# weak topology on a topology vector space

$$X$$ is a vector space. $$X'$$ is a family of linear functionals on $$X$$ and separate point on $$X$$ (for any point $$x\neq y$$ in $$X$$, there is a linear functional $$L$$ in the family $$X'$$ s.t. $$Lx\neq Ly$$) . The weak topology on $$X$$ is the weakest topology making all the element in $$X'$$ continuous. And $$X^*$$(denote the dual space of $$X$$) is the span of $$X'$$.

Now for any linear functional $$L$$ on $$X$$, the function ,$$p_L:X\rightarrow \mathbb{R},p_L(x)=|Lx|$$ is a seminorm on $$X$$, $$X'$$ separates the point implies that the family of semi-norms is a separating family. My teacher told me that the topology generated by $$X'$$ is exactly the same the topology generated by the family of semi-norms(why?) In other words, why the weak topology has a local base $$\mathcal{B}_0=\{V_{L_1,…,L_n;r_1,…r_n}|L_1,…,L_n\in X',r_1,…,r_n＞0\}$$ , where $$V_{L_1,…,L_n;r_1,…r_n}=\{x\in X:|L_ix|\leq r_i, 1\leq i\leq n\}$$.

My second question: why $$x_n\rightarrow x$$ in the weak topology iff $$L(x_n)\rightarrow L(x)$$ for all $$L\in X^*$$($$X^*$$ is the dual space of $$X$$).

Let $$X$$ be a set and $$(f_i)$$ be a family of scalar valued functions on $$X$$. Let $$\tau$$ be the weak topology generated by this family. If $$x_n \to x$$ then $$f_i(x_n) \to f_i(x)$$ for each $$i$$ because $$f_i$$ is continuous for $$\tau$$. This proves one way of your result since,for any finite linear combination $$g$$ of $$f_i$$'s also $$g(x_n) \to g(x)$$.
Conversely, suppose $$f_i(x_n) \to f_i(x)$$ for each $$i$$. Consider any basic neighborhood of $$x$$ in $$\tau$$: $$\{y: |f_{i_j}(y)-f_{i_j}(x)| . Since $$f_{i_j}(x_n) \to f_{i_j}(x)$$ for each $$j \leq N$$ it follows that the inequalities $$|f_{i_j}(x_n)-f_{i_j}(x)| hold for $$n$$ suffcientyl large. Hence $$x_n$$ lie in the basic neighborhood for $$n$$ suffcientyl large.
EDIT: Weak topology $$\tau$$ generated by $$(f_i)$$: Consider sets of the type $$\{y: |f_{i_j}(y)-f_{i_j}(x)| . Form all possible unions of these sets. You can check that this gives a topology $$\tau'$$ on $$X$$. It is clear that $$\tau' \subseteq \tau$$ because each of the sets $$\{y: |f_{i_j}(y)-f_{i_j}(x)| is open in $$\tau$$. It is also easy to check that each $$f_i$$ is continuous for $$tau'$$. By definition of $$\tau$$ it follows that $$\tau \subseteq \tau'$$.
• $\tau \subset \tau'$ is very hard for me to prove (How to prove $f$ is continuous on $\tau'$. Could you write more deatils? Thank you! Commented Jan 30, 2021 at 9:52
• $f_i$ is continuous on $\tau'$ by definition.Thank you! Commented Jan 30, 2021 at 10:37