# Problem 9 M.L Krasnov variational calculus

Warning: Finding extreme value of a multivariable function My question differs from this since I try to use the Hessian criterion so it is not a repeated question.

My question: Problem 9 M.L Krasnov.

Find extrema of $$f:U \to \mathbb{R}$$ where $$U=\{(x_1,\ldots,x_n)\in \mathbb{R}^n:x_{i}>0 \text{ for all }i\in \mathbb{N}\}$$ and $$f(x_1,x_2,\ldots,x_n)=x_1x_2^2\ldots x_n^n(1-x_1-2x_2-\ldots-nx_n).$$

Well this is my attempt: Critical points are:

$$\frac{\partial f}{\partial x_n}=nx_1x_2^2x_3^3(1-x_1-2x_2-\cdots-nx_n)-nx_1x_2^2\cdots x_n^n=0$$

so $$x_1=x_2=\cdots =x_n=\frac{2}{n^2+n+2}.$$

These are critical points of $$f$$.

Computing second partial derivatives for use Hessian in critical point $$x_0=\left(\frac{2}{n^2+n+2},\ldots, \frac{2}{n^2+n+2}\right)$$ and suppose no loss of generality $$i I get:

$$\frac{\partial^2 f}{\partial x_i^2}=i(i-1)\left(\frac{2}{n^2+n+2}\right)^{\frac{n(n+1)}{2}}-2i^2\left(\frac{2}{n^2+n+2}\right)^{\frac{n(n+1)}{2}-1}.$$

So computing the cross derivatives and substituting into $$x_1=x_2=\cdots=x_n=(\frac{2}{n^2+n+2})$$

$$\frac{\partial^2 f}{\partial x_j \partial x_i}=-ij\left(\frac{2}{n^2+n+2}\right)^{\frac{n(n+1)}{2}-1}.$$

I have tried this exercise for many days without any success, I plan to use Sylvester's criterion to show that this critical point is a local maximum.

Sylvester's criterion: The real-symmetric matrix $$A$$ is positive definite if and only if all the leading principal minors of $$A$$ are positive.

• If the $(\frac{2}{n^2+n+2})^{\frac{n(n+1)}{2}-1}$ is a constant that's not changing with $i$ and $j$, then removing it, you're effectively left with the matrix $A_{ij} = ij$, which is positive semi-definite, not positive definite. It definitely has $0$ as an eigenvalue. Mar 19, 2022 at 4:08

1. OP's function is $$f(x)~:=~\pi(x)(1-\sigma(x)),\qquad x~=~(x_1,\ldots,x_n)~\in~ \mathbb{R}^n_{+},\tag{1}$$ $$\pi(x)~:=~ \prod_{i=1}^n x_i^i, \qquad \sigma(x)~:=~\sum_{i=1}^n ix_i,\tag{2}$$ $$\frac{\partial \pi(x)}{\partial x_j}~=~ \frac{j \pi(x)}{ x_j},\qquad \frac{\partial \sigma(x)}{\partial x_j}~=~ j,\tag{3}$$ $$\frac{\partial f(x)}{\partial x_j} ~=~j\pi(x)\left( \frac{1-\sigma(x)}{x_j}-1\right) ~=~j\pi(x) \frac{1-\sigma(x)-x_j}{x_j}.\tag{4}$$
2. We see from eq. (4) that critical points can only be on the diagonal $$x_1~=~\ldots~=~x_n~=~1-\sigma(x).\tag{5}$$ It follows immediately that the only critical point is $$x_1~=~\ldots~=~x_n ~=\left(1+\sum_{i=1}^n i\right)^{-1} ~=\left(1+\frac{n(n+1)}{2}\right)^{-1} ~=~\frac{2}{n^2+n+2},\tag{6}$$ as OP correctly states.
4. Hessian $$\frac{\partial^2 f(x)}{\partial x_j\partial x_k} ~=~\frac{jk\pi(x)}{x_k}\left( \frac{1-\sigma(x)}{x_j}-1\right) - j\pi(x)\left(\frac{k}{x_j}+ \frac{1-\sigma(x)}{x_j^2}\delta_{jk} \right). \tag{7}$$ Hessian on-shell: $$\left. \frac{\partial^2 f(x)}{\partial x_j\partial x_k}\right| ~=~-j(k+\delta_{jk})\frac{\pi(x)}{x_j}.\tag{8}$$
5. To prove that the critical point (6) is a maximum, it is enough to prove that the matrix $$C_{jk}~=~j(k+\delta_{jk})~=~A_{jk}+B_{jk}\tag{9}$$ is a positive matrix. The second term $$B_{jk}~=~j\delta_{jk}\tag{10}$$ is clearly a positive matrix. Since (semi)positive matrices form a convex cone, it is enough to prove that the first term $$A_{jk}~=~jk\tag{11}$$ is a semi-positive matrix, i.e. $$\forall v~\in~\mathbb{C}^n:~~v^{\dagger}A v~\geq~0.\tag{12}$$ By scaling $$w_i=iv_i$$, this is equivalent to proving that the matrix $$D_{jk}~=~1,\tag{13}$$ or equivalently that the matrix $$P_{jk}~=~\frac{1}{n}D_{jk}~=~\frac{1}{n},\tag{14}$$ is a semi-positive matrix. But $$P^2~=~P~=~P^{\dagger}\tag{15}$$ is an orthogonal projection matrix, and hence semi-positive. $$\Box$$