7
$\begingroup$

A Banach space $H$ is said to have Schur's property if weak convergence of a sequence implies converge in norm. The most famous example of such a space is $\ell^1(\mathbb N)$, while $L^1[0,1]$ does not have this property.

My question is the following:

Is there a characterization of such spaces?

Is there a list of known examples, other than examples of the type $\ell^1(X)$?

Is it true that such space are not reflexive, when they are infinite dimensional?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
14
$\begingroup$

Rosenthal's $\ell_1$ theorem (Google) says that every bounded sequence in a Banach space contains a subsequence that is either weakly Cauchy or is equivalent to the unit vector basis of $\ell_1$. From this you get that a Banach space has the Schur property iff for every $\epsilon > 0$, every $\epsilon$ separated bounded sequence has a subsequence that is equivalent to the unit vector basis of $\ell_1$. That answers your first question.

The answer to the third question is, obviously, yes. Unit balls of reflexive spaces are weakly sequentially compact by the Eberlein-Smulian theorem.

As for the second question, there are many examples, but because of the characterization above, all are in some sense constructed from $\ell_1$. What are you looking for?

Improve this answer
$\endgroup$
0
3
$\begingroup$

(Turning my comment into an answer here):

Regarding the third question: Yes, reflexive spaces with the Schur property need to be finite-dimensional. To see that, two ingredients suffice, once you notice that the Schur property can be phrased as "weakly sequentially compact sets are sequentially compact":

  • You need to show that in an infinite-dimensional space $X$, the unit ball is never (sequentially) compact. This follows from Riesz's lemma, which guarantees the existence of a sequence $(x_n) \in X^{\mathbb N}$ with $|x_n - x_m| \ge \frac 12$ for $n \ne m$.
  • You need to show that the unit ball in a reflexive space is weakly sequentially compact.

The proofs for both claims are elementary, unlike the Eberlein–Šmulian theorem; they can be found e.g. (in German, I'm afraid) in

Werner, Dirk. Funktionalanalysis. (German) [Functional analysis] Third, revised and extended edition. Springer-Verlag, Berlin, 2000. xii+501 pp. ISBN: 3-540-67645-7 MR1787146

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ That weakly compact sets are weakly sequentially compact is the easy direction in the Eberlein-Smulian Theorem. You just observe that a weakly compact separable set lives in a separable subspace and hence (this is easy) is weakly metrizable. $\endgroup$
    – Bill Johnson
    Dec 2, 2016 at 20:52
  • $\begingroup$ I agree that this is the easier direction the Eberlein–Šmulian theorem. $\endgroup$
    – anonymous
    Dec 3, 2016 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.