EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 6, No. 4, 2013, 451-459 ISSN 1307-5543 – www.ejpam.com On Integrability of Trigonometric Series with Special Type of Coefficients Xhevat Z. Krasniqi Department of Mathematics and Informatics, Faculty of Education, University of Prishtina "Hasan Prishtina", Avenue "Mother Theresa" 5, 10000 Prishtina, Republic of Kosovo Abstract. In this paper some condition on integrability of cosine and sine trigonometric series with coefficients that keep their signs are obtained. The results extend some previous results of Telyakovskĭı. 2010 Mathematics Subject Classifications: 42A16, 42A20 Key Words and Phrases: Trigonometric series, quasi-convex sequence, bounded variation sequence, integrability 1. Introduction and Preliminaries Several mathematicians have studied the integrability conditions for trigonometric series with different types of coefficients. The first results pertaining to the trigonometric series of the form a0 2 + ∞ ∑ k=1 ak cos kx (1) ∞ ∑ k=1 ak sin kx (2) considered the case of monotone coefficients. Later, some authors investigated the series (1) with quasi-monotone coefficients (an+1 ≤ an(1+α/n), n≥ n0, α > 0). Many papers have been written on the series (1) when the sequence {ak} is a null- sequence and convex or quasi-convex, i.e. 42ak ≥ 0 or ∞ ∑ k=1 (k+ 1)|42ak|<∞, (3) where 42ak =4 � 4ak � , 4ak = ak − ak+1. Email address: xhevat.krasniqi@uni-pr.edu http://www.ejpam.com 451 c© 2013 EJPAM All rights reserved. Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 452 Furthermore, when {ak} is a null-sequence of bounded variation, i.e. ∑∞ k=1 |4ak| <∞, is also considered. We shall consider the series (1) and (2) whose coefficients tend to zero and satisfy any condition that provides their convergence on (0,π]. Let us denote their sums with f (x) and g(x) respectively. If the coefficients ak are quasi-convex, it is well-known that f is an integrable function on [0,π] (see [1]), and the estimation ∫ π 0 | f (x)|d x ≤ π ∞ ∑ k=1 (k+ 1)|42ak| is valid. In a similar direction, among others, S. A. Telyakovskĭı [6] obtained some estimates of the integrals of the following form ∫ π/` π/(m+1) |φ(x)|d x , 1≤ `≤ m, (`, m ∈ N), (4) expressed in terms of the coefficients ak, where he used null-sequences of bounded variation of second order ( ∑∞ k=1 |4 2ak| < ∞), instead of quasi-convex null-sequences. Here φ(x) is either f (x) or g(x). It is obvious that the condition ∞ ∑ k=1 |42ak|<∞ (5) is a weaker condition than the condition (3). The following definition is introduced in [4]: A sequence {ak} is of bounded variation of integer order p ≥ 0 if ∞ ∑ k=1 |4pak|<∞, (6) where 4pak =4 � 4p−1ak � =4p−1ak −4p−1ak+1, and we agree with 40ak = ak. In [4] an example is given to show that (6) is an effective generalization of the null sequences of bounded variation. This fact encouraged the present author to consider the series (1) with coefficients that satisfy the condition (6). The results are published in [2]. Also similar results the reader can find in [3, 4]. For an integer non-negative number r and a sequence {ak} we write4r ak = ak−ak+r and 42 r ak =4r � 4r ak � = ak − 2ak+r + ak+2r . Note that for r = 1 we obtain ordinary differences 4ak = ak − ak+1 and 42ak =4 � 4ak � = ak − 2ak+1+ ak+2. Let r ∈ N, k = 1,2, . . . , r, n= 0, 1,2, . . . , B0 0,r,k(x) = sin ((2k− r)x/2) 2 sin(r x/2) , x 6= 2mπ/r, m ∈ Z, Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 453 B0 n+1,r,k(x) = cos(k+ nr)x , BC1 n,r,0(x) = sin � (2n+ 1)r x 2 � 2sin � r x 2 � B1 n,r,k(x) = n ∑ m=0 B0 m,r,k(x), B2 n,r,k(x) = n ∑ m=0 B1 m,r,k(x), BC1 n,r,k(x) = sin � (2k+ (2n+ 1)r) x 2 � − sin � (2k− r) x 2 � 2sin � r x 2 � B 1 n+1,r,k(x) = n ∑ m=0 sin(k+mr)x = cos � (2k− r) x 2 � − cos � (2k+ (2n+ 1)r) x 2 � 2 sin � r x 2 � . The following definition is introduced in [5]: A sequence {an} keeps its sign if either an ≥ 0 for all n, or an ≤ 0 for all n. Also in the same paper are proved some lemmas formulated below. Lemma 1. Let r ∈ N, k = 1,2, . . . , r, n= 0,1, 2, . . . . (a) If the sequence {4r ak+nr} keeps its sign separately for each k, then the series (1) and (2) converge for almost all x. The function g(x +m2π r ) is almost everywhere representable in the form g(x +m 2π r ) = r ∑ k=1 cos � km 2π r � ∞ ∑ n=0 4r ak+nr B 1 n+1,r,k(x) + r−1 ∑ k=1 sin � km 2π r � ∞ ∑ n=0 4r ak+nr BC1 n,r,k(x). (b) If the sequence {42 r ak+nr} keeps its sign separately for each k, then f (x) is almost everywhere representable in the form f (x) = r ∑ k=1 ∞ ∑ n=0 42 r ak+(n−1)r B2 n,r,k(x). Lemma 2. Let r ∈ N, an→ 0 for n→∞ and 42,r an ≥ 0 for all n. Then 4r an ≥ 0 and an ≥ 0 for all n. The aim of this paper is to achieve some results, similar to those of Telyakovskĭı [6], for the series (1) and (2) with coefficients that satisfy the conditions: the sequences {4r ak+nr} and {42 r ak+nr} keep their sign separately for each k. We write g(u) = Or (h(u)), u→ 0, if there exists a positive constant Ar , that depends only on r, such that g(u)≤ Arh(u) in a neighborhood of the point u= 0. The constants Ar may be, in general, different in different estimates. Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 454 2. Main Results We begin with the following result regarding to the cosine series. Theorem 1. Let r ∈ N, k = 1, 2, . . . , r. If an → 0 as n→∞ and the sequence {42 r ak+nr} keeps its sign separately for each k, then the series (1) converges for almost all x, and for 1 ≤ ` ≤ m, the sum function f (x) satisfies ∫ π/` π/(m+1) | f (x)|d x =O m+ 1− ` m r ∑ k=1 `−1 ∑ n=0 n+ 1 ` |4r ak+(n−1)r | ! +O r ∑ k=1 ∞ ∑ n=` min(n+ 1− `, m+ 1− `)|42 r ak+(n−1)r | ! . Proof. The convergence for almost all x of the series (1) has been proved in Lemma 1(a). Moreover, from Lemma 1(b) the sum function f (x) is almost everywhere representable in the form f (x) = r ∑ k=1 ∞ ∑ n=0 42 r ak+(n−1)r B2 n,r,k(x). (7) Let i be a positive integer and x ∈ � π i+1 , π i � . With agreement that B2 −1,r,k(x)≡ 0 and using the equality r ∑ k=1 i−1 ∑ n=0 42 r ak+(n−1)r B2 n,r,k(x) = = r ∑ k=1 i−1 ∑ n=0 � 4r ak+(n−1)r −4r ak+nr � B2 n,r,k(x) = r ∑ k=1 i−1 ∑ n=0 4r ak+(n−1)r � B2 n,r,k(x)− B2 n−1,r,k(x) � −4r ak+(i−1)r B2 i−1,r,k(x) ! = r ∑ k=1 i−1 ∑ n=0 4r ak+(n−1)r B1 n,r,k(x)−4r ak+(i−1)r B2 i−1,r,k(x) ! , from (7) we have f (x) = r ∑ k=1 i−1 ∑ n=0 4r ak+(n−1)r B1 n,r,k(x) + ∞ ∑ n=i 42 r ak+(n−1)r � B2 n,r,k(x)− B2 i−1,r,k(x) � ! . It is obvious that |B1 n,r,k(x)| ≤ n+ 1, and since from (see [5, page 65]) B2 n,r,k(x) = sin2 � (k+ nr) x 2 � − sin2 � (k− r) x 2 � 2sin2 � r x 2 � Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 455 follows |B2 n,r,k(x)− B2 i−1,r,k(x)| ≤ 2 sin2 � r x 2 � , we have ∫ π/i π/(i+1) | f (x)|d x = O ( r ∑ k=1 i−1 ∑ n=0 |4r ak+(n−1)r | k+ 1 i(i+ 1) + ∞ ∑ n=i |42 r ak+(n−1)r | !) . Now if we take the summation, when i goes from ` to m, to the both sides of the above equality we get ∫ π/` π/(m+1) | f (x)|d x = O ( r ∑ k=1 m ∑ i=` i−1 ∑ n=0 |4r ak+(n−1)r | n+ 1 i(i+ 1) + m ∑ i=` ∞ ∑ n=i |42 r ak+(n−1)r | !) . (8) For the first term in the parentheses of the right-hand side of (8) we have m ∑ i=` i−1 ∑ n=0 |4r ak+(n−1)r | n+ 1 i(i+ 1) = = m ∑ i=` `−1 ∑ n=0 |4r ak+(n−1)r | n+ 1 i(i+ 1) + m ∑ i=`+1 i−1 ∑ n=` |4r ak+(n−1)r | n+ 1 i(i+ 1) = `−1 ∑ n=0 (n+ 1)|4r ak+(n−1)r | � 1 ` − 1 m+ 1 � + m−1 ∑ n=` (n+ 1)|4r ak+(n−1)r | � 1 n+ 1 − 1 m+ 1 � ≤ m+ 1− ` m `−1 ∑ n=0 n+ 1 ` |4r ak+(n−1)r |+ m ∑ n=` ∞ ∑ j=n |42 r ak+( j−1)r |. (9) But the second term in (9) can be written as m ∑ i=` ∞ ∑ n=i |42 r ak+(n−1)r |= m ∑ i=` m ∑ n=i |42 r ak+(n−1)r |+ m ∑ i=` ∞ ∑ n=i |42 r ak+(n−1)r | = m ∑ n=` (n+ 1− `)|42 r ak+(n−1)r | + (m+ 1− `) ∞ ∑ n=m+1 |42 r ak+(n−1)r |. (10) The proof of theorem follows from (8), (9) and (10). Now we shall prove an estimation of the integral in Theorem 1 only in terms of second order difference of the sequence {ak+(n−1)r}. Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 456 Corollary 1. If the coefficients of the series (1) satisfy conditions of the Theorem 1, then ∫ π/` π/(m+1) | f (x)|d x = O m+ 1− ` m r ∑ k=1 ∞ ∑ n=0 min � (n+ 1)2 ` , n+ 1, m � |42 r ak+(n−1)r | ! . Proof. To deduce the required estimation we use the identity 4r ak+(n−1)r = ∞ ∑ i=n � 42 r ak+(i−1)r � . We have `−1 ∑ n=0 n+ 1 ` |4r ak+(n−1)r | ≤ `−1 ∑ n=0 n+ 1 ` ∞ ∑ i=n |42 r ak+(i−1)r | = `−1 ∑ i=0 i ∑ n=0 n+ 1 ` |42 r ak+(i−1)r |+ ∞ ∑ i=` `−1 ∑ n=0 n+ 1 ` |42 r ak+(i−1)r | ≤ `−1 ∑ i=0 (i+ 1)2 ` |42 r ak+(i−1)r |+ ` ∞ ∑ i=` |42 r ak+(i−1)r |. (11) If k < m, then we can estimate the second term in the estimation of the Theorem 1 by means of the fact that n+ 1− `≤ n+ 1− ` n+ 1 m = m− ` m (n+ 1). Finally, from the above and (11) along with the estimate of the Theorem 1 we immediately obtain the required estimation. In the following we shall deal with trigonometric series of the form (2). Theorem 2. Let r ∈ N, k = 1, 2, . . . , r. If an → 0 as n→∞ and the sequence {4r ak+nr} keeps its sign separately for each k, then the series (2) converges for almost all x, and for 1 ≤ ` ≤ m, the sum function g(x) satisfies ∫ π/` π/(m+1) | f (x)|d x = r ∑ k=1 m ∑ i=` di,r,k k |ak+ir | +Or    m+ 1− ` m r ∑ k=1 `−1 ∑ n=1 n2 `2 |4r ak+nr |+ r ∑ k=1 m ∑ n=` ∞ ∑ j=n |42 r ak+ jr |    , where di,r,k := ln sin rπ 2i sin rπ 2(i+1) + cos kπ 2i(i+ 1) cos kπ(2i+ 1) 2i(i+ 1) . Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 457 Proof. Under assumptions of the theorem and Lemma 1(a) the series (2) converges for almost all x and for m = 0 the sum function g(x) is almost everywhere representable in the form g(x) = r ∑ k=1 ∞ ∑ n=0 4r ak+nr B 1 n+1,r,k(x). Let us denote ϕn,r,k(x) :=− cos(2k+ (2n+ 1)r) x 2 2sin � r x 2 � , ψn,r,k(x) := n ∑ s=0 ϕs,r,k(x) = sin(k+ nr)x + sin kx 4sin2 � r x 2 � . Let i ∈ N be such that i ≥ r. From definition of B 1 n+1,r,k(x) we can write g(x) = r ∑ k=1 i−1 ∑ n=0 4r ak+nr B 1 n+1,r,k(x) + r ∑ k=1 ∞ ∑ n=i 4r ak+nr B 1 n+1,r,k(x) = r ∑ k=1 i−1 ∑ n=0 4r ak+nr B 1 n+1,r,k(x) + r ∑ k=1 ∞ ∑ n=i 4r ak+nr cos(2k− r) x 2 2 sin � r x 2 � − r ∑ k=1 ∞ ∑ n=i 4r ak+nr cos(2k+ (2n+ 1)r) x 2 2 sin � r x 2 � = r ∑ k=1 ak+ir cos(2k− r) x 2 2 sin � r x 2 � + r ∑ k=1 i−1 ∑ n=0 4r ak+nr B 1 n+1,r,k(x) + r ∑ k=1 ∞ ∑ n=i 4r ak+nrϕn,r,k(x) := h0(x) + h1(x) + h2(x). (12) For x ∈ � π i+1 , π i � , i = 1,2, . . . , we have ∫ π/` π/(m+1) |h1(x)|d x ≤ m ∑ i=` ∫ π/i π/(i+1) r ∑ k=1 i−1 ∑ n=0 |4r ak+nr ||B 1 n+1,r,k(x)|d x = r ∑ k=1 m ∑ i=` `−1 ∑ n=0 |4r ak+nr | ∫ π/i π/(i+1) |B1 n+1,r,k(x)|d x + r ∑ k=1 m ∑ i=`+1 i−1 ∑ n=` |4r ak+nr | ∫ π/i π/(i+1) |B1 n+1,r,k(x)|d x = r ∑ k=1 `−1 ∑ n=0 |4r ak+nr | ∫ π/` π/(m+1) |B1 n+1,r,k(x)|d x Xh. Krasniqi / Eur. J. Pure Appl. Math, 6 (2013), 451-459 458 + r ∑ k=1 m−1 ∑ n=` |4r ak+nr | ∫ π/(n+1) π/(m+1) |B1 n+1,r,k(x)|d x . (13) Since |B1 n+1,r,k(x)| ≤ n ∑ m=0 (k+mr)x ≤ 4rn2 x , then from (13) it follows that ∫ π/` π/(m+1) |h1(x)|d x ≤Cr r ∑ k=1 `−1 ∑ n=1 n2|4r ak+nr | � 1 `2 − 1 (m+ 1)2 � + Cr r ∑ k=1 m−1 ∑ n=` n2|4r ak+nr | � 1 (n+ 1)2 − 1 (m+ 1)2 � ≤Cr m+ 1− ` m r ∑ k=1 `−1 ∑ n=1 n2 `2 |4r ak+nr |+ Cr r ∑ k=1 m ∑ n=` ∞ ∑ j=n |42 r ak+ jr |. (14) Now we shall estimate the integral of the function |h2(x)| for x ∈ � π i+1 , π i � . Indeed, the summation by parts gives h2(x) = lim p→∞ r ∑ k=1 p ∑ n=i 4r ak+nrϕn,r,k(x) = r ∑ k=1 lim p→∞ � p−1 ∑ n=i 42 r ak+nr n ∑ s=0 ϕs,r,k(x) −4r ak+ir i−1 ∑ s=0 ϕs,r,k(x) +4r ak+pr p ∑ s=0 ϕs,r,k(x) � = r ∑ k=1 ∞ ∑ n=i 42 r ak+nrψn,r,k(x)−4r ak+irψi−1,r,k(x) ! = r ∑ k=1 ∞ ∑ n=i 42 r ak+nr � ψn,r,k(x)−ψi−1,r,k(x) � , where ψn,r,k(x) are defined as above. So, we have ∫ π/` π/(m+1) |h2(x)|d x ≤Cr r ∑ k=1 m ∑ i=` ∫ π/i π/(i+1) ∞ ∑ n=i |42 r ak+nr | d x x2 ≤Cr r ∑ k=1 m ∑ i=` ∞ ∑ n=i |42 r ak+nr |. (15) REFERENCES 459 It is clear that for i ≥ r ∫ π/i π/(i+1) � � � � cos(2k− r) x 2 2 sin � r x 2 � � � � � d x = 1 2 ∫ π/i π/(i+1) � � � � cos kx cot � r x 2 � + sin kx � � � � d x ≤ 1 2 ∫ π/i π/(i+1) cot � r x 2 � d x + 1 2 ∫ π/i π/(i+1) sin kxd x ≤ 1 k ln sin rπ 2i sin rπ 2(i+1) + cos kπ 2i(i+ 1) cos kπ(2i+ 1) 2i(i+ 1) ! = di,r,k k . (16) Therefore from (16) we have ∫ π/` π/(m+1) |h0(x)|d x ≤ r ∑ k=1 m ∑ i=` |ak+ir | ∫ π/i π/(i+1) � � � � cos(2k− r) x 2 2sin � r x 2 � � � � � d x ≤ r ∑ k=1 m ∑ i=` di,r,k k |ak+ir |. (17) Finally, the proof of the theorem is an immediate result of relations (12), (14), (15) and (17). References [1] A. N. Kolmogorov. Sur l’ordre de grandeur des coefficients de la série de Fourier- Lebesgue. Bulletin de l’Academie Polonaise, 83–86, 1923. [2] Xh. Z. Krasniqi. Integrability of cosine trigonometric series with coefficients of bounded variation of order p. Applied Mathematics E-Notes, 11:61–66, 2011. [3] Xh. Z. Krasniqi. On the first derivative of the sums of trigonometric series with quasi- convex coefficients of higher order. Acta et Commentationes Universitatis Tartuensis de Mathematica, 14:53–63, 2010. [4] Xh. Z. Krasniqi. Integrability of double cosine trigonometric series with coefficients of bounded variation of second order. Commentationes Mathematicae, 51:125-139, 2011. [5] B. V. Simonov. Trigonometric series in the Orlicz–Lorentz spaces. Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 51:61–74, 2007. [6] S. A. Telyakovskĭı. Localizing the conditions of integrability of trigonometric series (Rus- sian). Theory of functions and differential equations, Collection of articles. In honor of the ninetieth birthday of Academician S. M. Nikolskii(Russian), Trudy Matematicheskogo Instituta Imeni V. A. Steklova, 210:264–273, 1995.