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Let $T$ be a compact operator in Hilbert space. Let $(x_n)$ be a bounded sequence such that $(T T^*x_n)$ is Cauchy. Then $T^*x_n$ is also Cauchy.

I'm quite lost here. I know that $TT^*$ is also compact, so we can find some convergent subsequence $TT^*x_{n_k}$, but this does not help in showing that $T^*x_n$ is also Cauchy. I tried using the fact that $x_n$ is bounded so $x_n$ has a weakly convergent subsequence, however, I couldn't make much progress from this. How may I go about to show this? I would greatly appreciate any help.

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Note that: $$ ||T(x_n-x_m)||^2 = \langle T(x_n-x_m),T(x_n-x_m)\rangle \\= \langle (x_n-x_m),T^*T(x_n-x_m)\rangle \leq ||x_n-x_m|| \cdot ||T^*T(x_n-x_m)|| $$ Since $x_n$ are bounded, say $||x_n|| <M$, this gives that $||x_n-x_m||<2M$, and then we know that $||T^*T (x_n-x_m)||$ goes to zero anyway. Hence, it follows that $||T(x_n-x_m)|| \to 0$, or that $\{Tx_n\}$ is Cauchy.

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  • $\begingroup$ Of course, the converse is also true. $\endgroup$ Commented Dec 8, 2016 at 6:47

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