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Gaussian decay gives an entire Fourier-Laplace transform and its growth bound

Statement

Assume countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $a > 0$ , $C \geq 0$ , and let $f : R^{n} \to C$ be measurable with $∣ f (x) ∣ \leq C e^{- π a ∣ x ∣^{2}}$ for almost every $x$ (so $f \in L^{1}$ ). Define $F (z) : = \int_{R^{n}} f (x) e^{- 2 π i x \cdot z} d x$ . Then the integral converges absolutely for every $z \in C^{n}$ and is independent of the representative of the class; every coordinate slice of $F$ is entire, with $\frac{\partial}{\partial z_{j}} F (z) = \int_{R^{n}} f (x) (- 2 π i x_{j}) e^{- 2 π i x \cdot z} d x;$ $F (x) = \hat{f} (x)$ for every $x \in R^{n}$ ; and $∣ F (z) ∣ \leq C a^{- n / 2} e^{π ∣ Im z ∣^{2} / a} (z \in C^{n}) .$

Facts & Assumptions

Given: An integer $n \geq 1$ , reals $a > 0$ and $C \geq 0$ , a measurable $f : R^{n} \to C$ with $∣ f (x) ∣ \leq C e^{- π a ∣ x ∣^{2}}$ for almost every $x$ , points $z \in C^{n}$ and $w \in C$ , and countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Countable choice is assumed; it is the hypothesis carried by the change-of-variables corollary and by the Gaussian integral identity used below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

The complex exponential is defined by its power series, satisfies $\exp (ζ + η) = \exp ζ \exp η$ and $∣ \exp ζ ∣ = e^{Re ζ}$ , and is entire with $\exp^{'} = \exp$ , so $(\exp ζ - 1) / ζ \to 1$ as $ζ \to 0$ through nonzero complex values (The complex exponential by its power series, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The complex exponential is entire and its complex derivative is itself).

[F3]

For every $ζ \in C$ , $∣ e^{ζ} - 1 ∣ \leq ∣ ζ ∣ e^{∣ ζ ∣}$ : the defining power series of The complex exponential by its power series is absolutely convergent, so the triangle inequality for series bounds $∣ e^{ζ} - 1 ∣ \leq \sum_{m \geq 1} ∣ ζ ∣^{m} / m! = e^{∣ ζ ∣} - 1 \leq ∣ ζ ∣ e^{∣ ζ ∣}$ .

[F4]

If measurable complex-valued $h_{n}, h$ satisfy $h_{n} \to h$ almost everywhere and $∣ h_{n} ∣ \leq G$ almost everywhere for one nonnegative measurable $G$ with $\int G < \infty$ , then $\int h_{n} \to \int h$ (Dominated convergence).

[F5]

The Lebesgue integral is complex-linear on $L^{1}$ , and integrable functions that agree almost everywhere have equal integrals (The Lebesgue integral is linear on $L^{1} (μ)$ , Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[F6]

For nonnegative measurable functions the integral is monotone and $\int c φ = c \int φ$ for $c > 0$ ; and $∣ \int h ∣ \leq \int ∣ h ∣$ (Monotonicity and nonnegative homogeneity of the nonnegative integral, The modulus of an integral is bounded by the integral of the modulus).

[F7]

The Gaussian Lebesgue integral and its translations: for every $b > 0$ , $\int_{R^{n}} e^{- π b ∣ x ∣^{2}} d x = b^{- n / 2}$ (Euclidean Gaussian transform with the 2π normalization at $ξ = 0$ ); and for $h \in L^{1}$ and $c \in R^{n}$ the translation $T (x) = x - c$ has $\det D T = 1$ , so $\int h (x - c) d x = \int h (y) d y$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

[F8]

For every real $u$ , $1 + u \leq e^{u}$ ( $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ ).

[F9]

The $L^{1}$ transform is $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ , absolutely convergent at every real $ξ$ (Fourier transform on complex L1 classes, The integral transform is representative independent).

[F10]

A function is holomorphic on an open subset of $C$ when complex differentiable at every point, and holomorphic on all of $C$ means entire (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

Proof

technique · direct

1.1F5givenchoose

Fix a representative satisfying the bound. By [F5] we may modify $f$ on the null set where $∣ f (x) ∣ > C e^{- π a ∣ x ∣^{2}}$ without changing any integral below; after this modification $∣ f (x) ∣ \leq C e^{- π a ∣ x ∣^{2}}$ holds for every $x$ , and $f$ remains measurable.

1.2F1F6F7F8algebra

Two exact integrals. (i) For $v \in R^{n}$ the algebraic identity $- π a ∣ x ∣^{2} + 2 π x \cdot v = π ∣ v ∣^{2} / a - π a ∣ x - v / a ∣^{2}$ and the scalar rule of [F6], the translation identity of [F7] applied to $y \mapsto e^{- π a ∣ y ∣^{2}}$ , and the Gaussian identity of [F7] give $\int_{R^{n}} e^{- π a ∣ x ∣^{2} + 2 π x \cdot v} d x = e^{π ∣ v ∣^{2} / a} \int_{R^{n}} e^{- π a ∣ x - v / a ∣^{2}} d x = e^{π ∣ v ∣^{2} / a} a^{- n / 2} .$ (ii) For $c \geq 0$ , the elementary inequality $2 π ∣ x ∣ c \leq \frac{π a}{4} ∣ x ∣^{2} + \frac{4 π}{a} c^{2}$ (from $2 A B \leq A^{2} + B^{2}$ with $A = \frac{\sqrt{π a}}{2} ∣ x ∣$ and $B = \frac{2 \sqrt{π}}{\sqrt{a}} c$ ), together with $∣ x ∣ e^{- π a ∣ x ∣^{2} / 2} \leq e^{1 / (2 π a)}$ — which follows from [F8] as $∣ x ∣ \leq e^{∣ x ∣}$ and $s - \frac{π a}{2} s^{2} \leq \frac{1}{2 π a}$ for $s \geq 0$ — gives $∣ x ∣ e^{- π a ∣ x ∣^{2} + 2 π ∣ x ∣ c} \leq e^{1 / (2 π a)} e^{(4 π / a) c^{2}} e^{- π a ∣ x ∣^{2} / 4}$ ; by [F6] and [F7] (with $b = a / 4$ ) this is integrable with $\int_{R^{n}} ∣ x ∣ e^{- π a ∣ x ∣^{2} + 2 π ∣ x ∣ c} d x \leq (\frac{4}{a})^{n / 2} e^{1 / (2 π a)} e^{(4 π / a) c^{2}} < \infty .$

2.1F2F5F6step 1.1step 1.2

Absolute convergence, growth bound, and representative independence. Put $v : = Im z$ . The exact modulus identity in [F2] and the assumed Gaussian bound give, almost everywhere, $∣ f (x) e^{- 2 π i x \cdot z} ∣ = ∣ f (x) ∣ e^{2 π x \cdot v} \leq C e^{- π a ∣ x ∣^{2} + 2 π x \cdot v} .$ Completing the square as in computation (i) of step 1.2 and applying [F6] yields $\int_{R^{n}} ∣ f (x) e^{- 2 π i x \cdot z} ∣ d x \leq C \int_{R^{n}} e^{- π a ∣ x ∣^{2} + 2 π x \cdot v} d x = C a^{- n / 2} e^{π ∣ v ∣^{2} / a} < \infty .$ Hence the integral defining $F$ converges absolutely at the arbitrary point $z$ , and [F6] gives the growth bound $∣ F (z) ∣ \leq C a^{- n / 2} e^{π ∣ Im z ∣^{2} / a}$ . If $\tilde{f} = f$ almost everywhere is another representative, the integrands agree almost everywhere; the same majorant makes both integrable, so [F5] gives equal integrals.

3.1F2givenstep 1.2step 2.1

The slice as a one-variable integral. Fix $j$ and complex numbers $z_{k}$ for $k \neq j$ , and let $v^{'}$ be the vector of their imaginary parts, with zero in coordinate $j$ . Set $g (x) : = f (x) e^{- 2 π i \sum_{k \neq j} x_{k} z_{k}}$ . Then $g$ is measurable and $∣ g (x) ∣ \leq C e^{- π a ∣ x ∣^{2} + 2 π x \cdot v^{'}}$ almost everywhere, so $g \in L^{1}$ by the completed-square calculation of step 1.2. For $w \in C$ put $h (w) : = \int_{R^{n}} g (x) e^{- 2 π i x_{j} w} d x$ . The addition law [F2] turns the integrand into $f (x) e^{- 2 π i x \cdot z^{'}}$ with $z^{'} = (z_{1}, \dots, z_{j - 1}, w, z_{j + 1}, \dots, z_{n})$ , so $h (w) = F (z^{'})$ ; step 2.1 gives absolute convergence at every $w$ .

4.1F2F5step 3.1

Difference quotients are integrals. Fix $w \in C$ and $s \in C ∖ {0}$ . For every $x$ the addition law [F2] gives $e^{- 2 π i x_{j} (w + s)} - e^{- 2 π i x_{j} w} = e^{- 2 π i x_{j} w} (e^{- 2 π i x_{j} s} - 1)$ , and both integrands are integrable by step 3.1; additivity and scaling from [F5] therefore yield $\frac{h (w + s) - h (w)}{s} = \int_{R^{n}} g (x) e^{- 2 π i x_{j} w} \frac{e^{- 2 π i x_{j} s} - 1}{s} d x .$

5.1F3F4step 1.2step 3.1step 4.1

Pointwise limit and an integrable majorant. Let $s_{n} \to 0$ be any sequence in $C ∖ {0}$ . By [F2] the quotients converge pointwise to $g (x) (- 2 π i x_{j}) e^{- 2 π i x_{j} w}$ . By [F3], whenever $∣ s_{n} ∣ \leq 1$ one has $∣ (e^{- 2 π i x_{j} s_{n}} - 1) / s_{n} ∣ \leq 2 π ∣ x_{j} ∣ e^{2 π ∣ x_{j} ∣}$ . The $n$ -th quotient integrand is therefore bounded in modulus by $2 π C ∣ x ∣ e^{- π a ∣ x ∣^{2} + 2 π ∣ x ∣ (∣ Im w ∣ + ∣ v^{'} ∣ + 1)},$ where $v^{'}$ is from step 3.1. Its integral is finite by computation (ii) of step 1.2. Passing to the tail of the sequence, dominated convergence [F4] gives $\lim_{n \to \infty} \frac{h (w + s_{n}) - h (w)}{s_{n}} = \int_{R^{n}} g (x) (- 2 π i x_{j}) e^{- 2 π i x_{j} w} d x,$ and the limit integral is absolutely convergent by the same majorant.

6.1F9F10step 5.1∎

Conclusion. Since the nonzero null sequence $s_{n} \to 0$ was arbitrary, step 5.1 shows that $h$ is complex differentiable at every $w \in C$ , with $h^{'} (w) = \int_{R^{n}} g (x) (- 2 π i x_{j}) e^{- 2 π i x_{j} w} d x$ ; being holomorphic on all of $C$ , the slice is entire by [F10], and rewriting the derivative gives the displayed formula for $\partial F / \partial z_{j}$ . At real $z = ξ \in R^{n}$ the defining integral is literally the $L^{1}$ transform formula, so $F (ξ) = \hat{f} (ξ)$ by [F9].

Depends on

Used by

Hardy's Gaussian uncertainty principle in ℝⁿ Theorem

Dependency tree · two levels

84 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Calder Sheagren, Uncertainty Principles with Fourier Analysis (University of Chicago REU 2017, author PDF) (standard reference, not scraped)
Mathilda Lindell, The Phragmén–Lindelöf Principle and Its Applications (Lund University bachelor's thesis 2025:K15) (standard reference, not scraped)