# Functions on the unit disk $f(0)=1$ and $Re f >0$ Prove the given inequality.

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.

• – Martin R Apr 17 '18 at 16:41
• I hadn't seen these links! Thanks for the reference! – Jade Apr 17 '18 at 17:51

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.