Summability of a function If $u\in W^{1,p}(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$ and $\xi$
is a smooth compactly supported function in $\Omega$,
is it true that $\xi u^{\beta-p+1} \in W^{1,p}_0$ if $\beta >p-1$?
(In the end my problem is to say if $u^{\beta-p+1} \in W^{1,p}$ (from this I know it follows the result).)
I think if $\Omega$ is not bounded we can't say, but if it is bounded, then we know
the function $u$ belongs to $L^r$ for $r<p$, but $\beta>p-1$ could be also greater than p.
Maybe if I add the hypotesys $u\in L^{\infty}$ I could conclude?
Thanks for any help.
 A: The presence of $\xi$ helps only by reducing the region of integration to a compact subset of $\Omega$. The question is equivalent to asking whether $u^{\beta-p+1}\in W^{1,p}$ locally. 
When $p>n$ we are in good shape. Indeed, $u$ has a continuous representative by the Morrey-Sobolev embedding which means that it is locally bounded away from both $0$ and $\infty$. The function $\phi(t)=t^{\beta-p+1}$ is Lipschitz on any interval $[\alpha,\beta]$ with $0<\alpha<\beta<\infty$. It is a standard fact that composition with a Lipschitz function preserves Sobolev spaces of first order. Thus, $\phi\circ u\in W^{1,p}$ locally. 
The assumption $p>n$ was needed only to have two-sided bounds on $u$. If you are willing to impose such bounds artificially, then any $p\ge 1$ works. Otherwise there are counterexamples. Indeed, $u(x)=|x|^r$ belongs to $W^{1,p}$ in a neighborhood of origin exactly when $p(r-1)>-n$, equivalently $pr>p-n$. If $p<n$, this may hold with negative $r$, but then raising $u$ to a sufficiently high power breaks the inequality.
