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Assume $ \Omega \subset \mathbb{R}^N$ is a smooth bounded domain. There is well known Hardy inequality that says

For any $ u \in W_0^{1,2}(\Omega) $, $N\geq3$ we have $$ \Lambda \int_{\Omega} \frac{u^2}{|x|^2} \, \mathrm{d}x \leq \int_{\Omega} |\nabla u|^2 \, \mathrm{d}x $$ Where $ \Lambda=\frac{(N-2)^2}{4} $ is optimal constant.

Brezis and Vazquez has extended this inequality as follow:

For any $ u \in W_0^{1,2}(\Omega) $, $N\geq3$ and $ 1 \leq p < \frac{2N}{N-2} $we have

$$ \int_{\Omega} |\nabla u|^2 \, \mathrm{d}x \geq \Lambda \int_{\Omega} \frac{u^2}{|x|^2} \, \mathrm{d}x + C \int_{\Omega} u^p \, \mathrm{d}x $$

See This Article.

From the above inequality we deduce that $$ \|\phi\|= \bigg( \int_{\Omega} |\nabla \phi|^2 \, \mathrm{d}x - \Lambda \int_{\Omega} \frac{\phi^2}{|x|^2} \, \mathrm{d}x \bigg)^{\frac{1}{2}} $$

defines a norm on the space $W_0^{1,2}(\Omega)$. Now Let $H(\Omega)$ be the compeletion of $C_0^{\infty}(\Omega)$ with respect to above norm.

Now since

$$ \bigg( \int_{\Omega} |\nabla \phi|^2 \, \mathrm{d}x - \Lambda \int_{\Omega} \frac{\phi^2}{|x|^2}\bigg)^{\frac{1}{2}} \leq \bigg( \int_{\Omega} |\nabla \phi|^2 \, \mathrm{d}x \bigg)^{\frac{1}{2}} $$ we can see that $ W_0^{1,2}(\Omega) \subset H(\Omega) $, and also converse inclusion fails.

Also the above inequality results that $ H(\Omega) $ is embedded in $ L^p(\Omega)$, for any $ 1 \leq p < \frac{2N}{N-2} $.

Now I want to investigate that, is this embedding compact for $ 1 \leq p < \frac{2N}{N-2} $ or not? I know some criterion for precompactness in $L^p$ spaces, but I can't use it to show compactness of embedding.

I am thanksed for any hint, or comments.

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    $\begingroup$ in the `improve Hardy inequality' you need to renormalize the $L^p$ term; ie. it should be $ \| u \|_{L^p}^2$. Take a look at this paper regarding $H(\Omega)$ , I think it says that some of the earlier papers regarding $H$ may have had some flaw??? arxiv.org/pdf/1102.5661v1.pdf Further here is a paper regarding improved $L^2$ Hardy inequaliteis : arxiv.org/abs/0805.0610 $\endgroup$
    – Math604
    Commented Oct 8, 2015 at 18:54
  • $\begingroup$ @Math604: thanks. Did you know any other article about improved case of rellich inequality and similar space H as completion with respect to norm $$ \bigg( \int_{\Omega} \Big((\Delta u)^2 - \dfrac{N^2(N-4)^2}{16} \dfrac{u^2}{|x|^4}\Big) \,\mathrm{d}x \bigg)^{\frac{1}{2}} $$ $\endgroup$
    – Hheepp
    Commented Oct 9, 2015 at 18:31
  • $\begingroup$ there should be a bunch of "improved Hardy-Rellich papers". Try googling that phrase (also Amir Moradifam wrote some papers on this in the radial case). Regarding the space $H$ I don't really know of many references.... Do you need to understand $H$ for a particular reason or ?? $\endgroup$
    – Math604
    Commented Oct 9, 2015 at 18:47
  • $\begingroup$ Thanks. I must give a seminar in the classroom in this regard. I want to minimize a function in this space and I need an compact embedding. $\endgroup$
    – Hheepp
    Commented Oct 9, 2015 at 18:52
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    $\begingroup$ Maybe you can argue directly. Let $ u_m$ be bounded by say $1$ in $H$ and use the improved imbedding to see that $u_m$ is bounded in $L^2(\Omega)$. Then directly from the inequality again we see that $ u_m$ should be bounded in $H^1_0_{loc}( overline{\Omega} \backslash \{0\} )$ (trying to say its bounded in $H^1$ except near origin). By diagonal argument there should be some $ u in H^1_{loc} (remove near origin) such that $ u_m \rightharpoonup u$ in $H^1_{loc}(away from origin). Now try and show that $ u_m$ converges in $L^1(\Omega)$ (so in $L^1$ it can't concentrate at origin). Then... $\endgroup$
    – Math604
    Commented Oct 9, 2015 at 20:01

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