Answer to Question #8203 in Functional Analysis for jim

Question

Show that K(x-a) = O(k-a) [this is the letter O no the number 0] for any K.

Answer 1

K(x-a) = O(k-a) x->x0 <=> exists C>0 and exists eps>0: |K(x-a)|<|C(x-a)|, xε(x0-eps,x0+eps). So, let's assume C:=K+1 and eps:=1 for any K.