Let \(\Delta=\{z\in\mathbb C:|z|<1\}\).
Let \(f : \mathbb{C} \to \mathbb{C}\) be meromorphic with a pole at infinity. Show that \(f\) must be a rational function.
Use the above to prove the following: if \(f : \Delta \to \mathbb{C}\) is holomorphic with a continuous extension to the boundary of \(\Delta\) such that \(|f(z)| = 1\) for all \(|z| = 1\), then \(f(z)\) is the restriction of a rational function.
Solution
1. A meromorphic function on the sphere is rational; in particular, part (a) holds.
Suppose \(f\) is meromorphic on \(\mathbb C\) and has either a pole or a removable singularity at infinity. Then it has no finite pole for \(|z|>R\), for some \(R>0\). Its finite poles form a locally finite set: the local meromorphic factorization at any point gives a neighborhood with at most one pole. Compactness of \(|z|\leq R\) therefore makes the list of poles finite, say \(a_1,\ldots,a_m\).
Let the principal part at \(a_j\) be \[ S_j(z)=\sum_{\ell=1}^{d_j}\frac{c_{j,\ell}}{(z-a_j)^\ell}. \] The Laurent expansion at infinity has the form \[ f(z)=P_\infty(z)+b_0+O(1/z),\qquad P_\infty(z)=\sum_{k=1}^{N}b_k z^k, \] where \(N=0\) and \(P_\infty=0\) are permitted in the removable case [@SS03]. Subtracting all these principal parts gives \[ G=f-P_\infty-\sum_{j=1}^m S_j. \] Every finite singularity of \(G\) is removable. After their removal \(G\) is entire, and \(G(z)\to b_0\) at infinity because each \(S_j(z)\to0\). It is bounded outside a disk by that limit and on the remaining compact disk by continuity. Liouville’s theorem gives \(G\equiv b_0\) [@SS03]. Thus \[ f(z)=b_0+P_\infty(z)+\sum_{j=1}^{m}\sum_{\ell=1}^{d_j} \frac{c_{j,\ell}}{(z-a_j)^\ell}, \] which is rational. This proves (a), and also the removable case at infinity that will be needed in (b).
2. The boundary condition in part (b) produces a meromorphic function on the sphere.
Let \(f\) be as in part (b). Define for \(|z|>1\) \[ F(z)=\frac{1}{\overline{f(1/\overline z)}}. \] The function \[ f^*(w)=\overline{f(\overline w)} \] is holomorphic on the unit disk: conjugating the Taylor coefficients of \(f\) gives its Taylor series. Thus on \(|z|>1\), the expression \(F(z)=1/f^*(1/z)\) is meromorphic. The boundary modulus excludes \(f\equiv0\).
Uniform continuity of \(f\) on \(\overline\Delta\) gives \(0<\eta<1/2\) such that \(|f(z)|>1/2\) whenever \(1-\eta<|z|\leq1\): compare \(f(z)\) to the unit-modulus value \(f(z/|z|)\). The reflected expression consequently has no pole for \(1<|z|<1/(1-\eta)\).
On \(|z|=1\), one has \(1/\overline z=z\), and the boundary hypothesis gives \[ \frac1{\overline{f(z)}}=f(z). \] Hence the inside definition \(F=f\) and the outside reciprocal-reflection have the same continuous boundary values on the unit circle.
Fix \(p\) on the unit circle and use the fractional coordinate \[ w=i\frac{1-z/p}{1+z/p},\qquad z=p\frac{i-w}{i+w}. \] These inverse holomorphic maps take \(p\) to zero and the unit circle locally to the real axis. Indeed, \[ 1-\left|\frac{i-w}{i+w}\right|^2 =\frac{4\operatorname{Im}w}{|i+w|^2}, \] so the upper and lower sides correspond to the inside and outside of the circle. Choose \(\varepsilon>0\) so the inverse maps \(|w|<\varepsilon\) into the pole-free collar just established. The pullback \(K(w)=F(p(i-w)/(i+w))\) is continuous there and holomorphic off the real axis.
For any closed triangle contained in \(|w|<\varepsilon\), cut it along the real axis into at most two polygons. Translate its upper piece by \(i\delta\) and its lower piece by \(-i\delta\). For sufficiently small \(\delta>0\) both translated polygons remain in the disk and lie in the respective open half-planes, so their boundary integrals of \(K\) are zero by Cauchy’s theorem [@SS03]. For either piece \(P\), \[ \left|\int_{\partial P}(K(w\pm i\delta)-K(w))\,dw\right| \leq\operatorname{length}(\partial P) \sup_{w\in\partial P}|K(w\pm i\delta)-K(w)|\longrightarrow0. \] The limit follows from uniform continuity on a compact neighborhood of the triangle. Degenerate pieces have zero integral by cancellation. Passing to the limit and adding the two integrals cancels the shared real segment, so the original triangle integral is zero. Morera’s theorem proves that \(K\), and hence \(F\) near \(p\), is holomorphic [@SS03]. As \(p\) was arbitrary, \(F\) is meromorphic on the whole plane.
At infinity, write \(f(w)=w^k h(w)\) near zero, where \(k\ge0\) and \(h(0)\ne0\). Then for large \(z\), \[ F(z)=\frac{z^k}{h^*(1/z)}, \] so infinity is removable when \(k=0\) and is a pole when \(k>0\). Hence \(F\) is meromorphic on the Riemann sphere.
3. The reflected extension is rational, proving part (b).
By step 1, the meromorphic sphere function \(F\) from step 2 is rational. On the unit disk its definition is exactly the original \(f\). Therefore \(f\) is the restriction to \(\Delta\) of a rational function, as required.