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Let $f,g \in L^{1}(\mathbb R)$ we define $f\ast g(x)= \int_{\mathbb R} f(x-y)g(y) dy $ for all most all $x,$ and denote $\text{supp} (f)$ the support of $f.$

Fact: If $A$ is the closure of $\{x+y: s\in \text{supp}(f), y \in \text {supp}(g)\},$ then $\text{supp}(f\ast g) \subset A.$

[Proof. If $z\notin A,$ then for any $y\in \text{supp}(g)$ we have $z-y \notin \text{supp}(f);$ hence $f(z-y)g(y)=0$ for all $y,$ so $z\notin \text{supp}(f\ast g).$]

My Question is: How to understand the above proof geometrically ? Bit rough: Why does the support of convolution of functions is larger than the support of individuals function ?

Thanks,

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Think of a special case: $f$ and $g$ are zero except on a small interval, like "two copies of the box-function, which is $1$ on $[-1/2, 1/2]$ and $0$ elsewhere".

Then $f \star g(u)$ is, roughly, "how much $f$ looks like $g$, flipped over and shifted by $u$.

In the case where $g$ is symmetric (i.e., an even function), the "flip over" can be ignored, and $f \star g(u)$ tells you "how much does $f$ look like a shifted copy of $g$ (shifted by $u$, that is).

For two box functions, when $u = 0$, they look a lot alike -- the functions are identical. As you increase $u$, one of the graphs moves to the right, and the overlap is smaller; the same goes for moving to the left. But you need to move $g$ one whole unit to the right (or left) to get to the point where the graphs of $f$ and $g$ don't overlap at all.

Hence the support of $f \star g$ ends up having width $2$ instead of $1$.

BTW, When I say that $f \star g (0)$ is "how much $g$ (flipped) and $f$ "look alike", I'm speaking informally. But you can see that the form of the integral is exactly the inner product of $f$ with "flipped $g$", for the usual inner product on $L^2$.

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  • $\begingroup$ In the OP's proof , it says if $z \not\in A$, then for any $y \in \text{supp}(g)$, we have $z-y \not\in \text{supp}(f)$. How do we know this? $\endgroup$ – Al Jebr Jan 20 '18 at 5:24

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