The series Tf|a(x), in this case, is the sum from i=0 to infinity of a-i-1(x-a)i = (x/a-1)i/a.
This is a geometric series that converges when |x/a-1|<1, meaning when x ∈ (0,2a).
Therefore, for any d we can pick a=d/2, u=0.
This solution, however, uses u=0, and the riddle explicitly asks for a positive u. An example with a positive u would be
f(x)=e-1/(x-1) for x>1 and f(x)=0 for x≤1.
In this function, Tf|a(x) equals f(x) in the range (1,2a-1) for a>1. Therefore, we can choose a=d/2+1 for any d.
(My thank-yous to Christian Blatter for pointing out to me that u=0 violates the wording of the riddle.)