T compact if and only if $T^*T$ is compact.

Question

I have an operator $T \in B(\mathcal{H})$. I need to prove that T is comapct if and only if $T^*T$ is compact.

One way is ok, because if $A$ or $B$ is compact then $AB$ is compact, so I get at once that if $T$ is compact then $T^*T$ is compact.

But how do I go the other way? If I assume that $T^*T$ is compact I am not quite sure how to see that $T$ is compact. If I assume for contradiction that $T$ is not compact I must also have that $T^*$ is not compact. If I knew that either $T$ or $T^*$ was invertible it would be ok, because then I could find a bounded subsequence that did not converge. But when I do not have invertibility I am not quite sure how to proceed.

Answer 1

Assume that $S := T^\ast T$ is compact. By the spectral theorem for compact self-serving operators, this implies that $|T|=\sqrt{S}$ is also compact.

But the polar decomposition theorem yields $T=V \cdot |T|$ for some partial isometry $V$. Since the compact operators form an ideal in the space of bounded operators, we are done.

Answer 2

If $S = T^{\ast}T $ is compact, then so is $p(S)$ for any polynomial p such that $p (0) = 0$. By taking a norm limit of such expressions you can conclude that $\sqrt {S} = |T|$ is compact. Now write $T = V|T|$ by the polar decomposition. Hence $T $ is compact.

Answer 3

Let $\{x_n \}$ is a bounded sequence with bound $M$. If $T^{\star}T$ is compact, then there exists $\{ x_{n_{k}}\}$ such that $\{ T^{\star}Tx_{n_{k}}\}$ converges. Then \begin{align} \|Tx_{n_k}-Tx_{n_j}\|^2 & =(T^{\star}Tx_{n_{k}}-T^{\star}Tx_{n_j},x_{n_k}-x_{n_j}) \\ & \le 2M\|T^{\star}Tx_{n_k}-T^{\star}Tx_{n_j}\|. \end{align} This forces $\{Tx_{n_{k}}\}$ to be a Cauchy sequence and, hence, to converge.