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 $f_n$


  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!

  • $\begingroup$ 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: $\endgroup$ – Giuseppe Negro Jan 23 at 12:24
  • $\begingroup$ @GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion? $\endgroup$ – kaithkolesidou Jan 23 at 12:26
  • $\begingroup$ math.stackexchange.com/a/1717522/8157 $\endgroup$ – Giuseppe Negro Jan 23 at 12:27
  • $\begingroup$ I think that what you did is OK. $\endgroup$ – Giuseppe Negro Jan 23 at 12:28
  • $\begingroup$ @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 $\endgroup$ – kaithkolesidou Jan 23 at 12:32

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.

  • $\begingroup$ Exactly what I was looking for! Thanks a lot for your help! $\endgroup$ – kaithkolesidou Jan 24 at 10:11

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