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So the Banach-Alaoglu theorem states:

Let $X$ be the dual space to some Banach separable space $Z$, i.e $X=Z^*$. Take $M$ a bounded subset of $X$. Then any sequence in $M$ has a weak-* convergent subsequence.

My question is:

If we know that $Y$ is contained in the dual space of some Banach separable space $Z$, then if $M$ is a bounded subset of $Y$, is there still a weak-* convergent subsequence in $Y$?

The reason I ask is: Let $\Gamma$ be a 2-dimensional compact manifold. I know that the Lebesgue-Bochner space $L^1(0,T,C_c(\Gamma))$ is Banach and separable. It is also known that the dual space of $C_c(\Gamma)$ is the space of Radon measures on $\Gamma$ with finite mass, denoted by $\mathcal M(\Gamma)$. However since $\mathcal M(\Gamma)$ is not separable we can not claim that the dual space of $L^1(0,T, C_c(\Gamma))$ is $L^{\infty}(0,T,\mathcal M(\Gamma))$. But we do know that $L^1(0,T, C_c(\Gamma))^*$ contains always $L^{\infty}(0,T,\mathcal M(\Gamma))$. I have a sequence of functions bounded in $L^{\infty}(0,T,L^1(\Gamma))$.

So I wonder if $Y=L^{\infty}(0,T,\mathcal M(\Gamma))$, $Z=L^1(0,T, C_c(\Gamma))$ and $M=L^{\infty}(0,T,L^1(\Gamma))$, can I deduce a weak-* convergent subsequence then?

It seems true to me but after a lot hours of studying, I would like if somebody else could verify this.

Moreover, if this is not valid, then what can we say in this case? Is there any other theorem more suitable for that case that I miss at the moment?

Any help is much appreciated.

Thanks in advance!

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  • $\begingroup$ What is $Y$? A vector subspace? Could you please specify? $\endgroup$ – pitariver Jan 23 at 16:44
  • $\begingroup$ @pitariver I just edited. Hope it's more clear now! $\endgroup$ – kaithkolesidou Jan 23 at 17:27
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Well, if $(y_n)$ is a bounded sequence in $Y$, you can say that there is a subsequence which weak*-converges to an element of $Z^*$. But if $Y$ is not weak*-closed in $Z^*$, there is no reason that the limit should be in $Y$. For instance, if $(y_n)$ itself weak*-converges to a point of $Z^*$ which is not in $Y$, then clearly no subsequence can converge to a point of $Y$. For an explicit example, you could take $Z=L^1([0,1])$ and $Y=C([0,1])\subset L^\infty([0,1])=Z^*$. You can then pick a sequence of bounded continuous functions which converge almost everywhere to a discontinuous function like $1_{[0,1/2]}$, and such a sequence will not have any subsequence which weak*-conveges in $Y$.

In your case, $L^{\infty}(0,T,\mathcal M(\Gamma))$ is weak*-dense in $L^1(0,T, C_c(\Gamma))^*$ so it is not weak*-closed. It follows (by Krein-Smulian) that there is a bounded sequence in $L^{\infty}(0,T,\mathcal M(\Gamma))$ which weak*-converges to an element of $L^1(0,T, C_c(\Gamma))^*$ which is not in $L^{\infty}(0,T,\mathcal M(\Gamma))$. Such a sequence will have no weak*-convergent subsequence within $L^{\infty}(0,T,\mathcal M(\Gamma))$.

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  • $\begingroup$ Thanks a lot for your detailed answer! $\endgroup$ – kaithkolesidou Jan 24 at 10:12

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