This is an old problem from Ph.D Qualifying Exam of Complex Analysis.
Let $f$ be a holomorphic function in the open disc $D(0,2)$ of radius 2 centered at the origin and suppose that $|f(z)|=1$ whenever $|z|=1$, and $f(0)=0$. Prove that $|f'(z)|\ge 1$ if $|z|=1$.
My attempt: By maximum modulus principle, $|f(z)|< 1$ when $|z|<1$. Therefore, by Schwarz lemma, $|f(z)|\le |z|$ if $|z|\le 1$. Since $|f(z)|=1$ when $|z|=1$, I guess something similar to the Mean Value Theorem would hold, but I have no idea how to figure it out. Does anyone have ideas?
Thanks in advance!
Best Answer
Let $z_0$ in the unit circle, since we know that $f$ is holomorphic :
$$f'(z_0)=\lim_{\lambda \to 0^+}\frac {f(z_0(1-\lambda))-f(z_0)}{-\lambda z_0}.$$
Let $1>\lambda >0$ :
$$\left|\frac {f(z_0(1-\lambda))-f(z_0)}{-\lambda z_0}\right|=\left|\frac {f(z_0(1-\lambda))-f(z_0)}{\lambda}\right| \ge\left|\frac {\left|f(z_0(1-\lambda))\right|-|f(z_0)|}{\lambda}\right| .$$
As you noted it, we can use Schwarz lemma. Hence $|f(z)| \le |z|$ for every $z$ in the open unit disc. Furthermore, $|f(z_0)|=1$, then :
$$\left|\frac {f(z_0(1-\lambda))-f(z_0)}{-\lambda z_0}\right|\ge\frac {\left|f(z_0)\right|-|f(z_0(1-\lambda))|}{\lambda}\ge\frac {1-|(1-\lambda)|}{\lambda}=1.$$
The quantity we are taking the limit from is always greater or equal to $1$ for $\lambda<1$, then the limit is greater or equal to $1$.