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.

  • 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

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).

  • $\begingroup$ 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

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