Schwarz Reflection

Tie’s Extra Questions: Spring 2015 (Reflection for harmonic functions) #complex/exercise/completed

  • Assume \(u\) is harmonic on open set \(O\) and \(z_n\) is a sequence in \(O\) such that \(u(z_n) = 0\) and \(\lim z_n \in O\). Prove or disprove that \(u\) is identically zero. What if \(O\) is a region?

  • Assume \(u\) is harmonic on open set \(O\) and \(u(z) = 0\) on a disc in \(O\). Prove or disprove that \(u\) is identically zero. What if \(O\) is a region?

  • Formulate and prove a Schwarz reflection principle for harmonic functions

cf. Theorem 5.6 on p.60 of Stein et al.

Hint: Verify the mean value property for your new function obtained by Schwarz reflection principle.

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 \({\mathbf{R}}\) but zero on \(i{\mathbf{R}}\).

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 \({\mathbf{R}}\), and that \(u\equiv 0\) on \({\mathbf{R}}\cap S^+\). Define \(S^- = \left\{{\overline{z} {~\mathrel{\Big\vert}~}z\in S^+}\right\}\), and \begin{align*} U(z) \coloneqq \begin{cases} U(z) & z\in S^+ \\ -U(\overline{z}) & z\in S^-. \end{cases} .\end{align*} Then \(U\) is a harmonic extension of \(u\) to \(S \coloneqq S^+ \cup(S^+ \cap{\mathbf{R}}) \cup S^-\). 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 \(w_0 = \overline{z_0}\), then \begin{align*} U(z_0) &\coloneqq-u(\overline{z_0}) \\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\overline{z_0 + re^{it} }} \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\overline{z_0} + re^{-it} } \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\overline{z_0 + re^{-it} }} \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi U(z_0 + re^{it} ) \,dt .\end{align*}

Reflection for the disc #complex/exercise/completed

  • State the standard Schwarz reflection principle involving reflection across the real axis.

  • Give a linear fractional transformation \(T\) mapping \({\mathbb{D}}\) to \({\mathbb{H}}\). Let \(g(z) = \overline{z}\), and show \begin{align*} (T^{-1} \circ g \circ T)(z) = 1/\overline{z} .\end{align*}

  • Suppose that \(f\) is holomorphic on \({\mathbb{D}}\), continuous on \(\overline{{\mathbb{D}}}\), and real on \(S^1\). Show that \(f\) must be constant.

Part 1: Let \(\Omega = \Omega^+ \cup I \cup\Omega^-\) be a region symmetric about \({\mathbf{R}}\). 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 \(F(z) = \overline{f(\overline{z})}\).

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}\overline{\qty{-i {z+1\over z-1} }} \\ &= T^{-1}\qty{i{\overline{z} + 1 \over \overline{z} - 1}} \\ &= {i\qty{\overline{z} + 1 \over \overline{z} - 1} - i \over i\qty{\overline{z} + 1 \over \overline{z} - 1} + i } \\ &= {(\overline{z} + 1) - (\overline{z} - 1) \over (\overline{z} + 1) + (\overline{z} - 1)} \\ &= {1\over \overline{z}} .\end{align*}

Part 3: Define \(h: {\mathbb{H}}\to \overline{{\mathbb{H}}}\) by \(h(z) = (T\circ f\circ T^{-1})(z)\). Under \(T^{-1}: {\mathbb{D}}\to {\mathbb{H}}\), we have \(T(S^1) = {\mathbf{R}}\), so \(h\) is a holomorphic function on \({\mathbb{H}}\) that is continuous and real-valued on \({\mathbf{R}}\). By the reflection principle, defining \(H(z) \coloneqq\overline{h(\overline{z})}\) for \(\Im(z) < 0\) yields an entire function \(H: {\mathbf{C}}\to {\mathbf{C}}\) Noting that for \(g(z) \coloneqq\overline{z}\), \(g=g^{-1}\), 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 \(F:{\mathbf{C}}\to {\mathbf{C}}\) defined by \begin{align*} F(z) = \begin{cases} f(z) & z\in {\mathbb{D}} \\ f_-(z) \coloneqq{1\over \overline{f\overline{z}}} & 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.

Spring 2020 HW 2, SS 2.6.15 (Constant on boundary and nonvanishing implies constant, using Schwarz) #complex/exercise/completed

Suppose \(f\) is continuous and nonvanishing on \(\overline{{\mathbb{D}}}\), and holomorphic in \({\mathbb{D}}\). Prove that if \({\left\lvert {z} \right\rvert} = 1 \implies {\left\lvert {f(z)} \right\rvert} = 1\), then \(f\) is constant.

Hint: Extend \(f\) to all of \({\mathbf{C}}\) by \(f(z) = 1/ \overline{f(1/\overline{z})}\) for any \({\left\lvert {z} \right\rvert} > 1\), and argue as in the Schwarz reflection principle.

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 \(T(S^1) = {\mathbf{R}}\), and since \(f(z)\in S^1\) when \(z\in S^1\), we have \(\tilde f({\mathbf{R}}) = {\mathbf{R}}\), i.e. this is a real-valued function on \({\mathbf{R}}\). So \(\tilde f\) extends holomorphically to \(\tilde F:{\mathbf{C}}\to CC\), and we can pull this back to a holomorphic extension of \(f\).

Extend \(f\) to \(F:{\mathbf{C}}\to {\mathbf{C}}\) by \(f(z) = 1/\overline{f(1/\overline{z})}\) for \(z\in {\mathbb{D}}^c\), which generally has poles at the points \(1/\overline{z_k}\) 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 \(F({\mathbf{C}}) \subseteq f\qty{\left\{{{\left\lvert {z} \right\rvert} \leq 1}\right\}} \cup\overline{ f\qty{\left\{{{\left\lvert {z} \right\rvert} \geq 1}\right\}}}\), which are both the continuous images of compact sets and thus compact and bounded. So \(F\) is a bounded entire function and thus constant.