I find hard to understand the proof for this theorem. Hope that some one can help me to clarify this. Thanks
Suppose $X$ is a vector space and $X'$ is a separating space of linear functionals on $X$. Then the $X'$-topology $\tau'$ makes $X$ into a locally convex space whose dual space is $X'$.
This is a theorem from Functional Analysis of Walter Rudin (page 64-65). We know that local base of $X'$ is of the form: $$V = \{x: |f_{i}(x)| < r_{i}; 1 \le i \le n\} (1)$$ for $f_{i} \in X'$, $r_{i} \gt 0$ for all $1 \le i \le n$
I can understand the part of proving $\tau'$ is a locally convex vector topology. In the part of proving $X'$ is dual space of $X$, suppose $f$ is a $\tau'$-continuous linear functional on $X$. Then $|f(x)| \le 1$ for all $x$ in some set $V$ of the form $(1)$ corresponding to $f_{i} \in X'$, $1 \le i \le n$.
Then the author states that if $f_{i}(x) = 0$ for all $i$, then $f(x) = 0$. I think about this for a long time but still don't understand how he state this fact. Please clarify this for me. I really appreciate.
If $f_i(x)=0$ then $nx\in V$ for all $n\in\mathbb N$ so that $|f(nx)|\le 1$. This implies $|f(x)|\le 1/n$ for all $n$ and thus $f(x)=0$.
Therefore, the map $g: \lbrace (f_1(x),\ldots,f_n(x)): x\in X\rbrace \to \mathbb C$, $(f_1(x),\ldots,f_n(x)) \mapsto f(x)$ is well defined and linear and has a linear extension $G: \mathbb C^n \to \mathbb C$ which is of the form $G(z_1,\ldots,z_n)= \sum a_k z_k$ for some $a_k\in \mathbb C$. Then you obtain $f= \sum a_k f_k \in X'$.