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The support-measure uncertainty inequality
Statement
Assume countable choice. Let be nonzero and let be Lebesgue measurable sets of finite measure such that almost everywhere on and almost everywhere on . Here denotes the continuous transform (The integral transform is representative independent), which is defined because forces . Then No regularity of or beyond measurability and finite measure is assumed, and no complex analysis is used.
Facts & Assumptions
Given: Countable choice (The Axiom of Countable Choice ()), a nonzero , and Lebesgue measurable sets of finite measure with almost everywhere off and almost everywhere off .
Countable choice is assumed; it is the hypothesis carried by the transform interface, the agreement theorem, and Plancherel below (The Axiom of Countable Choice ()).
For the transform is defined at every frequency, satisfies , 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).
Hölder's inequality with conjugate exponents : for measurable real and one has (Holder's inequality for integrals, including the endpoint cases); and are the quotient spaces of The space 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).
If , its bounded continuous transform represents the Plancherel transform almost everywhere (Agreement of the integral and L2 transforms), and Plancherel gives (Plancherel theorem).
Proof
The function is integrable and its transform is bounded. Since almost everywhere on , one has ; applying [F3] with and , whose norm is , gives . Hence , its transform is defined and continuous [F2], and for every .
Plancherel size from the two supports. As , the continuous transform represents almost everywhere [F4]; since almost everywhere off , also almost everywhere off . Therefore is represented by the function that vanishes off and equals on , and where the last inequality inserts the uniform bound of step 1.1 and is used.
Conclusion. Plancherel's isometry [F4] gives , so step 2.1 yields . Since is nonzero, , and dividing gives .
Depends on
- Complex Lp classes and Euclidean test-function conventions
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- The space $L^p(\mu)$ as the quotient by null functions
- The integral transform is representative independent
- Holder's inequality for integrals, including the endpoint cases
- Agreement of the integral and L2 transforms
- Plancherel theorem
- Riemann–Lebesgue lemma
- Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree
Used by
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Sources
- David L. Donoho and Philip B. Stark, Uncertainty Principles and Signal Recovery, SIAM J. Appl. Math. 49(3) (1989) 906–931 (standard reference, not scraped)