April 08, 2026
Abstract
For a wide range of functions \(W\colon\mathbb{N}\to\mathbb{N}\), we establish a general result for estimating weighted averages of the form \[\operatorname{\mathbb{E}}^{W}_{n \le N} f(\vartheta(n))= \frac{1}{W(N)}\sum_{n=1}^N (W(n)-W(n-1))f(\vartheta(n)),\] where \(f\colon \{1,\ldots,N\}\to\mathbb{C}\) is an arbitrary function, and \(\vartheta(n)\) is any arithmetic function that adheres to a certain Gaussian distribution condition.(For instance, one may take \(\vartheta(n)=\Omega(n)\), where \(\Omega(n)\) counts the number of prime factors of \(n\) with multiplicity, or \(\vartheta(n)=s_q(p_n)\), where \(s_q\) is the sum-of-digits function in base \(q\) and \(p_n\) denotes the \(n\)-th prime. Additional natural examples are discussed in the paper.) Building on our main theorem, we show that if \(h(n)\) is a function from a Hardy field with polynomial growth then \((h(\vartheta(n)))_{n\in\mathbb{N}}\) is uniformly distributed mod \(1\) if and only if one of the following (mutually exclusive) conditions is satisfied:
\(\lim_{x\to\infty} \frac{|h(x)-p(x)|}{x \log x}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\);
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{\sqrt{x}}=\infty\) for each \(p(x)\in \mathbb{Q}[x]\) and there exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x}<\infty\).
This leads to several novel applications. For example, it follows that \((\Omega(n)^c)_{n\in\mathbb{N}}\) is uniformly distributed mod \(1\) if and only if \(c\) is a non-integer greater than \(\frac{1}{2}\).
Let \(\mathbb{N}=\{1,2,3,\ldots\}\) be the set of positive integers. Given a sequence \(W \colon \mathbb{N} \to [0,\infty)\), we define its discrete derivative \(\Delta W\) by \[\Delta W(N) = \begin{cases} W(N) - W(N-1), & \text{if } N \ge 2, \\[6pt] W(1), & \text{if } N = 1. \end{cases}\] We also use second and third order discrete derivatives \(\Delta^2 W = \Delta(\Delta W)\) and \(\Delta^3 W = \Delta(\Delta^2 W)\).
Definition 1. Let \(\mathscr{W}\) denote the class of functions \(W\colon \mathbb{N}\to[0,\infty)\) that are eventually non-decreasing and satisfy \(\lim_{N \to \infty} W(N) = \infty\). For \(W \in \mathscr{W}\) and \(f \colon \{1, \ldots, N\} \to \mathbb{C}\), we define the weighted discrete average \[\label{eq:equivalence95of95derivatives950} \operatorname{\mathbb{E}}^{W}_{n \le N} f(n) = \frac{1}{W(N)} \sum_{n=1}^N \Delta W(n) \, f(n).\tag{1}\] When \(W(N) = N\), this reduces to the standard Cesàro average, which we denote by \[\operatorname{\mathbb{E}}_{n \le N} f(n) = \frac{1}{N} \sum_{n=1}^N f(n).\]
In addition to the averages introduced in 1 , we consider two averaging schemes whose weights do not arise from a single underlying function: the binomial mean \[\label{def95bin} \operatorname{\mathbb{E}}^{\text{bin}}_{n \le N} f(n) = \frac{1}{2^N} \sum_{n=1}^N \binom{N}{n} f(n),\tag{2}\] and its “parity-neutral” variant \[\label{def952bin} \operatorname{\mathbb{E}}^{2\text{bin}}_{n \le N} f(n) = \operatorname{\mathbb{E}}^{\text{bin}}_{n \le N} \!\left( \frac{f(2n) + f(2n+1)}{2} \right) = \frac{1}{2^{N+1}} \sum_{n=1}^{2N} \binom{N}{\big\lfloor \frac{n}{2} \big\rfloor} f(n).\tag{3}\]
Let \(f\colon \{1,\ldots,N\}\to\mathbb{C}\) be an arbitrary function. The purpose of this paper is to develop a new technique for estimating averages of the form \[\label{eqn95general95theta95average} \operatorname{\mathbb{E}}^{W}_{n \le N} f(\vartheta(n))\tag{4}\] for an extensive class of arithmetic functions \(\vartheta\colon\mathbb{N}\to\mathbb{N}\).
To describe the class of sequences \(\vartheta(n)\) to which our method applies, consider the probability density function of the Gaussian normal distribution with mean \(\mu\) and standard deviation \(\sigma\) given by \[g(x,\mu,\sigma) = \frac{1}{\sigma \sqrt{2\pi}} \, e^{-\frac{(x-\mu)^2}{2\sigma^2}},\qquad x\in\mathbb{R}.\] Let \(\mathscr{L}\subseteq\mathscr{W}\) denote the class of all sequences \(L\in\mathscr{W}\) for which \(\Delta L(n)\in\{0,1\}\) for all but finitely many \(n\in\mathbb{N}\); this requirement can be interpreted as a discrete analogue of sublinear growth. The scope of our main result includes all arithmetic functions \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) for which there exists \(L\in\mathscr{L}\) such that \[\label{gaussian95condition} \sum_{k\in \mathbb{N}} \bigg|\frac{|\{1\leqslant n\leqslant N: \vartheta(n)=k\}|}{N}- g(k,L(N),\sqrt{L(N)})\bigg|= {\mathrm o}_{N\to\infty}(1).\tag{5}\] This “Gaussian condition” roughly says that the distribution of \(\vartheta(n)\) is close in variation distance to a normal distribution with mean \(L(N)\) and standard deviation \(\sqrt{L(N)}\).
There are many types of naturally occurring functions that satisfy this condition. For example, this property is satisfied by various summatory functions arising in number theory:
Let \[\Omega(n)=\sum_{p^k} 1_{p\mid n}\] denote the number of prime factors of \(n\) counted with multiplicity. Then \(\Omega(n)\) satisfies condition 5 with \(L(N) = \lfloor \log\log(N) \rfloor\). This follows from [1] (see [2] for details).
Let \[\omega(n)=\sum_{p} 1_{p\mid n}\] be the number of prime factors of \(n\) counted without multiplicities. Then \(\omega(n)\) satisfies 5 with \(L(N) = \lfloor \log\log(N) \rfloor\). This can be derived from [1].
The function \(\Omega(q_n)\) satisfies 5 with \(L(N) = \lfloor \log\log(N) \rfloor\), where \(q_n\) is the \(n\)-th squarefree number. This follows from [1].
Given an integer \(q\geqslant 2\), let \(s_q(n)\) denote the sum of digits function in base \(q\), i.e., \[s_q(n)=\sum_{j\geqslant 0}\varepsilon_j(n),\quad\text{where}\quad n=\sum_{j\geqslant 0} \varepsilon_j(n) q^j.\] Then \(s_q(n)\) satisfies 5 with \(L(N) = \lfloor \log(N) \rfloor\); this can be derived from [3].
The function \(s_q(p_n)\) satisfies 5 with \(L(N) = \lfloor \log(N) \rfloor\), where \(p_n\) is the \(n\)-th prime number. In fact, a stronger assertion is proved in [4].
While our main results are stated below for general \(\vartheta(n)\) satisfying condition 5 , it is worth noting that they are already new when \(\vartheta(n)\) is taken to be any one of the functions \(\Omega(n), \omega(n), \Omega(q_n), s_q(n), s_q(p_n)\).
For technical reasons, in our main theorem we need to restrict our attention to weights in \(\mathscr{W}\) that exhibit suitable regularity at infinity. More precisely, we require that the weight function \(W\) has the property that \[\label{eqn95smoothish95weights} \lim_{N\to\infty} \frac{N\cdot \Delta^2 W(N)}{\Delta W(N)} \text{ exists in } \mathbb{R}\cup \{-\infty,\infty\}.\tag{6}\] This assumption is mild, as many natural weights satisfy 6 . For instance, if \(W\) is a function that belongs to a Hardy field (defined on page ) then 6 holds and hence \(W \in \mathscr{W}^*\) if and only if \(W \in \mathscr{W}\).
Henceforth, we use \(\mathscr{W}^*\) to denote the subclass of \(\mathscr{W}\) that satisfy 6 . The following is our main theorem.
Theorem 1. Let \(W\in\mathscr{W}^*\) and assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 for some \(L\in\mathscr{L}\).
Uniformly over all \(f\colon \mathbb{N}\to\mathbb{C}\) with \(\|f\|_\infty\leqslant 1\), \[\label{eq95main95cesaro95scale} \operatorname{\mathbb{E}}_{n\leqslant N}f(\vartheta(n)) =\operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}}f(n)+o_{N\to\infty}(1).\tag{7}\]
If \(\lim_{N\to\infty}\frac{\log(W(N))}{\log(N)}=0\) then uniformly over all \(f\colon \mathbb{N}\to\mathbb{C}\) with \(\|f\|_\infty\leqslant 1\), \[\label{eq95main95log95scale} \operatorname{\mathbb{E}}_{n\leqslant N}^W f(\vartheta(n)) =\operatorname{\mathbb{E}}_{n\leqslant N}^W\operatorname{\mathbb{E}}_{k\leqslant L(n)}^{2\text{bin}}f(k)+o_{N\to\infty}(1).\tag{8}\]
If \(\lim_{N\to\infty}\frac{\log(W(N))}{N}=0\) and \(\lim_{N\to\infty}\frac{\log(W\circ L)(N)}{\log(N)}=0\) then uniformly over all \(f\colon \mathbb{N}\to\mathbb{C}\) with \(\|f\|_\infty\leqslant 1\), \[\label{eq95main95loglog95scale} \operatorname{\mathbb{E}}_{n\leqslant N}^{W\circ L} f(\vartheta(n)) = \operatorname{\mathbb{E}}^{W}_{n\leqslant L(N)}f(n)+o_{N\to\infty}(1).\tag{9}\]
Let us point out some interesting consequences of 1 to illustrate its usefulness. Applied to \(\vartheta(n) = \Omega(n)\) and \(W(N) = N\), part [itm95main951] gives that the Cesàro average of \(f(\Omega(n))\) is asymptotically equal to the parity-neutral binomial mean of \(f(n)\), i.e., \[\label{eqn95Cesaro95for95Omega} \operatorname{\mathbb{E}}_{n\leqslant N}f(\Omega(n)) =\operatorname{\mathbb{E}}_{n\leqslant\lfloor\log\log N\rfloor}^{2\text{bin}}f(n)+o_{N\to\infty}(1).\tag{10}\] Moreover, part [itm95main953] shows that the double-logarithmic average of \(f(\Omega(n))\) corresponds to the Cesàro average of \(f(n)\), \[\label{eqn95double-log-averages95for95Omega} \operatorname{\mathbb{E}}_{n\leqslant N}^{\log\log} f(\Omega(n)) = \operatorname{\mathbb{E}}_{n\leqslant\lfloor\log\log N\rfloor}f(n)+o_{N\to\infty}(1).\tag{11}\] Analogous formulas hold when \(\Omega(n)\) is replaced by either \(\omega(n)\) or \(\Omega(q_n)\).
In a similar vein, when \(\vartheta(n)=s_q(n)\), then part [itm95main951] of 1 yields \[\label{eqn95Cesaro95for95sumofdigits} \operatorname{\mathbb{E}}_{n\leqslant N}f(s_q(n)) =\operatorname{\mathbb{E}}_{n\leqslant\lfloor\log N\rfloor}^{2\text{bin}}f(n)+o_{N\to\infty}(1),\tag{12}\] and part [itm95main953] gives \[\label{eqn95log-averages95for95sumofdigits} \operatorname{\mathbb{E}}_{n\leqslant N}^{\log} f(s_q(n)) = \operatorname{\mathbb{E}}_{n\leqslant\lfloor\log N\rfloor}f(n)+o_{N\to\infty}(1).\tag{13}\] The same formulas apply when \(s_q(n)\) is replaced by \(s_q(p_n)\).
As an application of 1, we obtain new results on the equidistribution of sequences modulo \(1\) and new results in ergodic theory on the convergence of ergodic averages along arithmetic functions. These are described in the following two subsections.
We say that two functions \(f\colon [a,\infty)\to\mathbb{R}\) and \(g\colon [b,\infty)\to\mathbb{R}\) are eventually identical if there exists \(c\geqslant\max\{a,b\}\) such that \(f(x)=g(x)\) for all \(x\in [c,\infty)\). This yields a natural equivalence relation on the set of all real-valued continuous functions which are defined for all sufficiently large real arguments. A germ at \(\infty\) of real-valued functions is an equivalence class under this relation. Note that the operations of pointwise addition and pointwise multiplication of real-valued continuous functions extend in a natural way to germs at \(\infty\). Under these operations, the set of all germs at \(\infty\) forms a commutative ring.
A Hardy field \(\mathcal{H}\) is any subfield of this ring that is closed under differentiation, in the sense that if the germ at \(\infty\) of a differentiable function belongs to \(\mathcal{H}\) then so does the germ at \(\infty\) of its derivative.
Typical examples of Hardy fields include the field of rational functions, and the field of logarithmico-exponential functions (i.e., the smallest field closed under compositions and containing all polynomials, \(\log(x)\), and \(\exp(x)\)). By abuse of language, we say a function \(f\colon [a,\infty)\to\mathbb{R}\) belongs to a Hardy field if its germ at \(\infty\) belongs to a Hardy field. Examples include functions such as \(x^c\) for \(c\in\mathbb{R}\), \(x\log(x)\), or \(\exp(\sqrt{\log x})\)). It is a classical fact that functions belonging to a Hardy field are eventually monotone. In particular, this means that highly oscillatory functions such as \(\sin(x)\) do not belong to any Hardy field. For more information on Hardy fields, see [5]–[7] or [8].
We are now ready to formulate one of the main applications of 1.
Theorem 2. Let \(h\) be a Hardy field function with polynomial growth (i.e., there exist \(c,d\geqslant 1\) such that \(|h(x)|\leqslant x^d\) for all \(x\in [c,\infty)\)). Assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 for some \(L\in\mathscr{L}\). The following are equivalent:
The sequence \((h(\vartheta(n)))_{n\in \mathbb{N}}\) is uniformly distributed mod \(1\).
One of the following two (mutually exclusive) conditions is satisfied:
\(\lim_{x\to\infty} \frac{|h(x)-p(x)|}{x \log x}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\);
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{\sqrt{x}}=\infty\) for each \(p(x)\in \mathbb{Q}[x]\) and there exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x}<\infty\).
By L’Hospital’s rule, condition [itm95ud95of95Hardy95functions95along95Omega95Cesaro95ii95a] in Theorem 2 is equivalent to the assertion \[\lim_{x\to\infty} \frac{|h'(x)-p(x)|}{\log x}=\infty,\quad \forall p(x)\in \mathbb{Q}[x],\] where \(h'\) denotes the derivative of \(h\). In light of Boshernitzan’s theorem [7], this is in turn equivalent to \(h'(n)\) being uniformly distributed mod 1. This leads us to the following corollary.
Corollary 1. Let \(h\) be a function from a Hardy field with polynomial growth. If \(h'(n)\) is uniformly distributed mod \(1\) then \(h(\Omega(n))\) is uniformly distributed mod \(1\). The same applies to the sequences \(h(\omega(n))\), \(h(\Omega(q_n))\), \(h(s_q(n))\), and \(h(s_q(p_n))\).
The reverse implication in 1 does not hold. For example, if \(h(x)=x^{\frac{2}{3}}\) then \(h'(n)=\frac{2}{3 \sqrt[3]{n}}\) is not uniformly distributed mod \(1\), yet \(h(\Omega(n))\) is uniformly distributed mod \(1\) due to condition [itm95ud95of95Hardy95functions95along95Omega95Cesaro95ii95b] in Theorem 2.
Here is another corollary that follows immediately from Theorem 2.
Corollary 2. Let \(c>0\). The sequence \(\Omega(n)^c\) is uniformly distributed mod \(1\) if and only if \(c\in \big(\frac{1}{2},\infty\big)\backslash\mathbb{N}\). The same applies to the sequences \(\omega(n)^c\), \(\Omega(q_n)^c\), \(s_q(n)^c\), and \(s_q(p_n)^c\).
The surprising conclusion that we can draw from 2 is that if \(\vartheta(n)\) satisfies 5 for some \(L\in\mathscr{L}\) then there are many functions \(h\) from a Hardy field such that \((h(n))_{n\in\mathbb{N}}\) is uniformly distributed mod \(1\), but \((h(\vartheta(n)))_{n\in\mathbb{N}}\) is not. However, it follows from part [itm95main953] of 1 that if one switches from Cesàro averages to averages weighted by \(L(N)\) then uniform distribution mod \(1\) along \(n\) and along \(\vartheta(n)\) become equivalent.
Theorem 3. Let \(h\) be a function from a Hardy field with polynomial growth. Then the following are equivalent:
\(\lim_{x\to\infty} \frac{|h(x)-p(x)|}{\log x}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\);
\(h(n)\) is uniformly distributed mod \(1\);
\(h(p_n)\) is uniformly distributed mod \(1\), where \(p_n\) denotes the \(n\)-th prime;
\(h(\Omega(n))\) is uniformly distributed mod \(1\) with respect to double-logarithmic averages. The same applies to the sequences \(h(\omega(n))\) and \(h(\Omega(q_n))\).
\(h(s_q(n))\) is uniformly distributed mod \(1\) with respect to logarithmic averages. The same applies to the sequence \(h(s_q(p_n))\).
The equivalence between [ud95of95Hardy95functions95along95Omega95loglog95i], [ud95of95Hardy95functions95along95Omega95loglog95ii], and [ud95of95Hardy95functions95along95Omega95loglog95iii] in 3 is the content of [9]. The equivalence between [ud95of95Hardy95functions95along95Omega95loglog95ii] and [ud95of95Hardy95functions95along95Omega95loglog95iv] follows from 11 . The equivalence between [ud95of95Hardy95functions95along95Omega95loglog95ii] and [ud95of95Hardy95functions95along95sq95log95v] follows from 13 .
It is worth mentioning that we don’t know whether it is possible to replace the double-logarithmic averages in part [ud95of95Hardy95functions95along95Omega95loglog95iv] with logarithmic averages. (However, we think that it is unlikely to be true.)
For the purposes of this paper, a measure preserving system will refer to a triple \((X,\mu,T)\) where \(X\) is a compact metric space, \(T\colon X\to X\) is a continuous map, and \(\mu\) is a Borel probability measure on \(X\) that is preserved under the transformation \(T\), meaning that \(\mu(T^{-1}A)=\mu(A)\) holds for all Borel sets \(A\subseteq X\).
Given a point \(x\in X\), the sequence \((T^nx)_{n\in\mathbb{N}}\) is called the orbit of \(x\) under \(T\). Let \(C(X)\) denote the space of all (complex-valued) continuous functions on \(X\). The system \((X,\mu,T)\) is ergodic if the orbit of \(\mu\)-almost every point is uniformly distributed in \(X\) with respect to \(\mu\), i.e., for any \(f \in C(X)\) we have \[\lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(T^{n}x) =\int_X f \;d\mu,\qquad \text{for}~\mu\text{-a.e.}~x\in X.\] We call \((X,\mu,T)\) uniquely ergodic if the orbit of every point is uniformly distributed in \(X\) with respect to \(\mu\), that is, for any \(f \in C(X)\), \[\lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(T^{n}x) =\int_X f \;d\mu,\qquad\forall x\in X.\] Finally, we say \((X,\mu,T)\) is non-atomic if the measure \(\mu\) is non-atomic.
The following theorem is our dynamical application of 1. It provides new insights into the behavior of orbits of the form \((T^{\vartheta(n)}x)_{n\in\mathbb{N}}\) in measure-preserving systems, establishing refined equidistribution properties and clarifying the distinctions between different variants of the ergodic theorem along arithmetic functions.
Theorem 4. Assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 for some \(L\in\mathscr{L}\).
For any uniquely ergodic measure preserving system \((X,\mu,T)\) and any \(f \in C(X)\) we have \[\lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(T^{\vartheta(n)}x) =\int_X f \;d\mu,\qquad\forall x\in X.\]
\(\vartheta(n)\) has the strong sweeping out property, i.e., for any non-atomic measure preserving system \((X,\mu,T)\) there exists a residual set of Borel sets \(B\) such that \[\begin{align} \limsup_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N 1_B(T^{\vartheta(n)}x) =1,\qquad \text{for}~\mu\text{-a.e.}~x\in X, \\ \liminf_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N 1_B(T^{\vartheta(n)}x) =0,\qquad \text{for}~\mu\text{-a.e.}~x\in X. \end{align}\]
If \(L(N)=\lfloor W(N)\rfloor\) for some function \(W\) from a Hardy field satisfying \[\lim_{N\to\infty}\frac{\log(W(N))}{\log(N)}=0,\] then for any ergodic measure preserving system \((X,\mu,T)\) and any \(f\in L^{\infty}(X,\mu)\) we have \[\begin{align} \lim_{N \to \infty} \mathbb{E}^{W}_{n\leqslant N}\, f(T^{\vartheta(n)}x) =\int_X f \;d\mu,\qquad \text{for}~\mu\text{-a.e.}~x\in X. \end{align}\]
When \(\vartheta(n)=\Omega(n)\) then part [itm95thm95erg951] of 4 was proved in [10], part [itm95thm95erg952] was shown in [2], and part [itm95thm95erg953] appeared in [11]. However, for other choices of \(\vartheta(n)\), such as \(\Omega(q_n)\), \(s_q(n)\), or \(s_q(p_n)\), 4 provides new results.
The paper is organized as follows. The proof of our main technical result, 1, is split across three sections. In 2 we prove the first part (formula 7 ). The proof relies on quantitative estimates for binomial coefficients and elementary results regarding equivalent methods of summation.
In 3 we provide a proof of the second part of 1 (formula 8 ). The principal idea is to show that 7 implies 8 , and the main ingredient in this derivation is 4, which is a result from an unpublished preprint of Michael Boshernitzan.
In 4, we first prove that \[\operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}_{k\leqslant n}^{\text{bin}}f(k)=\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\operatorname{\mathbb{E}}_{k\leqslant n}f(k) = \operatorname{\mathbb{E}}_{n\leqslant\lfloor{N/2\rfloor}}f(n)+o_{N\to\infty}(1),\] which is the content of 9. This result is then used to prove the third and final part of 1 (formula 9 ).
In 5, we give conditions for convergence of a sequence with respect to \(\operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}}\) averages and use these results to derive Theorem 2 from 1.
Finally, in 6 we provide a proof of 4.
We thank Tristán Radić for suggestion that condition 5 applies to the sequences \(s_q(n)\) and \(s_q(p_n)\).
The goal of this section is to prove the first part of 1. For the convenience of the reader, we state this part separately as a theorem.
Theorem 5. Let \(W\in\mathscr{W}^*\), \(L\in\mathscr{L}\), and assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 . Then uniformly over all \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\operatorname{\mathbb{E}}_{n\leqslant N}f(\vartheta(n)) = \operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}}f(n)+o_{N\to\infty}(1).\]
The main idea behind the proof of 5 is to first show that the Gausssian condition 5 implies \[\operatorname{\mathbb{E}}_{n\leqslant N}f(\vartheta(n))\approx \sum_{n\in\mathbb{N}} g(n,L(N),\sqrt{L(N)}) f(n),\] and then use the fact that the values of the normalized binomial coefficients \(\frac{1}{2^{M+1}}\binom{M}{\lfloor m/2 \rfloor}\) form a close approximation to the gaussian curve with mean \(M\) and standard deviation \(\sqrt{M}\), which ultimately gives \[\sum_{n\in\mathbb{N}} g(n,L(N),\sqrt{L(N)}) f(n) \approx \sum_{n\in\mathbb{N}} \frac{1}{2^{L(N)+1}}\binom{L(N)}{\lfloor n/2 \rfloor} f(n) = \operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}}f(n).\] The details rely on elementary yet technical computations, beginning with 1, which characterizes when weighted sums yield equivalent methods of summation.
Let \((\alpha_{n,N})_{n, N\in \mathbb{N}}\) and \((\beta_{n,N})_{n, N\in\mathbb{N}}\) be nonnegative doubly indexed sequences satisfying \[\lim_{N\to\infty}\sum_{n\in \mathbb{N}}\beta_{n,N}=\lim_{N\to\infty}\sum_{n\in \mathbb{N}}\alpha_{n,N}=1.\] We seek conditions ensuring that the averages weighted by \((\alpha_{n,N})\) and those weighted by \((\beta_{n,N})\) agree asymptotically, meaning that \[\label{eq:changing95weights} \sum_{n\in\mathbb{N}} \alpha_{n,N}\, f(n) = \sum_{n\in\mathbb{N}} \beta_{n,N}\, f(n) + o_{N\to\infty}(1).\tag{14}\] Rewriting equation (14 ) and using the triangle inequality, we have \[\label{eq:changing95weights952} \lim_{N\to\infty}\left|\sum_{n\in \mathbb{N}}\alpha_{n,N}f(n)-\sum_{n\in \mathbb{N}}\beta_{n,N}f(n)\right|\leqslant \|{f}\|_{\infty}\cdot\lim_{N\to\infty}\sum_{n\in \mathbb{N}}|\alpha_{n,N}-\beta_{n,N}|,\tag{15}\] and so it suffices to show that \(\lim_{N\to\infty}\sum_{n\in \mathbb{N}}|\alpha_{n,N}-\beta_{n,N}|=0\). To this end, we have the following lemma.
Lemma 1. Suppose that \((\alpha_{n,N})_{n,N\in \mathbb{N}}\) and \((\beta_{n,N})_{n,N\in \mathbb{N}}\) are nonnegative doubly indexed sequences, and \((I_N)_{N\in \mathbb{N}}\) is a sequence of intervals such that \[\label{eq:condition95in95changing95weights} \lim_{N\to\infty}\sum_{n\in \mathbb{N}}\beta_{n,N}=\lim_{N\to\infty}\sum_{n\in \mathbb{N}}\alpha_{n,N}= \lim_{N\to\infty}\sum_{n\in I_N}\alpha_{n,N}=1.\tag{16}\] Assume that there is a function \(E\colon\mathbb{N}\rightarrow \mathbb{R}\) which tends to \(0\) such that \(|1-\frac{\beta_{n,N}}{\alpha_{n,N}}|\leqslant E(N)\) for all \(N\in \mathbb{N}\) and \(n\in I_N\). Then uniformly over all \(f:\mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\sum_{n\in \mathbb{N}}\alpha_{n,N}f(n)=\sum_{n\in \mathbb{N}}\beta_{n,N}f(n)+o_{N\to\infty}(1).\]
Proof. Observe that \(\lim_{N\to\infty}\sum_{n\in I_N}\beta_n=1\), since \[\begin{align} & \lim_{N\to\infty} \sum_{n\in I_N}\beta_n = \lim_{N\to\infty} \sum_{n\in I_N}(\beta_n-\alpha_n) + \lim_{N\to\infty} \sum_{n\in I_N}\alpha_n\\ =& \lim_{N\to\infty} \sum_{n\in I_N}\alpha_n\left(\frac{\beta_n}{\alpha_n}-1\right) + 1 = \lim_{N\to\infty} E(N)+1 =1. \end{align}\] Then \[\begin{align} \sum_{n\in \mathbb{N}}|a_{n,N}-\beta_{n,N}| = \sum_{n\in I_N}|a_{n,N}-\beta_{n,N}|+o_{N\to\infty}(1) = \sum_{n\in I_N}\alpha_{n,N}\cdot \left|1-\frac{\beta_{n,N}}{\alpha_{n,N}}\right|+o_{N\to\infty}(1). \end{align}\]
Further,
\[\begin{align} \lim_{N\to\infty}\sum_{n\in I_N}\alpha_{n,N}\cdot \left|1-\frac{\beta_{n,N}}{\alpha_{n,N}}\right| &\leqslant\lim_{N\to\infty}\sum_{n\in I_N}\alpha_{n,N}\cdot E(N) \\ &=\lim_{N\to\infty}E(N)\cdot\sum_{n\in I_N}a_{n,N} \\ &= 0\cdot 1=0. \end{align}\] This means that \(\lim_{N\to\infty}\sum_{n\in \mathbb{N}}|\alpha_{n,N}-\beta_{n,N}|=0\) as desired. ◻
Remark 6. By an almost identical argument it can be shown that the condition \(|1-\frac{\beta_{n,N}}{\alpha_{n,N}}|\leqslant E(N)\) in the lemma above can be replaced by the assumption that \(|1-\frac{\beta_{n,N}}{\alpha_{n,N}}|\leqslant E(n)\), so long as \(\lim_{N\to\infty}\alpha_{n,N} = 0\) for each fixed \(n\in \mathbb{N}\).
Proof of 5. Recall that \(g(x,\mu,\sigma)\) denotes the probability density function of the Gaussian normal distribution. For \(n,N\in \mathbb{N}\), define
\(\alpha_{n,N} = \frac{1}{N}\cdot|\{1\leqslant m\leqslant N:\vartheta (m)=n\}|\),
\(\beta_{n,N} = g(n,L(N),\sqrt{L(N)})\),
\(\gamma_{n,N} = g(2\lfloor n/2 \rfloor,L(N),\sqrt{L(N)})\),
\(\delta_{n,N} = \frac{1}{2^{L(N)+1}}\binom{L(N)}{\lfloor n/2 \rfloor}\).
We will show that the values \[\operatorname{\mathbb{E}}_{n\leqslant N}f(\vartheta(n)),\;\sum_{n\in \mathbb{N}}\alpha_{n,N}f(n),\;\sum_{n\in \mathbb{N}}\beta_{n,N}f(n),\;\sum_{n\in \mathbb{N}}\gamma_{n,N}f(n),\;\sum_{n\in \mathbb{N}}\delta_{n,N}f(n),\;\operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}}f(n)\] are each equal up to a \(o_{N\to\infty}(1)\) term, uniformly over all \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\). First we can note the equalities \(\operatorname{\mathbb{E}}_{n\leqslant N}f(\vartheta(n))=\sum_{n\in \mathbb{N}}\alpha_{n,N}f(n)\) and \(\operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}}f(n) =\;\sum_{n\in \mathbb{N}}\delta_{n,N}f(n)\) hold by definition. Next, we have \(\sum_{n\in \mathbb{N}}\alpha_{n,N}f(n)=\sum_{n\in \mathbb{N}}\beta_{n,N}f(n)+o_{N\to\infty}(1)\) by equations (5 ) and (15 ).
The last two equalities will follow from Lemma 1. Consider \(|1-\frac{\beta_{n,N}}{\gamma_{n,N}}|\). When \(n\) is even, we have \(\beta_{n,N} = \gamma_{n,N}\) and so \(|1-\frac{\beta_{n,N}}{\gamma_{n,N}}|=0\). When \(n\) is odd, we have \(\gamma_{n,N} = \beta_{n-1,N}\) and so \[\frac{\beta_{n,N}}{\gamma_{n,N}} = \frac{\frac{1}{\sqrt{L(N)}\cdot \sqrt{2\pi}} \, e^{-\frac{(n-L(N))^2}{2L(N)}}}{\frac{1}{\sqrt{L(N)}\cdot \sqrt{2\pi}} \, e^{-\frac{((n-1)-L(N))^2}{2L(N)}}} = e^{\frac{2L(N)-2n+1}{2L(N)}} = e^{1-\frac{n}{L(N)}+\frac{1}{2L(N)}}.\]
Next, we will approximate a sum of the form \(\sum_{n=A}^B\beta_{n,N}\) with the corresponding integral \(\int_A^Bg(x,L(N),\sqrt{L(N)})dx\). However, we know that \[\int_A^Bg(x,L(N),\sqrt{L(N)})dx = \int_A^B \frac{1}{\sqrt{L(N)}\cdot \sqrt{2\pi}} \, e^{-\frac{(x-L(N))^2}{2L(N)}}dx = \int_{\frac{A-L(N)}{\sqrt{2L(N)}}}^{\frac{B-L(N)}{\sqrt{2L(N)}}} \frac{1}{\sqrt{\pi}} \, e^{-x^2}dx.\] From this it follows that we can put \(I_N = [L(N) - (L(N))^{3/5},L(N)+(L(N))^{3/5}]\) so that \(\sum_{n\in I_N}\beta_{n,N} \to 1\) as \(N\to\infty\). But for \(n\in I_N\) we have \[|1-e^{1-\frac{n}{L(N)}+\frac{1}{2L(N)}}|\leqslant 1- e^{1-\frac{L(N)-(L(N))^{3/5}}{L(N)}+\frac{1}{2L(N)}} = 1-e^{\frac{-1}{L(N)^{2/5}}+\frac{1}{2L(N)}}\to 0 \text{ as }N\to\infty.\] We can conclude that the hypothesis of Lemma 1 is satisfied and so \(\sum_{n\in \mathbb{N}}\beta_{n,N}f(n) = \sum_{n\in \mathbb{N}}\gamma_{n,N}f(n)+o_{N\to\infty}(1)\). For the last equality, we refer to a fact about the asymptotics of binomial coefficients, whose proof can be found in [12]. Namely, there is a function \(E\colon \mathbb{N}\rightarrow \mathbb{R}\) with \(\lim_{N\to\infty}E(N)=0\) such that for \(|N-2n| = O(N^{2/3})\), \[\left|1-\frac{\sqrt{\frac{2}{\pi N }}\cdot 2^N\cdot e^{\frac{(N-2n)^2}{2N}}}{\binom{N}{n}}\right|\leqslant E(N).\] Seeing as how \(\frac{\sqrt{\frac{2}{\pi L(N) }}\cdot 2^{L(N)}\cdot e^{\frac{(L(N)-2n)^2}{2L(N)}}}{\binom{L(N)}{n}} = \frac{\frac{1}{\sqrt{2\pi L(N) }}\cdot e^{\frac{(L(N)-2n)^2}{2L(N)}}}{\frac{1}{2^{L(N)+1}}\binom{L(N)}{n}} = \frac{\gamma_{2n,N}}{\delta_{2n,N}}\), it follows that \(|1-\frac{\gamma_{n,N}}{\delta_{n,N}}|\leqslant E(L(N))\) for all \(n\) in an interval of the form \([L(N)-O(L(N)^{2/3}),L(N)+O(L(N)^{2/3})]\). By Lemma 1, we get \(\sum_{n\in \mathbb{N}}\gamma_{n,N}f(n) = \sum_{n\in \mathbb{N}}\delta_{n,N}f(n) +o_{N\to\infty}(1)\), completing the proof. ◻
In this section, we will provide some background on weighted averages in order to derive equation (8 ) from equation (7 ). We will begin with some preliminary facts, the first of which is the Stolz-Cesàro Theorem
Theorem 7 (Stolz-Cesàro). Let \(A\colon \mathbb{N}\rightarrow \mathbb{C}\) and \(B\colon\mathbb{N}\rightarrow (0,\infty)\) be functions such that \(B\) is strictly increasing with \(\lim_{N\to\infty}B(n) = \infty\). Let \(\ell\in \mathbb{C}\). \[\label{eq:stolz95cesaro} \text{ If } \lim_{N\to\infty}\frac{\Delta A(N)}{\Delta B(N)} = \ell \text{ then } \lim_{N\to\infty}\frac{ A(N)}{ B(N)} = \ell.\tag{17}\]
Next is a standard lemma, say from [13].
Lemma 2. Let \(W\in \mathscr{W}\) and let \(f:\mathbb{N}\rightarrow \mathbb{C}\). Suppose that \(\lim_{n\to\infty}f(n) = \ell\). Then \(\lim_{N\to\infty}\operatorname{\mathbb{E}}^W_{n\leqslant N}f(n) = \ell\).
Proof. Apply Theorem 7 with \(A(N) = \sum_{n=1}^N\Delta W(n)f(n)\) and \(B(N) = W(N)\), so that \(\frac{\Delta A(N)}{\Delta B(N)} = \frac{\Delta W(N) f(N)}{\Delta W(N)} = f(N)\). ◻
When \(W\) grows fast enough, we have a converse to this statement.
Lemma 3. Let \(W\in \mathscr{W}\) satisfy \(\lim_{N\to\infty}\frac{\Delta W(N)}{W(N)}>0\), let \(f:\mathbb{N}\rightarrow \mathbb{C}\), and let \(\ell\in \mathbb{C}\). Suppose that \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n) = \ell\). Then \(\lim_{N\to\infty}f(N)=\ell\).
Proof. Note that \[\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n) =\frac{W(N-1)}{W(N)}\operatorname{\mathbb{E}}_{n\leqslant N-1}^Wf(n)+\frac{\Delta W(N)}{W(N)}f(N),\] and so \[f(N) = \left(\frac{\Delta W(N)}{W(N)}\right)^{-1}\left(\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n)-\frac{W(N-1)}{W(N)}\operatorname{\mathbb{E}}_{n\leqslant N-1}^Wf(n)\right).\] Observe that \(\frac{W(N-1)}{W(N)} = 1-\frac{\Delta W(N)}{W(N)}\), which after taking limits gives that \[\lim_{N\to\infty}f(N) = \frac{\ell-(1-\lim_{N\to\infty}\frac{\Delta W(N)}{W(N)})\cdot \ell}{\lim_{N\to\infty}\frac{\Delta W(N)}{W(N)}}=\ell.\] ◻
The next lemma can be attributed to Michael Boshernitzan, who has a variant for Hardy functions in an unpublished preprint [14]. The proof we provide here is adapted from Boshernitzan’s proof.
Lemma 4. Let \(W\in \mathscr{W^*}\) and suppose that \(\lim_{N\to\infty}\frac{\log W(N)}{\log(N)}=0\). Then uniformly over all \(f:\mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\label{eq:boshernitzan} \operatorname{\mathbb{E}}_{n\leqslant N}^W(\operatorname{\mathbb{E}}_{k\leqslant n}f(k)) = \operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n)+o_{N\to\infty}(1).\tag{18}\]
Proof. Recall the summation by parts formula1, which says that for sequences \((x_n),(y_n)\), \[\begin{align} \label{eq:summation95by95parts} \sum_{n=1}^N\Delta x_n \cdot y_n = x_{N}y_{N}-\sum_{n=1}^{N-1}x_n\cdot \Delta y_{n+1}. \end{align}\tag{19}\]
Let \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) satisfy \(\| f \|_{\infty}\leqslant 1\). Now we will take the expression for \(\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n)\) and manipulate it into the form \(\operatorname{\mathbb{E}}_{n\leqslant N}^W\operatorname{\mathbb{E}}_{k\leqslant n}f(k)+o_{N\to\infty}(1)\). Put \(F(n) = \sum_{k=1}^nf(k)\) so that \(\Delta F(n) = f(n)\) and \(\frac{F(N)}{N}=\operatorname{\mathbb{E}}_{n\leqslant N}f(n)\). Apply summation by parts to obtain \[\begin{align} &\operatorname{\mathbb{E}}^W_{n\leqslant N}f(n) =\frac{1}{W(N)}\sum_{n=1}^N\Delta W(n)\cdot f(n) =\frac{1}{W(N)}\sum_{n=1}^N\Delta W(n)\cdot \Delta F(n)\\ =& \frac{1}{W(N)}\left(\Delta W(N)\cdot F(N)- \sum_{n=1}^{N-1}F(n)\cdot \Delta^2 W(n+1) \right)\\ =& \frac{N\cdot \Delta W(N)}{W(N)}\cdot \frac{F(N)}{N}-\frac{1}{W(N)}\sum_{n=1}^{N-1}\Delta W(n)\cdot \frac{F(n)}{n}\cdot \frac{n\cdot \Delta^2W(n+1)}{\Delta W(n)}, \end{align}\] which means that \[\label{eq:boshernitizan95lem95eq} \operatorname{\mathbb{E}}^W_{n\leqslant N}f(n)=\frac{N \Delta W(N)}{W(N)} \operatorname{\mathbb{E}}_{n\leqslant N}f(n)-\frac{W(N-1)}{W(N)} \operatorname{\mathbb{E}}_{n\leqslant N-1}^W\left(\frac{n\cdot \Delta^2W(n+1)}{\Delta W(n)}\operatorname{\mathbb{E}}_{m\leqslant n}f(m)\right).\tag{20}\] All that is left is to show the following claims:
\(\lim_{N\to\infty}\frac{W(N-1)}{W(N)}=1\).
\(\lim_{N\to\infty}\frac{N\cdot \Delta^2W(N+1)}{\Delta W(N)}=-1\).
\(\lim_{N\to\infty}\frac{N\cdot \Delta W(N)}{W(N)}=0\).
For any bounded function \(g\colon \mathbb{N}\rightarrow \mathbb{C}\), \(\operatorname{\mathbb{E}}_{n\leqslant N}^Wg(n) = \operatorname{\mathbb{E}}_{n\leqslant N-1}^Wg(n)+o_{N\to\infty}(1)\).
From these claims, equation (20 ) becomes \[\begin{align} \operatorname{\mathbb{E}}^W_{n\leqslant N}f(n) =&o_{N\to\infty}(1)- (1+o_{N\to\infty}(1))\cdot \operatorname{\mathbb{E}}_{n\leqslant N-1}^W\left(\operatorname{\mathbb{E}}_{m\leqslant n}f(m)\cdot (-1+o_{n\to\infty}(1))\right)\\ =& \operatorname{\mathbb{E}}_{n\leqslant N-1}^W\left(\operatorname{\mathbb{E}}_{m\leqslant n}f(m)\right)+o_{N\to\infty}(1)\\ =& \operatorname{\mathbb{E}}_{n\leqslant N}^W\left(\operatorname{\mathbb{E}}_{m\leqslant n}f(m)\right)+o_{N\to\infty}(1). \end{align}\] as desired.
To prove the above claims, we will make use of Theorem 7 along with the fact that \(\lim_{N\to\infty}\frac{N\cdot \Delta^2 W(N)}{\Delta W(N)}\) exists since \(W\in \mathscr{W}^*\). First, we know that \(\lim_{N\to\infty}\frac{\log W(N)}{\log N}=0\) and so \(\lim_{N\to\infty}\frac{\log W(N)}{ N}=0\). Then, \[0=\lim_{N\to\infty}\frac{\log W(N)}{N} = \lim_{N\to\infty}\frac{\Delta \log W(N)}{\Delta (N)} = \lim_{N\to\infty}\log\left(\frac{W(N)}{W(N-1)}\right).\] Therefore \(\lim_{N\to\infty}\frac{W(N)}{W(N-1)} = 1\) and we have proven (1). Next, we have \[1=\lim_{N\to\infty}\frac{W(N+1)}{W(N)} = \lim_{N\to\infty}\frac{\Delta W(N+1)}{\Delta W(N)} =\lim_{N\to\infty}\frac{\Delta^2 W(N+1)}{\Delta^2 W(N)}\] and so to prove (2) and (3), it suffices to show that \[\lim_{N\to\infty}\frac{ N\cdot \Delta W(N+1)}{W(N)}=0 \qquad\text{and}\qquad\lim_{N\to\infty}\frac{N\cdot \Delta^2 W(N+1)}{\Delta W(N)}=-1.\] To this end, we will use the approximation \(\log(1+x)\approx x\) as \(x\to0\) and apply Theorem 7 several times to the limit \(\lim_{N\to\infty}\frac{\log W(N)}{\log N} = 0\): \[\begin{align} 0=&\lim_{N\to\infty}\frac{\log W(N+1)}{\log (N+1)} = \lim_{N\to\infty}\frac{\Delta \log W(N+1)}{\Delta \log (N+1)} = \lim_{N\to\infty}\frac{ \log \frac{W(N+1)}{W(N)}}{\log \frac{N+1}{N}} \\ =&\lim_{N\to\infty}\frac{ \log (1+\frac{\Delta W(N+1)}{W(N)})}{\log (1+\frac{1}{N})} = \lim_{N\to\infty}\frac{\frac{ \Delta W(N+1)}{W(N)}}{1/N}= \lim_{N\to\infty}\frac{ N\cdot \Delta W(N+1)}{W(N)}\\ =& \lim_{N\to\infty}\frac{ \Delta(N\cdot \Delta W(N+1))}{\Delta W(N)} = \lim_{N\to\infty}\frac{ N\cdot \Delta W(N+1)-(N-1)\Delta W(N)}{\Delta W(N)}\\ =&\lim_{N\to\infty}\frac{N\cdot \Delta^2 W(N+1)}{\Delta W(N)}+1.\label{eq:smoothish95weights95limit} \end{align}\tag{21}\] This shows that \(\lim_{N\to\infty}\frac{ N\cdot \Delta W(N+1)}{W(N)}=0\) and \(\lim_{N\to\infty}\frac{N\cdot \Delta^2 W(N+1)}{\Delta W(N)}=-1\).
Claim (4) follows from the fact that \[\begin{align} \operatorname{\mathbb{E}}_{n\leqslant N-1}^Wg(n)=& \frac{1}{W(N-1)}\sum_{n=1}^{N-1}\Delta W(n)g(n) \\ =&\frac{W(N)}{W(N-1)}\left(\frac{1}{W(N)}\sum_{n=1}^{N}\Delta W(n)g(n) -\frac{\Delta W(N)}{W(N)}g(N)\right)\\ =&(1+o_{N\to\infty}(1))\cdot\left(\operatorname{\mathbb{E}}_{n\leqslant N}^Wg(n)+o_{N\to\infty}(1) \right) = \operatorname{\mathbb{E}}_{n\leqslant N}^Wg(n)+o_{N\to\infty}(1). \end{align}\] This completes the proof. ◻
By applying \(\operatorname{\mathbb{E}}_{n\leqslant N}^W\) to both sides of equation (7 ) and invoking Lemma 4, we obtain equation (8 ) as an immediate corollary:
Corollary 3. Let \(W\in\mathscr{W}^*\), \(L\in\mathscr{L}\), and assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 . If \(\lim_{N\to\infty}\frac{\log(W(N))}{\log(N)}=0\) then uniformly over all \(f\colon \mathbb{N}\to\mathbb{C}\) with \(\|f\|_\infty\leqslant 1\), \[\operatorname{\mathbb{E}}_{n\leqslant N}^W f(\vartheta(n)) =\operatorname{\mathbb{E}}_{n\leqslant N}^W\operatorname{\mathbb{E}}_{k\leqslant L(N)}^{2\text{bin}}f(k)+o_{N\to\infty}(1).\]
The goal of this section is to prove formula 9 , which is the third and final part of 1. Let us state this result as a standalone theorem.
Theorem 8. Let \(W\in\mathscr{W}^*\), \(L\in \mathscr{L}\), and assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 . If \(\lim_{N\to\infty}\frac{\log (W\circ L)(N)}{\log(N)}=0\) then uniformly over all \(f\colon \mathbb{N}\to\mathbb{C}\) with \(\|f\|_\infty\leqslant 1\), \[\operatorname{\mathbb{E}}_{n\leqslant N}^{W\circ L} f(\vartheta(n)) = \operatorname{\mathbb{E}}^{W}_{n\leqslant L(N)}f(n)+o_{N\to\infty}(1).\]
We will derive 8 from 3; one of the key components in this derivation is the following theorem.
Theorem 9. Uniformly over all \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\label{eqn95BC95of95C951} \operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}_{k\leqslant n}^{\text{bin}}f(k)=\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\operatorname{\mathbb{E}}_{k\leqslant n}f(k) = \operatorname{\mathbb{E}}_{n\leqslant\lfloor{N/2\rfloor}}f(n)+o_{N\to\infty}(1).\tag{22}\]
From 9, we obtain the following immediate corollary.
Corollary 4. Uniformly over all \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\label{eqn95BC95of95C952} \operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}_{k\leqslant n}^{2\text{bin}}f(k)= \operatorname{\mathbb{E}}_{n\leqslant N}f(n)+o_{N\to\infty}(1).\tag{23}\]
Proof. Using the definition of \(\operatorname{\mathbb{E}}^{2\text{bin}}\) and invoking 9, we get \[\begin{align} \operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}_{k\leqslant n}^{2\text{bin}}f(k) &= \frac{1}{2} \operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}_{k\leqslant n}^{\text{bin}}f(2k) + \frac{1}{2} \operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}_{k\leqslant n}^{\text{bin}}f(2k+1) \\ &= \frac{1}{2} \operatorname{\mathbb{E}}_{n\leqslant\lfloor{N/2\rfloor}} f(2n) + \frac{1}{2} \operatorname{\mathbb{E}}_{n\leqslant\lfloor{N/2\rfloor}}f(2n+1)+o_{N\to\infty}(1) \\ &= \operatorname{\mathbb{E}}_{n\leqslant N} f(n) +o_{N\to\infty}(1), \end{align}\] as desired. ◻
It remains to prove 9. The leftmost equality in 22 follows from the following fact.
Lemma 5. For any function \(f:\mathbb{N}\rightarrow \mathbb{C}\) and any \(N\in \mathbb{N}\) we have \[\operatorname{\mathbb{E}}_{n\leqslant N}\operatorname{\mathbb{E}}^{\text{bin}}_{k\leqslant n}f(k) = \operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\operatorname{\mathbb{E}}_{k\leqslant n}f(k).\]
The proof of this lemma can be found in [13] by combining Proposition 3.4.4(e) with Example 3.4.7 and Definition 3.4.8. It remains to prove the rightmost equality in Theorem 9. Our strategy will be to show that \(\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\operatorname{\mathbb{E}}_{k\leqslant n}f(k)\) is close to an average of \(\operatorname{\mathbb{E}}_{k\leqslant n}f(k)\) for \(n\) near \(\lfloor N/2 \rfloor\), and that each term of this form is very close to \(\operatorname{\mathbb{E}}_{k\leqslant\lfloor N/2 \rfloor}f(k)\). The idea for this proof strategy, albeit phrased slightly differently, can be found in [15].
Lemma 6. Let \(f:\mathbb{N}\rightarrow \mathbb{C}\) be bounded and \(N,M \in \mathbb{N}\). Then \[|\operatorname{\mathbb{E}}_{n\leqslant N}f(n)-\operatorname{\mathbb{E}}_{n\leqslant M}f(n)|\leqslant\left(\frac{2|N-M|}{\min\{M,N\}+1}\right)\cdot \|{f}\|_{\infty}.\]
Proof. Without loss of generality, assume that \(N\geqslant M\). \[\begin{align} |\operatorname{\mathbb{E}}_{n\leqslant N}f(n)-\operatorname{\mathbb{E}}_{n\leqslant M}f(n)| = &\left|\frac{1}{N+1}\sum_{n=0}^{N}f(n)-\frac{1}{M+1}\sum_{n=0}^{M}f(n)\right|\\ =& \left|\sum_{n=0}^{M}\left(\frac{f(n)}{N+1}-\frac{f(n)}{M+1}\right)+\frac{1}{N+1}\sum_{n=M+1}^{N}f(n)\right|\\ =& \left|\frac{M-N}{N+1}\cdot \frac{1}{M+1}\sum_{n=0}^{M}f(n)+\frac{1}{N+1}\sum_{n=M+1}^{N}f(n)\right|\\ \leqslant& \frac{|M-N|}{N+1}\cdot \frac{1}{M+1}\sum_{n=0}^{M}|f(n)|+\frac{1}{N+1}\sum_{n=M+1}^{N}|f(n)| \\=& \frac{N-M}{N+1}\cdot \operatorname{\mathbb{E}}_{n\leqslant M}|f(n)|+\frac{1}{N+1}\sum_{n=M+1}^{N}|f(n)| \\\leqslant& \frac{N-M}{N+1}\cdot \|f\|_{\infty}+\frac{N-M}{N+1}\|f\|_{\infty}\\ =& \frac{2(N-M)}{N+1}\|f\|_{\infty}. \end{align}\] ◻
Corollary 5. Let \((A_N)_{N\in \mathbb{N}}\) and \((B_N)_{N\in \mathbb{N}}\) be positive integer-valued sequences with \(\lim_{N\to\infty}B_N = \infty\) and \(\lim_{N\to\infty}\frac{A_N}{B_N}= 1\). Then uniformly over all \(f:\mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\operatorname{\mathbb{E}}_{n\leqslant A_N}f(n) = \operatorname{\mathbb{E}}_{n\leqslant B_N}f(n)+o_{N\to\infty}(1).\]
Now for the proof of Theorem 9.
Proof of Theorem 9. Let \(X_1,\dots, X_N\) be i.i.d. random variables with \(\mathbb{P}(X_1=0)=\mathbb{P}(X_i=1)=1/2\), so that \(\sum_{i=1}^NX_i\sim \text{Bin}(N,1/2)\). The central limit theorem states that the sequence \(\frac{\sum_{i=1}^NX_i}{\sqrt{N}}\) converges in distribution to \(\mathcal{N}(0,1)\) as \(N\to\infty\). So for any \(A<B\in \mathbb{R}\), \[\lim_{N\to\infty}\mathbb{P}\left(A<\frac{\sum_{i=1}^NX_i}{\sqrt{N}}<B\right) = \mathbb{P}(A<\mathcal{N}(0,1)<B).\] But \[\lim_{N\to\infty}\mathbb{P}\left(A<\frac{\sum_{i=1}^NX_i}{\sqrt{N}}<B\right) =\mathbb{P}\left(A\sqrt{N}<{\sum_{i=1}^NX_i}<B\sqrt{N}\right)=\frac{1}{2^N}\sum_{n=N/2+A\sqrt{N}}^{N/2+B\sqrt{N}}\binom{N}{n}.\] Taking \(A\to-\infty\) and \(B\to\infty\) we have \(\mathbb{P}(A<\mathcal{N}(0,1)<B)\to 1\). It follows that whenever \(D\) is a function which tends to \(\infty\) faster than \(\sqrt{N}\), we have \(\frac{1}{2^N}\sum_{n=N/2-D(N)}^{N/2+D(N)}\binom{N}{n}\to 1\) as \(N\to\infty\).
For each \(N\in \mathbb{N}\), put \(I_{N} = [\lfloor N/2-N^{2/3} \rfloor,\lfloor N/2+N^{2/3} \rfloor]\). Then \(\lim_{N\to\infty}2^{-N}\cdot\sum_{n\in I_N}\binom{N}{n}=1\), and so \[\begin{align} \left|\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\operatorname{\mathbb{E}}_{k\leqslant n}f(k)-\frac{1}{2^N}\sum_{n\in I_N}\binom{N}{n}\operatorname{\mathbb{E}}_{k\leqslant n}f(k)\right| < \|f\|_{\infty}\cdot o_{N\to\infty}(1). \end{align}\]
Now we will apply Lemma 6. For any \(n\in I_N\) we have that \[\begin{align} |\operatorname{\mathbb{E}}_{k\leqslant\lfloor{N/2\rfloor}}f(k)-\operatorname{\mathbb{E}}_{k\leqslant n}f(k)| \leqslant&\left(\frac{2|\lfloor N/2 \rfloor-n|+1}{\min\{\lfloor N/2 \rfloor,n\}+1}\right)\cdot \|{f}\|_{\infty}\\ \leqslant&\left(\frac{2\lceil{N^{2/3}\rceil}+1}{\lfloor{N/2 - N^{2/3}\rfloor}+1}\right)\cdot \|{f}\|_{\infty}\\ \leqslant& \left(\frac{2}{\lfloor{N^{1/3}/2-1\rfloor}}+o_{N\to\infty}(1)\right)\cdot \|{f}\|_{\infty} = o_{N\to\infty}(1). \end{align}\] So \(|\operatorname{\mathbb{E}}_{k\leqslant\lfloor{N/2\rfloor}}f(k)-\operatorname{\mathbb{E}}_{k\leqslant n}f(k)|\) goes to \(0\) as \(N\to\infty\) uniformly for \(n\in I_N\). Hence, \[\begin{align} \frac{1}{2^N}\sum_{n\in I_N}\binom{N}{n}\operatorname{\mathbb{E}}_{k\leqslant n}f(k) =& \frac{1}{2^N}\sum_{n\in I_N}\binom{N}{n}\operatorname{\mathbb{E}}_{k\leqslant\lfloor N/2 \rfloor}f(k)+o_{N\to\infty}(1)\\ =& \operatorname{\mathbb{E}}_{k\leqslant\lfloor N/2 \rfloor}f(k)\cdot \left(\frac{1}{2^N}\sum_{n\in I_N}\binom{N}{n}\right)+o_{N\to\infty}(1). \end{align}\] In total we have that \[\begin{align} \operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\operatorname{\mathbb{E}}_{k\leqslant n}f(k) =&\frac{1}{2^N}\sum_{n\in I_N}\binom{N}{n}\operatorname{\mathbb{E}}_{k\leqslant n}f(k)+o_{N\to\infty}(1)\\ =& \operatorname{\mathbb{E}}_{k\leqslant\lfloor N/2 \rfloor}f(k)\cdot \left(1+o_{N\to\infty}(1)\right)+o_{N\to\infty}(1)\\ =&\operatorname{\mathbb{E}}_{k\leqslant\lfloor N/2 \rfloor}f(k)+o_{N\to\infty}(1), \end{align}\] which concludes the proof. ◻
The final ingredient in the proof of 8 is a discrete counterpart of the change-of-variables formula for integrals. We introduce this identity next and include a proof for completeness. Recall the standard formula, which states that \[\label{eqn95int95changeofvariable95formula} \int_{1}^{N} (W\circ s)'(x)\, f(s(x))\, dx \;=\; \int_{s(1)}^{s(N)} W'(y)\, f(y)\, dy.\tag{24}\] The following proposition is a discrete variant of 24 .
Proposition 10. Let \(W\in \mathscr{W}\), \(s\in\mathscr{L}\), and suppose that \(\lim_{N\to\infty}\frac{\Delta W(N)}{W(N)}=0\). Then uniformly over all \(f:\mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\operatorname{\mathbb{E}}^{W\circ s}_{n\leqslant N}(f(s(n)))= \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W}(f(k))+ o_{N\to\infty}(1).\]
For the proof of 10, we use the next lemma.
Lemma 7. Let \(s\in \mathscr{L}\) and define \(\hat{s}(k)= \max\{n:s(n)\leqslant k\}\). Let \(W\in \mathscr{W}\) and suppose that \(\lim_{N\to\infty}\frac{\Delta(W\circ \hat{s})(N)}{(W\circ \hat{s})(N)}=0\). Then uniformly over all \(f:\mathbb{N}\rightarrow \mathbb{C}\) with \(\| f \|_{\infty}\leqslant 1\), \[\operatorname{\mathbb{E}}^W_{n\leqslant N}(f(s(n)))= \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W\circ \hat{s}}(f(k))+ o_{N\to\infty}(1).\]
The statement of 10 follows simply by rephrasing Lemma 7 to remove any reference to \(\hat{s}\) by replacing \(W\) with \(W\circ s\). It remains to prove Lemma 7.
Proof of Lemma 7. First we will note that \(\hat{s}\) is a right inverse for \(s\), so that \(s(\hat{s}(k))=k\) for all \(k\in \mathbb{N}\). When \(N\in \mathbb{N}\) is equal to \(\hat{s}(M)\) for some \(M\in \mathbb{N}\), we have that \(\hat{s}(s(N)) = \hat{s}(s(\hat{s}(M))) = \hat{s}(M) = N\). So for \(N\) contained in the image of \(\hat{s}\), we can calculate that \[\begin{align} \operatorname{\mathbb{E}}^W_{n\leqslant N}(f(s(n)))=& \frac{1}{W(N)}\sum_{n=1}^N\Delta W(n) f(s(n)) =\frac{1}{W(N)}\sum_{k=1}^{s(N)}\sum_{n\leqslant N: s(n)=k}\Delta W(n)f(k). \end{align}\] But for each \(k\leqslant s(N)\), \(\{n\leqslant N:s(n)=k\}\) is the interval \(\{ \hat{s}(k-1)+1,\dots, \hat{s}(k)\}\), so \[\sum_{n\leqslant N: s(n)=k}\Delta W(n) = \sum_{n=\hat{s}(k-1)+1}^{\hat{s}(k)}\Delta W(n)=W(\hat{s}(k))-W(\hat{s}(k-1)) = \Delta (W\circ \hat{s})(k).\] Additionally, writing \(W(N) = W(\hat{s}(s(N)) = (W\circ \hat{s})(s(N))\), we have \[\frac{1}{W(N)}\sum_{k=1}^{s(N)}\sum_{n\leqslant N: s(n)=k}\Delta W(n)f(k) = \frac{1}{(W\circ \hat{s})(s(N))}\sum_{k=1}^{s(N)}\Delta (W\circ \hat{s})(k)\cdot f(k) = \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W\circ \hat{s}}f(k).\] It total, we have shown that \(\operatorname{\mathbb{E}}^W_{n\leqslant N}(f(s(n))) = \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W\circ \hat{s}}f(k)\) whenever \(N\) belongs to the image of \(\hat{s}\).
Now suppose that \(N\) does not belong to the image of \(\hat{s}\), so that we have \(\hat{s}(s(N)-1)<N<\hat{s}(s(N))\). Then \(\{n\leqslant N:s(n)=s(N)\}\) is the interval \(\{ \hat{s}(s(N)-1)+1,\dots, N\}\), so \[\sum_{n\leqslant N: s(n)=s(N)}\Delta W(n) = \sum_{n=\hat{s}(s(N)-1)+1}^{N}\Delta W(n)=W(N)-W(\hat{s}(s(N)-1))\] By the argument from the first half of the proof, we have \[\begin{align} &\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(s(n))=\frac{1}{W(N)}\sum_{n=1}^{\hat{s}(s(N))}\Delta W(n)f(s(n))-\frac{1}{W(N)}\sum_{n=N+1}^{\hat{s}(s(N))}\Delta W(n)f(s(n))\\ =& \frac{(W\circ \hat{s})(s(N))}{W(N)}\cdot \frac{1}{(W\circ \hat{s})(s(N))}\sum_{k=1}^{s(N)}\Delta (W\circ \hat{s})(k) f(k)-\frac{1}{W(N)}\sum_{n=N+1}^{\hat{s}(s(N))}\Delta W(n)f(s(n))\\ =& \frac{(W\circ \hat{s})(s(N))}{W(N)}\cdot \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W\circ \hat{s}}f(k)-\frac{1}{W(N)}\sum_{n=N+1}^{\hat{s}(s(N))}\Delta W(n)f(s(n)). \end{align}\] Now we claim that \(\lim_{N\to\infty} \frac{\Delta(W\circ \hat{s})(s(N))}{W(N)}=0\) since \[\lim_{N\to\infty} \frac{\Delta(W\circ \hat{s})(s(N))}{W(N)} \leqslant\lim_{N\to\infty} \frac{\Delta(W\circ \hat{s})(s(N))}{W(\hat{s}(s(N)))}=\lim_{N\to\infty} \frac{\Delta(W\circ \hat{s})(s(N))}{W(\hat{s}(s(N)))} =0\] by the fact that \(W\) is eventually increasing, \(\hat{s}(s(N))\geqslant N\), and our assumption that \(\lim_{N\to\infty}\frac{\Delta (W\circ \hat{s})(N)}{(W\circ \hat{s})(N)}=0\). Noting that \[\begin{align} \left|\frac{1}{W(N)}\sum_{n=N+1}^{\hat{s}(s(N))}\Delta W(n)f(s(n))\right|\leqslant& \frac{\| f \|_{\infty}}{W(N)}\sum_{n=N+1}^{\hat{s}(s(N))}\Delta W(n)\\ \leqslant&\frac{ \| f \|_{\infty}}{W(N)}\cdot(W(\hat{s}(s(N)))-W(N))\\ \leqslant&\| f \|_{\infty}\cdot \frac{ \Delta(W\circ \hat{s})(N)}{W(N)}=o_{N\to\infty}(1), \end{align}\] we have \[\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(s(n)) = (1+o_{N\to\infty}(1)) \cdot \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W\circ \hat{s}}f(k)-o_{N\to\infty}(1) = \operatorname{\mathbb{E}}_{k\leqslant s(N)}^{W\circ \hat{s}}f(k)+o_{N\to\infty}(1),\] which means that we are done. ◻
Remark 11. When \(s(n) = \lfloor q^{-1}(n) \rfloor\) for some increasing function \(q\colon \mathbb{R}\rightarrow \mathbb{R}\) with \(q(\mathbb{N}) \subseteq\mathbb{N}\) and \(\Delta q^{-1}(n)\leqslant 1\) for all \(n\in \mathbb{N}\), we have \(\hat{s}(n) = q(n)\).
Example 1. Take \(W(N) = N\) and \(s(N) = \lfloor \sqrt{N} \rfloor\) so that \(\hat{s}(N) = N^2\). Then for any bounded function \(f:\mathbb{N}\rightarrow \mathbb{C}\) we have that \[\begin{align} \operatorname{\mathbb{E}}_{1\leqslant n\leqslant N}f({\lfloor\sqrt{n}\rfloor}) = \frac{1}{N}\sum_{n=1}^Nf(\lfloor\sqrt{n}\rfloor )= &\frac{1}{(\lfloor\sqrt{N}\rfloor)^2}\sum_{n=1}^{\lfloor \sqrt{N}\rfloor}(2n+1)f(n)+o_{N\to\infty}(1)\\=& \operatorname{\mathbb{E}}_{1\leqslant n\leqslant\lfloor \sqrt{N}\rfloor}^{V}f(n)+o_{N\to\infty}(1) \text{ for } V(N) = N^2. \end{align}\]
Proof of 8. 3 gives us that \(\operatorname{\mathbb{E}}_{n\leqslant N}^{W \circ L}f(n)=\operatorname{\mathbb{E}}_{n\leqslant N}^{W \circ L}\operatorname{\mathbb{E}}_{k\leqslant L(n)}^{2\text{bin}}f(k)+o_{N\to\infty}(1)\). Now we will use 10 to obtain \[\operatorname{\mathbb{E}}_{n\leqslant N}^{W \circ L}\operatorname{\mathbb{E}}_{k\leqslant L(n)}^{2\text{bin}}f(k) =\operatorname{\mathbb{E}}_{n\leqslant L(N)}^W\operatorname{\mathbb{E}}_{k\leqslant n}^{2\text{bin}}f(k)+o_{N\to\infty}(1).\] We may apply 10 because \(W\in \mathscr{W}^*\) and \(\lim_{N\to\infty}\frac{\log(W(N))}{N}=0\), so that we may compare equations (6 ) and (21 ) to see that \(\lim_{N\to\infty}\frac{\Delta W(N)}{W(N)} = 0\).
Now apply 4, 4, and 4 again, so that we have \[\begin{align} \operatorname{\mathbb{E}}_{n\leqslant L(N)}^W\operatorname{\mathbb{E}}_{k\leqslant n}^{2\text{bin}}f(k) =& \operatorname{\mathbb{E}}_{n\leqslant L(N)}^W\operatorname{\mathbb{E}}_{k\leqslant n}\operatorname{\mathbb{E}}_{m\leqslant k}^{2\text{bin}}f(m)+o_{N\to\infty}(1)\\ =& \operatorname{\mathbb{E}}_{n\leqslant L(N)}^W\operatorname{\mathbb{E}}_{k\leqslant n}f(k)+o_{N\to\infty}(1)\\ =& \operatorname{\mathbb{E}}_{n\leqslant L(N)}^Wf(n)+o_{N\to\infty}(1), \end{align}\] completing the proof. ◻
The goal of this section is to prove Theorem 2 (or rather, an equivalent form which we formulate now). Note that part [itm95main951] of 1 tells us that \((h(\vartheta(n)))_{n\in \mathbb{N}}\) is uniformly distributed mod \(1\) with respect to regular Cesàro averages if and only if \((h(n))_{n\in \mathbb{N}}\) is uniformly distributed mod \(1\) with respect to \(\operatorname{\mathbb{E}}^{2\text{bin}}\) averages. This allows us to state Theorem 2 in the following equivalent way.
Theorem 12. Let \(h\) be a Hardy field function with polynomial growth. The following are equivalent:
The sequence \((h(n))_{n\in \mathbb{N}}\) is uniformly distributed mod \(1\) with respect to \(\operatorname{\mathbb{E}}^{2\text{bin}}\) averages.
One of the following two (mutually exclusive) conditions is satisfied:
\(\lim_{x\to\infty} \frac{|h(x)-p(x)|}{x \log x}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\);
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{\sqrt{x}}=\infty\) for each \(p(x)\in \mathbb{Q}[x]\) and there exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x}<\infty\).
The rest of this section is devoted to the proof of 12. We begin by recalling that Hardy functions are totally ordered by asymptotic growth rate. Thus, we can prove 12 by considering cases. Let \(h\) be a function of polynomial growth belonging to a Hardy field. Exactly one of the following statements is true.
There exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{\sqrt{x}}<\infty\),
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{\sqrt{x}}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\) and there exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x}<\infty\),
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{x}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\) and there exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x\log(x)}=0\),
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{x\log(x)}>0\) for all \(p(x)\in \mathbb{Q}[x]\) and there exists \(q\in \mathbb{Q}[x]\) such that \(0<\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x\log(x)}<\infty\),
\(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{x\log(x)}=\infty\) for all \(p(x)\in\mathbb{Q}[x]\).
It is evident that conditions (2) and (5) above are identical to conditions (b) and (a) in 12, respectively. We will show that if either of conditions (2) or (5) hold then \((h(n))_{n\in \mathbb{N}}\) is uniformly distributed mod \(1\) with respect to \(\operatorname{\mathbb{E}}^{2\text{bin}}\) averages, and we will also show that if any of conditions (1), (3), or (4) hold then \((h(n))_{n\in\mathbb{N}}\) is not uniformly distributed mod \(1\) with respect to \(\operatorname{\mathbb{E}}^{2\text{bin}}\) averages.
Given a function \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) and an interval of natural numbers \([a,b]\), we will find it convenient to use the notation \[\operatorname{\mathbb{E}}_{n\in [a,b]}f(n)= \frac{1}{b-a}\sum_{n=a}^bf(n).\]
First we will consider the cases where one of conditions (2) or (5) holds. It suffices to prove the following theorems.
Theorem 13. Let \(f:\mathbb{N}\rightarrow \mathbb{C}\) be bounded, let \(\ell\in \mathbb{C}\), and let \(W(x) = e^{\sqrt{x}}\). Consider the following statements.
There exists a function \(V\) belonging to a Hardy field satisfying \(\lim_{N\to\infty}\frac{\log(W(N))}{\log(V(N))}=0\) and \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\in [N-s(N),N]}f(n)=\ell\] for each \(s\colon \mathbb{N}\rightarrow \mathbb{N}\) with \(\lim_{N\to\infty}s(N)\cdot \Delta \log(V(N))= \infty\),
\(\lim_{N\to\infty}\operatorname{\mathbb{E}}^W_{n\leqslant N}f(n)=\ell\),
\(\lim_{N\to\infty}\operatorname{\mathbb{E}}^{2\text{bin}}_{n\leqslant N}f(n)=\ell\).
Then (i)\(\implies\)(ii)\(\implies\)(iii).
Theorem 14. Let \(h\) be a function with polynomial growth which belongs to a Hardy field and let \(W(x)=e^{\sqrt{x}}\). Suppose that \(\lim_{x\to\infty}\sqrt{x} |h'(x)-p(x)|=\infty\) for all \(p(x)\in \mathbb{Q}[x]\) and there is some \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}|h'(x)-q(x)|<\infty\). Then \[\lim_{N\to\infty}\operatorname{\mathbb{E}}^W_{n\leqslant N}e^{2\pi i k h(n)}=0\] for each \(k\in \mathbb{Z}\backslash\{0\}\).
Theorem 15. Let \(h\) be a function with polynomial growth which belongs to a Hardy field. Suppose that \[\label{eq:s95ud} \lim_{x\to\infty} \frac{|h'(x)-p(x)|}{\log x}=\infty \text{ for all }p(x)\in \mathbb{Q}[x].\tag{25}\] Then there exists a function \(V\in \mathscr{W}^*\) which belongs to a Hardy field and satisfies \[\lim_{N\to\infty}\frac{\sqrt{N}}{\log(V(N))}=0\] such that \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\in [N-s(N),N]}e^{2\pi i k h(n)}=0\] for all \(k\in \mathbb{Z}\backslash\{0\}\) and all \(s\colon\mathbb{N}\rightarrow \mathbb{N}\) with \(\lim_{N\to\infty}s(N)\cdot \Delta \log(V(N))=\infty\) and \(s(N)\leqslant N-1\) for all \(N\in \mathbb{N}\).
The first implication of Theorem 13 follows from this next theorem.
Theorem 16 ([16]). Suppose that \(V\in \mathscr{W}^*\) belongs to a Hardy field and satisfies \(\lim_{N\to\infty}\frac{\log(V(N))}{\log (N)}=\infty\) and \(\lim_{N\to\infty}\frac{\log(V(N))}{ N}=0\). Let \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) be bounded, and let \(\ell\in \mathbb{C}\). The following statements are equivalent:
\(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n) = \ell\) for each \(W\in \mathscr{W}^*\) which belongs to the same Hardy field as \(V\) and satisfies \(\lim_{N\to\infty}\frac{\log(W(N))}{\log(V(N))}=0\),
\(\lim_{N\to\infty} \operatorname{\mathbb{E}}_{n\in [N-s(N),N]}f(n) = \ell\) for all nondecreasing functions \(s\colon\mathbb{N}\rightarrow\mathbb{N}\) which satisfy \(\lim_{N\to\infty}s(N) \cdot \Delta \log(V(N))=\infty\) and \(s(N)\leqslant N-1\) for all \(N\in \mathbb{N}\).
The second implication in Theorem 13 follows from [13] and Lemma 8 below.
Theorem 17 ([13]). Let \(W\in \mathscr{W}\) and let \((\alpha_{n,N})_{n,N\in \mathbb{N}}\) be a nonnegative doubly indexed sequence such that \(\lim_{N\to\infty}\sum_{n\in \mathbb{N}}\alpha_{n,N}=1\). Then the following are equivalent:
For each function \(f\colon \mathbb{N}\rightarrow \mathbb{C}\) and each \(\ell\in \mathbb{C}\), if \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n) = \ell\) then \(\lim_{N\to\infty}\sum_{n\in \mathbb{N}}\alpha_{n,N}f(n) = \ell\).
\(\sup_{N\to\infty} \sum_{n\in \mathbb{N}}W(n)\left|\frac{\alpha_{n,N}}{\Delta W(n)}-\frac{\alpha_{n+1,N}}{\Delta W(n+1)}\right|<\infty\), and for each \(N\in \mathbb{N}\), \(\lim_{n\to\infty}\frac{\alpha_{n,N}}{\Delta W(n)} = 0\).
Lemma 8. Let \(f:\mathbb{N}\rightarrow \mathbb{C}\) be any function, let \(\ell\in \mathbb{C}\) and let \(W(x) = e^{\sqrt{x}}\). Suppose that \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^Wf(n)=\ell\). Then \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}}f(n) = \lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}f(n) = \ell\).
Proof. We will apply Theorem 17 with \(\alpha_{n,N} = \frac{1}{2^N}\binom{N}{n}\) and \(W(x) = e^{\sqrt{x}}\). It is clear that we have \(\lim_{n\to\infty}\frac{\alpha_{n,N}}{\Delta W(n)} = 0\) for each \(N\), since \(\alpha_{n,N}=0\) for all \(n>N\). Next, we will consider \[\begin{align} \label{eq:messy95sum} \sum_{n\in \mathbb{N}}W(n)\left|\frac{\alpha_{n,N}}{\Delta W(n)}-\frac{\alpha_{n+1,N}}{\Delta W(n+1)}\right| = \frac{1}{2^N}\sum_{n\in \mathbb{N}}\left|\frac{W(n)\binom{N}{n}}{\Delta W(n)}-\frac{W(n)\binom{N}{n+1}}{\Delta W(n+1)}\right| . \end{align}\tag{26}\]
Note that \(\frac{ W(n)}{ W(n+1)} = e^{\sqrt{n}-\sqrt{n+1}} = 1+O_{N\to\infty}(\Delta \sqrt{n+1}) = 1+O_{N\to\infty}(n^{-1/2})\). Put \(\eta(n)= \frac{W(n)\binom{N}{n}}{2^N\Delta W(n)}\) so that we have \[\begin{align} & \frac{1}{2^N}\sum_{n\in \mathbb{N}}\left|\frac{W(n)\binom{N}{n}}{\Delta W(n)}-\frac{W(n)\binom{N}{n+1}}{\Delta W(n+1)}\right| =\sum_{n\in \mathbb{N}}\left|\eta(n)-\eta(n+1)(1+O_{N\to\infty}(n^{-1/2}))\right| \\ \leqslant&\sum_{n\in \mathbb{N}}\left|\eta(n)-\eta(n+1)\right| +\sum_{n\in \mathbb{N}}\eta(n+1)\cdot O_{n\to\infty}(n^{-1/2}). \end{align}\] The second sum is bounded since \[\begin{align} &\sum_{n=1}^{\infty}\eta(n+1)\cdot O_{n\to\infty}(n^{-1/2}) = \sum_{n=1}^{\infty}\frac{1}{2^N}\binom{N}{n+1}\frac{W(n+1)}{\Delta W(n+1)}\cdot O_{n\to\infty} (n^{-1/2})\\ =& \sum_{n=1}^{\infty}\frac{1}{2^N}\binom{N}{n+1}\cdot O_{n\to\infty}(n^{1/2})\cdot O_{n\to\infty} (n^{-1/2}) = O_{N\to\infty}(1). \end{align}\] To bound the other sum above, note that the ratio \(\frac{\eta(n+1)}{\eta(n)}\sim \frac{N-n}{n+1}\) uniformly in \(N\). Since \(\frac{N-n}{n+1}\) is decreasing in \(n\) this shows that, when \(n\) is large, \(\eta(n)\) increases to its maximum and then decreases. So \(\sum_{n\in \mathbb{N}}\left|\eta(n)-\eta(n+1)\right| \leqslant 2\cdot \sup_{n\in \mathbb{N}}\eta(n)\). We can bound \(\sup_{n\leqslant N}\eta(n)\) by noting that \(\binom{N}{n}\leqslant\frac{2^N}{\sqrt{\pi N}}\) for all \(n\) and \(\frac{W(n)}{\Delta W(n)}\leqslant\frac{W(N)}{\Delta W(N)} = 2\sqrt{N}\cdot (1+o_{N\to\infty}(1))\) for all \(n\leqslant N\). In particular, \(\max_{n\leqslant N}\eta(n) = O_{N\to\infty}(1)\), and so we can conclude that \(\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}f(n)=\ell\) by Theorem 17.
In order to prove that \(\operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}}f(n)=\ell\), we can take \(\alpha_{n,N} = \frac{1}{2^{N+1}}\binom{N}{\lfloor n/2 \rfloor}\) and perform a similar calculation as above. ◻
Now that we have shown Theorem 13, we will consider Theorem 14. Suppose that \(\lim_{x\to\infty}\sqrt{x} |h'(x)-p(x)|=\infty\) for all \(p(x)\in \mathbb{Q}[x]\) and there is a \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}|h'(x)-q(x)|<\infty\). By L’Hôpital’s rule, this means that \(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{\sqrt{x}}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\) and there is \(Q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-Q(x)|}{x}<\infty\).
Let \(\alpha,\beta\in\mathbb{R}\backslash\mathbb{Q}\), \(k\in \mathbb{Z}\backslash\{0\}\), and let \(W(n)=e^{\sqrt{n}}\). Put \(r(n) = \beta(h(n)-Q(n))\) so that \(r(n)\) grows faster than \(\sqrt{n}\) and slower than \(n\). Consider \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k \alpha \lfloor r(n) \rfloor }.\] By Lemma 7 we have \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k \alpha \lfloor r(n) \rfloor }=\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant r(N)}^{W\circ r^{-1}}e^{2\pi i k \alpha n },\] but we know that \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{W\circ r^{-1}}e^{2\pi i k \alpha n }=0\) by Theorem 16 and the fact that \(\{n\alpha \}_{n\in \mathbb{N}}\) is well distributed mod 1. So we have reduced Theorem 14 to the following lemma.
Lemma 9. Let \(W(n) = e^{\sqrt{n}}\). Let \(r\colon \mathbb{N}\rightarrow \mathbb{N}\) and suppose that \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i \alpha \lfloor \beta r(n) \rfloor }=0\] for all \(\alpha,\beta\in \mathbb{R}\backslash\mathbb{Q}\). Then \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k r(n) }=0\] for all \(k\in \mathbb{Z}\backslash\{0\}\).
Proof. Let \(k\in \mathbb{Z}\backslash\{0\}\) and pick any \(\varepsilon>0\) with \(\varepsilon\not\in \mathbb{Q}\). Write \(r(n) = \varepsilon(\varepsilon^{-1}r(n)\text{ mod } 1)+\varepsilon\lfloor \varepsilon^{-1}r(n) \rfloor\). Since \(\varepsilon(\varepsilon^{-1}r(n)\text{ mod }1)\in [0,\varepsilon)\) for all \(n\), we have \[\begin{align} &\limsup_{N\to\infty}\left|\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i kr(n)}-\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k \varepsilon\lfloor \varepsilon^{-1}r(n) \rfloor }\right|\\ =&\limsup_{N\to\infty}\left|\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k \varepsilon\lfloor \varepsilon^{-1}r(n) \rfloor }e^{2\pi i k \varepsilon(\varepsilon^{-1}r(n)\text{ mod }1) }-\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k \varepsilon\lfloor \varepsilon^{-1}r(n) \rfloor }\right|\\ =&\limsup_{N\to\infty}\left|\operatorname{\mathbb{E}}_{n\leqslant N}^We^{2\pi i k \varepsilon\lfloor \varepsilon^{-1}r(n) \rfloor }(e^{2\pi i k \varepsilon(\varepsilon^{-1}r(n)\text{ mod } 1) }-1)\right|\\ =&\limsup_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^W|e^{2\pi i k \varepsilon(\varepsilon^{-1}r(n)\text{ mod } 1) }-1|\\ \leqslant& \limsup_{n\to\infty} |e^{2\pi i k \varepsilon(\varepsilon^{-1}r(n)\text{ mod } 1) }-1|< 2\pi |k|\varepsilon\to 0\text{ as }\varepsilon\to 0. \end{align}\] So we are done. ◻
Before giving a proof of Theorem 15, recall van der Corput’s trick.
Theorem 18 (van der Corput’s trick). Let \((x_n)_{n\in \mathbb{N}}\) be a bounded sequence of complex numbers and let \((I_N)_{N\in \mathbb{N}}\) be a sequence of intervals of natural numbers with \(|I_N|\to\infty\) as \(N\to\infty\). Suppose that for each \(j\in \mathbb{N}\), \(\operatorname{\mathbb{E}}_{n\in I_N}(x_{n+j}\overline{x_n}) \to 0\) as \(N\to\infty\). Then \(\operatorname{\mathbb{E}}_{n\in I_N} x_n \to 0\) as \(N\to\infty\).
18 is a special case of [17] when \(F_N=I_N\), \(G=\mathbb{Z}\), and \(H=\mathbb{C}\).
Corollary 6. Let \(h\colon \mathbb{R}\rightarrow \mathbb{R}\) be a Hardy function of polynomial growth and let \((I_N)_{N\in \mathbb{N}}\) be a sequence of intervals of natural numbers with \(|I_N|\to\infty\) as \(N\to\infty\). Suppose that \(\operatorname{\mathbb{E}}_{n\in I_N}e^{2\pi i h'(n)} \to 0\) as \(N\to\infty\). Then \(\operatorname{\mathbb{E}}_{n\in I_N} e^{2\pi i h(n)} \to 0\) as \(N\to\infty\).
Now we are ready to give a proof of 15.
Proof of 15. By replacing \(h(x)\) with \(k(h(x)-\int_0^x p(t)dt)\) if necessary, we can assume without loss of generality that \(k=1\) and \(p(x)=0\). Additionally, assume that \(h\) eventually increases to \(\infty\).
Let \(m\geqslant 1\) be such that \(\lim_{x\to\infty}\frac{h(x)}{x^m}=\infty\) and \(\lim_{x\to\infty}\frac{h(x)}{x^{m+1}}<\infty\). We proceed by considering cases.
Case 1: \(m=1\),
Case 2: \(m=2\)
Case 3: \(m\geqslant 3\).
We can begin by observing that Case 3 reduces to Case 2. Indeed, if \(\lim_{x\to\infty}\frac{h(x)}{x^m}=\infty\) and \(\lim_{x\to\infty}\frac{h(x)}{x^{m+1}}<\infty\), then \(\lim_{x\to\infty}\frac{h^{(m-2)}(x)}{x^2}=\infty\) and \(\lim_{x\to\infty}\frac{h^{(m-2)}(x)}{x^{3}}<\infty\) and so we may apply Case 2 to \(h^{(m-2)}\) and invoke Corollary 6.
Now we will turn our attention to Case 1. Suppose that \(m=1\). Applying L’Hôpital’s rule to equation (25 ) gives us that \(\lim_{x\to\infty}\frac{h'(x)}{\log x}=\lim_{x\to\infty}x\cdot h''(x)=\infty\). Let \[V(N) = e^{\int_{n=1}^N{\sqrt{h''(n)}}}\] so that \(\Delta\log(V(N)) = \sqrt{h''(N)}\cdot (1+o_{N\to\infty}(1))\) and \[\lim_{N\to\infty}\frac{\sqrt{N}}{\log(V(N))} = \lim_{N\to\infty}\frac{\Delta \sqrt{N}}{\Delta \log(V(N)) }=\lim_{N\to\infty}\frac{\frac{1}{2\sqrt{N}}}{\sqrt{h''(N)}} =\lim_{N\to\infty}\frac{1}{2\sqrt{N\cdot h''(N)} }=0.\]
Let \(s\colon \mathbb{N}\rightarrow \mathbb{N}\) be any function with \(s(N)\leqslant N-1\) for all \(N\in \mathbb{N}\) and \(\lim_{N\to\infty}s(N)\cdot \Delta \log(V(N)) = \infty\) so that \(\lim_{N\to\infty}s(N)\sqrt{h''(N)}=\infty\). We will show that \[\label{eq:goal} \lim_{N\to\infty}\frac{1}{s(N)}\sum_{n=N-s(N)}^{N}e^{2\pi i h(n)}=0.\tag{27}\]
If \(\lim_{N\to\infty}\frac{s(N)}{N}\in (0,1]\) then (27 ) holds because \((h(n))_{n\in \mathbb{N}}\) is uniformly distributed in the usual sense, so suppose that \(s(N) = o_{N\to\infty}(N)\).
Theorem 2.2 in [18] says that if \(h\) is a smooth function and \(I\) is an interval with \(\lambda \leqslant|h{''}(x)|\leqslant\alpha \lambda\) for \(x\in I\) then \[\label{eq:vdc952462} \frac{1}{|I|}\left|\sum_{n\in I}e^{2\pi i k h(n)}\right| \leqslant C\left(\alpha\;\lambda^{1/2}+|I|^{-1}\lambda^{-1/2}\right)\tag{28}\] for some uniform constant \(C>0\).
Put \(\lambda = h''(N)\) and \(\alpha = \frac{h''(N-s(N))}{h''(N)}\) and \(I= [N-s(N),N]\) for \(N\in \mathbb{N}\). We have that \(\lambda\to 0\) as \(N\to\infty\) and that \(\lim_{N\to\infty}\alpha<\infty\) since \(s(N) = o_{N\to\infty}(N)\) and \(h''(x)\) tends to \(0\) slower than \(x^{-1}\). So \(\alpha \cdot \lambda^{1/2}\to 0\) as \(N\to\infty\).
Additionally, \(|I|\cdot \lambda^{1/2} = s(N)\cdot \sqrt{h''(N)}\to \infty\) as \(N\to\infty\). So, the right-hand side of equation (28 ) tends to \(0\) as \(N\to\infty\), so it follows that \(\left|\operatorname{\mathbb{E}}_{n\in I}e^{2\pi i h(n)} \right|= \frac{1}{|I|}\left|\sum_{n\in I}e^{2\pi i h(n)}\right| \to 0\) as \(N\to\infty\). This completes the proof for Case 1.
Lastly, consider Case 2. In this case, we have \(\lim_{x\to\infty}\frac{h(x)}{x^2}=\infty\). Then by L’Hôpital’s rule, we also have \[\infty=\lim_{x\to\infty}\frac{h(x)}{x^{3/2}} = \lim_{x\to\infty}\frac{h'(x)}{\frac{3}{2}x^{1/2}} = \lim_{x\to\infty}\frac{h''(x)}{\frac{3}{4}x^{-1/2}} = \lim_{x\to\infty}\frac{h'''(x)}{\frac{-3}{8}x^{-3/2}} =\lim_{x\to\infty}\frac{-8}{3}h'''(x)x^{3/2} .\]
Let \[V(N) = e^{\int_{n=1}^N{\sqrt[3]{h'''(n)}}}\] so that \(\Delta\log(V(N)) = \sqrt[3]{h'''(N)}\cdot(1+o_{N\to\infty}(1))\) and \[\lim_{N\to\infty}\frac{\sqrt{N}}{\log(V(N))} = \lim_{N\to\infty}\frac{\Delta \sqrt{N}}{\Delta \log(V(N)) }=\lim_{N\to\infty}\frac{\frac{1}{2\sqrt{N}}}{\sqrt[3]{h'''(N)}} =\lim_{N\to\infty}\frac{1}{2\sqrt[3]{N^{3/2}\cdot h'''(N)} }=0.\]
Let \(s\colon \mathbb{N}\rightarrow \mathbb{N}\) be any function with \(s(N)\leqslant N-1\) for all \(N\in \mathbb{N}\) and \(\lim_{N\to\infty}s(N)\cdot \Delta \log(V(N)) = \infty\) so that \(\sqrt[4]{h'''(N)\cdot s(N)^3}\to \infty\). Theorem 2.6 in [18] says that if \(h\) is a smooth function and \(I\) is an interval with \(\lambda \leqslant|h'''(x)|\leqslant\alpha \lambda\) for \(x\in I\) then
\[\label{eq:vdc952466} \frac{1}{|I|}\left|\sum_{n\in I}e^{2\pi i h(n)}\right| \leqslant C\left(\alpha^{1/3} \lambda^{1/6}+\alpha^{1/4}|I|^{-1/4}+\lambda^{-1/4}|I|^{-3/4}\right)\tag{29}\] for some uniform constant \(C>0\).
Put \(\lambda = h^{'''}(N)\) and \(\alpha = \frac{h^{(3)}(N-s(N))}{h^{'''}(N)}\) and \(I=I(N) = [N-s(N),N]\) for \(N\in \mathbb{N}\). We have that \(\lim_{N\to\infty}\alpha<\infty\) and so \(\alpha^{1/3} \lambda^{1/6}+\alpha^{1/4}|I|^{-1/4}\to 0\) as \(N\to\infty\). Also, \(\lambda^{1/4}|I|^{3/4} = \sqrt[4]{ h^{(3)}(N)\cdot s(N)^3} \to \infty\) as \(N\to\infty\).
The right-hand side of equation (29 ) tends to \(0\) as \(N\) tends to \(\infty\), so it follows that \(\left|\operatorname{\mathbb{E}}_{n\in I_N}e^{2\pi i h(n)}\right| = \frac{1}{|I|}\left|\sum_{n\in I}e^{2\pi i h(n)}\right| \to 0\) as \(N\to\infty\). This concludes the proof of Case 2, and so we are done. ◻
It remains to show that if any of conditions (1), (3), or (4) hold then \(h(n)\) is not uniformly distributed mod \(1\) with respect to \(\operatorname{\mathbb{E}}^{2\text{bin}}\) averages.
Suppose that condition (1) holds. There is a Tauberian theorem in [19]2 which says that if \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\sum_{j=1 }^na_j = \ell\) and \(a_n = o(n^{-1/2})\) then \(\lim_{n\to\infty}\sum_{j=1}^na_j = \ell\). Taking \(a_n = (e^{2\pi i k h(2n)}-e^{2\pi i k h(2n-2)}+e^{2\pi i k h(2n+1)}-e^{2\pi i k h(2n-1)})/2\) gives that \(a_n = o(n^{-1/2})\) and \(\sum_{j=1}^na_j = (e^{2\pi i kh(2n)}+e^{2\pi i kh(2n+1)}- e^{2\pi i kh(0)}-e^{2\pi i kh(1)})/2\) which does not tend to \(0\) as \(n\to\infty\). So it follows that \[\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}}e^{2\pi i k h(n)}=\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}\sum_{j=1 }^na_j \neq 0.\]
Next, we can consider the case where condition (3) holds.
Lemma 10. Let \(I = [a,b]\) with \(a\) arbitrarily large and let \(h\) be a hardy function with \(\lim_{x\to\infty}\frac{|h(x)|}{x}=\infty\) and \(\lim_{x\to\infty}\frac{h(x)}{x^2}=0\). Let \(c\in [a,b]\), and let \(L_c(x) = h'(c)(x-c)+h(c)\) be the tangent line to \(h\) at \(x=c\). Then \(|e^{2\pi i h(n)}-e^{2\pi i L_c(n)}|\leqslant 2\pi |I|^2h''(a)\) for all \(n\in I\). Notably, this bound is independent of \(c\).
Proof. Recall the chord inequality, which says that \(|e^{i\theta}-e^{i\phi}|\leqslant|\theta-\phi|\) for all \(\theta,\phi\in \mathbb{R}\). So \[|e^{2\pi i h(n)}-e^{2\pi i L_c(n)}|\leqslant 2\pi |h(n)-L_c(n)| = 2\pi |h(n)-h(c)-h'(c)(n-c)|.\] By Taylor’s theorem \(|h(n)-h(c)-h'(c)(n-c)|\leqslant(n-c)^2h''(a)/2\leqslant|I|^2h''(a)\). ◻
Theorem 19. Let \(h\) be a Hardy function with polynomial growth and suppose that \(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{x}=\infty\) for all \(p(x)\in \mathbb{Q}[x]\) and there exists \(q(x)\in \mathbb{Q}[x]\) such that \(\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x\log(x)}=0\). Then \(\lim_{N\to\infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}e^{2\pi i h(n)}\neq 0\).
Proof. We will show that \(\limsup_{N\to\infty}|\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}e^{2\pi i h(n)}|=1\). Without loss of generality assume that \(q(x)=0\) and \(h\) increasing to \(\infty\).
Let \(\varepsilon>0\) and pick \(C>0\) such that for \(I = [N/2-C\sqrt{N}, N/2+C\sqrt{N}]\) we have \(\liminf_{N\to\infty}\frac{1}{2^N}\sum_{n\in I}\binom{N}{n}>1-\varepsilon\).
It follows that \[\label{3} \left|\frac{1}{2^N}\sum_{n=1}^N\binom{N}{n}e^{2\pi i h(n)}-\frac{1}{2^N}\sum_{n\in I}\binom{N}{n}e^{2\pi i h(n)}\right|<\varepsilon.\tag{30}\]
Note that \(h'\) eventually increases to \(\infty\). So, for infinitely many values of \(N\) we can find a real number \(c\in I\) such that \(h'(c)\in \mathbb{Z}\). Pick a large enough \(N\) such that there is a value \(c\in I\) with \(h'(c)\in \mathbb{Z}\). Let \(L_c(x) = h'(c)(x-c) + h(c)\) be the tangent line to \(h\) at \(x=c\). Note that \(e^{2\pi i L_c(n)}\) is constant for all integers \(n\in I\) and so \[\label{1} \left|\frac{1}{2^N}\sum_{n\in I}\binom{N}{n}e^{2\pi i L_c(n)}\right|=\frac{1}{2^N}\sum_{n\in I}\binom{N}{n}>1-\varepsilon.\tag{31}\]
By Lemma 10, \(|e^{2\pi i h(n)}-e^{2\pi i L_c(n)}|\leqslant 2\pi |I|^2h''(N/2-C\sqrt{N})\) for all \(n\in I\). In particular, \(|I|^2 = 4C^2N\) and by using L’Hôpital’s rule on the hypothesis of the theorem, we have that \(\lim_{x\to\infty}h''(x)\cdot x =0\). It follows that \(\lim_{N\to\infty}|I|^2h''(N/2-C\sqrt{N}) = 0\). Then \[\label{2} \lim_{N\to\infty}\left|\frac{1}{2^N}\sum_{n\in I}\binom{N}{n}e^{2\pi i h(n)}-\frac{1}{2^N}\sum_{n\in I}\binom{N}{n}e^{2\pi i L_c(n)}\right|=0.\tag{32}\]
Combining (30 ), (31 ), and (32 ) we get that \(\limsup_{N\to\infty}\left|\operatorname{\mathbb{E}}_{n\leqslant N}^{\text{bin}}e^{2\pi i h(n)}\right|>1-2\varepsilon\). ◻
Lastly, consider the case where condition (4) holds.
Lemma 11. Suppose that \(h\) is a Hardy function such that \(\lim_{x\to\infty}\frac{h(x)}{x\log(x)}>0\) and there exists \(0<\lim_{x\to\infty}\frac{h(x)}{x\log(x)}<\infty\). There exist infinitely many \(N\in\mathbb{N}\) such that \[\|h'(N)\|_{\mathbb{Z}}\leqslant h''(N),\] where \(\|x\|_{\mathbb{Z}}\) denotes the distance from \(x\) to the closest integer.
Proof. Let \(m\) be a large positive integer. \(h'\) eventually increases to \(\infty\) and so we can pick \(N\in\mathbb{N}\) such that \(h'(N)\) is smaller than \(m\), but \(h'(N+1)\) is bigger or equal than \(m\). By using the mean value theorem, we can note that the inequality \[|h'(N+1)-h'(N)|= |h''(c)|\leqslant h''(N) \text{ for some } c\in (N,N+1)\] holds for all but at most finitely many \(N\in\mathbb{N}\), since \(h''\) is eventually decreasing. Since \(m\) lies between \(h'(N)\) and \(h'(N+1)\), it follows that \[|m-h'(N)|\leqslant h''(N).\] This completes the proof. ◻
Lemma 12. Suppose that \(h\) is a Hardy function such that \(\lim_{x\to\infty}\frac{|h(x)-p(x)|}{x\log(x)}>0\) for all \(p(x)\in \mathbb{Q}[x]\) and there exists \(q\in \mathbb{Q}[x]\) such that \(0<\lim_{x\to\infty}\frac{|h(x)-q(x)|}{x\log(x)}<\infty\). Let \(k\in \mathbb{Z}\backslash\{0\}\). Then there exists a constant \(A\in \mathbb{R}\) such that for every \(\varepsilon>0\) and infinitely many \(N\in\mathbb{N}\), \[\label{eq:nlogn95approx} \operatorname{\mathbb{E}}^{2\text{bin}}_{n \le N} e^{2\pi i kh(n)}= e^{2\pi i kh(N)}\cdot \Bigg(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{2}+A\pi i x^2} ~\mathrm{d}x \Bigg)+{\mathrm O}(\varepsilon).\tag{33}\]
Proof. Without loss of generality, assume that \(h\) eventually increases to \(\infty\), \(k=1\), and \(q(x)=0\). Let \(\varepsilon>0\), pick \(C\in \mathbb{N}\) arbitrarily large and put \[I_N= \mathbb{Z}\cap [N-C\sqrt{N}, N+C\sqrt{N}]\qquad\text{and}\qquad \tilde{I}_N= \mathbb{Z}\cap [-C\sqrt{N},C\sqrt{N}].\] If \(C\) is chosen sufficiently large we have \[\label{eq:central95limit} \frac{1}{2^{N+1}}\sum_{n\in I_N}\binom{N}{\big\lfloor \frac{n}{2} \big\rfloor} \geqslant 1-\varepsilon\text{ for all sufficiently large } N.\tag{34}\] From the proof of Theorem 5, we get \[\begin{align} \operatorname{\mathbb{E}}^{2\text{bin}}_{n \le N} e^{2\pi i h(n)} &=\frac{1}{\sqrt{2\pi N}} \sum_{n\in I_N} e^{-\frac{\big(\frac{N-n}{\sqrt{N}}\big)^2}{2}} e^{2\pi i h(n)} + {\mathrm O}(\varepsilon)+o_{N\to\infty}(1) \\ &=\frac{1}{\sqrt{2\pi N}} \sum_{n\in \tilde{I}_N} e^{-\frac{\big(\frac{n}{\sqrt{N}}\big)^2}{2}} e^{2\pi i h(N+n)} + {\mathrm O}(\varepsilon)+o_{N\to\infty}(1). \end{align}\] Using a second-degree Taylor approximation for \(h(x)\) at the point \(x=N\), we have \[h(N+n)= h(N)+ h'(N)\cdot n + \frac{n^2}{2}\cdot h''(N) + {\mathrm O}\Big(n^3\cdot h'''(N)\Big).\] Let \(A = \lim_{N\to\infty}\frac{h(N)}{N\log(N)} = \lim_{N\to\infty}\frac{h''(N)}{1/N}\) and note that \(A\in (0,\infty)\). The \({\mathrm O}\Big(n^3\cdot h'''(N)\Big)\) term above tends to \(0\) since for \(n\in \tilde{I}_N\) we have, \[\begin{align} |n^3\cdot h'''(N)| \leqslant|N^{3/2}\cdot h'''(N)| =&\frac{A^{3/2}}{(h''(N))^{3/2}}\cdot (-h'''(N))\cdot (1+o_{N\to\infty}(1))\\=&2A^{3/2}\cdot\left(\frac{1}{(h''(N))^{1/2}}\right)'\cdot (1+o_{N\to\infty}(1)), \end{align}\] and \[\lim_{N\to\infty}\frac{\left(\frac{1}{(h''(N))^{1/2}}\right)'}{1}=\lim_{N\to\infty}\frac{\left(\frac{1}{(h''(N))^{1/2}}\right)}{N}=\sqrt{\lim_{N\to\infty}\frac{1/N^2}{h''(N)}}=\sqrt{\lim_{N\to\infty}\frac{\log(N)}{h(N)}}=0.\]
Using 11, we can choose \(N\) arbitrarily large such that \(\|h'(N)\|_{\mathbb{Z}}\leqslant h''(N)\) and hence \[\max_{n\in\tilde{I}_N}\{ \|h'(N) \cdot n\|_{\mathbb{Z}}\} \leqslant h''(N)\cdot \sqrt{N}.\] But we can also note that \[\lim_{N\to\infty}h''(N)\cdot \sqrt{N} =\lim_{N\to\infty} \frac{h''(N)}{1/\sqrt{N}} = \lim_{N\to\infty} \frac{h'(N)}{2\sqrt{N}}=\lim_{N\to\infty} \frac{h(N)}{4N^{3/2}/3}=0.\] Note that \(n^2h''(N) = A\cdot \frac{n^2}{N}\cdot(1+o_{N\to\infty}(1))\). Then uniformly over all \(n\in\tilde{I}_N\) we have \[\begin{align} e^{2\pi i h(N+n)}&= e^{2\pi i h(N)} \cdot e\Big(\frac{An^2}{2N}\cdot (1+o_{N\to\infty}(1))\Big)+{\mathrm o}_{N\to\infty}(1) \\ &=e^{2\pi i h(N)} \cdot e^{A\pi i \left(\frac{n}{\sqrt{N}}\right)^2}+{\mathrm o}_{N\to\infty}(1). \end{align}\] Combined with the above, we thus have \[\begin{align} \operatorname{\mathbb{E}}^{2\text{bin}}_{n \le N}e^{2\pi i h(n)} &= e^{2\pi i h(N)} \cdot \Bigg( \frac{1}{\sqrt{2\pi N}} \sum_{n\in \tilde{I}_N} e^{-\frac{\big(\frac{n}{\sqrt{N}}\big)^2}{2}} e^{A\pi i \big(\frac{n}{\sqrt{N}}\big)^2}\Bigg) + {\mathrm O}(\varepsilon)+o_{N\to\infty}(1). \end{align}\]
Observe that the above sum is a Riemann sum. More precisely, we have \[\frac{1}{\sqrt{2\pi N}} \sum_{n\in \tilde{I}_N} e^{-\frac{\big(\frac{n}{\sqrt{N}}\big)^2}{2}} e^{A\pi i \big(\frac{n}{\sqrt{N}}\big)^2} = \frac{1}{\sqrt{2\pi}}\int_{-A}^A e^{-\frac{x^2}{2}+A\pi i x^2} ~\mathrm{d}x + {\mathrm o}_{N\to\infty}(1).\] Adding back the tails of the integral gives another error in the order of \({\mathrm O}(\varepsilon)\), and so we have \[\operatorname{\mathbb{E}}^{2\text{bin}}_{n \le N}e^{2\pi i h(n)} = e^{2\pi i h(N)} \cdot \Bigg(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{2}+A\pi i x^2} ~\mathrm{d}x \Bigg)+{\mathrm O}(\varepsilon)+o_{N\to\infty}(1).\] ◻
Recall the classical fact that \[\int_{-\infty}^{\infty} e^{-\alpha x^2}\,dx = \sqrt{\frac{\pi}{\text{Re}(\alpha)}}\] when \(\text{Re}(\alpha)>0\). It readily follows from equation (33 ) that \[\limsup_{N\to\infty}\Big|\operatorname{\mathbb{E}}^{2\text{bin}}_{n \le N} e^{2\pi i kh(n)}\Big| >0\] when condition (4) holds. This completes the proof of Theorem 12.
The main ingredient in the proof of Theorem 4, aside from our main result 1, is the following proposition.
Proposition 20.
For any uniquely ergodic measure preserving system \((X,\mu,T)\) and any \(f \in C(X)\) we have \[\lim_{N \to \infty} \operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}} f(T^{n}x) =\int_X f \;d\mu,\qquad\forall x\in X.\]
For any non-atomic measure preserving system \((X,\mu,T)\) there exists a residual set of Borel sets \(B\) such that \[\begin{align} \limsup_{N \to \infty} \operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}} 1_B(T^{n}x) =1,\qquad \text{for}~\mu\text{-a.e.}~x\in X, \\ \liminf_{N \to \infty}\operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}} 1_B(T^{n}x) =0,\qquad \text{for}~\mu\text{-a.e.}~x\in X. \end{align}\]
Proof. We will give an “indirect” proof of 20 by combining existing results in the literature concerning ergodic averages along \(\Omega(n)\) with 1. It is worth mentioning that one could also prove 20 directly without mentioning results concerning \(\Omega(n)\).
By 1 we have \[\label{eqn95dynamical95omega95to952bin} \frac{1}{N} \sum_{n = 1}^N f(T^{\Omega(n)}x) =\operatorname{\mathbb{E}}_{n\leqslant\lfloor \log\log N\rfloor}^{2\text{bin}} f(T^{n}x) + {\mathrm o}_{N\to\infty}(1).\tag{35}\] By [10], for any uniquely ergodic system \((X,\mu,T)\) and any \(f \in C(X)\), \[\lim_{N\to\infty} \frac{1}{N} \sum_{n = 1}^N f(T^{\Omega(n)}x)=\int_X f \;d\mu,\qquad\forall x\in X.\] Since any subsequence of a convergent sequence is convergent, we also have \[\lim_{N\to\infty} \frac{1}{\exp(\exp(N))} \sum_{n = 1}^{\exp(\exp(N))} f(T^{\Omega(n)}x)=\int_X f \;d\mu,\qquad\forall x\in X.\] Combined with 35 , this proves that \[\lim_{N\to\infty} \operatorname{\mathbb{E}}_{n\leqslant N}^{2\text{bin}} f(T^{n}x) = \int_X f \;d\mu,\qquad\forall x\in X,\] and part [itm95prop95erg951] of 20 follows.
For part [itm95prop95erg952] we follow a similar strategy. By [2], we know that for any non-atomic measure preserving system \((X,\mu,T)\) there exists a residual set of Borel sets \(B\) such that \[\begin{align} \limsup_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N 1_B(T^{\Omega(n)}x) =1,\qquad \text{for}~\mu\text{-a.e.}~x\in X, \\ \liminf_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N 1_B(T^{\Omega(n)}x) =0,\qquad \text{for}~\mu\text{-a.e.}~x\in X. \end{align}\] By 35 , we obtain \[\begin{align} \limsup_{N \to \infty} \operatorname{\mathbb{E}}_{n\leqslant\lfloor \log\log N\rfloor}^{2\text{bin}} 1_B(T^{n}x) =1,\qquad \text{for}~\mu\text{-a.e.}~x\in X, \\ \liminf_{N \to \infty}\operatorname{\mathbb{E}}_{n\leqslant\lfloor \log\log N\rfloor}^{2\text{bin}} 1_B(T^{n}x) =0,\qquad \text{for}~\mu\text{-a.e.}~x\in X, \end{align}\] and hence part [itm95prop95erg952] follows. ◻
Proof of 4. Assume \(\vartheta\colon \mathbb{N}\to\mathbb{N}\) satisfies 5 for some \(L\in\mathscr{L}\). By 1, we have \[\label{eqn95dynamical95vartheta95to952bin} \frac{1}{N} \sum_{n = 1}^N f(T^{\vartheta(n)}x) =\operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}} f(T^{n}x) + {\mathrm o}_{N\to\infty}(1).\tag{36}\] As shown in 20, the conclusion of parts [itm95thm95erg951] and [itm95thm95erg952] of 4 holds with \(\frac{1}{N} \sum_{n = 1}^N f(T^{\vartheta(n)}x)\) replaced by \(\operatorname{\mathbb{E}}_{n\leqslant L(N)}^{2\text{bin}} f(T^{n}x)\). Then by 36 , it also holds for \(\frac{1}{N} \sum_{n = 1}^N f(T^{\vartheta(n)}x)\).
Finally, note that it also follows from 1 that \[\label{eqn95dynamical95vartheta95to95Cesaro} \mathbb{E}^{W}_{n\leqslant N}\, f(T^{\vartheta(n)}x) = \frac{1}{N} \sum_{n = 1}^N f(T^{n}x) + {\mathrm o}_{N\to\infty}(1).\tag{37}\] So part [itm95thm95erg953] of 4 follows from Birkhoff’s pointwise ergodic theorem. ◻
Vitaly Bergelson
The Ohio State University
vitaly@math.ohio-state.edu
Michael Reilly
The Ohio State University
reilly.201@osu.edu
Florian K.Richter
École Polytechnique Fédérale de Lausanne (EPFL)
f.richter@epfl.ch
This formula for summation by parts holds because of our convention that \(\Delta x_n = x_n-x_{n-1}\). A different convention for \(\Delta x_n\) would give a different summation by parts formula.↩︎
In [19] the limit \(\lim_{N\to\infty}\operatorname{\mathbb{E}}^{\text{bin}}_{n\leqslant N}\) is referred to as the \((E,1)\) method↩︎