I'm trying to prove Milman-Pettis's theorem.
Let $E$ be a uniformly convex Banach space. Then $E$ is reflexive.
Clearly, my attempt is not correct because I have not used the uniform convexity of $E$. Could you elaborate on where my logic fails?
Let denote by $E_w$ the set $E$ together with the weak topology $\sigma(E, E')$. Let denote $E''_w$ the set $E''$ together with the weak$^\star$ topology $\sigma(E'', E')$. Let $J:E_w \to E''_w, x \mapsto \hat x$ be the canonical injection. Then $J$ when restricted to its image is a homeomorphism which sends open/close/compact set to open/close/compact set.
By Kakutani's theorem, it's enough to prove that $B_E:= \{x\in E \mid |x| \le 1\}$ is compact in $\sigma(E, E')$. Because $J$ (restricted to its image) is a homeomorphism, it suffices to show that $J[B_E]$ is compact in the weak$^\star$ topology $\sigma(E'', E')$. By Banach–Alaoglu theorem, $B_{E''} := \{\varphi \in E'' \mid \| \varphi \| \le 1\}$ is compact in $\sigma(E'', E')$. Clearly, $J[B_E] \subseteq B_{E''}$, so it suffices to show that $J[B_E]$ is closed in $\sigma(E'', E')$. However, this is true because $B_E$ is closed in $\sigma(E, E')$ and $J$ (restricted to its image) is a homeomorphism.
Update: I added a proof that $J$ (restricted to its image) is a homeomorphism.
Let $F:= J[E]$. We denote by $F_w$ the set $F$ with the subspace topology $\tau$ that $\sigma(E'', E')$ induces on $F$. Let's prove that $J:E_w \to F_w$ is a homeomorphism. Clearly, $J$ is bijective.
Lemma: Let $(X, \tau)$ be a topological space, $A \subseteq X$, and $\tau_A$ the subspace topology of $A$. Let $a\in A$ and $(x_d)_{d \in D}$ is a net in $A$. Then $x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$.
Let $x,x_d\in E$. To prove that $J, J^{-1}$ are continuous, we need to prove that $x_d \to x$ in $\sigma(E, E')$ if and only if $\hat x_d \to \hat x$ in $\tau$. By our Lemma, we need to prove $x_d \to x$ in $\sigma(E, E')$ if and only if $\hat x_d \to \hat x$ in $\sigma(E'', E')$. This equivalence is indeed true because
-
$\hat x_d \to \hat x$ in $\tau$ if and only if $f(x_d) \to f(x)$ for all $f\in E'$.
-
$\hat x_d \to \hat x$ in $\sigma(E'', E')$ if and only if $\hat x_d (f) \to \hat x (f)$ for all $f\in E'$ if and only if $f(x_d) \to f(x)$ for all $f\in E'$.
Best Answer
Let $f\in X\to Y$ be a homeomorphism and $Y\subseteq Z$. If $S\subseteq X$ is closed, then $f(S)$ is closed in $Y$, since $f$ is a homeomorphism. But it need not be closed in $Z$. For example, take $X=Y=(-1,1)$, $f$ is the identity, and $Z=[-1,1]$; then $X$ is closed in $X$, but its image is not closed in $Z$.
You want to show $J(B_E)$ closed in $E_w''$, not in $\operatorname{im}{J}$. But you only show the latter.