# Continuous semi-norms on subspace

Suppose $X$ is a locally convex topological vector space, let $P$ be the set of all continuous semi-norms on $X$. Suppose $M$ is a subspace of $X$, denote $P|_M$ as the set of semi-norms in $P$ restricted on $M$. Is it true that $P|_M$ is the set of continuous semi-norms on $M$?

Yes, every continuous semi-norm $p$ on $M$ can be extended to a continuous semi-norm on $X$:
Let $U=\lbrace x\in M: p(x)<1\rbrace$ be the unit ball of $p$. Since $U$ is open in $M$ and $0\in U$ there is a convex $0$-neighbourhood $V$ in $X$ such that $V\cap M\subseteq U$. Now, let $W$ be the convex hull of $U\cup V$. Since $U$ and $V$ are convex we have $W=\lbrace \lambda u + (1-\lambda)v: \lambda\in [0,1], u\in U,v\in V\rbrace$ and this implies $W\cap M=U$: Indeed, given $x=\lambda u + (1-\lambda)v \in W\cap M$ with $u\in U$ and $v\in V$ we have $(1-\lambda)v =x-\lambda u \in M$ and hence $\lambda=1$ or $v\in M\cap V \subseteq U$. In either case we get $x\in U$. The inclusion $U\subseteq W\cap M$ is clear.
We now define $q$ as the Minkowski functional of $W$, i.e., $$q(x)=\inf\lbrace t>0: x\in tW\rbrace.$$ This is a continuous semi-norm on $X$ because $V\subseteq W$ and the relation $W\cap M=U$ implies $q|_M=p$.
If $X$ is not locally convex the statement is not true in general: There are Hausdorff topological vector spaces without any non-zero continuous semi-norms. Any finite-dimensional subspace is locally convex and no non-zero continuous semi-norm can be extended.