I. INTRODUCTION
Let ( ) be a linear space consisting of measurable functions for which ( ) is Lebesgue integrable on any finite segment of the real axis with norm
\[| | f | | _ {B _ {p}} = \{\overline{{M}} [ | f (x) | ^ {p} ] \} ^ {\frac{1}{p}} = \{\overline{{\lim _ {T \to \infty}}} \int_ {- T} ^ {T} | f (x) | ^ {p} d x \} ^ {\frac{1}{p}} < \infty ,\]
\[Definition of the norm for B_\infty | | f | | _ {B _ {\infty}} = v r a i \sup _ {- \infty < x < \infty} | f (x) | < \infty .\]
At , A. Bezikovich [1] or [2], introduced the following concept of -almost-periodic function.
Definition 1. A function is called almost-periodic in the sense of Bezikovich or -almost-periodic if there exists a sequence of finite trigonometric polynomials of the form
\[P_{n}(x) = \sum_{k=1}^{n} A_{k}(f)e^{i\lambda_{k}x},\]
for which the following condition holds
\[\lim _ {n \to \infty} | | f (x) - P _ {n} (x) | | _ {B _ {p}} = 0.\]
For each , a function is defined
\[a(f,\lambda) = \lim_{T\to\infty} \frac{1}{T} \int_{-T}^{T} f(x)e^{-i\lambda x}\,dx = M\{f(x)e^{-i\lambda x}\}.\]
It can differ from zero no more than on the countable set of values of . The numbers are called exponents Fourier transform or the spectrum of the function in question, and the numbers - Fourier coefficients. Thus, each function can be written a Fourier series
\[f(x)\sim\sum_{k}A_{k}(f)e^{i\lambda_{k}x}.\]
The space of uniform almost-periodic functions denote (see, for example, [3], [4]).
Denote by the finite difference of the th order of the function at the point with a step of , i.e.
\[\Delta_t^m f(x) = \sum_{r=0}^{m} (-1)^{m-r} \binom{m}{r} f(x + rt).\]
To determine the smoothness of the function, the quality of the structural characteristic of the function , , we will use the continuity module of the order k
\[\omega_{m}(f,h)_{B_p} = \sup_{|t|\leq h} \|\Delta_t^m f(x)\|_{B_p}, \quad h>0, m\in N.\]
Let be any partition of the function of interval . Given , and a positive , we write
\[V_{r,T}^{m}(f) = \left[\sup_{r=0}^{n-1}\sum|\Delta_{h_r}^mf(x_r)|^r\right]^{1/r},\]
where , . Then we call the value
\[V_{r}^{m}(f) = \varlimsup_{T\to\infty}\frac{1}{2T} V_{r,T}^{m}(f)\]
-the variation of function of order .
In [9]-[17] and others, some necessary and sufficient conditions for the absolute convergence of the Fourier series of almost-periodic functions in the sense of Bohr and Bezikovich were obtained.
J. Museliak [10] showed that if the spectrum and , , , then for the function the condition
\[\sum_ {k = 1} ^ {\infty} k ^ {\frac {1 - \frac {\beta}{2}}{\alpha - 1}} \omega_ {1} ^ {\beta} (f; \frac {1}{k}) _ {B _ {2}} < \infty ,\tag{1}\]
at , entails the convergence of the series
\[\sum_{n=1}^\infty |A_k(f)|^{\beta}.\]
N.P. Kuptsov [11] showed that for functions , condition (1) for , and replacing the value by provides absolute convergence of the series (2).
In the work of A.G. Pritula [12] it is proved that if for , , ( ), the condition is met
\[\sum_ {\nu = 1} ^ {\infty} (\frac {\lambda_ {2 ^ {\nu}}}{\lambda_ {2 ^ {\nu - 1}}}) ^ {\beta} \omega_ {1} ^ {\beta} (f, \frac {1}{\lambda_ {2 ^ {\nu}}}) _ {B _ {p}} 2 ^ {\nu (\gamma + \frac {q - \beta}{q})} < \infty ,\]
that
\[\sum_{k=1}^\infty |A_k|^{\beta} k^{\beta} < \infty.\]
In the case when , A. S. Jafarova and G. A. Mammadova [13] established the convergence of the series
\[\sum_ {k = 1} ^ {\infty} | A _ {k} | ^ {\beta} \varphi (k),\]
with some restrictions on the functions . Instead of the continuity modulus, they used the following value based on the Laplace transform
\[\Omega(f;H;\delta;\theta) = \delta \sup_{x} \left| \int_0^\infty e^{\!-\!\delta\theta} f(x-t) e^{i\theta t} dt \right|, \quad \delta > 0, \theta \in \mathbb{R}.\]
In the work of Yu.K. Khasanov [14], some sufficient, and in the case of monotonous decrease of the Fourier coefficients, the necessary conditions for the absolute convergence of the Fourier series of almost-periodic Bezikovich functions are established when the Fourier exponents have a single limiting point at infinity or at zero.
The results of this note are analogs of some results of [10], [14] and [15] for the class of almost-periodic Bezikovich functions.
0.1 Main results
1.1. The note discusses some new sufficient conditions for the absolute convergence of Fourier series of almost-periodic functions from the space when the spectrum has a single limit point at infinity, i.e.
\[\lambda_ {0} = 0; \quad \lambda_ {- k} = - \lambda_ {k}; \quad | \lambda_ {k} | < | \lambda_ {k + 1} |; \quad \lim _ {k \to \infty} \lambda_ {k} = \infty .\]
It is well known that for an arbitrary function having a Fourier series expansion
\[f (x) \sim \frac {a _ {0} (f)}{2} + \sum_ {k = 1} ^ {\infty} (a _ {k} (f) \cos \lambda_ {k} x + b _ {k} (f) \sin \lambda_ {k}) x,\tag{3}\]
где
\[a _ {0} (f) = M \{f (x) \},\]
\[a _ {k} (f) = M \{f (x) \cos \lambda_ {k} x \},\]
\[b _ {k} (f) = M \{f (x) \sin \lambda_ {k} x \} \quad (k = 1, 2, \ldots),\]
\[M\{g(x)\} = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} g(x) dx.\]
We prove the following theorem concerning the absolute convergence of series (4) with known coefficients (see, for example, [3], [10-15]).
Theorem 1. Let is a limited function. Suppose that the function is non-decreasing such that if and is also a non-decreasing function. If at is executed
\[\sum_ {\nu = 1} ^ {\infty} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} \omega^ {\beta} (f, 2 ^ {- \nu}) \omega_ {\Phi} ^ {\beta / 2} (f, 2 ^ {- \nu}) \Phi^ {- \beta / 2} [ \omega (f, 2 ^ {- \nu}) ] < \infty ,\tag{4}\]
where
\[\omega(f,h) = v r a i \sup_{x} \sup_{|\delta| \leq h} |f(x + \delta) - f(x)|,\]
\[\omega_ {\Phi} (f, h) = \sup _ {| \delta | \leq h} \overline {{M}} \{| f (x + \delta) - f (x) | \},\]
that row
\[\sum_ {k = 1} ^ {\infty} (| a _ {k} (f) | ^ {\beta} + | b _ {k} (f) | ^ {\beta})\tag{5}\]
it fits.
Proof We first prove an important inequality
\[\sum_ {n \in A _ {\nu}} (| a _ {k} (f) | ^ {2} + b _ {k} (f) | ^ {2}) \leq \frac {1}{2} M \{| f (x + 2 ^ {- \nu - 1}) - f (x - 2 ^ {- \nu - 1}) | ^ {2} \},\tag{6}\]
where
For any , consider the function
\[F _ {h} (x) = f (x + h) - f (x - h).\]
The coefficients of the Fourier function are defined as follows:
\[a_{0}(F_{h}) = M\{F_{h}(x)\} = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} [f(x+h)-f(x-h)]dx = \\= \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} [f(x)-f(x)]dx = 0.\]
\[a_{k}(F_{h}) = M\{F_{h}(x) \cos \lambda_{k} x\} = M\{[f(x+h)-f(x-h)] \cos \lambda_{k} x\} = \\= \lim_{T\to\infty} \frac{1}{2T} \int_{-T+h}^{T+h} f(t) \cos \lambda_{k}(t-h) dt - \lim_{T\to\infty} \frac{1}{2T} \int_{-T-h}^{T-h} f(t) \cos \lambda_{k}(t+h) dt = \\= \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} f(t+h)[\cos \lambda_{k} t \cos \lambda_{k} h + \sin \lambda_{k} t \sin \lambda_{k} h] dt - \\- \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} f(t-h)[\cos \lambda_{k} t \cos \lambda_{k} h - \sin \lambda_{k} t \sin \lambda_{k} h] dt = \\= 2 \sin \lambda_{k} h \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} f(t) \sin \lambda_{k} t dt = \\= 2 \sin \lambda_{k} h M\{f(t) \sin \lambda_{k} t\} dt = 2 b_{k} \sin \lambda_{k} h.\]
Similarly, we have repeating these calculations for the coefficients , we find
\[b_{k}(F_{h}) = -2a_{k} \sin \lambda_{k} h.\]
Then by virtue of Bessel's inequality we get
\[\sum_{k=1}^\infty (|a_k(f)|^2 + |b_k(f)|^2) \sin^2 \lambda_k h = \sum_{k=1}^\infty (|a_k(f) \sin \lambda_k h|^2 + |b_k(f) \sin \lambda_n h|^2) =\]
\[= \frac {1}{4} \sum_ {k = 1} ^ {\infty} (| 2 a _ {k} (f) \sin \lambda_ {k} h | ^ {2} + | 2 b _ {k} (f) \sin \lambda_ {k} h | ^ {2}) = \frac {1}{4} \sum_ {k = 1} ^ {\infty} (| a _ {n} (F _ {h}) | ^ {2} + | b _ {k} (F _ {h}) | ^ {2}) \leq\]
\[\leq \frac {1}{4} M \{| f (x + h) - f (x - h) | ^ {2} \}.\]
For , consider
\[2 ^ {\nu - 1} \pi h \leq \lambda_ {k} h < 2 ^ {\nu} \pi h.\]
Let's put and from the latter we get
\[2^\nu-1\pi2^{-\nu-1} \leq \lambda_k h < 2^\nu\pi2^{-\nu-1}\]
or
\[\frac {\pi}{4} \leq \lambda_ {k} h < \frac {\pi}{2}.\]
Hence,
\[\sin^ {2} \lambda_ {k} 2 ^ {- \nu - 1} \geq \frac{1}{2}.\]
Hence, after using a number of calculations, the inequality (6) follows
\[\sum_ {k \in A _ {\nu}} (| a _ {k} (f) | ^ {2} + | b _ {k} (f) | ^ {2}) \leq 2 \sum_ {k \in A _ {\nu}} (| a _ {k} (f) | ^ {2} + | b _ {k} (f) | ^ {2}) \sin^ {2} \lambda_ {k} 2 ^ {- \nu - 1} \leq\]
\[\leq 2 \sum_{k=1}^\infty (|a_k(f)|^2 + |b_k(f)|^2) \sin^2 \lambda_k 2^{-\nu-1} \leq \frac{1}{2} M |f(x+2^{-\nu-1}) - f(x-2^{-\nu-1})|^2.\]
Next, we denote by the function , which is non-decreasing. Since the function is bounded, then . So, if , then and the assumption of the limitation of the function will be superfluous.
Multiplying and dividing the right part (6) by the function , we have
\[\sum_{k\in A_{\nu}} (|a_k(f)|^2 + |b_k(f)|^2) \leq \frac{1}{2} M\{|f(x+2^{-\nu-1}) - f(x-2^{-\nu-1})|^2\} \frac{\Phi[\omega(f,2^{-\nu})]}{\Phi[\omega(f,2^{-\nu})]} \leq\]
\[\leq 2^{-1} \omega^{2}(f,2^{-\nu}) \Phi^{-1}[\omega(f,2^{-\nu})] \overline{M}\{\Phi[\omega(f,2^{-\nu})]\} = 2^{-1} \omega^{2}(f,2^{-\nu}) \omega_{\Phi}(f,2^{-\nu}) \Phi^{-1}[\omega(2^{-\nu})].\]
Let be a measure of sets in . Then using the Helder inequality from the last inequality we get
\[\begin{array}{r l} & \sum_{k\in A_{\nu}} (|a_k(f)|^2 + |b_k(f)|^2)^{\frac{\beta}{2}} \leq [m(A_v)]^{1-\frac{\beta}{2}} [\sum_{k\in A_{\nu}} (|a_k(f)|^2 + |b_k(f)|^2)]^{\frac{\beta}{2}} \leq \\& \qquad \leq [m(A_v)]^{1-\frac{\beta}{2}} [2^{-1} \omega^2(f,2^{-\nu}) \omega_{\Phi}(f,2^{-\nu}) \Phi^{-1}[\omega(f,2^{-\nu})]]^{\frac{\beta}{2}} = \\& \qquad = 2^{-\frac{\beta}{2}} [m(A_v)]^{1-\frac{\beta}{2}} \omega^\beta(f,2^{-\nu}) \omega_{\Phi}^{\frac{\beta}{2}}(f,2^{-\nu}) \Phi^{-\frac{\beta}{2}}[\omega(f,2^{-\nu})] \leq \\& \qquad \leq 2^{-\frac{\beta}{2}} [\mu(2^{\nu} \pi) - \mu(2^{\nu-1} \pi) + 1]^{1-\frac{\beta}{2}} \omega^\beta(f,2^{-\nu}) \omega_{\Phi}^{\frac{\beta}{2}}(f,2^{-\nu}) \Phi^{-\frac{\beta}{2}}[\omega(f,2^{-\nu})].\tag{7} \\end{array}\]
\[\leq 2 ^ {- \frac {\beta}{2}} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} \omega^ {\beta} (f, 2 ^ {- \nu}) \omega_ {\Phi} ^ {\frac {\beta}{2}} (f, 2 ^ {- \nu}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ].\tag{7}\]
So for
\[k _ {0} = \min _ {\lambda_ {k} \geq \pi} k\]
from inequality (7) we find that
\[\sum_ {k = k _ {0}} ^ {\infty} \left(| a _ {k} (f) | ^ {2} + | b _ {k} (f) | ^ {2}\right) ^ {\frac {\beta}{2}} \leq\]
\[\leq 2 ^ {- \frac {\beta}{2}} \sum_ {\nu = 1} ^ {\infty} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} \omega^ {\beta} (f, 2 ^ {- \nu}) \omega_ {\Phi} ^ {\frac {\beta}{2}} (f, 2 ^ {- \nu}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ].\tag{8}\]
By virtue of condition (4), the series (5) converges. Theorem 1 is proved.
Theorem 2. Let the function . If at the condition is met
\[\sum_ {\nu = 1} ^ {\infty} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} \omega_ {2} ^ {\beta} (f, 2 ^ {- \nu}) < \infty ,\]
where
\[\omega_ {2} (f, h) = [ \sup _ {| \delta | \leq h} M \{| f (x + \delta) - f (x) | ^ {2} \} ] ^ {\frac {1}{2}},\]
then row (5) converges.
Proof We write inequality (6) in the following form
\[\sum_ {k \in A _ {\nu}} (| a _ {k} (f) | ^ {2} + | b _ {k} (f) | ^ {2}) \leq \frac {1}{2} M \{| f (x + 2 ^ {- \nu - 1}) - f (x - 2 ^ {- \nu - 1}) | ^ {2} \} \leq\]
\[\leq \frac{1}{2} \sup_{|\delta|\leq 2^{-\nu}} M\{|f(x+\delta)-f(x)|^{2}\} = \frac{1}{2} \omega(f,2^{-\nu}).\]
Hence, using the inequality (7), we will have
\[\sum_ {k \in A _ {\nu}} (| a _ {k} (f) | ^ {2} + | b _ {k} (f) | ^ {2}) ^ {\frac {\beta}{2}} \leq 2 ^ {- \frac {\beta}{2}} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ] ^ {\frac {\beta}{2}} =\]
\[= 2^{-\frac{\beta}{2}} [ \mu(2^{\nu} \pi) - \mu(2^{\nu-1} \pi) + 1 ]^{1-\frac{\beta}{2}} \omega_2^\beta (f, 2^{-\nu})\]
Then, by the conditions of the theorem, it follows from the latter that it follows that
\[\sum_ {k = k _ {0}} ^ {\infty} (| a _ {k} (f) | ^ {2} + | b _ {k} (f) | ^ {2}) ^ {\frac {\beta}{2}} \leq 2 ^ {- \frac {\beta}{2}} \sum_ {\nu = 1} ^ {\infty} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} \omega_ {2} ^ {\beta} (f, 2 ^ {- \nu}) < \infty .\]
Theorem 2 is proved.
In the future, we will need the following auxiliary statement.
Lemma 1. If for a non-decreasing function at
\[V_{\phi,T}(f) = \sup_{\Pi} \sum_{k=1}^{n} \phi[\|f(x_k) - f(x_{k-1})\|_{B_2}]\]
where is an arbitrary division of the interval by the points and
\[V_{\phi}(f) = \overline{M}\{V_{\phi,T}(f)\} = \varlimsup_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} V_{\phi,T}(f) dx,\]
then for any h > 0 the following estimate is valid
\[\overline{M}\{\Phi\left[f(x+h)-f(x-h)\right]\} \leq 2hV_{\Phi}(f).\]
Proof Let . Suppose that for there exists such a number that for every the inequality holds
\[V_{\Phi,T+3h}(f) \leq 2 [ V_{\Phi}(f) + \varepsilon ] (T + 3h).\]
Indeed, by definition of the upper limit
\[\frac {1}{2 (T + 3 h)} V _ {\Phi , T + 3 h} (f) \leq V _ {\Phi} (f) + \varepsilon ,\]
hence the inequality (10).
For a fixed , we define such an interval in which the points will be
\[x _ {k} - x _ {k - 1} = 2 h \quad (k = 1, 2, \dots , n),\]
\[x _ {k} - x _ {k - 1} \geq 2 h (k = n).\]
Then, given the values of , for up to the limiting average value of the function we get
\[\begin{array}{r l} & {\frac {1}{2 T} \int_ {- T} ^ {T} \Phi [ | f (x + h) - f (x - h) | ] d x = \frac {1}{2 T} \sum_ {k = 1} ^ {n} \int_ {x _ {k - 1}} ^ {x _ {k}} \Phi [ | f (x + h) - f (x - h) | ] d x =} \\& {\qquad = \frac {1}{2 T} \int_ {0} ^ {2 h} \sum_ {k = 1} ^ {n} \Phi [ | f (x _ {k} + t) - f (x _ {k - 1} + t) | ] d t +} \\& {\qquad + \frac {1}{2 T} \int_ {0} ^ {x _ {k} - x _ {k - 1}} \Phi [ | f (x _ {k} + t + 2 h) - f (x _ {k - 1} + t) | ] d t \leq} \\& {\qquad \leq \frac {1}{2 T} \int_ {0} ^ {2 h} \sup _ {x _ {k} \in [ - T + 3 h, T + 3 h ]} \sum_ {k = 1} ^ {n} \Phi [ | f (x _ {k}) - f (x _ {k - 1}) | ] d t =} \\& {\qquad = \frac {1}{2 T} \int_ {0} ^ {2 h} V _ {\phi , T + h + 3 h} (f) d t = \frac {1}{2 T} V _ {\Phi , T + 3 h} \int_ {0} ^ {2 h} d t =} \\& {\qquad = \frac {1}{2 T} \int_ {0} ^ {2 h} V _ {\phi , T + h + 3 h} (f) d t = \frac {h}{T} V _ {\Phi , T + 3 h} \leq} \\& {\qquad \frac {2 h}{T} [ V _ {\phi} (f) + \varepsilon ] [ T + 3 h ] = (2 h + \frac {6 h ^ {2}}{T}) (V _ {\Phi} (f) + \varepsilon),} \end{array}\]
Hence, at , going to the limit, we get an estimate (9), which implies the validity of Lemma 1. ☐
Theorem 3. Let and a non-decreasing function is given such that and for u > 0, . If at and the condition is met
\[\sum_ {\nu = 1} ^ {\infty} [ \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 ] ^ {1 - \frac {\beta}{2}} 2 ^ {- \frac {\beta \nu}{2}} \omega^ {\beta} (f, 2 ^ {- \nu}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ] < \infty ,\tag{11}\]
then the series (5) converges.
Proof The theorem is proved using Theorem 1 and Lemma 1. Indeed, since
\[\omega_ {\Phi} (f, 2 ^ {- \nu}) = \sup _ {| \delta | \leq 2 ^ {- \nu}} \overline {{M}} \{\Phi [ | f (x + \delta) - f (x) | ] \} \leq\]
\[\leq 2 h V _ {\phi} (f) = \sup _ {| \delta | \leq 2 ^ {- \nu}} \sum_ {\nu = 1} ^ {\infty} \phi [ | f (x + 2 ^ {- \nu}) - f (x) | ] \leq 2 ^ {- \nu},\]
then substituting instead of into inequality (4), we get inequality (8), which proves theorem 3.
In the future, we will establish the convergence condition of series (5) for the existence of the spectrum . In this case, we need to prove the following
Lemma 2. If are positive numbers, then for any series
\[\sum_ {\nu = 1} ^ {\infty} 2 ^ {\chi \nu} a _ {2 ^ {\nu}}\]
and
\[\sum_ {k = 1} ^ {\infty} k ^ {\chi - 1} a _ {k},\]
either converge or diverge at the same time.
Note that this lemma is implicitly contained in (see page 13).
Indeed, due to the monotony of , for any we have
\[2 ^ {\chi \nu} a _ {2 ^ {\nu}} \leq 2 ^ {| \chi | + 1} \sum_ {k = 2 ^ {\nu - 1} + 1} ^ {2 ^ {\nu}} k ^ {\chi - 1} a _ {k} \leq 2 ^ {2 | \chi | + 1} 2 ^ {\chi (\nu - 1)} a _ {2 ^ {\nu - 1}},\]
and from here
\[\sum_ {k = 1} ^ {\infty} k ^ {\chi - 1} a _ {k} = a _ {1} + \sum_ {\nu = 1} ^ {\infty} \sum_ {s = 2 ^ {\nu - 1} + 1} ^ {2 ^ {\nu}} s ^ {\chi - 1} a _ {s}.\]
Lemma 3. If and , then the flat , we have
\[\sum_ {k = 1} ^ {\infty} k \Delta a _ {k} = + \infty .\]
Let's put
\[k ^ {\chi - 1} a _ {k} = \sum_ {\nu = 1} ^ {k} \nu \Delta a _ {\nu}.\]
By virtue of we have ( ). So all and does not decrease monotonically. We need to prove that . If this were not true, then
\[k ^ {\chi - 1} a _ {k} \uparrow a (a \neq + \infty).\]
Then , , and hence since
\[k ^ {\chi - 1} a _ {k} - (k - 1) ^ {\chi - 1} a _ {k - 1} = k \Delta a _ {k} = (a - \varepsilon_ {k}) - (a - \varepsilon_ {k - 1}) = \Delta \varepsilon_ {k - 1},\]
that
\[\Delta a _ {k} = \frac {\Delta \varepsilon_ {k - 1}}{k}.\]
By virtue of and we have
\[a _ {k} = \sum_ {\nu = k} ^ {\infty} \Delta a _ {\nu} = \sum_ {\nu = k} ^ {\infty} \frac {\Delta \varepsilon_ {\nu - 1}}{\nu} \leq\]
\[\leq \frac {1}{k} \sum_ {\nu = k} ^ {\infty} \Delta \varepsilon_ {\nu - 1} = \frac {\varepsilon_ {k - 1}}{k},\]
and therefore
But, applying to the sum representing , the Abel transform, we find
\[k ^ {\chi - 1} a _ {k} = \sum_ {\nu = 1} ^ {k} \nu \Delta a _ {\nu} = \sum_ {\nu = 1} ^ {k + 1} a _ {\nu} = a _ {1} + a _ {2} + \ldots + a _ {k + 1},\]
and since , and , then , which contradicts the condition
\[\sum_ {k = 1} ^ {\infty} a _ {k} = + \infty .\]
Let be a convergent series with . We believe
\[r _ {n} = \sum_ {k = 1} ^ {\infty} k ^ {\chi - 1} a _ {k}.\]
We will say that a series satisfies condition (A) if
\[r _ {n} = O (k ^ {\chi - 1} a _ {k}).\tag{12}\]
If the terms of the series decrease no slower than some geometric progression, that is, if
\[(k + 1) ^ {\chi - 1} a _ {k + 1} < \theta k ^ {\chi - 1} a _ {k}, \quad 0 < \theta < 1,\]
then it satisfies condition (A), but the reverse conclusion is, of course, incorrect, as at least such an example shows
\[(2 k - 1) ^ {\chi - 1} a _ {2 k - 1} - (2 k) ^ {\chi - 1} a _ {2 k} = \theta^ {k}, \quad k \in N; \quad 0 < \theta < 1.\]
We show that if a series satisfies condition (A), then whatever , it can be divided into a finite l series (l envy of ) so that the terms of each of them decrease no slower than the geometric progression with the denominator .
Indeed, condition (12) means that , where is constant. Let be given. Let's choose the number so that
\[l = [ \frac {c}{\theta} ].\tag{13}\]
Then by virtue of the monotonous decreasing of the numbers and by virtue of (13) we have
\[(l + 1) (k + l) ^ {\chi - 1} a _ {k + l} \leq \sum_ {\nu = k} ^ {\nu = k + l} \nu^ {\chi - 1} a _ {\nu} \leq \sum_ {\nu = k} ^ {\infty} \nu^ {\chi - 1} a _ {\nu} \leq c k ^ {\chi - 1} a _ {k},\tag{14}\]
and therefore it follows from (14) that
\[(k + l) ^ {\chi - 1} a _ {k + l} \leq \frac {c}{l + 1} k ^ {\chi - 1} a _ {k} \leq \theta k ^ {\chi - 1} a _ {k}.\]
Hence, all series
\[1 ^ {\chi - 1} a _ {1} + (1 + l) ^ {\chi - 1} a _ {1 + l} + (1 + 2 l) ^ {\chi - 1} a _ {1 + 2 l} + \dots ,\]
\[2 ^ {\chi - 1} a _ {2} + (2 + l) ^ {\chi - 1} a _ {2 + l} + (2 + 2 l) ^ {\chi - 1} a _ {2 + 2 l} + \dots ,\]
\[l ^ {\chi - 1} a _ {l} + (2 l) ^ {\chi - 1} a _ {2 l} + (3 l) ^ {\chi - 1} a _ {3 l} + \dots ,\]
it can be decomposed into a series
\[\sum_ {\nu = 1} ^ {\infty} \nu^ {\chi - 1} a _ {\nu},\]
which decreases no slower than the geometric progression with the denominator .
1.2. Let the spectrum be and for the condition is satisfied. Let's choose an increasing function such that then
\[y ^ {\rho} = x \frac {y ^ {\rho}}{\lambda (y)} = O (x) \quad (x = \lambda (y)),\]
\[y = O \{x ^ {\frac {1}{\rho}} \}.\]
\[\mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) \leq \mu (2 ^ {\nu} \pi) + 1 = O \{2 ^ {\frac {\nu}{\rho}} \},\]
and condition (4) can be replaced by the condition
\[\sum_ {\nu = 1} ^ {\infty} 2 ^ {\frac {(1 - \beta) \nu}{\rho}} \omega^ {\beta} (f, 2 ^ {- \nu}) \omega_ {\Phi} ^ {\beta / 2} (f, 2 ^ {- \nu}) \Phi^ {- \beta / 2} [ \omega (f, 2 ^ {- \nu}) ] < \infty ,\]
или согласно леммы 2
\[\sum_ {\nu = 1} ^ {\infty} k ^ {\frac {(1 - \beta) \nu}{\rho} - 1} \omega^ {\beta} (f, k ^ {- 1}) \omega_ {\Phi} ^ {\beta / 2} (f, k ^ {- 1}) \Phi^ {- \beta / 2} [ \omega (f, k ^ {- 1}) ] < \infty .\]
The statement of Lemma 2 is obtained from the fact that the function is non-decreasing. By virtue of Theorem 2, the following holds.
Theorem 4. Let when . And let for the function at takes place
\[\sum_ {k = 1} ^ {\infty} k ^ {\frac {1 - \frac {\beta}{2}}{\rho} - 1} \omega_ {2} ^ {\beta} (f, k ^ {- 1}) < \infty ,\tag{15}\]
where . Then the series (7) converges.
For and , the result is obtained in (see [8], p. 137), and for we get Bernstein (see [7], p. 231)
\[\sum_ {k = 1} ^ {\infty} \frac {1}{\sqrt {n}} \omega_ {2} (f, k ^ {- 1}) < \infty .\]
Now we denote for . The value of is called r-a variation of the function .
According to Theorem 3 and Lemma 2, the following statement holds
Theorem 5. Let when . If at the function has a finite -variation, besides at exists
\[\sum_ {k = 1} ^ {\infty} k ^ {\frac {1 - \frac {\beta}{2}}{\rho} - \frac {\beta}{2} - 1} \omega_ {n} ^ {\beta (1 - \frac {r}{2})} (f, k ^ {- 1}) < \infty ,\]
In the case when and we get , which for obtained by Varashkevich and Zygmunde (see [8], p. 138). For , we obtain the following statement about the absolute convergence of the series (3).
Theorem 6. Let and when . Suppose that and for . Then the series (3) absolutely converges.
This theorem for was obtained by Sigmund. If in Theorem 5 we obtain the known condition of absolute convergence of series (3) ([7], p. 231)
\[\sum_ {k = 1} ^ {\infty} \frac {1}{k} \sqrt {\omega (f , k ^ {- 1})} < \infty ,\]
which generalizes Sigmund's result.
1.3. Definition 3. A sequence of natural numbers
\[n _ {1} < n _ {2} < \ldots < n _ {k} < \ldots\]
are called lacunar if there exists such a q > 1 that
\[\frac {n _ {k + 1}}{n _ {k}} \geq (k = 1, 2, \ldots).\]
Now let the lacunar condition
\[\frac {\lambda_ {k + 1}}{\lambda_ {k}} > q > 1\]
be satisfied for the spectrum . Then the spectrum will be increasing. Let's choose such a function so that the function . was increasing and . Then also increasing and denoting we have
\[y = \log_ {q} x - \log_ {q} \frac {\lambda (y)}{q ^ {y - 1}} + 1,\]
then in the case of , we have
\[\begin{array}{c} \mu (2 ^ {\nu} \pi) - \mu (2 ^ {\nu - 1} \pi) + 1 = l o g _ {q} 2 - l o g _ {q} \frac {\lambda (y _ {2}) q ^ {y _ {1}}}{q ^ {y _ {2}} \lambda (y _ {1})} + 1 \leq \\\leq l o g _ {q} 2 + 1 = O \{1 \} \end{array}\tag{16}\]
By virtue of Lemma 2, condition (6) can be replaced by the condition
\[\sum_ {k = 1} ^ {\infty} \frac {1}{k} \omega^ {\beta} (f, \frac {1}{k}) \omega_ {\Phi} ^ {\frac {\beta}{2}} (f, \frac {1}{k}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, \frac {1}{k}) ] < \infty .\]
Theorem 7. Let the function be bounded at . And let the condition of Theorem 1 be satisfied for the function , for , . Then for each the series (5) converges.
Theorem 8. Let . Then if and for the condition is satisfied, then for the series (5) converges.
By virtue of Theorem 3 we get:
Theorem 9. Let and the function satisfy the conditions of Theorem 3. Let the function be bounded and . Then at the series (5) converges.
For proofs, it is sufficient to note that the condition
\[\sum_ {\nu = 1} ^ {\infty} 2 ^ {\frac {- \beta \nu}{2}} \omega^ {\beta} (f, 2 ^ {- \nu}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ] < \infty\tag{17}\]
follows when substituting (16) for (11). However, according to Lemma 2, the condition (17) is equivalent to the condition
\[\sum_ {\nu = 1} ^ {\infty} \frac {1}{k ^ {1 + \frac {\beta}{2}}} \omega^ {\beta} (f, \frac {1}{k}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ] < \infty ,\]
which follows from the limitations of the spectrum
\[\omega^ {\beta} (f, \frac {1}{k}) \Phi^ {- \frac {\beta}{2}} [ \omega (f, 2 ^ {- \nu}) ]\]
and is performed when .
When is a bounded function, theorems 7, 8, 9 at are weaker than Sidon's results stating that the lacunar series of periodic and bounded absolute functions converge (see [8], page 139), for almost-periodic functions [6].
