Suppose $D\subseteq\mathbb C$ is a bounded domain, with $\partial D$ smooth. Then suppose $g:\mathbb C\backslash\overline D\to\mathbb C$ is analytic and extends continuously to $\partial D$. If $\lim_{z\to\infty}g(z)$ is finite, show that $$ g(u)=\lim_{z\to\infty}(g(z))-\frac1{2\pi i}\int_{\partial D}\frac{g(z)dz}{z-u} $$ for all $u\in\mathbb C\backslash\overline D$, assuming that $\partial D$ is oriented as the boundary of $D$.

This problem is confusing to me, because the integral term looks like the result of the Cauchy Integral Formula. But we are considering points outside $\overline D$. Would anyone be able to give some intuition for how to solve this problem?


Recall the Cauchy integral formula:

Let $\Omega$ be a bounded domain with smooth boundary. Then for any $h \in \mathcal{O}(\Omega)$ which extends continuously to $\partial{\Omega}$, we have $$ \forall u \in \Omega,\ h(u) = \frac{1}{2\pi i} \int_{\partial{\Omega}} \frac{h(z)}{z-u} dz. $$

We can assume that $D$ is contained in the unit disk $\Delta$. Set $\Omega$ the image of $\mathbb{C}\backslash \overline{D}$ by $z \mapsto \frac{1}{z}$ and for all $z \in \Omega$, set $h(z) = g(\frac{1}{z})$. Notice that $\Omega$ is a bounded domain since $D \subset \Delta$. Alas, $h$ is not holomorphic on $\Omega$ but it is holomorphic on $\Omega\backslash\{0\}$. Since the boundary of this latter domain is not smooth, we replace it by $\Omega\backslash \Delta_{\varepsilon}$ where $\Delta_{\varepsilon}$ stands for the disc centered at the origin with radius $\varepsilon$, which is chosen small enough so that $\Delta_{\varepsilon} \subset \Omega$. Now you can apply the Cauchy integral formula above and let $\varepsilon$ go to zero to conclude.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.