Is every seminorm induced by a linear operator into a normed space?

Question

I'm reading an analysis textbook chapter on convex topological vector spaces, and there is this statement that (one of) the most common way(s) to define a topology on a vector space $X$ is by requiring the continuity of certain linear maps, (i.e via the semi-norms induced by given linear maps $T\colon X\to Y$ where $Y$ is a normed space).

This makes sense, but then there's this that bugs me: "Can any semi-norm on $X$ be shown to be induced by some linear map T, to some normed linear space?" I can't find a counter-example to this, or show that it is true. Any insight would be appreciated.

Answer 1

Suppose that $p$ is a seminorm on $X$. Let $$ K=\{x\in X:\ p(x)=0\}. $$ Let $Y=X/K$, i.e. $Y$ is the vector space of the classes $x+K$. On $Y$, define $$\|x+K\|=p(x).$$ This is well-defined, because if $x_1+K=x_2+K$, this means that $p(x_1-x_2)=0$, so by the reverse triangle inequality $$ |p(x_1)-p(x_2)|\leq p(x_1-x_2)=0 $$ and so $p(x_1)=p(x_2)$. It is now easy to check that $\|\cdot\|$ is a norm on $Y$. If $\pi$ is the quotient map $\pi(x)=x+K$, then $\pi$ is linear and $$ p(x)=\|\pi(x)\|, \ \ \ x\in X. $$

Answer 2

Let $p$ be a seminorm on a space $X$. Then $X/ \{x\in X\colon p(x)=0\}$ is a normed space under the norm $\|\pi(x)\|=p(x)$. Then the seminorm $p$ is implemented by the quotient map $\pi$.