TITLE: $T$ is compact operator Hilbert space
QUESTION [5 upvotes]: I'm getting troubles to show the following
Let $X$ be an infinite-dimensional separable Hilbert space and $T$ a self-adjoint operator. Assume $T^n$ is compact for some $n \in \mathbb{N}$. Prove that $T$ is compact.
I was thinking about using the spectral theorem. However, I don't see this clearly. Any hint would be amazing.
Thank you.

REPLY [2 votes]: Assume that $T^2$ is compact. To show that $T$ is compact, let $x_n \rightharpoonup 0$ weakly and we claim that $Tx_n \to 0$ strongly.
Since $T^2$ is compact, we have $\|T^2x_n\| \to 0$. Therefore
$$\|Tx_n\|^2 = \langle Tx_n, Tx_n\rangle = \langle T^2x_n, x_n\rangle \le \|T^2x_n\|\|x_n\| \to 0$$
since $(x_n)_n$ is bounded. Hence $\|Tx_n\| \to 0$ so $T$ is compact.
Now, if $T^n$ is compact, let $k \in \Bbb{N}$ such that $n \le 2^k$. Then $T^{2^k}$ is also compact so
$$T^{2^k} \text{ compact} \implies T^{2^{k-1}} \text{ compact} \implies \cdots \implies T^2 \text{ compact} \implies T \text{ compact}.$$