Prove that every proper holomorphic map \(f:\mathbb H\to\mathbb H\) can be written in the form \[ f(z)=a_0z+b_0+\sum_{i=1}^n\frac{a_i}{b_i-z}, \] with \(a_i\ge0\) and \(b_i\in\mathbb R\).
Solution
1. The map \(f\) extends to a rational map on the Riemann sphere and takesthe extended real line \(\widehat{\mathbb R}\) to itself.
Let \[ \phi(z)=\frac{z-i}{z+i} \] be a Cayley map from \(\mathbb H\) to \(\Delta\), and let \(\psi=\phi^{-1}\). Then \[ F=\phi\circ f\circ\psi:\Delta\to\Delta \] is proper. Every proper holomorphic self-map of \(\Delta\) is a finite Blaschke product, so \(F\) is one, and it extends rationally to \(\widehat{\mathbb C}\) and maps \(S^1\) to itself. Conjugating this extension by the Möbius maps \(\phi\) and \(\psi\) gives a rational extension of \(f\), and since \(\psi(S^1)=\widehat{\mathbb R}\), this extension maps \(\widehat{\mathbb R}\) to itself.
2. The rational extension satisfies\[ f(\overline z)=\overline{f(z)}. \]Consequently its finite poles off the real axis would occur in conjugatepairs.
For every real \(x\) at which \(f\) is finite, 1 gives \(f(x)\in\mathbb R\). Thus the two rational functions \[ f(z) \qquad\text{and}\qquad \overline{f(\overline z)} \] agree at infinitely many points, hence agree identically. Poles are therefore preserved by complex conjugation.
3. Every finite pole of \(f\) lies on \(\mathbb R\).
There are no poles in \(\mathbb H\), because \(f\) is holomorphic there. If a pole lay in the lower half-plane, then 2 would give its conjugate as a pole in \(\mathbb H\), a contradiction. Hence every finite pole is real.
4. Every finite pole \(b\) is simple, and its residue is negative.
By 2 the Laurent coefficients of \(f\) at the real point \(b\) are real. If \(b\) were a pole of order \(m\), write its leading term as \[ c(z-b)^{-m}, \qquad c\in\mathbb R\setminus\{0\}. \] For \(z=b+re^{i\theta}\) with \(0<\theta<\pi\) and \(r\) small, the leading contribution to the imaginary part is \[ -c\,r^{-m}\sin(m\theta). \] If \(m\ge2\), then \(\sin(m\theta)\) changes sign as \(\theta\) ranges over \((0,\pi)\), contradicting \(\operatorname{Im}f(z)>0\) throughout \(\mathbb H\). Therefore \(m=1\).
If \[ f(z)=\frac{c}{z-b}+O(1), \] then along \(z=b+iy\), \[ \operatorname{Im}\frac{c}{iy}=-\frac{c}{y}. \] Positivity for small \(y>0\) forces \(c<0\). Writing \(a=-c>0\), the polar part is \[ \frac{a}{b-z}. \]
5. Subtracting all finite polar parts gives a real polynomial:\[ P(z) =f(z)-\sum_{i=1}^n\frac{a_i}{b_i-z}, \qquad a_i>0,\quad b_i\in\mathbb R. \]
By <1>3–<1>4 these terms remove every finite pole of the rational function \(f\). The remainder is therefore a rational function with no finite poles, hence a polynomial. The conjugation symmetry in 2, shared by each polar term, implies \[ P(\overline z)=\overline{P(z)}, \] so \(P\) has real coefficients.
6. The polynomial \(P\) has degree at most one.
Suppose \[ P(z)=cz^m+O(z^{m-1}), \qquad c\in\mathbb R\setminus\{0\}, \] with \(m\ge2\). Along a ray \(z=Re^{i\theta}\) in the upper half-plane, the polar terms in 5 are \(O(R^{-1})\), while \[ \operatorname{Im}P(Re^{i\theta}) =cR^m\sin(m\theta)+O(R^{m-1}). \] As \(\theta\) ranges through \((0,\pi)\), \(\sin(m\theta)\) takes both signs. For a suitable fixed \(\theta\) and large \(R\), this would make \(\operatorname{Im}f(Re^{i\theta})<0\), impossible. Hence \(m\le1\).
7. Thus\[ P(z)=a_0z+b_0 \]with \(a_0\ge0\) and \(b_0\in\mathbb R\).
5. Along the positive imaginaryaxis,\[ f(iy)=a_0iy+b_0+O(y^{-1}) \qquad(y\to\infty). \]Since \(\operatorname{Im}f(iy)>0\), division by \(y\) and passage to the limitgive \(a_0\ge0\).
8. Combining the preceding steps gives\[ \boxed{ f(z)=a_0z+b_0+\sum_{i=1}^n\frac{a_i}{b_i-z}} \]with \(a_i\ge0\) and \(b_i\in\mathbb R\).
7. For finitepoles, 4 actually gives \(a_i>0\); allowing \(a_i=0\) does not change thestated form.