0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ – Charles Pehlivanian Jan 17 at 1:56
  • $\begingroup$ Maybe I'll ask in another post. $\endgroup$ – Charles Pehlivanian Jan 17 at 1:58
  • $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Jan 17 at 2:14
  • $\begingroup$ Thanks, this all works. I'm going to add additional conditions and post as a separate question. $\endgroup$ – Charles Pehlivanian Jan 17 at 5:17

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.