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Finite variance is not compact support

Statement refuted

Assume countable choice. Every nonzero $f \in L^{2} (R^{n}; C)$ with finite second moments in both domains vanishes almost everywhere outside a compact set, and so does $\hat{f}$ ; equivalently, finite variance forces compact support.

Facts & Assumptions

Given: Countable choice, an integer $n \geq 1$ , a real $a > 0$ , and the Gaussian $f (x) = e^{- π a ∣ x ∣^{2}}$ .

[F1]

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

[F2]

For every $t > 0$ the Gaussian $e^{- π t ∣ x ∣^{2}}$ is absolutely integrable with $\int_{R^{n}} e^{- π t ∣ x ∣^{2}} d x = t^{- n / 2}$ and $L^{1}$ Fourier transform $t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ (Euclidean Gaussian transform with the 2π normalization); every polynomial times a positive real Gaussian is absolutely integrable. For $L^{1} \cap L^{2}$ functions this integral transform represents the Plancherel transform (Agreement of the integral and L2 transforms); the one-dimensional Gaussian integral is $\int_{R} e^{- s^{2}} d s = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F3]

Complex $L^{1}$ change of variables: for the reflection $T (x) = - x$ , which is a $C^{1}$ diffeomorphism with $∣ \det D T ∣ = 1$ , one has $\int h (- x) d x = \int h (y) d y$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

[F4]

Means and variances for a nonzero $L^{2}$ function with finite second moments are as in Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments; the support-measure hypothesis that is not satisfied here is the one of The support-measure uncertainty inequality $∣ E ∣ ∣ F ∣ \geq 1$ .

[F5]

Compact subsets of $R^{n}$ are bounded (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), and every box has its product-of-side-lengths Lebesgue measure under Countable Choice (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included). Thus a compact set is contained in a finite-measure cube, whereas $R^{n}$ has infinite measure since it contains cubes $[- R, R]^{n}$ of measure $(2 R)^{n}$ for all $R > 0$ .

Counterexample

technique · direct

1.1F1F2given

The Gaussian and its transform are strictly positive and integrable. $f (x) = e^{- π a ∣ x ∣^{2}} > 0$ for every $x$ , $f$ is continuous and even, and by [F2] with $t = 2 a$ , $∥ f ∥_{2}^{2} = \int_{R^{n}} f^{2} = (2 a)^{- n / 2} < \infty$ , so $f \in L^{1} \cap L^{2}$ and [F2] identifies its integral and Plancherel transforms, with $\hat{f} (ξ) = a^{- n / 2} e^{- π ∣ ξ ∣^{2} / a} > 0$ for every $ξ \in R^{n}$ .

2.1F2F3F4givenstep 1.1

Finite second moments and vanishing means. By [F2] with $t = 2 a$ one has $∥ f ∥_{2}^{2} = (2 a)^{- n / 2} < \infty$ , and since $∣ \hat{f} (ξ) ∣^{2} = a^{- n} e^{- 2 π ∣ ξ ∣^{2} / a}$ , [F2] with $t = 2 / a$ gives $∥ \hat{f} ∥_{2}^{2} = (2 a)^{- n / 2} < \infty$ as well. By [F2], $∣ x ∣^{2} e^{- 2 π a ∣ x ∣^{2}}$ and $∣ ξ ∣^{2} a^{- n} e^{- 2 π ∣ ξ ∣^{2} / a}$ are integrable polynomial multiples of positive Gaussians. Thus both second moments are finite. The functions $x \mapsto x_{j} ∣ f (x) ∣^{2}$ and $ξ \mapsto ξ_{j} ∣ \hat{f} (ξ) ∣^{2}$ are odd in their $j$ -th coordinate while $f$ and $\hat{f}$ are even, so [F3] applied to the reflection $T (x) = - x$ shows that each of these integrals equals its own negative and hence vanishes; consequently both means are $0$ and the variances $V_{x} (f), V_{ξ} (f)$ of [F4] are finite.

3.1F4F5givenstep 1.1step 2.1∎

No finite-measure support in either domain. Let $E \subseteq R^{n}$ be measurable with $f = 0$ almost everywhere on $R^{n} ∖ E$ . Since $f > 0$ everywhere, the set where $f$ differs from $0$ inside $R^{n} ∖ E$ is $R^{n} ∖ E$ itself, so $R^{n} ∖ E$ is null and $E$ has full measure, $∣ E ∣ = \infty$ ; the same argument with the strictly positive transform $\hat{f}$ shows that no measurable $F$ of finite measure can carry $\hat{f} = 0$ almost everywhere off it. By [F5], compact sets have finite measure, so neither function can have compact support. Consequently the counterexample has finite second moments and finite variances in both domains but neither it nor its transform is supported ( $a . e .$ ) on a set of finite measure, so finite variance does not force compact support, and the support-measure hypothesis of [F4] is not implied by the variance hypotheses.

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Sources

Calder Sheagren, Uncertainty Principles with Fourier Analysis (University of Chicago REU 2017, author PDF) (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (arXiv:0903.3845) (standard reference, not scraped)