If $T^{2}$ is a compact operator then $T$ is compact

Question

Suppose $T$ is a bounded , self-adjoint operator on a Hilbert space such that $T^{2}$ is compact. Then prove that $T$ is compact. I proved it by continuous functional calculus but am looking for a solution which does not use continuous functional calculus. Thanks!

Answer 1

Suppose that $A$ is a bounded selfadjoint operator on a Hilbert space $\mathcal{H}$. Then, $$ \|Ax\|^{2}=(Ax,Ax)=(A^{2}x,x)\le \|A^{2}x\|\|x\|. $$ Let $\{ x_{n} \}$ be a bounded sequence in $\mathcal{H}$ which is bounded by $M$. Assuming that $A^{2}$ is compact gives a subsequence $\{ x_{n_{k}}\}$ such that $\{A^{2}x_{n_{k}}\}$ is convergent and, hence, Cauchy. Hence, $\{ Ax_{n_k}\}$ is also Cauchy because $$ \|Ax_{n_{k}}-Ax_{n_{j}}\|^{2} \le 2M\|A^{2}x_{n_{k}}-A^{2}x_{n_{j}}\|. $$ Therefore $\{ Ax_{n_{k}}\}$ also converges. So $A$ is compact if $A^{2}$ is compact.