Proof of A Convex Function Bounded Above on a Set Symmetric about a Point is Bounded Below on It

By claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<2$, so $2$ has a multiplicative inverse $2^{-1}$, and $0<2^{-1}$ by claim 7 of that lemma. From $0\le1$, claim 1 of Elementary Arithmetic in an Ordered Field, and the translation of claim 3 of that lemma we get $1\le 2$; multiplying by the nonnegative number $2^{-1}$, by claim 5 of Elementary Arithmetic in an Ordered Field, gives

$$2^{-1}\le 2^{-1}2=1 .$$

Moreover $1-2^{-1}=2\,2^{-1}-2^{-1}=(2-1)2^{-1}=2^{-1}$.

Let $x\in D$ and put $x'=x_{0}+(x_{0}-x)$, an element of $D$ by hypothesis, and hence of $C$. For every coordinate index $j$, using the vector space operations of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and $2\,2^{-1}=1$,

$$2^{-1}x_{j}+2^{-1}x'_{j}=2^{-1}\Bigl(x_{j}+(x_{0})_{j}+\bigl((x_{0})_{j}-x_{j}\bigr)\Bigr)=2^{-1}\bigl(2\,(x_{0})_{j}\bigr)=(x_{0})_{j},$$

so, since $1-2^{-1}=2^{-1}$,

$$x_{0}=2^{-1}x+\bigl(1-2^{-1}\bigr)x' .$$

As $0\le2^{-1}$ and $2^{-1}\le1$, Convex Real-Valued Function on a Convex Subset of $\mathbb{R}^n$ applies and gives

$$u(x_{0})\le 2^{-1}u(x)+2^{-1}u(x')=2^{-1}\bigl(u(x)+u(x')\bigr).$$

Since $u(x')\le M$, claim 3 of Elementary Order Arithmetic in an Ordered Field gives $u(x)+u(x')\le u(x)+M$, and multiplying by the nonnegative number $2^{-1}$, by claim 5 of Elementary Arithmetic in an Ordered Field, gives

$$u(x_{0})\le 2^{-1}\bigl(u(x)+M\bigr),$$

using claim 1 of Elementary Order Arithmetic in an Ordered Field for transitivity. Multiplying this inequality by the nonnegative number $2$, again by claim 5 of Elementary Arithmetic in an Ordered Field, yields

$$2\,u(x_{0})\le u(x)+M ,$$

and adding $-M$ to both sides, by claim 3 of Elementary Order Arithmetic in an Ordered Field, gives $2\,u(x_{0})-M\le u(x)$.