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Is there a pair of integrable functions $(f,g)$, where $f\in L^p$ and $g\in L^q$ where $p>0$ and $q>0$ may be unrelated, such that $f*g=f$ where $*$ stands for convolution?

I have considered the Fourier transform $\hat u$ assuming it is meaningful on any function $u$ under consideration. Assume $\hat f\hat g=\hat f$ holds. Then $\hat g(k)=1, \forall k\in\{k: \hat f(k)\neq 0\}$.

Hans
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1 Answers1

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Let $f$ be a Schwarz function such that $\hat f$ is supported on a compact set $K$. Let $g$ be another Schwarz function such that $\hat g(\xi)=1$ for all $\xi\in K$. Then $$ \widehat{f\ast g}=\hat f\,\hat g=\hat f\implies f\ast g=f. $$

Julián Aguirre
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  • Excellent. Thank you, Julian. The motivation for this example is the following. We do not need $\hat g=1$ for all $\xi$ but only on a compact set, we need the value elsewhere to be free for manipulation which is facilitated by the suppression by the vanishing $\hat f$. This allows $\hat g$ to be integrable and most importantly the Fourier transform to exist. Schwartz space is but one concrete condition to guarantee the existence of the Fourier transform, since Fourier transform is a linear isomorphism in the Schwartz space, and any smooth compactedly supported function is a Schwartz function. – Hans Dec 12 '16 at 20:26