$u\in L^2(\Omega)$ does this imply that $u^p\in L^2(\Omega)$?

Question

Suppose that a function $u:\Omega \rightarrow \mathbb{R}^n$ is such that $u \in L^2(\Omega)$. Does this imply that $u^p \in L^2(\Omega)$? if not can you give a counterexample?

Here $\Omega$ is an open bounded subset of $\mathbb{R}^n$ and $p > 2$ is a real number.

Answer 1

It does not. Indeed let $\Omega = (0,1)$, $u(x) = \frac{1}{\sqrt[3]{x}}$. Then $u \in L^2((0,1))$, but $u^3 = \frac{1}{x}$ and $\frac{1}{x}$ is clearly not square integrable.