Is the square of an $H^1$ function also $H^1$?

Question

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?

Answer 1

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 $f^2 \not\in H^1(\Omega)$.

Answer 2

Your question can be answered in the negative for any $\Omega\subset \mathbb R^n$, unless $n=1$. The rationale is that if your conjecture in the title did hold, then the weak derivative would satisfy a chain rule, i.e., $\frac{\partial (G\circ u)}{\partial x_i}(x)=(G'(u(x))\frac{\partial u}{\partial x_i}(x)$. Because $u\in H^1(\Omega)$, $\frac{\partial u}{\partial x_i}\in L^2(\Omega)$ and the only way $\frac{\partial (G\circ u)}{\partial x_i}$ may lie in $L^2(\Omega)$ is that $(G'\circ u)\in L^\infty$. But this cannot be satisfied in your case, since you are considering $G(s)=s^2$ and $u\not\in L^\infty(\Omega)$ in any dimension larger than $1$. However, for $n=1$ your conjecture does hold: again by the above argument, or simply because, according to Theorem 4.39 in the book by Adams-Fournier, $W^{1,p} (\mathbb R)$ is a Banach algebra for any $p>1$.