# Does midpoint-convex imply rationally convex?

Say that a function $$f$$ defined on $$\mathbb{Q}^n$$ is midpoint convex if $$f((x+y)/2) \le (f(x) + f(y))/2$$. Say that it is rationally convex if, for $$\lambda \in \mathbb{Q} \cap [0,1]$$ and $$\bar \lambda = 1-\lambda$$, we have $$f(\lambda x + \bar\lambda y) \le \lambda f(x) + \bar\lambda f(y)$$.

Clearly every rationally-convex function is midpoint-convex.

Question: Is a midpoint-convex function on $$\mathbb{Q}^n$$ necessarily rationally-convex?

Background: I'm looking for sufficient conditions for a function on $$\mathbb{Q}^n$$ to extend to a continuous function on $$\mathbb{R}^n$$. The following results are known and/or easy:

• Rationally-convex functions on $$\mathbb{Q}^n$$ are continuous; similarly real-convex functions on $$\mathbb{R}^n$$ are continuous. (I'm sticking to "proper" functions, with real values, for convenience.)

• Using the axiom of choice, there are rationally-convex functions on $$\mathbb{R}$$. (Take a $$\mathbb{Q}$$-basis of $$\mathbb{R}^n$$ and go from there.)

• Every rationally-convex function $$f$$ on $$\mathbb{Q}^n$$ extends continuously to a real-convex function. (Define the extension to be the $$\liminf$$ of the nearby rational values, and check the result is convex.) As a result, every rationally-convex function on $$\mathbb{Q}^n$$ extends to a continuous function on $$\mathbb{R}^n$$.

• Every measurable, midpoint-convex function on $$\mathbb{R}^n$$ is continuous. (Attributed by Wikipedia to Sierpinski, I have not read the reference. There are many prior questions on MathOverflow that are answered by this theorem.)

These all seem like it should be implying midpoint-convex implies rationally-convex, but I couldn't quite close the loop.

In the case I'm actually interested in, this is a side issue, since my function $$f$$ is additionally (rationally)-homogeneous, and homogeneous and midpoint-convex implies rationally-convex. But I was curious and could not find it in the literature.

• Midpoint convexity implies the convexity condition for dyadic $\lambda$. Why is it not possible to use the same kind of liminf-extension directly from dyadic numbers to $\mathbb{R}^n$ without the intermediate step of $\mathbb{Q}^n$? After all, dyadic numbers are also dense in $\mathbb{R}$ just like $\mathbb{Q}$ is. – Johannes Hahn Dec 3 '19 at 0:19
• @JohannesHahn I was not able to prove continuity (on the dyadics) with midpoint (aka dyadic) convexity. Thus I don't know how to show liminf-extension is an extension. – Dylan Thurston Dec 3 '19 at 1:16

Assume that $$g$$ defined on $$\mathbb{Q}^n$$ is midpoint convex. First we show that we can extend the midpoint inequality to arbitrary means:

$$g((x_1+\dots+x_m)/m)\leq (g(x_1)+\dots+g(x_m))/m$$ for any $$m\in\mathbb{Z}_{\geq1}$$

We can easily prove this for $$m=2^k$$ by using midpoint convexity $$k$$ times.

For general $$m\leq 2^{i}$$ you take $$x_1,\dots,x_m$$ plus $$2^i-m$$ copies of $$x'=(x_1+\dots+x_m)/m$$

This gives $$g(x')=g\left(\frac{(2^i-m)x'+x_1+\cdots+x_m}{2^i}\right) \le\frac{(2^i-m)g(x')+g(x_1)+\dots+g(x_m)}{2^i},$$ which implies $$g(x')=g((x_1+\dots+x_m)/m) \leq (g(x_1)+\dots+g(x_m))/m$$ for any $$m\in\mathbb{Z}_{\geq1}$$.

Finally, taking $$a$$ copies of $$x$$ and $$b$$ copies of $$y$$ we obtain $$g(\frac{ax+by}{a+b})\leq \frac{ag(x)+bg(y)}{a+b}=\left[\frac{a}{a+b}\right]g(x)+\left[\frac{b}{a+b}\right]g(y)$$ for $$a,b\in \mathbb{Z}_\geq0$$ not both zero and hence that $$g$$ is rationally convex.

• Ivan's proof feels like a classic! – Wlod AA Dec 3 '19 at 7:32
• @WlodAA Thank you! The underlying idea though is not mine - I learnt it from a striking proof of the AM-GM inequality which uses exactly the same substitution to go back from powers of 2. The AM-GM inequality can be proved from Jensen's inequality for convex functions hence the connection I guess. – Ivan Meir Dec 3 '19 at 12:30
• Beautiful, thank you! I'm curious to see that AM-GM proof. I guess you prove AM-GM for just 2 variables first and then go from there? – Dylan Thurston Dec 3 '19 at 15:27
• Ivan, I had an association with the old Cauchy's proof of am-gm. You played this theme very nicely, I have enjoyed your clean and simple proof. – Wlod AA Dec 3 '19 at 15:46
• @DylanThurston Yes, basically you just need the 2 variable case and it works out the same from there. I think I read it in "Inequalities" by Hardy, Littlewood and Polya. (I just checked it's there on p 17 of the 2nd edition!) – Ivan Meir Dec 3 '19 at 16:35