Supremum of $ \lbrace |f'(a)| : f : D_1(0) \rightarrow D_1(0) \text{ holomorphic with } f(a) = b \rbrace,$ Let $a, b \in D_1(0)$ (the open unit disk). I want to find the supremum of
$$ \lbrace |f'(a)| : f : D_1(0) \rightarrow D_1(0) \text{ holomorphic with } f(a) = b \rbrace,$$
and find out if that supremum is ever taken by some function. I tried to do something using the Schwarz lemma: It seems like using a function like $h(z) = f(z) - b$ could give a nice bound on $f(z)$, but it's not actually guaranteed that this would be a function $D_1(0) \rightarrow D_1(0)$, nor do I know how I would get from here to $f'$.
 A: Well, let $\mathcal{F}=\{f\in H(\mathbb{D},\mathbb{D}): f(a)=b\}$. Note that this is non-empty (consider a holomorphic automorphism of the unit disk mapping $a$ to $b$). since every function $f\in\mathcal{F}$ has absolute value at most 1, by Montel's theorem $\mathcal{F}$ is a normal family. Let $(f_n)\subset\mathcal{F}$ be a sequence such that $|f_n'(a)|\to\sup_{f\in\mathcal{F}}|f_n'(a)|$. Then, due to normality, we obtain a subsequence $(f_{n_k})$ such that $f_{n_k}\to f$ locally uniformly where $f:\mathbb{D}\to\mathbb{C}$ is a holomorphic function (by Weierstrass' theorem, locally uniform limits of sequences of holomorphic functions are also holomorphic).  Note that $f_{n_k}'(a)\to f'(a)$, again by Weierstrass' covnergence theorem. Thus $|f'(a)|=\sup_{g\in\mathcal{F}}|g'(a)|$. Now we have that $|f(z)|\leq1$ for all $z\in\mathbb{D}$. By the maximum modulus principle, if $|f(z)|=1$ for some $z$, then $f$ is constant, hence $f'=0$, so $f'(a)=0$ and thus $g'(a)=0$ for all $g\in\mathcal{F}$, a contradiction. Thus $|f(z)|<1$ and also $f(a)=\lim_{k}f_{n_k}(a)=b$, hence $f\in\mathcal{F}$.
To compute the supremum, we use the Schwarz-Pick lemma: if $f\in\mathcal{F}$, then
$$|f'(a)|\leq\frac{1-|b|^2}{1-|a|^2}:=s$$
so the supremum is at most $s$. On the other hand, set $\phi_1(z)=\frac{z-a}{1-\bar{a}z}$ and $\phi_2(z)=\frac{z+b}{1+\bar{b}z}$. These are holomorphic self-maps of the unit disk and the composition $h=\phi_2\circ\phi_1$ satisfies $h(a)=\phi_2(0)=b$, so $h\in\mathcal{F}$. Now
$h'(z)=\phi_2'(\phi_1(z))\phi_1'(z)$
so $h'(a)=\phi_2'(0)\phi_1'(a)=\frac{1-|b|^2}{1-|a|^2}=s$, so the supremum is greater than or equal to $s$. We conclude that the supremum, is precisely equal to $s=\frac{1-|b|^2}{1-|a|^2}$ and it actually is attained by this function $h$ we considered, i.e. it is a maximum.
Edit: thanks Martin R for your comment, the first part was not needed indeed:)
