1
$\begingroup$

Given $X$ is a topological vector space and $V$ is a convex, balanced neighbourhood of $0$ in X. Then for $x \in V$, the Minkowski functional $\mu_{V}(x)$ = inf $\{t>0: t^{-1}x\in V\}$ $<$ $1$.

The reason given for strict inequality is "$V$ is open". But I dont get it. I can observe the following. Since $x\in V$, $\mu_{V}(x)$ $\leq 1$ and since $V$ is balanced $tV$ $\subseteq V$ for all $0<t<1$. Let $x\in V$. As $V$ is open, there exists a neighbourhood $U$ of $x$ such that $x\in U \subset V$.

I dont know how to proceed with this. Please help!

$\endgroup$
2
$\begingroup$

Using continuity of the map $M\colon X\times \mathbf R\to x$ defined by $M(x,\lambda):=\lambda x$ (continuity for the product topology), we have that the map $t\mapsto tx$ is continuous, hence the set $$S:=\{t>0,t^{-1}x\in V\}$$ is an open subset of $(0,+\infty)$. Since $1$ belongs to $S$, the infimum of $S$ is necessarily strictly smaller than $1$.

$\endgroup$

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.