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Let $V$ be a complex vector space, which we endow with the finest linear topology. Then continuous dual $V'$ coincides with the algebraic dual $V^*$. We choose the weak-star topology $\sigma(V^*,V)$ on $V^*$. Consider the map $(-,-) \ : \ V^* \times V \to \mathbb C$ defined by $(\phi,v)=\phi(v)$. I have been able to show that it is separately continuous and jointly sequentially continuous. I would like to know if it is jointly continuous.

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No, whenever $V$ is infinite dimensional the evaluation isn't continuous with respect to the product of the weak$^*$ and the finest locally convex topology. Indeed, continuity at $0$ would give a finite set $E\subseteq V$, $\varepsilon >0$, and a $0$-neighbourhood $U$ in the finest locally convex topology such that $|f(u)|\le 1$ whenever $|f(e)|\le \varepsilon$ for all $e\in E$ and $u\in U$. Given $v\in V$ not contained in the linear span of $E$ you can choose $\delta>0$ such that $\delta v \in U$ and a linear functional $f$ which vanishes on $E$ with $|f(v)|\ge 2/\delta$ which yields a contradiction.

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  • $\begingroup$ Dear Jochen, I meant the finest $\mathbb C$-linear topology. I believe it is not locally convex unless dimension of $V$ is countable. Thus it is finer than the finest locally convex topology. I can't answer my question myself because I don't really understand what neighbourhoods of $0$ look like in this topology. It is possible that topology you are referring to would be more useful for me, though. Do you have any reference to some description of its properties? $\endgroup$ – Blazej Aug 21 '18 at 7:44
  • $\begingroup$ I just realized that confusion might have arisen because of the tag "locally convex spaces". I guess I added it because the dual of $V$ equipped with the weak-star topology is locally convex. $\endgroup$ – Blazej Aug 21 '18 at 7:49
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The argument in my answer also works for the finest linear topology. You just need that neighbourhoods of zero are absorbing.

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  • $\begingroup$ Oops, this should have been a comment. The device I am using now makes things difficult. $\endgroup$ – Jochen Aug 21 '18 at 8:13

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