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Spring 2018 - Problem 7

Cauchy's integral formula

Let $\func{F}{\C \times \C}{\C}$ be (jointly) continuous and holomorphic in each variable separately. Show that $z \mapsto F\p{z, z}$ is holomorphic.

Solution.

Fix $z_0 \in \C$ and let $z \in B\p{z_0, 1}$ . Observe that because $F$ is holomorphic in the first variable,

F\p{z, z} = \frac{1}{2\pi i} \int_{\partial B\p{z_0, 1}} \frac{F\p{\zeta, z}}{\zeta - z} \,\diff\zeta,

by Cauchy's integral formula. Since for each $\zeta$ , $z \mapsto F\p{\zeta, z}$ is still holomorphic, we get

F\p{\zeta, z} = \frac{1}{2\pi i} \int_{\partial B\p{z_0, 2}} \frac{F\p{\zeta, w}}{w - z} \,\diff{w}.

Combining these, we get

\begin{aligned} F\p{z, z} &= \frac{1}{\p{2\pi i}^2} \int_{\partial B\p{z_0, 2}} \int_{\partial B\p{z_0, 3}} \frac{F\p{\zeta, w}}{\p{\zeta - z}\p{w - z}} \,\diff{w} \,\diff\zeta \\ &= \frac{1}{\p{2\pi i}^2} \int_{\partial B\p{z_0, 2}} \int_{\partial B\p{z_0, 3}} \frac{F\p{\zeta, w}}{\zeta - w}\p{\frac{1}{w - z} - \frac{1}{\zeta - z}} \,\diff{w} \,\diff\zeta \\ &= \frac{1}{\p{2\pi i}^2} \int_{\partial B\p{z_0, 2}} \int_{\partial B\p{z_0, 3}} \frac{F\p{\zeta, w}}{\zeta - w}\sum_{n=0}^\infty \br{\frac{1}{w}\p{\frac{z - z_0}{w - z_0}}^n - \frac{1}{\zeta - z_0}\p{\frac{z - z_0}{\zeta - z_0}}^n} \,\diff{w} \,\diff\zeta \\ &= \frac{1}{\p{2\pi i}^2} \int_{\partial B\p{z_0, 2}} \int_{\partial B\p{z_0, 3}} \frac{F\p{\zeta, w}}{\zeta - w}\sum_{n=0}^\infty \p{\frac{1}{\p{w - z_0}^{n+1}} - \frac{1}{\p{\zeta - z_0}^{n+1}}} \p{z - z_0}^n \,\diff{w} \,\diff\zeta. \end{aligned}

If we can justify switching the sum and integral, then we are done. Observe that the sum converges uniformly, since $\abs{w - z_0}, \abs{\zeta - z_0} \geq 1$ . Also, $\abs{\zeta - w} \geq 1$ , by construction, and on $\partial B\p{z_0, 3}$ , we see that $w \mapsto F\p{\zeta, w}$ is bounded, so we may apply Fubini's theorem to swap the inner sum and integral. By the same argument, since $\zeta \mapsto F\p{\zeta, w}$ is a continuous hence bounded function on $\partial B\p{z_0, 2}$ , so we may swap again. Thus,

F\p{z, z} = \sum_{n=0}^\infty \p{\frac{1}{\p{2\pi i}^2} \int_{\partial B\p{z_0, 2}} \int_{\partial B\p{z_0, 3}} \frac{F\p{\zeta, w}}{\zeta - w} \p{\frac{1}{\p{w - z_0}^{n+1}} - \frac{1}{\p{\zeta - z_0}^{n+1}}} \,\diff{w} \,\diff\zeta} \p{z - z_0}^n,

so $F$ is holomorphic.