For an upper half-plane, $\mathbb{H} = \{z \in \mathbb{C} : Im(z) > 0 \}$ $f$ is a function holomorphic in $\mathbb{H}$ and we have $f(\mathbb{H}) \subset \mathbb{D}$ where $\mathbb{D}$ is an open unit disc. How large can $|f'(i)|$ be ?
My attempt : Cayley's transformation maps upper half plane to unit disc. Suppose $F$ be a set of all such $f$ now $f$ map will be of the form $ z \mapsto e^{i\theta}\frac{z+i}{z-i}$
Now I cant apply schwartz lemma to find $|f'(i)|$ as $0$ is not mapped to $0$ How should I proceed further ? am I on right track ?
Your ideas are correct. You can use Mobius transformations of the unit disk to map a point to $0$.
Let $f(z)=(z-i)/(z+i)$ be the Cayley transformation from the upper half plane to the unit disk that maps $i$ to $0$, and consider $h$ an arbitrary map from the upper half plane into the unit disk, such that $h(i)=a$. Then, move $a$ to $0$ using the Mobius transformation $g(z)=(z-a)/(1-\bar{a}z)$. Now, you can apply Schwarz's lemma to the composition $g\circ h\circ f^{-1}$, which yields $$ |(g\circ h\circ f^{-1})'(0)| \leq 1$$ Computing the derivative of $g$ and $f$, and rearranging yields $$|h'(i)|\leq (1-|a|^2)/2$$ where $|f'(i)|=1/2$. Note that the maximum of $|h'(i)|$ is $1/2$ and is attained when $a$ is $0$. In the case that the maximum is attained $(a=0)$, we have equality in Schwarz's lemma, and this implies that $h=e^{i\theta}f$ for some $\theta$.