# Optimizing a multivariate symmetric (permutation-invariant) function

Let $$\ell$$ and $$d$$ be two integers such that $$\ell \le d$$. I would like to find the global maxima of the following symmetric function $$f\colon (0,1]^n \to \mathbb{R}$$,

$$f(x_1, \ldots, x_n) := \sum_{\substack{S \subseteq [n] \\ |S| = \ell}} \sum_{\substack{z \in [d]^S \\ \sum_i z_i = d}} \prod_{i \in S} x_i \left(\ln\Bigl(\frac{e}{x_i^{1/z_i}}\Bigr)\right)^{z_i/2},$$

subject to $$x_1 + \dots + x_n = 1$$.

Note that this function has the same value under any permutation of the coordinates of $$x$$.

For example, when $$n = 3, d = 2, \ell = 2$$, we have $$f(x,y,z) = xy \sqrt{\ln(e/x) \ln(e/y)} + yz \sqrt{\ln(e/y) \ln(e/z)} + xz \sqrt{\ln(e/x) \ln(e/z)} .$$

When $$n=3, d=3, \ell=2$$, we have $$\begin{multline*} f(x,y,z) = xy \ln(e/\sqrt{x}) \sqrt{\ln(e/y)} + xy \sqrt{\ln(e/x)} \ln(e/\sqrt{y}) + yz \ln(e/\sqrt{y}) \sqrt{\ln(e/z)} + \\yz \sqrt{\ln(e/y)} \ln(e/\sqrt{z}) + xz \ln(e/\sqrt{x}) \sqrt{\ln(e/z)} + xz \sqrt{\ln(e/x)} \ln(e/\sqrt{z}) . \end{multline*}$$

The Purkiss Principle says that the point $$(1/n, \ldots, 1/n)$$ is a local maxima. I believe this is also the global maximum, but cannot think of a way to prove this.

One possible way is to show that this function is strictly concave, but it's not clear at all if this is true and I don't know how to show its Hessian is strictly negative-definite.

Is there an easy way to see this? or is my guess simply not true?

Edit: In the original question, the factor $$\ln\Bigl(\frac{e}{x_i^{1/z_i}}\Bigr)$$ was $$\ln\Bigl(\frac{e}{x_i}\Bigr)$$, which was clearly false as pointed out by @fedja.

• For $n=\ell=2$ and large $d$, the maximum is certainly not in the middle, which leaves little hope for the general case without additional assumptions. – fedja Dec 10 '18 at 20:58
• @fedja thanks. Indeed the original question did not make sense. I've now fixed the question accordingly. – nichehole Dec 14 '18 at 18:56

One can show that this function is Schur-concave using the Schur-Ostrowski criterion, which then implies the maximum is attained at the diagonal.

I will omit the calculations here.