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Uncertainty Principles for Fourier Analysis

1 · Prerequisites

2 · Summary

This page isolates three inequivalent notions of localisation of a function on $R^{n}$ and states precisely how much can be said about each. The variance formulation measures spread by the second moments of $∣ f ∣^{2}$ and $∣ \hat{f} ∣^{2}$ ; the support-measure formulation counts the measures of sets off which $f$ and $\hat{f}$ vanish; the Hardy formulation assumes Gaussian decay of both and exhibits a threshold at the critical product $a b = 1$ . Gaussian decay in both domains implies finite second moments, but the hypotheses are not interchangeable; the companion page supplies the Gaussian computations and the separating examples.

The variance part begins with the spatial and frequency centres and variances of a nonzero $L^{2}$ function with finite second moments, whose well-definedness is discharged by Cauchy-Schwarz and Plancherel. The Fourier characterization of $H^{1}$ identifies this domain with $f \in H^{1}$ and $x f \in L^{2}$ . Translation and modulation are then shown to centre the variance pair without changing the product, and the coordinate commutator estimate $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$ , with $D_{j} f$ the weak derivative, is proved by compact cutoffs and vanishing $L^{2}$ tails. Fourier differentiation and finite-tuple Cauchy-Schwarz convert that estimate into the summed $n$ -dimensional Heisenberg bound $∥ ∣ x ∣ f ∥_{2} ∥ ∣ ξ ∣ \hat{f} ∥_{2} \geq \frac{n}{4 π} ∥ f ∥_{2}^{2}$ on the same $H^{1}$ -with-finite-spatial-moment domain. The sharp theorem and its equality classification are owned by the functional-analysis track and remain quoted only on Schwartz functions; this pair extends the lower bound, not that classification.

The support-measure part proves $∣ E ∣ ∣ F ∣ \geq 1$ for a nonzero $L^{2}$ function supported on $E$ whose continuous $L^{1}$ transform is supported on $F$ , with no regularity beyond measurability and finiteness of the two measures. The complex-analytic route to compact-support rigidity is developed next: compact support makes the transform a function with entire coordinate slices by differentiation under the integral sign, Gaussian decay gives the same entire continuation together with the growth bound $∣ F (z) ∣ \leq C a^{- n / 2} e^{π ∣ Im z ∣^{2} / a}$ , and a separately holomorphic function vanishing on a real box is identically zero. Combining the continuation with the nonempty open complement of a compact frequency support yields the qualitative theorem: a nonzero $L^{1}$ function with compact support cannot have compactly supported transform.

The Hardy part states and proves the Gaussian uncertainty principle on $R^{n}$ by coordinate slices: entire continuation and one-variable rigidity give vanishing in the supercritical case and successive Gaussian factors in the critical case. The separate-holomorphy vanishing lemma then extends the critical factorization to complex arguments. The proof's complex-analytic cost is the entire growth-rigidity lemma with its two sector bounds and Phragmen-Lindelof argument, and the remark on proof cost records which part of the argument is real-variable and which is complex-analytic. The subcritical Gaussians show that the threshold is sharp: for $a b < 1$ every Gaussian $e^{- π c ∣ x ∣^{2}}$ with $a < c < 1 / b$ satisfies both Gaussian bounds, so no vanishing conclusion can hold. The page closes by contrasting the finite support-product bound of the unitary discrete Fourier transform, whose right side is $N$ and whose equality set is different, with the continuous inequalities, and by recording the sense in which the three localisation notions are not interchangeable.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableprecheck passjudge pass (gpt-6.1-sol)Open item page →

Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments

Definition

Assume countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $f \in L^{2} (R^{n}; C)$ (The space $L^{p} (μ)$ as the quotient by null functions) be nonzero with finite second moments, $\int_{R^{n}} ∣ x ∣^{2} ∣ f (x) ∣^{2} d x < \infty, \int_{R^{n}} ∣ ξ ∣^{2} ∣ \hat{f} (ξ) ∣^{2} d ξ < \infty,$ where $\hat{f}$ is the Plancherel transform of Plancherel theorem. Define the spatial mean $a = (a_{1}, \dots, a_{n})$ and spatial variance $V_{x} (f)$ , and the frequency mean $b = (b_{1}, \dots, b_{n})$ and frequency variance $V_{ξ} (f)$ , by $a_{j} : = \frac{1}{∥ f ∥_{2}^{2}} \int_{R^{n}} x_{j} ∣ f (x) ∣^{2} d x, V_{x} (f) : = \frac{1}{∥ f ∥_{2}^{2}} \int_{R^{n}} ∣ x - a ∣^{2} ∣ f (x) ∣^{2} d x,$ $b_{j} : = \frac{1}{∥ \hat{f} ∥_{2}^{2}} \int_{R^{n}} ξ_{j} ∣ \hat{f} (ξ) ∣^{2} d ξ, V_{ξ} (f) : = \frac{1}{∥ \hat{f} ∥_{2}^{2}} \int_{R^{n}} ∣ ξ - b ∣^{2} ∣ \hat{f} (ξ) ∣^{2} d ξ .$ These are the probability-normalised means and variances of the measures $∣ f (x) ∣^{2} d x / ∥ f ∥_{2}^{2}$ and $∣ \hat{f} (ξ) ∣^{2} d ξ / ∥ \hat{f} ∥_{2}^{2}$ . The integrals are read in the componentwise convention of Integrable real and complex functions, and their integrals, with $∣ f ∣^{2} = f \overline{f}$ and $x_{j}, ξ_{j}$ the real coordinate functions; the frequencies are measured in the $e^{- 2 π i x \cdot ξ}$ convention of Plancherel theorem and Complex Lp classes and Euclidean test-function conventions. No centring is asserted here: the mean is subtracted in the variance but the transformation property of the pair $(V_{x}, V_{ξ})$ is proved separately in Centring by translation and modulation preserves the variance product.

Well-definedness

Since $f \neq 0$ in $L^{2}$ , we have $∥ f ∥_{2} > 0$ . By the $L^{2}$ Cauchy–Schwarz inequality and the pairing convention of Complex completeness, density, and inner product: the consumer interface, applied to the functions $x \mapsto ∣ x_{j} ∣ ∣ f (x) ∣$ and $∣ f ∣$ , $\int_{R^{n}} ∣ x_{j} ∣ ∣ f (x) ∣^{2} d x \leq ∥ ∣ x_{j} ∣ f ∥_{2} ∥ f ∥_{2} \leq ∥ ∣ x ∣ f ∥_{2} ∥ f ∥_{2} < \infty,$ where $∣ x_{j} ∣ \leq ∣ x ∣$ and the hypothesis on the second moment bound the first factor. Each numerator $∣ \int x_{j} ∣ f ∣^{2} ∣$ is therefore finite, and the same estimate with $∣ x - a ∣^{2} \leq 2 ∣ x ∣^{2} + 2 ∣ a ∣^{2}$ makes the spatial variance numerator finite. On the frequency side Plancherel theorem gives $∥ \hat{f} ∥_{2} = ∥ f ∥_{2} > 0$ and, together with the assumed second moment of $\hat{f}$ , the same Cauchy–Schwarz estimate makes both frequency numerators finite. Hence all four quantities are well-defined finite real numbers and $a, b \in R^{n}$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The coordinate inequality $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$

Statement

Assume Countable Choice. Let $n \geq 1$ , $j \in {1, \dots, n}$ and $f \in H^{1} (R^{n}; C)$ with $x f \in L^{2} (R^{n}; C^{n})$ . Here $H^{1} = W^{1, 2}$ under the regular-distribution identification of Integer-order W^{k,2} and H^k agree with equivalent norms, and $D_{j} f \in L^{2}$ denotes the weak partial derivative. Then $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2} .$ By the Fourier characterization of $H^{1}$ , this is the coordinate estimate on the natural domain where both $∣ x ∣ f$ and $∣ ξ ∣ \hat{f}$ lie in $L^{2}$ . Both sides vanish when $f = 0$ ; for nonzero $f$ the right side is positive.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $j \in {1, \dots, n}$ , and $f \in H^{1} (R^{n}; C)$ with $x f \in L^{2} (R^{n}; C^{n})$ .

[A1]

Countable Choice is the assumption carried by the Sobolev and compact-support integration-by-parts interfaces below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The Fourier characterization identifies $H^{1}$ with $W^{1, 2}$ under the regular-distribution embedding, so each weak derivative $D_{k} f$ belongs to $L^{2}$ ; its Fourier-side weight is $⟨ ξ ⟩ \hat{f}$ (Integer-order W^{k,2} and H^k agree with equivalent norms).

[F2]

There is a real smooth cutoff $χ$ with $0 \leq χ \leq 1$ , $χ = 1$ on $∣ x ∣ \leq 1$ , $χ = 0$ on $∣ x ∣ \geq 2$ , and $D_{j} (χ (x / R)) = R^{- 1} (D_{j} χ) (x / R)$ (Explicit compactly supported smooth cutoffs).

[F3]

For $u \in W^{1, 2}$ and a smooth multiplier $ρ$ with bounded derivatives, $ρ u \in W^{1, 2}$ and $D_{j} (ρ u) = (D_{j} ρ) u + ρ D_{j} u$ (Weak Leibniz rule with a smooth factor).

[F4]

If $u, v \in W^{1, 2} (R^{n})$ and one factor is compactly supported as an almost-everywhere class, then $\int u D_{j} v = - \int v D_{j} u$ (Integration by parts for dual-exponent Sobolev functions).

[F5]

Complex $L^{2}$ satisfies Cauchy–Schwarz, so the product of two $L^{2}$ functions is integrable (Complex completeness, density, and inner product: the consumer interface).

[F6]

If integrable functions converge almost everywhere under a common integrable majorant, their integrals converge (Dominated convergence).

[F7]

The Sobolev test identity is bilinear and does not conjugate its test function (Integer-order Sobolev spaces and their norms, Complex Lp classes and Euclidean test-function conventions).

Proof

technique · cutoff integration by parts, followed by vanishing $L^2$ tails

1.1A1F1F7given

Since $f \in H^{1} = W^{1, 2}$ by [F1], each weak derivative $D_{k} f$ lies in $L^{2}$ . Conjugating the bilinear weak-derivative test identity for $f$ shows that $\overline{f} \in W^{1, 2}$ and $D_{k} \overline{f} = \overline{D_{k} f}$ : for each $ϕ \in C_{c}^{\infty}$ , apply the identity to $\overline{ϕ}$ and conjugate it.

2.1F2F3F4step 1.1given

For $R \geq 1$ set $χ_{R} (x) = χ (x / R)$ and $ρ_{R} (x) = x_{j} χ_{R} (x)$ . By [F2], $ρ_{R} \in C_{c}^{\infty}$ ; by [F3] and step 1.1, $v_{R} : = ρ_{R} \overline{f}$ belongs to $W^{1, 2}$ and is compactly supported, with $D_{j} v_{R} = (χ_{R} + x_{j} D_{j} χ_{R}) \overline{f} + x_{j} χ_{R} \overline{D_{j} f} .$ Applying [F4] to $u = f$ and $v = v_{R}$ gives $\int f D_{j} v_{R} = - \int v_{R} D_{j} f,$ hence $\int χ_{R} ∣ f ∣^{2} + \int x_{j} D_{j} χ_{R} ∣ f ∣^{2} = - 2 Re \int x_{j} χ_{R} \overline{f} D_{j} f .$ The last identity uses that $χ_{R}$ and $x_{j}$ are real-valued, so the two derivative terms are complex conjugates.

3.1F2F5F6step 2.1

Let $R \to \infty$ through positive integers. We have $χ_{R} (x) \to 1$ pointwise and $0 \leq χ_{R} \leq 1$ , so dominated convergence [F6] gives $\int χ_{R} ∣ f ∣^{2} \to ∥ f ∥_{2}^{2}$ . The derivative $D_{j} χ_{R}$ vanishes unless $R \leq ∣ x ∣ \leq 2 R$ , and $∣ x_{j} D_{j} χ_{R} ∣ \leq 2 ∥ D_{j} χ ∥_{\infty}$ there by [F2]; for every fixed $x$ this factor is eventually zero. Since $∣ f ∣^{2} \in L^{1}$ , [F6] gives $\int x_{j} D_{j} χ_{R} ∣ f ∣^{2} \to 0$ . Finally, $x_{j} \overline{f} D_{j} f \in L^{1}$ by [F5], because $x_{j} f, D_{j} f \in L^{2}$ ; dominated convergence with $∣ χ_{R} ∣ \leq 1$ yields $∥ f ∥_{2}^{2} = - 2 Re \int x_{j} \overline{f} D_{j} f .$

4.1F5step 3.1∎

By [F5] and $∣ Re z ∣ \leq ∣ z ∣$ , $∥ f ∥_{2}^{2} \leq 2 \int ∣ x_{j} f ∣ ∣ D_{j} f ∣ \leq 2 ∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} .$ Dividing by $2$ proves the inequality. If $f = 0$ both sides vanish; if $f \neq 0$ the lower bound is positive.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Compact support gives an entire Fourier-Laplace transform by slices

Statement

Let $n \geq 1$ and let $f \in L^{1} (R^{n}; C)$ vanish almost everywhere outside a compact set $K \subseteq R^{n}$ . Define $F (z) : = \int_{R^{n}} f (x) e^{- 2 π i x \cdot z} d x, z \in C^{n} .$ Then the integral converges absolutely for every $z$ and its value does not depend on the representative of the $L^{1}$ class; for every $j$ and every fixed values of the other $n - 1$ complex coordinates, the coordinate slice $w \mapsto F (z_{1}, \dots, z_{j - 1}, w, z_{j + 1}, \dots, z_{n})$ is entire on $C$ , with $\frac{\partial}{\partial z_{j}} F (z) = \int_{R^{n}} f (x) (- 2 π i x_{j}) e^{- 2 π i x \cdot z} d x;$ and $F (x) = f^{\land} (x)$ for every $x \in R^{n}$ , where $f^{\land}$ is the $L^{1}$ transform of Fourier transform on complex L1 classes.

Facts & Assumptions

Given: An integer $n \geq 1$ , a class $f \in L^{1} (R^{n}; C)$ vanishing almost everywhere outside a compact set $K \subseteq R^{n}$ , and points $z \in C^{n}$ and $w \in C$ ; here $x \cdot z = \sum_{k = 1}^{n} x_{k} z_{k}$ for $x \in R^{n}$ .

[F1]

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

[F2]

For an integrable complex function $u$ , $∣ \int u ∣ \leq \int ∣ u ∣$ (The modulus of an integral is bounded by the integral of the modulus).

[F3]

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).

[F4]

The $L^{1}$ transform $\hat{f} (ξ) = \int_{R^{n}} f (x) e^{- 2 π i x \cdot ξ} d x$ is absolutely convergent at every real $ξ$ , is unchanged by null-set modifications of the representative, and satisfies $∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ (Fourier transform on complex L1 classes, The integral transform is representative independent).

[F5]

A nonempty compact subset $K \subseteq R^{n}$ is bounded, so $M : = \sup_{x \in K} ∣ x ∣ < \infty$ and $S_{j} : = \sup_{x \in K} ∣ x_{j} ∣ \leq M$ for every $j$ ; if $K \neq \emptyset$ each $S_{j}$ is attained on $K$ by the extreme value theorem (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value). Only finiteness of $S_{j}$ is used below.

[F6]

Complex-valued functions are measurable when their components are; pointwise sums, products, and compositions with continuous functions of the coordinates of measurable complex-valued functions are measurable, and so is the modulus (Complex Lp classes and Euclidean test-function conventions).

[F7]

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

Proof

technique · direct

1.1givenF3F5F6

Fix a representative. If $K = \emptyset$ then $f = 0$ almost everywhere, so every integral below vanishes by [F3], giving $F \equiv 0$ , and all claims of the lemma hold trivially; assume $K \neq \emptyset$ and let $M$ be the bound of [F5]. Multiplying $f$ by the indicator of the closed set $K$ produces a measurable representative in $L^{1}$ that vanishes everywhere outside $K$ and agrees with $f$ almost everywhere, so no integral below changes ([F3, F6]); from here we use this representative.

1.2givenF1F3F5F6

Absolute convergence and representative independence. For $x \in K$ the exponential law and modulus formula [F1] give $∣ f (x) e^{- 2 π i x \cdot z} ∣ = ∣ f (x) ∣ e^{2 π x \cdot Im z} \leq ∣ f (x) ∣ e^{2 π M ∣ Im z ∣}$ , while for $x \notin K$ both sides vanish. Hence $\int_{R^{n}} ∣ f (x) e^{- 2 π i x \cdot z} ∣ d x \leq e^{2 π M ∣ Im z ∣} ∥ f ∥_{1} < \infty$ for the given $z$ , and since $z$ was arbitrary the integral defining $F$ converges absolutely at every point of $C^{n}$ . If $\tilde{f} = f$ almost everywhere is a second representative, the two integrands agree almost everywhere and are both integrable, so the two integrals agree by [F3]; thus the value is representative-independent.

2.1givenF1F6step 1.2

The slice as a one-variable integral. Fix $j$ and complex numbers $z_{k}$ for $k \neq j$ , and set $g (x) : = f (x) e^{- 2 π i \sum_{k \neq j} x_{k} z_{k}}$ . Then $g$ is measurable and $∣ g (x) ∣ \leq ∣ f (x) ∣ e^{2 π M ∣ Im z ∣}$ for every $x$ by the same bound as in step 1.2, so $g \in L^{1}$ ; for $w \in C$ define $h (w) : = \int_{R^{n}} g (x) e^{- 2 π i x_{j} w} d x$ . By the addition law [F1] the integrand equals $f (x) e^{- 2 π i x \cdot z^{'}}$ with $z^{'} = (z_{1}, \dots, z_{j - 1}, w, z_{j + 1}, \dots, z_{n})$ , so $h (w) = F (z^{'})$ and step 1.2 shows that the integral for $h (w)$ converges absolutely at every $w \in C$ .

3.1givenF1F3step 2.1

Difference quotients are integrals. Fix $w \in C$ and $s \in C ∖ {0}$ . For every $x$ the addition law [F1] 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 $g (x) e^{- 2 π i x_{j} (w + s)}$ and $g (x) e^{- 2 π i x_{j} w}$ are integrable because $g \in L^{1}$ and the exponential factors are bounded on the support of $g$ by step 2.1. Applying additivity and scaling from [F3] therefore yields $\frac{h (w + s) - h (w)}{s} = \int_{R^{n}} g (x) e^{- 2 π i x_{j} w} \frac{e^{- 2 π i x_{j} s} - 1}{s} d x .$

4.1F1F2F3F5step 2.1step 3.1

Uniform remainder estimate. Put $S : = S_{j} \leq M$ and $L : = \int g (x) (- 2 π i x_{j}) e^{- 2 π i x_{j} w} d x$ , which is absolutely convergent since $∣ x_{j} ∣ \leq S$ on the support of $g$ . If $S = 0$ , step 3.1 is identically zero and $L = 0$ . Otherwise, for any $ε > 0$ , [F1] gives $δ > 0$ such that $∣ (e^{ζ} - 1) / ζ - 1 ∣ < ε$ when $0 < ∣ ζ ∣ < δ$ . For $0 < ∣ s ∣ < δ / (2 π S)$ , setting $ζ = - 2 π i x_{j} s$ therefore bounds the difference between the quotient integrand and its limiting integrand by $2 π S ε e^{2 π S ∣ Im w ∣} ∣ g (x) ∣$ on $K$ , with zero remainder when $x_{j} = 0$ . By [F2] and linearity [F3], $∣ \frac{h (w + s) - h (w)}{s} - L ∣ \leq 2 π S ε e^{2 π S ∣ Im w ∣} ∥ g ∥_{1} .$ Since $ε$ is arbitrary, the complex difference quotient tends to $L$ directly, without selecting a sequence.

5.1F4F7givenstep 4.1∎

Conclusion. The limit in step 4.1 shows that $h$ is complex differentiable at every $w \in C$ with derivative $h^{'} (w) = \int_{R^{n}} g (x) (- 2 π i x_{j}) e^{- 2 π i x_{j} w} d x$ ; a function holomorphic on all of $C$ is entire by definition ([F7]). Rewriting the derivative integrand gives the displayed formula for $\partial F / \partial z_{j}$ . At real $z = ξ \in R^{n}$ the defining integral of $F$ is literally the $L^{1}$ transform formula, so $F (ξ) = \hat{f} (ξ)$ at every real $ξ$ by [F4].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

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]

[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].

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Entire rigidity under Gaussian growth and real-axis decay

Statement

Let $a > 0$ , $b > 0$ and let $F : C \to C$ be entire. Suppose there are $C_{1}, C_{2} \geq 0$ with $∣ F (x + i y) ∣ \leq C_{1} e^{π y^{2} / a}, ∣ F (x) ∣ \leq C_{2} e^{- π b x^{2}} (x, y \in R) .$ Then: (i) if $a b > 1$ , $F \equiv 0$ ; (ii) if $a b = 1$ , $F (z) = F (0) e^{- π z^{2} / a}$ for every $z \in C$ . All constants are absorbed into the two bounds; no further hypothesis on $F$ is imposed.

Facts & Assumptions

Given: Reals $a > 0$ , $b > 0$ , an entire $F : C \to C$ , constants $C_{1}, C_{2} \geq 0$ satisfying the two displayed bounds, and $C : = \max (C_{1}, C_{2})$ .

[F1]

The complex exponential is entire with $\exp^{'} = \exp$ , satisfies $\exp (ζ + η) = \exp ζ \exp η$ , and $∣ \exp ζ ∣ = e^{Re ζ}$ (The complex exponential is entire and its complex derivative is itself, $\exp (z + w) = \exp z \exp w$ , 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 $e^{i π} + 1 = 0$ ).

[F2]

Sums, scalar multiples and products of complex differentiable functions are complex differentiable with the usual rules, and a composition of complex differentiable maps is complex differentiable (Linearity, product, reciprocal, and quotient rules for complex derivatives, The chain rule for complex derivatives); holomorphy on all of $C$ means entire (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[F3]

On the slit plane $S = C ∖ {x \in R : x \leq 0}$ the principal logarithm is holomorphic, and for $w \in C$ the principal power $z \mapsto z_{p r}^{w} = \exp (w Log z)$ is holomorphic on $S$ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers, The principal logarithm is the normalised holomorphic branch on the slit plane); for real $s > 0$ one has $∣ z_{p r}^{s} ∣ = ∣ z ∣^{s}$ , and $[0, \infty) \to R$ , $t \mapsto t^{s}$ , is continuous (Real powers for positive bases, with the zero-base positive-exponent convention, Continuity and derivatives of positive-base real powers).

[F4]

Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

[F5]

Maximum modulus principle with boundary and infinity control: if $Ω \subseteq C$ is a domain, $G$ is holomorphic on $Ω$ and $M \geq 0$ is such that for every $ε > 0$ every boundary point of $Ω$ has a neighbourhood $V$ with $∣ G ∣ < M + ε$ on $V \cap Ω$ , while $∣ G ∣ < M + ε$ outside some large circle inside $Ω$ , then $∣ G ∣ \leq M$ on $Ω$ (Maximum modulus principle with boundary and infinity control).

[F6]

Every $z \neq 0$ has a polar form $z = r e^{i α}$ with $r = ∣ z ∣ > 0$ and $α = \arg z$ (Every nonzero complex number has a unique polar form $r (\cos θ + i \sin θ)$ with $r > 0$ and $- π < θ \leq π$ ); the Cartesian field laws ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ) and [F1] give the identities $Re (z^{2}) = (Re z)^{2} - (Im z)^{2}$ , $Im (z^{2}) = ∣ z ∣^{2} \sin 2 α$ and $(Re z)^{2} = ∣ z ∣^{2} \cos^{2} α$ ; by [F1] these give $∣ e^{\pm i δ z^{2}} ∣ = e^{\mp δ ∣ z ∣^{2} \sin 2 α}$ for real $δ > 0$ .

Proof

technique · direct

To prove the asserted cases (i) and (ii), it suffices to treat $a b \geq 1$ ; assume $b \geq 1 / a$ throughout the proof. This makes the real-axis bound for the critical function uniform on the two anchor rays used in the sector argument.

1.1F1F2F6given

The critical function. Put $Φ (z) : = e^{π z^{2} / a} F (z)$ . The map $z \mapsto π z^{2} / a$ is a polynomial, hence complex differentiable everywhere, and composing it with the entire exponential and multiplying by the entire $F$ shows that $Φ$ is entire ([F1, F2]). For real $x$ and $y$ , using $∣ e^{π z^{2} / a} ∣ = e^{π Re (z^{2}) / a}$ and the two hypotheses, $∣ Φ (x) ∣ = e^{π x^{2} / a} ∣ F (x) ∣ \leq C_{2} e^{- π (b - 1 / a) x^{2}} \leq C_{2}, ∣ Φ (i y) ∣ = e^{- π y^{2} / a} ∣ F (i y) ∣ \leq C_{1},$ and, since $Re (z^{2}) = (Re z)^{2} - (Im z)^{2}$ , $∣ Φ (z) ∣ = e^{π ((Re z)^{2} - (Im z)^{2}) / a} ∣ F (z) ∣ \leq C_{1} e^{π (Re z)^{2} / a} .$ In particular $∣ Φ (0) ∣ = ∣ F (0) ∣ \leq C_{2} \leq C$ .

1.2F1F2F3F6construct

Sector data. Fix $δ > 0$ and any $θ$ with $\arctan \frac{π}{2 a δ} < θ < \frac{π}{2}$ ; equivalently $π \cos^{2} θ / a < δ \sin 2 θ$ . Fix also $ε > 0$ small enough that $(2 + ε) θ < π$ , and put $σ_{ε} : = \cos \frac{(2 + ε) θ}{2} > 0, η_{ε} : = \frac{π}{2} - \frac{(2 + ε) θ}{2},$ defining two auxiliary functions on the sectors $S_{1} = {0 < \arg z < θ}$ and $S_{2} = {π - θ < \arg z < π}$ , both contained in the slit plane $S$ of [F3]: $h_{ε} (z) : = \exp ⁣ (i ε e^{i μ} z_{p r}^{2 + ε}), q_{δ} (z) : = e^{i δ z^{2}} (z \in S_{1}), q_{δ} (z) : = e^{- i δ z^{2}} (z \in S_{2}),$ with $μ : = η_{ε}$ on $S_{1}$ and $μ : = η_{ε} - (2 + ε) (π - θ)$ on $S_{2}$ . Since $z_{p r}^{2 + ε}$ is holomorphic on $S$ and exponentials and polynomials are entire, $h_{ε}$ and $q_{δ}$ are holomorphic on each sector, and so is $G_{ε} : = h_{ε} q_{δ} Φ$ ([F1, F2, F3]). For $z = r e^{i α}$ in the closure of either sector, $∣ z_{p r}^{2 + ε} ∣ = r^{2 + ε}$ and $∣ h_{ε} (z) ∣ = \exp ⁣ (- ε r^{2 + ε} \sin (μ + (2 + ε) α)) \leq e^{- ε σ_{ε} r^{2 + ε}} \leq 1,$ because $μ + (2 + ε) α \in [\frac{π}{2} - \frac{(2 + ε) θ}{2}, \frac{π}{2} + \frac{(2 + ε) θ}{2}]$ for $α \in [0, θ]$ on $S_{1}$ and for $α \in [π - θ, π]$ on $S_{2}$ . Also $∣ q_{δ} (z) ∣ = e^{\mp δ Im (z^{2})}$ with the sign making $∣ q_{δ} ∣ \leq 1$ on the sector.

2.1F1F3F6givenstep 1.1step 1.2

Boundary bounds. On the far ray $\arg z = θ$ of $S_{1}$ one has $Re z = r \cos θ$ and, by [F6], $∣ q_{δ} (z) ∣ = e^{- δ r^{2} \sin 2 θ}$ , so steps 1.1 and 1.2 give $∣ G_{ε} (z) ∣ \leq ∣ q_{δ} (z) ∣ ∣ Φ (z) ∣ \leq C_{1} \exp (r^{2} (π \cos^{2} θ / a - δ \sin 2 θ)) \leq C_{1} \leq C$ by the choice of $θ$ . On the far ray $\arg z = π - θ$ of $S_{2}$ one has $Re z = - r \cos θ$ and $∣ q_{δ} (z) ∣ = e^{- δ r^{2} \sin 2 θ}$ as well, so the same computation gives $∣ G_{ε} (z) ∣ \leq C_{1} \leq C$ there. On the anchor rays $\arg z = 0$ and $\arg z = π$ the identity [F6] gives $∣ q_{δ} (z) ∣ = 1$ , and the display of step 1.2 gives $∣ h_{ε} ∣ \leq 1$ ; hence $∣ G_{ε} (z) ∣ \leq C_{2} \leq C$ there, and at $z = 0$ one has $∣ G_{ε} (0) ∣ = ∣ F (0) ∣ \leq C$ .

3.1F1F3F5step 1.2step 2.1

Boundedness on the sectors. In the closure of either sector, steps 1.1 and 1.2 give $∣ G_{ε} (z) ∣ \leq C_{1} \exp (r^{2} (\frac{π}{a} + 2 δ) - ε σ_{ε} r^{2 + ε}) ⟶ 0 (r \to \infty),$ because $\cos^{2} α \leq 1$ and $∣ \sin 2 α ∣ \leq 1$ on the compact angular interval; this is the control at infinity. Step 2.1 gives $∣ G_{ε} ∣ \leq C$ on the boundary rays, and $∣ G_{ε} ∣$ is continuous on the closed sector (the principal power extends continuously from $S$ to the closure of each sector, and $∣ G_{ε} ∣$ is a finite product of continuous functions), so every boundary point has a neighbourhood $V$ with $∣ G_{ε} ∣ < C + ε^{'}$ on $V$ intersected with the sector. The maximum modulus principle [F5], applied to the domains $S_{1}$ and $S_{2}$ , therefore gives $∣ G_{ε} ∣ \leq C$ on both sectors.

4.1F1step 3.1

Removing the auxiliary and the sector truncation. Fix $δ > 0$ . The perturbation $h_{ε}$ tends to $1$ pointwise as $ε \to 0$ along any sequence with $(2 + ε) θ < π$ , so step 3.1 gives $∣ q_{δ} Φ ∣ \leq C$ on each sector $S_{1} (θ), S_{2} (θ)$ for every admissible $θ \in (\arctan \frac{π}{2 a δ}, \frac{π}{2})$ . Since $θ$ was arbitrary in that interval and the sectors with larger $θ$ contain those with smaller $θ$ , while $q_{δ}$ does not depend on $θ$ , taking $θ \to π / 2$ yields $∣ e^{i δ z^{2}} Φ (z) ∣ \leq C$ for every $z$ with $0 < \arg z < π / 2$ and $∣ e^{- i δ z^{2}} Φ (z) ∣ \leq C$ for every $z$ with $π / 2 < \arg z < π$ . Letting now $δ ↓ 0$ along any sequence, $∣ q_{δ} (z) ∣ \to 1$ at each fixed $z$ by [F1] (the exponent $i δ z^{2} \to 0$ ), so $∣ Φ (z) ∣ \leq C (0 < \arg z < π, z \neq 0) .$

5.1F2givenstep 4.1

The lower half-plane. The function $\tilde{Φ} (z) : = Φ (- z)$ is entire ([F2]) and satisfies the same three estimates as $Φ$ in step 1.1, since the bounds $∣ F (- x) ∣ \leq C_{2} e^{- π b x^{2}}$ , $∣ F (- i y) ∣ \leq C_{1} e^{π y^{2} / a}$ and the growth estimate only involve absolute values and $(Re (- z))^{2} = (Re z)^{2}$ . Applying steps 1.1–4.1 to $\tilde{Φ}$ gives $∣ Φ (- z) ∣ \leq C$ for $0 < \arg z < π$ , that is, $∣ Φ (w) ∣ \leq C$ for $- π < \arg w < 0$ .

6.1F4givenstep 1.1step 4.1step 5.1∎

Conclusion of the critical case and of (i). By steps 4.1 and 5.1 the entire function $Φ$ satisfies $∣ Φ ∣ \leq C$ off the coordinate axes, while on the axes step 1.1 gives $∣ Φ (x) ∣ \leq C_{2} \leq C$ and $∣ Φ (i y) ∣ \leq C_{1} \leq C$ directly. Hence $Φ$ is a bounded entire function, so $Φ$ is constant by [F4]; the constant is $Φ (0) = F (0)$ . Therefore, whenever $b \geq 1 / a$ , $F (z) = F (0) e^{- π z^{2} / a} (z \in C) .$ If $a b > 1$ then $b > 1 / a$ , so the display applies; evaluating at real $x$ gives $∣ F (0) ∣ e^{- π x^{2} / a} = ∣ F (x) ∣ \leq C_{2} e^{- π b x^{2}}$ , that is, $∣ F (0) ∣ \leq C_{2} e^{- π (b - 1 / a) x^{2}}$ for every real $x$ , and letting $x \to \infty$ forces $F (0) = 0$ and hence $F \equiv 0$ . If $a b = 1$ then the display is assertion (ii).

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Subcritical Gaussians show the Hardy threshold $a b = 1$ is sharp

Statement

Assume countable choice. Let $n \geq 1$ and let $a, b > 0$ with $a b < 1$ . Then $(a, 1 / b)$ is a nonempty open interval, and for every $c$ with $a < c < 1 / b$ the Gaussian $f_{c} (x) : = e^{- π c ∣ x ∣^{2}}$ satisfies $∣ f_{c} (x) ∣ \leq e^{- π a ∣ x ∣^{2}}, ∣ \hat{f_{c}} (ξ) ∣ \leq c^{- n / 2} e^{- π b ∣ ξ ∣^{2}} (x, ξ \in R^{n}) .$ In particular Hardy's two Gaussian bounds hold at every subcritical pair $a b < 1$ with a nonzero function, so the vanishing and classification conclusions genuinely require $a b \geq 1$ .

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), an integer $n \geq 1$ , reals $a, b > 0$ with $a b < 1$ , and a real $c$ with $a < c < 1 / b$ .

[F1]

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

[F2]

For every $t > 0$ , the Gaussians $e^{- π t ∣ x ∣^{2}}$ are absolutely integrable and have $L^{1}$ Fourier transform $t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ at every frequency; every polynomial times a positive real Gaussian is absolutely integrable (Euclidean Gaussian transform with the 2π normalization).

[F3]

The real exponential is strictly increasing on $R$ (The exponential function is strictly increasing); the real power $c^{s}$ of a positive base is a positive real number, and $s \mapsto c^{s}$ , $t \mapsto t^{s}$ are continuous on their domains (Real powers for positive bases, with the zero-base positive-exponent convention, Continuity and derivatives of positive-base real powers).

Proof

technique · direct

1.1F3given

The interval and the first bound. Since $b > 0$ , the inequality $a b < 1$ is equivalent to $a < 1 / b$ , so $(a, 1 / b)$ is a nonempty open interval and the given $c$ satisfies $c > a > 0$ . The function $f_{c}$ is continuous and hence measurable, with $f_{c} > 0$ and $f_{c} (0) = 1$ . For every $x \in R^{n}$ one has $- π c ∣ x ∣^{2} \leq - π a ∣ x ∣^{2}$ because $c \geq a$ , and the real exponential is strictly increasing [F3], so $0 < f_{c} (x) = e^{- π c ∣ x ∣^{2}} \leq e^{- π a ∣ x ∣^{2}}$ . Hence $∣ f_{c} (x) ∣ \leq e^{- π a ∣ x ∣^{2}}$ for every $x$ , and $f_{c} \neq 0$ .

2.1F1F2F3step 1.1

The transform and the second bound. By [F2] the Gaussian $f_{c}$ is absolutely integrable and its $L^{1}$ Fourier transform is $\hat{f_{c}} (ξ) = c^{- n / 2} e^{- π ∣ ξ ∣^{2} / c}$ for every $ξ \in R^{n}$ ; this is a positive real number. Since $c < 1 / b$ gives $b < 1 / c$ , one has $- π ∣ ξ ∣^{2} / c \leq - π b ∣ ξ ∣^{2}$ , and strict increase of the exponential [F3] gives $e^{- π ∣ ξ ∣^{2} / c} \leq e^{- π b ∣ ξ ∣^{2}}$ . The factor $c^{- n / 2}$ is positive by [F3]. Therefore $∣ \hat{f_{c}} (ξ) ∣ = c^{- n / 2} e^{- π ∣ ξ ∣^{2} / c} \leq c^{- n / 2} e^{- π b ∣ ξ ∣^{2}}$ for every $ξ$ .

3.1givenstep 1.1step 2.1∎

Conclusion. By steps 1.1 and 2.1 the nonzero Gaussian $f_{c}$ satisfies both Gaussian bounds of the subcritical pair $(a, b)$ whenever $a < c < 1 / b$ , and such $c$ exists at every pair with $a b < 1$ . Hence at every subcritical pair the two Gaussian hypotheses admit a nonzero solution, so the vanishing conclusion cannot hold below that threshold. Nor can the critical classification with rate $a$ hold: $f_{c} (x) / e^{- π a ∣ x ∣^{2}} = e^{- π (c - a) ∣ x ∣^{2}}$ equals $1$ at $0$ and is smaller at $x = (1, 0, \dots, 0)$ , so is nonconstant. By continuity it cannot be constant almost everywhere either.

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Separately holomorphic functions vanishing on a real box are zero

Statement

Let $n \geq 1$ and let $G : C^{n} \to C$ be separately holomorphic: for every $j$ and every fixed $w \in C^{n}$ , the one-variable map $z \mapsto G (w_{1}, \dots, w_{j - 1}, z, w_{j + 1}, \dots, w_{n})$ is entire on $C$ . If there are nondegenerate intervals $I_{1}, \dots, I_{n} \subseteq R$ with $G = 0$ on $I_{1} \times \dots \times I_{n}$ , then $G \equiv 0$ on $C^{n}$ .

Facts & Assumptions

Given: An integer $n \geq 1$ , a separately holomorphic $G : C^{n} \to C$ , and nondegenerate intervals $I_{1}, \dots, I_{n}$ (that is, each $I_{j}$ contains a nonempty open subinterval, so it has more than one point) with $G = 0$ on $I_{1} \times \dots \times I_{n}$ .

[F1]

Identity theorem: if two functions holomorphic on a complex domain $Ω \subseteq C$ agree on a set having an accumulation point in $Ω$ , then they agree on $Ω$ (Identity theorem for holomorphic functions).

[F2]

A function is entire when it is complex differentiable on all of $C$ , that is, holomorphic on the domain $C$ ; a separately holomorphic $G$ has every one-variable slice entire by hypothesis (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[F3]

If $J \subseteq R$ has nonempty interior and $p$ lies in that interior, then $p$ is an accumulation point of $J$ in $C$ : some open ball $(p - ε, p + ε)$ is contained in $J$ , and for every neighbourhood radius $r > 0$ , $p + \min (ε, r) / 2$ is a point of $J$ different from $p$ within that neighbourhood. Every coordinate slice of a separately holomorphic function is determined by the values it takes on such a set.

Proof

technique · induction on $n$

1.1baseF1F2F3given

Base case $n = 1$ . Here $G$ is an entire function of one variable and vanishes on the nondegenerate interval $I_{1}$ . Choose $p$ in the interior of $I_{1}$ ; by [F3], $p$ is an accumulation point in $C$ of the set where $G$ and the zero function agree, and both are entire [F2]. The identity theorem [F1] on the domain $C$ gives $G \equiv 0$ .

1.2ihgivenchoose

Inductive hypothesis and setup. Assume $n \geq 2$ and that the assertion holds for $n - 1$ variables. Choose $p \in int I_{1}$ , which is possible because $I_{1}$ is nondegenerate, and fix an arbitrary $z^{'} \in C^{n - 1}$ ; it remains to show $G (p, z^{'}) = 0$ .

2.1F2givenstep 1.2

The slice in the last $n - 1$ variables. The map $z^{'} \mapsto G (p, z^{'})$ on $C^{n - 1}$ is separately holomorphic, because each of its one-variable slices is a slice of $G$ with all other coordinates fixed, hence entire by [F2]. It vanishes on the box $I_{2} \times \dots \times I_{n}$ , whose factors are nondegenerate, so the induction hypothesis of step 1.2 applies and gives $G (p, z^{'}) = 0$ .

3.1step 2.1

Vanishing on a slab. The point $p \in int I_{1}$ in step 1.2 was chosen arbitrarily in the interior, so step 2.1 gives $G = 0$ on $int I_{1} \times C^{n - 1}$ .

4.1F1F2F3step 1.2step 3.1

The slice in the first variable. For the fixed $z^{'}$ of step 1.2, the one-variable map $w \mapsto G (w, z^{'})$ is entire by [F2] and vanishes on the nondegenerate interval $int I_{1}$ by step 3.1. Its zero set therefore has the accumulation point $p$ of [F3] inside the domain $C$ , and [F1] gives $G (w, z^{'}) = 0$ for every $w \in C$ .

5.1step 1.2step 4.1discharge-induction∎

Conclusion. Since $z^{'} \in C^{n - 1}$ was arbitrary in step 1.2, step 4.1 gives $G \equiv 0$ on $C^{n}$ , which discharges the induction step and completes the induction.

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The support-measure uncertainty inequality $∣ E ∣ ∣ F ∣ \geq 1$

Statement

Assume countable choice. Let $f \in L^{2} (R^{n}; C)$ be nonzero and let $E, F \subseteq R^{n}$ be Lebesgue measurable sets of finite measure such that $f = 0$ almost everywhere on $R^{n} ∖ E$ and $\hat{f} = 0$ almost everywhere on $R^{n} ∖ F$ . Here $\hat{f}$ denotes the continuous $L^{1}$ transform (The integral transform is representative independent), which is defined because $∣ E ∣ < \infty$ forces $f \in L^{1}$ . Then $∣ E ∣ ∣ F ∣ \geq 1.$ No regularity of $E$ or $F$ beyond measurability and finite measure is assumed, and no complex analysis is used.

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), a nonzero $f \in L^{2} (R^{n}; C)$ , and Lebesgue measurable sets $E, F$ of finite measure with $f = 0$ almost everywhere off $E$ and $\hat{f} = 0$ almost everywhere off $F$ .

[F1]

Countable choice is assumed; it is the hypothesis carried by the $L^{1}$ transform interface, the agreement theorem, and Plancherel below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

For $g \in L^{1} (R^{n}; C)$ the $L^{1}$ transform is defined at every frequency, satisfies $∣ \hat{g} (ξ) ∣ \leq ∥ g ∥_{1}$ , and is unchanged by null-set modifications of the representative (The integral transform is representative independent); it is continuous and vanishes at infinity (Riemann–Lebesgue lemma).

[F3]

Hölder's inequality with conjugate exponents $p = q = 2$ : for measurable real $u \in L^{2}$ and $v \in L^{2}$ one has $\int ∣ u v ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ (Holder's inequality for integrals, including the endpoint cases); $L^{1}$ and $L^{2}$ are the quotient spaces of The space $L^{p} (μ)$ as the quotient by null functions with the norms of Complex Lp classes and Euclidean test-function conventions, and integrable functions that agree almost everywhere have equal integrals (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[F4]

If $g \in L^{1} \cap L^{2}$ , its bounded continuous $L^{1}$ transform represents the Plancherel transform $F_{2} g$ almost everywhere (Agreement of the integral and L2 transforms), and Plancherel gives $∥ F_{2} g ∥_{2} = ∥ g ∥_{2}$ (Plancherel theorem).

Proof

technique · direct

1.1F1F2F3given

The function is integrable and its transform is bounded. Since $f = 0$ almost everywhere on $R^{n} ∖ E$ , one has $\int_{R^{n}} ∣ f ∣ = \int_{E} ∣ f ∣$ ; applying [F3] with $u = ∣ f ∣$ and $v = 1_{E}$ , whose $L^{2}$ norm is $∣ E ∣^{1 / 2} < \infty$ , gives $\int_{E} ∣ f ∣ \leq ∥ f ∥_{2} ∣ E ∣^{1 / 2} < \infty$ . Hence $f \in L^{1} \cap L^{2}$ , its $L^{1}$ transform $\hat{f}$ is defined and continuous [F2], and $∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1} \leq ∣ E ∣^{1 / 2} ∥ f ∥_{2}$ for every $ξ \in R^{n}$ .

2.1F4givenstep 1.1

Plancherel size from the two supports. As $f \in L^{1} \cap L^{2}$ , the continuous transform $\hat{f}$ represents $F_{2} f$ almost everywhere [F4]; since $\hat{f} = 0$ almost everywhere off $F$ , also $F_{2} f = 0$ almost everywhere off $F$ . Therefore $F_{2} f$ is represented by the function that vanishes off $F$ and equals $\hat{f}$ on $F$ , and $\int_{R^{n}} ∣ F_{2} f ∣^{2} = \int_{F} ∣ \hat{f} ∣^{2} \leq ∣ F ∣ \sup_{ξ \in R^{n}} ∣ \hat{f} (ξ) ∣^{2} \leq ∣ E ∣ ∣ F ∣ ∥ f ∥_{2}^{2},$ where the last inequality inserts the uniform bound of step 1.1 and $∣ F ∣ < \infty$ is used.

3.1F4givenstep 2.1∎

Conclusion. Plancherel's isometry [F4] gives $∥ F_{2} f ∥_{2}^{2} = ∥ f ∥_{2}^{2}$ , so step 2.1 yields $∥ f ∥_{2}^{2} \leq ∣ E ∣ ∣ F ∣ ∥ f ∥_{2}^{2}$ . Since $f$ is nonzero, $∥ f ∥_{2}^{2} > 0$ , and dividing gives $∣ E ∣ ∣ F ∣ \geq 1$ .

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The n-dimensional Heisenberg uncertainty inequality

Statement

Assume Countable Choice. Let $n \geq 1$ and let $f \in H^{1} (R^{n}; C)$ satisfy $x f \in L^{2} (R^{n}; C^{n})$ . Write $\hat{f} = F_{2} f$ for its Plancherel transform. Then $∥ ∣ x ∣ f ∥_{2} ∥ ∣ ξ ∣ \hat{f} ∥_{2} \geq \frac{n}{4 π} ∥ f ∥_{2}^{2} .$ The Fourier characterization of $H^{1}$ makes this exactly the domain where both spatial and Plancherel-frequency second moments are finite. For $f = 0$ both sides vanish. Equality cases on this full domain are not decided here; the published sharp equality theorem is stated for Schwartz functions.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $f \in H^{1} (R^{n}; C)$ with $x f \in L^{2} (R^{n}; C^{n})$ , its weak derivatives $D_{j} f$ , and its Plancherel transform $\hat{f} = F_{2} f$ .

[A1]

Countable Choice is the hypothesis carried by the Sobolev, multiplier, and Plancherel interfaces below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The integer-order Fourier characterization identifies $H^{1}$ with $W^{1, 2}$ and hence supplies every $D_{j} f \in L^{2}$ (Integer-order W^{k,2} and H^k agree with equivalent norms).

[F2]

For a weak derivative in $L^{2}$ , $F_{2} (D_{j} f) (ξ) = 2 π i ξ_{j} F_{2} f (ξ)$ almost everywhere (Distributional derivatives are polynomial Fourier multipliers).

[F3]

Plancherel is a complex-linear isometry on $L^{2}$ (Plancherel theorem).

[F4]

Cauchy–Schwarz for finite tuples in complex $L^{2}$ gives $\sum_{j = 1}^{n} a_{j} b_{j} \leq (\sum_{j} a_{j}^{2})^{1 / 2} (\sum_{j} b_{j}^{2})^{1 / 2}$ for nonnegative real $a_{j}, b_{j}$ (Complex completeness, density, and inner product: the consumer interface).

[F5]

The coordinate estimate $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$ holds on this $H^{1}$ -with-finite-spatial-moment domain (The coordinate inequality $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$ ).

Proof

technique · convert the coordinate estimate by Plancherel, sum, and apply finite-dimensional Cauchy–Schwarz

1.1A1F1F2F3F5given

Fix $j \in {1, \dots, n}$ . By [F1] the weak derivative $D_{j} f$ is in $L^{2}$ , and by [F2]–[F3] $∥ D_{j} f ∥_{2} = ∥ F_{2} (D_{j} f) ∥_{2} = 2 π ∥ ξ_{j} \hat{f} ∥_{2} .$ Applying the coordinate estimate [F5] and dividing by $2 π$ yields $∥ x_{j} f ∥_{2} ∥ ξ_{j} \hat{f} ∥_{2} \geq \frac{1}{4 π} ∥ f ∥_{2}^{2} .$

2.1F4step 1.1∎

Summing the inequalities of step 1.1 gives $\sum_{j = 1}^{n} ∥ x_{j} f ∥_{2} ∥ ξ_{j} \hat{f} ∥_{2} \geq \frac{n}{4 π} ∥ f ∥_{2}^{2} .$ By [F4] the left side is at most $(\sum_{j} ∥ x_{j} f ∥_{2}^{2})^{1 / 2} (\sum_{j} ∥ ξ_{j} \hat{f} ∥_{2}^{2})^{1 / 2} = ∥ ∣ x ∣ f ∥_{2} ∥ ∣ ξ ∣ \hat{f} ∥_{2},$ where the equalities follow by summing the coordinate integrals. This proves the asserted inequality, including $n = 1$ .

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Centring by translation and modulation preserves the variance product

Statement

Assume countable choice. Let $f \in L^{2} (R^{n}; C)$ be nonzero with finite second moments, spatial mean $a$ , frequency mean $b$ , and variances $V_{x} (f), V_{ξ} (f)$ as in Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments (Translation of a function on $R^{n}$ ). Define $g (x) : = e^{- 2 π i b \cdot x} f (x + a) .$ Then $g \in L^{2}$ is nonzero with finite second moments, its spatial mean is $0$ and its frequency mean is $0$ , and for every $j$ $∥ x_{j} g ∥_{2} = ∥ (x_{j} - a_{j}) f ∥_{2}, ∥ ξ_{j} \hat{g} ∥_{2} = ∥ (ξ_{j} - b_{j}) \hat{f} ∥_{2} .$ Consequently $V_{x} (g) = V_{x} (f)$ , $V_{ξ} (g) = V_{ξ} (f)$ , and $V_{x} (g) V_{ξ} (g) = V_{x} (f) V_{ξ} (f)$ .

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), a nonzero $f \in L^{2} (R^{n}; C)$ with finite second moments and means and variances $a, b, V_{x} (f), V_{ξ} (f)$ as in Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments, and $g (x) = e^{- 2 π i b \cdot x} f (x + a)$ .

[F1]

Countable choice is assumed; it is used by the change-of-variables interface and to select the $L^{2}$ -approximating sequence in step 2.2 (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

Complex $L^{1}$ change of variables: for a $C^{1}$ diffeomorphism $T$ with absolute Jacobian determinant $∣ \det D T ∣$ and $h \in L^{1}$ , $\int h (T x) ∣ \det D T (x) ∣ d x = \int h (y) d y$ ; the affine maps $x \mapsto x + a$ and $ξ \mapsto ξ + b$ have determinant $1$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions). Complex $L^{2}$ carries the norm $∥ \cdot ∥_{2}$ of Complex completeness, density, and inner product: the consumer interface.

[F3]

Translation and modulation: with $τ_{a} f (x) = f (x - a)$ and $M_{b} f (x) = e^{2 π i b \cdot x} f (x)$ , for $f \in S$ one has $\hat{τ_{a} f} (ξ) = e^{- 2 π i a \cdot ξ} \hat{f} (ξ)$ and $\hat{M_{b} f} (ξ) = \hat{f} (ξ - b)$ at every frequency (Translation, modulation, linear dilation and reflection laws, Translation of a function on $R^{n}$ ).

[F4]

Schwartz functions lie in $L^{1} \cap L^{2}$ ; Schwartz space is dense in $L^{2}$ , the integral Fourier transform of any $L^{1} \cap L^{2}$ function represents its Plancherel transform almost everywhere, and Plancherel is an isometry (Schwartz derivatives are integrable, Schwartz space is dense in L2, Agreement of the integral and L2 transforms, Plancherel theorem).

[F5]

Integrable functions that agree almost everywhere have equal integrals (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

Proof

technique · direct

1.1F2given

The centred function is admissible on the spatial side. Translation preserves null equivalence and the $L^{2}$ norm by [F2], while modulation has unit modulus, so $g \in L^{2}$ and $∥ g ∥_{2} = ∥ f ∥_{2} > 0$ . Substituting $y = x + a$ gives $\int ∣ x ∣^{2} ∣ g (x) ∣^{2} d x = \int ∣ y - a ∣^{2} ∣ f (y) ∣^{2} d y \leq 2 \int ∣ y ∣^{2} ∣ f (y) ∣^{2} d y + 2 ∣ a ∣^{2} ∥ f ∥_{2}^{2} < \infty .$ Also $\int ∣ x_{j} ∣ ∣ g (x) ∣^{2} d x < \infty$ by Cauchy--Schwarz from $g, x_{j} g \in L^{2}$ . Thus the spatial mean and variance of $g$ are defined; frequency-side finiteness is established in the frequency computation below.

2.1F2givenstep 1.1

Spatial side. Substituting $y = x + a$ and using $∣ g (x) ∣^{2} = ∣ f (x + a) ∣^{2}$ [F2] gives, for each $j$ , $\int x_{j} ∣ g (x) ∣^{2} d x = \int y_{j} ∣ f (y) ∣^{2} d y - a_{j} \int ∣ f (y) ∣^{2} d y = ∥ f ∥_{2}^{2} a_{j} - ∥ f ∥_{2}^{2} a_{j} = 0, \int x_{j}^{2} ∣ g (x) ∣^{2} d x = \int (y_{j} - a_{j})^{2} ∣ f (y) ∣^{2} d y .$ Hence the spatial mean of $g$ is $0$ , $∥ x_{j} g ∥_{2} = ∥ (x_{j} - a_{j}) f ∥_{2}$ , and $V_{x} (g) = ∥ g ∥_{2}^{- 2} \int ∣ x ∣^{2} ∣ g ∣^{2} = ∥ f ∥_{2}^{- 2} \int ∣ y - a ∣^{2} ∣ f (y) ∣^{2} d y = V_{x} (f)$ .

2.2F1F2F3F4F5givenstep 1.1

Frequency side. Choose $f_{k} \in S$ with $f_{k} \to f$ in $L^{2}$ , using [F4] and countable choice [F1], and set $g_{k} = M_{- b} (τ_{- a} f_{k})$ . By [F4], $f_{k} \in L^{1} \cap L^{2}$ ; [F2] shows translation and modulation preserve both spaces and their norms, so $g_{k} \in L^{1} \cap L^{2}$ . Translation and modulation preserve $L^{2}$ distances, hence $g_{k} \to g$ in $L^{2}$ ; Plancherel gives $F_{2} f_{k} \to F_{2} f$ and $F_{2} g_{k} \to F_{2} g$ . By [F3], for each $k$ the integral transforms satisfy ${\hat{g}}_{k} (ξ) = e^{2 π i a \cdot (ξ + b)} {\hat{f}}_{k} (ξ + b)$ , and [F4] identifies these transforms with their Plancherel classes. Translation and multiplication by this unit-modulus phase are isometries on $L^{2}$ by [F2], so passing to the norm limits proves the Plancherel-class identity $\hat{g} (ξ) = e^{2 π i a \cdot (ξ + b)} \hat{f} (ξ + b)$ almost everywhere. First, the affine change of variables $η = ξ + b$ gives $\int ∣ ξ ∣^{2} ∣ \hat{g} (ξ) ∣^{2} d ξ = \int ∣ η - b ∣^{2} ∣ \hat{f} (η) ∣^{2} d η < \infty,$ so the first moments are absolutely integrable by Cauchy--Schwarz. Using [F5] for representatives, the same substitution now yields $\int ξ_{j} ∣ \hat{g} (ξ) ∣^{2} d ξ = \int (η_{j} - b_{j}) ∣ \hat{f} (η) ∣^{2} d η = ∥ f ∥_{2}^{2} b_{j} - ∥ f ∥_{2}^{2} b_{j} = 0, \int ξ_{j}^{2} ∣ \hat{g} (ξ) ∣^{2} d ξ = \int (η_{j} - b_{j})^{2} ∣ \hat{f} (η) ∣^{2} d η .$ Thus $g$ has finite frequency second moments and mean $0$ , $∥ ξ_{j} \hat{g} ∥_{2} = ∥ (ξ_{j} - b_{j}) \hat{f} ∥_{2}$ , and $V_{ξ} (g) = ∥ \hat{g} ∥_{2}^{- 2} \int ∣ ξ ∣^{2} ∣ \hat{g} ∣^{2} = ∥ \hat{f} ∥_{2}^{- 2} \int ∣ η - b ∣^{2} ∣ \hat{f} (η) ∣^{2} d η = V_{ξ} (f)$ ; Plancherel and the unitary covariance give $∥ \hat{g} ∥_{2} = ∥ g ∥_{2} = ∥ f ∥_{2} = ∥ \hat{f} ∥_{2}$ .

3.1step 2.1step 2.2∎

Conclusion. Steps 2.1 and 2.2 give $V_{x} (g) = V_{x} (f)$ , $V_{ξ} (g) = V_{ξ} (f)$ and hence $V_{x} (g) V_{ξ} (g) = V_{x} (f) V_{ξ} (f)$ , and both centred means vanish.

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Hardy's Gaussian uncertainty principle in $R^{n}$

Statement

Assume countable choice. Let $n \geq 1$ and $a, b, C > 0$ and let $f : R^{n} \to C$ be measurable with $∣ f (x) ∣ \leq C e^{- π a ∣ x ∣^{2}}$ for almost every $x$ ; then $f \in L^{1}$ and $\hat{f}$ is its continuous $L^{1}$ transform (Fourier transform on complex L1 classes). Suppose $∣ \hat{f} (ξ) ∣ \leq C e^{- π b ∣ ξ ∣^{2}}$ for every $ξ \in R^{n}$ . Then: (i) if $a b > 1$ , $f = 0$ almost everywhere; (ii) if $a b = 1$ , there is $c \in C$ with $f (x) = c e^{- π a ∣ x ∣^{2}}$ for almost every $x$ ; necessarily $c = \hat{f} (0) a^{n / 2}$ with $\hat{f} (0) = \int f$ . Equality in (ii) is asserted almost everywhere only; no continuity of $f$ is assumed.

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), reals $a, b, C > 0$ , and a measurable $f : R^{n} \to C$ with $∣ f (x) ∣ \leq C e^{- π a ∣ x ∣^{2}}$ for almost every $x$ and $∣ \hat{f} (ξ) ∣ \leq C e^{- π b ∣ ξ ∣^{2}}$ for every $ξ \in R^{n}$ .

[F1]

Countable choice is assumed; it is the hypothesis carried by the entire continuation, the Gaussian transform and the uniqueness theorem below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

Gaussian decay gives an entire continuation: $f \in L^{1}$ , $F (z) : = \int_{R^{n}} f (x) e^{- 2 π i x \cdot z} d x$ converges absolutely for every $z \in C^{n}$ , has entire coordinate slices, satisfies $F (x) = \hat{f} (x)$ for real $x$ , and $∣ F (z) ∣ \leq C a^{- n / 2} e^{π ∣ Im z ∣^{2} / a}$ for every $z \in C^{n}$ (Gaussian decay gives an entire Fourier-Laplace transform and its growth bound).

[F3]

One-variable rigidity: if $a, b > 0$ , $C_{1}, C_{2} \geq 0$ and the entire $φ : C \to C$ satisfies $∣ φ (x + i y) ∣ \leq C_{1} e^{π y^{2} / a}$ and $∣ φ (x) ∣ \leq C_{2} e^{- π b x^{2}}$ for all real $x, y$ , then $φ \equiv 0$ when $a b > 1$ , and $φ (z) = φ (0) e^{- π z^{2} / a}$ for all $z \in C$ when $a b = 1$ (Entire rigidity under Gaussian growth and real-axis decay).

[F4]

A separately holomorphic $G : C^{n} \to C$ vanishing on a nondegenerate real box is identically zero (Separately holomorphic functions vanishing on a real box are zero).

[F5]

For every $t > 0$ the Gaussian $e^{- π t ∣ x ∣^{2}}$ is absolutely integrable with $L^{1}$ transform $t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ (Euclidean Gaussian transform with the 2π normalization).

[F6]

The $L^{1}$ transform is defined by $\hat{g} (ξ) = \int g (x) e^{- 2 π i x \cdot ξ} d x$ , so $\hat{g} (0) = \int g$ ; if $g, h \in L^{1}$ have equal transforms then $g = h$ almost everywhere; a scalar multiple has the correspondingly scaled transform (Fourier transform on complex L1 classes, Uniqueness of the L1 Fourier transform).

Proof

technique · apply one-variable rigidity on coordinate slices, then iterate the critical factors

1.1F1F2given

Entire continuation. By [F1, F2], $f \in L^{1}$ and its continuation $F$ has entire coordinate slices, $F (x) = \hat{f} (x)$ for real $x$ , and $∣ F (z) ∣ \leq C a^{- n / 2} e^{π ∣ Im z ∣^{2} / a}, ∣ F (x) ∣ \leq C e^{- π b ∣ x ∣^{2}} .$

2.1F3step 1.1

Coordinate rigidity. Fix $j$ and real coordinates $y_{k}$ for $k \neq j$ . The entire slice $φ (w) = F (y_{1}, \dots, y_{j - 1}, w, y_{j + 1}, \dots, y_{n})$ satisfies $∣ φ (u + i v) ∣ \leq C a^{- n / 2} e^{π v^{2} / a}, ∣ φ (u) ∣ \leq C e^{- π b \sum_{k \neq j} y_{k}^{2}} e^{- π b u^{2}} .$ Thus [F3] applies with $C_{1} = C a^{- n / 2}$ and $C_{2} = C e^{- π b \sum_{k \neq j} y_{k}^{2}}$ . If $a b > 1$ , every such slice is zero, so $F = 0$ on $R^{n}$ . If $a b = 1$ , every such slice satisfies $φ (w) = φ (0) e^{- π w^{2} / a}$ .

3.1F4step 2.1

Critical factorization. If $a b = 1$ , apply the slice identity in step 2.1 successively to coordinates $1, \dots, n$ of a real point $x$ , leaving the other coordinates real at each application. This gives $F (x) = F (0) \prod_{j = 1}^{n} e^{- π x_{j}^{2} / a} = F (0) e^{- π ∣ x ∣^{2} / a} .$ In fact the same formula holds for complex $z$ : the difference $F (z) - F (0) e^{- π \sum_{j} z_{j}^{2} / a}$ has entire coordinate slices and vanishes on $[0, 1]^{n}$ , so [F4] makes it identically zero. This includes $n = 1$ .

4.1F2F5F6step 2.1step 3.1∎

Fourier uniqueness and the constant. If $a b > 1$ , step 2.1 gives $\hat{f} = 0$ and [F6] yields $f = 0$ almost everywhere. If $a b = 1$ , [F5] says $g (x) = F (0) a^{n / 2} e^{- π a ∣ x ∣^{2}}$ is integrable with transform $F (0) e^{- π ∣ ξ ∣^{2} / a}$ , equal to $\hat{f}$ by step 3.1. By [F6], $f = g$ almost everywhere. Hence the scalar is $c = F (0) a^{n / 2} = \hat{f} (0) a^{n / 2}$ , and $\hat{f} (0) = \int f$ .

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A nonzero L1 function and its transform cannot both have compact support

Statement

Assume countable choice, used through the $L^{1}$ uniqueness theorem. Let $f \in L^{1} (R^{n}; C)$ and let $K_{1}, K_{2} \subseteq R^{n}$ be compact sets such that $f = 0$ almost everywhere on $R^{n} ∖ K_{1}$ and $f^{\land} = 0$ on $R^{n} ∖ K_{2}$ , where $f^{\land}$ is the continuous $L^{1}$ transform (Fourier transform on complex L1 classes). Then $f = 0$ almost everywhere. In particular, if $f$ is nonzero and $L^{1}$ with compact support, its transform cannot have compact support.

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), a function $f \in L^{1} (R^{n}; C)$ , compact sets $K_{1}, K_{2} \subseteq R^{n}$ with $f = 0$ almost everywhere on $R^{n} ∖ K_{1}$ and $\hat{f} = 0$ on $R^{n} ∖ K_{2}$ .

[F1]

Countable choice is assumed; it is the hypothesis carried by the $L^{1}$ uniqueness theorem below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

Compact support gives an entire continuation: $F (z) : = \int_{R^{n}} f (x) e^{- 2 π i x \cdot z} d x$ converges absolutely at every $z \in C^{n}$ , has entire coordinate slices, and satisfies $F (x) = \hat{f} (x)$ for every real $x$ (Compact support gives an entire Fourier-Laplace transform by slices).

[F3]

A separately holomorphic map on $C^{n}$ that vanishes on a nondegenerate real box $I_{1} \times \dots \times I_{n}$ is identically zero (Separately holomorphic functions vanishing on a real box are zero).

[F4]

If $g, h \in L^{1} (R^{n}; C)$ have equal $L^{1}$ transforms, then $g = h$ almost everywhere (Uniqueness of the L1 Fourier transform).

[F5]

A subset of $R^{n}$ is compact if and only if it is closed and bounded; in particular compact subsets are closed (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line). Consequently a nonempty open set $U \subseteq R^{n}$ contains a nondegenerate box: if $ξ \in U$ and the ball of radius $ε > 0$ around $ξ$ lies in $U$ , then $\prod_{j = 1}^{n} [ξ_{j} - ε / (2 \sqrt{n}), ξ_{j} + ε / (2 \sqrt{n})] \subseteq U$ , since every point of this box is at Euclidean distance at most $ε / 2$ from $ξ$ .

Proof

technique · direct

1.1F2given

The entire continuation of $f$ . By [F2] there is a function $F : C^{n} \to C$ with entire coordinate slices and $F (x) = \hat{f} (x)$ for every real $x$ ; hence $F$ is separately holomorphic and vanishes wherever $\hat{f}$ does.

1.2F5givenchoose

A box outside the compact frequency support. Since $K_{2}$ is bounded by [F5] and $n \geq 1$ , choose $R > 0$ such that $K_{2} \subseteq {∣ x ∣ \leq R}$ and take $ξ = (R + 1, 0, \dots, 0) \notin K_{2}$ . Since $K_{2}$ is closed, its complement is a nonempty open set; [F5] supplies a nondegenerate real box there. The argument also applies when $K_{2}$ is empty.

2.1F3givenstep 1.1step 1.2

Vanishing of the continuation. On the box from step 1.2, $\hat{f} = 0$ by hypothesis and $F = \hat{f}$ by step 1.1. Thus the separately holomorphic function $F$ vanishes on a nondegenerate real box. By [F3], $F \equiv 0$ on $C^{n}$ , so $\hat{f} \equiv 0$ on $R^{n}$ .

3.1F1F4step 2.1∎

Return to the original function. The $L^{1}$ functions $f$ and $0$ now have equal transforms, so [F1, F4] gives $f = 0$ almost everywhere. Consequently a nonzero compactly supported $L^{1}$ function cannot also have compactly supported transform.

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The sharp Heisenberg theorem is owned by functional analysis

Remarks

Assume countable choice, as does the cited theorem. The sharp Heisenberg uncertainty inequality and its equality classification are owned by functional analysis: the published theorem Heisenberg uncertainty and Gaussian equality of the functional-analysis track states that for $f \in S (R^{n})$ and $a, b \in R^{n}$ $∥ ∣ x - a ∣ f ∥_{2} ∥ ∣ ξ - b ∣ \hat{f} ∥_{2} \geq \frac{n}{4 π} ∥ f ∥_{2}^{2},$ with equality for nonzero $f$ exactly for $f (x) = c \exp (- λ ∣ x - a ∣^{2} / 2) \exp (2 π i b \cdot x)$ , $c \neq 0$ , $λ > 0$ . Both this inequality and its sharp equality classification are quoted here only on the Schwartz domain of the published source. Under this library's $e^{- 2 π i x \cdot ξ}$ convention its coordinate form on that domain is $∥ x_{j} f ∥_{2} ∥ ξ_{j} \hat{f} ∥_{2} \geq (4 π)^{- 1} ∥ f ∥_{2}^{2}$ .

This page does not extend the source's equality classification beyond Schwartz functions. Separately, the local cutoff argument in The coordinate inequality $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$ supplies the coordinate real-variable inequality on the natural $H^{1}$ domain with $x f \in L^{2}$ , Centring by translation and modulation preserves the variance product centres the variance formulation, and The n-dimensional Heisenberg uncertainty inequality records the summed $n$ -dimensional inequality on that same domain. Gaussian attainment in this convention is checked in The Gaussian attains equality in the Heisenberg inequality ↗.

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Proof cost and complex-analysis interface for Hardy uncertainty

Remarks

The complex-analysis cost of Hardy's Gaussian uncertainty principle in $R^{n}$ is the one-variable rigidity in Entire rigidity under Gaussian growth and real-axis decay. For the critical function $Φ (z) = e^{π z^{2} / a} F (z)$ , that lemma first bounds $q_{δ} Φ$ on two sectors of aperture $θ < π / 2$ , with $q_{δ} = e^{i δ z^{2}}$ on the first and $q_{δ} = e^{- i δ z^{2}}$ on the second. It then uses $h_{ε} (z) = \exp (i ε e^{i μ} z_{p r}^{2 + ε})$ , with the sector-dependent phase $μ$ specified in its step 1.2, so that $∣ h_{ε} (z) ∣ \leq e^{- ε σ_{ε} ∣ z ∣^{2 + ε}}$ uniformly in angle. Boundary and infinity control give the sector bound; removing the perturbations and applying Liouville gives critical rigidity. Coordinate slices then yield the higher-dimensional theorem.

The estimate $∣ F (x + i y) ∣ \leq C_{1} e^{π y^{2} / a}$ comes from the spatial Gaussian bound; the critical decay is $e^{- π x^{2} / a}$ . Tao's real-variable proof in the cited post proves a weaker, non-sharp threshold $a b > C_{0}$ . This describes that particular proof, not a limitation of all real-variable methods: the cited survey, §1, printed p. 2, also records a complete sharp real-variable proof.

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Finite support-product uncertainty for the unitary DFT

Statement

Let $N \geq 1$ , let $f \in C^{Z / N Z}$ be nonzero and let $F_{N}$ be the unitary discrete Fourier transform of The unitary discrete Fourier transform on $Z / N Z$ . Writing $supp f : = {x : f (x) \neq 0}$ and $supp F_{N} f : = {k : (F_{N} f) (k) \neq 0}$ , $∣ supp f ∣ \cdot ∣ supp F_{N} f ∣ \geq N .$ At $N = 1$ the bound is equality for every nonzero $f$ . No convergence or regularity hypothesis is involved.

Facts & Assumptions

Given: An integer $N \geq 1$ and a nonzero $f \in C^{Z / N Z}$ , with $S : = supp f$ , $T : = supp F_{N} f$ , the counting inner product $⟨ g, h ⟩ = \sum_{x \in Z / N} g (x) \overline{h (x)}$ and norm $∥ g ∥_{2}^{2} = \sum_{x} ∣ g (x) ∣^{2}$ of The counting inner product on $C^{Z / N Z}$ , and the unitary transform $(F_{N} g) (k) = N^{- 1 / 2} \sum_{x = 0}^{N - 1} g ([x]_{N}) e^{- 2 π i k x / N}$ of The unitary discrete Fourier transform on $Z / N Z$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[F1]

Finite sums in a commutative monoid are order-independent and linear with respect to scalar multiplication, and satisfy the triangle inequality $∣ \sum_{j} z_{j} ∣ \leq \sum_{j} ∣ z_{j} ∣$ ; the standard rules for real finite sums hold (A finite sum in a commutative monoid indexed by an arbitrary finite set, Laws of finite sums and finite products). The complex triangle inequality follows by induction on the number of summands from $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F2]

Cauchy–Schwarz for the counting inner product: $∣ ⟨ g, h ⟩ ∣ \leq ∥ g ∥_{2} ∥ h ∥_{2}$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors, The counting inner product on $C^{Z / N Z}$ ), where $∥ g ∥_{2}^{2} = ⟨ g, g ⟩ = \sum_{x} ∣ g (x) ∣^{2}$ .

[F3]

Finite Parseval: $⟨ F_{N} g, F_{N} h ⟩ = ⟨ g, h ⟩$ for all $g, h$ ; in particular $∥ F_{N} g ∥_{2} = ∥ g ∥_{2}$ (Finite Parseval and Plancherel identity for the unitary DFT).

[F4]

The modulus satisfies $∣ z ∣^{2} = z \overline{z} \geq 0$ and vanishes only at $z = 0$ ; complexes form a field (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

Proof

technique · direct

1.1F3F4given

The two supports. Since $f \neq 0$ there is a class with $f (x) \neq 0$ , so $S \neq \emptyset$ and $∥ f ∥_{2}^{2} = \sum_{x} ∣ f (x) ∣^{2} > 0$ by [F4]. Applying [F3] with $g = h = f$ and restricting the sum over all classes to the support $T$ , $\sum_{k \in T} ∣ (F_{N} f) (k) ∣^{2} = ∥ F_{N} f ∥_{2}^{2} = ∥ f ∥_{2}^{2} > 0,$ so $T \neq \emptyset$ as well.

1.2F1F2given

Pointwise bound on the support of the transform. For $k \in T$ , the triangle inequality and the normalisation $N^{- 1 / 2}$ of the transform give $∣ (F_{N} f) (k) ∣ \leq N^{- 1 / 2} \sum_{x \in S} ∣ f (x) ∣,$ the sum running only over $S$ because the remaining summands vanish. Cauchy–Schwarz [F2] applied on $Z / N Z$ to $u (x) = ∣ f (x) ∣$ and $v (x) = 1_{S} (x)$ gives $\sum_{x \in S} ∣ f (x) ∣ = ⟨ u, v ⟩ \leq ∥ u ∥_{2} ∥ v ∥_{2} = ∣ S ∣^{1 / 2} ∥ f ∥_{2}$ . Hence $∣ (F_{N} f) (k) ∣ \leq N^{- 1 / 2} ∣ S ∣^{1 / 2} ∥ f ∥_{2}$ for every $k \in T$ .

2.1F1F3step 1.1step 1.2

Summing over the support. Squaring the bound of step 1.2 and summing over the $∣ T ∣$ classes of $T$ gives $∥ F_{N} f ∥_{2}^{2} = \sum_{k \in T} ∣ (F_{N} f) (k) ∣^{2} \leq ∣ T ∣ N^{- 1} ∣ S ∣ ∥ f ∥_{2}^{2} .$ By Parseval [F3] the left side is $∥ f ∥_{2}^{2}$ , and step 1.1 gives $∥ f ∥_{2}^{2} > 0$ , so dividing yields $∣ S ∣ ∣ T ∣ \geq N$ .

3.1step 2.1given∎

The case $N = 1$ and conclusion. If $N = 1$ the group $Z / 1 Z$ has the single class $[0]$ , $N^{- 1 / 2} = 1$ and $e^{- 2 π i \cdot 0 \cdot 0} = 1$ , so $(F_{N} f) ([0]) = f ([0])$ ; hence $T = S = {[0]}$ for nonzero $f$ and $∣ S ∣ ∣ T ∣ = 1 = N$ , an equality. Together with step 2.1 this proves the claim for every $N \geq 1$ .

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Uncertainty principles measure different notions of localisation

Remarks

Three inequivalent notions of localisation are in play on this page and their hypotheses and conclusions are not interchangeable: the variance pair $(V_{x}, V_{ξ})$ of Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments with the product bound The n-dimensional Heisenberg uncertainty inequality; support measure with the product bound The support-measure uncertainty inequality $∣ E ∣ ∣ F ∣ \geq 1$ and the compact-support dichotomy; and Gaussian decay with the Hardy threshold Hardy's Gaussian uncertainty principle in $R^{n}$ , whose critical rigidity is isolated in Entire rigidity under Gaussian growth and real-axis decay. Gaussian decay implies finite second moments in both domains, but need not give compact support. Finite variance is not compact support — the Gaussian of Finite variance is not compact support ↗ has finite variances in both domains and full support — while Gaussian decay and its critical rigidity remain a stronger, distinct formulation. The finite product bound Finite support-product uncertainty for the unitary DFT is not the Heisenberg product: its right side is $N$ , not $(4 π)^{- 1}$ , and its equality set is different. No implication among these statements is asserted beyond the ones proved on this page.

5 · Examples, counterexamples and false statements

None yet.

Uncertainty Principles for Fourier Analysis

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Spatial and frequency centres and variances of an L2L2 function with finite second moments

Definition

Well-definedness

The coordinate inequality ∥xjf∥2∥Djf∥2≥12∥f∥22∥xj​f∥2​∥Dj​f∥2​≥21​∥f∥22​

Statement

Facts & Assumptions

Proof

Compact support gives an entire Fourier-Laplace transform by slices

Statement

Facts & Assumptions

Proof

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

Statement

Facts & Assumptions

Proof

Entire rigidity under Gaussian growth and real-axis decay

Statement

Facts & Assumptions

Proof

Subcritical Gaussians show the Hardy threshold ab=1ab=1 is sharp

Statement

Facts & Assumptions

Proof

Separately holomorphic functions vanishing on a real box are zero

Statement

Facts & Assumptions

Proof

The support-measure uncertainty inequality ∣E∣∣F∣≥1∣E∣∣F∣≥1

Statement

Facts & Assumptions

Proof

The n-dimensional Heisenberg uncertainty inequality

Statement

Facts & Assumptions

Proof

Centring by translation and modulation preserves the variance product

Statement

Facts & Assumptions

Proof

Hardy's Gaussian uncertainty principle in RnRn

Statement

Facts & Assumptions

Proof

A nonzero L1 function and its transform cannot both have compact support

Statement

Facts & Assumptions

Proof

The sharp Heisenberg theorem is owned by functional analysis

Remarks

Proof cost and complex-analysis interface for Hardy uncertainty

Remarks

Finite support-product uncertainty for the unitary DFT

Statement

Facts & Assumptions

Proof

Uncertainty principles measure different notions of localisation

Remarks

5 · Examples, counterexamples and false statements

Sources

Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments

The coordinate inequality $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$

Subcritical Gaussians show the Hardy threshold $a b = 1$ is sharp

The support-measure uncertainty inequality $∣ E ∣ ∣ F ∣ \geq 1$

Hardy's Gaussian uncertainty principle in $R^{n}$