Is it possible to prove that,


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.

  • $\begingroup$ 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

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.

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).

  • $\begingroup$ 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

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