Convolution Support
Theorem
Given two sets of real numbers $A, B$, we define $A + B$ as follows.
$$A + B := \left\{ a + b : \forall a \in A, \forall b \in \supp B \right\}$$
For two functions $f, g$, the following holds.
$$\supp f \ast g \subset \supp f + \supp g$$
Here, $\supp$ is the function’s support, and $\ast$ is the convolution.
Proof1
Assume $x \notin \supp f + \supp g$. Then, for any $y$ chosen, $f(y)g(x-y)=0$ occurs.
Case 1 $y \in \supp f$
In this case, $x - y \notin \supp g$ holds. If we assume $x - y \in \supp g$,
$$ \supp f + \supp g \ni (x - y) + y = x \notin \supp f + \supp g $$
this leads to a contradiction. Therefore, $x - y \notin \supp g$ and $g(x-y) = 0$ hold.
Case 2 $y \notin \supp f$
In this case, $f(y) = 0$ holds.
Therefore, if $x \notin \supp f + \supp g$, then $\displaystyle \int f(y)g(x - y) dy = f \ast g(x) = 0$, and so $x \notin \supp f \ast g$. Thus,
$$ \supp f \ast g \subset \supp f + \supp g $$
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