Weak convergence in a product space

Question

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

1. If $y_n \rightharpoonup y $, then $f(y_n)\rightharpoonup f(y) \text{ in } Y$;
2. $f$ 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 $f(x(\cdot))$

Answer 1

For $t\in I$ let $p_t:Y^I\to Y$ denote the projection on the $t$-th coordinate of the product, that is the evaluation map $x\mapsto x(t)$, and let $p^\intercal_t:Y^*\to (Y^I)^*$ be its transpose operator (thus for $x\in Y^I$ and $u\in Y^*$ one has $\langle p^\intercal_t u, x\rangle=\langle u, x(t)\rangle$.

Fact: The topological dual of $Y^I$ is generated as a linear space by the elements of the form $ p^\intercal_t u$ for $t\in I$ and $u\in Y^*$, and by linearity they are a sufficient set of test linear forms in order to check the weak convergence on $Y^I$.

Therefore $ f(x_n(\cdot)) \rightharpoonup f(x(\cdot))$ in $Y^I$ simply means that it weakly converges in $Y$ point-wise, that is, for any $t\in I$ and for any $u\in Y^*$ one has $\langle u, f(x_n(t))\rangle\to\langle u, f(x(t))\rangle$, which is true by assumption 1.

7/10/20 Detalis. Here is a direct proof of the mentioned Fact. Let $\lambda $ be a continuous linear form on $Y^I$. For $J\subset I$ let $P_J:Y^I\to Y^I$ denote the continuous linear projector given by the multiplication by the characteristic map of $J$, i.e. $x\mapsto \chi_Jx $. Since $Y^I$ has the product topology, the set $\{|\lambda|<1\}$, as any other nbd of the origin, contains a whole subspace $\text{ker}P_J$ for some finite subset $J\subset I$. Note that for all $x\in Y^I$ and for all $c\in\mathbb{R}$ we have $c(1-P_{ J})x\in \text{ker}P_J$, so $|\langle \lambda, c(1-P_{ J})x\rangle|<1$ as said; being $c$ arbitrary, this means $\langle \lambda, (1-P_{ J})x\rangle=0$, that is, $\langle \lambda, x \rangle=\langle \lambda, P_J x\rangle =\langle P_J^\intercal\lambda, x\rangle$, that is $\lambda= P_J^\intercal\lambda=\sum_{t\in J}P_t^\intercal\lambda $ as we wished to show.

(Note that today's $P_t:Y^I\to Y^I$ is slightly different to yesterday's $p_t:Y^I\to Y$ since now $\text{ran}\,P_t= \{x\in Y^I: \text{supp}(x) \subset\{t\}\}\sim Y$).