It's known that $L^1[0,1]\supset L^2[0,1]\supset L^3[0,1]\supset\cdots\supset L^\infty[0,1]$. How to show without using axiom of choice (nor Hahn-Banach Theorem) to show the unit ball in $L^\infty[0,1]$ is weakly closed in $L^3[0,1]$ ?
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Can you be more specific about what you mean by "weak-closedness"? – Lentes Sep 05 '17 at 04:34
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1I assume it means that the unit ball in $L^{\infty}([0, 1])$ as a subset of $L^3([0, 1])$ is closed in the weak topology on $L^3([0, 1])$. – Michael L. Sep 05 '17 at 04:35
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Yes. More precisely, any sequence $ – user74489 Sep 05 '17 at 04:42
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Thanks for your proof (suddenly deleted). I tried to prove it without use of Riesz representation theorm, so the use of Kakutani Theorem is a way. – user74489 Sep 05 '17 at 05:07
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Kakutani relies on Banach-Alaoglu, which is equivalent to the axiom of choice. I've undeleted the answer and can leave it, if you'd like. – Michael L. Sep 05 '17 at 05:09
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I also missed this. By the way, is Banach-Alaoglu Theorem equivalent to AC , or only Tyconoff's theorem, as you wrote below ? – user74489 Sep 05 '17 at 05:25
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1After doing a bit more reading, I've found that technically, Banach-Alaoglu is equivalent in ZF to the ultrafilter lemma, which is strictly weaker than the axiom of choice. – Michael L. Sep 05 '17 at 05:39
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Let's do this for general $L^p([0, 1])$, $1\leq p < \infty$. First, for $f\in L^{\infty}([0, 1])$, $$\|f\|_{L^p([0, 1])}\leq \|f\|_{L^{\infty}([0, 1])}$$ as an application of Hölder's inequality. Therefore, if $f_n\to f$ in $L^{\infty}([0, 1])$, then $f_n\to f$ in $L^p([0, 1])$, so closed subsets of $L^{\infty}([0, 1])$ are closed as subsets of $L^p([0, 1])$. Furthermore, the closed unit ball in $L^{\infty}([0, 1])$ is a subset of the closed unit ball of $L^p([0, 1])$ by the above inequality. As (1) the closed unit ball of $L^p([0, 1])$ is weakly compact by Kakutani's theorem, (2) strong closure is equivalent to weak closure for convex subsets of a Banach space, and (3) a closed subset of a compact set is compact in any topology, we have that the closed unit ball of $L^{\infty}([0, 1])$ is weakly compact in $L^p([0, 1])$. This is a stronger implication than the weak closedness of the closed unit ball of $L^{\infty}([0, 1])$ in $L^p([0, 1])$, as $L^p([0, 1])$ is Hausdorff in the weak topology (and therefore weakly compact sets are weakly closed).
Michael L.
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The proof of Alaoglu's theorem that you cite uses Tyconoff's theorem, which is known to be equivalent to the axiom of choice. – shalin Sep 05 '17 at 05:02
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Ah, true, good catch. I figured this was too easy to work without the AC hiding somewhere behind. – Michael L. Sep 05 '17 at 05:05
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Suppose that $f,f_n:[0,1] \to \Bbb R$ are a collection of measurable functions such that $\|f_n\|_{\infty} \leq 1$ for all $n$, and such that $\int_0^1 f_ng \stackrel{n \to \infty} {\longrightarrow} \int_0^1 fg$ for every $g \in L^{3/2}[0,1]$ (which means $f_n \to f$ weakly in $L^3$). We need to show that $\|f\|_{\infty} \leq 1$.
For positive functions $g \in L^{3/2}$, we clearly have $\int_0^1 fg \leq \int_0^1 g$, since $\int f_ng \leq \int |f_n|g \leq \int g$ for every $n$.
In particular, letting $g=1_A$, we get that $\int_A f \leq m(A)$ for every Borel set $A$ (here $m$ denotes Lebesgue measure).
Now for contradiction, assume $\|f\|_{\infty} > 1$. That means that $m(\{x:f(x)>1)+m(\{x:f(x)<-1\})>0$, so wlog, let's say that $m(\{x:f(x)>1\})>0$. Then letting $A:=\{x:f(x)>1\}$, we find that $\int_Af > \int_A 1=m(A)$, which contradicts the result of the preceding paragraph. Thus we conclude that $\|f\|_{\infty} \leq 1$.
shalin
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One should realize that this answer only shows the weak sequential closedness. – gerw Sep 05 '17 at 07:02
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@gerw Yes, you are right. My mistake. So, is the weak topology not first-countable? – shalin Sep 05 '17 at 07:07
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@gerw Actually, unless I'm mistaken one may show without using AC that the weak topology on $L^3$ is induced by the metric $$d(f,g) = \sum_n 2^{-n} \int_{[0,1]} |(f-g)h_n|$$ where ${h_n}$ is a dense subset of $L^{3/2}$. Thus it is first countable, so sequentially closed sets are closed. Let me know if that is incorrect, or if AC is somehow used. – shalin Sep 05 '17 at 07:27
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@Shalop: The weak topology is only metrizable on bounded subsets of $L^{3/2}$. However, the unit ball of $L^\infty$ is contained in a bounded subset of $L^{3/2}$, so it should be fine. – gerw Sep 05 '17 at 09:05
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