Let $D = \{ z \in \mathbb C : |z| < 1 \}$. Suppose $f : D \to \mathbb C$ is a non-constant analytic function such that $\mathrm{Re} f(z) \geq 0$ for all $z \in D$.

(a) Show that $\mathrm{Re} f(z) \neq 0$ for all $z \in D$.

(b) Using a Möbius transformation and Schwarz's lemma, prove that if $f(0) = 1$, then $$ |f(z)| \leq {1+|z| \over 1-|z|}. $$ What can be said if $f(0) \neq 1$?

(c) Show that if $f(0) = 1$, then $f$ also satisfies $$ |f(z)| \geq {1-|z| \over 1+|z|}. $$


(a) We can use the open mapping theorem.

(b) We can compose a Möbius transformation $m$ that sends the right-half plane to $D$ with $f$. Then $h = m \circ f$ satisfies the hypotheses of Schwarz's lemma by construction. Simple manipulations of the inequality from Schwarz's lemma give the result.

(c) Looking at part (a), we know that $\mathrm{Re} f(z) \neq 0$, and so $f(z) \neq 0$. Thus, we can use the function $g = 1/f$ in the statement of part (b).


What about when $f(0) \neq 1$? I don't think you can get anything from $f(z)-f(0)+1$ unless $\mathrm{Re} f(0) < 1$. And you can't in general rearrange the inputs of $f$ because $1$ isn't necessarily in the image of $f$. Also, I don't think you can get anything from $f(z)/f(0)$ unless $f(0)$ is real, because otherwise the range of $f(z)/f(0)$ could extend into the left-half plane.


If $f(0)\ne 1$, you can consider $\phi\circ f$ where $\phi$ is an automorphism of the right half plane such that $\phi(f(0))=1$. Namely, $$\phi(w) = \frac{w-i\operatorname{Im}f(0)}{\operatorname{Re}f(0)}$$

  • $\begingroup$ Should the numerator be $w- i \operatorname{Im} f(0)$? And is it important that $\phi$ be an automorphism? I don't see why it wouldn't work if $\phi$ was just any function that sends $f(0)$ to $1$, whose image is contained in the right-half plane when restricted to inputs in the right-half plane. I guess it needs to be analytic for Schwarz's lemma to apply. $\endgroup$ – A l'Maeaux Nov 1 '13 at 14:10
  • $\begingroup$ @DanielHirsbrunner Thanks for pointing out the missing $i$. If $\phi$ is not an automorphism, we can still get an inequality for $f$. But we won't get the sharp inequality for $f$. $\endgroup$ – user103402 Nov 2 '13 at 4:36

$$\bigg|\dfrac{f-f(0)}{f+f(0)}\bigg|\leqslant 1$$ (think why? hint: Re $f(z)>0$)

Schwarz's lemma dictates that if we consider $\dfrac{f-f(0)}{f+f(0)}$ as a single function, we have:


Solve it: $$\dfrac{1-|z|}{1+|z|}|f(0)|\leqslant|f(z)|\leqslant\dfrac{1+|z|}{1-|z|}|f(0)|$$ (now consider $|f|,|f(0)|,|z|$ as $x,y,z$ and solve it using the most elementary way)


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