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.

  • 1
    $\begingroup$ 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. $\endgroup$ Dec 3, 2019 at 0:19
  • $\begingroup$ @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. $\endgroup$ Dec 3, 2019 at 1:16

1 Answer 1


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.

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    $\begingroup$ Ivan's proof feels like a classic! $\endgroup$
    – Wlod AA
    Dec 3, 2019 at 7:32
  • $\begingroup$ @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. $\endgroup$
    – Ivan Meir
    Dec 3, 2019 at 12:30
  • $\begingroup$ 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? $\endgroup$ Dec 3, 2019 at 15:27
  • $\begingroup$ 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. $\endgroup$
    – Wlod AA
    Dec 3, 2019 at 15:46
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    $\begingroup$ @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!) $\endgroup$
    – Ivan Meir
    Dec 3, 2019 at 16:35

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