Let $K^*$ be a self-adjoint operator on $L^2(\mathbb{R}^3),$ and assume $(K^*)^4 := K^* \circ K^* \circ K^* \circ K^*$ is a Hilbert-Schmidt operator on $L^2(\mathbb{R}^3).$
Will $K^*$ be a compact operator on $L^2(\mathbb{R}^3)?$
I ask this question because I am reading the proof for Theorem 2.1 of the paper of Golse and Poupaud. The following is its proof.
I guess this is true. But I can't provide a rigorous proof here.
Heuristically, since $(K^∗)^4$ is a Hilbert-Schmidt operator, it's compact. Then, with its self-adjointness, $(K^∗)^4$ is similar to a diagonalized infinite matrix D (by spectral theorem) and its trace is finite. So $K^∗$ looks like a diagonalized matrix $\sqrt[4]{D}.$ (Here, we use the self-adjointness of $K^*$ to ensure $(K^∗)^4$ is nonnegative). Then we see the operator $K^∗$ can be approximated by truncating $\sqrt[4]{D}$ to finite matrices.
I am appreciated to any discussion and hint.
Let $T$ be a bounded self adjoint operator on a Hilbert space (to avoid the notation $K^*$ which looks like the adjoint of the operator $K$) and suppose that $T^n$ is compact for some positive integer $n$ (such as $n=4$). Then $T$ is compact.
The reason is as follows: $T^{2n}$ is positive and compact and hence so is $(T^{2n})^{1/p}$ for every $p>0$. Moreover $$ T = \lim_{p\to\infty} (T^{2n})^{1/p}T, $$ So $T$ is also compact.