I got stuck on this problem about weak convergence in normed space. This problem is exercise 9, page 263 in Functional Analysis of Erwin Kreyszig. So the question is basically on the title, let me state it clearly here:

Let $A$ be a set in a normed space $X$ such that every nonempty subset of $A$ contains a weak Cauchy sequence. Show that $A$ is bounded.

Here, a sequence $(x_n)$ is a weak Cauchy sequence if for every $f' \in X'$, the sequence $(f(x_n))$ is Cauchy in $\mathbb{R}$ or $\mathbb{C}$ ($X'$ is dual space of $X$: space of all bounded linear operators on $X$). I have no idea how to solve this problem, so I hope someone can give me a hint. Thanks so much.

  • $\begingroup$ do they mean any subset containing an infinite number of elements, then we can create a weak Cauchy sequence from the elements of the set, without repeating any of the element ? in that case it is obvious that an unbounded set contains a sequence that goes to $\infty$ in norm $\|.\|_{X}$, but some of those sequences can converge weakly, for example in a Hilbert space : $x_n = n^{1/4} e_n$ where $(e_n)$ is an orthonormal basis, then $x_n \overset{weakly}\to 0$ $\endgroup$ – reuns May 24 '16 at 17:11
  • $\begingroup$ I think he just means "nonempty", not necessarily "infinite"...But is it important, I asked because the condition is for "every" nonempty subset? $\endgroup$ – le duc quang May 24 '16 at 17:12
  • $\begingroup$ if the sequence is constant then it converges weakly obviously.. $\endgroup$ – reuns May 24 '16 at 17:15
  • $\begingroup$ @user1952009 There must have been a typo in your last comment - $(e_n)$ is certainly bounded. And no, $n^{1/4}e_n$ does not tend weakly to $0$. $\endgroup$ – David C. Ullrich May 24 '16 at 17:26
  • $\begingroup$ @user1952009 (Replying to a version of your comment with typos fixed:) No, $\sum|c_n|^2<\infty$ does not imply that $|c_n|\le Cn^{-1/2}$. $\endgroup$ – David C. Ullrich May 24 '16 at 17:35

If $x_n\in X$ and $||x_n||$ is unbounded then the Banach-Steinhaus Theorem, also known as the Uniform Boundedness Principle, shows that there exists $\Lambda\in X^*$ such that $\Lambda x_n$ is unbounded. So there is a subsequence $x_{n_j}$ with $|\Lambda(x_{n_j}|\to\infty$; that subsequence has no weakly Cauchy subsequence.

  • $\begingroup$ you use the uniform boundedness principle that saying that if $A$ is unbounded then $\sup_{x \in A} |y(x)|$ is unbounded for some $y \in X^*$, and hence any unbounded sequence doesn't converge weakly ? $\endgroup$ – reuns May 24 '16 at 17:45
  • $\begingroup$ Can I ask why you need to introduce complete of $X$ here? Sorry but I don't see where you use the completeness. $\endgroup$ – le duc quang May 24 '16 at 17:56
  • $\begingroup$ @user1952009 I don't quite see how to answer that except by what I wrote. A question that refers more spcifically to something I wrote might work better $\endgroup$ – David C. Ullrich May 24 '16 at 17:58
  • $\begingroup$ @leducquang No reason at all - I was thinking a tiny bit backwards. In fact $X^*$ is complete and $X$ is a collection of bounded linear fucntionals on $X^*$, which is all we use. Thanks. $\endgroup$ – David C. Ullrich May 24 '16 at 18:00
  • $\begingroup$ @DavidC.Ullrich: Thanks, but your previous version has a line that interests me. You said that $\overline{X}^* = X^*$. Is it true for all normed space $X$? So we have that if $f$ is a bounded linear functional on $X$, it is a bounded linear functional on $\overline{X}$, right? $\endgroup$ – le duc quang May 24 '16 at 18:03

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