# Every sequence has a weakly convergent subsequence with limit

Let $\mathbf{E}$ be a reflexive space and $\mathbf{A ⊂ E}$ be bounded and weakly closed. Show that $\mathbf{A}$ is sequentially compact, i.e., every sequence in $\mathbf{A}$ has a weakly convergent subsequence with limit in $\mathbf{A}$. This statement is a special case of the Eberlein-Smulian Thoerem, which you are not allowed to use in this task.

Hint: First assume that $\mathbf{E}$ is separable.

I just can solve it with this Eberlein Smulian Theorem and I think my other second approach is wrong... Has anybody a solution with an explication, why I should assume, that $\mathbf{E}$ is separable.

I also saw a proof of this here Every bounded sequence has a weakly convergent subsequence in a Hilbert space , but there they start with a Hilbert space? Hilbert space is reflexive but not the other way. Or is there no connection?

I think that the argument in the linked post should filter through fine (in fact, it seems constructed for general reflexive spaces).

A reflexive Banach space is separable if and only if its dual is. (The standard result is that $X^*$ separable implies $X$ separable.) Fix a countable dense subsequence $\{\phi_1,\phi_2,\ldots\}\subseteq X^*$, and let $\{a_1,a_2,\ldots\}\subseteq A$ be any sequence. Let them all be bounded by $M$ in norm, since $A$ was bounded. Consider $\{\phi_1(a_1),\phi_2(a_2),\ldots\}\subseteq \mathbb{K}$. This is a bounded sequence in a locally compact space (bounded because all of them have norm less than or equal to $\|\phi_1\|\cdot M$), so it has a convergent subsequence $\phi_1(a_{11}),\phi_2(a_{12}),\ldots$. Now for each $k+1$, similarly extract a convergent subsequence from $\phi_{k+1}(a_{1k}),\phi_{k_1}(a_{2k}),\ldots$. Do a standard diagonal argument and $a_{11},a_{22},\ldots$ is a sequence for which each functional $\phi_i$ converges. Moreover, because the $\phi_i$ were dense, for each $\phi\in X^*$, $\phi(a_{11}),\phi(a_{22}),\ldots$ is convergent. Define the functional $a$ on $X^*$ by $a(\phi)=\lim\limits_{n\to\infty} \phi(a_{nn})$. This is continuous (needs checking) and therefore by reflexivity is represented by some element $x\in X$. Then the sequence $\{a_{nn}\}\xrightarrow{w} x$, and because $A$ was $w$-closed, $x$ is in $A$.

For general spaces, fix your sequence $\{a_1,\ldots\}$ and let $Y$ be its closed linear span. Then $Y$ is separable, and it is a closed subspace of a reflexive space, so itself reflexive (A closed subspace of a reflexive Banach space is reflexive) and the dual is separable. But now we're in the old case, so we extract $x$ with $y^*(a_n)\rightarrow y^*(x)$ for each $y^*\in Y^*$. Now this convergence holds for any linear functional in $X^*$, because the restriction to $Y$ is a linear functional on $Y$.

Indeed, if $X$ is reflexive, then $\overline{B_1(0)} \subseteq X$ is weakly sequentially compact. The proof begins as follows:

Let $(x_n)_n \subseteq \overline{B_1(0)}$ be a sequence and define $Y := \overline{\mathrm{span}(\{x_k | k \in \mathbb{N}\})}$. Then:

• $Y$ is separable: you can approximate every finite linear combination by the same finite linear combination, but with coefficients from $\mathbb{Q}$ (or $\mathbb{Q} \times \mathbb{Q}$ when working over $\mathbb{C}$).

• $Y$ is reflexive: every closed linear subspace of reflexive spaces is reflexive as well.

Now apply the theorem that $\overline{B_1(0)} \subseteq X'$ is weak-$*$-ly sequentially compact.

I think that's why you were supposed to first consider $E$ to be separable.

In all cases, your claim directly follows now using the weak closedness assumption you have.