For some $r\in(0, 1)$, $f(0)=f(r)=f(-r)=0 $. Prove $|f(z)|\leq |z||\frac{z^2-r^2}{1-r^2z^2}|$.

Asked May 19 '21 at 13:31

Active May 19 '21 at 13:31

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Let $f(z)$ be analytical in the unit disk $\mathbb{D}=\{z:|z|<1\}$. Suppose $|f(z)|< 1$ when $|z|<1$ and for some $r\in(0, 1)$$f(0)=f(r)=f(-r)=0 $. Prove $|f(z)|\leq |z|\left|\dfrac{z^2-r^2}{1-r^2z^2}\right|$.

I think we should apply the Schwarz lemma here,but for now I can only prove $$ |f(z)|\leq\left|\dfrac{z-r}{1-rz}\right|, |f(z)|\leq\left|\dfrac{z+r}{1+rz}\right|, $$ which seem of no use. I do not know where $|z|$ of RHS comes from (the rest part are two maps in Aut($\mathbb{D}$)).

Appreciate any help or hint!

asked May 19 '21 at 13:31

user823011

1,335

That is a special case of Showing that $|f(z)| \leq \prod \limits_{k=1}^n \left|\frac{z-z_k}{1-\overline{z_k}z} \right|$. – Martin R May 19 '21 at 13:33
Note that if $z_k=0$ then the corresponding factor on the right becomes $|z|$, that's where it comes from! – Martin R May 19 '21 at 13:37
I got it! Thanks very much. – user823011 May 19 '21 at 13:38

For some $r\in(0, 1)$, $f(0)=f(r)=f(-r)=0 $. Prove $|f(z)|\leq |z||\frac{z^2-r^2}{1-r^2z^2}|$.

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