I'm trying to minimize the function $f(x) = \sum_{i=1}^{n}(\log(|x_i|))^2$ in the closed unit ball $B(0, 1) \subset \mathbb R^n$, where the function is defined to be $\infty$ if $x_i = 0$.
What I did
First notice that the function is symmetric with respect to sign, so to make things easier we can just work with non negative $x_i$'s and remove the absolute value. Thus we want to minimize $f(x) = \sum_{i=1}^{n}(\log(x_i))^2$ given constraints $x_i \geq 0$, $\sum_{i=1}^{n}x_i^2 \leq 1$.
Lets first look for a local minimum: $\nabla f(x) = \begin{pmatrix}\frac{2\log(x_1)}{x_1} \\ \vdots \\ \frac{2\log(x_n)}{x_n}\end{pmatrix} = 0$ implies $x_1 = x_2 = \dots = x_n = 1$ which is outside our domain - so no local minimum.
If any $x_i = 0$ then we definitely don't have a minimum since the function was defined to be $\infty$ there. Thus our minimum must be when $x_i > 0$ and $\sum_{i=1}^{n}x_i^2 = 1$
Using Lagrange multipliers, we get that we need to solve the system:
$\begin{cases}\frac{\log(x_i)}{x_i} = \lambda x_i \\ \sum_{i=1}^{n}x_i^2 = 1\end{cases}$
How do I solve this system? Obviously $x_1 = x_2 = \dots = x_n$ is a possible solution but maybe there are more...
