# Is Schwarz's Lemma true for squares?

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 $$0 since $$|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?

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 $$\mathbb{C}$$).

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 $$|f'(a)| \le 1$$.

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.

• 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 $g$? Commented May 25 at 10:21
• @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) Commented May 25 at 10:28