Let $f$ be a function defined on the unit square $R = [0,1]^2 \subseteq \mathbf{R}^2$ satisfying


* $f \geq 0$, $f(0,0) = 0$,
* $\frac{\partial{f}}{\partial{x}} \geq 0$, $\frac{\partial{f}}{\partial{y}} \leq 0$,
* $\frac{\partial^2{f} }{\partial{x}\partial{y}} \leq 0$.


The last condition is equivalent to the inequality $f(x\_1,y\_1) + f(x\_2,y\_2) \geq f(\min\left(x\_1,x\_2\right), \min\left(y\_1,y\_2\right)) + f(\max\left(x\_1,x\_2\right), \max\left(y\_1,y\_2\right))$ on any rectangle, which can be obtained by integrating $\frac{\partial^2{f} }{\partial{x}\partial{y}}$ over the rectangle.


If we label the vertices of the 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}


After change of variables 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.


Note that without the first bullet condition, the answer is no [Planar function inequality on parallelograms](https://mathoverflow.net/questions/381405/planar-function-inequality-on-parallelograms).


I'm pretty sure the answer is yes with the additional condition - proof?



Extending functional inequality from rectangles to parallelograms
Take for instance $f(x,y)=x^2$, that satisfies the assumptions. Since one has $f(x,y)+f(1-x,1-y)= 1-2x+2x^2 < 1=f(0,0)+f(1,1)$ for all $0<x<1$, it verifies the stated inequality for no parallelogram (inscribed in $R$, with vertices in $(0,0)$ and $(1,1)$) with non-vertical edges.


