1
$\begingroup$

Let $V \subset H$ be a dense and compact embedding. Let $$\lVert u_n\rVert_{L^\infty(0,T;H)} + \lVert u_n \rVert_{L^2(0,T;V)} < C$$ where $C$ is independent of $n$. It follows that eg. $u_n \rightharpoonup u$ in $L^2(0,T;V)$ and $u_n \rightharpoonup^* u$ in $L^\infty(0,T;H)$ for some $u$.

Does this imply $u_n \to u$ in $L^2(0,T;H)$ strongly?

I have seen this claim on page 11 of http://www.mat.unimi.it/users/rocca/cgquad2.ps (see equation 3.41) but I find it hard to believe. The reference cited there is a book by Lions in which I cannot find anything.

I would like a reference if thiis is true. Thanks.

$\endgroup$
2
  • 1
    $\begingroup$ Where does $u$ come from? A simple counterexample is obtained from letting the sequence oscillate between two vectors. $\endgroup$ – user1688 Mar 2 '15 at 18:07
  • $\begingroup$ @Corbennick please see edit. It comes from the first two bounds. $\endgroup$ – Charpe Mar 2 '15 at 18:13
1
$\begingroup$

As it is stated, this property does not hold: indeed consider the sequence of functinos $(u_n)_{n \in \mathbb{N}}$ defined for $t \in [0, T]$ by $$ u_n (t) = \sin (2n\pi t) v, $$ where $v \in V$ is a fixed vector. The sequence converges clearly weakly to $0$ in $L^2 (0, T; V)$. Also since $L^2 (0, T)$ is dense in $L^1 (0, T)$ and the squence $(\sin 2n\pi t)$ is bounded in $L^\infty (0, T)$, the sequence $(\sin 2n\pi t)_{n \in \mathbb{N}}$ converges weakly-* in $L^\infty (0, T)$, and thus the sequence $(u_n)_{n \in \mathbb{N}}$ converges weakly-* in $L^\infty (0, T, H)$. This sequence satisfies thus the assumptions, but not the conclusion.

Classical results on this topic involve an assumption of the type $(u_n')_{n \in\mathbb{N}}$ is bounded in $L^2 ([0, T], V')$ and imply the strong convergence (see for example Evans, Partial differential equations, 1998, section 5.9.2).

$\endgroup$
6
  • $\begingroup$ Thank you, but doesn't this contradict Theorem 1 of this paper? $\endgroup$ – Charpe Mar 3 '15 at 19:34
  • $\begingroup$ I am of course aware of the classical example involving the sine that weak convergence does not imply strong convergence. But I don't see why your example converges weakly in the $L^\infty(0,T;L^2)$ either. $\endgroup$ – Charpe Mar 3 '15 at 19:35
  • $\begingroup$ The theorem that you mention is quite different: it assumes that for almost every $t \in [0, T]$, the sequence $(u_n)$ converges weakly in $H$ and that it is equiintegrable in $L^2 (0, T; H)$. This is not a consequence of the weak-* $L^\infty (0, T; V)$ convergence. $\endgroup$ – Jean Van Schaftingen Mar 4 '15 at 7:19
  • $\begingroup$ Oh I see. Well, we certainly have the equiintegrability by Remark 1, point 2. I had thought that the uniform bound on $u_n$ in $L^\infty(0,T;L^2)$ gave $u_n(t) \rightharpoonup u(t)$ in $H=L^2$ a.e. $t$. $\endgroup$ – Charpe Mar 4 '15 at 8:05
  • $\begingroup$ @Charpe The uniform bound in $L^\infty (0, T; L^2)$ only implies that for almost every $t \in [0, T]$, a subsequence converges weakly; since the set $[0, T]$ is not contable, you cannot conclude by a diagonal argument. $\endgroup$ – Jean Van Schaftingen Mar 4 '15 at 9:20

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.