show that $uv\in W^{1,r}(\Omega)$

Question

Let $\Omega\subset\mathbb{R}^{n}$ is bounded,with $\partial\Omega\in C^{1},p,q\geq 1$,$u\in W^{1,p}(\Omega),v\in W^{1,q}(\Omega)$,show that $ uv\in W^{1,r}(\Omega)$.here $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-\frac{1}{n}$ when $\frac{1}{p}+\frac{1}{q}-\frac{1}{n}>0$,and $r=\min\{p,q\}$ when $\frac{1}{p}+\frac{1}{q}-\frac{1}{n}<0$ .

Answer 1

First you can prove (by approximating $u$ and $v$ by smooth functions) that $$ D(uv) = u Dv + vDu. $$ Assume first $p<n$ and $q<n$. Then using Sobolev embedding theorem, we know that $$ u\in L^{\frac{pn}{n-p}}(\Omega), \quad v\in L^{\frac{qn}{n-q}}(\Omega). $$ Using Hoelder inequality one can prove that $f\in L^p(\Omega)$, $g\in L^q(\Omega)$ implies $fg\in L^r(\Omega)$ for all $r$ such that $1/r \ge 1/p+1/q$ provided that $\Omega$ is has finite measure: $$ \|fg\|_{L^r(\Omega)} = \int_\Omega |f|^r|g|^r \le \|f\|_{L^p(\Omega)}^r \left(\int_\Omega |g|^{\frac{pr}{p-r}}\right)^{\frac{p-r}p} \le \|f\|_{L^q(\Omega)}^r \|g\|_{L^q(\Omega)}^r |\Omega|^{r(\frac1r-\frac1p-\frac1q)} $$

$$ %q'=q\frac{p-r}{pr}, 1/q'=\frac{pr}{qp-qr}, 1-1/q' = \frac{qp-qr-pr}{qp-qr}, %(1-1/q')(p-r)/p=\frac{qp-qr-pr}{qp}=r(1/r-1/p-1/q) %$$ This implies $$ uDv \in L^{r}(\Omega), \ vDu \in L^r(\Omega) $$ if $$ \frac1r \ge \max( \frac{n-p}{pn}+\frac1q , \frac{n-q}{qn}+\frac1p) = \frac1p+\frac1q-\frac1n. $$ Moreover, $uv\in L^r(\Omega)$ if $$ \frac1r\ge \frac{n-p}{pn}+\frac{n-q}{qn} = \frac1p+\frac1q-\frac2n $$ This completes the discussion if $p< n$ and $q<n$.

If $p>n$ and $q>n$ then $uv\in L^\infty(\Omega)$ and $D(uv)\in L^{\min(p,q)}(\Omega)$. If $p<n$ and $q>n$ then $uv\in L^p(\Omega)$ and $D(uv)\in L^r(\Omega)$ if $$ \frac1r\ge \max(\frac1p-\frac1n+\frac1q , \frac1p) = \frac1p = \frac1{\min(p,q)}. $$ Similarly, the argument works for $p>n$ and $q<n$.

The claim is false if $p=n$ or $q=n$ holds. Then $uv\in W^{1,r}(\Omega)$ for all $r<\min(p,q)$.