Equivalence of K-Functionals and Modulus of ... - Springer Link

c Allerton Press, Inc., 2008. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2008, Vol. 52, No. 8, pp. 1–11.  c E.S. Belkina and S.S. Platonov, 2008, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2008, No. 8, pp. 3–15. Original Russian Text 

Equivalence of K-Functionals and Modulus of Smoothness Constructed by Generalized Dunkl Translations E. S. Belkina* and S. S. Platonov** Petrozavodsk State University, pr. Lenina 33, Petrozavodsk, 185640 Russia Received July 26, 2006

Abstract—In a Hilbert space L2,α := L2 (R, |x|2α+1 dx), α > −1/2, we study the generalized Dunkl translations constructed by the Dunkl differential-difference operator. Using the generalized Dunkl translations, we define generalized modulus of smoothness in the space L2,α . Based on the Dunkl operator we define Sobolev-type spaces and K-functionals. The main result of the paper is the proof of the equivalence theorem for a K-functional and a modulus of smoothness. DOI: 10.3103/S1066369X0808001X Key words: Dunkl operator, generalized Dunkl translation, K-functional, modulus of smoothness.

1. INTRODUCTION AND STATEMENT OF MAIN RESULTS In the classical theory of approximation of functions on the axis R translation operators f (x) → f (x + y), x, y ∈ R, play the leading role. Thus, the differentiation operator is an infinitesimal translation operator, the Fourier transform is an expansion in eigenfunctions of the translation operator. The translation operator is used for the construction of modulus of continuity and smoothness which are the fundamental elements of direct and inverse theorems in the approximation theory. Various generalizations of translation operators allow one to state natural analogs of problems of the approximation theory. One of generalizations of translation operators is a group or a semigroup of operators in a Banach space. Various problems of the approximation theory in Banach spaces with a group or a semigroup of operators are studied in [1, 2]. Another generalization of translation operators are the so-called “generalized translation operators” (e.g., [3], Chap. I, § 1–2). These operators do not necessarily form a group or a semigroup; however, the generalized modulus of smoothness constructed with the help of them may appear to be more convenient (than the usual modulus of smoothness) for the study of the connection between the smoothness properties of a function and the best approximations of this function in weight functional spaces. See papers [4–8] and references therein for some results on the approximation of functions with the use of generalized modulus of continuity and smoothness. In many problems of the theory of approximation of functions the K-functionals play an important role. The study of the connection between the modulus of smoothness and K-functionals is one of the main problems in the theory of approximation of functions. For various generalized modulus of smoothness these problems are studied, for example, in [4, 6, 9]. Recently in mathematical papers a new class of generalized translations was described and put into use, namely, the generalized Dunkl translations. The generalized Dunkl translations are constructed on the base of certain differential-difference operators (the Dunkl operators) which are widely used in mathematical physics (e.g., [10, 11]). Here we consider only the Dunkl operators, whose rank equals 1 (i.e., on R1 ), see [11] for the general case. In this paper we study certain problems of the theory of approximation of functions on the entire axis R in the metrics of L2 with a certain weight. The main result is the proof of the theorem on the equivalence of a K-functional and the module of smoothness constructed by the generalized Dunkl translations. * **

E-mail: [email protected]. E-mail: [email protected].

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Let us describe the obtained results in detail. Hereinafter the symbol α stands for a real value such that α > −1/2. The Dunkl operator is a differential-difference operator D which satisfies the condition   df 1 f (x) − f (−x) Df (x) = (x) + α + . dx 2 x The action of the operator D is defined for all functions f ∈ C (1) (R). Note that any even function f (x) ∈ C (2) (R) satisfies the equality D2 f = Bf,

(1.1)

d2 dx2

d + 2α+1 where B = x dx is the differential Bessel operator. Let jα (x) be a normalized Bessel function of the first kind, i.e.,

2α Γ(α + 1) Jα (x) , xα where Jα (x) is a Bessel function of the first kind ([12], Chap. 7). The function jα (x) is infinitely differentiable and even; in addition, jα (0) = 1. We understand a generalized exponential function as the function jα (x) =

√

eα (x) := jα (x) + i cα x jα+1 (x),

(1.2)

where cα = (2α + 2)−1 , i = −1. The function y = eα (x) satisfies the equation Dy = iy with the initial condition y(0) = 1. In the limit case with α = −1/2 the generalized exponential function coincides with the usual exponential function eix . Using the correlation jα (x) = −

xjα+1 (x) 2(α + 1)

(it follows, for example, from [13], formula 8.472), we conclude that the function eα (x) admits the representation eα (x) = jα (x) − ijα (x).

(1.3)

For the main classes of functions defined on R (we assume that all functions are complex-valued) we use the following denotations: C stands for the set of continuous functions, Cc is the set of continuous functions with a compact support, C ∞ is the set of infinitely differentiable functions, D denotes the set of infinitely differentiable functions with a compact support. Let L2,α stand for the Hilbert space which consists of measurable functions f (x) defined on R (we consider the functions accurate to their values on a set of the null measure) with the finite norm  1/2 2 2α+1 |f (x)| |x| dx . f 2,α := Here and in the following formulas we calculate the integrals along the entire numerical axis (if other limits of integration are not specified). The scalar product in the Hilbert space L2,α obeys the formula  (f, g) := f (x) g(x) |x|2α+1 dx, f, g ∈ L2,α . By the partial integration one can verify the correlation (Df, g) = −(f, Df )

(1.4)

for any functions f, g ∈ D. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

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As usual, we endow the space D with a topology; this turns it into a topological vector space ([14], Chap. IV, § 4). Let D  stand for the set of generalized functions, i.e., linear continuous functionals on the space D. We denote the value of a functional f ∈ D  on a function ϕ ∈ D by f, ϕ. The space L2,α is embedded into D  , provided that for f ∈ L2,α and ϕ ∈ D we put  (1.5) f, ϕ := f (x) ϕ(x) |x|2α+1 dx. One can extend (in a natural way) the action of the Dunkl operator D onto the space of generalized functions D  , putting Df, ϕ := −f, Dϕ,

f ∈ D  , ϕ ∈ D.

(1.6)

In particular, the action of the operator Df is defined for any function f ∈ L2,α , but, generally speaking, Df is a generalized function. One can define the operator of the generalized Dunkl translation T y f (x) in various ways. For a function f (x) ∈ D one can define the operator of the generalized Dunkl translation u(x, y) = T y f (x) as a solution of the Cauchy problem (e.g., [15]) Dx u(x, y) = Dy u(x, y),

u(x, 0) = f (x),

(1.7)

where Dx and Dy are the Dunkl operators applied with respect to variables x and y, correspondingly. One can extend the operator T y by continuity from a linear subset D ⊂ L2,α onto the whole space L2,α (see Section 2); the extended operator is also denoted by T y . With the help of the generalized Dunkl translation for any function f (x) ∈ L2,α we define differences of the order m (m ∈ N = {1, 2, . . . }) with a step h > 0: h m Δm h f (x) = (I − T ) f (x);

here I is the unit operator. For any positive integer m we define the generalized module of smoothness of the mth order by the formula ωm (f, δ)2,α := sup Δm h f 2,α ,

δ > 0,

f ∈ L2,α .

0 0. For brevity, we denote m ). Km (f, t)2,α := K(f, t; L2,α ; W2,α

The following theorem establishes the equivalence of the modulus of smoothness and the K-functional. Theorem 1.1. One can find positive numbers c1 = c1 (m, α) and c2 = c2 (m, α) which satisfy the inequality c1 ωm (f, δ)2,α ≤ Km (f, δm )2,α ≤ c2 ωm (f, δ)2,α , where f ∈ L2,α , δ > 0. The proof of Theorem 1.1 adduced in Section 3 is the main result of this paper. In Section 2 we describe some properties (necessary in what follows) of the Dunkl transform and the generalized Dunkl translation. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

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2. THE DUNKL TRANSFORM AND THE GENERALIZED DUNKL TRANSLATION Let us describe some properties of the Dunkl transform and the generalized Dunkl translation . The Dunkl transform is the integral transform  F : f (x) → f(λ) = f (x) eα (λx) |x|2α+1 dx,

λ ∈ R.

The inverse Dunkl transform is defined by the formula  −1  F : g(λ) → f (x) = A g(λ) eα (−λx) |λ|2α+1 dλ, where

−2  A = 2α+1 Γ(α + 1) .

Let S stand for the space of rapidly decreasing functions defined on R, i.e., the set of all infinitely differentiable functions ϕ(x) which decrease as |x| → ∞, together with all their derivatives, more rapidly than any power of |x|−1 . One can endow (in a usual way) the space S with a topology, turning it into a locally convex space (e.g., [17], Chap. I, § 5). It is well-known [11] that the direct and inverse Dunkl transforms are mutually inverse automorphisms of the space S. The Dunkl transform satisfies the Parseval equality (f (x) ∈ S)   (2.1) |f (x)|2 |x|2α+1 dx = A |f(λ)|2 |λ|2α+1 dλ. One can extend the mapping f (x) → f(λ) by continuity up to the isomorphism of the Hilbert space L2,α onto itself. The extended mapping is also denoted by f (x) → f(λ) and is called the Dunkl transform; this does not affect the validity of formula (2.1). One can write it as follows: f 22,α = A f22,α .

(2.2)

In Section 1 we define the operator of the generalized Dunkl translation T y f (x) as a solution of the Cauchy problem (1.7). For any function f (x) ∈ C ∞ a solution of this Cauchy problem exists, is unique, and obeys the following explicit formula [15]:  π y fe (G(x, y, ϕ))he (x, y, ϕ) sin2α ϕ dϕ T f (x) = C 0   π o 2α fo (G(x, y, ϕ))h (x, y, ϕ) sin ϕ dϕ , (2.3) + 0

where

Γ(α + 1) , G(x, y, ϕ) = x2 + y 2 − 2|xy| cos ϕ, Γ(α + 1/2)Γ(1/2) he (x, y, ϕ) = 1 − sgn(xy) cos ϕ,

(x+y)he (x,y,ϕ) for (x, y) = (0, 0), o G(x,y,ϕ) h (x, y, ϕ) = 0 for (x, y) = (0, 0),   1 1 fe (x) = f (x) + f (−x) , fo (x) = f (x) − f (−x) . 2 2

C=

(2.4)

With the help of formula (2.3) one can also define the operator T y for a class of functions which is wider than C ∞ . In particular, the operator T y f is defined for any continuous function f . Further we show that formula (2.3) extends the operator T y up to a continuous operator in L2,α . RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

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Lemma 2.1. Any even or odd continuous function f (x) satisfies the inequality |T y f (x)|2 ≤ 2T y (|f (x)|2 ),

x, y ∈ R.

Proof. 1) If f (x) is an even function, then fe (x) = f (x). On the segment [0, π] we introduce the measure π dm(ϕ) = C(sin ϕ)2α dϕ, where C is the coefficient used in formula (2.3). Then dm(ϕ) = 1. Note that 0

(h (x, y, ϕ)) = 1 − 2 sgn(xy) cos ϕ + cos ϕ ≤ 2h (x, y, ϕ). e

2

2

e

Using formula (2.3) and the Cauchy–Bunyakowsky inequality, we obtain  π 2 y 2 e f (G(x, y, ϕ)) h (x, y, ϕ) · 1 dm(ϕ) |T f (x)| = 0  π   π  2 e 2 ≤ |f (G(x, y, ϕ))| (h (x, y, ϕ)) dm(ϕ) dm(ϕ) 0 0  π |f (G(x, y, ϕ))|2 he (x, y, ϕ) dm(ϕ) = 2T y (|f (x)|2 ). ≤2 0

2) Let f (x) be an odd function. Using the Cauchy–Bunyakowsky inequality, we obtain  π 2 y 2 o f (G(x, y, ϕ)) h (x, y, ϕ) · 1 dm(ϕ) |T f (x)| = 0  π   π  ≤ |f (G(x, y, ϕ))|2 (ho (x, y, ϕ))2 dm(ϕ) dm(ϕ) 0

0



π

= 0

(x + y)2 he (x, y, ϕ) dm(ϕ). (2.5) |f (G(x, y, ϕ))|2 he (x, y, ϕ) G2 (x, y, ϕ)

Let us prove that (x + y)2 he (x, y, ϕ) ≤ 2. G2 (x, y, ϕ)

(2.6)

Let xy > 0 (one can consider the case xy < 0 analogously), then sgn(xy) = 1 and |xy| = xy. Note that (x − y)2 (1 + cos ϕ) (x + y)2 (1 − cos ϕ) − 2 = ≥ 0, x2 + y 2 − 2xy cos ϕ x2 + y 2 − 2xy cos ϕ whence we get (2.6). Formulas (2.5) and (2.6) yield  π |f (G(x, y, ϕ))|2 he (x, y, ϕ) dm(ϕ) = 2T y (|f (x)|2 ). |T y f (x)|2 ≤ 2 0

Here we use the fact that |f (x)|2 is an even function. Lemma 2.2. Let g(x) be a continuous even function such that for all x ∈ [0, a + |y|] the following inequality is true: |g(x)| ≤ A (a, A, y ∈ R, a > 0, A > 0). Then all x ∈ [−a, a] meet the inequality |T y g(x)| ≤ 2A. Proof. Since g(x) is an even function, formula (2.3) yields  π y g(G(x, y, ϕ)) he (x, y, ϕ) (sin ϕ)2α dϕ. T g(x) = C 0

Note that 0 ≤ G(x, y, ϕ) ≤ |x| + |y|. Therefore, if x ∈ [−a, a], then G(x, y, ϕ) ∈ [0, a + |y|] and g(G(x, y, ϕ)) ≤ A. Since 0 ≤ he (x, y, ϕ) ≤ 2, we have  π  π y e 2α A h (x, y, ϕ) (sin ϕ) dϕ ≤ 2AC (sin ϕ)2α dϕ = 2A.  |T g(x)| ≤ C 0

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Lemma 2.3. Let f (x) be a continuous function such that |f (x)| ≤ A with any x ∈ [−a − |y|, a + |y|] (a > 0, A > 0). Then with x ∈ [−a, a] the following inequality is true: |T y f (x)| ≤ 4A.

(2.7)

Proof. Let us represent f (x) as the sum of even and odd functions: f (x) = fe (x) + fo (x), where fe (x) and fo (x) are defined by formulas (2.4). Functions |fe (x)|2 and |fo (x)|2 are even and on the segment [0, a + |y|] their values do not exceed A2 . According to Lemma 2.2, we have T y (|fe (x)|2 ) ≤ 2A2 , T y (|fo (x)|2 ) ≤ 2A2 with x ∈ [−a, a]. Using Lemma 2.1 and the inequality (u + v)2 ≤ 2(u2 + v 2 ), we obtain |T y f (x)|2 = |T y fe (x) + T y fo (x)|2 ≤ 2(|T y fe (x)|2 + |T y fo (x)|2 ) ≤ 4(T y (|fe (x)|2 ) + T y (|fo (x)|2 )) ≤ 4(2A2 + 2A2 ) = 16A2 , whence we get inequality (2.7). Corollary 2.1. Let a sequence of continuous functions fn (x) converge to a function f (x) uniformly on any segment I ⊂ R. Then for any y ∈ R the sequence of functions T y fn (x) converges to the function T y f (x) uniformly on any segment. Proof. The uniform convergence of T y fn (x) to T y f (x) on the segment [−a, a] is equivalent to the condition max |T y fn (x) − T y f (x)| → 0 as n → ∞. But in accordance with Lemma 2.3, max |T y fn (x) − |x|≤a

|x|≤a

T y f (x)| ≤ 4 max |fn (x) − f (x)|, and max |fn (x) − f (x)| → 0, because fn (x) converges to f (x) |x|≤a+|y|

|x|≤a+|y|

uniformly on the segment [−a − |y|, a + |y|].

Lemma 2.4. For any functions f (x) ∈ C and g(x) ∈ Cc the following equality is true:   (T y f (x))g(x)|x|2α+1 dx = f (x)(T −y g(x))|x|2α+1 dx, y ∈ R.

(2.8)

Proof. If f (x), g(x) ∈ D, then equality (2.8) is proved in [18], proposition 3.2. Let f (x) be an arbitrary continuous function, g(x) ∈ Cc . Assume that supp g ⊆ [−N, N ] (supp g is the support of the function g). Choose an arbitrary sequence of functions fn (x) ∈ D which converges to f (x) uniformly on each segment, and a sequence of functions gn (x) ∈ D such that supp gn ⊆ [−N, N ] and gn (x) converges to g(x) uniformly on the segment [−N, N ]. Then   y 2α+1 dx = fn (x)(T −y gn (x))|x|2α+1 dx. (2.9) (T fn (x))gn (x)|x| Corollary 2.1 implies that the sequence T y fn (x) converges to T y f (x) uniformly on any segment, and the sequence T −y gn (x) converges to T −y g(x) uniformly on the segment [−N − |y|, N + |y|]. Proceeding to the limit in Eq. (2.9) for n → ∞, we obtain (2.8). Let us make sure that

√ T y f 2,α ≤ 2 2f 2,α

(2.10)

with f ∈ Cc . If f (x) = fe (x) + fo (x), then T y f 2,α ≤ T y fe 2,α + T y fo 2,α . Let us estimate each term separately:   y 2 y 2 2α+1 dx ≤ 2 T y (|fe (x)|2 ) · 1 |x|2α+1 dx T fe 2,α = |T fe (x)| |x|   2 y 2α+1 dx = 2 |fe (x)|2 |x|2α+1 dx = 2fe 22,α . = 2 |fe (x)| (T 1) |x| RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

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Analogously, one can verify that T y fo 22,α ≤ 2fo 22,α . Here we use Lemma 2.1, correlation (2.8) for a particular case g(x) = 1, and the equality T y 1 = 1. Taking into account the fact that fe 2,α ≤ f 2,α and fo 2,α ≤ f 2,α , we obtain √ T y f 2,α ≤ T y fe 2,α + T y fo 2,α ≤ 2 2f 2,α . Inequality (2.10) implies that one can extend the operator T y by continuity from D up to a bounded operator in L2,α . The extended operator is also denoted by T y ; inequality (2.10) remains valid for it. Lemma 2.5. Let f ∈ L2,α , then the following equality is true for any y ∈ R: y f )(λ) = e (λy) f(λ);  (T α

(2.11)

here f → f is the Dunkl transform. Proof. For functions f ∈ D Eq. (2.11) is proved in [16], corollary 5.4. Since D is a dense subset in L2,α , formula (2.12) remains valid for f ∈ L2,α . Lemma 2.6. Let f (x) ∈ L2,α , then m (2.12) Δm h f 2,α ≤ 4 f 2,α . √ Proof. Using inequality (2.10), we obtain T h f 2,α ≤ 2 2f 2,α ≤ 3f 2,α , then Δ1h f 2,α = T h f − f 2,α ≤ T h f 2,α + f 2,α ≤ 4f 2,α , whence inequality (2.12) follows.

Recall (e.g., [12], item VIII.2) that a linear operator A in a Hilbert space H with a dense definition domain is said to be self-adjoint in essence, if its closure A is a self-adjoint operator. Note also that a self-adjoint in essence operator A meets the equality A = A∗ (i.e., the closure of the operator A coincides with the adjoint operator). Lemma 2.7. The operator T = iD, where D is the Dunkl operator, with the definition domain D ⊂ L2,α is self-adjoint in essence. Proof. Formula (1.4) implies that the operator T = iD is symmetric, i.e., (2.13)

(T f, g) = (f, T g) C ∞.

for all f, g ∈ D. Equalities (1.4) and (2.13) remain valid, when f ∈ D, g ∈ For a symmetric operator A the following criterion of the self-adjointness in essence ([19], P. 283) is known: A is self-adjoint in essence if and only if Ker (A∗ + i) = {0}. In order to prove that the operator T = iD is self-adjoint in essence, it suffices to verify that Ker (T ∗ + i) = {0}. Let a function g belong to the definition domain of the operator T ∗ and (T ∗ + i)g = 0.

(2.14)

Taking into account the symmetric property of the operator T , we obtain that Eq. (2.14) is equivalent to the following one (T + i)g = 0.

(2.15)

Dg = −g.

(2.16)

Eq. (2.15), in turn, is equivalent to Assume that g(x) = ge (x) + go (x), where ge (x) is the even part of the function g(x), go (x) is the odd one. Then Eq. (2.16) is equivalent to the system dge (x) = −go (x), dx (2α + 1) dgo (x) + go (x) = −ge (x). dx x RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

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Substituting go (x) from the first equation into the second one, we obtain (−B + 1)ge = 0, where B is the differential Bessel operator (see (1.1)). In [8], lemma 3.2, one proves that the equality (−B + 1)ge = 0 and the inclusion ge ∈ L2,α yield ge (x) = 0, therefore go (x) = 0, i.e., g(x) = 0. Corollary 2.2. If functions f and Df belong to the space L2,α (we understand the action of the operator D in the sense of the theory of generalized functions), then one can find a sequence of functions fn ∈ D such that fn → f and Dfn → Df in the space L2,α . Proof. Let T = iD. The definition of the adjoint operator T ∗ implies that f belongs to the definition domain of the operator T ∗ and g = T ∗ f if and only if f ∈ L2,α and g = i(Df ) in the sense of the theory of generalized functions. It remains to use the fact that the self-adjointness in essence implies that the closure of the operator T coincides with T ∗ . Consequently, one can find a sequence of functions fn ∈ D such that fn → f and T fn → T f in the space L2,α , whence we conclude that Dfn → Df in the space L2,α . m , then Lemma 2.8. If a function f belongs to the Sobolev space W2,α m f )(λ) = (−iλ)m f(λ).  (D

(2.17)

Proof. Let functions f and Df belong to the space L2,α . Let us prove that ( Df )(λ) = (−iλ)f(λ).

(2.18)

According to Corollary 2.2, one can find a sequence of functions fn ∈ D such that fn → f and Dfn → Df in the space L2,α . Therefore, it suffices to prove (2.18) for f ∈ D. Note that f(λ) = f (x), eα (λx) (see (1.5)), therefore, using equalities Deα (λx) = iλeα (λx) and (1.6), we obtain ( Df )(λ) = Df (x), eα (λx) = −f (x), Deα (λx) = −iλf (x), eα (λx) = −iλf(λ), which proves (2.18). Eq. (2.17) follows from (2.18). In the following lemma we obtain several bounds for functions eα (x); we use them below. Lemma 2.9. For x ∈ R the following inequalities are fulfilled: 1) |eα (x)| ≤ 1, and the equality is attained only with x = 0; 2) |1 − eα (x)| ≤ 2|x|; 3) |1 − eα (x)| ≥ c with |x| ≥ 1, where c > 0 is a certain constant which depends only on α. Proof. The function jα (x) admits the following integral representation ([13], formula 8.411):  π/2 (cos ϕ)2α cos(x sin ϕ) dϕ, jα (x) = c1

(2.19)

0

where



π/2

c1 = 0

−1 2 Γ(α + 1) . (cos ϕ)2α dϕ =√ π Γ(α + 1/2)

Using formulas (1.3) and (2.19), we obtain the integral representation  π/2   (cos ϕ)2α cos(x sin ϕ) + i sin ϕ sin(x sin ϕ) dϕ. eα (x) = c1 0

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Then



π/2

|eα (x)| ≤ c1

2α

(cos ϕ)



1/2 dϕ ≤ c1 cos (x sin ϕ) + sin ϕ sin (x sin ϕ) 2

2

9



π/2

2

0

(cos ϕ)2α dϕ = 1, 0

and the equality |eα (x)| = 1 is attained only with x = 0. Let us use representation (1.2) for the function eα (x) and the bounds (e.g., [7], lemma 3.5) |jα (x)| ≤ 1,

1 − jα (x) ≤ x2 /2.

(2.20)

Then |1 − eα (x)| ≤ |1 − jα (x)| + (2α + 2)−1 |x| |jα+1 (x)|. With |x| ≤ 1 formula (2.20) yields |x|2 1 |x| |x| + |x| ≤ + < 2|x|. 2 2α + 2 2 2α + 2 With |x| ≥ 1 inequality 1) gives |1 − eα (x)| ≤ 2 ≤ 2|x|, i.e., the inequality |1 − eα (x)| ≤ 2|x| is true with any x. The asymptotic formulas for the Bessel functions imply that jα (x) → 0 and jα (x) → 0 as x → ∞, therefore, taking into account Eq. (1.2), we conclude that eα (x) → 0 as x → ∞. Consequently, a number x0 > 0 exists such that with x ≥ x0 the inequality |eα (x)| ≤ 1/2 is true. Let m = minx∈[1,x0 ] |1 − eα (x)|. With x ≥ 1 we get the inequality |1 − eα (x)| ≥ c, where c = min{m, 1/2}. |1 − eα (x)| ≤

3. PROOF OF THEOREM 1 In what follows, f (x) is an arbitrary function of the space L2,α ; c, c1 , c2 , c3 , . . . are positive constants which, possibly, depend on m and α. For brevity, let  ·  =  · 2,α . Further we often use the Parseval Eq. (2.2). m , t > 0. The following inequality is true: Lemma 3.1. Let f ∈ W2,α

ωm (f, t)2,α ≤ c1 tm Dm f . h m Proof. Assume that h ∈ (0, t], Δm h f = (I − T ) f is the difference with the step h. Properties (2.11), (2.17) and the Parseval equality yield m Δm h f  = A(1 − eα (hλ)) f (λ),

Formula (3.1) implies the equality Δm h f

Dm f  = Aλm f(λ).

   (1 − eα (hλ))m m   = h A λ f (λ)  . m (hλ) m

(3.1)

(3.2)

According to Lemma 2.9, with all s ∈ R we have the inequality |(1 − eα (s))m s−m | ≤ c2 , where c2 = 2m . Relations (3.2) and (3.1) give m m m m m m Δm h f  ≤ c2 h Aλ f (λ) = c2 h D f  ≤ c2 t D f .

Calculating the supremum with respect to all h ∈ (0, t], we obtain ωm (f, t)2,α ≤ c2 tm Dm f . For any function f ∈ L2,α and any number ν > 0 let us define the function  ν f(λ)eα (λx)|λ|2α+1 dλ = F −1 (f(λ)χν (λ)), Pν (f )(x) := A −ν

where χν (λ) is the characteristic function of the segment [−ν, ν], F −1 is the inverse Dunkl transform. m , One can easily prove that the function Pν (f ) is infinitely differentiable and belongs to all classes W2,α m ∈ N = {1, 2, 3, . . . }. Lemma 3.2. For any function f ∈ L2,α the following inequality is true: f − Pν (f ) ≤ c3 Δm 1/ν f , RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

ν > 0.

10

BELKINA, PLATONOV

Proof. Let |1 − eα (t)| ≥ c with |t| ≥ 1 (see Lemma 2.9). Using the Parseval equality, we obtain    1 − χν (λ)  m   f − Pν (f ) = A(1 − χν (λ))f (λ) = A (1 − eα (λ/ν)) f (λ) . m (1 − eα (λ/ν))

(3.3)

Note that 1 1 − χν (λ) ≤ m. c λ∈R |1 − eα (λ/ν)|

sup

−m . Then relation (3.3) gives f − Pν (f ) ≤ c−m A(1 − eα (λ/ν))m f(λ) = c3 2Δm 1/ν f , where c3 = c

Corollary 3.1. f − Pν (f ) ≤ c3 ωm (f, 1/ν)2,α . Lemma 3.3. The following inequality is true: D m (Pν (f )) ≤ c4 ν m Δm 1/ν f ,

ν > 0, m ∈ N.

(3.4)

Proof. Using the Parseval equality and correlation (3.2), we obtain (Pν (f )) = Aλm χν (λ)f(λ) Dm (Pν (f )) = ADm   m χ (λ)   λ ν m   = A  (1 − eα (λ/ν))m (1 − eα (λ/ν)) f (λ). (3.5) Note that λm χν (λ) (λ/ν)m tm = ν m sup = ν m sup . m m m λ∈R |1 − eα (λ/ν)| |λ|≤ν |1 − eα (λ/ν)| |t|≤1 |1 − eα (t)|

sup Let

tm , m |t|≤1 |1 − eα (t)|

c4 = sup then formula (3.5) yields inequality (3.4).

Corollary 3.2. D m (Pν (f )) ≤ c4 ν m ωm (f, 1/ν)2,α . Proof of Theorem 1. 1◦ . Proof of the inequality c5 ωm (f, δ)2,α ≤ Km (f, δm )2,α .

(3.6)

m . Using Lemma 3.1 and inequality (2.12), we obtain Let h ∈ (0, δ], g ∈ W2,α m m m m m m m Δm h f  ≤ Δh (f − g) + Δh g ≤ 4 f − g + c1 h D g ≤ c6 (f − g + δ D g),

where c6 = max{4m , c1 }. Calculating the supremum with respect to h ∈ (0, δ] and the infimum with m , we obtain ω (f, δ) m respect to all possible functions g ∈ W2,α m 2,α ≤ c6 Km (f, δ )2,α , whence we get inequality (3.6). 2◦ . Proof of the inequality Km (f, δm )2,α ≤ c7 ωm (f, δ)2,α .

(3.7)

m , by the definition of a K-functional we have Since Pν (f ) ∈ W2,α

Km (f, δm )2,α ≤ f − Pν (f ) + δm Dm (Pν (f )).

(3.8)

Using Corollaries 3.1 and 3.2, let us proceed with inequality (3.8): Km (f, δm )2,α ≤ c3 ωm (f, 1/ν)2,α + c4 (δν)m ωm (f, 1/ν)2,α . Since ν is an arbitrary positive value, choosing ν = 1/δ, we obtain (3.7). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

EQUIVALENCE OF K-FUNCTIONALS

11

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