I'm a bit confuse with all theses notions. Let $E$ a normed vector space of infinite dimension (also Banach, but it's probably not important).

The theorem of Eberlin Smulian theorem says that : all bounded sequence that has a subsequence that converge weakly $\iff$ it's reflexive.

(In fact it just says implication, but the converse is also true)... anyway.

Q1) Does it mean that if $E$ is reflexive, then instead of the fact that the weak topology is not metrizable, the property

$C\subset E$ is compact $\iff$ $C$ is sequentially compact

hold ? Because in reflexive spaces, $\{x\in E\mid \|x\|\leq 1\}$ is compact. Eberlin Smulian theorem says that

$\{x\in E\mid \|x\|\leq 1\}$ is compact $\iff$ it's sequentially compact.

Can this be generalized for any compact $C$ ?

Q2) If a set is separable, we know that $\{x\in E\mid \|x\|\leq 1\}$ is metrizable for the weak topology. In particular, can we conclude from this that

If $E$ is separable, a set $C\subset E$ is compact $\iff$ it's sequentially compact.


1 Answer 1


Eberlein -Smulian is usually stated in the form:

Let $A$ be a subset of a Banach space $X$. Then, $A$ is weakly compact if and only if $A$ is weakly sequentially compact.

The form you gave is a consequence of this, as the unit ball is weakly compact if and only if the space is reflexive. But, since the form you gave says absolutely nothing about non-reflexive spaces, it is actually weaker than E-S.

Q1) The answer is true in general, by the (stronger) ES, and it does not require reflexivity.

Note that, if I remember right, the key for the proof is the following:

If $E$ is a separable Banach space, then $E*$ contains a countable total set. In this case, the weak topology becomes metrisable on weakly compact sets.

Now if $x_n$ is any sequence, the subspace spanned by $\{ x_n \}$ is separable, and the trick above shows that weak-compactness implies weak sequential compactness.

The other implication is a bit trickier, it usually is done by showing that if $A$ is weakly sequentially compact it is bounded and the weak* closure of $A$ is included in $E$.

  • $\begingroup$ Thank you for your answer. At the end, could we say that in a reflexive space, that set that are closed and bounded for strong topology are compact in weak topology and conversely ? It make sense ? $\endgroup$
    – Peter
    Commented Aug 30, 2018 at 22:13
  • $\begingroup$ @Peter Hopefully I won't say anything stupid: In a reflexive space the weak and weak* topology are equivalent, so bounded implies compact closure by Alaoglu Bourbacki. Now, with the closure it is trickier: weak closure implies closure and compactness implies strong compactness but the converses are not true in general. There is a theorem of Mazur though which says that for convex sets, closure is equivalent to weak closure, but this is true for convex sets. So you can apply this to balls for example. $\endgroup$
    – N. S.
    Commented Aug 30, 2018 at 22:26
  • $\begingroup$ I'm not sure to understand totally. Let $E$ a reflexive space. 1) Closed and bounded space (for the strong topology) are compact in the weak topology, that correct ? 2) Compacts in the weak topology are closed and bounded in the strong topology, that's correct? 3) I didn't get what you mean with the weak closure. And thank you :) $\endgroup$
    – Peter
    Commented Aug 31, 2018 at 8:46
  • $\begingroup$ Peter : I'm really not an expert, but I don't think that closed bounded in the strong topology is compact in the weak topology. Simply because a closed in the strong topology is not closed in the weak topology. But since convex and closed in strong topology are also closed in the weak topology, I guess that a closed, bounded and convex in the strong topology will be compact in the weak topology... $\endgroup$
    – Surb
    Commented Aug 31, 2018 at 9:52
  • $\begingroup$ @N.S.: Does Eberlein Smulian also work for the weak-* topology ? $\endgroup$
    – Peter
    Commented Sep 1, 2018 at 8:50

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