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By Raphael Salem
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Extra resources for Algebraic numbers and Fourier analysis (Heath mathematical monographs)
We have lim 'Yn(k) = 0 for = I n ~ 0, k=oo lim 'Yo(k) k= ~ o. 00 4. The derivative A£(X) exists for each x and each k, and is bounded (the bound may depend on k). Under these conditions, E is a set of uniqueness. 18 The Uniqueness of the Expansion in Trigonometric Series We shall first prove the following lemma. Let E be a closed set, X(x) a function vanishing for x E E and having an absolutely convergent Fourier series L 'Ynenix, and a bounded derivative X'(x). Let L cnenix be a trigonometric series converging to zero in every interval of the complementary set CE.
We fix u > small enough for P'(z) to have no zeros on the real axis in the interval ° o- u < z < 0 + u. , J1. being a positive number fixed as soon as (J' is fixed. If we take 0 real and I 0 I < u, P(O + 0) has the sign of 0 and is in absolute value not less than I 0 I J1.. , t We recall that we do not consider the number 1 as belonging to the class S (see Chapter I). t This proof, much shorter and simpler than the original one, has been communicated to me by Prof. Hirschman, during one of my lectures at the Sorbonne.
The situation is quite different iff is continuous, but singular. In this case "f(u) need not tend to zero, although there do exist singular functions for which "f(u) ~ 0 ([17J, and other examples in this chapter). The same remarks apply to the Fourier-Stieltjes coefficients cn • The problem which we shall solve here is the following one. Given a symmetrical perfect set with constant ratio of dissection ~, which we shall denote by E(~), we construct the Lebesgue function f connected with it, and we try to determine for what values of ~ the Fourier-Stieltjes transform (5) (or the Fourier-Stieltjes coefficient (4» tends or does not tend to zero as I u I (or I n J) increases infinitely.