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Let $\Gamma$ be a compact $C^2$ manifold and suppose that $f_n$ is a non negative sequence of functions such that ${\vert \vert f_n \vert \vert}_{L^{\infty}(0,T,L^1(\Gamma))} \le C$

I am interested in deducing convergence of $f_n$

MY ATTEMPTS:

  1. Since $L^1$ is not reflexive, $L^{\infty}(0,T,L^1(\Gamma))$ is also not reflexive and thus from the boundedness of $f_n$ I can't obtain a weak convergent subsequence.
  2. After that, I wondered if I could have a weak-* convergence so I thought the Banach-Alaoglu theorem. But again, this didn't work because I couldn't find the Banach and separable space whose dual is $L^{\infty}(0,T,L^1(\Gamma))$

2nd EDIT: I just came up with the following idea for which I need also verification:

Consider the space of continuous functions with compact support on $\Gamma$, i.e $C_c(\Gamma)$. Since $\Gamma$ is compact, we know that $C_c(\Gamma)$ is also Banach and separable. Its dual space is the space of (signed) Radon measures on $\Gamma$ with finite mass which is denoted by $\mathcal M(\Gamma)$.

If $L^{\infty}(0,T,\mathcal M(\Gamma))$ is contained in the dual space of $L^1(0,T,\mathcal C_c(\Gamma))$ then by Banach-Alaoglu theorem a weak-* convergent subsequence is obtained.

However I'm not completely sure if the duality argument that I used holds.

At this point I've been stuck. I would really appreciate any help or even hints.

Thanks in advance!

kaithkolesidou
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  • If it were $L^\infty(\Omega)$, where $\Omega$ is a $\sigma$-finite measure space, then $L^\infty$-boundedness would give you the existence of a subsequence that converges weakly-$\star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter: – Giuseppe Negro Jan 23 '19 at 12:24
  • @GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion? – kaithkolesidou Jan 23 '19 at 12:26
  • https://math.stackexchange.com/a/1717522/8157 – Giuseppe Negro Jan 23 '19 at 12:27
  • I think that what you did is OK. – Giuseppe Negro Jan 23 '19 at 12:28
  • @GiuseppeNegro My only doubt is if $\mathcal M(\Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot – kaithkolesidou Jan 23 '19 at 12:32
  • Unfortunately, $\mathcal{M}(\Gamma)$ is not separable. The Dirac measures form an uncountable family with $|\delta_x-\delta_y|=1$. – MaoWao Jan 23 '19 at 13:08
  • @MaoWao Provided that $\Gamma$ is compact and for positive measures, I think it holds... – kaithkolesidou Jan 23 '19 at 13:10
  • No. If there exists an uncountable family $(x_i)_{i\in I}$ in $X$ and $\delta>0$ such that $|x_i-x_j|\geq\delta$ for $i\neq j$, then $X$ cannot be separable. In $\mathcal{M}(\Gamma)$ the Dirac measures form such a family (unless $\Gamma$ is $0$-dimensional). – MaoWao Jan 23 '19 at 13:15
  • @MaoWao https://math.stackexchange.com/questions/71150/separability-of-the-set-of-positive-measures – kaithkolesidou Jan 23 '19 at 13:19
  • The answer in the link talks about the vague topology, while I was talking about the norm topology (which is the relevant topology for the duality $L^1(\Omega;X)^\prime\cong L^\infty(\Omega,X^\prime)$. – MaoWao Jan 23 '19 at 13:23
  • @MaoWao oh! you're right...thanks! is there any way to fix this duality argument? – kaithkolesidou Jan 23 '19 at 13:25
  • @MaoWao Ok, is $C_c(\Gamma)$ reflexive, right? In that case the duality argument would also hold – kaithkolesidou Jan 23 '19 at 13:32
  • No, $C(\Gamma)$ is not reflexive. The dual of a separable reflexive space is separable. In fact, I don't think the natural map $L^\infty(\Omega;\mathcal{M}(\Gamma))\to L^1(\Omega;C(\Gamma))^\prime$ is surjective. But it is probably better to post this as a separate question. – MaoWao Jan 23 '19 at 13:38
  • @MaoWao Ok, I think I won't insist more on this case...but one last question: even if $\Gamma$ is a 2-dimensional compact manifold, the space $C_c(\Gamma)$ is still no reflexive? – kaithkolesidou Jan 23 '19 at 14:19
  • No. Also $\mathcal{M}(\Gamma)$ does not have the Radon Nikodym property. – MaoWao Jan 23 '19 at 14:24
  • @MaoWao Fine...I stop! Thanks a lot for your time – kaithkolesidou Jan 23 '19 at 14:26

1 Answers1

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The dual space of $L^1(0,T; C(\Gamma))$ is $L^\infty_w(0,T;\mathcal M(\Gamma))$ and this space consists of weak-$*$ measurable functions, see, e.g., Theorem 10.1.16 in "Handbook of applied analysis" by Papageorgiou and Kyritsi-Yiallourou.

gerw
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