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Checking whether Hessian matrix of a function is positive semidefinite on a hypercube

Posted 1 day ago

Given parameters $w_1, \ldots, w_n > 0$ with $\sum_{j=1}^n w_j = 1$ ( $n\ge 2$), define function

$$ f(x_1, \ldots, x_{n-1})= \sum_{j=1}^{n} \frac{x_j}{1-x_j} \cdot \sum_{j=1}^n \frac{w_j}{1-w_j} - \sum_{j=1}^n \frac{x_j}{1-w_j} \cdot \sum_{j=1}^n \frac{w_j}{1-x_j} $$

with $$x_n = 1-\sum_{j=1}^{n-1}x_j$$ for $x_1, \dots, x_{n-1} \in (0,1)$.

For example, for $n=3$ the function is

$$ f(x_1, x_2)= \left ( \frac{x_1}{1-x_1} + \frac{x_2}{1-x_2}+ \frac{1- x_1-x_2}{x_1+x_2}\right) \left ( \frac{w_1}{1-w_1} + \frac{w_2}{1-w_2}+ \frac{w_3}{1-w_3}\right)- \left ( \frac{w_1}{1-x_1} + \frac{w_2}{1-x_2}+ \frac{w_3}{x_1+x_2}\right) \left ( \frac{x_1}{1-w_1} + \frac{x_2}{1-w_2}+ \frac{1- x_1-x_2}{1-w_3}\right)$$

Working on a conjecture in this MO question, for $n= 3, 4, 5$, I want to numerically check whether the Hessian matrix of $f(x_1, \ldots, x_{n-1})$ is positive semi-definite over the hypercube $(0,1)^{n-1}$ for any parameters $w_1, \ldots, w_n > 0$ with $\sum_{j=1}^n w_j = 1$.

I could manage the case $n=3$ using a desmos graph, which shows the answer is yes. Could you help me efficiently use Mathematica to handle the two remaining cases $n=4,5$?

I found a related question discussing how one can check whether a matrix is positive semi-definite by Mathematica.

POSTED BY: Amir A.
6 Replies
Posted 10 hours ago

Thanks! It was very helpful. In fact, the determinant of the Hessian matrix may have sharp drops near some of the extreme points of the hypercube, which cannot be detected by checking the Desmos figure.

POSTED BY: Amir A.

I think your plot of the determinant misses the behaviour near the origin. Just calculate the determinant at x=1/100, y=1/100, for example.

POSTED BY: Gianluca Gorni
Posted 18 hours ago

Thank you Gianluca for your kind consideration! However, it is not consistent with what I see from my method at the point. In this link, I depicted the two diagonal elements and the determinant of the Hessian matrix. You may see that all the three quantities are highly positive at the point you obtained (see the intersection of the green and orange lines). The reason may be that $x_3$ is used instead of $1-x_1-x_2$, or something else.

POSTED BY: Amir A.

Unless I have made a mistake, there seem to be counterexamples:

In[13]:= 
f[n_] := Sum[x[j]/(1 - x[j]), {j, n}]*Sum[w[j]/(1 - w[j]), {j, n}] -
    Sum[x[j]/(1 - w[j]), {j, n}]*Sum[w[j]/(1 - x[j]), {j, n}] /.
   {x[n] :> 1 - Sum[x[j], {j, n - 1}],
    w[n] :> 1 - Sum[w[j], {j, n - 1}]};
hessianMatrix = D[f[3], {Array[x, 2], 2}];
sol = FindInstance[Det[hessianMatrix] < 0 &&
   0 < x[1] < 1 && 0 < x[2] < 1 && 0 < w[1] < 1 && 0 < w[2] < 1 &&
   w[2] == 1/2 && w[1] + w[2] < 1,
  {x[1], x[2], w[1], w[2]}, Reals]
Eigenvalues[hessianMatrix /. sol[[1]]] // N

Out[15]= {{x[1] -> 5/608, x[2] -> 15/1216, w[1] -> 7/608, w[2] -> 1/2}}

Out[16]= {465210., -0.271271}
POSTED BY: Gianluca Gorni
Posted 1 day ago

I haven't made any progress using Mathematica to solve my problem yet. I could solve the problem for $n=3$ using a Desmos graph.

POSTED BY: Amir A.

Please provide your efforts in the form of the Wolfram Language code. This will make it easier for other members to help you. Check several methods available to include your code in the rules http://wolfr.am/READ-1ST

POSTED BY: EDITORIAL BOARD
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