# 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.

#complex/exercise/completed