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Spring 2019 - Problem 11

harmonic functions, Schwarz reflection principle

Let $u \in C^\infty\p{\R}$ be smooth $2\pi$ -periodic. Show that there exists a bounded holomorphic function $f_+$ in the upper half-plane $\Im{z} > 0$ and a bounded holomorphic function $f_-$ in the lower half-plane $\Im{z} < 0$ , such that

u\p{x} = \lim_{\epsilon\to0} \,\p{f_+\p{x + i\epsilon} - f_-\p{x - i\epsilon}}, \quad x \in \R.

Solution.

Since $u$ is $2\pi$ -periodic, then function $u\p{\frac{\log{z}}{i}}$ is a well-defined smooth function on $S^1$ . We may then solve the Dirichlet problem on the disk with boundary data $u\p{\frac{\log{z}}{i}}$ to get a harmonic function $v\p{z}$ on the disk which is continuous on $\cl{\D}$ . Since $\D$ is simply connected, there exists a holomorphic function $g$ on $\D$ such that $\Im{g} = v$ . Then $f\p{z} = g\p{e^{iz}}$ is a holomorphic function on the upper half-plane $\H$ such that for any $x \in \R$ ,

\Im{f\p{x}} = \Im{g\p{e^{ix}}} = v\p{e^{ix}} = u\p{x}.

Let $f_+ = \frac{f}{2i}$ , and by Schwarz reflection, $f_-\p{z} = \conj{f_+\p{\conj{z}}}$ is a holomorphic function on the lower half-plane. Then

f_+\p{x + i\epsilon} - f_-\p{x - i\epsilon} = \Im{f\p{x + i\epsilon}} \xrightarrow{\epsilon\to0} u\p{x},

which was what we wanted.