Bogdan Szal
A note on the uniform convergence and boundedness of a generalized class of sine series
Abstract. In this paper we essentially extend the Leindler’s results concerning the uniform convergence and boundedness of a certain class of sine series.
2000 Mathematics Subject Classification: 40A30, 42A10.
Key words and phrases: sine series, Fourier series, embedding relations, number sequences.
1. Introduction. Chaundy and Jolliffe [1] proved the following classical result (see also [6]).
Theorem 1.1 Suppose that bn ≥ bn+1and bn → 0. Then a necessary and sufficient condition for the uniform convergence of the series
(1)
X∞ n=1
bnsin nx
is nbn→ 0.
If a null sequence c := (cn) of positive numbers has the property X∞
n=m
|cn− cn+1| ≤ K (c) cm
for all m ∈ N, where K (c) is a constant depending only on c, then we call the sequence c an R+0BV - sequence; and briefly we write c∈ R+0BV S.
In [3] L. Leindler generalized above result to the class R+0BV S. Namely, he proved the following theorems.
Theorem 1.2 If a sequence b = (bn) belongs to the class R+0BV S, then the condi- tion nbn→ 0 as n → ∞ is both necessary and sufficient for the uniform convergence of series (1).
Theorem 1.3 If a sequence b = (bn) belongs to the class R+0BV S, then the con- dition nbn = O (1) is both necessary and sufficient for the uniform boundedness of the partial sums of series (1).
Theorem 1.4 Suppose that b∈ R+0BV S. Then a necessary and sufficient condi- tion for the series (1) to be the Fourier series of a continuous function is nbn→ 0.
A nonnegative sequence c is said to be a sequence of Group Bounded Variation (GBV S) if there exists a natural number N such that
X2m n=m
|cn− cn+1| ≤ K (c) max
m≤n≤m+Ncn
holds for all m. In [2] R. Le and S. Zhou proved that Theorem 1.1 is true if a sequence b ∈ GBV S.
Let γ := (γn) be a fixed sequence of positive numbers. We say that a null sequence c of real numbers belongs to the class γRBV S if the inequality
X∞ n=m
|cn− cn+1| ≤ K (c) γm holds for all m ∈ N.
In [4] L. Leindler generalized Theorem 1.2 and Theorem 1.3 to the class γRBV S but he proved that for the class γRBV S only a sufficient condition in those theorems is valid.
In order to formulate our new results we define another class of sequences (see [5]).
Definition 1.5 A null sequence c := (cn) of positive numbers is called of Mean Rest Bounded Variation, or briefly c ∈ MRBV S, if it has the property
(2) X∞
n=2m
|cn− cn+1| ≤ K (c) 1 m + 1
X2m n=m
cn
for all natural numbers m.
It is clear that
R0+BV S⊆ MRBV S ⊆ γRBV S.
In the present paper we show that R+0BV S 6= MRBV S and GBV S 6= MRBV S.
Moreover, we prove that Theorem 1.2, Theorem 1.3 and Theorem 1.4 are true if a sequence b belongs to MRBV S.
2. Main results. We have the following results:
Theorem 2.1 There exists a sequence d =: (dn) with the property that ndn → 0 as n → ∞, which belongs to the class MRBV S but it does not belong to the class R+0BV S and GBV S.
Theorem 2.2 If a sequence b = (bn) belongs to the class MRBV S, then the condi- tion nbn→ 0 as n → ∞ is both necessary and sufficient for the uniform convergence of series (1).
Theorem 2.3 If a sequence b = (bn) belongs to the class MRBV S, then the con- dition nbn = O (1) is both necessary and sufficient for the uniform boundedness of the partial sums of series (1).
Theorem 2.4 Suppose that b∈ MRBV S. Then a necessary and sufficient condi- tion for the series (1) to be the Fourier series of a continuous function is nbn→ 0.
Remark 2.5 By the embedding relation R+0BV S ⊆ MRBV S we can observe that Theorem 1.2, Theorem 1.3 and Theorem 1.4 are the corollaries of Theorem 2.2, Theorem 2.3 and Theorem 2.4, respectively.
3. Proofs of Theorems. In this section we shall prove our results.
3.1. Proof of Theorem 2.1. Denote by µm := 2m for m = 1, 2, 3, ... and define a sequence (dn) by the following formulas d1= 1 and
dn:= 1 + m + (−1)nm
(2µm)2m if µm≤ n < µm+1.
It is clear that ndn → 0 as n → ∞. Namely, for any n > 1 there exists a natural number m such that µm≤ n < µm+1. Hence
ndn≤ µm+11 + m + (−1)n+1m
(2µm)2m ≤ µm+1 1 + 2m
(2µm)2m ≤ 3µm+1
(2µm)2. Using the inequality
(3) 2s ≤ 2s (for s ≥ 1),
we have
ndn≤ 3µm+1
2µm2µm
= 3 2µm ≤ 3
2n2 . Since the sequence 3
2n2 → 0 as n → ∞, we obtain that ndn→ 0 as n → ∞.
Now we show that the sequence (dn) does not belong to the class R+0BV S.
Namely, for m ≥ 2 we have X∞
i=µm+1
|di− di+1| ≥
2µXm−3 i=µm+1
|di− di+1| ≥
2µXm−3 i=µm+1
(di− di+1)
= dµm+1− dµm+1−2
=
1
(2µm)2m− 1 + 2m (2µm)2m
= 2
(2µm)2 = 2mdµm+1
and since dµm+1= (2µm1)2m, the inequality X∞
i=n
|di− di+1| ≤ K (d) dn
does not hold, that is, (dn) does not belong to R+0BV S.
Next, we prove that the sequence (dn) does not belong to the class GBV S.
Let m ≥ 1. Then we get
2µm
X
i=µm
|di− di+1| ≥
µm+1X−2 i=µm
|di− di+1|
=
µm+1X−2 i=µm
1 + 2m
(2µm)2m− 1 (2µm)2m
=
µm+1X−2 i=µm
2
(2µm)2 = 2
(2µm)2(2m− 1) and since for any N = 0, 1, 2, ...
µm≤i≤µmaxm+Ndi= 1 + 2m (2µm)2m, the inequality
X2m n=m
|dn− dn+1| ≤ K (d) max
m≤n≤m+Ndn
does not hold, (dn) does not belong to GBV S.
Finally, we show that the sequence (dn) belongs to MRBV S.
Using the inequality (3) we have
(4) 2µs+1 ≥ 2µs+s+1 (for s ≥ 1).
Hence we get for r = 1, 2, 3, ... and k = 0, 1, 2, ..., 1 + 2 (k + r + 1)
(2µk+r+1)2(k + r + 1) ≤ 3
(2µk+r+1)2 ≤ 3
(2µk+r)22k+r2k+r+2
(5) ≤ 3
25
1
(2µk+r)2(k + r) ≤ 1
(2µk+r)2(k + r).
If n = 1, then by (3) and (5) we have X∞
i=2
|di− di+1| = X∞ r=1
µr+1X−2 i=µr
1 + 2r
(2µr)2r− 1 (2µr)2r
+
1
(2µr)2r − 1 + 2 (r + 1) (2µr+1)2(r + 1)
!
≤ X∞ r=1
µr+1X−2 i=µr
2
(2µr)2 + 1 (2µr)2r
≤X∞
r=1
3 (2µr)2µr
≤3 4
X∞ r=1
1 2r =3
4 ≤ 3 2
d1+ d2
2
and (2) obviously holds. For any n ≥ 2 there exist two numbers k = 1, 2, 3, ... and l = 0, 1, 2, ..., 2k− 1 such that n = µk+ l. Hence
X∞ i=2(µk+l)
|di− di+1| = X∞ i=µk+1+2l
|di− di+1| =
µk+2X−2 i=µk+1+2l
|di− di+1|+ dµk+2−1− dµk+2
+ X∞ r=1
µk+r+2X−2 i=µk+r+1
|di− di+1| + dµk+r+2−1− dµk+r+2
=
µk+2X−2 i=µk+1+2l
1 + 2 (k + 1)
(2µk+1)2(k + 1)− 1 (2µk+1)2(k + 1)
+
1
(2µk+1)2(k + 1)− 1 + 2 (k + 2) (2µk+2)2(k + 2)
+X∞
r=1
µk+r+2X−2 i=µk+r+1
1 + 2 (k + r + 1)
(2µk+r+1)2(k + r + 1)− 1
(2µk+r+1)2(k + r + 1)
+
1
(2µk+r+1)2(k + r + 1)− 1 + 2 (k + r + 2) (2µk+r+2)2(k + r + 2)
! . By (4) we have
X∞ i=2(µk+l)
|di− di+1| ≤
µk+2X−2 i=µk+1+2l
2
(2µk+1)2 + 1 (2µk+1)2(k + 1)
+X∞
r=1
µk+r+2X−2 i=µk+r+1
2
(2µk+r+1)2 + 1
(2µk+r+1)2(k + r + 1)
= 2
(2µk+1)2 2k+1− 2l − 1+ 1 (2µk+1)2(k + 1) +X∞
r=1
2
(2µk+r+1)2 2k+r+1− 2+ 1
(2µk+r+1)2(k + r + 1)
!
≤ 4
(2µk+1)2 2k− l
+ 1
(2µk+1)2(k + 1) +
X∞ r=1
2
(2µk+r+1)22k+r+1+ 1
(2µk+r+1)2(k + r + 1)
!
= 4
µk+1X−1 i=µk+l
1
(2µk+1)2 + 1 (2µk+1)2(k + 1)
+ X∞ r=1
2
(2µk+r+1)22k+r+1+ 1
(2µk+r+1)2(k + r + 1)
! . Using the inequalities (3), (4) and
2µk+r+1≥ 2µk+1+k+r+1 for r, k = 1, 2, 3..., we obtain
X∞ i=2(µk+l)
|di− di+1| ≤ 4
µk+1X−1 i=µk+l
1
(2µk)222(k+1) + 1 (2µk+1)2(k + 1)
+X∞
r=1
2
(2µk+1)222(k+r+1)2k+r+1+ 1
(2µk+1)222(k+r+1)(k + r + 1)
!
≤ 1
2k+1
µk+1X−1 i=µk+l
1
(2µk)2k + 1 (2µk+1)2(k + 1)
+ X∞ r=1
1
2r(2µk+1)2(k + 1)+ 1
(2µk+1)222(k+r+1)(k + r + 1)
!
≤ 1
2k+1
µk+1X−1 i=µk+l
1
(2µk)2k + 1 (2µk+1)2(k + 1)
+ 2
(2µk+1)2(k + 1) X∞ r=1
1 2r
= 1
2k+1
µk+1X−1 i=µk+l
1
(2µk)2k + 3 (2µk+1)2(k + 1)
= 1 2k+1
µk+1X−1 i=µk+l
1
(2µk)2k+ 3 2k+ 1
µk+1+l
X
i=µk+l
1 (2µk+1)2(k + 1)
≤ 1
2k+1
µk+1X−1 i=µk+l
1
(2µk)2k+ 6 2k+1
µk+1+l
X
i=µk+l
1 (2µk+1)2(k + 1)
= 1
2k+1
µk+1X−1 i=µk+l
1
(2µk)2k+ 6 2k+1
µk+1X−1 i=µk+l
+
µk+1+l
X
i=µk+1
1
(2µk+1)2(k + 1) By (4) we obtain
X∞ i=2(µk+l)
|di− di+1| ≤ 7 2k+1
µk+1X−1 i=µk+l
1
(2µk)2k+ 6 2k+1
µk+1+l
X
i=µk+1
1 (2µk+1)2(k + 1)
≤ 7
2k+1
µk+1X−1 i=µk+l
1 (2µk)2k+
µk+1+2l
X
i=µk+1
1 (2µk+1)2(k + 1)
.
Let for u ≤ v, A (u, v) and B (u, v) denote the sets of all even and odd integer from the set {u, u + 1, u + 2, ..., v}, respectively. Hence
X∞ i=2(µk+l)
|di− di+1| ≤ 7 2k+1
X
i∈A(µk+l,µk+1−1)
+ X
i∈B(µk+l,µk+1−1)
1
(2µk)2k
+
X
i∈A(µk+1,µk+1+2l)
+ X
i∈B(µk+1,µk+1+2l)
1
(2µk+1)2(k + 1)
≤ 7
2k+ l
X
i∈A(µk+l,µk+1−1)
+ X
i∈B(µk+l,µk+1−1)
1 + k + (−1)i+1k (2µk)2k
+
X
i∈A(µk+1,µk∗1+2l)
+ X
i∈B(µk+1,µk∗1+2l)
1 + (k + 1) + (−1)i+1(k + 1) (2µk+1)2(k + 1)
= 7
µk+ l
2(µk+l)
X
i=µk+l
di
and (2) is valid. This ends our proof.
3.2. Proof of Theorem 2.2. First we prove the necessity. Setting x = 4mπ , we get
(6)
X2m n=m
bnsin nx = X2m n=m
bnsin nπ
4m ≥sinπ 4
X2m n=m
bn.
If b ∈ MRBV S, by (2) with b in place of c, we have
b2m= X∞
n=2m
(bn− bn+1) ≤ X∞
n=2m
|bn− bn+1| ≤ K (b) 1 m + 1
X2m n=m
bn
and
b2m−1 = X∞
n=2m−1
(bn− bn+1) ≤ X∞
n=2m−1
|bn− bn+1| ≤ X∞ n=2m
|bn− bn+1|
≤ K (b) 1 m + 1
X2m n=m
bn. (7)
Hence, by (6) and (7), and taking into account that the series (1) converges uni- formly, we obtain that mb2m→ 0 and mb2m−1 → 0 as m → ∞, and this verifies the necessity of the condition nbn → 0 as n → ∞.
Now, we prove the sufficiency. Denote by
εn:= sup
k≥n
kbk and rn(x) :=
X∞ k=2n
bksin kx.
In view of the assumptions, we have that εn → 0 as n → ∞. We show that
(8) |rn(x)| ≤ Kεn
also holds. Since rn(kπ) = 0, it suffices to prove (8) for 0 < x < π.
Let N = N (x) be the integer such that
(9) π
N + 1 < x≤ π N. Then
rn(x) =
2(n+N )−1X
k=2n
bksin kx + X∞ k=2(n+N )
bksin kx = r(1)n (x) + rn(2)(x) .
Hence, by (9),
(10) rn(1)(x) ≤ x
2(n+N )−1X
k=2n
kbk ≤ 2xNεn ≤ 2πεn. Using the well known inequality
|Dn(x)| :=
Xn k=1
sin kx ≤ π
x for n = 1, 2, ...
and (2) and (9), by Abel’s transformation we have
r(2)n (x) ≤ X∞ k=2(n+N )
|bk− bk+1| |Dk(x)| + b2(n+N )
D2(n+N )−1(x)
≤π x
K (b) 1 n + N + 1
2(n+N )
X
k=n+N
bk+ b2(n+N )
≤ (N + 1)
K (b) 1 n + N + 1
2(n+N )
X
k=n+N
bk+ b2(n+N )
.
By (7) we get
rn(2)(x) ≤ 2K (b) N + 1 n + N + 1
2(n+N )
X
k=n+N
bk
(11) ≤ 2K (b) 1
n + N + 1
2(n+N )
X
k=n+N
kbk ≤ 2K (b) εn.
From the estimations (10) and (11) we obtain the uniform convergence of series (1) and thus the proof is complete.
3.3. Proof of Theorem 2.3. The proof of Theorem 2.3 goes analogously as the proof of Theorem 2.2. Now, we have
X2m n=m
bn≤ K.
Hence from (7) it follows that mb2m≤ K and mb2m−1 ≤ K.
In the proof of sufficiency, the only difference is that εn should be replaced by a positive constant.
3.4. Proof of Theorem 2.4. If nbn → 0 as n → ∞, by Theorem 2.2, we obtain that series (1) is uniformly convergent. From this and by Fej´er’s theorem we obtain that the series (1) is the Fourier series of a continuous function.
Now, we prove the necessity of the condition nbn→ 0. If series (1) is the Fourier series of a continuous function, then the (C, 1) −means
σn(x) = Xn k=1
bk
1 − k
n + 1
sin kx
of this series converges uniformly. In particular
(12) σ4m π
8m
→ 0 as m → ∞.
Using the inequality sin x ≥ 2πx in0,π2we have
(13) σ4m(x) = X4m k=1
bk
1 − k 4m + 1
sin kx ≥ X4m k=1
bk
1 − k 4m + 1
2kx π
for x ∈0,8mπ . Hence, by (13) and (7),
σ4m π 8m
≥ X4m k=1
bk
1 − k 4m + 1
k 4m
≥ 1 4m
X2m k=m
bk
1 − k 4m + 1
k≥ 1
8 X2m k=m
bk ≥ 1
8K (b)mb2m.
Hence by (12) we have that nbn → 0 as n → ∞, and the proof is complete.
References
[1] T. W. Chaundy and A. E. Jolliffe, The uniform convergence of a certain class of trigonomet- rical series, Proc. London Math. Soc. 15 (1916), 214-216.
[2] R. J. Le and S. P. Zhou, A new condition for the uniform convergence of certain trigonometric series, Acta Math. Hungar. 108 (1-2) (2005), 161-169.
[3] L. Leindler, On the uniform convergence and boundedness of a certain class of sine series, Analysis Math. 27 (2001), 279-285.
[4] L. Leindler, A note on the uniform convergence and boundedness of a new class of sine series, Analysis Math. 31 (2005), 269-275.
[5] L. Leindler, Integrability conditions pertaining to Orlicz space, J. Inequal. Pure and Appl.
Math. 8(2) (2007), Art. 38, 6 pp.
[6] A. Zygmund, Trigonometric series, Vol. I, University Press, Cambridge 1959.
Bogdan Szal
University of Zielona G´ora Faculty of Mathematics, Computer Science and Econometrics 65-516 Zielona G´ora, ul. Szafrana 4a, Poland
E-mail: B.Szal@wmie.uz.zgora.pl
(Received: 4.09.2007)
