# Functions of Boundedkth p-Variation and Continuity Modulus

Authors: Odalis Mejía; Pilar Silvestre
Journal: Journal of Function Spaces
 (2015)

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## Abstract

A scale of spaces exists connecting the class of functions of boundedkth p-variation in the sense of Riesz-Merentes with the Sobolev space of functions with p-integrable kth derivative. This scale is generated by the generalized functionals of Merentes type. We prove some limiting relations for these functionals as well as sharp estimates in terms of the fractional modulus of smoothness of order k-1/p.


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## Body

## 1. Introduction

Letf be a 1-periodic function on the real line. A set Π={x0,…,xn} of points such that x0<x1<⋯<xn, where xn=x0+1, will be called apartition. For a partition Π we denote by Π=maxjxj+1-xj its diameter. Let p>1, p′≔p/p-1, and 0≤α≤1/p′. For any Π, we set(1)vp,αf;Π=∑k=0n-1fxk+1-fxkpxk+1-xkαp1/p.We denote by Vp,α the class of all 1-periodic functions f such that vp,α(f)=supΠvp,αf;Π<∞, where the supremum is taken over all partitions of [0,1]; see, for example, [1].Ifα=0 we get the class Vp,0≔Vp of functions of bounded p-variation introduced by Wiener [2]. A celebrated result of Riesz (see, e.g., [3, 4]) states that the class Wp1 of absolutely continuous 1-periodic functions such that the first derivative belongs to the Lebesgue space Lp0,1 coincides with the class Vp,1/p′ and f′p=vp,1/p′(f).Another characterization of the classWp1 was given by Terehin in [5] in terms of the fractional continuity modulus of order 1-1/p, (2)ω1-1/pf;δ=supΠ≤δvpf;Π0<δ≤1,introduced by Wiener in [2]. If f∈Wp1, then ω1-1/pf;δ≤δ1/p′f′p, and conversely if ω1-1/p(f;δ)=O(δ1/p′), then f∈Wp1.The concept of functions of bounded second variation was introduced by de la Vallée Poussin in [6]. By arefined partition we mean a partition Π of the form(3)x0<y1≤z1<x1<y2≤z2<x2<⋯<yn≤zn<xn=x0+1.Motivated by the result of Riesz, Merentes in [7] introduced, for p>1, the class Vp,1/p′(2) of functions of bounded (p,1/p′) second variation as the class of those functions for which (4)
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′is finite. He proved a similar result to that of Riesz (see, e.g., [7, Theorem, page 122]) and gave a variational characterization of the Sobolev space Wp2 in terms of functions of bounded (p,1/p′) second variation, namely, a function belongs to Wp2 if and only if f∈Vp,1/p′(2) and vp,1/p′2f=f′′p. Observe that from the characterization of Riesz and Merentes it follows that vp,1/p′f′=vp,1/p′2f. The people interested in this topic can see [8].The above results motivated Barza and Silvestre to study the scale of spacesVp,α(2) between the class Vp(2) of functions of bounded second p-variation and the Sobolev space Wp2 (see [9]) as an extension of [1].Following [9, 10], let p>1 and 0≤α≤1/p′. Let Π be a partition of the form(5)x0=t1,1<t1,2<⋯<t1,k≤t1,k+1<⋯<t1,2k≤⋯≤t2,k+1<⋯≤⋯<t3,1<⋯<tj,1<⋯<tj,k≤tj,k+1<⋯<tj,2k≤⋯≤tm,1<⋯<tm,k≤tm,k+1<⋯<tm,2k=x0+1.Let for any k∈N(6)vp,αkf;Πp=∑j=1mftj,k+1,…,tj,2k-ftj,1,…,tj,kp·1tj,2k-tj,1αp,where the kth divided differences are defined by (7)ft0,t1,…,tk≔∑j=0kftjtj-t0⋯tj-tj-1tj-tj+1⋯tj-tk.We define the space Vp,α(k)(k∈N) as the space of 1-periodic functions f such that (8)vp,αkf=supΠvp,αkf;Π<∞,where the supremum is taken over all partitions of type (5). The number vp,α(k)(f) is called the (p,α)kth variation of f. For α=0, we denote it by vp(k)(f) and call it the kth p-variation of f. We have that Vp,1/p′(k) is the space RV(p,k) in [10]. The spaces Vp,α(k) form a scale between the space of functions of bounded kth p-variation and the Sobolev space Wpk. In [10] the authors extend the results of Riesz and Merentes by showing that f∈Wpk if and only if f∈Vp,1/p′(k)=RVp,k.Terehin studied in [5] properties of the function (9)ωk-1/pf;δ=sup0<h≤δω1-1/pΔhk-1f;h0<δ≤1,where Δhf(x)=f(x+h)-f(x), Δhkf(x)=ΔhΔhk-1f(x). The function ωk-1/p(f;δ) is called thefractional modulus of smoothness of orderk-1/p. A characterization of the space f∈Wpk in terms of ωk-1/p(f;δ) was also given in [5]: ωk-1/p(f;δ)=O(δk-1/p) if and only if f∈Wpk.We extend the results proved in [1, 9] to the case of functions of bounded kth p-variation. In Section 2 we prove some results which we will use in the proofs of our main theorems. In Section 3 we study sharp estimates of vp,α(k)(f) in terms of the fractional modulus of smoothness of order k-1/p, and we study limits in the scales of spaces Vp,α(k) for k∈N.


## 2. Preliminary Results

Recall that a functionf satisfying limδ→0+ω1-1/p(f;δ)=0 is called p-continuous and the class of p-continuous functions is denoted by Cp.Lemma 1 (see [11, Lemma  1]).
Letp>1 and f∈Cp such that f(r)∈Cp for k,r∈N. Then (10)ωk+r-1/pf;δ≤δrωk-1/pfr;δ0<δ≤1.Remark 2.
LetΠ be a partition as in (5). Since 0<α≤1/p′  (p′=p/p-1), then 0<αp<p-1 and since tj,2k-tj,1≤1, it follows that(11)tj,2k-tj,1αp>tj,2k-tj,1p-1.Then, vp,αkf;0,1≤vp,1/p′(k)(f;[0,1]). The space Vp,1/p′(k)[0,1] is the space RV(p,k)[0,1] in [10]. Then, RV(p,k)[0,1]⊂Vp,α(k)[0,1].Proposition 3.
Let1<p<∞, k≥2, and 0<α≤1/p′. If f∈Vp,α(k)[0,1], then fk-1 exists everywhere and is continuous on [0,1].Proof.
Suppose thatf∈Vp,α(k)[0,1]. Then, there exists a constant M>0 such that(12)M≥vp,αkf;0,1≥ftj,k+1,…,tj,2k-ftj,1,…,tj,kptj,2k-tj,1αp,for any partition of type (5) and 1≤j≤m.
Letx0∈(0,1). It follows by (12) that (13)tj,2k-tj,1αM1/p≥ftj,k+1,…,tj,2k-ftj,1,…,tj,k,for any partition of type (5) such that tj,k+1=x0 and 1≤j≤m. So that, Cauchy’s convergence criterion establishes the existence of the right-hand derivative f+(k-1)(x0) when 0≤x0<1. Similarly, the left-hand derivative f-(k-1)(x0) exists when 0<x0≤1. Let us consider the following partition (14)0≤t1,1<⋯<t1,k≤t1,k+1<⋯<t1,2k≤⋯<tj,k=x0=tj,k+1<⋯<tj,2k≤⋯<tm-1,2k≤tm,1<⋯<tm,2k.It follows that (15)ftj,k+1,…,tj,2k-ftj,1,…,tj,k≤M1/ptj,2k-tj,1α≤M1/ptj,2k-x0+x0-tj,1α≤2αM1/ptj,2k-x0α+x0-tj,1α,which it tends to zero as tj,1→x0- and tj,2k→x0+. Then, we obtain that f+(k-1)(x0)=f-(k-1)(x0) and so that, f(k-1)(x0) exists.
Let us considery∈[x0,1] and the following type of partition (16)0≤t1,1<⋯<t1,k≤t1,k+1<⋯≤tj,1=x0-k-1kh<tj,2=x0-k-2kh<⋯<tj,k=x0≤tj,k+1=y<tj,k+2=y+1kh<⋯<tj,2k=y+k-1kh≤⋯<tm,2k,where 0≤h<min{x0,1-y}. Then, by (12), (17)
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.Letting h→0, we obtain that (18)fk-1y-fk-1x0≤M1/py-x0α,from which the continuity of f(k-1) in (0,1) follows. Also, f(k-1) is right and left continuous in 0 and 1, respectively.Remark 4.
Let1<p<∞, k≥2, and 0<α≤1/p′. If f∈Vp,α(k)[0,1], then it follows that(19)ft0,t1,…,tk=fk-1ψk-1!for some ψ∈conv{t0,…,tk}. Indeed, by Proposition 3 we have that f(k-1) exists everywhere on [0,1] and it is continuous. Then, f∈C(k-1)[0,1]. The equality (19) follows by [10, (2.8)].Theorem 5.
Letp>1 and 0<α≤1/p′.(i)
Iff∈Vp,α(k)[0,1], then (20)vp,αkf;0,1≥1k-1!vp,αfk-1;0,1.
(ii)
Ifvp,α(f(k-1);[0,1])<∞, then (21)vp,αkf;0,1≤1k-1!vp,αfk-1;0,1.Proof.
(i) Iff∈Vp,α(k)[0,1], then vp,α(k)(f;[0,1])<∞ and f(k-1) exists everywhere on [0,1] by Proposition 3 and it is continuous. To prove the inequality, let Π={0=t0,…,tn=1} be a partition of [0,1] and take 0<h≤min0≤j≤m-1{tj+1-tj/2}. Define the partition Πj of the subinterval [tj,tj+1] as (22)tj=tj,1<tj,2<⋯<tj,k=tj+h≤tj,k+1=tj+1-h<⋯<tj,2k=tj+1.
Then, we have that(23)vp,αkf;0,1p≔supΠ⁡vp,αkf;Πp≥∑j=1mftj,k+1,…,tj,2k-ftj,1,…,tj,kptj,2k-tj,1αp=∑j=1mftj+1-h,…,tj+1-ftj,…,tj+hptj+1-tjαp.Then, by Remark 4, there exist ψj∈conv{tj+1-h,…,tj+1} and ψj-1∈conv{tj,…,tj+h} such that(24)vp,αkf;0,1p≥1k-1!p∑j=1mfk-1ψj-fk-1ψj-1ptj+1-tjαp.By letting h→0, since f(k-1) is continuous, it follows that (25)vp,αkf;0,1p≥1k-1!p∑j=1mfk-1tj+1-fk-1tjptj+1-tjαp.By taking then supremum over all Π’s, the inequality follows.
(ii) Givenɛ>0, there exists a partition of type (5) (26)0=t1,1<t1,2<⋯<t1,k≤t1,k+1<⋯<t1,2k≤⋯<tj,2k≤tm,1<⋯<tm,k≤tm,k+1<⋯<tm,2k=1such that (27)vp,αkf;0,1p-ɛ<∑j=1mftj,k+1,…,tj,2k-ftj,1,…,tj,kptj,2k-tj,1αp.Since vp,α(f(k-1);[0,1])<∞, it follows that f(k-1)∈Vp,α[0,1]. Moreover, as we can see in [9, Theorem  2.3], we have that (28)v12fk-2;0,1≤vp,α2fk-2;0,1≤vp,αfk-1;0,1.Moreover, by [9, Proposition  2.4], it follows that v1(2)(f(k-2);[0,1])=v1(f(k-1);[0,1]). Hence, f(k-1)∈BV[0,1]. Therefore, f(k-1) is Riemann integrable in [0,1]. Indeed, it can be written as a difference of two monotone functions and then, f(k-1) is Riemann integrable.
By using [10, Proposition  2.5], since f(k-1) is continuous, for ψj∈conv{tj,k+1,…,tj,2k} and ψj-1∈conv{tj,1,…,tj,k}, we have that (29)vp,αkf;0,1p-ɛ<∑j=1mftj,k+1,…,tj,2k-ftj,1,…,tj,kptj,2k-tj,1αp≤1k-1!p∑j=1mfk-1ψj-fk-1ψj-1pψj-ψj-1αp≤1k-1!pvp,αfk-1;0,1p.Letting ɛ tend to zero, the inequality follows.Proposition 6.
Letp≥1 and assume that f(k-1) exists everywhere. Then(30)vpkf≔vpkf;0,1=1k-1!vpfk-1;0,1≔1k-1!vpfk-1.(For simplicity, from now on we omit the domain [0,1] in the functionals vp,α(k).)Proof.
The equality follows as in Theorem5.Letf∈L1 and fh be the Steklov average of the function f (see, e.g., [1]) defined by (31)fhx≔1h∫0hfx+tdt0<h≤1.We define the kth Steklov average of f as the Steklov average of the (k-1)th Steklov average of f, (32)fh,…k,hx≔1h∫0hfh,…k-1,hx+tdt0<h≤1.Lemma 7.
Letp>1 and f∈Vp(k). Then(i)
(33)ωk-1/pfh;t≤ωk-1/pf;t0≤t≤1,(ii)
(34)fh,…k,hkp≤h-k-1/pωk-1/pf;h0<h≤1,(iii)
(35)vpkf-fh,…k,h≤6kk-1!ωk-1/pf;hhk-1.Proof (the argument is inductive).
(i) The inequality follows by the first inequality of [1, Lemma  2.2] and the definition of ωk-1/p(f;t). Indeed, since (Δδk-1fh)(x)=(Δδk-1f)h(x) and fh∈Vp(k) it follows that (36)ωk-1/pfh;t=sup0<δ≤tω1-1/pΔδk-1fh;δ=sup0<δ≤tω1-1/pΔδk-1fh;δ≤sup0<δ≤tω1-1/pΔδk-1f;δ≔ωk-1/pf;t.(ii) Let f∈Vp(k). We have that f′h=fh′, (37)ddxfh,hx=ddx1h∫0hfhx+tdt=∫0hfh′x+uduh=fh′hx,d2dx2fh,hx=ddxfh′hx=Δhfxhh′.In the computations in [9, Lemma  2.5], we find that(38)fh,h′′p≤h-2+1/pω1-1/pΔhf;h.We have that (39)ddxfh,h,hx=ddx1h∫0hfh,hx+tdt=∫0hfh,h′x+uduh=fh,h′hx=Δhfh,hxh=Δhfh,hxh,since (40)Δhfh,hx=1h∫0hΔhfhx+tdt=1h∫0hΔhfhx+tdt=Δhfh,hx.
Then, it follows that(41)d3dx3fh,h,hx=d2dx2Δhfh,hxh=Δhfh,hh′′x.
Therefore, by (38) (see [9, Lemma  2.5])(42)fh,h,h′′′p=d2dx2Δhfh,hxhp=1hΔhfh,h′′p≤h-3+1/pω1-1/pΔh2f;h≤h-3-1/pω3-1/pf;h,and the inequality holds for 3. Let us assume that(43)fh,…k-1,hk-1p≤h-k-1+1/pω1-1/pΔhk-2f;hand let us see that the inequality holds for k.
We have that(44)ddxfh,…k,hx=ddx1h∫0hfh,…k-1,hx+tdtx=Δhfh,…k-1,hxh=Δhfh,…k-1,hxh,dkdxkfh,…k,hx=dk-1dxk-1Δhfh,…k-1,hxh=Δhfh,…k-1,hxhk-1.Therefore, by (43), it follows that (45)fh,…k,hkp=Δhfh,…k-1,hxhk-1p=1hΔhfh,…k-1,hk-1p≤1hh-k-1+1/pω1-1/pΔhk-2Δhf;h≤h-k-1/pωk-1/pf;h.(iii) The inequality holds for 2 by [9]. We have that (46)vp3f-fh,h,h≤vp3f-fh,h+vp3fh,h-fh,h,h.
By Theorem5 and [9, Proposition  2.4] it follows that(47)vp3f-fh,h,h≤vp3f-fh,h+vp3fh,h-fh,h,h=12!vp2f′-fh,h′+vp2fh,h′-fh,h,h′≤12!2vp2fh′-fh,h′+vp2fh,h′-fh,h,h′,because vp2f′-fh,h′≤2vp(2)(fh′-fh,h′) since by [9, Lemma  2.5(iii)] vp(2)(f-fh,h)≤2vp(2)(fh-fh,h). Therefore, since in [9, Lemma  2.5(iii)] we see that vp(2)(f-fh)≤vp(2)(fh-fh,h), then we have that(48)vp2fh′-fh′h≤vp2fh′h-fh′h,h.Therefore,∗vp3f-fh,h,h≤12!2vp2fh′-fh,h′+vp2fh,h′-fh,h,h′=12!2vp2fh′-fh′h+vp2fh,h′-fh,h,h′≤12!2vp2fh′h-fh′h,h+vp2fh,h′-fh,h,h′=32!vp2fh,h′-fh,h,h′=32!vpfh,h′′-fh,h,h′′=32!vpΔhfhh′-Δhfhh,h′=32!vpΔhfhh′-Δhfhh′h=32!vpg-ghfor g=Δhfh′/h∈Vp. By [1, Lemma  2.2 (2.15)] it follows that(49)vpg-gh≤6ω1-1/pg;h,and then since Δhfh′=Δh2f/h, it follows that (50)vp3f-fh,h,h≤32!vpg-gh≤182!ω1-1/pΔhfh′h;h=182!ω1-1/pΔh2fh2;h≤182!1h2ω1-1/pΔh2f;h≤182!h-2ω3-1/pf;h.Then we see that the inequality holds for 3 as well.
In particular we have by [9, Lemma  2.5(iii)] that(51)vp2f-fh,h=vp2f-fh,…2,h≤2vpfh′-fh,h′=2vpfh,…2-1,h2-1-fh,…2,h2-1,and we have proven that(52)vp3f-fh,…3,h≤32!vpfh,…3-1,h3-1-fh,…3,h3-1.
Let us assume that the inequality holds fork-1 and let us take k. We can see that the inequality holds also for k. Indeed, by the same kind of argument than for 3 we see that (recall ∗)(53)vpkf-fh,…k,h≤vpkf-fh,…k-1,h+vpkfh,…k-1,h-fh,…k,h≤vpkf-fh,…k-1,h+vpfh,…k-1,hk-1-fh,…k,hk-1k-1!≤kk-1!vpfh,…k-1,hk-1-fh,…k-1,hk-1h≤6kk-1!ω1-1/pfh,…k-1,hk-1;h,since fh,…(k-1),h(k-1)∈Vp.
Recall that(54)Δhfh′=ΔhΔhfh.Then fh,h′′(x)=Δhfh′/h=Δh2f/h2. Hence, fh,h,h′′′(x)=Δh2f/h2h′=Δh3f/h3. Let us assume that this type of identity holds for k-2, that is, fh,…(k-2),h(k-2)=Δhk-2f/hk-2. By (54), we see that (55)fh,…k-1,hk-1=fh,…k-2,hk-2h′=Δhk-2fhk-2h′=ΔhΔhk-2f/hk-2h=Δhk-1fhk-1and the same identity holds for any k∈N.
Therefore, by (53) and the previous identity, we obtain that (56)vpkf-fh,…k,h≤6kk-1!ω1-1/pΔhk-1fhk-1;h≤6kk-1!ωk-1/pf;hhk-1.In the next proposition we give an exact description of Peetre’sK-functional for the pair (Vp(k),Wpk) for k∈N. By [12, page 172](57)Kf,t;Vpk,Wpk≔infg∈Wpkf-gVpk+tgkpt>0,with fVp(k)=f0+f′0+⋯+fk-10+vpkf for any f∈Vp(k).Proposition 8.
Letf∈Vp(k). For any t∈(0,1](58)1k-1!ωk-1/pf;tp′tp′k-1≤Kf,t;Vpk,Wpk≤k1+6k-1!ωk-1/pf;tp′tp′k-1.Proof.
Fort∈(0,1] let h=tp′ and g=fh,…(k),h. Since f∈Wpk, then g∈Wpk. Hence, by (ii) and (iii) in Lemma 7 it follows that(59)f0-g0+⋯+fk-10-gk-10+vpkf-g+h1/p′gkp≤k-1ωk-1/pf;hhk-1+6kk-1!ωk-1/pf;hhk-1+h1/p′h-k-1/pωk-1/pf;h≤kωk-1/pf;hhk-1+6kk-1!ωk-1/pf;hhk-1.Therefore,(60)f-gVpk+h1/p′gkp≤k1+6k-1!ωk-1/pf;hhk-1,and by taking infimum over all g∈Wpk, it follows that (61)Kf,t;Vpk,Wpk≤k1+6k-1!ωk-1/pf;tp′tp′k-1.
We prove now the left inequality. Letg∈Wpk. By [11, page 19], it follows that(62)ωk-1/pf;tp′tp′k-1≤ωk-1/pf-g;tp′tp′k-1+ωk-1/pg;tp′tp′k-1≤tp′k-1ω1-1/pf-gk-1;tp′tp′k-1+tp′k-1ω1-1/pgk-1;tp′tp′k-1=ω1-1/pfk-1-gk-1;tp′+tgkp≤vpfk-1-gk-1+tgkp=k-1!vpkf-g+tgkp,since g(k-1)∈Wp. Taking infimum over all g, we obtain that (63)ωk-1/pf;tp′tp′k-1≤k-1!Kf,t;Vpk,Wpk


## 3. Main Results

Proposition 9.
Letp>1 and f∈Wpk. Then (64)limδ→0+ωk-1/pf;δδk-1/p=fkp.Proof.
By Lemma1 and [1, Proposition  3.1] it follows that (65)limδ→0+ωk-1/pf;δδk-1/p≤limδ→0+ω1-1/pfk-1;δδ1-1/p=fkp.Now we prove the reverse inequality. Let Δδf(x)=f(x+δ)-f(x), Δδ2f(x)=Δδ(Δδf)(x), and Δδkf(x)=Δδ(Δδ(k-1)f)(x). Then(66)fkp≤fk-Δδfk-1δp+Δδfk-1δ-Δδ2fk-2δ2p+⋯+Δδk-1f′δk-1-Δδkfδkp+Δδkfδkp.We set ωk(f;δ)p=sup0<h≤δΔhkfp, which is the kth Lp-modulus of continuity; see [1, 9]. By Hölder’s inequality and Fubini’s theorem (67)fk-Δδfk-1δp=∫01fkx-1δ∫0δfkx+tdtpdx1/p≤1δ∫0δ∫01fkx-fkx+tpdxdt1/p≤1δ∫0δΔtfkppdt1/p≤1δ∫0δsup0≤t≤δΔtfkppdt1/p≤ωfk;δp.Moreover, it follows by Hölder’s inequality and Fubini’s theorem that (68)
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1
Δ
δ
f
k
-
1
δ
x
-
1
δ
2
∫
0
δ
Δ
δ
f
k
-
1
x
+
t
d
t
p
d
x
1
/
p
=
∫
0
1
1
δ
p
∫
0
δ
Δ
δ
f
k
-
1
δ
x
-
Δ
δ
f
k
-
1
x
+
t
δ
d
t
p
d
x
1
/
p
≤
∫
0
1
1
δ
p
∫
0
δ
Δ
δ
f
k
-
1
δ
x
-
Δ
δ
f
k
-
1
x
+
t
δ
p
d
t
·
∫
0
δ
d
t
p
-
1
d
x
1
/
p
=
1
δ
·
∫
0
δ
∫
0
1
Δ
δ
f
k
-
1
δ
x
-
Δ
δ
f
k
-
1
x
+
t
δ
p
d
x
d
t
1
/
p
≤
1
δ
·
∫
0
δ
Δ
t
Δ
δ
f
k
-
1
δ
p
p
d
t
1
/
p
≤
1
δ
·
∫
0
δ
s
u
p
0
≤
t
≤
δ
Δ
t
Δ
δ
f
k
-
1
δ
p
p
d
t
1
/
p
≤
1
δ
·
∫
0
δ
ω
Δ
δ
f
k
-
1
δ
;
δ
p
d
t
1
/
p
≤
ω
Δ
δ
f
k
-
1
δ
;
δ
p
.We repeat this type of argument to the other terms in the sum in the right hand side of inequality (66), except for the last one. For instance, in the last case we see that (69)Δδk-1f′δk-1-Δδkfδkp≤ωΔδk-1f′δk-1;δp.Since f(k), Δδf(k-1)/δ,…,Δδk-1f′/δk-1 belong to Lp, then by all the previous inequalities it follows that all the intermediate terms except the last one in the sum of the right hand side in (66) tend to zero as δ→0+. It was proved in [13] that ωk(f;δ)p≤δ1/pωk-1/p(f;δ). Then (70)Δδkfδkp≤ωkf;δpδk≤δ1/pωk-1/pf;δδk=ωk-1/pf;δδk-1/p,fkp≤limδ→0+ωk-1/pf;δδk-1/p.Theorem 10.
Let1<p<∞ and f be a 1-periodic function.(i)
Iff∈Wpk, then (71)lims→1/p′-1p′-s1/p·∫01t-s-k-1ωk-1/pf;tpdtt1/p=1p1/p·fkp.(ii)
Iflimt→0+ωk-1/p(f;t)/tk-1=0 and (72)lim¯s→1/p′-⁡1p′-s1/p·∫01t-s-k-1ωk-1/pf;tpdtt1/p<∞,thenf∈Wpk.Proof.
(i) Letf∈Wpk and 0<s<1/p′. By Proposition 9 we have that, for any ɛ>0, there exists δ>0 such that(73)fkpp-ɛ<ωk-1/pf;tptkp-1<fkpp+ɛ0<t<δ.Multiplying (73) by tp-sp-2 and then integrating from 0 to δ and subtracting f(k)pp from each side, it follows that(74)δp-sp-1-1fkpp-ɛδp-sp-1≤p-sp-1·∫0δt-s-k-1ωk-1/pf;tpdtt-fkpp≤δp-sp-1-1fkpp+ɛδp-sp-1.
On the other hand, sincef∈Wpk, then f∈Vp(k) and by Lemma 1 and Theorem 5 we have that (75)ωk-1/pf;t≤tk-1ω1-1/pfk-1;t≤tk-1vpfk-1≔tk-1vp,0fk-1≤tk-1k-1!vp,0kf=k-1!tk-1vpkf<∞.Hence, it follows that(76)p1p′-s∫δ1t-s-k-1ωk-1/pf;tpdtt≤p1p′-sδ-sp-1vpkfp.By adding up (74) and (76) we get (77)p1p′-s∫01t-s-k-1ωk-1/pf;tpdtt-fkpp≤1-δp-sp-1fkpp+ɛδp-1-sp+p1p′-sδ-sp-1vpkfp.As s→1/p′-, the right hand side tends to ɛ. Since ɛ>0 was arbitrary, by letting it tend to zero we get the first part of the theorem.
(ii) Applying the first part of the theorem tofh and Lemma 7(i), we have that there exists a constant C>0 such that (78)1pfhkpp=lims→1-1/p-1p′-s·∫01t-s-k-1ωk-1/pfh;tpdtt≤lim-s→1-1/p-⁡1p′-s·∫01t-s-k-1ωk-1/pf;tpdtt=Cand then (79)fhkpp=∫01fk-1x+h-fk-1xhpdx≤C0<h≤1.Since suphh-1Δhf(k-1)p<∞, then by a deep result of Hardy and Littlewood (see [14, Theorem  24]) it follows that f∈Wpk.In the following theorem we prove the continuity at the end points of the scale determined byvp,α(k)(f) in the same manner as in [1, 9].Theorem 11.
Let1<p<∞. If f is 1-periodic function, then(i)
limα→1/p′-vp,α(k)(f)=vp,1/p′(k)(f), and(ii)
iff∈Vp,a(k), for some a>0, then(80)limα→0+vp,αkf=vpkf.Proof.
The proof of (i) and (ii) is similar to the proof of [1, Theorem  3.4] (see also [9, Theorem  3.3]). It is enough to define for a partition Π of type (5), t1,1<⋯<t1,2k≤t2,1<⋯≤tm,k+1<⋯<tm,2k, Il=j:2-l-1<tj,2k-tj,1≤2-l, and(81)Slkf=∑l∈Ilftl,k+1,…,tl,2k-ftl,1,…,tl,kp1/p.By summation by parts, the result follows as in [9, Theorem  3.3].The conditionf∈Vp,a(k) for some a>0 is necessary for (80) to hold. If f∈Vp(k) has a discontinuous (k-1) derivative, then vp,αfk-1=∞ for any α>0. Therefore, vp,α(k)(f)=∞ for all α>0, while vp(k)(f)<∞.Terehin in [13] showed that, for any function f∈Cp, the modulus ω1-1/p(f;t) is equivalent to the nondecreasing function on [0,1] given by (82)ω1f;t=t1/p′inf0<u≤tω1-1/pf;uu1/p′.Clearly, t-1/p′ω1(f;t) is a decreasing function and (see, e.g., [1]) (83)ω1t≤ω1-1/pf;t≤21/p′ω1t0≤t≤1.Similarly, for any functionf such that limt→0+ωk-1/pf;t/tk-1=0, it is possible to prove that the following function ω defined by(84)ωt≔ωf;t=t1/p′inf0<u≤tωk-1/pf;uuk-1+1/p′is increasing and t-1/p′ω(t) is decreasing. Moreover, since ωk-1/pf;t/t(k-1)+1/p′≤2k-1/pωk-1/pf;s/s(k-1)+1/p′ for any s<t (see [15]), we have that(85)ωt≤ωk-1/pf;ttk-1≤2k-1/pωt0≤t≤1.In the next theorem we prove a sharp estimate ofvp,α(k)(f) in terms of ωk-1/p(f;t), which in particular implies the result of Merentes [7].Theorem 12.
Let1<p<∞, 0<α<1/p′, and f be a 1-periodic function in Vp(k). Assume that(86)Ip,αkf=∫01t-α-k-1ωk-1/pf;tpdtt<∞.Then, f∈Vp,α(k) and(87)vp,αkf≤Avpk+p′α1/p1p′-α1/pIp,αkf,where A is constant dependent of k and p.Proof.
Condition (86) implies that limt→0+⁡ωk-1/p(f;t)/tk-1=0. Without loss of generality (see [1, Theorem  4.1]), we may suppose that(88)limt→0+ωf;tt1/p′=∞.
Letωn=ω(f;2-n) and ω¯n=2n/p′ωf;2-n, n∈N. As in [1, page 5], we define the sequence of natural numbers nk. Set n0=0 and(89)nl+1=min⁡n;max⁡ωnωnl,ω¯nlω¯n≤12,l=0,1,….By the monotonicity properties of ω(f,t) we have that either(90)ωnl<4ωnl+1or(91)ω¯nl+1<4ω¯nl.Assume first that (90) holds. Let us define (92)glx=f2-nl,…k,2-nlx=2nl∫02-nlf2-nl,…k-1,2-nlx+tdt.For a fixed partition Π, we set(93)Jl=j:2-nl+1<tj,2k-tj,1≤2-nl.We define (94)
R
l
,
α
f
≔
∑
j
∈
J
l
f
t
j
,
k
+
1
,
…
,
t
j
,
2
k
-
f
t
j
,
1
,
…
,
t
j
,
k
p
1
t
j
,
2
k
-
t
j
,
1
α
p
1
/
p
,
R
l
f
≔
∑
j
∈
J
l
f
t
j
,
k
+
1
,
…
,
t
j
,
2
k
-
f
t
j
,
1
,
…
,
t
j
,
k
p
1
/
p
.For any j∈Jl, we have that gl∈C(k) and, by [10, Proposition  2.3], (95)glt0,…,tk=glkψk!ψ∈convt0,…,tk.Then, by Hölder’s inequality it follows that (96)gltj,k+1,…,tj,2k-gltj,1,…,tj,kp·1tj,2k-tj,1αp=glkψj′-glkψjpk!tj,2k-tj,1αp≤2-nlp-1-αpk!∫ψjψj′glkp,where ψj∈conv{tj,1,…,tj,k} and ψj′∈conv{tj,k+1,…,tj,2k}.
Thus, by Lemma7(ii) and by (85), it follows that (97)Rl,αgl≤2-nl1/p′-αglkp≤2-nl1/p′-α2nl/p′2k-1nlωk-1/pf;2-nl≤2-nl1/p′-α2nl/p′2k-1/pωnl.Then(98)Rl,αgl≤2nlα+kωnland moreover(99)Rl,αf-gl≤2nl+1αRlf-gl.By (98) and (99), it follows that (100)vp,αkf,Π≤2k∑l=0∞2nlαpωnlp1/p+∑l=0∞2nl+1αpRlf-glp1/p.We estimate the second term of the above sum. Obviously, Rl(f-gk)≤vp(k)(f-gl). By Lemma 7(iii), (85), and (90), it follows that (101)2nl+1αpRlf-glp≤2nl+1αpvpkf-glp≤2nl+1αp6kk-1!pωk-1/pf;2-nl2-nlk-1p≤2nl+1αp6kk-1!p2kp-1+2pωnl+1p.
Hence,(102)vp,αkf,Π≤2k∑l=0∞2nlαpωnlp1/p+∑l=0∞2nl+1αpRlf-glp1/p≤2k+2k-1/p+26kk-1!∑l=0∞2nlαpωnlp1/p.Assume now that (91) holds. Let us define (103)g^lx=f2-nl+1,…k,2-nl+1x=2nl+1∫02-nl+1f2-nl+1,…k-1,2-nl+1x+tdt.
Thus by Lemma7(ii) and (85) and by the definition of ω(f,t) and (91) (104)Rl,αg^l≤2-nl1/p′-α2nl+1k-1/pωk-1/pf;2-nl+1≤2-nl1/p′-α2nl+1k-1/p2k-1/pωnl+1≤4·2-nl1/p′-α2k-1/pω¯nl≤4·2-nl1/p′-α2k-1/p2nl/p′ωnl≤4·2nlα+kωnl.Then(105)Rl,αg^l≤4·2nlα+kωnland moreover(106)Rl,αf-g^l≤2nl+1αRlf-g^k.By (105) and (106), it follows that (107)vp,αkf,Π≤4·2k∑l=0∞2nlαpωnlp1/p+∑l=0∞2nl+1αpRlf-g^lp1/p.We estimate now the second term of the above sum. Obviously, we have Rlf-g^l≤vpkf-g^l. By Lemma 7(iii) and (85) we get that (108)2nl+1αpRlf-g^lp≤2nl+1αpvpkf-g^lp≤2nl+1αp6kk-1!pωk-1/pf;2-nl+12-nl+1k-1p≤2k6kk-1!p2nl+1αpωnl+1p.
Hence,(109)vp,αkf,Π≤4·2k∑l=0∞2nlαpωnlp1/p+∑l=0∞2nl+1αpRlf-g^lp1/p≤2k4+6kk-1!∑l=0∞2nlαpωnlp1/p.
By (102) and (109), (110)vp,αkf,Π≤2k+21+6kk-1!∑l=0∞2nlαpωnlp1/p.
Applying [1, Lemma  2.1] with q=p, β=pα and γ=1/p′, it follows that (111)
v
p
,
α
k
f
,
Π
≤
2
k
+
2
1
+
6
k
k
-
1
!
∑
l
=
0
∞
2
n
l
α
p
ω
n
l
p
1
/
p
≤
2
k
+
2
1
+
6
k
k
-
1
!
2
ω
0
p
+
2
p
+
2
p
′
p
p
α
p
p
′
-
α
p
∫
0
1
t
-
α
p
ω
t
p
d
t
t
1
/
p
≤
2
k
+
6
1
+
6
k
k
-
1
!
ω
0
+
p
′
α
1
/
p
1
p
′
-
α
1
/
p
∫
0
1
t
-
α
p
ω
t
p
d
t
t
1
/
p
,since p1/p≤2 and (p′)1/p≤p′.
On the other hand, by (85), (112)ω0≔ω1≤ωk-1/pf;11≤ω1-1/pfk-1;1≤vpfk-1=vpkf,and also (113)∫01t-αpωtpdtt1/p≤∫01t-α-k-1ωk-1/pf;tpdtt1/p=Ip,αkf.
Therefore,(114)vp,αkf,Π≤2k+61+6kk-1!·vpkf+p′α1/p1p′-α1/pIp,αkf.Remarks.If f∈Wpk, then, by Theorem 10(i), we have that (115)lims→1/p′-1p′-s1/p·∫01t-s-k-1ωk-1/pf;tpdtt1/p=1p1/p·fkp.Moreover, by the celebrated result of Riesz and Theorem 5, (116)vpkf=limα→0vp,αkf≤vp,1/p′kf=1k-1!vp,1/p′fk-1=1k-1!fkp.The upper limit as α→1/p′- of the right hand side of (87) is less than (117)2k+61+6kk-1!·1k-1!+p′1p′1/p1p1/pfkp.Also, by Theorem11(i), the left hand side of (87) tends to vp,1/p′(k)(f). Thus, taking limit as α→1/p′- in it, it follows that(118)vp,1/p′kf≤Cp,kfkp.This implies the result of Merentes [7] that states that Wpk⊂Vp,1/p′(k), and it shows that the order of the constant is optimal.Let us assume thatIp,α0(k)(f)<∞ for some 0<α0<1/p′. Since Ip,α(k)(f)≤Ip,α0(k)(f) for 0<α<α0, it follows that(119)limα→0+α1/pIp,αkf≤limα→0+α1/pIp,α0kf=0.Moreover, since Ip,α0(k)(f)<∞, by Theorem 12, f∈Vp,α0(k). Hence, vp,α(k)(f) tends to vp(k)(f) as α→0+ by Theorem 11(ii). Thus, the left hand side of (87) agree in behavior with the right hand side when α→0+ which means that the order of the constant is also optimal.


## 4. Further Results and Applications

Lemma 13.
Let1<p<∞, 0<α<1/p′, and f be a 1-periodic function. Assume that (120)Ip,αkf=∫01t-α-k-1ωk-1/pf;tpdtt1/p<∞.Then (121)Ip,αkf≤∫01t-αω1-1/pfk-1;tpdtt1/p=Ip,αfk-1≤Cp,α,kIp,αkf,where(122)Cp,α,k=C21-1/pk-1pαp+2kp-1Cp1/pif 1≤p≤2 and Cp,α,k=C21-1/p((k-1)p/αp+1/(p-αp-1))1/p if p≥2.Proof.
From the pointwise estimate of Lemma1, we have that (123)Ip,αkf≤∫01t-α-k-1tk-1ω1-1/pfk-1;tpdtt1/p=Ip,αfk-1.
We prove now the second inequality. SinceIp,α(k)(f)<∞, it follows that Ip,α(f(k-1))<∞. Then, f(k-1)∈Cp. We use the following Marchaud-Timan type inequality in [15, Theorem  4],(124)
ω
1
-
1
/
p
f
k
-
1
;
t
≤
C
k
-
1
∫
0
t
ω
k
-
1
/
p
f
;
u
u
k
d
u
+
t
1
-
1
/
p
∫
t
1
ω
k
-
1
/
p
f
;
u
s
u
s
k
-
1
/
p
+
1
d
u
1
/
s
,where C is an absolute constant independent on p and s=max⁡(p,2).
Assume first thats=p, thatis,p≥2. Let t>0. By raising (124) to p, then multiplying both sides of it by t-αp-1, and then integrating it from 0 to 1 we obtain (125)
∫
0
1
t
-
α
ω
1
-
1
/
p
f
k
-
1
;
t
p
d
t
t
≤
C
p
2
p
-
1
k
-
1
p
·
∫
0
1
t
-
α
p
-
1
∫
0
t
ω
k
-
1
/
p
f
;
u
u
k
d
u
p
d
t
+
∫
0
1
t
p
-
α
p
-
2
∫
t
1
ω
k
-
1
/
p
f
;
u
p
u
k
p
d
u
d
t
.
We estimate now the first term of the above sum by the use of the sharp weighted Hardy inequality (see, e.g, [16, Lemma  3.9]) and we get(126)∫01t-αp-1∫0tωk-1/pf;uukdupdt≤1αp∫01t-α-k-1ωk-1/pf;tpdtt.
For the second term, by Fubini’s theorem we have(127)∫01tp-αp-2∫t1ωk-1/pf;upukpdudt=1p-αp-1∫01t-αωk-1/pf;ttk-1pdtt.
Therefore, (126) and (127) imply the following estimate (128)∫01t-αω1-1/pfk-1;tpdtt≤Cp2p-1k-1pαp+1p-αp-1·∫01t-α-k-1ωk-1/pf;tpdtt.
Let nows=2, that is, 1<p≤2. By raising (124) to p, then multiplying both sides of it by t-αp-1, and then integrating it from 0 to 1 we get (129)
∫
0
1
t
-
α
ω
1
-
1
/
p
f
k
-
1
;
t
p
d
t
t
≤
C
p
2
p
-
1
k
-
1
p
·
∫
0
1
t
-
α
p
-
1
∫
0
t
ω
k
-
1
/
p
f
;
u
u
k
d
u
p
d
t
+
∫
0
1
t
p
-
α
p
-
2
∫
t
1
ω
k
-
1
/
p
f
;
u
2
u
2
k
-
1
/
p
+
1
d
u
p
/
2
d
t
.
The first term of the right-hand side of the inequality can be estimated as before (see (126)). The second term instead can be estimated by the fact that for any u>t the following inequality holds(130)ωk-1/pf;uuk-1/p2=ωk-1/pf;uuk-1+1/p′2≤2k-1/pωk-1/pf;ttk-1+1/p′2.Then, (131)∫01tp-αp-2∫t1ωk-1/pf;uuk-1/p2duup/2dt≤2kp-1Cp∫01t-α-k-1ωk-1/pf;updtt.Therefore,(132)∫01t-αω1-1/pfk-1;tpdtt≤Cp2p-1k-1pαp+2kp-1Cp·∫01t-α-k-1ωk-1/pf;tpdtt.Theorem12 can be derived by [1, Theorem  4.1] and the previous lemma, but the constant will not have right asymptotic behaviour. Indeed, since f∈Vp(k), then, as in Proposition 14, (133)vp,αkf≤Avpkf+cp,αCp,α,kk-1!Ip,αkf.Proposition 14.
Let1<p<∞, 0<α<1/p′, and f be a 1-periodic function such that f(k-1) belongs to Vp. Assume that(134)Ip,αkf=∫01t-α-k-1ωk-1/pf;tpdtt<∞.Then, f∈Vp,α(k) and(135)vp,αkf≤Avpkf+cp,αCp,α,kk-1!Ip,αkf,where A is an absolute constant and C(p,α,k) is in Lemma 13.Proof.
The result follows by [1, Theorem  4.1], Lemma 13, and Theorem 5. Indeed, since f(k-1)∈Vp and Ip,α(k)(f)<∞, then by [1, Theorem  4.1] it follows that f(k-1)∈Vp,α. Then, f∈Vp,α(k) by Theorem 5. By [1, Theorem  4.1] and Lemma 13, it follows that(136)vp,αfk-1≤Avpfk-1+cp,αIp,αfk-1≤Ak-1!vpkf+cp,αCp,α,kIp,αkf,where A is the absolute constant in [1, Theorem 4.1]. Finally, applying again Theorem 5, it follows that (137)vp,αkf≤Avpkf+cp,αCp,α,kk-1!Ip,αkf.The inequality is sharp in the sense that (134) cannot be removed.



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*Source: 253636-2015-08-23.xml*