A mean-value theorem for character sums
Let \(\lambda=(\lambda_ n)_{n\leq N}\) be a given finite sequence of complex numbers, and consider the mean value \[ S^*(\lambda)={1 \over \varphi(q)} \sum_{\chi\neq\chi_ 0} \bigl|\sum \lambda_ n \chi(n)\bigr|^ 2, \] where \(\chi\) runs over characters \(\pmod q\). The main result of the paper is an estimate for \(S^*(\lambda)\) if \(\lambda=\alpha* \beta* \gamma\) is a convolution of three sequences \(\alpha=(\alpha_ k)_{k\leq K}\), \(\beta=(\beta_ l)_{l\leq L}\), and \(\gamma=(\gamma_ m)_{m\leq M}\) with \(N=KLM\), \(\alpha_ k\) and \(\beta_ l\) arbitrary, and \(\gamma_ m=1\). Write generally \(\|\lambda\|=\sum_{(n,q)=1} |\lambda_ n|^ 2\). Then \[ S^*(\lambda) \ll \|\alpha\|\;\|\beta\|\;\|\gamma\|\bigl(1+q^{-3/4}(K+L)^{1/4} (KL)^{5/4}+q^{- 1}(KL)^{7/4}\bigr) q^ \varepsilon. \] The proof is surprisingly elementary, using only an estimate for the sum of a geometric progression. Fixing now a character \(\chi\) and putting \(\alpha_ k=\overline{\chi}(k)\), \(\beta_ l=\overline{\chi}(l)\), one obtains immediately an estimate for the character sum \(\sum_{m\leq M}\) on ignoring in \(S^*(\lambda)\) all the other terms except that corresponding to the specified character \(\chi\). For a suitable choice of \(K\) and \(L\), the resulting estimate is nontrivial for \(M\gg q^{5/11+\varepsilon}\); it is remarkable that the ``\(q^{1/2}\)- barrier'' is broken here. True, this has been done thirty years ago by D. A. Burgess, and the present argument fall short of superseding his results, but nevertheless two important aspects should be pointed out: firstly, the new method is elementary (unlike that of Burgess), and secondly, similar ideas apply even to character sums involving Fourier coefficients of cusp forms to yield quite new results; this variant of the method will be presented in a subsequent paper (see the following review). Authors' remark: In the middle of page 158 near the end of the proof of the Proposition, ``Then \(d_ 3d_ 4\Delta\mid l_ 1l_ 2\) so \dots'' should be replaced by ``Then \(d_ 3d_ 4\mid l_ 1l_ 2\) and \(\Delta\mid(d_ 3+d_ 4)l_ 1l_ 2\) so \dots''.
scientific article; zbMATH DE number 63019