Is the square of an $H^1$ function also $H^1$? Let $\Omega$ be a bounded domain with a smooth boundary, and let $u$ be in the Sobolev space $H^1(\Omega) = W^{1,2}(\Omega)$. In a paper that I am reading they state that a given function, $u$, that they are studying is in $H^1(\Omega)$ but then proceed to use $u^2$ like its in $H^1(\Omega)$ to get some estimates. I am wondering if this is really true. In the setting for the problem $H^1(\Omega)$ is compactly contained in $L^4(\Omega)$ so showing that $u^2$ is in $L^2(\Omega)$ is not a problem. However, I am at a loss for how to show that $\nabla u^2$ is in $H^1(\Omega)$. Does any one have any ideas or counterexamples? 

Hint: Look at functions of the form
$$f(x) = |x|^\alpha$$
or
$$f(x) = \big|\log|x|\big|^\alpha$$
and try to figure out for which exponents $\alpha$ they belong to $H^1(\Omega)$.
This should help to find an example $f \in H^1(\Omega)$ with $f^2 \not\in H^1(\Omega)$.
