Bounds for the moments of complex geometric sums

Question

Let $N,a,s,M$ be positive integers with $\gcd(a,N) = 1$. I'm wondering if a good upper bound is known for $$ S(s) = \sum_{k=0}^{N-1} \left| \sum_{m=0}^{M-1}e_N(mka)\right|^s, $$ where $e_N(x) := e^{2\pi i x/N}$. The cases with $s=1$ and $s=2$ are familiar: $S(1) = O(N \log M)$ and Parseval's identity gives $S(2) = MN$. For $s \geq 3$, I guess one could use induction together with Parseval to get $S(s) \leq M^{s-1}N$. But I wonder if one could do better than this for $s \geq 3$? Thanks.

Answer 1

Let's fix $s=4$ to get the idea: \begin{align*} S(4) &= \sum_{k=0}^{N-1} \bigg| \sum_{m=0}^{M-1} e_N(mka) \bigg|^4 \\ &= \sum_{k=0}^{N-1} \bigg( \sum_{m=0}^{M-1} e_N(mka) \bigg)^2 \bigg( \overline{\sum_{m=0}^{M-1} e_N(mka)} \bigg)^2 \\ &= \sum_{k=0}^{N-1} \bigg( \sum_{m=0}^{M-1} e_N(mka) \bigg)^2 \bigg( \sum_{m=0}^{M-1} e_N(-mka) \bigg)^2 \\ &= \sum_{k=0}^{N-1} \sum_{m_1=0}^{M-1} e_N(m_1ka) \sum_{m_2=0}^{M-1} e_N(m_2ka) \sum_{m_3=0}^{M-1} e_N(-m_3ka) \sum_{m_4=0}^{M-1} e_N(-m_4ka) \\ &= \sum_{m_1=0}^{M-1} \sum_{m_2=0}^{M-1} \sum_{m_3=0}^{M-1} \sum_{m_4=0}^{M-1} \sum_{k=0}^{N-1} e_N\big( (m_1+m_2-m_3-m_4)ka \big) \\ &= \sum_{m_1=0}^{M-1} \sum_{m_2=0}^{M-1} \sum_{m_3=0}^{M-1} \sum_{m_4=0}^{M-1} \begin{cases} N, &\text{if } (m_1+m_2-m_3-m_4)a \equiv 0\pmod N, \\ 0, & \text{otherwise} \end{cases} \\ &= \sum_{m_1=0}^{M-1} \sum_{m_2=0}^{M-1} \sum_{m_3=0}^{M-1} \sum_{m_4=0}^{M-1} \begin{cases} N, &\text{if } m_1+m_2 \equiv m_3+m_4 \pmod N, \\ 0, & \text{otherwise.} \end{cases} \end{align*} The counting problem for the number of nonvanishing terms is easy to understand, at least its order of magnitude: when $M\le N$ the order of magnitude is $M^3$, while when $M\ge N$ the order of magnitude is $M^4/N$. Therefore $$ S(4) = O\big( \max\{ M^3N, M^4\} \big), $$ and this can't be improved. (Note in particular that your evaluation $S(2) = MN$ holds only when $M\le N$.)

A similar argument show that $$ S(2k) = O\big( \max\{ M^{2k-1}N, M^{2k}\} \big) $$ for any positive integer $k$, and then $$ S(s) = O\big( \max\{ M^{s-1}N, M^s\} \big) $$ for any $s\ge2$ (even real $s$) by Hölder's inequality.