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$$(i)\qquad Tx(t)=\sum^\infty_{k=1}x\left(\frac{1}{k}\right)\frac{t^k}{k!}$$ 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 [itex](x_n)\subset B_1(0)[/itex], the sequence [itex](T_nx)[/itex] 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 [itex]T(C[0,1])[/itex] is equicontinuous, and then argue via arzela-ascoli to get compactness of the operator

- obviously they are somehow related to [itex]e^t[/itex]

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