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Build a Blaschke product such as $B^*(1)=\lim_{r\to 1}B(r)=0$?

We have $$B(z)=\prod_{n=1}^{\infty}\dfrac{|z_n|}{z_n}\dfrac{z_n-z}{1-\bar{z_n}z}$$ I Know that $|B^*|=1$ p.p. on $\partial D$ (which $D=\{z: |z|<1\})$

Perhaps $z_n=1-\dfrac{1}{n^2}$ be suitable but how can I show it?

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  • $\begingroup$ Exercise 13, Chapter 15 of Rudin's book (Real and Complex Analysis) $\endgroup$ – Mathmath Mar 8 '18 at 15:48
  • $\begingroup$ I need guidance to solve it! $\endgroup$ – Fair Mar 8 '18 at 16:12
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Here's a result that works for any bounded holomorphic function.

Thm: Let $0<a_1< a_2 < \cdots \to 1,$ and assume

$$\frac{a_{n+1}-a_n}{1-a_{n+1}}\to 0.$$

Suppose $f\in H^\infty(\mathbb D)$ and $f(a_n)\to 0.$ Then $\lim_{r\to 1^-} f(r)=0.$

Proof: Let $|f|\le M$ in $\mathbb D.$ Note that if $z\in D(a_n, a_{n+1}-a_n),$ then $d(z,\partial D) < 1-a_{n+1}.$ So Cauchy's estimates show that for any such $z,$

$$|f'(z)| \le \frac{M}{1-a_{n+1}}.$$

Thus for $r\in (a_n,a_{n+1}),$

$$|f(r)-f(a_n)| = |\int_{a_n}^r f' |\le \frac{M}{1-a_{n+1}}(r-a_n) \le \frac{M}{1-a_{n+1}}(a_{n+1}-a_n) \to 0.$$

The theorem follows, as does your result for a Blashke product with zeros at $a_n = 1-1/n^2, n=1,2,\dots$

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Let $a_n=1-n^{-2}$. Hence, for $r>a_n$ \begin{eqnarray*} |B(r)|&=&\prod_{n=1}^\infty\frac{r-a_n}{1-a_n r}\\ &\leq& \prod_{n=1}^{N-1}\frac{r-a_n}{1-a_n r} \text{ finite product}\\ &\leq& \prod_{n=1}^{N-1}\frac{r-a_n}{1-a_n } \text{ Bigger denominator}\\ \end{eqnarray*} For $r<a_{N}$, we have also have
\begin{eqnarray*}\displaystyle |B(r)| &\leq& \prod_{n=1}^{N-1}\frac{a_{N}-a_n}{1-a_n } \\ &\leq& \prod_{n=1}^{N-1} \left(1-\frac{n^2}{N^2}\right) \\ &\leq&e^{\ln \prod_{n=1}^{N-1} \left(1-\frac{n^2}{N^2}\right) }\\ &\leq&e^{\sum_{n=1}^{N-1}\ln \left(1-\frac{n^2}{N^2}\right) }\\ &\leq&e^{\sum_{n=1}^{N-1} \left( -\frac{n^2}{N^2}\right) }\,\, \, (\ln x\leq x-1)\\ \end{eqnarray*} and hence the result.

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