# Relaxing strict convexity

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.

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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