# Weak convergence for compact operators

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

• 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 – user682522 Jun 27 '19 at 11:14
• @AndiGoldberger Check the edit, is it now clear? – N. S. Jun 27 '19 at 11:26
• Does one not have to consider closure$(T(B))$ instead of simply $T(B)$? But the argument should be the same I guess? – user682522 Jun 27 '19 at 11:42
• @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 $T(B)$). – N. S. Jun 27 '19 at 12:00
• I use that the closure of the image of the closed ball is compact – user682522 Jun 27 '19 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?