A function $f\colon X\to \mathbb R$ is called convex if

$$\forall (x,y)\in X^2,\quad \forall t\in[0,1],\quad f(tx+(1-t)y)\leqslant tf(x)+(1-t)f(y).$$

Intuitively, it would seem that if we only impose that condition for $t=\frac 12$ we would get the same set of convex functions.

For a continuous function, I believe this is true (the following drawing convinced me):

enter image description here

My problem is that I was not able to find a counter-example, even for a discontinuous function.

I do believe there exists one though.

So what would be a function $\varphi$ which would verified (i) but not (ii) ?

$$(i) \quad \forall (x,y)\in X^2,\quad \varphi\left(\frac {x+y}2\right)\leqslant \frac{\varphi(x)+ \varphi(y)}2.$$

$$(ii) \quad \forall (x,y)\in X^2,\quad \forall t\in[0,1],\quad \varphi(tx+(1-t)y)\leqslant t \varphi(x)+(1-t) \varphi(y).$$


The phrase to look up is "midpoint convexity". You are correct that for continuous functions this implies convexity. In fact, it is true for Lebesgue measurable functions as well. Thus any counterexample will be non-constructive.

Here is one. Consider a Hamel basis $B$ of the real numbers over the rationals, let $b_0$ be one member of $B$, and let $f(x) = c_{b_0}$ where $x = \sum_{b \in B} c_b b$ is the unique expression of $x$ as a finite linear combination of members of $B$ with $c_b$ rational. Then $f$ is midpoint-convex, in fact $f((x+y)/2) = f(x)/2 + f(y)/2$, but $f$ is not convex.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.