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Let $f$ be bounded in $[a,b]$. Assume $f$ satisfies $$f\left(\frac{x_1 + x_2}{2}\right) \le\frac{f(x_1) + f(x_2)}{2}$$
for all $a\le x_1 \le x_2 \le b $ . Prove that $f$ is continous on $x$ when $a< x < b$.

I'm more concerned with how I can derive the prove of this question

Martins Bruveris
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lio
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  • I'm curious about why people put close votes on a problem like this. –  Mar 10 '17 at 07:48

2 Answers2

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A function that satisfies $$f\left(\frac{x_1 + x_2}{2}\right) \le\frac{f(x_1) + f(x_2)}{2}$$ is called midpoint convex, which is a slightly weaker notion than convexity.

While every convex function is midpoint convex, not every midpoint convex function is convex. Every convex function is contiuous. Also, every continuous, midpoint convex function is convex. However, not every midpoint convex function is contiunous. A counterexample is contructed here.

EDIT: I overlooked that the counterexample in the link is not bounded. So the problem is still open.

EDIT 2: According to this paper by Green and Gustin, Ostrowski showed in a paper in 1929, written in german, that every midpoint convex function, bounded above on a set of positive measure is continuous. On his part, Ostrowski refers to a paper by Jensen from 1906, written in French, where it is shown on p. 188 that every bounded midpoint convex function is continuous. Thus we are done.

Martins Bruveris
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$f(x)$ is bounded on $[a,b]$ and it is convex function on$[a,b]$ so it is continuous on $(a,b)$

xyz
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    Why is it convex? – Zain Patel Mar 10 '17 at 05:17
  • @xyz The question only presumes that $f$ is midpoint convex, which doesn't directly imply convexity unless you make additional assumptions such as continuity (see here or here for example), so your proposed answer is circular. – dxiv Mar 10 '17 at 07:57