1. Introduction
Let X=Lp or X=C, where Lp (1≤p≤∞) C is the class of all 2π–periodic real-valued functions, integrable in the Lebesgue sense with pth power when p≥1 and essentially bounded when p=∞ [continuous], over Q=[-π,π] with the norm (1)fLp≔f·Lp=∫Qftpdt1/pwhen 1≤p<∞,ess supt∈Qftwhen p=∞,fC≔f·C=supt∈Qftand consider the trigonometric Fourier series (2)Sfx≔a0f2+∑ν=1∞aνfcosνx+bνfsinνxwith the partial sums Skf.

Let A≔an,k be an infinite matrix of real numbers such that (3)an,k≥0 when k,n=0,1,2,…, limn→∞an,k=0, ∑k=0∞an,k=1.The A-transformation of Skf be given by (4)Tn,Afx≔∑k=0∞an,kSkfx n=0,1,2,….In this paper, we study the upper bounds of Tn,Af-fX by the second modulus of continuity of f in the space X defined by the formula (5)ω2f,δX≔sup0≤t≤δφxtX,where (6)φxt≔fx+t+fx-t-2fx.

The deviation Tn,Af-f with lower triangular infinite matrix such that(7)an,k≥0 when k=0,1,2,…,n,an,k=0 when k>n,(8)∑k=0nan,k=1, where n=0,1,2,…,was estimated in the sup norm ·C by Krasniqi [1, Theorem 3.4, p. 290] (see also [2]) as follows.

Theorem 1.
Let f∈C and (an,k) satisfy (8). Then, (9)Tn,Af-fC=Oωf,πnC+∑k=1nωf,π/kCk∑l=0k+1an,l+∑k=1nωf,πkC∑l=knan,l-an,l+2,where ωf,·C denotes the modulus of continuity of f and if ωf,·C is such that (10)∫uπt-2ωf,tCdt=OHu when u∈0,π,where H(u)≥0, we have (11)Tn,Af-fC=Oωf,πnC+Hπn∑k=0nan,k-an,k+2.Additionally, if H satisfies the condition (12)∫0uHtdt=OuHu when u∈0,π,then (13)Tn,Af-fC=OH∑k=0nan,k-an,k+2∑k=0nan,k-an,k+2,Tn,Af-fC=OHπn∑k=0nan,k-an,k+2.

In our theorems we will consider the r-differences an,k-an,k+r instead of an,k-an,k+2 considered above. We will formulate the general relation for r∈N like it formulated only for r=2 in [1, Theorem 3.4, p. 290].

2. Statement of the Results
Let us consider a function ω of modulus of continuity type on the interval [0,2π], that is, a nondecreasing continuous function having the following properties: ω0=0, ωδ1+δ2≤ωδ1+ωδ2 for any 0≤δ1≤δ2≤δ1+δ2≤2π.

Suppose that ω satisfies the condition(14)∫uπt-2ωtdt=OHu when u∈0,π,where H(u)≥0 is such that(15)∫0uHtdt=OuHu when u∈0,π.

Let (16)Xω=f∈X:ω2f,δX=Oωδ when δ∈0,2π.Our main results on the degrees of approximation are the following.

Theorem 2.
If f∈Xω, where ω satisfies condition (14) such that (15) holds and r∈N, then (17)Tn,Af-fX=OHπn+1πn+1+∑k=0∞an,k-an,k+r.Additionally, if a matrix A is such that(18)∑l=0n ∑k=lr+l-1an,k-1=O1,then (19)Tn,Af-fX=OHπn+1∑k=0∞an,k-an,k+r.

Theorem 3.
If f∈Xω, where ω satisfies condition (14) such that (15) holds and r∈N, then (20)Tn,Af-fX=OH∑k=0∞an,k-an,k+r∑k=0∞an,k-an,k+r.

Theorem 4.
Let f∈Xω, where ω satisfies condition (14) such that (15) holds and r∈N. If a matrix A is such that(21)∑k=0∞k+1an,k=On+1,then (22)Tn,Af-fX=Oωπn+1+Hπn+1∑k=0∞an,k-an,k+r.

Theorem 5.
If f∈X and a matrix A is such that (21) holds and r∈N, then (23)Tn,Af-fX=Oω2f,πn+1X+∑μ=1nω2f,π/μXμ∑k=0μ+1an,k+∑μ=1nω2f,πμX ∑k=μ∞an,k-an,k+r.

Corollary 6.
For a lower triangular infinite matrix A conditions (18) and (21) hold always and therefore for a lower triangular infinite matrix such that conditions (7) and (8) hold one can obtain the results from Theorems 2, 4, and 5 without the assumptions (18) and (21), where the mentioned results of Krasniqi follow as the special cases with r=2. Moreover, one can consider the essentially wider class of sequences than in the mentioned paper with the same degrees of approximation (see, e.g., [3, Theorem 2]).

Corollary 7.
From Theorem 5 it follows that if f∈Xω where ω satisfies condition (14) such that (15) holds and a matrix A is such that (21) and(24)∑k=m∞an,k-an,k+r≤K∑k=m/c∞an,kk,with some c>1 and r∈N, hold, then (25)Tn,Af-fX=OHπ/n+1n+1+∑k=1nan,kHπ/k+1k+1.

Remark 8.
The class of sequences defined by condition (24) was introduced by the second author in [4]. The similar classes were considered by Dyachenko and Tikhonov in [5] with r=1 (see also [6]).

Remark 9.
We note that instead of (14) and (15) one can use Bari-Stechkin conditions (26)∫0ut-1ωtdt=Oωu,∫u2πt-2ωtdt=Oωu when u∈0,2π.Then all results are true for t-1ωt instead of H(y) and Lemmas 12 and 14 are not necessary.

3. Auxiliary Results
We begin this section by some notations from [7]. Let for r=1,2,…(27)Dk,rt=sin2k+rt/22sinrt/2,D~k,rt=cos2k+rt/22sinrt/2.

It is clear by [8] that (28)Skfx=1π∫-ππfx+tDk,1tdt,Tn,Afx=1π∫-ππfx+t∑k=0nan,kDk,1tdt.Hence (29)Tn,Afx-fx=1π∫0πφxt∑k=0nan,kDk,1tdt.

Next, we present the known estimates.

Lemma 10 (see [<xref ref-type="bibr" rid="B10">8</xref>]).
If 0<t≤π then (30)Dk,1t≤π2t,D~k,1t≤π2tand, for any real t, we have (31)Dk,1t≤k+12.

Lemma 11 (see [<xref ref-type="bibr" rid="B6">7</xref>]).
Let r∈N, l∈Z, and a≔an⊂C. If x≠2lπ/r then for every m≥n(32)∑k=nmaksinkx=-∑k=nmak-ak+rD~k,rt+∑k=m+1m+rakD~k,-rt-∑k=nn+r-1akD~k,-rt,∑k=nmakcoskx=∑k=nmak-ak+rDk,rt-∑k=m+1m+rakDk,-rt+∑k=nn+r-1akDk,-rt.

Lemma 12 (see [<xref ref-type="bibr" rid="B1">9</xref>]).
If (14) and (15) hold then (33)∫0ut-1ωtdt=OuHu when u∈0,2π.

We additionally proved two slight changed estimates.

Lemma 13.
If (14) and (15) hold, then, for c≥1 and β>α>0,(34)∫αβt-1ωtdt=Oβ-αHcβ-α when β-α∈0,2π.

Proof.
By substitution of t=u+α and monotonicity of ωu/u (see Lemma 15) we obtain, for c≥1, (35)∫αβt-1ωtdt=∫0β-αωu+αu+αdu≤2∫0β-αωuudu≤2∫0cβ-αωuuduand the desired result follows from the Lemma 12.

Lemma 14.
If (14) and (15) hold, then, for b≥1,(36)∫uπt-2ωtdt=OHbu when u∈0,π.

Proof.
Using Lemma 13 with c=1, α=0 and (14) our result follows (37)∫uπt-2ωtdt=∫ubu +∫buπ t-2ωtdt≤1u∫ubuωttdt+OHbu≤1u∫0buωttdt+OHbu=1uObuHbu+OHbu=Ob+1Hbu,for b≥1.

Finally, we present very useful property of the modulus of continuity.

Lemma 15 (see [<xref ref-type="bibr" rid="B10">8</xref>]).
A function ω of modulus of continuity type on the interval [0,2π] satisfies the following condition: (38)δ2-1ωδ2≤2δ1-1ωδ1 for δ2≥δ1>0.

4. Proofs of the Results
Proof of Theorem <xref ref-type="statement" rid="thm1">2</xref>.
It is clear that for an odd r(39)Tn,Afx-fx=1π∑m=0r/2∫2mπ/r2mπ/r+π/rφxt∑k=0∞an,kDk,1tdt+1π∑m=0r/2-1∫2mπ/r+π/r2m+1π/rφxt∑k=0∞an,kDk,1tdt=I1x+I2xand for an even r(40)Tn,Afx-fx=1π∑m=0r/2-1∫2mπ/r2mπ/r+π/r +∫2mπ/r+π/r2m+1π/r φxt∑k=0∞an,kDk,1tdt=I1′x+I2x.Then, (41)Tn,Af-fX≤I1X+I1′X+I2X.Since, by Lemma 11, (42)∑k=0∞an,kDk,1t=∑k=0∞an,ksin2k+1t/22sint/2=12sint/2∑k=0∞an,ksinktcost/2+∑k=0∞an,kcosktsint/2=cost/22sint/2-∑k=0∞an,k-an,k+rD~k,rt-∑k=0r-1an,kD~k,-rt+12∑k=0∞an,k-an,k+rDk,rt+∑k=0r-1an,kDk,-rt,then(43)∑k=0∞an,kDk,1t≤12sint/2sinrt/2∑k=0∞an,k-an,k+r+∑k=0r-1an,k≤1sint/2sinrt/2∑k=0∞an,k-an,k+r.Hence, by Lemma 10, (44)I1X≤1π∑m=0r/2∫2mπ/r2mπ/r+π/rφxtX∑k=0∞an,kDk,1tdt=1π∑m=0r/2∫2mπ/r2mπ/r+π/rn+1 +∫2mπ/r+π/rn+12mπ/r+π/r φxtX∑k=0∞an,kDk,1tdt≤12∑m=0r/2∫2mπ/r2mπ/r+π/rn+1Oωttdt+1π∑m=0r/2∫2mπ/r+π/rn+12mπ/r+π/rOωtsint/2sinrt/2∑k=0∞an,k-an,k+rdt.Using Lemmas 13 and 14 with c=b=r and the estimates sint/2≥t/π for t∈0,π and sinrt/2≥rt/π-2m for t∈2mπ/r,2mπ/r+π/r, where m∈0,…,r/2, we obtain (45)I1X≤O1r2+1πrn+1Hπn+1+∑k=0∞an,k-an,k+r∑m=0r/2∫2mπ/r+π/rn+12mπ/r+π/rOωttrt/π-2mdt=O1πn+1Hπn+1+∑k=0∞an,k-an,k+r∑m=0r/2∫2mπ/r+π/rn+12mπ/r+π/rOωtrt/πt-2mπ/rdt=O1πn+1Hπn+1+∑k=0∞an,k-an,k+r∑m=0r/2∫π/rn+1π/rOωt+2mπ/rrt/πt+2mπ/rdt≤O1πn+1Hπn+1+r2+12πr∑k=0∞an,k-an,k+r∫π/rn+1π/rωtt2dt=O1πn+1Hπn+1+∑k=0∞an,k-an,k+rHπn+1.Analogously (46)I1′X=O1r2πrn+1Hπn+1+2πr∑k=0∞an,k-an,k+rHπn+1.Similarly, by Lemmas 10, 13, and 14 with c=b=r and the estimates sint/2≥t/π for t∈0,π and sinrt/2≥2m+1-rt/π for t∈2m+1π/r-π/r,2m+1π/r-π/rn+1, where m∈0,…,r/2-1, we get (47)I2X≤1π∑m=0r/2-1∫2mπ/r+π/r2m+1π/rφxtX∑k=0∞an,kDk,1tdt=1π∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1 +∫2m+1π/r-π/rn+12m+1π/r φxtX∑k=0∞an,kDk,1tdt≤12∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rOωttdt+1π∑k=0∞an,k-an,k+r∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1Oωtsint/2sinrt/2dt≤12∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rOωttdt+∑k=0∞an,k-an,k+r∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1Oωtrt/π2m+1π/r-tdt=12∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rOωttdt+∑k=0∞an,k-an,k+r∑m=0r/2-1∫π/rn+1π/rOω-t+2m+1π/rr/πt-t+2m+1π/rdt≤12∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rOωttdt+∑k=0∞an,k-an,k+rr22πr∫π/rn+1π/rOωtt2dt.Thus (48)I2X=O1πn+1Hπn+1+∑k=0∞an,k-an,k+rHπn+1.Collecting these estimates we obtain the first result.

Applying condition (18) we have (49)n+1∑k=0∞an,k-an,k+r-1=∑l=0n ∑k=0∞an,k-an,k+r-1≤∑l=0n ∑k=l∞an,k-an,k+r-1≤∑l=0n∑k=l∞an,k-an,k+r-1≤∑l=0n ∑k=lr+l-1an,k-1=O1and the second result also follows.

Proof of Theorem <xref ref-type="statement" rid="thm2">3</xref>.
Analogously, as in the proof of Theorem 2, we consider an odd r and an even r. Then, (50)Tn,Afx-fx=I1x+I2xor (51)Tn,Afx-fx=I1′x+I2x,respectively, and (52)Tn,Af-fX≤I1X+I1′X+I2X.Since An,r=∑k=0∞an,k-an,k+r≤2(53)I1X≤1π∑m=0r/2∫2mπ/r2mπ/r+π/rφxtX∑k=0∞an,kDk,1tdt=1π∑m=0r/2∫2mπ/r2mπ/r+1/rAn,r +∫2mπ/r+1/rAn,r2mπ/r+π/r φxtX∑k=0∞an,kDk,1tdt,I2X≤1π∑m=0r/2-1∫2mπ/r+π/r2m+1π/rφxtX∑k=0∞an,kDk,1tdt≤1π∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-1/rAn,r +∫2m+1π/r-1/rAn,r2m+1π/r φxtX∑k=0∞an,kDk,1tdt.Therefore, in the terms I1, I1′, and I2 we can estimate analogously as in the proof of Theorem 2 and thus we obtain the desired estimate.

Proof of Theorem <xref ref-type="statement" rid="thm3">4</xref>.
Similarly, as in the proof of Theorem 2, we consider an odd r and an even r. Then, (54)Tn,Afx-fx=1π∫0π/rn+1φxt∑k=0∞an,kDk,1tdt+1π∫π/rn+1π/rφxt∑k=0∞an,kDk,1tdt+1π∑m=1r/2∫2mπ/r2mπ/r+π/rφxt∑k=0∞an,kDk,1tdt+1π∑m=0r/2-1∫2mπ/r+π/r2m+1π/rφxt∑k=0∞an,kDk,1tdt=J1x+J2x+I1′′x+I2xor (55)Tn,Afx-fx=J1x+J2x+1π∑m=1r/2-1∫2mπ/r2mπ/r+π/r +∑m=0r/2-1∫2mπ/r+π/r2m+1π/r φxt∑k=0∞an,kDk,1tdt=J1x+J2x+I1′′′x+I2x,respectively. Therefore, (56)Tn,Af-fX≤J1X+J2X+I1′′X+I1′′′X+I2X.By Lemma 10 and (18), (57)J1X≤1π∫0π/rn+1φxtX∑k=0∞an,kDk,1tdt≤1π∑k=0∞k+1an,k∫0π/rn+1ωtdt=O1n+1∫0π/rn+1ωtdt≤O1πrωπrn+1=Oωπn+1.Further, by the same lemmas and conditions as above and Lemma 15, we obtain, with (58)κ=1when r is even,0when r is odd,that (59)I1′′X+J2X+I1′′′X≤1π∑m=1r/2-κ∫2mπ/r2mπ/r+π/r +∫π/rn+1π/r φxtX∑k=0∞an,kDk,1tdt=1π∑m=1r/2-κ∫2mπ/r2mπ/r+π/rn+1 +∑m=0r/2-κ∫2mπ/r+π/rn+12mπ/r+π/r φxtX∑k=0∞an,kDk,1tdt≤1π∑m=1r/2-κ∫2mπ/r2mπ/r+π/rn+1Oωt2sint/2dt+1π∑m=0r/2-κ∫2mπ/r+π/rn+12mπ/r+π/rOωtsint/2sinrt/2∑k=0∞an,k-an,k+rdt≤12∑m=1r/2-κ∫2mπ/r2mπ/r+π/rn+1Oωttdt+∑k=0∞an,k-an,k+r∑m=0r/2-κ∫2mπ/r+π/rn+12mπ/r+π/rOωttrt/π-2mdt≤∑m=1r/2-κOω2mπ/r2mπ/r∫2mπ/r2mπ/r+π/rn+1dt+∑k=0∞an,k-an,k+r∑m=0r/2-κ∫π/rn+1π/rOωt+2mπ/rrt/πt+2mπ/rdt≤2∑m=1r/2-κOω2π/r2π/rπrn+1+2πrr2+1∑k=0∞an,k-an,k+r∫π/rn+1π/rOωtt2dt=O1ωπn+1+∑k=0∞an,k-an,k+rHπn+1,I2X≤1π∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1 +∫2m+1π/r-π/rn+12m+1π/r φxtX∑k=0∞an,kDk,1tdt≤1π∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1Oωtsint/2sinrt/2∑k=0∞an,k-an,k+rdt+12∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rOωttdt≤∑m=0r/2-1 ∑k=0∞an,k-an,k+r∫2m+1π/r-π/r2m+1π/r-π/rn+1Oωtrt/π2m+1π/r-tdt+∑m=0r/2-1Oω2m+1π/r-π/rn+12m+1π/r-π/rn+1∫2m+1π/r-π/rn+12m+1π/rdt≤r22πr∑k=0∞an,k-an,k+r∫π/rn+1π/rOωtt2dt+2Oωπ/rπ/rπrn+1=O1∑k=0∞an,k-an,k+rHπn+1+ωπn+1.Thus our proof is complete.

Proof of Theorem <xref ref-type="statement" rid="thm4">5</xref>.
Let as above (60)Tn,Af-fX≤J1X+J2X+I1′′X+I1′′′X+I2X,J1X≤1π∫0π/rn+1φxtX∑k=0∞an,kDk,1tdt≤1π∑k=0∞k+1an,k∫0π/rn+1ω2f,tXdt=On+1∫0π/rn+1ω2f,tXdt≤O1πrω2f,πrn+1X=O1ω2f,πn+1X.Further, taking τm1=π/(rt-2mπ) and τ=π/rt, using Lemma 15, we obtain, with (61)κ=1when r is even,0when r is odd,that (62)I1′′X+J2X+I1′′′X≤1π∑m=1r/2-κ∫2mπ/r2mπ/r+π/r +∫π/rn+1π/r φxtX∑k=0∞an,kDk,1tdt=1π∑m=1r/2-κ∫2mπ/r2mπ/r+π/rn+1 +∑m=0r/2-κ∫2mπ/r+π/rn+12mπ/r+π/r φxtX∑k=0∞an,kDk,1tdt≤1π∑m=1r/2-κ∫2mπ/r2mπ/r+π/rn+1ω2f,tX2sint/2∑k=0∞an,k dt+1π∑m=0r/2-κ∫2mπ/r+π/rn+12mπ/r+π/rω2f,tX2sint/2∑k=0τm1an,k+ω2f,tXsint/2sinrt/2∑k=τm1∞an,k-an,k+rdt≤12∑m=1r/2-κ∫2mπ/r2mπ/r+π/rn+1ω2f,tXtdt+12r2+1∫π/rn+1π/rω2f,tXt∑k=0τan,k dt+2πrr2+1∫π/rn+1π/rω2f,tXt2∑k=τ∞an,k-an,k+rdt≤∑m=1r/2-κω2f,2mπ/rX2mπ/r∫2mπ/r2mπ/r+π/rn+1dt+12r2+1∑μ=1n∫μμ+1ω2f,π/rtXπ/rt∑k=0tan,kπdtrt2+2πrr2+1∑μ=1n∫μμ+1ω2f,π/rtXπ/rt2∑k=t∞an,k-an,k+rπdtrt2≤O1ω2f,πn+1X+O1∑μ=1nω2f,π/μXμ∑k=0μ+1an,k+O1∑μ=1nω2f,πμX∑k=μ∞an,k-an,k+r.Next, taking τm2=π/(-rt+2(m+1)π), we obtain (63)I2X≤1π∑m=0r/2-1∫2mπ/r+π/r2m+1π/rφxtX∑k=0∞an,kDk,1tdt≤1π∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1 +∫2m+1π/r-π/rn+12m+1π/r φxtX∑k=0∞an,kDk,1tdt≤1π∑m=0r/2-1∫2m+1π/r-π/r2m+1π/r-π/rn+1ω2f,tX2sint/2∑k=0τm2an,k+ω2f,tXsint/2sinrt/2∑k=τm2∞an,k-an,k+rdt+1π∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rω2f,tX2sint/2∑k=0∞an,k dt≤12∑m=0r/2-1∫π/rn+1π/rω2-t+2m+1π/rX-t+2m+1π/r∑k=0τan,k dt+∑m=0r/2-1∫π/rn+1π/rω2-t+2m+1π/rXr/πt-t+2m+1π/r∑k=τ∞an,k-an,k+rdt+12∑m=0r/2-1∫2m+1π/r-π/rn+12m+1π/rω2tXtdt≤r2∫π/rn+1π/rω2f,tXt∑k=0τan,k+2πrr2∫π/rn+1π/rω2f,tXt2∑k=τ∞an,k-an,k+rdt+∑m=0r/2-1ω2f,2m+1π/r-π/rn+1X2m+1π/r-π/rn+1∫2m+1π/r-π/rn+12m+1π/rdt≤O1∑μ=1nω2f,π/μXμ∑k=0μ+1an,k+O1∑μ=1nω2f,πμX∑k=μ∞an,k-an,k+r+O1ω2f,πn+1X.

Thus the result follows.

Proof of Corollary <xref ref-type="statement" rid="coro2">7</xref>.
Theorem 5 implies that (64)Tn,Af-fX=Oω2f,πn+1X+∑μ=1nω2f,π/μXμ∑k=0μ+1an,k+∑μ=1nω2f,πμX∑k=μ∞an,k-an,k+r.Since in (24) (65)∑μ=1nω2f,πμX∑k=μ∞an,k-an,k+r=O1∑μ=1nω2f,πμX∑k=μ/cμ-1an,kk+∑k=μnan,kk+∑k=n+1∞an,kk≤O1∑μ=1nω2f,πμX∑k=μ/cμ-1an,kk+O1∑μ=1nω2f,πμX∑k=μnan,kk+∑k=n+1∞an,kk≤O1c∑μ=1nω2f,π/μXμ∑k=0μ+1an,k+O1∑μ=1nω2f,πμX∑k=μnan,kk+∑k=n+1∞an,kkone has (66)Tn,Af-fX=O1ω2f,πn+1X+O11+c∑μ=1nω2f,π/μXμ∑k=0μ+1an,k+O1∑μ=1nω2f,πμX∑k=μnan,kk+O1∑k=n+1∞an,kk∑μ=1nω2f,πμX≤O1ω2f,πn+1X+O11+c∑μ=1nω2f,π/μXμ∑k=0μ-1an,k+2∑μ=1nω2f,π/μXμ+1an,μ+1+∑μ=1nω2f,π/μXμan,μ+O1∑μ=1nω2f,πμX∑k=μnan,kk+O1∑k=n+1∞an,kk∑μ=1nω2f,πμX≤O1ω2f,πn+1X+O11+c∑μ=0nω2f,π/μ+1Xμ+1∑k=0μan,k+O131+c+1∑μ=1nω2f,πμX ∑k=μnan,kk+O1∑k=n+1∞an,kk∑μ=1nω2f,πμX=O1ω2f,πn+1X+O1∑k=0nan,k∑μ=knω2f,π/μ+1Xμ+1+O1∑k=1nan,kk∑μ=1kω2f,πμX+O1∑k=n+1∞an,kk∑μ=1nω2f,πμX,If (14) and (15) hold, then (67)ω2f,πn+1X≤1n∑μ=1nω2f,πμX≤O14πHπ/n+1n+1,∑μ=knω2f,π/μ+1Xμ+1≤8∫π/n+2π/k+1ω2f,tXtdt≤O18πHπ/k+1k+1and therefore (68)Tn,Af-fX=OHπ/n+1n+1+O∑k=1nan,kHπ/k+1k+1+OHπn+1∑k=n+1∞an,kk.Since (69)∑k=n+1∞an,kk≤1n+1∑k=n+1∞an,k=O1n+1the result follows (cf. [6]).