3
$\begingroup$

Is it possible to prove that,

$$f(x,y)=\left(\frac{x^a(1-y)}{(1-x)}-\frac{y^{a}(1-x)}{(1-y)}\right)\frac{1}{x-y}$$

is convex in the following range: $0<y<x<1$, where $a\ge2$ is an integer parameter.

Numerical testing shows that it is definitely convex, but can someone help to prove it? If it is yet not convex I would be very thankful for a counter example.

The Hessian is:

$\frac{\partial^{2}f}{\partial x \partial y}=\frac{xy^{a}(a(x-y)+2y)+x^{a}y(ax-2x-ay)}{xy(x-y)^3}$

$\frac{\partial^{2}f}{\partial x \partial x}=\frac{(y-1) x^{a+2} \left(a^2 (y (y+4)+1)-3 a (y (y+4)+1)+2 \left(y^2+y+1\right)\right)-2 (a-2) a y \left(y^2-1\right) x^{a+1}-2 (a-3) (a-1) \left(y^2-1\right) x^{a+3}+(a-3) (a-2) (y-1) x^{a+4}+(a-1) a (y-1) y^2 x^a-2 x^5 y^a+6 x^4 y^a-6 x^3 y^a+2 x^2 y^a}{(x-1)^3 x^2 (x-y)^3}$

according to Mathematica.

Below you can find some developments but the problem is not solved yet.

$\endgroup$
  • $\begingroup$ could you compute the Hessian? $\endgroup$ – LinAlg Feb 5 at 21:35
  • $\begingroup$ I did, but i could not say much about it. Also thought about going through convexity definition, tried to split it as a sum and check each component as well...still, nothing. It seemed simple at first glance but it appears to be tricky. $\endgroup$ – Y.L Feb 5 at 21:43
  • $\begingroup$ please share the Hessian with us $\endgroup$ – LinAlg Feb 5 at 21:45
  • $\begingroup$ I did. Used Mathematica. $\endgroup$ – Y.L Feb 5 at 22:00
  • $\begingroup$ What's the story of $f$? Why is it desirable to have a convex $f$, and how did the parameter $a$ emerge? $\endgroup$ – Hanno Mar 29 at 12:53
1
$\begingroup$

This is a partial answer.

Let us write $$f(x,y) \:=\: \frac{x^a(1-y)^2-y^a(1-x)^2}{(1-x)(1-y)(x-y)}\tag{1}$$ and exploit that convex- or concavity is invariant under affine transformations.
Setting $\,u=1-y\,$ and $\,v=1-x\,$ leads to $$\varphi(u,v,a) \:=\: \frac{u^2(1-v)^a-v^2(1-u)^a}{uv(u-v)}\tag{2}$$ with $\,0<v<u<1$.
As a side note: If $\,u=v>0\,$ then both the denominator and the numerator vanish, hence these are removable discontinuities of this rational function. Furthermore, there's the symmetry $\,\varphi(v,u,a)=\varphi(u,v,a)\,$.

Expanding the $(1-\,?\,)^a$ terms in $(2)$, combining equal powers of $v$ and $u$, and simplifying then yields $$\varphi(u,v,a) \:=\: \frac 1u+\frac 1v -a +\sum_{k=2}^{a-1}(-1)^k\binom a{k+1} \underbrace{\frac{vu^k-uv^k}{u-v}}_{=\,\sum_{j=1}^{k-1}u^{k-j}v^j}\:.$$ Thus, $$\varphi(u,v,a=2) \:=\:\frac 1u+\frac 1v -2$$ is convex because each summand is itself convex, and $$\varphi(u,v,a=3) \:=\:\frac 1u+\frac 1v -3 +uv$$ is convex since its Hessian is $$\begin{pmatrix}\frac 2{u^3} & 1\\ 1 & \frac 2{v^3} \end{pmatrix} > 0$$ positive-definite.

Summary:
In the cases $a=2,3$ one gets that $f$ is a convex function.
The ansatz presented may pave the way for a complete answer
(which I currently do not see).

$\endgroup$
  • $\begingroup$ Thank you so much @Hanno. It helps. Do you think it is possible to proceed from your first two steps with induction? $\endgroup$ – Y.L Mar 31 at 17:27
  • $\begingroup$ @Y.L I got stuck in my analysis, and then opted to write a partial answer at least. But the challenge is still on my table, incl. "inductive trials". $\endgroup$ – Hanno Apr 1 at 8:18
  • $\begingroup$ trying to apply induction following this development one gets that $$\varphi(u,v,a+1)=\varphi(u,v,a)+\frac{(1-u)^{a}v-(1-v)^{a}u}{u-v}$$. $\endgroup$ – Y.L Apr 4 at 19:07
  • $\begingroup$ Unfortunately while assuming $\varphi(u,v,a)$ is convex, the added term on the RHS is concave. Do you see a way out? $\endgroup$ – Y.L Apr 4 at 19:53

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.