an analytic function being zero

Question

Let $f$ be an analytic function defined on the unit disc $D=\{z:|z|<1\}$. If $|f(z)|\leq 1-|z|$ for all $z\in D$ then show that $f$ is a zero function on $D$.

Please give only hints.

Answer 1

Hint: By the Cauchy integral formula for the derivatives, you have

$$ |f^{(n)}(0)| = \frac{n!}{2\pi} \left| \oint_{|z| = a} \frac{f(z)}{z^{n+1}}\, dz \right| \leq \frac{n!}{2\pi} \oint_{|z| = a} \frac{|f(z)|}{a^{n+1}} \, |dz| \leq n! \frac{1 - a}{a^{n}} $$

for all $0 < |a| < 1$.

Answer 2

Using a consequence of the maximum modulus theorem one has $\max_{z \in B} |f(z)| = \max_{z \in \partial B} |f(z)| \leq \max_{z \in \partial B} (1 - |z|) = 1 - r \hspace{1cm} \forall B = U(0,r)$ with $0 < r < 1$