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I'm trying to solve the following problem:

Suppose $f$ is analytic and bounded on the unit disk and $f (0) \ne 0$. Let ${s_n}$ be the zeros of $f$. Show that $\sum_n(1-|s_n|) < \infty$.

The hint given is to apply Jensen's formula to a disc of radius $R$ with $R < 1$. Clearly the hypotheses for the formula apply to f, but I'm having a hard time seeing how to relate it to the sum in question. It'd be great if someone could point me in the right direction on this.

Thanks!

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1 Answer 1

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Let $f$ be bounded by $M > 0$. For $0 < r \leq R < 1$ let $N_r$ be the finite multi-set of zeroes of $f$ on the disc $\Delta_r$ with radius $r$ (counting multiplicity). Then there is a holomorphic $h$, non-vanishing on $\Delta_r$ such that $|h(z)| = |f(z)|$ for $|z| = R$ and

$$ f(z) = h(z) R^{|N_r|} \prod_{\alpha \in N_r} \frac{\alpha - z}{R^2 - \overline{\alpha} z}. $$

By the maximum principle for $h$ the following inequality holds at $z=0$:

$$ R^{|N_r|}|f(0)| = |h(0)| \prod_{\alpha \in N_r} |\alpha| \leq \max_{|z|=R}|f(z)| \prod_{\alpha \in N_r} |\alpha| \leq M \prod_{\alpha \in N_r} |\alpha|. $$

Since this holds for all $R \in [r, 1)$ we can take the limit $R \to 1$ to get the inequality

$$ \prod_{\alpha \in N_r} |\alpha| \geq \frac{|f(0)|}{M} $$

for all $r \in (0, 1)$. In particular

$$ \lim_{r \to 1} \prod_{\alpha \in N_r}\left(1 - (1 - |\alpha|) \right) = \lim_{r \to 1} \prod_{\alpha \in N_r} |\alpha| \geq \frac{|f(0)|}{M} > 0. $$

This implies (by a well known correspondence) that

$$ \lim_{r \to 1} \sum_{\alpha \in N_r} (1 - |\alpha|) < \infty. $$

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