0
$\begingroup$

I wonder whether this fact is true or not (if a counter-example exists, please just give a hint on how to construct it!):

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. Consider a bounded sequence $\{f_n\}$ in $L^\infty\left([0,T];L^2(\Omega)\right)$. Suppose I also know that $\{f_n\}$ is in $L^2\left([0,T];L^2(\Omega)\right)$ and it also bounded there. Can I find some subsequence of $\{f_n\}$ that converges in $L^\infty\left([0,T];L^2(\Omega)\right)$?

Ok so I realized the question had an easy answer and thank you for the comments and answer! I decided to focus on the more important question which is Generally, how does one study convergence in $L^\infty\left([0,T];L^2(\Omega)\right)$?

I wish the answers would elaborate more on the weak$^*$-convergence suggested in the comments. Also inspired by the comments, what tools from the Banach valued Bochner spaces theory can one use to help answering questions of convergence?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ You can find a subsequence converging in the weak-$\ast$ sense in $L^\infty((0,T);L^2(\Omega))$. $\endgroup$
    – sharpend
    Commented Apr 8, 2020 at 14:38
  • 1
    $\begingroup$ I voted as off-topics, as this is an undergraduate question (not research-level). For your information: You don't need the whole machinery of Banach-valued Bochner spaces to realize that the answer is of course not! (As is clear from Nik Weaver's answer below, this fails even in the trivial real-valued setting $L^\infty(0,T)=L^\infty((0,T);\mathbb R)$) $\endgroup$
    – leo monsaingeon
    Commented Apr 8, 2020 at 16:53
  • 7
    $\begingroup$ As a rule-of-thumbs / personal advice: When considering such questions, first ask yourself whether the result is even true for standard $L^p$ spaces with real/complex values. If so, proceed with further investigation. If not, you've got your answer! $\endgroup$
    – leo monsaingeon
    Commented Apr 8, 2020 at 16:56
  • 1
    $\begingroup$ In particular, $L^\infty([0,T]; L^2(\Omega))$ contains isometrically embedded copies of $L^\infty([0,T])$ (consider $f(t) = \phi(t) h$ for $\phi \in L^\infty([0,T])$ and fixed $h \in L^2$ of norm 1) and of $L^2(\Omega)$ (consider functions constant in $t$). So you certainly need to make sure your conjectures hold in those cases. $\endgroup$
    – Nate Eldredge
    Commented Apr 8, 2020 at 20:04

1 Answer 1

Reset to default
5
$\begingroup$

It sounds like you are new to this subject, so there are some basic things for you to know. The first is that by night all Hilbert spaces are the same, so your $L^2(\Omega)$ can just be a generic $H$. Secondly, $L^\infty[0,T]$ is contained in $L^2[0,T]$ and the norm is smaller in $L^2$ up to a factor of $\sqrt{T}$, and nothing changes when you take values in $H$. So your sequence is automatically bounded in $L^2$, you don't need to assume that separately.

A third comment is that the standard sequence for counterexamples in $L^\infty[0,1]$ is $f_n =$ the function which is constantly $0$ on the intervals $[\frac{2k}{2^n}, \frac{2k+1}{2^n})$ and constantly 1 on the complementary intervals. There's clearly no subsequence that converges in sup norm, though the sequence does converge weak* to the function which is constantly $\frac{1}{2}$. As sharpend says in the comments, you will always be able to find a subsequence which converges in the weak* topology (but you have to pay attention to how you define "weak* topology" for Hilbert space-valued functions).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .