[August 2007, Problem \(\# 4]\)
State the standard Schwarz reflection principle involving reflection across the real axis.
Let \(H=\{z\in\mathbb C:\operatorname{Im}z>0\}\). Give, with justification, a linear fractional transformation \(T\) mapping \(\DD\) to \(H\). Let \(g(z) = \overline{z}\); show that \(T^{-1} \circ g \circ T(z) = \frac{1}{\overline{z}}\).
Suppose \(f\) is holomorphic on \(\DD\), continuous on \(\overline{\DD}\), and real on the unit circle. Prove that \(f\) must be constant.
Solution
1. Part (a), the Schwarz reflection principle across the real axis:let \(U\subseteq\CC\) be a domain symmetric under conjugation, and write\[ U_+=\{z\in U:\operatorname{Im}z>0\}. \]If \(F\) is holomorphic on \(U_+\), continuous on\[ U_+\cup(U\cap\RR), \]and real-valued on \(U\cap\RR\), then\[ \widetilde F(z) = \begin{cases} F(z),&\operatorname{Im}z\geq0,\\ \overline{F(\bar z)},&\operatorname{Im}z<0 \end{cases} \]defines a holomorphic function on all of \(U\).
2. Part (b): the linear fractional transformation\[ \boxed{ T(z)=i\frac{1+z}{1-z} } \]maps \(\DD\) biholomorphically onto the upper half-plane \(H\).
For \(z\in\DD\), \[ \begin{aligned} \operatorname{Im}T(z) &= \operatorname{Re}\frac{1+z}{1-z}\\ &= \frac{1-\abs{z}^2}{\abs{1-z}^2}\\ &> 0. \end{aligned} \] Thus \(T(\DD)\subseteq H\). Solving \[ w=i\frac{1+z}{1-z} \] for \(z\) gives \[ T^{-1}(w)=\frac{w-i}{w+i}, \] which maps \(H\) back into \(\DD\). Hence \(T\) is a biholomorphism from \(\DD\) onto \(H\).
3. With \(g(w)=\bar w\), one has on the Riemann sphere\[ \boxed{ T^{-1}\circ g\circ T(z)=\frac1{\bar z}. } \]
For finite \(z\neq0,1\), \[ \overline{T(z)} = -i\frac{1+\bar z}{1-\bar z}. \] Therefore \[ \begin{aligned} T^{-1}(\overline{T(z)}) &= \frac{-i\frac{1+\bar z}{1-\bar z}-i} {-i\frac{1+\bar z}{1-\bar z}+i}\\ &= \frac{-2i/(1-\bar z)} {-2i\bar z/(1-\bar z)}\\ &= \frac1{\bar z}. \end{aligned} \] Both sides are the same anti-Möbius transformation of the Riemann sphere, so the identity extends to the exceptional points as well, with \(0\) and \(\infty\) interchanged.
4. Part (c): the function \(f\) is constant.
Write \[ f=u+iv \] with real-valued functions \(u\) and \(v\). Since \(f\) is holomorphic on \(\DD\), the function \(v=\operatorname{Im}f\) is harmonic there. Since \(f\) is continuous on \(\overline{\DD}\) and real-valued on the unit circle, \(v\) is continuous on \(\overline{\DD}\) and \[ v=0 \] on \(\partial\DD\).
The maximum principle for harmonic functions applied to \(v\) gives \[ v\leq0 \] on \(\DD\), while applying it to \(-v\) gives \[ v\geq0. \] Hence \(v\equiv0\) on \(\DD\), so \[ f(\DD)\subseteq\RR. \] If \(f\) were nonconstant, the open mapping theorem would make \(f(\DD)\) open in \(\CC\), which is impossible for a subset of \(\RR\). Therefore \(f\) is constant.
5. Q.E.D.
Step 1 answers part (a), steps <1>2–<1>3 answer part (b), and step 4 proves part (c).