$X$ 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 $X$).


1 Answer 1


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 $\tau$: $\{y: |f_{i_j}(y)-f_{i_j}(x)| <r_i: 1 \leq j \leq N\}$. Since $f_{i_j}(x_n) \to f_{i_j}(x)$ for each $j \leq N$ it follows that the inequalities $|f_{i_j}(x_n)-f_{i_j}(x)| <r_i: 1 \leq j \leq N$ hold for $n$ suffcientyl large. Hence $x_n$ lie in the basic neighborhood for $n$ suffcientyl large.

HOWEVER there is a flaw in your approach. Use of sequences here is fine in metrizable case but not in the case of general topological vector spaces. For a proper proof you have to use nets, or work completely with neighborhoods.

EDIT: Weak topology $\tau$ generated by $(f_i)$: Consider sets of the type $\{y: |f_{i_j}(y)-f_{i_j}(x)| <r_i: 1 \leq j \leq N\}$. 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 $\{y: |f_{i_j}(y)-f_{i_j}(x)| <r_i: 1 \leq j \leq N\}$ 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 $\tau \subseteq \tau'$.

  • $\begingroup$ Thanks for you answer! But how can I prove my first question? This question puzzles me for a long time. $\endgroup$ Commented Jan 30, 2021 at 9:09
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    $\begingroup$ @ZeldovichYakov I have edited my answer. $\endgroup$ Commented Jan 30, 2021 at 9:20
  • $\begingroup$ $\tau \subset \tau'$ is very hard for me to prove (How to prove $f$ is continuous on $\tau'$. Could you write more deatils? Thank you! $\endgroup$ Commented Jan 30, 2021 at 9:52
  • $\begingroup$ $f_i$ is continuous on $\tau'$ by definition.Thank you! $\endgroup$ Commented Jan 30, 2021 at 10:37

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