# Derivative sequences and microlocal regularity

*Written by GPT-6.1 Sol (OpenAI), October 2026.  Public domain (CC0).*

A smooth function can have derivatives that grow too quickly for a chosen regularity class. A derivative sequence measures that growth. Its wavefront set then distinguishes frequency directions where the stronger class fails, even when ordinary wavefront is empty. Gaussian packets detect this information through estimates at every index with one common constant.

Basic references are [Analytic directions of holomorphic boundary values][A], [Pulling back and combining analytic singularities][P], and [Gaussian packets and homogeneous Fourier directions][G]. Boman [B] gives background on derivative-weight wavefront sets. The [complete proof](derivative-sequences-and-microlocal-regularity-formal.md) supplies the exact sequence convention, both directions of the packet criterion, all the needed operation bounds and the homogeneous Fourier interchange. Ordinary continuity and uniform class estimates are treated separately. [Cauchy bounds, root counts and analytic extensions][C], Lemmas 1.1 and 3.1, gives the exact coordinate estimates.

## 1. The sequence controls derivatives, not a Sobolev order

We use an increasing positive sequence satisfying
\[
 L_0=1,\qquad k\leq L_k,\qquad
 L_{k+1}\leq C_L L_k,\qquad M_k=L_k^k,\quad M_0=1.
 \tag{1}
\]
Membership in \(C^L\) means that on every compact set
\[
 |\partial^\alpha f|\leq A^{|\alpha|+1}M_{|\alpha|}
 \tag{2}
\]
for one constant \(A\) independent of the derivative index. The function is smooth. This is a derivative-growth condition, not a bound on finitely many Sobolev derivatives.

Three useful choices are:

| Sequence | Derivative weight | Class |
|---|---|---|
| \(L_k=k+1\) | \((k+1)^k\) | Analytic |
| \(L_k=(k+1)^s\), \(s\geq1\) | \((k+1)^{sk}\) | Gevrey of order \(s\) |
| \(L_k=2^k\) | \(2^{k^2}\) | A larger derivative class |

All satisfy (1). The factorial comparison
\[
 (k/e)^k\leq k!\leq k^k,\qquad
 (k+1)^k\leq e k^k
 \tag{3}
\]
makes the Gevrey weight equivalent to \((k!)^s\) up to exponential constants. The analytic case is exactly the factorial derivative criterion and the convergent Taylor series proved in [G], Corollary 3.2.

The fixed-shift estimate is central:
\[
 M_{N+c}\leq D_c^{N+1}M_N
                       \quad\text{for every fixed }c\geq0.
 \tag{4}
\]
Complete-proof Lemma 1.1 derives it from \(L_N\leq C_L^N\) and \(L_{N+c}\leq C_L^cL_N\). A fixed number of additional derivatives therefore changes the common exponential constant, while preserving the class weight.

The class is closed under products, sums, differentiation and composition with analytic coordinate maps. Complete-proof Proposition 1.2 proves these statements with the product rule and finite Taylor jets. It does not assume that a general \(C^L\) function extends holomorphically.

## 2. Class wavefront uses a common bounded localization

A pair \((x_0,\xi_0)\), \(\xi_0\ne0\), is absent from \(\operatorname{WF}_L(u)\) when there are a common neighborhood \(U\), a common frequency cone, and distributions \(w_N=u\) on \(U\), with common compact support and common finite order, such that
\[
 |\widehat w_N(\eta)|
       \leq A^{N+1}M_N(1+|\eta|)^{-N}
 \tag{5}
\]
throughout that cone. The constant is the same for every \(N\). A different compact distribution of unrelated order at each index would not meet this definition.

The comparisons are
\[
 \operatorname{WF}(u)\subseteq\operatorname{WF}_L(u)
                              \subseteq\operatorname{WF}_A(u).
 \tag{6}
\]
An analytic witness already has a weight bounded by \(M_N\). A class witness gives each desired ordinary inverse power after fixing a sufficiently large index. The same ordinary cutoff then serves every power. These are the two distinct arguments in complete-proof Lemma 2.1.

Controlled cutoffs may be chosen with a common plateau and support and
\[
 \|\partial^{\alpha+\beta}\chi_N\|_\infty
      \leq C_\beta(CN)^{|\alpha|}\quad(|\alpha|\leq N),
 \tag{7}
\]
while every fixed low derivative is uniform. They are smooth families with finite-index estimates; no compact cutoff in a quasianalytic class is asserted. Complete-proof Lemma 2.2 normalizes class witnesses using these cutoffs, permits simultaneous regular cones, and proves multiplication and differential decrease for \(C^L\) coefficients.

The base projection of class wavefront is exactly failure of local \(C^L\) membership. For the difficult direction, a finite directional cover gives one sequence estimated in every direction. Fourier inversion at index \(N=k+n+1\) then uses
\[
 \int_0^\infty r^{k+n-1}(1+r)^{-k-n-1}dr=\frac1{k+n}.
 \tag{8}
\]
The fixed shift (4) bounds the derivative of order \(k\) by \(C^{k+1}M_k\). This is complete-proof Theorem 2.3.

## 3. One Gaussian criterion works for every admissible sequence

Use complex-linear pairings and the transform
\[
 \widehat f(\eta)=\int e^{-iy\cdot\eta}f(y)\,dy,\qquad
 \widehat u(\phi)=u(\widehat\phi),
 \tag{9}
\]
with inverse factor \((2\pi)^{-n}\). For a tempered local extension,
\[
 T_hu(x,\xi)
       =u\big(e^{-|y-x|^2/(2h)}e^{-iy\cdot\xi/h}\big).
 \tag{10}
\]
The Gaussian has width \(\sqrt h\), while its frequency is \(\xi/h\).

Complete-proof Theorem 3.1 says that class absence is equivalent to
\[
 |T_hu(x,\xi)|\leq A^{N+1}M_Nh^N
   \quad(N\geq1,\ x\in U,\ \xi\in V,\ 0<h\leq h_0),
 \tag{11}
\]
on common neighborhoods, with \(V\) in an annulus away from zero. It is a uniform estimate over the neighborhoods and all indices.

The forward direction averages a compact class Fourier estimate against the Gaussian. Frequencies outside the regular cone and terms supported away from the tested positions produce exponential tails. Those tails have the bound \(C^{N+1}N^Nh^N\), which is at most the class bound.

The converse uses the finite inverse heat expansion from [G], Lemma 2.1. For an extension of Schwartz order \(s\), fix
\[
 K=N+s,\qquad u_N=\chi_{2K}u.
 \tag{12}
\]
For a frequency \(r\theta\), choose \(h=\rho/r\), \(\rho=|\xi_0|\), and the auxiliary amplitude
\[
 b=\sum_{j=0}^{K-1}\frac{(-h/2)^j}{j!}\Delta^j\chi_{2K}.
 \tag{13}
\]
The exact heat remainder has weighted norm \(C^{K+1}K^Kh^K\), and \(\|b\|_1\leq C e^{ChK^2}\). On \(r\geq DK\), the last exponential is at most \(C_1^{N+1}\). Apply (11) at the fixed additional index \(N+c\), \(c\geq n/2\), to account for the Gaussian normalization. Equation (4) absorbs this shift. The remainder has inverse power \(r^{-N}\) after paying the Schwartz order. Lower frequencies use the common compact-order bound.

The sequence \(u_N\) is fixed for each \(N\). The auxiliary amplitude may depend on the frequency. Complete-proof Section 3 gives the full inequalities and makes this distinction explicit.

For Gevrey order \(s\), the equivalent packet estimate is
\[
 |T_hu|\leq C e^{-c h^{-1/s}}.
 \tag{14}
\]
Optimizing the index proves one implication; maximizing \(r^{sN}e^{-cr}\) proves the other. The large sequence \(L_N=2^N\) has a different optimized rate, computed in Example 1.

## 4. Pullback, products and Fourier interchange

For an analytic map \(F:X\to Y\), the normal set contains target covectors killed by \(DF^T\). If it misses \(\operatorname{WF}_L(u)\), the actual ordinary pullback exists and
\[
 \operatorname{WF}_L(F^*u)
     \subseteq\{(x,DF(x)^T\eta):
                         (F(x),\eta)\in\operatorname{WF}_L(u)\}.
 \tag{15}
\]
For an analytic diffeomorphism the bound is an equality. Analytic submanifold restriction is the embedding case. These are complete-proof Theorem 5.1 and Corollary 5.2, using the full ordinary construction of [P].

If \(F:X\subset\mathbb R^b\to Y\subset\mathbb R^a\), the phase is \(F(x)\cdot\eta-x\cdot\xi\). A common compact input order \(m\) is handled by the index
\[
 J=N+m+a+2.
 \tag{16}
\]
Here \(a\) is the dimension of the input distribution's space, the target of the map \(F\). On the nonstationary part, the full analytic phase estimate of [P], Lemma 3.1, has weight \(J^J\leq M_J\). On the regular-input part, the class Fourier bound has weight \(M_J\) and the amplitude is bounded. The proof uses one weight on each part. It never multiplies them into \(M_J^2\). The shift (4) gives (15).

Adjoin zero over actual support, writing \(\operatorname{WF}_{L,0}\). The tensor and product bounds are
\[
 \begin{aligned}
 \operatorname{WF}_L(u\otimes v)
    &\subseteq
       (\operatorname{WF}_{L,0}(u)\times
                            \operatorname{WF}_{L,0}(v))\setminus0,\\
 \operatorname{WF}_L(uv)
    &\subseteq\{(x,\xi+\eta):
       (x,\xi)\in\operatorname{WF}_{L,0}(u),\
       (x,\eta)\in\operatorname{WF}_{L,0}(v)\}\setminus0 .
 \end{aligned}
 \tag{17}
\]
The product condition is absence of opposing nonzero class covectors. Complete-proof Section 6 proves it by analytic diagonal pullback. Zero components retain single-factor singularities.

For a projection proper on the source support,
\[
 \operatorname{WF}_L(\pi_*u)
    \subseteq\{(x,\xi):\xi\ne0,\
         (x,y;\xi,0)\in\operatorname{WF}_L(u)\text{ for some }y\}.
 \tag{18}
\]
Only zero fiber covectors contribute. The finite compact-fiber localization and full proper kernel corollary are complete-proof Section 7.

Finally, if \(u\) satisfies a finite Euler-chain equation of complex degree on the punctured space,
\[
 (E-\lambda)^{q+1}u=0,\qquad E=x\cdot\nabla_x,
 \tag{19}
\]
complete-proof Theorem 4.1 gives
\[
 (x,\xi)\in\operatorname{WF}_L(u)
     \quad\Longleftrightarrow\quad
 (\xi,-x)\in\operatorname{WF}_L(\widehat u),
                      \qquad x\ne0,\quad \xi\ne0.
 \tag{20}
\]
The all-dimension temperedness, full Fourier phase and both origin anomalies are the exact supplied proofs in [G]. Fixed polynomial and logarithmic factors are absorbed by (4). Arbitrary origin point terms are retained; their Fourier polynomials belong to every class.

## 5. Four worked examples

### Example 1: exponential sequence growth has an exact logarithmic packet rate.

Let \(L_N=2^N\), so \(M_N=2^{N^2}\). Its adjacent ratio is two, and \(N\leq2^N\) by induction. Its shift ratio is exactly
\[
 \frac{M_{N+c}}{M_N}=2^{2cN+c^2}.
 \tag{21}
\]
This is still an exponential constant in \(N\) for each fixed \(c\), so all the preceding theorems apply.

To optimize the packet bound, write \(a=\log2\) and
\(B=\log(1/(Ah))\). Its logarithm is bounded by
\[
 \log A+aN^2-BN
      =\log A-\frac{B^2}{4a}
               +a\left(N-\frac{B}{2a}\right)^2 .
 \tag{22}
\]
For sufficiently small \(h\), choose \(N=\lfloor B/(2a)\rfloor\geq1\). The last square is at most one, yielding
\[
 |T_hu|\leq 2A
       \exp\left(-\frac{(\log(1/(Ah)))^2}{4\log2}\right).
 \tag{23}
\]
Conversely, an estimate of this same form with arbitrary fixed prefactor \(C\) and scale \(B_0>0\) implies the all-index bound. Indeed put \(v=\log(1/(B_0h))\). Then
\[
 h^{-N}e^{-v^2/(4a)}
    =B_0^N e^{Nv-v^2/(4a)}
    \leq B_0^N e^{aN^2}=B_0^N2^{N^2}.
 \tag{24}
\]
Complete the square, or maximize the quadratic, to prove the inequality.

The quadratic coefficient \(1/(4\log2)\) in this equivalent estimate is prescribed by the sequence. Replacing it by an arbitrary smaller positive coefficient could change the derivative class; a linear shift of \(v\) cannot absorb a different quadratic coefficient.

For any fixed Gevrey order \(s\), \((N+1)^s\leq C_s2^N\). Thus
\[
 C^{\text{Gevrey }s}\subseteq C^{L_N=2^N},\qquad
 \operatorname{WF}_{L_N=2^N}\subseteq\operatorname{WF}_{\text{Gevrey }s}.
 \tag{25}
\]
A simple bound is \(C_s=\sup_{N\geq0}(N+1)^s2^{-N}<\infty\); its finiteness follows from decay of a polynomial times a geometric sequence. This larger class still consists of smooth functions. No assertion that it contains every smooth function follows from (25).

### Example 2: a smooth boundary has a sharp Gevrey threshold.

Consider
\[
 f(x)=\int_0^\infty e^{-\sqrt t}e^{itx}\,dt .
 \tag{26}
\]
All its derivative moments are finite, so differentiation under the integral is justified, and
\[
 f^{(k)}(0)=2i^k(2k+1)!.
 \tag{27}
\]
Substitute \(t=r^2\) and integrate \(r^{2k+1}e^{-r}\) by parts. Its derivatives everywhere satisfy
\[
 |f^{(k)}(x)|\leq2(2k+1)!
               \leq C^{k+1}(k+1)^{2k}.
 \tag{28}
\]
The fixed extra power of \(k+1\) is absorbed exponentially. Thus \(f\) is globally Gevrey of order two.

For every \(1\leq s<2\), the lower factorial bound in (3) makes (27) incompatible with \(A^{k+1}(k+1)^{sk}\), for any \(A\). So \(f\) does not belong locally to that smaller class at zero. It remains smooth there.

The integral is holomorphic for \(\operatorname{Im}z>0\), with uniform bound two. Its boundary is \(f\). The positive-polar analytic tube theorem [A], Theorem 4.1, excludes negative analytic covectors. Class inclusion (6) excludes negative covectors for every derivative class too.

It is analytic at every nonzero real \(x_0\). For a direct proof write
\[
 f(x)=2\int_0^\infty r e^{-r}e^{ixr^2}dr.
 \tag{29}
\]
Rotate the \(r\)-ray by \(\theta=\operatorname{sgn}(x_0)\pi/8\). The integrand is entire in \(r\). Its entire power series can be integrated term by term to obtain an entire primitive. Its integral round a finite sector boundary is therefore zero. On the large joining arc, the Gaussian phase has nonpositive real exponent and \(e^{-r}\) has modulus at most \(e^{-c|r|}\), so the arc integral tends to zero. The small joining arc tends to zero since the integrand is \(O(r)\). The rotated integral is
\[
 2e^{2i\theta}\int_0^\infty
       r e^{-e^{i\theta}r}e^{iz e^{2i\theta}r^2}dr .
 \tag{30}
\]
Near \(z=x_0\), the real part of \(iz e^{2i\theta}\) is strictly negative. A uniform bound \(Cr e^{-cr^2+Cr}\) justifies all compact complex derivatives, so (30) is a holomorphic continuation there.

Consequently, for the Gevrey weight of order \(s\),
\[
 \operatorname{WF}_{\text{Gevrey }s}(f)=
 \begin{cases}
   \{(0,\xi):\xi>0\},&1\leq s<2,\\
   \varnothing,&s\geq2.
 \end{cases}
 \tag{31}
\]
The lower ray follows from failure of class regularity, complete-proof Theorem 2.3, and positive conicity. The analytic fiber is the \(s=1\) case. Ordinary wavefront is empty because the function is smooth.

The whole Fourier transform, including its coefficient, is
\[
 \widehat f(\eta)=2\pi H(\eta)e^{-\sqrt\eta}.
 \tag{32}
\]
For a Schwartz test, interchange the integrals: \(\int e^{-\sqrt t}dt=2\) and the test transform is integrable. Fourier inversion [F] then gives the coefficient \(2\pi\).

The positive packet has an exact leading rate:
\[
 \lim_{h\downarrow0}e^{1/\sqrt h}T_hf(0,1)=2\pi e^{1/8}.
 \tag{32a}
\]
To prove it, interchange the defining integral with the Gaussian test, which is absolutely legitimate, and perform the full Gaussian Fourier integral. With \(p=h^{-1/2}\) and \(v=\sqrt h(t-h^{-1})\), this gives
\[
 e^pT_hf(0,1)=\sqrt{2\pi}
      \int_{-p}^{\infty}
        \exp\left(-\frac{v}{\sqrt{1+v/p}+1}-\frac{v^2}{2}\right)dv .
 \tag{32b}
\]
Rationalizing the square root gives the displayed exponent exactly. Extend the integrand by zero below \(-p\). It is bounded by \(e^{|v|-v^2/2}\), independently of \(p\), and converges pointwise to \(e^{-v/2-v^2/2}\). Dominated convergence and completion of the square yield (32a). This rate is incompatible even at the central pair with \(e^{-c h^{-1/s}}\) for \(s<2\), since \(1/s>1/2\). It is consistent with the full Gevrey-two neighborhood estimate already proved.

![Exact factorial derivative moments normalized by three Gevrey weights, beside the scaled positive Gaussian packet and its proved limit.](figures/factorial-growth-and-Gevrey-packet-rate.png)

*Figure 1. The function is \(f(x)=\int_0^\infty e^{-\sqrt t}e^{itx}dt\). Left: the exact moments have modulus \(2(2k+1)!\); the plotted values are \(|f^{(k)}(0)|^{1/k}/(k+1)^s\), at integer orders \(1\leq k\leq128\), for \(s=1,3/2,2\). Lines connect those integer samples. Right: \(p=1/\sqrt h\) takes values \(4,6,8,12,16,24,32,48\), and each point is the exact scaled integral \(e^pT_{p^{-2}}f(0,1)\), evaluated with 90-digit arithmetic. The dashed line is the proved limit \(2\pi e^{1/8}\), not a fitted value. The class threshold and the limit are proved in Example 2, equations (27)–(28) and (32a)–(32b), and Solution 3. Complete-proof Example 8.2 supplies the same moments for the convergence example.*

### Example 3: an angular Gevrey defect becomes a Fourier half-axis.

In the plane let \(\operatorname{Re}\lambda>-2\), and define
\[
 u_\lambda(x)=|x|^\lambda f\left(\frac{x_1}{|x|}\right).
 \tag{33}
\]
The angular function is bounded and smooth. The radial integrability condition at zero is exactly the stated inequality; at infinity the modulus has polynomial growth. Hence this defines a tempered distribution, globally homogeneous of degree \(\lambda\).

Away from zero the radial factor and its reciprocal are analytic. The angular map \(w=x_1/|x|\) is analytic. Its only class-singular input value is \(w=0\), where its differential is nonzero, so the transverse pullback exists. At \(x=(0,t)\), \(t\ne0\),
\[
 dw=\left(\frac1{|t|},0\right).
 \tag{34}
\]
Example 2 and the analytic pullback bound therefore allow only positive horizontal covectors there for \(1\leq s<2\). They give none elsewhere away from zero.

For equality use local analytic coordinates \((r,w)\):
\[
 x=(rw,\sigma r\sqrt{1-w^2}),\qquad
                       r>0,\quad \sigma=\operatorname{sgn}t.
 \tag{35}
\]
In these coordinates the distribution is \(r^\lambda f(w)\). Multiplication by the analytic nonzero radial factor preserves the class wavefront in both directions. Formula (55) of the complete proof gives the exact tensor-with-constant fiber, so the positive \(w\)-covector ray is present. Analytic coordinate covariance and (34) give the exact physical set
\[
 \{((0,t),(a,0)):t\ne0,\ a>0\}
                         \quad(1\leq s<2).
 \tag{36}
\]

The full homogeneous interchange sends it to
\[
 \{((a,0),(0,-t)):a>0,\ t\ne0\}.
 \tag{37}
\]
At every positive horizontal Fourier base point the class fiber has both nonzero vertical signs. At any other nonzero Fourier base it is empty. For \(s\geq2\), both nonzero-base class wavefront sets are empty, by the same theorem and the global angular Gevrey bound. Ordinary wavefront is empty on both punctured spaces. Analytic wavefront is the \(s=1\) version of (36)–(37), so smoothness here does not imply analytic regularity.

The dot product in every displayed source pair is zero. The packet rotation phase is therefore one at these pairs, while its full factor remains \(2\pi\) in dimension two. Origin terms are not inferred from the punctured description.

![Positive horizontal class covectors at two vertical source bases rotate to both vertical signs at one positive Fourier base.](figures/angular-class-defect-and-Fourier-covectors.png)

*Figure 2. For \(u_\lambda(x)=|x|^\lambda f(x_1/|x|)\), \(\operatorname{Re}\lambda>-2\), the displayed class is Gevrey of order \(1\leq s<2\). The marked physical bases are \(x_\pm=(0,\pm1)\), both with covector \(\xi=(1,0)\). Rotation \((x,\xi)\mapsto(\xi,-x)\) gives the same Fourier base \((1,0)\) with covectors \((0,-1)\) and \((0,1)\). Blue lines identify the full punctured class-singular base sets: the vertical physical axis and the positive horizontal Fourier half-axis. The dashed unit circle locates the two marked source points. Both outlined origins are excluded from interchange. Covectors use the Euclidean identification and display scale \(3/5\); coordinate scales are equal in each panel. At Gevrey orders \(s\geq2\), both punctured class wavefront sets are empty. See Example 3, Solution 5 and complete-proof Theorem 4.1.*

### Example 4: a proper kernel can retain a zero fiber covector.

Let
\[
 K(x,y)=f(x)\delta_0(y),\qquad a(y)=1+y.
 \tag{38}
\]
The kernel support is proper over \(x\), since a compact \(x\)-set sees only \(y=0\). The input \(a\) is analytic and belongs to every class. The actual kernel pairing gives
\[
 (\mathcal Ka)(x)=a(0)f(x)=f(x).
 \tag{39}
\]
For orders \(1\leq s<2\), its output class fiber at zero is positive by Example 2.

The contributing source covectors include \((0,0;\xi,0)\), \(\xi>0\). They really occur, despite the zero \(y\)-covector. Direct Gaussian pairing gives
\[
 T_hK((x,y),(\xi,\eta))=T_hf(x,\xi)e^{-y^2/(2h)}.
 \tag{40}
\]
At \(y=0\) this is exactly the packet of \(f\), independent of \(\eta\). A class absence neighborhood for \(K\) at \((0,0;\xi,0)\) would give one for \(f\) at \((0,\xi)\), contradicting (31). Thus these source covectors are present and survive projection.

If both tensor factors were required to have nonzero covectors, these would be discarded. The zero component over the delta's support in (51) is necessary. The ordinary kernel has only its vertical delta covectors near this point, since \(f\) is smooth. Its projected ordinary wavefront is empty. The class projection sees additional horizontal covectors because \(f\) is not in the smaller Gevrey class.

## 6. Exercises

The eight problems total 100 points. Complete solutions follow.

1. **Weights and their packet rate, 10 points.** Prove the fixed-shift bound for an arbitrary sequence satisfying (1). For \(L_N=2^N\), derive both directions of (23)–(24) and identify why an arbitrary quadratic coefficient would not express the same class. Prove the class and wavefront inclusions in (25).
2. **Keep the localization fixed, 12 points.** Explain how controlled cutoffs normalize a class witness without a compact \(C^L\) cutoff. In the Gaussian converse, derive the remainder bound with \(K=N+s\), and the main-term bound using index \(N+c\), \(c\geq n/2\). State which distribution is fixed before the frequency varies.
3. **Find a smooth class singularity, 12 points.** Prove (27)–(28), the failure of Gevrey membership below order two, and the exact fiber (31). Establish the analytic continuation (30), including both joining arcs, and the whole coefficient in (32).
4. **Retain the exact scaled Fourier identity, 14 points.** For a globally strict homogeneous distribution of complex degree \(\lambda\) in dimension \(n\), derive the exact identity relating \(T_h\widehat u(\xi,-x)\) to \(T_hu(x,\xi)\), including its complex phase and \(h\)-power. Explain how it preserves a class estimate through fixed index shifts. Describe the errors required for an arbitrary punctured Euler-chain extension.
5. **Transport an angular class defect, 12 points.** Verify every assertion in (33)–(37), including global temperedness, complex degree, analytic coordinates, the sign in (34), exact lower fibers and both Fourier covector signs. State which origin pairs are excluded.
6. **Project the kernel's horizontal covector, 12 points.** Derive (39)–(40) from tests. Prove the presence of the zero-fiber covectors and explain how they fit both the class tensor and proper projection bounds. Compare with the ordinary kernel covectors.
7. **Use one class weight in pullback and product, 14 points.** For \(F:X\subset\mathbb R^b\to Y\subset\mathbb R^a\), identify the role of each dimension in (16). Derive the two integral bounds (49)–(50) of the complete proof and their reduction to \(M_N\). Use analytic diagonal pullback to prove the product condition and retain the zero-component terms.
8. **Separate convergence from uniform class control, 14 points.** Let \(f_R\) be (62) of the complete proof. Prove convergence to \(f\) in the ordinary normal topology with empty ordinary cone. Explain why class absence is lost below Gevrey order two. Prove that common all-index packet bounds do pass to a distributional limit, and state the additional proper-support and ordinary-cone conditions required when taking limits of the admissible operations.

## 7. Complete solutions

### Solution 1

Iteration gives \(L_N\leq C_L^N\) and \(L_{N+c}\leq C_L^cL_N\). Thus
\[
 M_{N+c}\leq C_L^{c(N+c)}L_N^{N+c}
              \leq C_L^{2cN+c^2}M_N.
\]
For \(N=0\), take \(M_0=1\) and absorb \(C_L^{c^2}\). For positive \(N\), the same fixed \(D_c\) bounds the expression by \(D_c^{N+1}M_N\). Also \(N^N\leq M_N\) by (1).

For \(M_N=2^{N^2}\), the logarithm of (11)'s right side is
\(\log A+aN^2-N\log(1/(Ah))\), \(a=\log2\). Completing the square gives (22). The integer part of the minimizer has distance less than one from it, so the error in the exponent is at most \(a\). For \(h\) small enough the integer is positive. This proves (23); bounded remaining \(h\)'s can be included with a larger prefactor.

Conversely, assume the same quadratic denominator \(4a\) with scale \(B_0\). For every positive integer \(N\), the quadratic \(Nv-v^2/(4a)\) has maximum \(aN^2\). Equation (24) follows, hence a bound \(C B_0^N2^{N^2}h^N\). Increase one common \(A\) to absorb \(C\) and \(B_0\). This is exactly the packet criterion.

Changing \(a\) changes the coefficient of \(N^2\). A factor \(A^{N+1}\) changes its logarithm only by a linear function of \(N\), so it cannot absorb that change. Therefore an unspecified logarithmic Gaussian exponent is not a substitute for the stated weight.

Finally, \((N+1)^s/2^N\) is bounded for every fixed \(s\), since its logarithm is \(s\log(N+1)-N\log2\to-\infty\). Raising the bound to the power \(N\) shows that a Gevrey weight is at most \(C_s^N2^{N^2}\). Any Gevrey derivative or Fourier-witness estimate is therefore an estimate for the larger class. Absence for the smaller function class implies absence for the larger one, so the wavefront inclusion goes in the reverse direction stated in (25).

### Solution 2

The controlled families have fixed support and plateau, uniformly bounded low derivatives, and the mixed finite-index estimates (7). Their existence is the full convolution construction [A], Lemma 1.1. They need not belong to \(C^L\) as individual compact functions.

At a regular point let \(w_J\) be its class witness, and put \(J=N+c\). Choose \(\chi_J\) supported in the common agreement neighborhood, so \(\chi_Ju=\chi_Jw_J\). Its transform is their Fourier convolution. In the smaller cone, the near-convolution part uses the class estimate of \(w_J\) and the uniformly bounded \(L^1\)-norm of \(\widehat\chi_J\). The remaining frequencies are angularly separated, so the high cutoff derivative estimate absorbs the witness's common polynomial Fourier order. Complete-proof Lemma 2.2 writes both integrals fully. The resulting weight is a single \(M_J\), which (4) converts to \(C^{N+1}M_N\). Common low derivatives supply the common compact distributional order.

For the inverse heat argument, the exact remainder has weighted norm
\(C^{K+1}K^Kh^K\). Pairing it with a distribution of Schwartz order \(s\), after multiplying by the frequency exponential \(e^{-ir y\theta}\), costs at most \((1+r)^s\). Set \(K=N+s\) and \(h=\rho/r\). For \(r\geq1\), the resulting bound is
\[
 C^{K+1}K^K(\rho/r)^K(1+r)^s
     \leq C_1^{N+1}N^N r^{-N}
     \leq C_1^{N+1}M_N r^{-N}.
\]
The fixed excess powers of \(N\) and \(\rho\) are absorbed into the exponential constant.

The auxiliary amplitude has norm \(C e^{C\rho K^2/r}\). On \(r\geq DK\), this is at most \(C_2^{N+1}\). The exact packet integral has prefactor \((2\pi h)^{n/2}\), so division costs \(C r^{n/2}\). Apply the packet estimate at the index \(N+c\), \(c\geq n/2\), to obtain
\[
 C_3^{N+1}M_{N+c}r^{n/2-N-c}
        \leq C_4^{N+1}M_Nr^{-N}.
\]
At \(r<DK\) the common compact-order bound gives the result by \(N^N\leq M_N\). The compact localized distribution is \(u_N=\chi_{2(N+s)}u\), fixed for each \(N\). Only the amplitude used to estimate its transform depends on \(r\). That fixed sequence is exactly what the definition requires.

### Solution 3

For every fixed \(k\), \(\int_0^\infty t^ke^{-\sqrt t}dt\) converges. Substitution \(t=r^2\) gives
\[
 \int_0^\infty t^ke^{-\sqrt t}dt
       =2\int_0^\infty r^{2k+1}e^{-r}dr=2(2k+1)!.
\]
The last identity follows by repeated integration by parts, ending with \(\int_0^\infty e^{-r}dr=1\); all polynomial-exponential endpoint terms vanish. Dominated differentiation gives (27), and the absolute-value estimate gives the first part of (28).

For \(k\geq1\),
\((2k+1)!\leq(2k+1)^{2k+1}\leq C^{k+1}(k+1)^{2k}\).
The additional \(k+1\) is at most \(2^k\). Handle \(k=0\) by the value two. This proves global Gevrey-two membership. If \(s<2\), the lower factorial bound yields a ratio to any proposed \(A^{k+1}(k+1)^{sk}\) whose logarithm has leading term \((2-s)k\log k\). It tends to infinity. Local membership at zero is impossible.

For the continuation, the entire \(r\)-integrand in (29) can be integrated round a sector of angle \(\theta=\operatorname{sgn}(x_0)\pi/8\). On the large arc \(r=Re^{i\varphi}\), with \(\varphi\) between zero and \(\theta\), the phase modulus is
\(e^{-x_0R^2\sin(2\varphi)}\leq1\), and \(|e^{-r}|\leq e^{-R\cos\theta}\). The polynomial factor and arc length give a bound \(CR^2e^{-R\cos\theta}\to0\). The small arc is bounded by \(C\varepsilon^2\to0\). The integrand has an entire primitive by termwise integration of its entire power series. The finite sector integral is therefore zero, and the vanishing arcs yield (30) at \(x_0\).

At that center \(\operatorname{Re}(ix_0e^{2i\theta})=-|x_0|\sin(\pi/4)<0\). It remains at most a fixed negative number for \(z\) in a sufficiently small complex neighborhood. Every derivative under (30) is a polynomial in \(r\) times a bound \(e^{-cr^2+Cr}\), so it is integrable uniformly on compact complex subsets. The rotated integral is holomorphic there. Thus every nonzero real base point is analytic.

In the upper half-plane the original integral has modulus at most \(\int e^{-\sqrt t}dt=2\) and is holomorphic. Dominated convergence gives its boundary. The tube theorem excludes negative analytic covectors; class inclusion excludes negative class covectors. At zero the smaller classes fail, so the base-regularity theorem forces a nonempty class fiber. Closed positive conicity in dimension one makes it the full positive ray. For \(s\geq2\) global class membership makes wavefront empty. This proves (31).

For a Schwartz test \(\phi\), absolute integrability permits
\[
 \widehat f(\phi)=\int_0^\infty e^{-\sqrt t}
             \left(\int_{\mathbb R}e^{itx}\widehat\phi(x)dx\right)dt
       =2\pi\int_0^\infty e^{-\sqrt t}\phi(t)dt.
\]
Fourier inversion is [F]. This proves (32) as a whole tempered identity.

### Solution 4

Let \(g(y)=e^{-|y|^2/2}\) and \(h=t^{-2}\). Strict homogeneity gives
\[
 V_gu(tx,t\xi)=t^{n+\lambda}T_hu(x,\xi).
\]
The Fourier degree is \(\mu=-n-\lambda\), and the same identity for the transform is
\[
 V_g\widehat u(t\xi,-tx)=t^{n+\mu}T_h\widehat u(\xi,-x).
\]
The complete full Fourier packet identity [G], Lemma 4.1, gives
\[
 V_g\widehat u(t\xi,-tx)
      =(2\pi)^{n/2}e^{it^2x\cdot\xi}V_gu(tx,t\xi).
\]
Combine the three exact identities:
\[
 T_h\widehat u(\xi,-x)
       =(2\pi)^{n/2}h^{-n/2-\lambda}
                      e^{ix\cdot\xi/h}T_hu(x,\xi).
\]
Positive powers of \(h\) use the real logarithm. The phase has a plus sign, and its modulus is one. The modulus of the \(h\)-power is \(h^{-n/2-\operatorname{Re}\lambda}\).

A fixed inverse power is absorbed by applying the input estimate at index \(N+c\), with \(c\) larger than the positive part of \(n/2+\operatorname{Re}\lambda\), and then (4). The inverse identity is handled by a shift exceeding the opposite exponent. All neighborhoods remain common, and \(h_0\) can be reduced to at most one.

For a finite punctured Euler chain, strict scaling is replaced by its finite nilpotent logarithmic matrix. The full source identities are [G], Sections 5–7: the physical error consists of point jets and is bounded by \(t^b e^{-ct^2}\) away from the physical origin; the Fourier error consists of polynomials and has the same type of bound away from the covector origin. Maximizing \(t^{2N+b}e^{-ct^2}\) bounds either by \(C^{N+1}N^Nt^{-2N}\leq C^{N+1}M_Nt^{-2N}\). Matrix powers, logarithms and their inverses cost a fixed further shift. Keeping the entire vector and the differential-decrease union equality gives complete-proof Theorem 4.1. Discarding origin terms before this comparison would lose the full identities.

### Solution 5

The function \(f\) is bounded by two on the real line. Thus near zero the absolute radial integral of (33) is at most a constant times
\(\int_0^1r^{\operatorname{Re}\lambda+1}dr\), finite exactly under \(\operatorname{Re}\lambda>-2\). At infinity it is bounded by \(2r^{\operatorname{Re}\lambda}\), so sufficiently large Schwartz weights give temperedness. Positive dilation changes the argument \(x_1/|x|\) by nothing and multiplies the radial factor by \(t^\lambda\). The locally integrable change of variables proves strict global homogeneity, with no additional point term.

At a vertical point \(x=(0,t)\), the derivative of \(x_1/|x|\) in the first coordinate is \(1/|t|\), and its second derivative component in the first-order differential is zero. This is (34), with a positive sign for either sign of \(t\).

The coordinates (35) are analytic for \(r>0\), \(|w|<1\) and fixed \(\sigma\). Their inverse is \(r=|x|\), \(w=x_1/|x|\), so they form an analytic diffeomorphism near the vertical point. The pullback of \(u_\lambda\) is \(r^\lambda f(w)\). The radial multiplier and its reciprocal are analytic and nonzero there, so the multiplier lemma gives equality of wavefronts. The exact tensor-with-constant equality forces zero radial covector and the positive \(w\)-covector ray from Example 2. Coordinate covariance gives precisely the positive multiples of \(dw\), namely every \((a,0)\), \(a>0\), at \((0,t)\).

Elsewhere on the punctured plane the angular value is nonzero, except on those vertical axes. The function \(f\) is analytic at every such value. Its analytic composition with the angular map and multiplication by the radial factor are analytic, so no other nonzero-base class covector occurs. This proves the exact physical set (36) below order two. At orders at least two, \(f\) belongs to the class on every compact interval; analytic composition and the radial factor give empty class wavefront on the punctured plane.

The unique rotated pair from (36) is \(((a,0),(0,-t))\). Since \(t\) ranges over both nonzero signs, so does the target vertical covector. Every \(a>0\) occurs, and the inverse rotated-pair argument excludes all other nonzero target pairs. This proves (37). The theorem excludes the physical base \(x=0\) and the source covector \(\xi=0\); equivalently it excludes zero Fourier base and zero Fourier covector. It makes no statement determining possible origin terms of the full transform.

### Solution 6

For a compact output test \(\phi\), choose a compact fiber cutoff equal to one near zero. The kernel pairing is
\[
 K(\phi(x)a(y))=\int f(x)\phi(x)dx\;a(0).
\]
The cutoff is immaterial because the fiber delta is supported at zero. Since \(a(0)=1\), this is exactly the distribution of the function \(f\). A compact output set sees \(y=0\), so properness is explicit.

The Gaussian test splits into its two variable factors. The delta evaluates the second at zero, giving \(e^{-y^2/(2h)}\), with phase one and no dependence on \(\eta\). This proves (40).

Fix \(\xi>0\). If \(K\) were class-regular at \((0,0;\xi,0)\), its common Gaussian estimate could be evaluated at the fiber position zero and fiber covector zero. It would become precisely the same common estimate for \(f\) near \((0,\xi)\), contradicting (31). Thus the zero-fiber source covectors are present.

They lie in the tensor upper bound because the delta has zero adjoined over its actual support. Projection retains exactly these zero fiber components and gives the positive output fiber of \(f\). Requiring both components nonzero would omit this actual source contribution.

For ordinary wavefront, \(f\) is smooth and \(f(0)=2\ne0\). Near the source origin its ordinary kernel is a nonvanishing smooth multiplier times \(1_x\otimes\delta_0(y)\), so it has only covectors \((0,\eta)\), \(\eta\ne0\), there. Their fiber component is nonzero, so none is retained by projection. This agrees with the smooth output. The class horizontal covectors carry information absent from ordinary wavefront, without contradicting the ordinary projection theorem.

### Solution 7

The input Fourier variable \(\eta\) has dimension \(a\), because \(u\) lives on \(Y\). The output test variable \(x\) has dimension \(b\). Its phase gradient is \(DF(x)^T\eta-\xi\), with \(\xi\in\mathbb R^b\). The Fourier inversion constant in the pullback formula is therefore \((2\pi)^{-a}\).

In the bad input directions, the compact conic margin permits the exact analytic nonstationary estimate
\[
 |I_J|\leq C^{J+1}J^J(1+|\xi|+|\eta|)^{-J}.
\]
Use the common polynomial bound \(C(1+|\eta|)^m\). Substitution \(\eta=(1+|\xi|)z\) in the \(a\)-dimensional integral gives
\[
 C^{J+1}J^J(1+|\xi|)^{m+a-J}.
\]
The remaining \(z\)-integral is uniformly finite for \(J\geq m+a+3\). The same bound applies to small good frequencies because \(|\eta|\leq c_0|\xi|\) makes the gradient at least \(|\xi|/2\).

For large good input frequencies, use the class Fourier estimate \(C^{J+1}M_J(1+|\eta|)^{-J}\) and only the bounded amplitude \(|I_J|\leq C\). Radial integration above \(c_0|\xi|\) gives
\[
 C^{J+1}M_J(1+|\xi|)^{a-J}.
\]
Choose \(J=N+m+a+2\). The first exponent is \(-N-2\), and the second is no larger. Also \(J^J\leq M_J\). The fixed-shift estimate converts either weight to \(C^{N+1}M_N\). This is the required output class witness. Multiplying the phase weight and input class weight would produce an unjustified extra scale; they are used on different pieces.

For the product, the diagonal map has transpose \(D\Delta^T(\xi,\eta)=\xi+\eta\). Its normal covectors are nonzero pairs with \(\eta=-\xi\). Condition (56) excludes them from the class tensor upper bound, and ordinary inclusion gives the actual smooth product. The analytic pullback theorem supplies its class bound. Zero components over support in the tensor bound include \((\xi,0)\) and \((0,\eta)\), giving the single-factor terms in (57). They cannot be omitted even when only one factor is class-singular.

### Solution 8

Every fixed derivative of \(f-f_R\) has the uniform bound (63). Substitution \(t=r^2\) shows that the tail is \(2\int_{\sqrt R}^\infty r^{2k+1}e^{-r}dr\), which tends to zero. Thus there is smooth convergence on every compact set, with every derivative.

For a bounded set of compact tests, the order-zero estimate tends uniformly to zero, proving strong distributional convergence. For a fixed compact cutoff \(\chi\), the product rule shows every fixed derivative of \(\chi(f-f_R)\) tends uniformly to zero on its fixed support. Integrating its Fourier transform by parts through any prescribed order gives convergence of every rapid Fourier seminorm, in every cone. This is convergence in the ordinary normal topology with empty ordinary wavefront cone.

Each \(f_R\) is entire, so has empty class wavefront. The limit \(f\) has the positive fiber at zero for every Gevrey order below two, by the exact factorial moments and Example 2. Therefore this ordinary convergence does not preserve stronger class absence.

If instead compact extensions share support and order and satisfy the common bound (11), fix \(h,x,\xi,N\) and test distributional convergence against the Gaussian multiplied by a fixed support cutoff. Their packet values converge. The same inequality passes to the limit with all constants and neighborhoods unchanged. The packet theorem proves class absence of the limit. No exchange of the order limit with \(N\to\infty\) is needed: each inequality is passed separately, and the original common constant remains common.

For admissible operation limits, use the fixed ordinary wavefront cones satisfying the exact conditions in [P], Theorems 8.1–8.3, and the normal topology there. Pullback and proper projection are continuous. Tensor and nonopposing product are hypocontinuous and sequentially continuous, not asserted jointly continuous. Supply common class packet bounds on the compact regularity neighborhoods, common finite orders and fixed conic margins; complete-proof Proposition 8.3 propagates these bounds. Proper projection also needs a common closed source support whose projection is proper. These hypotheses identify the ordinary distributional limit with the required operation and then preserve its uniform class estimates. The example shows why individual class membership alone cannot replace those uniform estimates.

## References

[A] [Analytic directions of holomorphic boundary values][A], Lemma 1.1 and Theorem 4.1, with controlled smooth cutoff families and the positive-polar analytic tube estimate.

[P] [Pulling back and combining analytic singularities][P], Sections 1 and 8, with the full ordinary constructions, scalar composition law and exact normal continuity and hypocontinuity statements; Lemma 3.1 supplies the complete analytic phase iteration.

[G] [Gaussian packets and homogeneous Fourier directions][G], Lemma 2.1, Lemmas 4.1–6.1, Section 7 and Theorem 8.1, retaining full Gaussian constants, finite inverse heat errors and both origin anomalies.

[F] [Schwartz functions and Fourier inversion][F], Sections 1–5, with the whole distributional Fourier identities and compact-test estimates.

[C] [Cauchy bounds, root counts and analytic extensions][C], Lemma 1.1 for the polydisc coefficient estimate and Lemma 3.1 for the local complex extension of a real analytic map.

[B] Jan Boman, *Microlocal Quasianalyticity for Distributions and Ultradistributions*, Publications of the Research Institute for Mathematical Sciences **31** (1995), 1079–1095. [Primary article](https://ems.press/content/serial-article-files/40593), Section 1 for derivative-weight wavefront conventions. Its actual publisher terms are retained; the required proofs and original examples are supplied here.

Original exposition, examples and solutions are CC0. Linked prerequisite proofs and references retain their actual component authorship and terms.

[A]: ../../AN02-L186.html#complete-proof
[P]: ../../AN02-L188.html#complete-proof
[G]: ../../AN02-L189.html#complete-proof
[F]: https://github.com/KokunoYumeto/open-math-courses-public/blob/5949a0e862f3d87dbc078ae57149cf10c2b37dd2/docs/courses/AN-01/prerequisites/U011-free-foundations/schwartz-fourier-foundations-U017.md
[C]: ../../AN02-L045.html