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Fall 2020 - Problem 12

Cauchy's integral formula

Let $\Omega = \set{z \in \C \st -2 < \Im{z} < 2}$ . Show that there is a finite constant $C$ so that

\abs{f\p{0}}^2 \leq C \int_{-\infty}^\infty \abs{f\p{x + i}}^2 + \abs{f\p{x - i}}^2 \,\diff{x}

for every holomorphic $\func{f}{\Omega}{\D}$ for which the right-hand side is finite.

Solution.

Let $R > 0$ , and consider the rectangular contour $\gamma_R$ with vertices $R + i$ , $-R + i$ , $-R - i$ , and $R - i$ oriented counter-clockwise. Let $\gamma_1, \gamma_2, \gamma_3, \gamma_4$ be the right, top, left, and bottom edges, respectively. By Cauchy's integral formula,

\p{f\p{0}}^2 = \frac{1}{2\pi i} \int_\gamma \frac{\p{f\p{\zeta}}^2}{\zeta} \,\diff\zeta.

We calculate the integral over the edges separately:

\abs{\int_{\gamma_1} \frac{\p{f\p{\zeta}}^2}{\zeta} \,\diff\zeta} \leq \int_{-1}^1 \frac{1}{\abs{R + it}} \,\diff{t} \xrightarrow{R\to\infty} 0

and similarly for $\gamma_3$ . On $\gamma_2$ ,

\begin{aligned} \abs{\int_{\gamma_2} \frac{\p{f\p{\zeta}}^2}{\zeta} \,\diff\zeta} &\leq \int_{-R}^R \abs{\frac{\p{f\p{x + i}}^2}{x + i}} \,\diff{x} \\ &\leq \int_{-R}^R \frac{\abs{f\p{x + i}}^2}{\sqrt{x^2 + 1}} \,\diff{x} \\ &\leq \int_{-\infty}^\infty \abs{f\p{x + i}}^2 \,\diff{x}. \end{aligned}

and similarly for $\gamma_4$ . Hence, sending $R \to \infty$ , we have

\begin{aligned} \abs{f\p{0}}^2 &\leq \frac{1}{2\pi} \int_{-\infty}^\infty \abs{f\p{x + i}}^2 + \abs{f\p{x - i}}^2 \,\diff{x}, \end{aligned}

which was what we wanted to show.