Planar function inequality on parallelograms

Question

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.

Answer 1

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 $f(v_2) + f(v_4)\not\ge f(v_1) + f(v_3)$.