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 


*

*$\{Tx_n\}_{n\;\! \in \;\! \mathbb{N}}$ is (strongly) convergent to some $y \in X$;

*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 $\varphi \in X^\prime$.
 A: 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).
A: 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 $y$.
