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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:)

PPPiRi
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2 Answers2

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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 $\Bbb R$.

Ben Grossmann
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Well this is not a correct counterexample, but I suppose it may provide some ideas...

Consider Dirichlet function \begin{eqnarray} D(x)=\left\{ \begin{aligned} 1\ \ \ \ \ \ \ \ \ \ \ \ \ x\in \mathbb{Q}\\ 0\ \ \ \ \ x\in [0,1]-\mathbb{Q} \end{aligned} \right. \end{eqnarray} D(x) is defined on [0,1], and is nowhere continuous. Of course it's not a convex function.

When $x_1$, $x_2\in \mathbb{Q}$, we have $f(\dfrac{x_1+x_2}{2})=1$, and $\dfrac{f(x_1)+f(x_2)}{2}=1$. Similarly, for $x_1 \in \mathbb{Q}$, $x_2\in[0,1]-\mathbb{Q}$, we have left$=0$ and right=$\dfrac12$.

But for $x_1$, $x_2\in [0,1]-\mathbb{Q}$ we have left$=0$ or $1$ and right$=0$. The third situation does not satisfy (2). That's my problem, and I mistook it yesterday. I tried to improve it but failed.

Fish
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  • Notice that if (2) is satisfied, then for any $p\in \mathbb{Q}$, we have $f(px_1+(1-p)x_2)\leq pf(x_1)+(1-p)f(x_2)$ – Fish Jun 06 '16 at 16:28
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    What if $x=1/\pi, y = 1/3-1/\pi?$ – zhw. Jun 06 '16 at 17:04
  • Yes, you're right. I made a mistake for the situation when both $x_1$ and $x_2$ are not in $\mathbb{Q}$... – Fish Jun 07 '16 at 15:03
  • Thank you for answer. Thought your example include error, your answer give me some motivate and strategy to attack the problem! – PPPiRi Jun 08 '16 at 13:05
  • Well,you can refer to this: https://shreevatsa.wordpress.com/2010/06/29/convex-continuous-jensens/ – Fish Jun 08 '16 at 15:33