Radosława Kranz and Aleksandra Rzepka
On the approximation of conjugate functions from Lp by some special matrix means of conjugate
Fourier series
Abstract. The results corresponding to some theorems of W. Łenski and B. Szal are shown. From the presented pointwise results the estimates on norm approximation are derived. Some special cases as corollaries are also formulated.
2000 Mathematics Subject Classification: 42A24.
Key words and phrases: Degree of approximation, Fourier series, matrix means, con- jugate function.
1. Introduction. Let Lp (1 ¬ p < ∞) be the class of all 2π–periodic real–
valued functions integrable in the Lebesgue sense with p–th power over Q = [−π, π]
with the norm
kfk := kf(·)kLp = Z
Q
| f(t) |pdt
!1/p
. Consider the trigonometric Fourier series
Sf (x) := ao(f)
2 +
X∞ ν=1
(aν(f) cos νx + bν(f) sin νx) and the conjugate one
Sf (x) :=e X∞ ν=1
(aν(f) sin νx − bν(f) cos νx)
with the partial sums eSkf . We know that if f ∈ L1, then f (x) :=e −1
π Z π
0 ψx(t)1 2ctg t
2dt = lim
→0+f (x, ) ,e
where
f (x, ) :=e −1 π
Z π
ψx(t)1 2ctgt
2dt with
ψx(t) = f(x + t) − f(x − t), exists for almost all x and if ef ∈ L1, then eSf (x) = S ef (x) [3].
Let A := (an,k) and B := (bn,k) be infinite lower triangular matrices of real numbers such that
an,k 0 and bn,k 0 when k = 0, 1, 2, ...n, (1)
an,k = 0 and bn,k = 0 when k > n ,
(2)
Xn k=0
an,k = 1 and Xn k=0
bn,k= 1, where n = 0, 1, 2, ... .
Let the AB−transformation of ( eSkf ) be given by Ten,A,Bf (x) :=
Xn r=0
Xr k=0
an,n−rbr,kSekf (x) (n = 0, 1, 2, ...) .
As a measure of approximation of function ef and ef
·,n+1π
by eTn,A,Bf in the space Lp(1 ¬ p < ∞) we will use the pointwise modulus of continuity of f defined, for β 0, by the formula
e
wxf (δ)p,β :=
1 δ Zδ
0
|ψx(t)| sinβ t 2
p dt
1 p
,
and
e
ωLpf (δ)β:= sup
0<t¬δ
ψ·(t) sinβ t 2
Lp
.
The deviation Tn,A,Bf − f, where Tn,A,Bf defines the AB−transformation of (Skf ), was estimated by M. L. Mittal [2], with special matrix B. In the case of conjugate function, the deviation eTn,A,Bf − ef was considered by W. Łenski. and B.
Szal [1] as follows:
Theorem A Let f ∈ L1. If entries of our matrices satisfy the conditions an,n 1
n + 1, 1
s + 1 Xn r=0
an,r an,r for 0 ¬ s ¬ n and
|an,rbr,r−l− an,r+1br+1,r+1−l| an,r
(r + 1)2 for 0 ¬ l ¬ r ¬ n − 1
then
eTn,A,Bf(x) −fe x, π
n+ 1
Xn r=0
an,r+ Xr k=1
an,k
r+ 1
! "
1 r+ 1
Xr k=0
e wxf π
k+ 1
1,0
#
+ 1 n+ 1
Xn k=0
e wxf π
k+ 1
1,0
and under the additional condition 1
π
Z π/(n+1) 0
|ψx(t)|
t dt ewxf
π n + 1
1,0
,
eTn,A,Bf (x)− ef (x) Xn r=0
an,r+ Xr k=1
an,k
r + 1
! "
1 r + 1
Xr k=0
e wxf
π k + 1
1,0
#
+ 1 n + 1
Xn k=0
e wxf
π k + 1
1,0
for every natural n and all considered real x.
In this paper we shall consider the pointwise deviations eTn,A,Bf (·) − ef (·,n+1π ) and eTn,A,Bf (·) − ef (·). In the theorems we formulate the general and precise con- ditions for the matrices (an,r)nr=0 and (br,k)nk=0 and for the modulus of continuity type. Finally, we also give some results on the norm approximation.
We shall write K1 K2, if there exists a positive constant C, sometimes de- pending on some parameters, such that K1¬ CK2.
2. Statement of the results. Now we formulate our main results.
Theorem 2.1 Let f ∈ Lp (1 ¬ p < ∞). If the entries of matrices (an,r)nr=0 and (br,k)nk=0 satisfy the conditions (1), (2), and
(3)
Xn r=0
an,r n−rX
k=0
|bn−r,k− bn−r,k+1| 1 n + 1, then, for 0 ¬ β ¬ 1 when p = 1 and 0 ¬ β < 2 −1p when p > 1,
Ten,A,Bf (x)− ef
x, π
n + 1
(n + 1)β (
e wxf
π n + 1
p,β
+ 1
n + 1 Xn k=0
e wxf
π k + 1
1,β
)
holds for almost all considered x .
Let us consider the classes Lp( ewx)β =n
f ∈ Lp: ewxf (δ)p,β¬ ewx(δ)o , and
Lp(eω)β=n
f ∈ Lp: eωLpf (δ)β¬ eω (δ)o ,
with β 0 and 1 ¬ p < ∞, where ewx and eω are the functions of modulus of continuity type.
Theorem 2.2 Let f ∈ Lp( ewx)β (1 < p < ∞). If the entries of matrices (an,r)nr=0 and (br,k)nk=0 satisfy the conditions (1), (2), (3) and ewx satisfies
(4) (Z n+1π
0
"
e wx(t) t sinβ t2
#q dt
)1q
(n + 1)β+p1wex
π n + 1
,
(5)
(Z n+1π
0
|ψx(t)|
e wx(t)
p sinβp t
2dt )1p
(n + 1)−p1, for β 0, then
eTn,A,Bf (x)− ef (x) (n + 1)β−1 Xn k=0
e wx
π k + 1
,
holds for almost all considered x such that ef (x)exists.
Theorem 2.3 Let f ∈ Lp (1 ¬ p < ∞). If the entries of matrices (an,r)nr=0 and (br,k)nk=0 satisfy the conditions (1), (2) and (3), then
Ten,A,Bf (·) − ef
·, π n + 1
Lp
¬ (n + 1)β−1 Xn k=0
e ωLpf
π k + 1
β
,
holds for almost all considered x such that ef (x) exists and for 0 ¬ β ¬ 1 when p = 1, and 0 ¬ β < 2 −1p when p > 1 .
Theorem 2.4 Let f ∈ Lp(eω)β (1 < p < ∞). If the entries of matrices (an,r)nr=0 and (br,k)nk=0 satisfy the conditions (1), (2), (3) and eω satisfies
(6) (Z n+1π
0
"
e ω(t) t sinβ t2
#q
dt )1q
(n + 1)β+1pωe
π n + 1
, for β 0, then
eTn,A,Bf (·) − ef (·) Lp (n + 1)β−1 Xn k=0
e ω
π k + 1
, holds.
3. Corollaries.
Corollary 3.1 Under the assumptions of Theorem 2.1 we have
Ten,A,Bf (x)− ef
x, π
n + 1
(n + 1)β−
1 p
( n X
k=0
"
e wxf
π k + 1
p,β
#p)1p
.
Proof By Lemma 4.3 eTn,A,Bf (x)− ef (x) ¬
(n + 1)β
1 n + 1
Xn k=0
"
e wxf
π k + 1
p,β
#p!p1
+ 1
n + 1 Xn k=0
e wxf
π k + 1
1,β
¬ (n + 1)β−1p Xn k=0
"
e wxf
π k + 1
p,β
#p!1p
.
This completes the proof of Corollary 3.1.
Corollary 3.2 Under the assumptions of Theorem 2.1, Corollary 3.1 and Lemma 4.3 we obtain another estimation for norm approximation
Ten,A,Bf (·) − ef
·, π n + 1
Lp ¬ (n + 1)β−1p ( n
X
k=0
e ωLpf
π k + 1
β
!p)1p
.
If matrix A is defined by an,r= (r+1) ln(n+1)1 when r = 0, 1, 2, ...n and an,r= 0 when r > n, and matrix B is defined by br,k = r+11 when k = 0, 1, 2, ...r and br,k= 0 when k > r, then from Theorem 2.1 we obtain
Corollary 3.3 Let f ∈ Lp (1 ¬ p < ∞), then
eTn,A,Bf (x)− ef (x)
=
1 ln(n + 1)
Xn r=0
1 (r + 1)2
Xr k=0
Sekf (x)− ef (x)
= Ox(n + 1)β (
e wxf
π n + 1
p,β
+ 1
n + 1 Xn k=0
e wxf
π k + 1
1,β
) ,
for almost all considered x and 0 ¬ β < 1 −1p,when p > 1, and β = 0, when p = 1.
Proof For the proof we will show that (an,r) and (br,k) satisfy the assumptions of Lemma 4.2. The conditions (1) and (2) are satisfied evidently. Since
Xn r=0
1 (r + 1) ln(n + 1)
n−r−1X
k=0
1
n− r + 1− 1 n− r + 1
+ 1 n− r + 1
!
= 1 ln(n + 1)
Xn r=0
1 r + 1
1 n− r + 1
= 1
ln(n + 1) 1 n + 2
Xn r=0
1
r + 1+ 1 n− r + 1
¬ 1
ln(n + 1) 1
n + 22 ln(n + 1) 1 n + 1
the proof of Corollary 3.3 is complete.
4. Auxiliary results. We begin this section by some notations following A.
Zygmund [3, Section 5 of Chapter II].
It is clear that
Sekf (x) =−1 π
Z π
−π
f (x + t) fDk(t) dt and
Ten,A,Bf (x) =−1 π
Z π
−π
f (x + t) Xn r=0
Xr k=0
an,n−rbr,kDfk(t) dt, where
Dfk(t) = Xk ν=0
sin νt = cos2t− cos(k +12)t 2 sin2t . Hence
Ten,A,Bf (x)− ef
x, π
n + 1
= −1 π
Z n+1π
0 ψx(t) Xn r=0
Xr k=0
an,n−rbr,kDfk(t) dt
+1 π
Z π
n+1π
ψx(t) Xn r=0
Xr k=0
an,n−rbr,kDfk◦(t) dt, and
Ten,A,Bf (x)− ef (x) = 1 π
Z π 0 ψx(t)
Xn r=0
Xr k=0
an,n−rbr,kDf◦k(t) dt, where
(7) Df◦k(t) = cos(2k+1)t2 2 sin2t .
Now, we formulate some estimates for the conjugate Dirichlet kernels.
Lemma 4.1 (see [3]) If 0 < |t| ¬ π/2, then (8) fD◦k(t) ¬ π
2 |t| and fDk(t) ¬ π
|t|, and for any real t we have
(9) fDk(t) ¬1
2k (k + 1)|t| and fDk(t) ¬ k + 1.
Lemma 4.2 If (an,r)nr=0 and (br,k)rk=0 satisfy (1), (2) and (3), then
Xn r=0
an,n−r
Xr k=0
br,kDfk◦(t) τ2
n + 1 for n = 0, 1, 2, 3, ..., where τ = [π/t] ¬ n/2.
Proof Since Kn(t) :=
Xn r=0
an,n−r
Xr k=0
br,kDfk◦(t) = Xn r=0
an,r n−r
X
k=0
bn−r,kDf◦k(t) and using Abel’s transformation and from (7) we get
= Xn r=0
an,r
2 sin2t
"n−r−1 X
k=0
(bn−r,k− bn−r,k+1) Xk
l=0
cos(2l + 1) t 2 +bn−r,n−r
n−rX
l=0
cos(2l + 1) t 2
#
= 1
2 sin2t Xn r=0
an,r n−r
X
k=0
(bn−r,k− bn−r,k+1) Xk l=0
cos(2l + 1) t
2 .
Then
|Kn(t)| ¬ 1 sint2
Xn r=0
an,r n−r
X
k=0
|bn−r,k− bn−r,k+1|
Xk l=0
cos(2l + 1) t 2
. A simple calculation for the last sum gives
Xk l=0
cos(2l + 1) t 2
¬ 1
sint2. So
|Kn(t)| π2 t2
Xn r=0
an,r nX−r k=0
|bn−r,k− bn−r,k+1| and from (3) we obtain
|Kn(t)| τ2 n + 1.
The desired estimate is now evident.
Lemma 4.3 Let f ∈ Lp (1 ¬ p < ∞) and β 0, then
e wxf
π n + 1
p,β
( 1
n + 1 Xn k=0
"
e wxf
π k + 1
p,β
#p)1p
holds for every natural n and real x.
Proof By our assumption we get
e wxf
π n + 1
p,β
=
(n + 1 π
Z n+1π
0
ψx(u) sinβu 2 pdu
Xn k=0
2 (k + 1) (n + 1) (n + 2)
)p1
= ( 2
n + 1 Xn k=0
(k + 1) (n + 1) π (n + 2)
Z n+1π
0
ψx(u) sinβu 2 pdu
)1p
( 1
n + 1 Xn k=0
(k + 1) π
Z k+1π
0
ψx(u) sinβ u 2 pdu
)1p
=
( 1 n + 1
Xn k=0
"
e wxf
π k + 1
p,β
#p)1p
,
and this completes the proof.
4.1. Proof of Theorem 2.1. We start with Ten,A,Bf (x)− ef
x, π
n + 1
= −1 π
Z n+1π
0
ψx(t) Xn r=0
Xr k=0
an,n−rbr,kDfk(t) dt
+1 π
Z π
n+1π
ψx(t) Xn r=0
Xr k=0
an,n−rbr,kDfk◦(t) dt
= Ie1(x) + eI2◦(x) and Ten,A,Bf (x)− ef
x, π
n + 1
6
eI1(x) + eI2◦(x) . In case p = 1 and 0 ¬ β ¬ 1 from Lemma 4.1 and (9), we obtain
eI1(x) ¬ 1 2π
Xn r=0
Xr k=0
an,n−rbr,kk (k + 1) Z n+1π
0 |ψx(t)| tdt, and using the inequality sin2t πt (0 ¬ t ¬ π), we get
eI1(x) ¬ n (n + 1) π
Z n+1π
0 |ψx(t)|
sinβ t
2
t1−βdt
= n
π n + 1
1−β
e wxf
π n + 1
1,β (n + 1)βwexf
π n + 1
1,β
.
In case p > 1 and 0 ¬ β < 2−1p, the H¨older inequality for integrals, with 1p+1q = 1, Lemma 4.1 and (9) give
eI1(x) ¬ n (n + 1) 2π
Z n+1π
0 |ψx(t)| tdt
¬ n(n + 1)1−1p 2
(n + 1 π
Z n+1π
0
|ψx(t)| sinβ t 2
p
dt )1p
(Z n+1π
0
"
t sinβ t2
#q
dt )1q
(n + 1)2−1pwexf
π n + 1
p,β
(Z n+1π
0 t(1−β)qdt )1q
(n + 1)2−1pwexf
π n + 1
p,β
(n + 1)β−1−1q
= (n + 1)βwexf
π n + 1
p,β
.
Now we estimate the term eI2◦. Using Lemma 4.2 and the inequality sin2t πt
(0 ¬ t ¬ π), we obtain
eI2◦(x) ¬ 1 π
Z π
π n+1
|ψx(t)|
Xn r=0
Xr k=0
an,n−rbr,kDfk◦(t)
dt¬ π n + 1
Z π
π n+1
|ψx(t)|
t2 dt
¬ (n + 1)β−1 Z π
π n+1
|ψx(t)| sinβ t2
t2 dt, for β 0. Integrating by parts we get
eI2◦(x) (n + 1)β−1 (1
t2 Z t
0 |ψx(s)| sinβs 2ds
π
n+1π
+ 2 Z π
n+1π
1 t3
Z t
0 |ψx(s)| sinβs 2ds
dt
)
¬ (n + 1)β−1 ( 1
π2 Z π
0 |ψx(s)| sinβs 2ds + 2
Z π
n+1π
1
t2wexf (t)1,βdt )
(n + 1)β−1
e
wxf (π)1,β+Z n+1 1 wex
π u
1,βdu
¬ (n + 1)β−1 (
e
wxf (π)1,β+ Xn m=1
Z m+1
m wexf π u
1,βdu )
(n + 1)β−1 (
e
wxf (π)1,β+ Xn m=0
e wxf
π
m + 1
1,β
)
¬ (n + 1)β−1 Xn m=0
e wxf
π
m + 1
1,β
.
Collecting these estimates the desired result is proved. 4.2. Proof of Theorem 2.2. Let us usually
Ten,A,Bf (x)− ef (x) = −1 π
Z n+1π
0 ψx(t) Xn r=0
Xr k=0
an,n−rbr,kDfk◦(t) dt
+1 π
Z π
π n+1
ψx(t) Xn r=0
Xr k=0
an,n−rbr,kDf◦k(t) dt
= Ie1◦(x) + eI2◦(x)
and eTn,A,Bf (x)− ef (x) 6 eI1◦(x) + eI2◦(x) .
From Lemma 4.1 and (8), using the H¨older inequality for integrals, with 1p+1q = 1, and from (4) and (5) we obtain
eI1◦(x) ¬ 1 π
Z n+1π
0 |ψx(t)|
Xn r=0
Xr k=0
an,n−rbr,k
fDk◦(t) dt
Z n+1π
0
|ψx(t)|
t dt
(Z n+1π
0
|ψx(t)|
e wx(t)
p
sinβp t 2dt
)p1
(Z n+1π
0
"
e wx(t) t sinβ t2
#q dt
)1q
(n + 1)−1p(n + 1)β+1pwex
π n + 1
= (n + 1)βwex
π n + 1
¬ (n + 1)β−1 Xn k=0
e wx
π n + 1
for 1 < p < ∞. Using Lemma 4.2 for eI2◦ we have
eI2◦(x) ¬ 1 π
Z π
n+1π
|ψx(t)|
Xn r=0
Xr k=0
an,n−rbr,kDfk◦(t)
dt¬ π2 n + 1
Z π
n+1π
|ψx(t)|
t2 dt.
Analogously to the second part of the proof of Theorem 2.1, we obtain
eI2◦(x) (n + 1)β−1 Xn m=0
e wxf
π
m + 1
1,β
and eI2◦(x) (n + 1)β−1 Xn m=0
e wx
π
m + 1
.