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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

(2)

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



x(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

(3)

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

x(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 .

(4)

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

x(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.

(5)

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)β1p 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)β1p ( 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

!

(6)

= 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,kDfk(t) dt, where

(7) Dfk(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) fDk(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.

(7)

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,kDfk(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.

(8)

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

Xn r=0

Xr k=0

an,n−rbr,kk (k + 1) Z n+1π

0 x(t)| tdt, and using the inequality sin2t ­ πt (0 ¬ t ¬ π), we get

eI1(x) ¬ n (n + 1) π

Z n+1π

0 x(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)

Z n+1π

0 x(t)| tdt

(9)

¬ n(n + 1)1−1p 2

(n + 1 π

Z n+1π

0



x(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

x(t)|

Xn r=0

Xr k=0

an,n−rbr,kDfk(t)

dt¬ π n + 1

Z π

π n+1

x(t)|

t2 dt

¬ (n + 1)β−1 Z π

π n+1

x(t)| sinβ t2

t2 dt, for β ­ 0. Integrating by parts we get

eI2(x)  (n + 1)β−1 (1

t2 Z t

0 x(s)| sinβs 2ds

π

n+1π

+ 2 Z π

n+1π

1 t3

Z t

0 x(s)| sinβs 2ds

 dt

)

¬ (n + 1)β−1 ( 1

π2 Z π

0 x(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

.

(10)

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,kDfk(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 x(t)|

Xn r=0

Xr k=0

an,n−rbr,k

fDk(t) dt



Z n+1π

0

x(t)|

t dt

(Z n+1π

0

x(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

x(t)|

Xn r=0

Xr k=0

an,n−rbr,kDfk(t)

dt¬ π2 n + 1

Z π

n+1π

x(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

 .

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