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Suppose $U \subset \mathbb{R}^n$ is non-empty, bounded, open, connected with $C^1$ boundary. Is there any subspace of $L^2(U)$ that are closed under multiplication, that is under which conditions can we say that $$ u ,v \in L^2(U) \implies uv \in L^2(U).$$ I know that in general you can it's easy to prove that $$ u ,v \in L^2(U) \implies uv \in L^1(U)$$ by the Hölder inequality, but how could you prove that $uv \in L^2(U)$ with some conditions on $u$ and $v$?

Henrik Finsberg
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    If $u, v \in L^4(U)$, then $u , v \in L^2(U)$ (but not in $L^4(U)$. Is this what you are looking for? If $u, v \in L^\infty(U)$, then even $u , v \in L^\infty(U)$. – gerw Apr 05 '13 at 12:21
  • Do you have something in mind? For example: The space $C_0^\infty (\overline{U})$ is closed under multiplication. Also the space of constant functions satisfies it. – Tomás Apr 05 '13 at 12:28
  • Yes, actually the answer from gerw was just what I was looking for. Thanks :) – Henrik Finsberg Apr 05 '13 at 12:41
  • I know it doesn't answer the question, but it solves my problem. – Henrik Finsberg Apr 05 '13 at 13:37

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