Application of Schawarz lemma??

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 $D$.

We have $|f(z)|\leq 1-|z|$ for all $z$ which implies that $|f(z)|\leq 1$ for all $z$..

Let $a\in(0,1)$. $f$ is analytic on $\{z: |z|\leq a\}=D_a$. So, by cauchy integral formula, we have $$f(0)=\frac{1}{2\pi i}\int_{\partial D_a}\frac{f(w)}{w}dw.$$ Further, $$|f(0)|\leq \frac{1}{2\pi }\int_{\partial D_a}\frac{|f(w)|}{|w|}|dw|\leq\frac{1}{2\pi}\int_0^{2\pi}\frac{1-|w|}{|w|}ad\theta=\frac{1}{2\pi}\frac{1-a}{a}a2\pi=1-a.$$ So, $|f(0)|\leq 1-a$ for all $a\in(0,1)$. So, $f(0)=0$.

So, $f(0)=0$ and $|f(z)|\leq 1-|z|$. So, by Schwarz lemma, $|f(z)|\leq |z|$ for all $z\in D$...

Then i see that $|f(z)|\leq |z|$ and $|f(z)|\leq 1-|z|$ implies $2|f(z)|\leq 1-|z|+|z|=1$ implies $|f(z)|\leq \frac{1}{2}$

I thought i can repeat this and conclude that $|f(z)|\leq \frac{1}{n}$ for all $n$ and then conclude that $f$ is zero but i am getting $|f(z)|\leq \frac{1}{2}$ only and no better bound.

I am not able to proceed...

Answer 1

Here are hints:

1. For a function $f$ analytic on the open disc $D$, define $$M_f(r)=\max_{|z|=r}\left|f(z)\right|$$for $r\in\left[0,1\right). $Prove that $$M_f(r)=\max_{|z|\leq r}\left|f(z)\right|$$
2. Conclude that $M_f(r)$ is monotonically increasing on $\left[0,1\right)$.
3. When is $M_f(r)$ constant? I.e., can you classify all functions $f$ for which $M_f(r)$ is constant?
4. Use the fact that $\left|f(z)\right|\leq1-|z|$ to prove that $f$ is constant (using the above steps, of course).

I hope it helps.