Whitney-type theorems on extension of functions on Carnot groups

Siberian Mathematical Journal, Vol. 47, No. 4, pp. 601–620, 2006 c 2006 Vodop yanov S. K. and Pupyshev I. M. Original Russian Text Copyright 

WHITNEY-TYPE THEOREMS ON EXTENSION OF FUNCTIONS ON CARNOT GROUPS S. K. Vodop yanov and I. M. Pupyshev

UDC 517.518.23+512.813.52

Abstract: We generalize the classical Whitney theorem which describes the restrictions of functions of various smoothness to closed sets of a Carnot group. The main results of the article are announced in [1]. Keywords: Carnot group, Whitney theorem, extension of functions, traces of functions

1. Notations and Preliminaries A Carnot group [2] is a connected simply connected Lie group G whose Lie algebra is nilpotent and graded, i.e., representable as V = V1 ⊕ · · · ⊕ Vm ,

[V1 , Vj ] = Vj+1 , j < m,

[V1 , Vm ] = 0.

Let N be the topological dimension of G and let X1 , X2 , . . . , XN be left-invariant vector fields on G that constitute a basis for the Lie algebra V ; moreover, X1 , . . . , Xn1 is a basis for V1 , and Xn1 +···+ni−1 +1 , . . . , Xn1 +···+ni , 1 < i ≤ m, is the basis for Vi constituted by the commutators of order i − 1 of some basis vector fields of V1 . Here n1 = n, n2 , . . . , nm are the respective dimensions of V1 , . . . , Vm . If Xi ∈ Vdi then the number di is called the degree of Xi . Henceforth a vector field Xi of degree di = 1 is called a horizontal vector field.  The exponential mapping x = exp N i=1 xi Xi is a diffeomorphism of V onto G sending the objects of the algebra to the group. Let η1 , . . . , ηN be the coordinate functions defined as follows: ηi (x) = xi . In these coordinates, the family of dilations δt , where t > 0, is defined as δt (x) with the coordinates ηi (δt (x)) = tdi xi . A homogeneous norm ρ is a continuous function on G of the class C ∞ (G \ {0}) with the following properties: ρ(x) ≥ 0 and ρ(x) = 0 if and only if x = 0; ρ(x−1 ) = ρ(x); ρ(δt (x)) = tρ(x), t > 0; and ρ(xy) ≤ κ(ρ(x) + ρ(y)), κ ≥ 1. The estimate |ηi (x)| ≤ ρ(x)di is valid. Sometimes it is more convenient to work with the Carnot–Carath´eodory metric dCC (x, y) on the group G defined as follows: dCC (x, y) is the length of the shortest horizontal arc between x and y. By the Rashevski˘ı–Chow theorem, such an arc always exists (henceforth we denote it by γ(x, y)), and the metric dCC (x, y) is equivalent to the quasimetric ρ(y −1 x).  m The Hausdorff dimension of G is equal to Q = N i=1 di = i=1 ini . iN I If I = (i1 , . . . , iN ) is a multi-index then we denote by X the differential operator X I = X1i1 . . . XN , |I| = i1 +· · ·+iN , and d(I) = d1 i1 +· · ·+dN iN is referred to as the homogeneous degree of the multi-index. iN Obviously, |η I (x)| ≤ ρ(x)d(I) , where the expression η I = η1i1 . . . ηN is called the monomial of homogeneous degree d(I). A homogeneous polynomial of homogeneous degree d is a linear combination of monomials of the same homogeneous degree d. A polynomial of homogeneous degree d is a linear combination of monomials of homogeneous degree at most d. We use the symbol ei to denote the multi-index ei = (0, . . . , 1, . . . , 0), with 1 at the ith place. The letter C stands for (generally) different constants depending only on the exponent γ and the characteristics of G. Novosibirsk. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 47, No. 4, pp. 731–752, July–August, 2006. Original article submitted July 8, 2005. Revision submitted November 3, 2005. c 2006 Springer Science+Business Media, Inc. 0037-4466/06/4704–0601 

601

2. The Taylor Formula on Carnot Groups 2.1. Properties of differential operators. In [2] it was shown that the basis vector fields are representable as  ∂ ∂ + Pik , (2.1) Xi = ∂ηi ∂ηk dk >di

where Pik is a homogeneous polynomial of homogeneous degree dk − di , and the higher-order differential operators X J are representable as  K  ∂ J PJK , (2.2) X = ∂η |K|≤|J| d(K)≥d(J)

where PJK is a homogeneous polynomial of homogeneous degree d(K) − d(J). It is easy to validate the Leibniz product formula:  CI1 ,I2 X I1 f · X I2 g. X I (f · g) =

(2.3)

I1 +I2 =I

Also, we need the formula X I (X J f ) =



γIJS X S f,

(2.4)

d(S)=d(I)+d(J)

where the operators are rearranged by the rule Xj Xi = Xi Xj − [Xi , Xj ], with j > i; moreover, the homogeneous order of the operators is preserved. In fact, γIJS = 0 if |S| ≥ |I| + |J|, S = I + J, and γIJ,I+J = 1. So, we can rewrite (2.4) as  γIJS X S f. X I (X J f ) = X I+J f + d(S)=d(I)+d(J) |S| d(J), we have  K   J     ∂ ∂ J   + X |η=0 = PJK =  ∂η ∂η 0 0 |K|≤|J| d(K)=d(J)

 |K| 0 we have BCC (x, r) ⊂ Bρ (x, Dr) and Bρ (x, r) ⊂ BCC (x, Dr). Prove that for a given r-packing constituted by the balls BCC (xi , r), the balls Bρ (xi , Dr) constitute a Dr-packing. It is clear that i Bρ (xi , Dr) = G. Moreover, the balls BCC (xi , r/K) are pairwise disjoint; therefore, so are the inner balls Bρ (xi , r/KD). Hence, we obtain existence of an r-packing on G in the quasimetric ρ (with the constant c = KD2 ). Henceforth we denote by d the distance ρ, implying by the word “ball” an open ball with respect to ρ. Denote by Mk the r-packing constituted by the balls of radius r = 2k , with k an integer. Define the layers Ωk as follows: Ωk = {x ∈ c F : c2k ≤ d(x, F ) ≤ c2k+1 }, where k is an integer and c > 2 is some constant chosen below. It is obvious that c F = ∞ k=−∞ Ωk . Given k, consider the meeting the layer Ω ; i.e., the collection of balls W = balls in M k k k {B ∈ Mk : B ∩ Ωk = ∅}. Then c F . Show that we can choose the constant c so that K r ≤ d(x , F ) ≤ K r , B ∈ W . Let B = 1 i i 2 i i Bk ∈W k Bi ∈ Mk . Then ri = 2k . If Bi ∈ W then there is a point x ∈ Bi ∩ Ωk . Therefore, d(xi , F ) ≤ κ(d(x, F ) + ρ(x−1 xi )) ≤ κ(c2k+1 + 2k ) = κ(2c + 1)2k and

d(xi , F ) ≥ d(x, F )/κ − ρ(x−1 xi ) ≥ c2k /κ − 2k = (c/κ − 1)2k .

If we choose c so that (K1 + 1)κ ≤ c ≤ (K2 − κ)/2κ (which is possible, if we take K2 sufficiently large) then κ(2c + 1) ≤ K2 , c/κ − 1 ≥ K1 , and K1 ri ≤ (c/κ − 1)2k ≤ d(xi , F ) ≤ κ(2c + 1)2k ≤ K2 ri . In this case the balls B ∈ W are certainly disjoint from F (since if x ∈ Bi then d(x, F ) ≥ d(xi , F )/κ − ρ(x−1 xi ) ≥ (K1 ri /κ)−ri ≥ ri ) and cover c F . We exclude from the collection W the balls that lie entirely in the union of other balls in W . We now construct a collection of balls satisfying the conditions (1), (3), and (5) of the lemma. Validate (4). Let Bi ∩ Bj = ∅. Then there is a point x ∈ Bi ∩ Bj and −1 −1 ρ(x−1 i xj ) ≤ κ(ρ(x xi ) + ρ(x xj )) ≤ κ(ri + rj ).

We have

2 K1 ri ≤ d(xi , F ) ≤ κ(ρ(x−1 i xj ) + d(xj , F )) ≤ κ (ri + rj ) + κK2 rj .

Consequently, (K1 − κ2 )ri ≤ (κ2 + κK2 )rj . Since K1 ≥ 2κ2 , we have ri ≤ K3 rj , where K3 = (κ2 + κK2 )/(K1 − κ2 ). From symmetry reasons we obtain rj ≤ K3 ri . 608

Now, prove (2). Assume that a point x belongs to m balls B1 , . . . , Bm . Denote one of them by Bi . From (3) we find ri /K3 ≤ rj ≤ K3 ri for all j = 1, . . . , m. Let y ∈ Bj . Then −1 −1 −1 −1 2 ρ(x−1 i y) ≤ κ(ρ(xi x) + ρ(x y)) ≤ κri + κ (ρ(xj y) + ρ(xj x))

≤ κri + 2κ2 rj ≤ κ(2κK3 + 1)ri ; j = i.e., Bj ⊂ B(xi , K0 ri ) for all j = 1, . . . , m, where K0 = κ(2κK3 + 1). Consider the disjoint balls B B(xj , rj /K4 ). We have m 

j = mes mes B

j=1

m

j ≤ mes B(xi , K0 ri ) = (K0 ri )Q . B

j=1

m

Consequently, j=1 (rj /K4 )Q ≤ (K0 ri )Q and m(ri /K3 K4 )Q ≤ (K0 ri )Q . Thus, m ≤ (K0 K3 K4 )Q , and property (2) is proven. (6) follows from (4) and Lemma (3.7) of [2].  Introduce the notations. Given Bi ⊂ c F , denote by pi the point in F such that d(xi , F ) = ρ(p−1 i xi ).

(3.8)

For an arbitrary point x ∈ c F , we denote by δ(x) the distance from x to the set F ; i.e., δ(x) = d(x, F ). Lemma 4. The following hold: ρ(x−1 pi ) ∼ d(xi , F ) ∼ ri ∼ δ(x), ρ(y −1 pi ) ≤ Cρ(y −1 x),

x ∈ Bi ;

y ∈ F, x ∈ Bi .

(3.9) (3.10)

Here and in the sequel the notation A ∼ B means that there exist positive constants C1 and C2 depending only on the characteristics of G such that the inequality C1 A ≤ B ≤ C2 A holds. Proof. Establish (3.9). Let x ∈ Bi . Then from the properties of ρ we obtain −1 ρ(x−1 pi ) ≤ κ(ρ(x−1 xi ) + ρ(x−1 i pi )) ≤ κ(ri + d(xi , F )) ≤ κ(K1 + 1)d(xi , F ).

On the other hand, −1 −1 d(xi , F ) = ρ(x−1 i pi ) ≤ κ(ρ(x xi ) + ρ(x pi ))

≤ κ(ri + ρ(x−1 pi )) ≤ κ(K1−1 d(xi , F ) + ρ(x−1 pi )). Consequently, d(xi , F )(1 − κK1−1 ) ≤ κρ(x−1 pi ), whence we find that d(xi , F ) ≤ κK1 (K1 − κ)−1 ρ(x−1 pi ) (since K1 > κ). Thus, ρ(x−1 pi ) ∼ d(xi , F ) ∼ ri . Now, δ(x) = d(x, F ) ≤ κ(ρ(x−1 xi ) + d(xi , F )) ≤ κ(1 + K2 )ri . At the same time

d(xi , F ) ≤ κ(ρ(x−1 xi ) + d(x, F )) ≤ κ(ri + δ(x)).

Hence, δ(x) ≥ d(xi , F )/κ − ri ≥ (K1 /κ − 1)ri . Thus, ri ∼ δ(x) and (3.9) is proven. Prove (3.10). Take y ∈ F and x ∈ Bi . Then it follows from (3.9) that ρ(y −1 pi ) ≤ κ(ρ(y −1 x) + ρ(x−1 pi )) ≤ κ(ρ(y −1 x) + C  d(xi , F )). But

(3.11)

d(xi , F ) ≤ κ(ρ(x−1 xi ) + d(x, F )) ≤ κ(ri + ρ(y −1 x)) ≤ κ(K1−1 d(xi , F ) + ρ(y −1 x)). 609

Consequently,

ρ(y −1 x) ≥ κ−1 d(xi , F ) − K1−1 d(xi , F ) = (K1 − κ)(κK1 )−1 d(xi , F );

i.e., d(xi , F ) ≤ κK1 (K1 − κ)−1 ρ(y −1 x). From (3.11) we infer ρ(y −1 pi ) ≤ κ(1 + C  κK1 (K1 − κ)−1 )ρ(y −1 x) = Cρ(y −1 x), which proves (3.10).  3.4. The extension operator. This operator of a collection {fJ }d(J)≤k of functions on a set F to the whole group G is defined as follows:  x ∈ F; f0 (x), (3.12) f (x) = Ek ({fJ })(x) =  P (x, p )ϕ (x), x ∈ c F, i i i

where

 

P (x, y) =



d(L)≤k d(K)=d(L) |K|≤|L|

 η L (y −1 x) βLK fK (y) , L!

x ∈ G, y ∈ F,

 means that the sum is {ϕi } is a partition of unity, and the points pi are defined by (3.8). The sign taken over those balls Bi containing the point x for which d(xi , F ) ≤ 1. Remark 2. Since each of the functions ϕi belongs to the class C ∞ and the sum over i for each x contains at most N0 summands, the extended function f (x) belongs to the class C ∞ on c F and, hence, is continuous on c F . Remark 3. If k = 0 then the extension operator (3.12) has the form  f (x), x ∈ F; E0 (f )(x) =  f (p )ϕ (x), x ∈ c F. i i

i

Remark 4. If x ∈ Bi then δ(x) = d(x, F ) ∼ d(xi , F ); i.e., there exist constants C1 and C2 such that  is the total sum over all balls Bi containing the point x; • if δ(x) ≤ C1 then the sum  contains finitely many summands; • if C1 ≤ δ(x) ≤ C2 then the sum  is identically zero. • if δ(x) > C2 then the sum In the case C1 ≤ δ(x) ≤ C2 we have |X J f (x)| ≤ AJ M for all J. Indeed, let C1 ≤ δ(x) ≤ C2 . Then    X J f (x) = X J P (x, pi )ϕi (x) = i

i

(3.13)



CJ1 ,J2 PJ1 (x, pi )X J2 ϕi (x),

J1 +J2 =J

and |X J f (x)| ≤ C   ×





i

J1 +J2 =J







|PJ1 (x, pi )| · |X J2 ϕi (x)| ≤ C



d(J1 )+d(L)≤k d(K)=d(L) |K|≤|L|

≤C





i

610



 i

βLK







J1 +J2 =J  η L (p−1 x)   i

γKJ1 S fS (pi )

d(S)=d(J1 )+d(K)



J1 +J2 =J L,K,S

−d(J2 )

d(L) |fS (pi )|ρ(p−1 ri i x)

−d(J2 )

ri

≤ CM

L!



since the sum contains finitely many summands and −d(J2 )

d(L) ρ(p−1 ri i x)

∼ δ(x)d(L)−d(J2 ) ≤ C,

for C1 ≤ δ(x) ≤ C2 . Henceforth we consider only those x for which δ(x) ≤ C1 . 3.5. Proof of the extension theorem. Lemma 5. The following inequalities are valid: |f (x) − P (x, a)| ≤ AM ρ(a−1 x)γ , |X J f (x) − PJ (x, a)| ≤ AM ρ(a−1 x)γ−d(J) , |X J f (x)| ≤ AM,

x ∈ G, a ∈ F ;

(3.14)

x ∈ G, a ∈ F, d(J) ≤ k;

(3.15)

x ∈ G, d(J) ≤ k;

|X J f (x)| ≤ AM δ(x)γ−k−1 ,

x ∈ c F, d(J) = k + 1.

(3.16) (3.17)

Proof. Demonstrate (3.14). It obviouslyholds for x ∈ F with A = 1. Suppose that x ∈ c F and δ(x) ≤ C1 . Then from (3.7) and the fact that i ϕi (x) = 1 we obtain f (x) − P (x, a) = =−

   i



 i

d(L)≤k d(K)=d(L) |K|≤|L|

(P (x, pi ) − P (x, a))ϕi (x)

 η L (a−1 x) ϕi (x). βLK RK (a, pi ) L!

Hence, 



|f (x) − P (x, a)| ≤ C

d(L)≤k d(K)=d(L) |K|≤|L|

≤ CM



L,K

≤ CM





|RK (a, pi )| · |η L (a−1 x)| · |ϕi (x)|

i

ρ(a−1 pi )γ−d(K) ρ(a−1 x)d(L)

i

ρ(a−1 x)γ−d(L)+d(L) ≤ AM ρ(a−1 x)γ .

i

L,K

We have used the fact that ρ(a−1 pi ) ≤ Cρ(a−1 x) (inequality (3.10)), γ − d(K) = γ − d(L) > 0, and the sum contains finitely many summands. Prove (3.15). It obviously holds for x ∈ F with A = 1, if we use the fact that X J f (x) = fJ (x) for x ∈ F (we show this later). Suppose that x ∈ c F and δ(x) ≤ C1 . Then it follows from (3.5) that X J f (x) = X J +



  P (x, pi )ϕi (x) = PJ (x, pi )ϕi (x)

i





i

J1 +J2 =J J2 =0

i

CJ1 ,J2 PJ1 (x, pi )X J2 ϕi (x).

If ρ(a−1 x) ≤ βδ(x), where β > 1 is some constant, then we take b = a; if ρ(a−1 x) > βδ(x) then we choose an arbitrary point b ∈ F such that ρ(b−1 x) ≤ βδ(x). It is clear that in both cases ρ(b−1 x) ≤ βδ(x), ρ(b−1 x) ≤ ρ(a−1 x). 611

Since



≡ 1 and

i ϕi (x)



iX

J2 ϕ

≡ 0 for x ∈ c F and 0 < d(J2 ) ≤ k, from (3.7) we derive

i (x)

X J f (x) − PJ (x, a) = +



i

J1 +J2 =J J2 =0

d(J1 )+d(L)≤k

−







βLK

d(S)=d(J1 )+d(K)

d(K)=d(L) |K|≤|L|



 i

CJ1 ,J2 (PJ1 (x, pi ) − PJ1 (x, b))X J2 ϕi (x) = −





(P (x, pi ) − P (x, a))ϕi (x)

i



×





d(J)+d(L)≤k d(K)=d(L) |K|≤|L|



i

J1 +J2 =J J2 =0

CJ1 ,J2

 η L (b−1 x) X J2 ϕi (x) γKJ1 S RS (b, pi ) L!



βLK



d(S)=d(J)+d(K)

 η L (a−1 x) ϕi (x). γKJS RS (a, pi ) L!

Therefore,

≤C

i

J1 +J2 =J L,K,S J2 =0

+C ≤ CM

|X J f (x) − PJ (x, a)|  |RS (b, pi )| · |η L (b−1 x)| · |X J2 ϕi (x)|





 

 i

i

|RS (a, pi )| · |η L (a−1 x)| · |ϕi (x)|

L,K,S





−d(J2 )

γ−d(S) ρ(p−1 ρ(b−1 x)d(L) ri i b)

J1 +J2 =J L,K,S J2 =0

+CM

  i

γ−d(S) ρ(p−1 ρ(a−1 x)d(L) . i a)

L,K,S

−1 In the first sum γ − d(S) = γ − d(J1 ) − d(K) = γ − d(J1 ) − d(L) > 0 and ρ(p−1 i b) ≤ Cρ(b x) by (3.10). −1 Since ρ(b x) ≤ βδ(x), we have −d(J2 )

ri

∼ δ(x)−d(J2 ) ≤ Cρ(b−1 x)−d(J2 ) .

−1 In the second sum γ − d(S) = γ − d(J) − d(K) = γ − d(J) − d(L) > 0 and ρ(p−1 i a) ≤ Cρ(a x) by (3.10). Therefore,

|X J f (x) − PJ (x, a)| ≤ CM

 i

+CM

  i

L,K,S





J1 +J2 =J L,K,S J2 =0

ρ(a−1 x)γ−d(J)−d(L)+d(L) ≤ CM

ρ(b−1 x)γ−d(J1 )−d(L)+d(L)−d(J2 )  i





ρ(a−1 x)γ−d(J)

J1 +J2 =J L,K,S

≤ AM ρ(a−1 x)γ−d(J) , since ρ(b−1 x) ≤ ρ(a−1 x), γ − d(J) > 0, and the sum contains finitely many summands. Inequality (3.16) follows from (3.15) if, for x ∈ G, we choose a point a ∈ F such that ρ(a−1 x) ≤ C.  is taken over those balls Bi for which the distance from them to F is This is possible, since the sum bounded from above by some constant. 612

Prove (3.17). Let x ∈ c F , δ(x) ≤ C1 , and d(J) = k + 1. We have   CJ1 ,J2 PJ1 (x, pi )X J2 ϕi (x), X J f (x) = i

J1 +J2 =J J2 =0

J1 where J2 = 0, since X d(J) = k + 1. Choose a point a ∈ F such that  P J2≡ 0 for J1 = J and −1 ρ(a x) = δ(x). Since i X ϕi (x) ≡ 0 for x ∈ c F , from (3.7) we deduce

X J f (x) =

=−





i

J1 +J2 =J J2 =0



 i

d(S)=d(J1 )+d(K)



CJ1 ,J2





βLK

d(J1 )+d(L)≤k d(K)=d(L) |K|≤|L|

J1 +J2 =J J2 =0



×

CJ1 ,J2 (PJ1 (x, pi ) − PJ1 (x, a))X J2 ϕi (x)

 η L (a−1 x) X J2 ϕi (x) γKJ1 S RS (a, pi ) L!

and 

|X J f (x)| ≤ C

i

≤ CM

 i



|RS (a, pi )| · |η L (a−1 x)| · |X J2 ϕi (x)|

J1 +J2 =J L,K,S J2 =0





−d(J2 )

γ−d(S) ρ(p−1 ρ(a−1 x)d(L) ri i a)

J1 +J2 =J L,K,S J2 =0

≤ CM

 i

≤ CM



 i





δ(x)γ−d(J1 )−d(L)+d(L)−d(J2 )

J1 +J2 =J L,K,S J2 =0





δ(x)γ−d(J) ≤ AM δ(x)γ−k−1 .

J1 +J2 =J L,K,S J2 =0

−1 Here we have used the fact that ρ(a−1 x) = δ(x), ri ∼ δ(x), ρ(p−1 i a) ≤ Cρ(a x) by (3.10), γ − d(S) = γ − d(J1 ) − d(K) = γ − d(J1 ) − d(L) > 0, d(J) = k + 1, and the sum contains finitely many summands. The lemma is proven. 

Lemma 6. The function f = Ek ({fJ }) has bounded continuous derivatives X J f in G for all d(J) ≤ k; moreover, X J f (x) = fJ (x) for x ∈ F . Proof. Proceed by induction. The function f itself is bounded (by (3.16)) and continuous on c F and coincides with f0 on F . Prove continuity of f at a ∈ F . Take x ∈ G. Then it follows from (3.14) that |f (x) − f (a)| ≤ |f (x) − P (x, a)| + |f (a) − P (x, a)| ≤ AM ρ(a−1 x)γ + O(ρ(a−1 x)) → 0

as ρ(a−1 x) → 0.

(3.18)

Assume that the assertion is proven for |J| ≤ m, d(J) ≤ k: there exist bounded continuous derivatives X J f in G and X J f = fJ on F . Prove this for |I| = m + 1 and d(I) ≤ k. Let I = ei + J with |J| = m; i.e., X I = Xi X J . From the equality X J f (a) = fJ (a) for a ∈ F we find that (3.15) is also valid for x ∈ F . 613

It is clear that the continuous derivatives of all orders exist at x ∈ c F . Prove that the derivative  Xi X J f (a) = γei JS fS (a) d(S)=d(J)+di

exists at a ∈ F . Indeed,   −1 J t (X f (a · exp(tXi )) − fJ (a)) −



  γei JS fS (a)

d(S)=d(J)+di

≤ |t−1 (X J f (a · exp(tXi )) − PJ (a · exp(tXi ), a))|       γei JS fS (a) . +t−1 PJ (a · exp(tXi ), a) − fJ (a) − t

(3.19)

d(S)=d(J)+di

By (3.15), in the first summand we have |X J f (a · exp(tXi )) − PJ (a · exp(tXi ), a)| ≤ Cρ(exp(tXi ))γ−d(J) ∼ C|t|

γ−d(J) di

.

The second summand has order O(t). Thus, the derivative X I f (a) = Xi X J f (a) =



γei JS fS (a) = fei +J (a) = fI (a)

d(S)=d(J)+di

exists at a ∈ F , since γei J,ei +J = 1 and γei JS = 0 for all S = ei + J in our case. This follows from the fact that, by (2.4),  γei JS X S g for every function g. X ei +J g = Xi X J g = d(S)=d(J)+di

Use (3.15), to prove continuity of the derivative X I f at all points of F . Take a ∈ F . Then |X I f (x) − fI (a)| ≤ |X I f (x) − PI (x, a)| + |PI (x, a) − fI (a)| ≤ CM ρ(a−1 x)γ−d(I) + O(ρ(a−1 x)) → 0

(3.20)

as ρ(a−1 x) → 0. Boundedness of the derivative follows from (3.16). The lemma is proven.  Proof of Theorem 2. Let d(J) = k. Denote g(x) = X J f (x). We need the following inequalities: |g(x) − g(a)| ≤ AM ρ(a−1 x)γ−k , |Xi g(x)| ≤ AM δ(x)γ−k−1 ,

x ∈ G, a ∈ F ;

(3.21)

x ∈ c F, i = 1, . . . , n.

(3.22)

Prove (3.21). If x ∈ F or x ∈ c F , δ(x) ≤ C1 , then from (3.15) we infer |g(x) − g(a)| = |X J f (x) − fJ (a)| = |X J f (x) − PJ (x, a)| ≤ AM ρ(a−1 x)γ−k , since PJ (x, a) = fJ (a) for d(J) = k. If x ∈ c F and C1 ≤ δ(x) ≤ C2 then |g(x) − g(a)| ≤ |g(x)| + |g(a)| ≤ 2M ≤ CM δ(x)γ−k ≤ AM ρ(a−1 x)γ−k . Prove (3.22). If δ(x) ≤ C1 then from (3.17) we derive 614

  |Xi g(x)| = |X ei X J f (x)| = 



  γei JS X S f (x) ≤ C

d(S)=k+1



|X S f (x)| ≤ AM δ(x)γ−k−1 .

d(S)=k+1

If C1 ≤ δ(x) ≤ C2 then from (3.13) we obtain |Xi g(x)| ≤ CM ≤ AM δ(x)γ−k−1 . To prove the theorem, we need to establish that the functions g belong to the class Lip(γ − k, G). The fact that |g(x)| ≤ M  for some M  = A M follows from (3.16). We are left with showing that |g(x) − g(y)| ≤ M  ρ(y −1 x)γ−k for arbitrary x, y ∈ G. If x, y ∈ F then this obviously follows from the fact that {fJ } ∈ Lip(γ, F ) and in case x ∈ c F and y ∈ F , from (3.21). Let x, y ∈ c F . Denote L = {yz : ρ(z) ≤ bρ(y −1 x)}, where b is the constant from the Lagrange theorem on a Carnot group. Consider two cases: d(L, F ) > ρ(y −1 x) and d(L, F ) ≤ ρ(y −1 x). Let d(L, F ) > ρ(y −1 x). Then, by the Lagrange theorem (2.12), |g(x) − g(y)| ≤ Cρ(y −1 x)

sup ρ(z)≤bρ(y −1 x) i=1,...,n

≤ CAM ρ(y −1 x)

sup

ρ(z)≤bρ(y −1 x)

|Xi g(yz)|

δ(yz)γ−k−1 ≤ M  ρ(y −1 x)γ−k .

Here we have used (3.22) and the fact that δ(yz) = d(yz, F ) ≥ d(L, F ) > ρ(y −1 x); consequently (since γ − k − 1 ≤ 0), sup δ(yz)γ−k−1 ≤ ρ(y −1 x)γ−k−1 .

yz∈L

Now, let d(L, F ) ≤ ρ(y −1 x). In this case there exist points x = yz ∈ L and y  ∈ F such that ≤ ρ(y −1 x). Then

ρ((y  )−1 x )

ρ((y  )−1 x) ≤ κ(ρ((y  )−1 x ) + ρ((x )−1 x)) ≤ κ(ρ(y −1 x) + ρ((yz)−1 x)) ≤ κ(ρ(y −1 x) + ρ(z −1 y −1 x)) ≤ κρ(y −1 x) + κ2 (ρ(z) + ρ(y −1 x)) ≤ (κ + κ2 b + κ2 )ρ(y −1 x) and ρ((y  )−1 y) ≤ κ(ρ((y  )−1 x ) + ρ((x )−1 y)) ≤ κ(ρ(y −1 x) + ρ((yz)−1 y)) = κ(ρ(y −1 x) + ρ(z)) ≤ (κ + κb)ρ(y −1 x). Since y  ∈ F , by (3.22), |g(x) − g(y)| ≤ |g(x) − g(y  )| + |g(y) − g(y  )| ≤ AM ρ((y  )−1 x)γ−k + AM ρ((y  )−1 y)γ−k ≤ M  ρ(y −1 x)γ−k . The theorem is proven, if we observe that M  = CM for some constant C independent of the set F .  615

3.6. An extension theorem for the Lipschitz space with a continuity modulus of a more general form. Let ω(δ), 0 < δ < ∞, be a regular continuity modulus; i.e., a positive increasing continuous function of δ with the following properties: ω(0) = 0;

ω(δ)/δ decreases.

(3.23)

It follows from (3.23) that, for every C1 > 0, there is C2 > 0 such that ω(C1 δ) ≤ C2 ω(δ) for all δ > 0. Definition 5. Let k ≥ 0 be an integer. We say that a function f on G belongs to the Lipschitz space Lip(k + ω, G) if there is a constant M such that (for all d(J) ≤ k) |X J f (x)| ≤ M ; |RJ (x, y)| ≤ M ρ(y −1 x)k−d(J) ω(ρ(y −1 x)) for all x, y ∈ G,

(3.24)

where RJ is defined by (3.2). Definition 6. A collection {fJ }d(J)≤k of functions on F belongs to the Lipschitz space Lip(k + ω, F ) if there is a constant M such that (for all d(J) ≤ k) |fJ (x)| ≤ M ; |RJ (x, y)| ≤ M ρ(y −1 x)k−d(J) ω(ρ(y −1 x)) for all x, y ∈ F,

(3.25)

where RJ is defined by (3.4). Remark 5. For ω(δ) = δ γ−k , k < γ ≤ k + 1, the space Lip(k + ω, F ) coincides with Lip(γ, F ). Remark 6. For k = 0 condition (3.25) takes the form |f (x)| ≤ M ; |f (y) − f (x)| ≤ M ω(ρ(y −1 x)) for all x, y ∈ F. Theorem 3. There is a linear extension operator Ek mapping continuously Lip(k + ω, F ) into Lip(k + ω, G); moreover, the norm of the operator is bounded by a constant independent of F ⊂ G; i.e., if a collection {fJ }d(J)≤k belongs to Lip(k + ω, F ) then f = Ek ({fJ }) ∈ Lip(k + ω, G); moreover,

f Lip(k+ω,G) ≤ C {fJ } Lip(k+ω,F ) , where C is independent of F . This theorem is a generalization of Theorem 2 and its proof repeats almost verbatim that of Theorem 2. Therefore, we only state the basic auxiliary assertions. Lemma 7. If a function f is continuous and bounded in G and has the bounded continuous derivatives X J f up to the order k (i.e., d(J) ≤ k) and X J f ∈ Lip(ω, G) for d(J) = k then f ∈ Lip(k + ω, G). The lemma is proven by analogy with Lemma 1. Lemma 8. The following inequalities are valid: |f (x) − P (x, a)| ≤ AM ρ(a−1 x)k ω(ρ(a−1 x)), |X J f (x) − PJ (x, a)| ≤ AM ρ(a−1 x)k−d(J) ω(ρ(a−1 x)), |X J f (x)| ≤ AM,

x ∈ G, a ∈ F ;

(3.26)

x ∈ G, a ∈ F, d(J) ≤ k;

(3.27)

x ∈ G, d(J) ≤ k;

|X J f (x)| ≤ AM δ(x)−1 ω(δ(x)),

x ∈ c F, d(J) = k + 1.

(3.28) (3.29)

The lemma is proven by analogy with Lemma 5 (inequalities (3.14)–(3.17)), on using the properties (3.23) of ω(δ). 616

Lemma 9. The function f = Ek {fJ } has the bounded continuous derivatives X J f in G for all d(J) ≤ k; moreover, X J f (x) = fJ (x) for x ∈ F . The lemma is proven by analogy with Lemma 6, using the properties (3.23) of ω(δ). Proof of Theorem 3. Let d(J) = k. Denote g(x) = X J f (x). We need the following inequalities: |g(x) − g(a)| ≤ AM ω(ρ(a−1 x)), |Xi g(x)| ≤ AM δ(x)−1 ω(δ(x)),

x ∈ G, a ∈ F ;

x ∈ c F, i = 1, . . . , n.

(3.30) (3.31)

They are proven by analogy with the corresponding inequalities (3.21) and (3.22). To prove the theorem, we only have to establish that the functions g belong to the class Lip(ω, G); i.e., the inequality |g(x) − g(y)| ≤ M  ω(ρ(y −1 x)) holds for arbitrary x, y ∈ G. For x ∈ c F and y ∈ F this follows from (3.30). In the case of x, y ∈ c F the arguments repeat verbatim the proof of Theorem 2; the only difference is that we use (3.31) rather than (3.22) and the estimate in the case d(L, F ) > ρ(y −1 x) is obtained as follows: |g(x) − g(y)| ≤ Cρ(y −1 x)

sup ρ(z)≤bρ(y −1 x) i=1,...,n

|Xi g(yz)|

≤ CAM ρ(y −1 x) sup δ(ξ)−1 ω(δ(ξ)) ≤ M  ω(ρ(y −1 x)), ξ∈L

since δ(ξ) > ρ(y −1 x) and the function ω(δ)/δ decreases (see (3.23)).



4. Generalization of the Classical Whitney Theorem on Carnot Groups Theorem 4. Suppose that a collection {fJ }d(J)≤k of functions on F satisfies the following conditions: |fJ (x)| ≤ M , d(J) ≤ k, on every compact subset of F and RJ (x, y) = o(ρ(y −1 x)k−d(J) ) in the sense that, for arbitrary ε > 0 and x ¯ ∈ F , there is δ = δ(ε, x ¯ ) > 0 such that the inequality |RJ (x, y)| ≤ ερ(y −1 x)k−d(J) holds for all x, y ∈ F satisfying the conditions ρ(¯ x−1 x) < δ and ρ(¯ x−1 y) < δ. Then the operator Ek defined by (3.12) gives the extension of the collection {fJ } to the whole group G. The extended function f = Ek ({fJ }) belongs to the class C k (G); i.e., the continuous derivatives X J f exist for all J, where d(J) ≤ k; moreover, X J f |F = fJ . The proof of Theorem 4 is similar to those of Theorems 2 and 3. However, it is not a straightforward consequence of Theorem 3, since it imposes weaker conditions on {fJ }. This theorem is an analog of the classical Whitney theorem [4, 5]. The corresponding theorem for functions of the class C 1 on Heisenberg groups was proven in [6]. Some generalizations of the classical Whitney theorem in Rn were obtained by Fefferman [7]. Lemma 10. Let a ∈ F , x ∈ c F , and d(J) ≤ k. Then X J f (x) − PJ (x, a) = o(ρ(a−1 x)k−d(J) ) as x → a in the usual sense; i.e., for every ε > 0, there is ˜δ = ˜δ(ε, a) > 0 such that if ρ(a−1 x) < ˜δ then |X J f (x) − PJ (x, a)| ≤ ερ(a−1 x)k−d(J) . Proof. The lemma is proven by analogy with (3.15). It is obvious that the derivatives X J f (x)  exist and are continuous on c F . Since x → a, it suffices to consider the case δ(x) ≤ C1 ; i.e., the sum in (3.12) is the total sum over all balls Bi containing the point x. In this case from (3.5) we see    PJ (x, pi )ϕi (x) + CJ1 ,J2 PJ1 (x, pi )X J2 ϕi (x). X J f (x) = i

i

J1 +J2 =J J2 =0

617

If ρ(a−1 x) ≤ βδ(x), where β > 1 is some constant, then we take b = a; if ρ(a−1 x) > βδ(x) then we choose an arbitrary point b ∈ F such that ρ(b−1 x) ≤ βδ(x). It is clear that in both cases ρ(b−1 x) ≤ βδ(x) and ρ(a−1 x). ρ(b−1 x) ≤   Since i ϕi (x) ≡ 1 and i X J2 ϕi (x) ≡ 0 for x ∈ c F and 0 < d(J2 ) ≤ k, from (3.7) we obtain  (P (x, pi ) − P (x, a))ϕi (x) X J f (x) − PJ (x, a) = +

i





i

J1 +J2 =J J2 =0





×

CJ1 ,J2 (PJ1 (x, pi ) − PJ1 (x, b))X J2 ϕi (x) = − 

d(J1 )+d(L)≤k d(K)=d(L) |K|≤|L|

−



 i





βLK

d(S)=d(J1 )+d(K)



d(J)+d(L)≤k d(K)=d(L) |K|≤|L|

|X J f (x) − PJ (x, a)| ≤ C

d(S)=d(J)+d(K)

 i

+C



i

J1 +J2 =J J2 =0

CJ1 ,J2



 η L (a−1 x) ϕi (x), γKJS RS (a, pi ) L!

−d(J2 )

|RS (b, pi )|ρ(b−1 x)d(L) ri

J1 +J2 =J L,K,S J2 =0

  i



 η L (b−1 x) X J2 ϕi (x) γKJ1 S RS (b, pi ) L!



βLK



|RS (a, pi )|ρ(a−1 x)d(L) .

L,K,S

a) is chosen where δ = δ(ε/C, is a constant depending according to the condition of the theorem, C is the constant in (3.10), and C on k and the characteristics of G to be chosen later. By (3.10), Take an arbitrary ε > 0. Let x ¯ = a and ρ(a−1 x) < ˜δ =

ρ(a−1 pi ) < C˜δ ≤ δ,

δ κ(C+2κ) ,

ρ(a−1 b) ≤ κ(ρ(b−1 x) + ρ(a−1 x)) ≤ 2κρ(a−1 x) < 2κ˜δ ≤ δ,

ρ(b−1 pi ) ≤ κ(ρ(a−1 pi ) + ρ(a−1 b)) < κ(C + 2κ)˜δ = δ. It follows from the conditions of the theorem and (3.10) that ε ε ε |RS (a, pi )| ≤ ρ(a−1 pi )k−d(S) = ρ(a−1 pi )k−d(J)−d(L) ≤ C ρ(a−1 x)k−d(J)−d(L) C C C in the second sum and |RS (b, pi )| ≤

ε ε ε ρ(b−1 pi )k−d(S) = ρ(b−1 pi )k−d(J1 )−d(L) ≤ C ρ(b−1 x)k−d(J1 )−d(L) C C C −d(J2 )

in the first sum. Since ρ(b−1 x) ≤ βδ(x), we have ri |X J f (x) − PJ (x, a)| ≤ C

ε  C i

+C



∼ δ(x)−d(J2 ) ≤ Cρ(b−1 x)−d(J2 ) . Hence,



ρ(b−1 x)k−d(J1 )−d(L)+d(L)−d(J2 )

J1 +J2 =J L,K,S J2 =0

ε   ε ρ(a−1 x)k−d(J) = ερ(a−1 x)k−d(J) , ρ(a−1 x)k−d(J)−d(L)+d(L) ≤ C C C i

L,K,S

since ρ(b−1 x) ≤ ρ(a−1 x), k − d(J) ≥ 0, and the sum contains finitely many summands. The lemma is proven.  618

Proof of Theorem 4. We have to prove that the extended function f has continuous derivatives with d(J) ≤ k and X J f = fJ on the set F . Since f ∈ C ∞ (c F ), it suffices to prove existence and continuity of the derivatives at all points of F . We prove this by induction. By construction, the function f itself coincides with f0 on F . Prove its continuity at a ∈ F . Let x ∈ G and ρ(a−1 x) → 0. Then it follows from Lemma 10 that XJ f

|f (x) − f (a)| ≤ |f (x) − P (x, a)| + |f (a) − P (x, a)| ≤ o(ρ(a−1 x)k ) + O(ρ(a−1 x)) → 0 as ρ(a−1 x) → 0. Assume that f is already proven to have continuous derivatives X J f with |J| ≤ m, d(J) ≤ k, and J X f = fJ on F . The estimate for the remainder RJ in the condition of the theorem and the fact that X J f (x) = fJ (x) for x ∈ F imply that the estimate X J f (x) − PJ (x, a) = o(ρ(a−1 x)k−d(J) ) is also valid for x ∈ F . Consider X I f with |I| = m + 1, d(I) ≤ k. Assume that I = ei + J with |J| = m; i.e., X I = Xi X J . Then d(I) = di + d(J). Prove that the derivative



Xi X J f (a) =

γei JS fS (a)

d(S)=d(J)+di

exists at a ∈ F . Indeed, (3.19) is valid. In the first summand on the right-hand side of (3.19), using / F and the estimates for RJ if a exp(tXi ) ∈ F , we obtain Lemma 10 if a exp(tXi ) ∈ |X J f (a exp(tXi )) − PJ (a exp(tXi ), a)| = o(ρ(exp(tXi ))k−d(J) ) = o(|t|

k−d(J) di

).

The second summand has order O(|t|). Therefore,   −1 J t (X f (a exp(tXi )) − fJ (a)) −



  γei JS fS (a)

d(S)=d(J)+di

≤ o(|t|

k−d(J) −1 di

) + O(|t|) → 0 as |t| → 0,

since k − d(J) − di = k − d(I) ≥ 0. Thus, the derivative X I f (a) = Xi X J f (a) =



γei JS fS (a) = fei +J (a) = fI (a)

d(S)=d(J)+di

exists at a ∈ F . Here, as in Lemma 6, we have used (2.4). Prove continuity of the derivative X I f at all points of F . Take a ∈ F . By Lemma 10, |X I f (x) − fI (a)| ≤ |X I f (x) − PI (x, a)| + |PI (x, a) − fI (a)| ≤ o(ρ(a−1 x)k−d(I) ) + O(ρ(a−1 x)) → 0 as ρ(a−1 x) → 0, since k − d(I) ≥ 0. The second summand is estimated by analogy with Lemma 6 (see (3.20)). The theorem is proven.  619

References 1. Vodop yanov S. K. and Pupyshev I. M., “Whitney-type theorems on extension of functions on a Carnot group,” Dokl. Ross. Akad. Nauk, 406, No. 5, 586–590 (2006). 2. Folland G. B. and Stein E. M., Hardy Spaces on Homogeneous Groups, Princeton Univ. Press, Princeton, N.J. (1982). 3. Vodop yanov S. K. and Greshnov A. V., “On extension of functions of bounded mean oscillation from domains in a space of homogeneous type with intrinsic metric,” Siberian Math. J., 36, No. 5, 873–901 (1995). 4. Whitney H., “Analytic extensions of differentiable functions defined in closed sets,” Trans. Amer. Math. Soc., 36, 63–89 (1934). 5. Stein E. M., Singular Integrals and Differentiability Properties of Functions, Princeton University Press, Princeton, N.J. (1973). 6. Franchi B., Serapioni R., and Serra Cassano F., “Rectifiability and perimeter in the Heisenberg group,” Math. Ann., 321, No. 3, 479–531 (2001). 7. Fefferman C., “A sharp form of Whitney’s extension theorem,” Ann. Math. (2), 161, No. 1, 509–577 (2005). S. K. Vodop yanov; I. M. Pupyshev Sobolev Institute of Mathematics, Novosibirsk, Russia E-mail address: [email protected]; [email protected]

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