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

  • 2
    $\begingroup$ 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

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 $W_0^{1,p}(\Omega)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.