Continuity, sublinearity and strict sublevels of an open convex gauge

Write $S_x=\{t>0:x\in tU\}$. If $t\in S_x$ and $a\ge t$, then $x\mathrm{/}a=(t\mathrm{/}a)(x\mathrm{/}t)+(1-t\mathrm{/}a)0\in U$, so $a\in S_x$. If $a>p(x)$, it cannot be a lower bound of $S_x$; hence some $t\in S_x$ satisfies $t<a$, and then $a\in S_x$. This uses only the defining greatest-lower-bound property, not an assumption that the infimum is attained.

For $r>0$, $S_{rx}=r{S}_x$ by direct substitution, so $p(rx)=rp(x)$: multiplication by $r$ bijects lower bounds of $S_x$ with lower bounds of $r{S}_x$ and preserves their order. At $r=0$, both sides are zero. For $\eta >0$ put $a=p(x)+\eta$ and $b=p(y)+\eta$. These are positive admissible numbers, and $(x+y)\mathrm{/}(a+b)=\frac{a}{a+b}(x\mathrm{/}a)+\frac{b}{a+b}(y\mathrm{/}b)\in U\mathrm{.}$ Thus $p(x+y)\le p(x)+p(y)+2\eta$ for every $\eta >0$. If subadditivity failed by a positive gap $d$, take $\eta =d\mathrm{/}4$ to contradict this bound. Hence $p$ is sublinear.

If $p(x)<1$, choose $a$ with $p(x)<a<1$, for example $(p(x)+1)\mathrm{/}2$. It is admissible, and convexity with zero gives $x\in aU\subseteq U$. Conversely, if $x\in U$, continuity of $s\mapsto sx$ at $s=1$ and openness of $U$ give an $\eta >0$ with $(1+\eta )x\in U$. Thus $1\mathrm{/}(1+\eta )\in S_x$, and $p(x)\le 1\mathrm{/}(1+\eta )<1$. These prove both inclusions of the strict-sublevel identity.

Given $\epsilon >0$, the open zero-neighborhood $N=\epsilon (U\cap (-U))$ has $p(h)<\epsilon$ and $p(-h)<\epsilon$ for $h\in N$, by homogeneity and the strict-sublevel identity. Subadditivity gives $p(x+h)-p(x)\le p(h)$ and $p(x)-p(x+h)\le p(-h)$, hence $\mathrm{\mid }p(x+h)-p(x)\mathrm{\mid }<\epsilon$. Translating $N$ proves continuity at every $x$. This argument does not assert absolute domination by an asymmetric gauge.

Suppose $U$ is balanced. For $\mathrm{\mid }a\mathrm{\mid }=1$, balance gives $aU\subseteq U$ and $a^{-1}U\subseteq U$, whence $aU=U$ and $S_{ax}=S_x$. For $\lambda \mathrm{\ne }0$, write $\lambda =\mathrm{\mid }\lambda \mathrm{\mid }a$ with $\mathrm{\mid }a\mathrm{\mid }=1$, and obtain $p(\lambda x)=\mathrm{\mid }\lambda \mathrm{\mid }p(ax)=\mathrm{\mid }\lambda \mathrm{\mid }p(x)$. At $\lambda =0$ this follows from $p(0)=0$. Thus the finite nonnegative continuous sublinear $p$ is a seminorm.

For concrete boundary calculations on the real line, $U=(-1,1)$ gives $S_x=(\mathrm{\mid }x\mathrm{\mid },\mathrm{\infty })$ and $p(x)=\mathrm{\mid }x\mathrm{\mid }$, with $S_0=(0,\mathrm{\infty })$. On ${\mathbb{R}}^2$ the open strip $U=\{(x,y):\mathrm{\mid }x\mathrm{\mid }<1\}$ gives $p_U(x,y)=\mathrm{\mid }x\mathrm{\mid }$, since admissibility is exactly $t>\mathrm{\mid }x\mathrm{\mid }$; thus $p_U(0,1)=0$ despite $(0,1)\mathrm{\ne }0$. These computations show why no positive-definiteness conclusion is available. All claimed properties are established without HB or AC.