Volume 9 (2008), Issue 1, Article 28, 13 pp.
ON THE RATE OF STRONG SUMMABILITY BY MATRIX MEANS IN THE
GENERALIZED HÖLDER METRIC
BOGDAN SZAL
FACULTY OF M ATHEMATICS , C OMPUTER S CIENCE AND E CONOMETRICS
U NIVERSITY OF Z IELONA G ÓRA
65-516 Z IELONA G ÓRA , UL . S ZAFRANA 4 A , P OLAND
B.Szal@wmie.uz.zgora.pl
Received 10 January, 2007; accepted 23 February, 2008
Communicated by R.N. Mohapatra
A BSTRACT. In the paper we generalize (and improve) the results of T. Singh [5], with mediate
function, to the strong summability. We also apply the generalization of L. Leindler type [3].
Key words and phrases: Strong approximation, Matrix means, Special sequences.
2000 Mathematics Subject Classification. 40F04, 41A25, 42A10.
1. I NTRODUCTION
Let f be a continuous and 2π-periodic function and let
∞
a0 X
(1.1)
f (x) ∼
+
(an cos nx + bn sin nx)
2
n=1
be its Fourier series. Denote by Sn (x) = Sn (f, x) the n-th partial sum of (1.1) and by ω (f, δ)
the modulus of continuity of f ∈ C2π .
The usual supremum norm will be denoted by k·kC .
Let ω (t) be a nondecreasing continuous function on the interval [0, 2π] having the properties
ω (0) = 0,
ω (δ1 + δ2 ) ≤ ω (δ1 ) + ω (δ2 ) .
Such a function will be called a modulus of continuity.
Denote by H ω the class of functions
H ω := {f ∈ C2π ; |f (x + h) − f (x)| ≤ Cω (|h|)} ,
where C is a positive constant. For f ∈ H ω , we define the norm k·kω = k·kH ω by the formula
kf kω := kf kC + kf kC,ω ,
where
kf (· + h) − f (·)kC
,
ω (|h|)
h6=0
kf kC,ω = sup
018-07
2
B OGDAN S ZAL
and kf kC,0 = 0. If ω (t) = C1 |t|α (0 < α ≤ 1), where C1 is a positive constant, then
H α = {f ∈ C2π ; |f (x + h) − f (x)| ≤ C1 |h|α , 0 < α ≤ 1}
is a Banach space and the metric induced by the norm k·kα on H α is said to be a Hölder metric.
Let A := (ank ) (k, n = 0, 1, . . . ) be a lower triangular infinite matrix of real numbers satisfying the following condition:
(1.2)
ank ≥ 0 (k, n = 0, 1, . . . ) ,
ank = 0,
k>n
and
n
X
ank = 1.
k=0
Let the A−transformation of (Sn (f ; x)) be given by
(1.3)
tn (f ) := tn (f ; x) :=
n
X
ank Sk (f ; x)
(n = 0, 1, . . . )
k=0
and the strong Ar −transformation of (Sn (f ; x)) for r > 0 by
( n
) r1
X
r
Tn (f, r) := Tn (f, r; x) :=
ank |Sk (f ; x) − f (x)|
(n = 0, 1, . . . ) .
k=0
Now we define two classes of sequences ([3]).
A sequence c := (cn ) of nonnegative numbers tending to zero is called the Rest Bounded
Variation Sequence, or briefly c ∈ RBV S, if it has the property
(1.4)
∞
X
|cn − cn+1 | ≤ K (c) cm
k=m
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 a Head Bounded Variation
Sequence, or briefly c ∈ HBV S, if it has the property
(1.5)
m−1
X
|cn − cn+1 | ≤ K (c) cm
k=0
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 .
Therefore we assume that the sequence (K (αn ))∞
n=0 is bounded, that is, that there exists a
constant K such that
0 ≤ K (αn ) ≤ K
holds for all n, where K (αn ) denote the sequence of constants appearing in the inequalities
(1.4) or (1.5) for the sequence αn := (ank )∞
k=0 . Now we can give the conditions to be used later
on. We assume that for all n and 0 ≤ m ≤ n,
(1.6)
∞
X
|ank − ank+1 | ≤ Kanm
k=m
and
(1.7)
m−1
X
|ank − ank+1 | ≤ Kanm
k=0
hold if αn := (ank )∞
k=0 belongs to RBV S or HBV S, respectively.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
R ATE OF S TRONG S UMMABILITY BY M ATRIX M EANS
3
Let ω (t) and ω ∗ (t) be two given moduli of continuity satisfying the following condition (for
0 ≤ p < q ≤ 1):
p
(ω (t)) q
= O (1)
ω ∗ (t)
(1.8)
(t → 0+ ) .
In [4] R. Mohapatra and P. Chandra obtained some results on the degree of approximation
for the means (1.3) in the Hölder metric. Recently, T. Singh in [5] established the following two
theorems generalizing some results of P. Chandra [1] with a mediate function H such that:
Z π
ω (f ; t)
(1.9)
dt = O (H (u)) (u → 0+ ) , H (t) ≥ 0
t2
u
and
Z t
(1.10)
H (u) du = O (tH (t))
(t → O+ ) .
0
Theorem 1.1. Let A = (ank ) satisfy the condition (1.2) and ank ≤ ank+1 for k = 0, 1, . . . , n−1,
and n = 0, 1, . . . . Then for f ∈ H ω , 0 ≤ p < q ≤ 1,
h
p
(1.11) ktn (f ) − f kω∗ = O {ω (|x − y|)} q {ω ∗ (|x − y|)}−1
p
p
π − pq
π 1− q
q
×
H
ann n + ann
,
+ O ann H
n
n
if ω (f ; t) satisfies (1.9) and (1.10), and
h
i
p
(1.12) ktn (f ) − f kω∗ = O {ω (|x − y|)} q {ω ∗ (|x − y|)}−1
p
π 1− pq n π π o
p
π 1− q
q
×
ω
+ ann H
,
+ ann n H
+O ω
n
n
n
n
if ω (f ; t) satisfies (1.9), where ω ∗ (t) is the given modulus of continuity.
Theorem 1.2. Let A = (ank ) satisfy the condition (1.2) and ank ≤ ank+1 for k = 0, 1, . . . , n−1,
and n = 0, 1, . . . . Also, let ω (f ; t) satisfy (1.9) and (1.10). Then for f ∈ H ω , 0 ≤ p < q ≤ 1,
h
p
(1.13) ktn (f ) − f kω∗ = O {ω (|x − y|)} q {ω ∗ (|x − y|)}−1
p
oi
n
p
−p
+ O (an0 H (an0 )) ,
× (H (an0 ))1− q an0 n q + an0q
where ω ∗ (t) is the given modulus of continuity.
The next generalization of another result of P. Chandra [2] was obtained by L. Leindler in
[3]. Namely, he proved the following two theorems
Theorem 1.3. Let (1.2) and (1.9) hold. Then for f ∈ C2π
π π (1.14)
ktn (f ) − f kC = O ω
+ O ann H
.
n
n
If, in addition ω (f ; t) satisfies the condition (1.10), then
(1.15)
ktn (f ) − f kC = O (ann H (ann )) .
Theorem 1.4. Let (1.2), (1.9) and (1.10) hold. Then for f ∈ C2π
(1.16)
ktn (f ) − f kC = O (an0 H (an0 )) .
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
4
B OGDAN S ZAL
In the present paper we will generalize (and improve) the mentioned results of T. Singh [5]
to strong summability with a mediate function H defined by the following conditions:
Z π r
ω (f ; t)
(1.17)
dt = O (H (r; u))
(u → 0+ ) , H (t) ≥ 0 and r > 0,
t2
u
and
t
Z
H (u) du = O (tH (r; t))
(1.18)
(t → O+ ) .
0
We also apply a generalization of Leindler’s type [3].
Throughout the paper we shall use the following notation:
φx (t) = f (x + t) + f (x − t) − 2f (x) .
By K1 , K2 , . . . we shall designate either an absolute constant or a constant depending on the
indicated parameters, not necessarily the same at each occurrence.
2. M AIN R ESULTS
Our main results are the following.
Theorem 2.1. Let (1.2), (1.7) and (1.8) hold. Suppose ω (f ; t) satisfies (1.17) for r ≥ 1. Then
for f ∈ H ω ,
p
(2.1)
kTn (f, r)kω∗ = O {1 + ln (2 (n + 1) ann )} q
n
π o r1 (1− pq ) r−1
× ((n + 1) ann ) ann H r;
.
n
If, in addition ω (f ; t) satisfies the condition (1.18), then
p
(2.2)
kTn (f, r)kω∗ = O {1 + ln (2 (n + 1) ann )} q
r−1
× (ln (2 (n + 1) ann ))
r1 (1− pq )
ann H (r; ann )
.
Theorem 2.2. Under the assumptions of above theorem, if there exists a real number s > 1
such that the inequality

1
k −1
k −1
2X
 2X
s
1
s
k−1 s −1
(2.3)
≤ K1 2
(a )
ani
 k−1 ni 
k−1
i=2
i=2
for any k = 1, 2, . . . , m, where 2m ≤ n + 1 < 2m+1 holds, then the following estimates
n
π o r1 (1− pq ) (2.4)
kTn (f, r)kω∗ = O
ann H r;
n
and
(2.5)
kTn (f, r)kω∗
p
= O {ann H (r; ann )} (1− q )
1
r
are true.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
R ATE OF S TRONG S UMMABILITY BY M ATRIX M EANS
5
Theorem 2.3. Let (1.2), (1.6), (1.8) and (1.17) for r ≥ 1 hold. Then for f ∈ H ω
n
π o r1 (1− pq ) (2.6)
kTn (f, r)kω∗ = O
an0 H r;
.
n
If, in addition, ω (f ; t) satisfies (1.18), then
(2.7)
p
1
kTn (f, r)kω∗ = O {an0 H (r; an0 )} r (1− q ) .
Remark 2.4. We can observe, that for the case r = 1 under the condition (1.8) the first part of
Theorem 1.1 (1.11) and Theorem 1.2 are the corollaries of the first part of Theorem 2.1 (2.1)
and the second part of Theorem 2.3 (2.7), respectively. We can also note that the mentioned
estimates are better in order than the analogical estimates from the results of T. Singh, since
ln (2 (n + 1) ann ) in Theorem 2.1 is better than (n + 1) ann in Theorem 1.1. Consequently, if
nann is not bounded our estimate (2.7) in Theorem 2.3 is better than (1.13) from Theorem 1.2.
Remark 2.5. If in the assumptions of Theorem 2.1 or 2.3 we take ω (|t|) = O (|t|q ), ω ∗ (|t|) =
O (|t|p ) with p = 0, then from (2.1), (2.2) and (2.7) we have the same estimates such as (1.14),
(1.15) and (1.16), respectively, but for the strong approximation (with r = 1).
3. C OROLLARIES
In this section we present
some special cases of our results. From Theorems 2.1, 2.2 and 2.3,
β
∗
putting ω (|t|) = O |t| , ω (|t|) = O (|t|α ),
 rα−1
t
if αr < 1,



ln πt
if αr = 1,
H (r; t) =



K1
if αr > 1
where r > 0 and 0 < α ≤ 1, and replacing p by β and q by α, we can derive Corollaries 3.1,
3.2 and 3.3, respectively.
Corollary 3.1. Under the conditions (1.2) and (1.7) we have for f ∈ H α , 0 ≤ β < α ≤ 1 and
r ≥ 1,
 β
1+ r1 (1− α
α−β
)

O {ln (2 (n + 1) ann )}
{ann }
if αr < 1,





oα−β
n 1+α−β
π
kTn (f, r)kβ =
O
{ln
(2
(n
+
1)
a
)}
ln
a
if αr = 1,
nn
nn
ann




β
α−β
1
 
O {ln (2 (n + 1) ann )}1+ r (1− α ) {ann } αr
if αr > 1.
Corollary 3.2. Under the assumptions of Corollary 3.1 and (2.3) we have
 α−β

O
{a
}
if αr < 1,

nn



 n oα−β π
kTn (f, r)kβ =
O
ln ann ann
if αr = 1,





 O {ann } α−β
αr
if αr > 1.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
6
B OGDAN S ZAL
Corollary 3.3. Under the conditions (1.2) and (1.6) we have, for f ∈ H α , 0 ≤ β < α ≤ 1 and
r ≥ 1,
 α−β

O {an0 }
if αr < 1,





oα−β
n π
kTn (f, r)kβ =
O
ln an0 an0
if αr = 1,





 O {an0 } α−β
αr
if αr > 1.
4. L EMMAS
To prove our theorems we need the following lemmas.
Lemma 4.1. If (1.17) and (1.18) hold with r > 0 then
Z s r
ω (f ; t)
(4.1)
dt = O (sH (r; s)) (s → 0+ ) .
t
0
Proof. Integrating by parts, by (1.17) and (1.18) we get
Z π r
s Z s Z π r
Z s r
ω (f ; t)
ω (f ; u)
ω (f ; u)
dt = −t
du +
dt
du
2
t
u
u2
0
t
0
t
0
Z s
= O (sH (r; s)) + O (1)
H (r; t) dt
0
= O (sH (r; s)) .
This completes the proof.
Lemma 4.2 ([7]). If (1.2), (1.7) hold, then for f ∈ C2π and r > 0,

1 
r
n+1

[
]
4
 
r 
 X
.
(4.2)
kTn (f, r)kC ≤ O 
an,4k Ekr (f ) + E[ n+1 ] (f ) ln (2 (n + 1) ann )


4

 k=0

If, in addition, (2.3) holds, then
(4.3)

1 
r
n+1


[
]
2
 
 X
kTn (f, r)kC ≤ O 
an,2k Ekr (f ) 
.


 k=0

Lemma 4.3 ([7]). If (1.2), (1.6) hold, then for f ∈ C2π and r > 0,
(
) r1 
n
X
(4.4)
kTn (f, r)kC ≤ O 
ank Ekr (f )  .
k=0
Lemma 4.4. If (1.2), (1.7) hold and ω (f ; t) satisfies (1.17) with r > 0 then
(4.5)
n+1
[X
4 ]
an,4k ω
r
k=0
π
f;
k+1
π = O ann H r;
.
n
If, in addition, ω (f ; t) satisfies (1.18) then
(4.6)
n+1
[X
4 ]
k=0
an,4k ω
r
π
f;
k+1
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
= O (ann H (r; ann )) .
http://jipam.vu.edu.au/
R ATE OF S TRONG S UMMABILITY BY M ATRIX M EANS
7
Proof. First we prove (4.5). If (1.7) holds then
anµ − anm ≤ |anµ − anm | ≤
m−1
X
|ank − ank+1 | ≤ Kanm
k=µ
for any m ≥ µ ≥ 0, whence we have
anµ ≤ (K + 1) anm .
(4.7)
From this and using (1.17) we get
n+1
[X
4 ]
an,4k ω
k=0
r
π
f;
k+1
n
X
π
ω f;
≤ (K + 1) ann
k+1
k=0
Z n+1 π
r
≤ K1 ann
ω f;
dt
t
1
Z π r
ω (f ; u)
du
= πK1 ann
π
u2
n+1
π = O ann H r;
.
n
r
Now we prove (4.6). Since
(K + 1) (n + 1) ann ≥
n
X
ank = 1,
k=0
we can see that
(4.8)
n+1
[X
4 ]
k=0
an,4k ω
r
π
f;
k+1
≤
1
[ 4(K+1)a
]−1
Xnn
r
an,4k ω
k=0
π
f;
k+1
n
X
+
an,4k ω
1
k=[ 4(K+1)a
nn
r
]−1
π
f;
k+1
= Σ 1 + Σ2 .
Using again (4.7), (1.2) and the monotonicity of the modulus of continuity, we can estimate the
quantities Σ1 and Σ2 as follows
(4.9)
Σ1 ≤ (K + 1) ann
1
]−1
[ 4(K+1)a
Xnn
k=0
Z
ω
r
π
f;
k+1
1
4(K+1)ann
π
ωr f ;
dt
t
1
Z π
ω r (f ; u)
= πK2 ann
du
u2
4π(K+1)ann
Z π r
ω (f ; u)
≤ πK2 ann
du
u2
ann
≤ K2 ann
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
8
B OGDAN S ZAL
and
n
X
r
Σ2 ≤ K3 ω (f ; 4π (K + 1) ann )
(4.10)
k=[
1
4(K+1)ann
an,4k
]−1
≤ K3 (8π (K + 1))r ω r (f ; ann )
a nn
≤ K3 (32π (K + 1))r ω r f ;
Z ann 2r
ω (f ; t)
dt
≤ 2K3 (32π (K + 1))r
ann
t
2
Z ann r
ω (f ; t)
≤ K4
dt.
t
0
If (1.17) and (1.18) hold then from (4.8) – (4.10) we obtain (4.6). This completes the proof. Lemma 4.5. If (1.2), (1.7) hold and ω (f ; t) satisfies (1.17) with r ≥ 1 then
n
π o r1 π
1− r1
(4.11)
ω f,
ln (2 (n + 1) ann ) = O {(n + 1) ann }
ann H r;
.
n+1
n
If, in addition, ω (f ; t) satisfies (1.18) then
1
1
π
ln (2 (n + 1) ann ) = O {ln (2 (n + 1) ann )}1− r {ann H (r; ann )} r .
(4.12) ω f,
n+1
Proof. Let r = 1. Using the monotonicity of the modulus of continuity
π
π
ω f,
ln (2 (n + 1) ann ) ≤ 2ann ω f,
(n + 1)
n+1
n+1
Z n+1
π
dt
≤ 4ann ω f,
n+1
1
Z n+1 π
dt
≤ 4ann
ω f,
t
1
Z π
ω (f, u)
= 4πann
du
π
u2
n+1
and by (1.17) we obtain that (4.11) holds. Now we prove (4.12). From (1.2) and (1.7) we get
Z π(K+1)ann
π
π
1
ω f,
ln (2 (n + 1) ann ) ≤ K1 ω f,
dt,
π
n+1
n+1
t
n+1
Z
π(K+1)ann
K1
π
n+1
Z ann
ω (f, t)
ω (f, u)
dt ≤ 2K1 (K + 1) π
du
1
t
u
(K+1)(n+1)
Z ann
ω (f, u)
≤ K2
du
u
0
and by Lemma 4.1 we obtain (4.12).
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
R ATE OF S TRONG S UMMABILITY BY M ATRIX M EANS
9
Assuming r > 1 we can use the Hölder inequality to estimate the following integrals
)1− r1
) r1 (Z
(Z
Z π
π
π
r
1
ω (f, u)
ω (f, u)
du
du
du ≤
2
2
π
π
π
u2
u
u
n+1
n+1
n+1
) r1
1− r1 (Z π r
n+1
ω (f, u)
du
≤
π
u2
π
n+1
and
Z
ann
1
(K+1)(n+1)
)1− r1
) r1 (Z
ann
1
ω r (f, u)
du
du
1
1
u
u
(K+1)(n+1)
(K+1)(n+1)
r1
Z ann r
ω (f, u)
1− r1
≤ {ln (2 (n + 1) ann )}
du .
u
0
ω (f, u)
du ≤
u
(Z
ann
From this, if (1.17) holds then
) r1
1− r1 (Z π r
n+1
ω (f, u)
π
ω f,
ln (2 (n + 1) ann ) ≤ 4πann
du
π
n+1
π
u2
n+1
n
π o r1 1− r1
ann H r;
= O {(n + 1) ann }
n
and if (1.17) and (1.18) hold then
π
ω f,
ln (2 (n + 1) ann )
n+1
r1
Z ann r
ω (f, u)
1− r1
≤ 2K1 (K + 1) π {ln (2 (n + 1) ann )}
du
u
0
n
π o r1 1− r1
= O {ln (2 (n + 1) ann )}
ann H r;
.
n
This ends our proof.
Lemma 4.6. If (1.2), (1.6) hold and ω (f ; t) satisfies (1.17) with r > 0 then
n
π X
π
r
(4.13)
ank ω f ;
= O an0 H r;
.
k+1
n
k=0
If, in addition, ω (f ; t) satisfies (1.18), then
n
X
π
r
(4.14)
ank ω f ;
= O (an0 H (r; an0 )) .
k+1
k=0
Proof. First we prove (4.13). If (1.6) holds then
ann − anm ≤ |anm − ann |
≤
≤
n−1
X
k=m
∞
X
|ank − ank+1 |
|ank − ank+1 | ≤ Kanm
k=m
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
10
B OGDAN S ZAL
for any n ≥ m ≥ 0, whence we have
ann ≤ (K + 1) anm .
(4.15)
From this and using (1.17) we get
n
n
X
X
π
π
r
r
ω f;
≤ (K + 1) an0
ank ω f ;
k+1
k+1
k=0
k=0
Z n+1 π
≤ K1 an0
ωr f ;
dt
t
1
Z π r
ω (f ; u)
= πK1 an0
du
2
π
u
n+1
π = O an0 H r;
.
n
Now, we prove (4.14). Since
n
X
(K + 1) (n + 1) an0 ≥
ank = 1,
k=0
we can see that
h
n
X
ank ω
r
k=0
π
f;
k+1
≤
1
(K+1)an0
i
−1
X
ank ω
r
k=0
π
f;
k+1
n
X
+
k=
h
ank ω
1
(K+1)an0
r
i
−1
π
f;
k+1
.
Using again (1.2), (1.6) and the monotonicity of the modulus of continuity, we get
h
(4.16)
n
X
ank ω
k=0
r
π
f;
k+1
1
(K+1)an0
X
≤ (K + 1) an0
k=0
i
−1
ω
r
π
f;
k+1
n
X
r
+ K1 ω (f ; π (K + 1) ano )
k=
Z
h
1
(K+1)an0
ank
i
−1
1
(K+1)an0
≤ K2 an0
1Z
≤ K3 an0
π
an0
π
ωr f ;
dt + K1 ω r (f ; π (K + 1) ano )
t
ω r (f ; u)
r
du + ω (f ; an0 ) .
u2
According to
r
r
ω (f ; an0 ) ≤ 4 ω
r
an0 f;
≤ 2 · 4r
2
(1.17), (1.18) and (4.16) lead us to (4.14).
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
Z
an0
an0
2
ω r (f ; t)
dt ≤ 2 · 4r
t
an0
Z
0
ω r (f ; t)
dt,
t
http://jipam.vu.edu.au/
R ATE OF S TRONG S UMMABILITY BY M ATRIX M EANS
11
5. P ROOFS OF THE T HEOREMS
In this section we shall prove Theorems 2.1, 2.2 and 2.3.
5.1. Proof of Theorem 2.1. Setting
Rn (x + h, x) = Tn (f, r; x + h) − Tn (f, r; x)
and
gh (x) = f (x + h) − f (x)
and using the Minkowski inequality for r ≥ 1, we get
|Rn (x + h, x)|
(
) r1 ( n
) r1 X
n
X
= ank |Sk (f ; x + h) − f (x + h)|r
−
ank |Sk (f ; x) − f (x)|r k=0
k=0
( n
) r1
X
≤
ank |Sk (gh ; x) − gh (x)|r
.
k=0
By (4.2) we have
|Rn (x + h, x)|

1
r
n+1

[
]
4
X

r 
≤ K1
an,4k Ekr (gh ) + E[ n+1 ] (gh ) ln (2 (n + 1) ann )
4


 k=0

≤ K2

n+1

4 ]
[X


an,4k ω r gh ,
k=0
π
k+1
+ ω gh ,
π
n+1
1
r
r 

ln (2 (n + 1) ann )
.


Since
|gh (x + l) − gh (x)| ≤ |f (x + l + h) − f (x + h)| + |f (x + l) − f (x)|
and
|gh (x + l) − gh (x)| ≤ |f (x + l + h) − f (x + l)| + |f (x + h) − f (x)| ≤ 2ω (|h|) ,
therefore, for 0 ≤ k ≤ n,
π
ω gh ,
k+1
(5.1)
π
≤ 2ω f,
k+1
and f ∈ H ω
π
ω gh ,
k+1
(5.2)
≤ 2ω (|h|) .
From (5.2) and (1.2)
(5.3)
|Rn (x + h, x)| ≤ 2K2 ω (|h|)

n+1

4 ]
[X


an,4k + (ln (2 (n + 1) ann ))r
k=0
1
r




≤ 2K2 ω (|h|) (1 + ln (2 (n + 1) ann )) .
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
12
B OGDAN S ZAL
On the other hand, by (5.1),
(5.4)
|Rn (x + h, x)|
1

r
n+1

]
[
4
r

X
π
π
≤ 2K2
an,4k ω r f,
+ ω f,
ln (2 (n + 1) ann )
.


k+1
n+1

 k=0
Using (5.3) and (5.4) we get
kTn (f, r; · + h) − Tn (f, r; ·)kC
(5.5) sup
ω (|h|)
h6=0
p
p
(kRn (· + h, ·)kC ) q
= sup
(kRn (· + h, ·)kC )1− q
ω (|h|)
h6=0
p
≤ K3 (1 + ln (2 (n + 1) ann )) q
 1 (1− p )

r
q
n+1

r 
4 ]
[X

π
π
×
an,4k ω r f,
+ ω f,
ln (2 (n + 1) ann )
.


k+1
n+1
 k=0

Similarly, by (4.2) we have
(5.6)
kTn (f, r)kC

1
r
n+1

]
[
4
r
X

π
π
an,4k ω r f,
+ ω f,
ln (2 (n + 1) ann )
≤ K4


k+1
n+1

 k=0
≤ K4

n+1

4 ]
[X


×
an,4k ω r f,
k=0

n+1

4 ]
[X


k=0
an,4k ω r f,
π
k+1
π
k+1
+ ω f,
+ ω f,
π
n+1
π
n+1
1 p
r q
r 

ln (2 (n + 1) ann )


 1 (1− p )
r
q
r 

ln (2 (n + 1) ann )


p
≤ K5 (1 + ln (2 (n + 1) ann )) q
 1 (1− p )

r
q
n+1

r 
4 ]
[X

π
π
×
an,4k ω r f,
+ ω f,
ln (2 (n + 1) ann )
.


k+1
n+1
 k=0

Collecting our partial results (5.5), (5.6) and using Lemma 4.4 and Lemma 4.5 we obtain that
(2.1) and (2.2) hold. This completes our proof.
5.2. Proof of Theorem 2.2. Using (4.3) and the same method as in the proof of Lemma 4.4
we can show that
n+1
[X
2 ]
π π
r
an,2k ω f,
= O ann H r;
(5.7)
k+1
n
k=0
holds, if ω (t) satisfies (1.17) and (1.18), and
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/
R ATE OF S TRONG S UMMABILITY BY M ATRIX M EANS
(5.8)
n+1
[X
2 ]
k=0
an,2k ω
r
π
f,
k+1
13
= O (ann H (r; ann ))
if ω (t) satisfies (1.17).
The proof of Theorem 2.2 is analogously to the proof of Theorem 2.1. The only difference
being that instead of (4.2), (4.5) and (4.6) we use (4.3), (5.7) and (5.8) respectively. This
completes the proof.
5.3. Proof of Theorem 2.3. Using the same notations as in the proof of Theorem 2.1, from
(4.4) and (5.2) we get
( n
) r1
X
(5.9)
|Rn (x + h, x)| ≤ K1
ank Ekr (gh )
k=0
≤ K2
( n
X
ank ω r
k=0
π
gh ,
k+1
) r1
≤ 2K2 ω (|h|) .
On the other hand, by (4.4) and (5.1), we have
( n
X
r
|Rn (x + h, x)| ≤ K2
ank ω gh ,
(5.10)
k=0
≤ 2K2
( n
X
ank ω r f,
k=0
π
k+1
) r1
π
k+1
) r1
.
Similarly, we can show that
(5.11)
kTn (f, r)kC ≤ K3
( n
X
ank ω r f,
k=0
π
k+1
) r1
.
Finally, using the same method as in the proof of Theorem 2.1 and Lemma 4.6, (2.6) and (2.7)
follow from (5.9) – (5.11).
R EFERENCES
[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 function, Acta Math. Hungar.,
62 (1993), 21–23.
[3] L. LEINDLER, On the degree of approximation of continuous functions, Acta Math. Hungar., 104(12), (2004), 105–113.
[4] R.N. MOHAPATRA AND P. CHANDRA, Degree of approximation of functions in the Hölder metric, Acta Math. Hungar., 41(1-2) (1983), 67–76.
[5] T. SINGH, Degree of approximation to functions in a normed spaces, Publ. Math. Debrecen, 40(3-4)
(1992), 261–271.
[6] XIE-HUA SUN, Degree of approximation of functions in the generalized Hölder metric, Indian J.
Pure Appl. Math., 27(4) (1996), 407–417.
[7] B. SZAL, On the strong approximation of functions by matrix means in the generalized Hölder
metric, Rend. Circ. Mat. Palermo (2), 56(2) (2007), 287–304.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 28, 13 pp.
http://jipam.vu.edu.au/