Question 12.17: Let the function f : R² → R be given for (x, y) = x + ιy by ......
Let the function f : R² → R be given for (x, y) = x + ιy by
f(x+\imath y)=\frac{1}{\pi} \cdot \frac{1}{x+\imath y} .
Prove that
a) the function f is locally integrable on R², and so it defines a distribution on R²;
b) the function f is the fundamental solution for the Cauchy-Riemann operator
\bar{\partial}=\frac{1}{2}\left(\frac{\partial}{\partial x}+\imath \frac{\partial}{\partial y}\right).
a) Since we have |f(z)|=\frac{1}{\pi} \cdot \frac{1}{|z|} \text { for } z=x+\imath y \text {, it follows that } f \in L_{l o c}^1\left( \mathbf{R} ^2\right) \text {. }
b) We have to prove that
<\bar{\partial} f, \varphi>=\varphi(0) (12.35)
To that end, we start from the equality
<\bar{\partial}, \varphi>=-<f, \bar{\partial} \varphi>=-\frac{1}{2 \pi} \int_{ \mathbf{R} ^2}\left(\frac{\partial \varphi}{\partial x}+\imath \frac{\partial \varphi}{\partial y}\right) \frac{d x d y}{x+\imath y}
Changing the variables x=r \cos t, y=r \sin t, \varphi_1(r, t)=\varphi(r \cos t, r \sin t), we obtain
<\bar{\partial} f, \varphi>=-\frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty}\left(\frac{\partial \varphi_1}{\partial r}-\frac{\imath}{r} \cdot \frac{\partial \varphi_1}{\partial t}\right) d r d t
Therefore we have
<\bar{\partial} f, \varphi>=-\frac{1}{2 \pi} \int_0^{2 \pi}\left(\varphi(\infty, t)-\varphi_1(0, t)\right) d t+\frac{\imath}{2 \pi} \int_0^{2 \pi}\left(\varphi(r, 2 \pi)-\varphi_1(r, 0)\right) d r
By the compactness of the support of the function \varphi and hence also of \varphi_1 and the periodicity of \varphi_1, we obtain (12.35).