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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If I remember right, $L^p$ is an algebra for any $p \geq 1$... – Sean Roberson Nov 08 '16 at 21:18
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2You might consider functions of one variable and consider $\Omega=(0,1)$. – Michael Nov 08 '16 at 21:19
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2In particular, on $(0,1)$ consider $x^{-1/4}$ and its square. – Robert Israel Nov 08 '16 at 21:23
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2Take $f(x)=g(x)=x^{-1/4}$ 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. – Mark Viola Nov 08 '16 at 21:24
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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 $f \in L^2(\Omega) \setminus L^4(\Omega)$.
gerw
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