Jensen's Formula/Proof 3

Proof
, assume that $r = 0$.

Write:
 * $\map f z = \map {B_{\rho_1} } z \dotsm \map {B_{\rho_n} } z \map g z$

where $\map {B_\rho} z := \dfrac {z - \rho} {1 - z \overline \rho}$ is the Blaschke factor at $\rho$.

So:
 * $\map g z$ is holomorphic and nonzero in $D_r$

and:
 * each $\map {B_{\rho_i} } z$ is holomorphic and has a simple zero $\rho_i$ inside $D_r$.

It is sufficient to check the equality for each factor of $f$ in this expansion.

First let:
 * $\map h z = \map {B_{\rho_k} } z = \dfrac {z - \rho_k} {1 - z \overline {\rho_k} }$

for some $k \in \set {1, \ldots, n}$.

Note that:
 * $\size {\map h z} \equiv 1$ for $\size z = 1$

so the integral is $0$.

Also:
 * $\map {B_{\rho_k} } 0 = \rho_k$

so the equality holds in this case.

To show equality for $\map g z$, first observe that by the Cauchy's Residue Theorem:


 * $\ds \int_{\size z \mathop = r} \frac {\ln \map g z} z \rd z = 2 \pi i \ln \map g 0$

Therefore substituting $z = r e^{i \theta}$ we have:


 * $\ds 2 \pi i \ln \map g 0 = i \int_0^{2 \pi} \ln \map g {r e^{i \theta} } \rd \theta$

Comparing the imaginary parts of this equality we see that:


 * $\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \size {\map g 0}$

as required.