Multifractality and intermittency in the limit evolution of polygonal vortex filaments

Abstract

With the aim of quantifying turbulent behaviors of vortex filaments, we study the multifractality and intermittency of the family of generalized Riemann’s non-differentiable functions

$$\begin{aligned} R_{x_0}(t) = \sum _{n \ne 0} \frac{e^{2\pi i ( n^2 t + n x_0 ) } }{n^2}, \qquad x_0 \in [0,1]. \end{aligned}$$

These functions represent, in a certain limit, the trajectory of regular polygonal vortex filaments that evolve according to the binormal flow. When $x_0$ is rational, we show that $R_{x_0}$ is multifractal and intermittent by completely determining the spectrum of singularities of $R_{x_0}$ and computing the $L^p$ norms of its Fourier high-pass filters, which are analogues of structure functions. We prove that $R_{x_0}$ has a multifractal behavior also when $x_0$ is irrational. The proofs rely on a careful design of Diophantine sets that depend on $x_0$, which we study by using the Duffin–Schaeffer theorem and the Mass Transference Principle.

1 Introduction

Multifractality and intermittency are among the main properties expected in turbulent flows but, as usual in the theory of turbulence, it is challenging to analyze them rigorously. The motivation of this article is to quantify the multifractal and intermittent behavior of regular polygonal vortex filaments that evolve with the binormal flow. This evolution is represented, in a certain limit, by the function $R_{x_0}: {\mathbb {R}} \rightarrow {\mathbb {C}}$ defined by

$$\begin{aligned} R_{x_0}(t) = \sum _{n \ne 0} \frac{e^{2\pi i ( n^2 t + n x_0 ) } }{n^2}, \end{aligned}$$

(1)

for $x_0 \in [0,1]$ fixed. This function is one of the possible generalizations of the classic Riemann’s non-differentiable function, which is recovered when $x_0 = 0$, and it can also be seen as the solution to a periodic Cauchy problem for the free Schrödinger equation. In this article we study the multifractality and intermittency of $R_{x_0}$, which until now was unknown for $x_0 \ne 0$:

• When $x_0 \in {\mathbb {Q}}$, we completely describe the multifractality of $R_{x_0}$ by computing its spectrum of singularities (Theorem 1.1). We also compute the $L^p$ norms of its Fourier high-pass filters to deduce its intermittency exponents (Theorem 1.6) and show that $R_{x_0}$ is intermittent.

• When $x_0 \not \in {\mathbb {Q}}$, we give a result that proves multifractality (Theorem 1.3) and strongly suggests that the spectrum of singularities depends on the irrationality of $x_0$, and hence that it is different from when $x_0 \in {\mathbb {Q}}$.

The main novelty in this article is a careful design of Diophantine sets and the use of the Duffin–Schaeffer theorem and the Mass Transference Principle to compute their measure and dimension. When $x_0 \in {\mathbb {Q}}$, we use the partial Duffin–Schaeffer theorem as proved by Duffin and Schaeffer in [21], while when $x_0 \not \in {\mathbb {Q}}$ we need the full strength of the theorem as proved by Koukoulopoulos and Maynard [37]. We give an overview of these arguments in Sect. 2. Before that, we introduce the concepts of multifractality and intermittency in Sect. 1.1, we discuss the connection of $R_{x_0}$ and vortex filaments in Sect. 1.2 and we state our results in Sects. 1.3 and 1.4.

1.1 Multifractality and intermittency

The concepts of multifractality and intermittency arise in the study of three dimensional turbulence of fluids and waves, both characterized by low regularity and a chaotic behavior. These are caused by an energy cascade by which the energy injected in large scales is transferred to small scales. In this setting, large eddies constantly split in smaller eddies, generating sharp changes in the velocity magnitude. Moreover, this cascade is not expected to be uniform in space, and the rate at which these eddies decrease depends on their location.

Mathematically speaking, an option to measure the irregularity of the velocity v is to compute the local Hölder regularity, that is, the largest $\alpha = \alpha (x)$ such that $|v(x + h) - v(x)| \lesssim |h|^\alpha$ when $|h| \rightarrow 0$. The lack of uniformity in space suggests that the Hölder level sets $D_\alpha = \{ \, x : \, \alpha (x) = \alpha \, \}$ should be non-empty, and of different size, for many values of $\alpha$. In this context, the spectrum of singularities is defined as $d(\alpha ) = {\text {dim}}_{{\mathcal {H}}} D_\alpha$, where ${\text {dim}}_{{\mathcal {H}}}$ is the Hausdorff dimension, and the velocity v is said to be multifractal if $d(\alpha )$ takes values in multiple Hölder regularities $\alpha$.

On the other hand, intermittency is a measure of the likelihood of localized bursts or outlier events. One way to quantify it is by analyzing the structure functions $S_p(h) = \langle |v(x+h) - v(x)|^p \rangle$ of the velocity when the scale h tends to zero. More precisely, defining the flatness as

$$\begin{aligned} F_4(h) = \frac{S_4(h)}{S_2(h)^2}, \quad \text { for very small } h, \end{aligned}$$

(2)

we have small-scale intermittency¹ if $\lim _{h \rightarrow 0} F_4(h) = +\infty$. Assuming the typical power law

$$\begin{aligned} S_p(h) \simeq |h|^{\zeta _p}, \end{aligned}$$

(3)

it is usual to rephrase the definition of intermittency as $\zeta _4 - 2\zeta _2 < 0$ for the intermittency exponent²$\zeta _p$. This definition, and in particular (2), is inspired by the probabilistic concept of kurtosis,³ which quantifies how large the tails of the underlying probability distribution are. A large kurtosis implies fat tails, which suggests that outlier events are more likely than for a normal distribution, agreeing with the widespread idea of non-Gaussianity. More generally, moments $F_p(h) = S_p(h)/S_2(h)^{p/2}$ of order $p \ge 4$ can be used to measure the tails of a probability distribution (see [27, p. 124]) and therefore intermittency, so it is common in recent physics literature to measure $\zeta _p$ for different p (see [42] and references therein, also [2] for a numeric intermittent model). The intermittency condition is then rewritten as $\zeta _p - p \zeta _2 /2 < 0$, a behavior that corresponds to a sublinear $\zeta _p$.

1.2 $R_{x_0}$ as the trajectory of polygonal vortex filaments

The binormal flow is a model introduced by Da Rios⁴ in 1906 [19] as an approximation to the evolution of a vortex filament according to Euler equation and whose validity has been precisely and rigorously described theoretically by Fontelos and Vega in [26] in the setting of the Navier–Stokes equations. This model describes the motion of the filament ${\varvec{X}}: {\mathbb {R}} \times {\mathbb {R}} \rightarrow {\mathbb {R}}^3$, ${\varvec{X}} = {\varvec{X}}(x,t)$ by the equation ${\varvec{X}}_t = {\varvec{X}}_x \times {\varvec{X}}_{xx}$. Inspired by Jerrard and Smets [33], De la Hoz and Vega [20] observed numerically that if the initial filament ${\varvec{X}}_M(x,0)$ is a regular polygon with M corners at the integers $x \in {\mathbb {Z}}$, then the trajectory of the corners ${\varvec{X}}_M(0,t)$ is a plane curve which, identifying the plane with ${\mathbb {C}}$ and when M is large, looks like

$$\begin{aligned} \phi (t) = \sum _{n \in {\mathbb {Z}}} \frac{e^{2\pi i n^2 t} - 1}{n^2} = 2\pi it - \frac{\pi ^2}{3} + R_0(t). \end{aligned}$$

(4)

Moreover, let $\varvec{\chi }_M(x,0)$ be an infinite polygonal line that loops the polygon of M sides a finite but large number of times and ends in two half-lines, symmetrized at $x=0$. Banica and Vega rigorously proved in [4] that, under certain hypotheses, its binormal flow evolution $\varvec{\chi }_M(x,t)$ obtained in [3] satisfies

$$\begin{aligned} \lim _{M \rightarrow \infty } M \, \varvec{\chi }_M(x_0,t) = \phi _{x_0}(t):= \sum _{n \in {\mathbb {Z}}} \frac{e^{2 \pi i n^2 t} - 1}{n^2} \, e^{2\pi i n x_0}, \quad \forall x_0 \in [0,1]. \end{aligned}$$

(5)

We show in Figs. 1 and 2 the image of $\phi _{x_0}$ for some values of $x_0$. Like in (4), noticing that the Fourier series ${\sum _{n \ne 0} \frac{ e^{2\pi i n x} }{n^2}}$ is $2\pi ^2 \left( x^2 - x + \frac{1}{6} \right)$, we can write

$$\begin{aligned} \phi _{x_0}(t) =2\pi i t - 2 \pi ^2 \Big (x_0^2 - x_0 + \frac{1}{6}\Big ) + R_{x_0}(t), \end{aligned}$$

which shows that $\phi _{x_0}$ and $R_{x_0}$ have the same regularity as functions of t. In other words, $R_{x_0}$ captures the regularity of the limit trajectory of polygonal vortex filaments that evolve with the binormal flow. This connection motivates us to study the multifractality and intermittency of $R_{x_0}$.

Fig. 1

Image of $\phi _{x_0}$, $t \in [0,1]$, defined in (5), for some values of $x_0$

Fig. 2

The images of $\phi _{x_0}$, $t \in [0,1]$, for the values $x_0 = 0, 0.1, 0.2, 0.3, 0.4, 0.5$, from the rightmost to the leftmost

1.3 Definitions and notation

We now rigorously define the concepts discussed above.

1.3.1 Holder regularity

A function $f: {\mathbb {R}} \rightarrow {\mathbb {C}}$ is $\alpha$-Hölder at $t \in {\mathbb {R}}$, which we denote by $f \in {\mathcal {C}}^\alpha (t)$, if there exists a polynomial $P_t$ of degree at most $\alpha$ such that $|f(t+h) - P_t(h)| \le C |h|^\alpha$ for some constant $C > 0$ and for h small enough. In particular, if $0< \alpha < 1$, the definition above becomes

$$\begin{aligned} f \in {\mathcal {C}}^\alpha (t) \quad \Longleftrightarrow \quad |f(t+h) - f(t)| \le C |h|^\alpha , \quad \text { for } h \text { small enough}. \end{aligned}$$

The local Hölder exponent of f at t is $\alpha _f(t) = \sup \{ \, \alpha : \, f \in {\mathcal {C}}^\alpha (t) \, \}$. We say f is globally $\alpha$-Hölder if $f \in {\mathcal {C}}^\alpha (t)$ for all $t \in {\mathbb {R}}$.

1.3.2 Spectrum of singularities

The spectrum of singularities of f is

$$\begin{aligned} d_f(\alpha ) = {\text {dim}}_{{\mathcal {H}}} \{ \, t \,: \, \alpha _f(t) = \alpha \, \}, \end{aligned}$$

where ${\text {dim}}_{{\mathcal {H}}}$ is the Hausdorff dimension,⁵ and convene that $d(\alpha ) = -\infty$ if $\{ \, t : \, \alpha _f(t) = \alpha \, \} = \emptyset$.

1.3.3 Intermittency exponents

As discussed in (3), the exponents $\zeta _p$ of the structure functions $S_p(h)$ describe the behavior of the increments of functions in small scales. Here we take the analogous approach of studying the high-frequency behavior of functions. Let $\Phi \in C^\infty ({\mathbb {R}})$ be a cutoff function such that $\Phi (x) = 0$ in a neighborhood of the origin and $\Phi (x) = 1$ for $|x| \ge 2$. For a periodic function f with Fourier series $f(t) = \sum _{n \in {\mathbb {Z}}} a_n e^{2\pi i n t}$, define the high-pass filter by

$$\begin{aligned} P_{\ge N} f(t) = \sum _{n \in {\mathbb {Z}}} \Phi \Big ( \frac{n}{N} \Big ) \, a_n \, e^{2\pi i n t }, \qquad N \in {\mathbb {N}}. \end{aligned}$$

We treat the $L^p$ norms $\Vert P_{\ge N} f \Vert _p^p$ as the analytic and Fourier space analogues of the structure functions.⁶ Our analogous to the power law (3) is⁷

$$\begin{aligned} \eta _f(p) = \liminf _{N \rightarrow \infty } \frac{\log (\Vert P_{\ge N} f \Vert _p^p ) }{\log (1/N)}, \end{aligned}$$

(6)

which means that for any $\epsilon > 0$ we have $\Vert P_{\ge N} f \Vert _p^p \le N^{-\eta _f(p) + \epsilon }$ for $N \gg _\epsilon 1$, and that this is optimal in the sense that there is a subsequence $N_k \rightarrow \infty$ such that $\Vert P_{\ge N_k} f \Vert _p^p \ge N_k^{-\eta _f(p) - \epsilon }$ for $k \gg _\epsilon 1$. We define the ${\varvec{p}}$-flatness to be

$$\begin{aligned} F_p(N) = \frac{\Vert P_{\ge N} f \Vert _p^p}{\Vert P_{\ge N} f \Vert _2^p}, \qquad N \gg 1. \end{aligned}$$

The corresponding intermittency exponent⁸ is $\eta _f(p) - p\, \eta _f(2)/2$.

1.4 Results

To simplify notation, let us denote $\alpha _{R_{x_0}}(t) = \alpha _{x_0}(t)$, $d_{R_{x_0}}(\alpha ) = d_{x_0}(\alpha )$ and $\eta _{R_{x_0}}(p) = \eta _{x_0}(p)$ for our function $R_{x_0}$ defined in (1).

Since Weierstrass [47] announced⁹ Riemann’s non-differentiable function as the first candidate of a continuous and non-differentiable function in 1872, the regularity of $R_0$ has been studied by several authors. After Hardy [30] and Gerver [28, 29] proved that it is only almost nowhere differentiable (see also the simplified proof of Smith [45]), Duistermaat [22] launched the study of its Hölder regularity. Jaffard completed the picture in his remarkable work [31, Theorem 1] (see also [11] for a recent alternative proof) by computing

$$\begin{aligned} \alpha _0(t) = \frac{1}{2} + \frac{1}{2{\widetilde{\mu }}(t)}, \quad \text { for } t \not \in {\mathbb {Q}}, \end{aligned}$$

(7)

where ${{\widetilde{\mu }}}(t)$ is the exponent of irrationality of t restricted to denominators $q \not \equiv 2 \text { (mod } 4)$.¹⁰ He combined this with an adaptation of the Jarník–Besicovitch theorem to prove

$$\begin{aligned} d_0(\alpha ) = \left\{ \begin{array}{ll} 4\alpha - 2, & 1/2 \le \alpha \le 3/4, \\ 0, & \alpha = 3/2, \\ -\infty , & \text { otherwise.} \end{array} \right. \end{aligned}$$

Our first results concern the spectrum of singularities of $R_{x_0}$ for $x_0 \ne 0$.

Theorem 1.1

Let $x_0 \in {\mathbb {Q}}$. Then,

$$\begin{aligned} d_{x_0}(\alpha ) = \left\{ \begin{array}{ll} 4 \alpha - 2, & 1/2 \le \alpha \le 3/4, \\ 0, & \alpha = 3/2, \\ -\infty , & \text {otherwise.} \end{array} \right. \end{aligned}$$

Remark 1.2

• (a) To prove Theorem 1.1, we adapt the classical approach due to Duistermaat [22] and Jaffard [31] by carefully choosing subsets of the irrationals with novel Diophantine restrictions to disprove Hölder regularities. However, the arguments in [31] to compute their Hausdorff dimension do not suffice¹¹ when $x_0 \ne 0$. We solve this by using the Duffin–Schaeffer theorem and the Mass Transference Principle; see Sect. 2 for the outline of the argument.

• (b) Even if $d_{x_0} = d_0$ for all $x_0 \in {\mathbb {Q}}$, we think that $\alpha _{x_0}(t) \ne \alpha _0(t)$. However, Theorem 1.1 does not require computing $\alpha _{x_0}(t)$ for all $t \in {\mathbb {R}}$. A full description of the sets $\{ \, t : \, \alpha _{x_0}(t) = \alpha \, \}$ is an interesting and challenging problem because when $x_0 \ne 0$ it is not clear how to characterize the Hölder regularity $\alpha _{x_0}(t)$ in terms of some irrationality exponent like in (7). We do not pursue this problem here, which we leave for a future work.

Let now $x_0 \not \in {\mathbb {Q}}$. Let $p_n/q_n$ be its approximations by continued fractions, and define the exponents $\mu _n$ by $|x_0 - p_n/q_n| = 1/q_n^{\mu _n}$. Define the alternative¹² exponent of irrationality

$$\begin{aligned} \sigma (x_0) = \limsup _{n \rightarrow \infty } \, \{ \, \mu _n \,: \, q_n \not \in 4{\mathbb {N}} \, \}. \end{aligned}$$

(8)

This exponent always exists and $\sigma (x_0) \ge 2$ (see Proposition 5.2). Our result is the following.

Theorem 1.3

Let $x_0 \not \in {\mathbb {Q}}$. Let $2 \le \mu < 2\sigma (x_0)$, with $\sigma (x_0)$ as in (8). Then, for all $\delta > 0$,

$$\begin{aligned} \frac{1}{\mu } \le {\text {dim}}_{{\mathcal {H}}} \left\{ \, t \,: \frac{1}{2} + \frac{1}{4\mu } - \delta \le \alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu } \right\} \le \frac{2}{\mu }. \end{aligned}$$

(9)

Remark 1.4

• (a) We show in Fig. 3 a graphic representation of Theorem 1.3.

• (b) Theorem 1.3 shows that $R_{x_0}$ is multifractal when $\sigma (x_0) > 2$.

• (c) Theorem 1.3 would be strengthened to $1/\mu \le d_{x_0}(1/2 + 1/2\mu ) \le 2/\mu$ for $\mu < 2\sigma (x_0)$ if we could compute the dimension of some well-identified Diophantine sets, see Remark 5.4. This would give a nontrivial spectrum of singularities in an open interval for all $x_0 \not \in {\mathbb {Q}}$. We leave this for a future work.

• (d) The reasons to have an interval $(1/\mu , 2/\mu )$ for the dimension in (9) seem to us deeper in nature. Unlike the upper bound $2/\mu$, which follows from approximating t with rationals p/q with unrestricted $q\in {\mathbb {N}}$ and with error $q^{-\mu }$ (see the Jarník–Besicovitch theorem 2.2), the lower bound depends on the nature of $x_0$ which imposes restrictions to q. When $x_0 = P/Q \in {\mathbb {Q}}$, we require $q \in 4Q{\mathbb {N}}$, which still results in a set of dimension $2/\mu$. However, when $x_0 \not \in {\mathbb {Q}}$ we require q be restricted to an exponentially growing sequence (given by the denominators of the continued fraction approximations of $x_0$). This restriction is much stronger and gives a set of t of dimension $1/\mu$. These results follow from the Duffin–Schaeffer theorem and the Mass Transference Principle.

• (e) The theorem and its proof (see the heuristic discussion in Sect. 5.2.1) suggest that the spectrum of singularities may be $d_{x_0}(\alpha ) = 4\alpha - 2$ in the range $\frac{1}{2} + \frac{1}{4\sigma (x_0)} \le \alpha \le \frac{3}{4}$, and possibly something different outside of this range. In particular, we expect the segment of the spectrum in $5/8 \le \alpha \le 3/4$ to be present for all $x_0$.

Remark 1.5

Our results suggest that the trajectories of the binormal flow do not have a generic behavior in terms of regularity. Indeed, if $X_n$ is a sequence of independent and identically distributed complex Gaussian random variables, then the random function

$$\begin{aligned} S(t) = \sum _{n=1}^\infty X_n \frac{e^{2\pi i n^2 t}}{n^2} \end{aligned}$$

(10)

has¹³ almost surely $\alpha _S(t) = 3/4$ for all $t \in {\mathbb {R}}$ [34]. Hence the generic behavior of (10) is monofractal. In contrast, the fine structure of the linear phase $nx_0$ of $R_{x_0}$ causes a multifractal behavior.

Regarding intermittency, we compute the $L^p$ norms of the Fourier high-pass filters of $R_{x_0}$ and the intermittency exponents $\eta _{x_0}(p)$ when $x_0 \in {\mathbb {Q}}$, from which we deduce that $R_{x_0}$ is intermittent.

Fig. 3

A graphic representation of Theorem 1.3. We have a continuum of Whitney-type boxes parametrized by $\mu$ along the dashed diagonal line $d(\alpha ) = 4\alpha - 2$. The graph of $d_{x_0}(\alpha )$ has at least a point in each of those boxes

Theorem 1.6

Let $x_0 \in {\mathbb {Q}}$. Let $1< p < \infty$. Then,

$$\begin{aligned} \big \Vert P_{\ge N} R_{x_0} \big \Vert _p^p \simeq \left\{ \begin{array}{ll} N^{-\frac{p}{2} - 1}, & p > 4, \\ N^{-3} \, \log N, & p = 4, \\ N^{-3p/4}, & p < 4, \end{array} \right. \end{aligned}$$

(11)

and therefore

$$\begin{aligned} \eta _{x_0}(p) = \lim _{N \rightarrow \infty } \frac{\log (\Vert P_{\ge N} f \Vert _p^p ) }{\log (1/N)} = \left\{ \begin{array}{ll} p/2 + 1, & p > 4, \\ 3p/4, & p \le 4. \end{array} \right. \end{aligned}$$

Consequently, $\lim _{N \rightarrow \infty } F_p(N) = +\infty$ for $p \ge 4$. In particular, $R_{x_0}$ is intermittent.

Remark 1.7

• (a) The $p=4$ intermittency exponent in (11) is $\eta (4) - 2\eta (2) = 0$, but the fact that $\Vert P_{\ge N} R_{x_0}\Vert _4^4$ does not follow a pure power law makes $F_4(N) \simeq \log N$. For $p > 4$, we have $\eta (p) - p\eta (2)/2 = 1 - p/4 < 0$, so $R_{x_0}$ is intermittent in small scales when $x_0 \in {\mathbb {Q}}$.

• (b) The upper bound in (11) in Theorem 1.6 holds for all $x_0 \in [0,1]$. The theorem shows that this is optimal when $x_0 \in {\mathbb {Q}}$, but we do not expect it to be optimal when $x_0 \not \in {\mathbb {Q}}$. We suspect that the exact behavior, and hence $\eta _{x_0}(p)$, depends on the irrationality of $x_0$. We aim to study this question in a future work.

1.5 Related literature on the analytic study of Riemann’s non-differentiable function

Beyond the literature for the original Riemann’s function $R_0$, the closest work to the study of $R_{x_0}$ is by Oskolkov and Chakhkiev [40]. They studied the regularity of $R_{x_0}(t)$ almost everywhere as a function of two variables $(x_0,t)$, which is not fine enough to capture multifractal properties.

Alternatively, there are many works studying $R_{x_0}(t)$ as a function of $x_0$ with t fixed, motivated by the fact that $R_{x_0}$ is the solution to an initial value problem for the periodic free Schrödinger equation. From this perspective, Kapitanski and Rodnianski [35] studied the Besov regularity of the fundamental solution¹⁴ as a function of x with t fixed. This approach is also intimately related to the Talbot effect in optics which, as proposed by Berry and Klein [7], is approximated by the fundamental solution to the periodic free Schrödinger equation. Pursuing the related phenomenon of quantization,¹⁵ the geometry of the profiles of Schrödinger solutions have been studied for fixed t by Berry [6] and Rodnianski [43]. Following the numeric works of Chen and Olver [16, 17], this perspective has also been extended to the nonlinear setting and other dispersive relations by Chousonis et al. [18, 24] and Boulton, Farmakis and Pelloni [8, 9].

There is a literature for other natural generalizations of Riemann’s function, like

$$\begin{aligned} F(t) = \sum _{n=1}^\infty \frac{e^{2\pi i P(n) t}}{n^\alpha }, \quad P \text { a polynomial}, \quad \alpha > 1, \end{aligned}$$

For $P(n) = n^2$, Jaffard [31] gave his results for all $\alpha > 1$. Chamizo and Córdoba [13] studied the Minkowski dimension of their graphs. Seuret and Ubis [44] studied the non-convergent case $\alpha < 1$, using a local $L^2$ exponent. Chamizo and Ubis [14, 15] studied the spectrum of singularities for general polynomials P. Further generalizations concerning fractional integrals of modular forms were studied by Pastor [41].

1.6 Structure of the article

In Sect. 2 we discuss the general strategy we follow to prove our theorems, stressing the new ideas related to Diophantine sets with restrictions, the Duffin–Schaeffer theorem and the Mass Transference Principle. In Sect. 3 we prove preliminary results for the local Hölder regularity of $R_{x_0}$, in particular the behavior around rational points t. In Sect. 4 we compute the spectrum of singularities of $R_{x_0}$ when $x_0 \in {\mathbb {Q}}$ and prove Theorem 1.1. In Sect. 5 we prove Theorem 1.3. In Sect. 6 we prove Theorem 1.6 by computing the $L^p$ norms of the high-pass filters of $R_{x_0}$. The proofs of some auxiliary results are postponed to Appendices A and B to avoid breaking the continuity of the main arguments.

2 An overview on the general arguments and on Diophantine approximation

2.1 General argument

An important part of the arguments in this article relies on Diophantine approximation. We will work with both the exponent of irrationality

$$\begin{aligned} \mu (x)= & \sup \Big \{ \, \mu > 0 \,: \Big | x - \frac{p}{q} \Big | \le \frac{1}{q^\mu } \text { for infinitely many coprime pairs } (p,q) \in {\mathbb {N}} \times {\mathbb {N}} \, \Big \},\nonumber \\ \end{aligned}$$

(12)

and the Lebesgue and Hausdorff measure properties of the related sets

$$\begin{aligned} A_\mu= & \Big \{ \, x \in [0,1] \, \mid \, \Big | x - \frac{p}{q} \Big | \le \frac{1}{q^\mu } \text { for infinitely many coprime pairs } (p,q) \in {\mathbb {N}} \times {\mathbb {N}} \, \Big \},\nonumber \\ \end{aligned}$$

(13)

where the case $\mu = \infty$ is understood as $A_\infty = \bigcap _{\mu \ge 2} A_\mu$. In a somewhat hand-waving way, $\mu (x) = \mu$ means that $|x - p/q| \simeq 1/q^\mu$ infinitely often, which ceases to be true for any larger $\mu$.

With these concepts in hand, the classic way to study the regularity of $R_{x_0}$ (used by Duistermaat, Jaffard and subsequent authors) is to first compute the asymptotic behavior of $R_{x_0}$ around rationals. Using the Poisson summation formula we will get a leading order expression of the form

$$\begin{aligned} R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q}\Big ) \sim \frac{\sqrt{h}}{q}G_q \sim \frac{\sqrt{h}}{\sqrt{q}}, \end{aligned}$$

(14)

where $G_q$ includes a quadratic Gauss sum of period q, hence $|G_q| \sim \sqrt{q}$ whenever it does not cancel. This shows that in most rationals the regularity of $R_{x_0}$ is 1/2. Let now $t\not \in {\mathbb {Q}}$ with irrationality exponent $\mu (t) = \mu$. Then, essentially $|t - p/q| \simeq 1/q^{\mu }$, so choosing $h = t-p/q$ we get

$$\begin{aligned} R_{x_0}\Big ( t \Big ) - R_{x_0}\Big ( t - h \Big ) \sim \frac{\sqrt{h}}{\sqrt{q}} \sim h^{\frac{1}{2} + \frac{1}{2\mu }}. \end{aligned}$$

This suggests that $\alpha _{x_0}(t) = \frac{1}{2} + \frac{1}{2\mu }$. Combining this with the Jarnik–Besicovitch theorem, which says that $\dim _{{\mathcal {H}} } A_\mu = 2/\mu$, we get the desired $d(\alpha ) = 4\alpha - 2$ in the range $1/2 \le \alpha \le 3/4$.

This argument is essentially valid up to assuming $G_q \ne 0$ in (14). This, however, does not always hold. Apart from a parity condition on q coming from the Gauss sums (present already in previous works), an additional condition arises that depends on $x_0$. For example, if $x_0 = P/Q \in {\mathbb {Q}}$, this condition has the form of $Q \mid q$. In terms of the sets $A_\mu$, this means that we need to restrict the denominators of the approximations to a subset of the natural numbers. So let ${\mathcal {Q}} \subset {\mathbb {N}}$, and define

$$\begin{aligned} A_{\mu , {\mathcal {Q}}}= & \Big \{ \, x \in [0,1] \,: \, \Big | x - \frac{p}{q} \Big |\le \frac{1}{q^\mu } \text { for infinitely many coprime pairs } (p,q) \in {\mathbb {N}} \times {\mathcal {Q}} \, \Big \}.\nonumber \\ \end{aligned}$$

(15)

Clearly $A_{\mu , {\mathcal {Q}}} \subset A_\mu$, but a priori it could be much smaller. Does $A_{\mu , {\mathcal {Q}}}$ preserve the measure of $A_\mu$? Previous works need to work with situations analogue to $Q=2$, but here we need to argue for all $Q \in {\mathbb {N}}$. For that, at the level of the Lebesgue measure we will use the Duffin–Schaeffer theorem, while we will compute Hausdorff measures and dimensions via the Mass Transference Principle.

2.2 Lebesgue measure: Dirichlet approximation and the Duffin–Schaeffer theorem

Both the Dirichlet approximation theorem and the theory of continued fractions imply $A_2 = \mathbb {[}0,1] {\setminus } {\mathbb {Q}}$. However, neither of them give enough information about the sequence of denominators they produce, so they cannot be used to determine the size of the set $A_{2,{\mathcal {Q}} } \subset A_2$. The recently proved Duffin–Schaeffer conjecture gives an answer to this kind of questions.

Theorem 2.1

(Duffin–Schaeffer theorem [37]) Let $\psi : {\mathbb {N}} \rightarrow [0,\infty )$ be a function. Define

$$\begin{aligned} A_{\psi } = \Big \{ \, x \in [0,1] \,: \, \Big | x {-} \frac{p}{q} \Big | {\le } \psi (q) \text { for infinitely many coprime pairs } (p,q) {\in } {\mathbb {N}} {\times } {\mathbb {N}} \, \Big \}. \end{aligned}$$

Let $\varphi$ denote the Euler totient function.¹⁶ Then, we have the following dichotomy:

• (a) If $\sum _{q = 1}^\infty \varphi (q) \psi (q) = \infty$, then $|A_\psi | = 1$.

• (b) If $\sum _{q = 1}^\infty \varphi (q) \psi (q) < \infty$, then $|A_\psi | = 0$.

The relevant part of this theorem is (a), since (b) follows from the canonical limsup covering

$$\begin{aligned} & A_\psi \subset \bigcup _{ q = Q }^\infty \bigcup _{\begin{array}{c} 1 \le p \le q \\ \, (p,q) = 1 \end{array}} \Big ( \frac{p}{q} - \psi (q), \, \frac{p}{q} + \psi (q) \Big ), \quad \forall \, Q \in {\mathbb {N}} \quad \nonumber \\ & \qquad \Longrightarrow \quad |A_\psi | \le \sum _{q = Q}^\infty \varphi (q) \psi (q), \quad \forall \, Q \in {\mathbb {N}}. \end{aligned}$$

(16)

On the other hand, as opposed to the classic theorem by Khinchin¹⁷ [36, Theorem 32], the arbitrariness of $\psi$ allows to restrict the denominators to a set ${\mathcal {Q}} \subset {\mathbb {N}}$ just by setting $\psi (q) = 0$ when $q \not \in {\mathcal {Q}}$. In particular, $A_{\mu ,{\mathcal {Q}}} = A_\psi$ if we define $\psi (q) = \mathbbm {1}_{{\mathcal {Q}}}(q) / q^\mu$, where $\mathbbm {1}_{{\mathcal {Q}}}$ is the indicator function of the set ${\mathcal {Q}}$. Hence, the relevant sum for the sets $A_{\mu ,{\mathcal {Q}}}$ is

$$\begin{aligned} \sum _{q=1}^\infty \varphi (q) \psi (q) = \sum _{q \in {\mathcal {Q}}} \frac{\varphi (q)}{q^\mu }. \end{aligned}$$

In particular, it is fundamental to understand the behavior of the Euler totient function $\varphi$ on ${\mathcal {Q}}$.

The complete proof of the Duffin–Schaeffer theorem was given recently by Koukoulopoulos and Maynard [37, Theorem 1], but Duffin and Schaeffer [21] proved back in 1941 that the result holds under the additional assumption that there exists $c > 0$ such that

$$\begin{aligned} \sum _{q=1}^N \varphi (q) \, \psi (q) \ge c \sum _{q=1}^N q \, \psi (q), \quad \text { for infinitely many } N \in {\mathbb {N}}. \end{aligned}$$

(17)

In the setting of $A_{\mu ,{\mathcal {Q}}}$, this condition is immediately satisfied by sets ${\mathcal {Q}}$ for which there is a $c > 0$ such that $\varphi (q) > c \, q$ for all $q \in {\mathcal {Q}}$. Examples of this are:

• ${\mathcal {Q}} = {\mathbb {P}}$ the set of prime numbers, and

• ${\mathcal {Q}} = \{ \, M^n : \, n \in {\mathbb {N}} \, \}$ where $M \in {\mathbb {N}}$, that is, the set of power of a given number M.

It follows from our computations in Appendix A that the condition (17) is also satisfied by

• ${\mathcal {Q}} = \{ \, Mn : \, n \in {\mathbb {N}} \, \}$ where $M \in {\mathbb {N}}$, that is, the set of multiples of a given number M.

To prove Theorem 1.1 for $x_0 = P/Q$, we restrict the denominators to the latter set with $M = 4Q$; in particular, the 1941 result by Duffin and Schaeffer [21] suffices. However, in the case of $x_0 \not \in {\mathbb {Q}}$ we need to restrict the denominators to an exponentially growing sequence $q_n$ for which we do not know if (17) holds. Hence, in this case we need the full power of the result by Koukoulopoulos and Maynard [37]. This might give an indication of the difficulty to settle the case $x_0 \not \in {\mathbb {Q}}$.

2.3 Hausdorff dimension: the Jarník–Besicovitch theorem and the Mass Transference Principle

We mentioned that $A_2 = [0,1] {\setminus } {\mathbb {Q}}$, and it follows from the argument in (16) that $|A_\mu | = 0$ for $\mu > 2$. But how small is $A_\mu$ is when $\mu > 2$? A measure theoretic answer to that is the Jarník and Besicovitch theorem from the 1930s (see [25, Section 10.3] for a modern version).

Theorem 2.2

(Jarník–Besicovitch theorem) Let $\mu > 2$ and let $A_\mu$ be defined as in (13). Then, ${\text {dim}}_{{\mathcal {H}}} A_\mu = 2/ \mu$ and ${\mathcal {H}}^{2/\mu }(A_\mu ) = \infty$.

In this article we need to adapt this result to $A_{\mu ,{\mathcal {Q}}}$. First, using the Duffin–Schaeffer Theorem 2.1 we will be able to find the largest $\mu _0 \ge 1$ such that $|A_{\mu _0, {\mathcal {Q}}}| = 1$, so that $|A_{\mu , {\mathcal {Q}}}| = 0$ for all $\mu > \mu _0$. To compute the Hausdorff dimension of those zero-measure sets, we will use a theorem by Beresnevich and Velani, called the Mass Transference Principle [5, Theorem 2]. We state here its application to the unit cube and to Hausdorff measures.

Theorem 2.3

(Mass Transference Principle [5]) Let $B_n = B_n(x_n,r_n)$ be a sequence of balls in $[0,1]^d$ such that $\lim _{n \rightarrow \infty } r_n = 0$. Let $\alpha < d$ and let $B_n^\alpha = B_n(x_n,r_n^{\alpha /d})$ be the dilation of $B_n$ centered at $x_n$ by the exponent $\alpha$. Suppose that $X^\alpha := \limsup _{n \rightarrow \infty } B_n^\alpha$ is of full Lebesgue measure, that is, $|X^\alpha | = 1$. Then, calling $X:= \limsup _{n \rightarrow \infty } B_n$, we have ${\text {dim}}_{{\mathcal {H}}} X \ge \alpha$ and ${\mathcal {H}}^\alpha ( X ) = \infty$.

To illustrate the power of the Mass Transference Principle, let us explain how the Jarnik–Besicovitch Theorem 2.2 follows as a simple corollary of the Dirichlet approximation theorem. From the definition of $A_\mu$ we can write¹⁸

$$\begin{aligned} A_\mu = \limsup _{q \rightarrow \infty } \bigcup _{1 \le p \le q, (p,q) = 1} B\Big ( \frac{p}{q}, \, \frac{1}{q^\mu } \Big ). \end{aligned}$$

(18)

Choose $\alpha = 2/\mu$ so that $(A_\mu )^\alpha = A_{\mu \alpha } = A_2$, which by the Dirichlet approximation theorem has full measure. Then, the Mass Transference Principle implies ${\text {dim}}_{{\mathcal {H}}} A_\mu \ge 2/\mu$ and ${\mathcal {H}}^{2/\mu } (A_\mu ) = \infty$. The upper bound follows from the canonical cover of $A_\mu$ in (18), proceeding like in (16).

For $A_{\mu ,{\mathcal {Q}}}$, once we find the largest $\mu _0$ for which $|A_{\mu _0,{\mathcal {Q}}}|=1$ using the Duffin–Schaeffer theorem, we will choose $\alpha = \mu _0 / \mu$ so that the property $(A_{\mu ,{\mathcal {Q}}})^\alpha = A_{\mu \alpha , {\mathcal {Q}} } = A_{\mu _0, {\mathcal {Q}}}$ has full measure, and the Mass Transference Principle will then imply ${\text {dim}}_{{\mathcal {H}}} A_{\mu , {\mathcal {Q}}} \ge \mu _0/\mu$.

3 Preliminary results on the local regularity of $R_{x_0}$

In this section we carry over to $R_{x_0}$ regularity results that are by now classical for $R_0$. In Sect. 3.1 we prove that $R_{x_0}$ is globally $C^{1/2}$. In Sect. 3.2 we compute the asymptotic behavior of $R_{x_0}$ around rationals. In Sect. 3.3 we give a lower bound for $\alpha _{x_0}(t)$ that is independent of $x_0$.

3.1 A global Hölder regularity result

Duistermaat [22, Lemma 4.1.] proved that $R_0$ is globally $C^{1/2}(t)$. The same holds for all $x_0 \in {\mathbb {R}}$. We include the proof for completeness.

Proposition 3.1

Let $x_0 \in {\mathbb {R}}$. Then, $\alpha _{x_0}(t) \ge 1/2$ for all $t \in {\mathbb {R}}$. That is, $R_{x_0}$ is globally $C^{1/2}$.

Proof

For $h \ne 0$, let $N \in {\mathbb {N}}$ such that $\frac{1}{(N+1)^2} \le |h| < \frac{1}{N^2}$, and write

$$\begin{aligned} R_{x_0}(t+h) - R_{x_0}(t)= & \sum _{|n| \le N} \frac{e^{2\pi i n^2 t}\, e^{2\pi i n x_0} }{n^2} \Big ( e^{2\pi i n^2 h} - 1 \Big )\\ & + \sum _{|n| > N} \frac{e^{2\pi i n^2 t}\, e^{2\pi i n x_0} }{n^2} \Big ( e^{2\pi i n^2 h} - 1 \Big ). \end{aligned}$$

Since $|e^{ix} - 1| \le |x|$ for all $x \in {\mathbb {R}}$, we bound

$$\begin{aligned} \Bigg | \sum _{|n| \le N} \frac{e^{2\pi i n^2 t}\, e^{2\pi i n x_0} }{n^2} \Big ( e^{2\pi i n^2 h} - 1 \Big ) \Bigg |\le & \sum _{|n| \le N} \frac{ \big | e^{2\pi i n^2 h} - 1 \big |}{n^2}\\\le & 2|h|N < 2 |h| \frac{1}{\sqrt{|h|}} = 2 \sqrt{|h|}. \end{aligned}$$

For the other sum, we trivially bound $\big | e^{2\pi i n^2 h } - 1 \big | \le 2$ to get

$$\begin{aligned} \Bigg | \sum _{|n| > N} \frac{e^{2\pi i n^2 t}\, e^{2\pi i n x_0} }{n^2} \Big ( e^{2\pi i n^2 h} - 1 \Big ) \Bigg | \le 2\, \sum _{n=N+1}^\infty \frac{2}{n^2} \le \frac{4}{N} \le \frac{8}{N+1} \le 8 \sqrt{|h|}. \end{aligned}$$

Hence $\big | R_{x_0}(t+h) - R_{x_0}(t) \big | \le 10 |h|^{1/2}$. This holds for all t, so $R_{x_0} \in C^{1/2}(t)$ for all $t \in {\mathbb {R}}$. $\square$

3.2 Asymptotic behavior of $R_{x_0}$ around rational t

The building block for all results in this article is the behavior of $R_{x_0}$ around rationals, which we compute explicitly.

Proposition 3.2

Let $x_0 \in {\mathbb {R}}$. Let $p,q \in {\mathbb {N}}$ be such that $(p,q)=1$. Then,

$$\begin{aligned} R_{x_0}\left( \frac{p}{q} + h \right) - R_{x_0}\left( \frac{p}{q} \right)= & - 2\pi i h\\ & + \frac{\sqrt{|h|}}{q} \, \sum _{m \in {\mathbb {Z}}} G(p,m,q) \, F_\pm \left( \frac{x_0 - m/q}{\sqrt{h}} \right) , \quad \text { for } h \ne 0, \end{aligned}$$

where $F_\pm = F_+$ if $h > 0$ and $F_\pm = F_-$ if $h < 0$, and

$$\begin{aligned} G(p,m,q) = \sum _{r=0}^{q-1} e^{2\pi i \frac{pr^2 + mr}{q}}, \quad F_\pm (\xi ) = \int _{{\mathbb {R}}} \frac{e^{\pm 2\pi i x^2} - 1}{x^2} \, e^{2\pi i x \xi } \, dx. \end{aligned}$$

The function $F_\pm$ is bounded and continuous, $F_\pm (0) = 2\pi (-1 \pm i)$, and

$$\begin{aligned} F_\pm (\xi ) = (1 \pm i) \, \frac{e^{\mp \pi i \xi ^2/2}}{\xi ^2} + O\left( \frac{1}{\xi ^4} \right) = O\left( \frac{1}{\xi ^2} \right) , \quad \text { as } \quad \xi \rightarrow \infty . \end{aligned}$$

Proof

We follow the classical approach, which can be traced back to Smith [45], of using the Poisson summation formula. From the definition of $R_{x_0}$, complete first the sum to $n \in {\mathbb {Z}}$ to write

$$\begin{aligned} R_{x_0}\left( \frac{p}{q} + h \right) - R_{x_0}\left( \frac{p}{q} \right) = - 2\pi i h + \sum _{n \in {\mathbb {Z}}} \frac{e^{2\pi i n^2 h} - 1}{n^2} \, e^{2\pi i \frac{pn^2}{q} } \, e^{2\pi i n x_0}, \end{aligned}$$

where we must interpret the term $n=0$ as the value of $\frac{e^{2\pi i n^2 h} - 1}{n^2} \simeq 2\pi i h$ as $n \rightarrow 0$. Split the sum modulo q by writing $n = mq+r$ and

$$\begin{aligned} \sum _{n \in {\mathbb {Z}}} \frac{e^{2\pi i n^2 h} - 1}{n^2} \, e^{2\pi i \frac{pn^2}{q} } \, e^{2\pi i n x_0} = \sum _{r=0}^{q-1} e^{2\pi i \frac{p r^2}{q} } \, \sum _{m \in {\mathbb {Z}}} \frac{e^{2\pi i (mq+r)^2 h} - 1}{(mq+r)^2} \, e^{2\pi i ( mq + r) x_0}.\nonumber \\ \end{aligned}$$

(19)

Use the Poisson summation formula for the function

$$\begin{aligned} f(y) = \frac{e^{2\pi i (yq+r)^2 h} - 1}{(yq+r)^2} \, e^{2\pi i (yq + r) x_0}, \end{aligned}$$

for which, changing variables $(yq+r) \sqrt{|h|} = z$, we have

$$\begin{aligned} {\widehat{f}}(\xi )= & \frac{\sqrt{|h|}}{q} \, e^{2\pi i r \xi /q } \, \int \frac{e^{2\pi i {\text {sgn}}(h) z^2}-1}{z^2} \, e^{2\pi i \frac{z}{\sqrt{|h|}} (x_0 - \xi /q ) }\, dz\\= & \frac{\sqrt{|h|}}{q} \, e^{2\pi i r \xi /q } \, F_\pm \Bigg ( \frac{x_0 - \xi /q}{\sqrt{|h|}} \Bigg ). \end{aligned}$$

Therefore,

$$\begin{aligned} (19)= & \sum _{r=0}^{q-1} e^{2\pi i \frac{p r^2}{q} } \, \sum _{m \in {\mathbb {Z}}} \frac{\sqrt{|h|}}{q} \, e^{2\pi i r m/q } \, F_\pm \Bigg ( \frac{x_0 - m/q}{\sqrt{|h|}} \Bigg ) \\= & \frac{\sqrt{|h|}}{q} \, \sum _{m \in {\mathbb {Z}}} G(p,m,q) \, F_\pm \Bigg ( \frac{x_0 - m/q}{\sqrt{|h|}} \Bigg ). \end{aligned}$$

The properties for $F_\pm$ follow by integration by parts and the value of the Fresnel integral. $\quad \square$

The main term in Proposition 3.2 corresponds to $m \in {\mathbb {Z}}$ such that $x_0 - m/q$ is closest to 0. Define

$$\begin{aligned} \left\{ \begin{array}{l} m_q = {\text {argmin}}_{m \in {\mathbb {Z}}} \big |x_0 - \frac{m}{q} \big |, \\ x_q = x_0 - \frac{m_q}{q}, \end{array} \right. \quad \text { so that } \quad |x_q| = \Big | x_0 - \frac{m_q}{q} \Big | = {\text {dist}} \Bigg ( x_0, \frac{{\mathbb {Z}}}{q} \Bigg ) \le \frac{1}{2q}.\nonumber \\ \end{aligned}$$

(20)

Then, shifting the sum,

$$\begin{aligned} R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big )= & - 2\pi i h + \frac{\sqrt{|h|}}{q} \, G(p,m_q,q) F_\pm \Big ( \frac{x_q}{\sqrt{|h|}} \Big )\\ & + \frac{\sqrt{|h|}}{q} \, \sum _{m \ne 0} G(p,m_q + m, q) \, F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ). \end{aligned}$$

Let us now bound the sum as an error term. As long as $(p,q) = 1$, it is a well-known property of Gauss sums that $|G(p,m,q)| \le \sqrt{2q}$ for all $m \in {\mathbb {N}}$, so

$$\begin{aligned} \frac{\sqrt{|h|}}{q} \Bigg | \sum _{m \ne 0} G(p,m_q + m, q) \, F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg | \le 2\, \frac{\sqrt{|h|}}{\sqrt{q}} \, \sum _{m \ne 0} \Bigg | F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg |. \end{aligned}$$

Since $|x_q| \le 1/(2q)$ and $m \ne 0$, we have $|x_q - m/q| \simeq |m|/q$. This suggests separating two cases:

• If $q \sqrt{|h|} < 1$, we use the property $F_\pm (x) = O(x^{-2})$ to bound

  $$\begin{aligned} \sum _{m \ne 0} \Bigg | F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg | \lesssim \sum _{m \ne 0} \frac{|h|}{|x_q - m/q|^2} \simeq q^2 \, |h|\, \sum _{m \ne 0} \frac{1}{m^2} \simeq q^2 |h|. \end{aligned}$$

• If $q \sqrt{|h|} \ge 1$, we split the sum as

  $$\begin{aligned} \begin{aligned} \sum _{m \ne 0} \Bigg | F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg |&= \sum _{|m| \le q\sqrt{|h|}} \Bigg | F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg |\\&\quad + \sum _{|m| \ge q\sqrt{|h|}} \Bigg | F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg | \\&\le \sum _{|m| \le q\sqrt{|h|}} C + \sum _{|m| \ge q\sqrt{|h|}} \frac{|h|}{|x_q - m/q|^2} \\&\simeq q \sqrt{|h|} + q^2 |h| \sum _{|m| \ge q\sqrt{|h|}} \frac{1}{m^2} \simeq q \, \sqrt{|h|}. \end{aligned} \end{aligned}$$

These two bounds can be written simultaneously as

$$\begin{aligned} \sum _{m \ne 0} \Bigg | F_\pm \Bigg ( \frac{x_q - m/q}{\sqrt{|h|}} \Bigg ) \Bigg | \lesssim \min \big ( q \, \sqrt{|h|}, q^2 |h| \big ), \end{aligned}$$

where the underlying constant is universal. Multiply by $\sqrt{|h|} / \sqrt{q}$ to get the following corollary.

Corollary 3.3

Let $x_0 \in {\mathbb {R}}$. Let $p,q \in {\mathbb {N}}$ be such that $(p,q)=1$. Then,

$$\begin{aligned} & R_{x_0}\Bigg ( \frac{p}{q} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg ) = - 2\pi i h + \frac{\sqrt{|h|}}{q} \, G(p,m_q,q) F_\pm \Bigg ( \frac{x_q}{\sqrt{|h|}} \Bigg ) \\ & \quad + O\left( \min \left( \sqrt{q} \, h, q^{3/2}\, h^{3/2} \right) \right) , \end{aligned}$$

where the underlying constant of the O is independent of p, q and $x_0$.

Remark 3.4

The difference between $x_0 = 0$ and $x_0 \ne 0$ is clear from Corollary 3.3.

• If $x_0 = 0$, we have $x_q = 0 = m_q$ for all q. The main term is $|h|^{1/2} q^{-1} \, G(p,0,q) F_\pm (0)$, so there is a clear dichotomy: $R_0$ is differentiable at p/q if and only if $G(p,0,q) = 0$, which happens if and only if $q \equiv 2 \pmod {4}$; in all other rationals, $R_{x_0}$ is $C^{1/2}$.

• If $x_0 \ne 0$, it is in general false that $x_q = 0$, so to determine the differentiability of $R_{x_0}$ we need to control the magnitude of $F_\pm (x_q/\sqrt{|h|})$.

3.3 Lower bounds for the local Hölder regularity

We now give lower bounds for $\alpha _{x_0}(t)$ that do not depend on $x_0$. In Sect. 3.3.1 we work with $t \in {\mathbb {Q}}$, and in Sect. 3.3.2 with $t \not \in {\mathbb {Q}}$.

3.3.1 At rational points

There is a dichotomy in the Hölder regularity of $R_{x_0}$ at rational points.

Proposition 3.5

Let $x_0 \in {\mathbb {R}}$ and $t \in {\mathbb {Q}}$. Then, either $\alpha _{x_0}(t) = 1/2$ or $\alpha _{x_0}(t) = 3/2$.

Proof

Let $t = p/q$ with $(p,q)=1$. If q is fixed, we get $\min \left( \sqrt{q} \, |h|, q^{3/2}\, |h|^{3/2} \right) = q^{3/2} |h|^{3/2}$ for small enough |h|, so from Corollary 3.3 we get

$$\begin{aligned} R_{x_0}\Bigg ( \frac{p}{q} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg )= & - 2\pi i h + \frac{\sqrt{|h|}}{q} \, G(p,m_q,q) F_\pm \Bigg ( \frac{x_q}{\sqrt{|h|}} \Bigg )\nonumber \\ & + O\Big ( q^{3/2} h^{3/2} \Big ). \end{aligned}$$

(21)

Then, differentiability completely depends on the Gauss sum $G(p,m_q,q)$ and on $x_q$.

Case 1:

If $G(p,m_q,q) =0$, then $\big | R_{x_0}\big ( \frac{p}{q} + h \big ) - R_{x_0}\big ( \frac{p}{q} \big ) + 2\pi i h \big | \lesssim _q h^{3/2}$, so $\alpha _{x_0}(p/q) \ge 3/2$.

Case 2:

If $G(p,m_q,q) \ne 0$ and $x_q \ne 0$. Then, $|G(p,m_q,q)| \simeq \sqrt{q}$ and $\lim _{h \rightarrow 0} x_q/\sqrt{|h|} = \infty$, so $\big | F_\pm \big ( x_q/\sqrt{|h|} \big ) \big | \lesssim h/x_q^2$. Hence, $\alpha _{x_0}(p/q) \ge 3/2$ because

$$\begin{aligned} R_{x_0}\Bigg ( \frac{p}{q} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg )= & - 2\pi i h + O\Bigg ( \frac{\sqrt{h}}{\sqrt{q}} \frac{h}{x_q^2} + q^{3/2}h^{3/2} \Bigg )\\= & - 2\pi i h + O_q\big ( h^{3/2} \big ). \end{aligned}$$

Case 3:

If $G(p,m_q,q) \ne 0$ and $x_q = 0$, we have $|G(p,m_q,q)| \simeq \sqrt{q}$, so from (21) we get

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p}{q} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg ) \Bigg |\ge & \frac{\sqrt{|h|}}{q} |G(p,m_q,q)| |F_\pm (0)| \\ & + O_q( h ) \simeq \frac{\sqrt{h}}{\sqrt{q}} + O_q( h ) \gtrsim _q h^{1/2} \end{aligned}$$

for $h \ll _q 1$. Together with Proposition 3.1, this implies $\alpha _{x_0}(p/q) = 1/2$.

That Cases 1 and 2 actually imply $\alpha _{x_0}(t) = 3/2$ is a bit more technical; we postpone the proof to Proposition B.6 in Appendix B. $\square$

3.3.2 At irrational points

We give a lower bound $\alpha _{x_0}(t)$ that depends on the exponent of irrationality of t, but not on $x_0$.

Proposition 3.6

Let $x_0 \in {\mathbb {R}}$ and $t \in {\mathbb {R}} {\setminus } {\mathbb {Q}}$. Let $\mu (t)$ be the exponent of irrationality of t. Then, $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu (t)}$.

The proof of this result, which we include for completeness, closely follows the procedure by Chamizo and Ubis [15, Proof of Theorem 2.3].

Remark 3.7

Similar to what happens for $x_0 = 0$, where $\alpha _0(t) = 1/2 + 1/2{{\widetilde{\mu }}}(t) \ge 1/2 + 1/2\mu (t)$ (see (7)), we do not expect the bound in Proposition 3.6 to be optimal for all $t \not \in {\mathbb {Q}}$. However, it will be enough to compute the spectrum of singularities.

Proof

In view of Proposition 3.1, there is nothing to prove if $\mu (t) = \infty$, so assume $\mu (t) < \infty$. Let $p_n/q_n$ be the n-th approximation by continued fractions of t. Center the asymptotic behavior in Corollary 3.3 at $p_n/q_n$, and bound it from above by

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) \Bigg | \lesssim \frac{\sqrt{|h|}}{\sqrt{q_n}} + |h| + \min \Big ( \sqrt{q_n} \, h, q_n^{3/2}\, h^{3/2} \Big ), \end{aligned}$$

(22)

where we used that $|G(p_n, m_{q_n}, q_n)| \le \sqrt{2q_n}$ for all $n \in {\mathbb {N}}$ and $|F(x)| \lesssim 1$ for all $x \in {\mathbb {R}}$.

Let $h \ne 0$ be small enough. The sequence $| t - p_n/q_n|$ is strictly decreasing, so choose n such that

$$\begin{aligned} \left| t - \frac{p_n}{q_n} \right| \le |h| < \left| t - \frac{p_{n-1}}{q_{n-1}} \right| . \end{aligned}$$

(23)

Then, from (22), (23) and $|t - p_n/q_n + h| \le 2|h|$, we get

$$\begin{aligned} \begin{aligned}&\left| R_{x_0}\left( t + h \right) - R_{x_0}\left( t \right) \right| \\&\quad \le \left| R_{x_0}\left( \frac{p_n}{q_n} + t - \frac{p_n}{q_n} + h \right) - R_{x_0}\left( \frac{p_n}{q_n} \right) \right| \\&\qquad + \left| R_{x_0}\left( \frac{p_n}{q_n} + t - \frac{p_n}{q_n} \right) - R_{x_0}\left( \frac{p_n}{q_n} \right) \right| \\&\quad \lesssim \frac{\sqrt{|h|}}{\sqrt{q_n}} + |h| + \min \left( \sqrt{q_n} \, |h|, q_n^{3/2}\, |h|^{3/2} \right) . \end{aligned} \end{aligned}$$

(24)

Next we compute the dependence between $q_n$ and h. By the property of continued fractions

$$\begin{aligned} \frac{1}{q_n^{\mu _n}} = \Big | t - \frac{p_n}{q_n} \Big | \le \frac{1}{q_{n+1}q_n}, \end{aligned}$$

we get $1/q_n \le 1/q_{n+1}^{1/(\mu _n - 1)}$ for all $n \in {\mathbb {N}}$. Then, from (23) we get

$$\begin{aligned} \frac{1}{q_n^{\mu _n}} \le |h| < \frac{1}{q_{n-1}^{\mu _{n-1}}} \le \frac{1}{q_n^{\mu _{n-1} / (\mu _{n-1} - 1 ) }}. \end{aligned}$$

(25)

We now bound each term in (24) using (25).

• For the first term, by (25), $\sqrt{|h|}/\sqrt{q_n} \le |h|^{\frac{1}{2} + \frac{1}{2\mu _n}}.$

• The fact that $\mu _n \ge 2$ implies $\frac{1}{2} + \frac{1}{2\mu _n} \le \frac{3}{4}$, so $|h| \le |h|^{3/4} \le |h|^{\frac{1}{2} + \frac{1}{2\mu _n}}$ and the second term is absorbed by the first one.

• For the third term, we write the minimum as

  $$\begin{aligned} \min \Big (\sqrt{q_n} \, |h|, q_n^{3/2} \, |h|^{3/2}\Big ) = \left\{ \begin{array}{ll} \sqrt{q_n} \, |h|, & \text { when } |h| \ge 1/q_n^2, \\ q_n^{3/2} \, |h|^{3/2}& \text { when } |h| \le 1/q_n^2. \end{array} \right. \end{aligned}$$

  So we have two regions:

  • When $|h| \ge 1/q_n^2$, use (25) to bound

    $$\begin{aligned} \sqrt{q_n} \, |h| \le \frac{|h|}{|h|^{(\mu _{n-1} - 1)/2\mu _{n-1}}} = |h|^{\frac{1}{2} + \frac{1}{2\mu _{n-1}}}. \end{aligned}$$

  • When $|h| \le 1/q_n^2$, we directly have $q_n \le |h|^{-1/2}$, so

    $$\begin{aligned} q_n^{3/2} \, |h|^{3/2} = |h|^{3/2 - 3/4} = |h|^{3/4} \le |h|^{\frac{1}{2} + \frac{1}{2\mu _{n-1}}}, \end{aligned}$$

    where in the last inequality we used $\frac{1}{2} + \frac{1}{2\mu _{n-1}} \le \frac{3}{4}$ as before.

Gathering all cases, we get

$$\begin{aligned} | R_{x_0}(t + h) - R_{x_0}(t) | \le |h|^{\frac{1}{2} + \frac{1}{2\mu _n}} + |h|^{ \frac{1}{2} + \frac{1}{2\mu _{n-1}} }. \end{aligned}$$

From the definition of the exponent of irrationality $\mu (t) = \limsup _{n \rightarrow \infty } \mu _n$, for any $\delta > 0$ there exists $N_\delta \in {\mathbb {N}}$ such that $\mu _n \le \mu (t) + \delta$ for all $n \ge N_\delta$. Then, since $|h| < 1$, we have $|h|^{\frac{1}{2} + \frac{1}{2\mu _n}} \le |h|^{\frac{1}{2} + \frac{1}{2\mu (t) + 2\delta }}$ for all $n \ge N_\delta$. Renaming $\delta$, we get $N_\delta \in {\mathbb {N}}$ such that

$$\begin{aligned} | R_{x_0}(t + h) - R_{x_0}(t) | \le |h|^{\frac{1}{2} + \frac{1}{2\mu (t)} - \delta }, \quad \text { for all } \quad |h| \le \Bigg | t - \frac{p_{N_\delta }}{q_{N_\delta }} \Bigg |, \end{aligned}$$

so $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu (t)} - \delta$. Since this holds for all $\delta > 0$, we conclude that $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu (t)}$. $\square$

4 Proof of Theorem 1.1: spectrum of singularities when $x_0 \in {\mathbb {Q}}$

In this section we prove Theorem 1.1. Let us fix $x_0 = P/Q$ such that $(P,Q) = 1$. To compute the spectrum of singularities $d_{x_0}$, we first characterize the rational points t where $R_{x_0}$ is not differentiable, and then we give an upper bound for the regularity $\alpha _{x_0}(t)$ at irrational t.

4.1 At rational points t

In the proof of Proposition 3.5 we established that $R_{x_0}$ is not differentiable at $t=p/q$ if and only if $G(p,m_q, q) \ne 0$ and $x_q = {\text {dist}} (x_0,{\mathbb {Z}} /q ) = 0$. We characterize this in the following proposition.

Proposition 4.1

Let $x_0 = P/Q$ with ${\text {gcd}}(P,Q) = 1$, and let $p,q \in {\mathbb {N}}$ such that ${\text {gcd}}(p,q) = 1$. Then, $R_{x_0}$ is non-differentiable at $t = p/q$ if and only if

• $q = kQ$ with $k \equiv 0,1,3 \pmod {4}$, in the case $Q \equiv 1 \pmod {2}$.

• $q = kQ$ with $k \equiv 0 \pmod {2}$, in the case $Q \equiv 0 \pmod {4}$.

• $q = kQ$ with $k \in {\mathbb {Z}}$, in the case $Q \equiv 2 \pmod {4}$.

In all such cases, the asymptotic behavior is

$$\begin{aligned} R_{x_0}\left( \frac{p}{q} + h \right) - R_{x_0}\left( \frac{p}{q} \right)= & c \, e^{2\pi i \phi _{p,q,x_0}} \, F_\pm (0) \, \frac{\sqrt{|h|}}{\sqrt{q}} - 2\pi i h \nonumber \\ & + O\left( \min \left( \sqrt{q} \, h, q^{3/2}\, h^{3/2} \right) \right) . \end{aligned}$$

(26)

where $c=1$ or $c = \sqrt{2}$ depending on parity conditions of Q and q. In particular, $\alpha _{x_0}(t) = 1/2$.

Proof

In view of the proof of Proposition 3.5, we must identify the conditions for $G(p,m_q,q) \ne 0$ and $x_q = 0$. Since $x_q = {\text {dist}}(P/Q, {\mathbb {Z}} / q)$, we have $x_q = 0$ when there exists $m_q \in {\mathbb {Z}}$ such that

$$\begin{aligned} \frac{P}{Q} = \frac{m_q}{q} \quad \Longleftrightarrow \quad Pq = m_qQ. \end{aligned}$$

Since ${\text {gcd}}(P,Q) = 1$, then necessarily Q|q. Reversely, if $q = kQ$, then picking $m_q = kP$ we have $m_q/q = P/Q$. In short,

$$\begin{aligned} x_q = 0 \quad \Longleftrightarrow \quad q \text { is a multiple of } Q. \end{aligned}$$

So let $q = kQ$ for some $k \in {\mathbb {N}}$. Then, $m_q = kP$. Let us characterize the second condition $G(p,m_q,q) = G(p,kP, kQ) \ne 0$. It is well-known that

$$\begin{aligned} G(a,b,c) \ne 0 \quad \Longleftrightarrow \quad \text { either } \left\{ \begin{array}{ll} c \text { is odd, or } \\ c \text { is even and } \frac{c}{2} \equiv b \pmod {2}. \end{array} \right. \end{aligned}$$

(27)

We separate cases:

• Suppose Q is odd. Then, according to (27), we need either

  • kQ odd, which holds if and only if k is odd, or

  • kQ even, which holds if and only if k is even, and $kQ/2 \equiv kP \pmod {2}$. Since Q is odd and k is even, this is equivalent to $k/2 \equiv 0 \pmod {2}$, which means $k \equiv 0 \pmod {4}$.

  Therefore, if $q = kQ$, the Gauss sum $G(p,m_q,q) \ne 0$ if and only if $k \equiv 0,1,3 \pmod {4}$.

• Suppose $Q \equiv 0 \pmod {4}$. Since $q = kQ$ is even, by (27) we need $kQ/2 \equiv kP \pmod {2}$. Since Q is a multiple of 4, this is equivalent to $kP \equiv 0 \pmod {2}$. But since Q is even, then P must be odd. Therefore, k must be even. In short, if $q = kQ$, we have $G(p,m_q,q) \ne 0$ if and only if k is even.

• Suppose $Q \equiv 2 \pmod {4}$. Since $q = kQ$ is even, by (27) we need $kQ/2 \equiv kP \pmod {2}$. Now both Q/2 and P are odd, so this is equivalent to $k \equiv k \pmod {2}$, which is of course true. Therefore, if $q = kQ$, we have $G(p,m_q,q) \ne 0$ for all $k \in {\mathbb {Z}}$.

Once all cases have been identified, (26) follows from Corollary 3.3 and from the fact that if $G(p,m_q,q) \ne 0$ we have $|G(p,m_q,q)| = c\sqrt{q}$ with $c = 1$ or $c = \sqrt{2}$.

$\square$

4.2 A general upper bound for irrational t

We begin the study of $t \not \in {\mathbb {Q}}$ by giving a general upper bound for $\alpha _{x_0}(t)$ for $t \not \in {\mathbb {Q}}$. The proof uses an alternative asymptotic expression around rationals that we postpone to Appendix B.

Proposition 4.2

Let $x_0 \in {\mathbb {Q}}$ and $t \not \in {\mathbb {Q}}$. Then, $\alpha _{x_0}(t) \le 3/4$.

Proof

See Appendix B, Proposition B.3. $\quad \square$

4.3 Upper bounds depending on the irrationality of t

We now aim at an upper bound for $\alpha _{x_0}(t)$ that depends on the irrationality of t at the level of Proposition 3.6. The idea is to approximate t by rationals p/q where $R_{x_0}$ is non-differentiable, which we characterized in Proposition 4.1. To avoid treating different cases depending on the parity of Q, let us restrict¹⁹$q \in 4Q {\mathbb {N}}$, such that the three conditions in Proposition 4.1 are simultaneously satisfied and (26) holds.

Let $\mu \in [2, \infty )$. Define the classic Diophantine set

$$\begin{aligned} A_\mu = \left\{ \, t \in (0,1) {\setminus } {\mathbb {Q}} \,: \, \big | t - \frac{p}{q} \big | \le \frac{1}{q^\mu } \, \, \text { for i. m. coprime pairs } (p,q) \in {\mathbb {N}} \times {\mathbb {N}} \, \right\} \end{aligned}$$

and for $0< a < 1$ small enough define the restricted Diophantine set

$$\begin{aligned} A_{\mu ,Q} = \left\{ \, t \in (0,1) {\setminus } {\mathbb {Q}} \,: \, \big | t - \frac{p}{q} \big | \le \frac{a}{q^\mu } \, \, \text { for i. m. coprime pairs } (p,q) \in {\mathbb {N}} \times 4Q{\mathbb {N}} \, \right\} . \end{aligned}$$

For $\mu = \infty$ we define $A_\infty = \bigcap _{\mu \ge 2} A_\mu$ and $A_{\infty , Q} = \bigcap _{\mu \ge 2} A_{\mu , Q}$. Clearly, $A_{\mu ,Q} \subset A_\mu$. Our first step is to give an upper bound for $\alpha _{x_0}(t)$ for $t \in A_{\mu ,Q}$.

Proposition 4.3

Let $\mu \ge 2$ and $t \in A_{\mu ,Q}$. Then, $\alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu }$.

Proof

We begin with the case $\mu < \infty$. If $t \in A_{\mu ,Q}$, there is a sequence of irreducible fractions $p_n/q_n$ with $q_n \in 4Q{\mathbb {N}}$, for which we can use (26) and write

$$\begin{aligned} R_{x_0}\left( t \right) - R_{x_0}\Big ( \frac{p_n}{q_n} \Big ) = c \, e^{2\pi i \phi _{n,x_0}} \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}} - 2\pi i h_n + O\left( \min \left( \sqrt{q_n} \, h_n, q_n^{3/2}\, h_n^{3/2} \right) \right) ,\nonumber \\ \end{aligned}$$

(28)

where we absorbed F(0) into c and we defined $h_n$ and $\mu _n$ as

$$\begin{aligned} h_n = t - \frac{p_n}{q_n}, \quad |h_n| = \frac{1}{q_n^{\mu _n}} \le \frac{a}{q_n^\mu } < \frac{1}{q_n^\mu }. \end{aligned}$$

(29)

We now absorb the second and third terms in (28) in the first term. First, $\mu \ge 2$ implies $q_n^2 |h_n| \le 1$, so $\min ( \sqrt{q_n} \, |h_n|, q_n^{3/2}\, |h_n|^{3/2} ) = q_n^{3/2}\, |h_n|^{3/2}.$ Letting C be the universal constant in the O in (28),

$$\begin{aligned} C\, q_n^{3/2} |h_n|^{3/2} \le \frac{c}{4} \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \quad \Longleftrightarrow \quad q_n^2 |h_n| \le \frac{ c}{4C}, \end{aligned}$$

and since $q_n^2 |h_n| \le a q_n^{2 - \mu } \le a$, it suffices to ask $a \le c/(4C)$. Regarding the second term, we have

$$\begin{aligned} 2\pi |h_n| \le \frac{c}{4} \, \frac{ \sqrt{|h_n|}}{\sqrt{q_n} } \quad \Longleftrightarrow \quad q_n \, |h_n| \le \Big ( \frac{c}{8\pi } \Big )^2 \end{aligned}$$

This holds for large n because $q_n^2 |h_n| \le 1$ implies $q_n \, |h_n| \le 1/q_n$, and because $\limsup _{n \rightarrow \infty } q_n = \infty$ (otherwise $q_n$ would be bounded and hence the sequence $p_n/q_n$ would be finite). All together, using the reverse triangle inequality in (28) and the bound for $h_n$ in (29)

$$\begin{aligned} \Bigg | R_{x_0}\left( t \right) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) \Bigg | \ge \frac{c}{2} \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \ge \frac{c}{2} \, |h_n|^{\frac{1}{2} + \frac{1}{2\mu }}, \quad \forall n \gg 1. \end{aligned}$$

This means that $R_{x_0}$ cannot be better than ${\mathcal {C}}^{\frac{1}{2} + \frac{1}{2\mu }}$ at t, thus concluding the proof for $\mu < \infty$.

If $t \in A_{ \infty ,Q}$, by definition $t \in A_{\mu , Q}$ for all $\mu \ge 2$, hence we just proved that $\alpha _{x_0}(t) \le 1/2 + 1/(2\mu )$ for all $\mu \ge 2$. Taking the limit $\mu \rightarrow \infty$ we get $\alpha _{x_0}(t) \le 1/2$. $\square$

To prove Theorem 1.1, we need to compute ${\text {dim}}_{{\mathcal {H}}}\{ \, t : \, \alpha _{x_0}(t) = \alpha \, \}$ with prescribed $\alpha$. For that, we need to complement Proposition 4.3 by proving that for $t \in A_{\mu ,Q}$ we also have $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu }$. By Proposition 3.6, it would suffice to prove that $t \in A_{\mu ,Q}$ has irrationality $\mu (t) = \mu$. Unfortunately, when $\mu < \infty$ this need not be true. To fix this, for $2 \le \mu < \infty$ define the companion sets

$$\begin{aligned} B_\mu= & A_\mu {\setminus } \bigcup _{\epsilon> 0} A_{\mu + \epsilon }\\= & \Bigg \{ \, t \in A_\mu \, \mid \, \forall \epsilon > 0, \, \big | t - \frac{p}{q} \big | \le \frac{1}{q^{\mu + \epsilon }} \, \, \text { only for finitely many } \frac{p}{q} \, \Bigg \}, \end{aligned}$$

and

$$\begin{aligned} B_{\mu ,Q}= & A_{\mu ,Q} {\setminus } \bigcup _{\epsilon> 0} A_{\mu + \epsilon }\nonumber \\= & \Bigg \{ \, t \in A_{\mu ,Q} \, \mid \, \forall \epsilon > 0, \, \big | t - \frac{p}{q} \big | \le \frac{1}{q^{\mu + \epsilon }} \, \, \text { only for finitely many } \frac{p}{q} \, \Bigg \},\nonumber \\ \end{aligned}$$

(30)

which have the properties we need.

Proposition 4.4

Let $2 \le \mu < \infty$. Then,

• (i) $B_{\mu ,Q} \subset B_\mu \subset \{ \, t \in {\mathbb {R}} {\setminus } {\mathbb {Q}} : \, \mu (t) = \mu \, \}$.

• (ii) If $t \in B_{\mu ,Q}$, then $\alpha _{x_0}(t) = \frac{1}{2} + \frac{1}{2\mu }$.

• (iii) If $t \in A_{\infty ,Q}$, then $\alpha _{x_0}(t) = 1/2$.

Proof

(i) First, $B_{\mu ,Q} \subset B_\mu$ because $A_{\mu ,Q} \subset A_\mu$. The second inclusion is a consequence of the definition of the irrationality exponent in (12). Indeed, $t \in B_\mu \subset A_\mu$ directly implies that $\mu (t) \ge \mu$. On the other hand, for all $\epsilon > 0$, $t \in B_\mu$ implies $t \notin A_{\mu + \epsilon }$, so t can be approximated with the exponent $\mu + \epsilon$ only with finitely many fractions, and thus $\mu (t) \le \mu + \epsilon$. Consequently, $\mu (t) \le \mu$.

(ii) By (i), $t \in B_{\mu ,Q}$ implies $\mu (t) = \mu$, so by Proposition 3.6 we get $\alpha _{x_0} (t) \ge \frac{1}{2} + \frac{1}{2\mu }$. At the same time, $t \in B_{\mu ,Q} \subset A_{\mu ,Q}$, so Proposition 4.3 implies $\alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu }$.

(iii) It follows directly from Propositions 3.1 and 4.3. $\square$

Corollary 4.5

Let $2< \mu < \infty$. Then, for all $\epsilon > 0$,

$$\begin{aligned} B_{\mu ,Q} \subset \left\{ \, t \in (0,1) \,: \, \alpha _{x_0}(t) = \frac{1}{2} + \frac{1}{2\mu } \, \right\} \subset A_{\mu - \epsilon }. \end{aligned}$$

For $\mu = 2$ we have the slightly more precise

$$\begin{aligned} B_{2,Q} \subset \{ \, t \in (0,1) \,: \, \alpha _{x_0}(t) = 3/4 \, \} \subset A_2. \end{aligned}$$

For $\mu = \infty$,

$$\begin{aligned} A_{\infty ,Q} \subset \{ \, t \in (0,1) \,: \, \alpha _{x_0}(t) = 1/2 \, \} \subset A_\infty \cup {\mathbb {Q}}. \end{aligned}$$

Proof

Left inclusions follow from Proposition 4.4 for all $\mu \ge 2$, so we only need to prove the right inclusions. When $\mu = 2$, it follows from the Dirichlet approximation theorem, which states that ${\mathbb {R}} {\setminus } {\mathbb {Q}} \subset A_2$, and Proposition 3.5, in which we proved that if t is rational, then either $\alpha _{x_0}(t) = 1/2$ or $\alpha _{x_0}(t) \ge 3/2$. Thus, $\{ \, t \in (0,1) : \, \alpha _{x_0}(t) = 3/4 \, \} \subset (0,1) {\setminus } {\mathbb {Q}} \subset A_2$. Suppose now that $2< \mu < \infty$ and that $\alpha _{x_0}(t) = \frac{1}{2} + \frac{1}{2\mu }$. By Proposition 3.6, $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu (t)}$, so we get $\mu \le \mu (t)$. In particular, given any $\epsilon > 0$, we have $\mu - \epsilon < \mu (t)$, so $\big | t - \frac{p}{q} \big | \le 1/q^{\mu - \epsilon }$ for infinitely many coprime pairs $(p,q) \in {\mathbb {N}} \times {\mathbb {N}}$, which means that $t \in A_{\mu - \epsilon }$. Finally, for $\mu = \infty$, if $t \not \in {\mathbb {Q}}$ is such that $\alpha _{x_0}(t) = 1/2$, then by Proposition 3.6 we get $\mu (t) = \infty$, which implies that $t \in A_\mu$ for all $\mu \ge 2$, hence $t \in A_\infty$. $\quad \square$

Now, to prove Theorem 1.1 it suffices to compute ${\text {dim}}_{{\mathcal {H}}} A_\mu$ and ${\text {dim}}_{{\mathcal {H}}} B_{\mu ,Q}$.

Proposition 4.6

For $2 \le \mu < \infty$, ${\text {dim}}_{{\mathcal {H}}} A_\mu = {\text {dim}}_{{\mathcal {H}}} B_{\mu ,Q} = 2/\mu$. Also, ${\text {dim}}_{{\mathcal {H}}} A_\infty = 0$.

Form this result, whose proof we postpone, we can prove Theorem 1.1 as a corollary.

Theorem 4.7

Let $x_0 \in {\mathbb {Q}}$. Then, the spectrum of singularities of $R_{x_0}$ is

$$\begin{aligned} d_{x_0}(\alpha ) = \left\{ \begin{array}{ll} 4 \alpha - 2, & 1/2 \le \alpha \le 3/4, \\ 0, & \alpha = 3/2, \\ -\infty , & \text {otherwise.} \end{array} \right. \end{aligned}$$

Proof

Proposition 3.1 implies $d(\alpha ) = -\infty$ when $\alpha < 1/2$, while Propositions 3.5 and 4.2 imply that $d_{x_0}(3/2) = 0$ and $d_{x_0}(\alpha ) = -\infty$ if $\alpha > 3/4$ and $\alpha \ne 3/2$. When $1/2 \le \alpha \le 3/4$, it follows from Corollary 4.5, Proposition 4.6 and the periodicity of $R_{x_0}$. First, $d_{x_0}(1/2) \le {\text {dim}}_{{\mathcal {H}}} ( A_\infty \cup {\mathbb {Q}} ) = 0$ because ${\text {dim}}_{{\mathcal {H}}} {\mathbb {Q}} = {\text {dim}}_{{\mathcal {H}}} A_\infty = 0$. On the other hand, for $2 \le \mu < \infty$ we get

$$\begin{aligned} \frac{2}{\mu } \le d_{x_0} \left( \frac{1}{2} + \frac{1}{2\mu } \right) \le \frac{2}{\mu - \epsilon }, \quad \forall \epsilon > 0 \quad \Longrightarrow \quad d_{x_0}\left( \frac{1}{2} + \frac{1}{2\mu } \right) = \frac{2}{\mu }. \end{aligned}$$

which gives the result for $1/2 < \alpha \le 3/4$ by renaming $\alpha = 1/2 + 1/(2\mu )$. $\square$

Let us now prove Proposition 4.6.

Proof of Proposition4.6

We have $A_2 = (0,1) {\setminus } {\mathbb {Q}}$ by Dirichlet approximation, so ${\text {dim}}_{{\mathcal {H}}} A_2 = 1$. For $\mu > 2$ we have ${\text {dim}}_{{\mathcal {H}}} A_\mu = 2/\mu$ by the Jarnik–Besicovitch Theorem 2.2. Also, $A_\infty \subset A_\mu$ for all $\mu \ge 2$, so ${\text {dim}}_{{\mathcal {H}}} A_\infty \le 2/\mu$ for all $\mu \ge 2$, hence ${\text {dim}}_{{\mathcal {H}}} A_\infty = 0$. So we only need to prove that ${\text {dim}}_{{\mathcal {H}}} B_{\mu ,Q} = 2/\mu$ for $2 \le \mu < \infty$. Moreover,

$$\begin{aligned} B_{\mu ,Q} = A_{\mu ,Q} {\setminus } \bigcup _{\epsilon > 0} A_{\mu + \epsilon } \subset A_{\mu ,Q} \subset A_\mu , \end{aligned}$$

which implies $\dim _{{\mathcal {H}}} B_{\mu ,Q} \le \dim _{{\mathcal {H}}} A_\mu = 2/\mu$. Hence it suffices to prove that ${\text {dim}}_{{\mathcal {H}}} B_{\mu ,Q} \ge 2/\mu$. This claim follows from ${\mathcal {H}}^{2/\mu }( A_{\mu ,Q} ) > 0$. Indeed, we first remark that the sets $A_\mu$ are nested, in the sense that $A_{\sigma } \subset A_\mu$ when $\sigma > \mu$. We can therefore write

$$\begin{aligned} \bigcup _{\epsilon > 0} A_{\mu + \epsilon } = \bigcup _{n \in {\mathbb {N}}} A_{\mu + \frac{1}{n}}. \end{aligned}$$

By the Jarnik–Besicovitch Theorem 2.2, $\dim _{{\mathcal {H}}} A_{\mu + 1/n} = 2/(\mu + 1/n) < 2/\mu$, so ${\mathcal {H}}^{2/\mu }(A_{\mu + 1/n}) = 0$ for all $n \in {\mathbb {N}}$, hence

$$\begin{aligned} {\mathcal {H}}^{2/\mu } \Bigg ( \bigcup _{\epsilon > 0} A_{\mu + \epsilon } \Bigg ) = {\mathcal {H}}^{2/\mu } \Bigg ( \bigcup _{n \in {\mathbb {N}}} A_{\mu + \frac{1}{n}} \Bigg ) = \lim _{n \rightarrow \infty } {\mathcal {H}}^{2/\mu } \Big ( A_{\mu + \frac{1}{n}} \Big ) = 0. \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathcal {H}}^{2/\mu } \big ( B_{\mu ,Q} \big )= & {\mathcal {H}}^{2/\mu } \Bigg ( A_{\mu ,Q} {\setminus } \bigcup _{\epsilon> 0} A_{\mu + \epsilon } \Bigg ) = {\mathcal {H}}^{2/\mu } ( A_{\mu ,Q} ) - {\mathcal {H}}^{2/\mu } \Bigg ( \bigcup _{\epsilon > 0} A_{\mu + \epsilon } \Bigg )\\= & {\mathcal {H}}^{2/\mu } \left( A_{\mu ,Q} \right) , \end{aligned}$$

so ${\mathcal {H}}^{2/\mu }( A_{\mu ,Q}) > 0$ implies ${\mathcal {H}}^{2/\mu }( B_{\mu ,Q}) > 0$, hence $\dim _{{\mathcal {H}}} B_{\mu ,Q} \ge 2/\mu$.

Let us thus prove ${\mathcal {H}}^{2/\mu }( A_{\mu ,Q}) > 0$, for which we follow the procedure outlined in Sect. 2 with the set of denominators ${\mathcal {Q}} = 4Q{\mathbb {N}}$. We first detect the largest $\mu$ such that $A_{\mu ,Q}$ has full Lebesgue measure using the Duffin–Schaeffer Theorem 2.1. Define

$$\begin{aligned} \psi _{\mu ,Q}(q) = a\, \frac{ \mathbbm {1}_{ 4Q {\mathbb {N}}}(q)}{q^{\mu }}, \end{aligned}$$

where $a> 0$ comes from the definition of $A_{\mu ,Q}$ and $\mathbbm {1}_{ 4Q {\mathbb {N}}}(q)$ is the indicator function of $4Q{\mathbb {N}}$,

$$\begin{aligned} \mathbbm {1}_{ 4Q {\mathbb {N}}}(q) = \left\{ \begin{array}{ll} 1, & \text { if } 4 Q \, \mid \, q, \\ 0, & \text { otherwise.} \end{array} \right. \end{aligned}$$

Then, we have $A_{\mu ,Q} = A_{\psi _{\mu ,Q}}$, where

$$\begin{aligned} A_{\psi _{\mu ,Q}} = \Big \{ \, t \in [0,1] \,: \, \Big | t - \frac{p}{q} \Big | \le \psi _{\mu ,Q}(q) \, \, \text { for i. m. coprime pairs } (p,q) \in {\mathbb {N}} \times {\mathbb {N}} \, \Big \} \end{aligned}$$

has the form needed for the Duffin–Schaeffer Theorem 2.1. Indeed, the inclusion $\subset$ follows directly from the definition of $\psi _{\mu ,Q}$. For the inclusion $\supset$, observe first that if $t \in A_{\psi _{\mu ,Q}}$ with $\mu > 1$, then $t \not \in {\mathbb {Q}}$. Now, if a coprime pair $(p,q) \in {\mathbb {N}}^2$ satisfies $|t - p/q| \le \psi _{\mu ,Q}(q)$, then $q \in 4Q{\mathbb {N}}$ because otherwise we get the contradiction

$$\begin{aligned} 0 < \Big | t - \frac{p}{q} \Big | \le \psi _{\mu ,Q}(q) = a \, \frac{\mathbbm {1}_{4Q{\mathbb {N}}}(q)}{q^\mu } = 0. \end{aligned}$$

In this setting, the Duffin–Schaeffer theorem says that $A_{\mu ,Q}$ has Lebesgue measure 1 if and only if

$$\begin{aligned} \sum _{q=1}^\infty \varphi (q) \, \psi _{\mu ,Q}(q) = \frac{a}{(4Q)^\mu } \, \sum _{n=1}^\infty \frac{\varphi (4Qn)}{ n^\mu } = \infty , \end{aligned}$$

and has zero measure otherwise. Using this characterization, we prove now

$$\begin{aligned} | A_{\mu ,Q} | = \left\{ \begin{array}{ll} 1, & \mu \le 2, \\ 0, & \mu > 2, \end{array} \right. \end{aligned}$$

(31)

independently of a. To detect the critical $\mu =2$, trivially bound $\varphi (n) < n$ so that

$$\begin{aligned} \sum _{n=1}^\infty \frac{\varphi (4Qn)}{ n^\mu }< \sum _{n=1}^\infty \frac{4Qn}{ n^\mu } = 4Q\, \sum _{n=1}^\infty \frac{1}{ n^{\mu -1} } < \infty , \quad \text { if } \, \, \mu > 2. \end{aligned}$$

However, this argument fails when $\mu = 2$. What is more, denote by ${\mathbb {P}}$ the set of primes so that

$$\begin{aligned} \sum _{n=1}^\infty \, \frac{\varphi (4Qn)}{n^2}> \sum _{p \in {\mathbb {P}}, \, p > 4Q} \, \frac{\varphi (4Qp)}{p^2} \end{aligned}$$

If $p \in {\mathbb {P}}$ and $p > 4Q$, then $\gcd (p,4Q) = 1$ because $p \not \mid 4Q$ (for if $p \mid 4Q$ then $p \le 4Q$). Therefore, $\varphi (4Qp) = \varphi (4Q) \, \varphi (p) = \varphi (4Q) \, (p-1) > \varphi (4Q) \, p/2$, so

$$\begin{aligned} \sum _{n=1}^\infty \, \frac{\varphi (4Qn)}{n^2}> \frac{\varphi (4Q)}{2} \, \sum _{p \in {\mathbb {P}}, \, p > 4Q} \, \frac{1}{p} = \infty , \end{aligned}$$

because the sum of the reciprocals of the prime numbers diverges.²⁰ The Duffin–Schaeffer Theorem 2.1 thus implies that $| A_{2,Q}| = 1$ and, in particular, ${\text {dim}}_{{\mathcal {H}}} A_{2,Q} = 1$. From this we immediately get $| A_{\mu ,Q}| = 1$ when $\mu < 2$ because $A_{2,Q} \subset A_{\mu ,Q}$.

Once we know (31), we use the Mass Transference Principle Theorem 2.3 to compute the dimension of $A_{\mu ,Q}$ for $\mu > 2$. Write first

$$\begin{aligned} A_{\mu ,Q} = \limsup _{q \rightarrow \infty } \bigcup _{p \le q, \, (p,q) = 1} B \Bigg ( \, \frac{p}{q}, \psi _{\mu ,Q}(q) \Bigg ). \end{aligned}$$

Let $\beta = 2/\mu$ so that

$$\begin{aligned} \psi _{\mu ,Q}(q)^\beta = \Bigg (a \, \frac{\mathbbm {1}_{4Q{\mathbb {N}}}(q)}{q^{\mu }} \Bigg )^\beta = a^{\beta } \, \frac{\mathbbm {1}_{4Q{\mathbb {N}}}(q)}{q^{\mu \beta }} = a^{2/\mu } \, \frac{\mathbbm {1}_{4Q{\mathbb {N}}}(q)}{q^2} = \psi _{2,Q}(q), \end{aligned}$$

with a new underlying constant $a^{2/\mu }$. Therefore,

$$\begin{aligned} ( A_{\mu ,Q})^\beta:= & \limsup _{q \rightarrow \infty } \bigcup _{p \le q, \, (p,q) = 1} B \Bigg ( \, \frac{p}{q}, \psi _{\mu ,Q}(q)^\beta \Bigg )\\= & \limsup _{q \rightarrow \infty } \bigcup _{p \le q, \, (p,q) = 1} B \Bigg ( \, \frac{p}{q}, \psi _{2,Q}(q) \Bigg ) = A_{2,Q}. \end{aligned}$$

Observe that $\beta$ is chosen to be the largest possible exponent that gives $| ( A_{\mu ,Q})^\beta | = | (A_{\mu \beta ,Q})| = 1$. Since (31) is independent of a, we get $| (A_{\mu ,Q})^{2/\mu } | = | A_{2,Q} | = 1$, and the Mass Transference Principle Theorem 2.3 implies that ${\mathcal {H}}^{2/\mu } \big ( A_{\mu ,Q} \big ) = \infty$. The proof is complete. $\square$

5 Proof of Theorem 1.3: spectrum of singularities when $x_0 \not \in {\mathbb {Q}}$

In this section we work with $x_0 \not \in {\mathbb {Q}}$ and prove Theorem 1.3. Following the strategy for $x_0 \in {\mathbb {Q}}$, we first study the Hölder regularity at rational t in Sect. 5.1, and at irrational t in Sect. 5.2

5.1 Regularity at rational ${\varvec{t}}$

Let $t = p/q$ an irreducible fraction. With Corollary 3.3 in mind, we now have $x_q = {\text {dist}} ( x_0, {\mathbb {Z}}/q ) \ne 0$. Since q is fixed, $\lim _{h \rightarrow 0} x_q / |h|^{1/2} = \infty$, so $F_\pm (x) = O(x^{-2})$ implies $F_\pm ( x_q/\sqrt{|h|} ) \lesssim |h|/x_q^2$ when $h \rightarrow 0$. Also $|G(p,m_q,q)| \le \sqrt{2q}$ for all $m_q$. Hence,

$$\begin{aligned} \Bigg | \, R_{x_0}\Bigg ( \frac{p}{q} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg ) + 2\pi i h \, \Bigg | \lesssim \left( \frac{1}{\sqrt{q} \, x_q^2} + q^{3/2} \right) \, h^{3/2}. \end{aligned}$$

This regularity is actually the best we can get.

Proposition 5.1

Let $x_0 \in {\mathbb {R}} {\setminus } {\mathbb {Q}}$ and let $t \in {\mathbb {Q}}$. Then, $\alpha _{x_0}(t) = 3/2$.

We postpone the proof of $\alpha _{x_0}(t) \le 3/2$ to Proposition B.6. In any case, this means that when $x_0 \notin {\mathbb {Q}}$, $R_{x_0}$ is more regular at rational points than when $x_0 \in {\mathbb {Q}}$.

5.2 Regularity at irrational ${\varvec{t}}$

Let now $t \notin {\mathbb {Q}}$. Again, we aim at an upper bound for $\alpha _{x_0}(t)$ that complements the lower bound in Proposition 3.6. by approximating $t \not \in {\mathbb {Q}}$ by rationals $p_n/q_n$ and using the asymptotic behavior in Corollary 3.3. However, now $x_0 \not \in {\mathbb {Q}}$ implies $x_{q_n} \ne 0$, so we cannot directly assume $F_\pm (x_{q_n}/\sqrt{|h_{q_n}|}) \simeq F_\pm (0) \simeq 1$ anymore. Therefore, it is fundamental to understand the behavior of the quotient $x_{q_n}/\sqrt{|h_{q_n}|}$.

5.2.1 Heuristics

Let $q \in {\mathbb {N}}$ and define the exponents $\mu _q$ and $\sigma _q$ as usual,

$$\begin{aligned} x_q = {\text {dist}} \Bigg ( x_0, \frac{{\mathbb {Z}}}{ q} \Bigg ) = \frac{1}{q^{\sigma _q}}, \quad |h_q| = {\text {dist}} \Bigg ( t, \frac{{\mathbb {Z}}}{ q} \Bigg ) = \frac{1}{q^{\mu _q}}, \quad \Longrightarrow \quad \frac{x_q}{\sqrt{|h_q| }} = \frac{1}{q^{\sigma _q - \mu _q / 2}}. \end{aligned}$$

If $\sigma _q - \mu _q/2> c > 0$ holds for a sequence $q_n$, we should recover the behavior when $x_0 \in {\mathbb {Q}}$ because

$$\begin{aligned} \lim _{n \rightarrow \infty } \Bigg ( \sigma _{q_n} - \frac{\mu _{q_n} }{2} \Bigg ) \ge c > 0 & \quad \Longrightarrow \quad \lim _{n \rightarrow \infty } \frac{x_{q_n}}{\sqrt{|h_{q_n}|}} = 0\nonumber \\ & \quad \Longrightarrow \quad F_\pm \Bigg ( \frac{x_{q_n}}{\sqrt{|h_{q_n}|}} \Bigg ) \simeq F_\pm (0), \quad n \gg 1. \qquad \end{aligned}$$

(32)

The main term in the asymptotic behavior for $R_{x_0} (t) - R_{x_0}(p_n/q_n)$ in Corollary 3.3 would then be

$$\begin{aligned} \text { Main Term } = \frac{\sqrt{|h_{q_n}|}}{q_n} G(p_n,m_{q_n},q_n) F_\pm (0) \simeq \frac{\sqrt{|h_{q_n}|}}{\sqrt{q_n} } \simeq h_{q_n}^{\frac{1}{2} + \frac{1}{2\mu _{q_n}}} \end{aligned}$$

if we assume the necessary parity conditions so that $|G(p_n,m_{q_n},q_n)| \simeq \sqrt{q_n}$. Recalling the definition of the exponent of irrationality $\mu (\cdot )$ in (12), we may think of $\sigma _{q_n} \rightarrow \mu (x_0)$ and $\mu _{q_n} \rightarrow \mu (t)$, so these heuristic computations suggest that $\alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu (t)}$ for t such that $\mu (t) \le 2\mu (x_0)$. Since Proposition 3.6 gives $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu (t)}$, we may expect that

$$\begin{aligned} \alpha _{x_0}(t) = \frac{1}{2} + \frac{1}{2\mu (t)}, \quad \text { if } \quad 2 \le \mu (t) \le 2\mu (x_0), \end{aligned}$$

(33)

or at least for a big subset of such t. It is less clear what to expect when $\mu (t) > 2\mu (x_0)$, since (32) need not hold. Actually, if $\sigma _{q_n} - \mu _{q_n}/2< c < 0$ for all sequences, then since $F_\pm (x) = x^{-2} + O(x^{-4})$,

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{x_{q_n}}{\sqrt{|h_{q_n}|}}= & \lim _{n \rightarrow \infty } q_n^{\mu _{q_n}/2 - \sigma _{q_n}} = \infty \quad \Longrightarrow \quad F_\pm \Bigg ( \frac{x_{q_n}}{\sqrt{|h_{q_n}|}} \Bigg ) \simeq \frac{1}{q_n^{\mu _{q_n} - 2\sigma _{q_n}} }\nonumber \\= & |h_{q_n}|^{ 1 - \frac{2\sigma _{q_n}}{\mu _{q_n}} }, \end{aligned}$$

which in turn would make the main term in $R_{x_0} (t) - R_{x_0}(p_n/q_n)$ be

$$\begin{aligned} \text { Main Term } = \frac{\sqrt{h_{q_n}}}{q_n} G(p_n,m_{q_n},q_n) F_\pm \Bigg ( \frac{x_{q_n}}{\sqrt{|h_{q_n}|}} \Bigg ) \simeq h_{q_n}^{ \frac{1}{2} + \frac{1}{2\mu _{q_n} } } \, h_{q_n}^{ 1 - \frac{2\sigma _{q_n}}{\mu _{q_n}} } \simeq h_{q_n}^{\frac{3}{2} - \frac{4\sigma _{q_n} - 1}{2\mu _{q_n}}}, \end{aligned}$$

which corresponds to an exponent $\frac{3}{2} - \frac{4\mu (x_0) - 1}{2\mu (t)}$. Together with lower bound in Proposition 3.6, we would get $\frac{1}{2} + \frac{1}{2\mu (t)} \le \alpha _{x_0}(t) \le \frac{3}{2} - \frac{4\mu (x_0) - 1}{2\mu (t)}$, which leaves an open interval for $\alpha _{x_0}(t)$.

The main difficulty to materialize the ideas leading to (33) is that we need the sequence $q_n$ to generate good approximations of both $x_0$ and t simultaneously, which a priori may be not possible. In the following lines we show how we can partially dodge this problem to prove Theorem 1.3.

5.2.2 Proof of Theorem 1.3

Let $\sigma \ge 2$. Recalling the definition of the sets $A_{\mu , {\mathcal {Q}}}$ in (15), define

$$\begin{aligned} & A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} = \left\{ \, x \in [0,1] \,: \Bigg | x - \frac{b}{q} \Bigg | < \frac{1}{ q^\sigma } \right. \\ & \qquad \qquad \qquad \quad \left. \text { for infinitely many coprime pairs } (b,q) \in {\mathbb {N}} \times ({\mathbb {N}} {\setminus } 4{\mathbb {N}}) \, \right\} . \end{aligned}$$

We first prove that the restriction in the denominators²¹ does not affect the Hausdorff dimension.

Proposition 5.2

Let $\sigma \ge 2$. Then, ${\text {dim}}_{{\mathcal {H}}} A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} = 2/\sigma$. Moreover, $A_{2, \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} = (0,1) {\setminus } {\mathbb {Q}}$, hence $|A_{2, \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} | = 1$. If $\sigma > 2$, then ${\mathcal {H}}^{2/\sigma }(A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} ) = \infty$.

Proof

The proof for the upper bound for the Hausdorff dimension is standard. Writing

$$\begin{aligned} A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}= & \limsup _{ q \rightarrow \infty \, \, (q \not \in 4{\mathbb {N}}) } \bigcup _{ 1 \le b< q, \, (b.q)=1 } B\Bigg ( \frac{b}{q}, \frac{1}{q^\sigma } \Bigg )\\= & \bigcap _{Q = 1}^\infty \bigcup _{ q \ge Q, \, q \not \in 4{\mathbb {N}} } \Bigg ( \bigcup _{ 1 \le b < q, \, (b.q)=1 } B\Big ( \frac{b}{q}, \frac{1}{q^\sigma } \Big ) \Bigg ), \end{aligned}$$

we get an upper bound for the Hausdorff measures using the canonical cover

$$\begin{aligned} & A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} \subset \bigcup _{ q \ge Q, \, q \not \in 4{\mathbb {N}} } \Bigg ( \bigcup _{ 1 \le b < q } B\Big ( \frac{b}{q}, \frac{1}{q^\sigma } \Big ) \Bigg ), \quad \forall Q \in {\mathbb {N}} \quad \nonumber \\ & \quad \Longrightarrow \quad {\mathcal {H}} ^\beta (A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}) \le \lim _{Q \rightarrow \infty } \sum _{ q \ge Q } \frac{1}{q^{\sigma \beta - 1}}. \end{aligned}$$

(34)

Thus, ${\mathcal {H}}^\beta (A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}) = 0$ when $\sigma \beta - 1 > 1$, and consequently ${\text {dim}}_{{\mathcal {H}}} A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} \le 2/\sigma$.

For the lower bound we follow the procedure discussed in Sect. 2, though unlike in the proof of Proposition 4.6 we do not need the Duffin–Schaeffer theorem here. We first study the Lebesgue measure of $A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$. From (34) with $\beta = 1$, we directly get $|A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}|=0$ when $\sigma > 2$. When $\sigma = 2$, we get $A_{2, \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} = A_2 = (0,1) {\setminus } {\mathbb {Q}}$. Indeed, if $b_n / q_n$ is the sequence of approximations by continued fractions of $x \in (0,1) {\setminus } {\mathbb {Q}}$, two consecutive denominators $q_n$ and $q_{n+1}$ are never both even.²² This means that there is a subsequence $b_{n_k}/q_{n_k}$ such that $|x - b_{n_k}/q_{n_k}| < 1/q_{n_k}^2$ and $q_{n_k}$ is odd for all $k\in {\mathbb {N}}$. In particular, $q_{n_k} \not \in 4{\mathbb {N}}$, so $(0,1) {\setminus } {\mathbb {Q}} \subset A_{2, \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$. Hence,

$$\begin{aligned} |A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}| = \left\{ \begin{array}{ll} 1, & \sigma \le 2, \\ 0, & \sigma > 2, \end{array} \right. \end{aligned}$$

(35)

With this in hand, we use the Mass Transference Principle Theorem 2.3. For $\beta > 0$,

$$\begin{aligned} (A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}})^\beta= & \limsup _{ \begin{array}{c} q \rightarrow \infty \\ q \not \in 4{\mathbb {N}} \end{array} } \bigcup _{ 1 \le b< q, \, (b,q) = 1 } B\Bigg ( \frac{b}{q}, \Bigg ( \frac{1}{q^\sigma } \Bigg )^\beta \Bigg )\\ \quad= & \limsup _{ \begin{array}{c} q \rightarrow \infty \\ q \not \in 4{\mathbb {N}} \end{array} } \bigcup _{ 1 \le b < q, \, (b,q) = 1 } B\Bigg ( \frac{b}{q}, \frac{1}{q^{\sigma \beta }} \Bigg ) = A_{\sigma \beta , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}. \end{aligned}$$

Thus, choosing $\beta = 2/\sigma$ we get $(A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}})^{2/\sigma } = A_{2, \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$, hence by (35) we get $|(A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}})^{2/\sigma } | = 1$. The Mass Transference Principle implies ${\text {dim}}_{{\mathcal {H}}} A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} \ge 2/\sigma$ and ${\mathcal {H}}^{2/\sigma } (A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}) = \infty$. $\square$

Let $x_0 \in A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$. Then there exists a sequence of pairs $(b_n, q_n) \in {\mathbb {N}} \times ({\mathbb {N}} {\setminus } 4{\mathbb {N}})$ such that $|x_0 - b_n/q_n| < 1/q_n^\sigma$ and moreover $b_n/q_n$ are all approximations by continued fractions. Define

$$\begin{aligned} {\mathcal {Q}}_{x_0} = \{ \, q_n \,: \, n \in {\mathbb {N}} \, \} \end{aligned}$$

to be the set of such denominators. This sequence exists because:

• if $\sigma = 2$, there is a subsequence of continued fraction approximations with odd denominator, in particular with $q_n \not \in 4{\mathbb {N}}$.

• if $\sigma > 2$, by definition there exist a sequence of pairs $(b_n, q_n) \in {\mathbb {N}} \times ({\mathbb {N}} {\setminus } 4{\mathbb {N}})$ such that

  $$\begin{aligned} \Bigg | x_0 - \frac{b_n}{q_n} \Bigg | < \frac{1}{q_n^\mu } \le \frac{1}{2q_n^2}, \quad \text { for large enough } n \in {\mathbb {N}}. \end{aligned}$$

  By a theorem of Khinchin [36, Theorem 19], all such $b_n/q_n$ are continued fraction approximations of $x_0$.

Since all such $q_n$ are the denominators of continued fraction approximations, the sequence $q_n$ grows exponentially.²³ Following again the notation in (15) in Sect. 2, for $\mu \ge 1$ and $0< c < 1/2$, let²⁴

$$\begin{aligned} A_{\mu ,{\mathcal {Q}}_{x_0}}= & \left\{ \, t \in [0,1] \,: \Bigg | t - \frac{p}{q} \Bigg | < \frac{c}{q^\mu } \text { for infinitely many coprime pairs } (p,q) \in {\mathbb {N}} \times {\mathcal {Q}}_{x_0} \, \right\} . \end{aligned}$$

Proposition 5.3

For $\mu \ge 1$, ${\text {dim}}_{{\mathcal {H}}} (A_{\mu ,{\mathcal {Q}}_{x_0}} ) =1/\mu$.

Proof

As in the proof of Proposition 5.2, the upper bound follows from the limsup expression $A_{\mu ,{\mathcal {Q}}_{x_0}} = \limsup _{n \rightarrow \infty } \bigcup _{ 1 \le p \le q_n, \, (p,q_n) = 1 } B( p/q_n, c/q_n^\mu )$ and its canonical covering

$$\begin{aligned} A_{\mu ,{\mathcal {Q}}_{x_0}} \subset \bigcup _{n \ge N} \bigcup _{1 \le p \le q_n} B\Bigg ( \frac{p}{q_n}, \, \frac{c}{q_n^\mu } \Bigg ), \quad \forall N \in {\mathbb {N}} \;\Longrightarrow \; {\mathcal {H}}^\beta \big ( A_{\mu ,{\mathcal {Q}}_{x_0}} \big ) \le c^\beta \lim _{N \rightarrow \infty } \sum _{n = N}^\infty \frac{1}{q_n^{\mu \beta - 1}}.\qquad \end{aligned}$$

(36)

Since $q_n \ge 2^{n/2}$, the series converges if and only if $\mu \beta - 1 > 0$. Thus, ${\mathcal {H}}^\beta (A_{ \mu ,{\mathcal {Q}}_{x_0}}) = 0$ for all $\beta > 1/\mu$, hence ${\text {dim}}_{{\mathcal {H}}} (A_{\mu ,{\mathcal {Q}}_{x_0}}) \le 1/\mu$.

For the lower bound we follow again the procedure in Sect. 2. First we compute the Lebesgue measure of $A_{\mu ,{\mathcal {Q}}_{x_0}}$. From (36) with $\beta = 1$ we get $|A_{\mu ,{\mathcal {Q}}_{x_0}}| = 0$ if $\mu > 1$. When $\mu \le 1$, we need the full strength of the Duffin–Schaeffer theorem proved by Koukoulopoulos and Maynard [37] (see Theorem 2.1 in this paper). Indeed, we have $|A_{\mu ,{\mathcal {Q}}_{x_0}}| = 1$ if and only if $\sum _{n=1}^\infty \varphi (q_n)/q_n^\mu = \infty$, and otherwise $|A_{\mu ,{\mathcal {Q}}_{x_0}}| = 0$. If $\mu < 1$, we use one of the classic properties of Euler’s totient function, namely that for $\epsilon = (1-\mu )/2 > 0$ there exists $N \in {\mathbb {N}}$ such that $\varphi (n) \ge n^{1-\epsilon }$ for all $n \ge N$. In particular, there exists $K \in {\mathbb {N}}$ such that

$$\begin{aligned} \sum _{n=1}^\infty \frac{\varphi (q_n)}{q_n^\mu } \ge \sum _{n=K}^\infty \frac{\varphi (q_n)}{q_n^\mu } \ge \sum _{n=K}^\infty q_n^{1- \mu - \epsilon } \ge \sum _{n=K}^\infty 1 = \infty , \end{aligned}$$

so $|A_{\mu ,{\mathcal {Q}}_{x_0}}| = 1$ if $\mu < 1$. For $\mu =1$, none of these arguments work, and we need to know the behavior of $\varphi (q_n)$ for $q_n \in {\mathcal {Q}}_{x_0}$, of which we have little control. So independently of $c > 0$,

$$\begin{aligned} |A_{\mu ,{\mathcal {Q}}_{x_0}}| = \left\{ \begin{array}{ll} 1, & \mu < 1, \\ ?, & \mu = 1, \\ 0, & \mu > 1. \end{array} \right. \end{aligned}$$

(37)

Even not knowing $|A_{1,{\mathcal {Q}}_{x_0}}|$, the Mass Transference Principle Theorem 2.3 allows us to compute the Hausdorff dimension of $A_{\mu ,{\mathcal {Q}}_{x_0}}$ from (37). As usual, dilate the set with an exponent $\beta >0$:

$$\begin{aligned} (A_{\mu ,{\mathcal {Q}}_{x_0}})^\beta= & \limsup _{n \rightarrow \infty } \bigcup _{ 1 \le p \le q_n } B \Bigg ( \frac{p}{q_n}, \Bigg ( \frac{c}{q_n^\mu } \Bigg )^\beta \Bigg )\\= & \limsup _{n \rightarrow \infty } \bigcup _{1 \le p \le q_n } B \Bigg ( \frac{p}{q_n}, \frac{c^\beta }{q_n^{\mu \beta }} \Bigg ) = A_{\mu \beta ,{\mathcal {Q}}_{x_0}}, \end{aligned}$$

with a new constant $c^\beta$. Since (37) is independent of c, we have $|(A_{\mu ,{\mathcal {Q}}_{x_0}})^\beta | = |A_{\mu \beta ,{\mathcal {Q}}_{x_0}}| = 1$ if $\mu \beta < 1$, and the Mass Transference Principle implies ${\text {dim}}_{{\mathcal {H}}} A_{\mu ,{\mathcal {Q}}_{x_0}} \ge \beta$. Taking $\beta \rightarrow 1/\mu$, we deduce ${\text {dim}}_{{\mathcal {H}}} A_{\mu ,{\mathcal {Q}}_{x_0}} \ge 1/\mu$. $\quad \square$

As in Proposition 4.4 and in the definition of $B_{\mu ,{\mathcal {Q}}}$ in (30), to get information about $\alpha _{x_0}(t)$ for $t \in A_{\mu ,{\mathcal {Q}}_{x_0}}$ we need to restrict their exponent of irrationality. We do this by removing sets $A_{\mu + \epsilon }$ defined in (13). However, compared to Proposition 4.4 we have two fundamental difficulties:

• (a) The dimensions ${\text {dim}}_{{\mathcal {H}}} A_\mu = 2/\mu > 1/\mu = {\text {dim}}_{{\mathcal {H}}} A_{\mu ,{\mathcal {Q}}_{x_0}}$ do not match anymore.

• (b) Because do not know the Lebesgue measure of $A_{1,{\mathcal {Q}}_{x_0}}$ in (37), we cannot conclude that ${\mathcal {H}}^{1/\mu } (A_{\mu ,{\mathcal {Q}}_{x_0}}) = \infty$ if $\mu > 1$.

To overcome these difficulties, let $\delta _1, \delta _2 > 0$ and define the set

$$\begin{aligned} B_{\mu , {\mathcal {Q}}_{x_0}} ^{\delta _1, \delta _2} = \Big ( A_{\mu , {\mathcal {Q}}_{x_0}} {\setminus } A_{\mu + \delta _1,{\mathcal {Q}}_{x_0}} \Big ) {\setminus } \Bigg ( \bigcup _{\epsilon >0} A_{2\mu + \delta _2 + \epsilon } \Bigg ). \end{aligned}$$

Remark 5.4

(Explanation of the definition of $B_{\mu , {\mathcal {Q}}_{x_0}} ^{\delta _1, \delta _2}$) The role of $\delta _2$ is to avoid the problem (b) above, while $\delta _1$ has a technical role when controlling the behavior of $F_\pm (x_{q_n}/\sqrt{h_{q_n}})$ in (40). Last, we remove $A_{2\mu +\epsilon }$ instead of $A_{\mu + \epsilon }$ to avoid problem (a) and to ensure that $B_{\mu , {\mathcal {Q}}_{x_0}} ^{\delta _1, \delta _2}$ is not too small. The downside of this is that we can only get $\mu (t) \in [\mu , 2\mu + \delta _2]$ for the exponent of irrationality of $t \in B_{\mu , {\mathcal {Q}}_{x_0}} ^{\delta _1, \delta _2}$. If instead we worked with the set

$$\begin{aligned} {{\widetilde{B}}}_{\mu , {\mathcal {Q}}_{x_0}}^{\delta _1} = \Big ( A_{\mu , {\mathcal {Q}}_{x_0}} {\setminus } A_{\mu + \delta _1,{\mathcal {Q}}_{x_0}} \Big ) {\setminus } \Bigg ( \bigcup _{\epsilon >0} A_{\mu + \epsilon } \Bigg ) \end{aligned}$$

we would deduce $\mu (t) =\mu$ and therefore $\alpha _{x_0}(t) = 1/2 + 1/(2\mu )$. However, we do not know how to compute the dimension of ${{\widetilde{B}}}^{\delta _1}_{\mu , {\mathcal {Q}}_{x_0}}$.

Proposition 5.5

Let $\mu \ge 1$. Then,

• (a) ${\text {dim}}_{{\mathcal {H}}} B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}} = 1/\mu$.

• (b) If $t \in B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}}$, then $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{4\mu + 2\delta _2}$.

• (c) If $2 \le \mu < 2\sigma - \delta _1$ and $t \in B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}}$, then $\alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu }$.

Proof of Proposition 5.5

(a) The inclusion $B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}} \subset A_{\mu , {\mathcal {Q}}_{x_0}}$ directly implies ${\text {dim}}_{{\mathcal {H}}} B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}} \le 1/\mu$. We prove the lower bound following the proof of Proposition 4.6 in a few steps:

• (a.1) Since ${\text {dim}}_{{\mathcal {H}}} A_{\mu + \delta _1, {\mathcal {Q}}_{x_0} } = 1/(\mu + \delta _1) < 1/\mu$, we have ${\text {dim}}_{{\mathcal {H}}} ( A_{\mu , {\mathcal {Q}}_{x_0}} {\setminus } A_{\mu + \delta _1, {\mathcal {Q}}_{x_0} } ) = 1/\mu$.

• (a.2) The sets $A_\mu$ are nested, so by the Jarnik–Besicovitch Theorem 2.2

  $$\begin{aligned} {\text {dim}}_{{\mathcal {H}}} \Bigg (\bigcup _{\epsilon > 0} A_{2\mu + \delta _2 + \epsilon } \Bigg )= & \sup _{n \in {\mathbb {N}}} \left\{ {\text {dim}}_{{\mathcal {H}}} \Big ( A_{2\mu + \delta _2 + \frac{1}{n}} \Big ) \right\} \\= & \sup _{n \in {\mathbb {N}}} \frac{2}{2\mu + \delta _2 + \frac{1}{n}} = \frac{1}{\mu + \delta _2 / 2}. \end{aligned}$$

  Moreover, ${\mathcal {H}}^\gamma \big (\bigcup _{\epsilon > 0} A_{2\mu + \delta _2 + \epsilon } \big ) = \lim _{n \rightarrow \infty } {\mathcal {H}}^\gamma \big ( A_{2\mu + \delta _2 + 1/n} \big ) = 0$ for all $\gamma \ge 1/(\mu + \delta _2/2)$.

Take $\gamma$ such that $1/ (\mu + \delta _2 /2)< \gamma < 1/\mu$. From (a.1) we get ${\mathcal {H}}^{\gamma } ( A_{\mu , {\mathcal {Q}}_{x_0}} {\setminus } A_{\mu + \delta _1, {\mathcal {Q}}_{x_0} } ) = \infty$, and from (a.2) we hav e ${\mathcal {H}}^{\gamma } \big (\bigcup _{\epsilon > 0} A_{2\mu + \delta _2 + \epsilon } \big ) = 0$, so

$$\begin{aligned} {\mathcal {H}}^\gamma (B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}}) = {\mathcal {H}}^\gamma ( A_{\mu , {\mathcal {Q}}_{x_0}} {\setminus } A_{\mu + \delta _1, {\mathcal {Q}}_{x_0} } )- {\mathcal {H}}^\gamma \Bigg (\bigcup _{\epsilon> 0} A_{2\mu + \delta + \epsilon } \Bigg ) > 0. \end{aligned}$$

Consequently ${\text {dim}}_{{\mathcal {H}}} B^{\delta _1,\delta _2}_{\mu , {\mathcal {Q}}_{x_0}} \ge \gamma$, and taking $\gamma \rightarrow 1/\mu$ we conclude ${\text {dim}}_{{\mathcal {H}}} B^{\delta _1,\delta _2}_{\mu , {\mathcal {Q}}_{x_0}} \ge 1/\mu$.

(b) Let $t \in B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}}$. Then, $t \notin \bigcup _{\epsilon > 0} A_{2\mu + \delta _2 + \epsilon }$ implies $\mu (t) \le 2\mu + \delta _2$, where $\mu (t)$ is the exponent of irrationality of t. Combining this with Proposition 3.6 we get $\alpha _{x_0}(t) \ge \frac{1}{2} + \frac{1}{2\mu (t)} \ge \frac{1}{2} + \frac{1}{4\mu + 2\delta _2}$.

(c) Let $t \in B^{\delta _1, \delta _2}_{\mu , {\mathcal {Q}}_{x_0}}$. Since $t \in A_{\mu , {\mathcal {Q}}_{x_0}} {\setminus } A_{\mu + \delta _1, {\mathcal {Q}}_{x_0} }$, there is a subsequence of denominators $(q_{n_k})_k \subset {\mathcal {Q}}_{x_0}$ such that $c/q_{n_k}^{\mu + \delta _1} \le \big | t - p_{n_k}/q_{n_k} \big | < c/q_{n_k}^\mu$ for $k \in {\mathbb {N}}$. Define the errors $h_{n_k}$ and $x_{n_k}$, and the exponent $\mu _{n_k}$ as

$$\begin{aligned} h_{n_k} = t - \frac{p_{n_k}}{q_{n_k}}, \quad |h_{n_k}| = \frac{1}{q_{n_k}^{\mu _{n_k}}} \quad \text { and } \quad x_{n_k} = \Big | x_0 - \frac{b_{n_k}}{q_{n_k}} \Big | < \frac{1}{q_{n_k}^{\sigma }}. \end{aligned}$$

(38)

From the condition above, since $c < 1$, we immediately get that for any $\epsilon >0$,

$$\begin{aligned} \mu < \mu _{n_k} \le \mu + \delta _1 + \epsilon , \quad \forall k \gg _\epsilon 1. \end{aligned}$$

(39)

By the asymptotic expansion in Corollary 3.3, we have

$$\begin{aligned} R_{x_0}(t) - R_{x_0}\Bigg ( \frac{p_{n_k}}{q_{n_k}} \Bigg ) = \frac{|h_{n_k}|^{1/2}}{q_{n_k}} \, G(p_{n_k}, b_{n_k}, q_{n_k}) \, F_\pm \Bigg ( \frac{x_{n_k}}{\sqrt{h_{n_k}}} \Bigg ) - 2\pi i h_{n_k} + \text {Error}, \end{aligned}$$

where $\text {Error} = O\Big ( \min \big ( q_{n_k}^{3/2}\, h_{n_k}^{3/2}, q_{n_k}^{1/2} \, h_{n_k} \big ) \Big )$. Let us treat the elements in this expression separately.

• Since $q_{n_k} \not \in 4{\mathbb {N}}$, we have $|G(p_{n_k},b_{n_k}, q_{n_k})| \ge \sqrt{q_{n_k}}$ for $k \in {\mathbb {N}}$. Indeed, if $q_{n_k}$ is odd, then $|G(p_{n_k},b_{n_k}, q_{n_k})| = \sqrt{q_{n_k}}$. If $q_{n_k} \equiv 2 \pmod {4}$, then $b_{n_k}$ is odd, so $q_{n_k}/2 \equiv b_{n_k} \pmod {2}$ and hence $|G(p_{n_k},b_{n_k}, q_{n_k})| = \sqrt{2 q_{n_k}}$. Also, by (38) and (39),

  $$\begin{aligned} \frac{x_{n_k}}{\sqrt{|h_{n_k}|}} = x_{n_k} \, q_{n_k}^{\mu _{n_k}/2} < \frac{q_{n_k}^{\mu _{n_k}/2}}{q_{n_k}^\sigma } \le \frac{q_{n_k}^{ \frac{\mu }{2} + \frac{\delta _1}{2} + \frac{\epsilon }{2}}}{q_{n_k}^\sigma } = \frac{1}{q_{n_k}^{\sigma - \frac{\mu }{2} - \frac{\delta _1}{2} - \frac{\epsilon }{2}}}. \end{aligned}$$

  (40)

  Hence, if $2\sigma > \mu + \delta _1$, we can choose $\epsilon = \sigma - \mu /2 - \delta _1 / 2 > 0$ and we get

  $$\begin{aligned} \lim _{k \rightarrow \infty } \frac{x_{n_k}}{\sqrt{|h_{n_k}|}} \le \lim _{k \rightarrow \infty } \frac{1}{q_{n_k}^{\sigma - \mu /2 - \delta _1/2 - \epsilon / 2}} = \lim _{k \rightarrow \infty } \frac{1}{q_{n_k}^{(\sigma - \mu /2 - \delta _1/2)/ 2}} = 0. \end{aligned}$$

  Since $F_\pm$ is continuous, we get $| F_\pm ( x_{n_k}/|h_{n_k}|^{1/2} ) | \ge |F_\pm (0)| / 2 \simeq 1$ for all $k \gg 1$. Therefore,

  $$\begin{aligned} \text {Main term} = \Bigg | \frac{\sqrt{|h_{n_k}|}}{q_{n_k}} \, G(p_{n_k}, b_{n_k}, q_{n_k}) \, F \Bigg ( \frac{x_{n_k}}{|h_{n_k}|^{1/2}} \Bigg ) \Bigg | \simeq \frac{\sqrt{|h_{n_k}|}}{\sqrt{q_{n_k}}}, \quad \forall k \gg 1. \end{aligned}$$

• The term $2\pi i h_{n_k}$ is absorbed by the Main Term if $|h_{n_k}| \ll \sqrt{|h_{n_k}|}/\sqrt{q_{n_k}}$, which is equivalent to $|h_{n_k}| \ll 1/q_{n_k}$. If $\mu > 1$, we get precisely $|h_{n_k}| < c/q_{n_k}^\mu \ll 1/q_{n_k}$.

• Regarding the error term, we can write

  $$\begin{aligned} q_{n_k}^{1/2} |h_{n_k}| = \frac{\sqrt{|h_{n_k}|}}{\sqrt{q_{n_k}}} \, (q_{n_k}^2 |h_{n_k}|)^{1/2}, \quad q_{n_k}^{3/2} |h_{n_k}|^{3/2} = \frac{\sqrt{|h_{n_k}|}}{\sqrt{q_{n_k}}} \, q_{n_k}^2 |h_{n_k}|. \end{aligned}$$

  Since $\text {Error} \le C \, \min \big ( q_{n_k}^{3/2}\, |h_{n_k}|^{3/2}, q_{n_k}^{1/2} \, |h_{n_k}| \big )$ for some $C > 0$, the error is absorbed by the Main Term if $q_{n_k}^2\, |h_{n_k}| \le c$ for a small enough, but universal constant c. Choosing $c>0$ in the definition of $A_{\mu ,{\mathcal {Q}}_{x_0}}$, the condition $|h_{n_k}| \le c/q_{n_k}^\mu \le c/q_{n_k}^2$ is satisfied if $\mu \ge 2$.

Hence, if $2 \le \mu < 2\sigma - \delta _1$ and $t \in B^{\delta _1,\delta _2}_{\mu ,{\mathcal {Q}}_{x_0}}$, then $| R_{x_0}(t) - R_{x_0}( p_{n_k}/q_{n_k} ) | \gtrsim \sqrt{|h_{n_k}|}/\sqrt{q_{n_k}}$ for all $k \gg 1$. From (39) we have $1/\sqrt{ q_{n_k} } = |h_{n_k}|^{1/(2\mu _{n_k})} > |h_{n_k}|^{1/(2\mu )}$, so $| R_{x_0}(t) - R_{x_0}( p_{n_k}/q_{n_k} ) | \gtrsim |h_{n_k}|^{\frac{1}{2} + \frac{1}{2\mu }}$ for all $k \gg 1$, which implies $\alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu }$. $\square$

From Proposition 5.5 we can deduce the main part of Theorem 1.3.

Theorem 5.6

Let $\sigma \ge 2$ and let $x_0 \in A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$. Let $2 \le \mu < 2\sigma$. Then, for all $\delta > 0$,

$$\begin{aligned} \frac{1}{\mu } \le {\text {dim}}_{{\mathcal {H}}} \left\{ \, t \,: \frac{1}{2} + \frac{1}{4\mu } - \delta \le \alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu } \right\} \le \frac{2}{\mu }. \end{aligned}$$

Proof

Choose $\delta _2 > 0$ and any $\delta _1 < 2\sigma - \mu$. Hence, $2 \le \mu < 2\sigma - \delta _1$ and Proposition 5.5 implies

$$\begin{aligned} B^{\delta _1,\delta _2}_{\mu ,{\mathcal {Q}}_{x_0}} \subset \left\{ \, t \,: \frac{1}{2} + \frac{1}{4\mu + 2\delta _2} \le \alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu } \right\} . \end{aligned}$$

Since ${\text {dim}}_{{\mathcal {H}}} B^{\delta _1, \delta _2}_{\mu ,{\mathcal {Q}}_{x_0}} = 1/\mu$ and $\delta _2$ is arbitrary, we get the lower bound. Let us now prove the upper bound. If $\alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu }$, by Proposition 3.6 we get $\frac{1}{2} + \frac{1}{2\mu (t)} \le \alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu }$, hence $\mu (t) \ge \mu$. This implies $t \in A_{\mu -\epsilon }$ for all $\epsilon >0$, so by the Jarnik–Besicovitch Theorem 2.2 we get

$$\begin{aligned} {\text {dim}}_{{\mathcal {H}}} \left\{ \, t \,: \frac{1}{2} + \frac{1}{4\mu } - \delta \le \alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu } \right\} \le {\text {dim}}_{{\mathcal {H}}} A_{\mu - \epsilon } = \frac{2}{\mu - \epsilon } \end{aligned}$$

for all $\delta \ge 0$. We conclude by taking the limit $\epsilon \rightarrow 0$. $\square$

To get the precise statement of Theorem 1.3, we only need to relate the sets $A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$ with the exponent $\sigma (x_0) = \limsup _{n \rightarrow \infty } \{ \, \mu _n : \, q_n \not \in 4{\mathbb {N}} \, \}$ defined in (8). We proceed as follows. Since $\{A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}\}_{\sigma \ge 2}$ is a nested family and $A_{2, \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} = (0,1) {\setminus } {\mathbb {Q}}$, for every $x_0 \in (0,1) {\setminus } {\mathbb {Q}}$ there exists ${\widetilde{\sigma }}(x_0) = \sup \{ \, \sigma : \, x_0 \in A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}} \, \}$. Let us check that $\sigma (x_0) = {{\widetilde{\sigma }}} (x_0)$. Indeed, call ${{\widetilde{\sigma }}}(x_0) = {{\widetilde{\sigma }}}$.

• If ${{\widetilde{\sigma }}} > 2$. Then for $\epsilon > 0$ small enough there exists a sequence $b_k/q_k$ such that $q_k \not \in 4{\mathbb {N}}$ and $|x_0 - b_k/q_k|< 1/q_k^{{{\widetilde{\sigma }}} - \epsilon } < 1/(2q_k^2)$. By Khinchin’s theorem [36, Theorem 19], $b_k/q_k$ is an approximation by continued fraction, for which $|x_0 - b_k/q_k| = 1/q_k^{\mu _k} < 1/q_k^{{{\widetilde{\sigma }}} - \epsilon }$, and therefore $\mu _k \ge {{\widetilde{\sigma }}} - \epsilon$. This implies $\sigma (x_0) \ge {{\widetilde{\sigma }}} - \epsilon$ for all $\epsilon > 0$, hence $\sigma (x_0) \ge {{\widetilde{\sigma }}}$. On the other hand, for all approximations by continued fractions with $q_n \not \in 4{\mathbb {N}}$ with large enough n we have $|x_0 - b_n/q_n| = 1/q_n^{\mu _n} > 1/q_n^{{{\widetilde{\sigma }}} + \epsilon }$, hence $\mu _n \le {{\widetilde{\sigma }}} + \epsilon$. This holds for all $\epsilon > 0$, so $\sigma (x_0) \le {{\widetilde{\sigma }}}$.

• If ${{\widetilde{\sigma }}} = 2$, then $|x_0 - b_n/q_n| = 1/q_n^{\mu _n} > 1/q_n^{2 + \epsilon }$, hence $\mu _n \le 2+\epsilon$, for all approximations by continued fractions with $q_n \not \in 4{\mathbb {N}}$. Therefore, $\sigma (x_0) \le 2$. Since $\sigma (x_0) \ge 2$ always holds, we conclude. Therefore, let $x_0 \in (0,1) {\setminus } {\mathbb {Q}}$. Then, $x_0 \in A_{\sigma , \, {\mathbb {N}} {\setminus } 4{\mathbb {N}}}$ for all $\sigma < \sigma (x_0)$, so the conclusion of Theorem 5.6 holds for $2 \le \mu < 2\sigma$, for all $\sigma < \sigma (x_0)$. That implies that for every $\delta > 0$,

  $$\begin{aligned} \frac{1}{\mu } \le {\text {dim}}_{{\mathcal {H}}} \left\{ \, t \,: \frac{1}{2} + \frac{1}{4\mu } - \delta \le \alpha _{x_0}(t) \le \frac{1}{2} + \frac{1}{2\mu } \right\} \le \frac{2}{\mu }, \quad \text { for all } \quad 2 \le \mu < 2\sigma (x_0). \end{aligned}$$

6 Proof of Theorem 1.6—the high-pass filters when $x_0 \in {\mathbb {Q}}$

In this section we prove Theorem 1.6. For that, we compute the $L^p$ norms of the high-pass filters of $R_{x_0}$ when $x_0 \in {\mathbb {Q}}$. In Sect. 6.1 we define Fourier high-pass filters using smooth cutoffs, reduce the computation of their $L^p$ norms to the study of Fourier localized $L^p$ estimates, state such localized estimates and deduce Theorem 1.6 from them. We prove such localized estimates in Sect. 6.2.

6.1 High-pass filters and frequency localization

We begin with the definition of high-pass filters we use in the proofs. Let $\phi \in C^\infty$ a positive and even cutoff with support on $[-1,1]$ and such that $\phi (x) = 1$ on $x\in [-1/2,1/2]$. Let $\psi (x) = \phi (x/2) - \phi (x)$, and

$$\begin{aligned} \psi _{-1}(x) = \frac{\phi (x)}{\phi (x) + \sum _{i \in {\mathbb {N}}} \psi ( x/2^i )}, \quad \psi _k(x) = \frac{\psi ( x/2^k )}{\phi (x) + \sum _{i \in {\mathbb {N}}} \psi ( x/2^i )}, \quad \text { for } k \ge 0, \end{aligned}$$

so that we have the partition of unity $\sum _{k = -1}^\infty \psi _k(x) = 1$. For $k \ge 0$, $\psi _k$ is supported on $[-2^{k+1},-2^{k-1}] \cup [2^{k-1},2^{k+1}]$. Let f be a periodic function with Fourier series $f(t) = \sum _{ n \in {\mathbb {Z}}} a_n e^{2\pi i n t}$. With the partition of unity above, we perform a Littlewood–Paley decomposition

$$\begin{aligned} f(t) = \sum _{k = -1}^\infty P_kf(t), \quad \text { where } \quad P_kf(t) = \sum _{n \in {\mathbb {Z}}} \psi _k(n) a_n e^{2\pi i n t}. \end{aligned}$$

The Fourier high-pass filter at frequency $N \in {\mathbb {N}}$ is roughly $P_{\ge N} f(t) = \sum _{k \ge \log N} P_kf(t)$. Let us be more precise working directly with $R_{x_0}$, whose frequencies in t are squared. Let $N \in {\mathbb {N}}$ be large, and define $k_N$ to be the unique $k_N \in {\mathbb {N}}$ such that $2^{k_N} \le \sqrt{N} < 2^{k_N+1}$. We define the high-pass filter of $R_{x_0}$ at frequency N as

$$\begin{aligned} P_{\ge N} R_{x_0}(t) = \sum _{k \ge k_N} P_k R_{x_0}(t), \quad \text { where } \quad P_kR_{x_0}(t) = \sum _{n \in {\mathbb {N}} } \psi _k(n) \frac{e^{2\pi i (n^2t + nx_0)}}{n^2}.\nonumber \\ \end{aligned}$$

(41)

We first estimate $\Vert P_k R_{x_0} \Vert _p$ and then extend the result to estimate $\Vert P_{\ge N} R_{x_0} \Vert _p$.

Remark 6.1

At a first glance, using pure Littlewood–Paley blocks in the definition for high-pass filters in (41) may seem restrictive, since it is analogue to estimating high-frequency cutoffs only for a sequence $N_k \simeq 2^k \rightarrow \infty$. However, the estimates we give depend only on the $L^1$ norm of the cutoff $\psi$, so slightly varying the definition and support of $\psi$ does not affect the estimates. This is analogous to having a cutoff $\Phi (x/N)$ for a fixed $\Phi$ as we state in the introduction.

We now state the estimates for the frequency localized $L^p$ estimates. For the sake of generality, let $\Psi \in C^\infty$ be compactly supported outside the origin and bounded below in an interval of its support (for instance, $\psi$ defined above).

Theorem 6.2

Let $x_0 \in {\mathbb {R}}$. Then, for $N \gg 1$,

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi \big (\frac{n}{N}\big )\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \lesssim \left\{ \begin{array}{ll} N^{p-2}, & \text { when } p > 4, \\ N^2 \log N, & \text { when } p=4, \\ N^{p/2}, & \text { when } p < 4. \end{array} \right. \end{aligned}$$

(42)

When $p=2$, the upper bound is sharp, that is, $\big \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \big \Vert _{L^2(0,1)}^2 \simeq N$.

If $x_0 \in {\mathbb {Q}}$, then the upper bound is sharp. That is, if $x_0 = P/Q$ with $(P,Q)=1$, then

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi \big (\frac{n}{N}\big )\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \simeq _Q \left\{ \begin{array}{ll} N^{p-2}, & \text { when } p > 4, \\ N^2 \log N, & \text { when } p=4, \\ N^{p/2}, & \text { when } p < 4. \end{array} \right. \end{aligned}$$

(43)

Remark 6.3

All estimates in Theorem 6.2 depend on $\Vert \Psi \Vert _1$ due to Lemma 6.4.

We postpone the proof of Theorem 6.2 to Sect. 6.2. and use it now to compute the $L^p$ norms of the high-pass filters $\Vert P_{\ge N} R_{x_0} \Vert _p$ and therefore to prove Theorem 1.6.

Proof of Theorem 1.6

Denote the estimate for $x_0 \in {\mathbb {Q}}$ on (43) in Theorem 6.2 by

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \simeq G_p(N). \end{aligned}$$

(44)

First, use the triangle inequality in (41) to bound

$$\begin{aligned} \Vert P_{\ge N} R_{x_0} \Vert _p \le \sum _{k \ge k_N} \Vert P_k R_{x_0} \Vert _p = \sum _{k \ge k_N} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \psi _k(n) \, \frac{e^{2\pi i (n^2 t + nx_0)}}{n^2} \Bigg \Vert _p. \end{aligned}$$

Since $\psi _k$ is supported on $[2^{k-1}, 2^{k+1}]$, we can take the denominator $n^2$ out of the $L^p$ norm to get

$$\begin{aligned} \Vert P_{\ge N} R_{x_0} \Vert _p \lesssim \sum _{k \ge k_N} \frac{1}{2^{2k}} \, \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \psi _k(n) \, e^{2\pi i (n^2 t + nx_0)} \Bigg \Vert _p, \end{aligned}$$

for example using [23, Lemma 3.1, Corollary 3.2]. We can now use (44) to get²⁵

$$\begin{aligned} \Vert P_{\ge N} R_{x_0} \Vert _p \lesssim \sum _{k \ge k_N} \frac{G_p(2^k)^{1/p}}{2^{2k}} \simeq \frac{G_p(2^{k_N})^{1/p}}{2^{2k_N}}, \end{aligned}$$

(45)

where the last equality follows by direct calculation because the defintion of $G_p$ makes the series be geometric. For the lower bound, as long as $1< p < \infty$, the Mihklin multiplier theorem.²⁶ combined again with [23, Lemma 3.1, Corollary 3.2] and (44) gives

$$\begin{aligned} \Vert P_{\ge N} R_{x_0} \Vert _p \gtrsim \Vert P_{k_N}R_{x_0} \Vert _p \simeq \frac{1}{2^{2k_N}} \, \Bigg \Vert \sum _{n} \psi _{k_N}(n) \, e^{2\pi i (n^2 t + nx_0)} \Bigg \Vert _p \simeq \frac{G_p(2^{k_N})^{1/p}}{2^{2k_N}}.\nonumber \\ \end{aligned}$$

(46)

Joining (45) and (46) and recalling that $2^{k_N} \simeq \sqrt{N}$, we conclude that

$$\begin{aligned} \Vert P_{\ge N} R_{x_0} \Vert _p \simeq \frac{G_p(2^{k_N})^{1/p}}{2^{2k_N}} \simeq \left\{ \begin{array}{ll} N^{-1/2 - 1/p}, & p > 4, \\ N^{-3/4} \, (\log N)^{1/4}, & p = 4, \\ N^{-3/4}, & p < 4, \end{array}\right. \end{aligned}$$

from which we immediately get

$$\begin{aligned} \eta (p) = \lim _{N \rightarrow \infty } \frac{\log (\Vert P_{\ge N} R_{x_0} \Vert _p^p ) }{\log (1/N)} = \left\{ \begin{array}{ll} p/2 + 1, & p > 4, \\ 3p/4, & p \le 4. \end{array} \right. \end{aligned}$$

$\square$

6.2 Frequency localized $L^p$ norms

In this section we prove Theorem 6.2. The $L^2$ estimate, which holds for all $x_0$, follows from Plancherel’s theorem. For $p \ne 2$, we use the following well-known lemma, whose proof can be found in [10, Lemma 3.18] (see also [4, Lemma 4.4]).

Lemma 6.4

Let $\Psi \in C^\infty _0({\mathbb {R}})$. Let $N \in {\mathbb {N}}$ and $q \in {\mathbb {N}}$ such that $q \le N$. Let also $a\in {\mathbb {Z}}$ such that $(a,q) = 1$. Then,

$$\begin{aligned} \Bigg | t - \frac{a}{q} \Bigg |\le & \frac{1}{qN} \quad \Longrightarrow \quad \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi \left( \frac{n}{N} \right) \, e^{2 \pi i (n^2 t + n x)} \, \Bigg |\nonumber \\\lesssim & _{\Vert \Psi \Vert _1} \frac{N}{\sqrt{q} \, \left( 1 + N\, \sqrt{ |t - a/q| } \right) }. \end{aligned}$$

(47)

Moreover, there exist $\delta , \epsilon \le 1$ only depending on $\Psi$ such that if

$$\begin{aligned} q \le \epsilon N, \quad \Big | t - \frac{a}{q} \Big | \le \frac{\delta }{N^2}, \quad \Big | x - \frac{b}{q} \Big | \le \frac{\delta }{N} \end{aligned}$$

for some $b \in {\mathbb {Z}}$, then

$$\begin{aligned} \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi \left( \frac{n}{N} \right) \, e^{2 \pi i (n^2 t + n x)} \, \Bigg | \simeq _{\Vert \Psi \Vert _1} \frac{N}{\sqrt{q}}. \end{aligned}$$

We are now ready to prove Theorem 6.2.

Proof of Theorem 6.2

Let $x_0 \in {\mathbb {R}}$. For simplicity, we prove the $L^2$ estimate for a symmetric $\Psi$. Considering f as a Fourier series in t, by Plancherel’s theorem we write

$$\begin{aligned} \begin{aligned}&\Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi \big (\frac{n}{N}\big )\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^2(0,1)}^2 = \sum _{n =1}^\infty \left| \Psi \big (\frac{n}{N}\big )\,e^{2\pi i n \, x_0} + \Psi \big (-\frac{n}{N}\big )\,e^{-2\pi i n \, x_0} \right| ^2 \\&\quad = \sum _{n=1}^\infty \Psi \big (\frac{n}{N}\big )^2 \left| e^{2\pi i n x_0} + e^{-2\pi i n x_0} \right| ^2 \simeq \sum _{n=1}^\infty \Psi \big (\frac{n}{N}\big )^2 \cos ^2 (2\pi n x_0) \end{aligned} \end{aligned}$$

This sum is upper bounded by N by the triangle inequality. If $x_0$ is rational, say $x_0 = P/Q$, the bound from below follows²⁷ by summing only over multiples of Q in [N, 2N], so that

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi \big (\frac{n}{N}\big ) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^2(0,1)}^2 \gtrsim \sum _{ k = N/Q}^{2N/Q} \cos ^2 (2\pi kQ x_0) = \frac{N}{Q} \simeq _Q N. \end{aligned}$$

If $x_0$ is irrational, it is known that the sequence $(nx_0)_n$ is equidistributed in the torus, which means that for any continuous p-periodic function

$$\begin{aligned} \lim _{N \rightarrow \infty } \frac{1}{N} \sum _{n=1}^N f(nx_0) = \int _0^p f. \end{aligned}$$

In particular, since for $f(y) = \cos (4\pi y)$ we have $\int _0^{1/2} f(y)\, dy=0$, we get²⁸ for large N that

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi \big (\frac{n}{N}\big )\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^2(0,1)}^2 \gtrsim \sum _{n = N}^{2N} \cos ^2 (2\pi n x_0) \simeq N + \sum _{n = N}^{2N} \cos (4\pi n x_0) \simeq N. \end{aligned}$$

We now prove the upper bound (42) for any $x_0 \in {\mathbb {R}}$. The Dirichlet approximation theorem implies that any $t \in {\mathbb {R}} {\setminus } {\mathbb {Q}}$ can be approximated as follows:

$$\begin{aligned} \forall N \in {\mathbb {N}}, \quad \exists q \le N, \quad 1 \le a \le q \quad \text { such that } \quad \Big | t - \frac{a}{q} \Big | \le \frac{1}{qN}, \end{aligned}$$

which can be rewritten as ${\mathbb {R}} {\setminus } {\mathbb {Q}} \subset \bigcup _{q=1}^N \bigcup _{a=1}^q B \big ( \frac{a}{q}, \frac{1}{qN} \big )$ for all $N \in {\mathbb {N}}$. Therefore, for any $N \in {\mathbb {N}}$,

$$\begin{aligned} & \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p\nonumber \\ & \quad \le \sum _{q=1}^N \sum _{a=1}^q \int _{B( \frac{a}{q}, \frac{1 }{qN} )} \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg |^p\, dt. \end{aligned}$$

(48)

We split each integral according to the two situations in (47) in Lemma 6.4:

$$\begin{aligned} \begin{aligned}&\int _{|t - \frac{a}{q}|< \frac{1}{N^2} } \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi (n/N) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg |^p\, dt \\&\qquad + \int _{ \frac{1}{N^2}< |t - \frac{a}{q}|< \frac{1}{qN} } \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi (n/N) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg |^p\, dt \\&\quad \le \int _{|t - \frac{a}{q}|< \frac{1}{N^2} } \Bigg ( \frac{N}{\sqrt{q} } \Bigg )^p\, dt + \int _{\frac{1}{N^2}<|t - \frac{a}{q}| < \frac{1}{qN} } \Bigg ( \frac{1}{\sqrt{q} \, |t - \frac{a}{q}|^{1/2 } } \Bigg )^p\, dt \\&\quad \simeq \frac{N^{p-2}}{q^{p/2}} + \frac{1}{q^{p/2}} \, \int _{\frac{1}{N^2}}^{\frac{1}{qN}} \frac{1}{h^{p/2}} \, dh. \end{aligned} \end{aligned}$$

(49)

The behavior of that last integral changes depending on p being greater or smaller than 2.

• If $p < 2$,

  $$\begin{aligned} (49) \simeq \frac{N^{p-2}}{q^{p/2}} + \frac{1}{q^{p/2}} \left( \left( \frac{1}{qN} \right) ^{1-p/2} - \left( \frac{1}{N^2} \right) ^{1-p/2} \right) \le \frac{N^{p-2}}{q^{p/2}} + \frac{1}{q\, N^{1-p/2}}, \end{aligned}$$

  so

  $$\begin{aligned} (48) \le N^{p-2}\, \sum _{q=1}^N \sum _{a=1}^q \frac{1}{q^{p/2}} + \frac{1}{ N^{1-p/2}}\, \sum _{q=1}^N \sum _{a=1}^q \frac{1}{q} \lesssim N^{p/2}. \end{aligned}$$

• If $p=2$,

  $$\begin{aligned} (49) \simeq \frac{1}{q} \Big ( 1 + \int _{\frac{1}{N^2}}^{\frac{1}{qN}} \frac{dh}{h} \Big ) \lesssim \frac{1}{q} \left( 1 + \log (N^2) - \log (qN) \right) = \frac{1 + \log (N/q) }{q}, \end{aligned}$$

  hence

  $$\begin{aligned} (48) & \lesssim \sum _{q=1}^N \Big ( 1 - \log ( q/N ) \Big ) \simeq N - \int _1^N \log (x/N) \, dx \simeq N \Big ( 1 -\int _{\frac{1}{N}}^1 \log (y)\, dy \Big )\\ & \simeq N. \end{aligned}$$

• If $p > 2$,

  $$\begin{aligned} & (49) \simeq \frac{N^{p-2}}{q^{p/2}} + \frac{ \left( N^2 \right) ^{p/2 - 1} - (qN )^{p/2 - 1} }{q^{p/2}}\lesssim \frac{N^{p-2}}{q^{p/2}}\\ & \quad \quad \Longrightarrow \quad (48) \lesssim N^{p-2} \, \sum _{q=1}^N \frac{1}{q^{p/2 - 1}}. \end{aligned}$$

  This series converges if and only if $p > 4$, and more precisely,

  $$\begin{aligned} (48) \lesssim \left\{ \begin{array}{ll} N^{p-2}, & p > 4, \\ N^2 \, \log N, & p = 4, \\ N^{p-2} \, N^{2 - p/2} = N^{p/2}, & p < 4. \end{array}\right. \end{aligned}$$

  This concludes the proof of (42).

We now prove the lower bound in (43) for $x_0 \in {\mathbb {Q}}$. Let $x_0 = P/Q$ with $(P,Q) = 1$. Let $\delta , \epsilon > 0$ as given in Lemma 6.4, and let $N \in {\mathbb {N}}$ be such that $Q \le \epsilon N$. Bound the $L^p$ norm from below by

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \ge \int _{B\big ( \frac{a}{Q}, \frac{\delta }{N^2} \big )} \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi (n/N) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg |^p\, dt,\nonumber \\ \end{aligned}$$

(50)

where a is any $1 \le a \le Q$ such that $(a,Q)=1$. Use Lemma 6.4 with $q=Q$ and $b = P$, for which the condition $0 = |x_0 - P/Q| < \delta / N$ is satisfied trivially, and $|t - a/Q| < \delta / N^2$, which is valid on the domain of integration. Then, for $N \ge Q / \epsilon$,

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \gtrsim \int _{B\big ( \frac{a}{Q}, \frac{\delta }{N^2} \big )} \Bigg ( \frac{N}{\sqrt{Q}} \Bigg )^p\, dt \simeq \frac{N^p}{Q^{p/2}} \, \frac{\delta }{N^2} \simeq _Q N^{p-2}. \end{aligned}$$

In view of the upper bound in (42), this is optimal when $p > 4$. When $p \le 4$, we refine the bound in (50) as follows. Define the set

$$\begin{aligned} {\mathcal {Q}}_N = \{ \, q \in {\mathbb {N}} \,: \, Q \mid q \, \text { and } \, q \le \epsilon N \, \}, \end{aligned}$$

whose cardinality $\simeq \epsilon N /Q$ is as large as needed if $N \gg 1$. Observe that

$$\begin{aligned} B\Bigg ( \frac{a}{q}, \frac{\delta }{N^2} \Bigg ) \cap B\Bigg ( \frac{a'}{q'}, \frac{\delta }{N^2} \Bigg ) = \emptyset , \quad \forall q,q' \in {\mathcal {Q}}_N, \quad (a,q) = 1 = (a',q'), \end{aligned}$$

as long as $a/q \ne a'/q'$. Indeed, the distance from the centers is $\frac{|aq' - a'q|}{q \, q'} \ge \frac{1}{q \, q'} \ge \frac{1}{\epsilon ^2 N^2}$, while the radius is $\frac{\delta }{N^2} < \frac{1}{\epsilon ^2 N^2}$ (choosing a smaller $\delta > 0$ if needed). Hence the balls in the family $\{ B(a/q, \delta /N^2) : \, q \in {\mathcal {Q}}_N, \, (a,q)=1 \, \}$ are pairwise disjoint, and we can bound

$$\begin{aligned} & \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \nonumber \\ & \quad \gtrsim \sum _{q \in {\mathcal {Q}}_N} \, \sum _{a: (a,q) = 1 } \int _{B\left( \frac{a}{q}, \frac{\delta }{N^2} \right) } \Bigg | \sum _{n \in {\mathbb {Z}}} \Psi (n/N) \, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg |^p \, dt. \end{aligned}$$

(51)

For each of those integrals we have $q = Qn$ for some $n \in {\mathbb {N}}$. To use Lemma 6.4 we chose $b = Pn$ so that $0 = |x_0 - b/q| < \delta / N$, hence

$$\begin{aligned} (51) & \gtrsim \sum _{q \in {\mathcal {Q}}_N} \, \sum _{a: (a,q) = 1 } \int _{B\left( \frac{a}{q}, \frac{\delta }{N^2} \right) } \Big ( \frac{N}{\sqrt{q}} \Big )^p \, dt\nonumber \\ & \quad \simeq \delta \, N^{p-2}\, \sum _{q \in {\mathcal {Q}}_N} \, \frac{\varphi (q)}{q^{p/2}} \simeq \frac{N^{p-2}}{Q^{p/2}} \, \sum _{n=1}^{\epsilon N / Q} \, \frac{\varphi (Qn)}{n^{p/2}}. \end{aligned}$$

(52)

We estimate this sum in the following lemma, which we prove in Appendix A, Corollary A.5.

Lemma 6.5

Let $Q \in {\mathbb {N}}$. Then, for $N \gg 1$,

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (Qn)}{n^2} \simeq \log N, \quad \text { and } \quad \sum _{n=1}^N \frac{\varphi (Qn)}{n^\alpha } \simeq N^{2 - \alpha }, \quad \text { for } \quad \alpha < 2, \end{aligned}$$

where the implicit constants depend on Q and $\alpha$.

Using this lemma in (52), when $p < 4$ we get

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^p(0,1)}^p \simeq _{p,Q} \frac{N^{p-2}}{Q^{p/2}} \, \Big ( \frac{\epsilon N}{Q} \Big )^{2-\frac{p}{2}} \simeq _{p,Q} N^{p/2}. \end{aligned}$$

Similarly, when $p = 4$ we get

$$\begin{aligned} \Bigg \Vert \sum _{n \in {\mathbb {Z}}} \Psi (n/N)\, e^{2\pi i (n^2 \, t + n \, x_0)} \Bigg \Vert _{L^4(0,1)}^4 \simeq _{Q} \frac{N^2}{Q^2} \, \log \Big ( \frac{\epsilon N}{Q} \Big ) \simeq _{Q} N^2 \, \log N. \end{aligned}$$

Together with the upper bounds in (42), this completes the proof. $\square$

Appendices

Sums of Euler’s totient function

Sums of the Euler totient function play a relevant role in this article, especially in Lemma 6.5. In Sect. A.1 we state the classical results and briefly prove them for completeness. In Sect. A.2 we adapt these classical proofs to sums modulo Q that we need in this article. Throughout this appendix, $\varphi$ denotes the Euler totient function and $\mu$ denotes the Möbius function.²⁹

1.1 Sums of Euler’s totient function

Define the sum function

$$\begin{aligned} \Phi (N) = \sum _{n=1}^N \varphi (n), \quad N \in {\mathbb {N}}. \end{aligned}$$

Proposition A.1

For $N \gg 1$,

$$\begin{aligned} \Phi (N) = C N^2 + O\Big ( N \log N \Big ), \quad \text { where } \quad C = \frac{1}{2} \, \sum _{n=1}^\infty \frac{\mu (n)}{n^2} = \frac{3}{\pi ^2} \end{aligned}$$

Proof

By the Möbius inversion formula,

$$\begin{aligned} \Phi (N) = \sum _{n = 1}^N \varphi (n) = \sum _{n=1}^N n \bigg ( \sum _{d \mid n} \frac{\mu (d)}{d} \bigg ) = \sum _{n=1}^N \sum _{d \mid n} \frac{n}{d}\, \mu (d). \end{aligned}$$

Calling $n/d = d'$, the sum is in all natural numbers d and $d'$ such that $d d' \le N$. Therefore,

$$\begin{aligned} \Phi (N) = \sum _{d, d' \,: \, dd' \le N} d' \mu (d) = \sum _{d=1}^N \mu (d) \, \sum _{d'=1}^{\lfloor N/d \rfloor } d' = \sum _{d=1}^N \mu (d) \, \frac{ \lfloor N/d \rfloor \left( \lfloor N/d \rfloor + 1 \right) }{2}. \end{aligned}$$

For $x \in {\mathbb {R}}$, write $x = \lfloor x \rfloor + \{ x \}$, where $0 \le \{ x \} < 1$ is the fractional part of x. Then, direct computation shows that $\lfloor x \rfloor \left( \lfloor x \rfloor + 1 \right) = x^2 + O(x)$ when $x \ge 1$, so

$$\begin{aligned} \Phi (N) = \frac{1}{2} \, \sum _{d=1}^N \mu (d) \, \bigg ( \Big ( \frac{N}{d} \Big )^2 + O\Big (\frac{N}{d} \Big ) \bigg ) = \frac{N^2}{2} \sum _{d=1}^N \frac{\mu (d)}{d^2} + O\left( N \, \sum _{d=1}^N \frac{1}{d} \right) . \end{aligned}$$

The series $\sum _{d=1}^\infty \mu (d)/d^2$ is absolutely convergent, and its value is known to be $2C = 6/\pi ^2$, so write

$$\begin{aligned} \sum _{d=1}^N \frac{\mu (d)}{d^2} = 2C - \sum _{d=N+1}^\infty \frac{\mu (d)}{d^2} = 2C + O\bigg ( \sum _{d=N+1}^\infty \frac{1}{d^2} \bigg ) = 2C + O\Big ( \frac{1}{N} \Big ). \end{aligned}$$

Since $\sum _{d=1}^N 1/d \simeq \log N$, we get $\Phi (N) = C\, N^2 + O( N ) + O( N \log N) = CN^2 + O( N \log N)$. $\square$

As a Corollary of Lemma A.1 we obtain the analogue result for the sums weighted by $n^{-\alpha }$. Observe that when $\alpha > 2$ the sum is convergent.

Corollary A.2

Let $\alpha \le 2$. For $N \gg 1$,

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (n)}{n^2} \simeq \log N, \quad \text { and } \quad \sum _{n=1}^N \frac{\varphi (n)}{n^\alpha } \simeq N^{2 - \alpha }, \quad \text { if } \, \alpha < 2. \end{aligned}$$

Proof

Upper bounds immediately follow from $\varphi (n) \le n$. For lower bounds, assume first that $\alpha \ge 0$. From Proposition A.1 we directly get

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (n)}{n^\alpha } \ge \frac{1}{N^\alpha } \sum _{n=1}^N \varphi (n) = \frac{1}{N^\alpha } \Phi (N) \simeq N^{2- \alpha }, \end{aligned}$$

which is optimal when $\alpha < 2$. For the case $\alpha = 2$ we use the summation by parts formula³⁰ to get

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (n)}{n^2}= & \frac{\Phi (N)}{N^2} - \sum _{n=1}^{N-1} \Phi (n) \Bigg (\frac{1}{(n+1)^2 } - \frac{1}{n^2} \Bigg )\nonumber \\= & \frac{\Phi (N)}{N^2} + \sum _{n=1}^{N-1} \Phi (n) \frac{ 2n+1}{n^2 \, (n+1)^2 }. \end{aligned}$$

(53)

Restrict the sum to $\log N \le n \le N-1$, and combine it with $\Phi (n) \simeq n^2$ for $n \gg 1$ from Proposition A.1 to get

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (n)}{n^2} \gtrsim 1 + \sum _{n \ge \log N}^{N-1} \frac{ 1}{n} \simeq \log N - \log \log N \simeq \log N, \quad \text { for } \, N \gg 1. \end{aligned}$$

When $\alpha < 0$, restrict the sum to $n \in [N/2, N]$ and use $\Phi (N) = C N^2 + O(N\log N)$ in Proposition A.1 to get

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (n)}{n^\alpha } = \sum _{n=1}^N \varphi (n) \, n^{|\alpha |} \ge \Big (\frac{N}{2}\Big )^{|\alpha |} \, \sum _{n \ge N/2}^N \varphi (n) \simeq _{|\alpha |} \frac{\Phi (N) - \Phi (N/2)}{N^\alpha } \simeq N^{2 - \alpha }. \end{aligned}$$

$\square$

1.2 Sums of Euler’s totient function modulo ${\varvec{Q}}$

For $Q \in {\mathbb {N}}$, let

$$\begin{aligned} \Phi _Q(N) = \sum _{n=1}^N \varphi (Qn) \quad \text { when } \, N \gg 1, \end{aligned}$$

To estimate the behavior when $N \rightarrow \infty$ we adapt the proofs of Proposition A.1 and Corollary A.2.

Proposition A.3

Let $Q \in {\mathbb {N}}$. Then, $\Phi _Q(N) \le QN^2$, and there exists a constant $c_Q > 0$ such that

$$\begin{aligned} \Phi _Q(N) \ge c_Q N^2 + O_Q( N \log N). \end{aligned}$$

Consequently, $\Phi _Q(N) \simeq _Q N^2$ when $N \gg 1$.

Proof

The upper bound follows directly from $\varphi (n) < n$ for all $n \in {\mathbb {N}}$, so it suffices to prove the lower bound. For that, first restrict the sum to $n \le N$ such that $(Q,n) = 1$. By the multiplicative property of the Euler function, we get

$$\begin{aligned} \Phi _Q(N) \ge \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \varphi (Qn) = \varphi (Q) \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \varphi (n). \end{aligned}$$

(54)

The proof now follows the same strategy as in Proposition A.1. Use Möbius inversion to write

$$\begin{aligned} \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \varphi (n) = \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \Bigg ( n \sum _{d \mid n} \frac{\mu (d)}{d} \Bigg ) = \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \sum _{d \mid n} \, \frac{n}{d} \, \mu (d). \end{aligned}$$

Observe that if $(Q,n)=1$ and if we decompose $n = d \, d'$, then both d and $d'$ are coprime with Q. Conversely, if d and $d'$ are coprime with Q, then so is $n = d \, d'$. Thus,

$$\begin{aligned} \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \varphi (n) = \sum _{\begin{array}{c} d, d' \,: \, d \, d' \le N \\ (Q,d) = 1 = (Q,d') \end{array} } d' \, \mu (d) = \sum _{\begin{array}{c} d=1 \\ (Q,d) = 1 \end{array} }^N \mu (d) \left( \sum _{\begin{array}{c} d'=1 \\ (Q,d') = 1 \end{array} }^{\lfloor N/d \rfloor } d' \right) . \end{aligned}$$

(55)

In the following lemma we give a closed formula for the inner sum. We postpone its proof.

Lemma A.4

Let $Q \in {\mathbb {N}}$, $Q \ge 2$. Then,

$$\begin{aligned} S_Q = \sum _{\begin{array}{c} n = 1 \\ (Q,n) = 1 \end{array}}^{Q-1} n = \frac{Q \, \varphi (Q)}{2}, \quad \text { and } \quad S_{Q,k} = \sum _{\begin{array}{c} n = 1 \\ (Q,n) = 1 \end{array}}^{kQ - 1} n = \frac{Q \, \varphi (Q)}{2} \, k^2, \quad \forall k \in {\mathbb {N}}. \end{aligned}$$

Now, for every $d \le N$, find $k_d \in {\mathbb {N}}\cup \{ 0 \}$ such that $k_dQ \le \lfloor N/d \rfloor < (k_d+1)Q$, and write

$$\begin{aligned} \sum \limits _{\begin{array}{c} d'=1 \\ (Q,d') = 1 \end{array} }^{\lfloor N/d \rfloor } d'= & \sum _{\begin{array}{c} d'=1 \\ (Q,d') = 1 \end{array} }^{k_dQ - 1} d' + \sum _{\begin{array}{c} d' = k_dQ + 1 \\ (Q,d') = 1 \end{array} }^{\lfloor N/d \rfloor } d' = S_{Q,k_d} + O\Big ( (k_d+1)Q^2 \Big )\nonumber \\= & \frac{Q \, \varphi (Q)}{2} \, k_d^2 + O\Big ( (k_d+1)Q^2 \Big ). \end{aligned}$$

(56)

Since the definition of $k_d$ is equivalent to $\frac{1}{Q} \, \lfloor N/d\rfloor - 1 < k_d \le \frac{1}{Q} \, \lfloor N/d\rfloor$, we deduce that $k_d = \lfloor \frac{1}{Q} \lfloor N/d \rfloor \rfloor$. Consequently, since $\lfloor x \rfloor = x + O(1)$ and $\lfloor x \rfloor ^2 = x^2 + O(x)$, we get

$$\begin{aligned} k_d = \frac{N}{Qd} + O(1) \quad \text { and } \quad k_d^2 = \frac{N^2}{Q^2d^2} + \frac{1}{Q} \, O\Big ( \frac{N}{d} \Big ). \end{aligned}$$

(57)

Hence, from (56) and (57) we get

$$\begin{aligned} \sum _{\begin{array}{c} d'=1 \\ (Q,d') = 1 \end{array} }^{\lfloor N/d \rfloor } d' = \frac{ \varphi (Q)}{2Q}\, \frac{N^2}{ d^2} + O \left( \varphi (Q)\, \frac{N}{d} + Q\frac{N}{d} + Q^2 \right) = \frac{ \varphi (Q)}{2Q}\, \frac{N^2}{ d^2} + Q^2 \, O \bigg ( \frac{N}{d}\bigg ). \end{aligned}$$

We plug this in (55) to get

$$\begin{aligned} \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \varphi (n) = \frac{\varphi (Q)}{2Q} N^2 \sum _{\begin{array}{c} d=1 \\ (Q,d) = 1 \end{array} }^N \frac{\mu (d)}{ d^2} + O \Bigg ( Q^2 N \sum _{\begin{array}{c} d=1 \\ (Q,d) = 1 \end{array} }^N \frac{\mu (d)}{d} \Bigg ). \end{aligned}$$

The sum $\sum _{n=1}^\infty \mu (d)/d^2$ is absolutely convergent, and $c_Q:= \sum _{ d=1, \, (Q,d) = 1 }^\infty \mu (d)/d^2 > 0$ because

$$\begin{aligned} c_Q = 1 + \sum _{\begin{array}{c} d=2 \\ (Q,d) = 1 \end{array} }^\infty \frac{\mu (d)}{d^2} \quad \text { and } \quad \Bigg | \sum _{\begin{array}{c} d=2 \\ (Q,d) = 1 \end{array} }^\infty \frac{\mu (d)}{d^2} \Bigg | \le \frac{\pi ^2}{6} - 1 < 1. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{\begin{array}{c} d=1 \\ (Q,d) = 1 \end{array} }^N \frac{\mu (d)}{d^2} = c_Q - \sum _{\begin{array}{c} d=N+1 \\ (Q,d) = 1 \end{array} }^\infty \frac{\mu (d)}{d^2} = c_Q + O\Bigg ( \sum _{ d=N+1 }^\infty \frac{1}{d^2} \Bigg ) = c_Q + O(1/N). \end{aligned}$$

Together with $|\sum _{d=1, \, (Q,d)=1}^N \mu (d)/d| \lesssim \log N$, this implies

$$\begin{aligned} \sum _{\begin{array}{c} n=1 \\ (Q,n) = 1 \end{array} }^N \varphi (n)= & c_Q\, \frac{\varphi (Q)}{2Q} N^2 + O\Big ( \frac{\varphi (Q)}{Q} N \Big ) + O (Q^2 N \log N)\\= & c_Q\, \frac{\varphi (Q)}{2Q} N^2 + O_Q(N \log N). \end{aligned}$$

Together with (54) we conclude $\Phi _Q(N) \ge c_Q\, \frac{\varphi (Q)^2}{2Q} N^2 + O_Q(N \log N)$. $\quad \square$

Proof of Lemma A.4

We begin with $k=1$. When $Q=2$, we have $S_{2,1} = 1 = 2\, \varphi (2)/2$, so we may assume $Q \ge 3$. We first observe that $\varphi (Q)$ is even, because if Q has an odd prime factor p, then $\varphi (p) = p-1$, which is even, is a factor of $\varphi (Q)$. Otherwise, $Q = 2^r$ with $r \ge 2$, so $\varphi (Q) = 2^{r-1}$ is even. Now, the observation that $(Q,n) = 1 \, \Longleftrightarrow \, (Q,Q-n)=1$ implies

$$\begin{aligned} S_{Q,1} = \sum _{\begin{array}{c} n = 1 \\ (Q,n) = 1 \end{array}}^{\lfloor Q/2 \rfloor } n + \sum _{\begin{array}{c} n = \lfloor Q/2 \rfloor + 1 \\ (Q,n) = 1 \end{array}}^{Q - 1} n = \sum _{\begin{array}{c} n = 1 \\ (Q,n) = 1 \end{array}}^{\lfloor Q/2 \rfloor } \big (n + (Q - n) \big ) = Q \, \frac{\varphi (Q)}{2}. \end{aligned}$$

Let now $k \ge 2$, so that

$$\begin{aligned} \sum _{\begin{array}{c} n = (k-1)Q + 1 \\ (Q,n) = 1 \end{array}}^{kQ-1} n= & \sum _{\begin{array}{c} n = 1 \\ (Q,n) = 1 \end{array}}^{Q-1} \bigg ( n + (k-1)Q \bigg )\\= & S_{Q,1} + (k-1)Q\varphi (Q) = Q \varphi (Q) \Big ( k - \frac{1}{2} \Big ). \end{aligned}$$

Consequently,

$$\begin{aligned} S_{Q, k} = \sum _{\ell =1}^k \left( \sum _{\begin{array}{c} n = (\ell - 1) Q + 1 \\ (Q,n) = 1 \end{array}}^{\ell Q} n \right) = \sum _{\ell =1}^k Q \varphi (Q) \Bigg ( \ell - \frac{1}{2} \Bigg ) = \frac{Q\varphi (Q)}{2} k^2. \end{aligned}$$

$\square$

To conclude, we prove the estimates for the weighted sums that we needed in Lemma 6.5 as a corollary of Proposition A.3. As before, when $\alpha > 2$ the sums are absolutely convergent.

Corollary A.5

(Lemma 6.5) Let $Q \in {\mathbb {N}}$ and $\alpha \le 2$. For $N \gg 1$,

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (Qn)}{n^2} \simeq \log N, \quad \text { and } \quad \sum _{n=1}^N \frac{\varphi (Qn)}{n^\alpha } \simeq N^{2 - \alpha } \quad \text { for } \quad \alpha < 2. \end{aligned}$$

The implicit constants depend on Q, and also on $\alpha$ when $\alpha < 0$.

Proof

Upper bounds follow directly from $\varphi (n) \le n$. Lower bounds follow from Proposition A.3 with the same strategy as in the proof of Corollary A.2. If $\alpha \ge 0$, by Proposition A.3 we get

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (Qn)}{n^\alpha } \ge \frac{1}{N^\alpha }\, \Phi _Q(N) \simeq _Q N^{2-\alpha }, \quad \text { when } N \gg 1. \end{aligned}$$

When $\alpha = 2$, combine Proposition A.3 with summing by parts as in (53) to get

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (Qn)}{n^2} = \frac{\Phi _Q(N)}{N^2} + \sum _{n=1}^{N-1} \Phi _Q(n) \frac{ 2n+1}{n^2 \, (n+1)^2 } \gtrsim 1 + \sum _{n=\log N}^{N-1} \frac{ 1}{n } \simeq \log N. \end{aligned}$$

When $\alpha < 0$, choosing $\delta > 0$ small enough depending on Q, Proposition A.3 implies

$$\begin{aligned} \sum _{n=1}^N \frac{\varphi (Qn)}{n^\alpha } \ge _\alpha N^{|\alpha |} \sum _{n=\delta N}^N \varphi (Qn) = N^{|\alpha |} \Big ( \Phi _Q(N) - \Phi _Q(\delta N) \Big ) \simeq _{Q,\alpha } N^{|\alpha |} N^2 = N^{2-\alpha }. \end{aligned}$$

$\square$

Alternative asymptotic behavior of $R_{x_0}$ around rational t

Following Duistermaat [22], we give an alternative asymptotic behavior of $R_{x_0}$ around rationals that complements Corollary 3.3 and allows us to prove Propositions 3.5 and 4.2.

Proposition B.1

Let $x_0 \in {\mathbb {R}}$. Let $p,q \in {\mathbb {N}}$ be such that $(p,q)=1$. Let $x_q = {\text {dist}}(x_0, {\mathbb {Z}}/q)$. Let $h \ne 0$ and denote ${\text {sign}}(h) = \pm$ so that $h = \pm |h|$. If $x_q = 0$,

$$\begin{aligned} \begin{aligned}&R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h \\&\quad = 2\pi (- 1 \pm i) \frac{\sqrt{|h|}}{\sqrt{q}} \frac{ G(p,m_q,q)}{\sqrt{q}}\\&\qquad + 2 (1 \pm i) \, q^{3/2} |h|^{3/2} \sum _{m \ne 0} \, \frac{ G(p,m_q + m,q)}{\sqrt{q}} \, \frac{e^{- 2\pi i \frac{m^2}{4 q^2 h}}}{ m^2} + O\left( q^{7/2} h^{5/2} \right) , \end{aligned} \end{aligned}$$

(58)

If $x_q \ne 0$,

$$\begin{aligned} \begin{aligned}&R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h \\&\quad = 2 (1 \pm i) \, q^{3/2} |h|^{3/2} \,\sum _{m \in {\mathbb {Z}}} \, \frac{ G(p,m_q + m,q)}{\sqrt{q}} \, \frac{e^{- 2\pi i \frac{(m - qx_q)^2}{4 q^2 h}}}{ (m - qx_q)^2} \, \\&\qquad + O\left( q^{7/2} h^{5/2} \sum _{m \in {\mathbb {Z}}} \frac{1 }{(m - qx_q)^4} \right) . \end{aligned} \end{aligned}$$

(59)

Proof

From the definition $R_{x_0}(t) = \sum _{n \ne 0}e^{2\pi i ( n^2 t + n x_0 ) } /n^2$, we first write

$$\begin{aligned} \begin{aligned} R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h&= 2 \pi i h + \sum _{n \ne 0} \frac{e^{2\pi i n^2 h} - 1}{n^2}\, e^{2\pi i \frac{pn^2}{q}} \, e^{2\pi i n x_0} \\&= 2\pi i h \, \sum _{n \in {\mathbb {Z}}} \Big ( \int _0^1 e^{2\pi i n^2 h \tau } \, d\tau \Big ) \, e^{2\pi i \frac{pn^2}{q}} \, e^{2\pi i n x_0}. \end{aligned} \end{aligned}$$

Split the sum modulo q by writing $n = mq+r$ and use the Poisson summation formula to obtain

$$\begin{aligned} \begin{aligned}&R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h\\&\quad = 2\pi i h \, \sum _{r = 0}^{q-1} e^{2\pi i r^2 p/q} \, \sum _{m \in {\mathbb {Z}}} \Big ( \int _0^1 e^{2\pi i (mq+r)^2 h \tau } \, d\tau \Big ) \, e^{2\pi i (mq+r) x_0} \\&\quad = 2\pi i h \, \sum _{r = 0}^{q-1} e^{2\pi i r^2 p/q} \, \sum _{m \in {\mathbb {Z}}} \int \Big ( \int _0^1 e^{\pm 2\pi i (zq+r)^2 |h| \tau } \, d\tau \Big ) \, e^{2\pi i (zq+r) x_0} \, e^{-2\pi i m z} \, dz \\&\quad = \pm 2\pi i \frac{\sqrt{|h|}}{q} \,\sum _{m \in {\mathbb {Z}}} \, \sum _{r = 0}^{q-1} e^{2\pi i \frac{p r^2 + m r }{q}} \, \int _0^1 \int e^{\pm 2\pi i y^2 \tau } \, e^{2\pi i \frac{ y}{\sqrt{|h|}} ( x_0 - \frac{m}{q})} \, dy \, d\tau . \end{aligned} \end{aligned}$$

where we changed variables $(zq+r)^2 |h| = y^2$. Now complete the square to get

$$\begin{aligned} & R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h\nonumber \\ & \quad = \pm 2\pi i \frac{\sqrt{|h|}}{q} \,\sum _{m \in {\mathbb {Z}}} \, G(p,m,q) \, \int _0^1 \Big ( \int e^{\pm 2\pi i \tau \Big ( y \pm \frac{x_0 - m/q}{2 \tau \sqrt{|h|}} \Big )^2} \, dy \Big ) \, e^{\mp 2\pi i \frac{(x_0 - m/q)^2}{4\tau |h|} } d\tau \nonumber \\ & \quad = \pm 2\pi i \frac{1\pm i}{2} \, \frac{\sqrt{|h|}}{q} \,\sum _{m \in {\mathbb {Z}}} \, G(p,m,q) \, \int _0^1 \frac{1}{\sqrt{\tau }} \, e^{ \mp 2\pi i \frac{(x_0 - m/q)^2}{4\tau |h|} } d\tau . \end{aligned}$$

(60)

By changing variables, and defining $x_q = \min _{m\in {\mathbb {Z}}} |x_0 - m/q| = |x_0 - m_q/q|$ as in (20), we write

$$\begin{aligned} & R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h\nonumber \\ & = \pi (-1 \pm i) \, \frac{\sqrt{|h|}}{\sqrt{q}} \sum _{m \in {\mathbb {Z}}} \, \frac{ G(p,m_q + m,q)}{\sqrt{q}} \, \int _1^\infty \frac{1}{ \xi ^{3/2}} \, e^{- 2\pi i \frac{(x_q - m/q)^2}{4 h} \xi } d\xi . \end{aligned}$$

(61)

We now separate cases. If $x_q = 0$, the integral of the term $m = 0$ is $\int _1^\infty \xi ^{-3/2}d\xi =2$. In all other cases, that is, if either $x_q \ne 0$ or $m \ne 0$, integration by parts implies

$$\begin{aligned} \begin{aligned}&\int _1^\infty \frac{1}{ \xi ^{3/2}} \, e^{- 2\pi i \frac{(x_q - m/q)^2}{4 h} \xi } d\xi \\&\quad = \frac{2}{\pi i}\frac{ q^2 h}{(m - qx_q)^2} \left( \, e^{- 2\pi i \frac{(m - qx_q)^2}{4 q^2 h}} + \frac{3}{2} \, \int _1^\infty \frac{1}{ \xi ^{5/2}} \, e^{- 2\pi i \frac{(x_q - m/q)^2}{4 h} \xi } d\xi \right) \\&\quad = O\left( \frac{q^2 h}{(m-qx_q)^2} \right) . \end{aligned} \end{aligned}$$

What is more, integrating by parts again we obtain

$$\begin{aligned} \begin{aligned} \int _1^\infty \frac{1}{ \xi ^{3/2}} e^{- 2\pi i \frac{(x_q - m/q)^2}{4 h} \xi } d\xi&= \frac{2}{\pi i} \frac{q^2 h}{ (m - qx_q)^2} \left( e^{- 2\pi i \frac{(m - qx_q)^2}{4 q^2 h}} + O\Big ( \frac{ q^2 h}{(m - qx_q)^2} \Big ) \right) . \end{aligned} \end{aligned}$$

Combining these with (61) give the desired expressions. $\square$

Remark B.2

Computations for (60) are made rigorous to avoid convergence problems by writing

$$\begin{aligned} \sum _{n \in {\mathbb {Z}}} \frac{e^{2\pi i n^2 h} - 1}{n^2}\, e^{2\pi i n^2 p/q} \, e^{2\pi i n x_0} = \lim _{\epsilon \rightarrow 0} \sum _{n \in {\mathbb {Z}}} \frac{e^{2\pi i n^2 h(1+i\epsilon )} - 1}{n^2}\, e^{2\pi i n^2 p/q} \, e^{2\pi i n x_0}. \end{aligned}$$

Proposition B.1 will allow us to give upper bounds of $\alpha _{x_0}(t)$ for general t.

Proposition B.3

Let $x_0 \in {\mathbb {Q}}$ and $t \not \in {\mathbb {Q}}$. Then, $\alpha _{x_0}(t) \le 3/4$.

Proof

Set $x_0 = P/Q$ with $P,Q \in {\mathbb {N}}$ and $(P,Q) = 1$. Let $t \not \in {\mathbb {Q}}$ and let $p_n/q_n$ be its approximations by continued fractions. It is well-known³¹ that there is a subsequence of odd denominators $q_{n_k}$. Renaming that subsequence back to $q_n$, we may assume that all $q_n$ are odd. Consequently, $|G(p_n, m, q_n)| = \sqrt{q}$ for all $m, n \in {\mathbb {N}}$. As usual, let

$$\begin{aligned} h_n = t - \frac{p_n}{q_n}, \quad |h_n| < \frac{1}{q_n^2}, \quad x_{q_n} = \min _{m \in {\mathbb {Z}}} \Big | \frac{P}{Q} - \frac{m}{q_n} \Big | = \Big | \frac{P}{Q} - \frac{m_{q_n}}{q_n} \Big |, \end{aligned}$$

and we immediately deduce that either $x_{q_n} = 0$ or $1/Q \le q_n x_{q_n} \le 1/2$. We separate cases:

Case 1:

We have $x_{q_n} = 0$ for infinitely many $n \in {\mathbb {N}}$. Rename that subsequence and rewrite (58) as

$$\begin{aligned} & \Big | R_{x_0} \Big ( \frac{p_n}{q_n} + h_n \Big ) - R_{x_0}\Big ( \frac{p_n}{q_n} \Big ) + 2\pi i h_n \Big | \nonumber \\ & \quad = 2\pi \sqrt{2} \frac{\sqrt{|h_n|}}{\sqrt{q_n}} + O\left( q_n^{3/2} h_n^{3/2} \right) \simeq \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \left( 1 + O\big ( q_n^2 h_n \big ) \right) . \end{aligned}$$

(62)

Let $\delta > 0$ which we determine later. Separate cases again:

Case 1.1.:

Suppose that $\left| 1 + O\big ( q_n^2 h_n \big ) \right| \ge \delta$ for infinitely many $n \in {\mathbb {N}}$. Then,

$$\begin{aligned} \left| R_{x_0} (t) - R_{x_0}( t - h_n ) + 2\pi i h_n \right| \ge \delta \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \ge \delta \, |h_n|^{3/4}, \end{aligned}$$

because $q_n^2 |h_n| \le 1$. Hence $| R_{x_0} (t) - R_{x_0}( t - h_n )| \ge (\delta /2) \, |h_n|^{3/4}$ for infinitely many $n \in {\mathbb {N}}$, and consequently $\alpha _{x_0}(t) \le 3/4$.

Case 1.2.:

We have $\left| 1 + O\big ( q_n^2 h_n \big ) \right| < \delta$ for all large enough n. In that case, we evaluate (62) at a point closer to $p_n/q_n$. Let $\epsilon > 0$ and write (58) for $\epsilon h_n$, so that instead of (62) we get

$$\begin{aligned} \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i \epsilon h_n \Bigg | \simeq \sqrt{\epsilon } \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \left( 1 + \epsilon O\big ( q_n^2 h_n \big ) \right) . \end{aligned}$$

Since $q_n^2 |h_n| < 1$ and the constant underlying the big-O is universal, say C, choose $\epsilon \le 1/(2C)$, in such a way that

$$\begin{aligned} \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i \epsilon h_n \Bigg | \gtrsim \frac{\sqrt{\epsilon }}{2} \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}}. \end{aligned}$$

From this and (62), we write

$$\begin{aligned} \begin{aligned} \frac{\sqrt{\epsilon }}{2} \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}}&\lesssim \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) \Bigg | + 2\pi \epsilon |h_n| \\&\le \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}( t ) \Bigg | + \Bigg | R_{x_0} ( t ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) \Bigg | + 2\pi \epsilon |h_n| \\&\lesssim \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}( t ) \Bigg | \\&\quad + \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \left( 1 + O\big ( q_n^2 h_n \big ) \right) + 2\pi (1 + \epsilon ) |h_n| \\&\le \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}( t ) \Bigg | + 2\delta \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}}. \end{aligned} \end{aligned}$$

In the last line we used the hypothesis of Case 1.2 and $|h_n| \le \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \, \frac{1}{\sqrt{q_n}}$. Hence,

$$\begin{aligned} \Bigg | R_{x_0} \Bigg ( \frac{p_n}{q_n} + \epsilon h_n \Bigg ) - R_{x_0}( t ) \Bigg | \gtrsim \left( \frac{\sqrt{\epsilon }}{2} - C\delta \right) \frac{\sqrt{|h_n|}}{\sqrt{q_n}}, \end{aligned}$$

for some $C>0$. Fix $\sqrt{\epsilon } = 4C \delta$ small enough. Writing $p_n/q_n + \epsilon h_n = t - (1-\epsilon ) h_n$ and observing that $(1-\epsilon )|h_n| \simeq |h_n|$, we conclude that

$$\begin{aligned} & \Big | R_{x_0} \big ( t - (1- \epsilon ) h_n \big ) - R_{x_0}( t ) \Big |\\ & \quad \gtrsim \delta \, \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \ge \delta \, |h_n|^{3/4} \simeq | (1-\epsilon ) h_n |^{3/4}, \quad \text { for large enough } n. \end{aligned}$$

Hence $\alpha _{x_0}(t) \le 3/4$.

Case 2:

We have $x_{q_n} \ne 0$ for all large enough $n \in {\mathbb {N}}$, hence $1/Q \le q_n x_{q_n} \le 1/2$. We now use (59) which has no leading $h^{1/2}$ term. Rewrite it,³² assuming $1/Q \le qx_q \le 1/2$, as

$$\begin{aligned} \begin{aligned}&R_{x_0}\Bigg ( \frac{p}{q} + h \Bigg ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg ) + 2\pi i h \\&\quad = 2(1 \pm i) \, \frac{G(p,0,q)}{\sqrt{q}}\, \frac{\sqrt{|h|}}{\sqrt{q}} \, q^{2} |h| \, \left[ \sum _{m \in {\mathbb {Z}}} \, e^{2\pi i (4p)^{-1} \frac{(m_q + m)^2}{q}} \, \frac{e^{- 2\pi i \frac{(m - qx_q)^2}{4 q^2 h}}}{ (m - qx_q)^2} \, + O_Q\left( q^{2} h \right) \right] . \end{aligned} \end{aligned}$$

Define the auxiliary function

$$\begin{aligned} f_q(y) = \sum _{m \in {\mathbb {Z}}} \, e^{2\pi i (4p)^{-1} \frac{ m^2 + 2m_q m}{q}} \, \frac{e^{- 2\pi i (m^2 - 2 qx_q m) y }}{ (m - qx_q)^2}. \end{aligned}$$

(63)

Take absolute values and write

$$\begin{aligned} & \Bigg | R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Bigg ( \frac{p}{q} \Bigg ) + 2\pi i h \Bigg |\nonumber \\ & \quad = 2\sqrt{2} \frac{\sqrt{|h|}}{\sqrt{q}} \, q^{2} |h| \, \left| f_q\Bigg ( \frac{1}{4q^2 h} \Bigg ) + O_Q\left( q^{2} h \right) \right| . \end{aligned}$$

(64)

We now state the properties of this function, whose proof we postpone.

Lemma B.4

Let $q \in {\mathbb {N}}$, let $p \in {\mathbb {N}}$ be coprime with q and $f_q$ defined in (63). Then,

• (a) $f_q$ is periodic of period Q.

• (b) there exists $y_0^q \in [0,Q]$ depending on q (and on p) such that $|f_q(y_0^q) | \ge 5$.

• (c) The sequence defined by $y_k^q = y_0^q + kQ$ satisfies

  $$\begin{aligned} \lim _{k \rightarrow \infty } y_k^q = \infty , \quad \text { and } \quad |f_q(y_k^q)| \ge 5, \quad \forall k \in {\mathbb {N}}. \end{aligned}$$

Remark B.5

The dependence on p of the point $y_0^q$ is irrelevant for our purposes. Indeed, once we fix $t \not \in {\mathbb {Q}}$, we get the sequence of approximations $p_n/q_n$, hence each $q_n$ comes with one and only one $p_n$. Hence, we can assume that the sequence $f_{q_n}$ only depends on $q_n$.

We now evaluate (64) at $p_n/q_n$ and $h_n = t - p_n/q_n$ and we separate two cases:

Case 2.1.:

Suppose $\limsup _{n \rightarrow \infty } q_n^2 |h_n| > 0$, so that there exists $c > 0$ and a subsequence for which $c < q_n^2 |h_n| \le 1$. Then, from (64) we get

$$\begin{aligned} \Bigg | R_{x_0}( t ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i h_n \Bigg | \ge c \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \, \left| f_{q_n}\Bigg ( \frac{1}{4q_n^2 h_n} \Bigg ) + O_Q\left( q_n^2 h_n \right) \right| . \end{aligned}$$

Fix $\delta >0$ which we later determine. Proceeding like in Case 1, we separate two cases:

Case 2.1.1.:

Suppose $\left| f_{q_n}\Big ( \frac{1}{4q_n^2 h_n} \Big ) + O_Q\left( q_n^2 h_n \right) \right| \ge \delta$ for infinitely many n. Then,

$$\begin{aligned} \Bigg | R_{x_0}( t ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i h_n \Bigg | \ge c \delta \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \ge c\delta |h_n|^{3/4} \end{aligned}$$

for infinitely many n, which implies $\alpha _{x_0}(t) \le 3/4$.

Case 2.1.2.:

Suppose $\left| f_{q_n}\Big ( \frac{1}{4q_n^2 h_n} \Big ) + O_Q\left( q_n^2 h_n \right) \right| < \delta$ for all large enough n. Then, let $\epsilon _n$ be a sequence which we determine later, and define $\eta _n = \epsilon _n /q_n^2$. Observe that $\eta _n = \epsilon _n |h_n| / (q_n^2 |h_n|) \simeq \epsilon _n |h_n|$. Evaluate (64) at $\eta _n$ to get

$$\begin{aligned} \begin{aligned}&\Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i \eta _n \Bigg |\\&\quad = 2\sqrt{2} \frac{\sqrt{\eta _n}}{\sqrt{q_n}} \, q_n^2 \eta _n \, \left| f_q\Bigg ( \frac{1}{4q_n^2 \eta _n} \Bigg ) + O_Q\left( q_n^{2} \eta _n \right) \right| \\&\quad = 2\sqrt{2} \, \epsilon _n \, \frac{\sqrt{\eta _n}}{\sqrt{q_n}} \, \left| f_{q_n}\Bigg ( \frac{1}{4 \epsilon _n} \Bigg ) + O_Q\left( \epsilon _n \right) \right| . \end{aligned} \end{aligned}$$

Fix $k \in {\mathbb {N}}$ large enough and set $\epsilon _n = 1/(4y_k^{q_n})$. Then, by Lemma B.4 (c),

$$\begin{aligned} \Bigg | f_{q_n}\Bigg ( \frac{1}{4 \epsilon _n} \Bigg ) \Bigg | = \left| f_{q_n}( y_K^{q_n}) \right| \ge 5, \quad \forall n \text { large enough.} \end{aligned}$$

Since $\epsilon _n \simeq 1/(kQ)$, if $k \in {\mathbb {N}}$ is large enough we get $O_Q\left( \epsilon _n \right) \le C_Q \epsilon _n \le 1$. In particular, $|h_n| \simeq _Q k \eta _n$. Therefore,

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i \eta _n \Bigg | \ge \epsilon _n\, \frac{\sqrt{\eta _n}}{\sqrt{q_n}} \simeq \epsilon _n^{3/2} \frac{\sqrt{|h_n|}}{\sqrt{q_n}}. \end{aligned}$$

With this, and using the assumption of this case in (64), we write

$$\begin{aligned} \begin{aligned} \epsilon _n^{3/2} \frac{\sqrt{|h_n|}}{\sqrt{q_n}}&\lesssim \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg )\Bigg | + 2\pi \eta _n \\&\le \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}(t) \Bigg | + \Bigg | R_{x_0}(t) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg )\Bigg | + 2\pi \eta _n \\&\lesssim \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}(t) \Bigg | + \delta \frac{\sqrt{|h_n|}}{\sqrt{q_n}} + 2\pi ( |h_n| + \eta _n) \\&\lesssim \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}(t) \Bigg | + \delta \frac{\sqrt{|h_n|}}{\sqrt{q_n}}, \end{aligned} \end{aligned}$$

for large enough n, where in the last line we used $\eta _n \simeq |h_n|/k \le |h_n|$ and $|h_n| \le \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \, \frac{1}{\sqrt{q_n}} \ll \frac{\sqrt{|h_n|}}{\sqrt{q_n}}$. Since $\epsilon _n \simeq _Q 1/k$, set $\delta = 1/(c_Qk^{3/2})$ with some small enough $c_Q>0$ so that

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}(t) \Bigg | \gtrsim \Big ( \epsilon _n^{3/2} - C\delta \Big ) \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \gtrsim \delta \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \ge \delta |h_n|^{3/4}. \end{aligned}$$

Write $p_n/q_n + \eta _n = t - (h_n - \eta _n)$. Since $|h_n - \eta _n| \le 2|h_n|$, we get

$$\begin{aligned} \Bigg | R_{x_0}\Big ( t - (h_n - \eta _n) \Big ) - R_{x_0}(t) \Bigg | \ge \delta |h_n|^{3/4} \gtrsim \delta |h_n - \eta _n|^{3/4}, \quad \text { for large enough } n, \end{aligned}$$

which implies $\alpha _{x_0}(t) \le 3/4$.

Case 2.2.:

Suppose $\lim _{n \rightarrow \infty } q_n^2 |h_n| = 0$. In this case, the term $q_n^2 |h_n|$ in (64) tends to zero, which kills the desired $|h_n|^{3/4}$ that came from $\sqrt{h_n}/\sqrt{q_n}$. To counter that, define $\eta _n = \epsilon _n / q_n^2$ as in Case 2.1.2. By (64),

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i \eta _n \Bigg | = 2\sqrt{2} \epsilon _n \frac{\sqrt{\eta _n}}{\sqrt{q_n}} \, \left| f_{q_n}\Bigg ( \frac{1}{4\epsilon _n} \Bigg ) + O_Q\left( \epsilon _n \right) \right| . \end{aligned}$$

Fix $k \in {\mathbb {N}}$ large enough and set $\epsilon _n = 1/(4y_k^{q_n})$. Then,

$$\begin{aligned} \left| f_{q_n}\Bigg ( \frac{1}{4\epsilon _n} \Bigg ) \right| = \left| f_{q_n}\Big ( y_k^{q_n} \Big ) \right| \ge 5, \quad \text { and } \quad O_Q(\epsilon _n) \le C_Q \epsilon _n = \frac{C_Q}{4y_k^{q_n}} \simeq \frac{C_Q}{kQ} \le 1, \end{aligned}$$

so

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) + 2\pi i \eta _n \Bigg | \ge \epsilon _n \frac{\sqrt{\eta _n}}{\sqrt{q_n}}. \end{aligned}$$

With this and (64), we can write

$$\begin{aligned} \begin{aligned} \epsilon _n \frac{\sqrt{\eta _n}}{\sqrt{q_n}}&\le \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) \Bigg | + 2\pi \eta _n \\&\le \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) -R_{x_0}(t) \Bigg | + \Bigg | R_{x_0}(t) - R_{x_0}\Bigg ( \frac{p_n}{q_n} \Bigg ) \Bigg | + 2\pi \eta _n \\&\lesssim \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) -R_{x_0}(t) \Bigg | + \frac{\sqrt{|h_n|}}{\sqrt{q_n}} \, q_n^2 |h_n| + 2\pi (\eta _n + |h_n|). \end{aligned} \end{aligned}$$

Since $\lim _{n \rightarrow \infty } q_n^2 |h_n|=0$ implies $h_n = o(\eta _n)$, and $\eta _n = \frac{\sqrt{\eta _n}}{\sqrt{q_n}} \frac{\sqrt{\epsilon _n}}{\sqrt{q_n}}$, we get

$$\begin{aligned} \Bigg | R_{x_0}\Bigg ( \frac{p_n}{q_n} + \eta _n \Bigg ) -R_{x_0}(t) \Bigg | \gtrsim \Bigg ( \epsilon _n - q_n^2 h_n - \frac{\sqrt{\epsilon _n}}{ \sqrt{q_n} } \Bigg ) \frac{\sqrt{\eta _n}}{\sqrt{q_n}} \ge \frac{\epsilon _n}{2} \, \frac{\sqrt{\eta _n}}{\sqrt{q_n}} = \frac{\epsilon _n^{3/4}}{2} \, \eta _n^{3/4}. \end{aligned}$$

Write $p_n/q_n + \eta _n = t + (\eta _n - h_n)$. Recalling $\epsilon _n \simeq 1/(kQ)$ for all n, and since $h_n = o(\eta _n)$ implies $|\eta _n - h_n| \simeq \eta _n$, we conclude

$$\begin{aligned} \Bigg | R_{x_0}\Big ( t + (\eta _n - h_n) \Big ) -R_{x_0}(t) \Bigg | \ge \frac{\epsilon _n^{3/4}}{2} \, \eta _n^{3/4} \simeq _Q |\eta _n - h_n|^{3/4}, \end{aligned}$$

and therefore $\alpha _{x_0}(t) \le 3/4$. $\square$

We now prove Lemma B.4.

Proof of Lemma B.4

(a) Write first

$$\begin{aligned} 0 \ne qx_q = q \min _{m \in {\mathbb {Z}}} \Big |x_0 - \frac{m}{q} \Big | = q \, \Big |x_0 - \frac{m_q}{q} \Big | = \frac{1}{Q} \, | Pq - Q m_q | = \frac{m_q'}{Q}, \end{aligned}$$

where $m_q' = |Pq - Qm_q| \in {\mathbb {N}} {\setminus } \{0\}$. Hence, the variable y in (63) only appears in

$$\begin{aligned} e^{2\pi i (m^2 - 2qx_q m) \, y } = e^{2\pi i (Qm^2 - 2 m_q' m) \frac{y}{Q} }, \end{aligned}$$

which is Q-periodic. Hence $f_q$ has period Q.

(b) Split the sum in $f_p$ in the terms $m=0,1$ and the rest,

$$\begin{aligned} f_q(y) = \frac{1}{ (qx_q)^2} + e^{2\pi i (4p)^{-1} \frac{ 1 + 2m_q }{q}} \, \frac{e^{- 2\pi i (1 - 2 qx_q) y }}{ (1 - qx_q)^2} + \text {Error} \end{aligned}$$

where $1/Q \le qx_q \le 1/2$ implies

$$\begin{aligned} \begin{aligned} |\text {Error}|&= \Bigg | \sum _{m \ne 0,1} \, e^{2\pi i (4p)^{-1} \frac{ m^2 + 2m_q m}{q}} \, \frac{e^{- 2\pi i (m^2 - 2 qx_q m) y }}{ (m - qx_q)^2} \Bigg | \\&\le \sum _{m=2}^\infty \frac{1}{ (m - qx_q)^2} + \sum _{m=1}^\infty \frac{1}{ (m + qx_q)^2} \le \sum _{m=2}^\infty \frac{1}{ (m - 1/2)^2} + \sum _{m=1}^\infty \frac{1}{ m^2}\\&= \frac{\pi ^2}{2} - 4 + \frac{\pi ^2}{6} \le 3. \end{aligned} \end{aligned}$$

On the other hand, the phase in

$$\begin{aligned} e^{2\pi i (4p)^{-1} \frac{ 1 + 2m_q }{q}} \, e^{- 2\pi i (1 - 2 qx_q) y }. \end{aligned}$$

is continuous, decreasing, and Q-periodic. That implies that there exists $y_0^q \in [0,Q]$ such that $e^{2\pi i (4p)^{-1} \frac{ 1 + 2m_q }{q}} \, e^{- 2\pi i (1 - 2 qx_q) y_0^q } = 1$, and consequently,

$$\begin{aligned} |f_q(y_0^q)| \ge \frac{1}{ (qx_q)^2} + \frac{1}{ (1 - qx_q)^2} - 3 \ge \frac{1}{ (1/2)^2} + \frac{1}{ (1 - 1/2)^2} - 3 = 5 \end{aligned}$$

because in (0, 1) the function $1/x^2 + 1/(1-x)^2$ has a minimum in $x=1/2$.

(c) The fact that $f_q$ is Q-periodic implies that $|f_q(y_n^q)| = |f_q(y_0^q + nQ)| = |f_q(y_0^q )| \ge 5$. $\quad \square$

We now complete the proof of Proposition 3.5.

Proposition B.6

Let $x_0 \in {\mathbb {R}}$ and $t \in {\mathbb {Q}}$. If $\alpha _{x_0}(t) \ne 1/2$, then $\alpha _{x_0}(t) = 3/2$.

Proof

By Proposition B.1, $\alpha _{x_0}(t) = 1/2$ happens only if $x_q = 0$ and $G(p,m_q,q) \ne 0$.

• If $x_q = 0$ and $G(p,m_q,q) = 0$, then $x_0 \in {\mathbb {Q}}$ and $q \in 2{\mathbb {N}}$. From (58) and the fact that

  $$\begin{aligned} & G(p,m,q)\nonumber \\ & \quad = \left\{ \begin{array}{ll} e^{2\pi i (4p)^{-1} m^2/q} \, G(p,0,q), & q \text { odd,}\\ e^{2\pi i p^{-1} (m/2)^2/q} \, G(p,0,q), & q \equiv 0 \pmod {4} \text { and } m \text { even}, \\ e^{2\pi i p^{-1} ((m-1)/2)^2/q} e^{2\pi i p^{-1} ((m-1)/2)/q} \, G(p,1,q), & q \equiv 2 \pmod {4} \text { and } m \text { odd}, \end{array} \right. \nonumber \\ \end{aligned}$$

  (65)

  and $G(p,m,q) = 0$ otherwise, we have

  $$\begin{aligned} & R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h\\ & \quad = 2(1 \pm i)\, q^{3/2} |h|^{3/2} \,\sum _{m \text { odd}} \, \frac{ G(p,m_q + m,q)}{\sqrt{q}} \, \frac{e^{- 2\pi i \frac{m^2}{4 q^2 h}}}{ m^2} + O\left( q^{7/2} h^{5/2} \right) . \end{aligned}$$

  It suffices to find a sequence $y_k \rightarrow \infty$ such that $|g(y_k)| \ge c > 0$ for some $c > 0$, where

  $$\begin{aligned} g(y) = \sum _{m \text { odd}} \, \frac{ G(p,m_q + m,q)}{\sqrt{q}} \, \frac{e^{- 2\pi i m^2 y}}{ m^2}, \end{aligned}$$

  because that way, defining $h_k = 1/(4q^2 y_k)$, we get

  $$\begin{aligned} \Big | R_{x_0}\Big ( \frac{p}{q} + h_k \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h_k \Big | \gtrsim _q h_k^{3/2} |g(y_k)| - O(h_k^{5/2}) \gtrsim _q h_k^{3/2} \end{aligned}$$

  for all k large enough, hence $\alpha _{x_0}(t) \le 3/2$. So let us find that sequence $y_k$. According to (65), if $q \equiv 0 \pmod {4}$, by symmetry we can write

  $$\begin{aligned} \begin{aligned} g(y)&= \frac{G(p,0,q)}{\sqrt{q}} \, \sum _{m \ge 0 \text { odd}} \frac{e^{- 2\pi i m^2 y}}{ m^2} \Big ( e^{2\pi i p^{-1} \left( \frac{m_q + m}{2} \right) ^2 \frac{1}{q} } + e^{2\pi i p^{-1} \left( \frac{m_q - m}{2} \right) ^2 \frac{1}{q} } \Big ) \\&= 2 \, \frac{G(p,0,q)}{\sqrt{q}} \, e^{2\pi i p^{-1} \frac{m_q^2}{4q} } \, \sum _{m \ge 0 \text { odd}} \frac{e^{- 2\pi i m^2 (y - \frac{ p^{-1} }{4q} ) }}{ m^2} \cos \left( 2\pi \frac{p^{-1} m_q }{2q} \, m \right) . \end{aligned} \end{aligned}$$

  On the other hand, if $q \equiv 2 \pmod {4}$, then

  $$\begin{aligned} \begin{aligned} g(y)&= \frac{G(p,1,q)}{\sqrt{q}} \, \sum _{m \ge 0 \text { odd}} \frac{e^{- 2\pi i m^2 y}}{ m^2} \\&\quad \times \Big ( e^{2\pi i p^{-1} \Big [ \left( \frac{m_q + m - 1}{2} \right) ^2 + \frac{m_q + m - 1}{2} \Big ] \frac{1}{q} } + e^{2\pi i p^{-1} \Big [ \left( \frac{m_q - m - 1}{2} \right) ^2 + \frac{m_q - m - 1}{2} \Big ] \frac{1}{q} } \Big ) \\&= 2 \, \frac{G(p,1,q)}{\sqrt{q}} \, e^{2\pi i p^{-1} \frac{(m_q - 1)^2 + 2(m_q - 1)}{4q} } \, \sum _{m \ge 0 \text { odd}} \frac{e^{- 2\pi i m^2 (y - \frac{ p^{-1} }{4q} ) }}{ m^2} \, \cos \left( 2\pi \frac{ p^{-1} m_q }{2q} \, m \right) . \end{aligned} \end{aligned}$$

  Choose the sequence $y_k = p^{-1}/(4q) + k$ for $k \in {\mathbb {N}}$. Then, since $x_q = |x_0 - m_q/q| = 0$ implies $x_0 = m_q/q$, but also $x_0 = P/Q$ in its reduced form, we get

  $$\begin{aligned} |g(y_k)| \simeq \left| \sum _{m = 0}^\infty \frac{\cos \left( \pi \frac{p^{-1} P }{Q} \, (2m+1) \right) }{ (2m+1)^2} \right| , \quad \forall k \in {\mathbb {N}}. \end{aligned}$$

  (66)

  Define the Fourier series

  $$\begin{aligned} G(z) = \sum _{m = 0}^\infty \frac{\cos \left( (2m+1) \pi z \right) }{ (2m+1)^2} = \frac{ \pi ^2}{8} (1 - |2z|) \quad z \in (-1,1), \end{aligned}$$

  so that, after extending periodically to ${\mathbb {R}}$, in view of (66), we have $|g(y_n)| = |G(p^{-1}P/Q)|$ for all $n \in {\mathbb {N}}$. Observe that the only zeros of G are $(2m+1)/2$ for $m \in {\mathbb {Z}}$. We separate two cases again. If $q \equiv 0 \pmod {4}$, by (65) $m_q$ must be odd. Then $Qm_q = Pq$ implies $4 \mid Q$, hence both $p^{-1}$ and P are odd. We deduce $p^{-1}P/Q \ne (2m+1)/2$ for any $m \in {\mathbb {Z}}$, because otherwise $p^{-1}P = (2m+1) Q/2$ for some m, so $p^{-1}P$ would be even. If $q \equiv 2 \pmod {4}$, then $m_q$ is even and $Q(m_q/2) = P(q/2)$ implies that Q is odd. Hence $p^{-1}P/Q \ne (2m+1)/2$ for any $m\in {\mathbb {Z}}$. In both cases, this implies that $|g(y_k)| = |G(p^{-1}P/Q)| \ne 0$ for all k, which is what we wanted to prove.

• If $x_q \ne 0$, according to (59) we get

  $$\begin{aligned} \begin{aligned}&\Big | R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big )+ 2\pi i h\Big | \\&\quad \simeq \Big | (qh)^{3/2} \sum _{m \in {\mathbb {Z}}} \, \frac{ G(p,m_q + m,q)}{\sqrt{q}} \, \frac{e^{- 2\pi i \frac{(m - qx_q)^2}{4 q^2 h}}}{ (m - qx_q)^2} \, + O\left( q^{7/2} h^{5/2} \right) \Big | \end{aligned} \end{aligned}$$

  (67)

  because $0 < qx_q \le 1/2$. If q is odd, we use (65) and the definition of $f_q$ in (63) to write

  $$\begin{aligned} \Big |R_{x_0}\Big ( \frac{p}{q} + h \Big ) - R_{x_0}\Big ( \frac{p}{q} \Big ) + 2\pi i h \Big | \simeq q^{3/2} h^{3/2} \,\Big | f_q\Big ( \frac{1}{4q^2h} \Big ) + O\big ( q^2h\big ) \Big |.\nonumber \\ \end{aligned}$$

  (68)

  With the definition of $y_k^q$ in Lemma B.4, choose the sequence $h_k = 1/(4q^2y_k^q)$ that tends to zero and for which $|f_q(1/(4q^2h_k^q))| = |f_q(y^q_k)| \ge 5$. This and (68) show that $\alpha _{x_0}(t) = 3/2$. When q is even, by (65), the sum in (67) only has either even or odd terms. The main term is $m=0$ if even terms survive, and $m=1$ if odd terms survive, and crude estimates in the error suffice to conclude. $\quad \square$