Let $(X,\|.\|) $ be a separable Banach space.
Let $\{x_n\}$ be a sequence relatively weakly compact. We consider the sequence $\{y_n\}$ defined by : $$ y_n=\frac {1}{n}\sum_{k=1}^{n}{x_k} $$
Show that : $\{y_n\}$ is relatively weakly compact.
An idea please
As $(x_n)$ is weakly pre-compact, it is (norm-)bounded, hence so is $(y_n)$. By the Eberlein–Šmulian theorem, it is enough to show that every subsequence of $(w_n)$ of $(y_n)$ has a convergent subsequence.
In particular, without loss of generality we may take a subsequence of $(x_n)$ that is weakly convergent, and for convenience, still denote it by $(x_n)$. As translations are homeomorphisms under the weak topology, without loss of generality we may assume that $(x_n)$ is weakly null. Let $(y_{n_j})$ be the subsequence of $(y_n)$ corresponding the selected subsequnce of the original sequence $(x_n)$. We will show that $(y_{n_j})$ is weakly null.
We may define a bounded linear operator $T\colon \ell_1\to X$ by $Te_k = y_{n_k}$, where $(e_k)$ is the standard unit vector basis of $\ell_1$. Consider the adjoint operator $T^*\colon X^*\to \ell_1^* = \ell_\infty = (c_0)^{**}$. By the very definition,
$$\langle T^* f, (\xi_k) \rangle = \langle f, T(\xi_k)\rangle = \langle f, \sum_k \xi_k y_{n_k}\rangle = \sum_k \xi_k \langle f, y_{n_k}\rangle = \sum_k \sum_{i = 1}^{n_k} \frac{1}{n_k} \xi_k \langle f, x_i\rangle.$$
Thus,
$$\langle T^* f, e_k \rangle = \langle f, y_{n_k}\rangle = \sum_{i = 1}^{n_k} \frac{1}{n_k} \langle f, x_i\rangle\to 0$$
as $k\to \infty$ because $(x_i)$ is weakly null, hence $\langle f, x_i\rangle\to 0$ as $i\to\infty$ for every $f\in X^*$. Using this we may conclude that $ \sum_{i = 1}^{n_k} \frac{1}{n_k} \langle f, x_i\rangle\to 0$ as $k\to \infty$ because every subsequence of the sequence of Cesàro averages of $(\langle f, x_i\rangle)$ converges to 0 too.
We have thus proved that $T^*$ takes values in $c_0$, which means that it is weak*-to-weak convergent. As $(e_k)$ is weakly* precompact in the weak* topology of $\ell_1$ introduced by $c_0^* = \ell_1$, so is then the image of $(e_k)$ via $T^*$, namely $\{y_{n_j}\colon j\in \mathbb N\}$.