By definition, I know a locally convex space is a topological vector space whose topology is defined by a family of seminorms $\cal P$ such that $$\bigcap_{p\in{\cal P}}\{x\colon p(x)=0\}=\{0\}.$$ Also I can easily show that a locally convex space by above definition separates the points and conversely. So this two properties are equivalent.
Now I want to show weak operator topology (WOT) is locally convex.
By above definition, suppose for every $\xi , \eta \in H$, $p_{\xi,\eta}(x) =0 $, thus $|(x\xi,\eta)|=0$ for every $\xi,\eta$. Put $\eta=x\xi$ we have $\|x\xi\|=0$ for every $\xi$ and we can conclude $x=0$.
My problem is to show WOT is locally convex using the equivalent property.
Suppose $x,y\in B(H)$ and $p_{h,k}(x) = p_{h,k}(y)$ for every $h,k\in H$. We have $|(xh,k)|=|(yh,k)|$. There are $r_x,r_y$ with $|r_x|=|r_y|=1$ such that $|(xh,k)|=r_x(xh,k)$ and $|(yh,k)|=r_y(yh,k)$ and in the end we can see $r_x x=r_y y$.
In this way I can not conclude wot topology separates the points. Please help me to understand it. Thanks in advance.
These conditions are not equivalent. Take a non-zero operator $x$. Then certainly $x\neq -x$. However,
$$|(-xh, k)| = |-(xh,h)| = |(xh, k)|$$
for any $h,k\in H$.
Separation in the context of locally convex spaces means the following: