# Planar function inequality on parallelograms

Let $$f$$ be a function defined on the unit square $$R = [0,1]^2 \subseteq \mathbf{R}^2$$ which is convex and satisfies $$\frac{\partial{f}^2 }{\partial{x}\partial{y}} \leq 0$$. The last condition is equivalent to the inequality $$f(x_1,y_1) + f(x_2,y_2) \geq f(\min\left(x_1,x_2\right), \min\left(y_1,y_2\right)) + f(\max\left(x_1,x_2\right), \max\left(y_1,y_2\right))$$ which can be obtained by integrating $$\frac{\partial{f}^2 }{\partial{x}\partial{y}}$$ over the rectangle.

If we label the vertices of any given rectangle counterclockwise $$v_1, \dots, v_4$$, starting at the upper right, this is saying that $$f(v_2) + f(v_4) \geq f(v_1) + f(v_3)$$. Does this property also hold for parallelograms inscribed in $$R$$ with wlog $$v_1 = (1,1), v_3 = (0,0)$$?

For $$v_2 = (x_2,y_2)$$, $$v_4 = (x_1,y_1)$$, the linear function mapping $$R$$ to the parallelogram $$P$$ is

$$\begin{pmatrix} &x_1 & x_2 \\ &y_1 & y_2 \end{pmatrix}$$

The differential condition on $$P$$ can then be written

$$(x_1y_2 + y_1x_2)f_{xy} + x_1x_2f_{xx} + y_1y_2f_{yy}$$, I don't know that that is necessarily nonpositive.

E.g., let $$f(x,y):=(x-y)^2+x^2+y^2$$ for $$(x,y):=R=[0,1]^2$$, $$v_1:=(1,1)$$, $$v_2:=(0,t)$$, $$v_3:=(0,0)$$, and $$v_4:=(1,1-t)$$, where $$t\in(0,1/2)$$. Then $$f$$ is convex, $$\frac{\partial{f}^2 }{\partial{x}\partial{y}} \le 0$$, $$v_1v_2v_3v_4$$ is a parallelogram inscribed into $$R$$, but $$f(v_2) + f(v_4)\not\ge f(v_1) + f(v_3)$$.
• Thanks, that answers it. I wish to add conditions : $\frac{\partial{f}}{\partial{x}} \geq 0$ and $\frac{\partial{f}}{\partial{y}} \leq 0$. I think the parallelogram inequality holds in this case - proof? – Charles Pehlivanian Jan 17 at 1:56
• @CharlesPehlivanian : If you want these additional conditions to hold as well, add $nx-ny$ to $f(x,y)$ in the example, where $n>0$ is large enough. I guess $n=10$ will suffice. – Iosif Pinelis Jan 17 at 2:14