# Inequalities on kernels of compact operators

Suppose we have a $\sigma$-finite positive measure $\mu(v)$ on $\Bbb R^d$ and we have two positive kernels on $\Bbb R^d\times \Bbb R^d$ $k_1(v,u)>0$, $k_2(v,u)>0$. We define integral operators $$K_i[f](v) =\int_{\Bbb R^d}k_i(v,u)f(u)d\mu(u).$$

We know by some external argument that $K_2:L^2(d\mu)\to L^2(d\mu)$ is a compact operator; furthermore, we know that there exist two strictly positive constants $C_+$ and $C_-$ such that $$C_-k_2\le k_1\le C_+k_2.$$

I want to deduce from these conditions that $K_1$ is a compact operator, but I can't see an evident path to prove it. It's possible to show that $K_1$ is bounded by examining negative and positive parts of $f$.

It's important to say that neither $k_1$ nor $k_2$ satisfies the Hilbert-Schmidt criterion (i.e. they are not $L^2(d\mu(v)d \mu(u))$). We can additionally suppose that $K_i$ are symmetric.

So, is there a way to prove that $K_1$ is compact (or to prove that it can be non-compact)? Any help will be appreciated.

The answer was found with the help of shwedka at dxdy forum. The result holds and there's an even more general formulation.
Suppose that we have two spaces with separable measures $(X,dm)$ and $(Y,dn)$; suppose also that $q\in[1,+\infty)$, $p\in (1,+\infty)$. The operators $K_i$ are given by $$K_i[f](y) = \int_{X}k_i(y,x)f(x)dm(x).$$ If $$|k_1(x,y)|\le k_2(x,y)\quad (m,n)-a.e..$$ and $K_2:L^p(dm(x))\to L^q(dm(y))$ is compact, then $K_1:L^p(dm(x))\to L^q(dm(y))$ is compact, too. Note that we don't need positivity of $k_1$.
The proof, however, exceeds the format of math.SE. You can find it in the article by L.D. Pitt: L. D. Pitt, A compactness condition for linear operators of function spaces. J. Operator Theory 1 1979, 49-54.
You can also find a proof (in russian) in the book by Krasnoselskiy et al. "Integral operators in the spaces of integrable functions", ch.2, th. 5.6. (Красносельский и др. "Интегральные операторы в пространствах суммируемых функций" гл. 2, п.5.6)