In compact embedding Theorem, $u_0$ lies in $W^1,p(Omega)$?

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We know that $W^1,p(Omega)hookrightarrow L^q(Omega)$ if $Omegasubsetmathbb R^N$ is a bounded open and $partial Omega$ is $C^1$, $1leq p<N$ and $1leq q<p^*:=fracNpN-p$. Moreover, for every bounded sequence $(u_n)$ in $W^1,p(Omega)$ (unless subsequence), $u_nrightarrow u_0$ in $L^q(Omega)$. I want to $u_0in W^1,p(Omega)$, but I only know that $u_0in L^q(Omega)$. Is it true that $u_0in W^1,p(Omega)$? and how can I prove it?



My attempt:



Since $W^1,p(Omega)$ is reflexive, we obtain $u_nrightharpoonup u_0$ in $W^1,p(Omega)$. We conclue with compact embedding.



The problem in my attempt is that $W^1,p(Omega)$ is not reflexive for $p=1$.




functional-analysis analysis pde compactness sobolev-spaces




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  • For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional.
    – daw
    Jul 21 at 19:44










  • Even with this hint, I did not make it.
    – Raoní Cabral Ponciano
    Jul 22 at 18:22






  • 1




    @daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$
    – ktoi
    Jul 23 at 11:59














up vote
3
down vote

favorite
3












We know that $W^1,p(Omega)hookrightarrow L^q(Omega)$ if $Omegasubsetmathbb R^N$ is a bounded open and $partial Omega$ is $C^1$, $1leq p<N$ and $1leq q<p^*:=fracNpN-p$. Moreover, for every bounded sequence $(u_n)$ in $W^1,p(Omega)$ (unless subsequence), $u_nrightarrow u_0$ in $L^q(Omega)$. I want to $u_0in W^1,p(Omega)$, but I only know that $u_0in L^q(Omega)$. Is it true that $u_0in W^1,p(Omega)$? and how can I prove it?



My attempt:



Since $W^1,p(Omega)$ is reflexive, we obtain $u_nrightharpoonup u_0$ in $W^1,p(Omega)$. We conclue with compact embedding.



The problem in my attempt is that $W^1,p(Omega)$ is not reflexive for $p=1$.




functional-analysis analysis pde compactness sobolev-spaces




share|cite|improve this question



















  • For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional.
    – daw
    Jul 21 at 19:44










  • Even with this hint, I did not make it.
    – Raoní Cabral Ponciano
    Jul 22 at 18:22






  • 1




    @daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$
    – ktoi
    Jul 23 at 11:59












up vote
3
down vote

favorite
3









up vote
3
down vote

favorite
3






3





We know that $W^1,p(Omega)hookrightarrow L^q(Omega)$ if $Omegasubsetmathbb R^N$ is a bounded open and $partial Omega$ is $C^1$, $1leq p<N$ and $1leq q<p^*:=fracNpN-p$. Moreover, for every bounded sequence $(u_n)$ in $W^1,p(Omega)$ (unless subsequence), $u_nrightarrow u_0$ in $L^q(Omega)$. I want to $u_0in W^1,p(Omega)$, but I only know that $u_0in L^q(Omega)$. Is it true that $u_0in W^1,p(Omega)$? and how can I prove it?



My attempt:



Since $W^1,p(Omega)$ is reflexive, we obtain $u_nrightharpoonup u_0$ in $W^1,p(Omega)$. We conclue with compact embedding.



The problem in my attempt is that $W^1,p(Omega)$ is not reflexive for $p=1$.




functional-analysis analysis pde compactness sobolev-spaces




share|cite|improve this question











We know that $W^1,p(Omega)hookrightarrow L^q(Omega)$ if $Omegasubsetmathbb R^N$ is a bounded open and $partial Omega$ is $C^1$, $1leq p<N$ and $1leq q<p^*:=fracNpN-p$. Moreover, for every bounded sequence $(u_n)$ in $W^1,p(Omega)$ (unless subsequence), $u_nrightarrow u_0$ in $L^q(Omega)$. I want to $u_0in W^1,p(Omega)$, but I only know that $u_0in L^q(Omega)$. Is it true that $u_0in W^1,p(Omega)$? and how can I prove it?



My attempt:



Since $W^1,p(Omega)$ is reflexive, we obtain $u_nrightharpoonup u_0$ in $W^1,p(Omega)$. We conclue with compact embedding.



The problem in my attempt is that $W^1,p(Omega)$ is not reflexive for $p=1$.





functional-analysis analysis pde compactness sobolev-spaces





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 21 at 19:38









Raoní Cabral Ponciano

589




589











  • For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional.
    – daw
    Jul 21 at 19:44










  • Even with this hint, I did not make it.
    – Raoní Cabral Ponciano
    Jul 22 at 18:22






  • 1




    @daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$
    – ktoi
    Jul 23 at 11:59
















  • For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional.
    – daw
    Jul 21 at 19:44










  • Even with this hint, I did not make it.
    – Raoní Cabral Ponciano
    Jul 22 at 18:22






  • 1




    @daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$
    – ktoi
    Jul 23 at 11:59















For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional.
– daw
Jul 21 at 19:44




For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional.
– daw
Jul 21 at 19:44












Even with this hint, I did not make it.
– Raoní Cabral Ponciano
Jul 22 at 18:22




Even with this hint, I did not make it.
– Raoní Cabral Ponciano
Jul 22 at 18:22




1




1




@daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$
– ktoi
Jul 23 at 11:59




@daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$
– ktoi
Jul 23 at 11:59















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