# Why $\lim_{r\to 1^-} \int_0^{2\pi}\ln(|f(re^{i\vartheta})|)d\vartheta=-\infty$?

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 $$|z|=r$$.

• @MarkViola No, I mean $-\infty$. Once we get that, the proof is concluded since we reached a contradiction. – Pelota Dec 25 '20 at 0:36
• Apology. I thought the magnitude was taken on the logarithm, not its argument. – Mark Viola Dec 25 '20 at 3:43

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!

• 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? – Pelota Dec 25 '20 at 0:47
• 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. – FormulaWriter Dec 25 '20 at 1:04
• 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? – Pelota Dec 25 '20 at 1:08
• 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. – FormulaWriter Dec 25 '20 at 1:08