MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider points $a,b,c$ (not on a line) and $x_1,...,x_n$ in $\Bbb{R}^2$. I am looking for a necessary and sufficient condition in terms of the geometric configuration of these points such that for any convex function $f : \Bbb{R}^2 \longrightarrow \Bbb{R}$, $\frac{f(x_1)+\cdots+f(x_n)}{n} \leq \frac{f(a)+f(b)+f(c)}{3}$. (As a guess, I think the points $x_i$ must be in the triangle determined by $a,b,c$). If you know a reference or an idea please let me know.

share|cite|improve this question
up vote 6 down vote accepted

The necessary and sufficient conditions are: $(1/n)\sum x_j=(1/3)(a+b+c)$, and all $x_j$ lie inside the closed triangle $(a,b,c)$.

Proof of sufficiency. For every affine function, we have equality. Let $f$ be convex. Then there exists a unique affine function $g$ which matches $f$ at $a$, $b$ and $c$. And we have $f\leq g$ in the triangle $(a,b,c)$. Thus $$\sum f(x_j)/n\leq\sum g(x_j)/n=(g(a)+g(b)+g(c))/3=(f(a)+f(b)+f(c))/3.$$

Necessity. If some point is outside, it is easy to construct a piecewise affine function which is zero on the triangle and positive outside, so the inequality is violated. If $A=(x_1+...+x_n)/n$, does not coincide with $B=(a+b+c)/3$, then it is easy to see that there exists an affine function which violates the inequality: just take any affine $g$ such that $g(A)>g(B)$.

This has a straightforward generalization to any dimension.

share|cite|improve this answer
First, By the barycenter of $x_j$, you mean $\sum x_j$ or $\frac{\sum x_j}{n}$ ? Second, would you give a tip on how do you construct the affine function claimed in the last line of your answer. – user30300 Jul 28 '13 at 9:16
About barycenter: second expression. It is also called the center of gravity. About affine function: take any one which is 0 at A and 1 at B. – Alexandre Eremenko Jul 28 '13 at 11:34
@Alexandre: Reading the necessity part of you nice answer tells me you should edit the first line of your answer. I think you meant $\frac{\sum x_j}{n} = \frac{a+b+c}3$. – user37834 Jul 28 '13 at 15:13
Silvi: thanks, I corrected the error. – Alexandre Eremenko Jul 28 '13 at 16:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.