# Singularity in Jensen's Formula Proof

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?

$g$ 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.

• So do we just cancel and redefine the function at that point? I.e whatever the limit is after cancellation? – rubikscube09 Jul 13 '18 at 0:29
• As it is $g$ is not defined at the points $x=a_j$. It is understood that you do the cancellations so that $g$ is well defined and, in fact, holomorphic. – Kavi Rama Murthy Jul 13 '18 at 5:29
• Thank you. Last question, why is the numerator defined the way it is? – rubikscube09 Jul 13 '18 at 14:23
• To make $|g(z)|=|f(z)|$ for $|z|=R$. – Kavi Rama Murthy Jul 13 '18 at 23:10

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 $f$.

• Is there a cancellation occuring? Or am I missing some subtlety. – rubikscube09 Jul 13 '18 at 1:02
• @rubikscube09 the cancelation that is occurring is removing the root from $f$. Recall the fundamental theorem of algebra: every polynomial can be factored in the form $\prod (z-a_k)$. Now, it's a common theorem in complex analysis that we can do a similar thing with entire functions. Let $f$ have a root at $a$ of order $n$. Then $f = (z-a)^n h(z)$ in a small enough neighborhood of $a$, where $h$ is analytic and nonzero in the neighborhood. It's this result that gives the theorem in italics in my post. Think of it as factoring out the zeroes – Brevan Ellefsen Jul 13 '18 at 19:38
• And then of course $g$ isn't formally defined at the roots, but since the singularities are removable we simply redefine them. Think of the function $x/x$. Surely this function should be one at zero! And it turns out that when you remove the removable singularities this way you get a function analytic in some neighborhood around the removable singularity, just like we want – Brevan Ellefsen Jul 13 '18 at 19:42
• Finally, to answer your question below the other answer (why we choose the numerator the way we do), the way we construct $g$ is as follows: we start with $\prod\frac{f(z)}{z-a_k}$ (where we naturally redefine removable singularities) to kill off all the zeroes of $f$ (there are finitely many since the disk is compact, so this needs no further justification). We then realize that we also want $|g| = |f|$ on the circle of radius $R$, so we have to adjust our definition to make that work (using what amounts to a mobius transform to map the values, or you can simply check that the algebra works) – Brevan Ellefsen Jul 13 '18 at 19:45
• @rubikscube09 please look at my comments above where I have attempted to answer your questions. If you have any others please let me know and I'll try to help you! – Brevan Ellefsen Jul 13 '18 at 19:48