Let $\Omega \neq \emptyset$ and $\mu$ be a probability measure on $\Omega.$ Consider two reproducing kernels $k_1,k_2:\Omega \times \Omega \rightarrow \mathbb{R},$ such that they both represent the same element of $L^2(\Omega \times \Omega, \mu \otimes \mu)$ (in other words: they are both $L^2$ functions and are $\mu\otimes \mu$-a.s. equal). Asume furthermore that $\int_{\Omega} k_i(x,x) d\mu(x) < \infty.$
Denote the Reproducing Kernel Hilbert Space of $k_i$ by $H_i.$ Consider now the operators $$S_i: H_i \rightarrow L^2(\Omega, \mu), \quad f \mapsto f,$$ which we assume to be injective. It is easy to see that our assumptions guarantee the continuity of $S_i.$ My question is the following: suppose I know that $S_1$ is compact. Does it necessarily follow that $S_2$ is compact?
The most promising approach to proving that this is indeed the case seems to be using the fact that an operator between Hilbert Spaces is compact iff its image contains no closed infinitely-dimensional subspace.
I came upon the solution myself. It turns out that the conditions I imposed are enough to force the embeddings $S_i$ to be Hilbert-Schmidt operators. So in particular they are both compact.