Let $(X,\|.\|)$ be reflexive Banach space and $Y$ be a closed separable subspace of $X$ $\big((Y ,\|.\|)$is clearly a separable reflexive Banach space$\big)$, then the dual space $Y^*$ of $Y$ is separable. Let $\{y_n^*\}$ be a countable dense subset of $Y^*$.
Let $\{x_m\}$ be a bounded sequence in $X$, such that
$$
\langle y_n^*, x_m\rangle\underset{m}{\to }z_n\qquad \forall n
$$
With $z_n\in\mathbb{R}$.
We suppose that the sequence $\{x_m\}$ has a subsequece $\{x_{m_i}\}$ weakly convergente in $Y$ to an element $x_\infty$. Then
$$
\langle y_n^*, x_\infty\rangle=z_n\qquad \forall n\qquad (*)
$$
Since $\{y_n^*\}$ separates the points of $Y$, it follows from $(*)$ that every limit point of $\{x_m\}$ must equal $x_\infty$.
My problem
I don't understand why : we can conclude that $x_m$ converges weakly to $x_\infty$
This result was used in the article Infinite-Dimentional Extension of a Theorem of Komlos of Erik J.Balder, on pages 186-187. In the context of the article, the auther says that: "$\{s_n(t)\}$ converges weakly to a point $y_t$ in $Y$." But i don't understand why. An idea please.
Best Answer
By Banach-Alaoglu we know that any norm bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Thus any subsequence of $(x_m)$ must have have a weakly convergent subsequence, which by $(*)$ must have $x_\infty$ as a limit. Thus every subsequence has a further subsequence converging weakly to $x_\infty$, so we must have the original sequence converging weakly to $x_\infty$.
This holds due to the following fact: In any topological space, if any sequence $(x_n)$ satisfies the property that every subsequence contains a convergent subsequence converging to the same limit, then $(x_n)$ itself converges to that limit. This is easy to prove by contradiction.