Let $X$ be a smooth manifold. $X \times X$ is product manifold. $\mu$ is a Borel measure on $X$.
There are two aims
(i) Associate a $C^*$ norm to $C_c(X\times X)$, making it a $C^*$ algebra.
(ii) Show that this is in fact $*$-isomorphic to $K(L^2(X,\mu))$ the compact operators on $L^2(X,\mu)$.
$$ \pi:C_c(X\times X) \rightarrow B(L^2(X,\mu)), \quad \pi(f)(\xi)(x) = \int_{z \in X} f(x,z)\xi(z) \, d\mu (z) $$ This representation induces a norm on $C_c(X \times X)$.
I guess this is how (i) works. We have a
(a) a $*$-algebra homomoprhism.
This follows as $$\pi(f\ast g) \xi(x) = \int_z \int_yf(x,y)g(y,z) \xi(z) dy \, dz $$ $$ \pi(f) \pi(g)\xi (x) = \int_y f(x,y)\int_z g(x,z) \,dz dy = \int_y\int_z f(x,y)g(x,z) \xi(z) \, dy dz $$ (a') which is also injective: Firstly, $f|_x$ is a measurable function, so the integral makes sense. Secondly if $f|_x:X \rightarrow \Bbb C$ when tested against all $\xi \in L^2(X,\mu)$, yields $0$ integral, then $f|_x$ is zero. Hence $f$ is zero.
(b)Now this is a faithful representation. Hence we can give a norm on $C_c(X \times X)$ by $$||f||_1 = ||\pi(f)||_{op}$$ This is a $C^*$ norm. We take the completition.
(c) Now I wish to understand the compact operators on $L^2(X,\mu)$ for (ii). It suffices to show the image of $\pi$ is dense.
Remark: It seems that the problem is not well stated too, and the result seems to be a consequence of Hilbert Schimdt Operators - that the image lies in $K(L^2(X,\mu)$. That it is dense is not clear.
The source of this example is Example 4 at Page 18