Here is the problem:
Let $f$ be an analytic function defined on $\mathbb{H}$:={$z \in \mathbb{C}:Imz > 0 $}. Suppose that $|f(z)|<1$ for all $z \in \mathbb{H}$. Prove that for every $z \in \mathbb{H}, |f'(z)| \leqslant \frac{1}{2Imz}$
I used Cauchy's formula to calculate $f'$, but the radius of the contour circle can not be larger than $Imz$, since $f$ is only analytic on $\mathbb{H}$. And then I used that integral to approximate $|f'|$, I can only get $|f'| < \frac{2}{Imz}$.
Any hint is appreciated, thank you!
The contour used in the Cauchy integral formula need not be a circle. We have $$ f'(z) = \frac{1}{2\pi i} \int_{C} \frac{f(w)}{(z-w)^2} \, dw; $$ let us deform $C$ to be composed of a line of length $2R$ just above the real axis, and a large semicircle of radius $R$ in the upper half-plane. Then one can show that the integral along the semicircle is bounded by a multiple of $1/R$ (as one would in one of the standard proofs of Liouville's theorem), and so $$ \lvert f'(z) \rvert \leqslant \frac{1}{2\pi} \int_{-R+i\varepsilon}^{R+i\varepsilon} \frac{\lvert f(w) \rvert }{\lvert z-w \rvert^2} \, dw + \frac{K}{R}. $$ Taking $R \to \infty$ and $\varepsilon \to 0$, and using that $\lvert f(w) \rvert \leqslant 1$, we find $$ \lvert f'(z) \rvert \leqslant \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{dx}{\Im(z)^2+x^2}, $$ where $w=\Re(z)+x+0i$ on the contour. The remaining integral evaluates to $\pi/\Im(z)$, which gives the required result.