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 $f, g \in L^{1} (R^{n})$ be represented by Borel functions still denoted $f, g$ . For any complex-valued function $u$ on $R^{n}$ , use the support convention

$supp (u) : = \overline{{x \in R^{n} : u (x) \neq 0}},$

which extends the real-valued definition of The support of a function on $R^{n}$ and its compactly supported Riemann integral. Define $h (x) : = {\begin{cases} \int_{R^{n}} f (x - y) g (y) d y, & if the integral exists, \\ 0, & otherwise . \end{cases}$ Then

$supp (h) \subseteq \overline{supp (f) + supp (g)} .$

Facts & Assumptions

Given: Borel representatives $f, g \in L^{1} (R^{n})$ and the function $h$ defined above.

[L1]

The displayed integral defines convolution wherever it converges absolutely (Convolution of two functions on $R^{n}$ ).

[L2]

Support is defined by the closure of the nonzero set (The support of a function on $R^{n}$ and its compactly supported Riemann integral).

Proof

technique · direct

1.1L1L2givenchoose

Let [L1, L2, given, choose] $x \notin \overline{supp (f) + supp (g)}$ . Choose an open neighborhood $U$ of $x$ disjoint from that closure. For $z \in U$ and $y \in supp (g)$ , one has $z - y \notin supp (f)$ , so $f (z - y) = 0$ ; and if $y \notin supp (g)$ then $g (y) = 0$ . Hence $f (z - y) g (y) = 0$ for every $y$ and every $z \in U$ .

2.1L1L2step 1.1

Therefore $h (z) = 0$ for every $z \in U$ : 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] $\overline{supp (f) + supp (g)}$ lies outside $supp (h)$ , the support inclusion follows.

Depends on

Dependency tree · 2 levels

Sources

Walter Rudin, Real and Complex Analysis, 3rd ed.