My books talks about the conjugated space, but does it mean dual space? Are not the same thing?
I don't understand why in the dual space $E^\ast$ of $E$, the separation axiom $T_1$ is satisfied and the space is locally convex. The book says, if we take the functional $f_0 \in E^\ast$ and $f_0 \neq 0$, we can find an element $x_0 \in E$ such that $f_0(x_0) \neq 0$. Let $\epsilon=\frac{1}{2}|f_0(x_0)|$ and $A=\{x_0\}$, then $f_0 \notin U_{\epsilon, A}$; so $E^\ast$ is a $T_1$ space. ( I don't understand the logic of this proof )
To show the local convex it says that we can observe that for every $\epsilon > 0$ and for every $A \subset E$ the neighborhood $U_{\epsilon, A}$ is convex in $E$