_{|\gamma|<N}\frac1{\gamma!}\partial^\gamma_\eta a_1(x,\xi)D_y^\gamma g\big((x',1),\xi\big)\Big)\Big|\leq C(1+x_n)^{-\nu}(1+|\xi|)^{m_1+m_2-N-|\alpha|},
\] with \(C\) controlled by finitely many seminorms of \(a_1\) and
\(a_2\) (differentiation in \(x\) commutes with the multiplier, and a
derivative of the evaluation at \(y=(x',1)\), \(\eta=\xi\) is a sum of
derivatives in the two sets of variables, each controlled by the
estimate). Since \(\chi=1\) near 1,
\(D_y^\gamma g((x',1),\xi)=D^{\gamma'}_{x'}D^{\gamma_n}_s[a_2(x',sx_n,\xi',s\xi_n)]_{s=1}\).
So \(b_1\in S^{m_1+m_2}_+\) with the expansion (8.3), continuously in
\((a_1,a_2)\).

\emph{Step 2: the far part is residual.} The factor \(1-\chi(1-z_n)\)
vanishes near \(z_n=0\). Off \(z=0\) the inverse transform of
\(a_1(x,\cdot)\) is smooth, and for \(|z|\) bounded below its
derivatives are bounded by \(C_N(1+|z|)^{-N}(1+x_n)^{-N}\) (integrate by
parts in \(\xi\)). So \(A_1\) satisfies (6.2). By lacunarity \(A_1=0\)
for \(z_n\geq1\), and Taylor's formula at \(z_n=1\) gives \[
|\partial_x^\alpha\partial_z^\beta A_1(x,z)|\leq C|1-z_n|^N(1+|z|)^{-2N}(1+x_n)^{-N}\qquad(z_n\leq1).
\tag{8.5}
\] Let \(G(x,y,\xi)\) be the integrand of (8.2) without the exponential.
For \(y_n>0\) we have
\(\min(1,y_n)(1+|\xi|)\leq1+|\xi'|+y_n|\xi_n|\leq(1+y_n)(1+|\xi|)\). So
a derivative of \(a_2(y',x_ny_n,\xi',\xi_ny_n)\) of order \(\gamma\) in
\(\xi\) is bounded by \((1+|\xi|)^{m_2-|\gamma|}\) times a factor
\((1+x_n)^K(y_n+y_n^{-1})^K\); the powers of \(x_n\) come from
\(y_n\)-derivatives falling on the second argument. The factor \(A_1\)
absorbs all of this. Near \(y_n=0\) the factor \(y_n^{-K}\) is paid for
by \(y_n^N=|1-z_n|^N\) in (8.5). For large \(y_n\) the growth is paid
for by the decay in \(z_n=1-y_n\). The powers of \(1+x_n\) are paid for
by the decay of \(A_1\) in \(x_n\). Hence \(G\) is smooth across
\(y_n=0\), and \[
|\partial_x^\alpha\partial_y^\beta\partial_\xi^\gamma G|\leq C_N(1+|x'-y'|+|1-y_n|)^{-N}(1+x_n)^{-N}(1+|\xi|)^{m_2-|\gamma|}.
\] The phase is
\(e^{-i\langle x',\xi'\rangle-i\xi_n}e^{i\langle y,\xi\rangle}\), so
\(\xi^\kappa b_2=\int e^{-i\langle x'-y',\xi'\rangle-i(1-y_n)\xi_n}(-D_y)^\kappa G\,dy\).
Derivatives of \(b_2\) in \(x\) and \(\xi\) bring factors \(\xi'\)
(treated the same way) or \(x'-y'\), \(1-y_n\) (absorbed by the decay of
\(G\)). So \(b_2\in S^{-\infty}_+\), continuously in \((a_1,a_2)\).

\emph{Step 3: the product formula for residual \(a_1\) and compactly
supported \(a_2\).} Let \(a_1\in S^{-\infty}_{\mathrm{la}}\),
\(a_2\in S^{m_2}_{\mathrm{la}}\) with \(a_2=0\) for \(|x|\geq R\), and
\(u\in\mathcal S\). Then \(w=T_{a_2}u\) is a bounded function with
compact support, zero for \(x_n<0\). For \(x_n>0\) the kernel
\(K_{a_1}(x,\cdot)\) is a Schwartz function vanishing for \(y_n\leq0\)
(Theorem 6.2), so for every Schwartz extension \(W\) of
\(w|_{\mathbb R^n_+}\),
\(T_{a_1}W(x)=\int K_{a_1}(x,y)w(y)dy=(2\pi)^{-n}\int e^{i\langle x,\eta\rangle}a_1^\flat(x,\eta)\widehat w(\eta)d\eta\).
Here
\(\widehat w(\eta)=(2\pi)^{-n}\iint e^{i\langle y,\xi-\eta\rangle}a_2^\flat(y,\xi)\widehat u(\xi)\,d\xi\,dy\),
absolutely convergent. Combining the integrals (absolutely convergent
for fixed \(x\)), \[
T_{a_1}T_{a_2}u(x)=(2\pi)^{-n}\int e^{i\langle x,\xi\rangle}c(x,\xi)\widehat u(\xi)d\xi,\qquad c(x,\xi)=(2\pi)^{-n}\iint e^{-i\langle x-y,\xi-\eta\rangle}a_1^\flat(x,\eta)a_2^\flat(y,\xi)\,dy\,d\eta .
\] Put \(b(x,\xi)=c(x,\xi',\xi_n/x_n)\), so \(c=b^\flat\). The
substitutions \(\xi_n\mapsto\xi_n/x_n\), \(\eta_n\mapsto\eta_n/x_n\),
\(y_n\mapsto x_ny_n\) turn \(c\) into (8.4); the double integral
converges absolutely because \(a_1\) is residual and \(y\) stays in a
compact set. Now insert \(1=\chi(y_n)+(1-\chi(y_n))\). The first part is
the Gauss transform (8.1) of a residual symbol with compact
\(y\)-support, written as an absolutely convergent integral (as in
Proposition 7.1). In the second part, integrate in \(\eta\) first:
\((2\pi)^{-n}\int e^{i\langle x'-y',\eta'\rangle+i(1-y_n)\eta_n}a_1(x,\eta)d\eta\),
multiplied by \(1-\chi(y_n)=1-\chi(1-z_n)\) with \(z_n=1-y_n\), is
\(A_1(x,x'-y',1-y_n)\); what remains is (8.2). So
\(T_{a_1}T_{a_2}=T_{b_1+b_2}\) on \(\mathcal S\), in \(\mathbb R^n_+\).

\emph{Step 4: general \(a_2\).} Let
\(\vartheta\in C_0^\infty(\mathbb R^n)\) equal 1 near 0 and
\(a_{2,k}=\vartheta(x/k)a_2\). These are lacunary, bounded in
\(S^{m_2}_+\), and converge to \(a_2\) locally uniformly with all
derivatives. Since functions of \(x\) stand on the left,
\(T_{a_{2,k}}u=\vartheta(\cdot/k)T_{a_2}u\to T_{a_2}u\) in
\(\overline{\mathcal S}(\mathbb R^n_+)\); so
\(T_{a_1}T_{a_{2,k}}u\to T_{a_1}T_{a_2}u\) by Theorem 5.1. On the other
side, \(b_{2,k}\to b_2\) pointwise by dominated convergence. The near
parts \(b_{1,k}\) are Gauss transforms of symbols in \((y,\eta)\) that
stay bounded in \(S^{m_1}\) and converge locally smoothly; by the facts
on the Gauss transform in Section 1, the transforms converge locally
uniformly with all derivatives, so \(b_{1,k}\to b_1\) pointwise. All
\(b_k=b_{1,k}+b_{2,k}\) are bounded in \(S^{m_1+m_2}_+\) (here \(m_1\)
is any real number, since \(a_1\) is residual), so
\(T_{b_k}u(x)\to T_bu(x)\) for each \(x\in\mathbb R^n_+\) by dominated
convergence. Hence \(T_{a_1}T_{a_2}u=T_bu\).

\emph{Step 5: general \(a_1\).} Write \(a_1=(a_1)_\rho+r\) with
\(r=a_1-(a_1)_\rho\in S^{-\infty}_{\mathrm{la}}\) (Lemma 4.4); Step 4
applies to \(r\). Let \(a_{1,k}=a_1\psi(\xi/k)\in S^{-\infty}_+\),
\(\psi\in C_0^\infty\) equal to 1 near 0. Then
\((a_{1,k})_\rho\in S^{-\infty}_{\mathrm{la}}\) and
\((a_{1,k})_\rho\to(a_1)_\rho\) in \(S^{m_1+1}_+\). By Step 4,
\(T_{(a_{1,k})_\rho}T_{a_2}u=T_{b^{(k)}}u\), where \(b^{(k)}\) is built
from \((a_{1,k})_\rho\) and \(a_2\). As \(k\to\infty\), the left side
converges to \(T_{(a_1)_\rho}T_{a_2}u\) (Theorem 5.1, continuity in the
symbol), and \(b^{(k)}\) converges in \(S^{m_1+1+m_2}_+\) by Steps 1--2,
so the right side converges to \(T_bu\) with \(b\) built from
\((a_1)_\rho\). Bilinearity gives the formula for \(a_1\).

\emph{Step 6: conclusions.} \(T_bu=T_{a_1}T_{a_2}u\) depends only on
\(u|_{\mathbb R^n_+}\), so \(b\) is lacunary by Proposition 4.3. The
operator determines the symbol, so \(b\) does not depend on \(\chi\).
Continuity and the expansion come from Steps 1--2. \(\square\)

\textbf{Remark 8.2} (The truncated symbols do not converge in the symbol
topology). For the tangential-translation example in this paragraph
assume \(n\ge2\). In Step 4 the symbols \(b_k\) are bounded and converge
pointwise, but they need not converge to \(b\) in the Fréchet topology
of \(S^{-\infty}_+\), even when \(a_1\) is residual. Take
\(a_2=\theta(x_n)\), with \(\theta\in C_0^\infty(\mathbb R)\) equal to 1
on \([0,1]\), and \(a_1=e^{-x_n}\widehat h(\xi)\) with
\(0\leq h\in C_0^\infty(\{|z|<\tfrac12\})\), \(h\neq0\). Both are
lacunary, \(T_{a_2}\) is multiplication by \(\theta(x_n)\), and
\(T_{a_1}\) commutes with translations in \(x'\). If \(b_k\to b\) in
\(S^{-\infty}_+\), then \(T_{b_k}\to T_b\) in the operator norm on
\(L^2(\mathbb R^n_+)\), by the Schur bound of Proposition 6.4, which is
linear in a seminorm of the symbol. But let \(u_0\geq0\) be a bump near
\((0,\tfrac12)\), and let \(u\) be a translate of \(u_0\) in \(x'\) far
outside the support of \(\vartheta(\cdot/k)\). Then
\(\|(T_{b_k}-T_b)u\|=\|T_{a_1}(\theta u_0)\|>0\), independently of
\(k\). So only boundedness together with pointwise convergence is
available, and that is what Step 4 uses.

\textbf{Example 8.3} (A totally characteristic differential operator:
product, adjoint, jets). Let \(\theta\in C_0^\infty(\mathbb R)\) equal 1
on \([-1,2]\) and \(a(x,\xi)=\theta(x_n)\xi_n\). It lies in
\(S^1_{\mathrm{la}}\) (strongly lacunary, since \(\mathcal F_na\) is
supported at \(t=0\)), and \(T_a=\theta(x_n)x_nD_n\).

\emph{Product.} In (8.3) only \(\alpha=0\) and \(\alpha=e_n\)
contribute: \(a\,a=\theta^2\xi_n^2\), and
\(\partial_{\xi_n}a\cdot D_s[\theta(sx_n)s\xi_n]_{s=1}=-i\theta(\theta+x_n\theta')\xi_n\).
Directly,
\(\theta x_nD_n(\theta x_nD_nu)=\theta^2x_n^2D_n^2u-i\theta(\theta+x_n\theta')x_nD_nu\),
whose compressed symbol is the same. Where \(\theta=1\) this is
\((x_nD_n)^2=x_n^2D_n^2-ix_nD_n\), in agreement with (2.1).

\emph{Adjoint for a real cutoff.} If the cutoff is real valued, the
following calculation applies. The full complex-cutoff formula and proof
follow below. By (7.5), the term \(j=0\) is \(\theta\xi_n\), the term
\(j=1\) is
\(\partial_{\eta_n}D_{y_n}[\theta(x_ny_n)y_n\eta_n]_{y_n=1}=-i(\theta+x_n\theta')\),
and all later terms vanish. Directly,
\((\theta x_nD_n)^*=D_n\,x_n\theta=\theta x_nD_n-i(\theta+x_n\theta')\).
The expansion is exact here: the difference is a differential operator
with symbol in \(S^{-\infty}_+\), hence 0.

\emph{Jets.} In (5.2) only \(a_{kk}=\binom k1(-i)\theta(0)=-ik\) is
nonzero near the boundary, so
\(D_n^k(x_nD_nu)(x',0)=-ik\,D_n^ku(x',0)\); this is Leibniz' rule for
\(D_n^k(x_nw)\) at \(x_n=0\).

\AddToHookNext{cmd/@afterheading/after}{
\hypertarget{AN03-TC-ADJOINT-CUTOFF-001}{}
\begingroup
\expandafter\def\csname @currentHref\endcsname{AN03-TC-ADJOINT-CUTOFF-001}
\label{AN03-TC-ADJOINT-CUTOFF-001}
\endgroup
}

\subsubsection{Complex cutoffs in the differential
example}\label{complex-cutoffs-in-the-differential-example}

Example 8.3 allows a complex-valued cutoff. Its product and jet
calculations already hold for that original choice. The adjoint
calculation displayed there requires a real-valued cutoff; here is the
exact formula for the full stated class. With the same
\(\theta\in C_0^\infty(\mathbb R)\), equal to 1 on \([-1,2]\), retain
the original operator and its compression: \[
 \begin{aligned}
 B&=\theta(x_n)x_nD_n,\qquad
 a(x,\xi)=\theta(x_n)\xi_n,\\
 B^*&=D_n x_n\overline\theta\\
    &=\overline\theta x_nD_n
        -i(\overline\theta+x_n\overline{\theta'}),\\
 a^\dagger(x,\xi)&=\overline\theta(x_n)\xi_n
        -i(\overline\theta(x_n)+x_n\overline{\theta'(x_n)}).
 \end{aligned}
 \tag{CC1}
\] Both terms of the last symbol remain: the derivative of the full
coefficient \(x_n\overline\theta\) is
\(\overline\theta+x_n\overline{\theta'}\). It is in
\(S^1_{\mathrm{la}}\), with normal Fourier support at zero, including
the constant normal-frequency term.

For \(u,v\in\overline{\mathcal S}(\mathbb R^n_+)\), integration by parts
in \(t=x_n\) gives the complete pairing identity \[
 \begin{aligned}
 (Bu,v)
 &=\int_{\mathbb R^{n-1}}\int_0^\infty
       -i\theta(t)t\,\partial_tu(x',t)\,\overline{v(x',t)}\,dt\,dx'\\
 &=\int_{\mathbb R^{n-1}}\int_0^\infty
       u(x',t)\,i\big[(\theta(t)+t\theta'(t))\overline{v(x',t)}
                 +t\theta(t)\partial_t\overline{v(x',t)}\big]\,dt\,dx'\\
 &=(u,B^*v).
 \end{aligned}
 \tag{CC2}
\] The boundary contribution is
\(-i[\theta(t)t u(x',t)\overline{v(x',t)}]_{t=0}^{t=\infty}=0\): the
original factor \(t\) kills its value at zero, and the compact support
of \(\theta\) kills its value at infinity. Every integral is absolutely
convergent. For \(n=1\) the tangential integral is over \(\mathbb R^0\),
with measure one. Theorem 7.3 gives uniqueness of the adjoint symbol, so
(CC1) is its full symbol. Equivalently, (7.5) has zeroth term
\(\overline\theta\xi_n\), first term
\(-i(\overline\theta+x_n\overline{\theta'})\), and no later terms. When
\(\overline\theta=\theta\), this proves exactly the earlier display.

The omitted conjugation can change even the sign of the pairing. Choose
real cutoffs \(\theta_0,\eta\), with \(\theta_0=1\) on \([-1,2]\),
\(\theta_0=0\) on \((3,4)\), \(\operatorname{supp}\eta\subset(3,4)\),
and \(\eta=1\) on a nonempty open interval \(J\) about \(7/2\). Then
\(\theta=\theta_0+i\eta\) satisfies all the original hypotheses. On
\(J\), its original operator is \(B=t\partial_t\); (CC1) gives
\(B^*=-t\partial_t-1\), while the display without conjugation would give
\(B_{\mathrm{old}}^*=t\partial_t+1\). Take a nonzero real
\(\chi\in C_0^\infty(J)\), and a real tangential
\(\varphi\in C_0^\infty(\mathbb R^{n-1})\) with \(\|\varphi\|_2=1\),
using \(\varphi=1\) when \(n=1\). For \(u=v=\varphi(x')\chi(t)\), the
exact values are \[
 (Bu,v)=-\frac12\int_0^\infty\chi(t)^2\,dt,
 \qquad
 (u,B^*v)=-\frac12\int_0^\infty\chi(t)^2\,dt,
 \qquad
 (u,B_{\mathrm{old}}^*v)=\frac12\int_0^\infty\chi(t)^2\,dt.
 \tag{CC3}
\] Indeed \(\int t\chi'\chi=-\tfrac12\int\chi^2\), with both endpoint
terms zero. Thus conjugation is necessary for the original complex
class. Theorem 7.3 and the antidual extension in Theorem 9.1 already
retain this conjugation. Their proofs and actions require no change. The
product in Example 8.3, its boundary jets and the delta calculation in
Example 11.3 also retain their original formulas.

\subsection{9. Extension to
distributions}\label{extension-to-distributions}

By duality with the adjoints of Section 7, the operators act on
supported and on restricted tempered distributions.

\subsubsection{Supported and restricted
distributions}\label{supported-and-restricted-distributions}

\textbf{Theorem 9.1} (Extension to distributions). Let
\(a\in S^m_{\mathrm{la}}\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  For \(U\in\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\) and
  \(v\in\overline{\mathcal S}(\mathbb R^n_+)\) the pairing
  \((U,v)=U(\overline V)\), \(V\) any Schwartz extension of \(v\), is
  well defined, and it identifies
  \(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\) with the space of
  continuous antilinear functionals on
  \(\overline{\mathcal S}(\mathbb R^n_+)\).
\item
  The formula \[
(T_aU,v)=(U,T_{a^\dagger}v)\qquad(v\in\overline{\mathcal S}(\mathbb R^n_+))
\tag{9.1}
\] defines a continuous map
  \(T_a:\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\to\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\).
  For \(u\in\overline{\mathcal S}(\mathbb R^n_+)\) with zero extension
  \(u_0\), \(T_au_0\) is the zero extension of the function \(T_au\).
\item
  The restriction map
  \(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\to\overline{\mathcal S'}(\mathbb R^n_+)\)
  is surjective, and its kernel is \[
\{U\in\mathcal S':\operatorname{supp}U\subset\partial\mathbb R^n_+\}=\bigcup_{k\geq0}\dot{\mathcal S}'_k,\qquad \dot{\mathcal S}'_k=\{U\in\mathcal S'(\mathbb R^n):x_n^kU=0\}.
\tag{9.2}
\]
\item
  \(T_a\dot{\mathcal S}'_k\subset\dot{\mathcal S}'_k\) for every \(k\).
  Hence \(T_a\) induces a map
  \(\overline{\mathcal S'}(\mathbb R^n_+)\to\overline{\mathcal S'}(\mathbb R^n_+)\).
  Identifying \(\overline{\mathcal S'}(\mathbb R^n_+)\) with the
  antidual of \(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\), this map
  is again given by (9.1), now with
  \(v\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\).
\item
  Every element of \(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\),
  and every element of \(\overline{\mathcal S'}(\mathbb R^n_+)\), is a
  weak limit of a sequence in \(C_0^\infty(\mathbb R^n_+)\). So the
  action of \(T_a\) on either space is determined by its action on
  \(C_0^\infty(\mathbb R^n_+)\).
\end{enumerate}

\textbf{Proof.} (a) If two extensions differ by \(\varphi\), then
\(\varphi=0\) in \(\mathbb R^n_+\), and \(U(\overline\varphi)=0\) by
Lemma 3.1. Since \(|U(\overline V)|\leq Cp(V)\) for a Schwartz seminorm
\(p\) and every extension \(V\), \(|(U,v)|\leq C\bar p(v)\) with the
quotient seminorm, so the functional is continuous. Conversely, a
continuous antilinear \(\lambda\) on
\(\overline{\mathcal S}(\mathbb R^n_+)\) gives
\(U(\varphi)=\lambda(\overline\varphi|_{\mathbb R^n_+})\), which is
linear and continuous on \(\mathcal S\), vanishes on
\(C_0^\infty(\mathbb R^n_-)\) (so
\(\operatorname{supp}U\subset\overline{\mathbb R}{}^n_+\)), and
satisfies \((U,v)=\lambda(v)\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  \(T_{a^\dagger}\) is continuous on
  \(\overline{\mathcal S}(\mathbb R^n_+)\) (Theorem 5.1), so (9.1)
  defines a continuous map by (a). For
  \(u\in\overline{\mathcal S}(\mathbb R^n_+)\),
  \((T_au_0,v)=\int_{\mathbb R^n_+}u\,\overline{T_{a^\dagger}v}=(T_au,v)_{L^2(\mathbb R^n_+)}\)
  by Theorem 7.3(c).
\item
  Surjectivity. Let \(w=U|_{\mathbb R^n_+}\), \(U\in\mathcal S'\). There
  is a Schwartz seminorm \(p\) with \(|U(\varphi)|\leq p(\varphi)\). On
  the subspace
  \(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\subset\overline{\mathcal S}(\mathbb R^n_+)\)
  (restriction is injective on it), \(p(v)\) equals the corresponding
  sum of suprema over \(\mathbb R^n_+\), a continuous seminorm
  \(\bar p\) of \(\overline{\mathcal S}(\mathbb R^n_+)\) by Lemma
  3.2(a). The antilinear functional \(v\mapsto U(\overline v)\) on this
  subspace is bounded by \(\bar p\). By the Hahn--Banach theorem in
  seminorm form (Section 1), applied to the linear functional
  \(v\mapsto\overline{U(\overline v)}\), it extends to
  \(\overline{\mathcal S}(\mathbb R^n_+)\) with the same bound. By (a)
  the extension is some
  \(\tilde U\in\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\), and
  \(\tilde U=U\) on \(C_0^\infty(\mathbb R^n_+)\). So
  \(\tilde U|_{\mathbb R^n_+}=w\).
\end{enumerate}

Kernel. An element of \(\dot{\mathcal S}'\) that vanishes in
\(\mathbb R^n_+\) has support in \(\partial\mathbb R^n_+\); conversely
\(x_n^kU=0\) forces \(U=0\) on \(x_n\neq0\). Let
\(\operatorname{supp}U\subset\{x_n=0\}\). Being tempered, \(U\)
satisfies
\(|U(\varphi)|\leq C\sum_{|\alpha|,|\beta|\leq\mu}\sup|x^\alpha D^\beta\varphi|\)
for some \(\mu\). Let \(\theta\in C_0^\infty(\mathbb R)\) equal 1 on
\([-1,1]\) and vanish outside \([-2,2]\), and
\(\theta_\varepsilon(x)=\theta(x_n/\varepsilon)\). For
\(\varphi\in\mathcal S\), \((1-\theta_\varepsilon)x_n^{\mu+1}\varphi\)
vanishes near \(\operatorname{supp}U\), so
\(U(x_n^{\mu+1}\varphi)=U(\theta_\varepsilon x_n^{\mu+1}\varphi)\). A
derivative of order \(|\beta|\leq\mu\) of
\(\theta_\varepsilon x_n^{\mu+1}\varphi\) is a sum of terms of size
\(\varepsilon^{-i}\,\varepsilon^{\mu+1-j}\,|D^\gamma\varphi|\),
\(i+j+|\gamma|=|\beta|\), on \(|x_n|\leq2\varepsilon\); each is
\(O(\varepsilon)\), with the weights \(x^\alpha\) carried by
\(\varphi\). So \(U(x_n^{\mu+1}\varphi)=0\), that is,
\(U\in\dot{\mathcal S}'_{\mu+1}\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{3}
\item
  Let \(U\in\dot{\mathcal S}'_k\) and
  \(v\in\overline{\mathcal S}(\mathbb R^n_+)\). Then
  \((x_n^kT_aU,v)=(U,T_{a^\dagger}(x_n^kv))\). The jets of \(x_n^kv\) of
  order \(<k\) vanish, so by Theorem 5.1(d) those of
  \(T_{a^\dagger}(x_n^kv)\) do too, and Lemma 3.2(d) writes it as
  \(x_n^kh\), \(h\in\overline{\mathcal S}(\mathbb R^n_+)\). So
  \((x_n^kT_aU,v)=(x_n^kU,h)=0\). For the last assertion:
  \(\overline{\mathcal S'}(\mathbb R^n_+)\) is \(\mathcal S'\) modulo
  the distributions vanishing in \(\mathbb R^n_+\), and these are
  exactly the tempered distributions that annihilate the closed subspace
  \(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\) (one inclusion is
  Lemma 3.1 with the half spaces exchanged, the other holds because
  \(C_0^\infty(\mathbb R^n_+)\subset\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\)).
  With the Hahn--Banach theorem this identifies
  \(\overline{\mathcal S'}(\mathbb R^n_+)\) with the antidual of
  \(\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\). If
  \(U\in\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\) restricts to
  \(u\) and \(v\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\), then
  \(T_{a^\dagger}v\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\)
  (Theorem 5.1(d)) and
  \((T_aU,v)=(U,T_{a^\dagger}v)=(u,T_{a^\dagger}v)\).
\item
  Let \(U\in\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\). Choose
  \(\phi\in C_0^\infty(\mathbb R^n_+)\) with \(\int\phi=1\) and
  \(\theta\in C_0^\infty(\mathbb R^n)\) equal to 1 near 0, and put
  \(U_\varepsilon=\theta(\varepsilon x)\,(\phi_\varepsilon*U)\),
  \(\phi_\varepsilon=\varepsilon^{-n}\phi(\cdot/\varepsilon)\). Then
  \(U_\varepsilon\in C_0^\infty(\mathbb R^n_+)\), since
  \(\operatorname{supp}(\phi_\varepsilon*U)\subset\overline{\mathbb R}{}^n_++\operatorname{supp}\phi_\varepsilon\subset\mathbb R^n_+\).
  For \(\varphi\in\mathcal S\),
  \(U_\varepsilon(\varphi)=U(\check\phi_\varepsilon*(\theta(\varepsilon\cdot)\varphi))\),
  and
  \(\check\phi_\varepsilon*(\theta(\varepsilon\cdot)\varphi)\to\varphi\)
  in \(\mathcal S\). So \(U_\varepsilon\to U\) weakly. Restricting gives
  the statement for \(\overline{\mathcal S'}(\mathbb R^n_+)\). Both
  actions of \(T_a\) are weakly continuous, being transposes of
  continuous maps. \(\square\)
\end{enumerate}

In the ordinary calculus one works modulo smooth functions, the range of
operators of order \(-\infty\). Here one also loses the distributions
supported on the boundary when passing to
\(\overline{\mathcal S'}(\mathbb R^n_+)\): they are the kernel (9.2).

By (e), the composition formula \(T_{a_1}T_{a_2}=T_b\) of Theorem 8.1
holds on \(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\) and on
\(\overline{\mathcal S'}(\mathbb R^n_+)\) as well. Indeed, both sides
are weakly continuous, and by (b) they agree on
\(C_0^\infty(\mathbb R^n_+)\).

\subsubsection{Residual operators produce conormal
distributions}\label{residual-operators-produce-conormal-distributions}

\textbf{Lemma 9.2} (Bounded order implies conormality). Let
\(W\in\mathcal D'(\mathbb R^n)\). Suppose there is \(\mu\) such that
every \(D'^{\alpha'}(x_nD_n)^{\alpha_n}W\) has order at most \(\mu\) on
every compact set (the constants may depend on \(\alpha\) and the set).
Then \(W\in I^{\mu+n/4}(\mathbb R^n,\partial\mathbb R^n_+)\).

\textbf{Proof.} If \(w\) has order \(\leq\mu\) near the support of
\(\phi\in C_0^\infty\), then
\(|\widehat{\phi w}(\xi)|=|w(\phi e^{-ix\cdot\xi})|\leq C(1+|\xi|)^\mu\),
so \(\|\Pi_j(\phi w)\|_{L^2}^2\leq C2^{2j\mu}2^{jn}\) and
\(\phi w\in B^{-\mu-n/2}_{2,\infty}\). Products of first-order operators
whose principal symbols vanish on \(N^*(\partial\mathbb R^n_+)\) are, by
Hadamard's lemma and the commutation argument of Proposition 2.1(c),
finite sums of smooth functions times
\(D'^{\alpha'}(x_nD_n)^{\alpha_n}\); multiplication by smooth functions
preserves the local order. So all these products map \(W\) into
\(B^{-\mu-n/2}_{2,\infty,\mathrm{loc}}\). By the definition of conormal
distributions in Section 1, this says that \(W\in I^{\mu+n/4}\), since
\(-(\mu+n/4)-n/4=-\mu-n/2\). \(\square\)

\textbf{Theorem 9.3} (Conormal outputs). Let
\(a\in S^{-\infty}_{\mathrm{la}}\) and
\(U\in\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\). Then
\(\operatorname{supp}T_aU\subset\overline{\mathbb R}{}^n_+\) and
\(T_aU\in I^k(\mathbb R^n,\partial\mathbb R^n_+)\) for some \(k\). More
precisely, if
\(|(U,v)|\leq C\sum_{|\beta|+|\gamma|\leq\mu}q_{\beta,\gamma}(v)\), then
there is \(\mu'\), depending only on \(\mu\) and \(n\), such that every
\(D'^{\alpha'}(x_nD_n)^{\alpha_n}T_aU\) has order at most \(\mu'\) on
every compact set, and \(T_aU\in I^{\mu'+n/4}\).

\textbf{Proof.} The support statement is part of Theorem 9.1. Since
\(U\) is continuous on \(\overline{\mathcal S}(\mathbb R^n_+)\), a bound
of the stated form holds by Lemma 3.2(a). For
\(\varphi\in C_0^\infty(\mathbb R^n)\), using the formal adjoints
\((x_nD_n)^*=D_nx_n\) and (5.4), {\footnotesize
\[
\big(D'^{\alpha'}(x_nD_n)^{\alpha_n}T_aU,\varphi\big)=\big(U,T_{a^\dagger}D'^{\alpha'}(D_nx_n)^{\alpha_n}\varphi\big)=(U,T_{b_\alpha}\varphi),\qquad
b_\alpha=\xi'^{\alpha'}\big(\xi_n-i-i\xi_n\partial_{\xi_n}\big)^{\alpha_n}a^\dagger\in S^{-\infty}_{\mathrm{la}} .
\]
} By Theorem 5.1(a), applied in the fixed class
\(S^0_{\mathrm{la}}\supset S^{-\infty}_{\mathrm{la}}\), there are
\(\mu'\) (depending only on \(\mu\) and \(n\)) and a seminorm \(p\) with
\(\sum_{|\beta|+|\gamma|\leq\mu}q_{\beta,\gamma}(T_{b_\alpha}\varphi)\leq p(b_\alpha)\sum_{|\beta|+|\gamma|\leq\mu'}q_{\beta,\gamma}(\varphi)\).
For \(\varphi\) supported in a fixed compact set the right side is at
most \(C\,p(b_\alpha)\sum_{|\gamma|\leq\mu'}\sup|D^\gamma\varphi|\). So
the order is at most \(\mu'\), with constants depending on \(\alpha\)
only through \(p(b_\alpha)\). Lemma 9.2 finishes the proof. \(\square\)

In particular the wave front set of \(T_aU\) lies in the conormal bundle
of the boundary, since this holds for every element of
\(I^k(\mathbb R^n,\partial\mathbb R^n_+)\) (Section 1).

\subsection{\texorpdfstring{10. Boundedness on \(L^2\), Sobolev and
Besov
spaces}{10. Boundedness on L\^{}2, Sobolev and Besov spaces}}\label{boundedness-on-l2-sobolev-and-besov-spaces}

Operators of order 0 are bounded on \(L^2(\mathbb R^n_+)\). We prove
this first. Then we extend it to Sobolev spaces of integer order, using
the commutator identities and duality, and to all real orders and all
Besov exponents by interpolation.

\subsubsection{\texorpdfstring{Boundedness on
\(L^2\)}{Boundedness on L\^{}2}}\label{boundedness-on-l2}

\textbf{Theorem 10.1} (Boundedness on \(L^2\)). If
\(a\in S^0_{\mathrm{la}}\), then \(T_a\) extends to a bounded operator
on \(L^2(\mathbb R^n_+)\), with norm bounded in terms of finitely many
seminorms of \(a\).

\textbf{Proof.} \emph{Step A (order \(-n-2\)).} For
\(a\in S^{-n-2}_{\mathrm{la}}\), Proposition 6.4 and the proved Schur
test in Section1.1, whose raw-integral assertion applies here to
Lebesgue measure, give
\(\|T_au\|_{L^2(\mathbb R^n_+)}\leq C\|u\|_{L^2(\mathbb R^n_+)}\) for
\(u\in\overline{\mathcal S}(\mathbb R^n_+)\), since
\(T_au(x)=\int_{\mathbb R^n_+}K_a(x,y)u(y)dy\). Restrictions of Schwartz
functions are dense in \(L^2(\mathbb R^n_+)\).

\emph{Step B (doubling).} Suppose every operator with symbol in
\(S^{-2k}_{\mathrm{la}}\) is bounded, and let
\(a\in S^{-k}_{\mathrm{la}}\). For
\(u\in\overline{\mathcal S}(\mathbb R^n_+)\), Theorems 7.3 and 8.1 give
\(\|T_au\|^2=(T_{a^\dagger}T_au,u)=(T_cu,u)\) with
\(c=a^\dagger\#a\in S^{-2k}_{\mathrm{la}}\) (writing \(\#\) for the
composition symbol of Theorem 8.1). So
\(\|T_au\|^2\leq\|T_c\|\,\|u\|^2\).

\emph{Step C.} By Steps A--B, operators with symbols in
\(S^{-k}_{\mathrm{la}}\) are bounded for \(k\geq(n+2)/2\), then for
\(k\geq(n+2)/4\), and so on; after finitely many steps, for every
\(k>0\).

\emph{Step D (order 0).} Let \(a\in S^0_{\mathrm{la}}\) and
\(M>\sup|a|\). The function \(c_0=(M^2-|a|^2)^{1/2}-M\) lies in
\(S^0_+\): it is \(G(a,\overline a)\) with \(G\) smooth on a
neighbourhood of the closed range and \(G(0)=0\), so by the chain rule
every derivative is a sum of products containing at least one derivative
of \(a\) (or \(a\) itself, since \(|G(a)|\leq C|a|\)), which gives the
decay in \(x_n\). Let \(c=(c_0)_\rho\in S^0_{\mathrm{la}}\) (Lemma 4.4).
For \(u\in\overline{\mathcal S}(\mathbb R^n_+)\), \[
\|(M+T_c)u\|^2+\|T_au\|^2=M^2\|u\|^2+(T_ru,u),\qquad r=M(c+c^\dagger)+c^\dagger\#c+a^\dagger\#a,
\] using \(2\operatorname{Re}(T_cu,u)=(T_{c+c^\dagger}u,u)\),
\(\|T_cu\|^2=(T_{c^\dagger\#c}u,u)\) and
\(\|T_au\|^2=(T_{a^\dagger\#a}u,u)\). The symbol \(r\) is lacunary. Its
leading part, by Theorem 7.3(a) and (8.3), is \(2Mc_0+c_0^2+|a|^2\)
modulo \(S^{-1}_+\) (recall \(c-c_0\in S^{-\infty}_+\) and \(c_0\) is
real). This equals \((M+c_0)^2-M^2+|a|^2=0\). So
\(r\in S^{-1}_{\mathrm{la}}\), \(T_r\) is bounded by Step C, and
\(\|T_au\|^2\leq(M^2+\|T_r\|)\|u\|^2\). \(\square\)

\textbf{Matrix-valued symbols.} For \(a\) with values in
\(L(\mathbb C^p,\mathbb C^q)\) take \(M>\sup\|a\|\) and \[
 c_0=(M^2I_p-a^*a)^{1/2}-M I_p,
 \qquad
 C_M(A)=M\sum_{k=1}^\infty\binom{1/2}{k}
                (-A^*A/M^2)^k .
\] The full square-root series, including its constant term, is \[
 (M^2I_p-A^*A)^{1/2}
 =M\sum_{k=0}^\infty\binom{1/2}{k}(-A^*A/M^2)^k
 =M I_p+C_M(A).
\] The square root is positive; \(c_0\) is its displayed difference from
\(M I_p\). For \(\|A\|\le r<M\), the ordered power series and all its
real and imaginary entry derivatives converge uniformly. Its first term
is \(-A^*A/(2M)\); hence \(C_M(0)=0\) and its first derivative at zero
is zero. The square root itself equals \(M I_p\) there. The product and
chain rules, with the uniform derivative bounds on this ball, prove
\(c_0\in S^0_+\). The leading symbol of \(r\), with all identities
explicit, is \[
 M(c_0+c_0^*)+c_0^*c_0+a^*a
 =(M I_p+c_0)^2-M^2I_p+a^*a=0 .
\] Thus the same ordered proof applies, with the input and output vector
dimensions retained.

\subsubsection{Sobolev spaces on the half
space}\label{sobolev-spaces-on-the-half-space}

Let
\(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)=\{u\in H_{(s)}:\operatorname{supp}u\subset\overline{\mathbb R}{}^n_+\}\),
with the norm of \(H_{(s)}\), and let
\(\overline H_{(s)}(\mathbb R^n_+)\) be the space of restrictions, with
\(\|u\|_{\overline H_{(s)}}=\inf\{\|U\|_{(s)}:U=u\text{ in }\mathbb R^n_+\}\).
Define \(\dot B^s_{2,p}(\overline{\mathbb R}{}^n_+)\) in the same way as
\(\dot H_{(s)}\). We prove the three facts about these spaces that we
need.

\textbf{Proposition 10.2} (Sobolev spaces on the half space).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  \(C_0^\infty(\mathbb R^n_+)\) is dense in
  \(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\), and
  \(\overline{\mathcal S}(\mathbb R^n_+)\) is dense in
  \(\overline H_{(s)}(\mathbb R^n_+)\), for every real \(s\).
\item
  The sesquilinear form
  \((u,v)=(2\pi)^{-n}\int\widehat u\,\overline{\widehat V}\,d\xi\), for
  \(u\in\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\) and \(V\in H_{(-s)}\)
  any extension of \(v\in\overline H_{(-s)}(\mathbb R^n_+)\), is well
  defined. It identifies each of
  \(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\) and
  \(\overline H_{(-s)}(\mathbb R^n_+)\) isometrically with the antidual
  of the other. For
  \(u\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\),
  \(v\in\overline{\mathcal S}(\mathbb R^n_+)\) it equals
  \(\int_{\mathbb R^n_+}u\overline v\).
\item
  Let \(k\geq0\) be an integer. For
  \(u\in\dot H_{(k)}(\overline{\mathbb R}{}^n_+)\),
  \(\|u\|_{(k)}^2=\sum_{|\alpha|\leq k}\frac{k!}{\alpha!(k-|\alpha|)!}\|D^\alpha u\|^2_{L^2}\).
  For \(u\in\overline H_{(k)}(\mathbb R^n_+)\), with
  \(N_k(u)^2=\sum_{|\alpha|\leq k}\|D^\alpha u\|^2_{L^2(\mathbb R^n_+)}\),
  \[
C^{-1}N_k(u)\leq\|u\|_{\overline H_{(k)}}\leq C\,N_k(u).
\tag{10.1}
\]
\end{enumerate}

\textbf{Proof.} (a) Let \(u\in\dot H_{(s)}\). The translates
\(u_h=u(\cdot-he_n)\), supported in \(x_n\geq h\), converge to \(u\) in
\(H_{(s)}\) as \(h\downarrow0\), by dominated convergence on the Fourier
side. Mollifying with a kernel supported in \(\{|x|<h/2\}\) gives smooth
functions supported in \(x_n\geq h/2\), converging in \(H_{(s)}\); these
lie in \(H_{(\sigma)}\) for every \(\sigma\). Cutting off with
\(\theta(x/R)\) converges in \(H_{(k)}\) for integers \(k\geq s\)
(Leibniz' rule and dominated convergence), hence in \(H_{(s)}\). The
second statement holds because \(\mathcal S\) is dense in \(H_{(s)}\)
and restriction is continuous and onto.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  \(H_{(s)}\) and \(H_{(-s)}\) are each other's antiduals,
  isometrically, under this form: Cauchy--Schwarz with the weights
  \(\langle\xi\rangle^{\pm s}\), with equality for
  \(\widehat V=\langle\xi\rangle^{2s}\widehat u\). If \(V=0\) in
  \(\mathbb R^n_+\), then \((\varphi,V)=0\) for
  \(\varphi\in C_0^\infty(\mathbb R^n_+)\), and by (a) \((u,V)=0\) for
  all \(u\in\dot H_{(s)}\); so the form is well defined, and the
  annihilator of \(\dot H_{(s)}\) in \(H_{(-s)}\) is exactly
  \(\{V:V=0\text{ in }\mathbb R^n_+\}\). A continuous antilinear
  functional on the closed subspace \(\dot H_{(s)}\) extends with the
  same norm to \(H_{(s)}\) (orthogonal projection) and is then
  represented by some \(V\); two representatives differ by an element of
  the annihilator. So the antidual of \(\dot H_{(s)}\) is \(H_{(-s)}\)
  modulo the annihilator, that is \(\overline H_{(-s)}\), and the norms
  agree (the infimum over the coset is at most the norm of the
  norm-preserving extension). Conversely, a functional on the quotient
  \(\overline H_{(-s)}\) is a functional on \(H_{(-s)}\) vanishing on
  the annihilator; it is represented by \(u\in H_{(s)}\) orthogonal to
  the annihilator, and the double annihilator of the closed subspace
  \(\dot H_{(s)}\) is itself. The last statement is Plancherel.
\item
  The identity is Plancherel with
  \((1+|\xi|^2)^k=\sum_{|\alpha|\leq k}\frac{k!}{\alpha!(k-|\alpha|)!}\xi^{2\alpha}\).
  For (10.1), any extension \(U\) gives
  \(N_k(u)^2\leq\sum_{|\alpha|\leq k}\|D^\alpha U\|^2_{L^2(\mathbb R^n)}\leq C\|U\|^2_{(k)}\).
  For the other inequality, let \(c_1,\ldots,c_{k+1}\) solve the
  Vandermonde system \(\sum_{l=1}^{k+1}c_l(-l)^i=1\), \(i=0,\ldots,k\)
  (the nodes \(-1,\ldots,-(k+1)\) are distinct). For
  \(u\in\overline{\mathcal S}(\mathbb R^n_+)\) let \(Eu=u\) on
  \(x_n\geq0\) and \(Eu(x)=\sum_lc_lu(x',-lx_n)\) for \(x_n<0\). The
  normal derivatives of order \(i\leq k\) from both sides agree on
  \(x_n=0\), so \(Eu\in C^k\), and
  \(\|D^\alpha Eu\|_{L^2(\mathbb R^n_-)}\leq\sum_l|c_l|l^{\alpha_n-1/2}\|D^\alpha u\|_{L^2(\mathbb R^n_+)}\).
  Hence \(\|u\|_{\overline H_{(k)}}\leq\|Eu\|_{(k)}\leq CN_k(u)\) on the
  dense set \(\overline{\mathcal S}(\mathbb R^n_+)\), and by continuity
  of both sides everywhere. \(\square\)
\end{enumerate}

\subsubsection{Sobolev continuity at integer
orders}\label{sobolev-continuity-at-integer-orders}

\textbf{Theorem 10.3} (Integer orders). Let \(a\in S^0_{\mathrm{la}}\)
and \(k\in\mathbb Z\). Then \(T_a\) is bounded on
\(\dot H_{(k)}(\overline{\mathbb R}{}^n_+)\) and on
\(\overline H_{(k)}(\mathbb R^n_+)\). These bounded operators are the
restrictions of the maps of Theorem 9.1.

\textbf{Proof.} Iterating the commutator identities (5.1) gives, for
every \(\alpha\), \[
D^\alpha T_a=\sum_{|\beta|\leq|\alpha|}T_{c_{\alpha\beta}}D^\beta\quad\text{on }\overline{\mathcal S}(\mathbb R^n_+),\qquad c_{\alpha\beta}\in S^0_{\mathrm{la}},
\tag{10.2}
\] each \(c_{\alpha\beta}\) being a constant-coefficient combination of
\(x\)- and \(\xi_n\)-derivatives of \(a\), linear in \(a\). (Indeed
\(D_jT_c=T_cD_j-iT_{\partial_{x_j}c}-i\delta_{jn}T_{\partial_{\xi_n}c}D_n\),
and
\(\partial_{\xi_n}c\in S^{-1}_{\mathrm{la}}\subset S^0_{\mathrm{la}}\).)

\emph{Nonnegative \(k\), supported spaces.} For
\(u\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\),
\(T_au\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\) (Theorem
5.1(d)), and its derivatives on \(\mathbb R^n\) are the zero extensions
of the derivatives in \(\mathbb R^n_+\). By Proposition 10.2(c), (10.2)
and Theorem 10.1,
\(\|T_au\|_{(k)}^2\leq C\sum_{|\alpha|\leq k}\|D^\alpha T_au\|^2_{L^2(\mathbb R^n_+)}\leq C'\sum_{|\beta|\leq k}\|D^\beta u\|^2_{L^2}\leq C''\|u\|^2_{(k)}\).
By density (Proposition 10.2(a)) \(T_a\) extends to \(\dot H_{(k)}\).

\emph{Nonnegative \(k\), restricted spaces.} For
\(u\in\overline{\mathcal S}(\mathbb R^n_+)\), (10.1), (10.2) and Theorem
10.1 give
\(\|T_au\|_{\overline H_{(k)}}\leq CN_k(T_au)\leq C'N_k(u)\leq C''\|u\|_{\overline H_{(k)}}\);
then use density.

\emph{Negative \(k\).} Let \(k\geq0\). For
\(u\in\overline{\mathcal S}(\mathbb R^n_+)\) and
\(v\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\), Theorem 7.3 gives
\((T_au,v)=(u,T_{a^\dagger}v)\), so
\(|(T_au,v)|\leq\|u\|_{\overline H_{(-k)}}\|T_{a^\dagger}v\|_{(k)}\leq C\|u\|_{\overline H_{(-k)}}\|v\|_{(k)}\)
by the supported case for \(a^\dagger\in S^0_{\mathrm{la}}\). By
Proposition 10.2(a),(b),
\(\|T_au\|_{\overline H_{(-k)}}\leq C\|u\|_{\overline H_{(-k)}}\), and
density extends \(T_a\). In the same way, for
\(u\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)\) and
\(v\in\overline{\mathcal S}(\mathbb R^n_+)\),
\(|(T_au,v)|\leq\|u\|_{(-k)}\|T_{a^\dagger}v\|_{\overline H_{(k)}}\leq C\|u\|_{(-k)}\|v\|_{\overline H_{(k)}}\),
which bounds \(T_a\) on \(\dot H_{(-k)}\).

\emph{Consistency.} The maps of Theorem 9.1 are weakly continuous, the
spaces here embed continuously into
\(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\) or
\(\overline{\mathcal S'}(\mathbb R^n_+)\), and the two definitions agree
on the dense subspaces used above. \(\square\)

\subsubsection{All real orders and all Besov
exponents}\label{all-real-orders-and-all-besov-exponents}

\textbf{Lemma 10.4} (A mollifier supported in the half space). Let
\(\phi\in C_0^\infty(\mathbb R^n_+)\) with \(\int\phi=1\), and put
\(\psi=2\phi-\phi*\phi\). Then \(\psi\in C_0^\infty(\mathbb R^n_+)\),
\(\widehat\psi=1-(1-\widehat\phi)^2\), and for all
\(\zeta\in\mathbb R^n\) \[
|\widehat\psi(\zeta)|\leq C\min(1,|\zeta|^{-2}),\qquad|1-\widehat\psi(\zeta)|\leq C\min(1,|\zeta|^2).
\tag{10.3}
\] With \(\psi_\varepsilon(x)=\varepsilon^{-n}\psi(x/\varepsilon)\) and
\(|s|\leq\tfrac12\), \[
\int_0^1|\widehat\psi(\varepsilon\xi)|^2\varepsilon^{1-2s}d\varepsilon\leq C\langle\xi\rangle^{2s-2},\qquad
\int_0^1|1-\widehat\psi(\varepsilon\xi)|^2\varepsilon^{-3-2s}d\varepsilon\leq C\langle\xi\rangle^{2s+2}.
\tag{10.4}
\]

\textbf{Proof.}
\(\operatorname{supp}(\phi*\phi)\subset\operatorname{supp}\phi+\operatorname{supp}\phi\subset\mathbb R^n_+\).
\(\widehat\psi\) is a Schwartz function, and
\(|1-\widehat\phi(\zeta)|\leq C\min(1,|\zeta|)\) since
\(\widehat\phi(0)=1\); this gives (10.3). For (10.4) with
\(|\xi|\leq1\): the first integral is at most
\(C\int_0^1\varepsilon^{1-2s}d\varepsilon<\infty\) (as \(1-2s\geq0\)),
and the second at most
\(C|\xi|^4\int_0^1\varepsilon^{1-2s}d\varepsilon\). For \(|\xi|\geq1\)
substitute \(u=\varepsilon|\xi|\) and extend to \((0,\infty)\): the
integrals become \(|\xi|^{2s-2}\int_0^\infty\min(1,u^{-4})u^{1-2s}du\)
and \(|\xi|^{2s+2}\int_0^\infty\min(1,u^4)u^{-3-2s}du\), which converge
because \(1-2s>-1\), \(-3-2s<-1\) and \(-3+4-2s>-1\). \(\square\)

The quadratic vanishing of \(1-\widehat\psi\) at 0 is needed for
\(s\geq0\), and it cannot be had with \(\psi\geq0\): see Example 10.5.
That is why \(\psi\) is built from \(\phi\) in this way.

\textbf{Example 10.5} (A positive mollifier is not good enough). Let
\(0\leq\phi\in C_0^\infty(\mathbb R^n_+)\) with \(\int\phi=1\). Its
first moment \(m=\int x\phi\,dx\) has \(m_n>0\), and
\(\widehat\phi(\zeta)=1-i\zeta\cdot m+O(|\zeta|^2)\). For
\(\xi=\lambda e_n\) we get
\(|1-\widehat\phi(\varepsilon\xi)|\geq m_n\varepsilon\lambda/2\) when
\(\varepsilon\lambda\) is small, so
\(\int_0^1|1-\widehat\phi(\varepsilon\xi)|^2\varepsilon^{-3-2s}d\varepsilon=\infty\)
for \(s\geq0\). So the second inequality of (10.4) fails for \(\phi\),
and it fails for every \(\psi\geq0\) supported in \(\mathbb R^n_+\):
quadratic vanishing forces \(\int x_n\psi=0\), which is impossible when
\(\psi\geq0\) and \(x_n>0\) on the support. The function
\(\psi=2\phi-\phi*\phi\) of Lemma 10.4 takes negative values.

\textbf{Theorem 10.6} (Sobolev and Besov continuity). Let
\(a\in S^0_{\mathrm{la}}\), \(\sigma\in\mathbb R\) and
\(1\leq p\leq\infty\). Then \(T_a\) is bounded on
\(\dot B^\sigma_{2,p}(\overline{\mathbb R}{}^n_+)\). In particular:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  (\(p=2\)) \(T_a\) is bounded on
  \(\dot H_{(\sigma)}(\overline{\mathbb R}{}^n_+)\), and, by duality
  with \(a^\dagger\) (Proposition 10.2(b)), on
  \(\overline H_{(\sigma)}(\mathbb R^n_+)\), for every real \(\sigma\).
\item
  (\(p=\infty\)) \(T_a\) is bounded on
  \(\dot B^\sigma_{2,\infty}(\overline{\mathbb R}{}^n_+)\).
\end{enumerate}

\textbf{First proof, for \(p=2\) (continuous interpolation).} Write
\(\sigma=k+s\) with \(k\in\mathbb Z\), \(|s|\leq\tfrac12\), and let
\(u\in\dot H_{(\sigma)}(\overline{\mathbb R}{}^n_+)\). The pieces
\(\psi_\varepsilon*u\) and \(u-\psi_\varepsilon*u\) are supported in
\(\overline{\mathbb R}{}^n_+\), because
\(\operatorname{supp}\psi\subset\mathbb R^n_+\); they lie in
\(\dot H_{(k+1)}\) and \(\dot H_{(k-1)}\). By (10.4) and Fubini, \[
\int_0^1\Big(\|\psi_\varepsilon*u\|^2_{(k+1)}\varepsilon^{1-2s}+\|u-\psi_\varepsilon*u\|^2_{(k-1)}\varepsilon^{-3-2s}\Big)d\varepsilon\leq C\|u\|^2_{(\sigma)} .
\] Put \(v_\varepsilon=T_a(\psi_\varepsilon*u)\),
\(w_\varepsilon=T_a(u-\psi_\varepsilon*u)\) and
\(U=T_au=v_\varepsilon+w_\varepsilon\). By Theorem 10.3 the same
integral with \(v_\varepsilon,w_\varepsilon\) in place of the two pieces
is at most \(C'\|u\|^2_{(\sigma)}\). If
\(\tfrac12<\varepsilon\langle\xi\rangle<1\), then
\(|\widehat U(\xi)|^2\leq C\big(|\widehat{v_\varepsilon}(\xi)|^2(\varepsilon\langle\xi\rangle)^{2k+2}+|\widehat{w_\varepsilon}(\xi)|^2(\varepsilon\langle\xi\rangle)^{2k-2}\big)\).
Multiply by \(\varepsilon^{-1-2\sigma}\) and integrate over these
\(\varepsilon\) (all in \((0,1]\)): the left side becomes
\(c_\sigma\langle\xi\rangle^{2\sigma}|\widehat U(\xi)|^2\) with
\(c_\sigma=\int_{1/2}^1u^{-1-2\sigma}du>0\), and the right side is at
most the integrand of the previous display, evaluated for
\(v_\varepsilon,w_\varepsilon\) at the frequency \(\xi\). Integrating in
\(\xi\) gives \(\|U\|^2_{(\sigma)}\leq C''\|u\|^2_{(\sigma)}\).
\(\square\)

\textbf{Second proof, for all \(p\) (dyadic form).} Let \(\sigma=k+s\)
as before and \(u\in\dot B^\sigma_{2,p}(\overline{\mathbb R}{}^n_+)\).
Since \(B^\sigma_{2,p}\subset H_{(\sigma-\delta)}\) for every
\(\delta>0\) (the squares \(2^{2j(\sigma-\delta)}\|\Pi_ju\|^2_{L^2}\)
are at most \(2^{-2j\delta}\) times the square of the norm in
\(B^\sigma_{2,p}\), so they are summable) and \(s>-1\),
\(u\in\dot H_{(k-1)}\). For each \(j\geq0\) put
\(\varepsilon_j=2^{-j}\),
\(v_j=\psi_{\varepsilon_j}*u\in\dot H_{(k+1)}\),
\(w_j=u-v_j\in\dot H_{(k-1)}\). Then \(T_au=T_av_j+T_aw_j\) (the maps of
Theorems 9.1 and 10.3 agree), and, since \(\langle\xi\rangle\) is
comparable to \(2^j\) on \(A_j\), {\scriptsize
\[
2^{j\sigma}\|\Pi_jT_au\|_{L^2}\leq C\big(2^{j(s-1)}\|T_av_j\|_{(k+1)}+2^{j(s+1)}\|T_aw_j\|_{(k-1)}\big)\leq C'\big(2^{j(s-1)}\|v_j\|_{(k+1)}+2^{j(s+1)}\|w_j\|_{(k-1)}\big).
\]
} On \(A_l\), \(|\widehat\psi(2^{-j}\xi)|\leq C\min(1,2^{2(j-l)})\) and
\(|1-\widehat\psi(2^{-j}\xi)|\leq C\min(1,2^{2(l-j)})\) by (10.3). With
\(y_l=2^{l\sigma}\|\Pi_lu\|_{L^2}\) this gives {\scriptsize
\[
2^{j(s-1)}\|v_j\|_{(k+1)}\leq C\Big(\sum_l\big[2^{(j-l)(s-1)}\min(1,2^{2(j-l)})\big]^2y_l^2\Big)^{1/2},\quad
2^{j(s+1)}\|w_j\|_{(k-1)}\leq C\Big(\sum_l\big[2^{(j-l)(s+1)}\min(1,2^{2(l-j)})\big]^2y_l^2\Big)^{1/2}.
\]
} For \(|s|\leq\tfrac12\) both brackets are at most \(2^{-|j-l|/2}\):
for \(j\geq l\) they are \(2^{(j-l)(s-1)}\) and \(2^{(j-l)(s-1)}\); for
\(j<l\) they are \(2^{(j-l)(s+1)}\) and \(2^{(j-l)(s+1)}\). Since
\((\sum_lc_ly_l^2)^{1/2}\leq\sum_lc_l^{1/2}y_l\), we get
\(x_j:=2^{j\sigma}\|\Pi_jT_au\|_{L^2}\leq C\sum_l2^{-|j-l|/2}y_l\).
Convolution with the summable sequence \(2^{-|m|/2}\) is bounded on
\(\ell^p(\mathbb Z)\) for every \(1\leq p\leq\infty\), by the triangle
inequality for translates (Section 1), so
\(\|T_au\|_{B^\sigma_{2,p}}\leq C\|u\|_{B^\sigma_{2,p}}\). Finally
\(T_au\) is supported in \(\overline{\mathbb R}{}^n_+\). \(\square\)

The dyadic proof treats all \(1\leq p\leq\infty\) at once and contains
the case \(p=2\).

\subsection{11. Conormal distributions are
preserved}\label{conormal-distributions-are-preserved}

Let \(\mathcal P_b\) be the set of operators
\(P=\sum_{|\alpha|\leq M}c_\alpha(x)x_n^{\alpha_n}D^\alpha\) with
\(c_\alpha\in C^\infty_b(\mathbb R^n)\) and any \(M\). For
\(\kappa\in\mathbb R\) put \[
\mathcal A^\kappa=\big\{u\in\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+):\ Pu\in B^\kappa_{2,\infty}(\mathbb R^n)\text{ for every }P\in\mathcal P_b\big\},
\tag{11.1}
\] the distributions supported in the closed half space that are
conormal to the boundary uniformly at infinity.

\textbf{Lemma 11.1} (Exact compositions). Let
\(a\in S^m_{\mathrm{la}}\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  If \(P\in\mathcal P_b\) has order \(\leq M\), then
  \(PT_a=T_{p\star a}\) on \(\overline{\mathcal S}(\mathbb R^n_+)\),
  where \[
p\star a=\sum_\alpha c_\alpha(x)\prod_{j<n}(\xi_j+D_{x_j})^{\alpha_j}\;q_{\alpha_n}\big(\xi_n+x_nD_{x_n}+\xi_nD_{\xi_n}\big)\,a\in S^{m+M}_{\mathrm{la}},
\tag{11.2}
\] with \(q_k\) from (2.1).
\item
  Let \(M\geq0\) be even and \(Q(\xi)=|\xi|^M\). Then
  \(T_Q:=\sum_{|\beta|=M/2}\frac{(M/2)!}{\beta!}D'^{2\beta'}x_n^{2\beta_n}D_n^{2\beta_n}\in\mathcal P_b\)
  is the operator with compressed symbol
  \(Q^\flat=(|\xi'|^2+x_n^2\xi_n^2)^{M/2}\), and for
  \(f\in S^\mu_{\mathrm{la}}\) {\small
\[
T_fT_Q=T_{f\circ Q},\qquad f\circ Q=\sum_{|\beta|=M/2}\frac{(M/2)!}{\beta!}\,\xi'^{2\beta'}\xi_n^{2\beta_n}(1-i\partial_{\xi_n})^{2\beta_n}f\in S^{\mu+M}_{\mathrm{la}},\qquad f\circ Q-Qf\in S^{\mu+M-1}_+ .
\tag{11.3}
\]
}
\item
  If \(c\in S^\mu_{\mathrm{la}}\) and \(M\geq0\) is an even integer with
  \(M>\mu\), then \(c=f\circ Q+g\) with \(f,g\in S^0_{\mathrm{la}}\).
\end{enumerate}

\textbf{Proof.} (a) By (5.3),
\(D_{x_j}(e^{ix\cdot\xi}c^\flat)=e^{ix\cdot\xi}((\xi_j+D_{x_j})c)^\flat\)
for \(j<n\), and
\(x_nD_{x_n}(e^{ix\cdot\xi}c^\flat)=e^{ix\cdot\xi}((\xi_n+x_nD_{x_n}+\xi_nD_{\xi_n})c)^\flat\),
because \(x_n\xi_nc^\flat=(\xi_nc)^\flat\). So
\(D_jT_c=T_{(\xi_j+D_{x_j})c}\) and
\(x_nD_nT_c=T_{(\xi_n+x_nD_{x_n}+\xi_nD_{\xi_n})c}\), and
\(c_\alpha(x)T_c=T_{c_\alpha c}\). These symbol operators preserve
lacunarity and raise the order by at most one (the factor \(x_n\) is
absorbed by the decay in \(x_n\)). By (2.1),
\(x_n^{\alpha_n}D^\alpha=D'^{\alpha'}q_{\alpha_n}(x_nD_n)\), which gives
(11.2).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Expanding \((|\xi'|^2+x_n^2\xi_n^2)^{M/2}\) and quantizing on the left
  gives
  \(\operatorname{Op}(Q^\flat)=\sum\frac{(M/2)!}{\beta!}x_n^{2\beta_n}D^{2\beta}=T_Q\).
  By (5.5),
  \(T_fD'^{2\beta'}x_n^{2\beta_n}D_n^{2\beta_n}=T_{\xi'^{2\beta'}\xi_n^{2\beta_n}(1-i\partial_{\xi_n})^{2\beta_n}f}\).
  Expanding
  \((1-i\partial_{\xi_n})^{2\beta_n}f=f+(\text{terms with }\partial_{\xi_n})\)
  shows \(f\circ Q-Qf\in S^{\mu+M-1}_+\).
\item
  Let \(\chi\in C_0^\infty(\mathbb R^n)\) equal 1 near 0; then
  \((1-\chi)/Q\in S^{-M}\). Put \(g_0=c\). Given
  \(g_i\in S^{\mu-i}_{\mathrm{la}}\), put
  \(f_i=\big((1-\chi)g_i/Q\big)_\rho\in S^{\mu-i-M}_{\mathrm{la}}\)
  (Lemma 4.4) and \(g_{i+1}=g_i-f_i\circ Q\). Then \[
g_{i+1}=\chi g_i+\Big(\frac{(1-\chi)g_i}Q-f_i\Big)Q-\big(f_i\circ Q-f_iQ\big)\in S^{\mu-i-1}_{\mathrm{la}},
\] since the first two terms are in \(S^{-\infty}_+\) and the last is
  in \(S^{\mu-i-1}_+\) by (b); it is lacunary as a combination of
  lacunary symbols. After \(N\) steps with \(\mu-N\leq0\),
  \(c=\big(\sum_{i<N}f_i\big)\circ Q+g_N\), with
  \(\sum f_i\in S^{\mu-M}_{\mathrm{la}}\subset S^0_{\mathrm{la}}\) and
  \(g_N\in S^0_{\mathrm{la}}\). \(\square\)
\end{enumerate}

\textbf{Theorem 11.2} (Conormal distributions are preserved). Let
\(\kappa\in\mathbb R\) and \(k=-\kappa-n/4\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  \(I^k(\mathbb R^n,\partial\mathbb R^n_+)\cap\dot{\mathcal E}'(\overline{\mathbb R}{}^n_+)\subset\mathcal A^\kappa\subset I^k(\mathbb R^n,\partial\mathbb R^n_+)\cap\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\).
\item
  For every real \(m\) and every \(a\in S^m_{\mathrm{la}}\),
  \(T_a\mathcal A^\kappa\subset\mathcal A^\kappa\).
\item
  If \(a\in S^m_{\mathrm{la}}\) and
  \(u\in I^k(\mathbb R^n,\partial\mathbb R^n_+)\cap\dot{\mathcal E}'(\overline{\mathbb R}{}^n_+)\),
  then
  \(T_au\in I^k(\mathbb R^n,\partial\mathbb R^n_+)\cap\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\).
\end{enumerate}

\textbf{Proof.} (a) Let
\(u\in I^k\cap\dot{\mathcal E}'(\overline{\mathbb R}{}^n_+)\) and
\(P\in\mathcal P_b\). By (2.1),
\(P=\sum c_\alpha D'^{\alpha'}q_{\alpha_n}(x_nD_n)\) is a sum of words
in the tangent operators \(D_j\) (\(j<n\)) and \(x_nD_n\), times
\(C^\infty_b\) functions. By the definition of conormal distributions
(Section 1), these words map \(u\) into
\(B^{\kappa}_{2,\infty,\mathrm{loc}}\), since \(\kappa=-k-n/4\); and
multiplication by \(c_\alpha\) preserves that space. \(Pu\) has compact
support, so \(Pu=\vartheta Pu\in B^\kappa_{2,\infty}\) with
\(\vartheta\in C_0^\infty\) equal to 1 near \(\operatorname{supp}u\).
For the second inclusion, let \(L_1,\ldots,L_N\) be first-order
operators on \(\mathbb R^n\) whose principal symbols vanish on
\(N^*(\partial\mathbb R^n_+)\), and
\(\vartheta\in C_0^\infty(\mathbb R^n)\). By Hadamard's lemma and the
commutation argument of Proposition 2.1, \(\vartheta L_1\cdots L_N\) is
an element of \(\mathcal P_b\) (with compactly supported coefficients).
So \(\vartheta L_1\cdots L_Nu\in B^\kappa_{2,\infty}\) for
\(u\in\mathcal A^\kappa\), which is the definition of
\(I^k(\mathbb R^n,\partial\mathbb R^n_+)\).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Let \(u\in\mathcal A^\kappa\) and \(P\in\mathcal P_b\) of order
  \(M_P\). By Lemma 11.1(a), \(PT_a=T_c\) with
  \(c=p\star a\in S^{m+M_P}_{\mathrm{la}}\). Choose an even \(M>m+M_P\),
  \(M\geq0\), and write \(c=f\circ Q+g\) as in Lemma 11.1(c). Then
  \(PT_a=T_fT_Q+T_g\). These identities hold on
  \(\overline{\mathcal S}(\mathbb R^n_+)\), hence on
  \(\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)\): both sides are
  weakly continuous and agree on \(C_0^\infty(\mathbb R^n_+)\) (Theorem
  9.1(e)). They agree there because every element of \(\mathcal P_b\)
  commutes with extension by zero: if
  \(w\in\overline{\mathcal S}(\mathbb R^n_+)\) has zero extension
  \(w_0\), then \(x_n^{\alpha_n}D^\alpha w_0\) and the zero extension of
  \(x_n^{\alpha_n}D^\alpha w\) differ by terms
  \(x_n^{\alpha_n}\delta^{(l)}(x_n)\otimes g_l(x')\) with
  \(l<\alpha_n\), and these vanish. With Theorem 9.1(b), both sides
  therefore send \(\varphi\in C_0^\infty(\mathbb R^n_+)\) to the zero
  extension of the same function. Now \(T_Qu\in B^\kappa_{2,\infty}\)
  because \(T_Q\in\mathcal P_b\), it is supported in
  \(\overline{\mathbb R}{}^n_+\), and \(u\in B^\kappa_{2,\infty}\) (take
  \(P=1\)). By Theorem 10.6 with \(p=\infty\), \(T_f\) and \(T_g\) are
  bounded on \(\dot B^\kappa_{2,\infty}(\overline{\mathbb R}{}^n_+)\).
  So \(PT_au\in B^\kappa_{2,\infty}\) for every \(P\in\mathcal P_b\),
  that is, \(T_au\in\mathcal A^\kappa\).
\item
  follows from (a) and (b). \(\square\)
\end{enumerate}

The order \(m\) of \(a\) plays no role: conormal distributions are
infinitely regular in the directions of the totally characteristic
operators, so any loss of order can be moved onto the elliptic
b-operator \(T_Q\), which conormality controls. The exact formulas
(11.2)--(11.3) do not need the coefficients of \(P\) to decay in
\(x_n\), as the composition theorem would; this is what allows the
global class \(\mathcal A^\kappa\) in (b). Without conormality nothing
of this kind holds; see Example 11.3.

\textbf{Example 11.3} (Besov regularity alone is not preserved at
positive order). Let \(a\) be as in Example 8.3, of order 1, and
\(u=\delta(x_n-\tfrac32)\otimes\varphi(x')\) with
\(0\neq\varphi\in C_0^\infty(\mathbb R^{n-1})\). Then
\(u\in\dot B^{-1/2}_{2,\infty}(\overline{\mathbb R}{}^n_+)\) and \(u\)
has compact support, but \(u\) is not conormal to the boundary. The
exact distributional identity is \[
 T_au=x_nD_nu=-i\varphi(x')\big(\tfrac32\delta'(x_n-\tfrac32)-\delta(x_n-\tfrac32)\big),
 \qquad
 \widehat{T_au}(\xi',\xi_n)
 =\widehat\varphi(\xi')e^{-3i\xi_n/2}\big(\tfrac32\xi_n+i\big).
\]

\textbf{Editorial correction to the Fourier lower bound.} It holds on a
bounded tangential set where \(|\widehat\varphi|\) is bounded below,
rather than at every point of \(\{|\xi'|\le1\}\). Since Fourier
inversion and \(\varphi\neq0\) imply \(\widehat\varphi\not\equiv0\),
continuity gives a bounded set \(E\) of positive measure and a constant
\(c>0\) with \(|\widehat\varphi|\ge c\) on \(E\). In dimension one,
\(E=\mathbb R^0\), with its measure one and the nonzero scalar
\(\varphi\). For sufficiently large \(j\), \[
 E\times[\tfrac35\,2^j,\tfrac45\,2^j]\subset A_j,
 \qquad
 \|\Pi_jT_au\|_2^2
 \ge (2\pi)^{-n}c^2|E|
       \int_{(3/5)2^j}^{(4/5)2^j}
           \big(\tfrac94\xi_n^2+1\big)\,d\xi_n
 \ge C2^{3j}.
\] Thus \(2^{-j/2}\|\Pi_jT_au\|_2\ge C'2^j\), so
\(T_au\notin B^{-1/2}_{2,\infty}\). For the original input, integration
over \(A_j\) is bounded above by integration over \(|\xi_n|<2^j\) and
all tangential frequencies, giving \[
 \|\Pi_j u\|_2^2
 \le (2\pi)^{-n}2^{j+1}\|\widehat\varphi\|_2^2.
\] The order-zero annulus is finite as well, so the stated input
membership follows. It is not conormal to the boundary: it is singular
on \(x_n=3/2\) wherever \(\varphi\neq0\), whereas a boundary-conormal
distribution is smooth off \(x_n=0\). Conormality is what makes the
order irrelevant in Theorem 11.2.

\subsection{12. Residual operators need not gain
regularity}\label{residual-operators-need-not-gain-regularity}

An ordinary pseudodifferential operator of order \(-\infty\) maps every
Sobolev space into every other. For totally characteristic operators
this fails: the singularity of the kernel at the corner (Remark 6.3) can
prevent any gain.

\textbf{Theorem 12.1} (No gain of regularity). Let
\(a\in S^{-\infty}_{\mathrm{la}}\) with resolved kernel \(F\) (Theorem
6.2).

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\item
  Suppose that for some \(s<s'\) there is \(C\) with \[
|(T_au,v)|\leq C\|u\|_{(s)}\|v\|_{(-s')}\qquad(u,v\in C_0^\infty(\mathbb R^n_+)),
\tag{12.1}
\] which holds in particular if \(T_a\) maps
  \(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\) continuously into
  \(\dot H_{(s')}(\overline{\mathbb R}{}^n_+)\). Then \(F(x',y',0,r)=0\)
  for all \(x',y',r\).
\item
  There are \(a\in S^{-\infty}_{\mathrm{la}}\) with
  \(F(\cdot,\cdot,0,\cdot)\not\equiv0\). For these, \(T_a\) maps no
  \(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\) into any
  \(\dot H_{(s')}(\overline{\mathbb R}{}^n_+)\) with \(s'>s\).
\item
  There are also \(a\in S^{-\infty}_{\mathrm{la}}\), \(a\neq0\), for
  which \(T_a\) maps \(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\) into
  \(\dot H_{(s')}(\overline{\mathbb R}{}^n_+)\) for all \(s,s'\).
\end{enumerate}

\textbf{Proof.} (a) \emph{Test functions.} Let
\(\varphi,\vartheta\in C_0^\infty(\mathbb R^{n-1})\),
\(w,z\in C_0^\infty((0,\infty))\), and integers \(J,J'\geq0\). Put
\(u(y)=\varphi(y')D^J_{y_n}w(y_n)\),
\(v(x)=\vartheta(x')D^{J'}_{x_n}z(x_n)\), and
\(u_\varepsilon(y)=\varepsilon^{-1/2}u(y',y_n/\varepsilon)\),
\(v_\varepsilon(x)=\varepsilon^{-1/2}v(x',x_n/\varepsilon)\),
\(0<\varepsilon\leq1\). Then
\(\widehat{u_\varepsilon}(\xi)=\varepsilon^{1/2}\widehat u(\xi',\varepsilon\xi_n)\),
so \[
\|u_\varepsilon\|^2_{(\sigma)}=(2\pi)^{-n}\int\big(1+|\xi'|^2+\theta^2/\varepsilon^2\big)^\sigma|\widehat u(\xi',\theta)|^2\,d\xi'\,d\theta .
\] For \(\sigma\geq0\) the weight is at most
\(\varepsilon^{-2\sigma}(1+|\xi'|^2+\theta^2)^\sigma\). For \(\sigma<0\)
it is at most \(\varepsilon^{-2\sigma}|\theta|^{2\sigma}\), and
\(|\widehat u(\xi',\theta)|=|\widehat\varphi(\xi')||\theta|^J|\widehat w(\theta)|\),
so the integral is finite if \(2\sigma+2J>-1\). Hence
\(\|u_\varepsilon\|_{(\sigma)}\leq C\varepsilon^{-\sigma}\) when
\(J>-\sigma-\tfrac12\), and likewise for \(v_\varepsilon\). Choosing
\(J>-s-\tfrac12\) and \(J'>s'-\tfrac12\), (12.1) gives
\(|(T_au_\varepsilon,v_\varepsilon)|\leq C\varepsilon^{s'-s}\to0\).

\emph{The limit.} By Theorem 6.2(b),
\((T_au_\varepsilon,v_\varepsilon)=\iint K(x,y)u_\varepsilon(y)\overline{v_\varepsilon(x)}\,dy\,dx\).
Substitute \(x_n=\varepsilon X\), \(y_n=\varepsilon Y\). Since
\(\varepsilon K(x',\varepsilon X,y',\varepsilon Y)=2F\big(x',y',\varepsilon\tfrac{X+Y}2,r\big)/(X+Y)\)
with \(r=2(X-Y)/(X+Y)\), and \(F\) is bounded with decay in \(x'-y'\),
while \(X,Y\) stay in a compact subset of \((0,\infty)\), dominated
convergence gives \[
\lim_{\varepsilon\to0}(T_au_\varepsilon,v_\varepsilon)=\iint\kappa(x',y',X,Y)\,u(y',Y)\,\overline{v(x',X)}\,dx'\,dy'\,dX\,dY,\qquad \kappa=\frac{2F\big(x',y',0,\frac{2(X-Y)}{X+Y}\big)}{X+Y}.
\] So this integral vanishes for all choices above.

\emph{Conclusion.} Let
\(\kappa_{\varphi\vartheta}(X,Y)=\iint\kappa\,\varphi(y')\overline{\vartheta(x')}\,dx'\,dy'\),
a smooth function on \((0,\infty)^2\), homogeneous of degree \(-1\).
Integrating by parts in \(X\) and \(Y\), the vanishing says
\(\iint(\partial_X^{J'}\partial_Y^J\kappa_{\varphi\vartheta})\,w(Y)\overline{z(X)}\,dX\,dY=0\)
for all \(w,z\), so
\(\partial_X^{J'}\partial_Y^J\kappa_{\varphi\vartheta}=0\). Hence, for
fixed \(Y\), \(X\mapsto\partial_Y^J\kappa_{\varphi\vartheta}(X,Y)\) is a
polynomial of degree \(<J'\). But
\(\partial_Y^J\kappa_{\varphi\vartheta}(X,Y)=X^{-1-J}g(Y/X)\) with
\(g(\varrho)=\partial_\varrho^J[\kappa_{\varphi\vartheta}(1,\varrho)]\),
and \(\kappa_{\varphi\vartheta}(1,\varrho)\) is smooth on \([0,1]\) and
flat at \(\varrho=0\) (as \(\varrho\to0\), \(r\to2\), where \(F\)
vanishes to infinite order); so
\(\partial_Y^J\kappa_{\varphi\vartheta}(X,Y)\to0\) as \(X\to\infty\),
and the polynomial is 0. So
\(\partial_Y^J\kappa_{\varphi\vartheta}\equiv0\), and the same argument
in \(Y\) (now using \(r\to-2\)) gives
\(\kappa_{\varphi\vartheta}\equiv0\). As \(\varphi,\vartheta\) are
arbitrary and \(r=2(X-Y)/(X+Y)\) takes every value in \((-2,2)\),
\(F(x',y',0,r)=0\) for \(|r|<2\); for \(|r|\geq2\) it vanishes anyway.

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\setcounter{enumi}{1}
\item
  Let \(\theta(x_n)=e^{-x_n}\) and
  \(0\neq h\in C_0^\infty(\{|z|<\tfrac12\})\), and put
  \(a(x,\xi)=\theta(x_n)\widehat h(\xi)\). Then \(a\in S^{-\infty}_+\)
  and \(A(x,z)=\theta(x_n)h(z)\), which vanishes for
  \(z_n\geq\tfrac12\); so \(a\) is (strongly) lacunary. By (6.6),
  \(F(x',y',0,r)=2h\big(x'-y',\tfrac{2r}{2+r}\big)/(2+r)\), and
  \(2r/(2+r)\) runs through \((-\infty,1)\supset(-\tfrac12,\tfrac12)\)
  as \(r\) runs through \((-2,2)\); so
  \(F(\cdot,\cdot,0,\cdot)\not\equiv0\). By (a), no gain is possible.
\item
  Let \(\theta_1\in C_0^\infty((2,3))\), \(\theta_1\neq0\), and
  \(a(x,\xi)=\theta_1(x_n)\widehat h(\xi)\) with \(h\) as in (b). On
  \(\operatorname{supp}\theta_1\), \(x_n\) is bounded above and below,
  so \(a^\flat\) is an ordinary symbol of order \(-\infty\) on
  \(\mathbb R^n\times\mathbb R^n\), and \(T_a\) maps \(H_{(s)}\) into
  \(H_{(s')}\) for all \(s,s'\), by the Sobolev continuity of ordinary
  pseudodifferential operators (Section 1). Its outputs are supported in
  \(\{2\leq x_n\leq3\}\). Here \(F=0\) for \(t<1\). \(\square\)
\end{enumerate}

So a residual operator may or may not improve regularity: by (b) some
gain nothing at all, and by (c) others gain every amount. The reason for
(a) is dilation invariance. Near the corner the kernel is
\(F(x',y',0,r)/t\), homogeneous of degree \(-1\) in \((x_n,y_n)\).
Normal dilations \(u\mapsto\lambda^{1/2}u(x',\lambda x_n)\) preserve the
\(L^2\) norm but change the \(\dot H_{(s)}\) norms of normal
oscillations by different powers of \(\lambda\), so an operator that
commutes with them cannot gain derivatives unless it vanishes. Test
functions with vanishing normal moments make the argument work at every
\(s\): with generic bumps the norms \(\|u_\varepsilon\|_{(\sigma)}\) are
of size \(\varepsilon^{1/2}\) for \(\sigma<-\tfrac12\), and the estimate
then says nothing when \(s<s'<-\tfrac12\) or \(\tfrac12<s<s'\).

\subsection{13. Exercises}\label{exercises}

\textbf{Exercise 1.} Express \((x_nD_n)^3\) in the basis \(x_n^jD_n^j\),
and check the result on \(x_n^\lambda\) for \(x_n>0\).

\emph{Solution.} Write \(L=x_nD_n\). By (2.1), \(x_n^2D_n^2=L^2+iL\) and
\(x_n^3D_n^3=L(L+i)(L+2i)=L^3+3iL^2-2L\). Hence
\(L^3=x_n^3D_n^3-3iL^2+2L=x_n^3D_n^3-3ix_n^2D_n^2-x_nD_n\). Check:
\(D_nx_n^\lambda=-i\lambda x_n^{\lambda-1}\), so
\(L^3x_n^\lambda=(-i\lambda)^3x_n^\lambda=i\lambda^3x_n^\lambda\), while
the right side gives
\(i[\lambda(\lambda-1)(\lambda-2)+3\lambda(\lambda-1)+\lambda]x_n^\lambda=i\lambda^3x_n^\lambda\).

\textbf{Exercise 2.} Show that the pointwise product of two lacunary
symbols need not be lacunary, although by Theorem 8.1 the composition
symbol always is.

\emph{Solution.} With \(\mathcal Ff(t)=\int e^{-it\xi}f(\xi)d\xi\) we
have \(\mathcal F(fg)=(2\pi)^{-1}\mathcal Ff*\mathcal Fg\), so supports
add. Take \(0\neq g\in C_0^\infty((-0.95,-0.85))\),
\(G\in\mathcal S(\mathbb R)\) with \(\mathcal FG=g\),
\(0\neq\psi\in\mathcal S(\mathbb R^{n-1})\), and
\(a(x,\xi)=e^{-x_n}\psi(\xi')G(\xi_n)\in S^{-\infty}_{\mathrm{la}}\).
Then \(\mathcal F_n(a^2)\) is a multiple of \(g*g\), which is supported
in \((-1.9,-1.7)\) and is not zero (its Fourier transform is a multiple
of \(G^2\)). So \(a^2\) is not lacunary.

\textbf{Exercise 3.} Show that on
\(\overline{\mathcal S}(\mathbb R^n_+)\), for
\(a\in S^m_{\mathrm{la}}\), \[
[D_j,T_a]=T_{D_{x_j}a}\ (j<n),\qquad[x_nD_n,T_a]=T_{x_nD_{x_n}a},
\] so commutators with the generators of \(\operatorname{Diff}_b\) do
not raise the order, unlike \([D_n,T_a]\).

\emph{Solution.} By the proof of Lemma 11.1(a),
\(D_jT_a=T_{(\xi_j+D_{x_j})a}\) and
\(x_nD_nT_a=T_{(\xi_n+x_nD_{x_n}+\xi_nD_{\xi_n})a}\). By (5.4),
\(T_aD_j=T_{\xi_ja}\) and
\(T_ax_nD_n=T_{\xi_n(1-i\partial_{\xi_n})a}=T_{\xi_na+\xi_nD_{\xi_n}a}\).
Subtract. Since \(x_n\) is absorbed by the decay in \(x_n\),
\(x_nD_{x_n}a\in S^m_{\mathrm{la}}\). In contrast
\([D_n,T_a]=-iT_{\partial_{x_n}a}-iT_{\partial_{\xi_n}a}D_n\) by (5.1),
and \(D_n\) is not in \(\operatorname{Diff}_b\).

\textbf{Exercise 4.} For the symbol of Example 6.6, find the ratios
\(y/x\) at which the kernel can be nonzero, and show directly that \(F\)
is smooth for \(t\geq0\).

\emph{Solution.} \(h((x-y)/x)\neq0\) requires \(|1-y/x|<\tfrac12\), that
is \(\tfrac12<y/x<\tfrac32\). In the formula for \(F\),
\(h(2r/(2+r))\neq0\) requires \(-\tfrac12<2r/(2+r)<\tfrac12\), that is
\(-\tfrac25<r<\tfrac23\). On this interval \(2+r\geq\tfrac85\), so \(F\)
is a product of smooth functions of \((t,r)\) for \(t\geq0\), with
support in \(-\tfrac25\leq r\leq\tfrac23\); it vanishes near \(r=\pm2\).
The two descriptions agree, because \(y/x=(2-r)/(2+r)\) maps
\((-\tfrac25,\tfrac23)\) onto \((\tfrac12,\tfrac32)\).

\textbf{Exercise 5.} Let \(a\in S^m_{\mathrm{la}}\) and
\(u\in\overline{\mathcal S}(\mathbb R^n_+)\) with \(u(x',0)=0\). Show
that \((T_au)(x',0)=0\) and compute \(D_n(T_au)(x',0)\).

\emph{Solution.} By (5.2) with \(k=0\),
\((T_au)(x',0)=a_{00}(x',D')u(\cdot,0)=0\). With \(k=1\),
\(D_n(T_au)(x',0)=a_{10}(x',D')u(\cdot,0)+a_{11}(x',D')D_nu(\cdot,0)=a_{11}(x',D')\big(D_nu(\cdot,0)\big)\),
where \(a_{11}(x',\xi')=a(x',0,\xi',0)+(D_{\xi_n}a)(x',0,\xi',0)\). The
first term is the value of the symbol at the boundary with the normal
frequency compressed to 0; the second is a correction of order \(m-1\).

\subsection{14. Positive order on both original Sobolev
spaces}\label{positive-order-on-both-original-sobolev-spaces}

The following proof extends the zero-order bounds without changing the
compressed operator or its distribution action.

\subsubsection{The original theorem and
spaces}\label{the-original-theorem-and-spaces}

Let \(m\ge0\), \(s\in\mathbb R\), and let \(a\in S^m_{\mathrm{la}}\)
take values in \(L(\mathbb C^p,\mathbb C^q)\), with \(p,q\) fixed finite
positive integers. Use the original symbol estimates \[
 |\partial_\xi^\alpha\partial_x^\beta a(x,\xi)|
 \le C_{\alpha\beta\nu}
 (1+|\xi|)^{m-|\alpha|}(1+x_n)^{-\nu},
 \quad x_n\ge0,\quad \nu\ge0,
 \qquad
 \operatorname{supp}\mathcal F_n a\subset[-1,\infty).
 \tag{PS1}
\] The forward Fourier kernel is \(e^{-ix\cdot\xi}\), \(D=-i\partial\),
and {\small
\[
 a^\flat(x,\xi)=a(x,\xi',x_n\xi_n)\quad(x_n\ge0),\qquad
 a^\flat(x,\xi)=0\quad(x_n<0),\qquad
 T_a u=(2\pi)^{-n}\int e^{ix\cdot\xi}a^\flat(x,\xi)\widehat u(\xi)\,d\xi .
 \tag{PS2}
\]
} The full-space norm is
\(\|u\|_{(s)}^2=(2\pi)^{-n}\int(1+|\xi|^2)^s|\widehat u(\xi)|^2\,d\xi\).
The supported space is the closed subspace of this \(H_{(s)}\)
consisting of distributions supported in \(\{x_n\ge0\}\). The restricted
space is its original ambient restriction quotient, with the infimum
norm over \(H_{(s)}\) extensions. Neither space is replaced.

We prove \[
 T_a:\dot H_{(s)}(\overline{\mathbb R}{}^n_+;\mathbb C^p)
       \longrightarrow
       \dot H_{(s-m)}(\overline{\mathbb R}{}^n_+;\mathbb C^q),
 \qquad
 T_a:\overline H_{(s)}(\mathbb R^n_+;\mathbb C^p)
       \longrightarrow
       \overline H_{(s-m)}(\mathbb R^n_+;\mathbb C^q)
 \tag{PS3}
\] continuously. For each fixed \(m,s\), a finite sum \(p_{m,s}(a)\) of
the original symbol seminorms bounds both operator norms. The maps agree
with the original distribution action in Theorem 9.1.

\subsubsection{Exact integer-order
decomposition}\label{exact-integer-order-decomposition}

Fix the same normal convolution \(\rho\) as Lemma 4.4, including its
inverse Fourier convention, \(\widehat\rho=1\) near zero and
\(\operatorname{supp}\widehat\rho\subset(-1/2,1)\). For an integer
\(M\ge1\) and \(a\in S^M_{\mathrm{la}}\), put \[
 w_j=\frac{\xi_j a}{1+|\xi|^2},\quad
 a_j=(w_j)_\rho\quad(1\le j\le n),\qquad
 a_0=a-\sum_{j=1}^n\xi_j a_j
     =\frac{a}{1+|\xi|^2}
       +\sum_{j=1}^n\xi_j(w_j-a_j).
 \tag{PS4}
\] Every \(w_j\in S^{M-1}_+\). Lemma 4.4 gives
\(a_j\in S^{M-1}_{\mathrm{la}}\) and \(w_j-a_j\in S^{-\infty}_+\),
continuously with all the stipulated seminorms. Thus
\(a_0\in S^{M-2}_+\subset S^{M-1}_+\). It is lacunary because the first
definition in (PS4) is a difference of lacunary symbols. The residual
sum in the second definition is retained exactly; its separate summands
need not themselves be lacunary.

Multiplication of a symbol by \(x_n\) preserves every original order and
lacunarity: a weighted seminorm uses one extra \(x_n\)-decay seminorm,
and differentiating \(x_n\) produces only a derivative of \(a_j\). On
the original test spaces the exact left-quantization identity is \[
 T_a=\sum_{j<n}T_{a_j}D_j+
       T_{x_n a_n}D_n+T_{a_0}
     =\sum_{j<n}T_{a_j}D_j+
       x_n T_{a_n}D_n+T_{a_0}.
 \tag{PS5}
\] In the normal term the uncompressed derivative frequency is
\(\xi_n\); the left factor \(x_n\) supplies precisely the original
compression \(x_n\xi_n\). No derivative is commuted past this factor.

The zero-order theorem is the induction base, for every real \(s\), on
both spaces. Suppose the bound of order \(M-1\) has been proved for
every \(s\). Each ambient \(D_j:H_{(s)}\to H_{(s-1)}\) has norm at most
one, since \(|\xi_j|^2\le1+|\xi|^2\). Derivatives preserve supported
distributions and descend continuously to the restriction quotient.
Hence every derivative term in (PS5) maps index \(s\) to \(s-M\). The
term \(T_{a_0}\) maps index \(s\) to \(s-(M-1)\), which embeds into
index \(s-M\) with norm at most one. The same inequality descends to
quotient norms. This proves both integer bounds, with finite original
symbol seminorm control.

The identities initially hold on supported or restricted Schwartz test
functions. Candidate Proposition 10.2(a) gives density for every real
index, and Theorem 9.1 gives their weakly continuous distribution
actions. Consequently the bounded extensions retain precisely (PS5) and
the original distribution action; no boundary delta term is removed by
an arbitrary extension. The supported proof uses functions smooth and
flat on the boundary before taking density.

\subsubsection{A support-preserving exact change of Sobolev
index}\label{a-support-preserving-exact-change-of-sobolev-index}

Set \[
 \ell(\xi)=\sqrt{1+|\xi'|^2}+i\xi_n,\qquad
 -\frac{\pi}{2}<\arg\ell<\frac{\pi}{2},\qquad
 \Lambda_+^z=\mathcal F^{-1}\ell(\xi)^z\mathcal F,
 \quad \ell^z=\exp(z\log\ell).
 \tag{PS6}
\] In dimension one the first term is \(1\), with its zero-dimensional
Fourier factor. The full factor \(\ell\) is retained. In particular \[
 |\ell|^2=1+|\xi'|^2+\xi_n^2=1+|\xi|^2,\qquad
 |\ell^z|=(1+|\xi|^2)^{\operatorname{Re}z/2}
           e^{-\operatorname{Im}z\,\arg\ell}.
 \tag{PS7}
\] It follows by the original Plancherel norm that \[
 \|\Lambda_+^z u\|_{(r-\operatorname{Re}z)}
 \le e^{\pi|\operatorname{Im}z|/2}\|u\|_{(r)}.
 \tag{PS8}
\] For real \(z\) this is equality. The inverse is the literal
multiplier \(\Lambda_+^{-z}\), so the real-index map is an isometry onto
the full \(H_{(r-z)}\), and it will be an isometry of the supported
subspaces once support is proved. All multipliers have smooth
derivatives of polynomial growth; they act continuously on
\(\mathcal S,\mathcal S'\). On each bounded real strip their Schwartz
operator seminorms grow at most as a polynomial in
\(|\operatorname{Im}z|\) times \(e^{\pi|\operatorname{Im}z|/2}\). This
follows by differentiating the full \(\ell^z\): every derivative is a
finite sum of derivatives of \(\ell\), powers of \(\ell^{-1}\), and
polynomial factors in \(z\). Parameter derivatives add powers of
\(\log\ell\), bounded by any fixed positive power of \(1+|\xi|\), which
proves holomorphy on the test spaces.

Here is a proof of support, with the transform constants. For
\(\operatorname{Re}q>0\), the elementary Laplace identity gives \[
 \ell^{-q}=\frac1{\Gamma(q)}
       \int_0^\infty t^{q-1}
          e^{-t\sqrt{1+|\xi'|^2}}e^{-it\xi_n}\,dt .
 \tag{PS9}
\] It is valid first for positive real \(\ell\) by substitution in the
gamma integral and then for \(\operatorname{Re}\ell>0\) by holomorphy
and the identity theorem.

For clarity, the gamma denominator here is nonzero. Integration by parts
in the beta integral gives \[
 \int_0^1 u^{q-1}(1-u)^N\,du
 =\frac{N!}{q(q+1)\cdots(q+N)}.
\] Multiply by \(N^q\) and put \(t=Nu\). The integrand is bounded in
modulus by \(t^{\operatorname{Re}q-1}e^{-t}\) on \(0<t<N\), so dominated
convergence gives \(\Gamma(q)\). Rewriting the finite product, its limit
is \[
 \Gamma(q)=\frac1q\exp\!\left(
   -\gamma q+\sum_{k\ge1}
        \left[\frac qk-\log\left(1+\frac qk\right)\right]\right),
 \quad
 \gamma=\lim_{N\to\infty}\left(\sum_{k=1}^N\frac1k-\log N\right).
\] The latter real limit exists by integral comparison; the complex
series converges absolutely, with summands \(O(k^{-2})\). Each logarithm
is in the right half-plane branch, so the displayed value is nonzero.
This proves the division in (PS9), including complex \(q\).

Let \[
 E_t(x')=(2\pi)^{-(n-1)}
       \int e^{ix'\cdot\xi'}e^{-t\sqrt{1+|\xi'|^2}}\,d\xi'.
\] In dimension one \(E_t=e^{-t}\). The inverse Fourier kernel of (PS9)
is \[
 K_{-q}(x',x_n)=\frac1{\Gamma(q)}
        \int_0^\infty t^{q-1}E_t(x')\delta(x_n-t)\,dt .
 \tag{PS10}
\] This is a tempered distribution: near zero the tested integrand has
the integrable bound \(C t^{\operatorname{Re}q-1}\), and at infinity the
factor \(e^{-t}\), with finitely many test seminorms, makes it
integrable. The kernel is supported in \(x_n\ge0\). The tangential
Fourier factor is exactly \((2\pi)^{-(n-1)}\); inverse transformation of
\(e^{-it\xi_n}\) is the displayed delta, without an extra factor.

For arbitrary \(z\in\mathbb C\), choose an integer \(k\ge0\) with
\(k>\operatorname{Re}z\). Then \(\ell^z=\ell^k\ell^{-(k-z)}\). The
second multiplier has the kernel (PS10); the first is the finite
operator \((\sqrt{1+|D'|^2}+\partial_n)^k\). Tangential multipliers and
normal derivatives preserve normal support. Thus \(\Lambda_+^z\)
preserves support for every \(z\). More explicitly, its action on a
Schwartz function supported in the positive half space has that support
by (PS10) and the finite operator; the one-sided mollification and
compact cutoff approximation in Theorem 9.1(e) extends this conclusion
to every supported tempered distribution. All the operators are
continuous on \(\mathcal S'\), so the support is retained in the weak
limit. The same argument applies to \(-z\).

It follows that, for real \(r\), \[
 \Lambda_+^{-r}:\dot H_{(0)}\longrightarrow\dot H_{(r)},
 \qquad
 \Lambda_+^{r}:\dot H_{(r)}\longrightarrow\dot H_{(0)}
 \tag{PS11}
\] are inverse isometries, with exactly the original full-space norms.
This proves the support-preserving index change rather than assuming
that the multiplier \((1+|D|^2)^{r/2}\) preserves half-space support.

\subsubsection{The original analytic symbol family and the strip
estimate}\label{the-original-analytic-symbol-family-and-the-strip-estimate}

Fix \(0<m<M\), with \(M\) an integer, and retain \[
 b_z=a(1+|\xi|^2)^{(z-m)/2},\qquad
 A_z=(b_z)_\rho+(a-a_\rho),\qquad 0\le\operatorname{Re}z\le M.
 \tag{PS12}
\] The residual contribution is present for every \(z\); \(A_m=a\)
exactly. For each \(z\),
\(A_z\in S^{\operatorname{Re}z}_{\mathrm{la}}\). Frequency
differentiation of the displayed scalar factor yields finite polynomials
in \(z-m\), the full powers of \(1+|\xi|^2\), and the original frequency
coordinates. Hence each indicated symbol seminorm is bounded by
\(C(1+|\operatorname{Im}z|)^L p(a)\), uniformly on this real strip, with
a finite original seminorm \(p\). The convolution and residual bounds of
Lemma 4.4 retain the same property. Holomorphy holds in any fixed
slightly larger symbol order, such as \(S^{M+1}_+\): parameter
derivatives produce powers of \(\log(1+|\xi|^2)\), which the one extra
order bounds. No assertion of order-\(M\) holomorphy at its borderline
is needed.

For \(f,g\) in the original supported Schwartz spaces of dimensions
\(p,q\), respectively, set \[
 F_{f,g}(z)=
  \left(\Lambda_+^{s-z}T_{A_z}\Lambda_+^{-s}f,g\right)_{L^2}.
 \tag{PS13}
\] This is holomorphic on the strip, continuous on its closed boundary.
The preceding test-space estimates and Theorem 5.1 bound its growth
throughout the strip by \(C_{f,g}(1+|\operatorname{Im}z|)^L
 e^{\pi|\operatorname{Im}z|/2}\). On the two boundary lines, the proved
integer bounds, (PS8) and (PS11) give \[
 |F_{f,g}(i\tau)|
 \le C_0(1+|\tau|)^L e^{\pi|\tau|/2}
        p(a)\|f\|_2\|g\|_2,\qquad
 |F_{f,g}(M+i\tau)|
 \le C_M(1+|\tau|)^L e^{\pi|\tau|/2}
        p(a)\|f\|_2\|g\|_2 .
 \tag{PS14}
\] Take a single finite seminorm \(p\) large enough for both endpoints.
The zero-order norm depends continuously on finitely many original
symbol seminorms, as proved in Section 10. Rescaling a symbol by their
sum gives a linear bound in that sum; the integer induction uses only
continuous linear symbol operations. Thus the stated \(p(a)\) bounds are
homogeneous, including \(p(a)=0\).

For any fixed \(\varepsilon>0\), multiply \(F\) by
\(\exp(\varepsilon(z-m)^2)\). Its boundary bounds now have finite
constants \(C'_0,C'_M\), since
\((1+|\tau|)^L e^{\pi|\tau|/2-\varepsilon\tau^2}\) is bounded. Its
modulus tends to zero on the horizontal edges of large rectangles in the
strip. Dividing it by \(p(a)\|f\|_2\|g\|_2 C'_0\) and multiplying by
\(\exp[-(z/M)\log(C'_M/C'_0)]\) makes the two vertical-edge bounds at
most one. The maximum principle on these rectangles, followed by their
expanding limit, therefore gives \[
 |F_{f,g}(m)|
 \le (C'_0)^{1-m/M}(C'_M)^{m/M}
      p(a)\|f\|_2\|g\|_2 .
 \tag{PS15}
\] The zero-seminorm or zero-test-function case is immediate and does
not require dividing by zero. This is the required strip estimate with
its growth control proved.

Since \(A_m=a\), duality in the supported \(L^2\) space and density of
its Schwartz subspace show that \(\Lambda_+^{s-m}T_a\Lambda_+^{-s}\) is
bounded on supported \(L^2\). Conjugating by the exact isometries (PS11)
gives the supported bound (PS3) for this real \(m\), with finite
original seminorm control. Together with the integer case it covers
every \(m\ge0\) and every real \(s\). The bounded extension agrees with
the original supported distribution action by test-space density and its
weak continuity.

\subsubsection{The full restriction
quotient}\label{the-full-restriction-quotient}

Theorem 7.3 gives \(a^\dagger\in S^m_{\mathrm{la}}\), continuously and
conjugate-linearly, with the original reversed vector dimensions. Apply
the supported result at the index \(m-s\): \[
 T_{a^\dagger}:\dot H_{(m-s)}(\mathbb C^q)
      \longrightarrow\dot H_{(-s)}(\mathbb C^p).
 \tag{PS16}
\] For the supported/restricted duality of Proposition 10.2(b), \[
 |(T_a u,v)|=|(u,T_{a^\dagger}v)|
 \le C p_{m,s}(a)
       \|u\|_{\overline H_{(s)}}\|v\|_{\dot H_{(m-s)}} .
 \tag{PS17}
\] The antidual of \(\dot H_{(m-s)}\) is exactly \(\overline H_{(s-m)}\)
with its original quotient norm, so (PS17) proves the restricted bound,
without choosing or identifying a supported extension of the restricted
input. Density and Theorem 9.1(d) retain the original quotient
distribution action. This completes both assertions of (PS3).

For \(m<0\), the same zero-order membership proves boundedness at the
same index, but the argument above does not assert a gain of \(-m\).
Theorem 12.1 gives nonzero residual examples forbidding every such gain.
The original residual term, and this exact limitation, remain part of
the calculus.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={Figure PS-F1. PS6--PS11 keeps the full multiplier sqrt(1+\textbar xi'\textbar\^{}2)+i xi\_n, the original weighted norm, Gamma(q), the inverse Fourier factor and normal kernel support, with Re q\textgreater0. PS12--PS15 uses integer M\textgreater m and epsilon\textgreater0, retaining the residual in A\_z. The strip is a schematic with 0\textless m\textless M. All four spaces and arrows have their original vector dimensions; PS16--PS17 gives the restriction quotient. The figure source is figures/positive\_order\_halfspace.py.}]{../figures/positive_order_halfspace.png}}
\caption{Figure PS-F1. PS6--PS11 keeps the full multiplier
sqrt(1+\textbar xi'\textbar\^{}2)+i xi\_n, the original weighted norm,
Gamma(q), the inverse Fourier factor and normal kernel support, with Re
q\textgreater0. PS12--PS15 uses integer M\textgreater m and
epsilon\textgreater0, retaining the residual in A\_z. The strip is a
schematic with 0\textless m\textless M. All four spaces and arrows have
their original vector dimensions; PS16--PS17 gives the restriction
quotient. The figure source is figures/positive\_order\_halfspace.py.}
\end{figure}

\subsection{Where this leads}\label{where-this-leads}

\begin{itemize}
\tightlist
\item
  \emph{Operators of positive order.} An operator with symbol in
  \(S^m_{\mathrm{la}}\), \(m>0\), maps
  \(\dot H_{(s)}(\overline{\mathbb R}{}^n_+)\) continuously into
  \(\dot H_{(s-m)}(\overline{\mathbb R}{}^n_+)\) (Section 14). So the
  order bounds the loss of regularity, while by Theorem 12.1 an operator
  of negative order need not gain any.
\item
  \emph{The global calculus.} The class is invariant under changes of
  variables that preserve the boundary, so it makes sense on manifolds
  with boundary. Its kernels live on the product blown up at the corner,
  as in Section 6, and the calculus has principal symbols, wave front
  sets and a notion of ellipticity. This is Melrose's b-calculus. The
  free author version {[}Melrose{]} develops its index theory, and
  {[}Loya{]} surveys related Dirac-operator constructions.
\item
  \emph{Not treated here.} We do not discuss whether, conversely, the
  vanishing of the resolved kernel \(F\) on the front face makes a
  residual operator improve regularity. Nor do we treat continuity on
  the restricted Besov spaces \(\overline B^s_{2,p}(\mathbb R^n_+)\) for
  \(p\neq2\), which would need a duality theorem for dyadic Besov spaces
  on the half space. For \(p=2\) these are the spaces
  \(\overline H_{(s)}(\mathbb R^n_+)\) of Theorem 10.6.
\end{itemize}

\subsection{References}\label{references}

\begin{itemize}
\tightlist
\item
  {[}Loya{]} Paul Loya,
  \href{https://people.math.binghamton.edu/loya/papers/bpag-art.pdf}{\emph{Index
  theory of Dirac operators on manifolds with corners up to codimension
  two}}, freely accessible author-hosted version, listed on the
  \href{https://people.math.binghamton.edu/loya/research.html}{author's
  research page}.
\item
  {[}Melrose{]} Richard B. Melrose,
  \href{https://math.mit.edu/~rbm/tapsit.html}{\emph{The
  Atiyah--Patodi--Singer Index Theorem}, freely available author
  version}, with the \href{https://math.mit.edu/~rbm/tapsit1.ps}{first
  half} and \href{https://math.mit.edu/~rbm/tapsit2.ps}{second half}
  provided by the author.
\end{itemize}

\end{document}