Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.