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This is an exercise about the Schwarz lemma, to which the internet gave me no hint so far.

Let $\mathbb{D}=\{z : |z| < 1\}$, and let $f,g:\mathbb{D}\to\mathbb{D}$ be analytic, $1-1$ functions satisfying $$f(0)=g(0), \ f'(0)=g'(0).$$ Prove that $f \equiv g$ on $\mathbb{D}$.

My INCORRECT try: the Schwarz lemma applied to $f-g$, $f'-g'$ yields

$|f(z)-g(z)|\leq|z|, |f'(0)-g'(0)|\leq 1$,

$|f'(z)-g'(z)|\leq|z|, |f''(0)-g''(0)|\leq 1$.

But I don't know how to continue.

Why incorrect? Because I cannot apply the Schwarz lemma to $f-g$, since not necessarily $\forall z : |f(z)-g(z)| \leq 1$, and similarly not necessarily $\forall z : |f'(z)-g'(z)| \leq 1$.

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  • $\begingroup$ How do you know $f-g$ maps $D$ to itself? Merely the assumption that $f,g$ both map $D$ to itself doesn't give that, because you could have $f(z)=z,g(z)=-z$. Obviously that violates the second equality assumption, but that just means that somehow you need to be using the other assumptions from the start. $\endgroup$ – Ian Dec 9 '17 at 15:40
  • $\begingroup$ @Ian: you are right, I missed that not necessarily $|f(z)-g(z)| \leq 1$. $\endgroup$ – co.sine Dec 9 '17 at 15:47
  • $\begingroup$ @MyGlasses That's pretty standard notation: $f(z)=g(z)$ for all $z \in D$. $\endgroup$ – Ian Dec 9 '17 at 15:49
  • $\begingroup$ Anyway, you're going to want to use the Schwarz lemma but the only function you know vanishes at zero is $f-g$. The obvious way to use the equality assumptions to build an apparent counterexample would be to look at something like $f(z)-g(z)=kz^2$ for some $k$. (All I did was imagine the series expansion at $z=0$ and notice that the first two terms had to vanish.) That's breaking the 1-1ness, though. This example should already be telling you that actually 1-1 analytic functions from D to itself are quite restricted. So much so that they have a special term, "univalent". $\endgroup$ – Ian Dec 9 '17 at 15:52
  • $\begingroup$ @Ian You reminded me we had a theorem that all analytic functions from $D$ to $D$ which are invertible, are of the form $e^{i \theta} \frac{z-\alpha}{1-\bar{\alpha}z}$, i.e. univalent Möbius transformations. But in the exercise they are not said to be invertible, but only $1-$to$-1$. $\endgroup$ – co.sine Dec 9 '17 at 16:00
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This is false as stated. Start with $$F(z)=z,\quad G(z)=z+z^2.$$

Since $G'(0)\ne0$ there exists $\delta>0$ such that $G$ is injective in the disk $|z|<\delta$. Choose $R$ so that $|F(z)|<R$ and $|G(z)|<R$ for $|z|<\delta$ and set $$f(z)=F(\delta z)/R,\quad g(z)=G(\delta z)/R.$$

It's true if you assume in addition that $g$ is surjective; in that case you can apply the Schwarz Lemma to the function $f\circ g^{-1}$. (In fact it's enough to assume that $f(D)\subset g(D)$; then you can apply Schwarz to $g^{-1}\circ f$. In fact in that case you don't even need to assume that $g$ is injective; if you assume just $g'\ne0$ then analytic continuation gives the moral equivalent of $g^{-1}\circ f$, namely a function $h:D\to D$ with $h(0)=0$ and $g\circ h=f$.)

Edit The fact that $G'(0)\ne0$ implies that $G$ is injective in some disk $|z|<\delta$ is a well known fact from elementary complex analysis. We can give an ad hoc proof here without needing any complex analysis, showing that in fact we can take $\delta=1/2$. Suppose that $|z|<1/2$, $|w|<1/2$, $z\ne w$ and $G(z)=G(w)$. Then $$z-w=w^2-z^2=(w-z)(w+z).$$Since $z\ne w$ this implies that $w+z=-1$, and now the triangle inequality implies that $1\le|z|+|w|<1$.

Knowing that $\delta=1/2$ works allows us to give an explicit counterexample: $$f(z)=z/2,\quad g(z)=z/2+z^2/4.$$(There's a simple algebraic proof above that $g$ is injective in the unit disk. This fact will be obvious to readers who know a little complex analysis, if they note that $\Re g'(z)>0$ for $|z|<1$.)

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  • $\begingroup$ @MyGlasses Read the sentence immediately following the definition of $G$. $\endgroup$ – David C. Ullrich Dec 9 '17 at 16:29
  • $\begingroup$ @MyGlasses Nope. What part of the answer don't you understand? What seems most likely is you're omitting a hypothesis of Rogosinski's theorem - look it up and tell us exactly what it says. $\endgroup$ – David C. Ullrich Dec 9 '17 at 16:41
  • $\begingroup$ @MyGlasses $g$ is injective in the unit disk because $G$ is injective in the disk $|z|<\delta$. $\endgroup$ – David C. Ullrich Dec 9 '17 at 16:42
  • $\begingroup$ @MyGlasses See edit - the one slightly non-trivial part can be trivialized. $\endgroup$ – David C. Ullrich Dec 9 '17 at 17:05
  • $\begingroup$ @MyGlasses In fact we can just let $f(z)=z/2$, $g(z)=z/2+z^2/4$. Tell me what part of that you don't understand. $\endgroup$ – David C. Ullrich Dec 9 '17 at 17:21

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