Fall 2016 - Problem 12

Solution.

1. Let $\delta = \frac{1}{2} d\p{K, U^\comp}$. If $\delta = \infty$, then use $\delta = 1$ instead. Consider the grid of rectangles

   $$\mathcal{R} = \set{\br{n\delta, \p{n+1}\delta} \times \br{m\delta, \p{m+1}\delta} \st n, m \in \Z}$$

   which cover all of $\C$. Let

   $$V = \p{\bigcup_{R \in \mathcal{R} \mid R \cap K \neq \emptyset} R}^\itr,$$

   i.e., the interior of the union of these rectangles. This is a finite union, since $K$ is compact.

   Notice that $K \subseteq V$: if $z \in K$, then certainly there is $R \in \mathcal{R}$ such that $z \in R$. If $z \in R^\itr$, then clearly $z \in V$. Otherwise, if $z \in \partial R$, then there are two cases: if $z$ is not one of the vertices of $R$, then there is exactly one more rectangle $R'$ such that $z \in \partial R'$, and it's clear that $z \in \p{R \cup R'}^\itr$. In the other case, if $z$ is a vertex of $R$, then $z$ lies in exactly $4$ rectangles from $\mathcal{R}$, and by the same argument, we still have $z \in V$.

   Finally, $\cl{V} \subseteq U$ since $\delta$ was chosen to be small, and because $\cl{V}$ is a finite union of rectangles, its boundary comprises of finitely many closed line segments.

2. Let $V$ be as in (1). We claim that

   $$f\p{z} = \frac{1}{2\pi i} \int_{\partial V} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta$$

   for any $z \in K$: since $\partial V$ consists of finitely many closed line segments, we can extend all of these segments into lines, which gives a grid on $V$. In this grid, if $z$ lies in exactly one rectangle $R$, then by Cauchy's integral formula,

   $$\frac{1}{2\pi i} \int_{\partial V} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta = \frac{1}{2\pi i} \int_{\partial R} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta = f\p{z},$$

   since the integral will vanish on every other rectangle that doesn't contain $z$. If $z$ lies on the boundary of a rectangle, then $z$ is contained in the interior of a larger rectangle by the same argument as in (1), and the same representation holds.

   Let $r = \frac{1}{2} d\p{K, \partial V}$ which is positive since $K$ and $V^\comp$ are closed. Then

   $$\abs{\frac{f\p{\zeta}}{\zeta - z}} \leq \frac{1}{r} \sup_{\zeta \in \partial V} \abs{f\p{\zeta}} < \infty,$$

   so $\set{\zeta \mapsto \frac{f\p{\zeta}}{\zeta - z}}_{z \in K}$ is equicontinuous. Let $\delta > 0$ be small enough so that if $\abs{\zeta - w} < \delta$, then $\abs{\frac{f\p{\zeta}}{\zeta - z} - \frac{f\p{\zeta}}{w - z}} < \epsilon$.

   Since $\partial V$ comprises of finitely many closed line segments, there exist $\gamma_1, \ldots, \gamma_n$ such that $\partial V = \gamma_1 + \cdots + \gamma_n$. By uniform continuity of each of these, we may pick a partition $0 = t_0 < t_1 < \cdots < t_N = 1$ so fine so that $\abs{\gamma_j\p{t_{k+1}} - \gamma_j\p{t_k}} < \delta$ for each $j$. Approximating $f$ with Riemann sums, we get

   $$\begin{aligned} \abs{f\p{z} - \frac{1}{2\pi i} \sum_{j=1}^n \sum_{k=0}^{N-1} \frac{f\p{\zeta_k}}{\zeta_k - z}\p{\gamma_j\p{t_{k+1}} - \gamma_j\p{t_k}}} &= \abs{\frac{1}{2\pi i} \int_{\partial V} \frac{f\p{\zeta}}{\zeta - z} \,\diff\zeta - \frac{1}{2\pi i} \sum_{j=1}^n \sum_{k=0}^{N-1} \frac{f\p{\zeta_k}}{\zeta_k - z}\p{\gamma_j\p{t_{k+1}} - \gamma_j\p{t_k}}} \\ &= \abs{\frac{1}{2\pi i} \sum_{j=1}^n \sum_{k=0}^{N-1} \int_{t_k}^{t_{k+1}} \p{\frac{f\p{\gamma_j\p{t}}}{\gamma_j\p{t} - z} - \frac{f\p{\gamma_j\p{t_k}}}{\gamma_j\p{t_k} - z}}\gamma_j'\p{t} \,\diff{t}} \\ &\leq \frac{1}{2\pi} \sum_{j=1}^n \sum_{k=0}^{N-1} \int_{t_k}^{t_{k+1}} \epsilon \abs{\gamma_j'\p{t}} \,\diff{t} \\ &= \epsilon\sum_{j=1}^n \frac{\abs{\gamma_j}}{2\pi} \\ &= \epsilon \frac{\abs{\partial V}}{2\pi}, \end{aligned}$$

   where $\abs{\partial V}$ is the length of $\partial V$, which depends only on $K$ and is finite by construction. Hence, our bound is independent of $\epsilon$, so $f$ can be uniformly approximated by linear combinations of rational functions of the form $\frac{f\p{\zeta_k}}{\zeta_k - z}$, where $\zeta_k \in \partial V \setminus K$, which was what we wanted to show.