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 $m > 0$?
Best Answer
We assume $a<0<b$, $f$ 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