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Let $F:\mathbb{H}\rightarrow \mathbb{D}$ be holomorphic, where $\mathbb{H}$ is the upper half plane and $\mathbb{D}$ is the unit disc. Show that if $F(i)=0$, then $$|F(z)|\leq \left|\frac{z-i}{z+i}\right|$$ for all $z\in\mathbb{H}$.

I have trouble constructing an auxiliary function to apply the Schwarz Lemma or something, any hints?

Thanks

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  • $\begingroup$ You need a conformal $T\colon \mathbb{D}\to \mathbb{H}$ with $T(0) = i$. $\endgroup$ – Daniel Fischer Dec 15 '14 at 2:16
  • $\begingroup$ Look at the given right hand side for inspiration about how to construct the $T$ in Daniel's hint. $\endgroup$ – mrf Dec 15 '14 at 7:59
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Is it correct?:

Consider $G(z)=\dfrac{z-i}{z+1}$ and $g(z)=F\circ G^{-1}(z)$. We have $g(0)=F\circ G^{-1}(0)=F(i)=0$. By Schwarz's Lemma, $|g(z)|\leqslant |z|$ so $|F\circ G^{-1}(z)|\leqslant |z|$ and then $|F(z)|\leqslant |G(z)|=\left|\dfrac{z-i}{z+i}\right|$

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Well, the answer is in your question.

The conformal mapping from the unit disk to the upper half-plane is $G=\frac{z-1}{z+1}$. So, $F\circ G$ will satisfy the hypotheses of the Schwarz Lemma.

If you want to prove that this is your desired function, I suggest the chapter on conformal mappings in Bak/Newman.

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