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Full question: Let $f$ be analytic in $D = \{z : |z| < 1\}$ and suppose that $|f(z)|≤ M$ for all $z$ in $D$. If $f(z_k) = 0$ for $1 ≤ k ≤ n$ show that $|f(z)| ≤ M \prod_{k=1}^n$$|z − z_k| \over |1 − \bar z_k z|$ for $|z| < 1$.

I am new to complex analysis and schwarz lemma (which I believe is at play in this problem). If someone could point me in the right direction or provide a proof it would be greatly appreciated. I'm just not quite sure where to go with this problem.

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    Let $g(z)=f(z)\prod_{k=1}^n\frac{1 − \bar z_k z}{z-z_k}$; the hypothesis $f(z_k)=0$ implies $g$ is analytic in the unit disc and since $|\frac{1 − \bar z_k z}{z-z_k}|=1$ on the unit circle, one gets that $|g(z)| \le M$ too by maximum modulus (technically one needs to take $a_r$ the maximum of $|\prod_{k=1}^n\frac{1 − \bar z_k z}{z-z_k}|$ on $|z|=r$ and $a_r \to 1$ when $r \to 1$ so one gets $|g(z)| \le a_rM$ for any $|z| <r$ and let $r \to 1$.) – Conrad Oct 07 '21 at 02:01
  • @Conrad Wasn't aware this was a duplicate at first but your comment was excellent and I was able to get it just off that, thank you. –  Oct 07 '21 at 05:43
  • happy to be of help – Conrad Oct 07 '21 at 12:22

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