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In a recent complex analysis exam, we were asked which step(s) in the proof of the Riemann Mapping Theorem fail, when you replace every instance of the open unit ball $\mathbb{E}$ with the square $$\mathbb{S} = \{z\in\mathbb{C}:|Re(z)|, |Im(z)| < 1 \}$$

One error that arose is that the square root function is not a self map of $\mathbb{S}$. But what I thought was a problem was that we used Schwarz's Lemma, in particular

Given a holomorphic $f:\mathbb{E}\to \mathbb{E}$ that fixes $0$, we have $|f'(0)|\leq 1$

The proof for this goes by noting that we can find a holomorphic $g$ such that

$$f(z)= zg(z)$$ and then for $0<r<1$ since $|g|$ achieves it's maximum on $\overline{B(0,r)}$ at a point $z_r\in\partial B(0,1)$, we have

$$|g(0)|\leq \frac{|f(z_r)|}{|z_r|} \leq \frac{1}{r}$$

where this last inequality follows since the codomain of $f$ is $\mathbb{E}$, and $z$ in $\partial B(0,r)$ implies $|z| = r$. In the case of $\mathbb{S}$ however, it may happen that $|z_r| = r$, but our only obvious bound for the numerator is

$$|f_r(z)|< \sqrt{2}$$

So we get only that

$$|f'(0)| < \sqrt{2}$$

But of course, just because this proof doesn't work, doesn't mean the result is not true, but it is suspicious. Hence

Question: Prove or disprove Schwarz's Lemma for $\mathbb{S}$

If it is true, can the same be said for all simply connected open sets?

If it is false, is the disc the only geometry for which this result holds?

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3 Answers 3

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Yes, the result is true.

As the square is simply connected (and not all of $\mathbb{C}$), by Riemann's Theorem we know there exists some biholomorphic function $g:\mathbb{S}\to \mathbb{D}$ (that we may choose such that $g(0)=0$).

Take any function $f:\mathbb{S}\to\mathbb{S}$ such that $f(0)=0$.

Then, $h=g\circ f \circ g^{-1}$ verifies that $h(0)=0$ and $h(\mathbb{D})\subseteq \mathbb{D}$.

By the usual Schwarz's Lemma, $|h'(0)|\leq 1$.

But by the Chain Rule, $$h'(0)=g'(f(g^{-1}(0))\cdot f'(g^{-1}(0))\cdot [g^{-1}]'(0)=g'(0)\cdot f'(0)\cdot \dfrac{1}{g'(0)}=f'(0)$$

It follows that $|f'(0)|\leq 1$, as desired.

This can, of course, be generalized to any simply connected domain that contains $0$ (and is not all of $\mathbb{C}$).

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If $D \subsetneq \Bbb C$ is any simply connected domain and $f: D \to D$ a holomorphic function which fixes a point $a \in D$ then the Schwarz lemma can be applied to $$ g = \phi \circ f \circ \phi^{-1} $$ where $\phi$ is a conformal mapping from $D$ onto the unit disk with $\phi(a) = 0$.

It follows that $|g'(0)| \le 1$, which by the chain rule implies that $|f'(a)| \le 1$.

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One could say: if $D$ is equivalent ( biholomorphic) to a bounded domain, and $f\colon D\to D$, $f(z_0) = z_0$, then $|f'(z_0)\le 1$. We may reduce to $D$ is bounded, and (say) $z_0 =0$. Consider $f_n=f^n= f\circ \cdots \circ f$. We have $g(D) \subset D$, and $f_n(0) = 0$. Now, since $f_n ( D(0, r) ) \subset D(0, R)$, we have $f'_n(0)|\le \frac{R}{r}$. But notice that $f'_n(0) = (f'(0))^n$. We conclude that the sequence $(f'(0)^n)$ is bounded, so $|f'(0)|\le 1$.

Note that the proof works similarly for bounded domains of $\mathbb{C}^n$. We get that $Df(z_0)$ ( the Jacobian matrix) has all eigenvalues $|\cdot | \le 1$, so the determinant of the jacobian too. This is a part of a theorem of Cartan that also states ( like the Schwarz lemma for the disk) that if moreover $|\det Df(z_0)| =1$ then $f$ is an automorphish of $D$. A proof can be found in the book of Bochner-Martin.

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  • $\begingroup$ Thank you for the answer, I can see that the argument is very elegant, but I am having trouble following it, if possible, could you fill in some of the details, and quantifiers? For example, I am not familiar with the notation $D(0,r)$ (but can make an educated guess) and what is $r,R$ and $g$? $\endgroup$
    – Carlyle
    Commented May 25 at 10:21
  • $\begingroup$ @Carlyle: Hi, $g$ being defined on the domain $D$ and taking values in $D$. Now, I denote by $D(0, r)$ the disk of radius $r$ centered at $0$ ( somehow confusing). In any case, if $g$ is defined on at least a small disk ( contained in $D$) and being bounded on it ( since $D$ is bounded), then from the Cauchy estimates we get ineqs for the derivative. So that's why all of the compositions $f^n$ have the derivative at $0$ uniformly bounded. But ( chain rule) that means $(f'(0))^n$ a bounded sequence... Note that I could not prove such a thing for say smooth fns, ( no Cauchy estimates) $\endgroup$
    – orangeskid
    Commented May 25 at 10:28

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