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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?

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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 $|F(z)/z|\le 1$.

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