Singularity in Jensen's Formula Proof

Question

So Jensen's formula says that for $f$ holomorphic on $\Omega$ containing the closed disk $D_R$ (around $0$) that if $f(0) \neq 0 $ and $f \neq 0$ on $\partial D_R$ that if we count the zeros of $f$ $\{a_1, \cdots a,a_n\}$ with multiplicity, then: $$ \log|f(0)| = \sum_{ k= 1}^{N}\log\left(\frac{|a_k|}{R}\right) + \frac{1}{2\pi} \int_{0}^{2\pi} \log \left|f(Re^{i \theta}) \right| $$

My proof is as follows: For any analytic $g$ defined on $\overline{D_R} = \{z \in \mathbb{C} : |z| \leq R\}$ satisfying $g(z) \neq 0$ on $\partial D_R$ we have: $$ \frac{1}{|\partial D_R|}\int_{\partial D_R} \log |g(z)| dS(z) = \log|g(0)| $$ This is a real valued line integral, and the equality follows from the integral representation of harmonic functions. Expanding the average and reparameterizing gives: $$ \frac{1}{2\pi}\int_{0}^{2\pi} \log \left|g(Re^{i \theta})\right| \mathrm{d}\theta = \log |g(0)| $$ Define: $$ g(z) = f(z) \prod_{k = 1}^{n}\frac{R^2 + \overline{a_k}z}{R(z-a_k)} $$ A computation shows that $|g(z)| = |f(z)|$ on the circle $|z| = R$, and moreover, $g(z) \neq 0$ on $D_R$. So we conclude: $$ \log |g(0)| = \frac{1}{2\pi}\int_{0}^{2\pi} \log \left|g(Re^{i \theta})\right| \mathrm{d}\theta = \frac{1}{2\pi}\int_{0}^{2\pi} \log \left|f(Re^{i \theta})\right| \mathrm{d}\theta $$ Expanding with logarithms we see: $$ \log |g(0)| = \log\left|f(0)\prod_{k = 1}^{n}\frac{R^2 }{-R\cdot(a_k)}\right| = \log|f(0)| + \log\left|\prod_{k = 1}^{n}\frac{R^2}{|R\cdot a_k|}\right| = \log|f(0)| + \sum_{k = 1}^{n}\log\left|\frac{R}{a_k}\right| $$ Recall the equality: $$ \log |g(0)| = \frac{1}{2\pi}\int_{0}^{2\pi} \log \left|f(Re^{i \theta})\right| $$ Substituting and rearranging gives: $$ \log|f(0)| = \frac{1}{2\pi}\int_{0}^{2\pi} \log \left|f(Re^{i \theta})\right| - \sum_{k = 1}^{n}\log\left|\frac{R}{a_k}\right| = \frac{1}{2\pi}\int_{0}^{2\pi} \log \left|f(Re^{i \theta})\right| + \sum_{k = 1}^{n}\log\left|\frac{a_k}{R}\right| $$ as we desire.

My question is as follows: the $g$ we define has singularities at each of the roots $a_k$. What exactly does the product do? Why is it there for the integral? Why not just let $g =f$? I know we add a term to make sure we are not taking the log of $0$, but why that term? Is it to avoid some singularity?

Answer 1

$g$ does not have any singularities. For example if $a_j$ is a zero of order $1$ of $f$ then $f(z)=(z-a_j)h(z)$ for some analytic function $h$ in a neighborhood of $a_j$ and the factor $z-a_j$ cancels with the same factor in the denominator. Same argument holds for zeros of any order.

Answer 2

I am not sure you understand why we choose to define $g$ the way we do. Recall that:

whenever you have a zero of order $n$ of $f$ at $a_k,$ we have $f(z)(z-a_k)^{-n}$ nonzero and analytic in a small enough neighborhood about $a_k$.

We thus define $g$ the way we do to remove zeroes while preserving analyticity (along with correction factor making $|g| = |f|$ along the circle).

While I could mention some stuff about Mobius Transforms or Blashke Factors, the most important thing to realize is that the terms of the form $(z-a_k)$ in the denominator of the product are there to kill the corresponding zeroes of $f$.