Schwarz Reflection

Part 1: This is not true: take the holomorphic function \(f(z) = z\), then \(u(z) \coloneqq\Re(f(z)) = \Re(z)\) is harmonic on nonzero on \({\mathbb{R}}\) but zero on \(i{\mathbb{R}}\).

Part 2: Set \(f \coloneqq u_x + i u_y\), then \(f\) is holomorphic on \(O\). Since \(h\equiv 0\) on \({\mathbb{D}}_{\varepsilon}\subseteq O\), \(g\equiv 0\) on this disc. By the identity principle for holomorphic functions, \(g\equiv 0\) on \(O\). So \(h_x, h_y \equiv 0\), making \(h\) constant, and since \(h\equiv 0\) on \(U\) this forces \(h\equiv 0\) on \(O\).

Part 3: Let \(u\) be harmonic on \(S^+\), a region symmetric about \({\mathbb{R}}\), and that \(u\equiv 0\) on \({\mathbb{R}}\cap S^+\). Define \(S^- = \left\{{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu {~\mathrel{\Big\vert}~}z\in S^+}\right\}\), and \begin{align*} U(z) \coloneqq \begin{cases} U(z) & z\in S^+ \\ -U(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) & z\in S^-. \end{cases} .\end{align*} Then \(U\) is a harmonic extension of \(u\) to \(S \coloneqq S^+ \cup(S^+ \cap{\mathbb{R}}) \cup S^-\). To see that \(U\) is harmonic on \(S\), it suffices to check that \(U\) satisfies the mean value property on \(S\). Clearly this holds in \(S^+\), so for \(z_0\in S^+\) we have \begin{align*} U(z_0) &= u(z_0) \\ &= {1\over 2\pi} \int_{-\pi}^\pi u(z_0 + re^{it} )\,dt\\ &= {1\over 2\pi} \int_{-\pi}^\pi U(z_0 + re^{it} )\,dt\\ .\end{align*} So for \(w_0\in S^-\), write it as \(w_0 = \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu\), then \begin{align*} U(z_0) &\coloneqq-u(\mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu) \\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\mkern 1.5mu\overline{\mkern-1.5muz_0 + re^{it} \mkern-1.5mu}\mkern 1.5mu} \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu + re^{-it} } \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\mkern 1.5mu\overline{\mkern-1.5muz_0 + re^{-it} \mkern-1.5mu}\mkern 1.5mu} \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi U(z_0 + re^{it} ) \,dt .\end{align*}

Part 1: Let \(\Omega = \Omega^+ \cup I \cup\Omega^-\) be a region symmetric about \({\mathbb{R}}\). 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 \(F(z) = \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\).

Part 2: The map is \(T(z) = -i\qty{z+1\over z-1}\) with \(T^{-1}(z) = {z-i\over z+i}\), so \begin{align*} (T^{-1}\circ g \circ T)(z) &= T^{-1}\mkern 1.5mu\overline{\mkern-1.5mu\qty{-i {z+1\over z-1} }\mkern-1.5mu}\mkern 1.5mu \\ &= T^{-1}\qty{i{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1 \over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1}} \\ &= {i\qty{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1 \over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1} - i \over i\qty{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1 \over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1} + i } \\ &= {(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1) - (\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1) \over (\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1) + (\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1)} \\ &= {1\over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} .\end{align*}

Part 3: Define \(h: {\mathbb{H}}\to \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{H}}\mkern-1.5mu}\mkern 1.5mu\) by \(h(z) = (T\circ f\circ T^{-1})(z)\). Under \(T^{-1}: {\mathbb{D}}\to {\mathbb{H}}\), we have \(T(S^1) = {\mathbb{R}}\), so \(h\) is a holomorphic function on \({\mathbb{H}}\) that is continuous and real-valued on \({\mathbb{R}}\). By the reflection principle, defining \(H(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muh(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\) for \(\Im(z) < 0\) yields an entire function \(H: {\mathbb{C}}\to {\mathbb{C}}\) Noting that for \(g(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\), \(g=g^{-1}\), we can write \begin{align*} H \coloneqq g^{-1}\circ h \circ = h^{-1}\circ (T^{-1}\circ f \circ T)\circ g .\end{align*} We can then conjugate \(H\) by \(T\) to get a direct formula in terms of \(f\), and unwinding this yields the extension \(F:{\mathbb{C}}\to {\mathbb{C}}\) defined by \begin{align*} F(z) = \begin{cases} f(z) & z\in {\mathbb{D}} \\ f_-(z) \coloneqq{1\over \mkern 1.5mu\overline{\mkern-1.5muf\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\mkern-1.5mu}\mkern 1.5mu} & z\in {\mathbb{D}}^c \\ f(z) = f_i(z) & z\in S^1 \end{cases} .\end{align*} In particular, \(H\) is an entire bounded function and thus constant, making \(F\) constant as well and consequently \(f\) is constant.

First, note that the Schwarz reflection principle can be applied here: let \(T: {\mathbb{D}}\to {\mathbb{H}}\) be the Cayley map, and consider \(\tilde f \coloneqq T\circ f \circ T^{-1}: {\mathbb{H}}\to {\mathbb{H}}\). Now \(T(S^1) = {\mathbb{R}}\), and since \(f(z)\in S^1\) when \(z\in S^1\), we have \(\tilde f({\mathbb{R}}) = {\mathbb{R}}\), i.e. this is a real-valued function on \({\mathbb{R}}\). So \(\tilde f\) extends holomorphically to \(\tilde F:{\mathbb{C}}\to CC\), and we can pull this back to a holomorphic extension of \(f\).

Extend \(f\) to \(F:{\mathbb{C}}\to {\mathbb{C}}\) by \(f(z) = 1/\mkern 1.5mu\overline{\mkern-1.5muf(1/\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\) for \(z\in {\mathbb{D}}^c\), which generally has poles at the points \(1/\mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5mu\) for \(z_k\in {\mathbb{D}}\) zeros of \(f\). Since \(f\) is nonvanishing, \(F\) has no poles and thus defines an entire function. By definition of \(F\), we have \(F({\mathbb{C}}) \subseteq f\qty{\left\{{{\left\lvert {z} \right\rvert} \leq 1}\right\}} \cup\mkern 1.5mu\overline{\mkern-1.5mu f\qty{\left\{{{\left\lvert {z} \right\rvert} \geq 1}\right\}}\mkern-1.5mu}\mkern 1.5mu\), which are both the continuous images of compact sets and thus compact and bounded. So \(F\) is a bounded entire function and thus constant.