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Let $D$ denote the (open) unit disc. Suppose $f:D\to D$ is holomorphic and that for some distinct $a,b$ we have $f(a)=f(b)=0$. Show that for any $z\in D$ $$\vert f(z)\vert \le \Bigg\vert \frac{z-a}{1-\bar az} \Bigg\vert \cdot\Bigg\vert \frac{z-b}{1-\bar bz}\Bigg\vert \tag{$*$}$$

Attempt: For any $c$ in the unit disc, define $\phi_c(z)=\frac{z-c}{1-\bar cz}$ (it can be shown that $\phi_c$ maps $D$ to $D$ and is holomorphic). Let $n(z)=-z$. My first idea was to apply Schwarz' Lemma to the function $g(z)=f\circ n\circ \phi_{-a}\circ f\circ n\circ\phi _{-b}$, which is a map from $D$ to $D$ and sends $0$ to $f\circ n\circ \phi_{-a}\circ f\circ n(-b)=f\circ n\circ \phi_{-a}\circ f(b)=f\circ n\circ \phi _{-a}(0)=f\circ n(-a)=f(a)=0$. Schwarz' implies $\vert g(z)\vert \le \vert z\vert$, but this doesn't allow me to deduce $(*)$ since (a) there's no way to isolate $f$, and (b) even if $f$ could be isolated, the input in $f$ would not be $z$.

I'm not sure if this attempt has any hope. May I have some suggestions?

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Hint: Let $g(z) = \dfrac{f(z) (1-\overline{a}z)(1-\overline{b}z)}{(z-a)(z-b)}$. After removing the removable singularities, use the Maximum Modulus Principle.

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  • $\begingroup$ After removing the singularities, I extend $g$ to $a,b$ by defining $g(a)=\frac{f'(a)(1-\vert a \vert ^2)(1-\bar b a)}{a-b}$ and $g(b)=\frac {f'(b)(1-\vert b \vert ^2)(1-\bar a b)}{b-a}$. Admittedly, I am not exactly sure how Max Mod applies here. All I can conclude from it is that $f$ and $g$ do not attain maximum values on $D$ unless they are constant. $\endgroup$ Commented Feb 15, 2015 at 9:49
  • $\begingroup$ What happens as $|z| \to 1$? $\endgroup$ Commented Feb 15, 2015 at 20:10
  • $\begingroup$ As $\vert z \vert \to 1$, $f$ or $g$ may approach their supremum on which lies on $\bar D$. Even if $g$ approached its supremum, $(*)$ holds for all $z\in D$, not just for $z$ near the boundary. $\endgroup$ Commented Feb 15, 2015 at 23:31
  • $\begingroup$ And the supremum is at most ... $\endgroup$ Commented Feb 16, 2015 at 2:11
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Following Robert's hint, you can extend your claim to a generalized version of Schwarz's lemma: if $d_1,\ldots,d_n$ are pairwise distinct zeros of a holomorphic function $f:D\to\Bbb C$, then $$|f(z)|\leqslant \left| \frac{z-d_1}{1-\bar{d_1}z}\right|\cdots \left| \frac{z-d_n}{1-\bar{d_n}z}\right||f|_{D}$$

This gives what is known as Jensen's inequality: $|f(0)|\leqslant |d_1\cdots d_n||f|_D$, which is a special case of Jensen's formula.

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  • $\begingroup$ What does $\vert f \vert _D$ represent? $\endgroup$ Commented Feb 15, 2015 at 9:54
  • $\begingroup$ The supremum of f on the disk. $\endgroup$
    – Pedro
    Commented Feb 15, 2015 at 10:13

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