$u, v \in H^1(\Omega)$, then the product $uv \in W^{1,3/2}(\Omega)$ for $\Omega \subset \mathbb{R}^3$?

Question

I think this seems quite plausible, but cannot prove myself rigorously.

Let $\Omega \subset \mathbb{R}^3$ be a bounded region and $W^{1,2}(\Omega):=H^1(\Omega)$.

My guess is that if $u,v \in H^1(\Omega)$ then for the product $uv$, \begin{equation} uv \in W^{1,3/2}(\Omega) \text{ with } D(uv)=u Dv + (Du)v \end{equation}

This is because $H^1(\Omega) \subset L^6(\Omega)$ and $\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$. However, I cannot justify my guess rigorously..

Could anyone please help me?

Answer 1

As in the comments section already mentioned, one can show this by first working on a dense subclass and then using the theorem by Meyers and Serrin. For completeness and in the interest of future people who stumble upon the same question, I attach a short proof below.

Let first $u,v \in D:= C^\infty(\Omega)\cap H^1(\Omega)$ where $\Omega$ is bounded domain in $\mathbb{R}^3$. If we assume that the boundary of $\Omega$ is sufficiently smooth, then we may use Sobolev's embedding theorems to conclude that there exists some $C>0$ such that $\|f\|_6\leq C\|f\|_{H^1}$ for all $f\in H^1(\Omega)$. Using this and Hölder's inequality ($\frac{1}{6}+\frac{1}{2}=\frac{2}{3}$) we conclude

$\|uv\|_{3/2}\leq \|u\|_2 \|v\|_{6}\leq C \|u\|_{H^1}\|v\|_{H^1}$,

and similarly,

$\|(Du)v\|_{3/2}\leq \|Du\|_2 \|v\|_{6}\leq C \|u\|_{H^1}\|v\|_{H^1}$,

$\|u(Dv)\|_{3/2}\leq \|u\|_{3/2} \|Dv\|_{2}\leq C \|u\|_{H^1}\|v\|_{H^1}$,

which by the product rule $D(uv)=(Du)v+u(Dv)$ implies

$\|D(uv)\|_{3/2}\leq \|(Du)v\|_{3/2}+\|u(Dv)\|_{3/2}\leq 2C \|u\|_{H^1}\|v\|_{H^1}$.

We conclude

$\|uv\|_{W^{1, 3/2}}=\|uv\|_{3/2}+\|D(uv)\|_{3/2}\leq 3C \|u\|_{H^1}\|v\|_{H^1}$ for all $u,v \in D\quad \quad (\ast)$ .

Now, let $u$ be a general function in $H^1(\Omega)$ and $v\in D$. By the theorem of Meyer and Serrin $D$ is dense in $H^1(\Omega)$, i.e., there is a sequence $(u_n)\subseteq D$ with $u_n\to u$ $(n\to\infty)$ in $H^1(\Omega)$. By $(\ast)$, we have

$\|(u_n-u_m)v\|_{W^{1, 3/2}}\leq 3C \|u_n-u_m\|_{H^1}\|v\|_{H^1}$ for all $n,m\in \mathbb{N}$.

Thus, the sequence $(u_nv)_{n}$ is a Cauchy sequence in $W^{1,3/2}(\Omega)$ and hence converges to some $w\in W^{1, 3/2}(\Omega)$. By the theorem of Fischer-Riesz, there is a subsequence $(u_{n_k}v)_{k}$ which converges pointwise a.e. to $w$. Once again using the theorem of Fischer Riesz there is a subsequence $(u_{n_{k_\ell}})_\ell$ which converges pointwise a.e. to $u$. Hence the subsequence $(u_{n_{k_\ell}}v)_\ell$ converges pointwise a.e. to $w$ and to $uv$ and by the uniqueness of pointwise limits we conclude $w=uv$ a.e. Therefore $uv\in H^1(\Omega)$ and by $(\ast)$ we have $\|uv\|_{W^{1,3/2}}\leq 3 C \|u\|_{H^1}\|v\|_{H^1}$. We have shown

$\|uv\|_{W^{1, 3/2}}\leq 3C \|u\|_{H^1}\|v\|_{H^1}$ for all $u\in H^1(\Omega)$ and $v\in D$.

Repeating the same argument, but now keeping $u\in H^1(\Omega)$ fixed and approximating a general $v\in H^1(\Omega)$ by a sequence $(v_n)_n\subseteq D$ in $H^1(\Omega)$, we conclude that, for any $u,v\in H^1(\Omega)$ we have $uv\in W^{1,3/2}(\Omega)$ and $\|uv\|_{W^{1,3/2}}\leq 3 C \|u\|_{H^1}\|v\|_{H^1}$ as desired.