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Let $(X,\mathcal{P}_X)$ and $(Y,\mathcal{P}_Y)$ be locally convex topological vector spaces with topologies induced by the families of continuous seminorms $\mathcal{P}_X$ and $\mathcal{P}_Y$ respectively and let $A:X\to Y$ be a linear map.

I want to show that $A:X\to Y$ is continuous $\iff\forall q\in\mathcal{P}_Y$ there exists $p\in\mathcal{P}_X$ such that $q(A(x))\le p(x)$ for all $x\in X$.

I wanted to first show the $\Longleftarrow$ direction.

Via the pre-image

First I tried to do it by claiming that it suffices to show that $A$ is continuous at 0 in $X$. To this end, let $B_q(0,r)$, with $q\in\mathcal{P}_Y$ and $r>0$ be an open neighbourhood of $0$ in $Y$. Then we must show that the pre-image $A^{-1}(B_q(0,r))$ is open in $X$. That is, $A^{-1}(B_q(0,r))=\{x\in X:T(x)\in B_q(0,r)\}$...

However, this seemed a dead-end and I couldn't see how to relate it with the inequality.

Showing A is bounded

Next I considered that $A:X\to Y$ is continuous if and only if $A:X\to Y$ is bounded. And, in a locally convex topological space, $U\subset X$ is bounded if and only if every $p\in\mathcal{P}_X$ is bounded on $U$. Thus, since the topology on $Y$ is induced by $\mathcal{P}_Y$, the map $A:X\to Y$ is bounded if and only if for all $q\in\mathcal{P}_Y$ there exists $p\in\mathcal{P}_X$ such that $q(A(x))\le p(x)$...which would in fact prove the if and only if statement, however, I am not sure if this argument is complete.

I also thought of proving it via some definition of sequential continuity, e.g. $A$ continuous if and only if $x_n\to 0\implies Ax_n\to 0$, but it seems a little tricky to formulate in this seminorm setting and I don't know if it would be fruitful.

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  • $\begingroup$ Are the families of seminorms assumed to be scale-invariant ($p \in \mathcal{P} \land c > 0 \implies c\cdot p \in \mathcal{P}$) and upward filtering (for all $p_1,p_2 \in \mathcal{P}$ there is a $p_3 \in \mathcal{P}$ with $p_3 \geqslant \max \{ p_1,p_2\}$)? $\endgroup$ – Daniel Fischer Mar 12 '17 at 21:24
  • $\begingroup$ Also, for general locally convex spaces, boundedness of a linear map does not imply continuity. $\endgroup$ – Daniel Fischer Mar 12 '17 at 21:27
  • $\begingroup$ @DanielFischer I should have mentioned that the two families both form fundamental systems of seminorms so that we do indeed have the upward filtering. I am not sure of the "scale-invariance"; is that some kind of absorption condition? $\endgroup$ – Jason Born Mar 12 '17 at 21:52
  • $\begingroup$ Without the scale invariance, you need a constant factor, $q(Ax) \leqslant C_{p,q}\cdot p(x)$, like in normed spaces. (It would suffice that for every $p_1$ and $c$ there is a $p_2$ such that $p_2 \geqslant cp_1$.) $\endgroup$ – Daniel Fischer Mar 12 '17 at 21:54
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    $\begingroup$ It is not the case that "$A$ maps bounded sets to bounded sets" implies the continuity of $A$ for general locally convex spaces. Consider $Y$ an infinite-dimensional normed space, and $X$ the same space endowed with its weak topology. The identity maps bounded sets to bounded sets (every weakly bounded set is norm-bounded), but it is not continuous, since the norm topology is strictly finer than the weak topology. Similarly, sequential continuity does in general not imply continuity. $\endgroup$ – Daniel Fischer Mar 12 '17 at 22:51

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