On Smooth Mesoscopic Linear Statistics of the Eigenvalues of Random Permutation Matrices

Abstract

We study the limiting behavior of smooth linear statistics of the spectrum of random permutation matrices in the mesoscopic regime, when the permutation follows one of the Ewens measures on the symmetric group. If we apply a smooth enough test function f to all the determinations of the eigenangles of the permutations, we get a convergence in distribution when the order of the permutation tends to infinity. Two distinct kinds of limit appear: if $f(0)\ne 0$, we have a central limit theorem with a logarithmic variance; and if $f(0) = 0$, the convergence holds without normalization and the limit involves a scale-invariant Poisson point process.

1 Notation

In the present article, we denote by $A = \mathcal {O}(B)$ and by $A \ll B$ the fact that there exists a constant $C > 0$ such that $|A| \le CB$. The expressions $A = \mathcal {O}_{x,y} (B)$ and $A \ll _{x,y} B$ mean that there exists a quantity $C_{x,y} > 0$ depending only on x and y, such that $|A| \le C_{x,y} B$.

2 Introduction

The spectrum of random permutation matrices has been studied with much attention in the last few decades. On the one hand, working with matrices gives a different way to understand some of the classical properties satisfied by random permutations. On the other hand, the set of permutation matrices can be seen as a finite subgroup of the orthogonal group or the unitary group, and thus, an interesting problem consists in studying how similar are the spectral behaviors of random permutations and usual ensembles of random orthogonal or unitary matrices. For a random permutation matrix following one of the Ewens measures, the number of eigenvalues lying on a fixed arc of the unit circle has been studied in detail by Wieand [34] and satisfies a central limit theorem when the order n goes to infinity, with a variance growing like $\log n$. This rate of growth is similar to what is obtained for the Circular Unitary Ensemble and random matrices on other compact groups, for which a central limit theorem also occurs, as it can be seen in Costin and Lebowitz [11], Soshnikov [30] and Wieand [33]. A similar result has recently been proved by Bahier [6], on the number of eigenvalues lying on a mesoscopic arc, for a suitable modification of Ewens distributed permutation matrices, and the growth of the variance is also the same as for the CUE, i.e., the logarithm of n times the length of the interval. Some other results on the distribution of eigenvalues of matrices constructed from random permutations can be found in papers by Bahier [5], Evans [16], Najnudel and Nikeghbali [28], Tsou [31], Wieand [35].

The analogy between the permutation matrices and the CUE is not as strong when we consider smooth linear statistics of the eigenvalues. In this case, if we take a fixed, sufficiently smooth test function, it is known that the fluctuations of the corresponding linear statistics tend to a limiting distribution, without normalization, which is unusual for a limit theorem. In the CUE case, the distribution is Gaussian, as seen in Diaconis and Shahshahani [14], Johansson [22], Diaconis and Evans [13], and the variance is proportional to the squared $H^{1/2}$ norm of the test function. In the case of permutation matrices, the limiting distribution is not Gaussian anymore: its shape depends on the test function f and can be explicitly described in terms of f and a sequence of independent Poisson random variables. More detail can be found in Manstavicius [27], Ben Arous and Dang [7].

In the case of mesoscopic linear statistics, one also has a central limit theorem without normalization in the CUE case (see [29]). The behavior of mesoscopic linear statistics of other random matrix ensembles has also been studied: the Gaussian Unitary Ensemble (see [15]), more general Wigner matrices (see [19, 20]) and determinantal processes (see [23]), the Circular Beta Ensemble (see [25]), the thinned CUE, for which a random subset of the eigenvalues has been removed (see [8]). However, the smooth mesoscopic linear statistics of permutation matrices have not been previously studied. The main point of the present article is to show that they also satisfy some limit theorems.

The precise framework is given as follows. We fix a parameter $\theta > 0$, and we consider a sequence $(\sigma _n)_{n \ge 1}$, $\sigma _n$ following the the Ewens($\theta$) distribution on the symmetric group $\mathfrak {S}_n$, that is to say

$$\begin{aligned} \forall \sigma \in \mathfrak {S}_n, \ \mathbb {P} (\sigma _n = \sigma ) = \mathbb {P}_\theta ^{(n)} (\sigma ) = \frac{\theta ^{K(\sigma )}}{\theta (\theta + 1) \cdots (\theta + n-1 )}, \end{aligned}$$

where $K(\sigma )$ denotes the total number of cycles of $\sigma$ once decomposed as a product of cycles with disjoint supports. Note that the particular case $\theta = 1$ corresponds to the uniform distribution on $\mathfrak {S}_n$. The permutation matrix $M^{\sigma }$ associated with any element $\sigma$ of $\mathfrak {S}_n$ is defined as follows: for all $1\le i , j\le n$,

$$\begin{aligned} M^{\sigma }_{i,j} = \left\{ \begin{array}{ll} 1 &{} \text { if } i = \sigma (j) \\ 0 &{} \text {otherwise}. \end{array} \right. \end{aligned}$$

A key relationship between the cycle structure of $\sigma$ and the spectrum of the corresponding permutation matrix $M^{\sigma }$ appears in the expression of the characteristic polynomial of $M^{\sigma }$:

$$\begin{aligned} \forall x \in \mathbb {R}, \quad \det \left( I-xM^{\sigma }\right) = \prod \limits _{j=1}^n \left( 1-x^j\right) ^{a^{\sigma }_j}, \end{aligned}$$

where $a^{\sigma }_j$ denotes the number of j-cycles in the decomposition of $\sigma$ as a product of disjoint cycles. Indeed, $M^{\sigma }$ is similar to a block diagonal matrix where one adds one block equal to the companion matrix of $x^j - 1$ for each j-cycle of $\sigma$. As a consequence, the cycle structure of $\sigma$ is fully determined by the spectrum of $M^{\sigma }$, counted with multiplicity.

In this paper, we are interested in the mesoscopic behavior of smooth linear statistics of the spectrum of $M^{\sigma _n}$ when n goes to infinity. More precisely, we fix a function f from $\mathbb {R}$ to $\mathbb {C}$ which satisfies the following regularity conditions:

$$\begin{aligned} \left\{ \begin{array}{l} f \in \mathcal {C}^2 (\mathbb {R}) \\ f^\prime , f^{\prime \prime } \in L^1 (\mathbb {R}) \\ \exists M>0, \quad \exists \alpha >1, \quad \forall x\in \mathbb {R} , \quad \vert f(x) \vert \le \frac{M}{(1+\vert x \vert )^\alpha }. \end{array}\right. \end{aligned}$$

(1)

Moreover, we fix a sequence $(\delta _n)_{n \ge 1}$ in $\mathbb {R}_+^* = (0,+\infty )$ such that $\delta _n \rightarrow 0$ and $n \delta _n \rightarrow \infty$ when $n \rightarrow \infty$, which means that the corresponding scale is mesoscopic (small but large with respect to the average spacing between the eigenvalues of $M^{\sigma _n}$). In this article, we mainly study the following quantity:

$$\begin{aligned} X_{\sigma _n, \delta _n} (f) := \sum _{x \in \mathbb {R}, e^{i x} \in S(\sigma _n)} m_n \left( e^{i x}\right) f\left( \frac{x}{2 \pi \delta _n}\right) , \end{aligned}$$

where $S(\sigma _n)$ denotes the spectrum of $M^{\sigma _n}$ and $m_n \left( e^{ix}\right)$ is the multiplicity of $e^{ix}$ as an eigenvalue of $M^{\sigma _n}$. In other words, we sum the function f at the eigenangles of $M^{\sigma _n}$, divided by $2 \pi \delta _n$ and counted with multiplicity. Notice that all the determinations of the eigenangles are considered here, and the set of x involved in the sum is $2 \pi$-periodic. Notice that the sum giving $X_{\sigma _n, \delta _n} (f)$ is absolutely convergent, because of the assumption we make on the decay of f at infinity. We will also consider the version of the linear statistics where we restrict the sum to the determinations of the eigenangles which are in the interval $(-\pi , \pi ]$:

$$\begin{aligned} X'_{\sigma _n, \delta _n} (f) := \sum _{x \in (\pi , \pi ], e^{i x} \in S(\sigma _n)} m_n \left( e^{i x}\right) f\left( \frac{x}{2 \pi \delta _n}\right) . \end{aligned}$$

The two sums $X_{\sigma _n, \delta _n} (f)$ and $X'_{\sigma _n, \delta _n}(f)$ can be both considered as natural: in $X_{\sigma _n, \delta _n} (f)$, the arbitrary choice of the determination of the eigenangles is avoided, since all determinations are considered, and then, the situation is more symmetric, whereas in $X'_{\sigma _n, \delta _n}$, there is a bijection between eigenvalues and eigenangles, and considering the eigenangle in $(-\pi , \pi ]$ is a standard choice.

If f tends to zero sufficiently fast at infinity, it is natural to expect that $X_{\sigma _n, \delta _n} (f)$ and $X'_{\sigma _n, \delta _n}(f)$ are close to each other and then have the same asymptotic behavior. Our main theorems provide precise results in this direction.

In order to state these theorems, we need to introduce the Fourier transform of f, normalized as follows:

$$\begin{aligned} \hat{f} (\lambda ) := \int _{\mathbb {R}} f(x) e^{-2 i \pi x \lambda } \mathrm{d} x, \end{aligned}$$

and the two following functions from $\mathbb {R}_+^*$ to $\mathbb {C}$:

$$\begin{aligned} \Theta _f : x \mapsto \sum \limits _{k\in \mathbb {Z}} f(kx), \end{aligned}$$

and

$$\begin{aligned} \Xi _f : x \mapsto \Theta _f(x) - f(0) \mathbb {1}_{x > 1} - \frac{1}{x} \hat{f} (0). \end{aligned}$$

The series defining $\Theta _f$ is absolutely convergent because of the assumptions (1). Our main result can now be stated as follows:

Theorem 1.1

Let $(\delta _n)_{n \ge 1}$ be a positive sequence such that $\delta _n \underset{n \rightarrow \infty }{\longrightarrow } 0$ and $n \delta _n \underset{n \rightarrow \infty }{\longrightarrow } \infty$, and let f be a function from $\mathbb {R}$ to $\mathbb {C}$ satisfying the assumptions (1) given above.

• (i) If $f(0) \ne 0$, then we have the following asymptotics:

  $$\begin{aligned} \mathbb {E} \left( X_{\sigma _n, \delta _n} (f) \right) = n \delta _n \hat{f}(0) - \theta \log (\delta _n ) f(0) + \mathcal {O}_{f,\theta } (1), \end{aligned}$$

  and

  $$\begin{aligned} \mathrm {Var} \left( X_{\sigma _n, \delta _n} (f) \right) = - \theta \log (\delta _n ) f(0)^2 + \mathcal {O}_{f,\theta } (\sqrt{-\log (\delta _n ) }). \end{aligned}$$

  Moreover, the following central limit theorem holds:

  $$\begin{aligned} \frac{ X_{\sigma _n, \delta _n} (f) - \mathbb {E} \left( X_{\sigma _n, \delta _n} (f) \right) }{ \sqrt{ \mathrm {Var} \left( X_{\sigma _n, \delta _n} (f) \right) } } \overset{d}{\underset{n \rightarrow \infty }{\longrightarrow }} \mathcal {N}(0,1). \end{aligned}$$

• (ii) If $f(0) = 0$, then we have the following convergence in distribution:

  $$\begin{aligned} X_{\sigma _n, \delta _n} (f) - n \delta _n \hat{f}(0) \overset{d}{\underset{n\rightarrow \infty }{\longrightarrow }} \sum _{y\in \mathcal {X}} \Xi _f (y) , \end{aligned}$$

  (2)

  where $\mathcal {X}$ is a Poisson point process with intensity $\frac{\theta }{x} \mathrm {d}x$ on $(0,+\infty )$ and where the sum on $\mathcal {X}$ in the right-hand side is a.s. absolutely convergent.

• (iii) For any $\alpha > 1$ such that (1) is satisfied, the results given in (i) and (ii) are still true if we replace $X_{\sigma _n, \delta _n} (f)$ by $X^\prime _{\sigma _n, \delta _n} (f)$, as soon as $\delta _n = o(n^{-1/\alpha })$ when $n \rightarrow \infty$.

Regarding $X^\prime _{\sigma _n, \delta _n} (f)$, we obtain a more refined version of point (iii) of Theorem 1.1. In order to state this result conveniently, let us introduce two additional notations: for all positive integers j and all real numbers $x>0$,

$$\begin{aligned} \Theta _{f,j} (x) := \sum _{k=\lfloor -j/2 \rfloor + 1}^{\lfloor j/2 \rfloor } f(kx), \end{aligned}$$

and

$$\begin{aligned} \Xi _{f,j} (x) := \Theta _{f,j} (x) - f(0) \mathbb {1}_{x>1} - \frac{1}{x} \hat{f}(0). \end{aligned}$$

With this notation, all computations related to $X'_{\sigma _n, \delta _n}$ are similar to the computations related to $X_{\sigma _n, \delta _n}$, except that $\Theta _{f}$ and $\Xi _{f}$ are replaced by $\Theta _{f,\ell }$ and $\Xi _{f,\ell }$ in the contribution of a cycle of length $\ell$.

Theorem 1.2

Assume (1), $f(0) = 0$ and

$$\begin{aligned} x \int _{\vert u \vert > x} \vert f^{\prime \prime } (u) \vert \mathrm {d}u \underset{x\rightarrow +\infty }{\longrightarrow } 0. \end{aligned}$$

(3)

If $(\delta _n)_{n \ge 1}$ is a positive sequence such that $\delta _n \underset{n \rightarrow \infty }{\longrightarrow } 0$, $n \delta _n \underset{n \rightarrow \infty }{\longrightarrow } \infty$,

$$\begin{aligned} n\delta _n \int _{\vert u \vert > \frac{1}{\delta _n}} f(u) \mathrm {d}u \underset{n\rightarrow \infty }{\longrightarrow } 0, \end{aligned}$$

and

$$\begin{aligned} \log (n) f \left( \pm \frac{1}{2\delta _n} \right) \underset{n\rightarrow \infty }{\longrightarrow } 0, \end{aligned}$$

then

$$\begin{aligned} X^\prime _{\sigma _n, \delta _n} (f) - n \delta _n \hat{f}(0) = \sum _{\ell = 1}^n a_{n,\ell } \Xi _{f,\ell } \left( \frac{1}{\ell \delta _n}\right) \overset{d}{\underset{n\rightarrow \infty }{\longrightarrow }} \sum _{y\in \mathcal {X}} \Xi _f (y), \end{aligned}$$

where $\mathcal {X}$ is a Poisson point process with intensity $\frac{\theta }{x} \mathrm {d}x$ on $(0,+\infty )$.

Example 1.3

• If $f \in \mathcal {C}^2_c (\mathbb {R})$ (i.e., $\mathcal {C}^2$ and compactly supported on $\mathbb {R}$) and $f(0)=0$, then all the conditions of Theorem 1.2 are satisfied, for every choice of $\delta _n$ such that $\delta _n \underset{n \rightarrow \infty }{\longrightarrow } 0$ and $n \delta _n \underset{n \rightarrow \infty }{\longrightarrow } \infty$.

• If f satisfies (1), (3) and if $n\delta _n^\alpha \rightarrow 0$ (it is in particular the case if $\delta _n = n^{-\varepsilon }$ for any $\varepsilon \in \left( \frac{1}{\alpha }, 1 \right)$), then all the conditions of Theorem 1.2 are satisfied. Indeed,

  $$\begin{aligned} \left| n\delta _n \int _{\vert u \vert> \frac{1}{\delta _n}} f(u) \mathrm {d}u \right| \le n \delta _n \int _{\vert u \vert > \frac{1}{\delta _n}} \frac{M}{(1+ \vert u \vert )^\alpha } \mathrm {d}u \ll n \delta _n^\alpha \end{aligned}$$

  and

  $$\begin{aligned} \left| \log (n) f\left( \pm \frac{1}{2\delta _n} \right) \right| \ll \log (n) \delta _n^\alpha = o(n\delta _n^\alpha ). \end{aligned}$$

• If $f \in \mathcal {S} (\mathbb {R})$ (i.e. in the Schwartz space of $\mathbb {R}$), and if $\delta _n = n^{-\varepsilon }$ for any $\varepsilon \in \left( 0 , 1 \right)$, then all the conditions of Theorem 1.2 are satisfied.

The assumptions (1) of Theorem 1.1 are made in order to apply Proposition 1 and Lemma 1: we do not expect that our assumptions are optimal, but it is necessary to have some properties of regularity and decay of f. The extra assumptions in Theorem 1.2 are used in order to compare $X^\prime _{\sigma _n, \delta _n} (f)$ and $X_{\sigma _n, \delta _n} (f)$: it may also be possible that they can be relaxed, but they cannot be totally removed, as shown in the following counterexamples.

Counterexample 1.4

• If $f(x) = 1/(1+|x|)$, and if $\delta _n$ is such that $\delta _n \underset{n \rightarrow \infty }{\longrightarrow } 0$ and $n \delta _n \underset{n \rightarrow \infty }{\longrightarrow } \infty$, then $X_{\sigma _n, \delta _n} (f)$ is infinite, whereas in the expression of $X'_{\sigma _n, \delta _n} (f)$, a cycle of length $\ell$ gives a contribution of

  $$\begin{aligned} \sum _{k = \lfloor - \ell /2 \rfloor + 1}^{\lfloor \ell /2 \rfloor } \frac{1}{ 1 + |k|/(\ell \delta _n)}&= \mathcal {O}(1) + 2 \sum _{k = 1}^{\lfloor \ell /2 \rfloor } \frac{\ell \delta _n}{\ell \delta _n + k} \\&= 2 \ell \delta _n \left( \log \left( \frac{ \ell \delta _n + \ell /2}{\ell \delta _n + 1} \right) + \mathcal {O}(1) \right) + \mathcal {O}(1) \\&= 2 \ell \delta _n ( \log \ell - \log ( 1+ \ell \delta _n) + \mathcal {O}(1) ) + \mathcal {O}(1). \end{aligned}$$

  If $\ell \ge \delta _n^{-1}$, we get an estimate:

  $$\begin{aligned} 2 \ell \delta _n ( \log \ell - \log ( \ell \delta _n) + \mathcal {O}(1) ) + \mathcal {O}(1) = 2 \ell \delta _n \log (\delta _n^{-1} ) + \mathcal {O} (\ell \delta _n), \end{aligned}$$

  and if $\ell \le \delta _n^{-1}$, we get

  $$\begin{aligned} 2 \ell \delta _n ( \log \ell + \mathcal {O}(1) ) + \mathcal {O}(1) = 2 \ell \delta _n \log \ell + \mathcal {O}(1). \end{aligned}$$

  By the Feller coupling described in Section 3 (see the proof of Lemma  2), if $a_{n,\ell }$ is the number of $\ell$-cycles in $\sigma _n$, then the variables $(a_{n,\ell })_{1 \le \ell \le n}$ can be coupled with independent Poisson variables $(W_{\ell })_{1 \le \ell \le n}$, where $W_{\ell }$ has parameter $\theta /\ell$, in such a way that $a_{n, \ell } \le W_{\ell }$ for all $\ell \in \{1, \dots , n\}$, except for at most one value of $\ell$ for which $a_{n, \ell } = W_{\ell } +1$. Hence, the number of cycles smaller than or equal to $\delta _n^{-1}$ has expectation at most $\theta (1 + \log ( \delta _n^{-1}))$ and the sum of their lengths has expectation at most $1 + \theta \delta _n^{-1}$. Hence, as soon as $\omega (n) \ge 1$ goes to infinity at infinity, the number of small cycles is at most $\omega (n) \theta (1 + \log ( \delta _n^{-1}))$, and the sum of their length is at most $\omega (n) ( 1+ \theta \delta _n^{-1})$, with probability tending to 1 when $n \rightarrow \infty$. If we take $\omega (n)$ going to infinity slower than $n \delta _n$, we deduce that with probability going to 1, the sum of the lengths of the cycles larger than $\delta _n^{-1}$ is $n- o(n)$ with probability tending to 1. Hence, their contribution in $X'_{\sigma _n, \delta _n} (f)$ is equivalent to $2 n \delta _n \log (\delta _n^{-1} )$. The contribution of the cycles smaller than $\delta _n^{-1}$ is dominated by $\delta _n \log (\delta _n^{-1})$ times the sum of their lengths, plus the number of these cycles. With probability going to 1, this contribution is dominated by

  $$\begin{aligned} (\delta _n \log (\delta _n^{-1}) ( 1+ \delta _n^{-1}) + 1 + \log (\delta _n^{-1}) )\omega (n) \ll \log (\delta _n^{-1}) \omega (n) \end{aligned}$$

  if n is large enough, and if $\omega (n)$ goes to infinity slower than $n \delta _n$, we deduce that

  $$\begin{aligned} X'_{\sigma _n, \delta _n} (f) = (2 + o(1)) n \delta _n \log (\delta _n^{-1} ) \end{aligned}$$

  with probability tending to 1 when $n \rightarrow \infty$. This behavior at infinity does not correspond to what we get in the theorems.

• If $f(x) = x^2/(1+x^4)$, f satisfies the assumptions of Theorem 1.1 (ii), and then,

  $$\begin{aligned} X_{\sigma _n, \delta _n} - n \delta _n \hat{f}(0) \overset{d}{\underset{n \rightarrow \infty }{\longrightarrow }} \sum _{y \in \mathcal {X}} \Xi _f (y). \end{aligned}$$

  If we replace $X_{\sigma _n, \delta _n}$ by $X'_{\sigma _n, \delta _n}$, then we subtract at least n terms of the form $f(x/2 \pi \delta _n)$ for $\pi < x \le 3 \pi$ and then at least a quantity of order $n \delta _n^2$. If $n \delta _n^2$ tends to infinity when $n \rightarrow \infty$ (for example if $\delta _n = n^{-1/3}$), then

  $$\begin{aligned} X'_{\sigma _n, \delta _n} - n \delta _n \hat{f}(0) \overset{d}{\underset{n \rightarrow \infty }{\longrightarrow }} - \infty , \end{aligned}$$

  in the sense that

  $$\begin{aligned} \mathbb {P} [ X'_{\sigma _n, \delta _n} - n \delta _n \hat{f}(0) > -A] \underset{n \rightarrow \infty }{\longrightarrow } 0 \end{aligned}$$

  for any fixed $A > 0$. Hence, Theorem 1.1 (ii) does not extend to $X'_{\sigma _n, \delta _n}$.

2.1 Outline of the Paper

In Section 2, we give an expression of $X_{\sigma _n, \delta _n} (f)$ and provide preliminary results involving the regularity of f. In Section 3, we recall the construction of the Feller coupling, which is a suitable feature for dealing with numbers of cycles of equal size in random permutations. The Feller coupling provides a comparison between $X_{\sigma _n, \delta _n} (f)$ and a sum involving independent Poisson random variables. In Section 4, we combine this comparison and a central limit theorem in order to show the first part of Theorem 1.1. In Section 5, a comparison between the Fourier transforms of $X_{\sigma _n, \delta _n} (f)$ and the sum involving Poisson variables is then combined with the convergence of this sum toward the right-hand side of (2) in order to prove the second part of Theorem 1.1. The third part of Theorem 1.1 is then deduced by showing that $X_{\sigma _n, \delta _n} (f)$ and $X'_{\sigma _n, \delta _n} (f)$ are close to each other under the assumptions which are considered. Theorem 1.2 is proved in Section 6.

3 Expression of $X_{\sigma _n, \delta _n} (f)$ in Terms of $\Theta _f$ and $\Xi _f$

In the spectrum of $M^{\sigma _n}$, each cycle of length $\ell$ gives eigenangles equal to all multiples of $2 \pi / \ell$. The contribution of these eigenangles in the sum $X_{\sigma _n, \delta _n}(f)$ is:

$$\begin{aligned} \sum _{k \in \mathbb {Z}} f\left( \frac{k}{\ell \delta _n}\right) =\Theta _{f}\left( \frac{1}{\ell \delta _n}\right) . \end{aligned}$$

Then, if we denote by $a_{n,j}$ the number of j-cycles in the decomposition of $\sigma _n$ as a product of cycles with disjoint support, we get:

$$\begin{aligned} X_{\sigma _n, \delta _n}(f)&= \sum _{\ell = 1}^n a_{n,\ell } \Theta _f \left( \frac{1}{\ell \delta _n} \right) \\&= \sum _{\ell = 1}^n a_{n,\ell } \Xi _f \left( \frac{1}{\ell \delta _n} \right) + f(0) \sum _{\ell < \delta _n^{-1}} a_{n, \ell } + \hat{f}(0) \sum _{\ell = 1}^n \ell \delta _n a_{n, \ell }. \end{aligned}$$

Since the total number of elements of all cycles is n, we have

$$\begin{aligned} \sum _{\ell = 1}^n \ell a_{n, \ell } = n, \end{aligned}$$

and then,

$$\begin{aligned} X_{\sigma _n, \delta _n}(f) = n \delta _n \hat{f}(0) + \sum _{\ell = 1}^n a_{n, \ell } \Xi _f\left( \frac{1}{\ell \delta _n}\right) + f(0) \sum _{\ell < \delta _n^{-1}} a_{n, \ell }. \end{aligned}$$

(4)

Note that $\hat{f}(0)$ is the integral of f, and then, the term $n \delta _n \hat{f} (0)$ is what we would expect if the n eigenangles lying in $[0,2\pi )$ were uniformly distributed.

First note that under (1), the Poisson summation formula applies and gives for all $x>0$,

$$\begin{aligned} \Theta _{f} (x) = \frac{1}{x} \Theta _{\hat{f}} \left( \frac{1}{x} \right) . \end{aligned}$$

(5)

We now get the following asymptotic result on $\Theta _{f}$:

Proposition 1

Assume (1). Then,

• (i) $\Theta _f$ is continuous on $\mathbb {R}_+^*$ and converges at infinity to f(0) with rate dominated by $\frac{1}{x^\alpha }$ (where $\alpha$ is given by (1)).

• (ii) $\Theta _{\hat{f}}$ is continuous on $\mathbb {R}_+^*$ and converges at infinity to $\hat{f}(0)$ with rate dominated by $\frac{1}{x^2}$.

Proof

We prove the two items separately.

• Proof of (i) Since f is assumed to be continuous, the functions $f_k : x \mapsto f(kx)$ are clearly continuous on $(0,+\infty )$ for all $k\in \mathbb {Z}$. Moreover, for all $k\in \mathbb {Z}\setminus \{0\}$ and for all x in any compact set $[A,B] \subset (0,+\infty )$,

  $$\begin{aligned} \vert f_k (x) \vert \le \frac{M}{(1+ \vert kx \vert ^\alpha )} \le \frac{M}{\vert k \vert ^\alpha A^\alpha }; \end{aligned}$$

  hence, $\sum _k f_k$ converges uniformly on compact sets of $(0,+\infty )$. We deduce the continuity of $\Theta _f$. For the convergence of $\Theta _f$ to f(0) at infinity, we only have to notice that for all $x\ge 1$,

  $$\begin{aligned} \sum _{k\ne 0} \vert f(kx) \vert \le \frac{M}{x^\alpha } \sum _{k\ne 0} \frac{1}{\vert k \vert ^\alpha } \ll _{M, \alpha } \frac{1}{x^\alpha }. \end{aligned}$$

• Proof of (ii) It is clear that the functions $g_k : x \mapsto \hat{f}(kx)$ are continuous on $(0,+\infty )$ for all $k\in \mathbb {Z}$, and from two consecutive integrations by parts, it follows that for all $k\in \mathbb {Z}\setminus \{0\}$ and for all x in any compact set $[A,B] \subset (0,+\infty )$,

  $$\begin{aligned} \vert g_k (x) \vert&= \left| \frac{1}{(2 i \pi k x)^2} \int _{-\infty }^{+\infty } f^{\prime \prime } (y) \mathrm {e}^{-2i\pi k x y} \mathrm {d}y \right| \\&\le \frac{1}{4\pi ^2 k^2 A^2} \int _{-\infty }^{+\infty } \vert f^{\prime \prime } (y) \vert \mathrm {d}y; \end{aligned}$$

  hence, $\sum _k g_k$ converges uniformly on compact sets of $(0,+\infty )$. Note that there is no boundary term in the integration by parts, since by assumption, f goes to zero at infinity, and $f'$ and $f''$ are integrable, which implies that $f'$ also goes to zero at infinity. Now, for all $x\ge 1$,

  $$\begin{aligned} \sum _{k\ne 0} \left| \hat{f}(kx) \right| \le \frac{1}{x^2} \times \frac{1}{4\pi ^2} \int _{-\infty }^{+\infty } \left| f^{\prime \prime } (y) \right| \mathrm {d}y \sum _{k\ne 0} \frac{1}{k^2} \ll \frac{1}{x^2}, \end{aligned}$$

  and the proof is complete.

$\square$

From the proposition just above, we deduce the following lemma:

Lemma 1

If the function f satisfies the assumptions (1), then for all $x \in \mathbb {R}_+^*$,

$$\begin{aligned} |\Xi _f (x)| \ll _{f} \min \left( x, 1/x\right) . \end{aligned}$$

In particular,

$$\begin{aligned} \int _0^{\infty } \frac{|\Xi _f (x)|}{x} dx < \infty . \end{aligned}$$

Moreover, $\Xi _f (x)$ is continuous at any point of $\mathbb {R}_+^* \backslash \{1\}$, and also at 1 if $f(0) = 0$.

Proof

We have, for all $x\in (0,1]$,

$$\begin{aligned} \vert \Xi _f (x) \vert = \left| \Theta _f (x) - \frac{1}{x} \hat{f}(0) \right| = \frac{1}{x} \left| \Theta _{\hat{f}} \left( \frac{1}{x}\right) - \hat{f}(0) \right| \ll _f \frac{1}{x} \times x^2 = x, \end{aligned}$$

and for all $x \in (1, \infty )$,

$$\begin{aligned} \vert \Xi _f (x) \vert \le \left| \Theta _f (x) - f(0) \right| + \frac{1}{x} \vert \hat{f}(0) \vert \ll _{f,\alpha } \frac{1}{x^\alpha } + \frac{1}{x} \ll \frac{1}{x}. \end{aligned}$$

The continuity of $\Xi _f$ is an immediate consequence of the continuity of $\Theta _f$. $\square$

4 The Feller Coupling

In [18], Feller introduces a construction of a uniform permutation on the symmetric group, such that the cycle lengths are given by the spacings between successes in independent Bernoulli trials. This construction can be extended to general Ewens distributions and provides a coupling between the cycle counts of a random permutation and a sequence of independent Poisson random variables. The coupling procedure and many related results are proved in [2] and [7, Section 4]: in order to make this article self-contained, we reproduce here some of the discussion given in [7, Section 4]. We consider a sequence $(\xi _i)_{i \ge 1}$ of independent Bernoulli random variables, such that

$$\begin{aligned} \mathbb {P} [ \xi _i = 1] = \frac{\theta }{ \theta +i-1}, \; \; \mathbb {P} [ \xi _i = 0] = \frac{i-1}{ \theta +i-1}. \end{aligned}$$

For $1 \le \ell \le n$, let us denote the number of spacings of length $\ell$ in the sequence $1 \xi _2 \xi _3 \dots \xi _n 1$ by $c_\ell (n)$, i.e.:

$$\begin{aligned} c_\ell (n) = \sum _{i=1}^{n-\ell } \xi _i (1-\xi _{i+1}) \dots (1-\xi _{i + \ell - 1}) \xi _{i+\ell } + \xi _{n-\ell +1} (1- \xi _{n-\ell +2} ) \dots (1-\xi _n). \end{aligned}$$

Define

$$\begin{aligned} W_\ell := \sum _{i=1}^{\infty } \xi _i (1-\xi _{i+1}) \dots (1-\xi _{i + \ell - 1}) \xi _{i+\ell }. \end{aligned}$$

The following result is proved in [7, Theorem 4.1]):

Proposition 2

• (i) $(c_\ell (n))_{ 1 \le \ell \le n}$ has the same probability distribution as $(a_{n,\ell })_{1 \le \ell \le n}$, where $a_{n,\ell }$ denotes the number of $\ell$-cycles in the random permutation $\sigma _n$.

• (ii) $(W_{\ell })_{\ell \ge 1}$ are independent Poisson random variables, with $\mathbb {E} [ W_{\ell }] = \theta /\ell$.

We deduce the following lemma:

Lemma 2

For all $n \ge 1$, one can couple the numbers $a_{n,\ell }$ of $\ell$-cycles in the random permutation $\sigma _n$, with a sequence of independent Poisson variables $W_\ell$ of parameters $\theta /\ell$, in such a way that

$$\begin{aligned} \mathbb {E} \left( \left( \sum _{\ell \le n} \vert a_{n,\ell } - W_\ell \vert \right) ^2 \right) \le C(\theta ), \end{aligned}$$

where $C (\theta )$ is a constant which does not depend on n.

Proof

We choose a coupling such that that $a_{n,\ell } = c_{\ell }(n)$, which is possible by Proposition 2. We then have $a_{n,\ell } = c_{\ell }(n) \le W_{\ell }$, except for at most one value of $\ell$, for which $a_{n,\ell }$ may be equal to $W_{\ell } +1$. Hence,

$$\begin{aligned} \sum _{\ell \le n } \vert a_{n,\ell } - W_\ell \vert \le 2 + \sum _{\ell \le n } (W_{\ell } - a_{n,\ell } ). \end{aligned}$$

It is then enough to bound the $L^2$ norm of $G_n - H_n$ by a quantity depending only on $\theta$, for $G_n := \sum \limits _{j=1}^n a_{n,j}$, $H_n := \sum \limits _{j=1}^n W_j$. Such a bound is a consequence of [7, Lemma 4.8], in the case where $u_j = 1$ for $1 \le j \le n$. $\square$

The lemma proved here allows to compare the quantity $X_{\sigma _n, \delta _n} (f)$ with a linear combination of independent Poisson random variables, for which classical tools in probability theory can be used to prove limit theorems.

5 Proof of Theorem 1.1 (i)

We couple the variables $(a_{n, \ell })_{1 \le \ell \le n}$ with independent Poisson variables $(W_{\ell })_{\ell \ge 1}$ by using the Feller coupling, as in the previous section. From (4), we get

$$\begin{aligned} X_{\sigma _n, \delta _n}(f)= & {} n \delta _n \hat{f}(0) + f(0) \sum _{\ell< \delta _n^{-1}} W_{\ell } + f(0) \sum _{\ell < \delta _n^{-1}} (a_{n, \ell } - W_{\ell })\nonumber \\&+ \sum _{\ell = 1}^n a_{n, \ell } \Xi _f\left( \frac{1}{\ell \delta _n}\right) . \end{aligned}$$

(6)

In order to prove the first part of Theorem 1.1, we will show that the sum of the two first terms satisfies the same central limit theorem and that the two last terms are bounded in $L^2$. We first prove the following result:

Proposition 3

We have:

$$\begin{aligned}&\mathbb {E} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell< \delta _n^{-1}} W_{ \ell } \right] = n \delta _n \hat{f}(0) - \theta \log (\delta _n) f(0) + \mathcal {O}_{f,\theta }(1), \\&{\text {Var}} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right] = - \theta \log (\delta _n) f(0)^2 + \mathcal {O}_{f,\theta }(1), \end{aligned}$$

and

$$\begin{aligned} \frac{n \delta _n \hat{f}(0) + f(0) \sum _{\ell< \delta _n^{-1}} W_{\ell } - \mathbb {E} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell< \delta _n^{-1}} W_{\ell } \right] }{ \sqrt{ {\text {Var}} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right] }} \overset{d}{\underset{n \rightarrow \infty }{\longrightarrow }} \mathcal {N}(0,1). \end{aligned}$$

Proof

Since $(W_{\ell })_{\ell \ge 1}$ are independent Poisson random variables, $W_{\ell }$ with parameter $\theta /\ell$, we get

$$\begin{aligned} \mathbb {E} \left( \sum _{\ell< \delta _n^{-1}} W_\ell \right) = \mathrm {Var} \left( \sum _{\ell< \delta _n^{-1}} W_\ell \right) = \sum _{\ell < \delta _n^{-1}} \frac{\theta }{\ell } = \theta \log \left( \delta _n^{-1}\right) + \mathcal {O}_\theta (1), \end{aligned}$$

(7)

which gives the estimates of the proposition for the expectation and the variance. The central limit theorem is easily obtained by applying the Lindeberg–Feller criterion, since the variables $(W_{\ell })_{\ell \ge 1}$ are independent. $\square$

We then prove that the two last terms of (6) are bounded in $L^2$:

Proposition 4

We have the estimate:

$$\begin{aligned} \mathbb {E} \left[ \left( f(0) \sum _{\ell < \delta _n^{-1}} |a_{n, \ell } - W_{\ell }| + \sum _{\ell = 1}^n a_{n, \ell } \left| \Xi _f\left( \frac{1}{\ell \delta _n}\right) \right| \right) ^2 \right] = \mathcal {O}_{f, \theta }(1). \end{aligned}$$

Proof

By Lemma 2, it is enough to show

$$\begin{aligned} \mathbb {E} \left[ \left( \sum _{\ell = 1}^n a_{n, \ell } \left| \Xi _f\left( \frac{1}{\ell \delta _n}\right) \right| \right) ^2 \right] = \mathcal {O}_{f, \theta }(1). \end{aligned}$$

Moreover, we have $a_{n, \ell } \le W_{\ell }$ for all $\ell$ except at most one value, for which we may have $a_{n, \ell } = W_{\ell } + 1$. It is then enough to check

$$\begin{aligned} \mathbb {E} \left[ \left( \sup _{\mathbb {R}_+^*} |\Xi _f| + \sum _{\ell = 1}^n W_{\ell } \left| \Xi _f\left( \frac{1}{\ell \delta _n}\right) \right| \right) ^2 \right] = \mathcal {O}_{f, \theta }(1), \end{aligned}$$

or equivalently,

$$\begin{aligned} \mathbb {E} \left[ \sup _{\mathbb {R}_+^*} |\Xi _f| + \sum _{\ell = 1}^n W_{\ell } \left| \Xi _f\left( \frac{1}{\ell \delta _n}\right) \right| \right] = \mathcal {O}_{f, \theta }(1), \end{aligned}$$

and

$$\begin{aligned} {\text {Var}} \left( \sum _{\ell = 1}^n W_{\ell } \left| \Xi _f\left( \frac{1}{\ell \delta _n} \right) \right| \right) = \mathcal {O}_{f, \theta }(1). \end{aligned}$$

These estimates are implied by the estimate

$$\begin{aligned} \sup _{\mathbb {R}_+^*} |\Xi _f| + \sum _{\ell = 1}^n \frac{1}{\ell } \left( \left| \Xi _f\left( \frac{1}{\ell \delta _n} \right) \right| + \left| \Xi _f\left( \frac{1}{\ell \delta _n} \right) \right| ^2 \right) = \mathcal {O}_{f}(1), \end{aligned}$$

which is a direct consequence of Lemma 1. $\square$

It is now easy to deduce Theorem 1.1 (i) from the two propositions just above. The estimate of the expectation is immediate, and the estimate of the variance is directly deduced from the fact that

$$\begin{aligned} {\text {Var}} (A+B) = {\text {Var}} (A) + {\text {Var}} (B) + \mathcal {O} \left( {\text {Var}}^{1/2} (A) {\text {Var}}^{1/2} (B) \right) . \end{aligned}$$

For the central limit theorem, we know that

$$\begin{aligned} \frac{n \delta _n \hat{f}(0) + f(0) \sum _{\ell< \delta _n^{-1}} W_{\ell } - \mathbb {E} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell< \delta _n^{-1}} W_{\ell } \right] }{ \sqrt{ {\text {Var}} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right] }} \overset{d}{\underset{n \rightarrow \infty }{\longrightarrow }} \mathcal {N}(0,1). \end{aligned}$$

and

$$\begin{aligned} \frac{f(0) \sum _{\ell< \delta _n^{-1}} (a_{n, \ell } - W_{\ell }) + \sum _{\ell = 1}^n a_{n, \ell } \Xi _f\left( \frac{1}{\ell \delta _n}\right) }{ \sqrt{ {\text {Var}} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right] }} {\underset{n \rightarrow \infty }{\longrightarrow }} 0 \end{aligned}$$

in $L^2$, since the numerator is bounded in $L^2$ and the denominator tends to infinity with n. Hence,

$$\begin{aligned} \frac{\mathbb {E} \left[ f(0) \sum _{\ell< \delta _n^{-1}} (a_{n, \ell } - W_{\ell }) + \sum _{\ell = 1}^n a_{n, \ell } \Xi _f\left( \frac{1}{\ell \delta _n}\right) \right] }{ \sqrt{ {\text {Var}} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right] }} {\underset{n \rightarrow \infty }{\longrightarrow }} 0, \end{aligned}$$

and by Slutsky’s lemma,

$$\begin{aligned} \frac{ X_{\sigma _n, \delta _n} - \mathbb {E} [ X_{\sigma _n, \delta _n} ] }{ \sqrt{ {\text {Var}} \left[ n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right] }} \overset{d}{\underset{n \rightarrow \infty }{\longrightarrow }} \mathcal {N}(0,1). \end{aligned}$$

Since the variance estimates we know imply that

$$\begin{aligned} \frac{ \sqrt{ {\text {Var}} \left( n \delta _n \hat{f}(0) + f(0) \sum _{\ell < \delta _n^{-1}} W_{\ell } \right) }}{ \sqrt{ {\text {Var}} (X_{\sigma _n, \delta _n})}} \underset{n \rightarrow \infty }{\longrightarrow } 1, \end{aligned}$$

we are done.

6 Proof of Theorem 1.1 (ii) and (iii)

Let $A_n:= \sum _{\ell = 1}^n a_{n,\ell } \Xi _f \left( \frac{1}{\ell \delta _n}\right)$, $B_n:= \sum _{\ell = 1}^n W_\ell \Xi _f \left( \frac{1}{\ell \delta _n}\right)$ and $Z=\sum _{y\in \mathcal {X}} \Xi _f (y)$. Here, $a_{n,\ell }$ and $W_{\ell }$ are again related by the Feller coupling. Notice that the sum defining Z is a.s. absolutely convergent, since

$$\begin{aligned} \mathbb {E} \left[ \sum _{y\in \mathcal {X}} |\Xi _f (y)| \right] = \theta \int _0^{\infty } |\Xi _f (x)| \frac{dx}{x} < \infty \end{aligned}$$

by Lemma 1.

We are going to prove the result in two steps:

• (i) For all $t\in \mathbb {R}$, $\mathbb {E}\left( \mathrm {e}^{it B_n}\right) \underset{n \rightarrow \infty }{\longrightarrow } \mathbb {E}\left( \mathrm {e}^{it Z}\right)$.

• (ii) For all $t\in \mathbb {R}$, $\left| \mathbb {E}\left( \mathrm {e}^{it A_n}\right) - \mathbb {E}\left( \mathrm {e}^{it B_n}\right) \right| \underset{n \rightarrow \infty }{\longrightarrow } 0$.

(i): Let $t\in \mathbb {R}$. Using that the variables $W_j$ are independent,

$$\begin{aligned} \mathbb {E}(\mathrm {e}^{it B_n})&= \prod _{\ell = 1}^n \mathbb {E} \left( \mathrm {e}^{it W_\ell \Xi _f (1/\ell \delta _n)} \right) \\&= \prod _{\ell =1}^n \exp \left( \frac{\theta }{\ell } \left( \mathrm {e}^{it \Xi _f (1/\ell \delta _n)} -1 \right) \right) \\&= \exp \left( \theta \delta _n \sum _{\ell = 1}^n \frac{1}{\ell \delta _n} \left( \mathrm {e}^{it \Xi _f (1/\ell \delta _n)} -1 \right) \right) \\&= \exp \left( \theta \delta _n \sum _{1 \le \ell \le n, \, \ell \delta _n \in [1/R, R] } \frac{1}{\ell \delta _n} \left( \mathrm {e}^{it \Xi _f (1/\ell \delta _n)} -1 \right) \right) \\&\times \exp \left( \theta \delta _n \sum _{1 \le \ell \le n, \, \ell \delta _n \notin [1/R, R] } \frac{1}{\ell \delta _n} \left( \mathrm {e}^{it \Xi _f (1/\ell \delta _n)} -1 \right) \right) , \end{aligned}$$

for any $R > 1$. For fixed R and n large enough depending on R, the condition $1 \le \ell \le n$ can be discarded in the first exponential of the last product, since $\delta _n \rightarrow 0$ and $n \delta _n \rightarrow \infty$ when $n \rightarrow \infty$. The sum in the first exponential is then a Riemann sum, which by continuity of $\Xi _f$ (proved in Lemma 1: recall that $f(0) = 0$ in this section), shows that the exponential tends to

$$\begin{aligned} \exp \left( \theta \int _{1/R}^{R} \frac{1}{x} \left( \mathrm {e}^{it \Xi _f (1/x)} -1 \right) \mathrm {d}x \right) \end{aligned}$$

when n goes to infinity. On the other hand, by Lemma 1 the sum inside the second exponential is dominated by

$$\begin{aligned} \sum _{\ell \delta _n > R} \frac{1}{(\ell \delta _n)^2} + \sum _{\ell \delta _n < 1/R} 1 = \mathcal {O} \left( (R \delta _n)^{-1}\right) , \end{aligned}$$

and then the second exponential is

$$\begin{aligned} \exp \left( \mathcal {O}_{f, \theta , t} (1/R) \right) = 1 + \mathcal {O}_{f, \theta , t} (1/R). \end{aligned}$$

Now, if L is the limit of a subsequence $(\mathbb {E}(\mathrm {e}^{it B_{n_k}}))_{k \ge 1}$, then

$$\begin{aligned}&\exp \left( \theta \delta _{n_k} \sum _{1 \le \ell \le n_k, \, \ell \delta _{n_k} \notin [1/R, R] } \frac{1}{\ell \delta _{n_k}} \left( \mathrm {e}^{it \Xi _f (1/\ell \delta _{n_k})} -1 \right) \right) \\&\underset{k \rightarrow \infty }{\longrightarrow } L \exp \left( - \theta \int _{1/R}^{R} \frac{1}{x} \left( \mathrm {e}^{it \Xi _f (1/x)} -1 \right) \mathrm {d}x \right) . \end{aligned}$$

Since the left-hand side of the convergence is $1 + \mathcal {O}_{f, \theta , t} (1/R)$, we deduce that

$$\begin{aligned} L = ( 1 + \mathcal {O}_{f, \theta , t} (1/R) ) \exp \left( \theta \int _{1/R}^{R} \frac{1}{x} \left( \mathrm {e}^{it \Xi _f (1/x)} -1 \right) \mathrm {d}x \right) . \end{aligned}$$

Letting $R \rightarrow \infty$, we deduce

$$\begin{aligned} L = \exp \left( \theta \int _{0}^{\infty } \frac{1}{x} \left( \mathrm {e}^{it \Xi _f (1/x)} -1 \right) \mathrm {d}x \right) = \exp \left( \theta \int _{0}^{\infty } \frac{1}{y} \left( \mathrm {e}^{it \Xi _f (y)} -1 \right) \mathrm {d}y \right) , \end{aligned}$$

where the convergence of the integrals is insured by the integrability of $|\Xi _f(x)| dx/x$ given in Lemma 1. By Campbell’s theorem,

$$\begin{aligned} L = \mathbb {E}\left( \mathrm {e}^{it Z}\right) , \end{aligned}$$

i.e., $\mathbb {E}\left( \mathrm {e}^{it Z}\right)$ is the unique possible limit of a subsequence of $\left( \mathbb {E}\left( \mathrm {e}^{it B_{n}}\right) \right) _{n \ge 1}$. Since this sequence is bounded, we have proved (i).

(ii): Let $t\in \mathbb {R}$.

$$\begin{aligned} \left| \mathbb {E} \left( \mathrm {e}^{itA_n}\right) - \mathbb {E}\left( \mathrm {e}^{itB_n}\right) \right|&\le \vert t \vert \mathbb {E} \left( \vert A_n - B_n \vert \right) \\&\le \vert t \vert \sum _{\ell =1}^n \vert \Xi _f (1/ \ell \delta _n) \vert \mathbb {E} ( \vert a_{n,\ell } - W_\ell \vert ), \end{aligned}$$

where, by [7, Lemma 4.4],

$$\begin{aligned} \mathbb {E} ( \vert a_{n,\ell } - W_\ell \vert ) \le \frac{C(\theta )}{n} + \frac{\theta }{n} \Psi _n (\ell ) \ll _\theta \frac{1}{n} (1 + \Psi _n (\ell )) \end{aligned}$$

for some $C(\theta ) > 0$ depending only on $\theta$ and for

$$\begin{aligned} \Psi _n (\ell ) := \prod _{k=0}^{\ell -1} \frac{n-k}{ \theta + n - k - 1}. \end{aligned}$$

Let $(u_n)_{n \ge 1}$ and $(v_n)_{n \ge 1}$ be two sequences of positive integers such that $u_n < v_n$ for all n and also such that $u_n$, $\delta _n^{-1}/ u_n$, $v_n / \left( \delta _n^{-1}\right)$ and $n/v_n$ all go to infinity with n. On the one hand,

$$\begin{aligned}&\frac{1}{n} \sum _{\begin{array}{c} \ell < u_n \\ \text {or } \ell >v_n \end{array}} (1 + \Psi _n (\ell )) \vert \Xi _f (1/ \ell \delta _n) \vert \\&\qquad \le \left( \sup _{z \in (0, u_n \delta _n) \cup (v_n \delta _n, +\infty )} \vert \Xi _f (1/z) \vert \right) \frac{1}{n} \sum _{\ell =1 }^n (1 + \Psi _n (\ell )) \\&\qquad = \left( 1 + \frac{1}{\theta } \right) \sup _{z \in (0, u_n \delta _n) \cup (v_n \delta _n, +\infty )} \vert \Xi _f (1/z) \vert \\&\qquad \underset{n\rightarrow \infty }{\longrightarrow } 0, \end{aligned}$$

because $\Xi _f$ tends to zero at zero and at infinity, and by [7, Lemma 4.6],

$$\begin{aligned} \frac{\theta }{n} \sum _{\ell =1}^n \Psi _n (\ell ) = \sum _{\ell =1}^n \mathbb {P} [ J_n = \ell ] = 1, \end{aligned}$$

where

$$\begin{aligned} J_n = \min \{j \ge 1, \xi _{n-j+1} = 1 \}, \end{aligned}$$

$(\xi _j)_{j \ge 1}$ being independent Bernoulli variables, $\xi _j$ having parameter $\theta / (\theta + j -1)$. On the other hand, noticing that $\Psi _n (\ell )$ is monotonic with respect to $\ell$ (the direction of monotonicity of $\Psi _n(.)$ changes at $\theta = 1$) and tends to 1 when $n/\ell$ goes to infinity,

$$\begin{aligned}&\frac{1}{n} \sum _{u_n \le \ell \le v_n } (1 + \Psi _n (\ell )) \vert \Xi _f (1/ \ell \delta _n) \vert \\&\qquad \le \Vert \Xi _f \Vert _\infty \frac{v_n - u_n +1 }{n} \times (1 + \max (\Psi _n (u_n) , \Psi _n (v_n) )) \\&\qquad \underset{n\rightarrow \infty }{\longrightarrow } 0, \end{aligned}$$

since $n/v_n \rightarrow \infty$ and then $\max (\Psi _n (u_n) , \Psi _n (v_n) ) \rightarrow 1$ as $n \rightarrow \infty$.

We can now deduce Theorem 1.1 (iii). Since $M^{\sigma _n}$ has n eigenangles in each interval of length $2 \pi$, replacing $X_{\sigma _n, \delta _n}$ by $X'_{\sigma _n, \delta _n}$ changes the sum by at most

$$\begin{aligned} n \sum _{k \ne 0} \sup _{|x - 2 k \pi | \le \pi } |f(x/2 \pi \delta _n)| \ll _f n \sum _{k \ge 1} \delta _n^{\alpha } k^{-\alpha } \ll _{f, \alpha } n \delta _n^{\alpha }, \end{aligned}$$

quantity which, by the assumption made in (iii), tends to zero when $n \rightarrow \infty$. Using Slutsky’s lemma, we easily deduce that (i) and (ii) are preserved when we replace $X_{\sigma _n, \delta _n}$ by $X'_{\sigma _n, \delta _n}$.

7 Proof of Theorem 1.2

We first show the following lemma:

Lemma 3

For all $n\ge 1$,

$$\begin{aligned}&\frac{1}{n}\sum _{j=1}^n \Psi _n (j) = \frac{1}{\theta }, \end{aligned}$$

(8)

$$\begin{aligned}&\sum _{j=1}^n \frac{\Psi _n (j)}{j} = \sum _{j=1}^n \frac{1}{\theta + j-1}, \end{aligned}$$

(9)

and

$$\begin{aligned} \sum _{j=1}^n \frac{\Psi _n (j)}{j^2} \underset{n\rightarrow \infty }{\longrightarrow } \frac{\pi ^2}{6}. \end{aligned}$$

(10)

Proof

The equalities (8) and (9) are proved in [6, Lemma 9]. Let us now show (10). Let $(u_n)_{n \ge 1}$ be a sequence of positive integers such that $u_n$ and $n/u_n$ both tend to infinity with n. We split the sum into two as follows

$$\begin{aligned} \sum _{j=1}^n \frac{\Psi _n (j)}{j^2} = \sum _{j=1}^{u_n} \frac{\Psi _n (j)}{j^2} + \sum _{j=u_n + 1}^n \frac{\Psi _n (j)}{j^2}. \end{aligned}$$

By monotonicity of $\Psi _n (k)$ with respect to k, we have

$$\begin{aligned} \min \limits _{k \le u_n} \Psi _n (k) = \min (\Psi _n (1) , \Psi _n (u_n)) \end{aligned}$$

and

$$\begin{aligned} \max \limits _{k \le u_n} \Psi _n (k) = \max (\Psi _n (1) , \Psi _n (u_n)), \end{aligned}$$

where $\lim \limits _{n \rightarrow \infty } \Psi _n (1) = \lim \limits _{n \rightarrow \infty } \Psi _n (u_n) = 1$. Thus, as n goes to infinity, we have

$$\begin{aligned} \sum _{j=1}^{u_n} \frac{\Psi _n (j)}{j^2} = (1+ o (1)) \sum _{j=1}^{u_n} \frac{1}{j^2} = \frac{\pi ^2}{6} + o (1). \end{aligned}$$

Besides,

$$\begin{aligned} \sum _{j=u_n + 1}^n \frac{\Psi _n (j)}{j^2}&\le \max \limits _{k = u_n+1, \dots ,n} \Psi _n (k) \times \sum _{j=u_n + 1}^{+\infty } \frac{1}{j^2} \\&\ll _\theta \max (1, n^{1-\theta }) \times \frac{1}{u_n}, \end{aligned}$$

which tends to 0 as n goes to infinity if we take for instance $u_n := \lfloor \max (n^{1- \frac{\theta }{2}}, n^{1/2}) \rfloor$. $\square$

We now have the ingredients needed in order to prove Theorem 1.2. The equality

$$\begin{aligned} X^\prime _{\sigma _n, \delta _n} (f) - n \delta _n \hat{f}(0) = \sum _{\ell = 1}^n a_{n,\ell } \Xi _{f,\ell } \left( \frac{1}{\ell \delta _n}\right) \end{aligned}$$

is proved similarly as (4), after taking into account the fact that $f(0) = 0$. From Theorem 1.1, it is then enough to show

$$\begin{aligned} \left| \mathbb {E}(\mathrm {e}^{it A_n}) - \mathbb {E}(\mathrm {e}^{it C_n}) \right| \underset{n \rightarrow \infty }{\longrightarrow } 0 \end{aligned}$$

for all $t\in \mathbb {R}$, where $A_n:= \sum _{\ell = 1}^n a_{n,\ell } \Xi _f \left( \frac{1}{\ell \delta _n}\right)$ and $C_n:= \sum _{\ell = 1}^n a_{n,\ell } \Xi _{f,\ell } \left( \frac{1}{\ell \delta _n}\right)$. Let $t\in \mathbb {R}$.

$$\begin{aligned} \left| \mathbb {E}(\mathrm {e}^{it A_n}) - \mathbb {E}(\mathrm {e}^{it C_n}) \right|&\le \vert t \vert \mathbb {E} (\vert A_n - C_n \vert )\\&\le \vert t \vert \sum _{\ell = 1}^n \mathbb {E} (a_{n,\ell }) \left| \Xi _f \left( \frac{1}{\ell \delta _n}\right) - \Xi _{f,\ell } \left( \frac{1}{\ell \delta _n}\right) \right| \\&= \vert t \vert \sum _{\ell = 1}^n \frac{\theta \Psi _n (\ell )}{\ell } \left| \sum _{k=-\infty }^{\lfloor -\ell /2\rfloor } f \left( \frac{k}{\ell \delta _n} \right) + \sum _{k=\lfloor \ell /2\rfloor + 1}^{+\infty } f \left( \frac{k}{\ell \delta _n} \right) \right| . \end{aligned}$$

Here, we use the fact that the expectation of the number of $\ell$-cycles is equal to $n/\ell$ times the probability that 1 is in an $\ell$-cycle, i.e., by the Feller coupling,

$$\begin{aligned} \mathbb {E} [ a_{n,\ell } ]&= \frac{n}{\ell } \mathbb {P} [ \xi _n = \xi _{n-1} = \dots = \xi _{n+2 - \ell } = 0, \xi _{n+1 - \ell } = 1] \\&= \frac{n}{\ell } \frac{\theta }{n-\ell + \theta } \prod _{k= 1}^{\ell -1} \frac{n-k}{n-k + \theta } = \frac{\theta \Psi _n (\ell )}{\ell }. \end{aligned}$$

We now estimate the sum over the positive indices $\sum _{k=\lfloor \ell /2\rfloor + 1}^{+\infty } f \left( \frac{k}{\ell \delta _n} \right)$: the sum over the negative indices behaves identically. To do this, we use the Euler–MacLaurin formula at order 2: for all positive integers $p<q$, and for all functions $g \in \mathcal {C}^2 (\mathbb {R})$,

$$\begin{aligned} \sum _{k=p}^q g(k) = \int _p^q g(x) \mathrm {d}x + \frac{g(p) + g(q)}{2} + \frac{g^\prime (q) - g^\prime (p)}{12} + \mathcal {O} \left( \int _p^q \vert g^{\prime \prime } (x) \vert \mathrm {d}x \right) , \end{aligned}$$

so that if g, $g^\prime$ and $g^{\prime \prime }$ are integrable at $+\infty$, we have, letting q tend to infinity,

$$\begin{aligned} \sum _{k=p}^{+\infty } g(k) = \int _p^{+\infty } g(x) \mathrm {d}x + \frac{g(p)}{2} - \frac{g^\prime (p)}{12} + \mathcal {O} \left( \int _p^{+\infty } \vert g^{\prime \prime } (x) \vert \mathrm {d}x \right) . \end{aligned}$$

Applying this formula to $g(x) = f\left( \frac{x}{\ell \delta _n}\right)$ gives, with a change of variables into the integrals,

$$\begin{aligned} \sum _{k=p}^{+\infty } f\left( \frac{k}{\ell \delta _n} \right) = -\ell \delta _n F \left( \frac{p}{\ell \delta _n} \right) + \frac{f\left( \frac{p}{\ell \delta _n} \right) }{2} - \frac{f^\prime \left( \frac{p}{\ell \delta _n} \right) }{12 \ell \delta _n} + \mathcal {O} \left( \frac{1}{\ell \delta _n} \int _{\frac{p}{\ell \delta _n}}^{+\infty } \vert f^{\prime \prime } (u) \vert \mathrm {d}u \right) , \end{aligned}$$

where F is the antiderivative of f such that $F(+\infty ) =0$.

Then, with $p=\lfloor \ell /2\rfloor + 1$, using Taylor-Lagrange formula at order 3 on F, at order 2 on f and at order 1 on $f^\prime$, between $\frac{1}{2\delta _n}$ and $\frac{p}{\ell \delta _n} = \frac{1}{2\delta _n} + \frac{1-\{\ell /2\}}{\ell \delta _n}$, we get

$$\begin{aligned} \sum _{k=\lfloor \ell /2\rfloor + 1}^{+\infty } f\left( \frac{k}{\ell \delta _n} \right)&= -\ell \delta _n F \left( \frac{1}{2\delta _n} \right) + \left[ - (1-\{\ell /2 \}) + \frac{1}{2} \right] f \left( \frac{1}{2\delta _n} \right) \\&\quad +\frac{1}{\ell \delta _n} \left[ - \frac{1}{2} (1-\{\ell /2 \})^2 + \frac{1}{2}(1-\{\ell /2 \}) - \frac{1}{12}\right] f^\prime \left( \frac{1}{2\delta _n} \right) \\&\quad + \mathcal {O} \left( \frac{1}{\ell \delta _n} \int _{\frac{1}{2\delta _n}}^{+\infty } \vert f^{\prime \prime } (u) \vert \mathrm {d}u \right) . \end{aligned}$$

Consequently,

$$\begin{aligned} \left| \sum _{k=\lfloor \ell /2\rfloor + 1}^{+\infty } f\left( \frac{k}{\ell \delta _n} \right) \right|&\le \ell \delta _n \left| F \left( \frac{1}{2\delta _n} \right) \right| + \frac{1}{2} \left| f \left( \frac{1}{2\delta _n} \right) \right| \\&\quad + \frac{1}{12\ell \delta _n} \left| f^\prime \left( \frac{1}{2\delta _n} \right) \right| \\&\quad + \frac{1}{\ell } \times \mathcal {O} \left( \frac{1}{\delta _n} \int _{\frac{1}{2\delta _n}}^{+\infty } \vert f^{\prime \prime } (u) \vert \mathrm {d}u \right) , \end{aligned}$$

where we point out that the implicit constant involved in $\mathcal {O}$ does not depend on $\ell$. Finally, using (8), (9) and (10), it follows

$$\begin{aligned} \sum _{\ell = 1}^n \frac{\theta \Psi _n (\ell )}{\ell } \left| \sum _{k=\lfloor \ell /2\rfloor + 1}^{+\infty } f\left( \frac{k}{\ell \delta _n} \right) \right|&\le n \delta _n \left| F \left( \frac{1}{2\delta _n} \right) \right| \\&\quad + \left( \frac{\theta }{2}\log n + \mathcal {O}_\theta (1)\right) \left| f \left( \frac{1}{2\delta _n} \right) \right| \\&\quad + \left( \frac{\theta \pi ^2}{72} + o_\theta (1) \right) \frac{1}{\delta _n} \left| f^\prime \left( \frac{1}{2\delta _n} \right) \right| \\&\quad + \mathcal {O}_{\theta } \left( \frac{1}{\delta _n} \int _{\frac{1}{2\delta _n}}^{+\infty } \vert f^{\prime \prime } (u) \vert \mathrm {d}u \right) , \end{aligned}$$

which tends to 0 as $n\rightarrow +\infty$, under the hypothesis made on f and $\delta _n$.