about convex function counterexample Proposition) If a function $f : [a, b] \rightarrow \mathbb{R}$ satisfies two conditions that
(1) $f$ is continuous
(2) $f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}$ for every $x$, $y$
then $f$ is convex function.
I already know the proof of above proposition.
My question is how can I find counterexample without condition (1). 
That is, I want to find non-convex function $f$ satisfying condition (2).
Thank you:)

My guess is that you're not going to be able to come up with any nice construction that works as a counterexample.  In particular, I think you need to invoke the axiom of choice to create a suitable counterexample.
For example, any discontinuous solution to Cauchy's functional equation will work here.

In particular, I would say that (2) is just enough to tell you that $f$ acts as a convex function when restricted to any coset of $\Bbb Q$ as a subset of $\Bbb R$.
