2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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:

Proposition: A real or complex topological vector space is normable if and only if it is Hausdorff and possesses a convex bounded neighborhood of zero.

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).

$\endgroup$
  • $\begingroup$ Thank you. Yes, I even realized that $\mu_{B'}(v)\leq \mu_B(v)$, but it didn't cross my mind to just use this for the definition of the seminorm (or norm in the Hausdorff case). Do you perhaps know some literture, where I can look up that Proposition? $\endgroup$ – Mahdimatika May 30 at 10:55
  • $\begingroup$ Most books on topological vector space should have this proposition. One reference I myself use recently is "Topological vector spaces and their applications" by Bogachev and Smolyanov. Check theorem 1.5.7, page 40 in that book here. $\endgroup$ – YeZ May 30 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.