Minkowski functional $p_E$ is continuous if and only if $0\in E^0=\text{int}E$

Question

Let $X$ be a normed topological vector space.

Prove $p_E$ is continuous $\iff 0\in E^0$.

In the above $p_E(x)=\inf\{t\ge0: x\in tE\},$ with $E$ an absorbing set $E\subset X$ is the Minkowski functional and $E^0$ denotes the interior of $E$.

I already proved the $\Rightarrow$ direction: A previous calculation yielded $E^0= E_1$ whenever $p_E$ is continuous, where $E_1=\{x\in X:p_E(x)\lt 1\}$. Since $E$ is absorbing, $0\in E = E^0$.

Now for the converse, I'm trying to prove $p_E$ is continuous at $0$ hoping that continuity everywhere else will follow. Now since $0\in E^0$, and $E^0$ there is a ball $B_r(0)\subset E^0$ for some $r>0$. All elements of said ball are in $E^0$ so $p_E(x)<1$ for all $\|x\|<r$:

Given $\epsilon>0$ I know that $\| p_E(x)-p_E(0)\|=\|p_E(x)\|<1$ for all $x\in B_r(0)$, but I can't make the above less than $\epsilon$. I also havent used the convexity of the Minkowski functional.

Any help would be greatly appreciated.

Answer 1

1. continuity at $0$: Let $\epsilon>0$. Since $0 \in E^0$, the set $$U := \epsilon E := \{ \epsilon \cdot x; x \in E\}$$ is a neighborhood of $0$. Moreover, by the sublinearity of $p_E$, $$|p_E(x)| = p_E(x) = \epsilon p_E \underbrace{\left( \frac{x}{\epsilon} \right)}_{\in E} \leq \epsilon \cdot 1$$ for all $x \in U$. This shows that $p_E$ is continuous at $0$.
2. continuity everywhere else: Using that $p_E$ is sublinear, we find $$\begin{align*} p_E(x) &\leq p_E(x-y)+ p_E(y) \\ p_E(y) &\leq p_E(y-x) + p_E(x) \\ \Rightarrow |p_E(y)-p_E(x)| &\leq \max\{p_E(x-y),p_E(y-x)\} \tag{1} \end{align*}$$ for all $x,y \in X$. For fixed $\epsilon>0$ there exists (by step 1) a neighborhood $U$ of $0$ such that $p_E(U) \subseteq [0,\epsilon]$. The set $$V := U \cap (-U)$$ is still a neighborhood of $0$, satisfies $V=-V$ and $p_E(V) \subseteq [0,\epsilon]$. For $y \in V_x := x+V$, $(1)$ shows $$|p_E(x)-p_E(y)| \leq \epsilon,$$ i.e. $p_E$ is continuous at $x$. This finishes the proof.