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.

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

  • $\begingroup$ I think "inscribed into $D$" usually means "contained in $D$, with all vertices on the boundary of $D$". Is this what you mean by "inscribed" here? $\endgroup$ Jan 17, 2021 at 13:49
  • $\begingroup$ Note that the first bullet condition is irrelevant. $\endgroup$ Jan 17, 2021 at 14:24

1 Answer 1


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.

  • $\begingroup$ Yes, thanks, this does it. I haven't looked at counterexamples enough apparently. The property follows trivially if I also assume subadditivity of $f$ (or 2-subhomogeneity, since $f$ is convex), but then that and the first bullet are sufficient, no partial deriv constraints needed. I was looking to avoid a subadditivity assumption... $\endgroup$ Jan 17, 2021 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.