Consider the following integral operator $\mathcal{A}$ on [EDITED: continuous vector function $f=(f_i):\partial S\to{\mathbb R}^3$]: $$ ({\mathcal A}f)_j(x_0):=\int_{\partial S}\sum_{i=1}^3 f_i(x)G_{ij}(x,x_0)dS(x),\quad j = 1,2,3,\quad x_0\in \partial S, $$ where $$ G_{ij}(x,x_0)=\frac{\delta_{ij}}{r}+\frac{\hat{x}_i\hat{x}_j}{r^3},\quad\hat{x}=x-x_0, \quad r = |\hat{x}|, $$ $S\subset{\mathbb R}^3$ is closed, bounded, simply connected, $\partial S$ is Lyapunov.

This operator is from the representation of the solution of Stokes flow by boundary integral from *Boundary Integral and Singularity Methods for Linearized Viscous Flow* by C. Pozrikidis. The author has slightly mentioned the compactness of this linear operator without giving a proof. I'm interested in *how* to prove it. I have not seen this integral operator before. Those in the book I learned functional analysis from are all of 1 dimension. Thus the basic techniques seem not helpful here.

**Questions:**

- How should I deal with the compactness of [EDITED: $\mathcal A:C(\partial S\to{\mathbb R}^3)\to C(\partial S\to{\mathbb R}^3)$]?
- Can anybody come up with a reference about this kind of high-dimension integral operators?

[**Added**]Some thoughts:
It may be easier to investigate the compactness of the following operator:
$$
({\mathcal A_{ij}}g)(x_0):=\int_{\partial S}g(x)G_{ij}(x,x_0)dS(x),
$$
where $g\in C(\partial S\to{\mathbb R})$. Is the compactness of ${\mathcal A}_{ij}$ related to ${\mathcal A}$?