TITLE: Product of $L^2$ functions
QUESTION [5 upvotes]: 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 $ ?

REPLY [8 votes]: 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 $f \in L^2(\Omega) \setminus L^4(\Omega)$.