About principal values and Wirtinger derivative

Question

Let $K$ be a compact of the plane of Lebesgues measure 0 and $\Omega$ a domain containing $K$. Denote by $E$ the vector space of functions that are holomorphic on $\Omega - K$.

I'm interested in knowing whether we can define a $\overline{\partial}$ operator on $E$ in the sense of distributions . The problem is, elements of $E$ do not define a priori distributions since they need not be locally integrable. So the problem amounts to finding a standard way to make them into distributions.

In the case where $f \in E$ is a rational fraction with poles in $K$, the decomposition theorem asserts that it is a sum of $\frac{a_i}{(z-y_i)^{n_i}}$, and those can be made into distributions by taking principal values. Is there a way to generalize this to any $f \in E$ ?

To rephrase more precisely my initial question : is there an operator $L : E \rightarrow D'(\Omega)$ such that :

• $L = \overline{\partial}$ on $E \cap D'(\Omega)$

• $L$ is continuous on $\mathbb{C}(z)$ with respect to its topology

For example, $\frac{1}{z(z-\epsilon)} \rightarrow \frac{1}{z^2}$, and $\\overline{\partial} \frac{1}{z(z-\epsilon)} \rightarrow \frac{1}{\pi}\delta'$ in the sense of distributions, so $L\frac{1}{z^2}$ should be $\frac{1}{\pi}\delta'$, and more generally $L\frac{1}{z^n}$ should be $\frac{1}{\pi}\delta^{(n)}(-1)^{n-1}$.

Answer 1

You probably want $L$ to be continuous with respect to the topology of $E$, not just $C(z)$ And $E$ is equipped with its usual topology: uniform convergence on compact subsets of $\Omega\backslash K$.

If this is so, the extension is impossible. Even if your $K$ is one point. In this case a function $f\in E$ expands into a Laurent series $\sum a_nz^n$ and $\overline\partial$ must be an infinite sum of derivatives of $\delta$-function. This will not converge for a test function in the Schwartz space $D$.

But this defines a continuous linear functional on a much smaller space of germs of analytic functions in a neighborhood of $K$, that is a hyperfunction. In other words, in this case $\overline\partial$ does not exist in the sense of distributions but exists in the sense of hyperfunctions.