Suppose $f:D\to\mathbb{C}$ is a holomorphic function with $\text{Re}\ f(z)\geq 0$ for all $z\in D$ and $f(0)=1$.

($D$ is the open unit disc $\{|z|<1\}$)

Prove that:

(i) $\text{Re}\ f(z)>0$ for all $z\in D$.

(ii) $|f(z)|\leq\frac{1+|z|}{1-|z|}$ for all $z\in D$.

(iii) $|f(z)|\geq\frac{1-|z|}{1+|z|}$ for all $z\in D$.

So far I am only able to prove (ii), using Schwarz's Lemma. Thanks for any help!

Proof of (ii):

Let $g(z)=\frac{z-1}{z+1}$ which is known to map the right half plane to the open unit disk, so that $gf:D\to D$ is analytic and $gf(0)=g(1)=0$. This satisfies the Schwarz Lemma so that


Breaking open the modulus: \begin{align*} |f(z)|-1&\leq||f(z)|-1|\\ &\leq|f(z)-1|\\ &\leq|zf(z)+z|\\ &\leq|f(z)||z|+|z| \end{align*}

Rearranging gives $|f(z)|(1-|z|)\leq 1+|z|$ which gives $|f(z)|\leq\frac{1+|z|}{1-|z|}$ which is (ii).


Hint. For part (i), note that $\operatorname{Re}f(z)$ is harmonic, hence obeys maximum principle. For part (iii), remember that we've already proved $\operatorname{Re}f(z)>0$, hence $f(z)\not=0$. Now note that $1/f(z)$ satisfies the same property as $f(z)$, so we may apply the consequence in part (ii).

  • $\begingroup$ Very nice. Thanks. $\endgroup$ – yoyostein Dec 2 '16 at 14:19

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