$T$ is compact operator Hilbert space

Question

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.

Answer 1

Let $x_k\rightharpoonup x$, then for any fixed $n$ $T^nx_k\rightharpoonup T^nx$, in particular, for any fixed $n$ sequence $\{T^nx_k\}$ is bounded. Because $T^n$ is compact, we have $T^nx_k\to T^nx$. By self-ajointness and Cauchy-Schwarz we have $$|(T^{n-1}(x_k-x),T^{n-1}(x_k-x))|=|(T^{n}(x_k-x),T^{n-2}(x_k-x))|\leq$$$$\leq\|T^{n}(x_k-x)\|\|T^{n-2}(x_k-x)\|.$$ The first factor tends to zero, and the second is bounded, so $T^{n-1}x_k\to T^{n-1}x$. Continuing this procedure, we obtain that $Tx_k\to Tx$, which means $T$ is compact.

Answer 2

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 $$T^{2^k} \text{ compact} \implies T^{2^{k-1}} \text{ compact} \implies \cdots \implies T^2 \text{ compact} \implies T \text{ compact}.$$