State the standard Schwarz reflection principle involving reflection across the real axis.
Give a linear fractional transformation \(T\) mapping \(\DD\) to \(\HH\). Let \(g(z) = \bar z\), and show \[ (T^{-1} \circ g \circ T)(z) = 1/\bar z .\]
Suppose that \(f\) is holomorphic on \(\DD\), continuous on \(\bar \DD\), and real on \(S^1\). Show that \(f\) must be constant.
Solution
Part 1: Let \(\Omega = \Omega^+ \union I \union \Omega^-\) be a region symmetric about \(\RR\). If \(f\) is holomorphic on \(\Omega^+\) extending continuously to \(I\) and real valued on \(I\), then \(f\) extends to a holomorphic function \(F\) on all of \(\Omega\) defined on \(\Omega^-\) by \(F(z) = \bar{f(\bar{z})}\).
Part 2: The map is \(T(z) = -i\qty{z+1\over z-1}\) with \(\inverseof{T}(z) = {z-i\over z+i}\), so \[ (\inverseof{T} \circ g \circ T)(z) &= \inverseof{T}\bar{\qty{-i {z+1\over z-1} }} \\ &= \inverseof{T}\qty{i{\bar z + 1 \over \bar z - 1}} \\ &= {i\qty{\bar z + 1 \over \bar z - 1} - i \over i\qty{\bar z + 1 \over \bar z - 1} + i } \\ &= {(\bar z + 1) - (\bar z - 1) \over (\bar z + 1) + (\bar z - 1)} \\ &= {1\over \bar{z}} .\]
Part 3: Define \(h: \HH\to \bar{\HH}\) by \(h(z) = (T\circ f\circ \inverseof{T})(z)\). Under \(\inverseof{T}: \DD\to \HH\), we have \(T(S^1) = \RR\), so \(h\) is a holomorphic function on \(\HH\) that is continuous and real-valued on \(\RR\). By the reflection principle, defining \(H(z) \da \bar{h(\bar z)}\) for \(\Im(z) < 0\) yields an entire function \(H: \CC\to \CC\) Noting that for \(g(z) \da \bar{z}\), \(g=\inverseof{g}\), we can write \[ H \da \inverseof{g} \circ h \circ = \inverseof{h} \circ (\inverseof{T} \circ f \circ T)\circ g .\] We can then conjugate \(H\) by \(T\) to get a direct formula in terms of \(f\), and unwinding this yields the extension \(F:\CC\to \CC\) defined by \[ F(z) = \begin{cases} f(z) & z\in \DD \\ f_-(z) \da {1\over \bar{f\bar{z}}} & z\in \DD^c \\ f(z) = f_i(z) & z\in S^1 \end{cases} .\] In particular, \(H\) is an entire bounded function and thus constant, making \(F\) constant as well and consequently \(f\) is constant.