I found this question in an old Mathematical Olympiad:
Let $0<a<1$ be a real number, and let $f(z)$ be a complex polynomial such that $$|f(z)|\leq \frac{1}{|z-a|}$$ on the unit disk $|z|\leq 1$. Prove that $$|f(a)| \leq \frac{1}{1-a^2}.$$
My attempt: Since $f$ is analytic in $\{z\in\mathbb{C}:|z|\leq 1\}$, we have that $$|f(a)| \leq \max_{|z|=1}|f(z)|.$$ Because of the triangle inequality, for all $z\in \mathbb{C}$ with $|z|=1$ we have $$|z-a|\geq |z|-a = 1-a.$$ By applying the hypothesis, we get $$|f(a)| \leq \max_{|z|=1}|f(z)| \leq \max_{|z|=1}\frac{1}{|z-a|}\leq \frac{1}{1-a}.$$ Nevertheless, we know that $1-a^2>1-a$ since $a\in (0,1)$. Therefore, we cannot get the desired result from the above inequality.
What can I apply to complete the proof?
Best Answer
Let $F(z)=(z-a)f(z)$ and $G(z)=F(\frac {z+a} {1+az})$. [Recall that $ z \to \frac {z+a} {1+az}$ is an analytic bijection of the unit disk into itself whose inverse is $ z \to \frac {z-a} {1-az}$]. Now $|G(z)| \leq 1$ and $G(0)=F(a)=0$. By Schwarz Lemma we get $|G(z)| \leq |z|$. This can be written as $|F(z)|\leq |\frac {z-a}{1-az}|$. Hence, $|f(z)|\leq |\frac 1{1-az}|$ and putting $z=a$ finishes the proof.