Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $F: \mathbb{R^2} \rightarrow \mathbb{R}$ is convex. If $F$ is moreover strictly convex, then

$$ \frac{1}{2} F(x_1,y_1) + \frac{1}{2} F(x_2, y_2) - F\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = 0 $$

implies $(x_1,y_1) = (x_2,y_2)$. For the application I have in mind, it is enough to show that $x_1 = x_2$ or $y_1 = y_2$.

I have a book that seems to claim the strict convexity of $F$ is not necessary for this to be true: instead, it's sufficient that either

  1. $F(\cdot,y)$ is strictly convex for all $y$.
  2. $F(x,\cdot)$ is strictly convex for all $x$.

holds. Simple as it may seem, I cannot work out why this is true. Any ideas?

P.S.: This is not for homework.

share|improve this question
Looks false. Take $F(x,y)=(x+y)^2$, and the points $(x_1, y_1)=(2,0)$ and $(x_2, y_2) = (0,2)$. Are you sure this is exactly what is writtent in your book ? –  D. Thomine Jan 15 '12 at 19:33
Ah, thank you! No this isn't what is written, this is part of a larger problem. The author states something without giving a proof, and I thought this is what he is relying on. –  user18063 Jan 15 '12 at 20:22

Your Answer


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

Browse other questions tagged or ask your own question.