Interchange complex differentiation and integration Let $g\in C(\mathbb C) \cap L_1(\mathbb C)$. Then the following formula holds:
$$
 \int\limits_{\mathbb C} g(z) \frac{\partial}{\partial \bar z}\frac{1}{z-\lambda} \, dx dy = \pi g(\lambda), \quad z = x + iy.
$$
Now let $f$ be defined by the formula:
$$
 f(\lambda) = -\frac{1}{\pi} \int\limits_{\mathbb C} g(z) \frac{1}{z-\lambda} \, dxdy.
$$
Is it true that $\frac{\partial}{\partial \bar \lambda} f(\lambda) = g(\lambda)$ so that we can interchange differentiation and integration? If it is true how to show this?
 A: The formula
$$\int_{\mathbb{C}} g(z) \frac{\partial}{\partial \overline{z}}\frac{1}{z-\lambda}\, dxdy = \pi g(\lambda)\tag{1}$$
holds only by abuse of notation (one that I deeply abhor). What it expresses is that the distribution
$$C \colon \varphi \mapsto \frac{1}{\pi} \int_{\mathbb{C}} \frac{\varphi(z)}{z}\,dxdy$$
is a fundamental solution of the Cauchy-Riemann operator, $\frac{\partial}{\partial\overline{z}}C = \delta$, where $\delta$ is the Dirac distribution, and $\frac{\partial}{\partial \overline{z}}$ is the distributional derivative. Then $(1)$ is simply the statement of the fact that $\delta$ is the unit for convolution, $A\ast \delta = A$ whenever that makes sense.
The distribution $C$ is given by integration of the product of the test function with a locally integrable function, $h(z) = \frac{1}{\pi z}$, so for sufficiently good functions $g$, the convolution of $C$ and $T_g \colon \varphi \mapsto \int_\mathbb{C} g(z)\varphi(z)\,dxdy$ is defined and we have
$$C\ast T_g = T_h \ast T_g = T_{h\ast g},$$
that it is given by the convolution of $g$ and $h$.
If $g \in C(\mathbb{C})\cap L^1(\mathbb{C})$, it is "sufficiently good", and for the function
$$f(\lambda) = -\frac{1}{\pi}\int_\mathbb{C} g(z)\frac{1}{z-\lambda}\,dxdy = (h\ast g)(\lambda),$$
we have $$\frac{\partial f}{\partial \overline{z}} = g\tag{2}$$
at least in the sense of distributions. Whenever $f$ is continuously real differentiable, for example when $g$ is, then $(2)$ holds in the classical sense.
