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Functions of Generalized Bounded Variation Martin Lind

Faculty of Health, Science and Technology Mathematics DISSERTATION | Karlstad University Studies | 2013:11

Functions of Generalized Bounded Variation Martin Lind

DISSERTATION | Karlstad University Studies | 2013:11

Functions of Generalized Bounded Variation Martin Lind DISSERTATION Karlstad University Studies | 2013:11 ISSN 1403-8099 ISBN 978-91-7063-486-4 ©

The author

Distribution: Karlstad University Faculty of Health, Science and Technology Department of Mathematics and Computer Science SE-651 88 Karlstad, Sweden +46 54 700 10 00 Print: Universitetstryckeriet, Karlstad 2013

WWW.KAU.SE

ii

Abstract This thesis is devoted to the study of different generalizations of the classical conception of a function of bounded variation. First, we study the functions of bounded p-variation introduced by Wiener in 1924. We obtain estimates of the total p-variation (1 < p < ∞) and other related functionals for a periodic function f ∈ Lp ([0, 1]) in terms of its Lp modulus of continuity ω(f ; δ)p . These estimates are sharp for any rate of decay of ω(f ; δ)p . Moreover, the constant coefficients in them depend on parameters in an optimal way. Inspired by these results, we consider the relationship between the Riesz type generalized variation vp,α (f ) (1 < p < ∞, 0 ≤ α ≤ 1 − 1/p) and the modulus of p-continuity ω1−1/p (f ; δ). These functionals generate scales of spaces that connect the space of functions of bounded p-variation and the Sobolev space Wp1 . We prove sharp estimates of vp,α (f ) in terms of ω1−1/p (f ; δ). In the same direction, we study relations between moduli of p-continuity and q-continuity for 1 < p < q < ∞. We prove an inequality that estimates ω1−1/p (f ; δ) in terms of ω1−1/q (f ; δ). The inequality is sharp for any order of decay of ω1−1/q (f ; δ). Next, we study another generalization of bounded variation: the so-called bounded Λ-variation, introduced by Waterman in 1972. We investigate relations between the space ΛBV of functions of bounded Λ-variation, and classes of functions defined via integral smoothness properties. In particular, we obtain the necessary and sufficient condition for the embedding of the class Lip(α; p) into ΛBV . This solves a problem of Wang (2009). We consider also functions of two variables. Applying our one-dimensional result, we obtain sharp estimates of the Hardy-Vitali type p-variation of a bivariate function in terms of its mixed modulus of continuity in Lp ([0, 1]2 ). (2) Further, we investigate Fubini-type properties of the space Hp of functions of bounded Hardy-Vitali p-variation. This leads us to consider the symmetric mixed norm space Vp [ Vp ]sym of functions of bounded iterated p-variation. (2) (2) For p > 1, we prove that Hp 6⊂ Vp [ Vp ]sym and Vp [ Vp ]sym 6⊂ Hp . In other words, Fubini-type properties completely fail in the class of functions of bounded Hardy-Vitali type p-variation for p > 1.

iii

Basis of the thesis This thesis is mainly based on the following works. Published/accepted papers [1] M. Lind, Functions of bounded Λ-variation and integral smoothness, to appear in Forum Math., 15 pages. [2] M. Lind, Estimates of the total p-variation of bivariate functions, J. Math. Anal. Appl. 401(2013), no. 1, 218–231. [3] M. Lind, On fractional smoothness of functions related to p-variation, Math. Inequal. Appl. 16(2013), no. 1, 21–39. [4] V.I. Kolyada and M. Lind, On moduli of p-continuity, Acta Math. Hungar. 137 (2012), no. 3, 191–213. [5] V.I. Kolyada and M. Lind, On functions of bounded p-variation, J. Math. Anal. Appl. 356 (2009), no. 2, 582–604. Submitted papers [6] M. Lind, Fubini-type properties of bivariate functions of bounded pvariation, 8 pages.

iv

Acknowledgements I am deeply grateful to my supervisor Professor Viktor Kolyada for his guidance and encouragement. Throughout my studies, Viktor has always been willing to take time to discuss mathematics with me, generously sharing his ideas and deep knowledge of the subject. His ability to explain difficult topics in a clear way is remarkable, and I have benefited greatly from working with him. Moreover, I thank Professor Alexander Bobylev for his crucial support when I wanted to begin my doctoral studies. Further, I thank Martin Brundin for encouraging me to study mathematics in the first place. Thanks also to past and present colleagues at the Department of Mathematics, Karlstad University, especially Ilie Barza, Sorina Barza, Martin Kˇrepela and ˚ Asa Windfäll. The support of the Graduate School of Mathematics and Computing (FMB) is gratefully acknowledged.

Contents 1 Introduction

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2 Auxiliary statements 2.1 Lp -moduli of continuity . . . . . . . 2.2 Properties related to p-variation . . 2.2.1 Local p-variation . . . . . . 2.2.2 The modulus of p-continuity 2.3 On γ-moduli of continuity . . . . .

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13 13 16 16 19 22

3 Integral smoothness and p-variation 3.1 Auxiliary results . . . . . . . . . . . . . 3.2 Estimates of L∞ -norm and p-variation . 3.3 Estimates of the modulus of p-continuity 3.4 The classes Vpα . . . . . . . . . . . . . . 3.5 On classes Up . . . . . . . . . . . . . . .

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4 Fractional smoothness via p-variation 55 4.1 Approximation with Steklov averages . . . . . . . . . . . . . . 55 4.2 Limiting relations . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.3 Estimates of the Riesz-type variation . . . . . . . . . . . . . . 64 5 Embeddings within the scale Vp 5.1 Some known results and statement of problem 5.2 Auxiliary results . . . . . . . . . . . . . . . . 5.3 Embeddings of the space Vqω . . . . . . . . . . 5.4 Sharpness of the main estimate . . . . . . . . v

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71 71 73 80 84

vi 6 On 6.1 6.2 6.3

Contents functions of bounded Λ-variation 91 Auxiliary results . . . . . . . . . . . . . . . . . . . . . . . . . 92 Embedding of Lipschitz classes . . . . . . . . . . . . . . . . . 94 A Perlman-type theorem . . . . . . . . . . . . . . . . . . . . . 101

7 Multidimensional results 7.1 Auxiliary results . . . . . . . . . . . . 7.2 Estimates of the L∞ -norm . . . . . . . 7.3 Estimates of the Vitali type p-variation (2) 7.4 Fubini-type properties of Hp . . . . .

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105 105 109 112 122

Chapter 1 Introduction General description of the area The notion of total variation of a function was introduced by Jordan in 1881 in connection with his investigation of convergence of Fourier series. As is well-known, bounded variation is a very important concept with many applications, for example in the study Stieltjes integration and rectifiable curves. Subsequently, several extensions of Jordan’s notion of bounded variation were considered. Two well-known generalizations are the functions of bounded pvariation and the functions of bounded Φ-variation, due to Wiener [82] and L.C. Young [84] respectively. These notions of generalized bounded variation have attracted much interest, e.g., [83, 45, 19, 20, 50, 23, 24, 25]. For a more extensive list of works related to generalized bounded variation, see [15, Part IV]. In [83], L.C. Young obtained an estimate of the Lp -modulus of continuity of a function in terms of its total p-variation. Conversely, sharp estimates of the total p-variation of a function in terms of the Lp -modulus of continuity were first obtained by Terehin [71]. Soon after Jordan’s work, many mathematicians began to study notions of bounded variation for functions of several variables. In the multivariate case, there is no unique concept of bounded variation. One of the most known approaches is based on ideas of Tonelli (it can be expressed in a different but equivalent form in terms of partial moduli of continuity in L1 ). Another definition, due to Vitali, is based on mixed differences. This is more similar to the original definition of Jordan. Later, Hardy restricted Vitali’s class by adding certain conditions on the sections of the functions. For a 1

2

Chapter 1. Introduction

survey of classical notions of multivariate functions of bounded variation, see [1]. Further, Golubov [26] introduced the total p-variation for multivariate functions that for p = 1 corresponds to the Hardy-Vitali type total variation. In the seventies, Waterman [79, 80] considered a completely different extension of bounded variation for univariate functions, the so-called functions of bounded Λ-variation. These classes have been studied by many authors, see, e.g., the surveys [67, 81]. There are also extensions to multivariate functions, see [16] and the references given there. Functions of generalized bounded variation are important in several different areas of mathematics, such as Fourier analysis (e.g, [53, 79, 16, 43, 44]) and operator theory (e.g., [15, 9, 10]). In [15, Part IV], many works related to applications of the concept of p-variation in probability are listed (we mention also [68], which will be considered below). Further, [55] surveys the relevance of Vitali-type variation in certain areas of numerical analysis. Main objectives and methods 1. The main objective of this thesis is to study various properties of functions of generalized bounded variation. In particular, we consider the following: • sharp relations between spaces of generalized bounded variation and spaces of functions defined by integral smoothness conditions (e.g., Sobolev and Besov spaces); • optimal properties of certain scales of function spaces of fractional smoothness generated by functionals of variational type; • sharp embeddings within the scale of spaces of functions of bounded p-variation; • bivariate functions of bounded p-variation, in particular sharp estimates of total variation in terms of the mixed Lp -modulus of continuity, and Fubini-type properties. 2. Two central methods in our work are approximations with Steklov averages and a decomposition technique for moduli of continuity. We also develop a scheme for constructing relevant counterexamples that is based on a sort of accumulation of piecewise linear functions. Such constructions are used to prove the sharpness of our results.

3 Summary We shall give a summary of the thesis and some antecedent results. Chapter 2 contains general results and definitions, in particular related to p-variation and moduli of continuity. It is convenient to introduce these notions right now. Let f be a 1-periodic function on the real line R and let 1 ≤ p < ∞. Any set Π = {x0 , x1 , ..., xn } of points x0 < x1 < ... < xn ,

where xn = x0 + 1,

will be called a partition of a period (or simply a partition). We also denote kΠk = maxk (xk+1 − xk ). For any partition Π, set !1/p n−1 X p . vp (f ; Π) = |f (xk+1 ) − f (xk )| k=0

We say that f is a function of bounded p-variation (written f ∈ Vp ) if vp (f ) = sup vp (f ; Π) < ∞,

(1.0.1)

Π

where the supremum is taken over all partitions Π. For 1 < p < ∞, we also consider the function ω1−1/p (f ; δ) = sup vp (f ; Π) (0 ≤ δ ≤ 1),

(1.0.2)

kΠk≤δ

where the supremum is taken over all partitions Π with kΠk ≤ δ. As it was mentioned above, for p = 1, the definiton (1.0.1) was given by Jordan, and for p > 1, both (1.0.1) and (1.0.2) are due to Wiener [82]. Following Terehin [70], we call the function (1.0.2) the modulus of p-continuity of f . For p > 1, there are non-constant functions f such that lim ω1−1/p (f ; δ) = 0.

δ→0+

Such functions are called p-continuous, and the class of all p-continuous functions is denoted by Cp . For f ∈ Lp ([0, 1]) (1 ≤ p < ∞), the Lp -modulus of continuity of f is given by Z 1 1/p p ω(f ; δ)p = sup |f (x + h) − f (x)| dx (0 ≤ δ ≤ 1). (1.0.3) 0≤|h|≤δ

0

4

Chapter 1. Introduction

In Chapter 3, we study relations between bounded p-variation and integral smoothness of univariate functions. Terehin [71] obtained sharp estimates of the total p-variation of a function in terms of its second order modulus of continuity (see (3.2.19) below). The same result was obtained by Peetre [57] with the use of interpolation methods. In particular, the following was proved in [71]: if Z Jp (f ) = 0

1

t−1/p ω(f ; t)p

dt < ∞ (1 < p < ∞), t

(1.0.4)

then f is equivalent1 to a continuous 1-periodic function f¯ ∈ Vp , and vp (f¯) ≤ AJp (f ).

(1.0.5)

The first part of this statement was proved in 1958 by Geronimus [21]. More exactly, it was shown in [21] that any function f ∈ Lp ([0, 1]) satisfying (1.0.4) is equivalent a continuous 1-periodic function, and kf k∞ ≤ A(kf kp + Jp (f )),

(1.0.6)

where A is an absolute constant. However, it can easily be shown that the constant coefficients of (1.0.5) and (1.0.6) should depend on p. We prove that the following stronger versions of (1.0.5) and (1.0.6) hold: 1 kf k∞ ≤ A kf kp + 0 Jp (f ) , pp

(1.0.7)

1 vp (f¯) ≤ A ω(f ; 1)p + 0 Jp (f ) , pp

(1.0.8)

and

where A is an absolute constant (as usual, p0 = p/(p − 1)). We show that the asymptotic behaviour of the constant A/(pp0 ) in (1.0.7) and (1.0.8) is optimal in a sense. It was shown in [83, 71] that for 1 < p < ∞, ω(f ; δ)p ≤ δ 1/p ω1−1/p (f ; δ) (0 ≤ δ ≤ 1). 1

Two functions are said to be equivalent if they coincide almost everywhere.

(1.0.9)

5 Reverse estimates were obtained in [71]. There it was shown that if (1.0.4) holds and the function f is modified on a set of measure zero so as to become continuous, then we have the following inequality Z δ dt t−1/p ω(f ; t)p , (1.0.10) ω1−1/p (f ; δ) ≤ A pδ −1/p ω(f ; δ)p + t 0 where A is an absolute constant. Applying (1.0.8), we obtain that the constant coefficients in (1.0.10) can be improved. Namely, we show that the following estimate holds. For 1 < p < ∞, we have Z δ 1 dt (1.0.11) ω1−1/p (f ; δ) ≤ A δ −1/p ω(f ; δ)p + 0 t−1/p ω(f ; t)p pp 0 t for any δ ∈ (0, 1], where A is an absolute constant. Our main result here is the sharpness of the estimate (1.0.11). More exactly, we construct a function with an arbitrary prescribed order of the modulus of continuity in Lp ([0, 1]), for which the opposite inequality holds for all δ ∈ [0, 1]. Assume that 1 < p < ∞ and α ≥ 0. Let f be an 1-periodic function and let Π = {x0 , x1 , ..., xn } be a partition. Set vp,α (f ; Π) =

n−1 X |f (xk+1 ) − f (xk )|p k=0

(xk+1 − xk )αp

!1/p .

(1.0.12)

We denote by Vpα the class of all 1-periodic functions f such that vp,α (f ) = sup vp,α (f ; Π) < ∞,

(1.0.13)

Π

where Π runs over all partitions of a period (see [57], p. 114). Obviously, Vp0 = Vp and Vpβ ⊂ Vpα if 0 ≤ α < β. Denote by Wp1 the class of all 1-periodic, absolutely continuous functions f with f 0 ∈ Lp ([0, 1]). By a theorem of F. Riesz (see Theorem 2.2 below), we 1/p0 have Vp = Wp1 for any 1 1/p0 , then any function f ∈ Vpα is constant. We obtain the following sharp estimate of the Riesz-type variation vp,α (f ) in terms of ω(f ; δ)p .

6

Chapter 1. Introduction If 1 < p < ∞, 0 < α < 1/p0 , and 1

Z

t−αp−1 ω(f ; t)pp

Kp,α (f ) = 0

dt t

1/p < ∞,

(1.0.14)

then f is equivalent to a continuous function f¯ ∈ Vpα , and 0 vp,α (f¯) ≤ Aα−1/p (1/p0 − α)1/p Kp,α (f ),

(1.0.15)

where A is an absolute constant. Further, we show that the condition (1.0.14) is sharp for any rate of the decay of the modulus of continuity, and the order of the constant in (1.0.15) is optimal as α → 0 or α → 1/p0 . In Chapter 4, we continue our study of the modulus of p-continuity (1.0.2) and Riesz-type variation (1.0.13). One of the main results of the chapter is the following sharp estimate of the Riesz-type variation vp,α (f ) in terms of ω1−1/p (f ; δ) (similar to (1.0.15)). If 1 < p < ∞, 0 < α < 1/p0 and Z Ip,α (f ) =

1

[t−α ω1−1/p (f ; t)]p

0

dt t

1/p < ∞,

(1.0.16)

then f ∈ Vpα and 0 vp,α (f ) ≤ A vp (f ) + p0 α1/p (1/p0 − α)1/p Ip,α (f ) ,

(1.0.17)

where A is an absolute constant. We prove that the constant coefficient of (1.0.17) has optimal order as α → 0 or α → 1/p0 . Observe that an estimate of the type (1.0.17) follows immediately from (1.0.15) and (1.0.9). However, the constant obtained in this way is not optimal. We also show that the condition (1.0.16) is sharp for any rate of decay of the modulus of p-continuity. Further, we obtain several limiting relations for the functionals ω1−1/p (f ; δ) and vp,α (f ). In particular, we prove the following. If f ∈ Vp (1 1, we have the following result. Let 1 < p < ∞ and f ∈ Lp ([0, 1]2 ) and assume that ZZ du dv (uv)−1/p ω(f ; u, v)p < ∞. (1.0.29) u v 2 [0,1]

11 (2) Then there exists a function f¯ ∈ Vp such that f = f¯ a.e. and Z 1 1 dt vp(2) (f¯) ≤ A ω(f ; 1, 1)p + 0 t−1/p [ω(f ; 1, t)p + ω(f ; t, 1)p ] pp 0 t ZZ 1 dv du + , (1.0.30) (uv)−1/p ω(f ; u, v)p (pp0 )2 u v [0,1]2

where A is an absolute constant. None of the terms at the right-hand side of (1.0.30) can be omitted, and the constants 1/pp0 and 1/(pp0 )2 have optimal asymptotic behaviour as p → 1 and p → ∞. This is a two-dimensional analogue of (1.0.8). Let X, Y be spaces of functions defined on the real line, with norms k · kX and k · kY . A function f (x, y) is said to belong to the symmetric mixed norm space X [ Y ]sym if the functions x 7→ kfx kY

and y 7→ kfy kY

both belong to the space X (fx , fy denotes the sections of f , see (1.0.26)). The importance of mixed norm spaces that are invariant under permutations of variables was first demonstrated by Fournier [18]. Recall that by the Fubini-Tonelli theorem, we have Lp ([0, 1]) [ Lp ([0, 1]) ]sym = Lp ([0, 1]2 ) (0 < p ≤ ∞). Fubini-type properties for the scale of Lorentz spaces were studied in [36]. (2) In Chapter 7, we also investigate Fubini-type properties of the class Hp for p ≥ 1. For this, we consider the symmetric mixed norm space Vp [ Vp ]sym of functions of bounded iterated p-variation. That is, if f is a bivariate functions and we denote ϕ(x) = vp (fx ) and ψ(y) = vp (fy ), then f ∈ Vp [ Vp ]sym if and only if ϕ, ψ ∈ Vp . (2) For p = 1, it was proved in [1] that H1 ⊂ V1 [ V1 ]sym . This is a Fubini(2) type property of H1 (in one direction). We prove that Fubini-type properties (2) completely fail for Hp when p > 1. In other words, the following holds. For p > 1, Hp(2) 6⊂ Vp [ Vp ]sym

and Vp [ Vp ]sym 6⊂ Hp(2) .

Chapter 2 Auxiliary statements In this chapter we collect some general results which are used throughout this thesis.

2.1

Lp-moduli of continuity

Let Lp ([0, 1]n ) denote the set of all measurable functions on Rn that are 1-periodic in each variable and satisfy Z kf kp =

|f (x)|p dx

1/p < ∞,

[0,1]n

if p < ∞, and kf k∞ = ess supx∈[0,1]n |f (x)| < ∞ for p = ∞. Let f be a function on Rn that is 1-periodic in each variable. For h ∈ Rn , we denote ∆(h)f (x) = f (x + h) − f (x). (2.1.1) Recall that for f ∈ Lp ([0, 1]n ) (1 ≤ p < ∞), the Lp -modulus of continuity is defined by ω(f ; δ)p = sup k∆(h)f kp (0 < δ ≤ 1). 0≤|h|≤δ

A modulus of continuity is a continuous functions ω defined on the interval [0, 1] such that ω(0) = 0 and ω is nonincreasing and subadditive. Denote by Ω the class of all moduli of continuity. For any ω ∈ Ω, we have ω(2δ) ≤ 2ω(δ) (0 ≤ δ ≤ 1/2), 13

14

Chapter 2. Auxiliary statements

and it follows that ω(µ) 2ω(h) ≤ µ h

for 0 < h < µ ≤ 1.

(2.1.2)

Whence, if ω is not identically 0, then for all δ ∈ [0, 1], we have ω(δ) ≥ cω δ

(cω = ω(1)/2).

Thus, the best order of decay for any modulus of continuity is ω(δ) = O(δ). Clearly, if f ∈ Lp ([0, 1]n ), then ω(f ; ·)p ∈ Ω. It follows from Lebesgue’s differentiation theorem that ω(f ; ·)p is identically 0 if and only if f is equivalent to a constant. For the rest of this chapter, we shall only consider the case n = 1. Recall that Wp1 (1 ≤ p < ∞) denotes the class of all 1-periodic absolutely continuous functions f with f 0 ∈ Lp ([0, 1]). If f ∈ Wp1 (1 ≤ p < ∞), then ω(f ; δ)p ≤ kf 0 kp δ

(0 ≤ δ ≤ 1).

(2.1.3)

Moreover, we have the following theorem of Hardy and Littlewood [27]. Theorem 2.1. Let f ∈ Lp ([0, 1]) (1 ≤ p < ∞). The following statements are true. (i) For 1 0

ω(f ; δ)p . δ

(ii) For p = 1, we have ω(f ; δ)1 = O(δ) if and only if f is equivalent to a function g ∈ V1 . Also, v1 (g) = sup δ>0

ω(f ; δ)1 . δ

For 1 < p < ∞, we also have the following variational characterization of Wp1 , due to F. Riesz (see, e.g., [51]).

2.1. Lp -moduli of continuity

15

Theorem 2.2. Let 1 < p < ∞, then f ∈ Wp1 if and only if sup Π

n−1 X |f (xk+1 ) − f (xk )|p

!1/p < ∞,

(xk+1 − xk )p−1

k=0

where the supremum is taken over all partitions Π. In other words, Wp1 = Vp1/p

0

(1 < p < ∞).

The next lemma is well-known (see, e.g., [7]). For the sake of completeness, we give the proof. Lemma 2.3. Let f ∈ Lp ([0, 1]) (1 ≤ p < ∞), then ω(f ; δ)p ≤

3 δ

Z

δ

k∆(t)f kp dt,

δ ∈ (0, 1],

(2.1.4)

0

where ∆(t)f is given by (2.1.1). Proof. Fix δ > 0 and let h ∈ (0, δ]. Clearly, for any t ∈ (0, δ], we have k∆(h)f kp ≤ k∆(h − t)f kp + k∆(t)f kp . Integrating the previous inequality with respect to t in [0, δ] and using the fact that k∆(u)f kp = k∆(−u)f kp , we get Z

δ

k∆(t)f kp dt,

δk∆(h)f kp ≤ 3 0

and (2.3) follows. One consequence of (2.1.4) that we shall use below is the following estimate: Z 1 Z 1 dt dt t−1/p ω(f ; t)p ≤ 3 t−1/p k∆(t)f kp . (2.1.5) t t 0 0 For any function f ∈ Lp ([0, 1]) (1 ≤ p < ∞), set Z

1

Z

Ωp (f ) = 0

0

1

|f (x) − f (y)|p dxdy

1/p

16


We have Ωp (f ) ≤ ω(f ; 1)p ≤ 2Ωp (f ). Indeed, since f is 1-periodic, Z 1 Z dh Ωp (f ) = 0

1

|f (x) − f (x + h)|p dx 1/p

ω(f ; h)pp dh

≤

1/p

0 1

Z

(2.1.6)

≤ ω(f ; 1)p

0

On the other hand, denoting I =

R1 0

f (y)dy, we obtain

ω(f ; 1)p = ω(f − I; 1)p ≤ 2kf − Ikp p 1/p Z 1 Z 1 =2 ≤ 2Ωp (f ). f (y)dy dx f (x) − 0

0

1

Let f ∈ L ([0, 1]). For any 0 < h ≤ 1, let Z 1 h f (x + t)dt fh (x) = h 0

(2.1.7)

be the Steklov average of the function f. Lemma 2.4. If f ∈ Lp ([0, 1]), 1 ≤ p < ∞, then kf − fh kp ≤ ω(f ; h)p

(2.1.8)

kfh0 kp ≤ ω(f ; h)p /h.

(2.1.9)

and These inequalities are well-known and their proofs are immediate.

2.2 2.2.1

Properties related to p-variation Local p-variation

Let f be a 1-periodic function on the real line and let [a, b] ⊂ R be any interval. A partition of the interval [a, b] is a finite set of points Πa,b = {x0 , x1 , ..., xn } such that a = x0 < x1 < ... < xn = b.

2.2. Properties related to p-variation

17

For 1 ≤ p < ∞, the p-variation of f on [a, b] is defined as vp (f ; [a, b]) = sup Πa,b

n−1 X

!1/p p

|f (xk+1 ) − f (xk )|

,

k=0

where the supremum is taken over all partitions Πa,b of [a, b]. Observe that for vp (f ) defined by (1.0.1), we have vp (f ) = sup vp (f ; [a, a + 1]). a∈R

Let 1 ≤ p < ∞ and assume that f is monotone on the interval [a, b]. Then vp (f ; [a, b]) = |f (b) − f (a)|, (2.2.1) the proof of this is trivial. The next lemma is well-known (see, e.g., [45]). Lemma 2.5. Let f be a 1-periodic function on the real line, let 1 ≤ p < ∞ and [a, b] ⊂ R. For any c ∈ (a, b), there holds vp (f ; [a, c])p + vp (f ; [c, b])p ≤ vp (f ; [a, b])p ,

(2.2.2)

vp (f ; [a, c]) + vp (f ; [c, b]) ≤ vp (f ; [a, b]).

(2.2.3)

and Moreover, if f attains global extremum at c ∈ (a, b), then vp (f ; [a, b])p = vp (f ; [a, c])p + vp (f ; [c, b])p .

(2.2.4)

Proof. The inequality (2.2.2) is obvious. To prove (2.2.3), we define the functions g(x) = [f (x) − f (c)]χ[a,c] (x) and h(x) = [f (x) − f (c)]χ[c,b] (x). Then f (x) = g(x) + h(x) + f (c) on [a, b], and vp (f ; [a, b]) = vp (g + h; [a, b]) ≤ vp (g; [a, b]) + vp (h; [a, b]). On the other hand, since g(x) = 0 for x ∈ [c, b], and g(x) = f (x) − f (c) for x ∈ [a, c], we have vp (g; [a, b]) = vp (g; [a, c]) = vp (f ; [a, c]).

18


In the same way, vp (h; [a, b]) = vp (f ; [c, b]), and (2.2.3) follows. Finally, we prove (2.2.4). By (2.2.2), it is enough to show that the right-hand side of (2.2.4) is not smaller than the left-hand side. Let Πa,b = {x0 , x1 , ..., xn } be any partition of [a, b] and assume that c ∈ (xi , xi+1 ) for some i. Clearly, |f (xi+1 ) − f (xi )|p ≤ max{|f (xi+1 ) − f (c)|p , |f (c) − f (xi )|p }, and thus, n−1 X

|f (xk+1 ) − f (xk )|p ≤

k=0

X

|f (xk+1 ) − f (xk )|p +

k6=i

+ |f (xi+1 ) − f (c)|p + |f (c) − f (xi )|p ≤ vp (f ; [a, c])p + vp (f ; [c, b])p , and (2.2.4) follows. The next lemma is a consequence of (2.2.4). Lemma 2.6. Let 1 ≤ p < ∞ and assume that {fn } is a sequence of nonnegative 1-periodic functions such that • (supp fn ) ∩ [0, 1] = [an , bn ]

(n ∈ N);

• the intervals [an , bn ] (n ∈ N) are nonoverlapping; • fn (an ) = fn (bn ) = 0 (n ∈ N). Set f (x) =

∞ X

fn (x),

x ∈ R,

n=1

then vp (f ) =

∞ X

!1/p vp (fn )

p

.

(2.2.5)

n=1

Proof. Since fn ≥ 0 (n ∈ N), it follows that x = 0, 1 and x = an , bn (n ∈ N) are points of global minimum of f . Then, by (2.2.4), we have !1/p ∞ X p vp (f ) = vp (f ; [0, 1]) = vp (f ; [an , bn ]) . n=1

On the other hand, vp (f ; [an , bn ]) = vp (fn ; [an , bn ]) = vp (fn ).


2.2.2

19

The modulus of p-continuity

Recall the moduli of p-continuity (1.0.2). For 1 < p < ∞ and I ⊂ R, we can define the local modulus of p-continuity ω1−1/p (f, I; δ) in the obvious way. Clearly, analogues of (2.2.2)-(2.2.4) hold. In particular, if f attains global extremum at c ∈ (a, b), then ω1−1/p (f, [a, b]; δ)p = ω1−1/p (f, [a, c]; δ)p + ω1−1/p (f, [c, b]; δ)p ,

(2.2.6)

for any δ ∈ (0, 1]. It was proved in [70] that for 1 < p < ∞ and any n ∈ N 0

ω1−1/p (f ; nδ) ≤ n1/p ω1−1/p (f ; δ) (0 ≤ δ ≤ 1/n), where p0 = p/(p − 1). Thus, ω1−1/p (f ; µ) 0 ω1−1/p (f ; h) ≤ 21/p 0 1/p µ h1/p0

for 0 < h < µ ≤ 1.

(2.2.7)

It follows that if f is not a constant function, then ω1−1/p (f ; δ) ≥ cδ 1/p

0

0

(c = vp (f )/21/p ). 0

Moreover, the best order of decay ω1−1/p (f ; δ) = O(δ 1/p ) is attained if f ∈ Wp1 . Indeed, assume that f ∈ Wp1 (1 < p < ∞) and let Π = {x0 , x1 , ..., xn } be any partition with kΠk ≤ δ. Applying Hölder’s inequality, we have vp (f ; Π) =

n−1 Z X

xk

k=0

≤ δ 1/p

0

n−1 Z X k=0

Thus,

xk+1

xk+1

p !1/p f (t)dt 0

!1/p |f 0 (t)|p dt

0

= δ 1/p ||f 0 ||p ..

xk

0

ω1−1/p (f ; δ) ≤ δ 1/p kf 0 kp ,

0 ≤ δ ≤ 1.

(2.2.8) 1/p0

In [70] the converse was proved. That is, if ω1−1/p (f ; δ) = O(δ ), then f ∈ Wp1 (1 < p < ∞). The next result is due to Terehin [71] for periodic functions (the easier non-periodic case was proved in [83]). The argument presented in [71] is not sufficiently clear, therefore we give a complete proof.

20


Proposition 2.7. Let 1 < p < ∞ and let f ∈ Vp . Then ω(f ; δ)p ≤ δ 1/p ω1−1/p (f ; δ),

0 ≤ δ ≤ 1.

(2.2.9)

Proof. We shall first prove (2.2.9) for functions in Vp that attain global maximum. Let f ∈ Vp and suppose that f attains a global maximum at some point c. We may assume that c = 0, since both ω(f ; δ)p and ω1−1/p (f ; δ) are invariant with respect to translation. Fix δ ∈ (0, 1] and let k/m ∈ (0, δ] be a rational number. By the periodicity of f , we have kk∆(k/m)f kp = =

Z

k

|f (x + k/m) − f (x)|p dx

0 m−1 X Z (j+1)k/m j=0

Z =

k/m m−1 X

0

Z ≤

|f (u + k/m) − f (u)|p du

jk/m

|f (u + (j + 1)k/m) − f (u + jk/m)|p du

j=0 δ

ω1−1/p (f, [u, u + k]; δ)p du.

(2.2.10)

0

Since k ∈ N, it is a point of global maximum of f . By (2.2.6), we have for any u ∈ [0, δ] ω1−1/p (f, [u, u + k]; δ)p = ω1−1/p (f, [u, k]; δ)p + ω1−1/p (f, [k, u + k]; δ)p . Further, since f is 1-periodic, ω1−1/p (f, [k, k + u]; δ) = ω1−1/p (f, [0, u]; δ). Whence, by the two previous equations and (2.2.2), ω1−1/p (f, [u, u + k]; δ)p = ω1−1/p (f, [u, k]; δ)p + ω1−1/p (f, [0, u]; δ)p ≤ ω1−1/p (f, [0, k]; δ)p . Applying (2.2.6) repetedly and using that f is 1-periodic, we have ω1−1/p (f, [0, k]; δ)p = kω1−1/p (f, [0, 1]; δ)p ≤ kω1−1/p (f ; δ)p .


21

Whence, ω1−1/p (f, [u, u + k]; δ)p ≤ kω1−1/p (f ; δ)p for any u ∈ [0, δ]. By the previous inequality and (2.2.10), we have kk∆(k/m)f kpp ≤ k

Z

δ

ω1−1/p (f ; δ)p du ≤ kδω1−1/p (f ; δ)p ,

0

and it follows that ω(f ; δ)p ≤ δ 1/p ω1−1/p (f ; δ) for all functions f ∈ Vp that attains global maximum. Let now f be an arbitrary function in Vp and set M = sup f (x). Given any ε > 0, there is a point c such that f (c) > M − ε. As above, we may suppose that c = 0. Define the 1-periodic function fε by setting fε (x) = f (x) for x ∈ / Z and fε (x) = M for x ∈ Z. Then fε attains global maximum and thus (2.2.9) holds for fε . On the other hand, since f = fε almost everywhere, we get ω(fε ; δ)p = ω(f ; δ)p . Further, it is clear that vp (f − fε ) ≤ 2ε, and therefore ω1−1/p (fε ; δ) ≤ ω1−1/p (f ; δ) + 2ε (0 ≤ δ ≤ 1). Then ω(f ; δ)p = ω(fε ; δ)p ≤ δ 1/p ω1−1/p (fε ; δ) ≤ δ 1/p ω1−1/p (f ; δ) + 2δ 1/p ε, and since ε > 0 was arbitrary, the inequality follows. Remark 2.8. One can give a simple proof of the inequality (2.2.9) with a worse constant. Indeed, for any δ > 0 let 0 < h ≤ δ and take n ∈ N such that nh < 1 ≤ (n + 1)h. Then k∆(h)f kpp

Z =

n−1 hX

0

Z

|f (x + (j + 1)h) − f (x + jh)|p dx +

j=0 1

+

|f (x + h) − f (x)|p dx

nh

≤ 2δω1−1/p (f ; δ)p . Thus, ω(f ; δ)p ≤ 21/p δ 1/p ω1−1/p (f ; δ). Recall that fh denotes the Steklov average function (2.1.7).

22


Lemma 2.9. Let f ∈ Vp , 1 < p < ∞, then ω1−1/p (fh , δ) ≤ ω1−1/p (f ; δ)

(0 ≤ δ ≤ 1),

(2.2.11)

and 0

kfh0 kp ≤ h−1/p ω1−1/p (f ; h).

(2.2.12)

The inequality (2.2.11) is immediate and (2.2.12) follows from (2.1.9) and (2.2.9). Recall that Cp (1 < p < ∞) denotes the class of p-continuous functions, that is, functions such that lim ω1−1/p (f ; δ) = 0.

δ→0+

It is easy to see that if f ∈ Cp , then f ∈ Vp and f is continuous. Love [45] considered the following property of functions: for any ε > 0, there exist δ > 0 such that if {(ak , bk )} is any finite collection of nonoverlapping intervals with !1/p X

(bk − ak )

p

!1/p < δ,

k

then

X

p

|f (bk ) − f (ak )|

< ε.

k

For p = 1, the previous condition is just the definition of absolute continuity. For 1 < p < ∞, it is equivalent to f ∈ Cp . Thus, for 1 < p < ∞, we can view p-continuity as an intermediate property of functions, between absolute continuity and continuity. Let W be van der Waerden’s function, i.e., W (x) =

∞ X

2−n φ(2n x),

where φ(x) = inf |x − k|, x ∈ R.

n=1

k∈Z

Then W is nowhere differentiable and thus not absolutely continuous. At the same time, W ∈ Lip(α) for any 0 < α < 1, whence it follows that W ∈ Cp for all p > 1.

2.3

On γ-moduli of continuity

We shall use the following scale of functions.

2.3. On γ-moduli of continuity

23

Definition 2.10. Let 0 < γ ≤ 1. We let Ωγ denote the class of all continuous functions ω defined on [0, 1] such that ω(0) = 0, ω(t) is nondecreasing and ω(t)/tγ is nonincreasing. For historical remarks and some new information concerning conditions of this type (including the close relation to index numbers), we refer to the paper [61] and the references given there. For γ = 1, the class Ω1 is “almost” the same as the class of moduli of continuity (see, e.g., [14, p.41]), in the sense that for any modulus of continuity η, there is ω ∈ Ω1 such that ω(t) ≤ η(t) ≤ 2ω(t), t ∈ [0, 1]. Similarly, Terehin [72] proved that for γ = 1/p0 , the class Ω1/p0 “almost coincides” with the class of all moduli of p-continuity for functions in Cp . Indeed, let f ∈ Cp and set 0

ω ∗ (t) = t1/p inf

0 ωnk

or ω nk+1 −1 < 4ω nk

holds. By (2.3.4) and (2.3.5), this implies that for each k = 0, 1, ... we have at least one of the inequalities ωnk < 8ωnk+1

(2.3.9)

ω nk+1 < 8ω nk .

(2.3.10)

or Partitions (2.3.7) for moduli of continuity have been used for a long time, beginning from the works [3, 52, 75].

Chapter 3 Integral smoothness and p-variation In this chapter, we study relations between variational properties of functions and integral smoothness. Recall that by Hardy-Littlewood’s theorem (Theorem 2.1 above), f ∈ L1 ([0, 1]) coincides a.e. with a function g ∈ V1 if and only if ω(f ; δ)1 = O(δ). For p > 1, the class Vp does not admit any similar characterization in terms of Lp -modulus of continuity. It was shown in [83], [70] that ω(f ; δ)p ≤ vp (f )δ 1/p

(0 ≤ δ ≤ 1).

(3.0.1)

1/p

However, for p > 1, the condition ω(f ; δ)p = O(δ ) does not even imply that f ∈ L∞ ([0, 1]) (take e.g. f (x) = log(1/|x|), |x| ∈ (0, 1/2]). As mentioned in the Introduction, Terehin [71] proved that if Z 1 dt Jp (f ) = t−1/p ω(f ; t)p < ∞, t 0 then f is equivalent to a continuous function f¯ ∈ Vp , and vp (f¯) ≤ AJp (f ).

(3.0.2)

Simple arguments show that the constant coefficient in (3.0.2) should depend on p and vanish as p → 1 or p → ∞. For example, for any continuously differentiable function f the left-hand side in (3.0.2) is bounded whilst the right-hand side tends to infinity as p → 1 (if f is not a constant). Furthermore, if p → ∞, then the right-hand side of (3.0.2) tends to the Dini integral, whilst the left-hand side tend to the oscillation of f¯. 25

26

Chapter 3. Integral smoothness and p-variation

The outline of this chapter is as follows. First, we determine the optimal asymptotics of the constant coefficient A = A(p) in (3.0.2) as p → 1 and p → ∞. Using this version of (3.0.2) with improved constant, we obtain the estimate (1.0.11), which strengthens Terehin’s inequality (1.0.10). We also show that (1.0.11) is sharp in a strong sense (see Theorem 3.12 below). Next, we prove sharp estimates of the Riesz type variation (1.0.13) in terms of Lp -moduli of continuity. Finally, we give some remarks on certain spaces of functions defined in terms of local oscillations.

3.1

Auxiliary results

We will use the next lemma at several places in this work. Lemma 3.1. Let 0 < γ ≤ 1 and let ω ∈ Ωγ satisfy (2.3.6). Let 1 ≤ q < ∞ and 0 < β < qγ be given numbers. Then Z 1 ∞ X dt 2qγ+2 β(qγ − β) t−β ω(t)q . 2nk β ωnq k ≤ 2ω0q + qγ t 0 k=0 Proof. By the first inequality of (2.3.8), we have nk+1 −1

X

q 2nβ (ωnq − ωn+1 ) ≥ 2nk β (ωnq k − ωnq k+1 ) ≥ 2nk β−1 ωnq k ,

n=nk

for any k ≥ 0. This implies that ∞ X

2nk β ωnq k ≤ 2

k=0

∞ X

q 2nβ (ωnq − ωn+1 ).

(3.1.1)

n=0

Further, applying the second inequality of (2.3.8), we obtain nk X

2n(β−qγ) (ω qn − ω qn−1 ) ≥ 2nk (β−qγ) (ω qnk − ω qnk−1 )

n=nk−1 +1

≥ 2nk (β−qγ)−1 ω qnk , for any k ≥ 1. Thus, ∞ X k=1

2nk β ωnq k ≤ 2

∞ X n=1

2n(β−qγ) (ω qn − ω qn−1 ).

(3.1.2)

3.1. Auxiliary results Since

∞ X

27

q 2nβ (ωnq − ωn+1 ) = ω0q + (2β − 1)

n=0

∞ X

2(n−1)β ωnq ,

n=1

and (2β − 1)2(n−1)β ωnq ≤ β

2−n+1

Z

t−β ω(t)q

2−n

dt . t

Whence, ∞ X

q 2nβ (ωnq − ωn+1 ) ≤ ω0q + β

1

Z

t−β ω(t)q

0

n=0

dt . t

(3.1.3)

Further, ∞ X

2n(β−qγ) (ω qn − ω qn−1 ) = (1 − 2β−qγ )

∞ X

n=1

2n(β−qγ) (ω qn − ω q0 ),

n=1

and by (2.3.5), Z

2−n

t−β ω(t)q

2−n−1

dt ≥ t

ωn 2γ

2−n

q Z

= 2−qγ ω qn

t−β+qγ−1 dt

2−n−1 β−qγ

1−2 2n(β−qγ) . qγ − β

Hence, ∞ X

Z

2n(β−qγ) (ω qn − ω qn−1 ) ≤ 2qγ (qγ − β)

1

t−β ω(t)q

0

n=1

Denote S=

∞ X

2nk β ωnq k

Z and J =

1

t−β ω(t)q

0

k=0

2(ω0q

dt . t

(3.1.4)

dt . t

Then (3.1.1) and (3.1.3) imply that S ≤ + βJ). By (3.1.2) and (3.1.4), we have also that S ≤ ω0q + 2qγ+1 (qγ − β)J. Thus, we get qγS ≤ 2qγω0q + 2qγ+2 β(qγ − β)J.

28


Lemma 3.2. Let f ∈ Lp ([0, 1]) (1 ≤ p < ∞) and let ψh,µ (x) = fh (x) − fµ (x)

(h, µ ∈ (0, 1]).

Then 1−1/p

kψh,µ k∞ ≤ h

ω(f ; µ)p

and vp (ψh,µ ) ≤ 5h1−1/p ω(f ; µ)p

1 1 + h µ

1 1 + h µ

(3.1.5)

.

(3.1.6)

Proof. For any x we have Z 1 x+h |ψh,µ (x)| ≤ |f (t) − fµ (x)|dt h x Z x+h Z 1 1 x+h ≤ |f (t) − fµ (t)|dt + |fµ (t) − fµ (x)|dt. h x h x Further, for any t ∈ [x, x + h] x+h

Z

|fµ0 (u)|du.

|fµ (t) − fµ (x)| ≤ x

Applying Hölder’s inequality, we obtain |ψh,µ (x)| ≤ h−1/p

x+h

Z

1/p |f (u) − fµ (u)|p du

x

+ h1−1/p

Z

x+h

1/p |fµ0 (u)|p du .

(3.1.7)

x

It follows that kψh,µ k∞ ≤ h1−1/p kfµ0 kp + h−1/p kf − fµ kp . Applying Lemma 2.4, we obtain (3.1.5). We shall now prove (3.1.6). Let Π = {x0 , x1 , ..., xn } be an arbitrary partition and let K 0 , K 00 be defined by (4.1.2). If j ∈ K 0 , then, applying Hölder’s inequality, we have |ψh,µ (xj+1 ) − ψh,µ (xj )| ≤ |fh (xj+1 ) − fh (xj )| + |fµ (xj+1 ) − fµ (xj )| !1/p Z xj+1 Z xj+1 0 0 1−1/p 0 0 p ≤ (|fh (x)| + |fµ (x)|)dx ≤ h (|fh (x)| + |fµ (x)|) dx . xj

xj

3.1. Auxiliary results

29

Thus, !1/p V

0

X

≡

p

|ψh,µ (xj+1 ) − ψh,µ (xj )|

j∈K 0

≤ h1−1/p (kfh0 kp + kfµ0 kp ).

(3.1.8)

00

Further, let j ∈ K . We have |ψh,µ (xj+1 ) − ψh,µ (xj )| ≤ |ψh,µ (xj )| + |ψh,µ (xj+1 )|. Using (3.1.7), we get !1/p X

00

V ≡

p

|ψh,µ (xj+1 ) − ψh,µ (xj )|

j∈K 00

XZ

−1/p

≤h

XZ

+h

j∈K 00

1−1/p

+h

p

|f (t) − fµ (t)| dt

xj

j∈K 00

−1/p

+h

!1/p

xj+1 +h p

|f (t) − fµ (t)| dt

xj+1

XZ

XZ j∈K 00

xj +h

!1/p |fµ0 (t)|p dt

xj

j∈K 00

1−1/p

!1/p

xj +h

xj+1 +h

!1/p |fµ0 (t)|p dt

.

xj+1

Observe that [xj , xj + h] ⊂ [xj , xj+1 ) for any j ∈ K 00 . Thus, if i < j and i, j ∈ K 00 , then [xi , xi + h] ∩ [xj , xj + h] = ∅ and [ [xj , xj + h] ⊂ [x0 , xn ] (xn = x0 + 1). j∈K 00

Further, if i < j and i, j ∈ K 00 , then xi+1 + h ≤ xj + h < xj+1 . Thus, [xi+1 , xi+1 + h] ∩ [xj+1 , xj+1 + h] = ∅ and [ [xj+1 , xj+1 + h] ⊂ [x0 + h, xn + h]. j∈K 00

30


Taking into account these observations, we obtain V 00 ≤ 2h−1/p (kf − fµ kp + hkfµ0 kp ).

(3.1.9)

Using (3.1.8), (3.1.9), and Lemma 2.4, we have vp (ψh,µ ) ≤ h−1/p hkfh0 kp + 3hkfµ0 kp + 2kf − fµ kp ω(f ; h)p 2 3 ≤ h1−1/p + ω(f ; µ)p + . h h µ Applying (2.1.2), we obtain (3.1.6). The following result is well known (see, e.g., [2], [5, p. 346]). Lemma 3.3. Let 1 ≤ p < q < ∞. Then for any function f ∈ Lp ([0, 1]) and any δ ∈ [0, 1] Z δ dt (3.1.10) t1/q−1/p ω(f ; t)p , ω(f, δ)q ≤ A t 0 where A is an absolute constant. Corollary 3.4. If Jp (f ) < ∞ for some 1 < p < ∞, then for any p < q < ∞ Jq (f ) ≤ AqJp (f ),

(3.1.11)

where A is an absolute constant. We shall use the following Hardy type inequality (see [39]). ∞ Lemma 3.5. Let {λk }∞ k=0 and {αk }k=0 be non-negative sequences. Assume that

λk+1 ≥ dλk

(k = 0, 1, ...),

where d > 1.

Let 1 ≤ p < ∞. Set λ−1 = 0. Then ∞ X k=0

(λk − λk−1 )

∞ X j=k

!p !1/p αj

≤p

d d−1

1/p0

∞ X k=0

!1/p λk αkp

.

3.2. Estimates of L∞ -norm and p-variation

3.2

31

Estimates of L∞-norm and p-variation

Theorem 3.6. Let f ∈ Lp ([0, 1]), 1 < p < ∞. Assume that Z 1 dt t−1/p ω(f ; t)p < ∞. Jp (f ) = t 0 Then f is equivalent to a continuous function f¯ ∈ Vp . Moreover, 1 kf k∞ ≤ A kf kp + 0 Jp (f ) pp and

1 vp (f¯) ≤ A Ωp (f ) + 0 Jp (f ) , pp

(3.2.1)

(3.2.2)

(3.2.3)

where A is an absolute constant. Proof. Let ω ∈ Ω1 be an arbitrary modulus of continuity such that ω(f ; t)p ≤ ω(t),

t ∈ [0, 1],

lim ω(t)/t = ∞,

(3.2.5)

t−1/p−1 ω(t)dt < ∞.

(3.2.6)

t→0+

and Z

(3.2.4)

1

0

Let nk = nk (ω, 1) be defined by (2.3.7). Set ϕk (x) = 2nk

Z

x+2−nk

f (t)dt,

k = 0, 1, ....

(3.2.7)

x

R1 Since n0 = 0, we have ϕ0 (x) = 0 f (t)dt = I. By Lebesgue’s differentiation theorem, for almost all x ∈ [0, 1] f (x) = I +

∞ X (ϕk+1 (x) − ϕk (x)).

(3.2.8)

k=0

Set ψk = ϕk+1 − ϕk . Fix k ≥ 0. Assume that (2.3.9) holds. Applying Lemma 3.2 with h = 2−nk+1 and µ = 2−nk , we obtain kψk k∞ ≤ 2nk+1 /p+1 ω(f ; 2−nk )p

32


and vp (ψk ) ≤ 2nk+1 /p+4 ω(f ; 2−nk )p . Thus, by (3.2.4) and (2.3.9), kψk k∞ ≤ 2nk+1 /p+4 ωnk+1

(3.2.9)

vp (ψk ) ≤ 2nk+1 /p+7 ωnk+1 .

(3.2.10)

and Now we assume that (2.3.10) is true. Then, applying Lemma 3.2 with h = 2−nk and µ = 2−nk+1 , we obtain kψk k∞ ≤ 2nk (1/p−1)+1 2nk+1 ω(f ; 2−nk+1 )p and vp (ψk ) ≤ 2nk (1/p−1)+4 2nk+1 ω(f ; 2−nk+1 )p . Using (3.2.4) and (2.3.10), we have kψk k∞ ≤ 2nk /p+4 ωnk

(3.2.11)

vp (ψk ) ≤ 2nk /p+7 ωnk .

(3.2.12)

and It follows from (3.2.9), (3.2.11), and (3.2.6), that the series (3.2.8) converges uniformly on [0, 1]. Thus, f is equivalent to a continuous 1-periodic function. Moreover, ∞ X kf k∞ ≤ |I| + 32 2nk /p ωnk . (3.2.13) k=0

We may assume that f is continuous. Then by (3.2.8) vp (f ) ≤

∞ X

vp (ψk ).

k=0

Applying (3.2.10) and (3.2.12), we get vp (f ) ≤ 256

∞ X k=0

2nk /p ωnk .

(3.2.14)


33

By Lemma 3.1 with q = 1, β = 1/p, γ = 1, ∞ X

2nk /p ωnk ≤ 2ω0 +

k=0

8 pp0

Z

1

t−1/p−1 ω(t)dt.

Using (3.2.13), (3.2.14), and (3.2.15), we obtain 1 kf k∞ ≤ |I| + A ω(1) + 0 Dp,ω pp and

1 vp (f ) ≤ A ω(1) + 0 Dp,ω , pp

where Z Dp,ω =

(3.2.15)

0

(3.2.16)

(3.2.17)

1

t−1/p−1 ω(t)dt

0

and A is an absolute constant. Here ω is any modulus of continuity satisfying (3.2.4) – (3.2.6). If ω(f ; t)p satisfies (3.2.5), then we take ω(t) = ω(f ; t)p . Otherwise, we take ω(t) = ω(f ; t)p + εtγ , where 1/p < γ < 1 and ε is an arbitrary positive number. Clearly, (3.2.4) – (3.2.6) are satisfied. Let ε → 0. Then (3.2.16), (3.2.17), and (2.1.6) imply (3.2.2) and (3.2.3). Remark 3.7. The condition (3.2.1) can not be improved. Moreover, it was shown by Ul’yanov [74] that if ω ∈ Ω, 1 < p < ∞, and Z 1 dt t−1/p ω(t) = ∞, t 0 then there exists an essentially unbounded function f ∈ Lp ([0, 1]) such that ω(f ; t)p ≤ ω(t) (see also Theorem 3.12 and Theorem 3.18 below). Remark 3.8. The term Ωp (f ) on the right-hand side of (3.2.3) can not be omitted. Indeed, let f (x) = sin(2πx). Then for all p ≥ 1 we have ω(f ; t)p ≤ 2πt (t ∈ [0, 1]), and Z 1 1 dt 2π t−1/p ω(f ; t)p ≤ . 0 pp 0 t p Thus, the second term on the right-hand side of (3.2.3) tends to 0 as p → ∞. On the other hand, vp (f ) ≥ 21/p (p ≥ 1).

34

Chapter 3. Integral smoothness and p-variation Set for f ∈ Lp ([0, 1]) and δ ∈ [0, 1] 1/p Z 1 (2) p |f (x + h) − 2f (x) + f (x − h)| dx ω (f ; δ)p = sup . 0≤h≤δ

0

Terehin [71] proved that if f ∈ Lp ([0, 1]), 1 ≤ p < ∞, and Z 1 dt t−1/p ω (2) (f ; t)p < ∞, Jp(2) (f ) = t 0

(3.2.18)

then f is equivalent to a continuous 1-periodic function f¯ ∈ Vp and vp (f¯) ≤ AJp(2) (f ),

(3.2.19)

where A is an absolute constant. We have the following improvement of the previous estimate. Corollary 3.9. Let f ∈ Lp ([0, 1]), 1 < p < ∞ and assume that (3.2.18) holds. Then f is equivalent to a continuous 1-periodic function f¯ ∈ Vp and 1 (2) ¯ vp (f ) ≤ A Ωp (f ) + Jp (f ) . (3.2.20) p Proof. By Marchaud’s inequality ([14], p. 47) Z 1 (2) ω (f ; u)p ω(f ; t)p ≤ ct du + Ωp (f ) u2 t

(3.2.21)

for all 0 < t ≤ 1, where c is an absolute constant. Applying (3.2.21) and Fubini’s theorem, we get Z 1 Z 1 (2) ω (f ; u)p 0 −1/p Jp (f ) ≤ c p Ωp (f ) + t dudt u2 0 t Z 1 du = cp0 Ωp (f ) + u−1/p ω (2) (f ; u)p . u 0 This estimate and (3.2.3) imply (3.2.20). As above (see Remark 3.8), the term Ωp (f ) on the right-hand side of (3.2.20) can not be omitted. However, we have that Z 1 dt Ωp (f ) ≤ ω (2) (f ; 1/2)p ≤ 2 t−1/p ω (2) (f ; t)p . (3.2.22) t 1/2


35

Indeed, by the periodicity of f , p 1/p Z 1 Ωp (f ) ≤ 2 f (t)dt dx f (x) − 0 0 p !1/p Z 1/2 Z 1 =2 [f (x + u) + f (x − u)]du dx f (x) − 0 0 !1/p Z Z Z

1

1

1/2

≤ 21/p

|f (x + u) + f (x − u) − 2f (x)|p dx

du 0

0

≤ ω (2) (f ; 1/2)p ≤ 2

Z

1

t−1/p ω (2) (f ; t)p

1/2

dt . t

Inequality (3.2.19) follows from (3.2.20) and (3.2.22). We emphasize that, in comparison with (3.2.19), the right-hand side of (3.2.20) contains the factor 1/p. This factor may play an essential role. We shall consider trigonometric polynomials n

Tn (x) =

a0 X + (ak cos 2πkx + bk sin 2πkx). 2 k=1

(3.2.23)

Terehin [71] observed that (3.2.19) yields that for every trigonometric polynomial Tn of degree n and any 1 ≤ p < ∞ vp (Tn ) ≤ Apn1/p kTn kp ,

(3.2.24)

where A is an absolute constant. Oskolkov [53, 54] proved that the coefficient p on the right-hand side of (3.2.24) can be omitted. That is, for any trigonometric polynomial of degree n and any 1 ≤ p < ∞ vp (Tn ) ≤ An1/p kTn kp ,

(3.2.25)

where A is an absolute constant. Oskolkov’s proof was based on the use of interpolation methods. We note that (3.2.25) can be obtained from (3.2.20) or, more directly, from (3.2.3). Indeed, we have ω(Tn ; t)p ≤ min(tkTn0 kp , 2kTn kp ),

t ∈ [0, 1].

36


Thus, 1

Z

t−1/p ω(Tn ; t)p

Jp (Tn ) = 0

Z

dt ≤ kTn0 kp t

Z

1/n

t−1/p dt

0

1

+ 2kTn kp

t−1/p−1 dt ≤ p0 kTn0 kp n1/p−1 + 2pn1/p kTn kp .

1/n

By the Bernstein inequality [14, p. 97], kTn0 kp ≤ 2πnkTn kp , and we obtain Jp (Tn ) ≤ 2πpp0 n1/p kTn kp . We have also Ωp (Tn ) ≤ 2kTn kp . Applying these estimates and (3.2.3), we obtain (3.2.25). We give one more observation concerning the behaviour of the right-hand side of (3.2.3) as p → ∞. It is easy to see that if f ∈ Vq for some q ≥ 1, then lim vp (f ) = osc(f )

p→∞

(3.2.26)

where osc(f ) is the oscillation of f on [0, 1]. The essential oscillation ess osc(f ) of a measurable 1-periodic function f is defined as the difference ess sup f (x) − ess inf f (x). x∈[0,1]

x∈[0,1]

Proposition 3.10. Let f be a 1-periodic measurable function. Assume that Jp0 (f ) < ∞ for some 1 < p0 < ∞. Then lim

p→∞

Jp (f ) ≤ ess osc(f ). p

(3.2.27)

Proof. Set ω0 = ess osc(f ). Let p > p0 . Then ω(f ; t)p ≤ ω0 and for any 0 < h < 1 we have Z Z dt 1 1 −1/p ω0 1 −1/p−1 t ω(f ; t)p ≤ t dt ≤ ω0 h−1/p . p h t p h Further, applying Lemma 3.3 and Fubini’s theorem, we obtain Z Z Z 1 h −1/p dt A h −1/p t 1/p−1/p0 du dt t ω(f ; t)p ≤ t u ω(f ; t)p0 p 0 t p 0 u t 0 Z h du ≤A u−1/p0 ω(f ; u)p0 . u 0


37

Let ε > 0. Then there exists h > 0 such that Z h du A u−1/p0 ω(f ; u)p0 < ε. u 0 Thus, we have 1 Jp (f ) < ω0 h−1/p + ε p and therefore lim

p→∞

1 Jp (f ) ≤ ω0 + ε. p

This implies (3.2.27). Applying (3.2.26) and (3.2.27), we see that the behaviour of the righthand side of (3.2.3) as p → ∞ agrees with that of the left-hand side. Namely, the left-hand side of (3.2.3) tends to ω0 = ess osc(f ) and the upper limit of the right-hand side does not exceed Aω0 . Note that the right-hand side of (1.0.5) may tend to infinity as p → ∞. The results given above show that the constant factor A/(pp0 ) in (3.2.3) has an optimal order as p → ∞. We observe now that its order is optimal as p → 1, too. Indeed, assume that f ∈ Wq1 for some q > 1. Then, by (2.1.3), Z 1 Jp (f ) ≤ ||f 0 ||p t−1/p dt = p0 ||f 0 ||p for any 1 < p ≤ q. 0

Thus, lim

p→1

Jp (f ) ≤ ||f 0 ||1 = v(f ). p0

Further, for the first term on the right-hand side of (3.2.3) we have lim Ωp (f ) ≤ ω(f ; 1)1

p→1

and for the left-hand side lim vp (f ) = v(f ).

p→1

It follows that the factor A/(pp0 ) in (3.2.3) can not be replaced by any factor α(p) such that limp→1 p0 α(p) = 0. Indeed, otherwise the inequality v(f ) ≤ Aω(f ; 1)1 would be true for any f ∈ Wq1 .

38


3.3

Estimates of the modulus of p-continuity

Let C denote the class of all continuous 1-periodic functions on R. The modulus of continuity of a function f ∈ C is defined by ω(f ; δ) = sup |f (x) − f (y)|,

0 ≤ δ ≤ 1.

|x−y|≤δ

Let f ∈ Lp ([0, 1]) (1 < p < ∞) and let 1

Z

t−1/p ω(f ; t)p

Jp (f ) = 0

dt < ∞. t

(3.3.1)

Then we may assume that f ∈ C. Moreover, δ

Z

t−1/p ω(f ; t)p

ω(f ; δ) ≤ A 0

dt , t

(3.3.2)

where A is an absolute constant (see [2],[56]). It was proved in [2] that (3.3.2) is sharp for any order of the modulus of continuity ω(f ; t)p . Let ω ∈ Ω be a modulus of continuity and let 1 ≤ p < ∞. Denote by Hpω the class of all functions f ∈ Lp ([0, 1]) such that ω(f ; t)p ≤ ω(t),

t ∈ [0, 1].

(3.3.3)

The result obtained in [2] can be formulated in the following equivalent way. Let 1 < p < ∞. Then there exist positive constants c and c0 such that for any modulus of continuity ω and any δ ∈ (0, 1] c0 ξp,ω (δ) ≤ sup ω(f ; δ) ≤ cξp,ω (δ), f ∈Hpω

where Z ξp,ω (δ) =

δ

t−1/p ω(t)

0

dt . t

Thus, for each separate value of δ it is impossible to strengthen (3.3.2). Namely, for any δ ∈ (0, 1] there exists a function fδ ∈ Hpω such that ω(fδ ; δ) ≥ c0 ξp,ω (δ).

3.3. Estimates of the modulus of p-continuity

39

However, a function f ∈ Hpω fitting all values δ may not exist. Moreover, it was proved in [31] that the following refinement of (3.3.2) is true: Z δ

1

1/p Z δ dt t−1/p ω(f ; t)p t−p ω(f ; t)p dt ≤ cδ 1/p−1 t 0

(3.3.4)

for any δ ∈ (0, 1]. In particular, if ω(t) = t and f ∈ Hpω , then by (3.3.4) Z

1

t−p ω(f ; t)p dt < ∞.

0

At the same time, ξp,ω (δ) = p0 δ 1−1/p , and the latter integral would diverge if the inequality ω(f ; δ) ≥ c0 ξp,ω (δ) was true for all δ ∈ [0, 1]. In this section we study estimates of the modulus of p-continuity (1 < p < ∞). As we have already mentioned above, such estimates were first obtained by Terehin (see (1.0.10)). First we shall show that the constant coefficients in (1.0.10) can be improved. Theorem 3.11. Let f ∈ Lp ([0, 1]) (1 < p < ∞) and assume that Jp (f ) < ∞. Then f can be modified on a set of measure zero so as to become continuous and Z δ 1 dt ω1−1/p (f ; δ) ≤ A δ −1/p ω(f ; δ)p + 0 t−1/p ω(f ; t)p , (3.3.5) pp 0 t for any δ ∈ (0, 1], where A is an absolute constant. Proof. By Theorem 3.6, we may assume that f is continuous. Let 0 < δ ≤ 1. We have (see (2.1.7)) ω1−1/p (f ; δ) ≤ ω1−1/p (fδ ; δ) + vp (f − fδ ).

(3.3.6)

By (2.2.8) and (2.1.9), ω1−1/p (fδ ; δ) ≤ δ 1−1/p ||fδ0 ||p ≤ δ −1/p ω(f ; δ)p .

(3.3.7)

Further, by Theorem 3.6, 1 vp (f − fδ ) ≤ A Ωp (f − fδ ) + 0 Jp (f − fδ ) . pp

(3.3.8)

40


First, by (2.1.8) Ωp (f − fδ ) ≤ 2||f − fδ ||p ≤ 2ω(f ; δ)p

(3.3.9)

and ω(f − fδ ; t)p ≤ 2||f − fδ ||p ≤ 2ω(f ; δ)p

(0 < t ≤ 1).

(3.3.10)

Besides, we have ω(f − fδ ; t)p ≤ ω(f ; t)p + ω(fδ ; t)p . It is easy to see that ω(fδ ; δ)p ≤ ω(f ; t)p . Thus, ω(f − fδ ; t)p ≤ 2ω(f ; t)p . (0 < t ≤ 1). Using estimates (3.3.10) and (3.3.11), we get Z δ dt . Jp (f − fδ ) ≤ 2 pδ −1/p ω(f ; δ)p + t−1/p ω(f ; t)p t 0

(3.3.11)

(3.3.12)

Applying (3.3.6) – (3.3.9) and (3.3.12), we obtain (3.3.5). It is clear that ω(f ; δ) ≤ ω1−1/p (f ; δ) for any 1 < p < ∞. As we have observed, the estimate (3.3.2) for ω(f ; δ) can be strengthened (see (3.3.4)). Now we shall show that, in contrast to (3.3.2), the estimate (3.3.5) is sharp in the following strong sense. Theorem 3.12. There exists a constant A > 0 such that for any 1 < p < ∞ and any ω ∈ Ω satisfying the condition Z 1 dt (3.3.13) t−1/p ω(t) < ∞, t 0 there is a continuous function f ∈ Hpω for which the inequality Z δ 1 dt ω1−1/p (f ; δ) ≥ A δ −1/p ω(δ) + 0 t−1/p ω(t) pp 0 t

(3.3.14)

holds for all δ ∈ (0, 1]. Proof. First we assume that (2.3.6) holds. Let nk = nk (ω) (see (2.3.7)). We set εn = ωnk if n = nk for some k ∈ N and εn = 0 otherwise (n ∈ N). Then ∞ X n=ν

εn ≤ 2ων

and

ν X n=1

2n εn ≤ 2ν+1 ων

(3.3.15)

3.3. Estimates of the modulus of p-continuity

41

for any ν ∈ N. Indeed, let j be the least natural number such that nj ≥ ν. Then by the first inequality in (2.3.8) ∞ X

εn =

n=ν

∞ X

ωnk ≤ 2ωnj ≤ 2ων .

k=j

Similarly, denoting by s the greatest natural number such that ns ≤ ν and applying the second inequality in (2.3.8), we get ν X

2n εn =

n=1

s X

2nk ωnk ≤ 2ns +1 ωns ≤ 2ν+1 ων .

k=1

Let ε(t) = εn for t ∈ (2−n−1 , 2−n ] (n ∈ N, n ≥ 2), and ε(t) = 0 for 1/4 < t ≤ 1. Set Z 1/2 f (x) = ε(t)t−1−1/p dt (3.3.16) |x|

if |x| ≤ 1/2, and extend f to the real line with period 1 (this construction was used before in [31, 32]). We first estimate ω(f ; δ)p . Since f is 1-periodic, even and monotonically decreasing on [0, 1/2], we easily get that for any 0 < h ≤ 1/4, Z 1 Z 1/2 |f (x) − f (x + h)|p dx ≤ 4 |f (x) − f (x + h)|p dx 0 0 p ∞ ∞ Z 2−n Z x+h X X Jn (h). ε(t)t−1−1/p dt dx ≡ 4 =4 2−n−1

n=2

Let 2

−ν−1

−ν

0 and a sequence of continuous 1-periodic functions {fn } such that kfn k∞ ≤ Jp (˜ ω1 ), ω(fn ; t)p ≤ ω ˜ n (t) ≤ ω(t) + tγ

(0 ≤ t ≤ 1),

(3.3.22)

dt ω(t) , t

(3.3.23)

and for 2−s ≤ δ < 2−s+1 vp (fn ; Πs ) ≥ A δ

−1/p

1 ω(δ) + 0 pp

Z

δ −1/p

t 0

where Πs is the partition of [0, 1] by points 2−s j, (j = 0, 1, ..., 2s ). The functions fn are equibounded and equicontinuous (see (3.3.22) and (3.3.2)). Thus, by the Ascoli–Arzelà theorem, there exists a subsequence {fnk } that converges uniformly to some function f ∈ C. It follows from (3.3.22) that f ∈ Hpω . Furthermore, (3.3.23) implies (3.3.14) for all δ ∈ [0, 1]. s Remark 3.13. For 1 ≤ p, q < ∞ and 0 < s < 1, the Besov space Bp,q p consists of all functions f ∈ L ([0, 1]) such that

Z kf kbsp,q =

1

(t−s ω(f ; t)p )q

0

dt t

1/q < ∞.

For 1 ≤ p < ∞, the dyadic p-variation of a 1-periodic function f is given by  vp,k (f ) = 

k −1 2X

1/p |f ((j + 1)2−k ) − f (j2−k )|p 

(k ≥ 0).

j=0

It was shown in [68] that for 1 < p, q < ∞ and 1/p < s < 1, there are s constants c0 , c00 > 0 such that for any function f ∈ Bp,q , there holds 0

c kf kbsp,q ≤

∞ X

!1/q k(s−1/p)q

2

vp,k (f )

q

≤ c00 kf kbsp,q .

(3.3.24)

k=0

The result (3.3.24) has applications in probability. The left inequality of (3.3.24) follows easily from an alternative description of Besov spaces due to Ciesielski et al. (see [68] for references). The proof of the right inequality given in [68] is more complicated. We observe

46


that the right inequality can be obtained directly from (3.3.5). Indeed, since vp,k (f ) ≤ ω1−1/p (f ; 2−k ), we have ∞ X

2k(s−1/p)q vp,k (f )q ≤

k=0

∞ X

(2k(s−1/p) ω1−1/p (f ; 2−k ))q

k=0 ∞ X

≤ A

k(s−1/p)q

2

k=0

∞ X

!q j/p

2

−j

ω(f ; 2 )p

.

j=k

By the previous estimates and Hardy’s inequality (see, e.g., [39]) we have ∞ X

2k(s−1/p)q vp,k (f )q ≤ cp,q,s

k=0

3.4

∞ X

2ksq ω(f ; 2−k )qp ≤ cp,q,s kf kqbsp,q .

k=0

The classes Vpα

In this section we obtain sharp estimates of vp,α (f ) (see (1.0.12)) in terms of the modulus of continuity ω(f ; δ)p . Theorem 3.14. Let 1 < p < ∞ and 0 < α < 1/p0 . Let f ∈ Lp ([0, 1]) and assume that 1/p Z 1 dt < ∞. (3.4.1) t−αp−1 ω(f ; t)pp Kp,α (f ) = t 0 Then f is equivalent to a continuous function f¯ ∈ Vpα and vp,α (f¯) ≤ cp,α Kp,α (f ),

(3.4.2)

where 0

cp,α = Aα−1/p (1/p0 − α)1/p

(3.4.3)

and A is an absolute constant. Proof. By Theorem 3.6, we may assume that f is continuous (indeed, (3.4.1) implies (3.2.1)). Set ω(δ) = ω(f ; δ)p . As in the proof of Theorem 3.6, we may suppose that ω satisfies (2.3.6). Let {nk } be defined by (2.3.7). Fix a partition Π = {x0 , x1 , ..., xn } (xn = x0 + 1). Let σk = {j : 2−nk+1 < xj+1 − xj ≤ 2−nk } (k = 0, 1, ...).

3.4. The classes Vpα

47

For any function ϕ, set X |ϕ(xj+1 ) − ϕ(xj )|p (xj+1 − xj )αp j∈σ

Rk (ϕ) =

!1/p

k

and

!1/p Sk (ϕ) =

X

p

|ϕ(xj+1 ) − ϕ(xj )|

.

j∈σk

For an integer k ≥ 0 we set µ(k) = k if (2.3.9) holds and µ(k) = k + 1 if (2.3.9) does not hold (in the latter case, (2.3.10) holds). Let gk (x) = 2nµ(k)

Z

2

−nµ(k)

f (x + t) dt. 0

Applying Hölder’s inequality, we obtain Z p xj+1 |gk (xj+1 ) − gk (xj )|p 0 −αp g (t)dt = (x − x ) j+1 j k xj (xj+1 − xj )αp Z xj+1 |gk0 (t)|p dt ≤ (xj+1 − xj )p−1−αp xj Z xj+1 |gk0 (t)|p dt ≤ 2−nk (p−1−αp) xj

for any j ∈ σk . Thus, by (2.1.9), 0

0

Rk (gk ) ≤ 2−nk (1/p −α) ||gk0 ||p ≤ 2−nk (1/p −α) 2nµ(k) ωnµ(k) . If (2.3.9) holds, then µ(k) = k and Rk (gk ) ≤ 2nk (α+1/p) ωnk . If (2.3.9) does not hold, then (2.3.10) holds. In this case µ(k) = k + 1 and by (2.3.10) Rk (gk ) ≤ 2nk (α+1/p)+3 ωnk . We have also Rk (f − gk ) ≤ 2nk+1 α Sk (f − gk ).

48


Using these estimates and denoting γk = Sk (f − gk ), we obtain ∞ X

vp,α (f ; Π) ≤

!1/p p

Rν (gν )

+

ν=0 ∞ X

≤ 8

∞ X

!1/p Rν (f − gν )

ν=0 !1/p nν (αp+1)

2

ωnp ν

p

+

ν=0

∞ X

!1/p 2nν+1 αp γνp

. (3.4.4)

ν=0

We estimate the latter sum. Applying Abel’s transform, we have ∞ X

2nν+1 αp γνp =

ν=0

∞ X

γkp +

∞ ∞ X X (2nν+1 αp − 2nν αp ) γkp . ν=0

k=0

(3.4.5)

k=ν

Further, reasoning as in Theorem 3.6 (see (3.2.10) and (3.2.12)), we obtain ∞ X

γk ≤ vp (f − gk ) ≤ A

2nj /p ωnj .

j=µ(k)

If (2.3.9) holds, then µ(k) = k and ωnk ≤ 8ωnk+1 . If (2.3.9) does not hold, then µ(k) = k + 1. Thus, ∞ X

γk ≤ 8A

2nj /p ωnj

(k = 0, 1, ...).

(3.4.6)

j=k+1

On the other hand, γk ≤ Sk (f ) + Sk (gk ). By (2.2.8) and (2.1.9), 0

0

Sk (gk ) ≤ 2−nk /p kgk0 kp ≤ 2−nk /p ω nµ(k) , and we obtain, as above Sk (gk ) ≤ 2nk /p+3 ωnk .

(3.4.7)

Further, applying (3.3.5), we have ∞ X k=m

X

Sk (f )p =

|f (xj+1 ) − f (xj )|p ≤ ω1−1/p (f ; 2−nm )p

S j∈ ∞ k=m σk

" ≤A

p

nm /p

2

ωnm

1 + 0 pp

Z

2−nm −1/p

t 0

dt ω(t) t

#p .


49

Considering cases (2.3.9) and (2.3.10), we obtain that Z 2−nk 1 dt t−1/p ω(t) ≤ 2nk /p+3 ωnk + 2nk+1 /p+3 ωnk+1 . 0 −n pp 2 k+1 t Thus, ∞ X

p

Sk (f ) ≤

!p

∞ X

A

k=m

nk /p

2

ωnk

(m = 0, 1, ...).

k=m

Using this inequality, (3.4.6), and (3.4.7), we obtain ∞ X

γkp

γνp

≤

+2

∞ X

p

∞ X

p

Sk (gk ) +

k=ν+1

k=ν ∞ X

0

≤

A

! Sk (f )

p

k=ν+1

!p nk /p

2

ωnk

.

k=ν+1

Thus, applying (3.4.5), we get ∞ X

+

" 2nν+1 αp γνp

ν=0 ∞ X

nν αp

(2

p

≤A

nν−1 αp

−2

ν=1

)

∞ X

!p nk /p

2

k=0 ∞ X

ωnk

+ !p #

nk /p

2

ωnk

.

(3.4.8)

k=ν

First we assume that αp < 4. Set λk = 2nk αp . Then λk+1 ≥ 2αp λk . Applying Lemma 3.5 to the right-hand side of (3.4.8), we get !1/p !1/p ∞ ∞ X X nν+1 αp p nν (αp+1) p 2 γν 2 ωnν ≤ Ap,α , ν=0

ν=0

where Ap,α = Ap

2αp 2αp − 1

1/p0

0

0

≤ A1 p(αp)−1/p = A1 p1/p α−1/p ≤ 2A1 α−1/p

0

(A1 is an absolute constant). Let now αp ≥ 4. Then, by Hölder’s inequality !p−1 !p ∞ ∞ ∞ X X X nk /p nk (1+αp/2) p −nk αp0 /2 2 ωnk ≤ 2 ωnk 2 k=ν

k=ν

k=ν

p−1 ∞ X 8 2−nν αp/2 2nk (1+αp/2) ωnp k ≤ α k=ν

50


(we have used the condition αp0 < 1). Thus, applying (3.4.8), changing the order of summations, and taking into account that αp ≥ 4, we obtain !1/p !1/p ∞ ∞ ∞ X X X A nν αp/2 nk (1+αp/2) p nν+1 αp p 2 γν 2 2 ωnk ≤ 1/p0 α ν=0 ν=0 k=ν !1/p ∞ X A1 nν (αp+1) p ≤ 1/p0 2 ωnν . α ν=0 These estimates and (3.4.4) yield that ∞ X

A vp,α (f ; Π) ≤ 1/p0 α

!1/p nν (αp+1)

2

ωnp ν

.

(3.4.9)

ν=0

Applying Lemma 3.1 with q = p, β = αp + 1 and γ = 1, we have !1/p ∞ X nν (αp+1) p 2 ωnν ≤ 8[ω(1) + (αp + 1)1/p (1/p0 − α)1/p Kp,α (f )]. ν=0

Further, (αp + 1)1/p ≤ p1/p ≤ 2, and by (2.1.2) Z Z 1 ω(1)p (1/p0 − α) 1 p(1−α)−2 ω(1)p dt t dt = . (1/p0 − α) t−αp−1 ω(t)p ≥ p t 2 p2p 0 0 Thus, ∞ X

!1/p nν (αp+1)

2

ωnp ν

≤ 48(1/p0 − α)1/p Kp,α (f ).

ν=0

From here and (3.4.9) it follows that 0

vp,α (f ; Π) ≤ Aα−1/p (1/p0 − α)1/p Kp,α (f ), where A is an absolute constant. Remark 3.15. Assume that f ∈ Wp1 (1 < p < ∞). It was proved in [11] (see also [13]) that in this case Z 1 1/p 1/p dt 1 lim (1 − s)1/p [t−s ω(f ; t)p ]p = kf 0 kp . s→1− t p 0 Thus, if α → 1/p0 , the right hand side of (3.4.2) tends to cp kf 0 kp . This agrees with Theorem 2.2 of F. Riesz. Besides, it shows that the order of the constant cp,α in (3.4.2) as α → 1/p0 is optimal.


51

Remark 3.16. We observe that the order of the constant (3.4.3) as α → 0 also is optimal. Indeed, let 1 < p < ∞, 0 < α < 1/(2p0 ). Set f (x) = | sin πx|2α . Then vp,α (f ) ≥ vp (f ) ≥ 1. Further, it is easy to see that ω(f ; δ)p ≤ cp αδ 2α+1/p . Thus Z 1 1/p 0 tαp−1 dt α−1/p Kp,α (f ) ≤ cp α1/p ≤ cp . 0

This implies that the constant cp,α in (3.4.2) can not replaced by c˜p,α such 0 that limα→0 c˜p,α α1/p = 0. Now we shall show that for 0 < α < 1/p0 the condition (3.4.1) is sharp. Theorem 3.17. Let 1 < p < ∞ and 0 < α < 1/p0 . Assume that ω ∈ Ω is a modulus of continuity such that Z 1 dt (3.4.10) t−αp−1 ω(t)p = ∞. t 0 Then there exists a function f ∈ Hpω which is not equivalent to a function in Vpα . Proof. The condition (3.4.10) implies (2.3.6). We define the function f as in Theorem 3.12 (see (3.3.16)). Then we have the estimate (3.3.18). Let n ∈ N and ξk = 2−n+k−1 (k = 0, 1, ..., n). Then !p Z 2−n+k n−1 n−1 X X |f (ξk+1 ) − f (ξk )|p (n+1−k)αp p −1−1/p = 2 εn−k t dt (ξk+1 − ξk )αp 2−n+k−1 k=0 k=0 ≥ 2−p

n X

2j(αp+1) εpj .

j=1

This implies that vp,α (f ) ≥ 2−p

∞ X j=1

2j(αp+1) εpj = 2−p

∞ X

2nk (αp+1) ωnp k .

k=1

It remains to show that the series at the right-hand side diverges. If (2.3.9) holds, then Z 2−nk dt 8p 2nk+1 (αp+1) ωnp k+1 . t−(αp+1) ω(t)p ≤ −nk+1 t αp + 1 2

52

Chapter 3. Integral smoothness and p-variation If (2.3.10) holds, then Z

2−nk

t−(αp+1) ω(t)p

2−nk+1

dt 8p ≤ 2nk (αp+1) ωnp k . t p(1 − α) − 1

These estimates and (3.4.10) yield that ∞ X

2nk (αp+1) ωnp k = ∞.

k=1

3.5

On classes Up

It was recently shown in [10, 9] that the classes Vp play an important role in problems of boundedness of superposition operators. In [10, 9, 8] there were also studied classes Up defined in terms of local oscillations. We consider one counterexample concerning these classes. For any 1-periodic measurable and almost everywhere finite function f we denote by ω(f ; x, δ) the essential oscillation of f on the interval (x − δ, x + δ), that is ω(f ; x, δ) = ess sup f (y) − ess inf f (y), |y−x| 1 the inclusion (3.5.4) is strict. Indeed, if f0 is the 1-periodic extension of log(1/|t|), |t| ∈ (0, 1/2], to the real line, then f0 ∈ Lip(1/p; p) for any p > 1; however, f0 ∈ / Up since f0 is unbounded. With the use of wavelet decompositions of Besov spaces, it was shown in [9] that there exists a bounded function in Lip(1/p; p) \ Up . We observe that the latter result can be obtained from the following theorem proved by direct methods in [30]. Theorem 3.18. Let 1 0. Since (3.5.5) holds with ω(t) = t1/p , there is a bounded function f ∈ Lip(1/p; p) such that ω(f ; x, δ) ≥ 1 for any x ∈ R and any δ > 0. Clearly, f∈ / Up .

Chapter 4 Fractional smoothness of functions via p-variation Let 1 < p < ∞, recall the definition (1.0.13) of the class Vpα (0 ≤ α ≤ 1/p0 ). 1/p0 For α = 1/p0 , Theorem 2.2 states that a function f ∈ Vp if and only if f ∈ Wp1 . For α = 0, we clearly have Vp0 = Vp . Thus, Vpα (0 < α < 1/p0 ) form a scale of spaces of fractional smoothness between Vp and Wp1 . Another characterization of Wp1 is given by moduli of p-continuity. In0 deed, f ∈ Wp1 if and only if ω1−1/p (f ; δ) = O(δ 1/p ) (see Chapter 2). Obviously, if f ∈ Vpα (0 < α ≤ 1/p0 ), then ω1−1/p (f ; δ) = O(δ α ).

(4.0.1)

However, for 0 < α < 1/p0 , the condition (4.0.1) does not imply that f ∈ Vpα . On the other hand, it is in general impossible to improve (4.0.1). The main objectives of this chapter are twofold: (i) to obtain sharp relations between vp,α (f ) and moduli of p-continuity; (ii) to study limits in the scales generated by vp,α (f ) and ω1−1/p (f ; δ).

4.1

Approximation with Steklov averages

We shall prove that the modulus of p-continuity “controls” the error of approximation by Steklov averages in Vp . 55

56

Chapter 4. Fractional smoothness via p-variation

Lemma 4.1. Let 1 < p < ∞ and f ∈ Vp . Then vp (f − fh ) ≤ 6ω1−1/p (f ; h).

(4.1.1)

Proof. Let Π = {x0 , x1 , ..., xn } be any partition and set K 0 = {j : xj+1 − xj ≤ h},

K 00 = {0, 1, ..., n − 1} \ K 0

(4.1.2)

Set also gh = f − fh and !1/p 0

V =

X

p

|gh (xj+1 ) − gh (xj )|

j∈K 0

and

!1/p 00

V =

X

p

|gh (xj+1 ) − gh (xj )|

.

j∈K 00

Then vp (gh ; Π) ≤ V 0 + V 00 . By Minkowski’s inequality !1/p V

0

≤

X

p

|f (xj+1 ) − f (xj )|

j∈K 0

!1/p +

X

p

|fh (xj+1 ) − fh (xj )|

j∈K 0

≤ ω1−1/p (f ; h) + ω1−1/p (fh ; h). Using (2.2.11), we get V 0 ≤ 2ω1−1/p (f ; h).

(4.1.3)

00

We now estimate V . We have p X Z h 00 p −p (V ) = h [f (xj+1 ) − f (xj+1 + t) − f (xj ) + f (xj + t)]dt . j∈K 00

0

Applying the trivial inequality |a+b|p ≤ 2p (|a|p +|b|p ) and Hölder’s inequality, we obtain Z h"X 00 p p −1 (V ) ≤ 2 h |f (xj+1 + t) − f (xj+1 )|p + 0

j∈K 00

# +

X j∈K 00

p

|f (xj + t) − f (xj )|

dt.

4.1. Approximation with Steklov averages

57

For t ∈ [0, h] and j ∈ K 00 we have [xj , xj + t] ⊂ [xj , xj+1 ), and hence [xj , xj + t] ∩ [xi , xi + t] = ∅ for i, j ∈ K 00 , i 6= j. Moreover, since j ≤ n − 1 and j ∈ K 00 , we have that xj + t ≤ xj+1 ≤ xn . Thus, [ [xj , xj + t] ⊂ [x0 , xn ], j∈K 00

and X

|f (xj + t) − f (xj )|p ≤ ω1−1/p (f ; h)p

j∈K 00

for each t ∈ [0, h]. Furthermore, if i, j ∈ K 00 and i < j, then xi+1 + t ≤ xj + t ≤ xj+1 . Whence, [xi+1 , xi+1 + t] ∩ [xj+1 , xj+1 + t] = ∅, i < j, and [ [xj+1 , xj+1 + t] ⊂ [x0 + t, xn + t]. j∈K 00

Thus, X

|f (xj+1 + t) − f (xj+1 )|p ≤ ω1−1/p (f ; h)p

j∈K 00

for each t ∈ [0, h]. It follows that V 00 ≤ 21+1/p ω1−1/p (f ; h).

(4.1.4)

By (4.1.3) and (4.1.4) we obtain vp (f − fh ) ≤ 6ω1−1/p (f ; h). This completes the proof. Remark 4.2. Applying Lemma 4.1, we can show that the Peetre K-functional 0 K(f, t; Vp , Wp1 ) is equivalent to ω1−1/p (f ; tp ). Set kf kVp = |f (0)| + vp (f ) for f ∈ Vp . It is simple to show that k · kVp is a norm on Vp and that Vp is a Banach space with respect to this norm. As in [14, p.172], we define the K-functional for the pair (Vp , Wp1 ) by the equality K(f, t; Vp , Wp1 ) = inf 1 (kf − gkVp + tkg 0 kp ). g∈Wp

We emphasize that the second term on the right-hand side is only a seminorm on Wp1 . We shall now prove that 0

0

ω1−1/p (f ; tp ) ≤ K(f, t; Vp , Wp1 ) ≤ 8ω1−1/p (f ; tp ).

(4.1.5)

58

Chapter 4. Fractional smoothness via p-variation 0

Fix an arbitrary t ∈ (0, 1] and set h = tp . Let g = fh be the Steklov average (2.1.7), then g ∈ Wp1 . By (2.2.12) and (4.1.1), we have that 0

|f (0) − g(0)| + vp (f − g) + h1/p kg 0 kp ≤ 8ω1−1/p (f ; h). 0

Substituting h = tp above yields 0

kf − gkVp + tkg 0 kp ≤ 8ω1−1/p (f ; tp ), and therefore, 0

K(f, t; Vp , Wp1 ) ≤ 8ω1−1/p (f ; tp ). On the other hand, for any g ∈ Wp1 , we have by (2.2.8) that 0

0

0

ω1−1/p (f ; tp ) ≤ ω1−1/p (f − g; tp ) + ω1−1/p (g; tp ) ≤ vp (f − g) + tkg 0 kp . Taking infimum over all g ∈ Wp1 , we obtain that 0

ω1−1/p (f ; tp ) ≤ K(f, t; Vp , Wp1 ). Thus, (4.1.5) is proved.

4.2

Limiting relations

Let f ∈ Lp ([0, 1]) (1 < p < ∞). It was proved in [11] that if Z sup (1 − s) 0 0 such that for 0 < t < δ kf 0 kpp − ε
0 is arbitrary, the proof of (i) is complete. Let now f ∈ Cp . For any 0 < h < 1, let fh be the Steklov average of f given by (2.1.7). Then fh ∈ Wp1 and fh0 (x) = [f (x + h) − f (x)]/h a.e. Applying (4.2.3) to the function fh and using (2.2.11), we have Z 1 1 0 p dt kfh kp = lim0 (1/p0 − s) [t−s ω1−1/p (fh ; t)]p s→1/p − p t 0 Z 1 dt [t−s ω1−1/p (f ; t)]p = C < ∞. ≤ lim (1/p0 − s) s→1/p0 − t 0 On the other hand, kfh0 kpp = h−p

Z

1

|f (x + h) − f (x)|p dx.

0

Thus, Z

1

|f (x + h) − f (x)|p dx

1/p ≤ Ch,

h ∈ (0, 1].

0

Since f is continuous, Theorem 2.1 implies that f ∈ Wp1 . Remark 4.5. Milman [49] studied continuity properties of interpolation scales at the endpoints. In particular, it follows from his results that for any f ∈ Wp1 , lim (1 − s)

s→1−

1/p

Z

1 −s

(t 0

dt K(f, t; Vp , Wp1 ))p t

1/p

1/p 1 = kf 0 kp p

Together with (4.1.5), this provides another look on (4.2.3).

62


We shall also give some limiting relations for the functionals vp,α (f ) defined by (1.0.13). Theorem 4.6. Let f be an 1-periodic function and let 1 0, then lim vp,α (f ) = vp (f ).

(4.2.6)

α→0+

Proof. To prove (i), we first observe that vp,α (f ) ≤ vp,1/p0 (f ),

0 < α < 1/p0 .

Further, let Π = {x0 , x1 , ..., xn } be any partition. Then, since vp,α (f ) ≥

n−1 X |f (xk+1 ) − f (xk )|p

!1/p

(xk+1 − xk )αp

k=0

,

we get lim vp,α (f ) ≥ α→1/p0 −

n−1 X |f (xk+1 ) − f (xk )|p k=0

(xk+1 − xk )p−1

Taking supremum over all partitions, we obtain lim vp,α (f ) ≥ vp,1/p0 (f ). α→1/p0 −

Thus, (4.2.5) holds. We proceed to prove (ii). Since vp (f ) ≤ vp,α (f ) for any α > 0, it is sufficient to show that lim vp,α (f ) ≤ vp (f ).

α→0+

!1/p .

4.2. Limiting relations

63

For any partition Π = {x0 , x1 , ..., xn }, we set σk = {j : 2−k−1 < xj+1 − xj ≤ 2−k }, and

!1/p Sk (f ) =

X

p

|f (xj+1 ) − f (xj )|

.

j∈σk

Then α

vp,α (f ; Π) ≤ 2

∞ X

!1/p kαp

2

p

Sk (f )

.

(4.2.7)

k=0

Furthermore, by applying the Abel transform we have # "∞ ∞ ∞ ∞ X X X X kαp p p p kαp Sj (f ) − Sj (f ) 2 Sk (f ) = 2 k=0

j=k

k=0 ∞ X

=

j=k+1 ∞ X

∞ X

k=1

j=k

Sk (f )p + (1 − 2−αp )

k=0

2kαp

Sj (f )p .

It is easy to see that ∞ X

Sj (f )p ≤ vp,α0 (f )p 2−kα0 p .

j=k

Whence, for 0 < α < α0 ∞ X

2kαp Sk (f )p ≤ vp (f )p + vp,α0 (f )p αp

k=0

∞ X

2−k(α0 −α)p .

k=1

Thus, by (4.2.7) α

vp,α (f ) ≤ 2

vp (f ) + α

1/p

vp,α0 (f )

p

1/p !

2(α0 −α)p − 1

and it follows that lim vp,α (f ) ≤ vp (f ),

α→0+

which concludes the proof. Remark 4.7. The condition that f ∈ Vpα0 for some α0 > 0 in (ii) cannot be omitted. Indeed, if f ∈ Vp has a discontinuity at some point, then vp,α (f ) = ∞ for all α > 0 whence lim vp,α (f ) = ∞, while vp (f ) < ∞.

64

4.3


Estimates of the Riesz-type variation

In this section we obtain a sharp estimate of vp,α (f ) (see (1.0.13)) in terms of the modulus of p-continuity ω1−1/p (f ; δ). Theorem 4.8. Let 1 < p < ∞ and let 0 < α < 1/p0 . Assume that f ∈ Vp and that 1/p Z 1 −α p dt Ip,α (f ) = < ∞. (4.3.1) [t ω1−1/p (f ; t)] t 0 Then f ∈ Vpα and vp,α (f ) ≤ A[vp (f ) + cp,α Ip,α (f )],

(4.3.2)

where A is an absolute constant and cp,α = p0 α1/p (1/p0 − α)1/p .

(4.3.3)

Proof. The condition (4.3.1) implies that f ∈ Cp . Let ω ∗ (t) be given by (2.3.1) and take ω ∈ Ω1/p0 such that ω ∗ (t) ≤ ω(t), and

t ∈ [0, 1]

(4.3.4)

0

lim ω(t)/t1/p = ∞.

(4.3.5)

t→0+

We specify later how such ω can be obtained. As before, set ωn = ω(2−n ) 0 and ω n = 2n/p ωn . Let the natural numbers nk ≡ nk (ω, 1/p0 ), k = 0, 1, ..., be defined by (2.3.7). Set µ(k) = k if (2.3.9) holds and µ(k) = k + 1 if (2.3.10) holds, and define nµ(k)

Z

2

−nµ(k)

f (x + t)dt.

gk (x) = 2

0

Fix a partition Π = {x0 , x1 , ..., xn } and set σk = {j : 2−nk+1 < xj+1 − xj ≤ 2−nk }. For any function ϕ we define Rk (ϕ) =

X |ϕ(xj+1 ) − ϕ(xj )|p (xj+1 − xj )αp j∈σ k

!1/p

4.3. Estimates of the Riesz-type variation

65

and

!1/p X

Sk (ϕ) =

p

|ϕ(xj+1 ) − ϕ(xj )|

.

j∈σk

By Hölder’s inequality we have for j ∈ σk Z p xj+1 0 gk (t)dt xj Z xj+1 ≤ (xj+1 − xj )p−1−αp |gk0 (t)|p dt xj Z xj+1 ≤ 2−nk (p−1−αp) |gk0 (t)|p dt.

|gk (xj+1 ) − gk (xj )|p 1 = (xj+1 − xj )αp (xj+1 − xj )αp

xj

Thus, by (2.2.12) and (4.3.4), 0

Rk (gk ) ≤ 2−nk (1/p −α) kgk0 kp 0 0 ≤ 2−nk (1/p −α) 2nµ(k) /p ω1−1/p (f ; 2−nµ(k) ) 0

0

≤ 2−nk (1/p −α) 2nµ(k) /p ωnµ(k) . If µ(k) = k, then Rk (gk ) ≤ 2nk α ωnk . If µ(k) = k + 1, then ω nk+1 < 8ω nk and 0 (4.3.6) Rk (gk ) ≤ 2−nk (1/p −α) ω nk+1 ≤ 2nk α+3 ωnk . Thus, (4.3.6) holds for each k ∈ N. Further, Rk (f − gk ) ≤ 2nk+1 α Sk (f − gk ).

(4.3.7)

Applying (4.3.6) and (4.3.7), we get vp,α (f ; Π) ≤

∞ X

!1/p p

Rk (gk )

∞ X

+

k=0

≤8

∞ X k=0

2

ωnp k

Rk (f − gk )

k=0

!1/p nk αp

!1/p p

+

∞ X

!1/p nk+1 αp

2

p

Sk (f − gk )

.

(4.3.8)

k=0

We estimate the latter sum. Clearly, Sk (f − gk ) ≤ vp (f − gk ). Applying (4.1.1), we obtain Sk (f − gk ) ≤ 6ωnµ(k) .

66


If µ(k) = k, then ωnk < 8ωnk+1 and 2nk+1 α Sk (f − gk ) ≤ 2nk+1 α+3 ωnk ≤ 2nk+1 α+6 ωnk+1 . If µ(k) = k + 1, then 2nk+1 α Sk (f − gk ) ≤ 2nk+1 α+3 ωnk+1 . Thus ∞ X

!1/p nk+1 αp

2

p

Sk (f − gk )

≤ 64

k=0

∞ X

!1/p nk αp

2

ωnp k

.

k=0

It follows from the previous estimate and (4.3.8) that !1/p ∞ X nk αp p vp,α (f ; Π) ≤ 72 2 ωnk . k=0 0

Applying Lemma 3.1 with γ = 1/p , q = p and β = αp yields 1/p Z 1 p p 0 0 −αp p dt . vp,α (f ) ≤ 72 2ω0 + 2 pp α(1/p − α) t ω(t) t 0 Set Z Dp,α (ω) =

1

t−αp ω(t)p

0

dt t

1/p .

Since p1/p ≤ 2 and (p0 )1/p ≤ p0 , we obtain vp,α (f ) ≤ 300 ω0 + p0 α1/p (1/p0 − α)1/p Dp,α (ω) .

(4.3.9)

If there holds

ω ∗ (t) = ∞, t1/p0 then we take ω(t) = ω ∗ (t). In this case lim

t→0+

(4.3.10)

Dp,α (ω) ≤ Ip,α (f ) and ω0 ≤ vp (f ), by (2.3.2). Thus, (4.3.2) is proved in this case. If (4.3.10) does not hold, we take ωε (t) = ω ∗ (t) + εtγ where α < γ < 1/p0 . Then ωε ∈ Ω1/p0 for each ε > 0 and ωε satisfies (4.3.4) and (4.3.5). Furthermore, by (2.3.2) and a simple calculation we have Dp,α (ωε ) ≤ Ip,α (f ) + ε(p(γ − α))1/p


67

and ωε (1) ≤ vp (f ) + ε. Thus, we get from (4.3.9) that vp,α (f ) ≤ 300(vp (f ) + ε + p0 α1/p (1/p0 − α)1/p [Ip,α (f ) + ε(p(γ − α))1/p ]). Letting ε → 0 yields (4.3.2). Remark 4.9. Assume that f ∈ Wp1 (1 < p < ∞). By Theorem 4.4 lim (1/p0 − α)1/p Ip,α (f ) = p−1/p kf 0 kp .

α→1/p0 −

Further, vp (f ) ≤ kf 0 kp for f ∈ Wp1 . Thus, the upper limit as α → 1/p0 − of the right-hand side of (4.3.2) does not exceed Akf 0 kp (where A is an absolute constant). On the other hand, by Proposition 4.6, the left-hand side of (4.3.2) tends to vp,1/p0 (f ) as α → 1/p0 −. Thus, vp,1/p0 (f ) ≤ Akf 0 kp . This agrees with Theorem 2.2, and shows that the order of the constant (4.3.3) is optimal as α → 1/p0 −. Remark 4.10. Assume that Ip,α0 (f ) < ∞ for some 0 < α0 < 1/p0 . Since Ip,α (f ) ≤ Ip,α0 (f ) for 0 < α ≤ α0 , we get that lim α1/p Ip,α (f ) ≤ lim α1/p Ip,α0 (f ) = 0.

α→0+

α→0+

Thus, as α → 0+, the limit of the right-hand side of (4.3.2) does not exceed Avp (f ) (where A is an absolute constant). On the other hand, if Ip,α0 (f ) < ∞, then f ∈ Vpa0 by Theorem 4.8. By (4.2.6), vp,α (f ) → vp (f ) as α → 0+. Thus, the behaviour of the left-hand side of (4.3.2) agrees with the behaviour of the right-hand side as α → 0+. Remark 4.11. We shall study the relationship between Theorem 4.8 and Theorem 3.14. In particular, we compare the estimates (3.4.2) and (4.3.2). For 1 < p < ∞, 0 < α < 1/p0 and f ∈ Vp , we have Kp,α (f ) ≤ Ip,α (f ) ≤

C Kp,α (f ), α

(4.3.11)

where C is an absolute constant. Indeed, the left inequality is an immediate consequence of (2.2.9), while the right inequality follows from the estimate of Theorem 3.11 combined with Hardy’s inequality (see [37, p.7]).

68

Chapter 4. Fractional smoothness via p-variation By (3.4.2) and the left inequality of (4.3.11), we have vp,α (f ) ≤ Ac0p,α Ip,α (f ), 0

where A is an absolute constant and c0p,α = α−1/p (1/p0 − α)1/p . Observe that for small α > 0, the constant c0p,α is much larger than the constant cp,α given by (4.3.3). Indeed, c0p,α → ∞ as α → 0+, while cp,α → 0 as α → 0+. Thus, (4.3.2) with the sharp constant (4.3.3) cannot be obtained from (3.4.2). However, as was observed, the order of the constant in (3.4.2) as α → 0+ is optimal. Now we show that for 0 < α < 1/p0 the condition (4.3.1) is sharp. Theorem 4.12. Let 1 < p < ∞ and 0 < α < 1/p0 . Assume that ω ∈ Ω1/p0 is any modulus of p-continuity such that Z 1 dt (t−α ω(t))p = ∞. (4.3.12) t 0 Then there is a function f ∈ Vp such that ω1−1/p (f ; δ) ≤ ω(δ) but f ∈ / Vpα . Proof. Define ωn , ω n by (2.3.3) with γ = 1/p0 . The condition (4.3.12) implies 0 that ω(δ) 6= O(δ 1/p ), thus we may construct {nk }∞ k=0 by (2.3.7). For k = 1, 2, ..., set ξk = 2−nk , δk = 2−nk −2 and Ik = [ξk − δk , ξk + δk ]. Then Ik ⊂ (0, 1). Further, since nk+1 ≥ nk + 1, we have ξk+1 + δk+1 < ξk − δk and thus the intervals {Ik }k∈N are pairwise disjoint and ordered from the right to the left. For k ∈ N, define ϕk as a continuous 1-periodic function such that ϕk (x) = 0 for x ∈ [0, 1]\Ik , ϕk (ξk ) = ωnk , and ϕk is linear on [ξk −δk , ξk ] and [ξk , ξk +δk ]. Set ∞ X ϕk (x). f (x) = k=1 −s

We shall estimate ω1−1/p (f ; 2 ) for s ∈ N. Assume that nm ≤ s < nm+1 for some m ≥ 1. Clearly, there holds ω1−1/p (f ; 2−s ) ≤

∞ X

ω1−1/p (ϕk ; 2−s ).

k=1

For each k ≥ m + 1 we have the trivial estimate ω1−1/p (ϕk ; 2−s ) ≤ v1 (ϕk ) = 2ωnk .


69

Fix 1 ≤ k ≤ m. Observe that |ϕ0k (x)| = 2nk +2 ωnk ,

x ∈ (ξk − δk , ξk ) ∪ (ξk , ξk + δk ),

and ϕ0k (x) = 0,

x ∈ [0, 1] \ Ik .

By (2.2.8), we have 0

ω1−1/p (ϕk ; 2−s ) ≤ 2−s/p kϕ0k kp = 2−s/p

0

Z Ik

2(nk +2)p ωnp k dx

1/p

0

= 2−s/p +2−1/p ω nk . By (2.3.8), "

m X

−s/p0

−s

ω1−1/p (f ; 2 ) ≤ 4 2

ω nk +

#

∞ X

ωnk

k=m+1

k=1 0

≤ 8(2−s/p ω nm + ωnm+1 ). 0

Further, since nm ≤ s < nm+1 , we have ω nm ≤ ω s = 2s/p ωs , and ωnm+1 ≤ ωs . Thus, ω1−1/p (f ; 2−s ) ≤ 16ωs . This implies that ω1−1/p (f ; δ) ≤ 32ω(δ) for 0 ≤ δ ≤ 1. We shall prove that f ∈ / Vpα . For any N ∈ N, consider the points 0 < ξN − δN < ξN < ξN −1 − δN −1 < .... < ξ1 − δ1 < ξ1 < 1. Clearly vp,α (f ) ≥

N X |f (ξk ) − f (ξk − δk )|p k=1

!1/p α

=4

δkαp

N X

!1/p nk αp

2

ωnp k

k=1

Thus, α

vp,α (f ) ≥ 4

∞ X

!1/p nk αp

2

ωnp k

.

k=1

It remains to show that the series at the right-hand side diverges.

.

70

Chapter 4. Fractional smoothness via p-variation If (2.3.9) holds, then Z

2−nk

(t−α ω(t))p

2−nk+1

dt 8p nk+1 αp p ≤ 2 ωnk+1 . t αp

If (2.3.10) holds, then Z

2−nk

(t−α ω(t))p

2−nk+1

dt 8p ≤ 2nk αp ωnp k . t p − 1 − αp

These estimates and (4.3.12) yield that ∞ X k=1

2nk αp ωnp k = ∞.

Chapter 5 Embeddings within the scale Vp Let 1 < p < q < ∞ and let f ∈ Vq . By Jensen’s inequality, we have that ω1−1/q (f ; δ) ≤ ω1−1/p (f ; δ) (0 ≤ δ ≤ 1). The main objective of this chapter is to obtain sharp reverse inequalities, that is, estimates of ω1−1/p (f ; δ) in terms of ω1−1/q (f ; δ). Such problems for different scales of function spaces of fractional smoothness have been studied for a long time. We shall first discuss some previous results concerning Lr −moduli of continuity.

5.1

Some known results and statement of problem

One of the origins of embedding theory is the classical Hardy-Littlewood theorem on Lipschitz classes [28]. This theorem states that if ω(f ; δ)r = O(δ α ) (1 ≤ r < ∞, 0 < α ≤ 1), r < s < ∞, and θ = 1/r − 1/s < α, then ω(f ; δ)s = O(δ α−θ ). Problems on general relations between moduli of continuity in different norms and their sharpness were first posed and studied in the works by Ul’yanov [74] – [76]. Therein, the conception of sharpness was formulated in terms of necessary and sufficient conditions for embeddings of classes of functions Hrω = {f ∈ Lr ([0, 1]) : ω(f ; δ)r = O(ω(δ))}, 71

72

Chapter 5. Embeddings within the scale Vp

where ω is a given majorant (ω ∈ Ω1 ) (see Definiton 2.10). Ul’yanov [75] proved that Z 1 dt Hrω ⊂ Ls ⇐⇒ (t−θ ω(t))s < ∞, t 0

(5.1.1)

where 1 ≤ r < s < ∞, θ = 1/r − 1/s, ω ∈ Ω1 . Furthermore, he obtained a general estimate Z δ 1/s −θ s dt ω(f ; δ)s ≤ c (t ω(f ; t)r ) (0 ≤ δ ≤ 1), (5.1.2) t 0 where 1 ≤ r < s < ∞ and θ = 1/r − 1/s (see also [56] and references in [31]). Andrienko [3] proved that this estimate is sharp in the following sense Z δ 1/s ω η −θ s dt Hr ⊂ Hs ⇐⇒ µr,s,ω (δ) ≡ (t ω(t)) = O(η(δ)), (5.1.3) t 0 whatever be ω, η ∈ Ω1 and 1 ≤ r < s < ∞. We observe that this result can be expressed in an alternative form. Set ω

H r = {f ∈ Lr ([0, 1]) : ω(f ; δ)r ≤ ω(δ)}. It is easy to see that (5.1.3) is equivalent to the following statement: there is ω a constant c > 0 such that for every δ ∈ [0, 1] there exists a function fδ ∈ H r for which Z δ 1/s dt (t−θ ω(t))s ω(fδ ; δ)s ≥ c . (5.1.4) t 0 We stress that for different values of δ we get different functions fδ . If r > 1, it may not exist a single function f fitting all δ ∈ (0, 1] (see [31]). Moreover, it was proved in [31] that inequality (5.1.2) can be strengthened in a sense. One of the questions that we consider in this chapter can be formulated in the following way: if 1 < p < q < ∞ and f ∈ Vq , what is the necessary and sufficient condition on the rate of decay of ω1−1/q (f ; δ) in order to have f ∈ Vp ? Of course, this is related to the problem of finding estimates of ω1−1/p (f ; δ) in terms of ω1−1/q (f ; δ). We shall show that the answers to these questions are given by results that are formally analogous to (5.1.1) and (5.1.2). However, the analogy fails to be complete. We show that in contrast to (5.1.2), the corresponding inequality for moduli of p-continuity is sharp in a stronger sense. Namely, the extremal function (similar to the one in (5.1.4)) can be chosen independently of δ.


5.2

73

Auxiliary results

We shall use the following construction. Definition 5.1. Let I = [a, b] ⊂ [0, 1] be an interval and N ∈ N. Set h = (b − a)/N , and let ξj = a + jh for j = 0, 1, ..., N , and ξj∗ = a + (j + 1/2)h for j = 0, 1, ..., N − 1. The function F (I, N, H; x) is defined to be the continuous 1-periodic function such that F (x) = 0 for x ∈ [0, 1] \ I, F (ξj ) = 0 (j = 0, 1, ..., N ), F (ξj∗ ) = H (j = 0, 1, ..., N − 1), and F is linear on each of the intervals [ξj , ξj∗ ] and [ξj∗ , ξj+1 ] (j = 0, 1, ..., N − 1). Thus, the graph of F consists of N congruent isosceles triangles with height H and base h. Using (2.2.1) and Lemma 2.6, we have vr (F ) = (2N )1/r H and

(1 ≤ r < ∞),

0

kF 0 kr = 2h−1/r N 1/r H

(1 ≤ r < ∞).

(5.2.1) (5.2.2)

Lemma 5.2. Let 1 < p < q < ∞ and θ = 1/p − 1/q. Suppose that ω ∈ Ω1/q0 satisfies (2.3.6) and let nk = nk (ω) (k ∈ N) be defined by (2.3.7). Fix natural numbers ν < µ and a number γ ∈ (0, 1]. Let !1/q µ−1 X nk θq q σ= 2 ωnk . k=ν

Then there exists a nonnegative continuous 1-periodic function f such that: (i) (supp f ) ∩ [0, 1] = [0, α], where 1 α≤ 8

γ+

µ−1 X

! 2−nk

;

(5.2.3)

k=ν

(ii) for any 0 < h ≤ 1, ω1−1/q (f ; h) ≤ 16

µ−1 X

min(1, (2nk h)1−1/q )ωnk

(5.2.4)

k=ν

and ω1−1/p (f ; h) ≤ 16 min(σ, h1−1/p ω nµ−1 );

(5.2.5)

74


(iii) the following estimates from below hold: ω1−1/p (f ; 2−nν ) ≥ 21/q γ θ σ,

(5.2.6)

and ω1−1/p (f ; 2−nµ−1 ) ≥ 21/q γ θ σ

2nµ−1 θ ωnµ−1 σ

q/p .

(5.2.7)

Proof. Set1 Nk = [2nk q/p ωnq k σ −q γ] + 1 (k = ν, ..., µ − 1), αν = 0,

αm =

m−1 X

Nk 2−nk −3

(m = ν + 1, ..., µ),

k=ν

Ik = [αk , αk+1 ] for k = ν, ..., µ − 1. −1/q

Further, for each ν ≤ k ≤ µ − 1, let Hk = ωnk Nk F (Ik , Nk , Hk ; x), and µ−1 X f (x) = Fk (x).

and set Fk (x) =

k=ν

Clearly, (supp f ) ∩ [0, 1] = [0, αµ ] and 1 αµ ≤ 8

−q

σ γ

µ−1 X

nk (q/p−1)

2

k=ν

ωnq k

+

µ−1 X

! −nk

2

k=ν

1 = 8

γ+

µ−1 X

! 2

k=ν

This implies (i). Let now 0 < h ≤ 1. We have (see (5.2.1)) ω1−1/q (Fk ; h) ≤ vq (Fk ) = 21/q ωnk , and by (2.2.8) ω1−1/q (Fk ; h) ≤ h1−1/q kFk0 kq ≤ 16h1−1/q 2nk (1−1/q) ωnk . These estimates imply (5.2.4). 1

[x] denotes the integral part of a number x.

−nk

.


75

Further, each Fk is nonnegative and equals to 0 at the endpoints of the interval Ik = supp Fk . Moreover, the intervals Ik have disjoint interiors. Thus, taking into account Lemma 2.6 and (5.2.1), we get vp (f )p =

µ−1 X

vp (Fk )p = 2

k=ν

µ−1 X

Hkp Nk = 2

k=ν

µ−1 X

1−p/q

ωnp k Nk

.

k=ν

Since Nk ≤ 2nk q/p ωnq k σ −q + 1, we have, by applying the first inequality in (2.3.8) ! µ−1 µ−1 X X vp (f )p ≤ 2 σ p−q 2nk θq ωnq k + ωnp k ≤ 2(σ p + 2ωnp ν ) ≤ 6σ p . k=ν

k=ν

Thus, ω1−1/p (f ; h) ≤ vp (f ) ≤ 61/p σ,

0 < h ≤ 1.

Further, by (5.2.2) kf 0 kpp =

µ−1 X

kFk0 kpp =

k=ν

µ−1 X

(Hk 2nk +4 )p Nk 2−nk −3

k=ν

= 24p−3

µ−1 X

1−p/q nk (p−1)

Nk

2

ωnp k ≤ 24p−3 (A + B),

k=ν

where A=

µ−1 X

2nk (p−1) ωnp k

k=ν

and B = σ p−q

µ−1 X

2nk p(1−1/q) ωnp k (2nk θq ωnq k )1−p/q .

k=ν

By the second inequality in (2.3.8), we have that A≤

µ−1 X

2−nk (1−p/q) ω pnk ≤

µ−1 X

ω pnk ≤ 2ω pnµ−1 .

k=ν

k=ν

Further, applying Hölder’s inequality and (2.3.8), we obtain !p/q µ−1 !1−p/q µ−1 X X p−q nk θq q q B≤σ ω nk 2 ωnk ≤ 2p/q ω pnµ−1 . k=ν

k=ν

(5.2.8)

76


Thus, kf 0 kp ≤ 16ω nµ−1 , and by (2.2.8) we have that ω1−1/p (f ; h) ≤ 16h1−1/p ω nµ−1 ,

0 < h ≤ 1.

(5.2.9)

Estimates (5.2.8) and (5.2.9) imply (5.2.5). All extremal points of the functions Fk (k = ν, ..., µ − 1) subdivide [0, αµ ] into intervals with lengths smaller than 2−nν . This implies that ω1−1/p (f ; 2−nν )p ≥ 2

µ−1 X

Nk Hkp = 2

µ−1 X

1−p/q

Nk

ωnp k .

k=ν

k=ν

Since Nk ≥ 2nk q/p−1 ωnq k σ −q γ, we have ω1−1/p (f ; 2−nν )p ≥ 2p/q σ p−q γ 1−p/q

µ−1 X

2nk θq ωnq k = 2p/q γ 1−p/q σ p .

k=ν

This proves (5.2.6). Finally, subdividing [0, αµ ] into intervals of the length 2−nµ−1 −4 , we take only the terms related to the interval [αµ−1 , αµ ]. Thus, we obtain p ω1−1/p (f ; 2−nµ−1 )p ≥ 2Nµ−1 Hµ−1 ≥ 2p/q γ 1−p/q σ p−q 2nµ−1 θq ωnq µ−1 .

This implies (5.2.7). We shall also use van der Waerden type functions to prove the following statement. Lemma 5.3. Let 1 ≤ p < q < ∞ and θ = 1/p − 1/q. Suppose that ω ∈ Ω1/q0 satisfies (2.3.6). Then there exists a nonnegative continuous 1-periodic function ψ such that (supp ψ) ∩ [0, 1] = [0, 1/2], ω1−1/q (ψ; δ) ≤ ω(δ),

0 ≤ δ ≤ 1,

(5.2.10) (5.2.11)

and ω1−1/p (ψ; δ) ≥ Aδ −θ ω(δ), where A is an absolute constant.

0 < δ ≤ 1,

(5.2.12)


77

Proof. Let nk = nk (ω) (see (2.3.7)). Denote I = [0, 1/2],

Nk = 2nk ,

Hk = ωnk 2−nk /q .

Further, applying Definition 5.1, we set gk (x) = F (I, Nk , Hk ; x) (k ∈ N) and g(x) =

∞ X

gk (x).

k=1

Then g is a nonnegative, continuous and 1-periodic function satisfying (5.2.10). By (5.2.1), we have ω1−1/q (gk ; h) ≤ vq (gk ) = 21/q ωnk

(5.2.13)

for any 0 < h ≤ 1. Besides, |gk0 (x)| = 2nk +2 Hk = 4ω nk

(k ∈ N)

(5.2.14)

for almost all x ∈ I (gk0 (x) = 0 for x ∈ (0, 1) \ I). By (2.2.8), it follows that ω1−1/q (gk ; h) ≤ 4h1−1/q ω nk ,

0 < h ≤ 1.

(5.2.15)

Let 2−nj+1 < h ≤ 2−nj (j ∈ N). Then, by (5.2.13), (5.2.15), and (2.3.8), ω1−1/q (g; h) ≤ 4

∞ X

1−1/q

ωnk + h

k=j+1

j X

! ω nk

k=1

≤ 8(ωnj+1 + h1−1/q ω nj ) ≤ 16ω(h).

(5.2.16)

Now we estimate ω1−1/p (g; 2−m ) from below. We shall use the inequality ω1−1/p (gk ; h) ≤ 22−1/p h1−1/p ω nk

(0 < h ≤ 1),

(5.2.17)

which follows directly from (2.2.8) and (5.2.14). Fix an integer m ≥ 0. Let nµ ≤ m < nµ+1 . First we assume that ω nµ+1 < 8ω nµ .

(5.2.18)

78


Set hm = 2−m−2 and let Π be the partition of [0, 1] by the points xi = ihm (i = 0, 1, ..., 2m+2 ). Then gk (xi ) = 0 (i = 0, 1, ..., 2m+2 ) for all k ≥ µ + 1. Further, if µ ≥ 2, then by (5.2.17) and (2.3.8), µ−1 X

µ−1 X

1−1/p vp (gk ; Π) ≤ 22−1/p hm

k=1

ω nk ≤ 23−1/p h1−1/p ω nµ−1 m

k=1 0

≤ 21/p h1−1/p ω nµ . m On the other hand, the function gµ is linear on each of the intervals [xi , xi+1 ]. Thus, by (5.2.14), 0

vp (gµ ; Π) = 4ω nµ hm 2(m+1)/p = 21+1/p h1−1/p ω nµ . m Applying these estimates and (5.2.18), we obtain 0

0

0

1−1/p ω nµ ≥ 21/p −3 h1−1/p ω m = 2−1/p −3 2mθ ωm . vp (g; Π) ≥ 21/p hm m

This implies that ω1−1/p (g; 2−m ) ≥ 2mθ−4 ωm

(5.2.19)

for nµ ≤ m < nµ+1 , provided that (5.2.18) holds. If (5.2.18) does not hold, then 8ω nµ ≤ ω nµ+1

(5.2.20)

and thus (see (2.3.9), (2.3.10) above) 8ωnµ+1 > ωnµ .

(5.2.21)

Set tµ = 2−nµ+1 −2 and let Π0 be the partition of [0, 1] by points xi = itµ (i = 0, 1, ..., 2nµ+1 +2 ). Then gj (xi ) = 0 for all j ≥ µ + 2. Further, by (5.2.17) and (5.2.20), we have µ X

vp (gk ; Π0 ) ≤ 22−1/p t1−1/p µ

k=1

µ X

ω nk ≤ 23−1/p t1−1/p ω nµ µ

k=1

≤ 2−1/p t1−1/p ω nµ+1 = 2−2+1/p 2nµ+1 θ ωnµ+1 . µ On the other hand, vp (gµ+1 ; Π0 ) = 2nµ+1 θ+1/p ωnµ+1 .


79

Applying (5.2.21), we obtain 0

0

vp (g; Π0 ) ≥ 2−1/p 2nµ+1 θ ωnµ+1 ≥ 2−1/p −3 2nµ+1 θ ωm ≥ 2mθ−4 ωm . Thus, (5.2.19) is true for nµ ≤ m < nµ+1 also in the case when (5.2.18) does not hold. Now, the statement of the lemma follows from (5.2.16) and (5.2.19). Lemma 5.4. Let 1 < p < q < ∞ and θ = 1/p − 1/q. Assume that ω ∈ Ω1/q0 satisfies (2.3.7) and let nk = nk (ω) (k ∈ N). Then: (i) the series ∞ X

q 2mθq ωm

(5.2.22)

m=1

converges if and only if the series ∞ X

2nk θq ωnq k

(5.2.23)

k=1

converges; (ii) if rn =

∞ X

!1/q mθq

2

q ωm

∞ X

nk θq

(n = 0, 1, ...)

m=n

and ρ(ν) =

!1/q 2

ωnq k

(ν ∈ N),

k=ν

then rn ≤ c(2nθ ωn + ρ(ν + 1)) for

nν ≤ n < nν+1

where c = c(p, q) depends only on p and q. Proof. Denote nk+1 −1

Sk =

X m=nk

q 2mθq ωm .

(ν ∈ N),

(5.2.24)

80


Assume that (2.3.9) holds. Then nk+1 −1

Sk ≤ 8q ωnq k+1

X

2mθq ≤ c2nk+1 θq ωnq k+1 .

(5.2.25)

m=nk

If (2.3.9) does not hold, then (2.3.10) holds. In this case nk+1 −1

Sk =

∞ X

0

X

2−mq/p ω qm ≤ 8q ω qnk

m=nk

0

2−mq/p

m=nk 0

= cω qnk 2−nk q/p = c2nk θq ωnq k .

(5.2.26)

Estimates (5.2.25) and (5.2.26) yield the statement (i). Similarly, we have nν+1 −1

X

q 2mθq ωm ≤ c max(2nθq ωnq , 2nν+1 θq ωnq ν+1 ).

(5.2.27)

m=n

Since nν+1 −1

rnq =

X

q 2mθq ωm +

m=n

∞ X

Sk ,

k=ν+1

estimates (5.2.25) – (5.2.27) imply (5.2.24).

5.3

Embeddings of the space Vqω

Theorem 5.5. Let 1 < p < q < ∞ and θ = 1/p − 1/q. Assume that f ∈ Vq and that Z 1 dt (t−θ ω1−1/q (f ; t))q < ∞. (5.3.1) t 0 Then f ∈ Vp and Z ω1−1/p (f ; δ) ≤ 4

−θ

q dt

(t ω1−1/q (f ; t)) 0

for all δ ∈ [0, 1].

δ

t

1/q (5.3.2)

5.3. Embeddings of the space Vqω

81

Proof. Let Π = {x0 , x1 , ..., xn } be any partition of a period. Applying Hölder’s inequality with exponents q/p and (q/p)0 = 1/(pθ), we obtain n−1 X

− f (xj )|p (xj+1 − xj ) (xj+1 − xj )pθ j=0 !1/q n−1 X |f (xj+1 ) − f (xj )|q . (xj+1 − xj )qθ j=0

vp (f ; Π) =

≤

pθ |f (xj+1 )

!1/p

(5.3.3)

Denote σk (Π) = {j : 2−k−1 < xj+1 − xj ≤ 2−k } (k = 0, 1, ...). Set also

1/q

 Sk (Π) = 

X

q

|f (xj+1 ) − f (xj )|

j∈σk (Π)

if σk (Π) 6= ∅ and Sk (Π) = 0 otherwise. Then, by (5.3.3) we have that   ∞ X X |f (xj+1 ) − f (xj )|q  vp (f ; Π) ≤  (xj+1 − xj )qθ k=0 j∈σk (Π) !1/q ∞ X (k+1)θq q ≤ 2 Sk (Π) . (5.3.4) k=0

Clearly, Sk (Π) ≤ ω1−1/q (f ; 2−k ),

(5.3.5)

for any partition Π. Using (5.3.4), (5.3.5), and (2.2.7), we obtain vp (f ) ≤

∞ X

!1/q (k+1)θq

2

−k q

ω1−1/q (f ; 2 )

k=0

Z ≤ 4

1 −θ

q dt

(t ω1−1/q (f ; t)) 0

t

1/q .

Thus, f ∈ Vp . Further, let 2−ν < δ ≤ 2−ν+1 , ν ∈ N. Let Π be any partition with kΠk ≤ δ. Then σk (Π) = ∅ and Sk (Π) = 0 for k < ν. Thus, from (5.3.4)

82


and (5.3.5) ∞ X

vp (f ; Π) ≤

!1/q (k+1)θq

2

−k q

ω1−1/q (f ; 2 )

k=ν

Z ≤ 4

δ

(t−θ ω1−1/q (f ; t))q

0

dt t

1/q .

This implies (5.3.2). Now, we obtain the following embedding theorem for the classes Vqω . Theorem 5.6. Let 1 < p < q < ∞, θ = 1/p − 1/q, and ω ∈ Ω1/q0 . Then the embedding Vqω ⊂ Vp (5.3.6) holds if and only if 1

Z

(t−θ ω(t))q

0

dt < ∞. t

(5.3.7)

Proof. The sufficiency of (5.3.7) for embedding (5.3.6) follows immediately from Theorem 5.5. To prove the necessity, we assume that the integral on the left-hand side of (5.3.7) diverges. Then ω satisfies (2.3.6). Let the sequence nk = nk (ω) be defined by (2.3.7). By Lemma 5.4(i), ∞ X

2nk θq ωnq k = ∞.

k=1

Thus, there exists a strictly increasing sequence of natural numbers {νj } such that ν1 = 1 and  1/q νj+1 −1 X n θq q σj ≡  2 k ωnk  > 2j (5.3.8) k=νj

for all j ∈ N. For each j ∈ N, we apply Lemma 5.2 with ν = νj , µ = νj+1 , and γ = 2−j . Then we have σ = σj ; we will write αj∗ instead of αµ (µ = νj+1 ). By (5.2.3), ! ∞ ∞ ∞ X 1 X −j X −nk 1 ∗ ≤ . αj ≤ 2 + 2 (5.3.9) 8 j=1 2 j=1 k=1

5.3. Embeddings of the space Vqω

83

By fj we denote the function f defined in Lemma 5.2. Set β1 = 0,

j−1 X

βj =

∗ αm

for j ≥ 2,

m=1

and define the functions ϕj (x) = fj (x + βj ) (j ∈ N), and ϕ(x) =

∞ X

ϕj (x).

j=1

Observe that by (5.3.9), (supp ϕj )∩[0, 1] = [βj , βj+1 ] ⊂ [0, 1/2], and supp ϕj (j ∈ N) have disjoint interiors. Assume that h ∈ (0, 1] and estimate ω1−1/q (ϕ; h). By (5.2.4), we have ω1−1/q (ϕ; h) ≤

∞ X

ω1−1/q (ϕj ; h)

j=1

≤ 16

∞ νj+1 X X−1


j=1 k=νj

= 16

∞ X

min(1, (2nk h)1−1/q )ωnk .

k=1

Let k(h) be the greatest natural number k such that 2nk h ≤ 1. Then h > 2−nk(h)+1 . Applying (2.3.8), we have   k(h) ∞ X X 1−1/q ω nk + ωnk  ω1−1/q (ϕ; h) ≤ 16 h k=1 1−1/q

≤ 32(h

k=k(h)+1

ω nk(h) + ωnk(h)+1 ) ≤ 64ω(h).

Thus, ϕ ∈ Vqω . On the other hand, by (5.2.6) and (5.3.8), we have that vp (ϕ) ≥ vp (ϕj ) > 2j(1−θ) Thus, ϕ ∈ / Vp .

for any j ∈ N.

84


Remark 5.7. Let p = 1 and 1 < q < ∞. Then the integral on the left-hand side of (5.3.7) diverges for any non-trivial ω ∈ Ω1/q0 . It is easy to show that embedding Vqω ⊂ V1 ≡ V (5.3.10) holds if and only if ω(t) = O(t1−1/q ).

(5.3.11)

Wq1

Vqω

(see Chapter 2) and thus we have ⊂ Indeed, if (5.3.11) is true, then (5.3.10). On the other hand, if (5.3.11) does not hold, then the function ψ defined in Lemma 5.3 (for p = 1) belongs to Vqω , but does not belong to V.

5.4

Sharpness of the main estimate

In this section we complete the proof of the main result of this chapter. Theorem 5.8. Let 1 < p < q < ∞, θ = 1/p − 1/q. There exists a positive constant c(p, q) such that for any ω ∈ Ω1/q0 there is a function f for which ω1−1/q (f ; δ) ≤ ω(δ),

δ ∈ [0, 1],

(5.4.1)

and Z

δ

ω1−1/p (f ; δ) ≥ c(p, q)

−θ

q dt

(t ω(t)) 0

1/q

t

,

δ ∈ [0, 1].

(5.4.2)

Proof. If the integral on the left-hand side of (5.3.7) diverges, then the statement follows from Theorem 5.6. We suppose that (5.3.7) holds. First we assume that ω satisfies condition (2.3.6). Let nk = nk (ω) (see (2.3.7)). As above, we set !1/q ∞ X nk θq q ρ(ν) = 2 ωnk (ν ∈ N). k=ν

Let ν1 = 1 and νj+1 = min{ν ∈ N : ρ(ν) ≤ α0 ρ(νj )} (j ∈ N), where α0 = 2−6 . Then ρ(νj+1 ) ≤ α0 ρ(νj )

(5.4.3)

5.4. Sharpness of the main estimate

85

and ρ(νj+1 − 1) > α0 ρ(νj ) (j ∈ N).

(5.4.4)

For each j ∈ N we apply Lemma 5.2 with ν = νj , µ = νj+1 , and γ = 1. We denote by fj the function f defined in this lemma. Further, by (5.4.3), we have  1/q νj+1 −1 X σ ≡ σj =  2nk θq ωnq k  = [ρ(νj )q − ρ(νj+1 )q ]1/q ≥ (1 − α0q )1/q ρ(νj ). k=νj

Thus, σj ≤ ρ(νj ) ≤ 21/q σj .

(5.4.5)

Observe also that by (5.2.3), (supp fj ) ∩ [0, 1] ⊂ [0, 1/2]. Set ϕ(x) =

∞ X

fj (x).

j=1

Then (supp ϕ) ∩ [0, 1] ⊂ [0, 1/2].

(5.4.6)

Let h ∈ (0, 1]. By (5.2.4), ω1−1/q (ϕ; h) ≤

∞ X

ω1−1/q (fj ; h)

j=1

≤ 16

∞ νj+1 X X−1


j=1 k=νj

= 16

∞ X

min(1, (2nk h)1−1/q )ωnk .

k=1

Estimating the last sum exactly as in the proof of Theorem 5.6, we obtain ω1−1/q (ϕ; h) ≤ 64ω(h).

(5.4.7)

Now we estimate the modulus of p-continuity of the function ϕ from below. We shall denote nk = lj if k = νj

and nk = lj∗ if k = νj+1 − 1.

(5.4.8)

86


By (5.2.5), ω1−1/p (fj ; h) ≤ 16 min(σj , h1−1/p ω lj∗ )

(5.4.9)

for all h ∈ (0, 1] and any j ∈ N. Besides, by (5.2.6), (5.2.7), and (5.4.5), we have ω1−1/p (fj ; 2−lj ) ≥ ρ(νj ), (5.4.10) and

∗

−lj∗

ω1−1/p (fj ; 2

2lj θ ωlj∗

) ≥ ρ(νj )

!q/p

ρ(νj )

(5.4.11)

for any j ∈ N. Let h ∈ (0, 1]. By (5.4.9), (5.4.5), and (5.4.3), ∞ X

ω1−1/p (fj ; h) ≤ 16

j=m+1

∞ X

ρ(νj ) ≤ 32ρ(νm+1 )

(5.4.12)

j=m+1

for any m ∈ N. Also, by (5.4.9) and (2.3.8), m−1 X

ω1−1/p (fj ; h) ≤ 16h1−1/p

j=1

m−1 X

ω lj∗

j=1 ∗ ≤ 32h1−1/p ω lm−1 ≤ 8h1−1/p ω lm

(5.4.13)

for any m ≥ 2. Fix now s ∈ N, s ≥ 2. First we assume that ρ(νs ) ≥ β0 ρ(νs − 1) (β0 = 2−12q ).

(5.4.14)

By (5.4.12) and (5.4.13), for any h ∈ (0, 1] ω1−1/p (ϕ; h) ≥ ω1−1/p (fs ; h) −

X

ω1−1/p (fj ; h)

j6=s

≥ ω1−1/p (fs ; h) − 32ρ(νs+1 ) − 8h1−1/p ω ls . Taking here h = 2−ls and applying (5.4.10) and (5.4.3), we obtain 1 ω1−1/p (ϕ; 2−ls ) ≥ ρ(νs ) − 2ls θ+3 ωls . 2

(5.4.15)

By (5.4.14) and (5.4.4), for any νs−1 ≤ k < νs , ρ(νs ) ≥ β0 ρ(νs − 1) ≥ α0 β0 ρ(νs−1 ) ≥ α0 β0 ρ(k).

(5.4.16)


87

In what follows, we denote λ(k) = 2nk θ ωnk .

(5.4.17)

Applying (5.4.15) and (5.4.16), we get ω1−1/p (ϕ; 2−nk ) ≥ c1 ρ(k) − 8λ(νs ) (c1 > 0)

(5.4.18)

for any νs−1 ≤ k < νs , provided that (5.4.14) holds. Now we assume that ρ(νs ) < β0 ρ(νs − 1).

(5.4.19)

Then (see notation (5.4.17)) 1 λ(νs − 1) ≥ ρ(νs − 1) − ρ(νs ) ≥ (1 − β0 )ρ(νs − 1) ≥ ρ(νs − 1). 2 By (5.4.4), ρ(νs − 1) > α0 ρ(νs−1 ). Whence, we obtain λ(νs − 1) ≥

α0 ρ(νs−1 ). 2

(5.4.20)

∗

∗ Set hs = 2−ls−1 . We have ls−1 = nνs −1 (see (5.4.8)). Thus, it follows from (5.4.11) and (5.4.20) that q/p λ(νs − 1) α0 q/p ρ(νs−1 ). (5.4.21) ω1−1/p (fs−1 ; hs ) ≥ ρ(νs−1 ) ≥ ρ(νs−1 ) 2

We observe also that by (5.4.19) ρ(νs ) < β0 ρ(νs−1 ).

(5.4.22)

Now we apply (5.4.21), (5.4.12) and (5.4.13) (in the case s ≥ 3) for m = s−1. We obtain X ω1−1/p (ϕ; hs ) ≥ ω1−1/p (fs−1 ; hs ) − ω1−1/p (fj ; hs ) j6=s−1

≥

α q/p 0

2

ρ(νs−1 ) − 32ρ(νs ) − 8h1−1/p ω ls−1 . s

Taking into account (5.4.22), we have ω1−1/p (ϕ; hs ) ≥ c2 ρ(νs−1 ) − 8h1−1/p ω ls−1 , s

88


where c2 = 2−7q/p − 25−12q > 0. Let νs−1 ≤ k < νs . Then hs ≤ 2−nk and ω ls−1 ≤ ω nk . Thus, ω1−1/p (ϕ; 2−nk ) ≥ c2 ρ(k) − 8λ(k)

(5.4.23)

for νs−1 ≤ k < νs , provided that (5.4.19) holds. Let now ψ be the function defined by Lemma 5.3. Set F (x) = ϕ(x) + ψ(x + 1/2). By (5.4.7) and (5.2.11), ω1−1/q (F ; δ) ≤ cω(δ),

0 ≤ δ ≤ 1.

(5.4.24)

On the other hand, taking into account (5.4.6) and (5.2.10), and applying (2.2.6), we have that ω1−1/p (F ; δ)p = ω1−1/p (ϕ; δ)p + ω1−1/p (ψ; δ)p .

(5.4.25)

Let νs−1 ≤ k < νs . Then, by (5.2.12) ω1−1/p (ψ; 2−nk ) ≥ cλ(k), and ω1−1/p (ψ; 2−nk ) ≥ ω1−1/p (ψ; 2−nνs ) ≥ cλ(νs ) (c > 0). Moreover, at least one of the inequalities (5.4.18) or (5.4.23) is true. Thus, taking into account (5.4.25), we obtain that ω1−1/p (F ; 2−nk ) ≥ cρ(k), for any k ∈ N. Finally, let nk ≤ n < nk+1 . By (5.4.25) and (5.2.12), ω1−1/p (F ; 2−n ) ≥ ω1−1/p (ψ; 2−n ) ≥ c2nθ ωn . On the other hand, by (5.4.26), ω1−1/p (F ; 2−n ) ≥ ω1−1/p (F ; 2−nk+1 ) ≥ cρ(k + 1). Thus, ω1−1/p (F ; 2−n ) ≥ c(ρ(k + 1) + 2nθ ωn ) (c > 0).

(5.4.26)


89

Applying Lemma 5.4, we have that −n

ω1−1/p (F ; 2

)≥c

0

∞ X

!1/q mθq

2

(c0 > 0)

q ωm

m=n

for any integer n ≥ 0. This estimate and (5.4.24) yield that the theorem is true provided that (2.3.6) holds. Now we assume that (2.3.6) does not hold. Then ω(t) = O(t1−1/q ). Take 1 − 1/p < γ < 1 − 1/q and set ωn (t) = ω(t) + tγ /n (n ∈ N). Clearly, ωn satisfies (2.3.6) and (5.3.7). As we have proved, there exists a constant c > 0 and a sequence of continuous 1-periodic functions {fn } such that fn (0) = 0, ω1−1/q (fn ; δ) ≤ ωn (δ) ≤ ω1 (δ) for all δ ∈ [0, 1], n ∈ N, and Z

δ

(t−θ ω(t))q

ω1−1/p (fn ; δ) ≥ c 0

dt t

(5.4.27)

1/q (5.4.28)

for all δ ∈ [0, 1], n ∈ N. By (5.4.27) and Theorem 5.5, δ

Z

−θ

ω1−1/p (fn ; δ) ≤ 4

q dt

(t ω1 (t)) 0

1/q ,

t

δ ∈ [0, 1],

(5.4.29)

for all n ∈ N. By the compactness criterion in Cp (see [24]), there exist a subsequence {fnk } and a function f ∈ Vp such that f (0) = 0 and vp (f − fnk ) → 0 as k → ∞.

(5.4.30)

Since f (0) = fnk (0) = 0, it follows that {fnk } converges uniformly to f . Thus, by (5.4.27), ω1−1/q (f ; δ) ≤ lim ω1−1/q (fn ; δ) ≤ ω(δ), n→∞

δ ∈ [0, 1].

Thus, f satisfies (5.4.1). Besides, (5.4.28) and (5.4.30) imply that f satisfies (5.4.2). Remark 5.9. Recall that for 1 < p < q < ∞ and ω ∈ Ω1/q0 , we denote Z ρp,q,ω (δ) = 0

δ

(t−θ ω(t))q

dt t

1/q ,

θ = 1/p − 1/q.

90


It follows from Theorems 5.5 and 5.8 that we have c(p, q) ≤ sup f ∈V

ω q

inf

0 1/p in (6.0.1) is essential; for α ≤ 1/p, the class Lip(α; p) contains unbounded functions and (6.0.1) cannot hold. We also show that Vp (p ≥ 1) can be expressed in terms of spaces ΛBV . For p = 1, this result is due to Perlman [58].

6.1

Auxiliary results

For f ∈ Lip(α; p) (1 ≤ p < ∞, 0 < α ≤ 1), we denote kf kLip(α;p) = sup δ>0

ω(f ; δ)p . δα

(6.1.1)

Let 1 0 such that for the modified function f¯, cp,α sup δ>0

ω1−1/p (f¯; δ) ω1−1/p (f¯; δ) ≤ kf k ≤ sup . Lip(α;p) δ α−1/p δ α−1/p δ>0

(6.1.2)

These statements follow from (2.2.9) and (1.0.10). Thus, if p > 1, 1/p < α ≤ 1 and f is a continuous 1-periodic function, then ω(f ; δ)p = O(δ α ) if and only if

ω1−1/p (f ; δ) = O(δ α−1/p ).

(6.1.3)

The following is a slight generalization of the construction given by Definition 5.1. Definition 6.1. Let I = [a, b] ⊂ [0, 1] be an interval, N ∈ N and H = (H0 , H1 , ..., HN −1 ) ∈ RN be a vector with Hj ≥ 0 for 0 ≤ j ≤ N − 1. Set h = (b − a)/N , ξj = a + jh (j = 0, 1, ..., N ) and ξj∗ = a + (j + 1/2)h (j = 0, 1, ..., N − 1). The function F (x) = F (I, N, H; x) is defined to be the continuous 1-periodic function such that F (x) = 0 for x ∈ [0, 1] \ I, F (ξj ) = 0 (j = 0, 1, ..., N ), F (ξj∗ ) = Hj (j = 0, 1, ..., N − 1), and F is linear on each of the intervals [ξj , ξj∗ ] and [ξj∗ , ξj+1 ] (j = 0, 1, ..., N − 1). Thus, the graph of F consists of N isosceles triangles of heights Hj (j = 0, ..., N − 1) and bases h. Using (2.2.1) and Lemma 2.6, we have !1/p N −1 X p 1/p vp (F ) = 2 Hj (1 ≤ p < ∞). (6.1.4) j=0


93

It is also easy to see that −1/p0

kF 0 kp = 2h

N −1 X

!1/p Hjp

(1 ≤ p < ∞).

(6.1.5)

j=0

The next lemma is of a known type (cf. [60]). In particular, it can be proved in the same way as Lemma 2.4 in [35]. Lemma 6.2. Let {αk } ∈ l1 be a sequence of non-negative numbers and let θ > 1 and γ > 0. There exists a sequence {βk } of positive numbers such that αk ≤ βk , ∞ X

k ∈ N, ∞

βk ≤

X θ1+γ αk , γ (θ − 1)(θ − 1) k=1

θ−γ ≤

βk+1 ≤ θ, βk

k=1

and

k ∈ N.

We shall also use the following Hardy-type inequality (see [39]). Lemma 6.3. Let β > 0 and 1 < r < ∞ be fixed. Let {ak } be a sequence of nonnegative real numbers, and {νn } an increasing sequence of positive real numbers with ν0 = 1. Then there exists a constant cβ,r > 0 such that  1/r !1/r ∞ ∞ X X X X 2−nβ  ak  . (6.1.6) ≤ cβ,r 2−nβ ak n=0

n=1

1≤k≤νn

νn−1 ≤k≤νn

Finally, we formulate the next well-known result (see, e.g., [17, Ch.6]). Lemma 6.4. Let 1 < p < ∞. Then {xn } ∈ lp if and only if ∞ X

αn xn < ∞,

n=1 0

for all {αn } ∈ lp . Moreover, sup

∞ X

k{αn }kp0 ≤1 n=1

αn xn = k{xn }kp .

94

Chapter 6. On functions of bounded Λ-variation

6.2

Embedding of Lipschitz classes

We shall now prove the main results of this chapter. Recall that kf kLip(α;p) is given by (6.1.1). Theorem 6.5. Let Λ ∈ S be given and 1 0 depending only on α and p such that for any f ∈ Lip(α; p),  ∞ X vΛ (f ) ≤ cp,α kf kLip(α;p) 

n+1 2X

n=0

k=2n

1

p0 !r0 /p0

k α−1/p λk

1/r0 

.

(6.2.2)

Proof. In light of (6.1.2), we may without loss of generality assume that sup δ>0

ω1−1/p (f ; δ) = 1. δ α−1/p

(6.2.3)

Take an arbitrary sequence I = {Ij } of nonoverlapping intervals contained in a period. Denote σk (I) = {j : 2−k−1 < |Ij | ≤ 2−k } (k ≥ 0). Then we have V =

∞ X |f (Ij )| j=1

λj

=

∞ X X |f (Ij )| . λj k=0 j∈σk (I)

We shall estimate V . By Hölder’s inequality, we have

V

1/p0 X 1 p0   ≤ |f (Ij )|p   λ j k=0 j∈σk (I) j∈σk (I)  0  p0 1/p ∞ X X 1  . ≤ ω1−1/p (f ; 2−k )  λj k=0 ∞ X

1/p 



X

j∈σk (I)

(6.2.4)

6.2. Embedding of Lipschitz classes

95

Thus, by (6.2.3) and (6.2.4) 1/p0 X 1 p0  . V ≤ 2−k(α−1/p)  λ j k=0 

∞ X

(6.2.5)

j∈σk (I)

Let the sequence {δn } be defined by card

n [

! σk (I)

= 2n δn ,

k=0

where, card(A) denotes the number of elements of the finite set A. Set also δ−1 = 0. There exists an n0 ≥ 0 such that δn > 0 for all n ≥ n0 , and we may assume n0 = 0. We observe that k{δn }kl1 ≤ 4. Indeed, first note that ∞ X

2−k card(σk (I)) ≤ 2

∞ X

|Ij | ≤ 2.

j=0

k=0

On the other hand, for n ≥ 0, we have 2−n card(σn (I)) = δn − δn−1 /2. Whence, for any N ∈ N, we have N

N +1 X

(δn − δn−1 /2) = δN +1 +

n=0

N

1X 1X δn ≥ δn , 2 n=0 2 n=0

and consequently, k{δn }kl1 ≤ 4. Applying Lemma 6.2 with θ = 2 and γ = 1/2 to {δk } yields a sequence {βk } such that δk ≤ βk , 2−1/2 ≤

βk+1 ≤ 2 (k ∈ N) and k{βk }kl1 ≤ 64. βk

(6.2.6)

Set νk = 2k βk . By the first relation of (6.2.6), we have 2νk ≤ νk+2 ≤ 16νk

(k ∈ N).

(6.2.7)

Since card(σk (I)) ≤ 2k δk = νk , and {λj } is increasing, we have by (6.2.5) !1/p0 ∞ X X 1 p0 −k(α−1/p) V ≤ 2 . λj 1≤j≤ν k=0 k

96


Applying (6.1.6) to the right-hand side of the previous inequality, we get 1/p0  ∞ X X 1 p0  , V ≤ cp,α 2−k(α−1/p)  λj ν ≤j≤ν k=1 k−1

k

for some constant cp,α > 0. Since 2−k = βk /νk , we have   0 α−1/p p0 1/p ∞ X X βk 1   . V ≤ cp,α ν λ k j ν ≤j≤ν k=1 k−1

(6.2.8)

k

By using Hölder’s inequality with exponents r and r0 , and the second inequality of (6.2.6), we estimate the right-hand side of (6.2.8)   0 0 1/r0  p0 r /p ∞ X X 1   1/r −r0 (α−1/p)   V ≤ cp,α k{βk }klr  νk  λ j ν ≤j≤ν k=1 k−1

  ∞ X  ≤ 64cp,α  k=1

X νk−1 ≤j≤νk

1 j α−1/p λj

k

 0 0 1/r0 p0 r /p    .

(6.2.9)

By collecting the terms of the sum at the right-hand side of (6.2.9) in pairs, and using that aq + bq ≤ 2(a + b)q for any q ≥ 0 and a, b ≥ 0, we get   r0 /p0 1/r0 0 ∞ p 1 X  X   (6.2.10) V ≤ 128cp,α   . α−1/p λ j j ν ≤j≤ν k=0 2k

2k+2

For k ≥ 0, we define mk ≥ 0 as the greatest integer m such that 2m < ν2k . By (6.2.7), we have 2ν2k ≤ ν2k+2 , and thus, 2mk +1 < ν2k+2 . Consequently, mk+1 ≥ mk + 1 (k ≥ 0).

(6.2.11)


97

Further, by (6.2.7), we have ν2k+2 ≤ 16ν2k . Therefore, for all k ≥ 0, [ν2k , ν2k+2 ] ⊂ [2mk , 2mk +5 ]. Whence,

X ν2k ≤j≤ν2k+2

p0

1

≤

j α−1/p λj

mk +5 2X

j=2mk

p0

1

.

j α−1/p λj

Since the terms of the previous sum decrease, it follows that mk +5 2X

j=2mk

p0

1 j α−1/p λj

≤ 40

mk +1 2X

j=2mk

p0

1

,

j α−1/p λj

Consequently, by the previous inequality and (6.2.10),  r0 /p0 1/r0  mk +1 0 ∞ 2 p 1  X  X  V ≤ c0p,α   , α−1/p j λ j m j=2 k k=0 for some c0p,α > 0. By (6.2.11), for each k ≥ 0, the intersection of [2mk , 2mk +1 ] and [2mk+1 , 2mk+1 +1 ] consists of at most one point. Hence,   0 n+1 p0 !r0 /p0 1/r ∞ 2X X 1  . V ≤ c0p,α  j α−1/p λj n=0 j=2n This proves (6.2.2). The estimate (6.2.2) is sharp in a sense. Namely, we have the following result. Theorem 6.6. Let Λ ∈ S be given, 1 0 depending only on α and p such that ω1−1/p (g; δ) ≤ c0p,α δ α−1/p

(0 < δ ≤ 1),

and  vΛ (g) ≥ c00p,α 

∞ X

n+1 2X

n=1

k=2n

1 k α−1/p λk

p0 !r0 /p0

(6.2.12)

1/r0 

.

(6.2.13)

98


Proof. Let {δn } ∈ l1 be a fixed but arbitrary positive sequence with k{δn }kl1 ≤ 1. Applying Lemma 6.2 with γ = 1 and θ = 3/2 (the value of γ does not matter, it is only important that 1 < θ < 2) to the sequence {δn }, we obtain a positive sequence {βn } such that δn ≤ βn (n ∈ N), βn+1 3 2 < ≤ 3 βn 2

(n ∈ N) and L = k{βn }kl1 ≤ 9.

(6.2.14)

Subdivide the interval [0, 1] into non-overlapping intervals Jn (n ∈ N) with |Jn | = βn /L. For n ∈ N, denote 2n+1 X−1

Sn =

1 λk

k=2n

p0 !1/p0 ,

and (n)

Hk

−1/(p−1)

= (2−n βn )α−1/p λk (n)

(n)

0

Sn−p /p

(n)

for 2n ≤ k ≤ 2n+1 − 1. n

Let also Hn = (H2n , H2n +1 , ..., H2n+1 −1 ) ∈ R2 . Put Fn (x) = F (Jn , 2n , Hn ; x) (see Definition 6.1), and ∞ X Fn (x). g(x) = n=1

It is clear that

n+1

vΛ (g) ≥ 2

∞ 2X −1 (n) X H k

n=1 k=2n

λk

.

(6.2.15)

On the other hand, 2n+1 X−1 k=2n

(n)

Hk = (2−n βn )α−1/p Sn . λk

Thus, since δk ≤ βk for k ∈ N, we have n+1

∞ 2X −1 (n) X H k

n=1 k=2n

λk

=

∞ X

βnα−1/p

−np0 (α−1/p)

2

2n+1 X−1 k=2n

n=1 −1/p+α

≥ 2

∞ X n=1

δnα−1/p

2n+1 X−1 k=2n

1 λk

p0 !1/p0

1 k α−1/p λk

p0 !1/p0 .


99

By the previous inequality and (6.2.15), vΛ (g) ≥ 2−1/p+α

∞ X

n+1 2X

δnα−1/p

k=2n

n=1

p0 !1/p0

1

.

k α−1/p λk

(6.2.16)

We proceed to estimate ω1−1/p (g; δ). By the first relation of (6.2.14), we have 2−n−1 βn+1 3 1 ≤ ≤ < 1. 3 2−n βn 4 In particular, the sequence {2−n βn } is strictly decreasing and 2−n βn → 0 as n → ∞. Fix 0 < δ ≤ 1. If δ > 2−1 β1 , then we set m = 0. Otherwise, define m ∈ N to be the unique natural number such that 2−m−1 βm+1 < δ ≤ 2−m βm . By (2.2.8), we have ω1−1/p (g; δ) ≤ δ 1/p

0

m X

∞ X

kFn0 kp +

n=1

vp (Fn ).

(6.2.17)

n=m+1

(The first sum is taken as zero if m = 0). We shall estimate the terms at the (n) right-hand side of (6.2.17). It follows from (6.1.4) and the definition of Hk that !1/p 2n+1 X−1 (n) 1/p p vp (Fn ) = 2 (Hk ) = 21/p (2−n βn )α−1/p . (6.2.18) k=2n

Further, by (6.1.5), kFn0 kp

= 2

βn /L 2n

−1/p0

2n+1 X−1

!1/p

(n) (Hk )p

k=2n

0

= 2L1/p (2−n βn )α−1 .

(6.2.19)

By the estimate L ≤ 9, (6.2.17), (6.2.18) and (6.2.19), ω1−1/p (g; δ) ≤ m ∞ X X 0 ≤ 18δ 1/p (2−n βn )α−1 + 21/p (2−n βn )α−1/p . n=1

n=m+1

(6.2.20)

100


Since

2−n+1 βn−1 2−n βn

α−1

=

2βn−1 βn

α−1

=

βn 2βn−1

1−α ≤

1−α 3 < 1, 4

we get m X

(2−n βn )α−1 ≤ (2−m βm )α−1

∞ n(1−α) X 3 n=0

n=1

4

= cα (2−m βm )α−1

≤ cα δ α−1 .

(6.2.21)

Similarly, ∞ X

−n

(2

α−1/p

βn )

−m−1

≤ (2

α−1/p

βm+1 )

n=m+1

∞ n(α−1/p) X 3 n=0

4

≤ cp,α δ α−1/p .

(6.2.22)

Thus, by (6.2.17), (6.2.21) and (6.2.22), ω1−1/p (g; δ) ≤ c0p,α δ α−1/p

(0 < δ ≤ 1).

Denote Ln =

n+1 2X

k=2n

p0 !1/p0

1 α−1/p

∞ X

.

k α−1/p λk

Clearly, {δn } ∈ l1 is equivalent to {δn choose {δn } ∈ l1 such that 1 δnα−1/p Ln ≥ 2 n=1

(6.2.23)

} ∈ lr . By Lemma 6.4, we can

∞ X

!1/r0 r0

Ln

.

(6.2.24)

n=1

0

If {Ln } ∈ / lr , then we must interpret (6.2.24) in the sense that we may choose {δn } ∈ l1 such that the left-hand side of (6.2.24) is infinite. In any case, the function g constructed above with this choice of {δn } satisfies (6.2.12) and (6.2.13), by (6.2.23), (6.2.16) and (6.2.24). Remark 6.7. As was mentioned in the Introduction, Wang observed that the condition 1/(1−α) ∞ X 1 < ∞. (6.2.25) λn n=1

6.3. A Perlman-type theorem

101

is necessary for the embedding (6.0.1) to hold, and he then conjectured that (6.2.25) is also sufficient. However, by combining Theorems 6.5 and 6.6, we obtain that the necessary and sufficient condition for (6.0.1)) is ∞ X

n+1 2X

n=0

k=2n

p0 !r0 /p0

1 k α−1/p λk

< ∞,

where r, r0 are given by (6.2.1). Clearly, this disproves Wang’s conjecture. Remark 6.8. For 1 ≤ p < ∞, α = 1, we have Lip(1; p) = Wp1 . It is easy to show that the embedding Wp1 ⊂ ΛBV holds for all sequences Λ ∈ S. Remark 6.9. Recall that for 1 ≤ p < ∞ and ω ∈ Ω1 , we denote Hpω = {f ∈ Lp ([0, 1]) : ω(f ; δ)p = O(ω(δ))}, and H ω = {f ∈ C : ω(f ; δ)C = O(ω(δ))}, where ω(f ; δ)C is the modulus of continuity in C. The problem of finding the necessary and sufficient condition for the embedding Hpω ⊂ ΛBV with general ω ∈ Ω1 and 1 ≤ p < ∞ is still open. On the other hand, the necessary and sufficient condition for the embedding H ω ⊂ ΛBV was obtained independently by Belov [4] and Medvedeva [47, 48]. Later, Leindler [40, 41] generalized these results.

6.3

A Perlman-type theorem

Perlman [58] showed that V1 =

\

ΛBV.

Λ∈S

We shall prove a similar result for Vp . Let 1 < p < ∞, denote by Sp0 the class of all sequences Λ = {λn } ∈ S such that p0 ∞ X 1 < ∞. λn n=1 Then we have the following theorem.

102


Theorem 6.10. Let 1 < p < ∞. Then \ Vp = ΛBV. Λ∈Sp0

Proof. Let f be a given function and {In } an arbitrary sequence of nonoverlapping intervals contained in a period. Applying Hölder’s inequality, we have p0 !1/p0 ∞ ∞ X X |f (In )| 1 ≤ vp (f ) . λn λn n=1 n=1 Thus, if Λ ∈ Sp0 , then Vp ⊂ ΛBV . Whence, \ Vp ⊂ ΛBV. Λ∈Sp0

Let now f be a bounded function with f ∈ / Vp . Then there exists a sequence {Jn } of nonoverlapping intervals contained in a period such that ∞ X

|f (Jn )|p = ∞.

n=1 0

Since {|f (Jn )|} ∈ / lp , there exists {αn } ∈ lp such that ∞ X

αn |f (Jn )| = ∞,

n=1

by Lemma 6.4. We may assume that αn > 0 for all n ∈ N and that {|f (Jn )|} is ordered nonincreasingly. Let {αn∗ } be the nonincreasing rearrangement of {αn }, set λn = 1/αn∗ and Λ = {λn }. Since {|f (Jn )|} is nonincreasing, we have ∞ ∞ ∞ X X |f (Jn )| X ∗ = αn |f (Jn )| ≥ αn |f (Jn )| = ∞, (6.3.1) λn n=1 n=1 n=1 whence f ∈ / ΛBV . It remains to show that Λ ∈ Sp0 . Clearly Λ is a positive and nondecreasing sequence. Moreover, |f (I)| ≤ 2kf k∞ for any interval. Therefore, ∞ X 1 X |f (Jn )| 1 ≥ = ∞, λ 2kf k∞ n=1 λn n=1 n

6.3. A Perlman-type theorem

103 0

by (6.3.1). Whence, Λ ∈ S. Furthermore, since {αn } ∈ lp , p0 X ∞ ∞ ∞ X X 1 0 0 = (αn∗ )p = αnp < ∞. λ n n=1 n=1 n=1 Thus, {λn } ∈ Sp0 . In connection to Theorem 6.10, we mention that embeddings between ΛBV and other spaces of functions of generalized bounded variation were previously studied in, e.g., [4, 6, 59, 62]. Remark 6.11. A result similar to Theorem 6.10 can also be proved for classes VΦ of functions of bounded Φ-variation. Remark 6.12. We can apply Theorem 6.10 to prove that there is a sequence Λ ∈ S that satisfies (6.2.25) but still Lip(α; p) 6⊂ ΛBV (thus disproving Wang’s conjecture mentioned above). Note first that 1 < 1/α < p < ∞. By Theorem 5.6, there exists a function f such that ω1−1/p (f ; δ) = O(δ α−1/p ), and at the same time f ∈ / V1/α . In light of (6.1.3), this means exactly that there is a function f ∈ Lip(α; p) such that f∈ / V1/a . Theorem 6.10 states that V1/α =

\

ΛBV.

(6.3.2)

Λ∈S1/(1−α)

Observe that S1/(1−α) is the collection of all sequences in S that satisfies (6.2.25). Since f ∈ / V1/α , (6.3.2) implies that for some Λ ∈ S1/(1−α) , we have f∈ / ΛBV . But since f ∈ Lip(α; p), we have shown that there exists a Λ that satisfies (6.2.25) while the embedding (6.0.1) does not hold.

Chapter 7 Multidimensional results The main objectives of this chapter is to study some problems related to (2) (2) bounded p-variation of bivariate functions (i.e., the classes Vp , Hp defined in the Introduction). In particular, we shall investigate the following: • sharp estimates of the Hardy-Vitali type p-variation and L∞ -norm of a function in terms of its mixed Lp -modulus of continuity; (2)

• Fubini-type properties of the class Hp (p ≥ 1).

7.1

Auxiliary results

Recall that Ω denotes the class of all moduli of continuity (see Chapter 2). Let f ∈ Lp ([0, 1]2 ), as we remarked before, ω(f ; ·)p ∈ Ω. Further, it is easy to show that for any fixed v ∈ [0, 1], the function ω(f ; ·, v)p ∈ Ω. Thus, by (2.1.2), ω(f ; u2 , v)p ω(f ; u1 , v)p ≤2 , u1 u2

0 < u2 ≤ u1 ≤ 1.

(7.1.1)

Similar relations hold with respect to the second variable v for a fixed u ∈ [0, 1]. Let h ∈ R, we shall use the following notations. ∆1 (h)f (x, y) = f (x + h, y) − f (x, y)

(7.1.2)

∆2 (h)f (x, y) = f (x, y + h) − f (x, y).

(7.1.3)

and

105

106

Chapter 7. Multidimensional results

The mixed difference (1.0.27) can be written as an iterated difference ∆(s, t)f (x, y) = ∆1 (s)∆2 (t)f (x, y) = ∆1 (s)∆2 (t)f (x, y). From here, k∆(s, t)(∆1 (h)f )kp = k∆1 (s)∆1 (h)∆2 (t)f kp . Applying the triangle inequality, we obtain the second estimate of the next lemma (the first inequality is proved similarly). Lemma 7.1. Let f ∈ Lp ([0, 1]2 ) (1 ≤ p < ∞) and h ∈ R. Then ω(∆1 (h)f ; δ)p ≤ 2 min{ω(f ; δ)p , ω(f ; h)p },

(7.1.4)

ω(∆1 (h)f ; u, v)p ≤ 2 min{ω(f ; u, v)p , ω(f ; h, v)p }.

(7.1.5)

and Similar estimates also hold if we consider ∆2 (h)f . Let f ∈ Lp ([0, 1]2 ) (1 < p < ∞). We shall use the following notations Z Jp (f ) =

1

t−1/p ω(f ; t)p

0

dt , t

(7.1.6)

1

Z

t−1/p [ω(f ; t, 1)p + ω(f ; 1, t)p ]

Kp (f ) = 0

and

1

Z

Z

Ip (f ) = 0

1

(uv)−1/p ω(f ; u, v)p

0

dt , t

du dv . u v

(7.1.7)

(7.1.8)

Let f ∈ Lp ([0, 1]2 ) (1 < p < ∞), then we have 4 Ip (f ). p0

Kp (f ) ≤

(7.1.9)

Indeed, by (7.1.1) 1

Z

−1/p−1

Ip (f ) =

u 0

≥

Z

1 2

1

v

−1/p ω(f ; u, v)p

0

Z 0

1

u−1/p−1 ω(f ; u, 1)p du

v Z 1 0

dv du

v −1/p dv.


107

Thus, 1

Z

t−1/p ω(f ; t, 1)p

dt 2 ≤ 0 Ip (f ). t p

t−1/p ω(f ; 1, t)p

2 dt ≤ 0 Ip (f ), t p

0

Similarly, one shows 1

Z 0

and (7.1.9) follows. In the same way, one demonstrates that 4 Ip (f ). (p0 )2

ω(f ; 1, 1)p ≤

(7.1.10)

Denote by Lp0 ([0, 1]2 ) the subspace of Lp ([0, 1]2 ) that consists of functions f such that Z Z 1

1

f (t, y)dt = 0

f (x, t)dt = 0

0

for a.e. x, y ∈ R. Observe that every function f ∈ Lp ([0, 1]2 ) can be written as f (x, y) = f¯(x, y) + φ1 (x) + φ2 (y), a.e. (x, y) ∈ R2 , (7.1.11) p 2 ¯ where f ∈ L ([0, 1] ). Indeed, let 0

Z

1

f (x, t)dt,

φ1 (x) =

(7.1.12)

0

Z

1

ZZ f (t, y)dt −

φ2 (y) = 0

f (s, t)dsdt.

(7.1.13)

[0,1]2

Then the function f¯(x, y) = f (x, y) − φ1 (x) − φ2 (y) belongs to Lp0 ([0, 1]2 ). It was proved in [65] that if f ∈ Lp0 ([0, 1]2 ), then ω(f ; δ)p ≤ 3[ω(f ; δ, 1)p + ω(f ; 1, δ)p ],

0 ≤ δ ≤ 1.

Whence, it follows that if f ∈ Lp0 ([0, 1]2 ) (1 < p < ∞), then Jp (f ) ≤ 3Kp (f ).

(7.1.14)

108


If f (x, y) = g(x)h(y), then for all p ≥ 1, there holds vp(2) (f ) = vp (g)vp (h),

(7.1.15)

and u, v ∈ [0, 1].

ω(f ; u, v)p = ω(g; u)p ω(h; v)p ,

(7.1.16)

(2) Hp

Recall that when defining the class (see the Introduction), we require in addition to (1.0.25) also that the sections fx , fy ∈ Vp for all x, y ∈ R. In fact, it is sufficient to assume that there exists at least two values x0 , y0 ∈ R such that f (x0 , ·), f (·, y0 ) ∈ Vp . Indeed, assume that f (x0 , ·) ∈ Vp for some x0 ∈ R and let x ∈ R be fixed but arbitrary. Take any partition Π = {y0 , y1 , ..., yn } and set ∆f (x0 , yj ) = f (x, yj+1 ) − f (x, yj ) − f (x0 , yj+1 ) + f (x0 , yj ), for 0 ≤ j ≤ n − 1. By the Minkowski inequality, we have vp (fx ; Π) =

n−1 X

!1/p p

|f (x, yj+1 ) − f (x, yj )|

j=0

≤

n−1 X

!1/p p

|∆f (x0 , yj )|

+ vp (fx0 ).

j=0

Whence, vp (fx ) ≤ vp(2) (f ) + vp (fx0 ). A similar inequality holds for vp (fy ). The next result is due to Golubov [26]. Lemma 7.2. Assume that f ∈ L10 ([0, 1]2 ) and let Z xZ y F (x, y) = f (s, t)dsdt. 0

Then (2)

Z

1

0

Z

1

|f (x, y)|dxdy.

v1 (F ) = 0

(7.1.17)

0

Remark 7.3. The condition f ∈ L10 ([0, 1]2 ) is imposed to assure that F is 1-periodic in both variables.

7.2. Estimates of the L∞ -norm

109

We shall also need the following lemma, which is a special case of a Hellytype principle proved in [42]. (2)

Lemma 7.4. Let {fn } be a sequence of functions in H1 . Assume that there exist x0 , y0 ∈ R and M > 0 such that the estimate (2)

v1 (fn ) + v1 (fn (·, y0 )) + v1 (fn (x0 , ·)) + |fn (x0 , y0 )| ≤ M holds uniformly in n. Then there exists a subsequence {fnj } that converges (2) at every point to a function f ∈ H1 .

7.2

Estimates of the L∞-norm

Recall the notations (7.1.6) and (7.1.8). Potapov [63, 64, 65] obtained estimates of the L∞ -norm of a function in terms of its mixed Lp -modulus of continuity (see also [66]). However, the behaviour of the constant coefficients in these estimates were not investigated. In this section, we study this problem. Observe first that for f ∈ Lp ([0, 1]2 ) (1 < p < ∞), the condition Ip (f ) < ∞ alone is not sufficient to ensure that f ∈ L∞ ([0, 1]2 ). Indeed, if f (x, y) = g(x, y)+φ(x), then Ip (f ) = Ip (g), but φ is an arbitrary function (in particular, φ can be unbounded). Theorem 7.5. Let f ∈ Lp ([0, 1]2 ) (1 < p < ∞) and suppose that Jp (f ) < ∞

and Ip (f ) < ∞.

Then f is equal a.e. to a continuous function and # " 2 1 1 Ip (f ) , kf k∞ ≤ A kf kp + 0 Jp (f ) + pp pp0

(7.2.1)

(7.2.2)

where A is an absolute constant. Proof. Assume that (7.2.1) holds, we shall first prove the estimate (7.2.2). For each x ∈ [0, 1], we apply (3.2.2) to the x-section fx . Using also (2.1.5), we have Z 1 1 kfx k∞ ≤ A kfx kp + 0 v −1/p−1 k∆(v)fx kp dv , (7.2.3) pp 0

110


where ∆(v)fx (y) = f (x, y + v) − f (x, y). Put α(x) = kfx kp ,

βv (x) = k∆(v)fx kp ,

and 1 Φ(x) = α(x) + 0 pp

Z

(7.2.4)

1

v −1/p−1 βv (x)dv.

(7.2.5)

0

By (7.2.3) kf k∞ = ess sup kfx k∞ ≤ A ess sup Φ(x). 0≤x≤1

(7.2.6)

0≤x≤1

We shall estimate kΦk∞ . By (3.2.2) and (2.1.5), we have Z 1 1 kΦk∞ ≤ A kΦkp + 0 u−1/p−1 k∆(u)Φkp du , pp 0

(7.2.7)

where ∆(u)Φ(x) = Φ(x + u) − Φ(x). It follows easily from the definitions (7.2.4) that kαkp = kf kp , k∆(u)αkp ≤ ω(f ; u)p , (7.2.8) and kβv kp ≤ ω(f ; v)p ,

k∆(u)βv kp ≤ ω(f ; u, v)p .

(7.2.9)

We estimate both terms of (7.2.7), starting with kΦkp . By Minkowski’s inequality and the left inequalities of (7.2.8) and (7.2.9), we get p 1/p Z 1 Z 1 1 −1/p−1 v βv (x)dv dx kΦkp ≤ kf kp + 0 pp 0 0 1/p Z 1 Z 1 1 −1/p−1 p ≤ kf kp + 0 v βv (x) dx dv pp 0 0 1 ≤ kf kp + 0 Jp (f ). (7.2.10) pp We proceed to estimate k∆(u)Φkp . Put Z 1 v −1/p−1 βv (x)dv. I(x) = 0

Then, by Minkowski’s inequality and the right inequality of (7.2.8) 1 k∆(u)Ikp pp0 1 ≤ ω(f ; u)p + 0 k∆(u)Ikp . pp

k∆(u)Φkp ≤ k∆(u)αkp +

(7.2.11)

7.2. Estimates of the L∞ -norm

111

Further, since 1

Z

v −1/p−1 |βv (x + u) − βv (x)|dv,

|I(x + u) − I(x)| ≤ 0

we get after applying Minkowski’s inequality that 1

Z

Z

1

k∆(u)Ikp ≤ Z ≤

0 1

v −1/p−1 |∆(u)βv (x)|dv

1/p

p dx

0

v −1/p−1 k∆(u)βv kp dv.

(7.2.12)

0

By (7.2.11), (7.2.12) and the right inequality of (7.2.9), we have Z 1 1 v −1/p−1 ω(f ; u, v)p dv. k∆(u)Φkp ≤ ω(f ; u)p + 0 pp 0

(7.2.13)

Now, (7.2.2) follows from (7.2.6), (7.2.7), (7.2.10) and (7.2.13). We now prove that f agrees a.e. with a continuous function. To do this, it is sufficient to show that ω(f ; δ)∞ → 0 as δ → 0. Fix δ ∈ (0, 1], then ω(f ; δ)∞ ≤ sup k∆1 (h)f k∞ + sup k∆2 (h)f k∞ , 0≤h≤δ

0≤h≤δ

where ∆1 (h)f, ∆2 (h)f are defined by (7.1.2) and (7.1.3) respectively. For h ∈ (0, δ], we have by (7.2.2) that # " 2 1 1 k∆1 (h)f k∞ ≤ A k∆1 (h)f kp + 0 Jp (∆1 (h)f ) + Ip (∆1 (h)f ) . pp pp0 By using Lemma 7.1, (2.1.2) and (7.1.1), we get for any 0 < h ≤ δ Z δ dt Jp (∆1 (h)f ) ≤ c t−1/p ω(f ; t)p , t 0 and

Z δZ

1

dv du , v u 0 0 for some constant c that is independent of δ. It follows that Ip (∆1 (h)f ) ≤ c

(uv)−1/p ω(f ; u, v)p

lim( sup k∆1 (h)f k∞ ) = 0.

δ→0 0≤h≤δ

112


In exactly the same way, we can show that lim( sup k∆2 (h)f k∞ ) = 0.

δ→0 0≤h≤δ

Hence, limδ→0 ω(f ; δ)∞ = 0. This concludes the proof. Corollary 7.6. Let f ∈ Lp ([0, 1]2 ) (1 < p < ∞) and assume that Ip (f ) < ∞. Then there exist a continuous function g ∈ Lp0 ([0, 1]2 ) and univariate functions φ1 , φ2 such that f (x, y) = g(x, y) + φ1 (x) + φ2 (y), for a.e. (x, y) ∈ R2 . Proof. By (7.1.11), we have f (x, y) = f¯(x, y) + φ1 (x) + φ2 (y) for a.e. (x, y) ∈ R2 , where f¯ ∈ Lp ([0, 1]2 ). We shall prove that f¯ is equal a.e. to a continuous function g. Clearly Ip (f¯) = Ip (f ) < ∞, and since f¯ ∈ Lp0 ([0, 1]2 ), we also have Jp (f¯) < ∞, by (7.1.14) and (7.1.9). The result now follows from Theorem 7.5.

7.3

Estimates of the Vitali type p-variation

In this section we shall consider the relationship between mixed integral smoothness and the Vitali type p-variation. In the case p = 1, we have the following theorem. Theorem 7.7. Assume that f ∈ L1 ([0, 1]2 ) and that ω(f ; u, v)1 = O(uv). (2)

Then there exist a function g ∈ H1 that for a.e. (x, y) ∈ R2 ,

and univariate functions φ1 , φ2 such

f (x, y) = g(x, y) + φ1 (x) + φ2 (y). Moreover, (2)

ω(f ; u, v)1 . uv u,v>0

v1 (g) = sup

(7.3.1)

7.3. Estimates of the Vitali type p-variation

113

Proof. We may without loss of generality assume that f ∈ L10 ([0, 1]2 ). For n ∈ N, denote Z 1/n Z 1/n fn (x, y) = n2 f (x + s, y + t)dsdt. 0

0

We shall first prove that (2)

v1 (fn ) ≤ sup

u,v>0

ω(f ; u, v)1 . uv

(7.3.2)

Observe that Z

x

Z

y

D1 D2 fn (s, t)dsdt − fn (x, 0) − fn (0, y) + fn (0, 0)

fn (x, y) = 0

0

= Fn (x, y) − fn (x, 0) − fn (0, y) + fn (0, 0). Moreover, D1 D2 fn (s, t) = n2 ∆(1/n, 1/n)f (s, t). Thus, by (7.1.17), Z 1Z 1 (2) (2) v1 (fn ) = v1 (Fn ) = n2 |∆(1/n, 1/n)f (x, y)|dxdy 0

0

ω(f ; u, v)1 ≤ sup . uv u,v>0 This proves (7.3.2). Let E be the set of Lebesgue points of f . Since R2 \ E has Lebesgue measure 0, there exist (x0 , y0 ) ∈ E such that the sections E(x0 ) = {y ∈ R : (x0 , y) ∈ E} and E(y0 ) = {x ∈ R : (x, y0 ) ∈ E}, have full measure. That is, mes1 (R \ E(x0 )) = mes1 (R \ E(y0 )) = 0,

(7.3.3)

where mes1 denotes linear Lebesgue measure. For n ∈ N, define now gn (x, y) = fn (x, y) − fn (x, y0 ) − fn (x0 , y) + fn (x0 , y0 ). For each n ∈ N, we have gn (x, y0 ) = gn (x0 , y) = 0 for all x, y ∈ R. Thus, by (7.3.2), (2)

v1 (gn ) + v1 (gn (·, y0 )) + v1 (gn (x0 , ·)) + |gn (x0 , y0 )| = ω(f ; u, v)1 (2) (2) . = v1 (gn ) = v1 (fn ) ≤ sup uv u,v>0

(7.3.4)

114


By Lemma 7.4, there is a subsequence gnj that converges at all points to a (2) function g ∈ H1 . On the other hand, by (7.3.3) and Lebesgue’s differentiation theorem, for a.e. (x, y) ∈ R2 there holds g(x, y) = f (x, y) − f (x, y0 ) − f (x0 , y) + f (x0 , y0 ). Take φ1 (x) = f (x, y0 ) and φ2 (y) = f (x0 , y) − f (x0 , y0 ), then f (x, y) = g(x, y) + φ1 (x) + φ2 (y) for a.e. (x, y) ∈ R2 . We now prove (7.3.1). Since gnj converges to g at all points, it follows from (7.3.4) that for any net N , (2)

(2)

v1 (g; N ) = lim v1 (gnj ; N ) ≤ sup j

u,v>0

ω(f ; u, v)1 . uv

(2)

Thus, v1 (g) ≤ sup ω(f ; u, v)/uv. On the other hand, since f = g a.e., we have for any u, v ∈ [0, 1] (2)

ω(f ; u, v)1 = ω(g; u, v)1 ≤ v1 (g)uv, (2)

by (1.0.28). Whence, sup ω(f ; u, v)1 /uv ≤ v1 (g). This proves (7.3.1). Recall the notations (7.1.7) and (7.1.8). Theorem 7.8. Let f ∈ Lp ([0, 1]2 ) (1 < p < ∞) and assume that Ip (f ) < ∞. (2) Then there exists a continuous function g ∈ Hp and univariate functions φ1 , φ2 such that for a.e. (x, y) ∈ R2 , we have f (x, y) = g(x, y) + φ1 (x) + φ2 (y).

(7.3.5)

Moreover, "

vp(2) (g)

1 ≤ A ω(f ; 1, 1)p + 0 Kp (f ) + pp

1 pp0

2

# Ip (f ) ,

(7.3.6)

where A is an absolute constant. If f ∈ Lp0 ([0, 1]2 ), then we may take φ1 = φ2 = 0 in (7.3.5). Proof. By Corollary 7.6, there is a continuous function g ∈ Lp0 ([0, 1]2 ) such that f (x, y) = g(x, y) + φ1 (x) + φ2 (y)


115

for a.e. (x, y) ∈ R2 (if f ∈ Lp0 ([0, 1]2 ), then φ1 = φ2 = 0). We shall prove (2) that g ∈ Hp . Take any net N = {(xi , yj ) : 0 ≤ i ≤ m, 0 ≤ j ≤ n}, and set gi (y) = g(xi+1 , y) − g(xi , y),

0 ≤ i ≤ m − 1.

Clearly, vp(2) (g; N )

m−1 n−1 XX

=

!1/p p

|∆g(xi , yj )|

i=0 j=0 m−1 n−1 XX

=

!1/p p

|gi (yj+1 ) − gi (yj )|

i=0 j=0 m−1 X

≤

!1/p p

vp (gi )

.

(7.3.7)

i=0

By (3.2.3) and (2.1.5), we have for 0 ≤ i ≤ m − 1 Z 1 1 v −1/p−1 k∆(v)gi kp dv , vp (gi ) ≤ A Ωp (gi ) + 0 pp 0

(7.3.8)

where ∆(v)gi (y) = gi (y + v) − gi (y). Set Z 1 Ii = v −1/p−1 k∆(v)gi kp dv. 0

By (7.3.8), m−1 X i=0

!1/p vp (gi )p

 ≤ A

m−1 X

!1/p Ωp (gi )p

i=0

1 + 0 pp

m−1 X

!1/p  Iip

i=0

Denote gy,v (x) = g(x, y + v) − g(x, y). Since Z 1Z 1 Ωp (gi )p = |gi (y + v) − gi (y)|p dydv, 0

0

.

(7.3.9)

116


we have m−1 X

1

Z

p

Ωp (gi )

1 m−1 X

Z

= 0

i=0

0 1

Z

Z

= 0

i=0 1 m−1 X

0 1

Z

|gy,v (xi+1 ) − gy,v (xi )|p dydv

i=0 1

Z

vp (gy,v )p dydv.

≤ 0

|gi (y + v) − gi (y)|p dydv

0

Further, by (3.2.3) and (2.1.5), we have p Z 1 1 −1/p−1 p p t k∆(t)gy,v kp dt . vp (gy,v ) ≤ A Ωp (gy,v ) + pp0 0 Thus, m−1 X

!1/p

"Z

p

Z

0

i=0 1

Z

1 + 0 pp

1

≤A

Ωp (gi )

1

Z

1

Z

−1/p−1

t 0

0

Z

Ωp (gy,v )p =

1

Z

0

1/p +

0

1/p # p . k∆(t)gy,v kp dt dydv

1

|∆(h, v)g(x, y)|p dxdh,

1

1

Z

(7.3.10)

0

thus 0

Ωp (gy,v )p dydv

0

Observe that

Z

1

Ωp (gy,v )p dydv

1/p ≤ ω(g; 1, 1)p .

0

Next, by Minkowski’s inequality, !1/p Z 1 1/p #p Z 1 Z 1 "Z 1 −1/p−1 p t |gy,v (x + t) − gy,v (x)| dx dt dydv 0

0

0

0 1

Z

t−1/p−1

≤ 0

Z ≤ 0

1

Z 0

1

t−1/p−1

Z 0

Z 0

1

Z

1

|gy,v (x + t) − gy,v (x)|p dxdydv

1/p dt

0

1

ω(g; t, v)pp dv

1/p

Z dt ≤ 0

1

t−1/p−1 ω(g; t, 1)p dt


117

Thus, we have m−1 X

!1/p

1 ≤ A ω(g; 1, 1)p + 0 Kp (g) . pp

p

Ωp (gi )

i=0

(7.3.11)

Now we estimate the second term of (7.3.9). Applying Minkowski’s inequality, we obtain m−1 X

!1/p Iip

1

Z ≤

v

−1/p−1

0

i=0

m−1 X

!1/p k∆(v)gi kpp

dv.

(7.3.12)

i=0

Furthermore, m−1 X

Z

k∆(v)gi kpp

1

= 0

i=0

Z ≡

m−1 X

! p

|gi (y + v) − gi (y)|

dy

i=0 1

Sv (y)dy.

(7.3.13)

0

On the other hand, Sv (y) = =

m−1 X i=0 m−1 X

|g(xi+1 , y + v) − g(xi , y + v) − g(xi+1 , y) + g(xi , y)|p |gy,v (xi+1 ) − gy,v (xi )|p ,

i=0

where gy,v (x) = g(x, y + v) − g(x, y). Thus, by (3.2.3) and (2.1.5), for a fixed y ∈ [0, 1], we have the following estimate p Z 1 1 Sv (y) ≤ A Ωp (gy,v ) + 0 u−1/p−1 k∆(u)gy,v kp du pp 0 p Z 1 1 −1/p−1 ≤ 2p A Ωp (gy,v )p + u k∆(u)g k du . (7.3.14) y,v p pp0 0 Further, by (7.3.10), Z 0

1

Ωp (gy,v )p dy ≤ ω(g; 1, v)pp .

118


This inequality, (7.3.14) and Minkowski’s inequality yield " Z 1 1/p 0 Sv (y)dy ≤ A ω(g; 1, v)p + 0

1 + 0 pp

Z

1

u

−1/p−1

k∆(u)gy,v kpp dy

1/p

# du .

(7.3.15)

0

0

Since k∆(u)gy,v kpp =

1

Z

Z

1

|∆(u, v)g(x, y)|p dx,

0

we obtain from (7.3.13) and (7.3.15) !1/p m−1 X p k∆(v)gi kp ≤ i=0

1 ≤ A ω(g; 1, v)p + 0 pp 0

Z

1

u

−1/p−1

ω(g; u, v)p du .

0

Integrating this inequality with respect to v and taking into account (7.3.12), we have !1/p m−1 X p 1 ≤ A0 Kp (g) + 0 Ip (g) . Ii pp i=0 The above inequality together with (7.3.9) and (7.3.11) yield !1/p m−1 X p vp (gi ) ≤ i=0

"

1 ≤ A ω(g; 1, 1)p + 0 Kp (g) + pp 0

1 pp0

2

# Ip (g) .

(7.3.16)

The estimate (7.3.6) follows now from (7.3.7), (7.3.16), and the fact that ω(g; u, v)p = ω(f ; u, v)p . (2) To show that g ∈ Hp , we also need to demonstrate that there exist x, y ∈ R such that gx , gy ∈ Vp . By applying (3.2.3) and (2.1.5) to an arbitrary x-section gx , we get Z 1 1 vp (gx ) ≤ A kgx kp + 0 v −1/p−1 k∆(v)gx kp dv = AΦ(x). pp 0


119

It was shown in the proof of Theorem 7.5 that if Jp (g) and Ip (g) are finite, then Φ ∈ L∞ ([0, 1]). Now, since g ∈ Lp0 ([0, 1]2 ), we have Jp (g) ≤ 12Ip (g)/p0 , by (7.1.14) and (7.1.9). Thus, for a.e. x ∈ R, vp (gx ) ≤ AkΦk∞ < ∞. In the same way, we have gy ∈ Vp for a.e. y ∈ R. This concludes the proof. Below we shall demonstrate that the estimate (7.3.6) is sharp in a sense. For this, we use the following results. Let tn (x) = sin 2πnx, for n ∈ N. It is easy to show that we have n1/p ≤ vp (tn ) ≤ 2πn1/p

(7.3.17)

ω(tn ; δ)p ≤ 2π min(1, nδ).

(7.3.18)

and Remark 7.9. Let 1 < p ≤ 2, by (7.1.9) and (7.1.10), we have ω(f ; 1, 1)p +

8 1 Kp (f ) ≤ 0 2 Ip (f ) 0 p (p )

Whence, for 1 < p ≤ 2, the estimate (7.3.6) assumes the form vp(2) (f ) ≤

A Ip (f ). (p0 )2

(7.3.19)

The constant 1/(p0 )2 has the optimal order as p → 1. Indeed, let f (x, y) = (2) t1 (x)t1 (y), then f ∈ Hp for all p ≥ 1. By (7.3.17) and (7.1.15), we have (2) vp (f ) ≥ 1 for all p ≥ 1. On the other hand, by (7.3.18) and (7.1.16), we easily get that Ip (f ) ≤ 4π 2 (p0 )2 for p > 1. This shows that the constant coefficient 1/(p0 )2 at the right-hand side of (7.3.19) cannot be replaced with some cp such that limp→1 (p0 )2 cp = 0. Remark 7.10. Let p > 2, then 1 < p0 < 2 and the estimate (7.3.6) takes the form. 1 1 vp(2) (f ) ≤ A ω(f ; 1, 1)p + Kp (f ) + 2 Ip (f ) . (7.3.20) p p

120


We shall prove that the first term at the right-hand side of (7.3.20) cannot be omitted, and that the constant coefficients of the other two terms have the optimal asymptotic behaviour as p → ∞. (2) Take first f (x, y) = t1 (x)t1 (y). As above, vp (f ) ≥ 1 for all p > 1 and (2) thus limp→∞ vp (f ) ≥ 1. On the other hand, by (7.3.18) and (7.1.16), we have for all p > 2 the inequalities 1 16π 2 Kp (f ) ≤ p p

and

1 16π 2 Ip (f ) ≤ 2 , 2 p p

This shows that the term ω(f ; 1, 1)p of (7.3.20) cannot be omitted. We proceed to show the sharpness of the constant coefficients. For fixed but arbitrary 1 < p < ∞, let αp , βp be any coefficients such that vp(2) (f ) ≤ A [ω(f ; 1, 1)p + αp Kp (f ) + βp Ip (f )] ,

(7.3.21)

holds for some absolute constant A and all (continuous) functions f ∈ Lp ([0, 1]2 ) with Ip (f ) < ∞. In light of Theorem 7.8, we may assume that αp ≤ 1/p and βp ≤ 1/p2 . We shall prove that these decay rates are optimal, i.e., that limp→∞ pαp > 0 and limp→∞ p2 βp > 0. Let f (x, y) = tn (x)t1 (y), where n ∈ N is fixed but arbitrary. By (7.3.18) and (7.1.16), we have ω(f ; u, v)p ≤ 4π 2 v min(nu, 1). Simple calculations shows that there exists an absolute constant A > 0 such that Kp (f ) ≤ Apn1/p and Ip (f ) ≤ Apn1/p . On the other hand, by (7.3.17) (2) and (7.1.15), we have vp (f ) ≥ n1/p . Putting these estimates into (7.3.21) and taking into consideration that βp ≤ 1/p2 yield that for all p > 2 and all n ∈ N, we have 1 n1/p , n1/p ≤ A 1 + pαp + p where A is an absolute constant. Assume that limp→∞ pαp = 0. Then, given any ε > 0, we may choose r = r(ε) such that for all n ∈ N, there holds n1/r ≤ A(1 + εn1/r ). In particular, take ε = 1/(2A) and choose subsequently n ∈ N large enough to have n1/r > 2A. This gives the contradiction n1/r n1/r ≤ A 1 + < n1/r . 2A


121

Whence, limp→∞ pαp > 0. To show that limp→∞ p2 βp > 0, take f (x, y) = tn (x)tn (y), where n ∈ N is fixed but arbitrary. As above, we have ω(f ; u, v)p ≤ 4π 2 min(nv, 1) min(nu, 1). Then there exists an absolute constant A > 0 such that Kp (f ) ≤ Apn1/p and (2) Ip (f ) ≤ Ap2 n2/p . On the other hand, vp (f ) ≥ n2/p . Putting these estimates into (7.3.21) yields that for all n ∈ N and p > 2, n2/p ≤ A[1 + pαp n1/p + p2 βp n2/p ] where A > 0 is an absolute constant. Dividing by n1/p and taking into consideration that pαp ≤ 1, we see that n1/p ≤ A[2 + p2 βp n1/p ], for all p > 2 and all n ∈ N. From here, we can give a proof by contradiction of the inequality limp→∞ p2 βp > 0, as above. Remark 7.11. We shall consider trigonometric polynomials of two variables and degree (n, m): Tn,m (x, y) =

n X m X [aj,k cos 2πjx cos 2πky + bj,k cos 2πjx sin 2πky j=0 k=0

+ cj,k sin 2πjx cos 2πky + dj,k sin 2πjx cos 2πky].

(7.3.22)

Oskolkov [54] proved that for any trigonometric polynomial (7.3.22) of degree (n, m) and any 1 ≤ p < ∞, there holds vp(2) (Tn,m ) ≤ A(nm)1/p kTn,m kp ,

(7.3.23)

where A is an absolute constant. We can obtain (7.3.23) directly from (7.3.6). Indeed, take any trigonometric polynomial T of degree (n, m). The estimate ω(T ; u, v)p ≤ min(uvkD1 D2 T kp , 4kT kp ),

u, v ∈ [0, 1],

is immediate. By using (7.3.24), we get Z 1/nm Z −1/p Kp (T ) ≤ 2kD1 D2 T kp t dt + 4kT kp 0

(7.3.24)

1

t−1/p−1 dt

1/nm

≤ 2p0 (nm)1/p−1 kD1 D2 T kp + 4p(nm)1/p kT kp .

(7.3.25)

122


It is a simple consequence of Bernstein’s inequality (see [14, p. 97]) that kD1 D2 T kp ≤ 4π 2 nmkT kp .

(7.3.26)

By (7.3.25) and (7.3.26), we get Kp (T ) ≤ 12π 2 pp0 (nm)1/p kT kp .

(7.3.27)

Similarly, by (7.3.24), Z

1/n

Z

1/m

Ip (T ) ≤ kD1 D2 T kp u1/p v 1/p dvdu 0 0 Z 1 Z 1 (uv)−1/p−1 dvdu + 4kT kp 1/n

1/m

≤ (p0 )2 (nm)1/p−1 kD1 D2 T kp + 4p2 (nm)1/p kT kp . By the above estimate and (7.3.26), we have Ip (T ) ≤ 8π 2 (pp0 )2 (nm)1/p kT kp .

(7.3.28)

Now, (7.3.23) is derived from (7.3.6), the estimate ω(T ; 1, 1)p ≤ 4kT kp , (7.3.27) and (7.3.28).

7.4

(2)

Fubini-type properties of Hp

Recall that for p ≥ 1, the set Vp [ Vp ]sym of functions of bounded iterated p-variation consists of all functions f such that if ϕ(x) = vp (fx ) and ψ(y) = vp (fy ), then ϕ, ψ ∈ Vp . We observe first that Vp [ Vp ]sym is not a vector space. Proposition 7.12. There are two functions f and g such that for any 1 ≤ p < ∞, we have f, g ∈ Vp [ Vp ]sym but (f + g) ∈ / Vp [ Vp ]sym . Proof. Let f, g be functions that are 1-periodic in each variable, and defined as follows on [0, 1]2 . Let f (x, y) = 1 if y = x and f (x, y) = 0 otherwise. Set g(x, y) = 1 if y = x and x ∈ / Q, g(x, y) = −1 if y = x and x ∈ Q and

(2)

123

7.4. Fubini-type properties of Hp

g(x, y) = 0 otherwise. Then it is easy to see that for any x, y ∈ [0, 1], we have vp (fx ) = 21/p , vp (fy ) = 21/p . Since vp (fx ), vp (fy ) are constant functions, they are of bounded p-variation, that is, f ∈ Vp [ Vp ]sym . In the same way, we have g ∈ Vp [ Vp ]sym . On the other hand,  / Q,  2 if y = x and x ∈ 0 if y = x and x ∈ Q, (f + g)(x, y) =  0 otherwise. Then vp ([f + g]x ) = 21+1/p if x ∈ / Q and vp ([f + g]x ) = 0 for x ∈ Q. Clearly, the function x 7→ vp ([f + g]x ) ∈ / Vp . As was mentioned before, it was shown in [1] that (2)

H1 ⊂ V1 [ V1 ]sym .

(7.4.1)

The inclusion (7.4.1) is strict. In fact, we have the following result. Proposition 7.13. Let 1 ≤ p < ∞, then there is a function f ∈ Vp [ Vp ]sym (2) such that f ∈ / Hp . Proof. Define f on (0, 1]2 f (x, y) =

1 if 0 < x ≤ y ≤ 1 0 if 0 < y < x ≤ 1,

and extend to the whole plane by periodicity. It is clear that vp (fx ) = vp (fy ) = 21/p for all x, y. Thus, f ∈ Vp [ Vp ]sym for 1 ≤ p < ∞. On the other hand, fix n ∈ N and let Nn = {(xi , yj )}, where xi =

i n

and yj =

j + 1/2 , n

0 ≤ i, j ≤ n.

Then |∆f (xi , yi )|p = 1 (2)

(2)

for 0 ≤ i ≤ n − 1, whence, vp (f ; Nn ) ≥ n1/p . Thus, f ∈ / Hp .

124

Chapter 7. Multidimensional results (2)

We will now proceed to consider the embedding Hp p > 1. We will use the following function φ(x) = inf |x − k|, k∈Z

⊂ Vp [ Vp ]sym for

x ∈ R.

(7.4.2)

For each n ∈ N, denote φn (x) = φ(nx). It is easy to see that vp (φn ) = 21/p−1 n1/p .

(7.4.3)

gn (x) = φ(2n x − 1)χ[0,1] (2n x − 1) for x ∈ [0, 1].

(7.4.4)

Define and extend gn to a 1-periodic function. Restricted to [0, 1], gn is supported on [2−n , 2−n+1 ] and the graph of gn is an isosceles triangle with height 1/2. Lemma 7.14. Let {αn } be any sequence of real numbers, and define g(x) =

∞ X

αn gn (x),

n=1

where the functions gn are given by (7.4.4). Then, for 1 ≤ p < ∞, we have !1/p ∞ X 1/p p vp (g) ≤ 2 |αn | . (7.4.5) n=1

Proof. For n ∈ N, set fn (x) = αn gn (x). Clearly, the functions fn have pairwise disjoint supports. Moreover, it is easy to see that vp (fn ) = 21/p−1 |αn | (n ∈ N).

(7.4.6)

Assume first that all αn are nonnegative. Then the functions fn are nonnegative, and by Lemma 2.6 and (7.4.6), we have !1/p ∞ X vp (g) = 21/p−1 αnp . (7.4.7) n=1

αn0

When {αn } changes sign, we set Then αn0 , αn00 ≥ 0 for all n ∈ N, and g(x) =

∞ X n=1

αn0 gn (x) −

= max(αn , 0) and αn00 = − min(αn , 0).

∞ X n=1

αn00 gn (x) = h1 (x) − h2 (x).

(2)

125


Applying (7.4.7) to h1 , h2 , we obtain  vp (g) ≤ vp (h1 ) + vp (h2 ) = 21/p−1 

∞ X

!1/p (αn0 )p

+

∞ X (αn00 )p

n=1

!1/p  .

n=1

Since αn0 , αn00 ≤ |αn |, (7.4.5) follows. Theorem 7.15. For p > 1, we have Hp(2) 6⊂ Vp [ Vp ]sym . Proof. Let 1 < p < ∞ and set f (x, y) =

∞ X

2−k/p gk (x)φ(2k y),

(7.4.8)

k=1

where φ is given by (7.4.2) and gk (k ∈ N) by (7.4.4). We shall prove that (2) the function f defined by (7.4.8) belongs to Hp \ Vp [ Vp ]sym . (2) First, we show that f ∈ Vp . Fix any net N = {(xi , yj ) : 0 ≤ i ≤ m, 0 ≤ j ≤ n}. For each j ∈ {0, 1, ..., n − 1}, denote fj (x) = f (x, yj+1 ) − f (x, yj ). Since ∆f (xi , yj ) = fj (xi+1 ) − fj (xi ), we get m−1 X

|∆f (xi , yj )|p =

i=0

m−1 X

|fj (xi+1 ) − fj (xi )|p ≤ vp (fj )p .

i=0

Thus, vp(2) (f ; N )p ≤

n−1 X

vp (fj )p .

j=0

On the other hand, we note that fj (x) =

∞ X k=1

2−k/p [φ(2k yj+1 ) − φ(2k yj )]gk (x).

(7.4.9)

126


By Lemma 7.14, we have vp (fj )p ≤ 2

∞ X

2−k |φ(2k yj+1 ) − φ(2k yj )|p .

(7.4.10)

2−k |φ(2k yj+1 ) − φ(2k yj )|p .

(7.4.11)

k=1

Thus, by (7.4.9) and (7.4.10), vp(2) (f ; N )p ≤ 2

n−1 X ∞ X j=0 k=1

Set σl = {j : 2−l−1 < yj+1 − yj ≤ 2−l } for integers l ≥ 0. Subdividing the sum at the right-hand side of (7.4.11), we have vp(2) (f ; N )p ≤ 2

∞ XX ∞ X

2−k |φ(2k yj+1 ) − φ(2k yj )|p .

(7.4.12)

l=0 j∈σl k=1

We shall estimate the right-hand side of (7.4.12). Observe that |φ(2k yj+1 ) − φ(2k yj )| ≤ min(1, 2k (yj+1 − yj )).

(7.4.13)

Indeed, since φ is a nonnegative function, we have |φ(2k yj+1 ) − φ(2k yj )| ≤ kφk∞ = 1/2, and, at the same time, |φ(2k yj+1 ) − φ(2k yj )| ≤ 2k (yj+1 − yj )kφ0 k∞ = 2k (yj+1 − yj ). Fix l ≥ 0 and let j ∈ σl . Then, yj+1 − yj ≤ 2−l , and by (7.4.13), we have ∞ X

2−k |φ(2k yj+1 ) − φ(2k yj )|p ≤

k=1

∞ X

2−k min(1, 2k−l )p

k=1

= 2−lp

l X

2k(p−1) +

k=1

∞ X k=l+1

Since p > 1, it follows that there is a constant cp > 0 such that ∞ X k=1

2−k |φ(2k yj+1 ) − φ(2k yj )|p ≤ cp 2−l ,

2−k .

(2)

127


for all j ∈ σl . Consequently, for l ≥ 0, there holds ∞ XX 2−k |φ(2k yj+1 ) − φ(2k yj )|p ≤ cp 2−l card(σl ),

(7.4.14)

j∈σl k=1

where card(σl ) denotes the cardinality of the finite set σl . To sum up, by (7.4.12) and (7.4.14), we have ∞ X vp(2) (f ; N )p ≤ cp 2−l card(σl ) l=0 ∞ X X

≤ 2cp

(yj+1 − yj ) = 2cp .

l=0 j∈σl (2)

(2)

Thus, f ∈ Vp . To prove that f ∈ Hp , it suffices to show the existence of x0 , y0 ∈ R such that fx0 , fy0 ∈ Vp . For all x ∈ R we have f (x, 0) = 0 and thus f (·, 0) ∈ Vp . Similarly, f (1, y) = 0 for all y ∈ R, so f (1, ·) ∈ Vp . Thus, (2) f ∈ Hp . Now we demonstrate that f ∈ / Vp [ Vp ]sym . First, we observe that gn (2−k ) = 0 (n, k ∈ N). Thus, vp (fx ) = 0 for x = 2−k (k ∈ N). On the other hand, if x = (2−k+1 + 2−k )/2 (k ∈ N), then fx (y) = 2−k/p−1 φ(2k y), and by (7.4.3), we have vp (fx ) = 2−k/p−1 vp (φ2k ) = 21/p−2 . Clearly, the function x 7→ vp (fx ) does not belong to Vp . Thus, f ∈ / Vp [ Vp ]sym . It follows from Proposition 7.13 and Theorem 7.15 that Fubini-type prop(2) erties fail in Hp for p > 1. Remark 7.16. It is easy to see that for any p ≥ 1, we have Hp(2) ⊂ L∞ [ Vp ]sym .

(7.4.15)

Moreover, the function constructed to prove Theorem 7.15 shows that for p > 1, the exterior L∞ -norm of (7.4.15) cannot be replaced by a stronger Vq -norm. That is, Hp(2) 6⊂ Vq [ Vp ]sym ,

for p > 1 and q ≥ 1.

However, for p = 1 we have (7.4.1), which is much stronger than (7.4.15).

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Functions of Generalized Bounded Variation The classical concept of the total variation of a function has been extended in several directions. Such extensions find many applications in different areas of mathematics. Consequently, the study of notions of generalized bounded variation forms an important direction in the field of mathematical analysis. This thesis is devoted to the investigation of various properties of functions of generalized bounded variation. In particular, we obtain the following results: • sharp relations between spaces of generalized bounded variation and spaces of functions defined by integral smoothness conditions (e.g., Sobolev and Besov spaces); • optimal properties of certain scales of function spaces of fractional smoothness generated by functionals of variational type; • sharp embeddings within the scale of spaces of functions of bounded p-variation; • results concerning bivariate functions of bounded p-variation, in particular p sharp estimates of total variation in terms of the mixed L -modulus of continuity, and Fubini-type properties.

ISBN 978-91-7063-486-4 ISSN 1403-8099 DISSERTATION | Karlstad University Studies | 2013:11