Suppose $1<p<\infty$ and $\Omega$ is an open bounded set in $\mathbb R^n$ with nice boundary (say Lipschitz or even better). Let $(f_j)_j \subset W^{1,p}(\Omega)$ s.t. $f_j \rightharpoonup f$ weakly in $W^{1,p}(\Omega)$.

Is it true that $f_j \to f$ strongly in $L^p(\Omega)$?

For sure it is true that $f_j \rightharpoonup f$ and $\nabla f_j \rightharpoonup\nabla f$. Moreover, we should have the strong convergence of a subsequence thanks to reflexivity: $(f_j)_j$ is bounded hence is has a strong convergent subsequence in $L^p(\Omega)$ because the embedding $W^{1,p} \to L^p$ is (always) compact.


  • $\begingroup$ In other words, you would like to prove that passing to a subsequence in the Rellich theorem is useless? I think this is not reasonable... $\endgroup$ – Siminore Jan 19 '14 at 11:14
  • $\begingroup$ Thanks for your comment. I agree with you but I was a bit confused and worried, because sometimes (in my notes of the course I've attended) I have written "weak convergence in $W^{1,p}$ implies strong convergence of the function in $L^p$" and I didn't understand how this can be true. For sure, the teacher meant "up to a subsequence". By the way, is there any useful "characterization" of the weak convergence in $W^{1,p}$? Or some other conclusions we can derive from it? Thanks again. $\endgroup$ – Romeo Jan 19 '14 at 11:25

Hint: prove the following topological result

Assume that $\Omega$ is a metric space and $x_n\in\Omega$ is a sequence. Suppose that every subsequence of $x_n$ has a further subsequence, which converges to some fixed limit $x\in \Omega$. Then, $$x_n\to x$$

  • $\begingroup$ I know this result, but I do not know how to apply it here... maybe I should say: for every subsequence of $f_j$ there exists a strongly convergent subsequence in $L^p$ via Rellich and its limit must be $f$. Hence $f$ is the strong limit of $f_j$. Is this correct? Thanks. $\endgroup$ – Romeo Jan 20 '14 at 15:47
  • $\begingroup$ Yes, that's right and in fact you know how to apply the theorem. $\endgroup$ – Tomás Jan 20 '14 at 16:06
  • $\begingroup$ This seems very interesting. Thanks a lot. $\endgroup$ – Romeo Jan 20 '14 at 17:04

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