Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)
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The support of a convolution lies in the closure of the support sumset

Statement

Let f,g∈L1(Rn) be represented by Borel functions still denoted f,g. For any complex-valued function u on Rn, use the support convention

supp⁡(u):={x∈Rn:u(x)≠0}‾,

which extends the real-valued definition of The support of a function on Rn and its compactly supported Riemann integral. Define h(x):={∫Rnf(x−y)g(y) dy,if the integral exists,0,otherwise. Then

supp⁡(h)⊆supp⁡(f)+supp⁡(g)‾.

Facts & Assumptions

Given: Borel representatives f,g∈L1(Rn) and the function h defined above.

[L1]

The displayed integral defines convolution wherever it converges absolutely (Convolution of two functions on Rn).

[L2]

Support is defined by the closure of the nonzero set (The support of a function on Rn and its compactly supported Riemann integral).

Proof

technique · direct
1.1L1L2givenchoose

Let [L1, L2, given, choose] x∉supp⁡(f)+supp⁡(g)‾. Choose an open neighborhood U of x disjoint from that closure. For z∈U and y∈supp⁡(g), one has z−y∉supp⁡(f), so f(z−y)=0; and if y∉supp⁡(g) then g(y)=0. Hence f(z−y)g(y)=0 for every y and every z∈U.

2.1L1L2step 1.1

Therefore h(z)=0 for every z∈U: by step 1.1 the integrand [L1, L2, step 1.1] vanishes for every y, so the convolution integral exists and equals 0 at each such z. So x lies outside the support of h in the sense of [L2].

3.1step 2.1∎

Since every point outside [step 2.1] supp⁡(f)+supp⁡(g)‾ lies outside supp⁡(h), the support inclusion follows.

Depends on

Used by

Dependency tree · two levels

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Sources