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If $ \Omega$ is any open subset of $\mathbb{R}^n$, is it true that the product of two $L^2$ functions over $\Omega$ is also $L^2$ ? What about $L^p $ ?

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  • $\begingroup$ If I remember right, $L^p$ is an algebra for any $p \geq 1$... $\endgroup$ Nov 8, 2016 at 21:18
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    $\begingroup$ You might consider functions of one variable and consider $\Omega=(0,1)$. $\endgroup$
    – Michael
    Nov 8, 2016 at 21:19
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    $\begingroup$ In particular, on $(0,1)$ consider $x^{-1/4}$ and its square. $\endgroup$ Nov 8, 2016 at 21:23
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    $\begingroup$ Take $f(x)=g(x)=x^{-1/4}$ on $(0,1)$. Then $\int_0^1f^2(x)\,dx=\int_0^1 g^2(x)\,dx=2$. But clearly $\int_0^1 f^2(x)g^2(x)\,dx$ does not exist. $\endgroup$
    – Mark Viola
    Nov 8, 2016 at 21:24

1 Answer 1

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No. In fact, the following is easy to prove for $p \ge 2$: a function $f : \Omega \to \mathbb{R}$ belongs to $L^p(\Omega)$ if and only if $f^2 \in L^{p/2}(\Omega)$.

Now, you obtain a counterexample to your question by choosing a function $f \in L^2(\Omega) \setminus L^4(\Omega)$.

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  • $\begingroup$ Should it be $f\in L^2(\Omega)\backslash L^4(\Omega)$? $\endgroup$
    – geometricK
    Mar 22, 2022 at 23:12
  • $\begingroup$ @geometricK: Yes, of course. Thank you! $\endgroup$
    – gerw
    Mar 23, 2022 at 7:17

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