Inverse transform of $\int K(s,t) \rho(s) dS $

Question

Assume that $V$ is a n-dimensional surface, or the disjoint union such surfaces. Let $K(s,t)$ be a function $V^2\to \mathbb R$. For the sake of simplicity, it can be assumed that $K$ is continuous, differentiable etc.

We assume furthermore that for every differentiable function $\rho: V \to \mathbb R$, integrable on $V$, the relation $$\int_V K(s,t) \rho(s) dS(s) = 0 $$ holds for all $t$ if and only if $\rho = 0$ identically. This means that the transform $F$ defined by $$F(\rho)(t) = \int_V K(s,t)\rho(s) dS(s)$$ is 1-1.

Under these conditions, I have some reasons (see below) to believe that $F$ is invertible, with invert of the form $$F^{-1}(\rho)(s) = \int_V K'(s,t)\rho(t) dS(t),$$ where $K':V^2\to \mathbb R$.

What work has been done in this direction? Is there an available general theorem?

Motivation: Let us divide the surface $V$ into a finite number $n$ of small patches $S_i$. Then approximately, $$F(\rho)(t) = \int_V K(s,t)\rho(s) dS(s) = \sum_i K(s_i,t) \rho(s_i) S_i,$$ where $s_i$ is centered at $S_i$. Writing $x_i = \rho(s_i) S_i$, we see that For every $t$, we have a linear relation of the form $$F(\rho)(t) = \sum_i K(s_i, t) x_i.$$ In particular, choosing $t = s_j$, we have for every $j\in \{1,\ldots n\}$ $$F(\rho)(s_j) = \sum_i K(s_i, s_j) x_i.$$ This is a system of $n$ equations in $n$ variables $x_i$, where $K(s_i, s_j)$ can be viewed as a square matrix. Because $F$ is 1-1, it should be true that $K(s_i, s_j)$ is invertible, hence we have $$x_i = \sum_j K^{-1}(s_i, s_j) F(\rho)(s_j).$$ But since $x_i = \rho(s_i)S_i$ is "very small" (infinitesimal of order 1) while $F(\rho)(s_j)$ is finite, $K^{-1}$ must be very very small (infinitesimal of order 2), of the form $$(K'(s_i, s_j) S_i S_j)_{i,j},$$ with $K'$ a finite matrix. Hence we "can" rewrite the previous equation as $$\rho(s_i) = \sum_i K'(s_i, s_j) F(\rho)(s_j)S_j.$$ Switching to the continuum, this would amount to: $$\rho(s) = \int_V K'(s, t) F(\rho)(t) dS(t).$$ Well, that's very very informal of course, but this makes sense.

Answer 1

This is not a full answer but something for the special case where $V$ is a domain in $\mathbb{R}^d$. Such integral operators $$ F\rho(t) = \int_V K(s,t)\rho(s)ds $$ are usually compact: If $K\in L^2(V\times V)$, then $F$ is linear and bounded from $L^2(V)$ into itself and also compact. Consequently, the singular values of $F$ go to zero and hence, no continuous inverse exist and on top of that the Moore-Penrose pseudo-inverse is bounded only if the range of $F$ is of finite dimension.

To comment on your approach to invertibility: The matrices you get will be badly conditioned (i.e. have large and small singular values) and the condition number will diverge to infinity when you discretize finer and finer.