# 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.

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I can't understand the role of $\beta$. Of course you cannot expect a function from $L^2$ to belong to $L^\beta$ for any $\beta$. –  Siminore Sep 14 '12 at 13:36
How do you define $u^{\beta-p+1}$ -- is $u$ nonnegative (or strictly positive)? –  user31373 Sep 14 '12 at 17:27
@LVK: let's assume it is strictly positive. –  balestrav Sep 15 '12 at 20:36
@ Siminore: but if $\Omega$ is bounded we can say something, and that was my question: are there any other assumptions (the set is bounded, the function is essentially bounded) which can make us conclude something about other exponents of summability? –  balestrav Sep 15 '12 at 20:39

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.