Xhevat Z. Krasniqi
On the degree of approximation of continuous functions by matrix means related to partial sums
of a Fourier series
Abstract. In this paper we generalize some results on the degree of approximation of continuous functions by matrix means related to partial sums of a Fourier series, obtained previously by some other authors (please consult references cited in this paper).
2010 Mathematics Subject Classification: 42A24, 41A25..
Key words and phrases: Matrix transformation, Degree of approximation, Single Fourier series.
1. Introduction and the aim of the paper. Let f(x) be a 2π- periodic continuous function. Let Sn(f; x) denote the n-th partial sum of its Fourier series at x and let ω(δ) = ω(δ, f) denote the modulus of continuity of f.
Let A := (an,k) (k, n = 0, 1, . . . ) be a lower triangular infinite matrix of real numbers and let the A-transform of {Sn(f; x)} be given by
Tn,A(f) := Tn,A(f; x) :=
Xn k=0
an,kSk(f; x) (n = 0, 1, . . . ).
The deviation kTn,A(f) − fk was estimated by P. Chandra [1] and [2] for mo- notonic sequences {an,k}, where k · k denotes the supnorm. Later on, these results are generalized by L. Leindler [3] who considered the sequences of Rest Bounded Variation and of Head Bounded Variation.
A sequence c := {cn} of nonnegative numbers tending to zero is called of Rest Bounded Variation, or briefly c ∈ RBV S, if it has the property
X∞ n=m
|cn− cn+1| ¬ K(c)cm
for all natural numbers m, where K(c) is a constant depending only on c.
A sequence c := {cn} of nonnegative numbers will be called of Head Bounded Variation, or briefly c ∈ HBV S, if it has the property
m−1X
n=0
|cn− cn+1| ¬ K(c)cm
for all natural numbers m, or only for all m ¬ N if the sequence c has only finite nonzero terms, and the last nonzero term is cN.
Since Chandra’s and Leindler’s results are not connected directly to our results, here we shall not recall those, just for interested reader we would like to make mention that some generalizations of Leindler’s results are made by present author in [4].
Very recently B. Wei and D. Yu [5] have generalized Leindler’s results, and thus Chandra’s results, without assumptions that A ∈ RBV S or A ∈ HBV S. They verified there that Leindler’s results are consequences of their results. Before we recall their results we shall suppose that
an,k 0, Xn k=0
ank= 1, (1)
and ω(t) is such that Z π
u
t−2ω(t)dt = O(H(u)), (u → 0+), (2) where H(u) 0, and
Z t
0 H(u)du = O(tH(t)), (t → 0+). (3)
B. Wei and D. Yu’s results read as follows:
Theorem 1.1 Let (1) hold. Suppose that ω(t) satisfies (2). Then
kTn,A(f) − fk = O
ω(π/n) + Xn k=0
|4ank|H(π/n)
.
If, in addition, ω(t) satisfies (3), then
kTn,A(f) − fk = O
Xn
k=0
|4ank|H
Xn
k=0
|4ank|
,
kTn,A(f) − fk = O
Xn
k=0
|4ank|H(π/n)
.
Theorem 1.2 Let (an,k)satisfies (1). Then
kTn,A(f) − fk = O
ω(π/n) + Xn k=1
k−1ω(π/k)
k+1X
µ=0
anµ+ Xn k=1
ω(π/k) Xn µ=k
|4anµ|
.
Now and further for an arbitrary sequence {bn} we denote 4bn= bn− bn+1 and 4νbn = 4 4ν−1bn
, ν = 2, 3, . . . .
The aim of the present paper is to generalize above results obtaining more close estimations of the deviation kTn,A(f) − fk than those of B. Wei and D. Yu. Also, we shall show that their results are consequences of ours as a special case.
We emphasize here that throughout of this paper we write u = Oβ(v) if there exists a positive constant C, depending on β, such that u ¬ Cv.
2. Helpful lemmas. To prove the main results we need some auxiliary state- ments.
Lemma 2.1 ([1]) If (1.2) and (1.3) hold then Z π/n
0
ω(t)dt = O n−2H(π/n) .
Lemma 2.2 ([2]) If (1.2) and (1.3) hold then Z r
0 t−1ω(t)dt = O (rH(r)) , (r → +0).
Lemma 2.3 For any lower triangular infinite matrix (an,k), k, n = 0, 1, 2, . . . of nonnegative numbers, it holds uniformly in 0 < t ¬ π, that
Xn k=0
an,ksin
k +1
2
t = Oβ
Xτ
k=0
ank+1 t
Xn k=τ
Gnk;β
, (4)
where Gnk;β :=
Xβ j=1
|4jank|, β ∈ {1, 2, . . . }, and τ denotes the integer part of πt. It also holds that
Xn k=0
an,ksin
k + 1
2
t = Oβ
1 t
Xn k=0
Gnk;β
. (5)
Proof For arbitrary λn 0 and for n m 0 we have Λm,n :=
Xn k=m
λksin
k +1
2
t sint
2
= 1
2
λmcos mt −
n−1
X
k=m
4λkcos(k + 1)t − λncos(n + 1)t
= 1
2
λmcos mt −
n−1
X
k=m
4λkcos
k + 1
2
t cos t
2
+
n−1X
k=m
4λksin
k +1
2
t sin t
2 −λncos(n + 1)t
= 1
2
λmcos mt −
n−1
X
k=m
4λkcos
k + 1
2
t cos t
2 −λncos(n + 1)t
+1 22
4λmcos mt −
n−2X
k=m
42λkcos(k + 1)t − 4λn−1cos nt
. Repeating this transformation, in the same way β−times, we easy obtain
Λm,n := 1 2
λmcos mt −
n−1X
k=m
4λkcos
k +1
2
t cost
2 −λncos(n + 1)t
+1 22
4λmcos mt −
nX−2 k=m
42λkcos(k + 1)t − 4λn−1cos nt
+ · · · +
+1 2β
4β−1λmcos mt −
nX−β k=m
4βλkcos(k + 1)t − 4β−1λn−β+1cos(n − β)t
, where β ∈ {1, 2, . . . }.
Thus,
|Λm,n| ¬ 1
2 λm+
n−1X
k=m
|4λk| + λn
!
+1
22 |4λm| +
n−2
X
k=m
|42λk| + |4λn−1|
! + · · · +
+ 1
2β |4β−1λm| +
nX−β k=m
|4βλk| + |4β−1λn−β+1|
!
, (6)
where β ∈ {1, 2, . . . }.
But since (ank) is a lower triangular matrix, then
anm¬ Xn k=m
|4ank|, |4anm| ¬ Xn k=m
|42ank| , . . . , |4β−1anm| ¬ Xn k=m
|4βank| (7)
hold for n − β + 1 m 0.
Now by (6) and (7), supposing that n − β + 1 τ, we have
Xn k=0
anksin
k + 1
2
t
¬
Xτ k=0
ank+
Xn k=τ
anksin
k + 1
2
t
¬ Xτ k=0
ank+ Oβ
1 t
an0+
n−1X
k=τ
|4ank| + ann
+
|4an0| +
n−2
X
k=τ
|42ank| + |4ann−1|
+ · · · +
|4β−1an0| +
nX−β k=τ
|4βank| + |4β−1ann−β+1|
= Oβ
Xτ
k=0
ank+1 t
Xn k=τ
|4ank| + |42ank| + · · · + |4βank| ,
where Oβ depends on β. This completely proves (4).
By a similar technique we have
Xn k=0
anksin
k +1
2
t
= Oβ
1 t
an0+
n−1X
k=0
|4ank| + ann
+
|4an0| +
nX−2 k=0
|42ank| + |4ann−1|
+ · · · +
|4β−1an0| +
nX−β k=0
|4βank| + |4β−1ann−β+1|
= Oβ
1 t
Xn k=0
|4ank| + |42ank| + · · · + |4βank| ,
which completes (5), and with this the proof of the lemma.
3. Main Results. We establish the following.
Theorem 3.1 Let (an,k)satisfies conditions
an,k 0 and Xn k=0
ank= 1
2β− 1, β∈ {1, 2, . . . }. (8) Assume that ω(t) satisfies condition (2). Then
kTn,A(f) − fk = Oβ
ω(π/n) + Xn k=0
Gnk;βH(π/n)
. (9)
If, in addition, ω(t) satisfies (3), then
kTn,A(f) − fk = Oβ
Xn
k=0
Gnk;βH
Xn
k=0
Gnk;β
, (10)
kTn,A(f) − fk = Oβ
Xn
k=0
Gnk;βH(π/n)
. (11)
Proof First we set the notation
φx;β(t) := f (x + t) + f (x− t) − 2(2β− 1)f(x)
2 ,
and throughout we shall keep in mind that β ∈ {1, 2, . . . }.
Then we easy obtain
Tn,A(f; x) − f(x) = 2 π
Z π 0 φx;β(t)
2 sin t
2
−1 nX
k=0
an,ksin
k + 1
2
t dt. (12)
By (12) we have
kTn,A(f; x) − f(x)k ¬ 2 π
Z π/n 0
+Z π π/n
:= J1(n) + J2(n). (13)
According to (8) and the inequality | sin t| ¬ t for 0 ¬ t ¬ π/n, we have
Xn k=0
an,ksin
k + 1
2
t
¬ 2nt 2β− 1. Thus,
J1(n) = Oβ(n)Z π/n
0 ω(t)dt = Oβ(ω(π/n)). (14)
Also, by (5) and (2) we have
J2(n) = Oβ
Xn
k=0
Gnk;β
Z π π/n
t−2ω(t)dt = Oβ
Xn
k=0
Gnk;βH(π/n)
. (15)
Now (9) follows from (13)–(15).
Then according to (8), and Xn
k=0
Gnk;β = Xn k=0
Xβ j=1
|4jank| ¬ (2 + 22+ · · · + 2β) Xn k=0
ank= 2 < π,
we get
kTn,A(f; x) − f(x)k ¬ 2 π
Z Pn
k=0Gnk;β 0
+Z π
Pn
k=0Gnk;β
:= R1(n) + R2(n). (16)
It is obvious from (8) that
Xn k=0
an,ksin
k + 1
2
t
¬ 1
2β− 1. Thus, by Lemma 2.1 we have
R1(n) = Oβ(1)Z Pnk=0Gnk;β
0 t−1ω(t)dt = Oβ
Xn
k=0
Gnk;βH
Xn
k=0
Gnk;β
. (17)
Using (5) and (2) we obtain
R2(n) = Oβ
Xn k=0
Gnk;β
Z π
Pn k=0Gnk;β
t−2ω(t)dt
!
= Oβ
Xn
k=0
Gnk;βH
Xn
k=0
Gnk;β
. (18)
From (16), (17), and (18) follows (10).
Now we turn back to prove (11). Since ank= 0 for k > n, we deduce that
an`¬ Xn k=`
|4ank| ¬ Xn k=0
|4ank| + |42ank| + · · · + |4βank|
for ` = 0, 1, 2, . . . , n, which implies 1
2β− 1 = Xn
`=0
an`¬ (n + 1) Xn k=0
|4ank| + |42ank| + · · · + |4βank| ,
i.e. Xn
k=0
|4ank| + |42ank| + · · · + |4βank|
1
2(2β− 1)n. Whence, according to Lemma 2.2 we obtain
J1(n) = Oβ
1
nH(π/n)
= Oβ
Xn
k=0
Gnk;βH(π/n)
. (19)
Therefore, by (13), (15), and (19), (11) is proved.
Example 3.2 We give an example of a lower triangular infinite matrix A := (ank), k, n = 0, 1, 2, . . . , that satisfies conditions (1.5). Namely, it is clear that the matrix
A =
21β 0 0 0 · · · 0 . . .
21β 1
4β 0 0 · · · 0 . . .
21β 1 4β 1
8β 0 · · · 0 . . .
1 2β 1
4β 1 8β 1
16β · · · 0 . . . ... ... ... ... · · · ... ...
satisfies conditions ank 0 and for n large enough holds Xn
k=0
ank= 1
2β− 1, β∈ {1, 2, . . . }.
Also note that in particular case β = 1 and n large enough holds Pn
k=0ank= 1.
Theorem 3.3 Let (an,k)satisfies (8). Then
kTn,A(f) − fk = Oβ
ω(π/n) + Xn k=1
k−1ω(π/k)
k+1X
µ=0
anµ+ Xn k=1
ω(π/k) Xn µ=k
Gnµ;β
. (20) Proof According to (4) and the property of the monotonicity of ω(t), we have
J2(n) = 2 π
Z π π/n
φx;β(t)
2 sin t
2
−1 nX
k=0
an,ksin
k +1
2
t dt
= Oβ
Z π π/n
t−1ω(t)
Xτ
µ=0
anµ+1 t
Xn µ=τ
Gnµ;β
dt
!
= Oβ n−1
X
k=1
Z π/k
π/(k+1)t−1ω(t)
Xτ
µ=0
anµ+1 t
Xn µ=τ
Gnµ;β
!
dt
= Oβ
Xn k=1
k−1ω(π/k)
k+1X
µ=0
anµ+ Xn k=1
ω(π/k) Xn µ=k
Gnµ;β
!
. (21)
Combining (13), (14), and (21), we immediately obtain (20). The proof of the
theorem is completed.
Remark 3.4 Note that if we take s = 1 in our results we exactly obtain Theorems 1.1-1.2.
Remark 3.5 Also, if we take s = 1 and suppose that {ank} ∈ HBV S or {ank} ∈ RBV S, then our results imply Leindler’s results, and thus also results of P. Chandra [1]–[2] (see discussions in [5]).
References
[1] P. Chandra, On the degree of approximation of a class of functions by means of Fourier series, Acta Math. Hungar.52 (1988), 199–205.
[2] P. Chandra, A note on the degree of approximation of continuous functions, Acta Math.
Hungar.62 (1993), 21–23.
[3] L. Leindler, On the degree of approximation of continuous functions, Acta Math. Hungar.
104 (2004), 105–113.
[4] Xh. Z. Krasniqi, On the degree of approximation of continuous functions that pertains to the sequence-to-sequence transformation, Aust. J. Math. Anal. Appl., Vol.7, No. 2, Art. 13, (2011), 1–10.
[5] B. Wei and D. Yu, On the degree of approximation of continuous functions by means of Fourier series, Math. Commun.17 (2012), 211–219.
Xhevat Z. Krasniqi
University of Prishtina, Department of Mathematics and Computer Sciences Avenue Mother Theresa, 10000 Prishtina, Kosova
E-mail: xhevat.krasniqi@uni-pr.edu
(Received: 23.08.2012)