Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find $\max |f'(i)|$ and mapping for which it reaches a maximum, if $f:\mathbb{H} \to \mathbb{D}$ is analytic function and $f(i)=0.$

Notation: $\mathbb{H}=\{z\in \mathbb{C}: \operatorname{\mathfrak{Im}}z>0\}$ and $\mathbb{D}=\{z\in \mathbb{C}: |z|<1\}.$

My solution: Let find (bilinear) map $z=g(\zeta), \ \ \zeta \in \mathbb{D}$ from $\mathbb{D}$ to $\mathbb{H}$ such that $g(0)=i$. We find that $g(\zeta)=i \dfrac{\zeta + 1}{1-\zeta}, \ \ \zeta \in \mathbb{D}$. Then, look at function $$\omega=F(\zeta)=(f \circ g)(\zeta)=f(g(\zeta))=f\left(i \frac{\zeta + 1}{1-\zeta}\right), \ \ \zeta \in \mathbb{D}$$ Clearly, $F:\mathbb{D} \to \mathbb{D}$ is analytic and $F(0)=f(g(0))=f(i)=0$, so we can apply Schwarz lemma to conclude that $$|F'(0)|\leqslant 1 \iff|f'(i)g(0)|\leqslant 1 \iff 2 |f'(i)|\leqslant 1 \iff |f'(i)|\leqslant \frac{1}{2}.$$ If $|F'(0)|=1$ then we find that $F(\zeta)=e^{i\alpha}\zeta, \ \ \alpha \in [0,2\pi), \ \ \zeta \in \mathbb{D}$, so $f\left(i \dfrac{\zeta + 1}{1-\zeta}\right)=e^{i\alpha}\zeta,\ \ \zeta \in \mathbb{D},$ or $$f(z)=e^{i\alpha} \frac{z-i}{z+i}, \ \ z \in \mathbb{H}, \ \ \alpha \in [0,2\pi).$$

My question: From Schwarz's lemma we also get $|F(\zeta)|\leqslant |\zeta|, \ \ \zeta \in \mathbb{D}.$ What we can conclude from that?

share|improve this question

1 Answer 1

I don't think $|F(z)|\le|z|$ gives you anything new. Think of Schwarz's lemma this way: it says that the function $F(z)/z$ is at most $1$ in modulus. (In fact, that's how the proof of Schwarz's lemma goes, via the maximum modulus principle.) But the expression $F(z)/z$ isn't defined at $z=0$; you need to put in the value that makes the function continuous there, and that value is exactly $F'(0)$ (by l'Hôpital's rule, if you like). So the conclusion that $|F'(0)|\le1$ is a "special case" of the general conclusion $|F(z)/z|\le 1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.