# Continuity wrt locally convex topology implies continuity wrt to at least one generating seminorm?

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 $$f \quad\text{is (\tau_{\mathcal{P}},\tau)-continuous} \quad \Longrightarrow \quad f \quad\text{is (p,\tau)-continuous for at least one p\in\mathcal{P} ?}$$

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).