# convergence of bounded, holomorphic functions on the disk

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

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 $$0$$.