It is a well known result in functional analysis that a Banach space $X$ is reflexive if and only if the unit ball is weakly compact (compact in the weak topology). This result is also known as Kakutani's theorem. However so far to my knowledge all the proofs for this theorem use in a way or another the Banach-Alaoglu theorem. Is there a topological proof using the definition of open cover/finite subcover? Below I present such a 'proof' for at least in one direction of the statement and in this respect I would like to know whether:

- its 'proof' is mathematically correct
- does it appear elsewhere in the literature?

Theorem: Let $(X, \|\cdot\|)$ be a reflexive Banach space then the unit ball $\mathbb{B}:=\{x\in X| \|x\|\leqslant 1\}$ is weakly compact.

Proof: Suppose $(X,\|\cdot\|)$ is reflexive but $\mathbb{B}$ is not weakly compact. Note that the collection of open sets $U(f;\varepsilon):=\{x\in X| |f(x)|<\varepsilon\}$ where $\varepsilon>0$ and $f$ is a bounded linear functional on $X$ forms a subbasis $\mathcal{B}$ for the weak topology $T_W$. From Alexander subbasis theorem it follows that there is an open cover $\{U_i\}_{i\in I}$ of $\mathbb{B}$ in $\mathcal{B}$ which has no finite subcover. Let $(V_n)_{n\in\mathbb{N}}$ be a family of open sets ordered by inclusion, where $V_n:=\bigcup_{j=1}^nU_{i_j}$ and $\bigcup_{n\in\mathbb{N}}V_n=\bigcup_{i\in I}U_i$. Denote by $W_n:=\mathbb{B}\setminus V_n$ for each $n$. Since $\mathbb{B}$ is convex and closed then it is weakly closed; so $W_n$ is weakly closed for all $n$ and satisfies $W_{n+1}\subseteq W_n$. Moreover for each $n$ there exists $x_n\in W_n$. Since $\mathbb{B}$ is bounded then $(x_n)$ is bounded. Assumption $(X,\|\cdot\|)$ is reflexive implies that there is a subsequence $(x_{n_k})$ weakly converging to some element $x\in X$. Moreover $x\in \mathbb{B}$ since $\mathbb{B}$ is weakly closed. On the other hand by definition of $w-\limsup_nW_n$ (the set of all weak cluster points of sequences $(x_n)$ such that $x_n\in W_n$ for each $n$) the weak limit $w-\lim_kx_{n_k}=x$ lies in $w-\limsup_nW_n$. The chain inclusion $W_{n+1}\subseteq W_n$ implies $w-\limsup_nW_n=\bigcap_nW_n$ and so $x\in \bigcap_{n\in\mathbb{N}}W_n$. Then $\bigcap_{n\in\mathbb{N}}W_n\neq\emptyset$ together with $\bigcap_{n\in\mathbb{N}}W_n\subseteq \mathbb{B}$ yields that $$\mathbb{B}\supset \mathbb{B}\setminus \bigcap_{n\in\mathbb{N}}W_n=\mathbb{B}\setminus\bigcap_{n\in\mathbb{N}}(\mathbb{B}\setminus V_n)=\bigcup_{n\in\mathbb{N}}(\mathbb{B}\cap V_n)=\mathbb{B}$$ This would be impossible.

If these arguments are correct then they could apply in general to any bounded closed convex set.

notweakly compact. But in order to get a contradiction (or to prove the theorem) you have to show that itisweakly compact. Am I missing something? $\endgroup$ – Nate Eldredge Dec 5 '19 at 16:073more comments