Weak closedness in $L^p$ spaces

Question

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 $L^3[0,1]$ ?

Answer 1

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).

Answer 2

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 $\|f\|_{\infty} \leq 1$.