Let for $j=1,\dots, m$, $z_j$ be distinct points from the unit disk $|z|<1$ and let $$g(z)=-\sum_{k=1}^m \log \frac{(1-|z|^2)(1-|z_k|^2)}{|1-z\overline{z_k}|^2}.$$ It seems that $g$ has a unique minimum in in the unit disk. How to prove this?
2
-
1$\begingroup$ Well, one can prove that $\log|1 - z\bar w|^2 - \log(1 - |z|^2)$ is strictly convex, and so $g$ is strictly convex. Since $g$ goes to infinity near the boundary of the unit disk, it must have a unique minimum in the interior. $\endgroup$– Mateusz KwaśnickiCommented Sep 11 at 13:53
-
$\begingroup$ @Mateusz Kwaśnicki Maybe the function is not strictly convex? $\endgroup$– AlphaHarmonicCommented Sep 11 at 21:08
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This answer is an elaboration of Mateusz Kwaśnicki's comment. We could manually check that the summand is convex, but there is a nicer (or at least more geometric) way to handle this. Your function can be written as $$ g(z) = \sum_{k=1}^m \log \cosh^2 \frac{d(z,z_k)}{2}, $$ where $d(z,w)$ is the Poincaré metric. The Poincaré disk is a Hadamard manifold, so its metric is (geodesically) convex. Now $f(x) = \log \cosh^2 x$ is a nondecreasing strictly convex function of the real variable $x$ since $f'(x)=2 \tanh x \geq 0$ and $\DeclareMathOperator{\sech}{sech} f''(x) = 2 \sech^2 x > 0$, so each term of the sum is strictly convex. Therefore the function $g$ is strictly convex and has a unique global minimum by the observation in Mateusz's comment that $g(z) \to \infty$ as $|z| \to 1$.
