Let $E$ a Topological Vector Spaces, $T: E \longrightarrow \mathbb{K}$, linear. If for some $x \in E$, $Tx \neq 0$. Then, $T$ is open.
I think that it is sufficient to prove that $T(G) \subset \mbox{Int}(T(G))$, $G \in \tau_E$
Any help is appreciated
Thanks!