Consider $u_n$ a bounded sequence in $L^{p}(\Omega)$ where $\Omega \subset R^n$ open and bounded. Suppose that exists a subsequence $u_{n_j}$ that converges weakly to a function $u $ in $L^{p}(\Omega)$. Then $u_n$ converges weakly to $u$ ?

Is the affirmation is true?

Can someone give me a hint to prove this (or disprove)? I am trying to prove, but nothing.

  • 1
    $\begingroup$ Consider a sequence taking finitely many values. $\endgroup$ – Daniel Fischer Sep 3 '13 at 20:40

No, take $u_n:=(-1)^n$.

Notice that when $1\lt p\lt \infty$, the space $L^p(\Omega)$ is reflexive, hence for each bounded sequence, we can extract a weakly convergent subsequence (so the assumption in the OP always holds). But it does not mean that the whole sequence converges.


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