Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I read somewhere that if:

  • $N\geq 2$ is an integer,
  • $p\in ]1,N[$, $r>N/p$,
  • $m\in L^r(0,a)$ (with $a>0$) and $m>0$ a.e. in $(0,a)$,

then the weighted Sobolev space $W^{1,p^\prime} ((0,a),m^{-1/(p-1)})$, defined as the completion of $C_c^1(0,a)$ w.r.t. the norm: $$\begin{split} \| u\|_{W^{1,p^\prime}((0,a),m^{-1/(p-1)})} &:= \| u\|_{L^{p^\prime}((0,a),m^{-1/(p-1)})} +\| \dot{u}\|_{L^{p^\prime}((0,a),m^{-1/(p-1)})}\\ &= \left( \int_0^a |u|^{p^\prime}\ \frac{1}{m^{1/(p-1)}}\ \text{d} s\right)^{1/p^\prime} + \left( \int_0^a |\dot{u}|^{p^\prime}\ \frac{1}{m^{1/(p-1)}}\ \text{d} s\right)^{1/p^\prime}\; , \end{split}$$ compactly embeds into $L^{p^\prime} ((0,a), m^{-1/(p-1)})$... But I didn't succeed in doing the right computations to get the correct Sobolev inequality.

Does anyone knows how to prove $W^{1,p^\prime} ((0,a),m^{-1/(p-1)}) \hookrightarrow L^{p^\prime} ((0,a), m^{-1/(p-1)})$ compactly?

Any hint will be appreciated.

share|improve this question
    
I assume you want $m > 0$? –  Willie Wong Jun 6 '12 at 15:23
    
@WillieWong : I forgot to mention it, but yes: $m>0$ a. e. in $(0,a)$. –  Pacciu Jun 7 '12 at 15:34
    
@DavideGiraudo : In this context $p^\prime$ is the Hölder conjugate of $p$, i.e. $p^\prime := \frac{p}{p-1}$. I thought it was a standard notation... –  Pacciu Jun 25 '12 at 20:32

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.