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I'm not sure how I could use Schwarz's Lemma to solve the following problem from an old complex analysis prelim: Let $\mathbb{D}$ be the unit disk and suppose we have $f: \mathbb{D} \rightarrow \mathbb{D}$ an analytic function. If $f(0)=a \neq 0$, show that $f$ has no zero in the disk $\{z: |z|< |a|\}$.

Now if I recall clearly, the two main hypotheses of Schwarz's Lemma are that $|f(z)| \leq 1$ for $z \text{ in } \mathbb{D}$ and $f(0)=0$. The conclusion is that $|f(z)| \leq |z| \text{ } \forall z \in \mathbb{D}$ and $|f'(0)| \leq 1$ and if there is equality, then $f(z)= cz$ for some constant $c$ with $|c|=1$. I know how to prove Schwarz's Lemma, but I'm still having trouble tweaking it to fit the conditions of the given problem. Any suggestions could be helpful.

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The method that most often succeeds in applications of the Schwarz lemma is composition of the function in question with suitably chosen automorphisms of the unit disk, and then applying the Schwarz lemma to $T\circ f \circ S$, where $S$ and $T$ are the chosen automorphisms.

Here, we have $f(0)$ given as a starting point, so we can choose $S = \operatorname{id}$, and for $T$ we take an automorphism mapping $a = f(0)$ to $0$, a standard choice is

$$T(z) = \frac{z-a}{1-\overline{a}z}.$$

So if you look at $g = T\circ f$, then $g$ is a function satisfying the hypotheses of Schwarz' lemma, and you want to see that $g(z) \neq T(0)$ for $\lvert z\rvert < \lvert a\rvert$.

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We can use the Schwarz-Pick Theorem which states that for a holomorphic map $f : \mathbb{D} \to \mathbb{D}$ we have

$$\left|\frac{f(z_1) - f(z_2)}{1-\overline{f(z_2)}f(z_1)}\right| \leq \left|\frac{z_1 - z_2}{1-\overline{z_2}z_1}\right|.$$

Now suppose $f(z_1) = 0$ and set $z_2 = 0$.

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  • $\begingroup$ The Schwarz-Pick Theorem is proved by applying Schwarz's Lemma to a conjugation of $f$ by automorphisms of the disc as in Daniel Fischer's answer. $\endgroup$ – Michael Albanese Aug 17 '14 at 19:48

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