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W lodzimierz Lenski, Bogdan Szal

Pointwise strong and very strong approximation by matrix means of Fourier series

Abstract. We generalize and extend the some results of the paper [6]. Considering a wider class of function and more general means we obtain the results of the V.

Totik type [8, 9].

2000 Mathematics Subject Classification: 42A24, 41A25.

Key words and phrases: Strong and very strong approximation, Rate of pointwise summability, Matrix means.

1. Introduction. Let Lp (1 < p < ∞) [resp.C] be the class of all 2π–periodic real–valued functions integrable in the Lebesgue sense with p–th power [continuous]

over Q = [−π, π] and let X = Xp where Xp = Lp when 1 < p < ∞ or Xp = C when p = ∞. Let us define the norm of f ∈ Xp as

kfkXp = kf(x)kXp =



R

Q | f(x) |pdx1/p

when 1 < p <∞ , supx∈Q| f(x) | when p =∞.

Consider the trigonometric Fourier series

Sf (x) =ao(f) 2 +X

k=0

(ak(f) cos kx + bk(f) sin kx)

and denote by Skf the partial sums of Sf . Let

Hkq0,krf (x) :=

( 1 r + 1

Xr ν=0

|Skνf (x)− f (x)|q )1q

, (q > 0) .

(2)

where 0 ≤ k0< k1< k2< ... < kr (≥ r) , and

Hn,Aq f (x) :=

( n X

k=0

ank|Skf (x)− f (x)|q )1q

,

where A := (ank) (k, n = 0, 1, ...) be a lower triangular infinite matrix of real num- bers satisfying the following conditions:

(1) ank≥ 0 (k, n = 0, 1, ...) , ank= 0, k > n and Xn k=0

ank= 1.

Now we define two classes of sequences (see [4]).

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

(2) X

k=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 the Head Bounded Variation Sequence, or briefly c ∈ HBV S, if it has the property

(3)

m−1X

k=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 terms 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) denotes the sequence of constants appearing in the in- equalities (2) or (3) 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

(4)

X k=m

|ank− ank+1| ≤ Kanm

and

(5)

mX−1 k=0

|ank− ank+1| ≤ Kanm

hold if αn:= (ank)k=0 belongs to RBV S or HBV S, respectively.

(3)

As a measure of approximation by the above quantities we use the pointwise characteristic

wx,β(δ; f)Lp:=



 1 δ1+βp

Zδ

0

x(t)|

sin t

2

β!p

dt



1 p

,

where β ≥ 0 and

ϕx(t) = f (x + t) + f (x − t) − 2f (x) .

The quantity wx,β(δ; f)Lpis constructed on the base of the definition of the Lebesgue (Lp-points), introduced in [5] (cf.[1]) and every point x for which

1 δ1+βp

Zδ

0

|f (x ± t) − f (x)|

sin t

2

β!p

dt = o(1) as δ→ 0+

we will call the Lpβ-Lebesgue point.

We can observe that for f ∈ Xpeand ep ≥ p ≥ 1, by the Minkowski inequality, kw·,βf (δ)LpkX ep≤ ωX epf (δ) ,

where ωXf is the modulus of continuity of f in the space X = Xpedefined by the formula

ωXf (δ) := sup

0<|h|≤δ·(h)kX .

We shall write I1 I2 if there exists a positive constant C such that I1≤ CI2. Let us consider a function wx of modulus of continuity type on the interval [0, 2π], i.e. a nondecreasing continuous function having the following properties:

wx(0) = 0, wx1+ δ2) ≤ wx1) + wx2) for any 0 ≤ δ1≤ δ2≤ δ1+ δ2≤ 2π and let for β ≥ 0

Lp(wx)?β = {f ∈ Lp:

"

1 δ1+βp

Z δ

0x(t) − ϕx(t ± γ)|

sin t

2

β!p

dt

#1/p

 wx(γ) and wx,βf (δ)Lp wx(δ) , where γ, δ > 0} .

Since, for β > α ≥ 0,

wx,β(δ; f)Lp=



 δαp δβp

1 δ1+αp

Zδ

0



x(t)|

sint

2

αp sin2t β sint2 α

!p dt



1 p

≤πα 2β



δp(α−β) 1 δ1+αp

Zδ

0



x(t)|

sin t

2

αp

tp(β−α)dt



1 p

≤πα

2βwx,α(δ; f)Lp

(4)

and

"

1 δ1+βp

Z δ 0



x(t) − ϕx(t ± γ)| sins 2

βp

dt

#1/p

≤ πα 2β

"

1 δ1+αp

Z δ 0

|ϕx(t) − ϕx(t ± γ)| sins 2

αp dt

#1/p

,

we have Lp(wx)α⊂ Lp(wx)β. Let also

Lp(ω) = {f ∈ Lp: ωLpf (δ)≤ ω (δ)} , where ω is a modulus of continuity type function.

It is well-known that H0,rq f (x)− means tend to 0 (as r → ∞) at the Lp−point x of f ∈ Lp (1 < p ≤ ∞) . In [2] this fact was by G. H. Hardy, J. E. Littlewood proved as a generalization of the Fej´er classical result on the convergence of the (C, 1) -means of Fourier series. In [6] W. Lenski obtained an estimation of the Hkq0,krf (x) means as an approximation version of the Totik type (see [8, Na1]) generalization of the result of G. H. Hardy, J. E. Littlewood, considered by N. L.

Pachulia in [7]. Namely, he proved the following theorem:

Theorem 1.1 [6] If f ∈ Lp(wx)?0 (1 < p ≤ 2) and 0 < k0 < k1 < k2 < ... < kr

(≥ r), then

(6) Hkq0,krf (x) wx

 π

k0+ 1



log kr+ 1 r + 1/2, where p1+1q = 1.

In the present paper we shall generalize this result taking a wider class of func- tions and more general means. We also give some corollaries on norm approxima- tion.

2. Main results. We start with the following theorem:

Theorem 2.1 If f ∈ Lp(wx)?β (p > 1, β ≥ 0) and 0 < k0 < k1 < k2 < ... < kr

(≥ r), then

(7) Hkq0,krf (x) wx

 π

k0+ 1

 

1 + logkr+ 1/2 r + 1/2

 ,

where q is such that 1 < q (q − 1)−1 ≤ p ≤ q.

Remark 2.2 If β = 0 and p = q (q− 1)−1, then we have (6) as a corollary of (7).

Similarly, we have estimate for norms.

(5)

Theorem 2.3 If f ∈ Lp(ω) (p ≥ 2) and 0 < k0< k1< k2< ... < kr (≥ r), then

Hkq0,krf (·)

Lp ω

 π

k0+ 1

 

1 + logkr+ 1/2 r + 1/2

 , where q is such that 0 < q ≤ p.

Basing on Theorem 2.1 and 2.3, we can formulate the next theorems.

Theorem 2.4 Let (1) and (4) hold. If f ∈ Lp(wx)?β (p > 1, β ≥ 0), then

Hn,Aq f (x) ( n

X

k=0

ank

 wx

 π

k + 1

q)1q

,

where q is such that 1 < q (q − 1)−1 ≤ p ≤ q.

Theorem 2.5 Let (1) and (5) hold. If f ∈ Lp(wx)?β (p > 1, β ≥ 0), then

Hn,Aq f (x)



 [Xn+14 ]

k=1

an,4k

 wx

 π k + 1

q +

wx

 π

n + 1



log ((n + 1) ann)q





1 q

,

where q is such that 1 < q (q − 1)−1 ≤ p ≤ q.

Theorem 2.6 Let (1) and (4) hold. If f ∈ Lp(ω) (p ≥ 2), then

Hn,Aq f (·)

Lp ( n

X

k=0

ank

 ω

 π k + 1

q)1q , where q is such that 0 < q ≤ p.

3. Lemmas. With the notation

Φxf (δ, γ) :=1 δ

Z γ+δ γ

ϕx(t) dt, wxf (δ, γ)Lp:=

"

1 δ

Z γ+δ

γx(t)|pdt

#1/p

we can formulate the following lemma:

Lemma 3.1 If f ∈ Lp(wx)?β (p ≥ 1) , then

xf (δ, γ)| ≤ wxf (δ, γ)Lp ≤ πβ(1 + βp)1/p(wx(δ) + wx(γ)) for any positive γ, δ.

(6)

Proof The first inequality is evident, then we prove the second one only. Using integration by parts we obtain

wxf (δ, γ)Lp=

"

1 δ

Z δ

0x(t + γ)|pdt

#1/p

"

1 δ

Z δ

0x(t + γ) − ϕx(t)|pdt

#1/p

+

"

1 δ

Z δ

0x(t)|pdt

#1/p

=

1 δ

Z δ 0

|ϕx(t + γ) − ϕx(t)| sint2 βp

sin2t βp dt

1/p

+

1 δ

Z δ 0

|ϕx(t)| sin2t βp

sint2 βp dt

1/p

≤ πβ



"

1 δ

Z δ 0

1 tβp

d dt

Z t 0



x(s + γ) − ϕx(s)| sins 2 β

p

ds

 dt

#1/p

"

1 δ

Z δ 0

1 tβp

d dt

Z t 0



x(s)| sins 2 β

p

ds

 dt

#1/p

= πβ ("

1 δ

 1 tβp

Z t 0



x(s + γ) − ϕx(s)| sins 2 β

p ds

δ

0

βp δ

Z δ 0

1 t1+βp

Z t 0



x(s + γ) − ϕx(s)| sins 2

βp

dsdt

#1/p

+

"

1 δ

 1 tβp

Z t 0



x(s)| sins 2

βp

ds

δ

0

βp δ

Z δ 0

1 t1+βp

Z t 0



x(s)| sins 2 β

p

dsdt

#1/p

. Since f ∈ Lp(wx)?β, we have

wxf (δ, γ)Lp≤ πβ



"

(wx(γ))p+βp δ

Z δ 0

(wx(γ))pdt

#1/p

"

(wx(δ))p+βp δ

Z δ 0

(wx(t))pdt

#1/p

≤ πβ(1 + βp)1/p(wx(δ) + wx(γ))

and this ends of our proof. 

(7)

Lemma 3.2 If f ∈ Lp(ω) (p ≥ 1), then

·f (δ, γ)kLp≤ kw·f (δ, γ)LpkLp≤ 2 (ω (δ) + ω (γ)) and

"

1 δ1+βp

Z δ

0·(t) − ϕ·(t ± γ)| sin t 2

β!p

dt

#1/p Lp

≤ 2ω (γ)

for any γ, δ > 0 and β ≥ 0.

Proof The first inequality is proved in [6]. Therefore we prove the second one, only. If we change the order of integration, then we obtain

"

1 δ1+βp

Z δ

0·(t) − ϕ·(t ± γ)| sin t 2

β!p dt

#1/p

Lp

"

1 δ

Z δ

0·(t) − ϕ·(t ± γ)|pdt

#1/p Lp

≤ (1

δ Z δ

0

Z π

−πx(t) − ϕx(t ± γ)|pdx

 dt

)1/p

≤ (1

δ Z δ

0

[2ωLpf (γ)] dt )1/p

= 2ωLpf (γ)

and our lemma follows. 

Lemma 3.3 [N1, Theorem 5.20 II, Ch. XII] Suppose that 1 < q (q − 1)−1 ≤ p ≤ q and ξ = p1+1q − 1. If t−ξg (t)

∈ Lp then

(8)

(|a0(g)|q 2 +X

k=0

(|ak(g)|q+ |bk(g)|q) )1q





 Zπ

−π

t−ξg (t) pdt



1 p

.

4. Proofs of the Results. We prove the theorems on pointwise approxima- tion only. The estimations of the norm of our means are an immediate consequences of these which will be proved.

4.1. Proof of Theorem 2.1. Let as usual

Hkq0,kr(x) = ( 1

r + 1 Xr ν=0

1

π Z π

0

ϕx(t) Dkν(t) dt

q)1/q

≤ Ar+ Br+ Cr,

(8)

where

Ar= ( 1

r + 1 Xr ν=0

1 π

Z ν

0

ϕx(t) Dkν(t) dt

q)1/q

Br= ( 1

r + 1 Xr ν=0

1

π Z r

ν

ϕx(t) Dkν(t) dt

q)1/q

Cr= ( 1

r + 1 Xr ν=0

1

π Z π

r

ϕx(t) Dkν(t) dt

q)1/q

,

with Dkν(t) = sin((kν+12)t)

2 sin2t , δν= k π

ν+1/2 and γr= r+1/2π , and

Ar≤ π 2

( 1 r + 1

Xr ν=0

"

kν+ 1/2 π

Z ν

0x(t)| dt

#q)1/q .

Using integration by parts we obtain Z ν

0x(t)| dt = Z ν

0

x(t)| sin2t β sin2t β dt

 Z ν

0

1 tβ

d dt

Z t

0x(s)| sins 2 βds

 dt

=  1 tβ

Z t

0x(s)| sins 2

β

ds

ν

0

+βZ ν

0

1 t1+β

Z t

0x(s)| sins 2

β

dsdt

= 2δνwx,βf (2δν)L1+ βZ ν

0

wx,βf (t)L1dt

≤ 2δνwx,βf (2δν)Lp+ βZ ν

0

wx,βf (t)Lpdt

(9) ≤ 2δνwx(2δν) + βZ ν

0

wx(t) dt  δνwx(2δν) , whence

Ar ( 1

r + 1 Xr ν=0

[wx(2δν)]q )1/q

 wx0).

The quantities Brand Cr we will estimate by the Totik method [9].

(9)

At the begin, we divide the term Brinto the two parts and use (9). Thus

Br= ( 1

r + 1 Xr ν=0

1 π

Z r

ν

ϕx(t) Dkν(t) dt

q)1/q

≤ ( 1

r + 1

νX0−1 ν=0

+ Xr ν=ν0

! 1 π

Z r

ν

ϕx(t) Dkν(t) dt

q)1/q

≤ ( 1

r + 1

νX0−1 ν=0

1 π

Z r

ν

ϕx(t) Dkν(t) dt

q)1/q

+ ( 1

r + 1 Xr ν=ν0

1

π Z r

ν

ϕx(t) Dkν(t) dt

q)1/q

 ( 1

r + 1

νX0−1 ν=0

"

kν+ 1/2 π

Z ν

r

x(t)| dt

#q)1/q + Br,ν0

 wx0) + Br,ν0,

where the index ν0 is such that kν0−1 < r≤ kν0. Next, we divide the term Br,ν0

into the three following parts

Br,ν0. = ( 1

r + 1 Xr ν=ν0

1

π Z r

ν

ϕx(t) Dkν(t) dt

q)1/q

= 1

2 ( 1

r + 1 Xr ν=ν0

1 π

Z r

ν

+Z r−δν

δν

+Z r

r−δν

− Z ν

δν

!

ϕx(t) Dkν(t) dt

q)1/q

≤ Br,ν1 0+ Br,ν2 0+ Br,ν3 0, where the first term

B1r,ν0 = 1 2

( 1 r + 1

Xr ν=ν0

1 π

Z r

ν

+Z r−δν

δν

!

ϕx(t) Dkν(t) dt

q)1/q

= 1 2

( 1 r + 1

Xr ν=ν0

1 π

Z r

ν

x(t) Dkν(t) + ϕx(t − δν) Dkν(t − δν)] dt

q)1/q

≤ 1 2

( 1 r + 1

Xr ν=ν0

1 π

Z r

ν

x(t) − ϕx(t − δν)) Dkν(t) dt

q)1/q

(10)

+1 2

( 1 r + 1

Xr ν=ν0

1

π Z r

ν

ϕx(t − δν) (Dkν(t) + Dkν(t − δν)) dt

q)1/q

≤ 1 2

( 1 r + 1

Xr ν=ν0

Z r

ν

1

t |ϕx(t) − ϕx(t − δν)| dt

q)1/q

+1 2

( 1 r + 1

Xr ν=ν0

1 π

Z r

ν

ϕx(t − δν) 1

2 sin2t − 1 2 sint−δ2ν

!

sin(2kν+ 1) t

2 dt

q)1/q

 ( 1

r + 1 Xr ν=ν0

Z r

ν

1

t |ϕx(t) − ϕx(t − δν)| dt

q)1/q

+ ( 1

r + 1 Xr ν=ν0

δν

Z r

ν

x(t − δν)|

t (t− δν) dt

q)1/q

≤ ( 1

r + 1 Xr ν=ν0

Z r

ν

1

t |ϕx(t) − ϕx(t − δν)| dt

q)1/q

+ ( 1

r + 1 Xr ν=ν0

δν

Z r

δν

x(t)|

t2 dt

q)1/q

. Integrating by parts and applying our assumption we get

B1r,ν0 ( 1

r + 1 Xr ν=ν0

Z r

ν

1 t1+β

d dt

Z t

0x(s) − ϕx(s − δν)| sins 2 βds

 dt

q)1/q

+ ( 1

r + 1 Xr ν=ν0

δν

Z r

δν

1 t2+β

d dt

Z t

0x(s)| sins 2 βds

 dt

q)1/q

≤ ( 1

r + 1 Xr ν=ν0

"

1 (2γr)1+β

Z r

0x(s) − ϕx(s − δν)| sins 2

β

ds

#q)1/q

+ ( 1

r + 1 Xr ν=ν0

"

1 (2δν)1+β

Z ν

0x(s) − ϕx(s − δν)| sins 2

β

ds

#q)1/q

+ ( 1

r + 1 Xr ν=ν0

Z r

ν

1 t2+β

Z t

0x(s) − ϕx(s − δν)| sins 2

β

dsdt

q)1/q

+ ( 1

r + 1 Xr ν=ν0

"

δν

1 (2γr)2+β

Z r

0x(s)| sins 2 βds

#q)1/q

(11)

+ ( 1

r + 1 Xr ν=ν0

"

1 (δν)1+β

Z δν

0x(s)| sins 2

β

ds

#q)1/q

+ ( 1

r + 1 Xr ν=ν0

 δν

Z r

δν

1 t3+β

Z t

0x(s)| sins 2

β

dsdt

q)1/q

≤ wx0) + ( 1

r + 1 Xr ν=ν0



wxν)Z r

ν

1 tdt

q)1/q

( 1 r + 1

Xr ν=ν0

 δν

wx(2γr) 2γr

q)1/q

+ wx0) + ( 1

r + 1 Xr ν=ν0

 δν

Z r

δν

wx(t) t2 dt

q)1/q

 wx0)



1 + logkr+ 1/2 r + 1/2

 +

( 1 r + 1

Xr ν=ν0

[wxν)]q )1/q

+ ( 1

r + 1 Xr ν=ν0



wxν)Z r

δν

1 tdt

q)1/q

 wx0)

1 + logkr+ 1/2 r + 1/2

 . Consequently, by Lemma 3.1,

Br,ν2 0 =1 2

( 1 r + 1

Xr ν=ν0

1 π

Z r

r−δν

ϕx(t) Dkν(t) dt

q)1/q

≤1 4

( 1 r + 1

Xr ν=ν0

Z r

r−δν

1

t |ϕx(t)| dt

q)1/q

≤1 4

( 1 r + 1

Xr ν=ν0

 1

r− δν Z r

r−δν

x(t)| dtq)1/q

≤ 1 4

( 1 r + 1

Xr ν=ν0

"

δν

r− δν 1 δν

Z r−δνν

r−δν

x(t)| dt

#q)1/q

≤1 4

( 1 r + 1

Xr ν=ν0

 δν

r− δν[wx(2γr− δν) + wxν)]q)1/q

≤ 1 4

( 1 r + 1

Xr ν=ν0



δνwx(2γr− δν)

r− δν + wxν)

q)1/q

 ( 1

r + 1 Xr ν=ν0



δνwxν) δν

+ wxν)q)1/q

 wx0)

(12)

and

B3r,ν0 = 1 2

( 1 r + 1

Xr ν=ν0

1 π

Z ν

δν

ϕx(t) Dkν(t) dt

q)1/q

≤1 4

( 1 r + 1

Xr ν=ν0

Z ν

δν

x(t)|1 tdt

q)1/q

 ( 1

r + 1 Xr ν=ν0

[wxν)]q )1/q

 wx0) . Thus

Br wx0)

1 + logkr+ 1/2 r + 1/2

 . For the estimate of Cr we use our notation in the form Φxf (δ0, t) := δ10 Rt+δ0

t ϕx(u) du with δ0= k0+1/2π . Then

Cr= ( 1

r + 1 Xr ν=0

1 π

Z π r

ϕx(t)

 2 sint

2

−1

sin



kν+1 2

 t

 dt

q)1/q

≤ ( 1

r + 1 Xr ν=0

1 π

Z π r

 Φxf (δ0, t)− ϕx(t) 2 sint2

 sin



kν+1 2

 t

 dt

q)1/q

+ ( 1

r + 1 Xr ν=0

1

π Z π

r

Φxf (δ0, t) 2 sin2t sin

kν+1 2

 t

 dt

q)1/q

= Cr1+ Cr2 and, by inequality (8),

Cr1 1 (1 + r)1/q

(1 π

Z π r

xf (δ0, t)− ϕx(t)] 2 sin t

2

−1

t1−1/p−1/q

p

dt )1/p

 1

(1 + r)1/q

Z π r

xf (δ0, t)− ϕx(t)|p

t1+p/q dt

1/p

= 1

(1 + r)1/q (Z π

r

1 δ0t1/p+1/q

Z t+δ0

t

x(u) − ϕx(t)] du

p

dt )1/p

≤ 1

(1 + r)1/q (Z π

r

"

1 δ0t1/p+1/q

Z δ0

0x(u + t) − ϕx(t)| du

#p

dt )1/p

. Applying the generalized Minkowski inequality and the partial integration we obtain

Cr1 1 δ0(1 + r)1/q

Z δ0

0

Z π r

x(u + t) − ϕx(t)|p t1+p/q dt

1/p

du 1

δ0(1 + r)1/q

(13)

· Z δ0

0

Z π r

1 t1+p/q+βp

d dt

Z t 0



x(u + v) − ϕx(v)| sinv 2

βp

dv

 dt

1/p du

= 1

(1 + r)1/q 1 δ0

Z δ0

0

( 1

t1+p/q+βp Z t

0



x(u + v) − ϕx(v)| sinv 2

βp

dv

π

t=2γr

+ (1 + p/q + βp)Z π r

 1

t2+p/q+βp Z t

0



x(u + v) − ϕx(v)| sinv 2 β

p dv

 dt

1/p du.

Hence, since f ∈ Lp(wx)?β,we have

Cr1 1 (1 + r)1/q

1 δ0

Z δ0

0

 1

π1+p/q+βp Z π

0



x(u + v) − ϕx(v)| sinv 2

βp

dv



+

"

1 (2γr)1+p/q+βp

Z r

0



x(u + v) − ϕx(v)| sinv 2 β

p dv

#

+Z π r

1

t1+p/q(wx(u))pdt

1/p

du

≤ 1

(1 + r)1/q 1 δ0

Z δ0

0

(

[wx(u)]p 1

πp/q + 1

(2γr)p/q +Z π r

1 t1+p/qdt

!)1/p

du

 1

(1 + r)1/qwx0) (

1 + 1

r)p/q +Z π r

t−(1+p/q)dt )1/p

 1

(1 + r)1/qwx0)

( 1

r)p/q +t−p/q

−p/q

π

t=2γr

)1/p

≤ 1

(1 + r)1/qwx0)

( 1

r)p/q

 1 +q

p

)1/p

 wx0) .

For the estimate of Cr2 we will consider the two cases r ≤ k0 and r ≥ k0 . In the first case, using the partial integration and inequality (8), we obtain

Cr2= 1 2 (r + 1)1/q

( r X

ν=0

1 π

Z π r

Φxf (δ0, t) sin2t

d dt

cos kν+12 t kν+12

! dt

q)1/q

= 1

2π (r + 1)1/q



 Xr ν=0

"

Φxf (δ0, t) sint2

cos kν+12 t kν+12

#π

r

− Z π

r

d dt

 Φxf (δ0, t) sin2t

 cos kν+12 t kν+12 dt

q)1/q

(14)

 1 (r + 1)1/q

( r X

ν=0

"

Φxf (δ0, 2γr) sin γr

cos kν+122γr kν+12

+

Z π r

d dt

 Φxf (δ0, t) sin2t

 cos kν+12 t kν+12 dt

#q)1/q

and analogously

Cr2 1 (r + 1)1/q

((r + 1)1/qxf (δ0, 2γr)|

γr(k0+ 1)

+ 1

k0+ 1

Z π r

d

dt

 Φxf (δ0, t) sint2



t1−1/p−1/q

p

dt

1/p)

 1

γr(k0+ 1)wx0) + 1 k0+ 1

wx(2γr) 2γr

+ 1

(k0+ 1) (r + 1)1/q

Z π r

d

dt

 Φxf (δ0, t) sint2



t1−1/p−1/q

p

dt

1/p

 wx0) + 1 k0+ 1

wx(2δ0) 2δ0

+ 1

(k0+ 1) (r + 1)1/q

"Z π r

d

dtΦxf (δ0, t)

sin2t −Φxf (δ0, t) cos2t 2 sin2t2

!

t1−1/p−1/q

p

dt

#1/p

 wx0)

+ 1

(k0+ 1) (r + 1)1/q

"Z π r

1

δ0x0+ t) − ϕx(t)|

t1/p+1/q +wx0) + wx(t) t1+1/p+1/q

p

dt

#1/p

 wx0) + 1

(k0+ 1) (r + 1)1/q

· 1 δ0p

Z π r

t−1−p/q−βpd dt

Z t 0



x0+ u) − ϕx(u)| sinu 2 β

p

du

 dt

1/p

+ 1

(k0+ 1) (r + 1)1/q (Z π

r

wx0) t1+1/p+1/q

p

dt

1/p +Z π

r

wx(t) t1+1/p+1/q

p

dt

1/p)

 wx0)

+ 1

(k0+ 1) (r + 1)1/qδ0

 1

π1+p/q+βp Z π

0



x0+ u) − ϕx(u)| sinu 2

βp

du

− 1

(2γr)1+p/q+βp Z r

0



x0+ u) − ϕx(u)| sinu 2

βp

du

(15)

+ (1 + p/q + βp)Z π r

t−2−p/q−βp Z t

0



x0+ u) − ϕx(u)| sinu 2

βp

dudt

1/p

+ wx0) (k0+ 1) (r + 1)1/q

Z π r

t−p−1−p/qdt

1/p

+ 1

(k0+ 1) (r + 1)1/q

Z π r

wx(t) /t t1/p+1/q

p

dt

1/p

 wx0)+ wx0) (k0+ 1) (r + 1)1/qδ0

"

1 πp/q + 1

γrp/q

+ (1 + p/q + βp)Z π r

t−1−p/qdt

#1/p

+(γr)−1−1/qwx0)

(k0+ 1) (r + 1)1/q + (γr)−1/qwxr) (k0+ 1) (r + 1)1/qγr

 wx0) + wx0) (k0+ 1) (r + 1)1/qδ0

"

1 γrp/q

+ (1 + p/q + βp)(2γr)−p/q p/q

#1/p

+ wx0)

(k0+ 1) (r + 1)1/qr)1+1/q + wxr) /γr

(k0+ 1) (r + 1)1/qγr1/q

 wx0) + wx0)

(k0+ 1) δ0(r + 1)1/qγr1/q

+ wx0)

(k0+ 1) (r + 1)1/qγr1+1/q

 wx0) . In the case r ≥ k0we will divide the interval of integrability into two parts. Thus, by inequality (8), Lemma 3.1 and the partial integration,

Cr2= 1 2 (r + 1)1/q

( r X

ν=0

1

π Z π

r

Φxf (δ0, t) sin2t sin

kν+1 2

 t

 dt

q)1/q

 1

(r + 1)1/q (Z 0

r

Φxf (δ0, t)

sint2 t1−1/p−1/q

p

dt )1/p

+ 1

2 (r + 1)1/q ( r

X

ν=0

1 π

Z π 0

Φxf (δ0, t) sin2t

d dt

cos kν+12 t kν+12

! dt

q)1/q

≤ 1

(r + 1)1/q (Z 0

r

Φxf (δ0, t)

sin2t t1−1/p−1/q

p

dt )1/p

+ 1

2π (r + 1)1/q ( r

X

ν=0

"

Φxf (δ0, t) sint2

cos kν+12 t kν+12

#π

0

− Z π

0

d dt

 Φxf (δ0, t) sint2

 cos kν+12 t kν+12 dt

q)1/q

(16)

 1 (r + 1)1/q

(Z 0

r

wx0) + wx(t) t1/p+1/q

p

dt )1/p

+ Φxf (δ0, 2δ0) sin δ0

1 k0+12

+ 1

(k0+ 1) (r + 1)1/q ( r

X

ν=0

Z π 0

d dt

 Φxf (δ0, t) sin2t

 cos

kν+1 2

 t

 dt

q)1/q

. Using once more inequality (8) and Lemma 3.1

Cr2 wx0) (r + 1)1/q

(Z 0

r

t−1−p/qdt )1/p

+wx0) + wx(2δ0) (k0+ 1) δ0

+ 1

(k0+ 1) (r + 1)1/q

Z π 0

d

dt

 Φxf (δ0, t) sint2



t1−1/p−1/q

p

dt

1/p

 wx0)

+ 1

(k0+ 1)1+1/q

"Z π 0

d

dtΦxf (δ0, t)

sint2 −Φxf (δ0, t) cos2t sin2t2

!

t1−1/p−1/q

p

dt

#1/p . Similarly to the first case

Cr2 wx0) + 1 (k0+ 1)1+1/q

· 1 δp0

Z π 0

t−1−p/q−βpd dt

Z t 0



x0+ u) − ϕx(u)| sinu 2

βp

du

 dt

1/p

+ 1

(k0+ 1)1+1/q (Z π

0

wx0) t1+1/p+1/q

p

dt

1/p +Z π

0

wx(t) t1+1/p+1/q

p

dt

1/p)

 wx0) + 1 (k0+ 1)1+1/qδ0

 1

π1+p/q+βp Z π

0

|ϕx0+ u) − ϕx(u)| sinu 2 p

du

+ 1

(2δ0)1+p/q+βp Z 0

0



x0+ u) − ϕx(u)| sinu 2

βp

du

+ (1 + p/q + βp)Z π 0

t−2−p/q−βp

Z t 0



x0+ u) − ϕx(u)| sinu 2 β

p

du

 dt

1/p

+ wx0) (k0+ 1)1+1/q

Z π 0

t−p−1−p/qdt

1/p

+ 1

(k0+ 1)1+1/q

Z π 0

wx(t) /t t1/p+1/q

p

dt

1/p

 wx0) + wx0) (k0+ 1)1+1/qδ0

"

1 πp/q + 1

δp/q0 + (1 + p/q + βp)Z π 0

t−1−p/qdt

#1/p

+(δ0)−1−1/qwx0)

(k0+ 1)1+1/q +(δ0)−1/qwx0) (k0+ 1)1+1/qδ0

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