Argue, why given Operators are compact or not.

[jonathan krill]

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

$$(i)\qquad Tx(t)=\sum^\infty_{k=1}x\left(\frac{1}{k}\right)\frac{t^k}{k!}$$ 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 $(T_nx)$ 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 are somehow related to $e^t$

sorry read the sticky to late, pls some moderator move this to HW forum.

 

[PeterDonis]

Moved to homework. @jonathan krill have you made any attempts at a solution? You give some ideas for showing compactness, but have you tried any of them? If so, please post what you obtained. (Note that in homework threads you are supposed to fill out the homework template; part of that is showing your attempts at a solution.)

 

[jonathan krill]

made progress showing (i) is compact the $T_nx$ are polynomials of finite rank, so for sequences in $C[0,1]$ $T_nx$ is bounded hence $T_nx$ contains a convergent subsequence. hence $T$ is compact.

for (ii) One probably can show that $T_nx$ is not cauchy using $x(t)=t^n$