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Revision difference : proof of Schwarz lemma |
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| Define $g(z)=f(z)/z$. Then $g:\Delta\to\Complex$ is a holomorphic function. The Schwarz lemma is just an application of the maximal modulus principle to $g$. |
Define $g(z)=f(z)/z$. Then $g:\Delta\to\Complex$ is a holomorphic function. The Schwarz lemma is just an application of the maximal modulus principle to $g$. |
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| For any $1>\epsilon>0$, by the maximal modulus principle $\size{g}$ must attain its maximum on the closed disk $\left\{z:\size{z}\le 1-\epsilon\right\}$ at its boundary $\left\{z:\size{z}=1-\epsilon\right\}$, say at some point $z_{\epsilon}$. But then $\size{g(z)}\le \size{g(z_{\epsilon})} \le \frac{1}{1-\epsilon}$ for any $\size{z}\le 1-\epsilon$. Taking an infinimum as $\epsilon\to0$, we see that values of $g$ are bounded: $\size{g(z)}\le 1$. |
For any $1>\epsilon>0$, by the maximal modulus principle $\size{g}$ must attain its maximum on the closed disk $\left\{z:\size{z}\le 1-\epsilon\right\}$ at its boundary $\left\{z:\size{z}=1-\epsilon\right\}$, say at some point $z_{\epsilon}$. But then $\size{g(z)}\le \size{g(z_{\epsilon})} \le \frac{1}{1-\epsilon}$ for any $\size{z}\le 1-\epsilon$. Taking an infinimum as $\epsilon\to0$, we see that values of $g$ are bounded: $\size{g(z)}\le 1$. |
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| Thus $\size{f(z)}\le\size{z}$. Additionally, $f'(0)=g(0)$, so we see that $\size{f'(0)}=\size{g(0)}\le 1$. This is the first part of the lemma. |
Thus $\size{f(z)}\le\size{z}$. Additionally, $f'(0)=g(0)$, so we see that $\size{f'(0)}=\size{g(0)}\le 1$. This is the first part of the lemma. |
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Now suppose, as per the premise of the second part of the lemma, that $|g(w)|=1$ for some $w\in\Delta$. For any $r>\size{w}$, it must be that $\size{g}$ attains its maximal modulus (1) \emph{inside} the disk $\left\{z:\size{z}\le r\right\}$, and it follows that $g$ must be constant inside the entire open disk $\Delta$. So $g(z)\equiv a$ for $a=g(w)$ of modulus 1, and $f(z)=az$, as required.
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Now suppose, as per the premise of the second part of the lemma, that $|g(w)|=1$ for some $w\in\Delta$. For any $r>\size{w}$, it must be that $\size{g}$ attains its maximal modulus (1) \emph{inside} the disk $\left\{z:\size{z}\le r\right\}$, and it follows that $g$ must be constant inside the entire open disk $\Delta$. So $g(z)\equiv a$ for $a=g(w)$ of size 1, and $f(z)=az$, as required.
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