I am trying to understand the proof of Theorem 1 in this paper (see page 784 for the proof). I understand that by the finite intersection property there exists an element $u_0$ in the weak closure of $\bigcap_{F_0 \in S} V_{F_0}$, where $S$ denotes the directed set of finite-dimensional subspaces of $X$, ordered by inclusion.
My questions are:
- How is $u_{F_1}$ related to $u_0$?
- Why can we conclude from $\langle Tv, v- u_{F_1} \rangle \geq 0$ for all $v\in F$, that $\langle Tv, v - u_0\rangle$ for all $v\in F$ holds. (It looks to me that we are using something like $u_{F_1} \rightharpoonup u_0$ but $(u_F)_{F\in S}$ is no sequence since $S$ is not a countable set.)
I think he uses "net" instead of sequence, which allows the index to be non-countable. By net, we mean a mapping $x:I\to X$, where $I$ is a directed set, and $X$ is a topological space, which is usually denoted by $\{x_\alpha\}_{\alpha\in I}$. We say that $x_\alpha\to x$ if and only if for any neighborhhood $U$ of $x$, there exists a $\alpha\in I$ such that it holds that $x_\beta \in V$ for any $\beta \geq \alpha$.
There are many usual properties that net possesses. Here we use two of them.
(1) For a mapping $f:X\to Y$ between two topological spaces, $f$ is continuous at $x$ if and only if $f(x_\alpha)\to f(x)$ whenever for any net $x_\alpha$ such that $x_\alpha\to x$.
(2) $x\in \overline{S}$ if and only if there exists a net $\{x_\alpha\}\subset S$ such that $x_\alpha\to x$.
Obviously, $(Tv,v-v_1)$ is continuous in $v_1$ from weak topology into $\mathbb{R}$, if $u_0$ is the limit of some net $u_{F}$, then it is done by the continuity. Since $u_0$ lies in the weak closure of $V_{F_1}$, therefore it is the limit of some net.
For the details on net, you can refer to the book "infinite dimensional analysis" by Aliprantis.