0
$\begingroup$

Let $A$ denote the set of all holomorphic functions $f(z)$ on the open unit disk $D=\lbrace |z|<1 \rbrace$ such that $f(0)=1$ and $\text{Re }f>0$. Show that if $F \in A$, then it satisifies the inequality

$$\frac{1-|z|}{1+|z|} \leq |f(z)| \leq \frac{1+|z|}{1-|z|} $$

I am not sure how to prove this inequality. I know that the mapping from the unit disk to the right half plane is given by $\frac{1+z}{1-z}$, so I think that is where the upper bound comes from, although I am not sure how to prove that this is the tightest upper bound and I don't know how to get the lower bound.

$\endgroup$
1
$\begingroup$

Agree with @Martin R. Since your question only involves $\left|f\right|$, it could observe some simplification.

Let $\mathbb{D}$ be the unit disk, $\mathbb{H}$ be the right-half plane. Define \begin{align} \Phi:\mathbb{H}\to\mathbb{D},&&z\mapsto\frac{z-1}{z+1}. \end{align} Obviously, $\Phi(1)=0$. Therefore, \begin{align} \Phi\circ f:\mathbb{D}\to\mathbb{D},&&0\mapsto 0. \end{align} Thanks to this fact, Schwarz lemma applies, i.e., $$ \left|\Phi\circ f(z)\right|\le\left|z\right| $$ holds for all $z\in\mathbb{D}$. The above inequality is exactly $$ \left|\frac{f(z)-1}{f(z)+1}\right|\le\left|z\right|\iff\left|f(z)-1\right|\le\left|z\right|\left|f(z)+1\right|. $$ The rest of your task is to play with this last inequality. For one thing, $$ \left|f\right|-1\le\left|f-1\right|\le\left|z\right|\left|f+1\right|\le\left|z\right|\left(\left|f\right|+1\right), $$ which yields $$ \left|f\right|\le\frac{1+\left|z\right|}{1-\left|z\right|}. $$ For another, $$ 1-\left|f\right|\le\left|f-1\right|\le\left|z\right|\left|f+1\right|\le\left|z\right|\left(\left|f\right|+1\right), $$ which leads to $$ \left|f\right|\ge\frac{1-\left|z\right|}{1+\left|z\right|}. $$ These are exactly what you want.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.