1
$\begingroup$

Which of the operators $T:C[0,1]\rightarrow C[0,1]$ are compact? give argumentation.

$$(i)\qquad Tx(t)=\sum^\infty_{k=1}x\left(\frac{1}{k}\right)\frac{t^k}{k!}$$ and $$(ii)\qquad Tx(t)=\sum^\infty_{k=0}\frac{x(t^k)}{k!}$$

ideas for compactness of the operator:

  • the image of the closed unit ball is relatively compact under T.
  • for any sequence $(x_n)\subset B_1(0)$, the sequence $(Tx_n)$ contains a cauchy sequence.
  • show that the operator is not bounded, which is equivalent of it not being continuous which is necessary to be a compact operator
  • show $T(C[0,1])$ is equicontinuous, and then argue via arzela-ascoli to get compactness of the operator
  • obviously they ar e somehow related to $e^t$

i already solved the question for some T where the mapping is an integral, but i font get these two solved.

$\endgroup$
1
$\begingroup$

Hint for (ii): Note that if $x(t)=t^n,$ then $(Tx)(t) = \exp (t^n), n = 1,2,\dots$ If $n_1<n_2 < \cdots $ are spaced widely enough, then the $\exp (t^{n_k})$ should be widely spaced enough to prove the non-compactness of $T.$

$\endgroup$
  • $\begingroup$ $\exp (t^{n_k})$ are widely spaced... does this mean $T(C[0,1])$ is not relatively compact, hence $T$ is not compact? $\endgroup$ – Jonathan Krill Nov 29 '16 at 22:41
  • 1
    $\begingroup$ It shows $T(B(0,1))$ is not relatively compact. Hence ... $\endgroup$ – zhw. Nov 30 '16 at 0:16
2
$\begingroup$

$(i)$ is compact since it is a limit of finite dimensional operators. $T(x)=lim_nP_n$ where $P_n(x)=\sum_{i=1}^{i=n}x({1\over i}){t^i\over {i!}}$ and $p_n$ takes value in the space of polynomial of degree less than $n$.

$\endgroup$
  • $\begingroup$ OK, so $P_n(x)$ is a convergent sequence and therefore is cauchy, hence the operator is compact. right? Thank you already. $\endgroup$ – Jonathan Krill Nov 29 '16 at 19:52
  • 1
    $\begingroup$ @JonathanKrill I think the point is that all bounded linear operators of finite rank are compact, and any norm-limit of compact operators is compact $\endgroup$ – zhw. Nov 29 '16 at 21:33

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.