I have some lecture notes containing the following.

An analytic function $F$ in the disc is outer if there is an $h\ge0$ s.t $\int_{\partial \mathbb{D}} \mid log h \mid < \infty$ and $F(z)=e^{\int_{\partial \mathbb{D}} \frac{x+z}{x-z}log h(x)dx}$

In this case $F$ is the outer function whose modulus equal $h$ a.e on $\partial \mathbb{D}$.

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According to me this suggests that $F$ in the second part of the theorem is of the form

$F(z)=e^{\int_{\partial \mathbb{D}} \frac{x+z}{x-z}log(log \mid f \mid) dx}$.

This does not lead me to the conclusion of the second part of the theorem. If we have $F(z)=e^{\int_{\partial \mathbb{D}} \frac{x+z}{x-z}log \mid f \mid dx}$ then atlest I get $f/F=1$ a.e on $\partial \mathbb{D}$ which is mentioned as consequence right after this proof. And I feel somewhat closer to prove the $F_n$ part, with $\mid F_n \mid =\mid f \mid + \frac{1}{n}$

Should it be $\mid F \mid = \mid f \mid$ a.e on $\partial \mathbb{D}$ in the hypoteses of the second part?

  • $\begingroup$ Yes, it should most definitely be $|F|=|f|$ a.e. on $\partial\mathbb{D}$. Then the absence of the Blaschke product gives $|f(z)| \le |F(z)|$ in $\mathbb{D}$. $\endgroup$ – DisintegratingByParts Aug 5 '16 at 16:56

Here are two classical references for this theorem:

  • Rudin's Real and Complex Analysis, theorem 17.17 on page 344. The second part of your quoted theorem is labelled (5). Note that $U$ is used by Rudin to denote the open unit disc in $\mathbb{C}$ and $Q_f$ is used to denote the outer function of $f$ (i.e. the outer function with $|f|$ acting as the $h$ in your definition).
  • Hoffman's Banach Spaces of Analytic Functions, pages 61-62.

There is indeed a typo in the second part of your quoted theorem: instead of saying "if $F$ is the outer function with $F|=\log|f|$ m-a.e. on $\mathbb{T}$" it should say something like "if $F$ is the outer function of $f$" if you follow the terminology of Rudin. Of course, as you point out, this implies that $|F|=|f|$ m-a.e. on $\mathbb{T}$.


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