Uncertainty Principles for Fourier Analysis — Examples

1 · Prerequisites

2 · Summary

These examples execute the three localisation principles of the companion page at explicit parameters and separate their hypothesis classes. The Gaussian $e^{- π a ∣ x ∣^{2}}$ is the running example: it attains equality in the summed Heisenberg inequality, with spatial variance $n / (4 π a)$ and frequency variance $n a / (4 π)$ whose product is exactly $(n / 4 π)^{2}$ , so every constant in the library's $e^{- 2 π i x \cdot ξ}$ convention is checked rather than quoted. The counterexample on the same function shows that finite variance is not compact support: the Gaussian and its transform are strictly positive everywhere, so neither vanishes off a set of finite measure, and the support-measure hypothesis of the companion theorem cannot be deduced from the variance hypotheses.

For Hardy's theorem the three parameter regimes are tabulated. At the critical product $a b = 1$ the Gaussian $e^{- π a ∣ x ∣^{2}}$ satisfies both Gaussian bounds and the theorem classifies it as a scalar multiple of itself; in the subcritical regime $a b < 1$ the interval $(a, 1 / b)$ supplies a nonzero Gaussian satisfying both bounds, so no vanishing conclusion holds; in the supercritical regime $a b > 1$ the two bounds force $c \geq a$ and $c \leq 1 / b$ on any Gaussian witness, which is impossible, matching the theorem's conclusion that $f = 0$ almost everywhere.

The finite discrete Fourier transform is treated separately, because its product bound is not the Heisenberg product. The delta and the constant function are evaluated explicitly as two examples of extremisers of the support-product bound at every length $N$ , with the two identities coinciding at $N = 1$ , and the last witness shows that both supports cannot be singletons when $N > 1$ : the support-product bound would give $1 \geq N$ . The boundary case $N = 1$ , where the delta is the constant function and the configuration does occur, is included to show that the hypothesis $N > 1$ is essential.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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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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The Gaussian attains equality in the Heisenberg inequality

Example

Assume countable choice. Let $n \geq 1$ and $a > 0$ and $f (x) : = e^{- π a ∣ x ∣^{2}}$ . Then $∥ f ∥_{2}^{2} = (2 a)^{- n / 2}$ , the means of $∣ f ∣^{2}$ and of $∣ \hat{f} ∣^{2}$ are $0$ , and $V_{x} (f) = \frac{n}{4 π a}, V_{ξ} (f) = \frac{n a}{4 π},$ so $V_{x} (f) V_{ξ} (f) = \frac{n^{2}}{16 π^{2}}$ and $\sqrt{V_{x} (f)} \sqrt{V_{ξ} (f)} = \frac{n}{4 π}$ ; hence $f$ attains equality in The n-dimensional Heisenberg uncertainty inequality, and FA-23's equality family $c \exp (- λ ∣ x - x_{0} ∣^{2} / 2) \exp (2 π i b_{0} \cdot x)$ is realised with $c = 1$ , $λ = 2 π a$ , and $x_{0} = b_{0} = 0$ .

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), an integer $n \geq 1$ , $a > 0$ , and $f (x) = e^{- π a ∣ x ∣^{2}}$ , whose membership in $L^{1} \cap L^{2}$ , finite moments, and zero means are verified below; the variances use Spatial and frequency centres and variances of an $L^{2}$ function with finite second moments.

[F1]

Countable choice is assumed; it is the hypothesis carried by the Gaussian transform identity, the parameter differentiation, the reflection substitution and Plancherel 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 $L^{1}$ transform $t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ ; every polynomial times a positive real Gaussian is absolutely integrable (Euclidean Gaussian transform with the 2π normalization); 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]

Differentiation under the integral sign: if $f (x, t)$ is integrable in $x$ for every $t$ in an open interval, differentiable in $t$ for almost every $x$ , and the $t$ -derivative is measurable in $x$ with $∣ \partial_{t} f (x, t) ∣ \leq g (x)$ for an integrable $g$ and all $t$ , then $t \mapsto \int f (x, t) d x$ is differentiable with derivative $\int \partial_{t} f d x$ (Differentiation under the integral sign).

[F4]

Complex $L^{1}$ change of variables for the reflection $T (x) = - x$ ( $∣ \det D T ∣ = 1$ ): $\int h (- x) d x = \int h (x) d x$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions). Complex $L^{2}$ carries $∥ \cdot ∥_{2}$ and Cauchy–Schwarz (Complex completeness, density, and inner product: the consumer interface), and for $g \in L^{1} \cap L^{2}$ the integral transform represents its Plancherel transform (Agreement of the integral and L2 transforms), so $∥ \hat{g} ∥_{2} = ∥ g ∥_{2}$ (Plancherel theorem).

[F5]

The power function $t \mapsto t^{α}$ on $(0, \infty)$ is differentiable with derivative $α t^{α - 1}$ (Continuity and derivatives of positive-base real powers, Real powers for positive bases, with the zero-base positive-exponent convention); in particular $\frac{d}{d c} (2 c)^{- n / 2} = - n (2 c)^{- n / 2 - 1}$ .

[F6]

The Fourier characterization identifies the $L^{2}$ classes with $⟨ ξ ⟩ \hat{f} \in L^{2}$ as $H^{1} = W^{1, 2}$ under the regular-distribution embedding (Integer-order W^{k,2} and H^k agree with equivalent norms, Statement 1 and Proof step 1.3 at $k = 1$ ).

Verification

technique · direct

1.1F1F2F4given

Norm, transform, evenness and vanishing means. By [F2] with $t = 2 a$ , $∥ f ∥_{2}^{2} = \int e^{- 2 π a ∣ x ∣^{2}} d x = (2 a)^{- n / 2} < \infty$ , and $\hat{f} (ξ) = a^{- n / 2} e^{- π ∣ ξ ∣^{2} / a}$ , so $∥ \hat{f} ∥_{2}^{2} = a^{- n} \int e^{- 2 π ∣ ξ ∣^{2} / a} d ξ = (2 a)^{- n / 2}$ by [F2] with $t = 2 / a$ and [F4]. Both $f$ and $\hat{f}$ are even and strictly positive. 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, so by [F4] (reflection) each integral equals its own negative; since they are integrable by [F2] and the remark above, both vanish. Hence both means are $0$ , and the variances are the uncentred second moments divided by $(2 a)^{- n / 2}$ .

2.1F2F3F5step 1.1

Second moments and variances. Differentiating the identity $\int_{R^{n}} e^{- 2 π c ∣ x ∣^{2}} d x = (2 c)^{- n / 2}$ in the parameter $c > 0$ ([F2] with $t = 2 c$ , [F5]) is legitimate by [F3]: on an open neighbourhood with closure contained in $(0, \infty)$ the derivative $- 2 π ∣ x ∣^{2} e^{- 2 π c ∣ x ∣^{2}}$ is dominated by $2 π ∣ x ∣^{2} e^{- 2 π c_{0} ∣ x ∣^{2}}$ for a positive lower bound $c_{0}$ of that interval, which is integrable by [F2]. Hence for every $c > 0$ $\int_{R^{n}} ∣ x ∣^{2} e^{- 2 π c ∣ x ∣^{2}} d x = \frac{1}{2 π} n (2 c)^{- n / 2 - 1} = \frac{n}{4 π c} (2 c)^{- n / 2} .$ With $c = a$ and step 1.1 this gives $\int ∣ x ∣^{2} ∣ f ∣^{2} = \frac{n}{4 π a} ∥ f ∥_{2}^{2}$ , hence $V_{x} (f) = \frac{\int ∣ x ∣^{2} ∣ f ∣^{2}}{∥ f ∥_{2}^{2}} = \frac{n}{4 π a} .$ Next, $∣ \hat{f} (ξ) ∣^{2} = a^{- n} e^{- 2 π ∣ ξ ∣^{2} / a} = a^{- n} e^{- 2 π c ∣ ξ ∣^{2}}$ with $c = 1 / a$ , so the same identity gives $\int ∣ ξ ∣^{2} ∣ \hat{f} ∣^{2} d ξ = a^{- n} \frac{n a}{4 π} (2 / a)^{- n / 2} = \frac{n a}{4 π} (2 a)^{- n / 2} = \frac{n a}{4 π} ∥ \hat{f} ∥_{2}^{2}$ by step 1.1, and therefore $V_{ξ} (f) = \frac{\int ∣ ξ ∣^{2} ∣ \hat{f} ∣^{2}}{∥ \hat{f} ∥_{2}^{2}} = \frac{n a}{4 π} .$

3.1F4F6step 1.1step 2.1∎

Equality. By step 2.1, $\int (1 + ∣ ξ ∣^{2}) ∣ \hat{f} ∣^{2} < \infty$ , so [F4, F6] give $f \in H^{1}$ ; the finite spatial moment is also verified there. Hence the domain of the cited Heisenberg corollary is satisfied. Steps 1.1 and 2.1 give $∥ ∣ x ∣ f ∥_{2}^{2} = \frac{n}{4 π a} ∥ f ∥_{2}^{2}$ and $∥ ∣ ξ ∣ \hat{f} ∥_{2}^{2} = \frac{n a}{4 π} ∥ f ∥_{2}^{2}$ , hence $∥ ∣ x ∣ f ∥_{2} ∥ ∣ ξ ∣ \hat{f} ∥_{2} = ∥ f ∥_{2}^{2} \sqrt{\frac{n}{4 π a}} \sqrt{\frac{n a}{4 π}} = \frac{n}{4 π} ∥ f ∥_{2}^{2},$ so the inequality of The n-dimensional Heisenberg uncertainty inequality is an equality. The product of variances is $\frac{n}{4 π a} \cdot \frac{n a}{4 π} = \frac{n^{2}}{16 π^{2}}$ and its square root is $\frac{n}{4 π}$ . Finally the entire family $c \exp (- λ ∣ x - a_{0} ∣^{2} / 2) \exp (2 π i b_{0} \cdot x)$ of the published Heisenberg theorem contains $f$ at $c = 1$ , $λ = 2 π a$ , $a_{0} = b_{0} = 0$ .

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Critical, subcritical and supercritical Gaussian regimes for Hardy's theorem

Example

Assume countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as Hardy's Gaussian uncertainty principle in $R^{n}$ and the Gaussian-transform interface do. Let $a, b > 0$ and put $C_{0} : = \max {1, a^{- n / 2}}$ . At the critical product $a b = 1$ the Gaussian $f (x) = e^{- π a ∣ x ∣^{2}}$ satisfies the two Hardy bounds $∣ f (x) ∣ \leq e^{- π a ∣ x ∣^{2}}$ and $∣ \hat{f} (ξ) ∣ \leq a^{- n / 2} e^{- π ∣ ξ ∣^{2} / a}$ of Hardy's Gaussian uncertainty principle in $R^{n}$ with the common constant $C_{0}$ , and the theorem returns a scalar multiple of the same Gaussian. In the subcritical regime $a b < 1$ , every $c$ with $a < c < 1 / b$ gives the Gaussian $e^{- π c ∣ x ∣^{2}}$ satisfying the two bounds with the common constant $C_{c} : = \max {1, c^{- n / 2}}$ , so no vanishing conclusion holds (Subcritical Gaussians show the Hardy threshold $a b = 1$ is sharp). In the supercritical regime $a b > 1$ the theorem forces $f = 0$ , and no Gaussian satisfies both bounds for any positive constants $C, C^{'}$ : $∣ e^{- π c ∣ x ∣^{2}} ∣ \leq C e^{- π a ∣ x ∣^{2}}$ forces $c \geq a$ , while $∣ \hat{f_{c}} ∣ \leq C^{'} e^{- π b ∣ ξ ∣^{2}}$ forces $c \leq 1 / b$ , and $a \leq 1 / b$ is exactly $a b \leq 1$ .

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), an integer $n \geq 1$ , reals $a, b > 0$ , the common critical bound $C_{0} = \max {1, a^{- n / 2}}$ , and for $c > 0$ the Gaussian $f_{c} (x) : = e^{- π c ∣ x ∣^{2}}$ (Real powers for positive bases, with the zero-base positive-exponent convention for real powers).

[F1]

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

[F2]

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

[F3]

Subcritical sharpness: if $a b < 1$ , then $(a, 1 / b)$ is a nonempty open interval and every $c$ with $a < c < 1 / b$ satisfies $∣ f_{c} (x) ∣ \leq e^{- π a ∣ x ∣^{2}}$ and $∣ \hat{f_{c}} (ξ) ∣ \leq c^{- n / 2} e^{- π b ∣ ξ ∣^{2}}$ (Subcritical Gaussians show the Hardy threshold $a b = 1$ is sharp).

[F4]

Hardy's theorem: if $a, b, C > 0$ and a measurable $g$ satisfies $∣ g (x) ∣ \leq C e^{- π a ∣ x ∣^{2}}$ almost everywhere and $∣ \hat{g} (ξ) ∣ \leq C e^{- π b ∣ ξ ∣^{2}}$ everywhere, then $g = 0$ almost everywhere when $a b > 1$ , and $g (x) = g^{\land} (0) a^{n / 2} e^{- π a ∣ x ∣^{2}}$ almost everywhere when $a b = 1$ (Hardy's Gaussian uncertainty principle in $R^{n}$ ).

[F5]

Every unit cube has Lebesgue measure $1$ 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). Hence a full-measure subset of $R^{n}$ is unbounded: if it were bounded, a unit cube outside a ball containing it would be contained in its null complement, a contradiction.

Verification

technique · direct

1.1F2F4given

Critical regime. For $g = f_{a}$ we have $∣ g (x) ∣ = e^{- π a ∣ x ∣^{2}} \leq C_{0} e^{- π a ∣ x ∣^{2}}$ and, by [F2] with $t = a$ , $∣ \hat{g} (ξ) ∣ = a^{- n / 2} e^{- π ∣ ξ ∣^{2} / a} \leq C_{0} e^{- π b ∣ ξ ∣^{2}}$ since $b = 1 / a$ . Thus $g$ meets both Hardy bounds with one common constant $C_{0}$ . Since $a b = 1$ , [F4] returns $g (x) = \hat{g} (0) a^{n / 2} e^{- π a ∣ x ∣^{2}}$ almost everywhere, with $\hat{g} (0) = \int g = a^{- n / 2}$ ; that is the same Gaussian, so the classification is attained, not merely bounded.

1.2F3given

Subcritical regime. Suppose $a b < 1$ , equivalently $a < 1 / b$ . By [F3] the interval $(a, 1 / b)$ is nonempty and every $c \in (a, 1 / b)$ produces a nonzero Gaussian $f_{c}$ satisfying $∣ f_{c} ∣ \leq e^{- π a ∣ x ∣^{2}}$ and $∣ \hat{f_{c}} ∣ \leq c^{- n / 2} e^{- π b ∣ ξ ∣^{2}}$ . With $C_{c} = \max {1, c^{- n / 2}}$ both bounds hold with a single constant, so the Hardy hypotheses admit a nonzero function and no vanishing conclusion can be drawn.

1.3F1F2F4F5given

Supercritical regime. Suppose $a b > 1$ , equivalently $a > 1 / b$ . If $c > 0$ and constants $C, C^{'} > 0$ satisfied $e^{- π c ∣ x ∣^{2}} \leq C e^{- π a ∣ x ∣^{2}}$ almost everywhere, then $e^{π (a - c) ∣ x ∣^{2}} \leq C$ would hold on a set of full measure, and a set of full measure is unbounded by [F5]; were $c < a$ , the left side would tend to $+ \infty$ along that unbounded set, which is impossible for a finite constant, so $c \geq a$ . Similarly, by [F2], $∣ \hat{f_{c}} (ξ) ∣ = c^{- n / 2} e^{- π ∣ ξ ∣^{2} / c} \leq C^{'} e^{- π b ∣ ξ ∣^{2}}$ for every $ξ$ would give $e^{π (b - 1 / c) ∣ ξ ∣^{2}} \leq C^{'} c^{n / 2}$ for every $ξ$ ; if $b > 1 / c$ , the left side diverges as $∣ ξ ∣ \to \infty$ , contradicting the finite bound; thus $1 / c \geq b$ , that is $c \leq 1 / b$ . Both requirements together would give $a \leq c \leq 1 / b$ , hence $a b \leq 1$ , contrary to the hypothesis; so no Gaussian meets the two bounds, and by [F4] the only function satisfying them is $0$ almost everywhere.

2.1step 1.1step 1.2step 1.3∎

Tabulation. Steps 1.1–1.3 separate the three parameter regions: at $a b = 1$ the Gaussian $e^{- π a ∣ x ∣^{2}}$ is a nonzero solution classified as itself, for $a b < 1$ nonzero Gaussian solutions exist with rate $c \in (a, 1 / b)$ , and for $a b > 1$ no Gaussian solution exists and every solution vanishes almost everywhere.

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Delta and constant functions are finite DFT extremisers

Example

Let $N \geq 1$ , let $δ_{0} \in C^{Z / N Z}$ be the delta at the class of $0$ and let $1$ be the constant function $1$ . Then $F_{N} δ_{0} = N^{- 1 / 2} 1, F_{N} 1 = N^{1 / 2} δ_{0} .$ Hence $∣ supp δ_{0} ∣ = 1 = ∣ supp F_{N} 1 ∣$ and $∣ supp 1 ∣ = N = ∣ supp F_{N} δ_{0} ∣$ , so both functions have support product $N$ and attain equality in Finite support-product uncertainty for the unitary DFT. At $N = 1$ one has $δ_{0} = 1$ and the two identities coincide.

Facts & Assumptions

Given: An integer $N \geq 1$ , the functions $δ_{0}, 1 \in C^{Z / N Z}$ with $δ_{0} ([0]) = 1$ , $δ_{0} (x) = 0$ for $x \neq [0]$ , and $1 (x) = 1$ for all $x$ , the unitary transform $(F_{N} f) (k) = N^{- 1 / 2} \sum_{x = 0}^{N - 1} f ([x]_{N}) e^{- 2 π i k x / N}$ of The unitary discrete Fourier transform on $Z / N Z$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , The counting inner product on $C^{Z / N Z}$ ).

[F1]

Character orthogonality: for $N \geq 1$ and $k, ℓ \in Z$ , $\sum_{x = 0}^{N - 1} e^{2 π i (k - ℓ) x / N} = N$ when $k \equiv ℓ (m o d N)$ and $0$ otherwise; at $N = 1$ the congruence always holds and the sum is $1$ (Orthogonality of the characters $x \mapsto e^{2 π i k x / N}$ on $Z / N Z$ ).

[F2]

Rational powers: $N^{\pm 1 / 2}$ is the positive real number with $N^{- 1 / 2} N^{1 / 2} = 1$ and $(N^{- 1 / 2})^{2} = N^{- 1}$ ; the usual power laws hold (Rational powers $a^{r}$ of a positive base, Laws of rational exponents), and complex arithmetic is that of the field $C$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F3]

Finite support-product uncertainty: $∣ supp f ∣ \cdot ∣ supp F_{N} f ∣ \geq N$ for every nonzero $f$ , with $supp f = {x : f (x) \neq 0}$ (Finite support-product uncertainty for the unitary DFT).

Verification

technique · direct

1.1F2given

The transform of the delta. In the defining sum $(F_{N} δ_{0}) (k) = N^{- 1 / 2} \sum_{x = 0}^{N - 1} δ_{0} ([x]_{N}) e^{- 2 π i k x / N}$ every summand with $x \equiv̸ 0 (m o d N)$ vanishes by definition of $δ_{0}$ , and the summand at $x = 0$ equals $1$ . Hence $(F_{N} δ_{0}) (k) = N^{- 1 / 2}$ for every $k$ , that is, $F_{N} δ_{0} = N^{- 1 / 2} 1$ .

1.2F1F2given

The transform of the constant function. For a frequency representative $k \in Z$ , $(F_{N} 1) (k) = N^{- 1 / 2} \sum_{x = 0}^{N - 1} e^{- 2 π i k x / N}$ . Apply [F1] with its first parameter $0$ and second parameter $k$ : the sum is $N$ when $k \equiv 0 (m o d N)$ and $0$ otherwise. Hence $(F_{N} 1) (k) = N^{- 1 / 2} \cdot N = N^{1 / 2}$ at $k = [0]$ and $0$ elsewhere, that is, $F_{N} 1 = N^{1 / 2} δ_{0}$ .

2.1F2F3step 1.1step 1.2∎

Supports, equality, and the case $N = 1$ . By step 1.1 the support of $F_{N} δ_{0}$ is the support of the constant function $1$ , namely all $N$ classes, while $supp δ_{0} = {[0]}$ has one element; by step 1.2 the support of $F_{N} 1$ is ${[0]}$ , while $supp 1$ has $N$ elements. Multiplying gives $∣ supp δ_{0} ∣ \cdot ∣ supp F_{N} δ_{0} ∣ = N$ and $∣ supp 1 ∣ \cdot ∣ supp F_{N} 1 ∣ = N$ , so both attain the equality case of the bound [F3]. At $N = 1$ the group has the single class $[0]$ , so $δ_{0} = 1$ and $N^{- 1 / 2} = N^{1 / 2} = 1$ , and the two identities of steps 1.1 and 1.2 coincide.

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Both finite supports cannot be singletons when $N > 1$

Statement refuted

For every $N \geq 1$ there exists a nonzero $f \in C^{Z / N Z}$ with $∣ supp f ∣ = ∣ supp F_{N} f ∣ = 1$ .

Facts & Assumptions

Given: An integer $N \geq 1$ and the unitary discrete Fourier transform $F_{N}$ of The unitary discrete Fourier transform on $Z / N Z$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[F1]

For every nonzero $f \in C^{Z / N Z}$ the support product satisfies $∣ supp f ∣ \cdot ∣ supp F_{N} f ∣ \geq N$ (Finite support-product uncertainty for the unitary DFT).

[F2]

At $N = 1$ there is exactly one class, and the delta $δ_{0}$ at it is the constant function $1$ ; the example Delta and constant functions are finite DFT extremisers computes $F_{1} δ_{0} = δ_{0}$ , so $∣ supp δ_{0} ∣ = ∣ supp F_{1} δ_{0} ∣ = 1$ .

Counterexample

technique · direct

1.1F1given

Impossibility for $N > 1$ . Suppose $N > 1$ and a nonzero $f$ satisfied $∣ supp f ∣ = ∣ supp F_{N} f ∣ = 1$ . Then the left-hand side of the bound [F1] equals $1$ , so $1 \geq N$ , contradicting $N > 1$ . Hence no such $f$ exists for $N > 1$ .

1.2F2given

The case $N = 1$ . At $N = 1$ the group $Z / 1 Z$ has the single class $[0]$ , and by [F2] the delta $δ_{0}$ is the constant function $1$ with $F_{1} δ_{0} = δ_{0}$ ; both its support and the support of its transform equal the one-element set ${[0]}$ .

2.1step 1.1step 1.2∎

Conclusion. The universal claim fails already at $N = 2$ , where [F1] forces a support product of at least $2$ ; step 1.2 shows that the hypothesis $N > 1$ is essential, since the excluded configuration does occur at $N = 1$ and the bound of [F1] is exactly attained there.

Uncertainty Principles for Fourier Analysis — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Finite variance is not compact support

Statement refuted

Facts & Assumptions

Counterexample

The Gaussian attains equality in the Heisenberg inequality

Example

Facts & Assumptions

Verification

Critical, subcritical and supercritical Gaussian regimes for Hardy's theorem

Example

Facts & Assumptions

Verification

Delta and constant functions are finite DFT extremisers

Example

Facts & Assumptions

Verification

Both finite supports cannot be singletons when N>1N>1

Statement refuted

Facts & Assumptions

Counterexample

Sources

Both finite supports cannot be singletons when $N > 1$