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?


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?

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(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

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  • 1
    $\begingroup$ 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 '16 at 20:52
  • $\begingroup$ I agree that this is the easier direction the Eberlein–Šmulian theorem. $\endgroup$ – anonymous Dec 3 '16 at 20:01

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