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Let $f$ be holomorphic from the right half-plane $Re z>0$ to the unit disk. Prove that $$|f'(z)|\leq\frac{1-|f(z)|^2}{2 Re z},$$ for any $z:\ Re z>0$.

I do know that $1-|f(z)|^2$ is the square of the distance from $f(z)$ to $z$ the to unit circle, and $Re z$ is the distance from $z$ to the boundary of $Re z>0$. But how can we relate them by the derivative? Thank you.

Let us fix $z$, and consider $$w=\frac{\zeta-z}{\zeta+\bar z}.$$ Then $Re \zeta>0$ is transformed to $|w|=1$.

Consider $$F(w)=f(\zeta)=f(\frac{z+w\bar z}{1-w})$$, we obtain, by Cauchy integral formula, that $$F'(0)=\frac{1}{2\pi i}\int_{|\eta|=1}\frac{F(\eta)}{\eta^2}d\eta.$$ Now, $$F'(0)=f'(z)\cdot 2 Re z,$$ and then $$\frac{1}{2\pi i}\int_{|\eta|=1}\frac{F(\eta)}{\eta^2}d\eta=?$$

XLDD
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2 Answers2

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I do know that $1−|f(z)|^2$ is the square of the distance from $f(z)$ to the unit circle

As a matter of fact, it's not. In any case, conformal maps have their own idea about distance to the boundary, which only loosely agrees with the Euclidean notion of distance.

Schwarz-Pick theorem gives $|g'(0)|\le 1-|g(0)|^2$ for any holomorphic map of unit disk to itself. Consider the map from your post $$\varphi(\zeta)=\frac{\zeta-z}{\zeta+\bar z}$$ and observe that $\varphi(z)=0$ and $|\varphi'(z)|=1/(2\operatorname{Re} z)$. Applying the Schwarz lemma to $f\circ \varphi^{-1}$ yields $$|f'(z)|\, |\varphi'(z)|^{-1}\le 1-|f(z)|^2$$ and the conclusion follows.

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This is a version of the Schwartz-Pick theorem. See Preposition 1 in the article by M.Qazi and Q. Rahman.

user64494
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