1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.