4 fixed another typo, sorry!

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 $\varphi(x_n) \lim_n \mapsto varphi(x_n) = \Psi(\varphi)(y)$ for all $\varphi \in X^\prime$.

3 improved text

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 linear function, $\Psi: X^\prime \to X^\prime$, mapping any given $\varphi \in X^\prime$ to the unique $\psi \in X^\prime$ such that condition 1 2 above is satisfied, is actually a bi-Lipschitz isomorphism $\mathbf{X}^\prime \to \mathbf{X}^\prime$, so $\varphi(x_n) \mapsto \Psi(\varphi)(y)$ for all $\varphi \in X^\prime$.

2 punctualization

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 $\mathbf{X}^\prime \equiv (X^\prime, \|\cdot\|_{X^\prime})$, we are given that the linear function, $\Psi: X^\prime \to X^\prime$, mapping any given $\varphi \in X^\prime$ to the unique $\psi \in X^\prime$ such that condition 1 above is satisfied, is actually a bilipschitz bi-Lipschitz isomorphism $\mathbf{X}^\prime \to \mathbf{X}^\prime$, so $\varphi(x_n) \mapsto \Psi(\varphi)(y)$ for all $\varphi \in X^\prime$.

1