# Analytic $f: \mathbb{D} \to \mathbb{D}$, $f(0)=0$, and $f$ has five zeros in $\overline{\frac{1}{2}\mathbb{D}}$

Suppose $$f: \mathbb{D} \to \mathbb{D}$$ is a holomorphic function and $$f(0)=0$$. The function $$f$$ has a total of five zeros (counting multiplicities) in the closed half-disc $$\overline{\frac{1}{2}\mathbb{D}} = \{z \in \mathbb{C}: |z| \leq \frac{1}{2}\}$$. How large can $$|f'(0)|$$ be?

Schwarz Lemma $$f: \mathbb{D} \to \mathbb{C}$$ is analytic with $$f(0) = 0$$ and $$|f(z)| \leq 1$$ on $$\mathbb{D}$$, then $$|f(z)| \leq |z|$$ for all $$z \in \mathbb{D}$$ and $$|f'(0)| \leq 1$$

It would be almost indisputable that Schwarz lemma will be of use here. Now, given that $$f$$ has five zeros in the domain $$\overline{\frac{1}{2}\mathbb{D}}$$, it's no question that Rouche's theorem will likely also be of use here.

Since $$f$$ maps $$\mathbb{D}$$ to itself, then implicitly we also have $$|f(z)| \leq 1$$ for $$z$$ on $$\mathbb{D}$$. Then $$f$$ satisfies Schwarz's lemma, and so an upper bound for $$|f'(0)|$$ is $$1$$. Perhaps this bound can be improved.

But I have not used the fact that $$f$$ has five zeros in $$\overline{\frac{1}{2}\mathbb{D}}$$. Again, I know this must be Rouche's theorem, but I'm not sure how to apply it.

• The answer is $1/16$ - hint: the problem shows that $f=zB(z)e^{g(z)}$ where $B$ is a Blaschke product with at least four zeroes in the half disc and $\Re g\le 0$; when you take the derivative at $0$ only $B(0)e^{g(0)}$ counts etc Commented May 14 at 14:14

## 1 Answer

Let $$a_1, \ldots, a_4$$ denote the four zeros of $$f$$ with $$0 < |a_k| \le 1/2$$. Then $$f(z) = h(z) \prod_{k=1}^n \frac{z-a_k}{1-\overline{a_k} z}$$ with $$h: \Bbb D \to \Bbb D$$ and $$h(0) = 0$$. It follows that $$f'(0) = h'(0) \prod_{k=1}^n (-a_k) \, .$$ We have $$|h'(0)| \le 1$$ by the Schwarz lemma, so that $$|f'(0)| \le \prod_{k=1}^n |a_k| \le \frac{1}{2^4} = \frac{1}{16} \, .$$ The bound is sharp, equality holds if and only if $$f(z) = c z \prod_{k=1}^n \frac{z-a_k}{1-\overline{a_k} z}$$ with complex numbers $$c, a_1, \ldots, a_4$$ satisfying $$|c|=1$$ and $$|a_1|=|a_2|=|a_3|=|a_4| = 1/2$$.

• I actually have never seen the first equality. How does one arrive there? Elegant answer though.... Commented May 15 at 7:36
• @HyperbolicPDEfriend: For $|a| < 1$ is $T_a(z) = \frac{z-a}{1-\bar a z}$ an automorphism of the unit disk (sometimes called “Blaschke factor”). If $f$ maps the unit disk into itself with $f(a) = 0$ then $f(z)/T_a(z)$ has a removable singularity and still maps the unit disk into itself (by the maximum modulus principle). Here $h(z) = \frac{f(z)}{z\prod_{k=1}^4 T_{a_k}}(z)$ has removable singularities and maps the unit disk into itself. The product in the denominator is sometimes called “finite Blaschke product.” Commented May 15 at 9:39
• Thank you so much! Commented May 15 at 9:51
• Can you justify $f'(0)= h'(0)\displaystyle\prod_{k=1}^n (-a_k)$? It's not clear to me how this is the case. The way this is written makes it look like the derivative is multiplicative, which I'm sure is merely a coincidence Commented May 25 at 14:05
• @GrigorHakobyan: It is just the product rule applied to $f(z) = h(z)T_1(z) \cdots T_n(z)$. Since $h(0) = 0$, the result is $f'(0) = h'(0)T_1(0) \cdots T_n(0)$. Commented May 25 at 14:44