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!


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 $(T(B), \| \, \|)$.

  • $\begingroup$ thanks for that answer. I see how iii)->i) works, but could you elaborate on iii)->i) I do not see how this works out $\endgroup$ – user682522 Jun 27 at 11:14
  • $\begingroup$ @AndiGoldberger Check the edit, is it now clear? $\endgroup$ – N. S. Jun 27 at 11:26
  • $\begingroup$ Does one not have to consider closure$(T(B))$ instead of simply $T(B)$? But the argument should be the same I guess? $\endgroup$ – user682522 Jun 27 at 11:42
  • $\begingroup$ @AndiGoldberger Depends which definition you are using: the image of the open ball being pre-compact, or the image of the closed ball being compact... If it is the first, then you are right : you don't get $z \in B$, but the rest of the proof still works (excepting again that $T(z)$ may not be in $T(B)$). $\endgroup$ – N. S. Jun 27 at 12:00
  • $\begingroup$ I use that the closure of the image of the closed ball is compact $\endgroup$ – user682522 Jun 27 at 12:03

Suppose iii) is false. Then there is a subseqeunce $(n_k)$ and $a >0$ such that $\|Tx_{n_k}\| \geq a$ for all $k$. Now $x_{n_k} \to 0$ weakly which implies $T(x_{n_k}) \to 0$ weakly. [Any norm-norm continuous linear map is weak-weak continuous]. But i) implies that some subsequence of $(T(x_{n_k}))$ converges in the norm (because this sequence lies inside a compact set). The limit cannot be anything other than $0$ (by weak convergence). Do you see a contradiction now?


Your Answer

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