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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.

  • 0
    @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...2012-06-25

0 Answers 0