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

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