1.8. IfO < a < 1 and J Ixla dF(x) < 00, then f(t) -1 = o(ltla ) as t -+ 0.

For 1 < a < 2 the same result is true under the additional assumption that J x dF(x) = 0. [HINT: The case 1 < a < 2 is harder. Consider the real and imaginary parts of f (t) -1 separately and write the latter as 1 sintxdF(x) +

1 sintxdF(x).

The second is bounded by (Itl/E)a ~Xl>f/ltllxla dF(x) = o(ltla) for fixed E. In the first integral use sin tx = tx + O(ltxI3), 1 fXdF(X)=tl xdF(x), Ixl:::f/ltl Ixl>f/ltl 1 Itxl3 dF(x) < E3-a 1 00 Itxl a dF(x).]
Ixl:::f/ltl -00