TITLE: T compact if and only if $T^*T$ is compact.
QUESTION [12 upvotes]: 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.

REPLY [18 votes]: 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.