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I am studying Greene and Krantz' "Function theory in one complex variable". In section $13.4$, they want to prove that, given $f\in H^p, p\in (0,+\infty),$ if the radial limit function $\tilde f$ is zero on a set of positive measure then $f\equiv 0$. Their proof is quite short: if $f\not\equiv 0$, we can assume wlog $f(0)\neq 0$ and then by Jensen's formula (given $r<1:\forall \vartheta f(re^{i\vartheta})\neq 0$) $$-\infty<\ln(|f(0)|)\le\int_0^{2\pi}\ln(|f(re^{i\vartheta})|)d\vartheta/2\pi $$

I have a problem with the following passage:

As $r\to 1^-$ through such$^1$ values, the right hand side of this expression tends to $$\frac{1}{2\pi}\int_0^{2\pi}\ln(|\tilde{f}(e^{i\vartheta})|)d\vartheta=-\infty$$

I am unsure about one thing: Why is the limit equal to $\frac{1}{2\pi}\int_0^{2\pi}\ln(|\tilde f|)$?

I tried using dominated convergence and Fatou's lemma, to no avail.

Thanks for the help, and Happy Holidays!

Note: for completeness, the proof then concludes by noting that we reached a contradiction and thus $f$ cannot be $\not\equiv 0$.

$^1$: meaning $r$ such that $f$ is not zero on $|z|=r$.

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  • $\begingroup$ @MarkViola No, I mean $-\infty$. Once we get that, the proof is concluded since we reached a contradiction. $\endgroup$ – Pelota Dec 25 '20 at 0:36
  • $\begingroup$ Apology. I thought the magnitude was taken on the logarithm, not its argument. $\endgroup$ – Mark Viola Dec 25 '20 at 3:43
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Define $P_r(\vartheta)=\ln(|{f}(re^{i\vartheta})|)^+, N_r(\vartheta)=\ln(|{f}(re^{i\vartheta})|)^-$ so that: $$\ln(|{f}(re^{i\vartheta})|)= P_r(\vartheta) - N_r(\vartheta)$$

Show first that as $r\to 1^{-}$ the integral$\frac{1}{2\pi}\int_0^{2\pi} P_r(\vartheta) d\vartheta$ is bounded using $f\in H^p$.

After this, all is left is to show that $\frac{1}{2\pi}\int_0^{2\pi} N_r(\vartheta) d\vartheta \to +\infty$. This can be proven by showing that $\frac{1}{2\pi}\int_0^{2\pi} N_{r_k}(\vartheta) d\vartheta \to +\infty$ for every sequence $(r_k)_k$ satisfying $r_k\to 1^-$ as $k\to +\infty$.

So, consider a sequence $(r_k)_k$ satisfying $r_k\to 1^-$ as $k\to +\infty$. By Egorov's theorem applied to the sequence $exp(-{N_{r_k}})$ you can show that $exp(-{N_{r_k}})$ converges uniformly to $0$ on a subset of $(0,2\pi)$ of positive measure. This can be used to prove $\frac{1}{2\pi}\int_0^{2\pi} N_{r_k}(\vartheta) d\vartheta \to +\infty$.

Merry Christmas and Happy New Year!

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  • $\begingroup$ Thanks, this answer my question. Just a question thoug: in order to apply Egorov's theorem we need a.e. pointwise convergence, and so we should restrict ourself to a subsequence of $r_k$, right? $\endgroup$ – Pelota Dec 25 '20 at 0:47
  • $\begingroup$ Egorov's theorem requires a sequence of measurable functions, so we can't apply it directly to the family $(exp(-{N_{r}}))_r$ as $r\to 1^-$. This is why a sequence of radii $(r_k)_k$ is used. Does this answer your question? If not, let me know. $\endgroup$ – FormulaWriter Dec 25 '20 at 1:04
  • $\begingroup$ Egorov's theorem requires a sequence of measurable functions converging pointwise a.e. to the limit (at least the version I know of), while $\exp(-N_{r_k})$ is not granted to satisfy this property. Am I right? $\endgroup$ – Pelota Dec 25 '20 at 1:08
  • $\begingroup$ Ok, sorry if I misunderstood your question. A.e. pointwise convergence of $(exp(-{N_{r_k}}))_k$ is a consequence of a.e convergence of $((f(re^{i\vartheta}))_r$, which is one of your hypoteses. $\endgroup$ – FormulaWriter Dec 25 '20 at 1:08

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