Consider the following fragment from the book "Lectures on von Neumann algebras".
Why is the line $\varphi$ is $\sigma(\mathcal{E}, \mathcal{F})$-continuous $\implies$ there exist $\psi_1, \dots, \psi_n \in \mathcal{F}$ such that $|\varphi| \leq \sum_k p_{\psi_k}$ true? This is probably an easy consequence of the initial topology definition, but I can't see it.
Hints:
There is a neighborhood $0\in \mathscr N$ such that $|\varphi(x)|<1$ whenever $x\in \mathscr N$ and there are seminorms $\psi_1,\cdots, \psi_n$ and an $ \epsilon>0$ such that $B_{\epsilon}(\psi_1,\cdots, \psi_n)\subseteq \mathscr N.$ Then $|\varphi(x)|<1$ if $|\psi_i(x)|<\epsilon$ for $1\le i\le n.$
It follows that, for any integer $n,\ |\varphi(nx)|<1$ whenever $x\in \ker \psi_i.$ But this means that $x\in \ker \varphi$ and so $\bigcap \ker \psi_i\subseteq \ker \varphi,$ which in turn implies that $\varphi=\sum c_i\psi_i.$
By definition of initial topology and using that $\mathscr{F}$ is a subspace, we can find $\psi_1, \dots, \psi_n \in \mathscr{F}$ such that $$\bigcap_{k=1}^n \{x \in \mathscr{E}:|\psi_k(x)| < 2\}\subseteq \{x \in \mathscr{E} :|\varphi(x)| < 1\}.$$
In particular, consider the vector $z:=x/\sum_{k}|\psi_k(x)|$ when $\sum_k |\psi_k(x)| \neq 0$. Then $$|\psi_k(z)| = |\psi_k(x)|/\sum_k|\psi_k(x)| \le 1 < 2$$ and thus $$|\varphi(z)| < 1$$ I.e. $$|\varphi(x)| < \sum_k |\psi_k(x)| = \sum_k p_{\psi_k}(x)$$
If $\sum_k |\psi_k(x)|= 0$, then $x \in \bigcap_k \ker \psi_k$ and thus $x \in \ker \varphi$ so that we always have the desired inequality.
