Let $H_1,H_2$ be Hilbert spaces and $T\in L(H_1,H_2)$. Prove TFA:
i) $T$ is compact
ii) $T^*T$ is compact
iii) $\lim_{n\to\infty}\Vert Tx_n\Vert=0$ for every sequence $(x_n)_n$ which converges weakly to zero.
I have shown that $i)\Leftrightarrow ii)$, but I do not see how to handle $i)\Leftrightarrow iii)$
$i)\to iii)$, Let $T$ be compact and $x_n$ converge weakly to zero. Then $\sup_n \Vert x_n\Vert$ is bounded. So $Tx_n$ contains a convergent subsequence, say ${Tx{_n}}_{k}$, then ${x_n}_k$ has to converge to zero, too. Then $T{x_n}_k$ has to converge to zero, too. But now?
Any help is welcome!
Best Answer
Hint Since $\{ T x_n \}$ is bounded, it is weakly compact.
If $Tx_n \not\to 0$ then you can find a subsequence which is weak-bounded away from zero. By compactness this contains a convergent subsequence, which you showed it converges to $0$.
$(iii) \to (i)$ If $B$ is the unit ball, then every sequence $y_n \in T(B)$ can be written as $y_n=Tx_n$.
Use the fact that $x_n$ has a weak- convergent subsequence $x_{k_n} \to z \in B$ and hence $x_{k_n}-z \to 0$.
Then by (iii) $T(x_{k_n}) \to T(z)$ in $(T(B), \| \, \|)$.