# Argue, why these operators are compact or not.

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.

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 $T.$
• $\exp (t^{n_k})$ are widely spaced... does this mean $T(C[0,1])$ is not relatively compact, hence $T$ is not compact? – Jonathan Krill Nov 29 '16 at 22:41
• It shows $T(B(0,1))$ is not relatively compact. Hence ... – zhw. Nov 30 '16 at 0:16
$(i)$ 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 $n$.
• OK, so $P_n(x)$ is a convergent sequence and therefore is cauchy, hence the operator is compact. right? Thank you already. – Jonathan Krill Nov 29 '16 at 19:52