# Pointwise Analog of the Stečkin Approximation Theorem

Authors: Włodzimierz Łenski
Journal: Journal of Mathematics
 (2013)

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## Abstract

We show the pointwise version of the Stečkin theorem on approximation by de la Vallée-Poussin means. The result on norm approximation is also derived.


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## Body

## 1. Introduction

LetLp(1≤p<∞)[C] be the class of all 2π-periodic real-valued functions integrable in the Lebesgue sense with pth power (continuous) over Q=[-π,π], and let Xp=Lp when 1≤p<∞ or Xp=C when p=∞.Let us define the norms off∈Xp as
(1)∥f∥=∥f∥Xp=∥f(·)∥Xp∥f∥≔{{∫Q|f(x)|pdx}1/pwhen1≤p<∞,supx∈Q|f(x)|whenp=∞,∥f∥x,δ=∥f∥Xp,x,δ=∥f(·)∥Xp,x,δ:=sup0<h≤δ∥f(·)∥Xp,x,h∘:={sup0<h≤δ{12h∫x-hx+h|f(t)|pdt}1/pwhen1≤p<∞,sup0<h≤δ{sup0<|t|≤h|f(x+t)|}whenp=∞,
where
(2)∥f∥x,δ∘=∥f∥Xp,x,δ∘=∥f(·)∥Xp,x,δ∘≔{{12δ∫x-δx+δ|f(t)|pdt}1/pwhen1≤p<∞sup0<|h|≤δ|f(x+h)|whenp=∞(δ>0).
We note additionally that
(3)∥f∥C,x,0=∥f∥C,x,0∘=|f(x)|.
Consider the trigonometric Fourier series of f(4)Sf(x)=ao(f)2+∑k=0∞‍(ak(f)coskx+bk(f)sinkx)
with the partial sums  Skf.Let(5)σn,mf(x):=1m+1∑k=n-mn‍Skf(x)(m≤n=0,1,2,…).As a measure of approximation by the previous quantities, we use the pointwise characteristics(6)wxf(δ)=wxf(δ)Xp:=∥Δxf(·)∥Xp,x,δ={sup0<h≤δ{12h∫-hh‍|Δxf(t)|pdt}1/pwhen1≤p<∞,sup0<h≤δ{sup0<|t|≤h|Δxf(t)|}whenp=∞,
compared to [1], and
(7)wx∘f(δ)=wx∘f(δ)Xp:=∥Δxf(·)∥Xp,x,δ∘={{12δ∫-δδ‍|Δxf(t)|pdt}1/pwhen1≤p<∞,sup0<|t|≤δ|Δxf(t)|whenp=∞,
and also
(8)Ωxf(πn+1)=Ωxf(πn+1)Xp:=1n+1∑k=0n‍wxf(πk+1)Xp,Ωx∘f(πn+1)=Ωx∘f(πn+1)Xp=1n+1∑k=0n‍wx∘f(πk+1)Xp,
where
(9)Δxf(t):=f(x+t)-f(x),
constructed on the base of definition of Xp-points  (Lebesgue points (Lp-points) or  points  of continuity (C-points)). We also use the modulus of continuity of f in the space Xp defined by the formula
(10)ωf(δ)=ωf(δ)Xp:=sup0<|h|≤δ∥Δ·f(h)∥Xp
and its arithmetic mean
(11)Ωf(πn+1)=Ωf(πn+1)Xp=1n+1∑k=0n‍ωf(πk+1)Xp.We can observe that, forf∈Xp~ and p~≥p,
(12)∥w·.f(δ)Xp∥C≤ωf(δ)C,
whence
(13)∥Ω·.f(δ)Xp∥C≤Ωf(δ)C,∥w·.∘f(δ)Xp∥Xp≤ωf(δ)Xp,
whence
(14)∥Ω·.∘f(δ)Xp∥Xp≤Ωf(δ)Xp.Let us introduce one more measure of pointwise approximation analogical to the best approximation of functionf by trigonometric polynomials T  of the degree at most n(T∈Hn)(15)En(f)Xp:=infT∈Hn{∥f(·)-T(·)∥Xp};
Namely,
(16)En(f,x;δ)=En(f,x;δ)Xp:=infT∈Hn{∥f(·)-T(·)∥Xp,x,δ}={infT∈Hn{sup0<h≤δ[12h∫-hh‍|f(x+t)12h∫-hh‍-T(x+t)|pdt{sup0<h≤δ[12h∫-hh‍|f(x+t)}]1/p}when1≤p<∞,infT∈Hn{sup0<|h|≤δ|f(x+h)-T(x+h)|}whenp=∞,En∘(f,x;δ)=En∘(f,x;δ)Xp:=infT∈Hn{∥f(·)-T(·)∥Xp,x,δ∘}.
We will also use its arithmetic mean
(17)Fn,m(f,x)=Fn,m(f,x)Xp:=1m+1∑k=0m‍En(f,x;πk+1)Xp,Fn,m∘(f,x)=Fn,m∘(f,x)Xp:=1m+1∑k=0m‍En∘(f,x;πk+1)Xp.Using these characteristics, we will show the pointwise version of the Stečkin [2] generalization of the Fejér-Lebesgue theorem. As a corollary, we will obtain the mentioned original result of Stečkin on norm approximation as well as the result of Tanović-Miller [3].ByK we will designate either an absolute constant or a constant depending on some parameters, not necessarily the same of each occurrence.


## 2. Statement of the Results

At the beginning, we formulate the partial solution of the considered problem.Theorem 1 (see [4]).
Iff∈Xp, then, for any positive integer m≤n and all real x,
(18)|σn,mf(x)-f(x)|≤π2En-m∘(f,x,π2n-m+1)Xp+6Fn-m,m∘(f,x)Xp+∫π/(2n-m+1)π/(m+1)‍En-m∘(f,x,t)Xptdt+En-m∘(f,x;0)Xp,|σn,mf(x)-f(x)|≤(6+π2)Fn-m,m(f,x)Xp×[1+lnn+1m+1]+En-m(f,x;0)Xp.Now, we can present the main result on pointwise approximation.Theorem 2.
Iff∈Xp,  then, for any positive integer  m≤n and all real x,
(19)|σn,mf(x)-f(x)|≤K∑ν=0n‍Fn-m+ν,m(f,x)Xp+Fn-m+ν,ν(f,x)Xpm+ν+1+E2n(f,x;0)Xp.Remark 3.
Theorem2, in case f∈C, immediately yields the result of Stečkin [2]. The result of Stečkin type also holds if instead of  C  we consider the spaces Xp with  1<p<∞. Then, in the proof, we need the Hardy-Littlewood estimate of the maximal function.At everyXp-point x of f(20)Ωxf(γ)Xp=ox(1)asγ→0+
and thus from Theorem 1, we obtain the corollary which states the result of the Tanović-Miller type [3].Corollary 4.
Iff∈Xp,  then, for any positive integer  m≤n  at every Xp-point  x of  f,
(21)|σn,mf(x)-f(x)|=ox(1)[1+lnn+1m+1]asn→∞.


## 3. Auxiliary Results

In order to prove our theorems, we require some lemmas.Lemma 5.
IfTn is the trigonometric polynomial of the degree at most  n  of the best approximation of  f∈Xp  with respect to the norm  ∥·∥Xp, then it is also the trigonometric polynomial of the degree at most  n  of the best approximation of  f∈Xp  with respect to the norm  ∥·∥Xp,x,δ for any δ∈[0,π].Proof.
From the inequalities(22)∥En(f,·,δ)Xp∥Xp≥∥En∘(f,·,δ)Xp∥Xp∥En(f,·,δ)Xp∥Xp=∥∥f-Tn,δ∥Xp,·,δ∘∥Xp=∥f-Tn,δ∥Xp∥En(f,·,δ)Xp∥Xp≥∥f-Tn∥Xp=En(f)Xp,∥En∘(f,·,δ)Xp∥Xp≤∥∥f-Tn∥Xp,·,δ∘∥Xp∥En(f,·,δ)Xp∥Xp=∥f-Tn∥Xp=En(f)Xp,
where  Tn,δ and  Tn are the trigonometric polynomials of the degree at most  n  of the best approximation of  f∈Xp  with respect to the norms  ∥·∥Xp,x,δ∘ and ∥·∥Xp, respectively, we obtain relation
(23)∥f-Tn,δ∥Xp=∥f-Tn∥Xp=En(f)Xp,
whence   Tn,δ=Tn  for any δ∈[0,π] by uniqueness of the trigonometric polynomial of the degree at most  n  of the best approximation of  f∈Xp  with respect to the norm  ∥·∥Xp(e.g. see [5] p. 96). We can also observe that for such Tn and any h∈[0,δ],
(24)∥f-Tn∥Xp,x,h∘=En∘(f,x,h)Xp≤En(f,x,δ)Xp≤∥f-Tn∥Xp,x,δ.
Hence,
(25)En(f,x,δ)Xp=∥f-Tn∥Xp,x,δ
and our proof is complete.Lemma 6.
Ifn∈ℕ0 and δ>0, then En(f,x;δ)Xp is nonincreasing function of  n  and nondecreasing function of δ. These imply that for m,n∈ℕ the function Fn,m(f,x)Xp is nonincreasing function of  n and m simultaneously.Proof.
The first part of our statement follows from the property of the norm∥·∥x,δ and supremum. The second part is a consequence of the calculation
(26)Fn,m+1(f,x)XpFn,m(f,x)Xp=m+1m+2(1+En(f,x;π/(m+2))Xp∑k=0mEn(f,x;π/(k+1))Xp)≤m+1m+2(1+En(f,x;π/(m+1))Xp∑k=0mEn(f,x;π/(m+1))Xp)=m+1m+2(1+1m+1)=1.Lemma 7.
Letm,n,q∈ℕ0 such that  m≤n and q≥m+1. If  f∈Xp, then
(27)|σn+q,mf(x)-σn,mf(x)|≤KFn-m,m(f,x)Xp∑ν=0q-1‍1m+ν+1.Proof.
It is clear that(28)σn,mf(x)=1m+1∑k=n-mn‍1π∫-ππf(x+t)Dk(t)dt=1π∫-ππf(x+t)Vn,m(t)dt,
where
(29)Vn,m(t)=1m+1∑k=n-mn‍Dk(t),Dk(t)=sin((2k+1)t/2)2sin(t/2).
Hence, by orthogonality of the trigonometric system,
(30)σn+q,mf(x)-σn,mf(x)=1π∫-ππ‍[f(x+t)-Tn-m(x+t)]×(Vn+q,m(t)-Vn,m(t))dt=1π(m+1)∑k=n-mn‍∫-ππ[f(x+t)-Tn-m(x+t)]1π(m+1)×(Dk+q(t)-Dk(t))dt=1π(m+1)×∑k=n-mn‍∫-ππ[f(x+t)-Tn-m(x+t)]×sin((2k+2q+1)t/2)-sin((2k+1)t/2)2sin(t/2)dt=1π(m+1)∑k=n-mn‍∫-ππ[f(x+t)-Tn-m(x+t)]1π(m+1)×sin(qt/2)cos((2k+q1)t/2)sin(t/2)dt,with trigonometric polynomial Tn-m  of the degree at most  n-m  of the best approximation of  f.
Using the notations(31)I1=[-πq,πq],I2=[-πm+1,-πq]∪[πq,πm+1],I3=[-π,-πm+1]∪[πm+1,π],
we get
(32)∑‍=1π(m+1)×∑k=n-mn‍(∫I1+∫I2+∫I3)×[f(x+t)-Tn-m(x+t)]×sin(qt/2)cos((2k+q+1)t/2)sin(t/2)dt=∑1‍+∑2‍+∑3‍,∑1‍≤1π(m+1)∑k=n-mn‍∫I1|f(x+t)-Tn-m(x+t)|qdt=qπ∫I1|f(x+t)-Tn-m(x+t)|dt≤2En-m(f,x;πq)Xp.
We next evaluate the sums  ∑2‍ and ∑3‍ using the partial integrating and  Lemma 5. Thus,
(33)∑2‍≤∫I2|f(x+t)-Tn-m(x+t)|tdt=2[12t∫-tt‍|f(x+u)-Tn-m(x+u)|du]t=π/qt=π/(m+1)+2∫π/qπ/(m+1)‍1t[12t∫-tt‍|f(x+u)-Tn-m(x+u)|du]dt≤2En-m(f,x;πm+1)Xp+2∫-tt‍1tEn-m(f,x;t)Xpdt≤4En-m(f,x;πm+1)Xp[1+lnqm+1]≤4En-m(f,x;πm+1)Xp[1+∑ν=0q-1‍1m+ν+1],∑3‍≤1m+1×∫I3|f(x+t)-Tn-m(x+t)|t×|∑k=n-mn‍cos(kt+q+12t)|dt≤1m+1×∫I3|f(x+t)-Tn-m(x+t)|t×|2sin((n+1)t/2)cos((2n-m+q+1)t/2)2sin(t/2)|dt≤πm+1∫I3|f(x+t)-Tn-m(x+t)|t2dt=πm+1{2[12t∫-tt|f(x+u)-Tn-m(x+u)|du{12t∫-tt}]t=π/(m+1)t=π+4∫π/(m+1)π‍1t2[12t∫-tt‍|f(x+u)12t∫-tt‍-Tn-m(x+u)|du{12t∫-tt‍}]dt}≤πm+1{{+4∫π/(m+1)π‍}2En-m(f,x;π)Xp+4∫π/(m+1)π‍1t2En-m(f,x;t)Xpdt}=2πm+1{{+2∫1m+1En-m(f,x;)Xpπ^2/u^2πduu2}En-m(f,x;π)Xp+2∫1m+1En-m(f,x;π/u)Xpπ2/u2πduu2}=2πm+1{{+2π∑k=0m-1‍∫k+1k+2‍En-m(f,x;πu)Xp}En-m(f,x;π)Xp+2π∑k=0m-1‍∫k+1k+2‍En-m(f,x;πu)Xpdu}=2πm+1{{∑k=0m-1‍(πk+1)Xp}En-m(f,x;π)Xp+2π∑k=0m-1‍En-m(f,x;πk+1)Xp}≤2π+2/πm+1∑k=0m-1‍En-m(f,x;πk+1)Xp,
which proves Lemma 7.Before formulating the next lemmas, we define a new difference. Letm,n∈ℕ0  and  m≤n.  Denote that
(34)τn,mf(x):=(m+1){σn+m+1,mf(x)-σn,mf(x)}.Lemma 8.
Letm,n,μ∈ℕ0 such that  2μ≤m≤n. If  f∈Xp, then
(35)|τn,mf(x)-τn-μ,m-μf(x)|≤KμFn-μ+1,μ-1(f,x)Xplnmμ.Proof.
The proof follows by the method of Leindler [6]. Namely,
(36)τn,mf(x)-τn-μ,m-μf(x)=(∑k=n+m-2μ+2n+m+1‍-2∑k=n-μ+1n‍)[Skf(x)-f(x)],|τn,mf(x)-τn-μ,m-μf(x)|≤|(∑k=n+m-2μ+2n+m-μ+1‍-∑k=n-μ+1n‍)[Skf(x)-f(x)]|+|(∑k=n+m-μ+2n+m+1‍-∑k=n-μ+1n‍)[Skf(x)-f(x)]|=μ|σn+m-μ+1,μ-1f(x)-σn,μ-1f(x)|+μ|σn+m+1,μ-1f(x)-σn,μ-1f(x)|.
By Lemma 6, for  2μ≤m,
(37)|τn,mf(x)-τn-μ,m-μf(x)|≤KμFn-μ+1,μ-1(f,x)Xp[1+ln(n-μ+1)+μ-1μ]+KμFn-μ+1,μ-1(f,x)Xp[1+lnm+μ-1μ]≤KμFn-μ+1,μ-1(f,x)Xp[1+lnmμ],
and our proof is complete.Lemma 9.
Letm,n∈ℕ0 and m≤n. If  f∈Xp, then
(38)|τn,mf(x)|≤K∑k=n-mn‍Fk,k-n+m(f,x)Xp.Proof.
Our proof runs parallel with the proof of Theorem 1 in [2].
Ifm=0, then
(39)|τn,0f(x)|=|σn+1,0f(x)-σn,0f(x)|≤KFn,0(f,x)Xp,
and if  m=1, then
(40)|τn,1f(x)|≤2|σn+2,1f(x)-σn,1f(x)|≤KFn-1,1(f,x)Xp≤K[Fn-1,1(f,x)Xp+Fn-1,1(f,x)Xp]≤K[Fn-1,1(f,x)Xp+Fn,1(f,x)Xp]
by Lemmas 6 and 7.
Next, we construct the same decreasing sequence(ms) of integers that was given by Stečkin. Let
(41)m0=m,ms=ms-1-[ms-12](s=1,2,…),
where [y]  denotes the integral part of  y. It is clear that there exists smallest index  t≥1 such that  mt=1 and
(42)m=m0>m1>⋯>mt=1.
By the definition of the numbers  ms, we have
(43)ms≥ms-12,ms-1-ms=[ms-12]≥[ms-13](s=1,2,…,t)
whence
(44)mt-1=2,mt-1-mt=1,ms-1-ms≤ms≤3(ms-ms+1)(s=1,2,…,t-1)
follow.
Under these notations, we get the following equality:(45)τn,mf(x)=∑s=1t‍(τn-m+ms-1,ms-1f(x)-τn-m+ms,msf(x))+τn-m+mt,mtf(x),
whence, by  mt=1,
(46)|τn,mf(x)|≤∑s=1t‍|τn-m+ms-1,ms-1f(x)-τn-m+ms,msf(x)|+|τn-m+mt,mtf(x)|
follows.
It is easy to see that the terms in the sum∑s=1t‍, by Lemma 8, with μ=ms-1-ms and  m=ms-1 do not exceed
(47)K(ms-1-ms)Fn-m+ms+1,ms-1-ms(f,x)Xplnms-1ms-1-ms,where(s=1,2,…,t-1),
and by Lemma 7, we get
(48)|τn-m+1,1f(x)|≤2|σn-m+2,1f(x)-σn-m+1,1f(x)|≤KFn-m,1(f,x)Xp.
Thus,
(49)|τn,mf(x)|≤K∑s=1t-1‍3(ms-ms+1)Fn-m+ms+1,ms(f,x)Xpln3+KFn-m+2,m-2(f,x)Xp+KFn-m,1(f,x)Xp,
whence, by the monotonicity of  Fν,μ(f,x)Xp,
(50)|τn,mf(x)|≤K(∑s=1t-1‍∑ν=ms+1+1ms‍Fn-m+ν+1,ν(f,x)Xp+∑ν=02‍Fn-m+ν,m-ν-1(f,x)Xp)+KFn-m,1(f,x)Xp≤K∑ν=0m1+1‍Fn-m+ν,ν(f,x)Xp+KFn-m,1(f,x)Xp≤K∑ν=0m‍Fn-m+ν,ν(f,x)Xp+KFn-m,1(f,x)Xp≤K∑k=n-mn‍Fk,k-n+m(f,x)Xp+KFn-m,1(f,x)Xp.


## 4. Proofs of the Results

Proof of Theorem2.
The proof follows the lines of the proofs of Theorem 4 in [2] and Theorem in [6]. Therefore, let n>0 and m≤n be fixed. Let us define an increasing sequence (ns:s=0,1,…,t) of indices introduced by Stečkin in the following way. Set n0=n. Assuming that the numbers  n0,…,ns are already defined and  ns<2n, we define ns+1 as follows. Let  νs denote the smallest natural number such that (51)Fns-m+νs,ν(f,x)Xp≤12Fns-m,ν(f,x)Xp(ν=0,1,…,n).
According to the magnitude of  νs,  we define
(52)ns+1={ns-m+1forνs≤m,ns+νsform+1≤νs<2n+m-ns,2n+mforνs≥2n+m-ns.
If  ns+1<2n,  we continue the procedure, and if once  ns+1≥2n,  then we stop the construction and define  t:=s+1.
By the previous definition of(ns), we have the following obvious properties:
(53)t≥1,n=n0<n1<⋯<nt,2n≤nt≤2n+m,ns+1-ns≥m+1(s=0,1,…,t-1),
and relations
(54)Fns+1-m,ν(f,x)Xp≤12Fns-m,ν(f,x)Xpfors=0,1,…,t-2,12Fns-m,ν(f,x)Xp≤Fns+1-m-1,ν(f,x)Xpfors=0,1,…t-1,
whenever  ns+1-ns>m+1.
Let us start with(55)|σn,mf(x)-f(x)|=∑s=0t-1‍[|σns,mf(x)-f(x)|-|σns+1,mf(x)-f(x)|]+|σnt,mf(x)-f(x)|≤∑s=0t-1‍|σns+1,mf(x)-σns,mf(x)|+|σnt,mf(x)-f(x)|=∑s=0t-1‍|1m+1τns,mf(x)|+|σnt,mf(x)-f(x)|.
Using Theorem 1 and that 2n≤nt≤2n+m,  we get
(56)|σnt,mf(x)-f(x)|≤KFnt-m,m(f,x)Xp[1+lnnt+1m+1]+|f(x)-Tnt-m(x)|≤K∑ν=0n‍Fn-m+ν,m(f,x)Xpm+ν+1+|f(x)-Tnt-m(x)|≤K∑ν=0n‍Fn-m+ν,m(f,x)Xp+Fn-m+ν,ν(f,x)Xpm+ν+1+|f(x)-Tnt-m(x)|≤K∑ν=0n‍Fn-m+ν,m(f,x)Xp+Fn-m+ν,ν(f,x)Xpm+ν+1+Ent-m(f,x;0)Xp≤K∑ν=0n‍Fn-m+ν,m(f,x)Xp+Fn-m+ν,ν(f,x)Xpm+ν+1+E2n(f,x;0)Xp.
The estimate of the sum in the right hand side of (55) we derive from the following one(57)|1m+1τns,mf(x)|≤K∑ν=0ns+1-ns-1‍Fns-m+ν,m(f,x)Xp+Fns-m+ν,m(f,x)Xpm+ν+1.
We split the proof of this inequality in two parts. If  ns+1-ns=m+1, then by Lemma 9, (58)|1m+1τns,mf(x)|≤K1m+1∑k=ns-mns‍Fk,k-ns+m(f,x)Xp≤K∑ν=0ns+1-ns-1‍Fns-m+ν,ν(f,x)Xpm+ν+1.
If ns+1-ns>m+1, then, by Lemma 7,
(59)|1m+1τns,mf(x)|≤KFns-m,m(f,x)Xp∑ν=0ns+1-ns-1‍1m+ν+1,
and since  (1/2)Fns-m,m(f,x)Xp≤Fns+1-m-1,m(f,x)Xp, we have
(60)|1m+1τns,mf(x)|≤2KFns+1-m-1,m(f,x)Xp∑ν=0ns+1-ns-1‍1m+ν+1≤2K∑ν=0ns+1-ns-1‍Fns-m+ν,m(f,x)Xpm+ν+1.
Consequently,(61)∑s=0t-1‍|1m+1τns,mf(x)|≤2K∑s=0t-1‍∑ν=0ns+1-ns-1Fns-m+ν,m(f,x)Xp+Fns-m+ν,ν(f,x)Xpm+ν+1‍.
Since  ns+1-ns≤2n+m-n-1=n+m-1  for all  s≤t-1, changing the order of summation, we get
(62)∑s=0t-1‍|1m+1τns,mf(x)|≤2K∑ν=0n+m-1‍1m+ν+1×∑s:ns+1-ns>ν[Fns-m+ν,m(f,x)Xp+Fns-m+ν,ν(f,x)Xp]‍.
Using the inequality(63)Fns+1-m,ν(f,x)Xp≤12Fns-m,ν(f,x)Xpfor{ν=0,1,2,…,ns+1-ns-1,s=0,1,2,…,t-2,
we obtain
(64)∑s:ns+1-ns>ν‍[Fns-m+ν,m(f,x)Xp+Fns-m+ν,ν(f,x)Xp]=Fnp-m+ν,m(f,x)Xp+Fnp-m+ν,ν(f,x)Xp+∑s≥p+1:ns+1-ns>ν‍[Fns+1-m+ν,m(f,x)Xp+∑s≥p+1:ns+1-ns>ν‍+Fns+1-m+ν,ν(f,x)Xp]≤Fnp-m+ν,m(f,x)Xp+Fnp-m+ν,ν(f,x)Xp+∑s:s≥p+1‍Fns-m,m(f,x)Xp[Fns-m,m(f,x)Xp+∑s:s≥p+1‍Fns-m,m(f,x)Xp+Fns-m,ν(f,x)Xp]≤Fnp-m+ν,m(f,x)Xp+Fnp-m+ν,ν(f,x)Xp+2[Fnp+1-m,m(f,x)Xp+Fnp+1-m,ν(f,x)Xp]≤3[Fnp-m+ν,m(f,x)Xp+Fnp-m+ν,ν(f,x)Xp],
where  p  denotes the smallest index  s  having the property  ns+1-ns>ν. Hence,
(65)∑s=0t-1‍|1m+1τns,mf(x)|≤K∑ν=0n+m-1‍Fn-m+ν,m(f,x)Xp+Fn-m+ν,ν(f,x)Xpm+ν+1≤K∑ν=0n‍Fn-m+ν,m(f,x)Xp+Fn-m+ν,ν(f,x)Xpm+ν+1,
and our proof follows.



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*Source: 426347-2013-03-11.xml*