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.



Planar function inequality on parallelograms
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 $f(v\_2) + f(v\_4)\not\ge f(v\_1) + f(v\_3)$.


