Consider the following famous theorem by Robert C. James (1964):

Let $X$ be a Banach space over $\mathbb R$ and $C$ a non-empty, bounded, weakly closed subset. Then, $C$ is weakly compact if and only if every continuous linear real-valued functional on $X$ attains its supremum on $C$.

The “only if” part is trivial, while known proofs of the “if” part are notoriously involved. Some texts prove it only for the case in which $X$ is assumed to be separable (see, for example, Holmes, 1975, pp. 157–161). Some sources do not make this extra assumption, yet they tend to be still too involved for my taste and patience (see, for example, Pryce, 1966).

Using Holmes (1975), who treats only the separable case, I can prove the following statement:

If $C\subseteq X$ is non-empty, bounded, weakly closed, but not weakly compact, then there exists a separable norm-closed subspace $Y\subseteq X$ and a continuous real-valued functional $f:Y\to\mathbb R$ on it such that $C\cap Y$ is not empty and $f$ doesn't attain its supremum on $C\cap Y$.

My question is: Do you think the second statement can be used to prove the existence of a continuous real-valued functional $F:X\to\mathbb R$ that doesn't attain its supremum on $C$? Do you think trying extending $f$ onto the whole space by using the Hahn–Banach theorem, or Zorn's lemma, could lead to anything useful?


  • James, R. C. (1964): “Weakly Compact Sets,” Transactions of the American Mathematical Society 113, 129–140.
  • Holmes, R. B. (1975): Geometric Functional Analysis and Its Applications, New York: Springer-Verlag.
  • Pryce, J. D. (1966): “Weak Compactness in Locally Convex Spaces,” Proceedings of the American Mathematical Society 17, 148–155.
  • $\begingroup$ I have not thought about this precisely, but can't you use the Eberlein-Smulian theorem to reduce the general case to the separable case? $\endgroup$ – Yemon Choi Jan 17 '15 at 17:30
  • $\begingroup$ @YemonChoi You mean, the theorem stating that weak compactness and weak sequential compactness are equivalent in norm vector spaces? How do you think that could work? $\endgroup$ – triple_sec Jan 17 '15 at 17:35
  • $\begingroup$ Also posted on Math.SE: math.stackexchange.com/questions/1108029/… $\endgroup$ – Nate Eldredge Jan 17 '15 at 17:42
  • $\begingroup$ Hmm, I guess I was too hasty - I think my idea merely ends up giving the same statement that you give in your question (about a separable closed subspace $Y$, etc) $\endgroup$ – Yemon Choi Jan 17 '15 at 17:44

I got it. I just read Pryce (1966), which is more accessible than I had thought. The proofs by Pryce (1966) and by Holmes (1975) use almost identical ideas, it's just that the latter refrains from dealing with the non-separable case for some reason.

Both proofs start out with assuming that $C$ is non-empty, weakly closed, bounded, but not weakly compact. In this case, there exist sequences $\{x_n\}_{n\in\mathbb N}\subseteq C$ and $\{f_m\}_{m\in\mathbb N}\subseteq X'$ with the amusingly pathological property that $\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$ and $\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)$ both exist in $\mathbb R$ but they are not equal.

Holmes (1975) actually constructs these sequences in a transparent way, but Pryce (1966) does not, just gives a quite obscure reference. Pryce (1966) claims also that the family $\{f_m\}_{m\in\mathbb N}$ is equicontinuous, while the construction of Holmes (1975) states that this sequence is in the closed unit ball of $X'$. What I was intimidated about is that I hadn't recognized immediately the fact that any non-empty family of functionals in the closed unit ball of $X'$ is actually equicontinuous.

However, once I recognized this, I could delve into the proof by Pryce (1966) with more confidence and enthusiasm. By the way, the trick Pryce (1966) uses is to consider the separable (with respect to a certain seminorm-topology) subspace of $X'$ given by the linear span of $\{f_m\}_{m\in\mathbb N}$, while Holmes (1975) uses the assumption that $X$ is separable to ensure that the closed unit ball in $X'$ be metrizable in the weak* topology. The point in both cases is to ensure the existence of a subsequence of $\{f_m\}_{m\in\mathbb N}$ with certain properties.

I don't really understand why Holmes (1975) makes the separability assumption—certainly restrictive for a textbook that is supposed to make the proofs in published papers more accessible. The trick by Pryce (1966), ingenious as it is, is actually quite elementary and does not require the separability assumption.

At any rate, I suggest anyone interested in an accessible proof of James' theorem read first the section in Holmes (1975) that constructs the pair of sequences mentioned above, and then immediately switch to Pryce (1966) and put those sequences into action. The end result will be the construction of a linear functional in $X'$ that fails to attain its supremum on $C$!

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    $\begingroup$ When I have taught James' theorem I have cheated by only proving the separable version (since in applications that is usually all that is needed and the non separable version looked to me too difficult and time consuming). But I'll rethink that in light of your comment. $\endgroup$ – Bill Johnson Jan 18 '15 at 15:10
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    $\begingroup$ Why is an economist interested in the non separable case of James' theorem? $\endgroup$ – Bill Johnson Jan 18 '15 at 15:11
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    $\begingroup$ @BillJohnson http://en.wikipedia.org/wiki/Physics_envy $\endgroup$ – triple_sec Jan 18 '15 at 16:11

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