Proof of A Convex Function is Lipschitz on a Ball around an Interior Point

Since $x_{0}$ is an interior point of $C$, claim 1 of Interior Points in the Metric Topology are Exactly the Centres of Contained Closed Balls provides $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and $\bar{B}_{d_{E}}(x_{0},\varepsilon)\subseteq C$; this $\varepsilon$ is fixed for the remainder of the proof.

For $i\in[n]$ let $e^{(i)}\in\mathbb{R}^{n}$ be the point whose $i$th coordinate is $1$ and whose other coordinates are $0$. All sums with a numerical index range are the finite sums of $\mathbb{R}$, and $2=1+1$.

Step 1: a coordinate-sum bound. Put $c=\sum_{k=1}^{n}1$. Each summand is nonnegative by claim 1 of Elementary Arithmetic in an Ordered Field, so claim 6 of Properties of Finite Sums gives $1\le c$; with $0<1$ from claim 6 of Elementary Order Arithmetic in an Ordered Field and claim 2 of that lemma we get $0<c$, so $c$ has an inverse $c^{-1}$ with $0<c^{-1}$ by claim 7 of that lemma. For every $y\in\mathbb{R}^{n}$, claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $|y_{i}|\le\lVert y\rVert$ for every $i\in[n]$, so comparing the two sums termwise by Comparison and Absolute Value Bounds for Finite Sums of Real Numbers and using claim 3 of Properties of Finite Sums,

Step 2: the coordinate neighbours lie in $C$. Fix $i\in[n]$. By claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and claim 7 of Properties of Finite Sums, applied to the family $j\mapsto e^{(i)}_{j}e^{(i)}_{j}$, which vanishes off $i$,

the dot product being that of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$. Writing $t=\lVert e^{(i)}\rVert$, the elementary field identities of Zero Products and Elementary Identities in a Field turn $t\,t=1$ into $(t-1)(t+1)=0$, so $t=1$ or $t=-1$ by that lemma; and claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $1=|e^{(i)}_{i}|\le t$, which excludes $t=-1$, since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, hence $-1<0$ by the sign reversal of claim 4 of that lemma, hence $-1<1$ by claim 1. Hence $\lVert e^{(i)}\rVert=1$.

By claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and the value $|\varepsilon|=|-\varepsilon|=\varepsilon$ from Absolute Value in an Ordered Field, both $\lVert\varepsilon e^{(i)}\rVert$ and $\lVert-\varepsilon e^{(i)}\rVert$ equal $\varepsilon$. Since $d_{E}(x_{0},x_{0}+v)=\lVert v\rVert$ for every $v\in\mathbb{R}^{n}$, the points $x_{0}+\varepsilon e^{(i)}$ and $x_{0}-\varepsilon e^{(i)}$ lie in $\bar{B}_{d_{E}}(x_{0},\varepsilon)$, hence in $C$; and $d_{E}(x_{0},x_{0})=0\le\varepsilon$ puts $x_{0}$ in $C$ as well.

Step 3: an upper bound at those points. Let $g:[n]\to\mathbb{R}$ be the family

and put $M=|u(x_{0})|+\sum_{i=1}^{n}g_{i}$. Absolute values are nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field, so each $g_{i}$ is nonnegative by claim 2 of Elementary Arithmetic in an Ordered Field; hence claim 5 of Properties of Finite Sums gives $0\le\sum_{i=1}^{n}g_{i}$ and claim 6 of that lemma gives $g_{i}\le\sum_{j=1}^{n}g_{j}$ for every $i\in[n]$. Using $w\le|w|$ from claim 3 of Properties of the Absolute Value in an Ordered Field and the additions of claim 3 of Elementary Order Arithmetic in an Ordered Field,

where in each chain the omitted terms are nonnegative and claim 1 of Elementary Order Arithmetic in an Ordered Field supplies transitivity.

Step 4: bounds on the coordinate-sum ball. Let

By Steps 2 and 3 the hypotheses of A Convex Function is Bounded Above near a Point by its Values at Coordinate Neighbours hold with $r=\varepsilon$ and with this $M$, so $P\subseteq C$ and $u(w)\le M$ for every $w\in P$.

The point $x_{0}$ lies in $P$, since every term of its defining sum is $|0|=0$ and claim 7 of Properties of Finite Sums makes the sum $0$. If $x\in P$ and $x'=x_{0}+(x_{0}-x)$, then $x'_{i}-(x_{0})_{i}=-\bigl(x_{i}-(x_{0})_{i}\bigr)$, so $|x'_{i}-(x_{0})_{i}|=|x_{i}-(x_{0})_{i}|$ by claim 4 of Properties of the Absolute Value in an Ordered Field together with $|-1|=1$; hence $x'\in P$. So A Convex Function Bounded Above on a Set Symmetric about a Point is Bounded Below on It applies with $D=P$ and gives

Since $u(x_{0})\le M$, multiplying by the nonnegative number $2$ by claim 5 of Elementary Arithmetic in an Ordered Field and adding $-M$ by claim 3 of Elementary Order Arithmetic in an Ordered Field give $m\le M$, so $0\le M-m$.

Step 5: the radius. Put $h=\varepsilon\,2^{-1}$ and $\rho=h\,c^{-1}$; here $0<2^{-1}$ and $0<h$ and $0<\rho$ by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field, and $h+h=\varepsilon(2^{-1}+2^{-1})=\varepsilon$. Also $h\le\varepsilon$, since $\varepsilon-h=h$ is nonnegative. If $x\in\bar{B}_{d_{E}}(x_{0},\rho)$, then $\lVert x-x_{0}\rVert\le\rho$, so by $(1)$ and claim 5 of Elementary Arithmetic in an Ordered Field,

In particular $\bar{B}_{d_{E}}(x_{0},\rho)\subseteq P\subseteq C$.

Step 6: the Lipschitz estimate. Put $L=c\,h^{-1}(M-m)$, a product of nonnegative numbers and hence nonnegative by claim 5 of Elementary Arithmetic in an Ordered Field. Let $x,y\in\bar{B}_{d_{E}}(x_{0},\rho)$.

If $x=y$, then $u(y)-u(x)=0$ and $\lVert y-x\rVert=0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so both sides of the asserted inequality are $0$.

Suppose $x\ne y$ and put $\delta=\sum_{i=1}^{n}|y_{i}-x_{i}|$. The summands are nonnegative, so $0\le\delta$ by claim 5 of Properties of Finite Sums; and if $\delta$ were $0$ the second part of that claim would force $|y_{i}-x_{i}|=0$, hence $y_{i}=x_{i}$, for every $i\in[n]$, contradicting $x\ne y$. So $0<\delta$ and $\delta^{-1}$ exists with $0<\delta^{-1}$.

Put $\mu=h\,\delta^{-1}$ and $z=y+\mu\,(y-x)$. By claim 4 of Properties of the Absolute Value in an Ordered Field, by $|\mu|=\mu$, and by claim 3 of Properties of Finite Sums,

The triangle inequality of claim 5 of Properties of the Absolute Value in an Ordered Field gives $|z_{i}-(x_{0})_{i}|\le|z_{i}-y_{i}|+|y_{i}-(x_{0})_{i}|$ for every $i\in[n]$, so Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, claim 2 of Properties of Finite Sums and the bound $(2)$ applied to $y$ give

so $z\in P$.

Put $\lambda=\delta\,(\delta+h)^{-1}$, which is defined and nonnegative because $0<\delta+h$. From $\delta\le\delta+h$ and claim 5 of Elementary Arithmetic in an Ordered Field, multiplying by the nonnegative number $(\delta+h)^{-1}$ gives $\lambda\le1$; and $1-\lambda=h\,(\delta+h)^{-1}$, since $\delta(\delta+h)^{-1}+h(\delta+h)^{-1}=(\delta+h)(\delta+h)^{-1}=1$. For every $i\in[n]$, using $\delta\mu=h$,

and multiplying by $(\delta+h)^{-1}$ gives $\lambda z_{i}+(1-\lambda)x_{i}=y_{i}$. Hence $y=\lambda z+(1-\lambda)x$.

The points $x$, $y$ and $z$ all lie in $P$, on which $m\le u\le M$ by Step 4, so Increment Bound for a Convex Function through an Extended Point applies with $D=P$ and gives $u(y)-u(x)\le\lambda(M-m)$.

It remains to bound $\lambda$. From $h\le\delta+h$ and $0\le\lambda$, claim 5 of Elementary Arithmetic in an Ordered Field gives $\lambda h\le\lambda(\delta+h)=\delta$, and multiplying by the nonnegative number $h^{-1}$ gives $\lambda\le\delta\,h^{-1}$. Multiplying by the nonnegative number $M-m$, then using $(1)$ with $y-x$ in place of $y$ and multiplying by the nonnegative number $h^{-1}(M-m)$, and finally using claim 1 of Elementary Order Arithmetic in an Ordered Field,

Interchanging the roles of $x$ and $y$ leaves $\delta$ unchanged, because $|x_{i}-y_{i}|=|y_{i}-x_{i}|$, and leaves $\lVert y-x\rVert$ unchanged, by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ with the factor $-1$; the same argument therefore gives $u(x)-u(y)\le L\lVert y-x\rVert$. Claim 6 of Properties of the Absolute Value in an Ordered Field now yields