Is operator $T_n$ a compact operator?

$$T_n:l_2\rightarrow l_2$$

$$T_nx=(\underbrace{0,0,\ldots,0,}_{n\text{ zeros}}, x_1,x_2,x_3,\ldots)\text{ where }x=(x_1,x_2,x_3,\ldots)\in l_2,\ \sum^\infty_{k=1} |x_k|^2\lt\infty$$

could you please help.

  • 1
    $\begingroup$ Well, to start, what is the definition of compact that you have learned? $\endgroup$ – Christopher A. Wong May 28 '14 at 21:53

This operator is not compact. Recall that any compact operator takes a bounded sequence to another sequences that has a convergent subsequence. If $\{ e_n \}_{n =1}^\infty$ is an orthonormal basis, then $T_n$ takes this basis to another orthonormal set, which clearly does not have a limit point. So $T_n$ is not compact.


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.