To show $\bar{D}u=f$ where $u(z)=\frac{1}{2\pi i} \int_{\Omega} \frac{f(\lambda)}{\lambda-z} d\lambda \wedge d\bar{\lambda}$,

Let $\Omega \subset \mathbb{C}$ be a bounded open set. Suppose $f \in C^{1 }(\Omega)$, $f$ is bounded and $$u(z)=\frac{1}{2\pi i} \int_{\Omega} \frac{f(\lambda)}{\lambda-z} d\lambda \wedge d\bar{\lambda}, (z \in \Omega)$$

I need to show that $u \in C^{1}(\Omega)$ and $\bar{D}u=f$

I start off to show that $u \in C^{1}(\Omega)$. Since the integral can be written as $$u(z)=\frac{1}{2\pi i} \int_{\mathbb{C}} \frac{f(z+\lambda)}{\lambda} d\lambda \wedge d\bar{\lambda}, (z \in \Omega)$$ where I am defining $f=0$ outside the region $\Omega$.

Since $f$ is bounded I can differentiate under the integral sign. (I am unable to give a mathematical rigor to it) So $f \in C^1({\Omega})$.

Now I fix $a \in \Omega$.After this I want to use the theorem : $$\text{Let \Omega be a bounded region in \mathbb{C}, with smooth oriented boundary \partial\Omega. If u \in C^1(\bar{\Omega}), then we have u(a)=\dfrac{1}{2\pi i}\int_{\partial \Omega}\dfrac{u(\lambda)d\lambda}{\lambda-a}-\dfrac{1}{2\pi i}\int_{\Omega}\dfrac{\bar{D}u(\lambda)d\bar{\lambda}\wedge d\lambda}{\lambda-a} }$$

The book I follow suggests to choose $\psi \in C^1(\mathbb{C})$, with support in $\Omega$, so that $\psi =1$ in a neighborhood of $V$ of $a$. (how does one come up with this??)

Then if $f$ is replaced by $(1-\psi)f$ in the original equation, the resulting integral is holomorphic in $V$ (because in $V$, the integral is zero??)

Hence $f$ can be replaced by $\psi f$ in the computation of $(\bar{D}u)(a)$ (why??). We obtain $$(\bar{D}u)(a)=\frac{1}{2\pi i} \int_{\mathbb{C}} \frac{\bar{D}(\psi f)(a+\lambda)}{\lambda} d\lambda \wedge d\bar{\lambda}$$ $$=\frac{1}{2\pi i} \int_{\mathbb{C}} \frac{\bar{D}(\psi f)(\lambda)}{\lambda-a} d\lambda \wedge d\bar{\lambda}=(\psi f)(a)$$

I guess this follows from the theorem I mentioned and that the boundary term cancels out because $\mathbb{C}$ has no boundary (there should be a minus sign as well ).

Finally $(\psi f)(a)=f(a)$ and we are done. (But this is only in $V$ and after that we are using identity theorem??)

Thanks for the help!!

Note that $$\bar{D_j}=\frac{\partial}{\partial \bar{z_j}}=\frac{1}{2}(\frac{\partial}{\partial x_j}+i\frac{\partial}{\partial y_j})$$

• So you want to prove the Residue Theorem? There are a lot of good references out there but I would suggest starting with the wiki en.wikipedia.org/wiki/Residue_theorem – Mark Sep 30 '15 at 17:53
• @Mark I dont see how this is Residue Theorem. Would you explain?? – tattwamasi amrutam Sep 30 '15 at 18:06
• What is $\overline{D}$? – Cameron Williams Oct 1 '15 at 17:07
• The Cauchy-Pompeiu formula might be of interest. – Cameron Williams Oct 1 '15 at 17:25
• @CameronWilliams I am myself confused here. – tattwamasi amrutam Oct 1 '15 at 17:29