Given a function $f: Y\longrightarrow Y$ ($Y$ is a Banach space). Assume that $f$ satisfies:

- If $y_n \rightharpoonup y $, then $f(y_n)\rightharpoonup f(y) \text{ in } Y$;
- $f$ is weakly compact;

Notation: $I:=[0,T]$, $T>0$ and $X:=\left(C(I;Y), \Vert\cdot\Vert_\infty \right)$

The question is to show that for any bounded sequence $(x_n)_{n\in \mathbb{N}}$ in $X$, there exists a subsequence, for simplicity we note the subsequence by $(x_n)_{n\in \mathbb{N}}$, such that $$ f(x_n(\cdot)) \rightharpoonup \color{red} {f(x(\cdot))}. \text{ (As a weak limit on the product space $Y^{I}$)} $$

$\bf{Hint}:$

i) There's a result Dobrokov’s Theorem says that for a bounded sequence in $X$, $x_n\rightharpoonup x$ iff $x_n(t)\rightharpoonup x(t)$, for each $t\in I$.

ii) $\{f(x_n(t))\}$ is relatively weakly compact for each $t\in I$, by 2.

iii) By (ii) and Tychonoff’s theorem: $\{f(x_n(\cdot)),\, n\in\mathbb{N}\}=\prod\limits_{t\in I} \{f(x_n(t)),\, n\in\mathbb{N}\}$ is relatively weakly compact in the product space $Y^{I}$

iv) Thus, there exists a subsequence, for simplicity we note the subsequence by $(f(x_n(\cdot))_{n\in \mathbb{N}}$, such that $ f(x_n(\cdot)) \rightharpoonup g(\cdot).$

$\bf{Problem}:$ To show that $g(\cdot)$ is exactly $f(x(\cdot))$