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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 $m > 0$?

2 Answers2

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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

Riemann
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Asinomás
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    Why didn't I think of that! Thanks! –  Apr 21 '20 at 21:07
  • I was wrong, but we can multiply it by $-1$. – Asinomás Apr 21 '20 at 21:10
  • But multiplying by -1 does not preserve midpoint convexity? –  Apr 21 '20 at 21:10
  • I also don't know why they assume that the limit of $f(x_n)$ exists – Asinomás Apr 21 '20 at 21:10
  • They can assume that because $f$ is bounded, so $f(x_n)$ has a convergent subsequence. –  Apr 21 '20 at 21:11
  • Oh ok, that makes sense – Asinomás Apr 21 '20 at 21:11
  • But don't we have a problem? If $f$ is midpoint convex, then $-f$ is not necessarily midpoint-convex. I think absolute value $x \mapsto |x|$ is a counterexample. –  Apr 21 '20 at 21:12
  • Yeah, my reasoning is not sound you are right. – Asinomás Apr 21 '20 at 21:13
  • I think we cannot assume it, but as long as the limit is nonzero we can do the same thing – Asinomás Apr 21 '20 at 21:19
  • Oh nvm ${}{}{}{}$ – Asinomás Apr 21 '20 at 21:22
  • I believe the proof fails if $m < 0$. But look at the comment above. I think that solves it. –  Apr 21 '20 at 21:22
  • Hi. Did you use the assumption that $a<0 <b$? I can't see where it is used –  Apr 22 '20 at 17:37
  • Also, I don't see how we can suppose wlog that $f(x) > 0$. If $f$ is continuous at the left, but not at the right in $x_0$, then you won't get that $|f(-x)| > \epsilon$ for example. –  Apr 22 '20 at 17:44
  • It isn't used, but I need for $0$ to be inside of the domain because it is the point of discontinuity. If you change $0$ with something else the numbers get a bit messy (they aren't $x,2x,4x$ etc – Asinomás Apr 22 '20 at 17:44
  • Thanks for the swift reply. What about my second comment above? As it now stands, the proof only holds for positive functions. –  Apr 22 '20 at 17:45
  • $f(-x)>\epsilon$ because $f(x) + f(-x)\geq 0$, not because of continuity. – Asinomás Apr 22 '20 at 17:45
  • I see! Thanks! I will go through it in detail again. –  Apr 22 '20 at 17:46
  • In fact this idea can be used to restore the original proof you had. Instead of taking $a_n$ you take $x_0-a_n$ and then take a convergent subsequence of it. ( that is the reason why you can take $m>0$ in the original proof you supplied). – Asinomás Apr 22 '20 at 17:47
  • I find this proof a lot easier to understand to be honest. Thanks again for that! –  Apr 22 '20 at 17:48
  • Sure, no problem ! – Asinomás Apr 22 '20 at 17:48
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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 $$\lim_{k\to\infty}f(y_{_{n_k}})=\limsup_{n\to\infty}f(y_n)\geq-m>0.$$

Riemann
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