## Discontinuous Midpoint Convex Functions

I’ve have provided a cleaner proof of Lemma 3 than the one presented in my previous post. I have also included a stronger result, which shows that a midpoint convex function is either continuous everywhere or discontinuous everywhere.

In what follows, we assume that ${E}$ is an open convex set of some normed space. We can actually replace normed space by metric vector space, which does not change the proof.

Lemma 1 If ${f:E\rightarrow\mathbb{R}}$ is midpoint convex and discontinuous at ${x_{0}\in E}$, then ${\limsup_{x\rightarrow x_{0}}f(x)=+\infty}$.

Proof: Replacing ${f}$ by ${g:=f(\cdot-x_{0})}$, we may assume that ${x_{0}=0}$. Replacing ${f}$ by ${f-f(0)}$, we may assume that ${f(0)=0}$. Set ${m=\limsup_{x\rightarrow 0}f(x)}$, and let ${(x_{n})_{n=1}^{\infty}}$ be a sequence converging to ${0}$ so that ${f(x_{n})\rightarrow m}$. I claim that ${m>0}$. Otherwise, by midpoint convexity,

$\displaystyle 0=f(0)=2f\left(\dfrac{x_{n}-x_{n}}{2}\right)\leq f(x_{n})+f(-x_{n})\Rightarrow 0<-m\leq\liminf_{n\rightarrow\infty} f(-x_{n})$

But ${-x_{n}\rightarrow 0}$, which contradicts the definiion of ${m}$. Discarding finitely many terms, we may assume that the sequence ${(2x_{n})_{n=1}^{\infty}}$ is contained in ${E}$. By midpoint convexity,

$\displaystyle 2f(x_{n})=2f\left(\dfrac{0+2x_{n}}{2}\right)\leq f(0)+f(2x_{n})=f(2x_{n})$

Taking the liminf of both sides, we obtain

$\displaystyle 2m\leq\liminf_{n\rightarrow\infty}f(2x_{n})$

Suppose we have shown that ${2^{k}m\leq\liminf f(2^{k}x_{n})}$. Discarding finitely many terms, we may assume that the sequence ${(2^{k+1}x_{n})_{n=1}^{\infty}}$ is contained in ${E}$. By midpoint convexity,

$\displaystyle 2f(2^{k}x_{n})=2f\left(\dfrac{0+2^{k+1}x_{n}}{2}\right)=f(0)+f(2^{k+1}x_{n})=f(2^{k+1}x_{n})$

Taking the liminf of both sides, we obtain

$\displaystyle 2^{k+1}m\leq 2\left(\liminf f(2^{k}x_{n})\right)\leq\liminf f(2^{k+1}x_{n})$

By induction, we obtain ${\liminf f(2^{k}x_{n})\geq 2^{k}m}$ for all ${k\in\mathbb{N}}$. $\Box$

Theorem 2 Let ${f:E\rightarrow\mathbb{R}}$ be a midpoint convex function defined on a convex open set ${U}$. If ${f}$ is discontinuous at a point ${x_{0}\in U}$, then ${f}$ is unbounded on every open convex subset of ${U}$. Hence, ${f}$ is discontinuous everywhere.

Proof: Replacing ${f}$ by the midpoint convex function ${g=f(\cdot-x_{0})}$, we may assume that ${x_{0}=0}$. By the same argument, we may assume that ${E}$ is symmetric. By Lemma 1, there exists a sequence ${x_{n}\rightarrow 0}$ and ${f(x_{n})\rightarrow +\infty}$. Let ${z\in E}$ be an arbitrary point. Since ${E}$ is open, there exists a neighborhood of ${z}$ which contains almost all terms of the sequence ${(z+2x_{n})_{n=1}^{\infty}}$. Relabeling, if necessary, we may assume that it contains the entire sequence. Observe that ${z+2x_{n}\rightarrow z}$.

$\displaystyle 2f(x_{n})=2f\left(\dfrac{z+2x_{n}-z}{2}\right)\leq f(z+2x_{n})+f(-z)$

Taking the liminf of both sides, we see that ${f(z+2x_{n})\rightarrow+\infty}$. Since ${z}$ was arbitrary, we conclude that ${f}$ is unbounded in every neighborhood of its domain. $\Box$