Proof of Eberlein–Smulian Theorem for a reflexive Banach spaces

Question

Looking for the proof of Eberlein-Smulian Theorem.

Searching for the proof is what I break with this morning. Some of my friends recommend Haim Brezis (Functional Analysis, Sobolev Spaces and Partial Differential Equations). After I search the book, I only found the statement of the theorem, is the proof very difficult to grasp? Why is Haim Brezis skip it in his book?

Please I need a reference where I can find the proof in detail.

---

Theorem:(Eberlein-Smul'yan Theorem) A Banach space $E$ is reflexive if and only if every (norm) bounded sequence in $E$ has a subsequence which converges weakly to an element of $E$.

---

Answer 1

• I think Megginson's book An Introduction to Banach Space Theory (GTM 183) is worth having look at. You can preview some parts of the book at Google Books. (I believe the whole section 2.8 Weak Compactness might be interesting for you.)

• Albiac–Kalton, Topics in Banach Space Theory (GTM 233), Corollary 1.6.4, page 24.

• Whitley, An elementary proof of the Eberlein–Šmulian theorem, Mathematische Annalen 172 (2), 1967, 116–118. Freely available from GDZ.

• Zeidler, Nonlinear functional analysis and its applications. II/A: Linear monotone operators, Springer, 1990, Theorem 21.D

• Theo Buhler, Dietmar A. Salamon, Functional Analysis (GSM 191), Theorem 3.4.1, page 134

I made this answer CW, so that other people can add further references if they think it's suitable.

Answer 2

Kôsaku Yosida, Functional Analysis, Springer 1980, Chapter V, Appendix, section 4. (This appears to be the 6th edition).

Answer 3

This is a direct consequence of Banach-Alaoglu's theorem.

First note that if $X$ is reflexive then so is $X^{\ast}.$ Hence weak and weak$^{\ast}$-topology on $X^{\ast \ast}$ coincide. So by Banach-Alaoglu's theorem it follows that any bounded closed ball in $X^{\ast \ast}$ is weakly compact. Consequently any bounded sequence in $X^{\ast \ast}$ has a weakly convergent subsequence. Now since $X$ is reflexive, it follows that $X$ is isometrically isomorphic to $X^{\ast \ast}$ and hence any bounded sequence in $X$ has a weakly convergent subsequence as well.