I want to show that a sequence of holomorphic, zero-free functions on the disk converges uniformly to zero on compact subsets of the disk if $|f_n| < 1$ and $\lim_{n \rightarrow \infty} f_n(0) = 0$.

My idea was to use Liouville's theorem to argue that $f_n$ must be constant , but I don't have a entire function. Is there a way to use Liouville for functions that are just holomorphic on the disk?


1 Answer 1


Each $f_n$ need not be zero, as we see from the example $f_n(z)=\frac{42+z}{2019n}$. On the other hand, it is still true that $(f_n)$ converges locally uniformly (converges uniformly on compact subsets) to the zero function.

Here is (possibly not the simplest) proof: Write $\mathbb{D}$ for the unit disk and let $f_n : \mathbb{D} \to \mathbb{D}$ be a sequence of holomorphic functions such that $f_n(0) \to 0$ as $n\to\infty$.

  • For each subsequence $(f_{n_k})$ of $(f_n)$, there exists a further subsequence $(g_{j})$ which converges locally uniformly on $\mathbb{D}$. Indeed, let $0 < r_1 < r_2 < \cdots $and $r_n \uparrow 1$. Then for each $n$, the bound

    $$ \forall z \in \overline{B(0,r_n)}, \qquad \left| f_n'(z) \right| = \left| \frac{1}{2\pi i} \int_{|\xi|=r_{n+1}} \frac{f(\xi)}{(z-\xi)^2} \, \mathrm{d}\xi \right| \leq \frac{r_{n+1}}{(r_{n+1} - r_n)^2} $$

    shows that $(f_n)$ is equicontinuous on each $\overline{B(0,r_n)}$, and so, we can emply Arzela-Ascoli theorem and diagonal argument to extract a (further) subsequence $(g_j)$ of $(f_{n_k})$ which converges uniformly on each $\overline{B(0,r_n)}$.

  • Since each $g_j$ is zero-free and converges locally uniformly, by Hurwitz's theorem, its limit is either identically zero or also zero-free. Since the latter option is impossible, it follows that the limit is identically zero.

  • We have shown that: for each subsequence of $(f_n)$, there is a further subsequence which converges locally uniformly to $0$. This implies that $(f_n)$ itself converges locally uniformly to $0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .