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Let $D$ denote the unit open disk in $\Bbb C$. Let $f$ be an analytic function defined in $D$, and suppose $f(0) \neq 0$, $|f(z)| \leq 1 (z \in D)$. If $z_1,z_2,...$ are the zeros of $f$ in $D$, I want to show that $ \sum _k (1- |z_k|) \leq - \ln f(0)$.

I think Jensen's formula seems useful, which asserts that:

For $0 \leq R <1$, $\sum _{|z_k| \leq R} \ln (\frac{R}{|z_k|}) = \frac{1}{2 \pi} \int _0 ^{2 \pi} \ln |f(Re^{i \theta})| d \theta - \ln |f(0)| $.

But I don't see how to apply this. Any hints?

blancket
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    Look here or here. – Daniel Fischer Dec 12 '19 at 14:31
  • @Daniel Fischer♦ Thanks but I think they are a little bit different with mine, and I don't see how to obtain an answer of mine from those. – blancket Dec 12 '19 at 16:10
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    They're not really different, but I see how that's not obvious. First step, can you see that the right hand side of Jensen's formula is $\leqslant -\log \lvert f(0)\rvert$? – Daniel Fischer Dec 12 '19 at 16:34
  • @Daniel Fischer♦ Yes, by the upper bound of $|f|$ is 1. – blancket Dec 12 '19 at 16:55
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    Right. So $$\sum_{\lvert z_k\rvert \leqslant R} \log \frac{R}{\lvert z_k\rvert} \leqslant -\log \lvert f(0)\rvert$$ for every $0 < R < 1$. And every term in that sum is non-negative, so for all $0 < R_1 \leqslant R_2 < 1$ we have $$\sum_{\lvert z_k\rvert \leqslant R_1} \log \frac{R_2}{\lvert z_k\rvert} \leqslant -\log \lvert f(0)\rvert,.$$ Any idea what a useful next step might be? – Daniel Fischer Dec 12 '19 at 17:00
  • @Daniel Fischer♦ First let $ R_2$ to 1, and then let $R_1$ to 1, right? – blancket Dec 13 '19 at 01:28
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    Right. (You can do it at once, letting $R\to 1$, but in my experience people feel safer doing such things in two steps at first.) Then it remains to write $\lvert z_k\rvert = 1 - (1-\lvert z_k\rvert)$. – Daniel Fischer Dec 13 '19 at 12:08

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