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

Statement

Assume countable choice. Let f∈L2(Rn;C) be nonzero and let E,F⊆Rn be Lebesgue measurable sets of finite measure such that f=0 almost everywhere on Rn∖E and f^=0 almost everywhere on Rn∖F. Here f^ denotes the continuous L1 transform (The integral transform is representative independent), which is defined because ∣E∣<∞ forces f∈L1. Then ∣E∣ ∣F∣≥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 (ACω)), a nonzero f∈L2(Rn;C), and Lebesgue measurable sets E,F of finite measure with f=0 almost everywhere off E and f^=0 almost everywhere off F.

[F1]

Countable choice is assumed; it is the hypothesis carried by the L1 transform interface, the agreement theorem, and Plancherel below (The Axiom of Countable Choice (ACω)).

[F2]

For g∈L1(Rn;C) the L1 transform is defined at every frequency, satisfies ∣g^(ξ)∣≤∥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∈L2 and v∈L2 one has ∫∣uv∣≤∥u∥2∥v∥2 (Holder's inequality for integrals, including the endpoint cases); L1 and L2 are the quotient spaces of The space Lp(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∈L1∩L2, its bounded continuous L1 transform represents the Plancherel transform F2g almost everywhere (Agreement of the integral and L2 transforms), and Plancherel gives ∥F2g∥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 Rn∖E, one has ∫Rn∣f∣=∫E∣f∣; applying [F3] with u=∣f∣ and v=1E, whose L2 norm is ∣E∣1/2<∞, gives ∫E∣f∣≤∥f∥2∣E∣1/2<∞. Hence f∈L1∩L2, its L1 transform f^ is defined and continuous [F2], and ∣f^(ξ)∣≤∥f∥1≤∣E∣1/2∥f∥2 for every ξ∈Rn.

2.1F4givenstep 1.1

Plancherel size from the two supports. As f∈L1∩L2, the continuous transform f^ represents F2f almost everywhere [F4]; since f^=0 almost everywhere off F, also F2f=0 almost everywhere off F. Therefore F2f is represented by the function that vanishes off F and equals f^ on F, and ∫Rn∣F2f∣2=∫F∣f^∣2≤∣F∣ sup⁡ξ∈Rn∣f^(ξ)∣2≤∣E∣ ∣F∣ ∥f∥22, where the last inequality inserts the uniform bound of step 1.1 and ∣F∣<∞ is used.

3.1F4givenstep 2.1∎

Conclusion. Plancherel's isometry [F4] gives ∥F2f∥22=∥f∥22, so step 2.1 yields ∥f∥22≤∣E∣∣F∣∥f∥22. Since f is nonzero, ∥f∥22>0, and dividing gives ∣E∣∣F∣≥1.

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