Let $f\in L^2(\mathbb{R}^n)$ and $g\in \mathcal{S}'(\mathbb{R}^n)$, where $\mathcal{S}'$ denotes the tempered distribution space, if $f=g$ in $\mathcal{S}'$, could we conclude $g=f$ in $L^2$?
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This question makes no sense since you did not assume $g\in L^2$. – Anne Bauval Mar 30 '24 at 07:31
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Quick beginner guide for asking a well-received question + please avoid "no clue" questions. – Anne Bauval Mar 30 '24 at 07:32
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I mean that could we prove $g\in L^2$? The similar problem is that we have $f=g$ in $L^1$, if $f$ is a continuous function, but we can not say $g$ is also continuous. – user1776247 Mar 30 '24 at 08:48
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It is not clear what you mean by "prove $g\in L^2$". You are assuming it when writing $\exists f\in L^2$ s.t. $g=f$ in $\mathcal S'$. – Anne Bauval Mar 30 '24 at 10:29
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Once clarified, your question would be more or less a duplicate of https://math.stackexchange.com/questions/4652345 – Anne Bauval Mar 30 '24 at 10:33
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@AnneBauval Thank you very much! – user1776247 Mar 30 '24 at 14:31