Compact resolvent

Question

Given that the operator $$ Hf(x) = -xf''(x) + (x - 1)f'(x) $$ on the Hilbert space $L^2([0,\infty),e^{-x}dx)$ possesses, for each $n \in \mathbb{N}$, an eigenvalue $\lambda_n = n$ with eigenvector the Laguerre polynomial of degree $n$, how can I deduce that the self adjoint operator $$ Kf(x) = Hf(x) + \frac{1}{3} \sin(x)f(x) $$ has a compact resolvent?

Answer 1

If you have a selfadjoint operator with compact resolvent, then bounded perturbation of it will have compact resolvent too. This can be seen like this: let me use the notation $Bf(x)=\frac{1}{3}\sin x f(x)$, then

$$(\lambda-H-B)= (I-B(\lambda-H)^{-1})(\lambda-H)$$.

Because of selfadointness, if $\lambda\in i\mathbb{R}$ is large enough (i.e., $\Im\lambda>\|B\|$), then $\|B(\lambda-H)^{-1}\|<1$, and hence

$$(\lambda-H-B)^{-1} = (\lambda-H)^{-1}(I-B(\lambda-H)^{-1})^{-1},$$

which is the product of a compact and a bounded operator.