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I'm working through a proof and the setting is that I have a convex, balanced, open subset $U$ of a topological vector space $V$.

The claim that I can't verify is made briefly: the Minkowski functional, $p_{C} := inf\{t\geq 0 : x\in tC\}$ is continuous. This is supposed to follow from the fact that $\{x\in V:p_{C}(x) < 1\}$ is open. But this I cannot verify.

Can anyone point out the logic link I am missing?

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Your definition of Minkovski's functional is incorrect. –  Norbert Jan 26 '12 at 21:38

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up vote 3 down vote accepted

I think your definition of the functional is incorrect (for once, taking an infimum over a vector space is not well-defined, and it is clearly not a fuctional, as the image of $\inf\lbrace x\in V\mid x\in tC\rbrace$ is not in $\mathbb R$). I assume you meant $p_C(x):=\inf\lbrace t\in \mathbb R\mid x\in tC\rbrace$.

Let $B=\lbrace x\in V\mid p_C(x)<1\rbrace$, we aim to show that in the case where $C$ is open we have that $B=C$ is open.

The fact that $B\subseteq C$ is immediate: if $x\in B$ then $p_C(x)<1$ implying that for any $t\ge 1,\: x\in tC$ (since $C$ is convex). We get that $x\in 1C=C$.

To the converse direction, let $x\in C$ be arbitrary. Since $C$ is open we have an open neighbourhood $x\in U\subseteq C$. Seeing that the action of multiplyication by a scaler is continuous, the pre-image of $U$ is open in $\mathbb C\times V$, and containes the element $\langle 1,x\rangle$. Thus, there exists some $\delta>0$ and an open ball $B(1,\delta)$ such that $B(1,\delta)\cdot x\subseteq C$.

In particular, this implies that $(1+\delta/2)x\in C$, and so $p_C(x)\le\frac{1}{1+\delta/2}<1$ and hence $x\in B$.

Once you have that, showing that $p_C$ is continuous is indeed immediate, I'll leave it to you for now, not to spoil all the fun :-)

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