Holomorphic images of disks

What's new 2020-11-28

Consider a disk ${D(z_0,r) := \{ z: |z-z_0| < r \}}$ in the complex plane. If one applies an affine-linear map ${f(z) = az+b}$ to this disk, one obtains

$\displaystyle f(D(z_0,r)) = D(f(z_0), |f'(z_0)| r).$

For maps that are merely holomorphic instead of affine-linear, one has some variants of this assertion, which I am recording here mostly for my own reference:

Theorem 1 (Holomorphic images of disks) Let ${D(z_0,r)}$ be a disk in the complex plane, and ${f: D(z_0,r) \rightarrow {\bf C}}$ be a holomorphic function with ${f'(z_0) \neq 0}$ .

(i) (Open mapping theorem) ${f(D(z_0,r))}$ contains a disk ${D(f(z_0),\varepsilon)}$ for some ${\varepsilon>0}$ .

(ii) (Bloch theorem) ${f(D(z_0,r))}$ contains a disk ${D(w, c |f'(z_0)| r)}$ for some absolute constant ${c>0}$ and some ${w \in {\bf C}}$ . (In fact there is even a holomorphic right inverse of ${f}$ from ${D(w, c |f'(z_0)| r)}$ to ${D(z_0,r)}$ .)

(iii) (Koebe quarter theorem) If ${f}$ is injective, then ${f(D(z_0,r))}$ contains the disk ${D(f(z_0), \frac{1}{4} |f'(z_0)| r)}$ .

(iv) If ${f}$ is a polynomial of degree ${n}$ , then ${f(D(z_0,r))}$ contains the disk ${D(f(z_0), \frac{1}{n} |f'(z_0)| r)}$ .

(v) If one has a bound of the form ${|f'(z)| \leq A |f'(z_0)|}$ for all ${z \in D(z_0,r)}$ and some ${A>1}$ , then ${f(D(z_0,r))}$ contains the disk ${D(f(z_0), \frac{c}{A} |f'(z_0)| r)}$ for some absolute constant ${c>0}$ . (In fact there is holomorphic right inverse of ${f}$ from ${D(f(z_0), \frac{c}{A} |f'(z_0)| r)}$ to ${D(z_0,r)}$ .)

Parts (i), (ii), (iii) of this theorem are standard, as indicated by the given links. I found part (iv) as (a consequence of) Theorem 2 of this paper of Degot, who remarks that it “seems not already known in spite of its simplicity”. The proof is simple:

Proof: (Proof of (iv)) Let ${w \in D(f(z_0), \frac{1}{n} |f'(z_0)| r)}$ , then we have a lower bound for the log-derivative of ${f(z)-w}$ at ${z_0}$ :

$\displaystyle \frac{|f'(z_0)|}{|f(z_0)-w|} > \frac{n}{r}$

(with the convention that the left-hand side is infinite when ${f(z_0)=w}$ ). But by the fundamental theorem of algebra we have

$\displaystyle \frac{f'(z_0)}{f(z_0)-w} = \sum_{j=1}^n \frac{1}{z_0-\zeta_j}$

where ${\zeta_1,\dots,\zeta_n}$ are the roots of the polynomial ${f(z)-w}$ (counting multiplicity). By the pigeonhole principle, there must therefore exist a root ${\zeta_j}$ of ${f(z) - w}$ such that

$\displaystyle \frac{1}{|z_0-\zeta_j|} > \frac{1}{r}$

and hence ${\zeta_j \in D(z_0,r)}$ . Thus ${f(D(z_0,r))}$ contains ${w}$ , and the claim follows. $\Box$

The constant ${\frac{1}{n}}$ in (iv) is completely sharp: if ${f(z) = z^n}$ and ${z_0}$ is non-zero then ${f(D(z_0,|z_0|))}$ contains the disk

$\displaystyle D(f(z_0), \frac{1}{n} |f'(z_0)| r) = D( z_0^n, |z_0|^n)$

but avoids the origin, thus does not contain any disk of the form ${D( z_0^n, |z_0|^n+\varepsilon)}$ . This example also shows that despite parts (ii), (iii) of the theorem, one cannot hope for a general inclusion of the form

$\displaystyle f(D(z_0,r)) \supset D(f(z_0), c |f'(z_0)| r )$

for an absolute constant ${c>0}$ .

Part (v) is implicit in the standard proof of Bloch’s theorem (part (ii)), and is easy to establish:

Proof: (Proof of (v)) From the Cauchy inequalities one has ${f''(z) = O(\frac{A}{r} |f'(z_0)|)}$ for ${z \in D(z_0,r/2)}$ , hence by Taylor’s theorem with remainder ${f(z) = f(z_0) + f'(z_0) (z-z_0) (1 + O( A \frac{|z-z_0|}{r} ) )}$ for ${z \in D(z_0, r/2)}$ . By Rouche’s theorem, this implies that the function ${f(z)-w}$ has a unique zero in ${D(z_0, 2cr/A)}$ for any ${w \in D(f(z_0), cr|f'(z_0)|/A)}$ , if ${c>0}$ is a sufficiently small absolute constant. The claim follows. $\Box$

Note that part (v) implies part (i). A standard point picking argument also lets one deduce part (ii) from part (v):

Proof: (Proof of (ii)) By shrinking ${r}$ slightly if necessary we may assume that ${f}$ extends analytically to the closure of the disk ${D(z_0,r)}$ . Let ${c}$ be the constant in (v) with ${A=2}$ ; we will prove (iii) with ${c}$ replaced by ${c/2}$ . If we have ${|f'(z)| \leq 2 |f'(z_0)|}$ for all ${z \in D(z_0,r/2)}$ then we are done by (v), so we may assume without loss of generality that there is ${z_1 \in D(z_0,r/2)}$ such that ${|f'(z_1)| > 2 |f'(z_0)|}$ . If ${|f'(z)| \leq 2 |f'(z_1)|}$ for all ${z \in D(z_1,r/4)}$ then by (v) we have

$\displaystyle f( D(z_0, r) ) \supset f( D(z_1,r/2) ) \supset D( f(z_1), \frac{c}{2} |f'(z_1)| \frac{r}{2} ) \supset D( f(z_1), \frac{c}{2} |f'(z_0)| r )$

and we are again done. Hence we may assume without loss of generality that there is ${z_2 \in D(z_1,r/4)}$ such that ${|f'(z_2)| > 2 |f'(z_1)|}$ . Iterating this procedure in the obvious fashion we either are done, or obtain a Cauchy sequence ${z_0, z_1, \dots}$ in ${D(z_0,r)}$ such that ${f'(z_j)}$ goes to infinity as ${j \rightarrow \infty}$ , which contradicts the analytic nature of ${f}$ (and hence continuous nature of ${f'}$ ) on the closure of ${D(z_0,r)}$ . This gives the claim. $\Box$