Let $H$ and $V$ be bounded self-adjoint operators. If $V$ is compact then $E_{H+V}(J)-E_{H}(J)$ is a compact operator.

Question

I read the following claim: Let $H$ and $V$ be bounded self-adjoint operators. If $V$ is compact then $E_{H+V}(J)-E_{H}(J)$ is a compact operator for any Borel set $J \subseteq \mathbb{R}$. Here $E_A(\cdot)$ is the spectral measure of $A$. Can you help me to show it?

I think that if $E_{H+V}(J)-E_{H}(J)= E_{V}(J)$, then the claim will be true just because $V$ is compact. But I'm no sure if this equality holds, and I'm not sure $E_{H+V}(J)-E_{H}(J)$ is a projection.

Edit: J is assumed to be bounded.

Answer 1

As stated, this is not true.

For instance on any infinite-dimensional Hilbert space, take $H=0$ and $V$ an injective compact operator. Take $J=(0,\infty)$. Then $E_H(J)=0$, and $E_{H+V}(J)=E_V(J)=I$, which is not compact since it is an infinite projection.

Although in this example it is, there is no reason for $E_{H+V}(J)-E_H(J)$ to be a projection in general.