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I have read that for a separable complete locally convex space $E$, any sequentially continuous linear map $f:E'\to \mathbb{K}$ is continuous (where $E'$ is equipped with the weak*-topology).

Is there any similar result for a linear map $f:E'\to E'$?

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Let $E$ be a separable complete locally convex space, $f:(E',w^*)\to (E',w^*)$ a sequentially continuous linear map and $ev_x:E'\ni x'\mapsto ev_x(x')=x'(x)\in \mathbb K$ the evaluation map. Then:

\begin{align} u_n\xrightarrow{w^*} u\quad&\Longrightarrow\quad f(u_n)\xrightarrow{w^*} f(u)\\ &\Longrightarrow\quad f(u_n)[x]\xrightarrow{} f(u)[x],\quad\forall \ x\in E\\ &\Longrightarrow\quad ev_x\circ f(u_n)\xrightarrow{} ev_x\circ f(u),\quad\forall \ x\in E \end{align} This shows that $ev_x\circ f:(E',w^*)\to \mathbb K$ is sequentially continuous for all $x\in E$ and thus, by the result in the post, $ev_x\circ f:(E',w^*)\to \mathbb K$ is continuous for all $x\in E$. It follows that the map $f:(E',w^*)\to (E',w^*)$ is continuous.

So the answer is yes, there is a similar result:

For a separable complete locally convex space $E$, any sequentially continuous linear map $f:E'\to E'$ is continuous (where $E'$ is equipped with the weak*-topology).

[Solution based on this solution of @LeBtz]

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