Let $E$, $F$, $G$ be locally convex topological vector spaces. Let $\mathcal{G}$ (resp. $\mathcal{H}$) be a family of bounded subsets of $E$ (resp. $F$). The $\mathcal{G}$-$\mathcal{H}$-topology on the space of separately continuous bilinear forms on $E \times F$ into $G$, $\mathscr{B}(E,\,F;\,G)$, is the topology of uniform convergence on the products $G \times H$ of $\mathcal {G} \times \mathcal {H}$.
A remark in the book says, that $\mathscr{B}(E,\,F;\,G)$ provided with such topology is Hausdorff, if $G$ is Hausdorff and if $\mathcal{G}$ and $\mathcal{H}$ are total families (families whose union is a total subset of $E$ resp. $F$). A subset is total, if its linear hull is dense.
We denote by $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ the space of separately continuous bilinear forms on weak duals of $E$ and $F$ into $\mathbb{C}$, with the $\mathcal{G}$-$\mathcal{H}$-topology, where $\mathcal{G}$ and $\mathcal{H}$ are taken to be the families of all equicontinuous subsets of $E'$ and $F'$ respectively.
In my opinion, $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ has to be Hausdorff, since $\mathbb{C}$ is Hausdorff and the union of all equicontinuous subsets of $E'$ is $E'$ (the same holds for $F$).
But another book says, that the space $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ is Hausdorff if and only if both $E$ and $F$ are Hausdorff.
My question is: under which conditions is the space $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ Hausdorff.
EDIT:
The basis of neighborhood of $0$ in $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ is defined by taking the sets $$ \mathcal{U}(A',\, B',\, V) = \{ \Phi \in \mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma}) \colon \Phi(A',\, B') \subseteq V \big\}$$ where $A'$ (resp. $B'$) varies over $\mathcal{G}$ (resp. $\mathcal{H}$) and $V$ over a basis of neighborhoods of $0$ in $\mathbb{C}$.
Since $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ is a topological vector space, if we want to show, that $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ is Hausdorff, we have to proof, that for every $\Phi \in \mathscr{B}(E'_{\sigma},\, F'_{\sigma})$, such that $\Phi \neq 0$, there is a neighborhood $U$ of $0$ in $\mathscr{B}_{\varepsilon}(E'_{\sigma},\, F'_{\sigma})$ such that $\Phi \notin U$.
Let $\Phi \in \mathscr{B}(E'_{\sigma},\, F'_{\sigma})$ be nonzero. That means that there are $x' \in E'$ and $y' \in F'$ such that $\Phi(x',\, y') \neq 0$. Since $\mathbb{C}$ is Hausdorff, there is a neighborhood $V$ of $0$ in $\mathbb{C}$ such that $\Phi(x',\, y') \notin V$. Hence we have $$ \Phi \notin \mathcal{U}(x',\, y',\, V)$$ The set $\mathcal{U}(x',\, y',\, V)$ is a neighborhood of zero in $\mathscr{B}_{\epsilon}(E'_{\sigma},\, F'_{\sigma})$. Hence the space $\mathscr{B}_{\epsilon}(E'_{\sigma},\, F'_{\sigma})$ is Hausdorff.
Is my proof right?