The Whitney extension problem for Zygmund spaces and Lipschitz selections in hyperbolic jet-spaces
aa r X i v : . [ m a t h . F A ] M a y The Whitney extension problem for Zygmund spacesand Lipschitz selections in hyperbolic jet-spaces By Pavel ShvartsmanDepartment of Mathematics, Technion - Israel Institute of Technology,32000 Haifa, Israele-mail: [email protected]
Abstract
We study a variant of the Whitney extension problem [28, 29] for the space C k Λ mω ( R n ) of functions whose partial derivatives of order k satisfy the generalizedZygmund condition. We identify C k Λ mω ( R n ) with a space of Lipschitz mappings froma metric space ( R n +1+ , ρ ω ) supplied with a hyperbolic metric ρ ω into a metric space( P k + m − × R n +1+ , d ω ) of polynomial fields on R n +1+ equipped with a hyperbolic-typemetric d ω . This identification allows us to reformulate the Whitney problem for C k Λ mω ( R n ) as a Lipschitz selection problem for set-valued mappings from ( R n +1+ , ρ ω )into a certain family of subsets of P k + m − × R n +1+ .
1. Introduction
Let m be a non-negative integer. We let Ω m denote the class of non-decreasing continu-ous functions ω : R + → R + such that ω (0) = 0 and the function ω ( t ) /t m is non-increasing.Given non-negative integers k and m and ω ∈ Ω m we define the space C k Λ mω ( R n ) as follows:a function f ∈ C k ( R n ) belongs to C k Λ mω ( R n ) if there exists a constant λ > α , | α | = k , and every x, h ∈ R n | ∆ mh ( D α f )( x ) | ≤ λω ( k h k ) . Here as usual ∆ mh f denotes the m -th difference of a function f of step h , i.e., the quantity∆ mh f ( x ) := m X i =0 ( − m − i (cid:18) mi (cid:19) f ( x + ih ) .C k Λ mω ( R n ) is normed by k f k C k Λ mω ( R n ) := X | α |≤ k sup x ∈ R n | D α f ( x ) | + X | α | = k sup x,h ∈ R n | ∆ mh ( D α f )( x ) | ω ( k h k ) . (1.1) Math Subject Classification
Key Words and Phrases
Whitney’s extension problem, smooth functions, Zygmund space, finiteness,hyperbolic metric, jet-space, set-valued mapping, Lipschitz selection
1n particular, for m = 1 and ω ∈ Ω the space C k Λ ω ( R n ) coincides with the space C k,ω ( R n ) consisting of all functions f ∈ C k ( R n ) whose partial derivatives of order k satisfythe Lipschitz condition (with respect to ω ): | D α f ( x ) − D α f ( y ) | ≤ ω ( k x − y k ) , x, y ∈ R n . In turn, the space Λ mω ( R n ) := C Λ mω ( R n ) , ω ∈ Ω m , coincides with the generalized Zygmundspace of bounded functions f on R n whose modulus of smoothness of order m , ω m ( · ; f ),satisfies the inequality ω m ( t ; f ) ≤ λω ( t ) , t ≥ . In particular, the space Λ ω ( R n ) with ω ( t ) = t is the classical Zygmund space Z ( R n ) ofbounded functions satisfying the Zygmund condition: there is λ > x, y ∈ R n | f ( x ) − f (cid:0) x + y (cid:1) + f ( y ) | ≤ λ k x − y k . (See, e.g. Stein [27].)Throughout the paper we let S denote an arbitrary closed subset of R n .In this paper we study the following extension problem. Problem 1.1
Given non-negative integers k and m , a function ω ∈ Ω m , and an arbitraryfunction f : S → R , what is a necessary and sufficient condition for f to be the restrictionto S of a function F ∈ C k Λ mω ( R n ) ? This is a variant of a classical problem which is known in the literature as the WhitneyExtension Problem [28, 29]. It has attracted a lot of attention in recent years. We refer thereader to [4]-[7], [8]-[14], [1, 2] and [30, 31] and references therein for numerous results inthis direction, and for a variety of techniques for obtaining them.This note is devoted to the phenomenon of “finiteness” in the Whitney problem for thespaces C k Λ mω ( R n ). It turns out that, in many cases, Whitney-type problems for differentspaces of smooth functions can be reduced to the same kinds of problems, but for finitesets with prescribed numbers of points. For the space Λ ω ( R n ), this phenomenon has been studied in the author’s papers [21,20, 22]. It was shown that a function f defined on S can be extended to a function F ∈ Λ ω ( R n ) with k F k Λ ω ( R n ) ≤ γ = γ ( n ) provided its restriction f | S ′ to every subset S ′ ⊂ S consisting of at most N ( n ) = 3 · n − points can be extended to a function F S ′ ∈ Λ ω ( R n ) with k F S ′ k Λ ω ( R n ) ≤ . (Moreover, the value 3 · n − is sharp [22].)This result is an example of “the finiteness property” of the space Λ ω ( R n ). We call thenumber N appearing in formulations of finiteness properties “the finiteness number”. In his pioneering work [29], Whitney characterized the restriction of the space C k ( R ) , k ≥ , to an arbitrary subset S ⊂ R in terms of divided differences of functions. An applicationof Whitney’s method to the space C k,ω ( R ) implies the finiteness property for this spacewith the finiteness number N = k + 2.The restriction of the space C k Λ mω ( R ) to an arbitrary subset S ⊂ R has been charac-terized by Jonsson [16] ( m is arbitrary, k = 0, ω ( t ) = t m − ), Shevchuk [18, 19] ( m, ω arearbitrary, k = 0), Galan [15] (the general case). These results imply the finiteness propertyfor C k Λ mω ( R ) with the finiteness number N = m + k + 1.For the space C ,ω ( R n ) the finiteness property (with the same finiteness number N ( n ) =3 · n − ) has been proved in [7], see also [4]. 2n impressive breakthrough in the solution of the Whitney problem for C k,ω -spaces hasrecently been made by Fefferman [8]-[14]. In particular, one of his remarkable results statesthat the space C k,ω ( R n ) possesses the finiteness property for all k, n > , see [8, 10]. (Anupper bound for the finiteness number N ( k, n ) is N ( k, n ) ≤ dim P k , see Bierstone, Milman[3], and Shvartsman [26]. Here P k stands for the space of polynomials of degree at most k defined on R n . Recall that dim P k = (cid:0) n + kk (cid:1) .)Thus for m > m = 2 and k > Problem 1.2
Whether the space C k Λ mω ( R n ) possesses the finiteness property? In this paper we develop an approach to Problem 1.1 which allows us to reformulatethis problem as a purely geometric question about the existence of Lipschitz selections ofset-valued mappings defined on metric spaces with certain hyperbolic structure.
2. Extensions of Zygmund functions and Lipschitz selections
We will demonstrate this approach for the case of the Zygmund space Z m ( R n ) := C Λ mω ( R n ) where ω ( t ) = t m − . Thus Z m ( R n ) is defined by the finiteness of the norm k f k Z m ( R n ) := sup x ∈ R n | f ( x ) | + sup x,h ∈ R n | ∆ mh f ( x ) |k h k m − . (2.1)The crucial ingredient of our approach is an isomorphism between the space Z m ( R n ) | S and a space of Lipschitz mappings from the set S × R + into the product P m − × R n +1+ equipped with certain metrics.This isomorphism is motivated by a description of the restrictions of C k Λ mω -functions interms of local approximations which we present in Section 3. Let us formulate this resultfor the space Z m ( R n ).We will assume that all cubes in this paper are closed and have sides which are parallelto the coordinate axes. It will be convenient for us to measure distances in R n in theuniform norm k x k := max {| x i | : i = 1 , ..., n } , x = ( x , ..., x n ) ∈ R n . Thus every cube Q = Q ( x, r ) := { y ∈ R n : k y − x k ≤ r } is a “ball” in the metric space ( R n , k · k ) of “radius” r centered at x . We let x Q := x denotecenter of Q and r Q := r its “radius”. Given a constant λ >
0, we let λQ denote the cube Q ( x, λr ).We let K ( S ) := { Q ( x, r ) : x ∈ S, r > } denote the family of all cubes centered in S . By K we denote the family of all cubes in R n ;thus K = K ( R n ). Theorem 2.1 (A) Let f : S → R . Suppose that there exists a function F ∈ Z m ( R n ) such that F | S = f . Then there exists a constant < λ ≤ C k F k C k Λ mω ( R n ) and a family ofpolynomials { P Q ∈ P m − : Q ∈ K ( S ) } such that: x ∈ S and every cube Q = Q ( x, r ) , r > , we have P Q ( x ) = f ( x ); (2). For every Q ∈ K ( S ) with r Q ≤ and every β, | β | ≤ m − , | D β P Q ( x Q ) | ≤ λ r −| β | Q ;(2.2) (3). For every Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K ( S ) , and every α, | α | ≤ m − , wehave | D α P Q ( x ) − D α P Q ( x ) | ≤ λ (max { r , r } + k x − x k ) m − −| α | (2.3) · ln (cid:18) { r , r } + k x − x k min { r , r } (cid:19) . (B) Conversely, suppose that there exists a constant λ > and a family of polynomials { P Q ∈ P m − : Q ∈ K ( S ) } such that conditions (2) and (3) are satisfied. Then for every x ∈ S there exists the limit f ( x ) = lim x Q = x, r Q → P Q ( x ) . (2.4) Moreover, there exists F ∈ Z m ( R n ) with k F k Z m ( R n ) ≤ Cλ such that F | S = f .Here C is a constant depending only on m and n . This result is a particular case of Theorems 3.2 and 3.7 proven in Section 3. It canbe considered as a certain version of the classical Whitney extension theorem [28] for theZygmund space Z m ( R n ) where the Taylor polynomials are replaced by corresponding ap-proximation polynomials P Q , Q ∈ K ( S ).Now our aim is to transform inequalities (2.3) into a certain Lipschitz condition for themapping K ( S ) ∋ Q → ( P Q , Q ) ∈ P m − × K ( S ). This is the crucial point of the approach.We equip the family K (of all cubes in R n ) with the distance ρ : K × K → R + definedby the following formula: if Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K , and Q = Q , then ρ ( Q , Q ) := ln (cid:18) r , r ) + k x − x k min( r , r ) (cid:19) , (2.5)and ρ ( Q , Q ) := 0 whenever Q = Q . We prove that ρ is a metric on K ; moreover, themetric space ( K , ρ ) can be identified (up to a constant weight) with the classical Poinˆcareupper half-space model of the hyperbolic space H n +1 , see Remark 2.2.Now, for every α, | α | ≤ m −
1, we will rewrite every inequality in (2.3) in such a waythat its right-hand side will be precisely equal to ρ ( Q , Q ). By (2.3), we have1 λ | D α P Q ( x ) − D α P Q ( x ) | min { r , r } m − −| α | ≤ (cid:18) max { r , r } + k x − x k min { r , r } (cid:19) m − −| α | · ln (cid:18) { r , r } + k x − x k min { r , r } (cid:19) = (cid:0) e ρ ( Q ,Q ) − (cid:1) m − −| α | ρ ( Q , Q ) . Put ψ α ( t ) := t ( e t − m − −| α | , ψ − α denote the inverse to ψ α . Then, by the latter inequality,1 λ | D α P Q ( x ) − D α P Q ( x ) | min { r , r } m − −| α | ≤ ψ α ( ρ ( Q , Q )) , so that ψ − α (cid:18) λ | D α P Q ( x ) − D α P Q ( x ) | min { r , r } m − −| α | (cid:19) ≤ ρ ( Q , Q ) . Of course, the same inequality holds for x instead of x . Taking the maximum over x , x , and over all α with | α | ≤ m −
1, we obtain I (cid:0) λ P Q , λ P Q (cid:1) ≤ ρ ( Q , Q )(2.6)where I ( P Q , P Q ) := max i =1 , , | α |≤ m − ψ − α (cid:18) | D α P Q ( x i ) − D α P Q ( x i ) | min { r , r } m − −| α | (cid:19) . (2.7)In general the quantity I ( · , · ) does not satisfy the triangle inequality on the set P m − × K := { T = ( P, Q ) : P ∈ P m − , Q ∈ K} . However, after a simple, but important modification the function I ( · , · ) transforms into ametric on P m − × K .Namely, let us add the quantity ρ ( Q , Q ) to the maximum in the left-hand side of(2.7). The function obtained we denote by δ . Thus for every T = ( P , Q ) , T = ( P , Q ) ∈P m − × K we put δ ( T , T ) := max { ρ ( Q , Q ) , I ( P , P ) } . (2.8)Given γ ∈ R and T = ( P, Q ) ∈ P m − × K we put γ ◦ T := ( γP, Q ) . Now inequality (2.6) is equivalent to the inequality δ (cid:0) λ ◦ T , λ ◦ T (cid:1) ≤ ρ ( Q , Q ) . (2.9)(Actually, (2.9) is equality, but it will be more convenient for us to work with inequalitiesrather than equalities).The function δ generates the standard geodesic metric d on P m − × K defined as follows:given T, T ′ ∈ P m − × K we put d ( T, T ′ ) := inf M − X i =0 δ ( T i , T i +1 )(2.10)where the infimum is taken over all finite families { T , T , ..., T M } ⊂ P m − × K such that T = T and T M = T ′ .Our main result, Theorem 4.2, being applied to the case k = 0, ω ( t ) = t m − , states thatfor every T, T ′ ∈ P m − × K the following inequality d ( T, T ′ ) ≤ δ ( T, T ′ ) ≤ d ( e n ◦ T, e n ◦ T ′ )5olds. (Of course, the first inequality is trivial and follows from definition (2.10).) Thisresult allows us to reformulate Theorem 2.1 as follows: f ∈ Z m ( R n ) | S ⇔ there exists λ > T ( Q ) = ( P Q , Q ) from ( K ( S ) , ρ ) into ( P m − × K ( S ) , d ) such that(i) for every Q ∈ K ( S ) with r Q ≤ β, | β | ≤ m − , we have | D β P Q ( x Q ) | r | β | Q ≤ λ ;(2.11)(ii) for every Q , Q ∈ K ( S ) d (cid:0) λ ◦ T ( Q ) , λ ◦ T ( Q ) (cid:1) ≤ ρ ( Q , Q );(2.12)(iii) for every x ∈ S we have f ( x ) = lim x Q = x,r Q → P Q ( x ) . (2.13)Inequality (2.12) motivates us to introduce a Lipschitz-type space LO ( K ( S )) of map-pings T : K ( S ) → T := P m − × K ( S )defined by the finiteness of the following “seminorm” k T k LO ( K ( S )) := inf { λ > k λ − ◦ T k Lip( K ( S ) , T ) ≤ } . Also, inequality (2.11) motivates us to define a “norm” k T k ∗ := sup {| D β P Q ( x Q ) | r | β | Q : Q ∈ K ( S ) , r Q ≤ , | β | ≤ m − } . By LO ( K ( S )) we denote a subspace of LO ( K ( S )) defined by the finiteness of the “norm” k T k ∗ LO ( K ( S )) := k T k ∗ + k T k LO ( K ( S )) . See Section 5 for details.Thus one can identify the space Z m ( R n ) | S with “limiting values” lim x Q = x,r Q → P Q ( x ) ofmappings Q ∈ K ( S ) → T ( Q ) = ( P Q , Q ) from the space LO ( K ( S )). Remark 2.2
By the formula:
K ∋ Q = Q ( x, r ) ⇔ y = ( x, r ) ∈ R n +1+ (2.14)we identify the family K of all cubes in R n with the upper half-space R n +1+ R n +1+ := R n × R + = { y = ( y , ..., y n , y n +1 ) ∈ R n +1 : y n +1 > } . This identification and the metric (2.5) generate a metric ̺ on R n +1+ defined by the followingformula: for z i = ( x i , r i ) ∈ R n +1+ , i = 1 , ,̺ ( z , z ) := ln (cid:18) r , r ) + k x − x k min( r , r ) (cid:19) , (2.15)if z = z , and ̺ ( z , z ) := 0 whenever z = z .6iven z = ( z , ..., z n +1 ) ∈ R n +1+ we put ¯ z := z = ( z , ..., − z n +1 ). Also, by k z k , wedenote the Euclidean distance in R n +1+ . We recall that the Poincar´e metric on R n +1+ isdefined by the formula ρ H ( z , z ) := ln k z − ¯ z k + k z − z k k z − ¯ z k − k z − z k . (2.16)This metric is the Riemannian metric for which the line element ds is given by ds := p dx + ... + dx n + dx n +1 x n +1 . It determines the classical Poincar´e upper half-space model of the hyperbolic space H n +1 :=( R n +1+ , ρ H ). It can be readily seen that ̺ ( z , z ) ∼ ρ H ( z , z ) , z , z ∈ R n +1+ . (2.17)This equivalence, (2.15) and (2.5) show that the metric space ( K , ρ ) can be identified(up to a constant weight) with the hyperbolic space H n +1 .In view of this remark and identification (2.14), one can interpret the equality (2.13) asthe restriction to R n of the mapping R n +1+ ∋ z → ( P z , z ) ∈ P m − × R n +1+ . This enables usto identify the Zygmund space Z m ( R n ) with the restriction to R n of all Lipschitz mappingsof the form T ( z ) = ( P z , z ) , z ∈ R n +1+ , defined on the hyperbolic space H n +1 and takingtheir values in the metric space ( P m − × R n +1+ , d ) .In Section 3 and 4 we develop this approach for the general case of the space C k Λ mω ( R n )with k >
0. This enables us to reformulate the extension Problem 1.1 as a geometricalproblem of the existence of Lipschitz selections of certain set-valued mappings from R n +1+ into a family of subsets in P k + m − × R n +1+ . We will discuss a generalization of Problem1.1 raised by C. Fefferman [11] (for the space C k,ω ( R n )) and show how this problem canbe reduced to the Lipschitz selection problem for certain jet-spaces generated by functionsfrom C k Λ mω ( R n ).We observe that the Lipschitz selection method has been used for proving the finitenessproperty of the spaces Λ ω ( R n ) [22], C ,ω ( R n ) [7] and C k Λ ω ( R n ) [6] (a jet-version). In [26]we used the same technique to prove a certain weak version of the finiteness property ofthe space C k,ω ( R n ). Acknowledgment.
I am greatly indebted to Michael Cwikel for helpful suggestionsand remarks.
3. The space C k Λ mω ( R n ) and local polynomial approximations Our notation is fairly standard. Throughout the paper
C, C , C , ... will be genericpositive constants which depend only on k, m, n . These constants can change even in asingle string of estimates. The dependence of a constant on certain parameters is expressed,for example, by the notation C = C ( k, m, n ). We write A ≈ B if there is a constant C ≥ A/C ≤ B ≤ CA . 7e let P ℓ = P ℓ ( R n ) , ℓ ≥ , denote the space of all polynomials on R n of degree at most ℓ . Finally, given k -times differentiable function f and x ∈ R n , we let T kx ( f ) denote theTaylor polynomial of f at x of degree at most k : T kx ( f )( y ) := X | α |≤ k α ! ( D α f )( x )( y − x ) α , y ∈ R n . Finally, we put L := k + m − . Theorem 3.1
Given a family of polynomials { P x ∈ P k : x ∈ S } there is a function F ∈ C k Λ mω ( R n ) such that T kx ( F ) = P x for every x ∈ S if and only if there is a constant λ > and a family of polynomials { P Q ∈ P L : Q ∈ K ( S ) } such that(1). For every cube Q ∈ K ( S ) we have T kx Q ( P Q ) = P x Q ; (2). For every Q ∈ K ( S ) with r Q ≤ and every α, | α | ≤ k, sup Q | D α P Q | ≤ λ ; (3). For every Q, Q ′ ∈ K ( S ) , such that Q ′ ⊂ Q we have sup Q ′ | P Q ′ − P Q | ≤ λ ( r Q ′ + k x Q − x Q ′ k ) k ω ( r Q ) . Moreover, inf {k F k C k Λ mω ( R n ) : T kx ( F ) = P x , x ∈ S } ≈ inf λ with constants of equivalence depending only on k, m and n . For the homogeneous space C k Λ mω ( R n ) (normed by the second item in (1.1)) a variantof this theorem has been proved in [6]. The present result can be obtained by a simplemodification of the method of proof suggested in [6]. Theorem 3.2
If a function F ∈ C k Λ mω ( R n ) , then there exists a family of polynomials { P Q ∈ P L : Q ∈ K ( S ) } such that:(1). For every cube Q ∈ K ( S ) we have T kx Q ( P Q ) = T kx Q ( F ); (2). For every Q ∈ K ( S ) with r Q ≤ and every α, | α | ≤ k, and β, | β | ≤ L − | α | , | D α + β P Q ( x Q ) | ≤ C k F k C k Λ mω ( R n ) r −| β | Q ;8 Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K ( S ) , and every α, | α | ≤ L, we have | D α P Q ( x ) − D α P Q ( x ) | ≤ C k F k C k Λ mω ( R n ) · (max { r , r } + k x − x k ) L −| α | r + r + k x − x k Z min { r ,r } ω ( t ) t m dt. Here C is a constant depending only on k, m and n . The proof of the theorem is based on the following auxiliary lemmas.
Lemma 3.3
Let ω ∈ Ω m . Assume that a family of polynomials { P Q ∈ P L : Q ∈ K ( S ) } and a constant λ > satisfy the inequality sup Q ′ | P Q ′ − P Q | ≤ λ ( r Q ′ + k x Q − x Q ′ k ) k ω ( r Q ) , (3.1) for every Q = Q ( x Q , r Q ) , Q ′ = Q ( x Q ′ , r Q ′ ) ∈ K ( S ) such that Q ′ ⊂ Q and r Q ≤ r Q ′ .Then for every Q, Q ′ ∈ K ( S ) , Q ′ ⊂ Q , and every α, | α | ≤ L, we have sup Q ′ | D α P Q ′ − D α P Q | ≤ Cλ r Q Z r Q ′ ω ( t ) t | α |− k dtt (3.2) where C = C ( k, m, n ) .Proof. Let x ∈ Q ′ . We put Q i := Q ( x Q ′ , i r Q ′ ) , i = 0 , ..., ℓ, where ℓ := h ln r Q r Q ′ i + 1. Then r Q ℓ ≤ ℓ − r Q ′ ≤ r Q < ℓ r Q ′ = r Q ℓ . Since 2 r Q ≤ r Q ℓ := 2 ℓ r Q ′ and x Q ′ ∈ Q ∩ Q ℓ , we have Q ⊂ Q ℓ , r Q ≤ r Q ℓ ≤ r Q . (3.3)By Markov’s inequality,sup Q i | D α ( P Q i − P Q i +1 ) | ≤ C r −| α | Q i sup Q i | P Q i − P Q i +1 | where C = C ( k, m, n ). Since r Q i +1 = 2 r Q i and x Q i +1 = x Q i , by (3.1) we obtainsup Q i | D α ( P Q i − P Q i +1 ) | ≤ Cλr −| α | Q i ( r kQ i ω ( r Q i +1 ) = Cλr k −| α | Q i ω (2 r Q i ) . Since x ∈ Q ′ ∩ Q i , we have | D α P Q i ( x ) − D α P Q i +1 ( x ) | ≤ Cλr k −| α | Q i ω (2 r Q i ) , i = 0 , ..., ℓ − . | D α P Q ′ ( x ) − D α P Q ( x ) | = | D α P Q ( x ) − D α P Q ℓ ( x ) | ≤ ℓ − X i =0 | D α P Q i ( x ) − D α P Q i +1 ( x ) | so that | D α P Q ′ ( x ) − D α P Q ( x ) | ≤ Cλ ℓ − X i =0 r k −| α | Q i ω (2 r Q i ) . (3.4)Recall that Q ⊂ Q ℓ and r Q ℓ ≤ r Q , so that by (3.1) and Markov’s inequality | D α P Q ( x ) − D α P Q ℓ ( x ) | ≤ sup Q | D α P Q − D α P Q ℓ |≤ Cr −| α | Q sup Q | P Q − P Q ℓ |≤ Cλ ( r Q + k x Q − x Q ′ k ) k r −| α | Q ω ( r Q ℓ ) . But x Q ′ ∈ Q ′ ⊂ Q so that k x Q − x Q ′ k ≤ r Q . Therefore by (3.3) | D α P Q ( x ) − D α P Q ℓ ( x ) | ≤ | α | k Cλr k −| α | Q ℓ ω ( r Q ℓ )= 4 | α | k Cλ (2 r Q ℓ − ) k −| α | ω (2 r Q ℓ − )= C ( k, m ) λr k −| α | Q ℓ − ω ( r Q ℓ − ) . Since ω ∈ Ω m , we have ω (2 t ) ≤ m ω ( t ) so that by (3.3) | D α P Q ′ ( x ) − D α P Q ( x ) | ≤ | D α P Q ′ ( x ) − D α P Q ℓ ( x ) | + | D α P Q ℓ ( x ) − D α P Q ( x ) |≤ Cλ ℓ − X i =0 r k −| α | Q i ω (2 r Q i ) ! + C ( k, m ) λr k −| α | Q ℓ − ω ( r Q ℓ − ) ≤ C ( k, m ) λ ℓ − X i =0 r k −| α | Q i ω ( r Q i ) . Since ω is non-decreasing, for every α, | α | 6 = k, and every a, b, < a < b , we have Z ba ω ( t ) t | α |− k dtt ≥ ω ( a ) Z ba t | α |− k dtt = ( | α | − k ) − ω ( a )( a k −| α | − b k −| α | )so that ω ( a ) a | α |− k ≤ ( | α | − k ) b | α |− k b | α |− k − a | α |− k Z ba ω ( t ) t | α |− k dtt . Hence r k −| α | Q i ω ( r Q i ) ≤ ( | α | − k )(2 r Q i ) | α |− k (2 r Q i ) | α |− k − r | α |− kQ i Z r Qi r Qi ω ( t ) t | α |− k dtt = ( | α | − k )2 | α |− k | α |− k − Z r Qi r Qi ω ( t ) t | α |− k dtt .
10n a similar way we show that ω ( r Q i ) ≤ Z r Qi r Qi ω ( t ) dtt . Thus for every α, | α | ≤ L , we have r k −| α | Q i ω ( r Q i ) ≤ C ( k, m ) Z r Qi +1 r Qi ω ( t ) t | α |− k dtt . Finally, we obtain | D α P Q ′ ( x ) − D α P Q ( x ) | ≤ C λ ℓ − X i =0 r k −| α | Q i ω ( r Q i ) ≤ C λ ℓ − X i =0 Z r Qi +1 r Qi ω ( t ) t | α |− k dtt = C λ Z r Qℓ r Q ω ( t ) t | α |− k dtt ≤ C λ Z r Q r Q ′ ω ( t ) t | α |− k dtt . It can be easily seen that Z r Q r Q ω ( t ) t | α |− k dtt ≤ C Z r Q r Q ω ( t ) t | α |− k dtt ≤ C Z r Q r Q ′ ω ( t ) t | α |− k dtt so that | D α P Q ′ ( x ) − D α P Q ( x ) | ≤ C λ Z r Q r Q ′ ω ( t ) t | α |− k dtt ≤ C (1 + C ) λ Z r Q r Q ′ ω ( t ) t | α |− k dtt proving the lemma. ✷ Lemma 3.4
Let λ > and let { P Q ∈ P L : Q ∈ K ( S ) } be a family of polynomials satisfyingthe conditions of Lemma 3.3. Then for every Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K ( S ) , andevery α, | α | ≤ L, we have | D α P Q ( x ) − D α P Q ( x ) | ≤ Cλ (max { r , r } + k x − x k ) L −| α | r + r + k x − x k Z min { r ,r } ω ( t ) t m dt where C = C ( k, m, n ) .Proof. Put ˜ r := r + r + k x − x k and e Q := Q ( x , ˜ r ) = Q ( x , r + r + k x − x k ) . Then I := | D α P Q ( x ) − D α P Q ( x ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X | β |≤ p − α β ! [ D β ( D α ( P Q − P e Q ))]( x )( x − x ) β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X | β |≤ L − α (cid:12)(cid:12)(cid:12) D α + β ( P Q − P e Q )( x ) (cid:12)(cid:12)(cid:12) k x − x k β .
11y Lemma 3.3 | D α + β P Q ( x ) − D α + β P e Q ( x ) | ≤ Cλ Z rr ω ( t ) t | α | + | β |− k dtt for every α, β, | α | + | β | ≤ L . Hence I ≤ Cλ X | β |≤ L −| α | (cid:18)Z rr ω ( t ) t | α | + | β |− k dtt (cid:19) k x − x k | β | = Cλ Z rr ω ( t ) t m X | β |≤ L −| α | k x − x k | β | x L −| α |−| β | dt. Since k x − x k ≤ ˜ r = r + r + k x − x k , we obtain I ≤ C λ ˜ r L −| α | Z rr ω ( t ) t m dt ≤ C λ ˜ r L −| α | Z r + r + k x − x k )min { r ,r } ω ( t ) t m dt. Since ω ( t ) /t m is non-increasing, for every 0 < a < b ≤ b we have1 b − a Z b a ω ( t ) t m dt ≤ b − a Z b a ω ( t ) t m dt so that r + r + k x − x k ) Z min { r ,r } ω ( t ) t m dt ≤ r + r + k x − x k ) − min { r , r } r + r + k x − x k − min { r , r } r + r + k x − x k Z min { r ,r } ω ( t ) t m dt = min { r , r } + 2 max { r , r } + 2 k x − x k max { r , r } + k x − x k r + r + k x − x k Z min { r ,r } ω ( t ) t m dt ≤ r + r + k x − x k Z min { r ,r } ω ( t ) t m dt. Hence I ≤ C λ ( r + r + k x − x k ) L −| α | r + r + k x − x k Z min { r ,r } ω ( t ) t m dt ≤ · L C λ (max { r , r } + k x − x k ) L −| α | r + r + k x − x k Z min { r ,r } ω ( t ) t m dt. The lemma is proved. ✷ Proof of Theorem 3.2.
We put P x := T kx ( F ) , x ∈ S. { P Q ∈ P L : Q ∈ K ( S ) } satisfying conditions (1)-(3) of this theorem with λ = C ( k, m, n ) k F k C k Λ mω ( R n ) .Then equality (1) of Theorem 3.2 immediately follows from that of Theorem 3.1. Inturn, condition (2) of Theorem 3.1 and Markov’s inequality imply inequality (2) of Theorem3.2. Finally, by Lemma 3.4, condition (3) of Theorem 3.1 implies inequality (3) of Theorem3.2. ✷ Definition 3.5
We say that a continuous function ω : R + → R + is quasipower if there isa constant C ω > t Z ω ( s ) dss ≤ C ω ω ( t )for all t > Example 3.6
Every function ω ( t ) = tϕ ( t ) where ϕ is a non-decreasing function is quasi-power (with C ω = 1). In fact t Z ω ( s ) dss = t Z ϕ ( s ) ds ≤ tϕ ( t ) = ω ( t ) . Theorem 3.7
Let ω ∈ Ω m be a quasipower function. Suppose that a family of polynomials { P Q ∈ P L : Q ∈ K ( S ) } and a constant λ > satisfy the following conditions:(1). For every Q ∈ K ( S ) with r Q ≤ and every α, | α | ≤ k, and β, | β | ≤ L − | α | , wehave | D α + β P Q ( x Q ) | ≤ λ r −| β | Q ; (2). For every two cubes Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K ( S ) , and every α, | α | ≤ L, | D α ( P Q − P Q )( x ) | ≤ λ (max { r , r } + k x − x k ) L −| α | r + r + k x − x k Z min { r ,r } ω ( t ) t m dt. (3.5) Then for every x ∈ S there exists the limit P x = lim x Q = x, r Q → T kx ( P Q ) . (3.6) Moreover, there exists a function F ∈ C k Λ mω ( R n ) with k F k C k Λ mω ( R n ) ≤ Cλ such that P x = T kx ( F ) , x ∈ S, and for every Q = Q ( x, r ) ∈ K ( S ) and | α | ≤ k we have | D α T kx ( F )( x ) − D α P Q ( x ) | ≤ Cλ r k −| α | ω ( r ) . (3.7) Here C is a constant depending only on k, m, n and the constant C ω (see Definition 3.5). roof. By condition (2), for every α, | α | ≤ L, and every two cubes Q := Q ( x, r ) and Q := Q ( x , ˜ r ) ∈ K ( S ) with r ≤ ˜ r ≤ r, we have | D α P Q ( x ) − D α P Q ( x ) | ≤ λ ˜ r L −| α | r +˜ r Z r ω ( t ) t m dt ≤ L −| α | λr L −| α | r Z r ω ( t ) t m dt. Since ω ( t ) /t m is non-increasing and L := k + m −
1, we have | D α P Q ( x ) − D α P Q ( x ) | ≤ L −| α | λr L −| α | ω ( r ) r m r Z r dt = 2 k + m −| α | λr k −| α | ω ( r ) . (3.8)Consider now two cubes Q ′ = Q ( x, r ′ ) and Q ′′ = Q ( x, r ′′ ) , r ′ < r ′′ . Put ℓ := [ln( r ′′ /r ′ )] and r i := 2 i r ′ , Q i := Q ( x, r i ) , i = 0 , , ... Thus r ℓ := 2 ℓ r ′ ≤ r ′′ < ℓ +1 r ′ =: r ℓ +1 . Hence, by (3.8), | D α P Q ′ ( x ) − D α P Q ′′ ( x ) | ≤ ℓ − X i =0 | D α P Q i ( x ) − D α P Q i +1 ( x ) | + | D α P Q ℓ ( x ) − D α P Q ′′ ( x ) |≤ k + m −| α | λ ( ℓ − X i =0 r k −| α | i ω ( r i ) ! + r k −| α | ℓ ω ( r ℓ ) ) = 2 k + m −| α | λ ℓ X i =0 r k −| α | i ω ( r i ) . But Z r i +1 r i t k −| α | ω ( t ) dtt ≥ ω ( r i ) ln( r i +1 /r i ) = (ln 2) ω ( r i ) , whenever | α | = k , and Z r i +1 r i t k −| α | ω ( t ) dtt ≥ ω ( r i ) k − | α | (cid:16) r k −| α | i +1 − r k −| α | i (cid:17) = 2 k −| α | − k − | α | )2 k −| α | r k −| α | i ω ( r i ) , if | α | 6 = k . Thus, for every α and every i, ≤ i ≤ ℓ, we have r k −| α | i ω ( r i ) ≤ C ( k, m ) Z r i +1 r i t k −| α | ω ( t ) dtt . Hence, | D α P Q ′ ( x ) − D α P Q ′′ ( x ) | ≤ Cλ Z r ℓ r t k −| α | ω ( t ) dtt ≤ Cλ Z r ′′ r ′ t k −| α | ω ( t ) dtt . (3.9) 14onsider now the case | α | ≤ k . Recall that, by the assumption, ω is a quasipowerfunction, so that Z r ′′ r ′ ω ( t ) dtt ≤ C ω ω (2 r ′′ ) ≤ m C ω ω ( r ′′ ) . Therefore, by (3.9), for every α, | α | = k, we have | D α P Q ′ ( x ) − D α P Q ′′ ( x ) | ≤ Cλω ( r ′′ )with C = C ( k, m, C ω ).If | α | < k , we obtain | D α P Q ′ ( x ) − D α P Q ′′ ( x ) | ≤ Cλ Z r ′′ r ′ t k −| α | ω ( t ) dtt ≤ Cλω (2 r ′′ ) Z r ′′ r ′ t k −| α | dtt = 1 k − | α | Cλω (2 r ′′ ) (cid:0) (2 r ′′ ) k −| α | − ( r ′ ) k −| α | (cid:1) ≤ k − | α | Cλ (2 r ′′ ) k −| α | ω (2 r ′′ ) . Thus for every α, | α | ≤ k, we have | D α P Q ′ ( x ) − D α P Q ′′ ( x ) | ≤ Cλ ( r ′′ ) k −| α | ω ( r ′′ ) , (3.10)where C = C ( k, m, C ω ).Since ω ( r ) → r →
0, there exists the limit p α ( x ) := lim x Q = x, r Q → D α P Q ( x ) . (3.11)We put P x ( y ) := X | β |≤ k p β ( x ) β ! ( y − x ) β . Thus P x ∈ P k and D α P x ( x ) = p α ( x ) , | α | ≤ k. (3.12)Since P k is finite dimensional, by (3.12) P x = lim x Q = x, r Q → T kx P Q . proving (3.6).Prove that the family of polynomials { P x ∈ P k : x ∈ S } satisfies the conditions ofTheorem 3.1, i.e., there exists a family of polynomials { e P Q ∈ P L : Q ∈ K ( S ) } such that:(a). T kx Q ( e P Q ) = P x Q for every Q ∈ K ( S );(b). sup Q | D α e P Q | ≤ Cλ for every cube Q ∈ K ( S ) with r Q ≤ α, | α | ≤ k ;15c). For every Q, Q ′ ∈ K ( S ), such that Q ′ ⊂ Q we havesup Q ′ | e P Q ′ − e P Q | ≤ Cλ ( r Q ′ + k x Q − x Q ′ k ) k ω ( r Q ) . Put e P Q := P Q + P x − T kx P Q , Q = Q ( x, r ) ∈ K ( S ) . (3.13)Then D α e P Q ( x ) = D α P x ( x ) , | α | ≤ k,D α P Q ( x ) , | α | > k. (3.14)In particular, the condition (a) is satisfied.Observe that tending r ′ to 0 in (3.10) we obtain | D α P x ( x ) − D α P Q ( x ) | ≤ Cλr k −| α | ω ( r )(3.15)for every cube Q = Q ( x, r ) ∈ K ( S ) and every α, | α | ≤ k . This proves (3.7).Let us prove (b). Note that for every α, | α | ≤ k , by property (1) of the theorem (with β = 0) we have | D α P Q ( x Q ) | ≤ λ for every Q ∈ K ( S ) with r Q ≤
1. Therefore by (3.11) and (3.12) | D α P x ( x ) | = | lim r → D α P Q ( x,r ) ( x ) | ≤ λ, x ∈ S. Hence, by (3.14), for every β such that | α + β | ≤ k we have | D α + β e P Q ( x Q ) | = | D α + β P x Q ( x Q ) | ≤ λ. In turn, if | α + β | > k , by (3.14) and condition (1) of the theorem | D α + β e P Q ( x Q ) | = | D α + β P Q ( x Q ) | ≤ λr −| β | Q . Therefore for every x ∈ Q we have | D α e P Q ( x ) | = | X | β |≤ L −| α | β ! D α + β e P Q ( x Q )( x − x Q ) β |≤ X | β |≤ L −| α | | D α + β e P Q ( x Q ) | k x − x Q k | β | = X | β |≤ k −| α | | D α + β e P Q ( x Q ) | k x − x Q k | β | + X k −| α | < | β |≤ L −| α | | D α + β e P Q ( x Q ) | k x − x Q k | β | ≤ X | β |≤ k −| α | λr | β | Q + X k −| α | < | β |≤ L −| α | λr −| β | Q r | β | Q . r Q ≤
1, we obtain | D α e P Q ( x ) | ≤ C ( k, m, n ) λ, x ∈ Q, proving (b).Let us prove (c). Put ¯ r := r + r ′ + k x ′ − x k and K := Q ( x ′ , r + r ′ + k x ′ − x k ) = Q ( x ′ , ¯ r ) . Then clearly, Q ′ ⊂ Q ⊂ K , and also r ≤ ¯ r = r + r ′ + k x ′ − x k ≤ r ′ + 2 r ≤ r. By (3.15) | D α e P Q ′ ( x ′ ) − D α e P K ( x ′ ) | ≤ Cλ ¯ r k −| α | ω (¯ r ) ≤ Cλr k −| α | ω ( r ) , | α | ≤ k. (3.16)On the other hand , by condition (2) of the theorem, for every α, | α | ≤ L , we have | D α P K ( x ′ ) − D α P Q ( x ′ ) | ≤ λ (max { ¯ r, r } + k x ′ − x k ) L −| α | ¯ r + r + k x ′ − x k Z min { ¯ r,r } ω ( t ) t m dt = λ (¯ r + k x ′ − x k ) L −| α | ¯ r + r + k x ′ − x k Z r ω ( t ) t m dt. Since r ≤ ¯ r ≤ r, k x ′ − x k ≤ r , we obtain | D α P K ( x ′ ) − D α P Q ( x ′ ) | ≤ L −| α | λr L −| α | r Z r ω ( t ) t m dt ≤ L −| α | λr L −| α | ω ( r ) r m r Z r dt so that | D α P K ( x ′ ) − D α P Q ( x ′ ) | ≤ k + m λr k −| α | ω ( r ) , | α | ≤ L. (3.17)Observe also that by (3.15) for all α with | α | ≤ k we have | D α P Q ( x ′ ) − D α e P Q ( x ′ ) | ≤ Cλr k −| α | Q ω ( r Q ) = Cλr k −| α | ω ( r ) . (3.18)Hence, | D α e P Q ′ ( x ′ ) − D α e P Q ( x ′ ) | ≤ | D α e P Q ′ ( x ′ ) − D α P K ( x ′ ) | + | D α P K ( x ′ ) − D α P Q ( x ′ ) | + | D α P Q ( x ′ ) − D α e P Q ( x ′ ) |
17o that by (3.16), (3.17) and (3.18) we obtain | D α e P Q ′ ( x ′ ) − D α e P Q ( x ′ ) | ≤ Cλr k −| α | ω ( r ) for all α, | α | ≤ k. (3.19)Now consider the case k < | α | ≤ L . By (3.14) and (3.9) we have | D α e P Q ′ ( x ′ ) − D α e P K ( x ′ ) | = | D α P Q ′ ( x ′ ) − D α P K ( x ′ ) |≤ Cλ r Z r ′ ω ( t ) t | α |− k dtt ≤ Cλω (2¯ r ) r Z r ′ t k −| α | dtt ≤ Cλ ( | α | − k ) − ( r ′ ) k −| α | ω (2¯ r ) . Since ¯ r ≤ r , we obtain | D α e P Q ′ ( x ′ ) − D α e P K ( x ′ ) | ≤ Cλ m | α | − k ( r ′ ) k −| α | ω ( r ) = C λ ( r ′ ) k −| α | ω ( r ) . On the other hand, by (3.17) and (3.14) | D α e P K ( x ′ ) − D α e P Q ( x ′ ) | = | D α P K ( x ′ ) − D α P Q ( x ′ ) | ≤ Cλ r k −| α | ω ( r ) . Hence | D α e P Q ′ ( x ′ ) − D α e P Q ( x ′ ) | ≤ | D α e P Q ′ ( x ′ ) − D α e P K ( x ′ ) | + | D α e P K ( x ′ ) − D α e P Q ( x ′ ) |≤ Cλ (( r ′ ) k −| α | ω ( r ) + r k −| α | ω ( r )) ≤ Cλ ( r ′ ) k −| α | ω ( r )) , k < | α | ≤ L. We have proved that for every two cubes Q ′ = Q ( x ′ , r ′ ) and Q = Q ( x, r ) such that Q ′ ⊂ Q , we have | D α e P Q ′ ( x ′ ) − D α e P Q ( x ′ ) | ≤ Cλ r k −| α | ω ( r ) , | α | ≤ k, ( r ′ ) k −| α | ω ( r )) , k < | α | ≤ L. (3.20)To estimate sup Q ′ | e P Q ′ − e P Q | we let e Q denote the cube e Q := Q ( x, r ′ + k x − x ′ k ). Since Q ′ ⊂ Q , we have r ′ + k x − x ′ k ≤ r so that Q ′ ⊂ e Q ⊂ Q . Also, by (3.14), D α e P e Q ( x ) = D α P x ( x ) = D α e P Q ( x )so that by (3.20) | D α e P e Q ( x ′ ) − D α e P Q ( x ) | ≤ Cλ , | α | ≤ k, ( r ′ + k x − x ′ k ) k −| α | ω ( r ) , k < | α | ≤ L. y ∈ e Q we have | e P e Q ( y ) − e P Q ( y ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X | α |≤ L α ! (cid:16) D α e P e Q ( x ) − D α e P Q ( x ) (cid:17) ( y − x ) α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X k< | α |≤ L α ! (cid:16) D α e P e Q ( x ) − D α e P Q ( x ) (cid:17) ( y − x ) α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X k< | α |≤ L (cid:12)(cid:12)(cid:12) D α e P e Q ( x ) − D α e P Q ( x ) (cid:12)(cid:12)(cid:12) k y − x k | α | ≤ Cλ X k< | α |≤ L ( r ′ + k x − x ′ k ) k −| α | ω ( r ) r | α | e Q ≤ C λ ( r ′ + k x − x ′ k ) k ω ( r ) . Thus sup Q ′ | e P e Q − e P Q | ≤ sup e Q | e P e Q − e P Q | ≤ Cλ ( r ′ + k x − x ′ k ) k ω ( r ) . Let us estimate sup Q ′ | e P Q ′ − e P e Q | . By (3.20) for every α, | α | ≤ L, | D α e P Q ′ ( x ′ ) − D α e P e Q ( x ) | ≤ Cλ r k −| α | e Q ω ( r e Q ) , | α | ≤ k, ( r ′ ) k −| α | ω ( r e Q ) , k < | α | ≤ L. Hence, for each y ∈ Q ′ we have | e P Q ′ ( y ) − e P e Q ( y ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X | α |≤ L α ! (cid:16) D α e P Q ′ ( x ′ ) − D α e P e Q ( x ′ ) (cid:17) ( y − x ′ ) α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X | α |≤ L (cid:12)(cid:12)(cid:12) D α e P Q ′ ( x ′ ) − D α e P e Q ( x ′ ) (cid:12)(cid:12)(cid:12) k y − x ′ k | α | = X | α |≤ k (cid:12)(cid:12)(cid:12) D α e P Q ′ ( x ′ ) − D α e P e Q ( x ′ ) (cid:12)(cid:12)(cid:12) k y − x ′ k | α | + X k< | α |≤ L (cid:12)(cid:12)(cid:12) D α e P Q ′ ( x ′ ) − D α e P e Q ( x ′ ) (cid:12)(cid:12)(cid:12) k y − x ′ k | α | ≤ Cλ X | α |≤ k ( r e Q ) k −| α | ω ( r e Q )( r ′ ) k −| α | + Cλ X k< | α |≤ L ( r ′ ) k −| α | ω ( r e Q )( r ′ ) | α | . Since r ′ ≤ r e Q = r ′ + k x − x ′ k , we obtain | e P Q ′ ( y ) − e P e Q ( y ) | ≤ Cλ r k e Q ω ( r e Q ) = Cλ ( r ′ + k x − x ′ k ) k ω ( r e Q )proving that sup Q ′ | e P Q ′ − e P e Q | ≤ Cλ ( r ′ + k x − x ′ k ) k ω ( r e Q ) . r e Q = r ′ + k x − x ′ k ≤ r , we havesup Q ′ | e P Q ′ − e P Q | ≤ sup Q ′ | e P Q ′ − e P e Q | + sup Q ′ | e P e Q − e P Q |≤ Cλ ( r ′ + k x − x ′ k ) k ω ( r e Q ) + Cλ ( r ′ + k x − x ′ k ) k ω ( r ) ≤ Cλ ( r ′ + k x − x ′ k ) k ω ( r ) . Theorem 3.7 is completely proved. ✷ C k Λ mω ( R n ) as a space of Lipschitz mappings The point of departure for our approach is the inequality (3.5) of Theorem 3.7. Thisinequality motivates the definition of a certain metric on the set P L × K = { T = ( P, Q ) : P ∈ P L , Q ∈ K} . This allows us to identify the restriction C k Λ mω ( R n ) | S with a space of Lipschitz mappingsfrom K ( S ) (equipped with a certain hyperbolic-type metric) into P L × K .Given v > α, | α | ≤ L, we define a function ϕ α ( · ; v ) on R + by letting ϕ α ( t ; v ) := t L −| α | Z v + tv ω ( s ) s m ds. (4.1)(Recall that L := k + m − ϕ − α ( · ; v ) we denote the inverse to the function ϕ α ( · ; v )(i.e., the inverse to the function ϕ α with respect to the first argument). Since for every v > ϕ α ( · ; v ) is strictly increasing, the function ϕ − α ( · ; v ) is well-defined.Thus for every u ≥ ϕ − α ( u ; v ) L −| α | v + ϕ − α ( u ; v ) Z v ω ( s ) s m ds = u. (4.2)In particular, v + ϕ − α ( u ; v ) Z v ω ( s ) s m ds = u, | α | = L. (4.3)Now fix two elements T = ( P , Q ) , T = ( P , Q ) ∈ P L × K where Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K and P i ∈ P L , i = 1 ,
2. Put∆( T , T ) := max { max { r , r } + k x − x k , (4.4) max | α |≤ L, i =1 , ϕ − α ( | D α ( P − P )( x i ) | ; min { r , r } ) } δ ω ( T , T ) := min { r ,r } +∆( T ,T ) Z min { r ,r } ω ( s ) s m ds , (4.5)if T = T , and δ ω ( T , T ) := 0, if T = T .Observe that definition (4.4) and equality (4.3) imply the following explicit formula for δ ω ( T , T ) , T = T , : δ ω ( T , T ) := max r + r + k x − x k Z min { r ,r } ω ( s ) s m ds, max | α | = L | D α P − D α P | , max | α | Given a family of polynomials { P Q ∈ P L : Q ∈ K ( S ) } and a constant λ > the following two statements are equivalent:(i). For every two cubes Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K ( S ) , and every α, | α | ≤ L, | ( D α P Q − D α P Q )( x ) | ≤ λ (max { r , r } + k x − x k ) L r + r + k x − x k Z min { r ,r } ω ( t ) t m dt. (4.8) (Recall that this is inequality (3.5) of Theorem 3.7).(ii). Let T : K ( S ) → P L × K be a mapping defined by the formula T ( Q ) := ( P Q , Q ) , Q ∈K ( S ) . Then δ ω ( λ − ◦ T ( Q ) , λ − ◦ T ( Q )) ≤ ρ ω ( Q , Q ) , Q , Q ∈ K ( S ) . (4.9) 21Recall that λ ◦ T := ( λP, Q ) provided T = ( P, Q ) ∈ P L × K and λ ∈ R .) Proof. Put A i := | ( D α ( λ − P Q ) − D α ( λ − P Q ))( x i ) | , i = 1 , . By definition (4.2) inequality (4.8) can be reformulated as follows: A ≤ ϕ α (max { r , r } + k x − x k ; min { r , r } ) . Hence ϕ − α ( A ; min { r , r } ) ≤ max { r , r } + k x − x k . Changing the order of cubes in this inequality and taking the maximum over all α, | α | ≤ L, we conclude that (4.8) is equivalent to the following inequality:max | α |≤ L, i =1 , ϕ − α ( A i ; min { r , r } ) ≤ max { r , r } + k x − x k . In turn, by definition (4.4), this inequality is equivalent to the next one:∆ (cid:0) λ ◦ T ( Q ) , λ ◦ T ( Q ) (cid:1) = max | α |≤ L, i =1 , { max { r , r } + k x − x k , ϕ − α ( A i ; min { r , r } ) }≤ max { r , r } + k x − x k . Since the function t → R v + tv ω ( s ) s m ds is strictly increasing, the inequality∆( λ − ◦ T ( Q ) , λ − ◦ T ( Q )) ≤ max { r , r } + k x − x k is equivalent to δ ω ( λ − ◦ T ( Q ) , λ − ◦ T ( Q )) := min { r ,r } +∆( λ − ◦ T ( Q ) ,λ − ◦ T ( Q )) Z min { r ,r } ω ( s ) s m ds ≤ min { r ,r } +max { r ,r } + k x − x k Z min { r ,r } ω ( s ) s m ds = r + r + k x − x k Z min { r ,r } ω ( s ) s m ds = ρ ω ( Q , Q ) . The claim is proved. ✷ Let us define a metric on P L × K as a geodesic metric generated by the function δ ω .Given T, T ′ ∈ P L × K we putd ω ( T, T ′ ) := inf M − X i =0 δ ω ( T i , T i +1 )(4.10)where the infimum is taken over all finite families { T , T , ..., T M } ⊂ P L × K such that T = T and T M = T ′ .Observe several elementary properties of d ω . In particular, as we have noted above, thefunction ρ ω : K × K → R + is a metric on K , see Remark 4.4. This property of ρ ω and (4.5)22nd (4.4) immediately imply the following inequality: for every P i ∈ P L , Q i ∈ K , i = 1 , , we have d ω (( P , Q ) , ( P , Q )) ≥ ρ ω ( Q , Q ) . In turn, this inequality and (4.7) imply the following:d ω (( P, Q ) , ( P, Q )) = ρ ω ( Q , Q ) , P ∈ P L , Q i ∈ K , i = 1 , . The main result of the section is the following Theorem 4.2 For every T, T ′ ∈ P L × K we have d ω ( T, T ′ ) ≤ δ ω ( T, T ′ ) ≤ d ω ( e n ◦ T, e n ◦ T ′ ) . The proof of this theorem relies on a series of auxiliary lemmas. Lemma 4.3 For every b , b , ..., b ℓ > , a , a , ..., a ℓ − ≥ and c , c , ..., c ℓ − ≥ we have max b + b ℓ + ℓ − P i =0 a i Z min { b ,b ℓ } ω ( t ) t m dt, min { b ,b ℓ } + ℓ − P i =0 c i Z min { b ,b ℓ } ω ( t ) t m dt ≤ ℓ − X i =0 max b i + b i +1 + a i Z min { b i ,b i +1 } ω ( t ) t m dt, min { b i ,b i +1 } + c i Z min { b i ,b i +1 } ω ( t ) t m dt . Proof. We put s − = a − = c − := 0 ,s i := max { max { b i , b i +1 } + a i , c i } , i = 0 , ..., ℓ − , and I := [min { b , b ℓ } , min { b , b ℓ } + max { max { b , b ℓ } + ℓ − X i =0 a i , ℓ − X i =0 c i } ] . Then the inequality of the lemma is equivalent to the following one: Z I ω ( t ) t m dt ≤ ℓ − X i =0 min { b i ,b i +1 } + s i Z min { b i ,b i +1 } ω ( t ) t m dt. (4.11)To prove this inequality we put I i := [min { b , b ℓ } + i − X j = − s j , min { b , b ℓ } + i X j = − s j ] , i = 0 , ..., ℓ − . | I i | = s i and l − X j = − s j = l − X j = − max { max { b j , b j +1 } + a j , c j }≥ max { l − X j = − max { b j , b j +1 } + l − X j = − a j , l − X j = − c j }≥ max { max { b , b ℓ } + l − X j = − a j , l − X j = − c j } . Thus ℓ − [ i =0 I i ⊃ I so that Z I ω ( t ) t m dt ≤ ℓ − X i =0 Z I i ω ( t ) t m dt. (4.12)Put A i := [min { b i , b i +1 } , min { b i , b i +1 } + s i ] , i = 0 , ..., ℓ − . Then | A i | = | I i | = s i . But the left end of the segment I i is bigger than the left end of thesegment A i . In fact,min { b , b ℓ } + i − X j = − s j ≥ s i − = max { max { b i − , b i } + a i − , c i − } ≥ b i ≥ min { b i , b i +1 } . Thus the segment A i is a shift of I i to the left. Since ω ( t ) /t m is non-increasing, this implies Z I i ω ( t ) t m dt ≤ Z A i ω ( t ) t m dt = min { b i ,b i +1 } + s i Z min { b i ,b i +1 } ω ( t ) t m dt. This inequality and inequality (4.12) imply (4.11). The lemma is proved. ✷ Remark 4.4 We put in Lemma 4.3 ℓ = 2 , c = c = c := 0 and get b + b + a + a Z min { b ,b } ω ( t ) t m dt ≤ b + b + a Z min { b ,b } ω ( t ) t m dt + b + b + a Z min { b ,b } ω ( t ) t m dt. (4.13)This inequality easily implies the triangle inequality for the function ρ ω : K × K → R + defined by (4.6). In fact, for every cubes Q i = Q ( x i , r i ) ∈ K , i = 0 , , , we have ρ ω ( Q , Q ) ≤ r + r + k x − x k Z min { r ,r } ω ( s ) s m ds ≤ r + r + k x − x k + k x − x k Z min { r ,r } ω ( s ) s m ds 24o that by (4.13) ρ ω ( Q , Q ) ≤ r + r + k x − x k Z min { r ,r } ω ( s ) s m ds + r + r + k x − x k Z min { r ,r } ω ( s ) s m ds = ρ ω ( Q , Q ) + ρ ω ( Q , Q ) . Lemma 4.5 Let { P , P , ..., P ℓ } be a finite subfamily of P L and let { x , x , ..., x ℓ } be asubset of R n .Then for every α, | α | ≤ L, we have | D α ( P − P ℓ )( x ) | ≤ e n max | β |≤ L −| α | ℓ − X i =0 | D α + β ( P i − P i +1 )( x i ) | ! · ℓ − X i =0 k x i − x i +1 k ! | β | . Proof. We have | D α ( P i − P i +1 )( x ′ ) | = | X | β |≤ L −| α | β ! D α + β ( P i − P i +1 )( x i ) · ( x ′ − x i ) β |≤ X | β |≤ L −| α | β ! | D α + β ( P i − P i +1 )( x i ) | · k x ′ − x i k | β | so that | D α ( P − P ℓ )( x ′ ) | ≤ ℓ − X i =0 | D α ( P i − P i +1 )( x ′ ) |≤ ℓ − X i =0 X | β |≤ L −| α | β ! | D α + β ( P i − P i +1 )( x i ) | · k x ′ − x i k | β | = X | β |≤ L −| α | β ! ℓ − X i =0 | D α + β ( P i − P i +1 )( x i ) | · k x ′ − x i k | β | ! . Hence, | D α ( P − P ℓ )( x ′ ) | ≤ X | β |≤ L β ! max | β |≤ L −| α | ℓ − X i =0 | D α + β ( P i − P i +1 )( x i ) | · k x ′ − x i k | β | ≤ e n max | β |≤ L −| α | ℓ − X i =0 | D α + β ( P i − P i +1 )( x i ) | · k x ′ − x i k | β | . It remains to note that k x ′ − x i k = k x − x i k ≤ ℓ − X j =0 k x j − x j +1 k and the lemma follows. ✷ emma 4.6 For every v, R, t > and every α, β, | α + β | ≤ L, we have v + ϕ − α ( R | β | t ; v ) Z v ω ( s ) s m ds ≤ max v + R Z v ω ( s ) s m ds , v + ϕ − α + β ( t ; v ) Z v ω ( s ) s m ds . Proof. Put s := ϕ − α + β ( t ; v ) and q := max { R, s } . Since R ≤ q , we obtain R | β | q L −| α + β | ≤ q L −| α | , so that R | β | q L −| α + β | v + q Z v ω ( s ) s m ds ≤ q L −| α | v + q Z v ω ( s ) s m ds . By definition, see (4.1), ϕ α ( u ; v ) := u L −| α | Z v + uv ω ( s ) s m ds , u > , so that the latter inequality can be written in the following form: R | β | ϕ α + β ( q ; v ) ≤ ϕ α ( q ; v ) . Since s ≤ q and ϕ α + β is increasing, this inequality implies the following R | β | ϕ α + β ( s ; v ) ≤ ϕ α ( q ; v ) . But ϕ α + β ( s ; v ) = t so that R | β | t ≤ ϕ α ( q ; v ). Since ϕ − α is increasing, we have ϕ − α ( R | β | t ; v ) ≤ q = max { R, ϕ − α + β ( t ; v ) } proving the lemma. ✷ Lemma 4.7 For every b , b , ..., b ℓ > , u , u , ..., u ℓ − ≥ and every α, | α | ≤ L, we have min { b ,b ℓ } + ϕ − α ( ℓ − P i =0 u i ;min { b ,b ℓ } ) Z min { b ,b ℓ } ω ( t ) t m dt ≤ ℓ − X i =0 max b i + b i +1 Z min { b i ,b i +1 } ω ( t ) t m dt, min { b i ,b i +1 } + ϕ − α ( u i ;min { b i ,b i +1 } ) Z min { b i ,b i +1 } ω ( t ) t m dt . Proof. Put A := ϕ − α ℓ − X i =0 u i ; min { b , b ℓ } ! and c i := ϕ − α ( u i ; min { b i , b i +1 } ) , i = 0 , ..., ℓ − . A ≤ B := ℓ − X i =0 c i = ℓ − X i =0 ϕ − α ( u i ; min { b i , b i +1 } ) . Hence I := min { b ,b ℓ } + ϕ − α ( ℓ − P i =0 u i ;min { b ,b ℓ } ) Z min { b ,b ℓ } ω ( t ) t m dt = min { b ,b ℓ } + A Z min { b ,b ℓ } ω ( t ) t m dt ≤ min { b ,b ℓ } + B Z min { b ,b ℓ } ω ( t ) t m dt = min { b ,b ℓ } + ℓ − P i =0 c i Z min { b ,b ℓ } ω ( t ) t m dt so that by Lemma 4.3 (with a i = 0 , i = 0 , ..., ℓ − 1) we obtain I ≤ ℓ − X i =0 max b i + b i +1 Z min { b i ,b i +1 } ω ( t ) t m dt , min { b i ,b i +1 } + c i Z min { b i ,b i +1 } ω ( t ) t m dt . This proves the lemma under the assumption A ≤ B .Suppose that A > B . By identity (4.2) ϕ − α ℓ − X i =0 u i ; min { b , b ℓ } !! L −| α | min { b ,b ℓ } + ϕ − α ℓ − P i =0 u i ;min { b ,b ℓ } ! Z min { b ,b ℓ } ω ( t ) t m dt = ℓ − X i =0 u i so that A L −| α | min { b ,b ℓ } + A Z min { b ,b ℓ } ω ( t ) t m dt = ℓ − X i =0 u i . Hence I = min { b ,b ℓ } + A Z min { b ,b ℓ } ω ( t ) t m dt = A | α |− L A L −| α | min { b ,b ℓ } + A Z min { b ,b ℓ } ω ( t ) t m dt = A | α |− L ℓ − X i =0 u i . Since A > B and | α | ≤ L , we obtain I = A | α |− L ℓ − X i =0 u i ≤ B | α |− L ℓ − X i =0 u i . (4.14)Again, by identity (4.2) for every i = 0 , ..., ℓ − , we have u i = ( ϕ − α ( u i ; min { b i , b i +1 } )) L −| α | min { b i ,b i +1 } + ϕ − α ( u i ;min { b i ,b i +1 } ) Z min { b i ,b i +1 } ω ( t ) t m dt 27o that by (4.14) I ≤ ℓ − X i =0 (cid:18) ϕ − α ( u i ; min { b i , b i +1 } ) B (cid:19) L −| α | min { b i ,b i +1 } + ϕ − α ( u i ;min { b i ,b i +1 } ) Z min { b i ,b i +1 } ω ( t ) t m dt . But ϕ − α ( u i ; min { b i , b i +1 } ) ≤ B := ℓ − X j =0 ϕ − α ( u j ; min { b j , b j +1 } )for every i = 0 , ..., ℓ − , so that I ≤ ℓ − X i =0 min { b i ,b i +1 } + ϕ − α ( u i ;min { b i ,b i +1 } ) Z min { b i ,b i +1 } ω ( t ) t m dt . The lemma is proved. ✷ Proof of Theorem 3.1 . The inequality d ω ( T, T ′ ) ≤ δ ω ( T, T ′ ) trivially follows from defi-nition (4.10) of the metric d ω .In turn, the inequality δ ω ( T, T ′ ) ≤ d ω ( e n ◦ T, e n ◦ T ′ ) is equivalent to the followingstatement: Let { T i = ( P i , Q i ) ∈ P L × K : i = 0 , ..., ℓ } where Q i = Q ( x i , r i ), be a subfamily of P L × K such that T = T, T ℓ = T ′ . Put I := min { r ,r ℓ } +∆( T ,T ℓ ) Z min { r ,r ℓ } ω ( t ) t m dt , (4.15)and A := ℓ − X i =0 min { r i ,r i +1 } +∆( e n ◦ T i ,e n ◦ T i +1 ) Z min { r i ,r i +1 } ω ( t ) t m dt . Then I ≤ A. (4.16)For the sake of brevity we put v := min { r , r ℓ } . Recall that∆( T , T ℓ ) := max | α |≤ L { max { r , r ℓ } + k x − x ℓ k ,ϕ − α ( | D α ( P − P ℓ )( x ) | ; v ) , ϕ − α ( | D α ( P − P ℓ )( x ℓ ) | ; v ) } . Given multiindex α, | α | ≤ L, we put I α := v + ϕ − α ( | D α ( P − P ℓ )( x ) | ; v ) Z v ω ( t ) t m dt , J α := v + ϕ − α ( | D α ( P − P ℓ )( x ℓ ) | ; v ) Z v ω ( t ) t m dt . Then by (4.15) I := v +∆( T ,T ℓ ) Z v ω ( t ) t m dt = max | α |≤ L v + k x − x ℓ k Z v ω ( t ) t m dt , I α , J α . (4.17)Prove that v + k x − x ℓ k Z v ω ( t ) t m dt ≤ A. (4.18)In fact, v + ℓ − P i =0 k x i − x i +1 k Z v ω ( t ) t m dt = min { r ,r ℓ } + ℓ − P i =0 k x i − x i +1 k Z min { r ,r ℓ } ω ( t ) t m dt ≤ r + r ℓ + ℓ − P i =0 k x i − x i +1 k Z min { r ,r ℓ } ω ( t ) t m dt so that applying Lemma 4.3 (with c i = 0 , i = 0 , ..., ℓ − , ) we obtain v + ℓ − P i =0 k x i − x i +1 k Z v ω ( t ) t m dt ≤ ℓ − X i =0 r i + r i +1 + k x i − x i +1 k Z min { r i ,r i +1 } ω ( t ) t m dt. By definition (4.4) for every i = 0 , ..., ℓ − , we havemax { r i , r i +1 } + k x i − x i +1 k ≤ ∆( e n ◦ T i , e n ◦ T i +1 )so that r i + r i +1 + k x i − x i +1 k = min { r i , r i +1 } + max { r i , r i +1 } + k x i − x i +1 k≤ min { r i , r i +1 } + ∆( e n ◦ T i , e n ◦ T i +1 ) . Hence r i + r i +1 + k x i − x i +1 k Z min { r i ,r i +1 } ω ( t ) t m dt ≤ min { r i ,r i +1 } +∆( e n ◦ T i ,e n ◦ T i +1 ) Z min { r i ,r i +1 } ω ( t ) t m dt (4.19)which implies the following inequality v + ℓ − P i =0 k x i − x i +1 k Z v ω ( t ) t m dt ≤ ℓ − X i =0 min { r i ,r i +1 } +∆( e n ◦ T i ,e n ◦ T i +1 ) Z min { r i ,r i +1 } ω ( t ) t m dt = A. (4.20) 29hus v + k x − x ℓ k Z v ω ( t ) t m dt ≤ v + ℓ − P i =0 k x i − x i +1 k Z v ω ( t ) t m dt ≤ A. proving (4.18).Now prove that I α ≤ A, | α | ≤ L. (4.21)To this end given multiindex γ and i = 0 , ..., ℓ − , we put U γ,i := | D γ ( e n P i − e n P i +1 )( x i ) | (4.22)and U γ := ℓ − X i =0 U γ,i . We also set R := ℓ − X i =0 k x i − x i +1 k . Then by Lemma 4.5 | D α ( P − P ℓ )( x ) | ≤ max | β |≤ L −| α | ℓ − X i =0 | D α + β ( e n P i − e n P i +1 )( x i ) | ! · ℓ − X i =0 k x i − x i +1 k ! | β | = max | β |≤ L −| α | R | β | U α + β . Hence I α ≤ v + ϕ − α „ max | β |≤ L −| α | R | β | U α + β ; v « Z v ω ( t ) t m dt = max | β |≤ L −| α | v + ϕ − α ( R | β | U α + β ; v ) Z v ω ( t ) t m dt so that by Lemma 4.6 I α ≤ max | β |≤ L −| α | v + R Z v ω ( t ) t m dt , v + ϕ − α + β ( U α + β ; v ) Z v ω ( t ) t m dt . (4.23)By (4.20) v + R Z v ω ( t ) t m dt = v + ℓ − P i =0 k x i − x i +1 k Z v ω ( t ) t m dt ≤ A. (4.24) 30rove that v + ϕ − α + β ( U α + β ; v ) Z v ω ( t ) t m dt ≤ A. (4.25)For the sake of brevity we put v i := min { r i , r i +1 } , i = 0 , ..., ℓ − . Then by Lemma 4.7 v + ϕ − α + β ( U α + β ; v ) Z v ω ( t ) t m dt = min { r ,r ℓ } + ϕ − α + β ( ℓ − P i =0 U α + β,i ;min { r ,r ℓ } ) Z min { r ,r ℓ } ω ( t ) t m dt ≤ ℓ − X i =0 max r i + r i +1 Z v i ω ( t ) t m dt , v i + ϕ − α + β ( U α + β,i ; v i ) Z v i ω ( t ) t m dt . By (4.19) r i + r i +1 Z v i ω ( t ) t m dt = r i + r i +1 Z min { r i ,r i +1 } ω ( t ) t m dt ≤ min { r i ,r i +1 } +∆( e n ◦ T i ,e n ◦ T i +1 ) Z min { r i ,r i +1 } ω ( t ) t m dt. (4.26)In turn, by (4.22) and definition (4.4) ϕ − α + β ( U α + β,i ; v i ) = ϕ − α + β ( | D γ ( e n P i − e n P i +1 )( x i ) | ; min { r i , r i +1 } ) ≤ ∆( e n ◦ T i , e n ◦ T i +1 )so that v i + ϕ − α + β ( U α + β,i ; v i ) Z v i ω ( t ) t m dt ≤ min { r i ,r i +1 } +∆( e n ◦ T i ,e n ◦ T i +1 ) Z min { r i ,r i +1 } ω ( t ) t m dt. Hence v + ϕ − α + β ( U α + β ; v ) Z v ω ( t ) t m dt ≤ ℓ − X i =0 min { r i ,r i +1 } +∆( e n ◦ T i ,e n ◦ T i +1 ) Z min { r i ,r i +1 } ω ( t ) t m dt = A, proving (4.25).Now inequality (4.21) follows from (4.23), (4.24) and (4.25). In the same way we provethat J α ≤ A, | α | ≤ L. Finally, this inequality, (4.25),(4.18) and (4.17) imply the required inequality (4.16).Theorem 3.1 is proved. ✷ We present several results related to calculation of the function δ ω , see (4.4) and (4.5),and the metric d ω , see (4.10). 31et T = ( P , Q ) , T = ( P , Q ) ∈ P L × K , where Q = Q ( x , r ) , Q = Q ( x , r ) ∈ K and P i ∈ P L , i = 1 , 2. Fix a point y ∈ R n and put∆( T , T ; y ) := max { max { r , r } + k x − x k , (4.27) max | α |≤ L ϕ − α ( | D α ( P − P )( y ) | ; min { r , r } ) } . Thus, we define ∆( T , T ; y ) by replacing in (4.4) the points x i , i = 1 , , by y .We also put δ ω ( T , T ; y ) := min { r ,r } +∆( T ,T ; y ) Z min { r ,r } ω ( s ) s m ds , (4.28)for T = T , and δ ω ( T , T ; y ) := 0, whenever T = T . Clearly,∆( T , T ) = max { ∆( T , T ; x ) , ∆( T , T ; x ) } , and δ ω ( T , T ) = max { δ ω ( T , T ; x ) , δ ω ( T , T ; x ) } . Proposition 4.8 For every y, z ∈ R n and every T i = ( P i , Q i ) ∈ P L × K , where Q i = Q ( x i , r i ) ∈ K , i = 1 , , we have δ ω ( T , T ; z ) ≤ δ ω ( γ ◦ T , γ ◦ T ; y ) , where γ = max (cid:26) , e n k y − z k L (max { r , r } + k x − x k ) L (cid:27) . Proof. Fix a multiindex α, | α | ≤ L, and put e P := P , e P = P , e P := P , and ˜ x := z, ˜ x = y, ˜ x := y. Let us apply Lemma 4.5 to polynomials { e P , e P , e P } and points { ˜ x , ˜ x , ˜ x } . We have | D α ( P − P )( z ) | = | D α ( e P − e P )(˜ x ) |≤ e n max | β |≤ L −| α | X i =0 | D α + β ( e P i − e P i +1 )(˜ x i ) | ! · X i =0 k ˜ x i − ˜ x i +1 k ! | β | = e n max | β |≤ L −| α | | D α + β ( P − P )( y ) | · k z − y k | β | . Put U α + β := | D α + β ( γP − γP )( y ) | . Then | D α ( P − P )( z ) | ≤ max | β |≤ L −| α | | D α + β ( P − P )( y ) |· e n (cid:18) k z − y k max { r , r } + k x − x k (cid:19) | β | (max { r , r } + k x − x k ) | β | ≤ max | β |≤ L −| α | | D α + β ( γP − γP )( y ) | · (max { r , r } + k x − x k ) | β | = max | β |≤ L −| α | (max { r , r } + k x − x k ) | β | U α + β . 32e also put v := min { r , r } , A := max { r , r } . Then I α := v + ϕ − α ( | D α ( P − P )( z ) | ; v ) Z v ω ( s ) s m ds ≤ max | β |≤ L −| α | v + ϕ − α (( A + k x − x k ) | β | U α + β ; v ) Z v ω ( s ) s m ds, so that by Lemma 4.6 I α ≤ max | β |≤ L −| α | v + A + k x − x k Z v ω ( s ) s m ds, v + ϕ − α + β ( U α + β ; v ) Z v ω ( s ) s m ds . But by (4.27) A + k x − x k = max { r , r } + k x − x k ≤ ∆( γ ◦ T , γ ◦ T ; y ) , and ϕ − α + β ( U α + β ; v ) = ϕ − α + β ( | D α + β ( γP − γP )( y ) | ; min { r , r } ) ≤ ∆( γ ◦ T , γ ◦ T ; y ) . Hence I α ≤ v +∆( γ ◦ T ,γ ◦ T ; y ) Z v ω ( s ) s m ds = δ ω ( γ ◦ T , γ ◦ T ; y ) , proving the proposition. ✷ Proposition 4.8 and Theorem 4.2 imply the following Corollary 4.9 Let θ ≥ . For every T i = ( P i , Q i ) ∈ P L ×K , where Q i = Q ( x i , r i ) ∈ K , i =1 , , and every point y ∈ R n such that k y − x k ≤ θ ( r + r + k x − x k ) we have δ ω ( γ − ◦ T , γ − ◦ T ; y ) ≤ δ ω ( T , T ) ≤ δ ω ( γ ◦ T , γ ◦ T ; y ) , and δ ω ( γ − ◦ T , γ − ◦ T ; y ) ≤ d ω ( T , T ) ≤ δ ω ( γ ◦ T , γ ◦ T ; y ) . Here γ = γ ( n, θ ) is a constant depending only on n and θ . For instance, by this corollary, δ ω ( γ − ◦ T , γ − ◦ T ; x i ) ≤ d ω ( T , T ) ≤ δ ω ( γ ◦ T , γ ◦ T ; x i ) , i = 1 , , (4.29)or δ ω (cid:18) γ − ◦ T , γ − ◦ T ; x + x (cid:19) ≤ d ω ( T , T ) ≤ δ ω (cid:18) γ ◦ T , γ ◦ T ; x + x (cid:19) , with γ = γ ( n ) depending only on n . 33et us present one more formula for calculation of δ ω . Given v > α, | α | ≤ L, we put h ( t ; v ) := v + t Z v ω ( s ) s m ds , t > , and ψ α ( u ; v ) := (cid:16) t (cid:2) h − ( t ; v ) (cid:3) L −| α | (cid:17) − ( u ) , u > . Thus t (cid:2) h − ( t ; v ) (cid:3) L −| α | = ψ − α ( t ; v ) , t > . (4.30) Proposition 4.10 For every T i = ( P i , Q i ) ∈ P L × K , where Q i = Q ( x i , r i ) ∈ K , i = 1 , , we have δ ω ( T , T ; x ) = max | α |≤ L r + r + k x − x k Z min { r ,r } ω ( s ) s m ds , ψ α ( | D α ( P − P )( x ) | ; min { r , r } ) . Proof. We put v := min { r , r } , U α := | D α ( P − P )( x ) | , and I α := v + ϕ − α ( U α ; v ) Z v ω ( s ) s m ds . Recall that δ ω ( T , T ; x ) := max | α |≤ L r + r + k x − x k Z v ω ( s ) s m ds , I α . Prove that I α = ψ α ( U α ; v ) . (4.31)Recall that by (4.2) ϕ − α ( u ; v ) L −| α | v + ϕ − α ( u ; v ) Z v ω ( s ) s m ds = u (4.32)so that I α = ϕ − α ( U α ; v ) L −| α | ϕ − α ( U α ; v ) L −| α | v + ϕ − α ( U α ; v ) Z v ω ( s ) s m ds = U α ϕ − α ( U α ; v ) L −| α | . Hence ϕ − α ( U α ; v ) = (cid:18) U α I α (cid:19) L −| α | 34o that by (4.32) (cid:18) U α I α (cid:19) L −| α | ! L −| α | v + ( UαIα ) L −| α | Z v ω ( s ) s m ds = U α . We obtain U α I α v + ( UαIα ) L −| α | Z v ω ( s ) s m ds = U α which implies I α = v + ( UαIα ) L −| α | Z v ω ( s ) s m ds = h (( U α /I α ) L −| α | ; v ) . Hence h − ( I α ; v ) = ( U α /I α ) L −| α | , so that I α [ h − ( I α ; v )] L −| α | = U α . By (4.30) this implies ψ − α ( I α ; v ) = U α proving (4.31) and the proposition. ✷ In particular, by this proposition and (4.29), for every T i = ( P i , Q i ) ∈ P L × K , where Q i = Q ( x i , r i ) ∈ K , i = 1 , , we haved ω (cid:16) γ ◦ T , γ ◦ T (cid:17) ≤ max | α |≤ L r + r + k x − x k Z min { r ,r } ω ( s ) s m ds, ψ α ( | D α ( P − P )( x ) | ; min { r , r } ) ≤ d ω ( γ ◦ T , γ ◦ T )where γ = γ ( n ) depends only on n .Let us present two examples. Example 4.11 Consider the case of the Zygmund space Z m ( R n ) := Λ mω ( R n ) with ω ( t ) = t m − , see (2.1). In this case h ( t ; v ) := v + t Z v s ds = ln (cid:18) tv (cid:19) , t > , so that h − ( s ; v ) = v ( e s − ψ α ( u ; v ) := (cid:2) sv m − −| α | ( e s − m − −| α | (cid:3) − ( u ) = (cid:2) s ( e s − m − −| α | (cid:3) − (cid:16) uv m − −| α | (cid:17) . Thus in this case for every T i = ( P i , Q i ) ∈ P L × K , where Q i = Q ( x i , r i ) ∈ K , i = 1 , , wehave δ ω ( T , T ) = max | α |≤ m − ,i =1 , (cid:26) ln (cid:18) { r , r } + k x − x k min { r , r } (cid:19) , (cid:2) s ( e s − m − −| α | (cid:3) − (cid:18) | D α ( P − P )( x i ) | min { r , r } m − −| α | (cid:19)(cid:27) , 35o that we again obtain the formula (2.8) for δ ω . Example 4.12 Consider the space C k,ω ( R n ) = C k Λ ω ( R n ) with ω ( t ) = t . This is theSobolev space W k +1 ∞ ( R n ) of bounded functions f ∈ C k ( R n ) whose partial derivatives oforder k satisfy the Lipschitz condition: | D α f ( x ) − D α f ( y ) | ≤ λ k x − y k , x, y ∈ R n , | α | = k. Since m = 1 and L = k , we have h ( t ; v ) := v + t Z v ds = t, t > , and ψ α ( u ; v ) := (cid:2) ss k −| α | (cid:3) − ( u ) = u k +1 −| α | . Thus, in this case for every T i = ( P i , Q i ) ∈ P k × K , where Q i = Q ( x i , r i ) ∈ K , i = 1 , , wehave δ ω ( T , T ) = max | α |≤ m − ,i =1 , n max { r , r } + k x − x k , | D α ( P − P )( x i ) | m −| α | o . A function δ ω of such a kind and the metric d ω generated by δ ω have been studied in[26].Now by Claim 4.1 and Theorem 4.2, the results of Theorem 3.2 and Theorem 3.7 canbe formulated in the following form. Theorem 4.13 Let F ∈ C k Λ mω ( R n ) and let λ := k F k C k Λ mω ( R n ) . There exists a family ofpolynomials { P Q ∈ P L : Q ∈ K ( S ) } such that:(1). T kx Q ( P Q ) = T kx Q ( F ) for every cube Q ∈ K ( S ) ;(2). For every Q ∈ K ( S ) with r Q ≤ and every α, | α | ≤ k, and β, | β | ≤ L , | D α + β P Q ( x Q ) | ≤ Cλ r −| β | Q ; (3). The mapping T ( Q ) := ( P Q , Q ) , Q ∈ K ( S ) , satisfies the Lipschitz condition d ω (( Cλ ) − ◦ T ( Q ) , ( Cλ ) − ◦ T ( Q )) ≤ ρ ω ( Q , Q ) , Q , Q ∈ K ( S ) . (4.33) Here C is a constant depending only on k, m and n . Theorem 4.14 Let ω ∈ Ω m be a quasipower function. Assume that a mapping T ( Q ) = ( P Q , Q ) , Q ∈ K ( S ) , from K ( S ) into P L × K and a constant λ > satisfy the following conditions:(1). For every Q ∈ K ( S ) with r Q ≤ and every α, | α | ≤ k, and β, | β | ≤ L − | α | , wehave | D α + β P Q ( x Q ) | ≤ λ r −| β | Q ;36 Q , Q ∈ K ( S )d ω ( λ − ◦ T ( Q ) , λ − ◦ T ( Q )) ≤ ρ ω ( Q , Q ) . (4.34) Then there exists a function F ∈ C k Λ mω ( R n ) with k F k C k Λ mω ( R n ) ≤ Cλ such that for every x ∈ S T kx ( F ) = lim x Q = x, r Q → T kx ( P Q ) . Moreover, for every Q = Q ( x, r ) ∈ K ( S ) and every α , | α | ≤ k we have | D α T kx ( F )( x ) − D α P Q ( x ) | ≤ Cλ r k −| α | ω ( r ) . Here the constant C depends only on k, m, n and the constant C ω . Recall that the metric ρ ω is defined by formula (4.6) and the constant C ω is defined inDefinition 3.5.Theorem 4.13 and Theorem 4.14 enable us to describe the space C k Λ mω ( R n ) and itsrestrictions to subsets of R n as a certain space of Lipschitz mappings defined on subsets ofthe metric space K ω := ( K , ρ ω )and taking their values in the metric space T ω := ( P L × K , d ω ) . Let ( M , ρ ) be a metric space and let Lip( M , T ω ) be the space of Lipschitz mappingsfrom M into P L × K equipped with the standard Lipschitz seminorm k T k Lip( M , T ω ) := sup x,y ∈M , x = y d ω ( T ( x ) , T ( y )) ρ ( x, y ) . We introduce a Lipschitz-type space LO ( M , T ω ) of mappings T : M → P L × K defined bythe finiteness of the following seminorm: k T k LO ( M , T ω ) := inf { λ > k λ − ◦ T k Lip( M , T ω ) ≤ } . (4.35)Thus T ∈ LO ( M , T ω ) whenever there exists a constant λ > λ − ◦ T belongs to the unit ball of the Lipschitz space Lip( M , T ω ). In other words, thequantity k · k LO ( M , T ω ) is the standard Luxemburg norm with respect to the unit ball ofLip( M , T ω ) (and the “multiplication” operation ◦ ) and LO ( M , T ω ) is the correspondingOrlicz space determined by this norm, see, e.g [17].We call the space LO ( M , T ω ) the Lipschitz-Orlicz space. We use it to define a second“norm”: given a mapping T ( z ) = ( P z , Q z ) , z ∈ M , we put k T k ∗ := sup {| D α + β P z ( x Q z ) | r | β | Q z : z ∈ M , r Q z ≤ , | α | ≤ k, | β | ≤ L − | α |} , and k T k LO ( M , T ω ) := k T k ∗ + k T k LO ( M , T ω ) . (4.36)We let LO ( M , T ω ) denote the subspace of LO ( M , T ω ) of “bounded” Lipschitz mappingsdefined by the finiteness of the “norm” (4.36).Given closed subset S ⊂ R n we consider the family of cubes K ( S ) (i.e, the cubes centeredat S ) as a metric space equipped with the metric ρ ω , i.e., as a subspace of the metric space K ω = ( K , ρ ω ). Now Theorem 4.13 and Theorem 4.14 imply the following37 heorem 4.15 (a). For every function F ∈ C k Λ mω ( R n ) there exists a mapping T ( Q ) = ( P Q , Q ) , Q ∈ K ( S ) , from LO ( K ( S ) , T ω ) with k T k LO ( K ( S ) , T ω ) ≤ C k F k C k Λ mω ( R n ) such that T kx ( P Q ) = T kx ( F ) forevery cube Q = Q ( x, r ) ∈ K ( S ) .(b). Conversely, let ω ∈ Ω m be a quasipower function. Assume that a mapping T ( Q ) =( P Q , Q ) , Q ∈ K ( S ) , belongs to LO ( K ( S ) , T ω ) . Then there exists a function F ∈ C k Λ mω ( R n ) with k F k C k Λ mω ( R n ) ≤ C k T k LO ( K ( S ) , T ω ) such that for all x ∈ ST kx ( F ) = lim x Q = x, r Q → T kx ( P Q ) . (4.37) Moreover, for every Q = Q ( x, r ) ∈ K ( S ) and α , | α | ≤ k , | D α T kx ( F )( x ) − D α P Q ( x ) | ≤ C k T k LO ( K ( S ) , T ω ) r k −| α | ω ( r ) . (4.38) Here C is a constant depending only on k, m, n and the constant C ω . 5. Lipschitz selections of polynomial-set valued mappings The ideas and results presented in Section 4 show that even though Whitney’s problemfor C k Λ mω ( R n ) deals with restrictions of k -times differentiable functions, it is also a problemabout Lipschitz mappings defined on subsets of K and taking values in a very non-linearmetric space T ω = ( P L ×K , d ω ). More specifically, the Whitney problem can be reformulatedas a problem about Lipschitz selections of set-valued mappings from K ( S ) into 2 T ω .We recall some relevant definitions: Let X = ( M , ρ ) and Y = ( T , d ) be metric spacesand let G : M → T be a set-valued mapping, i.e., a mapping which assigns a subset G ( x ) ⊂ T to each x ∈ M . A function g : M → T is said to be a selection of G if g ( x ) ∈ G ( x ) for all x ∈ M . If a selection g is an element of Lip( X, Y ) then it is said to bea Lipschitz selection of the mapping G . (For various results and techniques related to theproblem of the existence of Lipschitz selections in the case where Y = ( T , d ) is a Banachspace, we refer the reader to [23, 24, 25] and references therein.)In [11] C. Fefferman considered the following version of the Whitney problem: Let { G ( x ) : x ∈ S } be a family of convex centrally-symmetric subsets of P k . How can we decide whether there exist F ∈ C k,ω ( R n ) and a constant A > such that T kx ( F ) ∈ A ⊚ G ( x ) for all x ∈ S ? Here A ⊚ G ( x ) denotes the dilation of G ( x ) with respectto its center by a factor of A .Let P x ∈ P k be the center of the set G ( x ). This means that G ( x ) can be representedin the form G ( x ) = P x + σ ( x ) where σ ( x ) ⊂ P k is a convex family of polynomials whichis centrally symmetric with respect to 0. It is shown in [11] that, under certain conditionson the sets σ ( x ) (the so-called condition of Whitney’s ω -convexity), the finiteness propertyholds. The approach described in Section 4, see Example 4.12, and certain ideas related toLipschitz selections in Banach spaces [24], allows us to give an upper bound for a finitenessnumber in Fefferman’s theorem [11]: N ( k, n ) = 2 min { ℓ +1 , dim P k } , where ℓ = max x ∈ S dim σ ( x ), see [26].This improvement of the finiteness number follows from Fefferman’s result [11] and thefollowing 38 heorem 5.1 ([26]) Let G be a mapping defined on a finite set S ⊂ R n which assigns aconvex set of polynomials G ( x ) ⊂ P k of dimension at most ℓ to every point x of S . Supposethat, for every subset S ′ of S consisting of at most min { ℓ +1 , dim P k } points, there exists afunction F S ′ ∈ C k,ω ( R n ) such that k F S ′ k C k,ω ( R n ) ≤ and T kx ( F S ′ ) ∈ G ( x ) for all x ∈ S ′ .Then there is a function F ∈ C k,ω ( R n ) , satisfying k F k C k,ω ( R n ) ≤ γ and T kx ( F ) ∈ G ( x ) f or all x ∈ S. Here γ depends only on k, n and card S . We use the rather informal and imprecise terminology “ C k,ω ( R n ) has the weak finitenessproperty” to express the kind of result where γ depends on the number of points of S . Theweak finiteness property also provides an upper bound for the finiteness constant wheneverthe strong finiteness property holds. For instance, Fefferman’s theorems in [11] reduce theproblem to a set of cardinality at most N ( k, n ) while the weak finiteness property decreasesthis number to 2 min { l +1 , dim P k } .In [26], Theorem 1.10, we show that, in turn, the “weak finiteness” theorem, is equivalentto a certain Helly-type criterion for the existence of a certain Lipschitz selection of the set-valued mapping G ( x ) = ( G ( x ) , x ) , x ∈ S . An analog of this result for set-valued mappingsfrom K ω ( S ) := ( K ( S ) , ρ ω ) into 2 P L ×K where m = 1 and ω ( t ) = t , is presented in Theorem5.4 below.Let us see how these ideas and results can be generalized for the space C k Λ mω ( R n ) with m > 1, and what kind of difficulties appear in this way. We will consider the followinggeneral version of the problem raised in [11]. Problem 5.2 Let { G ( x ) : x ∈ S } be a family of convex closed subsets of P k . How can wedecide whether there exist a function F ∈ C k Λ mω ( R n ) such that T kx ( F ) ∈ G ( x ) for all x ∈ S ?In particular, if G ( x ) = { P ∈ P k : P ( x ) = f ( x ) } , x ∈ S, where f is a function definedon S , this problem is equivalent to the Whitney Problem 1.1 for C k Λ mω ( R n ).First, let us show that Problem 5.2 is equivalent to a Lipschitz selection problem forset-valued mappings from K ( S ) ω := ( K ( S ) , ρ ω ) into a certain family of subsets of T ω :=( P L × K , d ω ). To this end, given a cube Q = Q ( x, r ) ∈ K ( S ) and λ > H λ ( Q )denote the set of all polynomials P ∈ P L satisfying the following condition: There exists e P ∈ G ( x ) such that for every α , | α | ≤ k , | D α e P ( x ) − D α P ( x ) | ≤ λ r k −| α | ω ( r ) . (5.1)In particular, H ( Q ) := { P ∈ P L : T kx ( P ) ∈ G ( x ) } , Q = Q ( x, r ) ∈ K ( S ) . (5.2)Clearly, H λ ( Q ) is a convex closed subset of P L .By H λ we denote the set-valued mapping from K ( S ) into 2 P L ×K defined by the followingformula: H λ ( Q ) := ( H λ ( Q ) , Q ) , Q ∈ K ( S ) . heorem 5.3 (a) Suppose that F ∈ C k Λ mω ( R n ) and T kx ( F ) ∈ G ( x ) for every x ∈ S .Then the set-valued mapping H has a selection T ∈ LO ( K ( S ) , T ω ) with k T k LO ( K ( S ) , T ω ) ≤ C k F k C k Λ mω ( R n ) .(b) Conversely, let ω ∈ Ω m be a quasipower function. Suppose that there exists λ > such that H λ has a selection T ∈ LO ( K ( S ) , T ω ) with k T k LO ( K ( S ) , T ω ) ≤ λ . Then there exists F ∈ C k Λ mω ( R n ) with k F k C k Λ mω ( R n ) ≤ Cλ such that T kx ( F ) ∈ G ( x ) for every x ∈ S .Here C is a constant depending only on k, m, n and the constant C ω .Proof. (a) By part (a) of Theorem 4.15 there exists a mapping T : K ( S ) → P L ×K of theform T ( Q ) = ( P Q , Q ) , Q ∈ K ( S ) , satisfying the following conditions: T ∈ LO ( K ( S ) , T ω ), k T k LO ( K ( S ) , T ω ) ≤ C k F k C k Λ mω ( R n ) , and T kx ( P Q ) = T kx ( F ) for every cube Q ( x, r ) ∈ K ( S ).But T kx ( F ) ∈ G ( x ) so that T kx ( P Q ) ∈ G ( x ) as well, proving that P Q ∈ H , see (5.2).Thus, the mapping T is a selection of the set-valued mapping H , and part(a) of the theoremis proved.(b) Since T is a selection of H λ , it can be written in the form T ( Q ) = ( P Q , Q ) , Q ∈ K ( S ) , P Q ∈ P L . By part Theorem 4.15, part (b), there exists F ∈ C k Λ mω ( R n ) with k F k C k Λ mω ( R n ) ≤ Cλ suchthat (4.37) and (4.38) are satisfied.Prove that T kx ( F ) ∈ G ( x ) , x ∈ S. Since T is a selection of H λ , for each cube Q = Q ( x, r ) ∈ K ( S ) we have P Q ∈ H λ ( Q ), so that, by (5.1), there exists e P ∈ G ( x ) such that | D α e P ( x ) − D α P Q ( x ) | ≤ λ r k −| α | ω ( r ) , | α | ≤ k. We have | D α T kx ( F )( x ) − D α e P ( x ) | ≤ | D α T kx ( F )( x ) − D α P Q ( x ) | + | D α P Q ( x ) − D α e P ( x ) |≤ | D α T kx ( F )( x ) − D α P Q ( x ) | + λ r k −| α | ω ( r ) , | α | ≤ k. Since k T k LO ( K ( S ) , T ω ) ≤ λ , by (4.38), | D α T kx ( F )( x ) − D α e P ( x ) | ≤ Cλ r k −| α | ω ( r ) for all | α | ≤ k. (5.3)Clearly, given x ∈ S the quantity k P k := max {| D α P ( x ) : | α | ≤ k } presents anequivalent norm on the finite dimensional space P k . By inequality (5.3), the distance from T kx ( F ) to G ( x ) in this norm tends to 0 as r → 0. Since G ( x ) is closed, T kx ( F ) ∈ G ( x )proving the theorem. ✷ As we have noted above, a geometrical background of the weak finiteness property,Theorem 5.1, is a certain Helly-type criterion for the existence of a Lipschitz selection. Letus formulate a version of this result for set-valued mappings defined on finite families ofcubes in R n . Theorem 5.4 Let m = 1 and let ω ∈ Ω be a quasipower function. Let K ⊂ K be a finiteset of cubes in R n and let H ( Q ) = ( H ( Q ) , Q ) , Q ∈ K, be a set-valued mapping such thatfor each Q ∈ K the set H ( Q ) ⊂ P k is a convex set of polynomials of dimension at most ℓ .Suppose that there exists a constant A > such that, for every subset K ′ ⊂ K consistingof at most min { ℓ +1 , dim P k } elements, the restriction H| K ′ has a selection h K ′ ∈ LO ( K ′ , T ω ) with k h K ′ k LO ( K ′ , T ω ) ≤ A .Then H , considered as a map on all of K , has a Lipschitz selection h ∈ LO ( K, T ω ) with k h k LO ( K, T ω ) ≤ γA . Here the constant γ depends only on k, n and card K . 40e recall that for m = 1 we have L := k + m − k so that T ω := ( P L × K , d ω ) =( P k , d ω ). The proof of this result follows precisely the same scheme as in [26].It would be very useful to have such a criterion for arbitrary m > C k Λ mω ( R n ). However, the straightforward application of the method of proof given in [26]to this case meets certain difficulties. In particular, one of the crucial ingredients of theproof in [26] is “consistency” of the metrics ρ ( x, y ) := k x − y k and ρ ( x, y ) := ω ( k x − y k )in the following sense: for every x , x , x , x ∈ R n the inequality ρ ( x , x ) ≤ ρ ( x , x )imply the inequality ρ ( x , x ) ≤ ρ ( x , x ) (and vise versa).For instance, a corresponding analog of this property for the space Z m ( R n ), see (2.1),is “consistency” of the distance ρ ( Q , Q ) := max { r , r } + k x − x k , Q i = Q ( x i , r i ) , i = 1 , , defined on the family K of all cubes in R n , and the metric ρ defined by (2.5). However,in general, such a “consistency” does not hold. For example, consider the family { Q i = Q (0 , r i ) , i = 1 , , ... } of cubes in R n with r i = 2 − i . 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