Fall 2020 - Problem 10

Solution.

1. Because $K$ is compact, there exists $R > 0$ such that $K \subseteq B\p{0, R}$. Thus,

   $$\begin{aligned} \abs{F\p{z}} &\leq \iint_{B\p{0,R}} \frac{1}{\abs{w - z}} \,\diff\mu\p{w} \\ &= \iint_{B\p{-z,R}} \frac{1}{\abs{w}} \,\diff\mu\p{w} && \p{w \mapsto w - z} \\ &\leq \iint_{B\p{0,R}} \frac{1}{\abs{w}} \,\diff\mu\p{w} \\ &< \infty, \end{aligned}$$

   so $F$ is bounded. For continuity, let $z, z' \in \C$ and observe

   $$\abs{F\p{z} - F\p{z'}} \leq \iint_{B\p{0,R}} \abs{\frac{1}{z - w} - \frac{1}{z' - w}} \,\diff\mu\p{w}.$$

   Since the integrand is $L^1_{\mathrm{loc}}$ by the triangle inequality, we may apply dominated convergence and send $z \to z'$. Next, observe that for $z_0 \in \C \setminus K$, we have $\delta = d\p{z_0, K} > 0$, so it $\abs{z - z_0} \leq \frac{\delta}{2}$, we get

   $$F\p{z} = \iint_K \frac{1}{w - z} \,\diff\mu\p{w} = \iint_K \frac{1}{w - z_0} \sum_{n=0}^\infty \p{\frac{z - z_0}{w - z_0}}^n \,\diff\mu\p{w},$$

   and this converges uniformly by the Weierstrass M-test, as

   $$\abs{\frac{z - z_0}{w - z_0}} \leq \frac{\delta/2}{\delta} = \frac{1}{2}.$$

   Thus,

   $$F\p{z} = \sum_{n=0}^\infty \p{\iint_K \frac{1}{\p{w - z_0}^{n+1}} \,\diff\mu\p{w}} \p{z - z_0}^n$$

   for $z \in B\p{z_0, \frac{\delta}{2}}$, so $F$ is analytic.

2. We follow the hint (where we count the $z_k \in \C$ with multiplicity) and let

   $$G\p{z} = \frac{F\p{z} - a}{\prod_{j=1}^n \p{z - z_j}}.$$

   There are only finitely many of these, or else they must accumulate to some $z_0$. By assumption and continuity of $F$, we have $z_0 \in \C \setminus K$, which implies that $F$ is identically $a$, which is impossible.

   Since $F$ is bounded, it follows that if $n \neq 0$, then $G\p{\infty} = 0$. By construction, we also have $G\p{z} \neq 0$ for $z \in \C$, so $G\p{z} = e^{H\p{z}}$ for some $H$ holomorphic in $\C \setminus K$ and continuous on $\C$. Since $G\p{\infty} = 0$, we see that $\widetilde{G}\p{z} = G\p{1/z} = e^{\widetilde{H}\p{z}}$ is holomorphic near $0$ with $G\p{0} = 0$. By the argument principle, for $\delta > 0$ small enough, we get

   $$1 \leq \frac{1}{2\pi i} \int_{\partial B\p{0,\delta}} \frac{\widetilde{G}'\p{z}}{\widetilde{G}\p{z}} \,\diff{z} = \frac{1}{2\pi i} \int_{\partial B\p{0,\delta}} \widetilde{H}'\p{z} \,\diff{z} = 0,$$

   which is impossible. Hence, $F\p{z} \neq a$ for $z \in \C \setminus K$ to begin with.