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Bonsoir/bonjour à toutes et à tous.

The title has it all, but... We know (as a consequence of the Eberlein-Šmulian theorem) that any bounded sequence, $\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$, in a (real or complex) reflexive normed (and hence Banach) space, $\mathbf{X} \equiv (X, \|\cdot\|)$, contains a weakly convergent subsequence. Now, drop the assumption that $\mathbf{X}$ is complete and strengthen the hypotheses on $\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$ by replacing boundedness with cauchyness. Then the question comes:

Question. What are sufficient conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (real or complex) normed space (which is not supposed to be complete)? Of course, let's rule out trivially tautological conditions such as "the existence of a weakly convergent subsequence", "the (strong) convergence of the sequence", ...

I'm aware that the question may sound a little weird at face value - especially looking at the case in which $\mathbf{X}$ is a dense proper (normed) subspace of a Banach space, $\mathbf{Y}$, and $\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$ is a sequence in $X$ which is convergent in $\mathbf{Y}$ but not in $\mathbf{X}$. Nevertheless, the particular problem (*) on which I am working, implies additional conditions on $\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$ that may still force, as I hope, the existence of a weakly convergent subsequence - and hence, in my very particular case, the (strong) convergence of the sequence -, and this is basically why I am posing the question.


(*) Let me give some details on the problem, in the case that they may be useful to know. These include the existence of a bounded linear transformation, $T: \mathbf{X} \to \mathbf{X}$, such that

  1. $\{Tx_n\}_{n\;\! \in \;\! \mathbb{N}}$ is (strongly) convergent to some $y \in X$;
  2. for all $\varphi \in X^\prime$, there is a unique $\psi \in X^\prime$ for which $\varphi = \psi \circ T$.

Here, as you can guess, $X^\prime$ is the continuous dual of $\mathbf{X}$. Also, equipping $X^\prime$ with its usual norm, $\|\cdot\|_{X^\prime}: X^\prime \to \mathbb{R}: \varphi \mapsto \sup_{x \;\! \in \;\! X, \|x\| \;\! \le \;\! 1} |\varphi(x)|$, and letting $\mathbf{X}^\prime \equiv (X^\prime, \|\cdot\|_{X^\prime})$, the function, $\Psi: X^\prime \to X^\prime$, mapping any given $\varphi \in X^\prime$ to the unique $\psi \in X^\prime$ such that condition 2 above is satisfied, is actually a bi-Lipschitz isomorphism $\mathbf{X}^\prime \to \mathbf{X}^\prime$, so $\lim_n \varphi(x_n) = \Psi(\varphi)(y)$ for all $\varphi \in X^\prime$.

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2 Answers

up vote 1 down vote accepted

Your condition (2) is that $T^*$ is a surjective isomorphism, so $T$ induces a surjective isomorphism on the completion of $X$. For a counterexample, let $T$ be the right shift on $\ell_2(Z)$ restricted to an appropriate dense subspace. What subspace? Well, it must be dense, so throw in the unit vector basis. Throw in some natural vector $y$ which you want to be the limit of $Tx_n$; $y=\sum_{k=1}^\infty 2^{-k!} e_k$ should be fine. You need for $T$ to map the subspace back into itself, so throw in $T^k y$ for $k=1,2,...$. Let $X$ be the linear span of all the vectors thrown in. Set $x_n= \sum_{k=0}^n 2^{-(k+1)!} e_k$. Then $x_n$ converges to a point not in $X$ but $Tx_n$ converges to $y$.

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So nice, I would say! Thank you, BJ, this is very useful; though I had hoped for a positive answer, you just let me realise something that I had completely overlooked. I know, I know, I shouldn't fall in love with an idea, and continue forgetting the golden rule: when you can't prove something, break it down and try to seek a counterexample - in the worst case, you will get further insight on your problem. Thanks again! –  Salvo Tringali Sep 14 '11 at 22:16
    
For the sake of curiosity, aren't all of your isomorphisms already surjective by definition? :) –  Salvo Tringali Sep 14 '11 at 22:23
    
In normed space theory, "isomorphism" usually means "linear homeomorphism" into the target space. –  Bill Johnson Sep 14 '11 at 23:27
    
This is why I asked, I guess. It is all clear now. Thanks! –  Salvo Tringali Sep 15 '11 at 7:21
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I don't understand your question. If a norm Cauchy sequence has a weak cluster point, then it norm converges to the weak cluster point. Put another way, a norm Cauchy sequence is weakly Cauchy and hence weak* converges in the bidual to something; if that something is back in the space, the sequence is weakly convergent to it and, since the sequence is norm Cauchy, must norm converge to it (since e.g. tails of the sequence are contained in arbitrarily small balls and the weak limit must be in the closed convex hull of every tail of the sequence).

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@BJ. In fact, I'd like to prove that a certain Cauchy sequence (in a possibly incomplete normed space) is norm convergent. I just know that the sequence satisfies a few conditions (as given in the OP). I'm wondering if all of them together imply some well-known (but still not so well-known to me) sufficient condition for the existence of a weakly convergent subsequence, indeed whether or not all of them are sufficient to establish the existence of a weakly convergent subsequence (which would be enough). Also, I see by now that I phrased the question in the worst of all possible ways... –  Salvo Tringali Sep 13 '11 at 18:10
    
I'm especially looking at the existence of a weakly convergence subsequence since it seems to me the most natural way to go - as the conditions in the problem are stated in terms of the functionals in the (continuous) dual space. Does this (make more sense and) somehow clarify the question? –  Salvo Tringali Sep 13 '11 at 18:23
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