# Midpoint convex bounded function is continuous

I'm reading a proof of the following statement:

Let $$f: (a,b) \to \mathbb{R}$$ be a midpoint convex (i.e. $$f(1/2(x+y)) \leq 1/2f(x) + 1/2f(y)$$) function that is bounded. Then $$f$$ is continuous.

Here is a proof that I found here: Proving continuity of $f$

To prove that a bounded midpoint convex is continuous, argue by contradiction. Supose $$f$$ is discontinuous at $$x_0\in(a,b)$$. Without loss of generality we may assume $$x_0=0$$, $$f(x_0)=0$$.

First step. There exists a sequence $$\{x_n\}\subset(a,b)$$, such that $$\lim_{n\to\infty}x_n=0$$ and $$\lim_{n\to\infty}f(x_n)=m\ne0$$. We may assume that $$m>0$$.

Second step. The sequence $$\{2\,x_n\}$$ also converges to $$0$$ and $$f(x_n)=f\Bigl(\frac{0+2\,x_n}2\Bigr)\le\frac{f(0)+f(2\,x_n)}2\implies f(2\,x_n)\ge2\,f(x_n)\implies\liminf f(2\,x_n)\ge2\,m.$$ Iteration shows that $$\liminf f(2^k\,x_n)\ge2^k\,m,$$ which is impossible since $$f$$ is bounded.

Question: Why can we assume $$m > 0$$?

• Consider the function $f+2|m|$ which is also mid point convex. Apr 21 '20 at 21:14
• Thanks @copper.hat
– user745578
Apr 21 '20 at 21:16
• but that function wont satisfy $f(x_0)=0$ Apr 21 '20 at 21:21
• @JorgeFernándezHidalgo Good point. I didn't think of that.
– user745578
Apr 21 '20 at 21:25
• I found a simple proof of the theorem in question if you would like Apr 21 '20 at 21:28

We assume $$a<0, $$f$$ is discontinuous at $$0$$ and $$f(0)=0$$.

If not we can solve it by shifting the function horizontal and vertically and stretching horizontally.

Notice that since $$f$$ is discontinuous at $$0$$ there is an $$\epsilon>0$$ such that we can find $$x$$ as small as possible such that $$|f(x)|>\epsilon$$

Notice that if $$f(x) < 0$$ we have that $$f(-x)>0$$ because $$f(x)+f(-x)\ge 0$$.

So without loss of generality $$f(x)>0$$ and now we have $$f(x)\leq f(2x)/2\leq f(4x)/4\dots$$

This tells us $$f(2^nx)\geq 2^n\epsilon$$.

Of course we need to make $$x$$ small so that $$2^nx$$ is inside the interval

• Why didn't I think of that! Thanks!
– user745578
Apr 21 '20 at 21:07
• I was wrong, but we can multiply it by $-1$. Apr 21 '20 at 21:10
• But multiplying by -1 does not preserve midpoint convexity?
– user745578
Apr 21 '20 at 21:10
• I also don't know why they assume that the limit of $f(x_n)$ exists Apr 21 '20 at 21:10
• They can assume that because $f$ is bounded, so $f(x_n)$ has a convergent subsequence.
– user745578
Apr 21 '20 at 21:11

Suppose $$\lim_{n\to\infty}f(x_n)=m<0.$$ Let $$y_n=-x_n$$, then $$0=f(0)=f\left(\frac{x_n+y_n}{2}\right)\leq\frac{f(x_n)+f(y_n)}{2},$$ so $$f(y_n)\geq-f(x_n)$$, then $$\limsup_{n\to\infty}f(y_n)\geq\limsup_{n\to\infty}(-f(x_n)) =\lim_{n\to\infty}(-f(x_n))=-m>0.$$ So there exists subsequences $$\{f(y_{_{n_k}})\}$$ such that $$\lim_{k\to\infty}f(y_{_{n_k}})=\limsup_{n\to\infty}f(y_n)\geq-m>0.$$