Recent content by 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

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...