# Cauchy path integral as a linear operator: kernel and image?

Let $$\mathcal O(\Omega)$$ be the algebra of functions holomorphic on the open set $$\Omega\subset\mathbb C$$. For $$\gamma$$ a simple compact curve in $$\mathbb C$$ consider the linear operator given by path-integrating against the Cauchy kernel $$\Gamma : \mathcal O(\mathbb C) \longrightarrow \mathcal O(\mathbb C\setminus \gamma) \\ f\longmapsto \left(z\mapsto \int_\gamma \frac{f(w)}{w-z}\mathrm{d}w\right)$$

In the trivial case where $$\gamma$$ is closed then $$\Gamma$$ is injective (apply Cauchy formula to all $$z$$ in the bounded connected component of $$\mathbb C\setminus \gamma$$) and its range consists in those functions that are zero outside $$\gamma$$ and the restriction of an entire function inside.

My question: what happens if $$\gamma$$ is not closed (i.e. $$\mathbb C\setminus \gamma$$ connected)? Is it possible to characterize the range of $$\Gamma$$, besides the obvious $$O(|z|^{-1})$$ towards infinity?

(In this case $$f$$ may only be a germ near $$\gamma$$ of a holomorphic function, but keep it simple if necessary.)

I'd find it hard to believe that injectivity would fail (although I don't quite have an argument).

If this question has already a well-documented answer (which is fairly probable but I didn't know which keywords to look for), please accept my apologies. A pointer to a reference would be great!

Such an integral over a simple (non-closed) curve is called the Cauchy type integral.

The curve $$\gamma$$ is oriented, so for a function $$F$$ defined in $$C\backslash\gamma$$ and $$z\in\gamma$$ different from an endpoint, we can talk of the right limit $$F^+(z)$$ and left limit $$F^-(z)$$. Then we have Sokhotski formula $$F^-(z)-F^+(z)=f(z),\quad z\in\Gamma$$ As in addition $$F(z)\to0,\; z\to\infty$$, this shows that your map is injective. The image consists of those analytic functions in $$C\backslash\gamma$$ which tend to $$0$$ as $$z\to\infty$$, have limits on both sides of $$\gamma$$ and the jump between these limits is an entire function (analytic in $$C$$). (At the end-points, the limits do not exist, and we have logarithmic singularities.)

With some regularity of $$\gamma$$ and $$f$$ (non necessary analytic) we have $$F^+(z)=F(z)-\frac{1}{2}f(z),\quad F^-(z)=F^*(z)+\frac{1}{2}f(z),$$ where $$F^*(z)$$ is the principal value of your integral. The above formula is obtained by subtracting these two relations. But when $$f$$ is entire, the curve can be slightly deformed (preserving its ends), and you get rid of the regularity condition by approximating it with a smooth (or piecewise-linear) curve with the same ends.

If the curve happens to be closed then $$F^+=0$$ and we recover the result for the closed curve which you cite.

It is interesting that the curve $$\gamma$$ plays little role in the description of the image: functions from the image have analytic continuation along any curve starting and ending at $$\infty$$ and not passing through the endpoints of $$\gamma$$.

References. This material is standard and is included to every Russian undergraduate Complex Variables text, but for some strange reasons none of them is translated into English. A good reference in English which comes to my mind is a translation from the Japanese:

MR1026013
Kaneko, A. Introduction to hyperfunctions. Kluwer, Dordrecht, Tokyo, 1988.