# Sufficient conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (merely) normed space

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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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