Proof of A Convex Function is Bounded Above near a Point by its Values at Coordinate Neighbours

Let $x\in\mathbb{R}^{n}$ satisfy the stated inequality, write $y_{i}=x_{i}-(x_{0})_{i}$ for $i\in[n]$ and put $s=\sum_{i=1}^{n}|y_{i}|$, so that $s\le r$. Let $S$ be the successor map of Natural Numbers and put $N=S(n)$; by claim 3 of Basic Properties of Initial Segments of the Natural Numbers we have $[N]=[n]\cup\{N\}$ with $N\notin[n]$, so a family indexed by $[N]$ may be prescribed separately on $[n]$ and at $N$.

Since $0<r$, claim 7 of Elementary Order Arithmetic in an Ordered Field gives $0<r^{-1}$ for the multiplicative inverse $r^{-1}$ of $r$.

The family of points. For $i\in[n]$ set $\sigma_{i}=1$ if $0\le y_{i}$ and $\sigma_{i}=-1$ otherwise, so that $\sigma_{i}|y_{i}|=y_{i}$ in either case, by Absolute Value in an Ordered Field. Define $v:[N]\to\mathbb{R}^{n}$ by

$$v_{k}=x_{0}+(r\sigma_{k})\,e^{(k)}\ \ (k\in[n]),\qquad v_{N}=x_{0}.$$

Each $v_{k}$ with $k\in[n]$ is $x_{0}+r\,e^{(k)}$ or $x_{0}-r\,e^{(k)}$, so every value of $v$ lies in $C$ and satisfies $u(v_{k})\le M$, by hypothesis.

The weights. Define $t:[N]\to\mathbb{R}$ by

$$t_{k}=|y_{k}|\,r^{-1}\ \ (k\in[n]),\qquad t_{N}=1-s\,r^{-1}.$$

For $k\in[n]$ we have $0\le|y_{k}|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so $0\le t_{k}$ by claim 5 of Elementary Arithmetic in an Ordered Field. Multiplying $s\le r$ by the nonnegative number $r^{-1}$, again by claim 5 of Elementary Arithmetic in an Ordered Field, gives $s\,r^{-1}\le r\,r^{-1}=1$, so $0\le t_{N}$ by claim 3 of Elementary Order Arithmetic in an Ordered Field.

By the recursion in claim 1 and by claim 3 of Properties of Finite Sums,

$$\sum_{k=1}^{N}t_{k}=\Bigl(\sum_{k=1}^{n}|y_{k}|\,r^{-1}\Bigr)+t_{N}=s\,r^{-1}+\bigl(1-s\,r^{-1}\bigr)=1,$$

so $t$ is a system of convex weights of length $N$.

The combination is $x$. Fix $j\in[n]$. For $k\in[n]$ the $j$th coordinate of $v_{k}$ is $(x_{0})_{j}+r\sigma_{k}$ if $k=j$ and $(x_{0})_{j}$ otherwise, while the $j$th coordinate of $v_{N}$ is $(x_{0})_{j}$. Hence, writing $w:[N]\to\mathbb{R}$ for the family with $w_{j}=r\sigma_{j}$ and $w_{k}=0$ for every $k\in[N]$ with $k\ne j$, we have $(v_{k})_{j}=(x_{0})_{j}+w_{k}$ for every $k\in[N]$. By claims 2 and 3 of Properties of Finite Sums, by the value of $\sum_{k=1}^{N}t_{k}$ computed above, and by claim 7 of Properties of Finite Sums applied to the family $k\mapsto t_{k}w_{k}$, which vanishes off $j$,

$$\Bigl(\sum_{k=1}^{N}t_{k}v_{k}\Bigr)_{j}=\sum_{k=1}^{N}t_{k}\bigl((x_{0})_{j}+w_{k}\bigr)=(x_{0})_{j}\sum_{k=1}^{N}t_{k}+\sum_{k=1}^{N}t_{k}w_{k}=(x_{0})_{j}+t_{j}\,r\sigma_{j},$$

the convex combination being that of Convex Combination of Finitely Many Points of $\mathbb{R}^n$. Finally

$$t_{j}\,r\sigma_{j}=|y_{j}|\,r^{-1}r\,\sigma_{j}=\sigma_{j}|y_{j}|=y_{j},$$

so the $j$th coordinate of the combination is $(x_{0})_{j}+y_{j}=x_{j}$. As $j\in[n]$ was arbitrary, $x=\sum_{k=1}^{N}t_{k}v_{k}$.

Conclusion. All values of $v$ lie in the convex set $C$, so $x$ lies in $C$ by claim 3 of Small Cases, Reduction, and Membership for Convex Combinations. By Jensen's Inequality for Finite Convex Combinations,

$$u(x)\le\sum_{k=1}^{N}t_{k}\,u(v_{k}).$$

For every $k\in[N]$ we have $u(v_{k})\le M$ and $0\le t_{k}$, so $t_{k}u(v_{k})\le t_{k}M$ by claim 5 of Elementary Arithmetic in an Ordered Field. Comparing the two sums termwise by Comparison and Absolute Value Bounds for Finite Sums of Real Numbers and then using claim 3 of Properties of Finite Sums,

$$\sum_{k=1}^{N}t_{k}\,u(v_{k})\le\sum_{k=1}^{N}t_{k}M=M\sum_{k=1}^{N}t_{k}=M,$$

whence $u(x)\le M$ by claim 1 of Elementary Order Arithmetic in an Ordered Field.