An Inequality between two holomorphic function on $\mathbb{D}$

Question

Suppose $f:\mathbb{D} \rightarrow \mathbb{D}$ be a holomorphic function and $f(z_k) = 0 , k = 1,2,...,n$. How can I prove following inequality? $$|f(z)| \leq \Pi_{k=1}^{n}|\frac{z-z_k}{1-\bar{z_k}z}|$$ and if equality holds for som $z \not= z_k$ then what can I say about $f$?

Answer 1

Let $g(z)=\Pi_{k=1}^{n}\frac{z-z_k}{1-\bar{z_k}z}$; it is well known and easy to prove (as $g$ is a finite Blaschke product) that $|g(z)| \to 1$ (uniformly) as $|z| \to 1$ inside the disc

(this follows from $|g(z)|=1$ on the unit circle and $g$ analytic in a small neighborhood of the unit disc and incidentally, the converse is true but harder to prove and is known as Rado theorem)

Then for any $\epsilon >0$, $|g(z)| \ge 1-\epsilon$ for $r_{\epsilon} \le |z| <1$, so $\frac{1}{|g(z)|} \le 1+2\epsilon$ for $r_{\epsilon} \le |z| <1$ as $\frac{1}{1-\epsilon} \le 1+2\epsilon$ for $\epsilon$ small enough positive.

But now $\frac{f}{g}$ is holomorphic by hypothesis and by maximum modulus for $\epsilon >0$ there is $r_{\epsilon}$ (which goes to $1$ with $\epsilon \to 0$, so that $|\frac{f}{g}| \le 1+2\epsilon, |z|=r_{\epsilon}$, hence for $|z|\le r_{\epsilon}$ too. This clearly means $|\frac{f}{g}| \le 1$ inside the unit disc, so the inequality is proved.

Now if we have equality at some point which is not $z_k$, it follows that $|\frac{f}{g}|=1$ there (note that at $z_k$ we cannot conclude that since we cannot divide by zero!), hence by maximum modulus again $\frac{f}{g}$ is a unit modulus constant, so $f(z)=\lambda \Pi_{k=1}^{n}\frac{z-z_k}{1-\bar{z_k}z}, |\lambda|=1$