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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 ${\mathcal A}$?

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    $\begingroup$ What sould the domain and range spaces be? $\endgroup$
    – TaQ
    Feb 5, 2012 at 18:48
  • $\begingroup$ @TaQ: Good comment! I've edited my question. $\endgroup$
    – user14319
    Feb 5, 2012 at 23:04
  • $\begingroup$ Obviously, by Ascoli's theorem, compactness of the (original) operator in question follows if one can prove that for every $\varepsilon\in{\mathbb R}^+$ there is $\delta\in{\mathbb R}^+$ such that $\int_{\partial S}|G_{ij}(x,x_1)-G_{ij}(x,x_0)|dS(x) < \varepsilon$ whenever $x_0,x_1\in\partial S$ with $|x_1-x_0| < \delta$. (I do not know what does it mean that the boundary is Lyapunov.) $\endgroup$
    – TaQ
    Feb 6, 2012 at 20:17

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