I want to show, that a locally hausdorff space is seminormable, if and only if there exists a bounded and open set. My proof goes as follows:
Let $B \subset V$ be an open and bounded 0-neighborhood, $V$ a topological vector space and $\mathcal{T}_{ V} $ a topological local base. I claim that
\begin{align}\label{EQ:**}
B\cap \mathcal{T}_{ V} =\left\{ B\cap U\mid U\in \mathcal{T}_{ V} \right\}
\end{align} is also a local base, making $V$ a locally bounded TVS.
proof
The inclusion $B\cap U \subset U$ makes $B\cap \mathcal{T}_{ V} $ finer then $\mathcal{T}_{ V} $. On the other hand, $B\cap U$ is open with respect to $\mathcal{T}_{ V}$ so both topologies are equivalent.
Having shown this, let $B$ be a bounded 0-neighbourhood and $V$ be locally convex. We can assume $B$ to be absolutely convex, since the convex hull, as well as the balanced convex hull stay bounded and that $B$ is absorbing (every 0-neighborhood is). Now setting
$$
\left\| v \right\| := \sup\limits_{ B'\in \mathcal{B}} \left\{ \mu _{ B'} (v)\right\}
$$
should give the desired norm, where $\mathcal{B}=\mathcal{T}_{ V}\cap B $ and $\mu _{ B'} (v)=\inf\limits_{}\left\{ t\mid v\in tB'\right\} $ is the Minkowski functional. It is bounded, because of $\mu _{ B'} (v)\leq \mu _{ B} (v)$ for all $B'=B\cap U$, $U \in \mathcal{T}_{ V} $.
Is this proof correct?
Best Answer
The proposition you seek to prove is a version of a slightly more general one that is sometimes accredited to Kolmogorov. The more general proposition reads:
Dropping the Hausdorff assumption, the proposition remains true if "normable" is replaced by "seminormable".
Your proof seems to be in line with a standard one, except the supremum appearing in definition of the norm. Given that $B$ is a bounded absolutely convex neighborhood of zero, it suffices to consider the Minkowski functional $\mu_{B}$ (in your notation), which is a desired seminorm (or norm if the Hausdorff assumption is imposed).