1
$\begingroup$

Here discrepancy is from $(2.4)$ in https://www.ricam.oeaw.ac.at/files/people/siambook_nied.pdf given by 'The (extreme) discrepancy $D_N(P) = D_N(x_l,\dots,X_N)$ of the point set $P$ of $N$ points in $\mathbb Z^s$ is defined by $D_N(P) = D_N(\mathcal J; P)$, where $\mathcal J$ is the family of all subintervals of $\mathcal I^s$ of the form $\prod_{i=1}^s[ u_í, v_i)$ where $$D_N(\mathcal B; P) = \sup_{B\in\mathcal B}\Bigg|\frac{\sum_{i=1}^nc_{ B}(x_i)}N — \lambda_s(B)\Bigg|$$ with $c_{B}(x_i)$ being indicator function while $\lambda_s(B)=\lim_{N\rightarrow\infty}\frac1N\sum_{i=1}^Nc_{B}(x_i)$ with $\mathcal B$ being a nonempty family of Lebesgue-measurable subsets of $\overline{\mathcal I}^s$ ($\overline{\mathcal I}=[0,1]$ and ${\mathcal I}=[0,1)$). '

Given a vector $(v_1,\dots,v_n)\in\mathbb Z^n$ if $D$ is the discrepancy of the fractional parts $\big(\{\frac{mv_1}p\},\frac{mv_2}p\},\dots,\{\frac{mv_n}p\}\big)$ where $p$ is a prime and $m\in\{1,\dots,p\}$ then we know that we can find an $m$ such that $\big(\{\frac{mv_1}p\},\frac{mv_2}p\},\dots,\{\frac{mv_n}p\}\big)\in\mathcal I_1\times\dots\times\mathcal I_n$ where intervals $\mathcal I_i\subseteq(0,1)$ satisfy condition $\prod_{i=1}^n|\mathcal I_i|\geq D$ holds and in particular $$|\mathcal I_1|=\dots=|\mathcal I_n|=D^{1/n}+\epsilon$$ is possible at any $\epsilon>0$.

If $(v_1,\dots,v_n)=(a_1,b_1)\otimes(a_2,b_2)\otimes\dots\otimes(a_t,b_t)$ (note $n=2^t$) where each pair $a_i,b_j$ is coprime, each pair $a_i,a_j$ is coprime and each pair $b_i,b_j$ is coprime holds with $$p^{1/n}+1<a_i,b_j<2p^{1/n}$$ then is there an $m\in\mathbb Z$ such that $$|\mathcal I_1|=\dots=|\mathcal I_n|=D^{1/t}+\epsilon$$ is possible at any $\epsilon>0$ (even though $n=2^t$ we only have $t$ constituent vectors and each coordinate has only $t$ degrees of freedom for tensor product sequence)?

Note discrepancy of tensor product sequence is at most $p^{-t/n}$ (obtained from $5.12$ in https://www.ricam.oeaw.ac.at/files/people/siambook_nied.pdf by using $$\rho(V,p)=\min_{h:h\cdot V = 0 \bmod p} r(h)$$$$D_N(P)\rho(V,p)=O((\log N)^n)$$ where $V=(v_1,\dots,v_n)$ and $r(h)=\prod_{i=1}^n\max(1,|h_i|)$ (page $103$) and fact that tightness of Bombieri-Vaaler bound for this tensor sequence gives $\max(1,|h_i|)$ roughly $p^{-t/n(n-1)}$.

If $|\mathcal I_1|=\dots=|\mathcal I_n|=p^{-1/n}+\epsilon$ is possible then we can meet Dirichlet pigeonhole bound in Difference between Dirichlet Pigeonhole and Exponential sums bound in particular situation?.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.