# Weak convergence in a product space

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 $$f(x(\cdot))$$

• Is it a homework problem? – Nik Weaver Oct 6 at 22:41
• It is in a fixed-point paper, I reformulate it to this form. – Malik Amine Oct 7 at 10:48
• It looks like you got a good answer from Pietro. Just FYI, the word "hint" in your question makes it look like an assigned problem, and may hinder people from taking the question seriously. I'd suggest a different word choice in the future... – Nik Weaver Oct 7 at 14:28

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 $$\text{ran}\,P_t= \{x\in Y^I: \text{supp}(x) \subset\{t\}\}\sim Y$$).
• If you look at it, it is just a general fact about projectors ($P^2=P$): If $x=Pu$ for some $u$, then also $Px=P^2u=Pu=x$. I edited and changed a bit the notation, since using the same letter for different objects may be confusing, – Pietro Majer Oct 12 at 19:48