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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy