# upper bound for derivatives of analytic functions on upper half-plane

Let $f$ be analytic mapping of the upper half-plane into the unit disc. Given $f(i)=\alpha$, how does one rigorously obtain an upper bound for $|f'(i)|$?

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We can assume that $f(i)=i$ without loss of generality (scale and rotate). Then we use Cayley's transform $h(z):=\frac{z-i}{z+i}$, from the upper half-plane to the unit disc, and its inverse $h^{—1}(z)=-i\frac{z+1}{z-1}$. Let $g(z):=h\circ f\circ h^{-1}$ from the unit disk to itself. We have $g(0)=0$ and by Schwarz lemma, $|g'(0)|\leq 1$. Hence we have, after having used the chain rule, $$|f'(i)|\leq\frac 1{|h'(i)|\cdot |(h^{-1})'(0)|}.$$
no, it does not (give a better estimate) :-). Btw, shouldn't that read $(h^{-1})'(0)$ in the denominator? –  user20266 Aug 4 '12 at 12:16
There are Cauchy's estimates for derivatives of analytic functions: if $f$ is holomorphic in a neighourhood of the closed ball $\bar{B}(z_0)$, then, for every $z\in B(z_0)$ you have an estimate for the $k$-th derivative, $$|f^{(k)}(z)| \le k!\frac{r}{d(z)^{k+1}}|f|_{\partial B}, \,\, d(z)= \mbox{dist}(z, \partial B(z_0))$$ You can apply this with $z_0=i$, any ball around $i$ of radius less than $1$ and the knowledge that $|f|\le 1$ (this is, in particuar, a bound for $f$ at the boundary of the ball).