Use the Little Picard theorem to prove that there is a constant \(M>0\) such that every holomorphic map \[ f:\Delta\to\mathbb C\setminus\{0,1\} \] satisfies \[ \|f'\|\le M, \] where the derivative is measured from the hyperbolic metric on \(\Delta\) to the spherical metric on the range.
Hint: if the derivative were unbounded over such maps, rescale to obtain a nonconstant entire map \(\mathbb C\to\mathbb C\setminus\{0,1\}\).
Solution
Suppose, to the contrary, that no universal bound exists. Rescaling at points of maximal derivative produces a nonconstant entire function omitting both \(0\) and \(1\), contradicting Little Picard.
1. We may choose holomorphic maps\[ f_n:\Delta\to\mathbb C\setminus\{0,1\} \]for which the hyperbolic-to-spherical derivative norm attains a maximum\[ M_n\longrightarrow\infty. \]
Write \[ D_f(z) =\frac{|f'(z)|(1-|z|^2)}{1+|f(z)|^2}. \] This is the derivative norm from the hyperbolic metric \(|dz|/(1-|z|^2)\) on \(\Delta\) to the spherical metric \(|dw|/(1+|w|^2)\).
If no common bound exists, choose maps and points with \(D_f\) arbitrarily large. Replacing such a map \(f\) by \[ z\longmapsto f(rz), \qquad r<1 \] with \(r\) sufficiently close to \(1\) preserves an arbitrarily large value of the derivative norm at the corresponding interior point. The new map is holomorphic on a neighborhood of \(\overline\Delta\), and its factor \(1-|z|^2\) forces \(D_f(z)\to0\) as \(|z|\to1\). Thus \(D_f\) attains its maximum at an interior point. In this way we obtain \(f_n\) with maxima \(M_n\to\infty\).
2. After moving a maximizing point to \(0\) and rescaling, there are maps\[ g_n:\Delta(M_n)\to\mathbb C\setminus\{0,1\} \]such that\[ g_n^\#(0)=1 \]and, for \(|z|<M_n\),\[ g_n^\#(z) \le\frac{1}{1-|z|^2/M_n^2}, \]where\[ g^\#=\frac{|g'|}{1+|g|^2} \]is the Euclidean-to-spherical derivative.
Let \(p_n\) maximize \(D_{f_n}\), and choose an automorphism \[ A_n\in\operatorname{Aut}(\Delta) \] with \(A_n(0)=p_n\). Since the hyperbolic metric is invariant under disk automorphisms, for \[ F_n=f_n\circ A_n \] we have \[ D_{F_n}(0)=M_n, \qquad D_{F_n}(u)\le M_n \quad(u\in\Delta). \] At the origin, the hyperbolic factor is \(1\), so \[ F_n^\#(0)=M_n. \]
Now define \[ g_n(z)=F_n(z/M_n), \qquad |z|<M_n. \] Then \[ g_n^\#(0)=\frac{F_n^\#(0)}{M_n}=1. \] For \(u=z/M_n\), the maximality inequality says \[ F_n^\#(u)(1-|u|^2)\le M_n. \] Hence \[ g_n^\#(z) =\frac{F_n^\#(u)}{M_n} \le\frac{1}{1-|z|^2/M_n^2}. \]
3. A subsequence converges spherically and locally uniformly on \(\mathbb C\) to a nonconstant meromorphic map\[ g:\mathbb C\to\widehat{\mathbb C}. \]
Fix \(R>0\). For all sufficiently large \(n\), \(R<M_n/2\), and 2 gives \[ g_n^\#(z)\le\frac{1}{1-1/4}=\frac43 \qquad(|z|\le R). \] By Marty’s theorem, a family of meromorphic functions with locally uniformly bounded spherical derivative is normal, so \((g_n)\) is normal on every fixed disk. A diagonal subsequence converges spherically and locally uniformly on all of \(\mathbb C\) to a meromorphic map \(g\).
Moreover spherical derivatives converge at the normalized basepoint under this locally uniform convergence, so \[ g^\#(0)=\lim_{n\to\infty}g_n^\#(0)=1. \] Thus \(g\) is nonconstant.
4. The limit \(g\) omits \(0\), \(1\), and \(\infty\).
Every \(g_n\) is holomorphic and takes values in \[ \widehat{\mathbb C}\setminus\{0,1,\infty\}. \] By Hurwitz’s theorem, a value omitted by every member of a spherically locally uniformly convergent sequence of meromorphic functions is also omitted by a nonconstant limit. Therefore \[ g(\mathbb C)\cap\{0,1,\infty\}=\varnothing. \] In particular, omission of \(\infty\) means that \(g\) is an entire function \[ g:\mathbb C\to\mathbb C\setminus\{0,1\}. \] It is nonconstant by 3.
5. This contradicts Little Picard, so a universal \(M\) exists.
The Little Picard theorem says that a nonconstant entire function can omit at most one complex value. The entire function produced in 4 omits both \(0\) and \(1\), a contradiction. Hence the assumed unbounded sequence cannot exist. Therefore there is a constant \(M>0\) such that every holomorphic \[ f:\Delta\to\mathbb C\setminus\{0,1\} \] satisfies \[ \|f'\|\le M. \]