# A holomorphic function with infinitely many zeros in the unit disc

Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_1, z_2, z_3, \dotsc, z_n, \dotsc$ are its zeros ($\vert z_k \vert$ $\lt1$ ),then $$\sum_{k=1}^\infty (1-\vert z_k \vert) \lt \infty$$ [Hint:Use Jensen's formula.]

Since Jensen's formula can be used when $f$ vanishes nowhere on the circle $C_R$. I notice that there exist an increasing sequence $r_n$ for $\lim_{n\to \infty} r_n = 1$, and $f$ vanishes nowhere on each $C_{r_n}$.
Suppose $f(0) \neq 0$, then use Jensen's formula on each circle $C_r$ and get $$\sum_{k=1}^{n_r} \log \vert z_k \vert = \log \vert f(0) \vert + n_r \cdot \log r - \frac{1}{2\pi} \int_{0}^{2\pi} \log \vert f(re^{i\theta}) \vert \,\mathrm{d}\theta,$$
where $n_r$ denotes the numbers of zeros inside the disc $C_r$. But I don't know how to estimate the limit of $n_r \log r$ as $r$ tends to $1$.

• Maybe my complex analysis is weak here, but it appears that $f$ is zero if it has infinitely many zeroes. If $f$ is bounded, it can be continued on an open set $U$, such that $B_1(0) \Subset U$, i.e. $\bar{B_1(0)} \subset U$. But the zero-set of $f$ has an accumulation point in $\bar{B_1(0)}$. Therefore $f \equiv 0$ in $B_1(0)$ by the identity theorem. Did I do a mistake here? – red_trumpet Aug 25 '18 at 13:31
• If $f$ is holomorphic in an open disc that vanishes on a sequence of distinct points with a limit point in the disc.Then $f$ is identically $0$. However, for this question, the limit point might on the unti circle. – J.Guo Aug 25 '18 at 13:40
• Yeah, that's why I wanted to expand $f$ on an open which contains the closure of the unit circle. But actually I'm not sure anymore if this is possible. – red_trumpet Aug 25 '18 at 13:42
• Consider function $f(z)=\sum_{n=0}^{\infty}z^{2^n}$ for $z$ in the unit disc. This function can extends continuously to the unit circle, but cannot be analytically continued past the unit circle.(For nowhere differentiable in its real part). – J.Guo Aug 25 '18 at 13:53
• Sorry,I made a mistake here, function might be $f(z)=\sum_{n=0}^{\infty}2^{-n\alpha}z^{2^n}$ for any $\alpha$ is real and positive.And you can check the condition when z is real. – J.Guo Aug 25 '18 at 14:01

Don't pull out the $r$ from $\log(|z_k|/r)$. Instead look at it this way: Define $$\log^+(t)=\begin{cases}\log(t),&(t>1), \\0,&(0<t\le1).\end{cases}$$
Note that $$\sum_k\log^+(r/|z_k|)=-\sum_{|z_k|<r}\log(|z_k|/r).$$ So (assuming wlog that $f(0)\ne0$) Jensen implies that $$\lim_{r\to1}\sum_k\log^+(r/|z_k|)<\infty.$$
Applying the Monotone Convergence Theorem to that sum shows that $$\sum_k\log(1/|z_k|)<\infty.$$
Or a more elementary version of the same argument: Say $|f|\le c$. Fix $N$. If $r$ is close enough to $1$ that $|z_k|<r$ for $k=1,\dots, N$ then Jensen shows that $$\sum_{k=1}^N\log(|z_k|/r) \ge\sum_{|z_k|<r}\log(|z_k|/r)\ge\log|f(0)|-\log(c).$$Since $N$ is fixed we can let $r\to1$: $$\sum_{k=1}^N\log|z_k|\ge\log|f(0)|-\log(c).$$So $\sum_{k=1}^\infty\log|z_k|>-\infty$.