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I just look through a webpage from "mathoverfolw", on https://mathoverflow.net/questions/136681/the-convolution-of-integrable-functions-is-continuous

Let me remark that it is sufficient that one of the functions is bounded (the convolution of an $L^{1}$-function and an $L^{\infty}$-function on a unimodular group is always continuous.

How to prove it?

Are there other more discriminated methods about "continuity of convolution"?

When a convolution is continuous on locally compact Hausdorff topological groups?

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David Chan
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1 Answers1

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Let $G$ be a unimodular locally compact group. If $f\in L^1(G)$ and $g\in L^\infty (G)$. Then, $$f*g(x)-f*g(a)=\int_G (f(xy)-f(ay))g(y^{-1})dy$$ Therefore, $$|f*g(x)-f*g(a)|\leq \int_G|f(xy)-f(ay)||g(y^{-1})|dy \\ \leq \|g\|_{\infty}\int_G|f(xa^{-1}y)-f(y)|dy\\ =\|g\|_{\infty}\|L_{ax^{-1}}f-f\|_1$$ Where $L_z$ is the left translation by $z\in G$. Since the left translation is continuous map from $G$ into $L^p(G),1\leq p<\infty$. Hence $\|L_{ax^{-1}}f-f\|_1\to 0$ as $ax^{-1}\to 1$. Therefore, if $x\to a$ then $f*g(x)\to f*g(a)$, that if $f*g$ is continuous.

Yasser YKT
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