Automatic continuity and weighted composition operators between spaces of vector-valued differentiable functions

25 Pages·0.3 MB·English

by  Jesus Araujo

AUTOMATIC CONTINUITY AND WEIGHTED COMPOSITION
OPERATORS BETWEEN SPACES OF VECTOR-VALUED
1
DIFFERENTIABLE FUNCTIONS
0
0
2
JESU´S ARAUJO

n
a
J

9
1 Abstract. LetE andF beBanachspaces. ItisprovedthatifΩandΩ′ are
open subsets of Rp and Rq, respectively, and T is a linear biseparating map

] between twospaces ofdifferentiablefunctions An(Ω,E)andAm(Ω′,F),then
A p=q, n=m, and there exist a diffeomorphism h of class Cm from Ω′ onto
F Ω,andamapJ :Ω′→L(E,F)ofclasss−Cn suchthatforeveryy∈Ω′ and
. everyf ∈An(Ω,E),(Tf)(y)=(Jy)(f(h(y))). InparticularE andF areiso-
h
morphicasBanachspacesand,asaconsequence, alllinearbiseparatingmaps
t
a arecontinuous forusualtopologies inthespaces ofdifferentiablefunctions.
m
[ 1. Introduction

2 It is well known that an algebraic link between spaces of continuous functions
v mayleadtoatopologicallinkbetweenthespacesonwhichthefunctionsaredefined.
5
Forinstance,itturnsoutthatifthereexistsaringisomorphismT :C(X)→C(Y),
0
then the realcompactifications of X and Y are homeomorphic ([GJ, pp. 115-118]).
2
2 Also if h is the resultant homeomorphism from the realcompactification of Y onto
1 that of X, then Tf =f ◦h for every f ∈C(X), so we have a complete description
0 ofit. As aresult, whenX andY arerealcompact,we deduce thatifboth spacesof
0
continuousfunctionsC(X)andC(Y)areendowedwiththecompact-opentopology,
/
h then every ring isomorphism between them is continuous. In this result, the key
t
a point is that every ring isomorphism sends maximal ideals into maximal ideals.
m This implies that a good description of maximal ideals lead to the definition of a
map from Y onto X.
:
v
Of course, the pattern above have been succesfully applied to many other alge-
i
X bras of functions. However the situation becomes more complicated if we consider
r spaces of functions which take values in arbitrary Banach spaces. In this context
a and unlike algebra or ring homomorphisms, we can still use mappings satisfying
the property kTfkkTgk ≡ 0 if and only if kfkkgk ≡ 0. These maps are called
biseparating, and coincide with disjointness preserving mappings whose inverses
preserve disjointness too ([AK]). In general these maps turn out to be efficacious
substitutes for homomorphisms. Indeed, in [A1], we prove that the existence of a
biseparating mapping between a large class of spaces of vector-valued continuous
functions A(X,E) and A(Y,F) (E, F are Banach spaces) yields homeomorphisms
between some compactifications (and even the realcompactifications) of X and Y.
The automatic continuity of a linear biseparating mapping is also accomplished in
some cases (see [A2, A3]). Related results have also been given recently, for some
Research partially supported by the Spanish Direccio´n General de Investigacio´n Cient´ıfica y
T´ecnica(DGICYT,PB98-1102).
1
2 JESU´S ARAUJO
other families of scalar-valued functions, for instance, in [ABN, FH, J] and [JW].
In this paper, we go a step beyond and work in a context which does not seem to
have made its way into the literature yet, namely, linear operators between spaces
of differentiable functions taking values in arbitrary Banach spaces.
As for spaces (indeed algebras) of scalar-valued differentiable functions, Myers
([My]) showed that the structure of a compact differentiable manifold of class Cn
is determined by the algebra of all real-valued functions on M of class Cn. In this
line,Pursell([P])checkedthattheringstructureofinfinitelydifferentiablefunctions
defined on an open convex set of Rn determines such set up to a diffeomorphism.
HomomorphismsbetweenalgebrasofdifferentiablefunctionsdefinedonrealBanach
spaces have been studied by Aron, G´omez and Llavona ([AGL]); in this paper a
descriptionofhomomorphismsisgivenandthe automaticcontinuityisobtainedas
a corollary in quite a general setting, which includes in particular the case when
the Banach spaces are finite-dimensional. Further, Guti´errez and Llavona ([GL])
pursuedthestudyofcontinuoushomomorphismsbetweenalgebrasofdifferentiable
functions defined on open subsets of real Banach spaces, obtaining a description
in many cases. On the other hand, automatic continuity results for algebras of
differentiable functions have been also given for instance in [BCL], [Lr], [KN] and
[NRV]. For classical results and techniques in the study of automatic continuity,
seealso[D]and[S]. Finally,Molnarstatesinthefinalremarksof[Ml]thatitwould
seem to be of some importance to study some of the results above for separating
mappings instead of algebra or ring isomorphism. In his opinion, that is certainly
more difficult to carry out.
Our main goal here is to investigate the behaviour of biseparating maps when
defined between spaces of vector-valued differentiable functions. Let E and F be
Banach spaces. We shall prove that if Ω and Ω′ are open subsets of Rp and Rq,
respectively,andT isalinearbiseparatingmapbetweentwospacesofdifferentiable
functionsAn(Ω,E)andAm(Ω′,F),thenp=q,n=m,andthereexistadiffeomor-
phism h of class Cm from Ω′ onto Ω, and a map J :Ω′ →L(E,F) of class s−Cn
such that for every y ∈ Ω′ and every f ∈ An(Ω,E), (Tf)(y) = (Jy)(f(h(y))).
In particular E and F are isomorphic as Banach spaces and, as a consequence,
all linear biseparating maps are continuous for usual topologies in the spaces of
differentiable functions.
In Section 2 (as well as in the rest of Section 1), we assume that E is a real
Banach space.
If λ = (λ ,λ ,... ,λ ) is a p-tuple of non-negative integers, we set |λ| := λ +
1 2 p 1
λ +...+λ . If Ω is an open subset of Rp, then Cn(Ω,E) consists ofthe E-valued
2 p
functions f in Ω whose partial derivatives
∂λ1+λ2+...+λpf
∂λf :=
∂xλ1∂xλ2...∂xλp
1 2 p
exist and are continuous for each λ = (λ ,λ ,... ,λ ) ∈ Λ, where Λ := {λ ∈
1 2 p
(N∪{0})p :|λ|≤n}. Itiswellknownthat,iffor Lk(Rn,E)we denotethe spaceof
continuous k-R-linear maps of Rn into E, then Cn(Ω,E) coincides with the space
of maps f : Ω → E such that the differential Dkf : Ω → Lk(Rn,E) exists and is
continuous for each k =0,... ,n.
Fromnow onwe willassume that n≥1,and thatΩ⊂Rp is a (nonempty)open
set. Also, for C ⊂ Rp, clRpC denotes its closure in Rp. Ccn(Ω,R) will denote the
AUTOMATIC CONTINUITY 3
subringofCn(Ω,R),respectively,offunctionswithcompactsupport. Ontheother
hand, if x=(x ,x ,... ,x ) belongs to Rp, we set |x|:=max |x |.
1 2 p j j
2. Some previous results
Lemma 2.1. Suppose that a ,a ,... ,a ∈ Cn(Ω,E), and that f : Ω×R → E is
0 1 k
a polynomial in t defined as
k
f(x,t)= a (x)ti
i
i=0
X
for every (x,t)∈Ω×R. Then f ∈Cn(Ω×R,E).
Proof. It is immediate from the fact that all partialderivatives up to order n exist
and are continuous.
Lemma 2.2. Suppose that f,g ∈Cn(Ω,E) and k :Ω→R satisfy
f(x)=k(x)g(x)
for every x∈Ω. If g(x)6=0 for every x∈Ω, then k ∈Cn(Ω,R).
Proof. Fix x ∈Ω. We are going to prove that k is of class Cn in a neighborhood
0
of x . First we take f′ ∈E′ such that f′(g(x ))6=0. Since f′ and g are continuous,
0 0
then there exists an open neighborhood U of x such that
0
f′(g(x))6=0
for every x∈U.
Now
f′(f(x))=k(x)f′(g(x))
for every x∈U. This implies that, for every x∈U,
f′(f(x))
k(x)= ,
f′(g(x))
which is the quotient of two real-valued functions of class Cn.
This proves that k is of class Cn.
It is well known that, when it exists at a point a∈Ω, the differential Dkf(a)∈
Lk(Rp,E) is a symmetric form of degree k.
Now, given a map f ∈ Cn(Ω,E), we define its Taylor polynomial function of
degree n at a∈Ω as
1
T (x) := f(a)+Df(a)(x−a)+ D2f(a)(x−a,x−a)+...
a
2
1
+ Dnf(a)(x−a,x−a,... ,x−a).
n!
n
The proof of the following result is straightforward.
| {z }
Lemma 2.3. Suppose that f ∈ Cn(Ω,E), a ∈ Ω, and 1 ≤ k ≤ n. Then the k-th
derivative of the Taylor polynomial function T of degree n of f is equal to the
a
Taylor polynomial function of degree n−k of Dkf at a.
The following theorem, known as Whitney’s extension theorem, can be found,
for instance, in [Fr, Theorem 3.1.14].
4 JESU´S ARAUJO
Theorem 2.4. Suppose n ∈ N, A is a closed subset of Rp, and to each a ∈ A
corresponds a polynomial function
P :Rp →E
a
with degree P ≤n. Whenever C ⊂A and δ >0 let ρ(C,δ) be the supremum of the
a
set of all numbers
DiP (b)−DiP (b) ·|a−b|i−n·(n−i)!
a b
corresponding to i=(cid:13)0,... ,n and a,b∈C(cid:13) with 0<|a−b|≤δ.
(cid:13) (cid:13)
If ρ(C,δ) → 0 as δ → 0+ for each compact subset C of A, then there exists a
map g :Rp →E of class Cn such that
Dig(a)=DiP (a)
a
for i=0,... ,n and a∈A.
Proposition 2.5. Let p ≥ 2. For s ∈ R, s > 0, consider the following compact
subsets of Rp:
p
A:={(x ,x ,... ,x )∈Rp : x2 ≤x2/9,|x |≤s},
1 2 p i 1 1
i=2
X
A+ :={(x ,x ,... ,x )∈A:x ≥0},
1 2 p 1
and
A− :={(x ,x ,... ,x )∈A:x ≤0}.
1 2 p 1
Suppose that Ω is an open subset of Rp containing A and that f belongs to
Cn(Ω,E). If
∂λf(0,0,... ,0)=0
for every λ ∈ Λ, then there exists a function f ∈ Cn(Ω,E) with compact support
+
such that, for λ∈Λ,
∂λf (x)=∂λf(x)
+
for every x∈A+, and
∂λf (x)=0
+
for every x∈A−.
Proof. Suppose that for x ∈ A, T stands for the polynomial function of degree
x
n given in the Taylor formula for f at x. Now for a ∈ A−, we consider as P
a
the polynomial identically zero, and for a∈A+, we consider as P the polynomial
a
T . As it is seen for instance in [Ch, Theorem 2.71] or [Ln, p. 350], if r > 0 and
a
|b−a|<r, we have that
|b−a|n
kT (b)−f(b)k≤ sup kDnf(x)−Dnf(a)k.
a
n!
|x−a|≤r
Now it is easy to see that, by Lemma 2.3, for i∈{1,2,... ,n},
|b−a|n−i
DiT (b)−Dif(b) ≤ sup kDnf(x)−Dnf(a)k.
a
(n−i)!
|x−a|≤r
(cid:13) (cid:13)
This pr(cid:13)oves that if |b−a|(cid:13)<r,
DiT (b)−Dif(b) ·|b−a|i−n·(n−i)!≤ sup kDnf(x)−Dnf(a)k
a
|x−a|≤r
(cid:13) (cid:13)
(cid:13) (cid:13)
AUTOMATIC CONTINUITY 5
fori∈{0,1,...,n}. Thenwedefineτ(r)asthesupremumofthesetofallnumbers
sup kDnf(x)−Dnf(a)k
|x−a|≤r
for x,a ∈ A, which is a real number, because A is compact. Clearly, since Dnf is
continuous,if r tends to zero,τ(r) tends to zero. Now supposethata andb belong
to A and 0<|b−a|≤r. Then we have the following possibilities:
• a,b∈A+. Then we have that, by Lemma 2.3, for i∈{0,1,...,n},
DiP (b)=Dif(b)
b
and consequently
DiP (b)−DiP (b) = DiT (b)−Dif(b) .
a b a
• a,b∈/ A+. Th(cid:13)en (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
DiP (b)−DiP (b) =0.
a b
• a ∈/ A+, b ∈ A+. N(cid:13)ote that since we (cid:13)are assuming by hypothesis that
Dif(0,0,... ,0)=0,th(cid:13)enDiP (b)=0for(cid:13)alli∈{0,1,... ,n}. Consequently
0
DiP (b)−DiP (b) = DiP (b)
a b b
(cid:13) (cid:13) = (cid:13)DiP0(b)(cid:13)−DiPb(b)
(cid:13) (cid:13) (cid:13) (cid:13)
= (cid:13)DiT0(b)−Dif(b) (cid:13).
(cid:13) (cid:13)
• a∈A+, b∈/ A+. Then we have that (cid:13) (cid:13)
(cid:13) (cid:13)
DiP (b)−DiP (b) = DiT (b)
a b a
(cid:13) (cid:13) ≤ (cid:13)DiTa(b)(cid:13)−Dif(b) + Dif(b)
(cid:13) (cid:13) (cid:13) (cid:13)
= (cid:13)DiTa(b)−Dif(b)(cid:13)+(cid:13)Dif(b)(cid:13)−DiT0(b) .
(cid:13) (cid:13) (cid:13) (cid:13)
Note that in the third and fort(cid:13)h cases above, |a|,(cid:13)|b|≤(cid:13) |b−a|≤r. This(cid:13)implies
(cid:13) (cid:13) (cid:13) (cid:13)
that in these two cases
DiP (b)−DiP (b) ·|b−a|i−n·(n−i)! ≤ sup kDnf(x)−Dnf(a)k
a b
|x−a|≤r
(cid:13)(cid:13) (cid:13)(cid:13) + sup kDnf(x)−Dnf(0)k.
|x|≤r
On the other hand it is easy to see that in the other two cases
DiP (b)−DiP (b) ·|b−a|i−n·(n−i)!≤ sup kDnf(x)−Dnf(a)k.
a b
|x−a|≤r
(cid:13) (cid:13)
Th(cid:13)is facts imply that,(cid:13)if ρ is definedas inTheorem2.4,ρ(A,r)≤2τ(r). Also,it
isclearthatifC is acompactsubsetofA, thenρ(C,r)≤ρ(A,r). Consequentlyby
Theorem 2.4, we have that there exists f ∈Cn(Ω,E) such that, given any λ∈Λ,
0
∂λf (x)=∂λf(x)
0
for every x∈A+, and
∂λf (x)=0
0
foreveryx∈A−. Alsoitisclearthatifwetakeg ∈Cn(Ω,R)suchthatg ≡1on
0 c 0
an open neighborhoodof A, then f :=g f ∈Cn(Ω,E) satisfies the requirements
+ 0 0
of the theorem.
6 JESU´S ARAUJO
3. Biseparating maps: a first approach
In previous sections, we assumed that E was a real Banach space. Imagine now
that it is a complex Banach space instead. It is clear that in any case, it can also
beviewedasarealspace,andinthis senseweconsiderdefinedthespaceAn(Ω,E).
It is immediate that all results given before hold for them.
But being also complex ensures that An(Ω,E) is both real and complex as a
linear space, and consequently we can consider both real and complex linear maps
fromAn(Ω,E)into someother vectorspaces. This is the reasonwhy,unlike sofar,
in this section we assume that E and F are K-Banach spaces, where K=R or C.
In this section (and in the following ones) we also assume that our spaces are
defined according to one of the following two situations.
• Situation 1. Ω and Ω′ are (not necessarily bounded) open subsets of Rp
and Rq, respectively (p,q ∈ N). In this case, An(Ω,E) = Cn(Ω,E) and
Am(Ω′,F)=Cm(Ω′,F) (n,m≥1).
• Situation 2. Ω and Ω′ are bounded open subsets of Rp and Rq, respectively
(p,q ∈ N). In this case, An(Ω,E) = Cn(Ω¯,E) and Am(Ω′,F) = Cm(Ω¯′,F)
(n,m≥1). Here Cn(Ω¯,E) (respectively Cm(Ω¯′,F)) denotes the subspace of
Cn(Ω,E)(respectivelyCm(Ω′,F))ofthosefunctionswhosepartialderivatives
up to order n (respectively m) admit continuous extension to the boundary
of Ω (respectively Ω′).
For a function f ∈An(Ω,E) (respectively,in Am(Ω′,F)), we denote by c(f) the
cozero set of f, that is, the set {x∈Ω:f(x)6=0}.
Asforthespacesoflinearfunctions,wewilldenotebyL′(E,F)andbyB′(E,F)
the sets of (not necessarily continuous) linear maps and bijective linear maps from
E into F, respectively. L(E,F) and B(E,F) will denote the spaces of continuous
linear maps and bijective continuous linear maps from E into F.
Definition 3.1. A map T : An(Ω,E) → Am(Ω′,F) is said to be separating if it
is additive and c(Tf)∩c(Tg) = ∅ whenever f,g ∈ An(Ω,E) satisfy c(f)∩c(g) =
∅. Besides T is said to be biseparating if it is bijective and both T and T−1 are
separating.
Equivalently, we see that an additive map T : An(Ω,E) → Am(Ω′,F) is sepa-
rating if k(Tf)(y)kk(Tg)(y)k = 0 for all y ∈ Ω′ whenever f,g ∈ An(Ω,E) satisfy
kf(x)kkg(x)k=0 for all x∈Ω.
Let Ω :=Ω, Ω′ :=Ω′ when we are in Situation 1, and Ω :=Ω¯, Ω′ :=Ω¯′ if we
1 1 1 1
are in Situation 2. A point x ∈ Ω is said to be a support point of y ∈ Ω′ if, for
1 1
every neighborhood U of x in Ω , there exists f ∈ An(Ω,E) satisfying c(f) ⊂ U
1
such that (Tf)(y)6=0.
In a much more general context, (not necessarily linear) biseparating maps are
studied in [A1]. Some general results are given concerning conditions which allow
to link certain topological spaces when dealing with a biseparating map. Suppose
in particular that A is a subring of the space C(Ω ,R) of all real-valued functions
1
on Ω1. Clearly each map f :Ω1 →R admits a continuous extension fβΩ1 :βΩ1 →
R∪{∞}to the Stone-CˇechcompactificationofΩ (which obviouslycoincides with
1
Ω whenweareinSituation2above). Assumenowthatthespaceofallextensions
1
ofelementsofAseparatethepointsofβΩ ;following[A1],wesaythatAisstrongly
1
regular when given x ∈ βΩ and a nonempty closed subset K of βΩ which does
0 1 1
AUTOMATIC CONTINUITY 7
not contain x0, there exists f ∈ A such that fβΩ1 ≡ 1 on a neighborhood of x0
and fβΩ1 ≡ 0 on K. Now, let us denote by C(Ω1,E) the space of all continuous
E-valued functions on E. In particular, in [A1, Corollary 2.9], it is proven that if
A(Ω ,E)⊂C(Ω ,E) andB(Ω′,F)⊂C(Ω′,F)are anA-module anda B-module,
1 1 1 1
respectively,where A⊂C(Ω ,R)andB ⊂C(Ω′,R)arestronlyregularrings,then
1 1
there exists a homeomorphismh fromΩ′ onto Ω wheneverthere is a biseparating
1 1
map from A(Ω ,E) onto B(Ω′,F).
1 1
Applied to our situation, it is well known that if A = Cn(Ω ,R) and B =
1
Cm(Ω′,R),thenAandB arestronglyregularrings(seeforinstance[W,Corollary
1
1.2]). Onthe other hand, itis straightforwardto see that An(Ω,E) andAm(Ω′,F)
are an A-module and a B-module, respectively. As a consequence, the existence
of a biseparating map from An(Ω,E) onto Am(Ω′,F) ensures the existence of a
homeomorphism h from Ω′ onto Ω . As it is also shown in [A1], this map h sends
1 1
each point in Ω′ into its support point in Ω , and is called support map for T. It
1 1
turns out that the support map for T−1 is h−1 ([A1, Theorem 2.8]). Rephrasing
Lemma 2.4 in [A1], we have the following property.
Lemma 3.1. If y ∈ Ω′ and f ∈ An(Ω,E) vanishes on a neighborhood of h(y),
1
then (Tf)(y)=0.
Lemma 3.2. SupposethatΩcontainstheorigin,andthatT :An(Ω,E)→Am(Ω′,F)
is a biseparating map. Assume also that f ∈An(Ω,E) satisfies that for all λ∈Λ,
∂λf(0,0,... ,0)=0.
If (0,0,... ,0)∈Ω is the support point of y ∈Ω′, then (Tf)(y)=0.
Proof. First suppose that p>1 and that the closedball of center 0 and radius s is
containedinΩ. IfwetakeA+,A− andf asinProposition2.5,thenf andf−f
+ + +
belongtoAn(Ω,E)andsatisfyf (x)=0and(f−f )(x)=0foreveryx∈A− and
+ +
foreveryx∈A+ respectively. We havethat(Tf)(y)=(Tf )(y)+(T(f−f ))(y).
+ +
Also, since for any neighborhood U of the origin there exists an open subset V of
U such that f (x) = 0 for every x ∈ V, then we have that, taking into account
+
that the support map for T−1 is h−1, by Lemma 3.1, (Tf )(h−1(x)) = 0. Since
+
h : Ω′ → Ω is a homeomorphism, we deduce that (Tf )(y) = 0 and, in the same
+
way, (T(f −f ))(y)=0. We conclude that (Tf)(y)=0.
+
Consider now the case when p=1. Note that if f ∈An(Ω,E) satisfies f(0)=0
and 0 = f′(0) = ··· = f(n)(0), then it is clear that fξ and fξ belong
(−∞,0) (0,+∞)
to An(Ω,E) (where ξ stands for the characteristic function of A) and, as above,
A
(T(fξ ))(y)=0=(T(fξ ))(y). We conclude that (Tf)(y)=0.
(0,+∞) (−∞,0)
Next lemma is a first attempt to describe all biseparating maps, and does not
take into account some important details which will be discussed in Section 5. In
thiswaywecharacterizeallbiseparatinglinearmapsfromAn(Ω,E)ontoAm(Ω′,F)
asweightedcompositionbijectivemaps. Noticethatweassumenocontinuityprop-
erties on T. In fact, we will suppose that our spaces An(Ω,E) and Am(Ω′,F) are
not endowed with any topologies.
Lemma 3.3. Suppose that T : An(Ω,E) → Am(Ω′,F) is a K-linear biseparating
map. Then p = q, n = m, and there exist a diffeomorphism h of class Cn from
Ω′ onto Ω and a map J : Ω′ → B′(E,F) such that for every y ∈ Ω′ and every
f ∈An(Ω,E),
(Tf)(y)=(Jy)(f(h(y))).
8 JESU´S ARAUJO
Proof. First,the existenceofthe homeomorphismh(the supportmap)betweenΩ′
and Ω implies that p=q (see for instance [Fk, p. 120]).
Note that if f ∈An(Ω,E) and y ∈Ω′ satisfy
∂λf(h(y))=0
for all λ∈Λ, then by Lemma 3.2 we have that
(Tf)(y)=0.
Now take y ∈Ω′ and fix e∈E, e6=0. If #Λ stands for the cardinal of Λ, then
wecandefinealinearmapS :R#Λ →F asfollows. Given(a ,... ,a )∈R#Λ,
y λ1 λ#Λ
we consider any f ∈An(Ω,R) such that
∂λf(h(y))=a
λ
for every λ∈Λ. Then we define
S (a ,... ,a ):=(Tfe)(y).
y λ1 λ#Λ
The map S is linear and, as we have seen above, does not depend on the function
y
f we choose. This implies that it is well defined.
Then it is easy to see that there exist functions α from Ω′ into F, λ∈Λ, such
λ
that for every y ∈Ω′ and every f ∈An(Ω,R),
(Tfe)(y)= α (y)∂λf(h(y)). (3.1)
λ
λ∈Λ
X
From now on we consider i∈{1,2,... ,p} fixed. Also x ∈An(Ω,R) will be the
i
projection on the i-th coordinate for every point in Ω.
Next we define some functions
α0,α1,... ,αn+1
i i i
from Ω′×R into F. For every y ∈Ω′ and t∈R,
α0(y,t) := (Te)(y),
i
α1(y,t) := (Tx e)(y)−α0(y,t)t,
i i i
2!α2(y,t) := (Tbx2e)(y)−α0(y,t)t2−2α1(y,t)t,
i i i i
3!α3(y,t) := (Tx3e)(y)−α0(y,t)t3−3α1(y,t)t2−6α2(y,t)t,
i i i i i
and in general, for k ∈{1,2,...,n,n+1}
k!αk(y,t) := (Txke)(y)−α0(y,t)tk−kα1(y,t)tk−1−k(k−1)α2(y,t)tk−2
i i i i i
− ...−k!αk−1(y,t)t.
i
Claim 1. For l ∈{0,1,... ,n+1}, l!αl is a polynomial in t whose coefficients
i
are a linear combination of Te,Tx e,... ,Txle. Moreover, for y ∈ Ω′ fixed, the
i i
degree of the polynomial l!αl(y,t) is at most l. If we also assume that (Te)(y)6=0,
i
then the degree of l!αli(y,t) is lband its leading coefficient is equal to (−1)lα0i(y,t)
(notice that this term does not depend on t). b
We are going to prove it by applying induction on l. It is clear that this is true
for l = 0. Suppose that this relation also holds for l ∈ {0,1,2,... ,k} for some
AUTOMATIC CONTINUITY 9
k ≤n. We are going to see that it holds for l=k+1. We have that
(k+1)!αk+1(y,t) := (Txk+1e)(y)−α0(y,t)tk+1
i i i
− (k+1)α1(y,t)tk−(k+1)kα2(y,t)tk−1
i i
− (k+1)k(k−1)α3(y,t)tk−2
i
− ...
− (k+1)!αk(y,t)t,
i
whichimplies that itisa polynomialint and,for fixedy ∈Ω,its coefficientforthe
term tk+1 is α0(y,t)(−1+(k+1)− k+1 + k+1 −...−(k+1)(−1)k), which is
i 2 3
equal to α0(y,t)(−(1−1)k+1+ k+1 (−1)k+1), that is, to (−1)k+1α0(y,t).
i k+1(cid:0) (cid:1) (cid:0) (cid:1) i
Thus the claim is proved.
(cid:0) (cid:1)
Now, by Lemma 2.1, we have that for every k ∈{0,1,... ,n+1}, αk belongs to
i
Cm(Ω′×R,F).
Next define α :=Te, and for k ∈{1,2,... ,n}, α :=α :Ω′ →F, where
0 k λ
λ=(0,0,... ,0,k,0,... ,0).
b
i
Also, let h stand for the i-th coo|rdina{tze fun}ction of h.
i
Claim 2. For every k ∈{0,1,... ,n}, and for every y ∈Ω′,
α (y)=αk(y,h (y)).
k i i
First we have from Equation 3.1 that for k ∈{0,1,... ,n} and y ∈Ω′,
(Txke)(y)= α (y)∂λxk(h(y)),
i λ i
λ∈Λ
X
which can be written as
(Txke)(y) = α (y)hk(y) (3.2)
i 0 i
+ kα (y)hk−1(y)
1 i
+ ...
+ k!α (y)h (y)
k−1 i
+ k!α (y),
k
because
∂λxk(h(y))=0
i
whenever λ∈Λ, λ6=(0,0,... ,j,0,... ,0), j ∈{0,1,... ,k}.
i
On the other hand, it is clear that α (y) = (Te)(y) = α0(y,h (y)) for every
0 i i
y ∈ Ω′. Also suppose|tha{tzk <} n and that α (y) = αj(y,h (y)) for every j ∈
j i i
b
10 JESU´S ARAUJO
{0,1,...,k} and every y ∈Ω′. Then, by Equation 3.2, for y ∈Ω′
(k+1)!α (y) = (Txk+1e)(y)
k+1 i
− α (y)hk+1(y)
0 i
− (k+1)α (y)hk(y)
1 i
− ...
− (k+1)!α (y)h (y),
k i
which coincides with (k+1)!αk+1(y,h (y)), and the claim is proved.
i i
On the other hand, notice that in the same way as we obtain Equation 3.2, we
have
(Txn+1e)(y) = α (y)hn+1(y) (3.3)
i 0 i
+ (n+1)α (y)hn(y)
1 i
+ ...
+ (n+1)!α (y)h (y)
n i
for every y ∈Ω′.
Claim 3. Suppose that y ∈ Ω′ satisfies α (y ) 6= 0. Then for every open
0 0 0
neighborhood U of y , there exists a nonempty open subset W′ of U where h is of
0 i i
class Cm.
First we define F0 :Ω′×R→F as F0 :=(n+1)!αn+1, that is,
i i i
F0(y,t) = (Txn+1e)(y)
i i
− α0(y,t)tn+1
i
− (n+1)α1(y,t)tn
i
− (n+1)nα2(y,t)tn−1
i
− ...
− (n+1)!αn(y,t)t,
i
for every y ∈Ω′, t∈R.
Then, if j ∈{1,2,...,n+1}, we define Fj :Ω′×R→F as
i
∂jF0
Fj(y,t)= i (y,t),
i ∂tj
for all y ∈Ω′, t∈R.
Noticethatfromthe definitionofF0,Claim2andEquation3.3,wededucethat
i
F0(y,h (y))=0
i i
foreveryy ∈Ω′. Also,aswestatedinClaim1,thecoefficientsofF0asapolynomial
i
of degree n+1 in t are linear combinations of
(Te)(y),(Tx e)(y),... ,(Txn+1e)(y),
i i
andconsequently,byLemma2.1,fork ∈{1,2,... ,n},FkbelongstoCm(Ω′×R,F).
i
Taking into account thbat Fn+1(y,t) = (n+1)!(−1)n+1α (y) for every (y,t) ∈
i 0
Ω′ ×R, and the fact that α (y ) 6= 0, there exists k ∈ {0,1,... ,n} such that
0 0 0
Fk0(y,h (y)) = 0 for every y in a neighborhood of y and Fk0+1(y,h (y)) takes a
i i 0 i i
valuedifferentfrom0forsomey ineveryneighborhoodofy . SupposethenthatU
0