text:
\begin{document}
	\title{On Bounded Remainder Sets and \\ Strongly Non-Bounded Remainder Sets for Sequences $(\{a_n\alpha\})_{n\geq 1}$ }
	\author{Lisa Kaltenböck\footnote{The author is supported by the Austrian Science Fund (FWF), Project F5507-N26, which is a part of the Special Research Program Quasi-Monte Carlo Methods: Theory and Applications.}, Gerhard Larcher\footnote{The author is supported by the Austrian Science Fund (FWF), Project F5507-N26, which is a part of the Special Research Program Quasi-Monte Carlo Methods: Theory and Applications and Project I1751-N26.}}

	\date{\vspace{-5ex}}
	\maketitle

\begin{abstract}
	We give some results on the existence of bounded remainder sets (BRS) for sequences of the form $(\{a_n\alpha\})_{n\geq 1}$, where $(a_n)_{n\geq 1}$ - in most cases - is a given sequence of distinct integers. Further we introduce the concept of strongly non-bounded remainder sets (S-NBRS) and we show for a very general class of polynomial-type sequences that these sequences cannot have any S-NBRS, whereas for the sequence $(\{2^n\alpha\})_{n \geq 1}$ every interval is an S-NBRS.
\end{abstract}

\section{Introduction and Statement of Results}
Throughout this paper we will use $\{x\}$ to denote the fractional part of a number $x$ and $\|x\| = \min(\{x\},1-\{x\})$ to denote the difference between $x$ and the nearest integer.
\begin{definition}
	\label{def:ud}
	An infinite sequence $(x_n)_{n \geq 1}$ in $[0,1)$ is said to be uniformly distributed modulo 1 if for every interval $[a,b) \subseteq [0,1)$ we have
	\begin{equation*}
	\lim\limits_{N \rightarrow \infty}\frac{\#\{1 \leq n \leq N: x_n \in [a,b)\}}{N} = b-a.
	\end{equation*}
\end{definition}
A most common measure for the quality of the uniform distribution of a sequence is the discrepancy $D_N$ of the sequence:

\begin{definition}
	For a sequence $(x_n)_{n \geq 1}$ in $[0,1)$ and any subinterval $[a,b) \subseteq [0,1)$, let
	\begin{equation*}
	D_N([a,b)) = \frac{1}{N}\left(\#\{1 \leq n \leq N: x_n \in [a,b)\} - N(b-a)\right)
	\end{equation*}
	be the ($N$-th) discrepancy function of $[a,b)$.

	Then, the ($N$-th) discrepancy of $(x_n)$ is given by
	\begin{equation*}
		D_N = \sup_{[a,b)} |D_N([a,b))|.
	\end{equation*}
\end{definition}

A famous result of W. M. Schmidt \cite{SchmidtVII} gives the following general lower bound for the discrepancy of an arbitrary sequence in $[0,1)$:
\begin{thmm}[Schmidt]
	There exists a constant $c > 0$ such that for any infinite sequence $\xn$ in $\ui$ it holds that
	\begin{equation*}
	D_N \geq c \frac{\log N}{N}
	\end{equation*}
	for infinitely many $N$.
\end{thmm}
This result, however, does not contain any lower bound for the absolute value $|D_N([a,b))|$ of the discrepancy function of a concrete fixed interval $[a,b)$. Such a - essentially best possible - bound was given by Tijdeman and Wagner in \cite{Tijdeman1980}.
\begin{thmm}[Tijdeman, Wagner]
	For every sequence $\xn$ we have for almost all $b \in \ui$ that
	\begin{equation*}
	|D_N([0,b))| \geq \frac{1}{400}\frac{\log N}{N}
	\end{equation*}
	for infinitely many $N$.
\end{thmm}
The other extremal case are sets $[a,b)$ of bounded remainder.
\begin{definition}
	Let $(x_n)_{n\geq 1}$ be a sequence in $[0,1)$ and let $[a,b) \subseteq [0,1)$. $[a,b)$ is called bounded remainder set (BRS) if there exists some constant $c$ such that
	\begin{equation*}
	|D_N([a,b))| \leq c \frac{1}{N}
	\end{equation*}
	for all $N \in \N$.
\end{definition}
Note that for all $\ab$, $|D_N([a,b))| \geq \frac{1}{4N}$ for infinitely many $N$.
It was shown by Schmidt \cite{SchmidtVI} that for every sequence $(x_n)_{n \geq 1}$ in $[0,1)$ the set of intervals $[0,b)$ which are BRS is countable.

For some of the most well-known and commonly studied sequences $(x_n)_{n \geq 1}$ in $[0,1)$ the bounded remainder sets are explicitly known. For example for the one-dimensional Kronecker sequence $(\{n\alpha\})_{n \geq 1}$, where $\alpha$ is a fixed irrational, by Kesten \cite{Kesten1966} the following was shown:
\begin{thmm}[Kesten]
	\label{thm:Kesten}
	An interval $\ab \subseteq \ui$ is a BRS for the sequence $\na$ if and only if
	\begin{equation*}
	b - a = \{j\alpha\}
	\end{equation*}
	for some $j \in \Z$.
\end{thmm}
Kesten has given an elementary proof. Proofs and generalizations of this result based on ergodic theory and topological dynamics were given for example by Furstenberg, Keynes and Shapiro \cite{Furstenberg1973}, Petersen \cite{Petersen1973} and Oren \cite{Oren1982}.

Theorem \ref{thm:Kesten} shows that the collection of BRS of $\na$ is "dense" in $\ui$, meaning that for all $\varepsilon > 0$ and all $x,y \in \ui$ there is a BRS $\ab$ with $|a - x| < \varepsilon$ and $|b - y| < \varepsilon$. We say: \textit{the sequence has a dense collection of BRS}.

A deep result based on the techniques from ergodic theory was given by Liardet \cite{Liardet1987} in the case of "polynomial Kronecker sequences" of higher degree:
\begin{thmm}[Liardet]
	\label{thm:Liardet}
	Let $p(x) \in \R[x]$ of degree $d \geq 2$ and with irrational leading coefficient. Then the only BRS of the sequence $(\{p(n)\})\lc$ are the empty set and $\ui$.
\end{thmm}
So for example the sequence $(\{n^2\alpha\})_{n \geq 1}$ or more generally a sequence $(\{f(n)\alpha\})_{n\geq 1}$ with $f(x) \in \Z[x]$ of degree $d \geq 2$, does not have any non-trivial BRS. To our best knowledge until now there is no elementary proof of this fact.
A further nice result was given by Tijdeman and Voorhoeve in \cite{Tijdeman1981}, where they in some sense classify a collection $S$ of intervals in $\ui$ such that there exists a sequence $\xn$ having all elements of $S$ as BRS.
The notion of BRS was also studied in higher dimensions (in general and for concrete sequences) and for continuous motions for example in \cite{Hellekalek1984}, \cite{Grepstad2014}, \cite{Liardet1987}, \cite{Rauzy}, \cite{Ferenczi} or \cite{GrepstadLarcher}.

Our investigations of this paper started when we were asked whether the sequence $\xn = (\{2^n\alpha\})\lc$ has non-trivial BRS or not.
This question can be answered very easily even in the more general case of Pisot numbers $\beta$ as bases.
Here, we restrict ourselves to $\alpha$ whose $\beta$-expansion is normal to base $\beta$ and thus for which can be guaranteed that $(\{\beta^n \alpha\})_{n \geq 1}$ is uniformly distributed, as was shown by Bertrand-Mathis \cite{Bertrand}. For the special case that $\beta > 1$ is an integer, the normality of $\beta$ and the uniform distribution of $(\{\beta^n \alpha\})_{n \geq 1}$ are equivalent.
\begin{thm}
	\label{thm:brs_normal_alpha}
	Let $\beta > 1$ be a Pisot number and let $\alpha$ be normal to base $\beta$. Then the sequence $(\{\beta^n \alpha\})_{n \geq 1}$ does not have non-trivial bounded remainder sets.
\end{thm}
As will be seen from the proof of this theorem in Section \ref{sec:proofs_1}, for these sequences all intervals $[a,b) \subset \ui$ with $a \neq 0$ and $b \neq 1$ (if $\beta$ is an integer this restriction can be omitted) even are non-bounded remainder sets in a very strong sense.
Namely, all of these intervals are \textit{strongly non-bounded remainder sets}:
\begin{definition}
	Let $\xn$ be a sequence in $\ui$. An interval $\ab \subset \ui$ is said to be a \textit{strongly non-bounded remainder set (S-NBRS)} if for all $K \in \N$ there are $K$ successive elements of $\xn$ which all are contained in $\ab$ or which all are not contained in $\ab$.
\end{definition}
Of course the concepts of BRS and S-NBRS can be defined quite analogously in higher dimensions.
It is clear that an S-NBRS is not a BRS, Lemma \ref{lma:consecutive_elements} relates the two notions. Earlier we have asked for an elementary proof of the result of Liardet (Theorem \ref{thm:Liardet}), that certain polynomial sequences do not have (non-trivial) BRS. A first idea to approach such a proof would be to try to proceed like in the proof of Theorem \ref{thm:brs_normal_alpha}, i.e., to try to show, that for such sequences any interval is an S-NBRS, like in the case of the sequence $(\{\beta^n\alpha\})\lc$. In Theorem \ref{thm:sn-brs_polynomial} we show in a very general setting that such an approach has to fail.
\begin{thm}
	\label{thm:sn-brs_polynomial}
	Let $\bm{p}(n) = (p^{(1)}(n), p^{(2)}(n), \dots, p^{(s)}(n))$, where all $p^{(i)}(n)$, $i \leq s$ are real polynomials with the property, that for each point $\bm{h} \in \Z^s, \bm{h} \neq 0$, the polynomial $\langle \bm{h}, \bm{p}(n) \rangle$ has at least one non-constant term with irrational coefficient. Then the sequence $(\{\bm{p}(n)\})\lc$ does not have any s-dimensional S-NBRS.
\end{thm}

Let us now consider BRS for sequences of the form $(\{a_n\alpha\})\lc$ where $\an$ is a given strictly increasing sequence of integers, and where $\alpha$ again is an irrational number. Distribution properties of such sequences from different points of view (uniform distribution, discrepancy, law of iterated logarithm, Poissonian pair correlation) were studied in a multitude of papers (see for example for a survey \cite{Drmota:Sequences_Discrepancies_Applications}, or for more recent references \cite{aistleitner_larcher_2016} or \cite{Aistleitner2017}).

We already know: if $a_n = n$, then we have a dense collection of BRS, if $a_n = f(n)$ with $f(x) \in \Z[x]$ of degree $d \geq 2$, or if $a_n = \beta^n$ for some Pisot number $\beta$, then we do not have any BRS.
In a first impression one could think that the existence of BRS for sequences $(\{a_n\alpha\})\lc$ relies on a "not too strong" growth rate of the sequence $\an$. For example one could have the idea, that sequences $(\{a_n\alpha\})\lc$ with $\an$ lacunary never can have BRS. This indeed, of course, is wrong. A very simple counterexample is the sequence $(\{a_n\alpha\})\lc$ with $a_n = n + q_n$ with $q_n$ a best approximation denominator of $\alpha$. Since $\|q_n\|$, i.e. the distance of $q_n$ to the nearest integer, always is very small, the sequence $(\{a_n\alpha\})\lc$ behaves quite similar to the sequence $\na$ and from this observation we can deduce that $(\{a_n\alpha\})\lc$ has a dense collection of bounded remainder sets. However, in general, $(\{a_n\alpha\})\lc$ and $\na$ do not have the same collection of BRS.
\begin{thm}
	\label{thm:lacunary_brs}
	\begin{enumerate}
		\item Let $\alpha$ be irrational and let $q_n$ be the denominator of the $n$-th convergent from the continued fraction expansion of $\alpha$. Then every interval of the form $[\{k_1\alpha\},\{k_2\alpha\})$ with $k_1,k_2 \in \N$ and $\{k_1\alpha\} < \{k_2\alpha\}$ is a BRS for the uniformly distributed sequence $(\{(q_n+n)\alpha\})\lc$.
		\item In general, the sequence $(\{(q_n+n)\alpha\})\lc$ does not have the same collection of BRS as the sequence $\na$.
	\end{enumerate}
\end{thm}
Of course BRS for a sequence $(\{a_n\alpha\})\lc$ for an (ultimately) increasing $\an$ only can exist if $a_n$ grows at least linearly. This is a simple fact but for the sake of completeness we formulate it as a theorem:
\begin{thm}
	\label{thm:slow_growth_rate}
	Let $\an$ be an increasing sequence of integers and let $\ana$ be uniformly distributed. If
	\begin{equation*}
	\lim_{n\rightarrow \infty}\frac{a_n}{n} = 0,
	\end{equation*}
	then every interval $\ab \subset \ui$ of positive measure is an S-NBRS.
\end{thm}
For every other growth rate, however, it is possible to achieve even a dense collection of BRS.
\begin{thm}
	\label{thm:arbitrary_growth_rate_Brs}
	Let $\varphi(n)$ be any strictly increasing sequence with $\varphi(n) \geq 2n$ for all $n \in \N$. Then for all irrational $\alpha$ with bounded continued fraction coefficients there is a constant $L$ and a strictly increasing sequence of integers $\an$, with $\varphi(n) \leq a_n \leq L\varphi(n)$ for all $n \geq 1$ such that $\ana$ is uniformly distributed and such that $\ana$ has a dense collection of BRS.
\end{thm}
In Theorem \ref{thm:lacunary_brs} and in Theorem \ref{thm:arbitrary_growth_rate_Brs} we have given examples of sequences $\ana$ with dense collections of BRS for always one choice of $\alpha$. But it is also possible for any countable set of irrationals $\alpha$ to give an integer sequence $\an$ such that $\ana$ has a dense collection of BRS for all of these $\alpha$.
\begin{thm}
	\label{thm:countably_many_alpha}
	Let $\mathcal{A}$ be a countable set of irrational numbers. Then there is a lacunary sequence $\an$ of integers such that $\ana$ is uniformly distributed and such that $\ana$ has a dense collection of BRS for all $\alpha \in \mathcal{A}$.
\end{thm}
In Section \ref{sec:proofs_1} we will give the proofs of the Theorems \ref{thm:brs_normal_alpha}, \ref{thm:lacunary_brs}, \ref{thm:slow_growth_rate}, \ref{thm:arbitrary_growth_rate_Brs}, and \ref{thm:countably_many_alpha} and Section \ref{sec:proof_2} is dedicated to the proof of Theorem \ref{thm:sn-brs_polynomial}.

\section{Proofs of Theorems \ref{thm:brs_normal_alpha}, \ref{thm:lacunary_brs}, \ref{thm:slow_growth_rate}, \ref{thm:arbitrary_growth_rate_Brs}, and \ref{thm:countably_many_alpha}}
\label{sec:proofs_1}

The first lemma characterizes the relation between bounded remainder sets and \linebreak
S-NBRS and will be used to prove Theorem \ref{thm:brs_normal_alpha}.

\begin{lemma}
	\label{lma:consecutive_elements}
	Let $(x_n)\lc$ be a sequence in $[0,1)$.
	\begin{enumerate}
		\item If the interval $[a,b) \subset [0,1)$ is an S-NBRS, then $[a,b)$ is not a BRS.
		\item If $(x_n)\lc$ has a dense collection of BRS, then there do not exist any S-NBRS.
	\end{enumerate}
\end{lemma}

\begin{proof}
	For the proof of the first part consider intervals $[a,b) \subset [0,1)$ which are bounded remainder sets for an arbitrary sequence $(x_n)\lc$. By definition, there exists a constant $c$ independent of $N$ such that
	\begin{equation*}
	|\#\{1 \leq n \leq N: x_n \in [a,b)\} - (b-a) N| \leq c
	\end{equation*}
	for all $N \in \N$.
	Let $k \in \N, k > \frac{2c}{1-b+a}$ and first suppose that there exists an $N$ such that
	\begin{equation*}
	\#\{N < n \leq N+k: x_n \in [a,b)\} = k.
	\end{equation*}
	Considering this particular $N$ we have that
	\begin{align*}
	\#\{1 \leq n \leq N+k&: x_n \in [a,b)\} - (b-a)(N+k) \\
	&= \#\{1 \leq n \leq N: x_n \in [a,b)\}\\
	&\quad + \#\{N + 1 \leq n \leq N+k: x_n \in [a,b)\}
	- (b-a)(N+k) \\
	&\geq -c + k -(b-a)k \\
	&> c,
	\end{align*}
	which is a contradiction.
	On the other hand, if some $k > \frac{2c}{b-a}$ successive elements are not contained in $\ab$ we have
	\begin{align*}
	\#\{1 \leq n \leq N+k&: x_n \in [a,b)\} - (b-a)(N+k) \\
	&= \#\{1 \leq n \leq N: x_n \in [a,b)\}	- (b-a)(N+k) \\
	&\leq c -(b-a)k < -c,
	\end{align*}
	Therefore, an S-NBRS cannot be a BRS.

	For the second part of the lemma let $(x_n)\lc$ be a sequence with a dense collection of bounded remainder sets and consider $[a,b) \subset [0,1)$.
	Then there exists an interval $[\underline{a},\underline{b}) \subseteq [a,b)$ which is a BRS for $(x_n)\lc$. Since there exists $K$ such that at most $K$ consecutive elements of $(x_n)\lc$ are not contained in $[\underline{a},\underline{b})$ this also holds for the interval $\ab$.
	Moreover, if $a \neq 0$ and $b \neq 1$ there also exists $[\bar{a},\bar{b}) \supset \ab$, which is a BRS and thus the number of consecutive elements of $\xn$ in $[\bar{a},\bar{b})$ as well as in $\ab$ is bounded.
	In case that $a = 0$ or $b = 1$ we consider the complement of $\ab$. Since the number of successive elements of $\xn$ which are not contained in $\ui \setminus \ab$ is bounded, so is the number of successive elements which are contained in $\ab$.
	Therefore, no interval $[a,b)$ can be an S-NBRS.
\end{proof}

\begin{refproof}[Proof of Theorem \ref{thm:brs_normal_alpha}.]
	We exclude the unit interval $[0,1)$ from our considerations because it trivially fulfills the properties of a bounded remainder set.

	Let $\beta > 1$ be a Pisot number and let $\alpha \in \ui$ with expansion
	\begin{equation*}
	\alpha = \sum_{i = 1}^{\infty} \alpha_i\beta^{-i},
	\end{equation*}
	where $\alpha \in \{0,1, \dots, \lceil\beta\rceil -1\}$ and such that $\sum_{i > j} \alpha_i\beta^{-i} < \beta^{-j}$ for all $j \geq 0$. Moreover, we choose $\alpha$ to be normal to base $\beta$.
	At first, consider intervals of the form $\ab$, where $a \neq 0$ and $b \neq 1$ and define $\varepsilon := \min\left(\frac{\|a\|}{2}, \frac{\|b\|}{2}\right)$.

	Since $\beta$ is a Pisot number, $\beta^n$ is a good approximation of an integer, i.e. $\|\beta^n\|$ converges to zero at an exponential rate. Therefore, $\sum_{n = 0}^{\infty}\|\beta^n\| < c$ for some constant $c$ and by $m$ we denote the first index such that
	\begin{equation*}
	\sum_{n = m}^{\infty} \|\beta^n\| < \frac{1}{\lceil\beta\rceil -1} \varepsilon.
	\end{equation*}

	Furthermore, consider the $\beta$-expansion of $\varepsilon$, i.e.
	\begin{equation*}
	\varepsilon = \sum_{i = 1}^{\infty}\varepsilon_i\beta^{-i},
	\end{equation*}
	with coefficients $\varepsilon_i \in \{0,1, \dots,\lceil\beta\rceil -1\}$. Let $j$ denote the index of the first coefficient for which $\varepsilon_j \neq 0$.
	A sufficient condition for  $\sum_{i = n+1}^{\infty}\alpha_i\beta^{n-i} < \varepsilon$ is then given by $\alpha_{n+1} = \dots = \alpha_{n+j} = 0$.

	From the fact that $\alpha$ is normal to base $\beta$, i.e., every admissible finite block $d$ of integers $d_i \in \{0,1, \dots, \lceil\beta\rceil -1\}$ occurs with the right frequency in the $\beta$-expansion of $\alpha$, the occurrence of arbitrary long blocks consisting only of zeros in the base $\beta$ expansion of $\alpha$ follows. In other words, for all $k \in \N$ there exists an index $N$ for which it holds that

	\begin{equation*}
	\alpha_{N+1} = \alpha_{N+2} = \dots = \alpha_{N + m + k + j} = 0.
	\end{equation*}
	For the elements $\{\beta^n\alpha\}$ with $N + m < n \leq N + m + k$ it holds that $\alpha_{n-i} = 0$ for $i = 0, 1, \dots, m - 1$ which implies that
	\begin{equation*}
	\|\sum_{i = 0}^{n-1}\alpha_{n-i}\beta^{i}\| = \|\sum_{i = m}^{n-1}\alpha_{n-i}\beta^{i}\|
	\leq \sum_{i = m}^{n-1}\alpha_{n-i}\|\beta^{i}\| < \varepsilon.
	\end{equation*}
	Additionally, we have $\alpha_i = 0$ for $i = n +1, \dots, n + j$ and $N + m < n \leq N + m + k$. Hence,
	\begin{equation*}
	\sum_{i = n+1}^{\infty}\alpha_i\beta^{n-i} < \varepsilon.
	\end{equation*}
	Since $\varepsilon < 1/4$ it holds that
	\begin{align*}
	\|\beta^n\alpha\|
	&= \|\sum_{i = 0}^{n-1}\alpha_{n-i}\beta^{i} + \sum_{i = n+1}^{\infty}\alpha_i\beta^{n-i}\| \\
	&\leq \|\sum_{i = 0}^{n-1}\alpha_{n-i}\beta^{i}\| + \sum_{i = n+1}^{\infty}\alpha_i\beta^{n-i} \\
	&< 2\varepsilon = \min(\|a\|,\|b\|)
	\end{align*}
	for $N + m < n \leq N + m + k$.
	Therefore, for any $k \in \N$ there exist $k$ consecutive elements $\{\beta^n\alpha\}$ which are not contained in the interval $\ab$, which therefore is an S-NBRS.

	Note that if $\beta$ is an integer it holds that $\{\sum_{i = 0}^{n-1}\alpha_{n-i}\beta^{i}\} = 0$ and
	\begin{equation*}
	\{\beta^n\alpha\} = \sum_{i = n+1}^{\infty}\alpha_i\beta^{n-i}.
	\end{equation*}
	Thus, the occurrence of arbitrary long blocks of zeros implies the existence of arbitrarily many successive elements which are contained in some interval $[0,a)$. It follows that in this case even intervals of the form $[0,a)$ or $[b,1)$ are S-NBRS.

	Nevertheless, for arbitrary base $\beta$ it remains to show that intervals $[0,a)$ or $[b,1)$ cannot be BRS as well. From the previous considerations we know that the number of consecutive elements of $(\{\beta^n\alpha\})\lc$ which are not contained in some interval $\ab$ with $0 < a < b <1$ is unbounded, i.e. for any  $k$ there exists an $N$ such that $\{\beta^n\alpha\} \in [0,a)\cup[b,1)$ for all $N < n \leq N + k$. Now consider a strictly increasing sequence of such integers $(k_i)$ with corresponding indices $N_i$ and let $\varepsilon = \frac{b-a}{3}$.

	Then, let
	\begin{equation*}
	A_ik_i = \#\{N_i < n \leq N_i + k_i : \{\beta^n\alpha\} \in [0,a)\}
	\end{equation*}
	and
	\begin{equation*}
	B_ik_i = \#\{N_i < n \leq N_i + k_i : \{\beta^n\alpha\} \in [b,1)\}.
	\end{equation*}
	Since $A_i + B_i = 1$ and $1 - (b-a) + 2\varepsilon < 1$ it either holds that
	\begin{equation*}
	A_i > a + \varepsilon \qquad \text{for infinitely many i},
	\end{equation*}
	or
	\begin{equation*}
	B_i > 1-b + \varepsilon \qquad \text{for infinitely many i}.
	\end{equation*}
	W.l.o.g. let us assume the first case and suppose that there exists a constant $c$ such that
	\begin{equation*}
	ND_N([0,a)) = |\#\{1 \leq n \leq N: \{\beta^n\alpha\} \in [0,a)\} - N a| \leq c
	\end{equation*}
	for all $N \in \N$. We use this property for $N_i$ and obtain
	\begin{align*}
	(N_i + k_i)D_{N_i + k_i}([0,a)) &= \#\{1 \leq n \leq N_i + k_i : \{\beta^n\alpha\} \in [0,a)\} - (N_i+k_i)a\\
	&\geq A_i k_i - a k_i -c \\
	&> \varepsilon k_i -c.
	\end{align*}
	Since $\varepsilon k_i$ is unbounded, so is $(N_i + k_i)D_{N_i + k_i}([0,a))$, and $[0,a)$ as well as $[a,1)$ therefore cannot be BRS.
\end{refproof}

The next lemma will be of use in the proofs of Theorem \ref{thm:lacunary_brs}, \ref{thm:arbitrary_growth_rate_Brs} and \ref{thm:countably_many_alpha}.
\begin{lemma}
	\label{lma:dcbrs_ud}
	Let $(x_n)_{n \geq 1}$ be a sequence in $[0,1)$ which has a dense collection of bounded remainder sets. Then $(x_n)_{n \geq 1}$ is uniformly distributed modulo 1.
\end{lemma}

\begin{proof}
	Let $a,b \in (0,1)$. Since $(x_n)_{n \geq 1}$ has a dense collection of bounded remainder sets, for every $\varepsilon > 0$ there exist $\underline{a}, \bar{a}, \underline{b}, \bar{b}\in (0,1)$ such that
	\begin{align*}
	\underline{a} \leq a \leq \bar{a}, \quad  a - \underline{a} &\leq \varepsilon, \quad \bar{a} - a \leq \varepsilon, \\
	\underline{b} \leq b \leq \bar{b}, \quad  b - \underline{b} &\leq \varepsilon, \quad \bar{b} - b \leq \varepsilon,
	\end{align*}
	and $[\underline{a}, \bar{b})$, $[\bar{a},\underline{b})$ are bounded remainder sets.
	From
	\begin{align*}\\
	\#\{1 \leq n \leq N: x_n \in [\underline{a},\bar{b})\}
	&\geq \#\{1 \leq n \leq N: x_n \in [a,b)\} \\
	&\geq \#\{1 \leq n \leq N: x_n \in [\bar{a},\underline{b})\}
	\end{align*}
	for all $N \in \N$, it follows that
	\begin{align*}
	\lim\limits_{N \rightarrow \infty} \frac{\#\{1 \leq n \leq N: x_n \in [\underline{a},\bar{b})\}}{N}
	&\geq \lim\limits_{N \rightarrow \infty} \frac{\#\{1 \leq n \leq N: x_n \in [a,b)\}}{N} \\
	&\geq \lim\limits_{N \rightarrow \infty} \frac{\#\{1 \leq n \leq N: x_n \in [\bar{a},\underline{b})\}}{N}.
	\end{align*}
	By the definition of bounded remainder sets it holds that
	\begin{equation*}
	\lim\limits_{N \rightarrow \infty} \frac{\#\{1 \leq n \leq N: x_n \in [\underline{a},\bar{b})\}}{N} = \bar{b} - \underline{a} \leq b - a + 2\varepsilon,
	\end{equation*}
	and analogously
	\begin{equation*}
	\lim\limits_{N \rightarrow \infty} \frac{\#\{1 \leq n \leq N: x_n \in [\bar{a},\underline{b})\}}{N} = \underline{b} - \bar{a} \geq b - a - 2\varepsilon.
	\end{equation*}
	Since $\varepsilon$ was arbitrary small, we have that
	\begin{equation*}
	\lim\limits_{N \rightarrow \infty} \frac{\#\{1 \leq n \leq N: x_n \in [a,b)\}}{N} = b - a,
	\end{equation*}
	and therefore $(x_n)_{n \geq 1}$ is uniformly distributed.
\end{proof}

\begin{refproof}[Proof of Theorem \ref{thm:lacunary_brs}.]
	For the proof of the first part, our goal is to show that there exists an index $k$ such that
	\begin{equation}
	\label{eq:equality_of_numbers}
	\#\{k \leq n \leq N: \{a_n\alpha\} \in J\} = \#\{k \leq n \leq N: \{n\alpha\} \in J\}
	\end{equation}
	for all $N \in \N$ where $J = [\{k_1\alpha\},\{k_2\alpha\})$ with $k_1,k_2 \in \N$ and $\{k_1\alpha\} < \{k_2\alpha\}$ and where $a_n = n + q_n$ with $q_n$ the denominator from the $n$-th convergent of the continued fraction expansion of $\alpha$.
	Since $J$ is a BRS for $(\{n\alpha\})_{n \geq 1}$ this implies that
	\begin{equation*}
	|\#\{1 \leq n \leq N: \{a_n\alpha\} \in J\} - \lambda(J)N| \leq c + k
	\end{equation*}
	for all $N \in \N$ with the right side of the inequality independent of $N$ and where $\lambda(J) = \{k_2\alpha\} - \{k_1\alpha\}$.

	Equation (\ref{eq:equality_of_numbers}) is fulfilled if we are able to show that the point $\{a_n\alpha\}, n \geq k$ is a suitable approximation of $\{n\alpha\}$. More precisely, if the distance between $\{a_n\alpha\}$ and $\{n\alpha\}$ is smaller than the distance between the boundaries of $J$ and $\{n\alpha\}$.

	Let us begin with the distance between $\{a_n\alpha\}$ and $\{n\alpha\}$. This distance is given by
	\begin{align*}
	\| a_n\alpha - n\alpha \| &= \| (q_n + n)\alpha - n\alpha \| \\
	&= \| q_n\alpha \|.
	\end{align*}
	On the other hand, the distance between $\{n\alpha\}$ and the boundaries of $J$ can be estimated by the minimal distance between any two elements $0, \{\alpha\}, \{2\alpha\},...,\{n\alpha\}$ if $\{k_1\alpha\}$ and $\{k_2\alpha\}$ are contained in those points, i.e. if $n > \max(k_1, k_2)$. For this minimal distance it holds that
	\begin{align*}
	\min_{0 \leq l_1 < l_2 \leq n} \|l_2\alpha - l_1\alpha \|
	&= \min_{0 \leq l_1 < l_2 \leq n} \| \underbrace{(l_2-l_1)}_{\in \{1, \dots, n\}}\alpha  \| \\
	&= \min_{1 \leq l \leq n}\|l\alpha\|.
	\end{align*}

	Since the convergents of the continued fraction expansion of $\alpha$ provide best rational approximations, it holds that
	\begin{equation*}
	\|q_n\alpha\| < \min_{1 \leq l \leq n}\|l\alpha\|,
	\end{equation*}
	if $q_n > n$ which is certainly the case for $n \geq 4$.

	Summing up, we have to choose the index $k$ as $k = \max(k_1+1,k_2+1,4)$ such that the boundaries of $J$ are contained in the considered points and to guarantee the approximation property. For this choice of $k$, equation (\ref{eq:equality_of_numbers}) holds and $J$ therefore is a bounded remainder set for the sequence $(\{a_n\alpha\})_{n\geq 0})$. The uniform distribution of this sequence directly follows with Lemma \ref{lma:dcbrs_ud} since the intervals $J$ form a dense collection of BRS.

	\ \\

	For the proof of the second part, we consider $\alpha = [0;1,1,\bar{2}]$ and show that some interval $J = [a, a+ \alpha)$, which is a BRS for $\na$ is not a BRS for the sequence $(\{(q_n + n)\alpha\})_{n\geq 1})$. More precisely, we show the existence of an $a \in \R$, obviously $a \neq \{j\alpha\}$ for any $j \in \Z$, for which
	\begin{equation}
	\label{eq:infinit}
	\{n\alpha\} \notin J \text{ but } \{(q_n + n)\alpha\} \in J \qquad \text{for infinitely many } n \in \N
	\end{equation}
	and
	\begin{equation}
	\label{eq:finite}
	\{n\alpha\} \in J \text{ but } \{(q_n + n)\alpha\} \notin J \qquad \text{for only finitely many } n \in \N.
	\end{equation}
	This would imply that $|\#\{1 \leq n \leq N: \{(q_n + n)\alpha\} \in J\} - \alpha N|$ is unbounded and that $J$ is not a BRS for $(\{(q_n + n)\alpha\})_{n\geq 0})$ whereas it is a BRS for $\na$.

	Note that because of the special form of $\alpha$, the denominator $q_n$ of the $n$-th convergent of $\alpha$, is even if $n$ is even and odd if $n$ is odd.
	Moreover, it is a well known fact that $p_n/q_n$ is smaller than $\alpha$ for every
	even value of $n$ and greater for every odd value of $n$, which implies that $0 < \{q_n\alpha\} < 1/2$ for $n$ even and $1/2 < \{q_n\alpha\} < 1$ for $n$ odd. From the proof of the first part we know that the distance between $\{n\alpha\}$ and $\{(q_n+n)\alpha\}$ is given by $\|q_n\alpha\|$.
	Hence,
	\begin{align*}
	\{n\alpha\} &< \{(q_n+n)\alpha\} \quad \text{if } n \text{ is even},\\
	\{n\alpha\} &> \{(q_n+n)\alpha\} \quad \text{if } n \text{ is odd}.
	\end{align*}

	To begin with, we want to construct $a$ such that there exists a subsequence $(\{n_i\alpha\})_{i \geq 1}$ for which it holds that
	\begin{equation}
	\label{eq:a}
	\{n_i\alpha\} < a \text{ and } a \leq \{(q_{n_i}+n_i)\alpha\} < a + \alpha.
	\end{equation}
	This would guarantee property (\ref{eq:infinit}).
	Therefore, choose any even $n_1 \in \N$ such that $\{(q_{n_1}+n_1)\alpha\} + \alpha < 1$ and with $\{q_{n_1}\alpha\} < \alpha$. The first condition is necessary because from (\ref{eq:a}) we know that $a \in (\{n_1\alpha\}, \{(q_{n_1}+n_1)\alpha\}]$ and it therefore guarantees that $a < a + \alpha$. The condition on $\{q_{n_1}\alpha\}$ ensures that $\{(q_{n_i}+n_i)\alpha\} < a + \alpha$.

	Now, we want to choose $n_2$ such that $\{n_2\alpha\}$ lies between $\{n_1\alpha\}$ and $\{(q_{n_1}+n_1)\alpha\}$. It follows from the approximation property of $q_n$ that $n_2 = n_1 + q_{n_1+2}$ is a suitable choice. In fact, of the first $n_2$ elements of $\na$, $\{n_2\alpha\}$ is the one closest to $\{n_1\alpha\}$. Moreover, $n_2$ again is an even number which implies that
	\begin{equation*}
	\{n_1\alpha\} < \{n_2\alpha\} < \{(q_{n_2}+n_2)\alpha\} < \{(q_{n_1}+n_1)\alpha\}.
	\end{equation*}
	Therefore, by (\ref{eq:a}) it has to hold that $a \in (\{n_2\alpha\}, \{(q_{n_2}+n_2)\alpha\}]$. In the same manner, $n_2$ now can be used to construct $n_3$ and in general we obtain
	\begin{equation*}
	n_{i+1} = n_i + q_{n_i + 2}.
	\end{equation*}
	This subsequence satisfies
	\begin{equation*}
	\{n_i\alpha\} < \{n_{i+1}\alpha\} < \{(q_{n_{i+1}}+n_{i+1})\alpha\} < \{(q_{n_i}+n_i)\alpha\}
	\end{equation*}
	for all $i \in \N$.
	Hence, equation (\ref{eq:a}) is fulfilled if we choose
	\begin{equation*}
	a = \lim\limits_{n \rightarrow \infty} \{(q_{n_i}+n_i)\alpha\}
	\end{equation*}
	and it holds that
	\begin{equation*}
	\{n_i\alpha\} \notin [a,a+\alpha) \quad \text{ but } \quad \{(q_{n_i} + n_i)\alpha\} \in [a,a+\alpha).
	\end{equation*}
	\begin{figure}[h]
		\centering
		\includegraphics[width=1\textwidth]{counterexample4.pdf}
		\caption{Illustration of $n_1$ and $n_2$}\label{fig:counterexample}
	\end{figure}

	It remains to check property (\ref{eq:finite}).
	First, we want to show that there is no point $\{n\alpha\}$ with $n > n_1$ for which it holds that
	\begin{equation*}
	\{n\alpha\} > a \quad \text{ and } \quad \{(q_{n}+n)\alpha\} < a.
	\end{equation*}
	This can only be the case if $n$ is odd.
	Let $i \in \N$ be the index such that $n_i < n < n_{i+1}$.
	It holds that $\{n\alpha\} > \{n_{i+1}\alpha\}$. Additionally, $\{n\alpha\} > \{(q_{n_i}+n_i)\alpha\}$ and $\{(q_n + n)\alpha\} > \{(q_{n_i}+n_i)\alpha\}$ because otherwise
	$\{n\alpha\}$ and $\{(q_n + n)\alpha\}$ would be a better approximation of $\{n_{i+1}\alpha\}$ than $\{(q_{n_i}+n_i)\alpha\}$ which cannot be the case since $q_{n_i}+n_i > q_n + n$ and $q_{n_i}$ is a best approximation denominator.
	Therefore, for all $n \in \N$, $n > n_1$ it holds that
	\begin{equation}
	\label{eq:first}
	\{n\alpha\} > a \implies \{(q_{n}+n)\alpha\} > a.
	\end{equation}

	Secondly, we also have to check the right boundary of the interval $J$, that means we want to show that there are only finitely many $n$ with the property
	\begin{equation}
	\label{eq:a+alpha}
	\{n\alpha\} < a + \alpha \quad \text{ and } \quad \{(q_{n}+n)\alpha\} \geq a + \alpha.
	\end{equation}
	Note that if (\ref{eq:a+alpha}) holds then $n$ has to be even and that the distance between $\{n\alpha\}$ and $\{(q_{n}+n)\alpha\}$ is given by 	$\{q_n\alpha\}$. Therefore, the distance between $\{n\alpha\}$ and $a + \alpha$ is at most $\{q_n\alpha\}$. If $a > \{q_n\alpha\}$, which is certainly fulfilled for $n$ large enough (e.g. if $q_n > n_1$), we also obtain information about the distance of $\{(n-1)\alpha\}$ to the left boundary of $J$, namely
	\begin{equation}
	\label{eq:left_boundary}
	0 < a - \{(n-1)\alpha\} < \{q_n\alpha\}.
	\end{equation}
	For those $n$, we again have to consider two cases, depending on the position of \linebreak
	$\{(n-1)\alpha\}$.
	The first one is the case $\{(n-1)\alpha\} < \{n_1\alpha\}$. This however implies that
	\begin{equation*}
	a - \{(n-1)\alpha\} > a - \{n_1\alpha\}.
	\end{equation*}
	Since $a - \{n_1\alpha\} > \{n_2\alpha\} - \{n_1\alpha\} = \{q_{n_1 + 2}\alpha\}$ (this can be easily seen from Figure \ref{fig:counterexample}) we have that
	\begin{equation*}
	\{q_n\alpha\} > \{q_{n_1 + 2}\alpha\}.
	\end{equation*}
	Both, $n$ and $n_1 + 2$ are even numbers, therefore this implies that $n \leq n_1$.

	For the second case, $\{(n-1)\alpha\} > \{n_1\alpha\}$, observe that the points $\{n_i\alpha\}$ can get arbitrarily close to $a$. Therefore, for all $n > n_1$ which fulfill (\ref{eq:a+alpha}) there is an $i \in \N$ such that $\{n_i\alpha\} < \{(n-1)\alpha\} < \{n_{i+1}\alpha\}$.
	As in the previous case, from the right side of (\ref{eq:left_boundary}) and because $a - \{n_{i+1}\alpha\} > \{n_{i+2}\alpha\} - \{n_{i+1}\alpha\} = \{q_{n_{i+1}+2}\alpha\}$ it follows that
	\begin{equation*}
	\{q_n\alpha\} > \{q_{n_{i+1}+2}\alpha\},
	\end{equation*}
	which implies that $n \leq n_{i+1}$.
	Since of the first $n_{i+1}$ elements of $\na$, $\{n_{i+1}\alpha\}$ is the closest one to $\{n_{i}\alpha\}$, additionally it has to hold $n > n_{i+1}$. Thus, such an $n$ cannot exist.

	Summing up, for all $n \in \N$, $n > n_1$ it holds that
	\begin{equation}
	\label{eq:second}
	\{n\alpha\} < a + \alpha \implies \{(q_n + n)\alpha\} < a + \alpha
	\end{equation}
	By putting (\ref{eq:first}) and (\ref{eq:second}) together we obtain that
	\begin{equation*}
	\{n\alpha\} \in [a, a+ \alpha) \implies \{(q_n +n)\alpha\} \in [a, a+ \alpha)
	\end{equation*}
	for all $n > n_1$. Thus, property (\ref{eq:finite}) holds and the proof is complete.
\end{refproof}

\begin{refproof}[Proof of Theorem \ref{thm:slow_growth_rate}.]
	Let $\ab \subset \ui$.
	The condition that the increasing integer sequence $(a_n)\lc$ grows slower than linearly implies that for all $k \in \N$ there exists $N_k \in \N$ such that $a_{N_k+1} = a_{N_k + 2} = \dots = a_{N_k + k}$.
	Thus, for each $k$ the elements $\{a_n\alpha\}$, with $N_k < n \leq N_k + k$ are either all contained or not contained in $\ab$ which therefore is an S-NBRS.
\end{refproof}

\begin{refproof}[Proof of Theorem \ref{thm:arbitrary_growth_rate_Brs}.]
	Let $\alpha$ have bounded continued fraction coefficients, i.e.
	\linebreak $\alpha = [0;\alpha_1,\alpha_2,...]$ with $\alpha_i \leq L-1$ for all $i \in \N$ and let $q_n$ denote the $n$-th best approximation denominator of $\alpha$.
	Consider the sequence $a_n = q_{n'} + n$ where for each $n$ the parameter $n'$ is chosen such that
	\begin{equation*}
	q_{n' - 1} + n < \varphi(n) \leq q_{n'} + n.
	\end{equation*}
	It holds that
	\begin{align*}
	\varphi (n) \leq  a_n = q_{n'} + n
	&= \alpha_{n'}q_{n' -1} + q_{n' - 2} + n \\
	&\leq (L-1)q_{n' -1} + q_{n' - 2} + n \\
	&\leq Lq_{n' -1} + n\\
	&< L (\varphi(n) - n) + n \\
	&< L \varphi(n).
	\end{align*}
	Moreover, we already know from the proof of the first part of Theorem \ref{thm:lacunary_brs} that $(\{a_n\alpha\})_{n \geq 1}$ has a dense collection of BRS if the distance between $\{a_n\alpha\}$ and $\{n\alpha\}$, which is given by $\|\{q_{n'}\alpha\}\|$, is smaller than $\min_{1 \leq l \leq n} \|\{l\alpha\}\|$ for all $n \in \N$.
	Therefore, from the approximation property of $q_{n'}$ and from the fact that
	\begin{equation*}
	q_{n'} \geq \varphi(n) - n > n
	\end{equation*}
	it follows that $(\{a_n\alpha\})_{n \geq 1}$ has a dense collection of BRS. Thus, by Lemma \ref{lma:dcbrs_ud} $\ana$ is also uniformly distributed.
\end{refproof}

The proof of Theorem \ref{thm:countably_many_alpha} basically follows the same idea as the proof of the first part of Theorem \ref{thm:lacunary_brs}, $a_n$ is again of the form $q_n + n$. However, with the difference that $q_n$ now does not any more denote the denominator of the $n$-th convergent from the continued fraction expansion of some single $\alpha$.
Instead, we will use the simultaneous version of Dirichlet's Approximation Theorem (see e.g. \cite{Hardy1975}): \newline
For all irrational $\alpha_1, ...., \alpha_k$ and all $N \in \N$ there exist integers $p_1,...,p_k$ and $1 \leq q \leq N$ such that
\begin{equation*}
\left|\alpha_i - \frac{p_i}{q}\right| \leq \frac{1}{q N^{1/k}}.
\end{equation*}
In particular, for any $\alpha_1, ..., \alpha_k$ and any $\varepsilon > 0$ it is possible to find an integer $q$ such that $\| q\alpha_i \|$ is less than $\varepsilon$ for every $i = 1, \dots, k$.

\begin{refproof}[Proof of Theorem \ref{thm:countably_many_alpha}.]
	Let $\mathcal{A} = \{\alpha_1, \alpha_2, \dots\}$ with $\alpha_i$, $i \in \N$ irrational. We show that there exists a sequence $(a_n)_{n \geq 1}$ such that for all $i = 1, 2, \dots$ and all $k_1,k_2 \in \N_0$ with $\{k_2\alpha_i\} > \{k_1\alpha_i\}$, the interval
	\begin{equation*}
	J_i = [\{k_1\alpha_i\},\{k_2\alpha_i\}),
	\end{equation*}
	is a BRS for the sequence $(\{a_n\alpha_i\})_{n \geq 1}$.

	The minimal pairwise distance between any two of the first $k$ elements of $(\{n\alpha\})_{n \geq 1}$ is certainly at least $\min_{1 \leq l \leq k} \| l\alpha \|$.
	Since we now have to deal with countably many $\alpha_i$ we define
	\begin{equation*}
	\varepsilon_k := \min_{1 \leq n,i \leq k} \| n\alpha_i \|.
	\end{equation*}

	From the simultaneous version of Dirichlet's Approximation Theorem, for all of those $\varepsilon_k$ there exists a $q_k$ such that
	\begin{equation*}
	\| \{q_k\alpha_i\} \| < \varepsilon_k
	\end{equation*}
	for all $i  = 1,...,k$.
	Since $k$ was arbitrary, this approach defines an increasing integer sequence $(q_k)_{k \geq 1}$.
	It can be easily shown that for any $i \in \N$, intervals of the form $J = [\{k_1\alpha_i\},\{k_2\alpha_i\})$ with $k_1, k_2 \in \N$ are bounded remainder sets for the sequence $(\{(q_n + n)\alpha_i\})_{n \geq 1}$:

	Consider again the sequences $(\{n\alpha_i\})_{n \geq 1}$ and $(\{(q_n + n)\alpha_i\})_{n \geq 1}$. Then, for all $k \geq \max(k_1 +1,k_2 + 1,i)$ it holds that the boundaries of $J$ are contained in the first $k$ elements of $(\{n\alpha\})$. Moreover, the minimal distance of the element $\{k\alpha_i\}$ to the boundaries of $J$, which is greater than $\varepsilon_k$, is larger than the distance between $\{k\alpha_i\}$ and $\{(q_k +k)\alpha_i\}$, which is given by $\| q_k\alpha_i \|$. Therefore, it holds for every $i \in \N$ that
	\begin{equation*}
	\#\{K \leq n \leq N: \{a_n\alpha_i\} \in J\}
	= \#\{K \leq n \leq N: \{n\alpha_i\} \in J\},
	\end{equation*}
	for all $N \in \N$ where $K = \max(k_1 +1,k_2 + 1,i)$. This implies that
	\begin{equation*}
	|\#\{1 \leq n \leq N: \{a_n\alpha_i\} \in J\} - \lambda(J)N| \leq c
	\end{equation*}
	for all $N \in \N$ and with a constant $c$ independent of $N$. The intervals $J$ form a dense collection of BRS such that by Lemma \ref{lma:dcbrs_ud} the sequences $(\{a_n\alpha_i\})_{n \geq 1}$ are uniformly distributed.
\end{refproof}

\section{Proof of Theorem \ref{thm:sn-brs_polynomial}}
\label{sec:proof_2}

This section is dedicated to the proof of Theorem \ref{thm:sn-brs_polynomial} for which two auxiliary results are needed. The proofs of the lemmas as well as of the theorem basically follow the same idea, nevertheless, it is necessary to perform them separately.

\begin{lemma}
	\label{thm:different_alpha_sequence}
	Let $q \in \N$, $\alpha_1, \alpha_2, \dots, \alpha_q$ be irrational and linearly independent over $\mathbb{Q}$. Let $(\bm{x}_n)\lc = (\{n\alpha_1\}, \{n \alpha_2\}, \dots, \{n\alpha_q\})\lc$.
	Then, for all intervals $[\bm{0},\bm{b}) = [0,b_1)\times [0, b_2) \times \dots \times [0, b_q) \subset [0,1)^q$ of positive measure there exists $K \in \N$ such that at least one of $K$ consecutive elements of $(\bm{x}_n)\lc$ is contained in $[\bm{0},\bm{b})$.
\end{lemma}

\begin{proof}
	Let $(\bm{x}_n)\lc = (\{n\alpha_1\}, \{n \alpha_2\}, \dots, \{n\alpha_q\})\lc$, where $\alpha_1, \dots, \alpha_q$ are irrational and linearly independent over $\mathbb{Q}$.
	To begin with, we partition the interval $[0,1)^q$ into $l^q$ subintervals of volume $l^{-q}$, i.e. intervals of the form
	\begin{equation*}
	\mathcal{I}(j_1, \dots, j_q) = \left[\frac{j_1}{l},\frac{j_1+1}{l}\right) \times \dots \times \left[\frac{j_q}{l},\frac{j_q+1}{l}\right),
	\end{equation*}
	with $j_1, \dots, j_q \in \{0, 1, \dots, l-1\}$ where $l$ is chosen such that
	\begin{equation}
	\label{eq:supset_b}
	[\bm{0}, \bm{b}) \supseteq \left[0,\frac{2}{l} \right)^q.
	\end{equation}

	Since $(\bm{x}_n)\lc$ is uniformly distributed, each interval of the form $\mathcal{I}(j_1, \dots, j_q)$ contains elements of $(\bm{x}_n)\lc$. For each $\bm{j} = (j_1, \dots, j_q) \in \{0, 1, \dots, l-1\}^q$, by $K_{\bm{j}}$ we denote the index of the first element of $(\bm{x}_n)\lc$ which lies in $\mathcal{I}(\bm{j})$.
	Now define
	\begin{equation*}
	K_{\bm{\alpha},\bm{b}} =\max_{\bm{j} \in \{0,1, \dots, l-1\}^q}K_{\bm{j}}.
	\end{equation*}
	Then, assume that for any $N \in \N$ we have
	\begin{equation}
	\label{eq:elements}
	x_{N+k} \notin [\bm{0},\bm{b})
	\end{equation}
	for all $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$. Note that if $c$ is a constant, it holds that $\{x + c\} \in [a,b)$ is equivalent to $\{x \} \in [a - c, b - c) \pmod 1$ where we use the notation
	\begin{equation*}
	[a,b) \pmod 1 :=
	\begin{cases}
	[a \pmod 1, b \pmod 1) \quad &\text{if } a \pmod 1 > b \pmod 1 \\
	[a \pmod 1, 1) \cup [0, b \pmod 1) \quad &\text{if } a \pmod 1 < b \pmod 1.
	\end{cases}
	\end{equation*}
	Therefore, $\bm{x}_{N + k} \notin [\bm{0}, \bm{b})$ implies that
	\begin{equation*}
	\bm{x}_k \notin [-\bm{x}_N, \bm{b} -\bm{x}_N) \pmod 1
	\end{equation*}
	for all $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$.
	Because of \ref{eq:supset_b} the interval $[-\bm{x}_N, \bm{b} -\bm{x}_N) \pmod 1$ fully covers at least one interval of the form $\mathcal{I}(j_1, \dots, j_q)$ which has to contain at least one of the first $K_{\bm{\alpha},\bm{b}}$ points of $(\bm{x}_n)\lc$. Thus, we have a contradiction to (\ref{eq:elements}).
\end{proof}

\begin{lemma}
	\label{lma:combination}
	Let $p,q \in \N$, $\alpha_1, \alpha_2, \dots, \alpha_q$ be irrational and linearly independent over $\mathbb{Q}$. Let $(\bm{x}_n)\lc$ be the $pq$-dimensional sequence whose components have the form $\{n^{p'}\alpha_{q'}\}$ with $1 \leq p' \leq p$ and $1 \leq q' \leq q$.
	Then, for all intervals $[\bm{0},\bm{b}) \subset [0,1)^{pq}$ there exists $K \in \N$ such that at least one of $K$ consecutive elements of $(\bm{x}_n)\lc$ is contained in $[\bm{0},\bm{b})$.
\end{lemma}

\begin{proof}
	The elements of the sequence $(\bm{x}_n)\lc$ are given by
	\begin{equation*}
	\bm{x}_n = (\{n\alpha_1\}, \{n^2\alpha_1\}, \dots, \{n^p\alpha_1\}, \dots, \{n\alpha_q\}, \{n^2\alpha_q\}, \dots, \{n^p\alpha_q\}).
	\end{equation*}
	We prove the result by induction on $p$. By Lemma \ref{thm:different_alpha_sequence} the assertion holds for $p = 1$ and arbitrary $q \in \N$, which proves the base case.

	Now suppose that for all $\bm{b}' \in [0,1)^{(p-1)q}$ there exists $L \in \N$ such that at least one of $L$ consecutive elements of $(\bm{x}_n')\lc$, where
	\begin{equation*}
	\bm{x}'_n = (\{n\alpha_1\}, \{n^2\alpha_1\}, \dots, \{n^{p-1}\alpha_1\}, \dots, \{n\alpha_q\}, \{n^2\alpha_q\}, \dots, \{n^{p-1}\alpha_q\})
	\end{equation*}
	is contained in $[\bm{0},\bm{b}')$.
	For the inductive step, we have to show that for all $\bm{b} \in [0,1)^{pq}$ there exists $K \in \N$ such that at least one of $K$ consecutive elements of $(\bm{x}_n)\lc$ lies in  $[\bm{0},\bm{b})$.

	We start by partitioning the interval $[0,1)^{pq}$ into $l^{pq}$ subintervals of volume $l^{-pq}$, i.e. intervals of the form
	\begin{equation*}
	\mathcal{I}(j_1^{(1)}, \dots, j_q^{(p)}) = \left[\frac{j_1^{(1)}}{l},\frac{j_1^{(1)}+1}{l}\right) \times \dots \times \left[\frac{j_q^{(p)}}{l},\frac{j_q^{(p)}+1}{l}\right),
	\end{equation*}
	with $j_{q'}^{(p')} \in \{0, 1, \dots, l-1\}$ for $q' \leq q$ and $p' \leq p$.
	Moreover, $l$ is chosen such that
	\begin{equation*}
	b_1^{(1)}, \dots, b_q^{(p)} \geq \frac{2R + 2}{l}
	\end{equation*}
	with $R = \sum_{j = 1}^{p - 1}\binom{p}{j}$
	in order to guarantee that
	\begin{equation}
	\label{eq:supset}
	[\bm{0}, \bm{b}) = [0,b_1^{(1)}) \times \dots \times [0,b_q^{(p)}) \supseteq \left[0,\frac{2R + 2}{l} \right)^{pq}.
	\end{equation}

	Since for each lattice point $\bm{h} \in \Z^{pq}, \bm{h} \neq 0$, the polynomial
	\begin{equation*}
	\langle \bm{h}, (n\alpha_1, \dots, n^p\alpha_1, \dots, n\alpha_q, \dots, n^p\alpha_q) \rangle
	\end{equation*}
	has at least one non-constant term with irrational coefficient,
	the sequence $(\bm{x}_n)\lc$ is uniformly distributed (see e.g. Theorem 6.4 in \cite{Kuipers:Uniform_Distribution_of_Sequences}). Hence, each interval of the form $\mathcal{I}(j_1^{(1)}, \dots, j_q^{(p)})$ contains elements of $(\bm{x}_n)\lc$ and for each $\bm{j} = (j_1^{(1)}, \dots, j_q^{(p)}) \in \{0, 1, \dots, l-1\}^{pq}$, by $K_{\bm{j}}$ we denote the index of the first element of $(\bm{x}_n)\lc$ which lies in $\mathcal{I}(\bm{j})$.
	As before, define
	\begin{equation*}
	K_{\bm{\alpha},\bm{b}} = \max_{\bm{j} \in \{0,1, \dots, l-1\}^{pq}}K_{\bm{j}}
	\end{equation*}
	and note that $K_{\bm{\alpha},\bm{b}} > l^q$.
	By the induction hypothesis, the assertion of Lemma \ref{lma:combination} is true for $p -1$ and any $\bm{b}' \in [0,1)^{(p-1)q}$. We consider the special case $\bm{b}' = (K_{\alpha,\bm{b}}^{-p}, \dots, K_{\alpha,\bm{b}}^{-p}) \in \ui^{(p-1)q}$ and conclude that there exists $L = L(\alpha_1, \dots, \alpha_{q}, \bm{b})$ such that in any $L$ consecutive integers there is at least one integer $M$ with the property
	\begin{equation}
	\label{eq:2V_smaller_2}
	\|M^{p'}\alpha_{q'}\| \leq K_{\bm{\alpha},\bm{b}}^{-p},
	\end{equation}
	for all $q' \leq q$ and $p' \leq p-1$.
	Furthermore, define $K = L + K_{\bm{\alpha},\bm{b}}$ and and suppose that for some $N \in \N$ it holds that
	\begin{equation*}
	\label{eq:considered_points}
	\bm{x}_{n} \notin [\bm{0},\bm{b})
	\end{equation*}
	for $n = N+1, \dots, N+K$.
	More specifically, we consider the elements $\bm{x}_{M + k}$ for $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$ where $N +1 \leq M \leq N + L$ and $M$ satisfies inequalities (\ref{eq:2V_smaller_2}).
	For the components of these elements with $p' = 1$ it holds that
	\begin{equation*}
	\{(M+k)\alpha_{q'}\} = \{\underbrace{M\alpha_{q'}}_{=: \xi_{q'}^{(1)}} + k\alpha_{q'}\} \notin [0,b_{q'}^{(1)}),
	\end{equation*}
	for all $q' \leq q$.
	This is equivalent to
	\begin{equation*}
	\{k\alpha_{q'}\} \notin [- \xi_{q'}^{(1)},b_{q'}^{(1)} - \xi_{q'}^{(1)}) \pmod 1.
	\end{equation*}
	For the components with $p' = 2$ we have
	\begin{equation*}
	\{(M+k)^2\alpha_{q'}\} = \{\underbrace{M^2\alpha_{q'}}_{=:\xi_{q'}^{(2)}} + 2Mk\alpha_{q'} + k^2\alpha_{q'}\} \notin [0,b_{q'}^{(2)}).
	\end{equation*}
	Since $\{2Mk\alpha_{q'} + k^2\alpha_{q'}\}$ is either $\{k^2\alpha_{q'}\} + \|2Mk\alpha_{q'}\|$ or $\{k^2\alpha_{q'}\} - \|2Mk\alpha_{q'}\|$ and with the estimate
	\begin{align*}
	\|2Mk\alpha_{q'}\| &\leq \|M\alpha_{q'}\|2k \\
	& \leq \|M\alpha_{q'}\|2K_{\bm{\alpha},\bm{b}} \\
	& \leq \frac{2K_{\bm{\alpha},\bm{b}}}{K_{\bm{\alpha},\bm{b}}^p} < \frac{2}{l},
	\end{align*}
	we can conclude that
	\begin{equation*}
	\{k^2\alpha_{q'}\} \notin \left[- \xi_{q'}^{(2)} + \frac{2}{l}, b_{q'}^{(2)} - \xi_{q'}^{(2)} - \frac{2}{l}\right) \pmod 1.
	\end{equation*}
	We continue in this manner for all values of $p'$ until we arrive at
	\begin{equation*}
	\{(M+k)^p\alpha_{q'}\} = \{\underbrace{M^p\alpha_{q'}}_{=:\xi_{q'}^{(p)}} + \sum_{j = 1}^{p -1}\binom{p}{j}M^{p-j}k^j\alpha_{q'} + k^p\alpha_{q'}\} \notin [b_{q'}^{(p)},1)
	\end{equation*}
	for all $q' \leq q$.
	We apply the estimate
	\begin{align*}
	\|\binom{p}{j}M^{p-j}k^j\alpha_{q'}\| &\leq \|M^{p-j}\alpha_{q'}\|\binom{p}{j}k^j \\
	& \leq \|M^{p-j}\alpha_{q'}\|\binom{p}{j}K_{\bm{\alpha},\bm{b}}^{p-1} \\
	& \leq \binom{p}{j}K_{\bm{\alpha},\bm{b}}^{-1} \leq \binom{p}{j}l^{-1},
	\end{align*}
	for all $j = 1, \dots, p - 1$. Hence,
	\begin{equation*}
	\{k^p\alpha_{q'}\} \notin \left[- \xi_{q'}^{(p)} + \sum_{j = 1}^{p -1}\binom{p}{j}l^{-1},
	b_{q'}^{(p)} - \xi_{q'}^{(p)} - \sum_{j = 1}^{p -1}\binom{p}{j}l^{-1}\right) \pmod 1.
	\end{equation*}

	Finally, we use (\ref{eq:supset}) and obtain
	\begin{align*}
	\{k\alpha_{q'}\} &\notin [- \xi_{q'}^{(1)}, (2R+2)l^{-1} - \xi_{q'}^{(1)})  \pmod 1 =: \mathcal{I}_{q'}^{(1)}, \\
	\{k^2\alpha_{q'}\} &\notin \left[- \xi_{q'}^{(2)} + 2l^{-1}, (2R+2)l^{-1} - \xi_{q'}^{(2)} -2l^{-1}\right)  \pmod 1 =: \mathcal{I}_{q'}^{(2)}, \\
	&\vdots\\
	\{k^p\alpha_{q'}\} &\notin \left[- \xi_{q'}^{(p)} + \sum_{j = 1}^{p -1}\binom{p}{j}l^{-1},
	(2R+2)l^{-1} - \xi_{q'}^{(p)} - \sum_{j = 1}^{p -1}\binom{p}{j}l^{-1}\right) \pmod 1 =: \mathcal{I}_{q'}^{(p)}
	\end{align*}
	for all $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$ and all $q' \leq q$.
	However, because of the definition of $R$ each of these one-dimensional intervals is of length greater or equal to $2/l$. The $pq$-dimensional interval $\mathcal{I}_1^{(1)} \times \dots \times \mathcal{I}_q^{(p)}$ therefore fully contains at least one of the intervals $\mathcal{I}(\bm{j})$. Since $K_{\bm{\alpha},\bm{b}}$ was constructed such that each interval $\mathcal{I}(\bm{j})$ at least contains one of the elements $(\bm{x}_n)_{1 \leq n \leq K_{\bm{\alpha},\bm{b}}}$ we have a contradiction. Therefore, for each $N$ at least one element of $(\bm{x}_{n})_{N+1 \leq n \leq N + K}$ lies in $[\bm{0},\bm{b})$.
\end{proof}

Finally, we are able to give a proof of Theorem \ref{thm:sn-brs_polynomial}. We will use the notation $\{(x^{(1)}, \dots, x^{(s)})\} = (\{x^{(1)}\}, \dots, \{x^{(s)}\})$.

\begin{refproof}[Proof of Theorem \ref{thm:sn-brs_polynomial}.]
	Let $\bm{p}(n)$ fulfill the assumptions of the theorem and let $q$ be the highest degree of the polynomials $p^{(i)}(n)$, $i = 1, \dots, s$. Then, each of these polynomials is of the form $p^{(i)}(n) = \alpha_0^{(i)} + n\alpha_1^{(i)} + \dots + n^q\alpha_q^{(i)}$ with coefficients $\alpha_{q'}^{(i)} \in \R$, $q' \leq q$. Furthermore, we split the set of coefficients into three subsets:
	\begin{align*}
	A_1 &= \{\alpha_{q'}^{(i)}: \alpha_{q'}^{(i)} \in \mathbb{Q}, i = 1, \dots, s, q' \leq q\},\\
	A_2 &= \{\alpha_{q'}^{(i)}: \alpha_{q'}^{(i)} \in \R \setminus \mathbb{Q}, i = 1, \dots, s, q' \leq q \text{ and all } \alpha_{q'}^{(i)} \text{ are linearly independent over } \mathbb{Q} \},\\
	A_3 &= \{\alpha_{q'}^{(i)}: \alpha_{q'}^{(i)} \in \R \setminus \mathbb{Q}, i = 1, \dots, s, q' \leq q \text{ and } \alpha_{q'}^{(i)} = \sum_{\alpha \in A_2} c(\alpha_{q'}^{(i)}, \alpha)\alpha\}.
	\end{align*}
	Note that this partition is not unique. Using these sets we can now define for each single $i$:
	\begin{align*}
	Q_1^{(i)} &= \{0 < q_1 \leq q: \alpha_{q_1}^{(i)} \in A_1 \}, \\
	Q_2^{(i)} &= \{0 < q_2 \leq q: \alpha_{q_2}^{(i)} \in A_2 \}, \\
	Q_3^{(i)} &= \{0 < q_3 \leq q: \alpha_{q_3}^{(i)} \in A_3 \},
	\end{align*}
	for all $i = 1, \dots, s$.
	For each $i$, the set $Q_2^{(i)} \cup Q_3^{(i)}$ contains at least one element. We have
	\begin{equation*}
	p^{(i)}(n) = \underbrace{\sum_{q_1 \in Q_1^{(i)}} n^{q_1}\alpha_{q_1}^{(i)}}_{=: P_1^{(i)}} +
	\underbrace{\sum_{q_2 \in Q_2^{(i)}} n^{q_2}\alpha_{q_2}^{(i)}}_{=: P_2^{(i)}} +
	\underbrace{\sum_{q_3 \in Q_3^{(i)}} n^{q_3}\alpha_{q_3}^{(i)}}_{=: P_3^{(i)}} + \alpha_0^{(i)}.
	\end{equation*}
	First, we consider some interval $\abm \subset \ui^s$ and show that the maximal number of successive elements of $(\bm{p}(n))\lc$ which are not contained in $\abm$ is bounded.
	Therefore, let $D$ denote the least common multiple of the denominators of all $\alpha \in A_1$. We partition $[0,1)^s$ into intervals
	\begin{equation*}
	\mathcal{I}(\bm{j}) = \left[\frac{j_1}{l},\frac{j_1+1}{l}\right) \times \dots \times \left[\frac{j_s}{l},\frac{j_s+1}{l}\right),
	\end{equation*}
	with $\bm{j} = (j_1, \dots, j_s) \in \{0, 1, \dots, l-1\}^s$. The parameter $l$ is chosen such that
	\begin{equation*}
	b^{(i)} - a^{(i)} \geq \frac{2 R^{(i)} + 2}{l},
	\end{equation*}
	for $i = 1, \dots, s$ in order to obtain
	\begin{equation}
	\label{eq:interval_contained}
	\abm \supseteq \left[\bm{a}, \bm{a} + \frac{2 R^{(i)} + 2}{l}\right).
	\end{equation}
	The constants $R^{(i)}$ are defined as
	\begin{equation*}
	R^{(i)} := \sum_{q_2 \in Q_2^{(i)}}\sum_{j = 1}^{q_2-1}\binom{q_2}{j}D^j +
	\sum_{q_3 \in Q_3^{(i)}}\sum_{j = 1}^{q_3-1}\binom{q_3}{j}D^j\sum_{\alpha \in A_2}c(\alpha_{q_3}^{(i)},\alpha).
	\end{equation*}
	Consider now the sequence $(\{P_2^{(1)}(Dn) + P_3^{(1)}(Dn)\}, \dots, \{P_2^{(s)}(Dn) + P_3^{(s)}(Dn)\})\lc$ which, by the assumption of the theorem, is uniformly distributed. Therefore, each of the intervals $\mathcal{I}(\bm{j})$ contains points of the sequence and by $K_{\bm{j}}$ we denote the index of the first element in the respective interval. As before, we define
	\begin{equation*}
	K_{\bm{\alpha}, \bm{b}} = \max_{\bm{j} \in \{0,1,\dots,l-1\}^s}K_{\bm{j}}.
	\end{equation*}
	Now consider the sequence whose components are given by the sequences $n^{q'}\alpha$ with $\alpha \in A_2$ and $q' \leq q$. Since all $\alpha \in A_2$ are linearly independent, it follows from Lemma \ref{lma:combination} that there exists $L \in \N$ such that any $L$ integers contain at least one integer $M$ which fulfills
	\begin{equation}
	\label{eq:inequality_p(n)}
	\| M^{q'}\alpha\| < K_{\bm{\alpha}, \bm{b}}^{-q}
	\end{equation}
	for all $q' \leq q$, $\alpha \in A_2$.
	We use this $L$ to define $K = L + DK_{\bm{\alpha},\bm{b}}$ and show that for each $N \in \N$ at least one of the elements
	\begin{equation*}
	\{\bm{p}(N + 1)\}, \dots, \{\bm{p}(N + K)\}
	\end{equation*}
	lies in $\abm$, i.e. the number of consecutive elements of $(\bm{p}(n))\lc$ which are not contained in $\abm$ is bounded by $K$.
	The elements $\{\bm{p}(N + 1)\}, \dots, \{\bm{p}(N + L)\}$ contain an element $\{\bm{p}(M)\}$, where $M$ satisfies (\ref{eq:inequality_p(n)}). Therefore, we particularly consider the elements $\{\bm{p}(M + Dk)\}$ for $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$. Suppose that
	\begin{equation}
	\label{eq:assertion}
	\{\bm{p}(M + Dk)\} \notin \abm
	\end{equation}
	for all $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$ and use
	\begin{equation*}
	p^{(i)}(M + Dk)
	= P_1^{(i)}(M + Dk) + P_2^{(i)}(M + Dk) + P_3^{(i)}(M + Dk) + \alpha_0^{(i)}.
	\end{equation*}
	Since $D$ is the least common multiple of the denominators of $\alpha \in A_1$, it holds that
	\begin{equation*}
	P_1^{(i)}(M + Dk) = P_1^{(i)}(M) + z
	\end{equation*}
	for some $z \in \Z$.
	For $P_2^{(i)}$ we have
	\begin{align*}
	P_2^{(i)}(M + Dk)
	&= \sum_{q_2 \in Q_2^{(i)}}(M + Dk)^{q_2}\alpha_{q_2}^{(i)} \\
	&= \underbrace{\sum_{q_2 \in Q_2^{(i)}} M^{q_2}\alpha_{q_2}^{(i)}}_{ = P_2^{(i)}(M)} +
	\sum_{q_2 \in Q_2^{(i)}}\sum_{j = 1}^{q_2-1}\binom{q_2}{j} M^{q_2-j}(Dk)^j\alpha_{q_2}^{(i)} +
	\underbrace{\sum_{q_2 \in Q_2^{(i)}}(Dk)^{q_2}\alpha_{q_2}^{(i)}}_{= P_2^{(i)}(Dk)}.
	\end{align*}
	Similarly, for $P_3^{(i)}$ we obtain
	\begin{align*}
	P_3^{(i)}(M + Dk)
	&= \underbrace{\sum_{q_3 \in Q_3^{(i)}} M^{q_3}\alpha_{q_3}^{(i)}}_{ = P_3^{(i)}(M)} +
	\sum_{q_3 \in Q_3^{(i)}}\sum_{j = 1}^{q_3-1}\binom{q_3}{j} M^{q_3-j}(Dk)^j\alpha_{q_3}^{(i)} +
	\underbrace{\sum_{q_3 \in Q_3^{(i)}}(Dk)^{q_3}\alpha_{q_3}^{(i)}}_{= P_3^{(i)}(Dk)},
	\end{align*}
	where the second term can be written as
	\begin{equation*}
	\sum_{q_3 \in Q_3^{(i)}}\sum_{j = 1}^{q_3-1}\binom{q_3}{j} \sum_{\alpha \in A_2} c(\alpha_{q_3}^{(i)},\alpha)M^{q_3-j}(Dk)^j\alpha.
	\end{equation*}
	By the inequalities in (\ref{eq:inequality_p(n)}) it holds that
	\begin{equation*}
	\| M^{q'-j}k^j\alpha \| \leq \| M^{q'-j}\alpha \| K_{\alpha,\bm{b}}^{q} < l^{-1},
	\end{equation*}
	for all $q' \leq q$, $j \leq q'$ and $\alpha \in A_2$.
	Using the fact that $\{x\} - c\| y \| \leq  \{ x + cy \} \leq \{x\} + c\| y \|$, where $c$ is a constant,
	the definition of $R^{(i)}$, (\ref{eq:interval_contained}) and (\ref{eq:assertion}), we obtain with $\xi^{(i)} = P_1^{(i)}(M) + P_2^{(i)}(M) + P_3^{(i)}(M) + \alpha_0^{(i)}$  that
	\begin{equation*}
	\{P_2^{(i)}(Dk) + P_3^{(i)}(Dk) \} \notin [a^{(i)} - \xi^{(i)} + R^{(i)}l^{-1}, a^{(i)} + (2R^{(i)} + 2)l^{-1} - \xi^{(i)} - R^{(i)}l^{-1}) \pmod 1
	\end{equation*}
	for all $ i = 1, \dots s$ and all $k = 1, \dots, K_{\bm{\alpha},\bm{b}}$.
	The length of each of those intervals is $2/l$, therefore, the resulting $s$-dimensional interval covers at least one of the intervals $\mathcal{I}(\bm{j})$. Since $K_{\bm{\alpha},\bm{b}}$ was chosen such that any $\mathcal{I}(\bm{j})$ contains at least one element of
	$(\{P_2^{(1)}(Dk) + P_3^{(1)}(Dk)\}, \dots, \{P_2^{(s)}(Dk) + P_3^{(s)}(Dk)\})_{1\leq k \leq K_{\bm{\alpha},\bm{b}}}$ this is a contradiction to the construction of $K_{\bm{\alpha},\bm{b}}$.
	Thus, the maximal number of successive elements of $(\{\bm{p}(n)\})\lc$ which are not contained in some interval $\abm \subset \ui^s$ is bounded by some constant $K$. Obviously, the same holds for the number of successive elements of $(\{\bm{p}(n)\})\lc$ in $\ui^s \setminus \abm$.

	In order to prove that the sequence $(\{\bm{p}(n)\})\lc$ does not have any S-NBRS, it remains to show that the number of successive elements in $\abm \subset \ui^s$ is also bounded. This can be easily seen since either
	\begin{equation*}
	\abm \subseteq [\bm{0},\bm{b}) \subset \ui^s \setminus [\bm{b},\bm{1}),
	\end{equation*}
	or
	\begin{equation*}
	\abm \subseteq [\bm{a},\bm{1}) \subset \ui^s \setminus [\bm{0},\bm{a}).
	\end{equation*}
	We already know that the number of consecutive elements in $\ui^s \setminus [\bm{b},\bm{1})$ as well as in $\ui^s \setminus [\bm{0},\bm{a})$ is bounded, therefore this also holds for the interval $\abm$.
\end{refproof}

\bibliographystyle{plain}
\bibliography{literatur}

\textbf{Author’s Addresses:} \\
Lisa Kaltenböck and Gerhard Larcher, Institut für Finanzmathematik und Angewandte Zahlentheorie, Johannes Kepler Universität Linz, Altenbergerstraße 69, A-4040 Linz, Austria. \\ \\
Email: lisa.kaltenboeck(at)jku.at, gerhard.larcher(at)jku.at
\end{document}


TITLE: Dynamical systems: Is there always an attractor?
QUESTION [4 upvotes]: I am just starting out with dynamical systems, and I am wondering if you can have a dynamical system with out any attractors (if so, can you give an example).
Explanations or references backing up answers are appreciated.

REPLY [5 votes]: There are several systems that don't have any attractors. E.g. all unstable systems are without attractors, as illustrated by
$$x_1' = −2 x_1 − 6 x_2 + 8, \; x_2' = 8 x_1 + 4x_2 − 12,$$ as shown in on page 19 of this note (provided by Moo in a comment)


TITLE: v, an extension of a premeasure will always be so v(E) <=u(E)
QUESTION [0 upvotes]: I don't understand the part of the proof that shows that v(E) is always less than or equal to u(E). It appears to me that they are arguing backwards but I don't know how to conclude v(E) ≤ u(E).

REPLY [0 votes]: If $\nu$ extends $\mu_0$ from $\mathscr{A}$ to $\mathscr{M}$ and $A_i \in \mathscr{A} \Rightarrow \nu(A_i) = \mu_0(A_i)$.
Therefore $\sum_{i=1}^{\infty}\nu(A_i) = \sum_{i=1}^{\infty}\mu_0(A_i)$. For all families $\{A_i\}_{i=1}^{\infty}$  of elements of $\mathscr{A}$ (i)
$E \subseteq \cup_{i=1}^{\infty} A_i \Rightarrow \nu(E) \leq \nu(\cup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty}\nu(A_i)$ by the additivity of $\nu$ (ii)
(i), (ii) $\Rightarrow \nu(E) \leq  \sum_{i=1}^{\infty}\mu_0(A_i)$. This again holds for all families $\{A_i\}_{i=1}^{\infty}$  of elements of $\mathscr{A}.$
I.e. $\nu(E)$ is a lower bound for $\sum_{i=1}^{\infty}\mu_0(A_i)$ for these families (iii)
As a lower bound it must be smaller or equal to the infimum:
(iii) $\Rightarrow \nu(E) \leq inf\{\sum_{i=1}^{\infty}\mu_0(A_i) | E \subseteq \cup_{i=1}^{\infty} A_i \wedge A_i \in \mathscr{A} \hspace{.2cm} \forall  i \geq 1\} = \mu^*(E) = \mu(E)$


TITLE: Magnetic potential in spherical coordinates
QUESTION [2 upvotes]: In a current-free region, we have $\nabla \times \mathbf{B} = 0$, allowing us to write $\mathbf{B} = -\nabla V$ for some scalar function $V$. We also have $\nabla \cdot \mathbf{B} = 0$, meaning that the function $V$ must satisfy the Laplace equation:
$$
\nabla^2 V = 0
$$
The solution to this equation is well known in spherical coordinates, and it is given by (in complex form),
$$
V(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^l A_l^m \frac{1}{r^{l+1}} Y_l^m(\theta,\phi) + B_l^m r^l Y_l^m(\theta,\phi)
$$
where $Y_l^m(\theta,\phi)$ are the spherical harmonics. If we consider only the part of the potential due to internal sources, then
$$
V^{int}(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^l A_l^m \frac{1}{r^{l+1}} Y_l^m(\theta,\phi)
$$
Now, many authors throw away the $l=0, m=0$ term, starting the summation at $l=1$, in other words setting $A_0^0 = 0$. One example is Campbell's Introduction to Geomagnetic Fields (2003).
My question is why is it allowed to discard the $l=0$ term? It seems to be related to the absence of magnetic monopoles. However the equation $\nabla \cdot \mathbf{B} = 0$ already accounts for no magnetic monopoles, and so the $l=0$ part of the potential already satisfies this condition. So there must be another reason why authors set $A_0^0 = 0$, however I cannot find a satisfactory explanation.

REPLY [2 votes]: $\nabla\cdot{\bf B}$ does not equal zero for the $l=0$ term.  $\nabla\cdot{\bf B}=4\pi\delta({\bf r}$) (in Gaussian units).\


text:
\begin{document}

\newcommand{\Gal}{{\rm Gal}}
\newcommand{\Pic}{{\rm Pic}}

\addtolength{\textwidth}{0mm}
\addtolength{\hoffset}{-0mm}
\addtolength{\textheight}{0mm}
\addtolength{\voffset}{-0mm}

\title{On finiteness of curves with high canonical degree on  a surface}

\author{Ciro Ciliberto, Xavier Roulleau}
\address{Universit\`a degli Studi di Roma Tor Vergata, Via della
  Ricerca Scientifica 1, 00133, Rome,
  Italy.}\email{cilibert@axp.mat.uniroma2.it}

  \address{Laboratoire de Math\'ematiques et Applications de lÕuniversit\'e de Poitiers,
  T\'el\'eport 2 -- BP 30179, Boulevard Marie et Pierre Curie, 86962 Futuroscope Chasseneuil Cedex,
  France.}\email{Xavier.Roulleau@math.univ-poitiers.fr}

\begin{abstract} The \emph{canonical degree} of a curve $C$ on a surface $X$ is
$K_X\cdot C$. Our main result, Theorem \ref {thm:MAIN}, is that on a surface of general type there are only finitely many curves
with negative self--intersection and sufficiently large canonical degree. Our proof strongly relies on results by Miyaoka. We extend our result both to surfaces not of general type and to non--negative curves, and give applications, e.g. to finiteness of negative curves on a general blow--up of $\mathbb P^ 2$ at $n\geq 10$ general points (a result related to  \emph{Nagata's Conjecture}).  We finally discuss a conjecture by Vojta concerning the asymptotic behaviour of the ratio between the canonical degree and the geometric genus of a curve varying on a surface.
The results in this paper go in the direction of understanding the \emph{bounded negativity} problem.
\end{abstract}
\maketitle

\section{Introduction.}

Let $C$ be a projective curve on a smooth projective complex surface $X$. By \emph{curve} we mean an irreducible, reduced 1--dimensional scheme. We denote by $g=g(C)$
its geometric genus and by $p=p_{a}(C)$ its arithmetic genus, i.e.
$C^{2}+K\cdot C=2p_{a}-2$, where $K=K_{X}$ is a canonical divisor of
$X$. We set $\delta=\delta(C)=p-g$.
We call a curve \emph{negative} if $C^{2}<0$. The \emph{canonical degree} of $C$ is $k_C=K\cdot C$, often simply denoted by $k$.
If $g(C)\neq 1$, we  set
\[
\beta_C=\frac{k_C}{g(C)-1},
\]
 often simply denoted by $\beta$.
For a surface $X$ we  set
\[a_X=3c_{2}(X)-K_{X}^{2},\]
often simply  denoted by $a$.  If the Kodaira
dimension $\kappa=\kappa(X)$ is non--negative, one has $a\geq 0$.

The main result of this paper concerns negative curves with high canonical degree:

\begin{theorem}
\label{thm:MAIN} (A) Let $C$  be a negative curve
not isomorphic to $\mathbb{P}^{1}$ on a surface $X$ with $\kappa\geq0$.
If $a=0$, there are no rational curves on $X$, i.e.
either $a\not = 0$ or $g>0$. Moreover
\begin{equation}\label{eq:1}
k_C\leq3(g-1)+\frac{3}{4}a+\frac{1}{4}\sqrt{9a^{2}+24a(g-1)}.
\end{equation}
Furthermore, if $g>1$ then
 \begin{equation}\label{eq:3}
\beta_C \leq3+\frac{3}{4}a+\frac{1}{4}\sqrt{9a^{2}+24a}\leq4+\frac{3}{2}a.
\end{equation}
If, in addition, $\beta=3+\epsilon>3$, then
\begin{equation}\label{eq:4}
g\leq1+\frac{3a(\epsilon+1)}{2\epsilon^{2}}.
\end{equation}
(B) Suppose $\kappa(X)=2$. Then for any $\epsilon>0$
there are at most finitely many negative curves $C$ on $X$ such that
 $k_C\geq(3+\epsilon)(g-1)>0$.
\end{theorem}

From (B) it follows that:

\begin{cor}
Let $X$ be a surface of general type. There  is a function $B(\epsilon), $ defined for $\epsilon \in ]0,\infty[$ such that for all negative curves $C$ we have
\[
k_C\leq(3+\epsilon)(g-1)+B(\epsilon) \,\,\, \text{and}\,\,\, -C^{2}\leq(1+\epsilon)(g-1)+B(\epsilon).
\]

\end{cor}

On some smooth quaternionic Shimura surfaces $X$ there are infinitely many totally geodesic curves
(see \cite {Chinburg}). Such a curve $C$ is
also called a \emph{Shimura curve} and satisfies $k_C=4(g-1)$. We thus obtain the following corollary,
which was one of the main result of
\cite{BR}:

\begin{cor} On a Shimura surface, there exist finitely many (may be none) negative
Shimura curves.
\end{cor}

The proof of Theorem \ref {thm:MAIN}, contained in \S \ref {sec:proof}, strongly relies on a result by
Miyaoka's (see  \cite[Cor 1.4]{Miyaoka} stated as Theorem \ref {thm:(Miyaoka)} below).
In particular, the inequality \eqref{eq:1} is very similar to  \cite[formula (3)]{Miyaoka}, which has a slightly lower growth in $g$, but applies only to minimal surfaces.

In \S \ref {sec:nagata} we make an extension of Theorem \ref {thm:MAIN} which works also in the case $\kappa=-\infty$, and we prove a finiteness result for negative curves on a general blow--up of $\mathbb P^ 2$ at $n\geq 10$ general points. This is  a bounded negativity result which is
reminiscent of the famous \emph{Nagata's Conjecture}, predicting that there is no negative curve on such a surface except for $(-1)$--rational curves.

In \S \ref {sec:pos}, using again   Miyaoka's result, we prove a boundedness theorem  for non--negative curves of high canonical degree. In \S \ref {sec:voita} we discuss a conjecture by Vojta concerning the asymptotic behaviour of $k_C/(g-1)$ when $C$ varies among all curves on a surface. We introduce an invariant related to Vojta's conjecture and we prove a bound for it.

The results in this paper go in the direction of understanding \emph{bounded negativity} (see \cite {BR}). The \emph{Bounded Negativity Conjecture} (BNC)
predicts that on a surface of general type over $\mathbb{C}$
the self--intersection of negative curves is bounded below. Nagata's conjecture, which we mentioned above, is also a sort of bounded negativity assertion.  As a general reference on both bounded negativity and Nagata's conjecture, see \cite{Harbourne}. Also Vojta's conjecture is related to bounded negativity, as we discuss in \S \ref {sec:voita}.
As a consequence of Theorem \ref {thm:MAIN}, we have the following information on negative curves:

\begin{cor}
Suppose BNC fails, so that there
exists a sequence $(C_{n})_{n \in \mathbb N}$ of negative curves of genus $g_{n}$
with $\lim g_{n}=\infty$ and $\lim C_{n}^{2}=-\infty$. Then
\[
\limsup_n\frac{K\cdot C_{n}}{g_{n}-1}\leq3.
\]
\end{cor}

In conclusion, the authors would like to thank B. Harbourne and J. Ro\'e for useful exchanges of ideas about the application to Nagata's Conjecture in \S \ref {sec:nagata}.

\section{The proof of the main theorem} \label{sec:proof}

Our proof relies on the following result by Miyaoka's
(see  \cite[Cor 1.4]{Miyaoka}):

\begin{thm}
\label{thm:(Miyaoka)} Let $C$ be curve on a surface $X$ with $\kappa\geq 0$.  Then for all $\alpha\in[0,1]$,
we have:
\begin{equation}
{\alpha^{2}}(C^{2}+3k_C-6g+6)-4\alpha(k_C-3g+3)+2a\geq0.\label{eq:Miyaoka 1}
\end{equation}
Suppose $C$ is not isomorphic to $\mathbb{P}^{1}$ and
$k_C>3(g-1)$. Then
\begin{equation}
2(k_C-3g+3)^{2}-a(C^{2}+3k_C-6g+6)\leq0.\label{eq:Miyaoka 1bis}
\end{equation}
Suppose  in addition $K^{2}>0$. Then
\begin{equation}
(\frac{c_{2}}{K^{2}}-1)k_C^{2}+(4(g-1)+a)k_C-2(g-1)(3(g-1)+a)\geq(\frac{c_{2}}{K^{2}}-\frac{1}{3})[k_C^{2}-C^{2}K^{2}]\geq0.\label{eq:Miyaoka 2}
\end{equation}
\end{thm}
\begin{proof}
Inequality \eqref {eq:Miyaoka 1} is \cite[Thm 1.3, (i)]{Miyaoka} and \eqref {eq:Miyaoka 1bis} is \cite[Thm 1.3, (ii)]{Miyaoka}.
As for \eqref {eq:Miyaoka 2}
this is \cite[Thm 1.3, (iii)]{Miyaoka}, which is stated there under the
assumption that $X$ is minimal of general type and $C\not\simeq\mathbb{P}^{1}$.
However Miyaoka's argument works more generally  under the weaker assumption $K^{2}>0$.
\end{proof}

We are now ready for the:
\begin{proof}[Proof of Theorem \ref{thm:MAIN}]
Let us prove (A). Let $C$ be a negative curve on $X$
 not isomorphic to $\mathbb{P}^{1}$.  Then
$-aC^{2}\ge  0$, with equality if and only if $a=0$. If $k_C\leq3(g-1)$, there is nothing
to prove. Let us suppose $k=k_C>3(g-1)$ and set $\mathfrak g=g-1$. By
\eqref{eq:Miyaoka 1bis}, one has
\begin{equation}\label{eq:p}
P(k):=2(k-3\mathfrak g)^{2}-a(3k-6\mathfrak g ) \leq 0,
\end{equation}
with strict inequality if $a>0$.

If $a=0$ and $g=0$, the polynomial $P$ is positive, thus this cannot occur. In the remaining cases,  $k_C$ is less than or equal to the largest root of  $P$, whence we get  \eqref {eq:1}.

 Suppose $g>1$ and let $\epsilon>0$ be such that $k_C=(3+\epsilon)(g-1)$. By \eqref {eq:1} we obtain
\begin{equation}\label{eq:big}
\epsilon(g-1)\leq\frac{3}{4}a+\frac{1}{4}\sqrt{9a^{2}+24a(g-1)},
\end{equation}
hence
\[
\epsilon\leq\frac{3a}{4(g-1)}+\frac{1}{4(g-1)}\sqrt{9a^{2}+24a(g-1)}\leq\frac{3}{4}a+\frac{1}{4}\sqrt{9a^{2}+24a}\leq\frac{3}{2}a+1.
\]
which yields \eqref {eq:3}.
On the other hand, \eqref {eq:big} reads
\[
4\epsilon(g-1)-3a\leq\sqrt{9a^{2}+24a(g-1)},
\]
and by squaring one gets \eqref {eq:4}, finishing  the proof of (A).

Next we prove (B). Let $\beta_0>3$. By \eqref {eq:3} and \eqref {eq:4}, negative curves with $\beta>\beta_0$
have bounded genus $g$, therefore by \eqref {eq:1} also $k_C$ is bounded, hence the arithmentic genus $p$ is bounded.

Suppose $K$ is big. By \cite[Cor. 2.2.7]{Lazarsfeld} there exist
$m\in\mathbb{N}^{*}$, an ample divisor $A$ and an effective divisor
$Z$ such that
\[
mK\equiv A+Z.
\]
Since $Z$ is effective, the set of integers $Z\cdot C$, when $C$ varies among
negative curves, is bounded from below, therefore the degree $A\cdot C=(mK-Z) \cdot C$ of these curves with respect to the ample
divisor $A$ is bounded. Hence, by results of Chow--Grothendieck \cite{Grothendieck},
\cite[Lecture 15]{Mumford}, one has only finitely many components of the Hilbert scheme
containing points corresponding to such curves.
Since they are negative, these components contain only one curve, proving the assertion.
\end{proof}

\section{Surfaces not of general type} \label {sec:nagata}

We want to deduce from Theorem \ref {thm:MAIN} a result valid for any smooth surface.

\begin{thm} \label{prop:nongen}Let $Y$ be any smooth projective surface. Let $\eta\in \Pic(Y)$ be such that
$\vert K_{Y}+\eta\vert$ is big and $\vert 2\eta \vert$ contains a base point free pencil.
Let $\beta_{0}>3$. Then there are finitely many negative curves  $D$ on $Y$ such that
\begin{equation}\label{eq:nongen}
k_D \geq\beta_{0}(g-1)+\frac{\beta_{0}-2}{2}D\cdot \eta.
\end{equation} \end{thm}

\begin{proof}
Under the hypotheses there is a smooth curve $B\equiv 2\eta$ intersecting
all negative curves of $X$ only at smooth points with intersection  multiplicity 1. Let us make a double cover $f:X\to Y$ branched along $B$.  Then for all negative curve $D$ of $X$, $C=f^ *(D)$ is irreducible, negative and  $g(C)=2g(D)-1+\eta\cdot D$ by Hurwitz formula.  Since
$f_*(K_X)=K_Y\oplus (K_{Y}+\eta)$, then $\kappa(X)=2$ and
we finish by applying  (B) of Theorem \ref {thm:MAIN} to
$X$ and to $C=f^{*}(D)$.
\end{proof}

As an application, we take $Y_n$ to be the plane blow-up at $n$ general points.
Then ${\rm Pic}(Y_n)\cong \mathbb Z^ {n+1}$
 generated by the classes of the pull--back $L$ of a line and of minus the exceptional divisors $E_1,\ldots, E_{n}$ over the blown--up points. We write $D=(d,m_{1},\dots,m_{n})$ to denote the
class of a curve with components $d,m_{1},\dots,m_{n}$ with respect to this basis.
We may use exponential notation to denote repeated  $m_i$'s.
Thus $-K=(3, 1^ {n})$.

 \begin{prop}\label {prop:nagata}  Fix  $\beta_0>3$. There are finitely many irreducible curves of class $D=(d,m_{1},\dots,m_{n})$  on $Y_n$ such that
\begin{equation}\label{eq:nagata}
\frac {D^ 2}d \leq \frac {2-\beta_0}  {\beta_0} \big (1+ \frac M d\big )\,\,\, \text {where} \,\,\,\,\, M=\sum_{i=1}^ n m_i.
\end{equation}
\end{prop}
\begin{proof} We apply Theorem \ref {prop:nongen}, by taking $\eta=4L$. Indeed
$-K+\eta=(1,-1^ {n})$ is big.  \end{proof}

Recall that
\[
\epsilon_n=\inf \{\frac d M, \,\,\, \text {for all effective} \,\,\, D=(d,m_{1},\dots,m_{n}),\,\,\, \text {such that}\,\,\, M >0 \}
\]
is the \emph{Seshadri constant} of $Y_n$. Nagata's Conjecture (see \cite {Nagata}) is equivalent to say that $\epsilon _n=1/\sqrt n$ if $n\ge 10$ (see \cite {HarbourneRoe}).

\begin{rem} Proposition  \ref {prop:nagata} can be seen as weak form of Nagata's Conjecture.
Indeeed, let us look at the \emph{homogenous case} $D=(d,m^ n)$ with $n\geq 10$. Nagata's Conjecture predicts that, if the $n$ blown--up points are in very general position, there is no  irreducible such curve  with $D^ 2<0$ (see \cite {CM, Nagata}), i.e., with
$d<\sqrt n m$. The conclusion of Proposition \ref {prop:nagata} is not absence of curves, but finiteness of their set, under a stronger assumption than Nagata's. Let us look at the difference between the two assumptions. In the $(m,d)$--plane \eqref {eq:nagata} applies   to pairs $(m,d)$ in the first quarter below the hyperbola  with equation
\begin{equation} \label{eq:hyper}
\beta_0d^ 2+d(\beta_0-2)-n\beta_0m^ 2+(\beta_0-2)nm= 0
\end{equation}
drawn in black in Figure \ref {fig:1}. One of its asymptotes (drawn in blue) is parallel to the \emph {Nagata line} $d=\sqrt n m$ (drawn in  green).

\begin{figure}[h]

        \begin{center}

           \begin{overpic}[scale=0.50]{Picture.png}

                \end{overpic}
        \end{center}
                 \caption{The hyperbola, its asymptote and the Nagata line}
       \label{fig:1}
    \end{figure}

Since for all effective divisor $D=(d,m_1,\ldots,m_n)$ one has $d/M\ge \epsilon_n$, one has
approximations $d/M\ge e_n$ to Nagata's conjecture for any lower approximation $e_n$ of $\epsilon_n$.
The best known in general is the one in \cite  {HarbourneRoe}
\begin{equation}\label{eq:hr}
\epsilon _n \geq e_n=\sqrt {\frac 1n \big (1 - \frac 1 {f(n)}\big )}
\end{equation}
where $f(n)$ is, for most $n$, an explicitly given quadratic function of $n$ (see \cite [Corollary 1.2.3] {HarbourneRoe}). For $n=10$ in the homogeneous case the best result is $e_{10}=228/721$ (see \cite{Petra}).

The hyperbola \eqref{eq:hyper} meets the line $d=e_n m$, therefore Proposition \ref{prop:nagata} gives some information in an unlimited region where the above approximations to Nagata do not work.
\end{rem}

\begin{rem} Proposition \ref{prop:nagata} implies that there are finitely many irreducible curves of class $D=(d,m_{1},\dots,m_{n})$  on $Y_n$ such that
\begin{equation}\label{eq:sesh}
\frac {D^ 2}d \leq \frac {2-\beta_0}  {\beta_0} \big (1+ \frac 1 \epsilon_n \big),
\end{equation}
where $\epsilon_n$ can be replaced by $e_n$ in Harbourne--Ro\'e's approximation \eqref {eq:hr}.
This result is not surprising. Indeed, J. Ro\'e pointed out to us an easy argument which shows that  there is no irreducible curve of class $D=(d,m_{1},\dots,m_{n})$  on $Y_n$ such that
\[
\frac {D^ 2} d< - \frac 1 {n\epsilon_n}
\]
which is better than \eqref {eq:sesh}, and the difference
\[
\frac {\beta_0-2}  {\beta_0} \big (1+ \frac 1 \epsilon_n \big)- \frac 1 {n\epsilon_n}
\]
 tends to $\frac {\beta_0-2}  {\beta_0}\sim \frac 1 3$ for $n\to \infty$. \end{rem}

\section{A boundedness result for non--negative curves}\label{sec:pos}

With the usual notation, for a curve $C$ on the surface $X$ with $C^ 2\neq 0$, we set  $x_C:=\frac{\delta_C}{C^{2}}$, with the usual convention that the index $C$ can be dropped if there is no ambiguity.

\begin{thm}\label{thm:MAIN2}
Consider real numbers  $x_{0}>\frac{1}{2}$ and $\beta_{0}>3$. Let $C$ be a curve
on $X$, with $\kappa(X)\geq 0$,  satisfying the following conditions:\\
(1)  $C^{2}>0$, $k_C=\beta(g-1)$ with $\beta>\beta_{0}$ and $g>1$;\\
(2) $x_C>x_{0}$.\\
Then
\begin{equation}
g\leq a\frac{(\beta-2)}{(\beta-3)^{2}}\frac{3x_{0}-1}{2x_{0}-1}+1\label{eq:positif 1},
\end{equation}
\begin{equation}
k_C\leq a\frac{(\beta-2)}{\beta(\beta-3)^{2}}\frac{3x_{0}-1}{2x_{0}-1},\label{eq:positif 2}
\end{equation}
\begin{equation}
k_C\leq 2(g-1)  +a\frac {(\beta-2)^ 2}{(\beta-3)^ 2}\cdot \frac{3x_{0}-1}{2x_{0}-1}.\label{eq:positif 3}
\end{equation}
If $\kappa(X)=2$, then the Hilbert scheme of curves  on $X$ satisfying
 (1) and (2) has finitely many irreducible components.\end{thm}
\begin{proof} One has
\[

k_C-2(g-1)=2\delta-C^ 2=(\beta-2)(g-1)>0.

\]
Hence
by \eqref{eq:Miyaoka 1}, we have
\begin{equation}
P(\alpha):=\alpha^{2}(3\delta-C^{2})+\alpha\frac{2(\beta-3)}{\beta-2}(C^{2}-2\delta)+a\geq0\,\label{eq:Positive}
\end{equation}
 for $\alpha \in [0,1]$. Since the
 coefficient of the leading term of  $P$ is positive, the minimum of $P(\alpha)$
 is attained for
\[
\alpha_{0}=\frac{(\beta-3)(2\delta-C^{2})}{(\beta-2)(3\delta-C^{2})}.
\]
Since $\beta>3$, we have $\alpha_{0}\in ]0,1[$, and, by \eqref {eq:Positive} we have
\[
P(\alpha_0)=-\frac{(\beta-3)^{2}(2\delta-C^{2})^{2}}{(\beta-2)^{2}(3\delta-C^{2})}+a\geq0
\]
Thus
\[
\frac{a}{\mu}\geq\frac{(2\delta-C^{2})^{2}}{(3\delta-C^{2})}\,\,\, \text {where}\,\,\, \mu=\frac{(\beta-3)^{2}}{(\beta-2)^{2}},
\]
hence
\[
\frac{a}{\mu}\cdot \frac{3\delta-C^{2}}{2\delta-C^{2}}\geq2\delta-C^{2}.
\]
We have
\[
\frac{3\delta-C^{2}}{2\delta-C^{2}}=\frac{3x-1}{2x-1}<\frac{3x_{0}-1}{2x_{0}-1}
\]
because $\frac{3x-1}{2x-1}$ is decreasing for $x>x_{0}>\frac{1}{2}$, hence
\[
(\beta-2)(g-1)=k_C-2(g-1)=2\delta-C^{2}\leq\frac{a}{\mu}\cdot \frac{3x_{0}-1}{2x_{0}-1},
\]
which implies \eqref {eq:positif 1}, \eqref {eq:positif 2} and \eqref {eq:positif 3}.
Moreover both $g$ and  $k_C$ are bounded from above and, if $\kappa(X)=2$, we conclude with the same argument at the end of the proof of Theorem \ref {thm:MAIN}.
\end{proof}

\begin{cor}
Let be $\beta_{0}>3$ and let $(C_{n})_{n\in \mathbb  N}$ be a sequence of curves
on
$X$ with $\kappa=2$ such that $k_{C_{n}}>\beta_{0}(g(C_{n})-1)$, $C_{n}^{2}>0$ and $\lim g(C_{n})=\infty$.
Then
\begin{equation}\label{eq:limsup}
\lim_n \frac {\delta_{C_n} }  {C_{n}^{2}} = \frac{1}{2},
\end{equation}
moreover  $\lim_n \frac {g(C_n)} {\delta_{C_{n}}}=\lim_n \frac{K\cdot C_n}{\delta_n}=0.$
\end{cor}
\begin{proof}
Let $C$ be a curve with $k_C=(3+\epsilon)(g-1)$, $\epsilon >0$. Since $(1+\epsilon)(g-1)=2\delta-C^2$, we get $\frac{\delta}{C^2}-\frac{1}{2}=(1+\epsilon)\frac{g-1}{2C^2}\geq 0$. Therefore $\lim\inf_n \frac{\delta_{n}}{C_{n}^{2}} \geq \frac{1}{2}$. On the other hand,  by Theorem \ref{thm:MAIN2} we obtain $\lim\sup_n \frac{\delta_{n}}{C_{n}^{2}} \leq \frac{1}{2}$. The remaining limits are readily computed.
\end{proof}

\begin{example}
For Shimura curves on Shimura surfaces, we have $K\cdot C=4(g-1)$ and, if there is one,
there are infinitely many of them, with the genus going to infinity.
\end{example}

\section{On a conjecture by Vojta}\label{sec:voita}

The results in \S \ref {sec:pos} are reminiscent of the following conjecture (see \cite {Autissier}), which predicts that curves of bounded geometric
genus on a surface of general type form a bounded family:
\begin{conjecture}\label{conj:1}
Let $X$ be a smooth projective surface. There exist
constants $A,\, B$ such that for any curve $C$ we have
\[
k_C\leq A(g-1)+B.
\]

\end{conjecture}
If this conjecture is satisfied for  $X$ with $\kappa(X)=2$, then $X$
 contains finitely many curves of genus $0$ or
$1$. This is known to hold for minimal surfaces
with big cotangent bundle (see \cite{Bogomolov, Deschamps}).

A stronger version of Conjecture \ref {conj:1} is the following conjecture by Vojta (see again \cite {Autissier}):
\begin{conjecture}\label{conj:2}
For any real number $\epsilon>0$, we can take $A=4+\epsilon$ in Conjecture \ref {conj:1}
(and $B=B(\epsilon)$  a function of $\epsilon$).
\end{conjecture}

An even stronger, more recent version, predicts that $A=2+\epsilon$ (see \cite {Autissier}).

\begin{rem}
If $C$ is a smooth curve on $X$, then $k_C=2(g-1)-C^2$, therefore if BNC holds, then Vojta's conjecture holds for smooth curves with $A=2$. This suggests a close relationship between Vojta's conjecture and BNC.
\end{rem}

Miyaoka proves in \cite {Miyaoka} that Conjecture \ref {conj:1} also holds if  $K^{2}>c_{2}$ and he gives explicit values for $A$ and
$B$,  but they are far away from the ones predicted by Conjecture \ref {conj:2}.
Moreover Miyaoka proves that $k_C\leq 3(g-1)$
for (smooth) compact ball quotient surfaces on which the equality is attained by an
infinite number of curves, i.e.,  Shimura curves, if they
exists.

In \cite{Autissier} one proves
that for surfaces whose universal cover is the bi--disk, one has
\[
k_C\leq4(g-1).
\]
This is sharp since for Shimura curves on Shimura
surfaces, one has $k_C=4(g-1)$.

For $X$ a surface, we define
\[
\Lambda_{X}=\sup_{(C_{n})_{n\in\mathbb{N}}} \big \{ \limsup_{n}\frac{K\cdot C_{n}}{g_{n}-1}\big \}
\]
where $(C_{n})_{n\in\mathbb{N}}$ varies among all sequences of curves
$C_{n}$ in $X$ of genus $g_{n}=g(C_{n})>1$ with $\lim_{n}g_{n}=\infty$.
Conjecture \ref {conj:1} says that $\Lambda_X<\infty$.

If $X$ has trivial canonical bundle, then $\Lambda_X=0$. Apart form this  case,
and the aforementioned cases  studied in  \cite{Autissier, Miyaoka},  nothing is known about $\Lambda_X$.
The following result gives us a  piece of information:

\begin{thm}\label{thm:voita} Let $X$ be a surface of general type. Let $L$ be a very ample divisor on $X$, and let $\gamma$ be the arithmetic genus of curves in $\vert L\vert$. Then
\[
\Lambda_X\geq \frac {K\cdot L}{L^ 2+\gamma-1}>0.
\]
\end{thm}

\begin{proof}  Look at the surface $X$ embedded in $\mathbb P^ r$, with $r\geq 3$, via $L$.
Then take a general projection $\pi: X\to \mathbb P^ 2$. Consider a general rational curve  of degree $n$ in $\mathbb P^ 2$ and let $C_n$ be its pull--back via $\pi$. Then $C_n\in \vert nL\vert$.

By Hurwitz formula, the ramification divisor $R$ of $\pi$ is such that $R\equiv K+3L$. So Hurwitz formula again, implies that the geometric  genus $g_n$ of $C_n$
satisfies
\[
2g_n-2=nL\cdot K+(3n-2)L^ 2.
\]
Therefore
\[
\frac {K\cdot C_n}{g_n-1}= \frac {L\cdot K}{L^ 2+\gamma-1- \frac {L^ 2}n}
\]
and this proves the assertion.\end{proof}

\begin{example} Suppose that, in the setting of  Theorem \ref {thm:voita}, one has $K=mL$, with $m>0$. Then
\[
\Lambda_X\geq	 \frac {2m}{m+3}.
\]
So there are sequences $(X_n)_{n\in \mathbb N}$ of surfaces, e.g., complete intersections of increasing degree in projective space, with $m\to \infty$, and therefore $\Lambda_{X_n}\to 2$ (from below).
\end{example}

config: arxiv


TITLE: Extreme value verification using Bordered Hessian matrix
QUESTION [0 upvotes]: Determine the minimum value of $$bcx+cay+abz$$ subject to the conditions $$xyz=abc$$
Here is the solution in the book (using the Lagrange multipliers)
$$F = bcx+cay+abz+k(xyz-abc)$$
After solving it, the minimum value given in the book, comes out at $x=a, y=b, z=c$
I was trying to verify whether this answer corresponds to minimum or maximum value using bordered Hessian matrix (let's call it $H$).
$$H = \begin{bmatrix}0&bc&ca&ab\\bc&0&-c&-b\\ac&-c&0&-a\\ab&-b&-a&0\end{bmatrix}$$
$$H = -3a^2b^2c^2$$
$$H < 0$$
This implies $x=a, y=b, z=c$ correspond to maximum value
Where am I going wrong?

REPLY [3 votes]: Even in the unconstrained case it is not sufficient to only check the determinant of the Hessian, and one needs to calculate the determinant of the minors in order to classify exterma.
In the constrained case things get more complicated, as you are only allowed to move on the constrained surface $xyz=abc$ (in this instance). It means that you need to discard eigenvalues associated with directions out of this surface.
The way to do it is outlined in Wikipedia (in the following $m$ is the number of constraints)

Specifically, sign conditions are imposed on the sequence of leading principal minors (determinants of upper-left-justified sub-matrices) of the bordered Hessian, for which the first $2m$ leading principal minors are neglected... A sufficient condition for a local minimum is that all of these minors have the sign of $\left(–1\right)^{m}$.

Applying this in your case, $m=1$ so we discard the first $2m=2$ minors and we need to calculate only
$$H_{3}=\begin{vmatrix}0&bc&ca\\bc&0&-c\\ac&-c&0\end{vmatrix}=-2abc^{3}<0$$
$$H_{4}=\begin{vmatrix}0&bc&ca&ab\\bc&0&-c&-b\\ac&-c&0&-a\\ab&-b&-a&0\end{vmatrix}=-3a^{2}b^{2}c^{2}<0$$
Both of which has a sign of $\left(-1\right)^{m}=-1$, so this results in a minima.


TITLE: Antisymmetric matrices in effective field theory
QUESTION [4 upvotes]: I'm trying to construct a nonlinear $d$-dimensional version E&M as an effective field theory. Let $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ be the field strength. The most general action I can write down will be a generic function of the following? Treating $F_\mu^\nu$ as a $d$-dimensional antisymmetric matrix, I can create Lorentz invariant combinations by taking
$$\text{tr}(F^2),~~\text{tr}(F^4),~~\text{tr}(F^6),\dots,\text{tr}(F^{2n}).$$
My question is: What is the largest value of $n$ such that the above traces are all independent of one another?

REPLY [4 votes]: The number of independent invariants of electromagnetic field in $d$ dimensions is $\lfloor \frac d 2 \rfloor $, where $\lfloor \,. \rfloor$  is the floor function.
The number could be guessed because for a general (non-null) field  there would be a Lorentz transform of the field strength tensor (see e.g. here) which would have at most $\lfloor \frac d 2 \rfloor $ independent real parameters:
$$
F'_{\mu\nu} = \Lambda \Lambda F  = \begin{bmatrix}
  \begin{matrix}0 & B_1 \\ -B_1 & 0\end{matrix} &  0 & \cdots & 0 \\
  0 & \begin{matrix}0 & B_2 \\ -B_2 & 0\end{matrix} & & 0 \\
  \vdots & & \ddots & \vdots \\
  0 & 0 & \cdots & \begin{matrix}0 & B_r\\ -B_r & 0\end{matrix} \\
    & & & & \begin{matrix}0 \\ & \ddots \\ & & 0 \end{matrix}
\end{bmatrix}
$$
So, for  $\text{tr}(F^n)$ with $n>d$ there should be recurrence relation expressing the trace through traces of lower powers.
Explicit construction of algebraic invariants of field strength tensor in arbitrary dimensions and the discussion of algebraically special cases could be found here:

Sokolowski, L. M., Occhionero, F., Litterio, M., & Amendola, L. (1993). Classical electromagnetic radiation in multidimensional space-times. Annals of Physics, 225(1), 1-47, doi:10.1006/aphy.1993.1050.


TITLE: Unimodular matrix definition?
QUESTION [4 upvotes]: I'm a bit confused. Based on Wikipedia:

In mathematics, a unimodular matrix M is a square integer matrix
  having determinant +1, 0 or −1. Equivalently, it is an integer matrix that is invertible over the integers.

So determinant could be +1, 0 or −1. But a matrix is invertible only if determinat is non-zero! In fact, from Wolfram:

A unimodular matrix is a real square matrix A with determinant det(A) = -1|+1.

Which is right answer?

REPLY [6 votes]: Well spotted. In a case like this, it's a good idea to check the article's history (using the "View history" link at the top). In the present case, the error was introduced only two days ago by an anonymous user in this edit (which I just reverted).


TITLE: Find the difference equation for {2, 4, 16, 256, ...}
QUESTION [0 upvotes]: Write a difference equation to represent the change during the nth interval as a function of the previous term in the sequence.
b. {2,4,16, 256,...}
I know that an= 22n but I can't figure out how to find Δan = c*an.
I'm lost. I'd really appreciate some help.

REPLY [1 votes]: The ratio $a_n/a_{n-1}$ is always $a_{n-1}$. So the term difference is
$$
\Delta a_n \equiv a_n-a_{n-1} = a_{n-1}^2 - a_{n-1} = ( a_{n-1} -1) a_{n-1}
$$
If you insist on writing this as $c \cdot a_n$ the form would be
$$
\Delta a_n = \frac{a_{n-1}-1}{a_{n-1}} a_n
$$