The open convex gauge is sublinear and recovers its set

Write $p=p_U$. The gauge definition gives $p(x)\ge 0$. For every $t>\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$ one has $x\mathrm{/}t\in B(0,r)\subseteq U$, so $p(x)\le t$. If $p(x)>\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$, their midpoint is such a $t$ smaller than $p(x)$, impossible. Hence $p(x)\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$.

For $a>0$, the condition $s\in S_{ax}$ is equivalent to $s\mathrm{/}a\in S_x$, so $S_{ax}=a{S}_x$. The number $ap(x)$ is a lower bound of this set. Conversely, for every $\epsilon >0$, choose $t\in S_x$ with $t<p(x)+\epsilon \mathrm{/}a$; then $at\in S_{ax}$ and $at<ap(x)+\epsilon$. The infimum criterion gives $p(ax)=ap(x)$. For $a=0$, both sides are zero by $p(0)=0$.

If $s>p(x)$, choose $s^{\mathrm{\prime }}\in S_x$ with $s^{\mathrm{\prime }}<s$ using the infimum criterion with $\epsilon =s-p(x)$. Since $0<s^{\mathrm{\prime }}\mathrm{/}s<1$, convexity and $0\in U$ give $x\mathrm{/}s=(s^{\mathrm{\prime }}\mathrm{/}s)(x\mathrm{/}{s}^{\mathrm{\prime }})+(1-s^{\mathrm{\prime }}\mathrm{/}s)0\in U$. Thus every $s>p(x)$ is admissible.

If $p(x)<1$, choose $s\in S_x$ with $s<1$ using the infimum criterion with $\epsilon =1-p(x)$. Then $0<s<1$ and $x=s(x\mathrm{/}s)+(1-s)0\in U$ by convexity.

Given $\epsilon >0$, set $s=p(x)+\epsilon$ and $t=p(y)+\epsilon$. Both are positive and admissible by step 1.3. Convexity gives $(x+y)\mathrm{/}(s+t)=(s\mathrm{/}(s+t))(x\mathrm{/}s)+(t\mathrm{/}(s+t))(y\mathrm{/}t)\in U$. Hence $p(x+y)\le s+t=p(x)+p(y)+2\epsilon$. If the desired inequality failed with positive difference $d$, taking $\epsilon =d\mathrm{/}4$ would give $d\le d\mathrm{/}2$. Therefore $p(x+y)\le p(x)+p(y)$. Together with step 1.2, this is sublinearity on the real space.

Conversely, let $x\in U$. For $x=0$, $p(x)=0<1$. If $x\mathrm{\ne }0$, openness gives $\eta >0$ with $B(x,\eta )\subseteq U$. Put $d=\eta \mathrm{/}(2\mathrm{\parallel }x\mathrm{\parallel })>0$. Since $\mathrm{\parallel }(1+d)x-x\mathrm{\parallel }=\eta \mathrm{/}2$, $(1+d)x\in U$, so $1\mathrm{/}(1+d)\in S_x$ and $p(x)\le 1\mathrm{/}(1+d)<1$. This proves both inclusions in the asserted set equality.

Subadditivity gives $p(x)-p(y)\le p(x-y)\le \mathrm{\parallel }x-y\mathrm{\parallel }\mathrm{/}r$ and, with $x,y$ interchanged, $p(y)-p(x)\le p(y-x)\le \mathrm{\parallel }x-y\mathrm{\parallel }\mathrm{/}r$. These two real inequalities yield the Lipschitz bound. When $x=y$ both differences are zero; no use of $p(-v)=p(v)$ occurs.