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If $a_n\in\mathbb{C}$ are complex number such that $|a_n|<1$ and $\sum_{n}(1-|a_n|)<\infty$, then I know that following Blaschke product define an analytic function on the open unit disk $\mathbb{D}$. $$B(z) = \prod_{n=1}^\infty\frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1}\tag{1}$$. Then, $\frac{1}{2\pi}\int_{-\pi}^{\pi} |B(re^{i\theta})|d\theta\rightarrow 1$, as $r\rightarrow 1$.

It seems that I am supposed to use either one of Jensen's formula or Poisson's formula to prove this claim, but I am not able to finish it. For a different approach, I have tried writing $$\log |B(z)| = \sum_{n} \log\Big|\dfrac{|a_n|}{a_n}\dfrac{z-a_n}{\overline a_nz-1}\Big| = \sum_{n}\log\Big|1 - \dfrac{1-|a_n|}{1-z\overline a_n}\Big|$$, and try to estimate each summand, but still finding no success.

Can anybody give me a hint?

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    $\begingroup$ Rudin, "real and complex analysis", 3rd ed., thm. 11.32 and thm. 15.24. Proves a stronger result, but it is a little involved. It may help you to find a proof. $\endgroup$ – user90189 May 8 '16 at 1:43
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For a function f holomorphic in a neighbourhood of zero we have $$ \log|f(0)| = \frac 1{2\pi} \int_{-\pi}^\pi\log|f(re^{i\theta})|\, d\theta $$ Taking exp and applying Jensen's inequality, we get the inequality there for $|f|$ (the so called mean value inequality).

The proof goes similarly to the proof of Thm 15.24 in Rudin's book. Take $$ B_N(z) := \prod_{n=N}^\infty \frac {|a_n|}{a_n} \frac{z-a_n}{\bar a_nz-1}, $$ plug it into the mean value inequality and notice that for any $\epsilon>0$ we can choose $r$ so close to $1$, that $$ |B_N(0)| \le \frac 1{2\pi} \int_{-\pi}^\pi|B_N(re^{i\theta})|\, d\theta \le (1+\epsilon) \cdot \frac 1{2\pi} \int_{-\pi}^\pi|B(re^{i\theta})|\, d\theta \le (1+\epsilon). $$ (This holds because $|B/B_N|$ is continuous and equals $1$ on the unit circle.)

Now, take $r\to1$, then $N\to \infty$ (see that $|B_N(0)|\to 1$), and finally $\epsilon \to 0$.


Remark: If we already knew that a Blaschke product has a radial limit almost everyhere on the unit circle, then the proof would be merely an application of Lebesgue's monotone convergence theorem (because an integral mean is increasing as a function of $r$).

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