It's homework time again. Consider a function $f\in C^k_c(\mathbb{C})$, i.e. $C^k$ functions with compact support. Then the function $$u(z) = -\frac{1}{2\pi i} \iint_{\mathbb{C}} \frac{f(\zeta)}{\zeta -z } d\zeta \wedge d\bar{\zeta}$$

is in $C^k(\mathbb{C})$ and solves $$\frac{\partial u}{\partial \bar{z}}=f$$

I want to show that $u(z)$ goes to $0$ as $|z| \rightarrow \infty$. I guess I have to use that $f$ has compact support somehow. But how to start? Do I have to use the Cauchy integral formula? What if I made a change of variables like

$$u(z) = -\frac{1}{2\pi i} \iint_{\mathbb{C}} \frac{f(\zeta+z)}{\zeta} d\zeta \wedge d\bar{\zeta}$$

Would that do anything?


Hint: Let $K$ be the support of $f.$ Set $R = \max \{|z|:z\in K\}.$ Use the fact that $f$ is bounded on $K$ to see that there is a constant $C$ such that for $|z|>R,$

$$|u(z)| \le C\int_K \frac{1}{|\zeta-z|}\, dA(z) \le C\int_K \frac{1}{|z|-R}\, dA(z).$$

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  • $\begingroup$ Thanks! One little question: what is $dA(z)$ in your definition? Just another notation for the Lebesgue measure? $\endgroup$ – Jakob Elias Nov 4 '17 at 0:47
  • $\begingroup$ Yes, area measure. $\endgroup$ – zhw. Nov 4 '17 at 16:06
  • $\begingroup$ I don't quite see how we can replace $\frac{1}{|\zeta-z|}$ by $\frac{1}{|z|-R}$. Could you perhaps say a few words about it? $\endgroup$ – Jakob Elias Nov 4 '17 at 19:42
  • $\begingroup$ Nvm, I figured it out :) $\endgroup$ – Jakob Elias Nov 6 '17 at 8:30

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