Proof of Properties of the Minkowski Gauge

Write $T_{x}$ for the set of The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum, so that $p_{U}(x)=\inf T_{x}$ for $x\in X$ by The Minkowski Gauge of an Open Convex Neighbourhood of the Origin (i). Throughout, balls and openness in $X$ refer to the norm metric $d_{X}(u,v)=\lVert u-v\rVert$ (The Norm Metric of a Real Normed Space (i)), which is a metric, in particular symmetric, by The Norm Metric of a Real Normed Space (ii). Two elementary facts about infima are used below: for $\varepsilon>0$ the number $\inf T_{x}+\varepsilon$ is not a lower bound of $T_{x}$, so some element of $T_{x}$ is smaller than it; and a real number that is at most every $\varepsilon>0$ is at most $0$.

Homogeneity. Let $x\in X$ and $\lambda>0$. Since $(\lambda t)^{-1}(\lambda x)=t^{-1}x$ for $t>0$ (Vector Space over a Field (v)), a real $t$ lies in $T_{x}$ exactly when $\lambda t$ lies in $T_{\lambda x}$; that is,

$$T_{\lambda x}=\{\lambda t:t\in T_{x}\}.$$

Hence $\lambda p_{U}(x)$ is a lower bound of $T_{\lambda x}$; and if $b$ is any lower bound of $T_{\lambda x}$, then $\lambda^{-1}b$ is a lower bound of $T_{x}$, so $\lambda^{-1}b\le p_{U}(x)$, that is, $b\le\lambda p_{U}(x)$. Thus $\lambda p_{U}(x)$ is the greatest lower bound of $T_{\lambda x}$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), and $p_{U}(\lambda x)=\lambda p_{U}(x)$. For $\lambda=0$ we have $0x=0_{X}$ (Elementary Identities in a Vector Space (ii)) and $t^{-1}0_{X}=0_{X}\in U$ for every $t>0$ (Elementary Identities in a Vector Space (iii)), so every real $\varepsilon>0$ lies in $T_{0_{X}}$. Then $0$ is a lower bound of $T_{0_{X}}$, and every lower bound $b$ satisfies $b\le\varepsilon$ for all $\varepsilon>0$, hence $b\le0$; so, by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv), $p_{U}(0x)=p_{U}(0_{X})=0=0\cdot p_{U}(x)$. This is Sublinear Functional (ii).

Subadditivity. Let $x,y\in X$ and $\varepsilon>0$, and put $s=p_{U}(x)+\varepsilon$ and $t=p_{U}(y)+\varepsilon$; both are positive, as $p_{U}\ge0$ by The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum (iii). Some element of $T_{x}$ is smaller than $s$, so $s\in T_{x}$ by The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum (ii); likewise $t\in T_{y}$. The identity

$$\frac{1}{s+t}(x+y)=\frac{s}{s+t}\Bigl(\frac{1}{s}x\Bigr)+\frac{t}{s+t}\Bigl(\frac{1}{t}y\Bigr)$$

holds by Vector Space over a Field (vii) and Vector Space over a Field (v), and its right-hand side is a convex combination of $s^{-1}x\in U$ and $t^{-1}y\in U$, so it lies in $U$ by Convex Subset of a Real Vector Space (i). Hence $s+t\in T_{x+y}$, and

$$p_{U}(x+y)\le s+t=p_{U}(x)+p_{U}(y)+2\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $p_{U}(x+y)\le p_{U}(x)+p_{U}(y)$, which is Sublinear Functional (i). Together with homogeneity, Sublinear Functional (iii) shows that $p_{U}$ is sublinear, proving (i).

Bounds. Let $r>0$ with $B(0_{X},r)\subseteq U$, and let $x\in X$. First, $p_{U}(x)\ge0$ by The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum (iii). By The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum (i) every real $t>r^{-1}\lVert x\rVert$ lies in $T_{x}$, so $p_{U}(x)\le r^{-1}\lVert x\rVert+\varepsilon$ for every $\varepsilon>0$, whence $p_{U}(x)\le r^{-1}\lVert x\rVert$. This proves (ii).

The sublevel set. Let $x\in U$. By Open Subset of a Metric Space (i) fix $\rho>0$ with $B(x,\rho)\subseteq U$, and put $\theta=\rho/(\lVert x\rVert+1)>0$. By Vector Space over a Field (viii) and Norms and Real Normed Spaces (iii),

$$\lVert(1+\theta)x-x\rVert=\theta\lVert x\rVert<\rho,$$

so $(1+\theta)x\in B(x,\rho)\subseteq U$ by Open Ball in a Metric Space (ii). Hence $(1+\theta)^{-1}\in T_{x}$, and since an infimum is a lower bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), $p_{U}(x)\le(1+\theta)^{-1}<1$. Conversely, let $p_{U}(x)<1$. Then some $s\in T_{x}$ satisfies $s<1$, so $1\in T_{x}$ by The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum (ii); that is, $x=1^{-1}x\in U$ (Vector Space over a Field (vi)). This proves (iii).