1 Introduction

The theory of almost periodic functions initiated by Bohr around a hundred years ago (see [1,2,3]) has been very widely developed. It is connected with the fact that such functions have applications in many fields. It is worth to mention here that almost periodic patterns, which correspond to almost periodic measures, describe the structure of quasicrystals (see e.g., [4] for more details).

There are many other classes of almost periodic functions which constitute generalizations of Bohr almost periodic functions (see e.g., [5,6,7] and [8]). In our opinion, among those generalizations, Levitan almost periodic functions (briefly: LAP functions) and almost periodic functions in view of the Lebesgue measure (briefly: \(\mu\)-a.p. functions) seem to be very interesting. Let us recall that in general, \(\mu\)-a.p. functions do not have to be continuous or even locally integrable in the Lebesgue sense and this fact leads to many problems while investigating them. As regards LAP functions let us recall that they do not have to be bounded or even do not have to satisfy the condition

$$\begin{aligned} \limsup _{T \rightarrow +\infty } \quad \frac{1}{2T}\int _{-T}^{T}|f(t)|dt<+\infty , \end{aligned}$$

although they are continuous. Therefore, an investigation of this class of functions is quite difficult. The interested reader can found a comparison of these two above-mentioned classes of almost periodic functions in the paper [9].

There is a rich literature concerning almost periodic as well as almost automorphic solutions to some type of differential equations (see e.g., [10,11,12,13,14,15] and [16]). Levitan almost periodic solutions and almost periodic in view of the Lebesgue measure solutions to linear differential equations of the first order have been investigated, for example, in the papers [17] and [9], respectively. In these papers an important role has played the investigation of the behavior of these classes of functions under convolution. In particular, the authors of these papers have examined the asymptotic behavior and the convolutions with functions

$$\begin{aligned} f(x)=\frac{1}{2+\cos {(x)}+\cos {(\alpha x)}}\quad \text {for }x\in {\mathbb {R}}, \end{aligned}$$
(1)

where \(\alpha \notin {\mathbb {Q}}\). Let us recall that this function is a classical example of continuous unbounded LAP function (it is also an example of \(\mu\)-a.p. function).

As it was shown in the papers [17] and [9], the investigation concerning the asymptotic behavior of function (1) as well as its convolution with the function \(g_{\lambda } :{\mathbb {R}} \rightarrow {\mathbb {R}}\) (\(\lambda <0\)), given by the formula

$$\begin{aligned} g_\lambda (x)= {\left\{ \begin{array}{ll} e^{\lambda x} \quad &{} \hbox { for}\ x\ge 0,\\ 0 \quad &{} \hbox { for}\ x<0. \end{array}\right. } \end{aligned}$$
(2)

plays an important role in proving of the results concerning the solutions to linear differential equations in the space of LAP functions or in the space of \(\mu\)-a.p. functions. The main goal of this paper is to continue that investigation (conducted also in the paper [18]). In particular, we prove some estimations for functions (1), where \(\alpha\) is an irrational number of finite irrationality measure or a Liouville number. These results are contained in Sect. 3. Next, in Sect. 4, we focus on the set of all irrational numbers for which the convolution of function (1) with function (2) exists as well as on its complement. We provide its characterization from the topological and set-theoretical point of view. In particular, it is a set with the cardinality of the continuum which is dense in \({\mathbb {R}}\) and its complement in \({\mathbb {R}}\) is of the second Baire category.

2 Preliminaries

At the beginning of this section we collect basic definitions and results concerning Levitan almost periodic functions and almost periodic functions in view of the Lebesgue measure.

One of several equivalent definitions of Levitan almost periodic functions uses the notion of an \([N, \varepsilon ]\)- almost period of a function which is defined as follows.

Definition 1

([7]) A real number \(\tau\) is said to be an \([N, \varepsilon ]\)- almost period of a function \(f:{\mathbb {R}} \rightarrow {\mathbb {R}}\) (briefly: \([N, \varepsilon ]\)- a.p.) if \(|f(x+\tau )-f(x)|< \varepsilon\) for \(|x|< N\).

Using the notion of an \([N, \varepsilon ]\)- almost period of a function one can define Levitan almost periodic function as follows.

Definition 2

([7]) A continuous function \(f:{\mathbb {R}} \rightarrow {\mathbb {R}}\) is said to be Levitan almost periodic if for every \(N, \varepsilon >0\) there exist \(\lambda _1,\ldots , \lambda _p \in {\mathbb {R}}\) and \(\delta >0\) such that every number \(\tau\) which satisfies inequalities \(|\lambda _r \tau |<\delta\) (mod \(2\pi\)) for \(r=1, 2, \ldots , p\), is an \([N, \varepsilon ]\)- a.p. of the function f.

Remark 1

Inequality \(|x|<\delta\) (mod \(2\pi\)) means that there exists \(k\in {\mathbb {Z}}\) such that \(-\delta<x-2k\pi <\delta\).

Let \(\Sigma\) be the \(\sigma\)-algebra of subsets of \(\mathbb {R}\) which are measurable in the Lebesgue sense, \(\mu\) - be the Lebesgue measure on \(\Sigma\), and let \(\mathcal {X}\) be the space of all \(\Sigma\)-measurable functions \(f :\mathbb {R}\rightarrow \mathbb {R}\). For \(f, g \in \mathcal {X}\) we set \(f=g \Leftrightarrow f(x)=g(x)\) for \(\mu\)-a.e. \(x\in {\mathbb {R}}\).

For a function \(f:\mathbb {R}\rightarrow \mathbb {R}\), by \(f_{\tau }\) for \(\tau \in {\mathbb {R}}\), we denote the function \(f_{\tau }:\mathbb {R}\rightarrow \mathbb {R}\), defined by the formula \(f_{\tau }(x)=f(x+\tau )\), for \(x\in {\mathbb {R}}\).

Now, we recall the notion of \((\varepsilon , \eta )\)-almost period of a function \(f \in \mathcal {X}\) and the definition of an almost periodic function in view of the Lebesgue measure.

Definition 3

([19]) Let \(f \in \mathcal {X}\). If for \(\varepsilon , \eta >0\) we have \(D(\eta ; f_{\tau }, f)\le \varepsilon\), then the real number \(\tau\) is said to be \((\varepsilon , \eta )\)-almost period (briefly: \((\varepsilon , \eta )\)-a.p.) of the function f.

By \(E\{\varepsilon , \eta ; f\}\) we will denote the set of all \((\varepsilon , \eta )\)-a.p. of the function f, that is

$$\begin{aligned} E\{\varepsilon , \eta ; f\}{:}{=}\bigl \{\tau \in {\mathbb {R}}:\sup _{u \in \mathbb {R}}\mu \bigl (\{x \in [u, u+1]:|f(x+\tau )-f(x)|\ge \eta \}\bigr )\le \varepsilon \bigr \}. \end{aligned}$$

Definition 4

([19]) A function \(f\in \mathcal {X}\) is said to be almost periodic in view of the Lebesgue measure \(\mu\), if for arbitrary numbers \(\varepsilon , \eta >0\), the set \(E\{\varepsilon , \eta ; f\}\) is relatively dense. By \(\widetilde{M}\) we will denote the set of all \(\mu\)-a.p. functions.

One can define almost periodic functions in view of the Lebesgue measure in the following equivalent way.

Definition 5

([5]) Fix \(d>0\). A function \(f \in \mathcal {X}\) is said to be \(\mu\)-a.p. or measurably almost periodic, if for any \(\varepsilon >0\), the set

$$\begin{aligned} E_{d}\{\varepsilon ; f\}{:}{=}\bigl \{\tau \in {\mathbb {R}}:{\forall u\in {\mathbb {R}} \quad \mu \bigl (\{x \in [u, u+d]:|f(x+\tau )-f(x)|\ge \varepsilon \}\bigr )<\varepsilon d}\bigr \} \end{aligned}$$

is relatively dense.

It is well-know that if f is a bounded \(\mu\)-a.p. function, then f is Stepanov almost periodic. Moreover, every bounded and uniformly continuous \(\mu\)-a.p. function is Bohr almost periodic (see [8]).

For a real number x, the symbols \(\bigl \lfloor x\bigr \rfloor\) and \(\{x\}\) will denote its integer part (éntier) and its fractional part, respectively. The symbol ||x|| will denote the Euclidean distance of x to the nearest integer. It is easy to check that \(||x||=|\{x+\frac{1}{2}\}-\frac{1}{2}|\). Now, we are going to define the relations “\(\ll\), \(\gg\), \(\approx\).”

Definition 6

Let us consider functions \(f,g:X\rightarrow \mathbb {R^+}\), where X denotes \(\mathbb {R}\), \(\mathbb {R}^+\) or \(\mathbb {N}\). We say that \(f(x)\ll g(x)\) if

$$\begin{aligned} \exists _{C>0}\exists _{x_1\in X}\forall _{x>x_1}\,\,f(x)\le Cg(x). \end{aligned}$$

We write

$$\begin{aligned} g\gg f \quad \text {when}\quad f\ll g. \end{aligned}$$

Finally

$$\begin{aligned} f\approx g\quad \text {if and only if}\quad f\ll g \quad \text {and}\quad f\gg g. \end{aligned}$$

From the above definition it follows that "\(\approx\)" is an equivalence relation. Moreover, for \(f \approx g\), if the limit, upper limit and lower limit of f at \(+\infty\) is equal to 0 or \(+\infty\), then the limit, upper limit or lower limit of g is equal to 0 or \(+\infty\), respectively.

Now, let us recall some definitions and results from the number theory.

Theorem 1

([20]) If \(\alpha\) is an irrational algebraic number of degree n, then there exists a constant \(c>0\) such that for every \(p\in {\mathbb {Z}}\) and \(q\in {\mathbb {N}}\) we have

$$\begin{aligned} \biggl |\alpha -\frac{p}{q}\biggr |>\frac{c}{q^n}. \end{aligned}$$

Definition 7

Let \(\alpha\) be a real number, and \({\textbf {R}}\) be the set of positive real numbers \(\nu\) for which

$$\begin{aligned} 0<\biggl |\alpha -\frac{p}{q}\biggr |<\frac{1}{q^{\nu }} \end{aligned}$$

has (at most) finitely many solutions \(\frac{p}{q}\) for \(p,q\in {\mathbb {Z}}\), \(q\ne 0\). Then, the irrationality measure, sometimes called the Liouville-Roth constant or the irrationality exponent, is defined as the threshold at which the Liouville Approximation Theorem kicks in and \(\alpha\) is no longer approximable by rationals numbers, that is

$$\begin{aligned} \nu (\alpha ):=\inf _{\nu \in {\textbf {R}}} \nu . \end{aligned}$$

If the set \({\textbf {R}}\) is empty, then \(\nu (\alpha )\) is defined to be \(\nu (\alpha )=+\infty\), and \(\alpha\) is called a Liouville number.

Now, let us recall a classical definition of a Liouville number.

Definition 8

A real number \(\alpha\) is said to be a Liouville number, if for every \(n\in {\mathbb {N}}\) there exist \(p, q\in {\mathbb {Z}}, q>1\) such that

$$\begin{aligned} 0<\biggl |\alpha -\frac{p}{q}\biggr |<\frac{1}{q^n}. \end{aligned}$$

Theorem 2

(Roth, [20]) If \(\alpha\) is an irrational algebraic number, then for every \(\varepsilon >0\) there exists a constant \(C=C(\alpha , \varepsilon )\) such that for all \(p\in {\mathbb {Z}}\), \(q\in {\mathbb {N}}\) holds

$$\begin{aligned} \bigl |\alpha -\frac{p}{q}\bigr |\ge \frac{C}{q^{2+\varepsilon }}. \end{aligned}$$

Remark 2

It is known that if \(\alpha\) is a real number, then

  • \(\nu (\alpha )=1\) if \(\alpha\) is rational;

  • \(\nu (\alpha )=2\) if \(\alpha\) is an irrational algebraic number;

  • \(\nu (\alpha )\ge 2\) if \(\alpha\) is transcendental.

Now, we are going to collect basic definitions and facts concerning continued fractions which will be needed in the sequel. As usual, every infinite sequence of real numbers \(\langle a_{0}; a_{1}, a_{2}, \ldots \rangle\) will be called infinite continued fraction, provided \(a_{j}\ge 1\) \((j=1, 2, \ldots )\). The numbers \(a_{1}, a_{2}, \ldots\) are said to be quotients of the continued fraction, while the number \(r_{k}=[a_{0}; a_{1}, \ldots , a_{k}]\) (that is, the value of the finite continued fraction \([a_{0}; a_{1}, a_{2}, \ldots, a_{n}]\)) is said to be its kth convergent. If all the quotients are positive integers and \(a_{0}\) is an integer, then such a continued fraction is said to be an arithmetic continued fraction. One can easily prove that the limit \(\lim _{n \rightarrow \infty }[a_{0}; a_{1} \ldots , a_{n}]\) exists; it is said to be the value of the continued fraction \(\langle a_{0}; a_{1}, a_{2}, \ldots \rangle\) and it is denoted by \([a_{0}; a_{1}, a_{2}, \ldots ]\).

For better calculation of convergents of continued fractions we recall the definition of two sequences of polynomials \((P_{n})_{n=-1}^{\infty }\) and \((Q_{n})_{n=-1}^{\infty }\) which coefficients are nonnegative integers.

Definition 9

Set

$$\begin{aligned} P_{-1}=1, \qquad Q_{-1}=0, \qquad P_{0}(x_{0})=x_{0}, \qquad Q_{0}(x_{0})=1, \end{aligned}$$

and

$$\begin{aligned} P_{k+1}(x_{0}, \ldots , x_{k+1})=x_{k+1}P_{k}(x_{0}, \ldots , x_{k})+P_{k-1}(x_{0}, \ldots , x_{k-1}), \end{aligned}$$

and

$$\begin{aligned} Q_{k+1}(x_{0}, \ldots , x_{k+1})=x_{k+1}Q_{k}(x_{0}, \ldots , x_{k})+Q_{k-1}(x_{0}, \ldots , x_{k-1}) \end{aligned}$$

for \(k\in \mathbb {N}_{0}\).

Lemma 1

([20]) If \(\alpha\) is the value of an infinite continued fraction \(\langle a_{0}; a_{1}, a_{2}, \ldots \rangle\), and \(r_{n}=\frac{P_n}{Q_n}\) is its nth convergent, then

$$\begin{aligned} \frac{1}{2Q_{n}Q_{n+1}}<\Bigl |\alpha -\frac{P_{n}}{Q_{n}}\Bigr |<\frac{1}{Q_{n}Q_{n+1}}<\frac{1}{Q_{n}^2}. \end{aligned}$$

If the continued fraction \(\langle a_{0}; a_{1}, a_{2}, \ldots \rangle\) is arithmetic, then the fraction \(\frac{P_{n}}{Q_{n}}\) is irreducible.

Definition 10

([20]) We say that a rational number \(\frac{a}{b}\) (\(b>0\); a and b are coprime) is a best rational approximation of a real number \(\alpha\), if for all fractions \(\frac{a'}{b'}\) with positive integers denominators which are less than b, the following inequality holds

$$\begin{aligned} |b\alpha -a|<|b'\alpha -a'|. \end{aligned}$$

Theorem 3

([20]) Let \(\langle a_{0}; a_{1}, a_{2}, \ldots \rangle\) be an arithmetic infinite continued fraction of irrational value \(\alpha\). Then, every rational number, being the best rational approximation of \(\alpha\), is equal to some convergent of this fraction and, conversely, for \(k\ge 1\), k-th convergent of this fraction is the best rational approximation of the number \(\alpha\).

Theorem 4

([20]) Every irrational number can be expressed in just one way as an arithmetic continued fraction.

3 Asymptotic behavior of some almost periodic functions

The theorem given below describes the asymptotic behavior of the function of form (1), where \(\alpha\) is not a Liouville number.

Theorem 5

(cf. [21]) If \(\alpha\) is an irrational number such that \(\nu (\alpha )=\nu <+\infty\), then for every \(\varepsilon >0\)

$$\begin{aligned} \frac{1}{2+\cos {(x)}+\cos {(\alpha x)}} \ll x^{2(\nu -1)+\varepsilon }. \end{aligned}$$

Proof

Let \(\varepsilon >0\). For \(x>\frac{1}{2}\pi\) let

$$\begin{aligned} P(x):=\biggl \lfloor \frac{x\alpha }{\pi }+\frac{1}{2}\biggr \rfloor \qquad \text {and} \qquad Q(x):=\biggl \lfloor \frac{x}{\pi }+\frac{1}{2}\biggr \rfloor . \end{aligned}$$

Moreover, let \(d(x):=\max \{d_1(x), d_2(x)\}\), where

$$\begin{aligned} d_1(x):=|\alpha x-P(x)\pi | \qquad \text {and} \qquad d_2(x):=|x-Q(x)\pi |. \end{aligned}$$

From the definition we have \(Q(x)\ge 1\) and \(0 < d(x)\le \frac{\pi }{2}\). Moreover,

$$\begin{aligned} | \alpha Q(x)\pi -P(x)\pi |&\le | \alpha Q(x)\pi - \alpha x| + |\alpha x- P(x)\pi |\\&= |\alpha |d_2(x)+d_1(x) \ll d(x), \end{aligned}$$

which implies that

$$\begin{aligned} \biggl |\alpha -\frac{P(x)}{Q(x)}\biggr | \ll \frac{d(x)}{Q(x)}. \end{aligned}$$

Moreover,

$$\begin{aligned} \frac{1}{Q(x)^{\nu +\varepsilon }} \ll \biggl |\alpha -\frac{P(x)}{Q(x)}\biggr |. \end{aligned}$$

Then,

$$\begin{aligned} \frac{1}{d(x)}\ll \bigl (Q(x)\bigr )^{\nu +\varepsilon -1}. \end{aligned}$$
(3)

On the other hand, because \(Q(x)\ge 1\) and \(0 \le d_2(x) \le \frac{\pi }{2}\), so we infer that

$$\begin{aligned} \frac{1}{2}Q(x)\le Q(x)-\frac{1}{2}\le \frac{x}{\pi } \end{aligned}$$

and therefore we have

$$\begin{aligned} Q(x)\ll x. \end{aligned}$$
(4)

Moreover,

$$\begin{aligned} 1+\cos (y+k\pi )\ge 1-\cos (|y|)\qquad \text {for }y\in \bigl [-\frac{\pi }{2}, \frac{\pi }{2}\bigr ], k\in {\mathbb {Z}} \end{aligned}$$

and

$$\begin{aligned} 1-\cos (y)\ge \frac{y^2}{2}-\frac{y^4}{24}\qquad \text {for }y\in {\mathbb {R}}. \end{aligned}$$

By the definition of \(d_1\) and \(d_2\), we have

$$\begin{aligned} 2+\cos {(x)}+\cos {(\alpha x)}&\ge 1-\cos {(d_2(x))}+1-\cos {(d_1(x))} \\&\ge 1-\cos {(d(x))}\ge \frac{(d(x))^2}{2}-\frac{(d(x))^4}{24}\gg d(x)^2. \end{aligned}$$

Hence, by (3), (4) and the above inequalities, we get

$$\begin{aligned} \frac{1}{2+\cos {(x)}+\cos {(\alpha x)}}\ll \frac{1}{d(x)^2}\ll \bigl (Q(x)\bigr )^{2(\nu +\varepsilon -1)}\ll x^{2(\nu +\varepsilon -1)}. \end{aligned}$$

Since \(\varepsilon >0\) is arbitrary, the theorem is proved. \(\square\)

In the proof of the next result we will need the following

Lemma 2

For any function \(h:\mathbb {R}^+\rightarrow \mathbb {R}\) for which \(\lim \limits _{x\rightarrow +\infty }h(x)=+\infty\), there exists a strictly increasing bijection \(g:\mathbb {R}^+\rightarrow \mathbb {R}^+\) such that:

  • there exists \(x_0\in \mathbb {R^+}\) such that \(g(x)\le h(x)\) for \(x>x_0\);

  • \(\lim \limits _{x\rightarrow +\infty }g(x)=+\infty\);

  • \(g(x)\le x\) for all \(x>0\);

  • for all \(x, y\in \mathbb {R}^+\) if \(|x-y|\le 2\), then \(|g(x)-g(y)|\le 3\).

Proof

Since the function h tends to \(+\infty\), for every \(n\in \mathbb {N}\) there exists such a point \(x_n\in \mathbb {R}^+\) that for \(x\ge x_n\), it holds \(h(x)\ge n+1\). Let \(p_0=0\) and

$$\begin{aligned} p_{n}=\max \{p_{n-1}+1, x_{n}\} \end{aligned}$$

for \(n>0\). Obviously \(\lim \limits _{n\rightarrow \infty }p_n=+\infty\) and \(h(x)\ge n+1\) for \(x\ge p_n\), \(n\in {\mathbb {N}}\) . For every \(x\in \mathbb {R}^+\) there exists exactly one \(n\in \mathbb {N}_0\) for which \(p_n<x\le p_{n+1}\). Let

$$\begin{aligned} g(x)=n+\frac{x-p_n}{p_{n+1}-p_n}, \qquad \text {for }p_n<x\le p_{n+1}. \end{aligned}$$

It is clear that this function is a strictly increasing bijection from \(\mathbb {R}^+\) to \(\mathbb {R}^+\) and if \(p_n<x\le p_{n+1}\), then \(g(x)\le n+1\). Because \(p_{n+1}\ge p_n+1\), so \(p_n\ge n\) and therefore

$$\begin{aligned} g(x)=n+\frac{x-p_n}{p_{n+1}-p_n}\le p_n+\frac{x-p_n}{1}=x. \end{aligned}$$

Moreover, for every \(x>p_1\), there exists \(n\ge 1\) such that \(p_n<x\le p_{n+1}\). Then,

$$\begin{aligned} h(x)\ge n+1\ge g(x). \end{aligned}$$

Now, let us notice that if \(|x-y|\le 2\) for some \(x, y>0\), then there exists \(n\in \mathbb {N}_0\) for which \(p_n<x\le p_{n+3}\) and \(p_n<y\le p_{n+3}\). Because g is increasing, then

$$\begin{aligned} |g(x)-g(y)|\le g(p_{n+3})-g(p_n)= 3, \end{aligned}$$

which completes the proof. \(\square\)

Now, we are going to prove

Theorem 6

For any function \(f:\mathbb {R}^+\rightarrow \mathbb {R}^+\) such that \(\lim \limits _{x\rightarrow +\infty }f(x)=+\infty\) there exists a Liouville number \(\alpha \in \mathbb {R}\) for which

$$\begin{aligned} \frac{1}{2+\cos x+\cos (x\alpha )}\ll x^2f(x). \end{aligned}$$

Proof

Let \(h(x)=\min \{\log _{10} x, \frac{\log _{10} f(\pi x)}{2}\}\). Take a function g satisfying all the assumptions of Lemma 2. Since g is a bijection, obviously there exists the inverse function \(g^{-1}\). Let us define \(\phi _1=1\) and

$$\begin{aligned} \phi _{n+1}=\phi _n+[g^{-1}(\phi _n)]+1. \end{aligned}$$

The sequence \((\phi _n)\) is a strictly increasing sequence of positive numbers; in particular \(\phi _n\ge n\) for every \(n\in \mathbb {N}_+\). Thus, the series \(\sum \limits _{k=1}^{\infty }10^{-\phi _k}\) is convergent. Denote its sum by \(\alpha\). Because \(g(x)\le h(x)\le \log _{10} x\) for \(x>x_0\), so \(x\ge 10^{g(x)}\) and therefore for \(x> x_0\) we have \(g^{-1}(x)\ge 10^x\).

$$\begin{aligned} 0<\{10^{\phi _n}\alpha \}=\sum \limits _{k=n+1}^{\infty }10^{-\phi _k+\phi _n}<2\cdot 10^{\phi _n-\phi _{n+1}}\approx 10^{-([g^{-1}(\phi _n)]+1)}\ll 10^{-10^{\phi _n}}\ll \frac{1}{(10^{\phi _n})^s} \end{aligned}$$

for any \(s\in \mathbb {N}\). Therefore, \(\alpha\) is a Liouville number.

Let us notice that for any \(x, y\in \mathbb {R}\) there exist \(m, n\in \mathbb {Z}\) for which

$$\begin{aligned} ||x||+||y||=|x-m|+|y-n|\ge |(x+y)-(m+n)|\ge ||x+y||, \end{aligned}$$

so \(||x+y||=||x+y||+||-y||-||y||\ge ||x||-||y||\). Let \(10^{\phi _n}<q<10^{\phi _{n+1}}\) for some odd \(q>0\). If \(q<\frac{1}{4}\cdot 10^{\phi _{n+1}-\phi _n}\), then

$$\begin{aligned} ||q\cdot \sum \limits _{k=1}^{n}10^{-\phi _k}||\ge 10^{-\phi _n} \end{aligned}$$

and

$$\begin{aligned} 0<q\cdot \sum \limits _{k=n+1}^{\infty }10^{-\phi _k}<\frac{1}{4}\cdot 10^{\phi _{n+1}-\phi _n}\cdot 2 \cdot 10^{-\phi _{n+1}}=\frac{1}{2}\cdot 10^{-\phi _n}<\frac{1}{2}, \end{aligned}$$

and therefore

$$\begin{aligned} ||q\alpha ||\ge ||q\cdot \sum \limits _{k=1}^{n}10^{-\phi _k}||-||q\cdot \sum \limits _{k=n+1}^{\infty }10^{-\phi _k}||\ge \frac{1}{2}\cdot 10^{-\phi _n}> \frac{1}{2q}. \end{aligned}$$

Further, if \(q\ge \frac{1}{4}\cdot 10^{\phi _{n+1}-\phi _n}\), then

$$\begin{aligned} ||q\cdot \sum \limits _{k=1}^{n+1}10^{-\phi _{k}}||\ge 10^{-\phi _{n+1}} \end{aligned}$$

and

$$\begin{aligned} 0<q\cdot \sum \limits _{k=n+2}^{\infty }10^{-\phi _k}<10^{\phi _{n+1}}\cdot 2\cdot 10^{-\phi _{n+2}}=2\cdot 10^{\phi _{n+1}-\phi _{n+2}}<\frac{1}{2}. \end{aligned}$$

Because \(g(x)\le x\) for every \(x>0\), so \(g^{-1}(x)\ge x\) and

$$\begin{aligned} \phi _{n+2}-\phi _{n+1}=[g^{-1}(\phi _{n+1})]+1>g^{-1}(\phi _{n+1})\ge \phi _{n+1}, \end{aligned}$$

and therefore \(\phi _{n+2}-\phi _{n+1}\ge \phi _{n+1}+1\). Therefore,

$$\begin{aligned}&||q\alpha ||\ge ||q\cdot \sum \limits _{k=1}^{n+1}10^{-\phi _k}||-||q\cdot \sum \limits _{k=n+2}^{\infty }10^{-\phi _k}||\ge 10^{-\phi _{n+1}}-2\cdot 10^{\phi _{n+1}-\phi _{n+2}}\\&\quad \ge 10^{-\phi _{n+1}}-\frac{1}{5}\cdot 10^{-\phi _{n+1}}> \\&\quad \frac{1}{2}\cdot 10^{-\phi _{n+1}}=4\cdot 10^{\phi _{n}-\phi _{n+1}}\cdot \frac{1}{8}\cdot 10^{-\phi _n}\ge \frac{1}{8q}\cdot 10^{-\phi _n}. \end{aligned}$$

Let us notice that

$$\begin{aligned} \phi _n=g(g^{-1}(\phi _n))<g([g^{-1}(\phi _n)]+1)=g(\phi _{n+1}-\phi _n), \end{aligned}$$

so

$$\begin{aligned} ||q\alpha ||>\frac{1}{8q}\cdot 10^{-\phi _n}>\frac{1}{8q}\cdot 10^{-g(\phi _{n+1}-\phi _n)}. \end{aligned}$$

It is easily possible to establish that \(k<\frac{1}{4}\cdot 10^k\) for positive integers \(k\in {\mathbb {N}}\). Because the function g is increasing, so

$$\begin{aligned} ||q\alpha ||>\frac{1}{8q}\cdot 10^{-g(\phi _{n+1}-\phi _n)}>\frac{1}{8q}\cdot 10^{-g(\frac{1}{4}\cdot 10^{\phi _{n+1}-\phi _n})}\ge \frac{1}{8q}\cdot 10^{-g(q)}. \end{aligned}$$

Every odd \(q>10\) occurs between two subsequent terms of the sequence \((10^{\phi _n})\). Since \(\frac{1}{8q}\cdot 10^{-g(q)}<\frac{1}{2q}\), so for every odd \(q>10\) we have \(||q\alpha ||>\frac{1}{8q}\cdot 10^{-g(q)}\). Let us notice that

$$\begin{aligned}&\sqrt{2+\cos (2\pi x)+\cos (2\pi x\alpha )}=\sqrt{1+\cos (2\pi x)+1+\cos (2\pi x\alpha )}\\&\quad =\sqrt{2\cos ^2(\pi x)+2\cos ^2(\pi x\alpha )}\approx |\cos (\pi x)|+|\cos (\pi x\alpha )|\\&\quad =|\cos (\pi \{ x\})|+|\cos (\pi \{ x\alpha \})|\approx \bigl |\{x\}-\frac{1}{2}\bigr |+\bigl |\{x\alpha \}-\frac{1}{2}\bigr |=:t(x). \end{aligned}$$

The graphs of two components of the function t consist of segments. In the first case their slope is equal to 1 and \(-1\), ; in the second one: \(\alpha\) and \(-\alpha\). Therefore, the graph of the function t consists of segments, the slopes of which are equal to \(1+\alpha ; 1-\alpha ; -1+\alpha ;-1-\alpha\). Let us consider local minima of this function. They have to appear at the points at which the function is not differentiable, that is at points \(\frac{z}{2}\) or \(\frac{z}{2\alpha }\) for \(z\in \mathbb {Z}\), because for other points there exists a neighborhood which is a segment of nonzero slope. One can easily notice that because \(\alpha <1\), so there are not minima at points \(\frac{z}{2\alpha }\), and that at points \(\frac{z}{2}\) minima and maxima occur alternately. Let \(x_n=\frac{(2n+1)}{2}\) for \(n\in {\mathbb {N}}\). Obviously the function t is continuous and attains minima at points \(x_n\), so t(x) is greater or equal to the value at one of the subsequent terms of the sequence \((x_n)\), between which x lies. For every \(x>x_1\) let us assign the number u(x) such that \(u(x_n)=x_n\) for every \(n\in \mathbb {N}\) and for \(x_n<x<x_{n+1}\), \(u(x)=x_n\) if \(t(x_n)<t(x_{n+1})\), and \(u(x)=x_{n+1}\) in other cases. For every \(x\ge \frac{11}{2}\) we have \(2u(x)\ge 11\) and 2u(x) is odd, so

$$\begin{aligned} \frac{1}{| \{x\}-\frac{1}{2}|+|\{x \alpha \}-\frac{1}{2}|} \le \frac{1}{|\{u(x)\cdot \alpha \}-\frac{1}{2}|}\le \frac{2}{||2u(x)\cdot \alpha ||}\le 16 \cdot 2u(x)\cdot 10^{g(2u(x))}, \end{aligned}$$

where the second inequality above follows from the inequality \(||2x||\le 2|\{x\}-\frac{1}{2}|\), for every \(x\in \mathbb {R}\). Because \(|2u(x)-2x|\le 2\), so by the property of the function g it follows that \(|g(2u(x))-g(2x)|\le 3\), and therefore

$$\begin{aligned} \frac{1}{|\{x\}-\frac{1}{2}|+|\{x\alpha \}-\frac{1}{2}|}\ll x\cdot 10^{g(2x)}, \end{aligned}$$

Moreover, for \(x>x_0\) we have \(10^{2g(x)}\le 10^{2h(x)}\le f(\pi x)\), so

$$\begin{aligned} \frac{1}{2+\cos x+\cos {(x\alpha )}}\approx \biggl (\frac{1}{|\{\frac{x}{2\pi }\}-\frac{1}{2}|+|\{\frac{x\alpha }{2\pi }\}-\frac{1}{2}|}\biggl )^2\ll x^2\cdot 10^{2g(\frac{x}{\pi })}\ll x^2 f(x), \end{aligned}$$

which ends the proof. \(\square\)

4 Convolution of some almost periodic functions

Now, we will consider the convolution of Levitan almost periodic functions (which are also \(\mu\)-almost periodic functions) with the function \(g_{\lambda } :{\mathbb {R}} \rightarrow {\mathbb {R}}\) (\(\lambda <0\)), defined in (2).

First, let us notice that Theorem 5 implies the existence of the convolution of function (1), where \(\alpha\) is not a Liouville number, with functions \(g_\lambda\). Indeed, if \(\nu (\alpha )=\nu\), then by the symmetry of the function f, we have

$$\begin{aligned} \int _{-\infty }^{0}f(t)e^{-\lambda t}d t=\int _{0}^{+\infty }f(t)e^{\lambda t}d t\le \int _{0}^{\frac{\pi }{2}}f(t)e^{\lambda t}d t+C\int _{\frac{\pi }{2}}^{+\infty }t^{2(\nu -1)+1}e^{\lambda t}d t<+\infty , \end{aligned}$$

for some constant \(C>0\). In particular, this means the existence of the convolution of function (1) for irrational algebraic numbers \(\alpha\), with functions \(g_\lambda\).

Moreover, as a simple consequence of Theorem 6 we get

Corollary 1

For any \(\lambda <0\) there exists a Liouville number \(\alpha \in \mathbb {R}\) for which

$$\begin{aligned} \int \limits _{0}^{+\infty }\frac{e^{\lambda x}}{2+\cos x+\cos (x\alpha )} dx<+\infty . \end{aligned}$$

Proof

Obviously

$$\begin{aligned} \lim \limits _{x\rightarrow +\infty }\frac{e^{\frac{-\lambda x}{2}}}{x^2}=+\infty . \end{aligned}$$

By Theorem 6, there exists a Liouville number \(\alpha \in \mathbb {R}\) such that

$$\begin{aligned} \frac{1}{2+\cos x+\cos (x\alpha )}\ll x^2 \cdot \frac{e^{\frac{-\lambda x}{2}}}{x^2}=e^{\frac{-\lambda x}{2}}, \end{aligned}$$

so

$$\begin{aligned} 0<\frac{e^{\lambda x}}{2+\cos x+\cos (x\alpha )}\ll e^{\frac{\lambda x}{2}}. \end{aligned}$$

The observation that the integral

$$\begin{aligned} \int \limits _{0}^{+\infty }e^{\frac{\lambda x}{2}} dx=-\frac{2}{\lambda } \end{aligned}$$

is finite, ends the proof. \(\square\)

In what follows we will need the following

Remark 3

Let w be a generalized trigonometric polynomial which is not a constant function, that is

$$\begin{aligned} w(x)=\sum _{n=1}^{k}\bigl (a_{n}\sin (\lambda _{n}x)+b_{n}\cos (\lambda _{n}x)\bigr ), \end{aligned}$$

for some \(a_{n},b_{n}, \lambda _{n}, \in \mathbb {R}\) for \(1\le n \le k\). By the Lagrange Mean Value Theorem, we know that there exists a constant \(L>0\) such that for \(x\in {\mathbb {R}}, h>0\), we have

$$\begin{aligned} |w(x+h)|\le |w(x)|+Lh. \end{aligned}$$

Since the function w has an analytic extension to the whole plane and it is not a constant function, it has countably many zeros on the plane. Hence, for almost all \(x\in {\mathbb {R}}\) the quotient \(\frac{1}{w(x)}\) is well defined. Thus, putting

$$\begin{aligned} f(x)= {\left\{ \begin{array}{ll} \dfrac{1}{w(x)}, \quad &{} \hbox { if}\ w(x)\ne 0, \\ 0, \quad &{} \hbox { if}\ w(x)=0, \end{array}\right. } \end{aligned}$$

for \(u\in {\mathbb {R}}\), we can write

$$\begin{aligned} \int _{u}^{u+1}|f(t)|d t=\int _{0}^{1}|f(u+t)|d t\ge \int _{0}^{1}\frac{1}{|w(u)|+Lt}d t. \end{aligned}$$
(5)

Theorem 7

The set \(\bigcup _{\lambda <0}S_\lambda\), where

$$\begin{aligned} S_\lambda =\{\alpha \in {\mathbb {R}} \setminus {\mathbb {Q}} :\frac{1}{2+\cos ({\cdot }) +\cos (\alpha {\cdot })}*g_\lambda \quad \text {exists}\}, \end{aligned}$$

is of the first Baire category. Moreover, \(\bigcap _{\lambda <0}S_\lambda\) and \(S_{\lambda _0}\), for \(\lambda _0<0\), are of the first Baire category. Thereby \(\bigcap _{\lambda <0}S_\lambda '\), \(S_{\lambda _0}'\), for \(\lambda _0<0\), and \(\bigcup _{\lambda <0}S_\lambda '\), where \(S_{\lambda }'={\mathbb {R}} \setminus S_{\lambda }\), are of the second Baire category.

Proof

The proof is similar to that one for Liouville’s numbers (see [22]). Obviously the existence of the convolution \(f*g_\lambda\) is equivalent to the condition

$$\begin{aligned} \int _{-\infty }^{0}f(t)e^{-\lambda t}dt<+\infty . \end{aligned}$$

If the convolution does not exists for some \(\lambda <0\), then it does not also exist for \(\lambda<\lambda '<0\), because

$$\begin{aligned} +\infty =\int _{-\infty }^{0}f(t)e^{-\lambda t}dt\le \int _{-\infty }^{0}f(t)e^{-\lambda ' t}dt. \end{aligned}$$

Hence,

$$\begin{aligned} \bigcap _{\lambda <0}S_\lambda '=\bigcap _{k=1}^\infty S_{-k}'. \end{aligned}$$

Let

$$\begin{aligned} U_n=\bigcup _{q\ge n}\bigcup _{p\in {\mathbb {Z}}}\left( \frac{2p+1}{2q-1}-\frac{1}{\Phi (2q-1)}, \frac{2p+1}{2q-1}+\frac{1}{\Phi (2q-1)}\right) , \end{aligned}$$

for \(n \in {\mathbb {N}}\) and

$$\begin{aligned} U=\bigcap _{n=1}^\infty U_n= \bigcap _{n=1}^\infty \bigcup _{q\ge n}\bigcup _{p\in {\mathbb {Z}}}\left( \frac{2p+1}{2q-1}-\frac{1}{\Phi (2q-1)}, \frac{2p+1}{2q-1}+\frac{1}{\Phi (2q-1)}\right) , \end{aligned}$$

where

$$\begin{aligned} \Phi (n)=\frac{n\pi \sqrt{e^{e^{n^2}}-1}}{\sqrt{2}}. \end{aligned}$$

The sets \(U_n\) are dense and open, because every number \(\frac{2p+1}{2q-1}, p\in {\mathbb {Z}}, q\in {\mathbb {N}}\) belongs to this set. Indeed, multiplying the numerator and the denominator by a sufficiently big odd number, one gets a quotient with an arbitrarily big denominator. Therefore, the complements of the sets \(U_n\) are nowhere dense. Because

$$\begin{aligned} U'={\mathbb {R}} \setminus U={\mathbb {R}} \setminus (\bigcap _{n=1}^\infty U_n)=\bigcup _{n=1}^\infty U_n', \end{aligned}$$

so \({\mathbb {R}} \setminus U\) is a set of the first category.

Now, we are going to establish that \(U\setminus {\mathbb {Q}}\subset \bigcap _{k=1}^\infty S_{-k}'=\bigcap _{\lambda <0}S_\lambda '\). Let \(\alpha \in U\setminus {\mathbb {Q}}\) and let \(w(x)=2+\cos x +\cos (\alpha x)\). Then, there exists a sequence \((p_n)_{n\in \mathbb N}\) of integers and an increasing sequence \((q_n)_{n\in {\mathbb {N}}}\) of positive integers such that

$$\begin{aligned} |\alpha -\frac{2p_n+1}{2q_n-1}|<\frac{\sqrt{2}}{(2q_n-1)\pi \sqrt{e^{e^{(2q_n-1)^2}}-1}}. \end{aligned}$$

It means that

$$\begin{aligned} |\alpha (2q_n-1)\pi -(2p_n+1)\pi |<\frac{\sqrt{2}}{ \sqrt{e^{e^{(2q_n-1)^2}}-1}}. \end{aligned}$$

Let us denote

$$\begin{aligned} d_n=|\alpha (2q_n-1)\pi -(2p_n+1)\pi |, \end{aligned}$$

for \(n \in {\mathbb {N}}\). Then,

$$\begin{aligned} d_n^2<\frac{2}{ e^{e^{(2q_n-1)^2}}-1}. \end{aligned}$$

For every \(y\in {\mathbb {R}}\) we have

$$\begin{aligned} 1-\cos (y)\le \frac{y^2}{2}. \end{aligned}$$

Therefore, for every \(k\in {\mathbb {N}}\) and sufficiently large n, we have

$$\begin{aligned}&w((2q_n-1)\pi )=1+ \cos (\alpha (2q_n-1)\pi )=1-\cos (d_n))\le \frac{d_n^2}{2} \\&\quad \le \frac{1}{ e^{e^{(2q_n-1)^2}}-1}<\frac{1}{ e^{e^{k (2q_n-1)\pi }}-1}<\frac{|\alpha |+1}{ e^{e^{k (2q_n-1)\pi }}-1}. \end{aligned}$$

By inequality (5), we have

$$\begin{aligned} \int _{(2q_n-1)\pi }^{(2q_n-1)\pi +1}\frac{1}{w(t)}dt\ge \int _{0}^{1}\frac{1}{w((2q_n-1)\pi )+(|\alpha |+1)t}d t= \frac{1}{|\alpha |+1}\ln \left( 1+\frac{|\alpha |+1}{w((2q_n-1)\pi )}\right) , \end{aligned}$$

so for every \(k\in {\mathbb {N}}\) and sufficiently large n, we have

$$\begin{aligned}&\int _{(2q_n-1)\pi }^{(2q_n-1)\pi +1}\frac{e^{-k t}}{w(t)}d t \ge e^{-k ((2q_n-1)\pi +1)}\int _{(2q_n-1)\pi }^{(2q_n-1)\pi +1}\frac{1}{w(t)}d t\\&\quad \ge \frac{e^{-k ((2q_n-1)\pi +1)}}{(|\alpha |+1)}\ln (1+\frac{(|\alpha |+1)}{w((2q_n-1)\pi )})\ge \frac{e^{-k}}{(|\alpha |+1)}. \end{aligned}$$

Hence,

$$\begin{aligned} \int _{-\infty }^{0}\frac{e^{k t}}{w(t)}d t=\int _{0}^{+\infty }\frac{e^{-k t}}{w(t)}d t\ge \sum _{n=1}^{+\infty } \int _{(2q_n-1)\pi }^{(2q_n-1)\pi +1}\frac{e^{-k t}}{w(t)}d t=+\infty . \end{aligned}$$

We have established that \(U\setminus {\mathbb {Q}}\subset \bigcap _{\lambda <0}S_\lambda '\). Hence \(\bigcup _{\lambda <0}S_\lambda \subset \mathbb (R \setminus U) \cup {\mathbb {Q}}\). Thus, \(\bigcup _{\lambda <0}S_{\lambda }\) is a set of the first category. \(\square\)

We will also need the following result which, roughly speaking, shows how one can construct irrational numbers \(\alpha\) such that functions of form (1), do not satisfy at infinity a given asymptotic.

Theorem 8

([17]) For any function \(g:{\mathbb {R}} \rightarrow (0, +\infty )\) and any \(a\in {\mathbb {R}}, \epsilon >0\) there exists \(\alpha\) such that

$$\begin{aligned} |a- \alpha |<\epsilon \quad \text {and}\quad \limsup _{x \rightarrow +\infty }\frac{g(x)}{2+\cos {x}+\cos {(\alpha x})}=+\infty . \end{aligned}$$

The result below shows how one can construct irrational numbers \(\alpha\) for which the convolution of a function of form (1) with the function \(g_\lambda\) does not exist.

Theorem 9

For every \({a\in {\mathbb {R}}}\) and every \({\epsilon >0}\) there exists \(\alpha \in {\mathbb {R}}\setminus {\mathbb {Q}}\) such that for all \(\lambda <0\)

$$\begin{aligned} |a- \alpha |<\epsilon \quad \text {and}\quad \int _{-\infty }^0 \frac{e^{-\lambda t}}{2+\cos {t}+\cos {(\alpha t})}d t=+\infty . \end{aligned}$$

In other words, the set

$$\begin{aligned} \bigcap _{\lambda<0} S_\lambda '=\bigl \{\alpha \in {\mathbb {R}}\setminus {\mathbb {Q}} :\frac{1}{2+\cos {\cdot }+\cos {(\alpha \cdot })}* g_\lambda \quad \hbox { does not exist for all}\ \lambda <0\bigr \} \end{aligned}$$

is dense in \({\mathbb {R}}\). Thereby for \(\lambda _0<0\) the sets \(S_{\lambda _0}'\) and \(\bigcup _{\lambda <0}S_\lambda '\) are also dense in \({\mathbb {R}}\).

Remark 4

It is well-known that the set of all irrational algebraic numbers is dense in \({\mathbb {R}}\). Moreover, by Theorem 2 (cf. also Remark 2), \(\nu (\alpha )=2\), for any irrational algebraic number \(\alpha\). By Theorem 5, for such numbers \(\alpha\) there exists the convolution with the function \(g_{\lambda } :{\mathbb {R}} \rightarrow {\mathbb {R}}\) (\(\lambda <0\)). Therefore, the sets \(\bigcap _{\lambda<0}S_\lambda , S_{\lambda _0}, \bigcup _{\lambda <0}S_\lambda\), for \(\lambda _0<0\) are dense in \({\mathbb {R}}\).

Proof

Fix \(a\in {\mathbb {R}}, \epsilon >0\). Let

$$\begin{aligned} g(x)=\frac{L}{e^{ e^{x^2}}-1} \quad \hbox { for}\ x\in {\mathbb {R}}, \end{aligned}$$

where \(L=|a|+1+\epsilon\). Let us observe that for any \(\lambda <0\) there exists \(x_0\) such that for \(x>x_0\) we have

$$\begin{aligned} g(x)\le \frac{L}{e^{ e^{-\lambda x}}-1}. \end{aligned}$$

By Theorem 8, there exists \(\alpha\) such that \(|\alpha -a|<\epsilon\) and the sequence \((x_n)_{n=1}^{\infty }\) such that \(x_n\ge 0\), \(x_{n+1}-x_n\ge 1\), for \(n\in {\mathbb {N}}\) and

$$\begin{aligned} \frac{g(x_n)}{w(x_n)}\ge 1, \end{aligned}$$

where \(w(x)=2+\cos (x)+\cos (\alpha x)\). Then, by the parity of the function w, we get

$$\begin{aligned} \int _{-\infty }^{0}\frac{e^{-\lambda t}}{w(t)}d t=\int _{0}^{+\infty }\frac{e^{\lambda t}}{w(t)}d t\ge \sum _{n=1}^{+\infty } \int _{x_n}^{x_n+1}\frac{e^{\lambda t}}{w(t)}d t=+\infty , \end{aligned}$$

because, by inequality (5) for sufficiently large n, we have

$$\begin{aligned}&\int _{x_n}^{x_n+1}\frac{e^{\lambda t}}{w(t)}d t \ge e^{\lambda (x_n+1)}\int _{x_n}^{x_n+1}\frac{1}{w(t)}d t\ge e^{\lambda (x_n+1)}\int _{0}^{1}\frac{1}{w(x_n)+Lt}d t\\&\quad =\frac{e^{\lambda (x_n+1)}}{L}\ln (1+\frac{L}{w(x_n)})\ge \frac{e^{\lambda (x_n+1)}}{L}\ln (1+\frac{L}{g(x_n)})\ge \frac{e^\lambda }{L}. \end{aligned}$$

\(\square\)

Remark 5

Denote by L the set of all Liouville numbers. At the beginning of the Sect. 4 it was noticed that

$$\begin{aligned} {\mathbb {R}} \setminus (L\cup {\mathbb {Q}}) \subset \bigcap _{\lambda <0}S_\lambda . \end{aligned}$$

Thus,

$$\begin{aligned} \bigcup _{\lambda <0}S_\lambda '\subset L. \end{aligned}$$

Therefore, for \(\lambda _0<0\)

$$\begin{aligned} \mu (\bigcap _{\lambda<0}S_\lambda ')\le \mu (S_{\lambda _0}')\le \mu ( \bigcup _{\lambda <0}S_\lambda ')\le \mu (L)=0\quad (\text {see } [20]), \end{aligned}$$

where \(\mu\) denotes the Lebesgue measure on \({\mathbb {R}}\).

Now, we are going to focus on cardinality of the sets \(\bigcap _{\lambda <0}S_\lambda\) and \(\bigcap _{\lambda <0}S_\lambda '\).

Theorem 10

\(\bigcap _{\lambda <0}S_\lambda\) is a set with the cardinality of the continuum. Moreover, the set \(\bigcap _{\lambda <0}S_\lambda '\) is also with the cardinality of the continuum. Thereby \(\bigcup _{\lambda <0}S_\lambda\) and \(\bigcup _{\lambda <0}S_\lambda '\) are also with the cardinality of the continuum.

Proof

We will show that if denominators of an arithmetic continued fraction \([a_0, a_1, a_2,\ldots ]\) are bounded, then this number satisfies the Liouville’s Theorem for irrational algebraic numbers of degree 2. Indeed, let \(M>0\) be an upper bound of terms of the sequence \((a_n)\). Let us consider arbitrary \(q\ge 1\). Let n be an index such that \(Q_n\le q<Q_{n+1}\). Since the convergents of a continued fraction are its best approximations, for arbitrary \(p\in {\mathbb {Z}}\), we have

$$\begin{aligned} |\alpha -\frac{p}{q}|\ge |\alpha -\frac{P_{n+1}}{Q_{n+1}}|\ge \frac{1}{2Q_{n+1}Q_{n+2}}. \end{aligned}$$

Moreover,

$$\begin{aligned}&2Q_{n+1}Q_{n+2}= 2Q_{n+1}(a_{n+2}Q_{n+1}+Q_n)\le 2(a_{n+2}+1)Q_{n+1}^2 \\&\quad = 2(a_{n+2}+1)(a_{n+1}Q_{n}+Q_{n-1})^2\\&\quad \le 2(a_{n+2}+1)(a_{n+1}+1)Q_{n}^2\le 2(a_{n+2}+1)(a_{n+1}+1)^2q^2\le 2(M+1)^3q^2. \end{aligned}$$

Therefore, for such numbers the convolution exists (for arbitrary \(\lambda <0\)). Obviously, the cardinality of such sequences is continuum, so the cardinality of the set \(\bigcap _{\lambda <0}S_\lambda\) is continuum, too. We are going to establish that the cardinality of the set

$$\begin{aligned} \bigcap _{\lambda <0}S_\lambda '=\bigcap _{k=1}^\infty S_{-k}'. \end{aligned}$$

is also continuum. Let

$$\begin{aligned} g(x)=\frac{1}{e^{e^{x^2}}-1}. \end{aligned}$$

Now, let \(\alpha\) be constructed according to Theorem 8 for the function g. Then, there exists a sequence \((x_n)_{n=1}^{\infty }\) such that \(x_n\ge 0\), \(x_{n+1}-x_n\ge 1\), for \(n\in {\mathbb {N}}\) and

$$\begin{aligned} \frac{g(x_n)}{w(x_n)}\ge 1, \end{aligned}$$

where \(w(x)=2+\cos (x)+\cos (\alpha x)\). Moreover, for every \(k\in {\mathbb {N}}\) and sufficiently large n, we have

$$\begin{aligned} w(x_n) \le \frac{1}{ e^{e^{x_n^2}}-1}<\frac{1}{ e^{e^{k x_n}}-1}<\frac{|\alpha |+1}{ e^{e^{k x_n}}-1}. \end{aligned}$$

Then, by the parity of the function w, we get

$$\begin{aligned} \int _{-\infty }^{0}\frac{e^{k t}}{w(t)}d t=\int _{0}^{+\infty }\frac{e^{-k t}}{w(t)}d t\ge \sum _{n=1}^{+\infty } \int _{x_n}^{x_n+1}\frac{e^{- kt}}{w(t)}d t=+\infty , \end{aligned}$$

because for every \(k\in {\mathbb {N}}\) and sufficiently large n, we have

$$\begin{aligned}&\int _{x_n}^{x_n+1}\frac{e^{- kt}}{w(t)}d t \ge e^{- k(x_n+1)}\int _{x_n}^{x_n+1}\frac{1}{w(t)}d t\ge e^{-k (x_n+1)}\int _{0}^{1}\frac{1}{w(x_n)+(|\alpha |+1)t}d t\\&\quad = \frac{e^{- k(x_n+1)}}{|\alpha |+1}\ln (1+\frac{|\alpha |+1}{w(x_n)})\ge \frac{e^{-k}}{|\alpha |+1}. \end{aligned}$$

By the construction, it follows that in an arbitrary neighborhood of any real number, the cardinality of the set of such constructed numbers is continuum, because having the denominators \(a_1,\ldots , a_n\), the subsequent one is constructed as any odd number greater than a certain quantity depending on \(a_1,\ldots , a_n\). Thus, at every step there are denumerably many possibilities of choice. Moreover, there are also denumerably many stages of choice. Hence, the cardinality of the set of such constructed numbers is continuum. Therefore, the cardinality of the set \(\bigcap _{\lambda <0}S_\lambda '\) is continuum. \(\square\)