Let \(f:\Delta\to\Delta\) be a proper analytic map, and let \(F:\overline\Delta\to\overline\Delta\) be its continuous extension. Suppose \(f(0)=0\). Show that for every continuous function \(\phi:S^1\to\mathbb R\), the functions \(\phi\) and \(\phi\circ F\) have the same average over \(S^1\).
Hint: use the harmonic extension of \(\phi\).
Solution
Let \(u:\overline\Delta\to\mathbb R\) be the harmonic extension of \(\phi\).
1. The function \(u\circ f\) is harmonic on \(\Delta\) and extendscontinuously to \(\overline\Delta\) with boundary values \(\phi\circ F\).
Since \(u\) is harmonic and \(f\) is holomorphic, \[ \Delta(u\circ f)=|f'|^2(\Delta u)\circ f=0, \] so \(u\circ f\) is harmonic.
A proper disk map is a finite Blaschke product, hence its continuous extension satisfies \(F(S^1)\subset S^1\). Since \(u|_{S^1}=\phi\), \[ (u\circ F)|_{S^1}=\phi\circ F. \] Continuity of \(u\) and \(F\) on the closed disk gives the claimed continuous extension.
2. The average of \(\phi\) over \(S^1\) equals \(u(0)\).
By the mean-value property for harmonic functions, \[ u(0)=\frac1{2\pi}\int_0^{2\pi}\phi(e^{it})\,dt. \]
3. The average of \(\phi\circ F\) over \(S^1\) equals \((u\circ f)(0)\).
Apply the mean-value property to the harmonic function \(u\circ f\) from 1: \[ (u\circ f)(0) =\frac1{2\pi}\int_0^{2\pi}\phi(F(e^{it}))\,dt. \]
4. The two averages are equal.
Since \(f(0)=0\), \[ (u\circ f)(0)=u(f(0))=u(0). \] Combining 2 and 3 therefore gives \[ \frac1{2\pi}\int_0^{2\pi}\phi(F(e^{it}))\,dt =\frac1{2\pi}\int_0^{2\pi}\phi(e^{it})\,dt. \]