Gaussian regularization and quadratic stationary phase

This reading develops Gaussian regularization, Fourier calculus on Schwartz functions, and quadratic stationary phase with every coefficient and an explicit finite-order remainder. It includes smooth Hessian and parameter families, nonstationary estimates, and an exact Gaussian example with a reproducible illustration.

The parameter-Morse companion supplies the smooth signed normal form used in the stationary-phase and critical-manifold reading. Mathematical sources and their conventions are described below.

Original programme exposition: CC0 to the extent rights exist. Cited human works are external sources and retain their own terms.

F0. The exact earlier programme proofs

The following earlier arguments provide the elementary tools used below. The proof index gives the statement, proof locator and lower inputs for each named result.

Finite sums, multi-indices, Euclidean scalar products and the determinant have their usual algebraic definitions. The human-proof references identify the external foundational statements and proofs, including those from Jiří Lebl’s freely accessible author edition.

Integrals in this module are compact Riemann integrals or absolutely convergent improper integrals, as defined and proved in P17–P18. In particular, every global interchange in Q3–Q6 satisfies the product-majorant hypotheses of P18.4.

Set

The second equality will be proved in Q4, before it is used. A function belongs to $\mathcal S(\mathbb R^n)$ when it is smooth and $\sup_x(1+|x|)^m|\partial^\alpha a(x)|<\infty$ for every integer $m\geq0$ and every multi-index $\alpha$.

Q1. The limit and differentiation rules actually used here

Suppose continuous functions $f_t$ converge to $f_0$ uniformly on every compact set as $t\to0$, and $|f_t(x)|\leq g(x)$ for one integrable $g$, including $t=0$. Then $\int f_t\to\int f_0$.

Proof. Given $\delta>0$, choose a cube $K$ so that $\int_{K^c}g<\delta$. The difference of the integrals on $K^c$ is at most $2\delta$. On $K$ it is at most $|K|\sup_K|f_t-f_0|$, which tends to zero. Taking a limit superior and then letting $\delta$ decrease to zero proves the claim.

For a parameter derivative, suppose $f(t,x)$ and $\partial_t f(t,x)$ are continuous for $t$ in a neighbourhood of $t_0$, the integral at $t_0$ exists absolutely, and $|\partial_t f(t,x)|\leq g(x)$ there. The fundamental theorem of calculus gives

These quotients are bounded by $g$ and converge uniformly on compact sets to $\partial_t f(t_0,x)$. The first part proves differentiation under the integral. It also proves continuity of that derivative when the same local bound holds. Repeating the argument proves any finite order for which the corresponding bounds are available. This proof uses F0, not a general unproved appeal to dominated convergence. $\square$

Q2. A complex Gaussian without an analytic-continuation import

For $z\in\mathbb C$ with $\operatorname{Re}z>0$ and real $\eta$,

Here $\sqrt z$ is the square root with positive real part.

Proof. The integral and all derivatives used below converge absolutely: on a compact subset of the right half-plane their absolute values are bounded by a polynomial in $|x|$ times $e^{-c x^2}$, for some $c>0$. Such bounds are integrable: differentiation shows that $r^m e^{-cr^2/2}$, for $r\geq0$, has a finite maximum, so the remaining factor is bounded by $C e^{-cr^2/2}$, and hence by $C e^{-cr/2}$ for $r\geq1$. The latter has a finite integral by the fundamental theorem. Q1 therefore applies.

First let $J=\int e^{-x^2/2}\,dx>0$. Squaring, using absolute iteration and then planar polar coordinates, gives

Thus $J=\sqrt{2\pi}$. P21.5 in change-of-variables-prerequisite-completions.md proves the polar Jacobian, the compact-sector substitution, and the limits at the origin, missing ray and infinity used in this step.

For fixed $z$ join 1 to $z$ by $z_t=1+t(z-1)$, $0\leq t\leq1$. Its real part is positive. Put $J(t)=\int e^{-z_t x^2/2}\,dx$. Integration of $\partial_x(xe^{-z_t x^2/2})$ over $[-R,R]$, followed by $R\to\infty$, gives $\int x^2e^{-z_t x^2/2}\,dx=J(t)/z_t$, because the boundary terms vanish exponentially. Consequently

The chosen root $s(t)=\sqrt{z_t}$ is differentiable and satisfies $2s(t)s'(t)=z'_t$. One can see this without a complex-analysis theorem by writing $z_t=r_t e^{i\theta_t}$, with $-\pi/2<\theta_t<\pi/2$, and setting $s(t)=\sqrt{r_t}e^{i\theta_t/2}$. P16.3 in exponential-prerequisite-completions.md proves the smooth angle, the unique positive-real-part root and this differentiation rule. The product rule now gives $(sJ)'=0$; hence $J(1)=\sqrt{2\pi}\,z^{-1/2}$.

Finally, with $z$ fixed write the left side of (G) as $F(\eta)$. The integral of $\partial_x e^{-zx^2/2+i\eta x}$ vanishes, so $\int xe^{-zx^2/2+i\eta x}\,dx=(i\eta/z)F(\eta)$. Thus $F'(\eta)=-(\eta/z)F(\eta)$, and $\partial_\eta(e^{\eta^2/(2z)}F(\eta))=0$. The value at zero already computed proves (G). $\square$

Q3. Schwartz estimates and the signs in the Fourier rules

If $a\in\mathcal S$, then $\widehat a\in\mathcal S$, and

These maps are continuous for the defining Schwartz seminorms.

Proof. Polynomial multiples of derivatives of $a$ are integrable. For example use the rectangular shells between $[-2^j,2^j]^n$ and $[-2^{j+1},2^{j+1}]^n$. P18.3 proves their integral bounds directly from finite rectangle partitions. Decay $(1+|x|)^{-m}$, $m>n$, reduces the bound to a convergent geometric series. Q1 proves every differentiation in the first formula. For the second, integrate first in a single coordinate on $[-R,R]$. Its boundary terms tend to zero by rapid decrease; the remaining integrals converge absolutely. Iteration proves the multi-index formula. Applying these identities together shows that each $\xi^\beta\partial_\xi^\alpha\widehat a$ is the Fourier transform of a finite linear combination of polynomial multiples of derivatives of $a$. Its supremum is bounded by their $L^1$ norms. The rectangular-shell estimate bounds each such norm by finitely many input seminorms, proving both rapid decrease and continuity.

A useful quantitative consequence is, for integers $M>N+n/2$,

Indeed (D) gives $(1+|\xi|^2)^M\widehat a=\widehat{(1-\Delta)^M a}$; take absolute values and integrate. The same rectangular-shell bound proves finiteness of $C_{n,N,M}$, with its stated threshold unchanged. $\square$

Q4. Fourier inversion proved before stationary phase

For every $a\in\mathcal S$, the inversion identity in (F) holds.

Proof. For $\varepsilon>0$ define

Insert the definition of $\widehat a$. The double absolute integral is bounded by $\|a\|_1\int e^{-\varepsilon|\xi|^2/2}\,d\xi<\infty$, so F0-INT permits interchange. Applying (G) in every coordinate yields

By positive dilation and Q2, $\int K_\varepsilon=1$. Since $a$ has bounded gradient, the fundamental theorem along a line segment gives $|a(x-v)-a(x)|\leq\|\nabla a\|_\infty|v|$, with the Euclidean norm for the gradient. Therefore

On the Fourier side the multiplier converges uniformly on compact sets to 1 and is bounded by 1. Q3 supplies integrability of $\widehat a$, so Q1 shows that $A_\varepsilon(x)$ tends to the right side of (F). This proves inversion without using stationary phase. Applying the same argument to $\partial^\alpha a$ also gives every derivative of (F). $\square$

Q5. Diagonalizing a real symmetric matrix

A real symmetric $n\times n$ matrix $H$ has an orthonormal basis of real eigenvectors. If $H$ is invertible, all its eigenvalues are nonzero. Define $\operatorname{sgn}H$ to be their number of positive entries minus their number of negative entries.

Proof. In dimension 0 the empty orthonormal basis proves the assertion. For $n\geq1$, the function $v\mapsto v^THv$ attains a maximum on the unit sphere, by F0-COMP. Let $v$ be a maximizing unit vector. For $w\perp v$, differentiate this function along $p(t)=(v+tw)/\sqrt{1+t^2|w|^2}$ at $t=0$. The product, chain and positive-root derivative rules give $p'(0)=w$. The scalar Fermat theorem at this maximum therefore gives $2w^THv=0$. Set $\lambda=v^THv$. The vector $w=Hv-\lambda v$ is perpendicular to $v$, and the preceding identity gives $|w|^2=w^THv=0$. Thus $Hv=\lambda v$. If $w\perp v$, then $v^THw=(Hv)^Tw=0$; hence $v^\perp$ is invariant under $H$. P9.4 in differential-prerequisite-completions.md constructs an orthonormal basis of this complement and proves that the restriction is a symmetric $(n-1)$-dimensional matrix. Apply induction to that matrix, starting with dimension 0 or 1. This gives an orthogonal matrix $U$ with $U^THU=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. If any $\lambda_j=0$, its eigenvector lies in the kernel, and conversely. Since $U^TU=I$, determinant multiplication gives $(\det U)^2=1$ and $\det(U^THU)=\det H$. In the permutation formula for a diagonal matrix only the identity permutation can have a nonzero term; its term is $\prod_j\lambda_j$. Hence $\det H=\prod_j\lambda_j$. $\square$

Q6. Exact quadratic identity, with a justified regularization

Let $H$ be real, symmetric and invertible, $h>0$, $B=H^{-1}$, and $a\in\mathcal S(\mathbb R^n)$. Then

Both integrals displayed here are absolutely convergent because the amplitudes are Schwartz. The bare quadratic exponential is not treated as an absolutely integrable function.

Proof. Multiply the integrand on the left by $e^{-\varepsilon|x|^2/2}$, $\varepsilon>0$, and insert Q4. The double absolute integral is finite, bounded by $(2\pi)^{-n}\|\widehat a\|_1\int e^{-\varepsilon|x|^2/2}\,dx$. Thus its order can be exchanged. Diagonalize $H$ using Q5 and make the orthogonal change $x=Uy$, $\eta=U^T\xi$. The inner integral is the product of (G) with $z_j=\varepsilon-i\lambda_j/h$. Its value is

Since $\operatorname{Re}(1/z_j)>0$, the modulus of the exponential is at most 1, while $|z_j|^{-1/2}\leq(h/|\lambda_j|)^{1/2}$. This supplies an $\varepsilon$-independent integrable bound after multiplication by $|\widehat a(\xi)|$.

As $\varepsilon\downarrow0$, with each root continued through the right half-plane,

The convergence is uniform for $\xi$ in compact sets. Q1 therefore takes the limit on the Fourier side of (RG), giving the right side of (QI). On the original side it applies with bound $|a|$, removing the regularization. The product of the root phases is precisely $e^{i\pi\operatorname{sgn}H/4}$. $\square$

Original text: public domain (CC0).

The root limit and its $\pi/4$ phase in this proof are established explicitly in P16.3, including both signs of $\lambda_j$.

Q7. Every coefficient and an explicit remainder

Put $\Delta_B=\sum_{j,k}B_{jk}\partial_j\partial_k$. For every integer $N\geq1$,

where

for any integer $M>N+n/2$. Here $\|B\|$ is its Euclidean operator norm. The error in the unnormalized integral is $|C_H(h)R_N|$, of order $h^{n/2+N}$ for fixed $H,a$.

Proof. Repeatedly apply the fundamental theorem of calculus to the function $t\mapsto e^{it}$, or inductively integrate its remainder once. This gives, for real $t$,

For completeness, the inductive integral form starts with $f(1)-f(0)=\int_0^1 f'(s)ds$. Integration by parts changes $\int_0^1 (1-s)^{N-1}f^{(N)}(s)/(N-1)!\,ds$ into $-f^{(N-1)}(0)/(N-1)!+ \int_0^1(1-s)^{N-2}f^{(N-1)}(s)/(N-2)!\,ds$; descending to the first identity proves the formula for $f(s)=e^{its}$.

Apply (T) with $t=-h\xi^TB\xi/2$ in (QI). The remainder integrates absolutely by Q3, giving the first inequality (R). Moreover, (D) and Q4 give

Thus $(-ih/2)^k$ becomes $(ih/2)^k$, establishing the sign in (SP). Finally $|\xi^TB\xi|\leq\|B\||\xi|^2$, and (W) proves the second inequality. $\square$

Q8. What is uniform when the Hessian varies

Let $p$ range over an open parameter set, let $H(p)$ be smooth, real symmetric and invertible, and let $a(p,\cdot)$ be smooth with values in $\mathcal S$: every parameter derivative has locally bounded Schwartz seminorms. On every compact parameter subset, the normalized integral in (QI), divided by $C_{H(p)}(h)$, extends smoothly through $h=0$ as a function of real $h$. Its Taylor coefficients are the ones in (SP), and the normalized remainder is $O(h^N)$ after any fixed number of parameter derivatives.

Proof. Matrix inversion is smooth where the determinant is nonzero, by the cofactor formula, so $B(p)$ and all derivatives are bounded on a compact parameter subset contained in the open set. For real $h$ define the normalized expression directly by

Each $h$-derivative multiplies the integrand by a power of the quadratic polynomial $-i\xi^TB(p)\xi/2$. A parameter derivative either differentiates the Fourier amplitude or contributes a derivative of this polynomial multiplied by $h$. Repeated product rules therefore give a finite sum of polynomial factors in $\xi$, with bounded coefficients for $p$ in a compact set and $|h|\leq1$, times a Fourier transform of a parameter derivative of $a$. Q3 makes all these factors uniformly integrable. Q1 proves smoothness and permits all the derivatives just described.

At $h=0$, Q4 and (D) give $\partial_h^k A(p,0)=(i/2)^k\Delta_{B(p)}^k a(p,0)$. Apply the integral Taylor formula proved in Q7 to $h\mapsto\partial_p^\alpha A(p,h)$. Its $N$th $h$-derivative is uniformly bounded by the preceding integral estimates. The remainder is bounded by $C_{N,\alpha}|h|^N$, uniformly on the compact parameter subset. This proves the asserted parameter estimate. It is not an estimate for unnormalized parameter derivatives of an additional oscillatory factor $e^{i\phi(p)/h}$; that factor must be kept separate. $\square$

Q9. The region with no stationary point

Suppose $\phi$ is smooth and real on an open neighbourhood of the compact support of $a\in C_c^\infty$, and $|\nabla\phi|\geq c>0$ on such a neighbourhood. Set $v=\nabla\phi/|\nabla\phi|^2$ there and $Ta=-\operatorname{div}(va)$. Then, for $h>0$ and each integer $N\geq0$,

The right side is finite and independent of $h$.

Proof. The identity $v\cdot\nabla e^{i\phi/h}=(i/h)e^{i\phi/h}$ follows by the chain rule. Integrating it against $a$ and using integration by parts in each coordinate gives (NS) for $N=1$. Each $T^j a$ has support inside $\operatorname{supp}a$, because derivatives and multiplication do not enlarge support. Thus the same argument applies repeatedly. There are no boundary terms: one can choose the integration box outside the compact support. The final bound follows from the absolute-value bound on the integral and $|e^{i\phi/h}|=1$. For a family with a common compact support, a uniform lower bound on $|\nabla\phi|$, and bounded derivatives of $\phi,a$ to the required finite order, the same finite product-rule expansion bounds $\|T^Na\|_1$ uniformly. A parameter derivative of the oscillatory factor costs a power of $h^{-1}$; applying (NS) sufficiently many extra times still proves decay to every order for each fixed derivative order. $\square$

E1. An exact model that checks the normalization and the error

Take $n=1$, $H=1$, $a(x)=e^{-x^2/2}$. Q2 gives

The roots in this equality have the branches specified in Q2: their arguments add to $-\pi/2+\arctan h$, which lies in $(-\pi/2,0)$. Hence no sign ambiguity is hidden in the factorization.

Repeated integration by parts gives the Gaussian moments $\int\xi^{2N}e^{-\xi^2/2}d\xi=(2N-1)!!\sqrt{2\pi}$: the boundary term in $\int\xi^{2N-1}\xi e^{-\xi^2/2}$ vanishes and each step reduces the exponent by two; the base case is Q2. For the derivatives of the amplitude, use $a'=-xa$. Its $(m-1)$st derivative at zero, for $m\geq2$, gives $a^{(m)}(0)=-(m-1)a^{(m-2)}(0)$. Together with $a(0)=1$, $a'(0)=0$, induction gives $a^{(2k)}(0)=(-1)^k(2k)!/(2^kk!)$. Consequently Q7 reads

For example the first three terms are $1-ih/2-3h^2/8$. The bound holds for every $h>0$; convergence of an infinite Taylor series for large $h$ is neither asserted nor used.

Figure 1. The left panel samples the real and imaginary parts of the exact normalized integral (EX) and of $1-ih/2-3h^2/8$. The right panel samples $|R_N(h)|/h^N$ for $N=1,2,3$; the horizontal dashed lines are the proved constants $\binom{2N}{N}/4^N$ in (ER), not fitted estimates. The horizontal variable is $0.01\leq h\leq1$. The sampled curves illustrate Q7 and E1; the proof of the bound is (T), (QI), and the moment calculation above. Reproducible source: figures/draw_quadratic_remainder.py. The human-source route for the Gaussian and quadratic calculation is Guillemin–Sternberg, author draft, §§14.1–14.5; the displayed example and figure are written for this reconstruction.

Mathematical sources and notation

• Victor Guillemin and Shlomo Sternberg, Semi-classical analysis, author-posted draft dated 13 January 2010, §§14.1–14.5, printed pp. 459–466, discuss Gaussian evaluation, regularization, the quadratic Fourier identity and the Taylor remainder. Section 13.14.3, pp. 455–457, gives the Morse lemma and its parameter variant, treated in the Morse companion. Section 14.8, p. 472, relates stationary phase to Fourier inversion.
• Richard Melrose, Tempered distributions and the Fourier transform, author-posted Chapter 1, §1.7, printed pp. 21–23, gives the Fourier normalization, Schwartz estimates and inversion; see also the MIT 18.157 course page.
• Jiří Lebl, Basic Analysis I–II, version 6.3 (15 May 2026), freely accessible author edition, supplies foundational calculus and compactness results. The human-proof references give exact statement and proof locators; the local prerequisite companions supply the additional finite-dimensional and integral arguments.

The transform convention is (F), with inverse factor $(2\pi)^{-n}$. Q2 derives the complex Gaussian by differential identities along a path in the right half-plane. Q4 proves inversion by a Gaussian approximate identity before the quadratic identity uses it. Q6 specifies the regularization, root phases and absolute majorants, and Q7 gives the finite Taylor estimate in this convention. E1 checks the normalization and remainder in an exact model. These arguments use the named elementary and integral inputs in F0; they do not import a distribution kernel theorem.

The source PDFs are external readings. Their authorship and reuse terms remain with their respective works. The independently written programme exposition, worked example and reproducible figure are CC0 to the extent rights exist.