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Let $(V,\mathcal{P})$ be a locally convex topological vector space, that is $V$ is a (real) vector space endowed with the locally convex topology $\tau_{\mathcal{P}}$ induced by the given family $\mathcal{P}$ of seminorms $p : V \rightarrow \mathbb{R}_0^+$. Let $(W, \tau)$ be any other topological space.

My question is as follows: Given a map $f : V \rightarrow W$, do we then have that $$ f \quad\text{is $(\tau_{\mathcal{P}},\tau)$-continuous} \quad \Longrightarrow \quad f \quad\text{is $(p,\tau)$-continuous for at least one $p\in\mathcal{P}$ ?} $$

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Take $W=(V,\mathcal P)$ and $f$ the identity map on $V$. Since the topology of $W$ is finer than the topology of $(V,p)$, then $f^{-1}$ is continuous. If $f$ is continuous, then it must be a homeomorphism, so the two topologies of $(V,\mathcal P)$, $(V,p)$ must coincide (since $f$ is just the identity). This is only possible if $p$ controls all the other norms of $\mathcal P$, that is, $(V,p)$ is a (semi-)normed space from the beginning.

So the answer is: what you said never holds, unless $(V,\mathcal P)$ is a (semi-)normed space (that is, if the topology can be induced by a single semi-norm). If you assume that your space is Hausdorff, then of course, it must be a normed space.

Edit: actually, the precise statement you wrote is not true even for normable spaces, (that is, topological vector spaces whose topology can be induced by one single norm). Consider $V=\mathbb R^2$ with the usual topology: it is a l.c. t.v.s. with topology induced by the family of seminorms $p_1((x,y))=|x|$, $p_2((x,y))=|y|$, but the identity map $f\colon (\mathbb R^2,p_j)\to (\mathbb R^2,\mathcal P)$ is not continuous for $j=1,2$. It is possible to give an analogous example in infinite dimensions, where $p_1$, $p_2$ are both strict norms (in finite dimensions this is not possible, since all norms are equivalent).

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    $\begingroup$ The example in the edit is a good reason to consider directed sets of seminorms. $\endgroup$
    – Jochen
    Commented Nov 12, 2023 at 12:39

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