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Spring 2020 - Problem 9

inhomogeneous Cauchy integral formula

Let $f \in L^1\p{\C} \cap C^1\p{\C}$ . Show that the integral

u\p{z} = -\frac{1}{2\pi} \iint_\C \frac{f\p{\zeta}}{\zeta - z} \,\diff\lambda\p{\zeta}, \quad z \in \C,

defines a $C^1$ function on the whole complex plane that satisfies

\p{\pder{}{x} + i\pder{}{y}}u\p{x + iy} = f\p{x + iy}.

In this problem, $\diff\lambda\p{\zeta}$ denotes (planar) Lebesgue measure on $\C$ and $C^1$ is meant in the real-variable sense.

Solution.

Recall that by an application of Green's theorem, we have for any $g \in C_{\mathrm{c}}^1\p{\C}$ that

g\p{z} = -\frac{1}{\pi} \iint_\C \pder{g}{\conj{\zeta}}\p{\zeta} \frac{1}{\zeta - z} \,\diff\lambda\p{\zeta}.

To extend this to our $f$ , let $\set{K_n}_n$ be a compact exhaustion of $\C$ , e.g., $K_n = \cl{B\p{0, n}}$ . Let $\set{\phi_n}_n \in C_{\mathrm{c}}^1\p{\C}$ be a bump function on $K_n$ , i.e., $0 \leq \phi_n \leq 1$ and $\phi_n = 1$ on a neighborhood of $K_n$ . Let $\psi_1 = \phi_1$ and $\psi_n = \phi_n - \phi_{n-1}$ for $n \geq 2$ so that $\psi_n$ is $0$ on a neighborhood of $K_{n-1}$ . Since $K_n \subseteq K_{n+1}$ for each $n$ , it follows that this sum is locally finite. Hence, at $z \in K_N$ , we have

\sum_{n=1}^\infty \psi_n\p{z} = \phi_1\p{z} + \sum_{n=2}^N \p{\phi_n\p{z} - \phi_{n-1}\p{z}} = \phi_N\p{z} = 1,

so $\set{\psi_n}_n$ is a locally finite partition of unity. Thus,

\begin{aligned} u\p{z} &= -\frac{1}{2\pi} \iint_\C \sum_{n=1}^\infty \frac{\psi_n\p{\zeta}f\p{\zeta}}{\zeta - z} \,\diff\lambda\p{\zeta} \\ &= -\frac{1}{2\pi} \iint_\C \sum_{n=1}^\infty \frac{\psi_n\p{\zeta + z}f\p{\zeta + z}}{\zeta} \,\diff\lambda\p{\zeta} && \p{\zeta \mapsto \zeta - z} \\ &= -\frac{1}{2\pi} \sum_{n=1}^\infty \iint_\C \frac{\psi_n\p{\zeta + z}f\p{\zeta + z}}{\zeta} \,\diff\lambda\p{\zeta}, \end{aligned}

since after the change of variables, the sum is finite for each $z \in \C$ . Again by local finiteness, because $\psi_n, f$ are $C^1$ , it follows that $u$ is also $C^1$ . Similarly, local finiteness allows us to interchange the derivative and the sum, and because $f \in L^1\p{\C}$ and each $\psi_n$ is compactly supported, we may differentiate under the integral sign and apply change of variables freely:

\begin{aligned} \pder{u}{\conj{z}}\p{z} &= -\frac{1}{\pi} \sum_{n=1}^\infty \pder{}{\conj{z}} \iint_\C \frac{\psi_n\p{\zeta + z}f\p{\zeta + z}}{\zeta} \,\diff\lambda\p{\zeta} \\ &= -\frac{1}{\pi} \sum_{n=1}^\infty \iint_\C \pder{\p{\psi_nf}}{\conj{\zeta}}\p{\zeta + z} \frac{1}{\zeta} \,\diff\lambda\p{\zeta} \\ &= -\frac{1}{\pi} \sum_{n=1}^\infty \iint_\C \pder{\p{\psi_nf}}{\conj{\zeta}}\p{\zeta} \frac{1}{\zeta - z} \,\diff\lambda\p{\zeta} && \p{\zeta \mapsto \zeta + z} \\ &= \sum_{n=1}^\infty \psi_n\p{z} f\p{z} && \p{\psi_n f \in C_{\mathrm{c}}^1} \\ &= f\p{z}, \end{aligned}

since we had a partition of unity.