关于破弧上外尔和的积分中值定理
An Integral Mean Value Theorem for Weyl Sums over Broken Arcs
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中文总结 AI 辅助
本文针对破弧上的外尔和，通过丢番图逼近、Vinogradov主值定理及改进的外尔差分法，突破幂的奇偶限制得到积分中值的新估计，完成相关证明。
中文摘要 AI 辅助
本文研究指数和 $S(\alpha)=\sum_{u\in I}e(\alpha u^k)$ 的积分中值，其中 $I$ 是长度为 $2N^\theta$ 的短区间，$\theta<1$；破弧 $\mathfrak{m^*}$ 是满足以下条件的次弧 $\mathfrak{m}$ 的子集：$\mathfrak{m}=\bigcap_{j\leq k-1}\left\{\alpha:\forall q<(\log N)^A,h<q,(h,q)=1,\Big|\alpha-\frac{h}{q}\Big|>\frac{1}{qN^{(k-j-1/2)\theta}}\right\}$，且 $\mathfrak{m}$ 的测度至少为 $c>0$。设 $m$ 为足够大的数，$N$ 相对于 $m$ 足够大，当 $k>\log m$ 且 $\frac{\log k}{\log m}<1/2$ 时，我们得到估计式：$\int_{\mathfrak{m^*}}\bigg|\sum\limits_{{N_1}<u<{N_2}}{e(zu^k)}\bigg|^{m}\mathrm{d}z\ll_cN^{\theta m(\frac{2k+1}{2k+2}+o(1))}$。该估计突破了幂的奇偶限制。为得到该界，我们首先通过丢番图逼近和Vinogradov主值定理对几乎所有 $\alpha$ 建立强估计，再结合该结果将差分步划分为大范围与小范围，构造出比经典方法更优的精细外尔差分法，利用概率版本计算每个和的重数，最终完成证明。
英文摘要
In this article, we research the mean value of integral of exponential sum $S(α)=\sum_{u\in I}e(αu^k)$, where $I$ is a short interval whose length is $2N^θ,θ<1$, and the broken arc $\mathfrak{m^*}$ is a subset of a following minor arcs \[ \mathfrak{m}=\bigcap_{j\leq k-1}\left\{α:\forall q<(\log N)^A,h<q,(h,q)=1,\Big|α-\frac{h}{q}\Big|>\frac{1}{qN^{(k-j-1/2)θ}}\right\} \] which has measure at least $c>0$. By setting $m$ is a sufficiently large number, $N$ is be sufficiently large in terms of $m$. When $k>\log m$ and $\frac{\log k}{\log m}<1/2$ we set the following estimate: \[ \int_{\mathfrak{m^*}}\bigg|\sum\limits_{{N_1}<u<{N_2}}{e(zu^k)}\bigg|^{m}\mathrm{d}z\ll_cN^{θm(\frac{2k+1}{2k+2}+o(1))} \] We can find this estimate moving beyond the even-odd restriction of powers. To get this bound, we first set a strong estimate for almost $α$ by Diophantine approximation and Vinogradov's main value theorem. Then, combining this result, we construct a refined Weyl differencing argument by partitioning the differences step into large and small range, which significantly outperforms the classical one. By the version of probability, we can calculate the multiplicity of each sum. Put them together and we can complete the proof.
