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This question comes from Rudin's book "principles of mathematical analysis" chapter 4,exercise 24,on page 101.

The original question is:

Assume that f is a continuous real function defined in $(a,b)$ such that $$f(\frac{x+y}{2})\leqslant\frac{1}{2}f(x)+\frac{1}{2}f(y)$$ for all $x,y\in (a,b)$.Prove that f is convex.

I have solved this question.But when I am reading the definition of convex function,I find that convex function is not always continuous.So I want to ask if there exists a discontinuous function which satisfies $f(\frac{x+y}{2})\leqslant\frac{1}{2}f(x)+\frac{1}{2}f(y)$ but is not convex? Thanks!

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marked as duplicate by Brahadeesh, Calvin Khor, Parcly Taxel, Arnaud D., José Carlos Santos Oct 26 '18 at 16:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ All convex functions on $(a,b)$ are continuous. $\endgroup$ – Kabo Murphy Oct 26 '18 at 12:04
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Any additive function, i.e., one with $$\tag1f(x+y)=f(x)+f(y)$$ for all $x,y$ will have $$f\left(\frac{x+y}2\right) =f\left(\frac x2\right)+f\left(\frac y2\right)=\frac12\left(f\left(\frac x2\right)+f\left(\frac x2\right)+f\left(\frac y2\right)+f\left(\frac y2\right)\right)=\frac12\left(f(x)+f(y)\right).$$ Once you abandon continuity, there are many solutions to $(1)$ - and they are not convex either. In fact, they are so discontinuous that they are unbounded in every open interval.

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