Powers of bounded Sobolev functions

Question

I felt that the claim

$$u\in L^\infty(\Omega)\cap W^{1,p}_{loc}(\Omega)\implies u^k\in W^{1,p}_{loc}(\Omega),$$

for $1<p<\infty$ and $k\geq 1$, should be true but I find it hard to believe. The proof I had in mind went as follows. Since, $u\in L^\infty$, all its powers are also bounded and hence in $L^p$. For the derivative, we find a sequence of smooth functions $u_n$ converging to $u$ in $W^{1,p}(\Omega')$ for $\Omega'\subset\subset\Omega$. Then, it holds true that for any $\phi\in \mathcal{D}(\Omega)$ compactly supported in $\Omega'$, we have $$\int_{\Omega'}u_n^k \phi_{x_i}dx=-k\int_{\Omega'}u_n^{k-1}(u_n)_{x_i}\phi\,dx.$$

Now, since both integrands converge pointwise almost everywhere for a subsequence, we can apply the Dominated convergence theorem to conclude the expected limits.

Where am I wrong? If it is true, does anyone know of a reference?

Answer 1

You can easily prove that is true by just computing the weak derivative of $u^k$. However, a more general result holds. In fact, $L^\infty(\Omega)\cap W^{1,p}(\Omega)$, $1\leq p\leq \infty$, is an algebra:

If $u,v\in L^\infty(\Omega)\cap W^{1,p}(\Omega)$ with $1\leq p\leq \infty$, then $u v \in L^\infty(\Omega)\cap W^{1,p}(\Omega)$.

What you ask is the particular case in which $u=v$. The proof is easy and can be found in Brezis's book (Prop.9.4, p.269).