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Weak convergence of a function in Sobolev spaces implies weak convergence of positive and negative parts of that function or not?

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    $\begingroup$ Isn't the sequence $(f_n)$, where $f_n(x)=\sin nx$ for $n$ even, $f_n=0$ for $n$ odd a counterexample, or am I missing something? $\endgroup$ Apr 22 '16 at 9:03
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Yes, this is true for $W_0^{1,p}(\Omega)$ for open and bounded $\Omega \subset \mathbb R^n$ and $1 < p < \infty$. If $\Omega$ has some regularity, it also works for $W^{1,p}(\Omega)$.

The argument is quite easy:

If $u_n \rightharpoonup u$ in $W_0^{1,p}(\Omega)$, you get $u_n \to u$ in $L^p(\Omega)$ by compact embedding. Hence, $u_n^+ \to u^+$ in $L^p(\Omega)$. Further, $u_n^+$ is bounded in $W_0^{1,p}(\Omega)$ by Stampacchia's lemma and (using the reflexivity of $W_0^{1,p}(\Omega)$) it converges (along a subsequence) weakly to some $w$ in $W_0^{1,p}(\Omega)$. Again by compactness, you have $u_n^+ \to w$ in $L^p(\Omega)$ which gives $w = u^+$. By a subsequence-subsequence argument, you get $u_n^+ \to u^+$ in $W_0^{1,p}(\Omega)$.

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  • $\begingroup$ I see. Actually, if $p>\frac1n$ (so that $W^{1,p}(\Omega)$ is an algebra), your argument works even if we replace the positive part by every continuous function of $u$. Namely $u_r\rightharpoonup u$ implies $f(u_r)\rightharpoonup f(u)$. My error was that I wanted to use the fact, in a Hilbert space, that if in addition $\|u_r\|\rightarrow\|u\|$, then the convergence is strong. In $L^2$ this amounts to $u_r^2\rightharpoonup u^2$, but not in $H^1$ !! $\endgroup$ Apr 22 '16 at 20:34
  • $\begingroup$ @DenisSerre: I do not get that point, $f(u)$ might even fail to belong to $W^{1,p}(\Omega)$: $\Omega = (0,1)$, $p = 2$, $u(x) = x$ and $f(t) = \sqrt{t}$. Then, $\nabla(f(u))(x) = x^{-1/2}/2$, which is not square-integrable. It might work if $f$ is Lipschitz. $\endgroup$
    – gerw
    Apr 23 '16 at 6:36
  • $\begingroup$ Of course. I just wrote too fast by saying "$f$ continuous". I meant "$f$ locally Lipschitzian". $\endgroup$ Apr 23 '16 at 12:42

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