14
$\begingroup$

(a) Let $\mathbb{D}$ denote the unit disk. Is there an analytic function $f \colon \mathbb{D} \to \mathbb{D}$ with $f(0) = 1/2$ and $f′(0) = 3/4?$ Either find such a function $f$ or explain why it does not exist.

(b) Answer the same question for $f(0) = 1/2$ and $f′(0) = 4/5.$

It seems like I could use the Schwarz lemma, but that is not working out so well. Any suggestions? Thanks.

$\endgroup$
1
  • $\begingroup$ Typically one composes the function with $(z-1/2)/(1-(\overline{1/2})z)$; the composition preserves the image as inside the unit circle but now sends $0$ to $0$. Now you can use the Schwarz lemma and so on. $\endgroup$ Jul 15, 2014 at 22:12

1 Answer 1

14
$\begingroup$

The "Schwarz-Pick" variant of Schwarz's lemma, proved, as Greg Martin's comment suggests, by composing with Möbius transformations of the unit disk, exactly answers your question: a holomorphic function $f\colon \mathbb{D} \to \mathbb{D}$ must satisfy

$$\frac{|f'(z)|}{1-|f(z)|^2} \le \frac{1}{1-|z|^2}$$

In case (a), the Schwarz-Pick inequality is saturated for $z=0$, so basically the only possible function is the Möbius transformation $f(z) = \frac{2z+1}{z+2}$ taking $0$ to $\frac{1}{2}$, which has derivative $\frac{3}{4}$ there. In case (b), the inequality forbids such a function.

$\endgroup$
2
  • $\begingroup$ but here your f'(0) should be 3/4.. $\endgroup$
    – Shri hari
    Apr 16, 2022 at 18:16
  • $\begingroup$ The answer should be $f(z)=\frac{2z+1}{z+2}$. $\endgroup$ Apr 19, 2022 at 8:35

You must log in to answer this question.

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