The Whitney extension problem and Lipschitz selections of set-valued mappings in jet-spaces

0.23 MB·English

by  Pavel Shvartsman

The Whitney extension problem and
Lipschitz selections of set-valued mappings in jet-spaces
By Pavel Shvartsman
Department of Mathematics, Technion - Israel Institute of Technology,
6 32000 Haifa, Israel
0
0 e-mail: [email protected]
2

n
a
J
Abstract
9
2
We study a variant of the Whitney extension problem [21, 22] for the space

] Ck,ω(Rn). We identify Ck,ω(Rn) with a space of Lipschitz mappings from Rn into
A
the space P ×Rn of polynomial fields on Rn equipped with a certain metric. This
F k
identification allows us to reformulate the Whitney problem for Ck,ω(Rn) as a Lip-
.
h
schitz selection problem for set-valued mappings into a certain family of subsets of
t
a P ×Rn. We prove a Helly-type criterion for the existence of Lipschitz selections for
m k
such set-valued mappings defined on finite sets. With the help of this criterion, we
[
improve estimates for finiteness numbers in finiteness theorems for Ck,ω(Rn) due to

1 C. Fefferman [8, 10, 11].
v
1
1
7
1
0
6
0
/
h 1. The main problem and main results
t
a
m Let ω : R → R be a continuous concave function satisfying ω(0) = 0. We let
+ +
: C˙k,ω(Rn) denote the (homogeneous) space of all functions f : Rn → R with continuous
v
i derivatives of all orders up to k, for which the seminorm
X
ar |Dαf(x)−Dαf(y)|
kfk := sup
C˙k,ω(Rn) x,y∈Rn,x6=y ω(kx−yk)
|α|=k
X
is finite. By Ck,ω(Rn) we denote the Banach subspace of C˙k,ω(Rn) defined by the norm
kfk := sup |Dαf(x)|+kfk .
Ck,ω(Rn) C˙k,ω(Rn)
x∈Rn
|α|≤k
X
Throughout the paper we let S denote an arbitrary closed subset of Rn.
Math Subject Classification 46E35,49K24,52A35, 54C60,54C65
Key Words and Phrases Whitney’s extension problem, smooth functions, finiteness, metric, jet-space,
set-valued mapping, Lipschitz selection
1
In this paper we study the following extension problem.
Problem. Given a positive integer k and an arbitrary function f : S → R, what
is a necessary and sufficient condition for f to be the restriction to S of a function F ∈
Ck,ω(Rn)?
This is a variant of a classical problem which is known in the literature as the Whitney
Extension Problem [21, 22]. It has attracted a lot of attention in recent years. We refer
the reader to [3]-[6], [8]-[14], [1, 2] and [23, 24] and references therein for numerous results
in this direction, and for a variety of techniques for obtaining them.
This note is devoted to the phenomenon of “finiteness” in the Whitney problem. It
turns out that, in many cases, Whitney-type problems for different spaces of smooth func-
tions can be reduced to the same kinds of problems, but for finite sets with prescribed
numbers of points.
For the space C1,ω(Rn) (with ω(t) = tp, 0 < p ≤ 1,) and for the Zygmund space, this
phenomenon has been studied in the author’s papers [16, 17]. The case of an arbitrary ω
was treated in joint papers with Yu. Brudnyi [3, 6]. It was shown that a function f defined
on S can be extended to a function F ∈ C1,ω(Rn) with kFk ≤ γ = γ(n) provided
C1,ω(Rn)
its restriction f| to every subset S′ ⊂ S consisting of at most N(n) = 3·2n−1 points can
S′
be extended to a function F ∈ C1,ω(Rn) with kF k ≤ 1. (Moreover, the value
S′ S′ C1,ω(Rn)
3·2n−1 is sharp [17, 6].)
This result is an example of “the finiteness property” of the space C1,ω(Rn). We call
the number N appearing in formulations of finiteness properties “the finiteness number”.
In his pioneering work [22], H. Whitney characterized the restriction of the space
Ck(R),k ≥ 1, to an arbitrary subset S ⊂ R in terms of divided differences of functions.
An application of Whitney’s method to the space Ck,ω(R) implies the finiteness property
for this space with the finiteness number N = k +2.
An impressive breakthrough in the solution of the Whitney problem for Ck,ω-spaces
has recently been made by C. Fefferman [8]-[14]. In this paper we will consider two of his
remarkable results related to the finiteness property and its generalizations for the space
Ck,ω(Rn). Here is the first of them:
Theorem 1.1 (C. Fefferman [8, 10]). There is a positive integer N = N(k,n) such that
the following is true: Suppose we are given a function ω, a set S ⊂ Rn, and functions
f : S → R and ξ : S → R . Assume that, for any S′ ⊂ S with at most N points, there
+
exists a function F ∈ Ck,ω(Rn) with kF k ≤ 1, and
S′ S′ Ck,ω(Rn)
|F (x)−f(x)| ≤ ξ(x) for all x ∈ S′.
S′
Then there exists F ∈ Ck,ω(Rn), with kFk ≤ γ and
Ck,ω(Rn)
|F(x)−f(x)| ≤ γ ·ξ(x), x ∈ S.
Here γ = γ(k,n) is a constant depending only on k and n.
In particular, if the function ξ is chosen to be identically zero, Theorem 1.1 shows that
the space Ck,ω(Rn) possesses the finiteness property for all k,n ≥ 1.
An upper bound for the finiteness number N(k,n) given in [8, 10] is
N(k,n) ≤ (dimP +1)3·2dimPk. (1.1)
k
2
Here P stands for the space of polynomials of degree at most k defined on Rn. (Recall
k
that dimP = n+k .)
k k
Our first result, Theorem 1.2, states that the expression bounding N(k,n) in (1.1) can
(cid:0) (cid:1)
be replaced by a considerably smaller expression which depends on dimP exponentially.
k
Theorem 1.2 Theorem 1.1 holds with the finiteness number N(k,n) = 2dimPk.
Remark 1.3 In the spring of 2005 I learned that E. Bierstone and P. Milman obtained an
improvement of estimate (1.1) to a bound that is exponential. Recently P. Milman kindly
drew my attention to the fact that their result gives precisely the estimate 2dimPk for the
spaces Ck(Rn) and Ck,ω(Rn).
In fact there are many different versions of the Whitney extension problem. These
versions arise when one considers a possibly different space of smooth functions on Rn and
a possibly different collection of given information about the function on the set S. In
his classical paper [21], Whitney solved a version for the space Ck(Rn) in the case where
the given information about the function includes its values and the values of all of its
partial derivatives of all orders up to k on the set S. Using Whitney’s extension method G.
Glaeser [15] proved a similar result for the space Ck,ω(Rn). Let us recall its formulation.
Given a k-times differentiable function f and x ∈ Rn, we let Tk(f) denote the Taylor
x
polynomial of f at x of degree at most k:
1
Tk(f)(y) := (Dαf)(x)(y −x)α , y ∈ Rn.
x α!
|α|≤k
X
Theorem 1.4 (Whitney-Glaeser). Given a family of polynomials {P ∈ P : x ∈ S} there
x k
is a function F ∈ Ck,ω(Rn) such that Tk(F) = P for every x ∈ S if and only if there is a
x x
constant λ > 0 such that for every α,|α| ≤ k we have
|DαP (x)| ≤ λ for all x ∈ S, (1.2)
x
and
max{|Dα(P −P )(x)|,|Dα(P −P )(y)|} ≤ λkx−ykk−|α|ω(kx−yk), (1.3)
x y x y
for all x,y ∈ S. Moreover,
inf{kFk : Tk(F) = P ,x ∈ S} ≈ infλ
Ck,ω(Rn) x x
with constants of equivalence depending only on k and n.
Observe that Theorem 1.4 can be interpreted as a finiteness theorem with the finiteness
number N = 2. In fact, the inequalities (1.2) and (1.3) depend on at most 2 (arbitrary)
points of S so that the sufficiency part of this result can be reformulated as follows: There
is a function F ∈ Ck,ω(Rn) with kFk ≤ γ(k,n) satisfying Tk(F) = P , x ∈ S,
Ck,ω(Rn) x x
provided for every two-point set S′ ⊂ S there exists a function F ∈ Ck,ω(Rn) with
S′
kF k ≤ 1 such that Tk(F ) = P , x ∈ S′.
S′ Ck,ω(Rn) x S′ x
In [11] C. Fefferman considered a version of the Whitney problem in which the family
of polynomials {P ∈ P : x ∈ S} is replaced by a family {G(x) : x ∈ S} of convex
x k
3
centrally-symmetric subsets of P . He raised the following question: How can we decide
k
whether there exist F ∈ Ck,ω(Rn) and a constant A > 0 such that
Tk(F) ∈ A⊚G(x) for all x ∈ S ?
x
Here A⊚G(x) denotes the dilation of G(x) with respect to its center by a factor of A.
Let P ∈ P be the center of the set G(x). This means that G(x) can be represented
x k
in the form G(x) = P +σ(x) where σ(x) ⊂ P is a convex family of polynomials which is
x k
centrally symmetric with respect to 0. It is shown in [11] that, under certain conditions on
the sets σ(x), the finiteness property holds. We say that a set σ(x) ⊂ P is “Whitney ω-
k
convex” (with Whitney constant A) at x ∈ Rn if the following two conditions are satisfied:
(i). σ(x) is closed, convex and symmetric with respect to 0;
(ii). Suppose P ∈ σ(x), Q ∈ P and δ ∈ (0,1]. Assume that P and Q satisfy the
k
estimates
|∂βP(x)| ≤ ω(δ)δk−|β| and |∂βQ(x)| ≤ δ−|β|
for all |β| ≤ k. Then Tk(P ·Q) ∈ Aσ(x). (See [11], p. 579.)
x
Theorem 1.5 ([11]) Given integers k,n ≥ 1 there is a constant N = N(k,n) for which the
following holds: Foreach x ∈ S, suppose we are given a polynomial P ∈ P , and a Whitney
x k
ω-convex set σ(x) with Whitney constant A. Suppose that for every subset S′ of S with
cardinality at most N there exists a function F ∈ Ck,ω(Rn) such that kF k ≤ 1
S′ S′ Ck,ω(Rn)
and
Tk(F ) ∈ P +σ(x) for all x ∈ S′.
x S′ x
Then there exists a function F ∈ Ck,ω(Rn), satisfying kFk ≤ γ and
Ck,ω(Rn)
Tk(F) ∈ P +γ ·σ(x) , x ∈ S. (1.4)
x x
Here γ depends only on k,n and the Whitney constant A.
A particular case of this result for σ(x) = {P ∈ P : DαP(x) = 0, |α| ≤ k − 1} and
k
ω(t) = tp, 0 < p ≤ 1, with the finiteness number N = 3·2(n+kk−2) has been proved in [5].
Analogously to Theorem 1.2, our second result in this paper gives an explicit upper
bound for a finiteness number.
Theorem 1.6 Theorem 1.5 holds with the finiteness number
N(k,n) = 2min{ℓ+1,dimPk},
where ℓ = max dimσ(x).
x∈S
Remark 1.7 Of course ℓ necessarily satisfies ℓ ≤ dimP = n+k .
k k
(cid:0) (cid:1)
In fact both of our new estimates for finiteness numbers are corollaries of the following
theorem which is the main result of this paper.
4
Theorem 1.8 Let G be a mapping defined on a finite set S ⊂ Rn which assigns a convex
set of polynomials G(x) ⊂ P of dimension at most ℓ to every point x of S. Suppose that,
k
for every subset S′ of S consisting of at most 2min{ℓ+1,dimPk} points, there exists a function
F ∈ Ck,ω(Rn) such that kF k ≤ 1 and Tk(F ) ∈ G(x) for all x ∈ S′. Then there
S′ S′ Ck,ω(Rn) x S′
is a function F ∈ Ck,ω(Rn), satisfying kFk ≤ γ and
Ck,ω(Rn)
Tk(F) ∈ G(x) for all x ∈ S.
x
Here γ depends only on k,n and cardS.
Comparing this result with Theorem 1.5, let us note that here there are no restrictions
on G. Moreover, here Tk(F) belongs to G(x) itself and not merely to its dilation as in
x
(1.4). However the price of that we have to pay to obtain such a general result is that we
have to permit the constant γ (controlling the Ck,ω-norm of the function F) to depend on
the number of points of S.
We can use the rather informal and imprecise terminology “Ck,ω(Rn) has the weak
finiteness property” to express the kind of result obtained in Theorem 1.8 where γ depends
on the number of points of S. The fact that such a weak finiteness property holds, strongly
suggeststhatwecanreasonablyhopetoestablishananalogous“strongfinitenessproperty”,
by which we mean a result with γ depending only on k and n. Such a result may possibly
require some additional very mild conditions to be imposed on the mapping G.
The weak finiteness property also provides an upper bound for the finiteness constant
whenever the strong finiteness property holds. For instance, Fefferman’s Theorems 1.1 and
1.5 reduce the problem to a set of cardinality at most N(k,n) while the weak finiteness
propertydecreases thisnumber to2dimPk (asinTheorem1.2)orto2min{l+1,dimPk} (Theorem
1.6).
We prove Theorem 1.8 in Section 4. The proof is based on an approach presented in
Sections 2 and 3.
The crucialingredient inthisapproachisanisomorphism between thespaceCk,ω(Rn)|
S
and a certain space of Lipschitz mappings from S into the product P ×Rn equipped with
k
a certain metric d . We define d and study its properties in Section 2. One of these
ω ω
properties, which is obtained in Proposition 2.5, is a useful formula for calculating d ,
ω
namely
d (T,T′) ≈ max{ω(kx−x′k),ϕ (|Dα(P −P′)(x)|),ϕ (|Dα(P −P′)(x′)|)}
ω α α
|α|≤k
where T = (P,x) and T′ = (P′,x′) are any two elements of P × Rn, and ϕ :=
k α
ω((sk−|α|ω(s))−1). We refer to the set
P ×Rn := {(P,x) : P ∈ P ,x ∈ Rn}
k k
as the space of (potential) k-jets. This name and also the definition of d are motivated by
ω
the Whitney-Glaeser extension theorem 1.4.
Given T = (P,x) ∈ P ×Rn and λ ∈ R we define λ ◦ T := (λP,x). Then inequality
k
(1.3) of the Whitney-Glaeser extension theorem can be reformulated as follows:
d (λ−1 ◦T ,λ−1 ◦T ) ≤ ω(kx−yk), T := (P ,x),T := (P ,y) ∈ P ×S. (1.5)
ω x y x x y y k
5
We define a metric on S by setting r (x,y) := ω(kx−yk) for all x,y ∈ S and we let
ω
S be the metric space S := (S,r ). We also consider P × Rn as a metric space with
ω ω ω k
respect to d , i.e., we set T := (P × Rn,d ). Let Lip(S ,T ) denote the space of
ω k,n k ω ω k,n
Lipschitz mappings from S (equipped with the metric r ) into P ×Rn (with the metric
ω k
d ). Inequality (1.5) motivates us to equip this space with a “norm” by setting
ω
kTk := inf{λ : kλ−1 ◦Tk ≤ 1}. (1.6)
LO(S) Lip(Sω,Tk,n)
We call k ·k the Lipschitz-Orlicz norm. We use it to define a second “norm” by
LO(S)
setting
kTk∗ := max sup|DαP (x)|+kTk (1.7)
LO(S) x LO(S)
|α|≤k x∈S
and we introduce the subspace Lip(S ,T ) of Lip(S ,T ) of “bounded” Lipschitz map-
ω k,n ω k,n
pings T(x) = (P ,z ), x ∈ S, defined by the finiteness of the “norm” (1.7).
x x
Now the Whitney-Glaeser extension theorem implies the following
Proposition 1.9 Given a family of polynomials {P ∈ P : x ∈ S}, there is a function
x k
F ∈ Ck,ω(Rn) such that Tk(F) = P for every x ∈ S if and only if the mapping T(x) :=
x x
(P ,x),x ∈ S, belongs to Lip(S ,T ). Moreover,
x ω k,n
inf{kFk : Tk(F) = P ,x ∈ S} ≈ kTk∗
Ck,ω(Rn) x x LO(S)
with constants of equivalence depending only on k and n.
Applying this proposition to S = Rn we obtain an interesting isomorphism between
Ck,ω(Rn) and a certain subfamily of Lip(Rn,T ). Namely, every function F ∈ Ck,ω(Rn)
ω k,n
gives rise to a Lipschitz mapping from Rn := (Rn,ω(k· k)) into P × Rn defined by the
ω k
formula T(x) := (Tk(F),x),x ∈ Rn. On the other hand, every Lipschitz mapping from Rn
x ω
toP ×Rn oftheformT(x) := (P ,x),x ∈ Rn,generatesafunctionF(x) := P (x),x ∈ Rn,
k x x
such that F ∈ Ck,ω(Rn) and Tk(F) = P ,x ∈ Rn.
x x
Let us restate this more concisely: The mapping
Ck,ω(Rn) ∋ F 7→ T(x) := (Tk(F),x) ∈ Lip(Rn,T )
x ω k,n
and its inverse mapping
Lip(Rn,T ) ∋ T(x) := (P ,x) 7→ F(x) := P (x) ∈ Ck,ω(Rn)
ω k,n x x
provide an isomorphism between Ck,ω(Rn) and the subfamily of Lip(Rn,T ) consisting
ω k,n
of all elements of the form T(x) := (P ,x), x ∈ Rn. Moreover, Proposition 1.9 states that
x
this isomorphism in some sense “preserves restrictions”.
The above ideas and results are presented in Section 2. They show that even though
Whitney’s problem deals with restrictions of k-times differentiable functions, it is also a
problem about Lipschitz mappings defined on subsets of Rn and taking values in a very
non-linear metric space T = (P ×Rn,d ). More specifically, the Whitney problem can
k,n k ω
be reformulated as a problem about Lipschitz selections of set-valued mappings from S into
2Tk,n. We study this problem in Section 3. We remark that the Lipschitz selection method
6
has already been used to obtain a solution to the Whitney problem for the space C1,ω(Rn),
see [17, 19, 6].
We recall some relevant definitions: Let X = (M,ρ) and Y = (T,d) be metric spaces
and let G : M → 2T 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 be
a Lipschitz selection of the mapping G. (For various results and techniques related to the
problem of the existence of Lipschitz selections in the case where Y = (T ,d) is a Banach
space, we refer the reader to [18, 19, 20] and references therein.)
It turns out that Theorem 1.8, the “weak finiteness” theorem, is equivalent to the
following Helly-type criterion for the existence of a Lipschitz selection.
Theorem 1.10 Let S ⊂ Rn be a finite set and let G(x) = (G(x),x),x ∈ S, be a set-
valued mapping such that for each x ∈ S the set G(x) ⊂ P is a convex set of polynomials
k
of dimension at most ℓ. Suppose that there exists a constant K > 0 such that, for every
subsetS′ ⊂ S consisting of atmost 2min{ℓ+1,dimPk} points, the restriction G|S′ has a Lipschitz
selection g ∈ Lip(S′,T ) with kg k ≤ K. Then G, considered as a map on all of
S′ k,n S′ LO(S′)
S, has a Lipschitz selection g ∈ Lip(S,T ) with kgk ≤ γK, where the constant γ
k,n LO(S)
depends only on k,n and cardS.
The proof of this result relies on some methods and ideas developed for the case of set-
valuedmappingswhichtaketheirvaluesinBanachspaces, see, e.g. Shvartsman[18,19,20].
In particular, an analog of Theorem 1.10 for Banach spaces has been proved in [19]. Our
strategy will beto adapt that proof to the case of the metric space T = (P ×Rn,d ). As
k,n k ω
inthecaseofBanachspacesouradaptedproofwillbebasedonHelly’sintersection theorem
[7] and a combinatorial result about a structure of finite metric graphs (Proposition 3.1).
Acknowledgment. I am greatly indebted to Michael Cwikel, Charles Fefferman and
Naum Zobin for interesting discussions and helpful suggestions and remarks.
2. Ck,ω(Rn) as a space of Lipschitz mappings
The point of departure for our approach is inequality (1.3) of the Whitney extension
theorem. This inequality motivates the definition of a certain special metric on the set
P × Rn which allows us to identify the restriction Ck,ω(Rn)| with a space of Lipschitz
k S
mappings from S into P ×Rn.
k
Observe that without loss of generality we may assume that ω is a strictly increasing
concave function on R . (In fact, for every positive concave ω : R → R there is a
+ + +
concave strictly increasing function ω∗ such that ω∗ ≤ ω. Therefore ω˜ := ω + ω∗ is a
concave strictly increasing function satisfying ω ≤ ω˜ ≤ 2ω.)
Now let us define a metric on P × Rn. To this end given multiindex α,|α| ≤ k, we
k
define a function ϕ : R → R by letting
α + +
ϕ := ω((sk−|α|ω(s))−1) (2.1)
α
for |α| < k and ϕ (t) := t for |α| = k. Since for |α| < k the function sk−|α|ω(s) is strictly
α
increasing, the inverse function ψ := (sk−|α|ω(s))−1 is well-defined so that the function
α
7
ϕ = ω(ψ ) is well-defined as well. It can be also readily seen that
α α
1
ϕ′ (t) = .
α ψk−|α|(t)+(k −|α|)ψk−|α|−1(t)ω(ψ (t))/ω′(ψ (t))
α α α α
Since ψ and ω are non-decreasing and ω′ is non-increasing, ϕ′ is non-increasing, so that
α α
ϕ is a concave function.
α
Fix two k-jets T = (P ,x ),T = (P ,x ) ∈ P ×Rn and put
0 0 0 1 1 1 k
δ (T ,T ) := max{ω(kx −x k),ϕ (|Dα(P −P )(x )|),ϕ (|Dα(P −P )(x )|}. (2.2)
ω 0 1 0 1 α 0 1 0 α 0 1 1
|α|≤k
Clearly,
δ ((P,x ),(P,x )) = ω(kx −x k) for every P ∈ P , x,y ∈ Rn. (2.3)
ω 0 1 0 1 k
Recall that λ◦T := (λP,x) where T = (P,x) ∈ P ×Rn and λ ∈ R. In these settings
k
inequality (1.3) of the Whitney-Glaeser extension theorem means the following
δ (λ−1 ◦T ,λ−1 ◦T ) ≤ ω(kx−yk), T := (P ,x),T := (P ,y).
ω x y x x y y
Now we define a metric d on P ×Rn by letting
ω k
m−1
d (T,T′) := inf δ (T ,T ) (2.4)
ω ω i i+1
i=0
X
where the infimum is taken over all finite families {T ,T ,...,T } ⊂ P × Rn such that
0 1 m k
T = T and T = T′.
0 m
In particular, since ω is subadditive, by this definition for every x,y ∈ Rn
d ((P,x),(Q,y)) ≥ ω(kx−yk), P,Q ∈ P ,
ω k
and by (2.3)
d ((P,x),(P,y)) = ω(kx−yk) for every P ∈ P . (2.5)
ω k
The main result of this section is the following
Theorem 2.1 For every T,T′ ∈ P ×Rn we have
k
d (T,T′) ≤ δ (T,T′) ≤ d (en ◦T,en ◦T′).
ω ω ω
A proof of the theorem is based on a series of auxiliary lemmas.
Lemma 2.2 For every t ,t > 0 and every multiindexes α,β such that |α|+ |β| ≤ k we
1 2
have
|β|
ϕ (t t ) ≤ max{ω(t ),ϕ (t )}.
α 1 2 1 α+β 2
8
Proof. If |α| = k, then β = 0 so that nothing to prove. Therefore we will assume that
|α| < k. In this case by (2.1) ϕ (t|β|t ) = ω(u) where u := (sk−|α|ω(s))−1(t|β|t ). Hence
α 1 2 1 2
t|β|t = uk−|α|ω(u).
1 2
Suppose that |α|+|β| < k. Then ϕ (t ) = ω(v) where v := (sk−|α|−|β|ω(s))−1(t ) so
α+β 2 2
that t = vk−|α|−|β|ω(v).
2
Put w := max{t ,v}. Then
1
max{ω(t ),ϕ (t )} = max{ω(t ),ω(v)} = ω(max{t ,v}) = ω(w). (2.6)
1 α+β 2 1 1
Hence
t|β|t = t|β|vk−|α|−|β|ω(v) ≤ w|β|wk−|α|−|β|ω(w) = wk−|α|ω(w).
1 2 1
But t|β|t = uk−|α|ω(u) so that
1 2
uk−|α|ω(u) = t|β|t ≤ wk−|α|ω(w).
1 2
Since tk−|α|ω(t) is an strictly increasing function, this implies u ≤ w. Recall also that
|β|
ϕ (t t ) = ω(u). Then by (2.6)
α 1 2
|β|
ϕ (t t ) = ω(u) ≤ ω(w) = max{ω(t ),ϕ (t )}.
α 1 2 1 α+β 2
It remains to consider the case |α|+ |β| = k. In this case by the definition of ϕ we
α
have ϕ (t ) = t . If sup ω(t) ≤ t , then
α+β 2 2 t>0 2
|β|
ϕ (t t ) = ω(u) ≤ t = ϕ (t )
α 1 2 2 α+β 2
and the lemma follows. If t < sup ω(t) then there is v > 0 such that t = ω(v). This
2 t>0 2
shows that equality t = vk−|α|−|β|ω(v) holds for the case |α|+|β| = k as well. The lemma
2
2
is proved.
Lemma 2.3 Let Q ∈ P and let a,b ∈ Rn. Then for every multiindex α,|α| ≤ k, we have
k
1
|DαQ(b)| ≤ |Dα+βQ(a)|·kb−ak|β|.
β!
|β|≤k−|α|
X
Proof. Since DαQ is a polynomial of degree at most k−|α|, Taylor’s formula for DαQ
at a gives
1
DαQ(b) = Dα+βQ(a)(b−a)β
β!
|β|≤k−|α|
X
2
which immediately implies the required inequality of the lemma.
Lemma 2.4 Let T = (P,x),T′ = (P′,x′) ∈ P ×Rn and let
k
{T = (P ,x ) : i = 0,1,...,m}
i i i
be a finite family of elements of P × Rn such that T = T, T = T′. Then for every
k 0 m
α,|α| ≤ k,
m−1
|Dα(P −P′)(x)| ≤ en max |Dα+β(P −P )(x )|·kx−x k|β|.
i i+1 i i
|β|≤k−|α|
i=0
X
9
Proof. By Lemma 2.3
1
|Dα(P −P )(x)| ≤ |Dα+β(P −P )(x )|·kx−x k|β|
i i+1 i i+1 i i
β!
|β|≤k−|α|
X
so that
m−1
|Dα(P −P′)(x)| ≤ |Dα(P −P )(x)|
i i+1
i=0
X
m−1
1
≤ |Dα+β(P −P )(x )|·kx−x k|β|
i i+1 i i
β!
i=0 |β|≤k−|α|
X X
m−1
1
= |Dα+β(P −P )(x )|·kx−x k|β|.
i i+1 i i
β!
|β|≤k−|α| i=0
X X
Hence
m−1
1
|Dα(P −P′)(x)| ≤ max |Dα+β(P −P )(x )|·kx−x k|β|
i i+1 i i
 β!|β|≤k−|α|
|β|≤k−|α| i=0
X X
 m−1
≤ en max |Dα+β(P −P )(x )|·kx−x k|β|
i i+1 i i
|β|≤k−|α|
i=0
X
2
proving the lemma.
We are in a position to prove Theorem 2.1.
Proof of Theorem 2.1. The first inequality follows from definition (2.4). Let us prove
the second inequality. Consider a family {T = (P ,x ) : i = 0,1,...,m} ⊂ P ×Rn such
i i i k
that T = T := (P,x),T = T′ := (P′,x′). Thus P ∈ P , x ∈ Rn for every i = 0,...,m
0 m i k i
and P = P,P = P′,x = x,x = x′. Let us prove that
0 m 0 m
m−1
δ (T,T′) ≤ δ (en ◦T ,en ◦T ). (2.7)
ω ω i i+1
i=0
X
Let us fix a multiindex α, |α| ≤ k and estimate ϕ (|Dα(P −P′)(x)|). By Lemma 2.4
α
m−1
|Dα(P −P′)(x)| ≤ en max |Dα+β(P −P )(x )|·kx−x k|β|.
i i+1 i i
|β|≤k−|α|
i=0
X
Put P := enP . Then the latter inequality implies
i i
m−1
e |Dα(P −P′)(x)| ≤ max |Dα+β(P −P )(x )|·kx−x k|β|.
i i+1 i i
|β|≤k−|α|
i=0
X
e e
Since kx−x k ≤ m−1kx −x k we obtain
i i=0 i i+1
P m−1 |β| m−1
|Dα(P −P′)(x)| ≤ max kx −x k · |Dα+β(P −P )(x )| .
i i+1 i i+1 i
|β|≤k−|α| ! !
i=0 i=0
X X
e e
10