I know that, if $f,g \in L^1(\mathbb{R})$, then, $\widehat{f*g}=\widehat{f}\widehat{g}$ (Classical Convolution's theorem) . If $f,g\in L^2(\mathbb{R})$. Is this still valid?
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2See https://math.stackexchange.com/a/118463/27978 – copper.hat Dec 18 '23 at 05:34