functions in endowed with the norm k kCk( ) defined as supremum of the derivatives of order ... be expressed in coordinates by saying that an object is Ck if we can find ... they do di er and some statements such as that certain operators acting on ... this way. If k < 1, we cannot apply the implicit function theorem to conclude.
Decomposition theorems for some groups of di eomorphisms and regularity of the composition operator. R. de la Llave yz Dept. Math. U. Texas at Austin Austin TX 78712 R. Obaya Dept. Matematica aplicada a la Ingeniera Escuela Superior de Ingenieros Industriales Univ. de Valladolid 47011 Valladolid SPAIN Abstract
We study the algebraic structure of several di eomorphism groups in Sn . We show that all di eomorphisms can be written as a nite product of di eomorphisms which have certain symmetries. Since the natural group operation among di eomorphisms is the composition and Holder regularities appear naturally in intermediate steps, we present an appendix in which optimal results about regularity of the composition operator are obtained when the spaces of di eomorphisms are given the topology of the classical Holder norms. This preprint is avaliable from the Math Physics preprints archive. Send e-mail to details. y Supported by N.S.F. grants z e-mail address: llavemath.utexas.edu
mp arcmath.utexas.edu for
1
1 Introduction The goal of this paper is to prove several decomposition theorems for di eomorphism groups. These theorems, roughly, state that any di eomorphism in the group can be written as a nite product of di eomorphisms lying in a certain subgroup enjoying certain properties such as symmetry. There are several motivations for the study of this theorems. For example, in [6] it is shown that one theorem of this type can be used to solve the inverse problem for scattering of geodesic elds in surfaces of genus one. It can also be argued that these theorems are analogues of the usual factorization theorems in Lie algebras and that stronger versions could be useful in the problem of computing representations of di eomorphism groups [Bi]. If we consider these theorems as in nite dimensional versions of factorization theorems for Lie groups, one rst diculty is that for di eomorphism groups, the corresponding Lie algebra, the Lie algebra of vector elds, is considerably worse behaved than those of nite dimensional Lie groups. For example, the exponential is not surjective in any neighborhood of the identity. Hence, it is not possible to apply the usual implicit function theorem in Banach spaces to obtain results in a small enough neighborhood of the origin using the \in nitesimal results" using the Lie algebra. Nevertheles, we will show that it is possible to obtain similar results to those in the nite dimensional case by using appropriate \hard" implicit function theorems of the Nash{Moser type. This seems to be a natural generalization of the systematic methods of nite dimensional Lie group theory to groups of diffeomorphisms and we hope that they have wider applicabilty. e.g. when the groups of di eomorphisms are required to preserve a certain geometric structure. In a rst chapter, we will consider decomposition theorems of the Langer { Singer type for di eomorphisms on Sn. Roughly, we show that 2n+1 subgroups of di eomorphisms consisting of di eomorphisms which commute with re ection generate the whole group of di eomorphisms. That is, any di eomorphism can be written as the composition of a nite number of factors, each of which commutes with re actions across planes of codimension one. Moreover, the factors are essentially one dimensional. If we assume that the factor is di erentiable suciently often, we can conclude that the factors are also severl times di erentiable. the factors are 2
The method we use can obtain the previously known theorems for diffeomorphisms in Tn and obtain results for di eomorphisms of Sn. From the point of view of the theory of representations, it would be useful to have global analogues of the local decomposition theorems [Bi]. We produce some expamples that show that such global results would be dicult to obtain.
2 Acknowledgements The work of R.L. has been supported by NSF. grants. Also R.L wants to ackwoledge the hospitality enjoyed at Valladolid during several visits. Letters from J. Langer and F. Bien discussing the implications of these theorems were very useful for us.
3 Decomposition theorem for spheres.
3.1 Notation and statement of results
We willnconsider the standard embedding of the sphere in Euclidean space o Pn?1 n ? 1 n ? 1 2 S = x (x1 ; ; xn) 2 IR j i=1 xi = 1 If x is a point in Sn?1, we will denote by x the hyperplane passing through the origin and consisting of vectors orthogonal to x, by Mx; we will denote the one-dimensional maximum circle going from x to ?x and passing through the point and we will refer to those sets as meridians . We will also denote by Px; the subsets of Sn?1 whose points have a constant projection
along the line joining the origin and the point x. The sets of these form, will be referred to as parallels. Finally, we denote by x, the re ection across the plane x. If k is an integer, and is a compact set in IRd we will say that f : ! IR is C k if it has continuous derivatives of order k. C k ( ), the space of C k functions in endowed with the norm k kC k ( ) de ned as supremum of the derivatives of order up to k is a Banach space. If k is not an integer, we de ne as usual
kf kC k ( ) = max(sup kDk f (x) ? Dk f (y)k=jx ? yjk?[k]; kf kC [k]( ) ) (1) x6=y
3
To cover Lipschitz conditions as well, we introduce the notation k? = (k ? 1) + 1?.In this notation, 1? will be bigger than any number in [0; 1) but smaller than 1. These notions of regularity can be lifted to geometric objects that can be expressed in coordinates by saying that an object is C k if we can nd coordinate patches that cover the manifold on which the expression is C k . Furthermore, if we x a set of coordinate patches that cover the manifold we can talk about the C k distance between two objects by declaring it to be the maximum of the C k distance between the coordinate expressions. A moment's re ection will show that the statement that a geometric object is C k is independent of the coordinate patches chosen (that is, if the coordinate expressions in one patch are C k so are in all others). Unfortunately, the C k distances do depend on the patch. Even if it is easy to show that the distances obtained using two patches are equivalent, they do di er and some statements such as that certain operators acting on C k functions are contractions require speci c choices of coordinate patches. For our purposes, this is not a shortcoming since many analytical operations that we will need to perform require taking coordinates anyway. We should however point out that, when k is an integer, there is a geometrically natural notion of C k distance in Riemannian geometry based on the notion of jets. (See e.g. [4] Ch. 2) For non-integer k, since the de nition of Holder distance 1 requires comparing the values at di erent points, there are no geometrically natural notions of C k distance. Nevertheless, if we select coordinate charts covering the manifold, we can de ne a distance. In a compact manifold, even if the choice of di erent charts leads to di erent metrics, the resulting metrics are equivalent. We will denote by Di k (M; N ), the set of di eomorphisms of M into N and, when M = N we will simply write Di k (N ). When k is nite k 1, these sets can be given the structure of Banach manifolds as follows. By using charts, we can de ne a norm on the space of C k vector elds on N that makes it a Banach space. Given a di eomorphism f 2 Di k (M; N ) and a C k vector eld v, the map Exp(f; v) : M ! N de ned by Exp(f; v)(x) = expf (x) v(f (x)) { where exp denotes the Riemannian geometry exponential associated to a C 1 metric { is a C k mapping. The implicit function theorem shows that for suciently small v, this mapping is also a di eomorphism and, 4
moreover, any di eomorphism in a C k neighborhood of f can be written in this way. If k < 1, we cannot apply the implicit function theorem to conclude that Exp(f; v) is a di eomorphism, nevertheless, it is possible to show that Exp covers a neighborhood of f in the space of C k mappings. (Similar argument show that Di 1(M ) is a Frechet manifold). In Di k (M ) it is natural to de ne the group operation of composition. Unfortunately, when k is not an integer, this operation will be discontinuous and when k is an integer, even when composition is continuous, it will not be di erentiable. We refer to appendix A for a sharp results about the di erentiability of composition in Holder spaces. We will denote by Di k0 (Sn?1 ) the connected component of the identity of Di k (Sn?1 ) and the subset of Di k (Sn?1) commuting with x. Symkx;0 will be the connected component containing the identity. We also have to consider families of di eomorphisms depending of a parameter. We will denote the parameters by subindices. We will say that f is a C r family of di eomorphisms when for every xed 2 , where IRm is an open set or a compact manifold, the map x ! f(x) is a C r di eomorphism and, moreover (; x) ! f(x) is also C r . The composition of di eomorphisms can be extended in a natural way to families. We denote by g f the family (; x) ! g(; f(x)). Remark.It is also possible to think of families as mappings from M to N such that the rst variable remains unaltered. This point of view makes it clear that many results about composition proved for C r mappings are true for C r families. Notice also that the notation f is somewhat ambiguous since it denotes at the same time the family and the mapping for a concrete value of . The notation is, however, standard use and it does not lead to confusion. Remark. Notice that the map ! C k (M; N ) de ned by ! f is only 0 C . There are many C 0 mappings ! C k (M; N ) which are not C k families. The main theorem in the paper will be Theorem 3.1 For any n 2 IN and any k 7, k 2 IN, we can nd points x1 ; ; x2n in such a way that any f 2 Diff0k (Sn?1) can be written f = f1 f2 fN with each fi 2 Symkx?j 2 for some j . For the cases of n = 1, k = 1 this theorem was proved in [6]. Also for n = 1 and nite di erentiability a di erent proof was given in [7]. The latter proof also provided with an analogue for Tn. 5
We remark that as a corollary of the proof we will obtain that the di eomorphisms appearing as factors can be identi ed with families of di eomorphisms of the interval.
3.2 Proof of the theorem
We start by observing that it suces to prove the result in an arbitrary small neighborhood of the identity. In e ect, given any open neighborhood U of the identity, it is possible to write any di eomorphism in the connected component of the identity as the composition of a nite number of di eomorphisms in U . We will establish the local theorem by induction on the dimension. The result for n = 1 is stated explicitly in [7] Our rst goal will be to decompose di eomorphisms of Sn into simpler di eomorphisms that preserve lower dimensional spheres, so that we can apply induction. It will turn out that the induction will proceed in such a way that we prove the theorem for each parallel and we consider the results as a di eomorphism parametrized by the height. We will start by proving a parametrized version of the well known fragmentation lemma. Lemma 3.1 Let M be a compact manifold, n 2 IN and fKigni=1 be a collection of compact subsets of M such that n
\
i=1
Ki = ; :
Then, we can nd a neighborhood U of the identity in the space of C l and a map F0 that, to each family f in U , associates C l families f1 ; ; fn, such that i) f = f1 fn . ii) fi jKi = Id . iii) If for some x, f (x) = x for all , then fi (x) = x for i = 1; : : : ; n and all values of .
6
The mapping F0 is continuous when we give its domain and its range the topology of C l families. Remark. Properties i)-iii) do not determine F0 uniquely. So that the claim of continuity refers only to the one we construct. Proof. We will assume that U is small enough so that f(x) is whitin the injectivity radius of the exponential. We will also need that some intermediate steps in our construction lie in this domain. Since those are nite number of conditions, it will be easy to check that they are satis ed for a non-trivial neighborhood. We will discuss rst the case n = 2. We can nd a C 1-function ' such that 'jK1 1 ; 'jK2 0 :
If f can be represented as the exponential of the vector eld F = E ?1(f ), we see that f2 = E (F ') will be the identity restricted to K2 . Moreover, since f (x) = x () E ?1 (f)(x) = 0, the set of xed points of f2 is bigger than those of f. Provided that f is suciently close to the identity { depending only on the K 's and ' { then f2 jK1 = f and f2jK2 = Id . Hence, setting f1 = f (f2)?1, the claims of the lemma are satis ed. Notice that since the choice of ' depends only on the compacts K1 and K2 the continuity follows. Remark. Notice that the previous argument works even in the case that one of the compacts is the empty set! To prove the general case, we will proceed by induction in the number of compacts set n. We will assume that it is true when n n0 , n0 2, and will prove it for n = n0 + 1 . If K1 ; : : : ; Kn0 +1 are compact sets with empty intersection, we can nd 0 V = ;. open sets V1 ; : : : ; Vn0+1 such that Ki Vi, and \ni=1 i n 0 Set Ln0 +1 = Kn0 +1, L0 = \i=1 V i : Since Ln0 +1 \ L0 = ; we apply the previous result with n0 = 2 to obtain families f0 , fn0+1 which are the identity on L0 , Ln0 +1 respectively and with not less xed points than f such that f = f0 fn0+1. 0 V . They have We can now consider the compact sets Li = Ki ? \ni=1 i n n 0 0 empty intersection because \i=1Ki \i=1 Vi , so we can apply the induction 7
hypothesis to f0 to write f0 = f1 fn0 . Each one of the fi will be the identity on a neighborhood of Kn0 +1. This gives us the representation f = f1 fn0 that verify all the conditions of the statement. The continuity of F0 is also a consequence of the induction process. Lemma 3.2 Let x be a point in Sn, ?x the antipodal point and U , U 0 two open neighborhoods around x, ?x respectively. Then, for every r 2 IN [f1g, we can nd a neighborhood U of the identity in the space of C r families of di eomorphisms in such a way that for any family f in U , such that fjU = Id, fjU 0 = Id we can nd C r families fh, fv verifying: i) f = fh fv . ii) fv Mx; = Mx; . iii) fhPx; = Px; . iv) If x 2 Sn and f(x) = x then fh (x) = fv (x) = x. In particular fhjU [U 0 = fv jU [U 0 = Id. Moreover, if we restrict fh, fv to be in a neighborhood of the identity, they are unique, and the mapping F1 that assigns to each f the fh, fv is continuous. We will henceforth refer to di eomorphisms and families that preserve the parallels as horizontal and those that preserve the meridians as vertical. Proof. This lemma says that any family can be factored into a family that preserves the parallels and another one that preserves the meridians. Obviously, we should set fv (p) to be the point in the meridian of p and on the parallel of f (p). Then, fh would move along the parallels so as to get to f (p). More formally, using the notation for parallels and meridians introduced in 1, we have: fv (p) = Mx;p \ Px;f(p) fh(p) = Mx;f((fv )?1 (p)) \ Px;p Using the uniform transversality of the meridian and parallel foliations outside a neighborhood of the poles, the usual implicit function theorem, establishes that fv (p) fh(p) depend jointly C r on p and on the parameters . 8
Remark. If f ('; ) = (' + v('; ); ) denotes a vertical di eomorphism in a small neighborhood of the identity with v 2 C r (Tm ), r 1, then f?1 is also a vertical di eomorphism f?1('; ) = (' + v('; ); ) and v 2 C r (Tm )
The induction in the dimension to prove the main theorem 3.1 will proceed as follows. For " suciently small, we will consider the sets Ki" = fx 2 Sn = jxij "g. Using Lemma 3.1 we will be able to write any di eomorphism f as composition of di eomorphisms fi, each of which is the identity restricted to Ki". Furthermore, for fi we will pick a polar axis going through the points xi = 1, xj = 0, j 6= i and using Lemma 3.2 will write it fi = fih fiv . If fi was suciently close to the identity then fih, fiv would be the identity in Ki". We will show how to factor fiv and will apply the induction hypothesis to fih. To apply the induction hypothesis, we will think of fih as a family of maps sending Sn?1 into Sn?1 , the parameter being the height of the parallel. Notice how we are lead to consider families of di eomorphisms even if we start with a single one. With hindsight, this is the reason why we formulated Lemma 3.1 and Lemma 3.2 in terms of families. Clearly, if the induction result obtains in its conclusions that the factors are a smooth family, this will imply that the factors are really di eomorphisms in Sn. We now proceed to establish the factorization of fiv into symmetric diffeomorphisms. To simplify the notation and without any loss of generality, we will assume that i = 1 and that the axis is going through the points x = (0; : : : ; 1). We also can think about the height as a periodic variable and consider the parameters in a set =Tm?2 .
Lemma 3.3 We can nd two points y1, y2 in the sphere such that there is a neighborhood U of the identity in the space of C r r 7 families of di eomorphisms such that for every family fv 2 U and satisfying 9
i) fv Mx+; = Mx+;
8.
ii) fv jK1" = Id,
then, we can nd two C r?2 families fv;1 , fv;2 satisfying a) fv;1 Mx+; = fv;2 Mx+; = Mx+;
8.
b) fv;1 y1 = y1 fv;1 .
fv;2 y2 = y2 fv;2 . c) fv = fv;1 fv;2 .
Proof. We will take polar coordinates in Sn ? K1" de ned by: xn+1 xn xn?1 ... x3 x2 x1
= cos 2' = sin 2' cos 2w1 = sin 2' sin 2w1 cos 2w2 ... = sin 2' sin 2w1 sin 2wn?2 = sin 2' sin 2w1 sin 2wn?2 cos 2 = sin 2' sin 2w1 sin 2wn?2 sin 2
for ', w1 ; : : : ; wn?2 2 (0; 1=2), 2 (0; 1). Notice that if jx1 j ", then jxn+1j 1 ? "0 so that ' is de ned univocally, and, since jxnj2 + jxn+1j2 1 ? "2, w1 is likewise outside of the critical values of sin and the same argument is applicable to w2 ; : : : ; wn?2. In others words, this polar coordinate system is a good coordinate system admitting a di erentiable inverse de ned in a open set V Tn+1. We observe that when we express a vertical di eomorphism in these coordinates, the value of ' changes but the others coordinates remain unaltered. Similarly, a re ection across the plane y1 with y1 = (1; 0; : : : ; 0) is expressed as a change in sign for the coordinate while all the others remain xed. More generally, a re ection across y2 with y2 = (cos ; sin ; 0; : : : ; 0) can be expressed by sending into 2 ? and leaving all the other variables unchanged. 10
If we take coordinates ('; ; w) the family fv will be written as ('; ; w) ! (' + v ('; ; w); ; w) and we will look for fv;1 fv;2 that can be written as ('; ; w) ! (' + vi ('; ; w); ; w) with
v1 ('; ; w) = v1 ('; ?; w) (2) v2 ('; ; w) = v2 ('; ? + 2; w) The equation that v1 v2 have to solve is, as an straightforward calculation shows: v('; ; w) = v1 (' + v2 ('; ; w); ; w) + v2 ('; ; w) (3) Notice that there is an open set of values of w for which the point correspond ('; ; w) is in K1" hence v ('; ; w) = 0. For these values of w, (3) admits the trivial solution v1 ('; ; w) = v2 ('; ; w) = 0. This implies that fv;1 = Id, fv;2 = Id for the points in the sphere satisfying this condition. Clearly, its solutions can be extended by the identity to all points of the sphere in K1" even for those for which our coordinate system does not work. Under conditions of proximity to the identity of fv , fv;1 , fv;2 these will be solutions of our original problem. Since, except for these considerations w does not enter either (3) or in (2) we can consider these variables as parameters in the problem in the same way as . We will, hence assume that refers both to the parameters in the family and to w and omit w from the notation. In the enunciate of the theorem y1, y2 were left at our choice. The choices of { the angle between the planes passing through the origin and y1, y2 respectively{ that are acceptable for the theorem are those satisfying the so-called Diophantine conditions. Those are conditions of the form ?1 p ? q Cq2+ (4)
11
It is well known that for > 0, the numbers satisfying inequalities of this form have full measure. For = 0, the set of numbers satisfying these inequalities is called the of constant type numbers. (4) is equivalent to having a bounded continued fraction expansion. All irrational numbers that satisfy a quadratic equation with integer coecients are constant type. In particular constant type numbers are dense. Even if all Diophantine 's would lead to a theorem of the type we want and for small with the conditions of di erentiability required in the statement, constant type numbers will lead to the sharpest di erentiability conclusions and we will, from now on assume we have chosen a constant type number. The actual choice will a ect the sizes of the neighborhoods but not the di erentiability properties. Proceeding heuristically for the moment, the importance of the Diophantine properties becomes apparent if we consider the linearized equations obtained by expanding formally (3) in the unknowns and keeping only the linear terms. We obtain:
v('; ) = v1 ('; ) + v2 ('; ) (5) We expand in Fourier coecients in v ('; ) = k v^;k (')e2ik and analogously for v1 , v2 and obtain that (5) is equivalent to: P
1 (') + v^2 (') v^;k (') = v^;k ;k 1 Similarly, the symmetry conditions for v , v2 can be expressed as:
(6)
1 (') v^;1 ?k (') = v^;k (7) v^;2 ?k (')e4i = v^;2 ?k (') When k 6= 0, then we can group the equations for k, ?k in (6) and, using (7) we obtain: 1 (') + v^2 (') v^;k (') = v^;k ;k 1 2 v^;?k (') = v^;?k (') + v^;k (')e?4i When k 6= 0, these equations admit the solutions: 1 (') = (^ v^;k v;k (')e?4ik ? v^;?k ('))=(e?4ik ? 1) (8) 2 (') = (^ v^;k v;k (') ? v^;?k ('))=(e?4ik ? 1)
12
When k = 0, there are many solutions. We just choose: v^;1 0(') = v^;2 0(') = 21 v^;0(') (9) Notice that, if were rational, the denominators would become zero, that is, there would be no solution for the equation. Even if is irrational the denominators would become arbitrarily close to zero. Nevertheless, if is a number of constant type, we have the bounds
j(e?2ik ? 1)?1j C 0jkj
(10)
that allow us to control the new Fourier expansions. If we think of (3) as an equation in spaces of functions, (5) would be the derivative { in some sense that will have to be speci ed { at zero. The calculation shows that the derivative, even if invertible is not bounded, hence the usual implicit function theorem in Banach spaces does not apply. We also emphasize that the calculation above shows only existence of the inverse of the derivative zero. If we had started making perturbations around a non-zero, v1 , v2 we would have been lead to an equation that cannot be readily analyzed in terms of Fourier components. The above heuristic discussion suggests that, among the several hard implicit function theorems can cope with unbounded inverses, the one that will be useful for this problem is that of Zehnder [13] which requires existence of inverse only at one point, but which also requires a group structure for the equation. We continue the proof by recalling Zehnder's theorem, which we have taken from [13] in the version we are going to apply.
De nition 3.1 We say (X ) 0 is a family of regular spaces when X are real Banach spaces with norm k k and, whenever for 0 we have: X0 X X 0 X1 = X 0 kuk kuk 0 8u 2 X 0 \
Furthermore, it is possible to de ne on them a C 1-regularization, that is, a family (St )t>0 of linear mappings St : X0 ! X1 and constants C ( ; 0)
13
satisfying:
i) tlim !1kSt u ? uk0 = 0 8u 2 X0 ii) kStuk t ? 0 C ( ; 0)kuk 0 8u 2 X 0 0 0 iii) kStu ? uk 0 t 0 ? C ( ; 0)kuk 8u 2 X 0 0 We now consider several families of regular spaces and denote by the same symbol their norms which are probably di erent. However the confusion is not possible because the kind of vector that we are evaluating indicate the norm we are using.
Theorem 3.2 Let X , Y , Z be families of regular spaces. Let F : X0 Y0 ! Z0 be such that F (x0 ; y0) = 0 and continuous on a set B0 . (For any , we will write B = fx; y 2 X0 Y0 = kx ? x0 k 1 ; ky ? y0k 1g.) Assume (H.1.) F (x; ) : Y0 ! Z0 is twice di erentiable and
kd2F (x; y)k M0 ; kd22F (x; y)k M0 8x; y 2 B0 : (H.2.) F is uniformly Lipschitz in X0. That is, for all (x; y) 2 B0, (x0 ; y) 2 B0 kF (x; y) ? F (x0; y)k0 M0 kx ? x0 k0 : (H.3.) The triple (F; x0; y0) is of order 1. That is (H.3.1.) (x0 ; y0) 2 X1 Y1. (H.3.2.) F (B0 \ X Y ) Z 0 . (H.3.3.) There exist constants M , 1 such that if (x; y) 2 X Y \ B1 satis es kx ? x0 k K; ky ? y0 k K , then kF (x; y)k K M . (H.4.) There exist an approximate right-inverse of loss , 1 . That is, for every (x; y) 2 B there exist a linear mapping (x; y) 2 L(Z ; Y ? ) such that: (H.4.1.) k(x; y)zk0 M0 kzk . 14
(H.4.2.) kd2F (x; y) (x; y) ? 1)zk0 M0 kF (x; y)k kzk . Then if " is a small positive number and = 2 + ", we can nd an open neighborhood D = fx 2 X = kx ? x0 k < Cg and a mapping : D ! Y such that a) F (x; (x)) = 0. b) k (x) ? y0 k C?1 kx ? x0 k . In particular (D \ X1) Y1 . Moreover, if (x; y) depends continuously on (x; y), then the mapping : D ! Y is continuous. To apply the theorem to solve (3) we have to choose an appropriate functional and appropriate spaces of families functions. We will consider Banach spaces of C r -families of mappings. Let 2 IR, we stand by Cr (Tm) the space of functions v 2 C r (Tm) which verify:
v ('; ) = v ('; ? + 2); (; '; ) 2 Tm : Let us x s 0 and a small parameter " > 0. We take X = Z = C s+ (Tm ), Y = C0s+ +2+"(Tm) Cs+ +2+"(Tm ) with the corresponding Holder norms and de ne the operators F : X Y ?! Z (v; v1 ; v2 ) ?! v1 (' + v2 ('; ); ) + v2 ('; ) ? v ('; ) We will show that F verify all the conditions of the previous statement around (0; 0; 0). In our case Zehnder's theorem is not anymore than a hidden inverse function theorem. Most of its conditions are deduced in the Appendix A. The hypothesis (H.1.) (H.2.) (H.3.) are immediate consequences of XXXXXX . The most crucial hypothesis is that about existence of an approximate right-inverse of the linearized operator, which we now discuss. We will start the construction of such approximate inverse (0; 0; 0). To study its di erentiability properties, we will use the following result proved in Russmann [10]. 15
Lemma 3.4 Let be a constant type number and h(z) =
1
X
k=?1
h^ k eikz
an analytic complex function in Br = fz 2 C = jImzj < rg with mean value zero and jh(z)j M 8z 2 Br . If we take
u(z) =
1
X
k=?1
h^ k eikz e2ik?1
there is a constant K such that if r0 < r then
ju(z)j rK?Mr0
for jImzj < r0:
(11)
Notice that F : X Y ! Z is di erentiable with respect the Y 0s components and d2F (0; 0; 0)(v1 ; v2 ) = v1 + v2 in all the cases. If we take a periodic analytic function v('; ) de ned in the strip Urm and consider the pair v1 ; v2 de ned by the equations (8), (9) that verify v = v1 + v2 then both functions are analytic in Urm0 for r0 < r and satis es a inequality similar to (11) with may be some other constants. (Notice that if ?4ik ? 1) = 1 + 1=(e?4ik ? is a constant type number, so is ?2, e?4ik =(eP P 2 ikx ^ 1) and that sup=xr j k hk e j = sup=xr j k h^ ?k e2ikxj ) From lemma 3.4, it is possible to deduce regularity properties of the inverse di erence operator in the classes of nitely di erentiable functions. The key is that the di erentiability properties of functions can be read of from the size of analytic approximations on thin strips { see [8] { The natural spaces on which to consider such characterizations are the r spaces ( a modern survey of the properties of these spaces, including the characterizations by approximation properties is [12] We just remark that, when r 2= IN, r = C r . When r 2 IN, C r r and the inclusion is strict. Notice, however that the de nition of C r functions makes sense on Banach spaces ( for 0 r 1? it makes sense in metric spaces ) whereas the characterization by analytic approximations is useful only for functions on IRn. A detailed argument along these lines can be found in [8] p. 528, [13] We have: 16
Lemma 3.5 Let be a constant type number and h(z) =
1
X
k=?1
h^ k eikz
a C r function, r 2= IN r > 1, with mean value zero. The function: 1 X h^ k eikz u(z) = 2ik?1 k=?1 e
satis es:
jjujjC r?1 K jjhjjC r
where K is a constant depending only on . Using 3.5 to estimate the solution of the approximate inverse that we are considering, we obtain: Lemma 3.6 Let > 3 and . For every v 2 Z there exist functions (v1; v2) 2 Y ? such that v = v1 + v2. Moreover the map
: Z ?! Y ? v ?! (v1; v2) is continuous.
Now a approximate right-inverse in a complete neighborhood of (0; 0; 0) is required. An important observation is that one exact right-inverse can be obtained thanks to the group action. See Hamilton [3], pg 198], Zehnder [13], pg. 133]. We proceed heuristically in the following. Consider for instance a smooth manifold M, which locally can be identi ed with a Banach space, and a di erentiable action : G M ! M where G is an in nite dimensional group. Assume that i) F : G ! M is de ned as F (g) = (g; m0) for m0 2 M. ii) dF (id) : G = TidG ! Tm0 M is surjective with right inverse . If Lg : G ! G denotes the left multiplication by g on G and g : M ! M denotes the action of g on M then g F (h) = F Lg (h) 17
(12)
for every g; h 2 G , which yields
dg (m0) dF (id) = dF (g) dLg (id) (13) Hence the map (g) = dLg (id) dg?1 (F (g)) de nes a right-inverse of dF (g). We now make precise the spaces and the domains in which the above calculations are valid.
Lemma 3.7 Let and take (v0 ; v1 ; v2 ) 2 B . There exists (v1 ; v2 ) 2 L(Z ; Y ? ) such that d2F (v0 ; v1 ; v2 ) (v1 ; v2 )(v) = v for every v 2 Z . Proof. Let us consider F2 : Y ! Z , (v1 ; v2 ) ! F (0; v1 ; v2 ). It is obvious
that d2F (v0 ; v1 ; v2 ) = dF2(v1 ; v2 ). We can realize the sets X ; Y ; Z as the corresponding tangent spaces at the identity of groups of di eomorphisms in Tm , where the functional equation de ned in lemma (3.3,c) is the composition. Let (v1 ; v2 ) 2 Y and v = F2 (v1 ; v2 ). In our problem the group operation and di erentiable action which appear in (13) are de ned by the relations : Z ?! Z 1 ' + v2 ('; ); ) + v2 ('; ); )+ v ?! vv(('' ++ vv2 (('; ); ) + v2 ('; ):
and
L : Y ? ?! Y ? (v1 (' + w1 ('; ); ) + w1 ('; ); (w1 ; w2 ) ?! w 2 2 2 (' + v ('; ); ) + v ('; )): It follows from proposition (A.8) that is continuously di erentiable and d(v) 2 L(Z ; Z ) and d(v)(w) = w(' + v2 ('; ); )+ @ v1 (' + v (' + v2 ('; ); )w (' + v2 ('; ); ): @' 18
Similarly, L is continuously di erentiable, dL(0) 2 L(Y ? ; Y ? ) and 1 ('; )w1 ('; ) + w1 ('; ); w2 (' + v 2 ('; ); )) dL(0)(w1 ; w2 ) = ( @v @' The map (v1 ; v2 ) = dL(0) d(v) 2 L(Z ; Y ? ) and de nes a right-inverse of dF2. We have veri ed all the hypotheses of theorem (3.2). Therefore, given v 2 C s(Tm),s > 6 close to 0 there are (v1 ; v2 ) 2 C0s?1?"(Tm ) Cs?1?"(Tm ) such that F (v; v1 ; v2 ) = 0, and hence the conclusions of the lemma (3.3) hold. Therefore, we have shown that in a C r neighborhood of the identity r > 6 all di eomorphisms can be factored. Unfortunately, the set of C r di eomorphisms is not a group unless r 2 IN. This is the reason why the statement of 3.1 is made using only C 7 regularity.
A Di erentiability of composition in Holder spaces A.1 Introduction
In this appendix we collect some theorems and counterexamples that clarify the di erentiability properties of composition in the classical spaces of Holder functions. We have emphasized the methods based on elementary estimates, triangle inequality , mean value theorem and the like. Therefore, the results presented here will be valid in a great generality - e.g., open sets of Banach spaces. In nite dimensions, it is sometimes possible to obtain similar results using characterizations of Holder spaces by approximation properties by analytic functions [11], [12], [9]. As intermediate steps we will also prove elementary versions of interpolation inequalities that work in in nite dimensional domains with good geometric properties. Since the composition operator is the basic operator of dynamical systems, it is clear that the study of this operator acting on di erent functional spaces should be very useful. The systematic study of di erentiability properties of the composition operator in the spaces of continuously di erentiable 19
functions can be found in [5]. In order to deduce good properties from this operator uniform continuity of the factors is required. Irwin formulates his version on compact sets where continuity automatically implies uniform continuity. In this appendix, we have studied only classical Holder spaces. Since Holder character assures uniform continuity we consider maps de ned on open subsets. Notice that some of the results for these spaces, even on compact sets, are di erent from the straightforward extensions of those for continuously di erentiable functions. Unfortunately, the literature abounds with confusions caused by this fact. Notice also that, as stated in 1, many of the results discussed here carry over to manifolds.
A.2 Basic de nitions and notation
Let E; F be Banach spaces. We will denote the space of bounded linear functions from E to F by L(E; F ) and in general Ln(E; F ) = L(E; Ln?1(E; F )). Let U E be an open set. A function f : U ! F is Frechet di erentiable at x0 if we can nd a bounded linear function df (x0) such that k f (x0 + h) ? f (x0) ? df (x0)h k = 0 lim h!0 khk We say that f is di erentiable in U if f is di erentiable at every point x0 2 U . We say that f is of class C 1 if it is di erentiable and the mapping df : U ! L(E; F ) that to each x0 2 U associates df (x0) is continuous. Proceeding by induction, we de ne dnf to be the di erential of dn?1f 1 (d f = df ) and we say that a function f is n times continuously di erentiable, df : U ! L(E; F ) is (n ? 1) times continuously di erentiable. We will denote by Ln(E1; : : : ; En; F ) the space of n-multilinear functions having arguments in E1; : : : ; En and values in F . It is standard that we can identify Ln(E; F ) with the space of bounded multilinear functions with n arguments in E taking values in F . We will think of derivatives as multilinear functions. We will say that f 2 C n(U; F ) if f is n times continuously di erentiable and k dnf (x) k is uniformly bounded by x 2 U . In that case, we denote k f kC n = sup sup k dk f (x) k : x
0kn
20
Notice that, according to our de nition there could be functions which are n times continuously di erentiable but not in C n(U; F ): If f 2 C n(U; F ), V is an open subset of F with f (U ) V and g 2 n C (V; G) then g f 2 C n(U; G). Moreover by a repeated application of the chain rule we can nd constants cn;j1;:::;jn such that
dn(g f )(x) =
n
X
k=1
ck;j1;:::;jn dk g(f (x))fdj1 f (x); : : : ; djk f (x)g
Remark. Weaker notions of di erentiability spaces have been proposed and the one we consider here is sometimes called "strongly di erentiable functions." Moreover, if the manifold under consideration is not compact, e.g., IRd or a bounded subset of an in nite dimensional Banach space - there may be functions which are continuously di erentiable but whose derivative is unbounded. In those cases, besides the topology induced by the supremum norm we consider here, there are other natural topologies, e.g., the Whitney topology associated to uniform convergence on compact sets. This topology is an inductive limit of the ones we consider here and, hence, the results of continuity can be immediately lifted. Since this topology does not come from a norm only weak concepts of di erentiability can be introduced for composition and similar operators, when we consider such topologies. De nition A.1 If r = n + with n 2 IN and 0 < 1? we say that f 2 C r (U; F ) if f 2 C n(U; F ) and moreover we have n n H (dnf ) = sup k d fk(xx)??ydkf (y) k < 1 x6=y With the introduction of k f kC r = sup(k f kC n ; H (dnf )) C r (U; F ) is a Banach space. Notice that with this de nition, we have k f kC r sup(k f kC 0 ; k df kC r?1 ) this makes it easy to carry out several inductions and hence, this de nition is preferable to other equivalent norms. 21
Remark. If U = U1 [ U2 , U1 \ U2 6= ;, we can write C r (U ) = C r (U1 )
C r (U2) and it is possible to study C r (Ui) separately. Similarly, the study of composition operators can be broken up into the study of the components. So, we will henceforth assume that the domains in which we study the functions are connected by arcs.
A.3 Geometry of domains and Interpolation properties The following are immediate. Proposition A.1 Let 0 < 1?; and n 2 IN.Then i) C n+ (U ) C n+ (U ) C n (U ). ii) C n+ (U )is a closed subset of C n+ (U ). Remark. For most spaces useful in analysis the inclusion in Proposition A.1 is strict. It suces to consider jxj +n on [0,1] which is in C n+ but not on C n+ . In the following lemmas we develop some interpolation properties of the norms that we will use later in the discussions of the composition operator.
Lemma A.1 Let r = n + ; s = n + ; t = n + with 0 < < 1 and = (t ? s)=(t ? r) . There is a constant Mr;t such that if f 2 C t (U; F );
then
(14) k f kC s Mr;t k f kC r k f k1C?t Proof: Notice that 1 ? = (s ? r)=(t ? r) and s = r + (1 ? )t. We
will establish the claim using induction on n. We will rst describe the case n=0 If we de ne for 2 (0; 1?]: H (f ) = sup k fk(xx)??yf k(y ) k we have 1? H = sup k fk(xx)??yf k(y ) k sup k fk(xx)??yfk( y(1)?k) = Hr (f )Ht (f )1?: 22
From here it is obvious that k f kC s Mr;t k f kC r k f k1C?t where ( < 1? Mr;t = 12 ifif 00 < = < 1?
For n 1 using the inductive hypothesis we nd k df kC s?1 Mr;t k df kC r?1 k df k1C?t?1 Mr;t k f kC r k f k1C?t and thus k f kC s Mr;t k f kC r k f k1C?t Notice that ln k f kC r is a convex , therefore continuous, function in each open interval (n; n + 1) with n 2 IN.
Proposition A.2 If 0 < 1? and U is separable then bounded sets
of C n+ (U ) are compact in C n+ (U ). Proof. For = 0 this is a consequence of Ascoli-Arzela theorem. For > 0 we use the result for = 0 and the interpolation formula (14). Notice that in Proposition (A.1) we have not claimed that if r = n + < s = m + then C s C r . The fact that this is not unfortunately true in general can be seen in the following. Example A.1 Let D = fz 2 C=jzj < 1g, S = fz 2 C==z = 0; 0 0; > 0 such that jf (x; y) ? f (x0 ; y0)j kj(x; y) ? (x0 ; y0)j Hence, in the space U C 1(U ) 6 C (U ) for any 0 < 1?.
23
In many respects this example is a "pathology." Its origins are clear if one tries to use the mean value theorem.
f (x) ? f (y) =
Z
1
0
df ( (s)) _ (s)ds
(15)
where is a path joining x; y. This immediately gives jf (x) ? f (y)j k f kC 1 distU (x; y) where the symbol distU (x; y) denotes the in mum of the lengths of paths joining x to y. In the space U of the example this could be much larger than jx ? yj. Clearly, if we want to use arguments based on the mean value formula (15) we have to restrict ourselves to spaces in which distU (x; y) is commensurate to j(x; y)j. When we consider problems involving composition, we have to ensure that the intermediate ranges also have this property.
De nition A.2 balanced We say that an open set U is balanced if there exists a constant kU such that dU (x; y) kU k x ? y k 8x; y 2 U . This property depends on the shape of U . For example, a convex set of a Banach space is balanced or more generally, a connected set which is a nite union of convex sets, is balanced. Notice that, as stated, the de nition only makes sense for connected sets. This is not a shortcoming in view of the fact that , as remarked in the introduction, it suces to consider spaces of function on connected sets. Notice also that a subset of a balanced subset needs not be balanced. Unfortunately, C 1 functions could transform balanced domains into unbalanced ones.
Example A.2 Let D+ = fz 2 C=jzj < 1; =z > 0g the map z ! z2 transforms the balanced open D+ on the unbalanced U de ned in the example (A.1) The following results are useful in the representation of the sucesive derivatives of the composition. We omit their proofs which are routine
Proposition A.3 Let l 2 L(F; G). Then l? : C r (U; F ) ! C r (U; G) de ned by l? f = l f is linear and bounded. Moreover, k l? k=k l k. 24
Proposition A.4 Let r; s; t 2 IR, r t; s t. Let b 2 L2(F; G; H ) . Then the map b? : C r (U; F ) C s (U; G) ! C t (U; G) de ned by b? (f; g)(x) = b(f (x); g(x)) is continuous and bilinear. Moreover k b? k 2 k b k. We specially employ these relations when b : L(F; G) L(G; H ) ?! L(F; H ) (l; m) ?! l m represents the composition of linear operators. If F = K then L(F; G) = G, L(F; H ) = H . We will stand by g f the map b? (f; g) de ned as b : C r (U; G) C s(U; L(G; H )) ?! C t(U; H ) (f; g) ?! [x ! g(x)(f (x))] Lemma A.2 Let F; G be Banach spaces U F; V G be balanced subsets. Let r = n + ; s = m + with m; n 2 IN; ; 2 [0; 1? ]: If f 2 C r (U; F ), f (U ) V; g 2 C s(V; G) then g f belongs to C t(U; G) where we can take t as follows: i) If n = m = 0; t = rs Moreover i.1) k g f kC rs k g kC s k f ksC r + k g kC 0 ii) If max(r; s) 1; t = min(r; s) Moreover ii.1) if r 1; 0 < s 1?
k g f kC s k g kC s k f ksC 1 kUs + k g kC 0 ii.2) If 0 < r 1?; s 1
k g f kC r k g kC 1 k f kC_ r kV + k g kC 0 ii.3) If r 1; s 1; 9Mt such that
k g f kC t Mt k g kt (1+ k f ktt ) 25
Proof. If n = m = 0; we have k g f (x) ? g f (y) kk g kC k f (x) ? f (y) k k g kC k f kC k x ? y k This establishes i), i.1). The cases ii.1) and ii.2) can be obtained in a similar way. To verify the last case, we assume that s = r = t = k + with k; 2 IN k 1 and 2 [0; 1?]. We proceed by induction in k. For k = 1, d(g f ) = dg f df . Applying Proposition [A.4] and the i) of Proposition [A.2] we obtain
k dg f kC kU k dg kC k f k C 1 + k dg kC 0
M k g kC 1+ (1+ k f k1+ C 1+ )
Let us assume that the above inequality holds until the integer k ? 1 and consider t = k + . Then by induction we have that d(g f ) = dg f df 2 C t?1 (U ) and
k d(g f ) kC t?1 2 k dg f kC t?1 k df kC t?1 2M k dg kC t?1 (1+ k df ktC?t?1 1 ) k df kC t?1 since Mt (1 + jxjt) M (1 + jxjt?1)jxj for some Mt > 1, we obtain
k g f kC t sup(k g f kC 0 ; k d(g f ) kC t?1 ) Mt k g kC t (1+ k f ktC t ) This completes the proof of Lemma A.2 The following examples show that the hypotheses in proposition (A.2) are sharp.
Example A.3 Let > 0 and f : [?1; 1] ! IR; x ! jxj i) If 0 < ; 1?, then f f 2 C ([?1; 1]; IR) if and only if . ii) If =2 62 IN then f Id 2 C ([?1; 1]; IR) if and only if ( ? 1) + 1?.
26
As a corollary of proposition (A.2) we obtain that if U is a balanced open set then the formula (14) also holds for t = n + 1 taking (
kU if 0 < < = 1 Mr;t = max(2 ; kU ) if 0 = < = 1 However, unless we impose some condition on the domain, these inequalities are not possible for exponents with di erent number of derivatives as shown in the following example.
Example A.4 Let U = f(x; y) 2 IR2=x > 0; y > 1; xy < 1g. We take ' as a C 1 cut-o function with k '0 kC 0 4 which takes the value 1 on [? 31 ; 13 ] and zero outside of [?1; 1]. We de ne on U the following functions fn(x; y) = log(1 + x)'(y ? n) For n 2 IN big enough we nd @Fn (x; y ) = 1 '(y ? n) @x 1+x @Fn (x; y ) = log(1 + x)'0 (y ? n) @y @ 2 Fn = ? 1 '(y ? n) @x2 (1+x)2 @ 2 Fn = log(1 + x)'00 (y ? n) @y2 @ 2 Fn = 1 '0 (y ? n) @x@y 1+x Observe that since the support of '(y ? n) is centereed around y = n, which in the domain we are considering implies x 1=n we have: j @F@xn j = O(1); j @F@yn j = O(1=n); @ 2 F j = O(1) j @@x2 F2n j = O(1); j @@y2F2 j = O(1=n); j @x@y jjF jjC 0 = O(1=n) This implies that k F kC 2 remains bounded and also jjF jjC 0 converges to
zero. Since @F@xn evaluated at x = zn1 ; y = n does not converge to zero, we conclude that there cannot be an inequality of the form
jjFnjjC 1 M jjFn jj1C=02jjFnjj1C=22 27
More generally, we observe that for any s 2 [1?; 1 + 1?) jjFnjjC s does not tend to zero. For any r 2 [0; 1?) jjFnjjC r tends to zero and for any t 2; jjFnjjC t remains bounded. Nevertheless, as it is well known, interpolation inequalities for spaces of functions de ned on IRn can be proved those ranges of regularities (see e.g. [11]). This shows that it should be neccessary to include some conditions in the domain U if interpolation inequalities are to hold.
Remark We observe that there is an abstract argument (see [13] ) that
shows that if there are smoothing operators, then interpolation inequalities hold. Hence, we conclude that it is impossible to de ne smooting operators for Holder spaces of functions in U . Notice that U in Example A.4 is a subset of IR2 and that there are interpolation formulas for C (IR2 ) when is not an integer (see e.g. [Kr ] ). Hence, the fact that there are interpolation formulas is for C spaces is not inherited by subsets. This is somewhat surprising since a naive argument would suggest that we can apply a Whitney extension theorem to extend our function F in U to a function in IR2 , apply interpolation to the extended functions and then restrict. The reason why this naive argument is wrong is that the extension operators used in Whitney theorem depend on the regularity considered. In [11] VI, 2 one can nd a construction of extension operators k which extend C for k < k + 1. Example A.4 shows that it is impossible to nd a unique extension operator working simultaneously in C 1 C 2 when the integer parts of 1 and 2 are not the same. This situation is in contrast with regularities measured in other scales of spaces - e.g. Sobolev spaces Lpk where it is possible to nd extension operators that work for all the scale of spaces simultaneously ( see. [11] VI 3). In the following, we will develop son conditions that guarantee that one can obtain interpolation inequalities. We do not know whether condition A.3 on U is neccessary for the validity of interpolation inequalities in U . We do not know conditions which ensure that there is a unique extension operator for all C regularities. 28
De nition A.3 Let U a balanced open set and 0 < 1?. We say that U is -regular if there is a constant M0 such that for every x0 2 U and h 2 E with k h k= 1 we can nd 2 C 1+ ([0; 1]; U ) with (0) = x0 , 0 (0) = h with = 1 or ?1 and k kC 1+ M0 . We say that U is regular if it is -regular for some .
Notice that the union of -regular subsets is also -regular. The relevance of regularity lies in the following lemma. Lemma A.3 Assume that U is regular. Let r; s; t be positive numbers with 0 r < s < t and = (t ? s)=(t ? r) . There is a constant Mr;t such that if f 2 C t (U; F ); then
k f kC s Mr;t k f kC r k f k1C?t
(16)
Proof: We suppose that U is a -regular open. First we consider r = n ? 1 + ; s = n; t = n + with . Let n = 1. Let z 2 C t(IR+; F ) and assume that there is t0 2 IR+ with 0 k z kC 0 =k z0 (t0 ) k. Substituting z by az(t) if neccessary we can assume that k z0 (t0 ) k= 0 k z kC 0 = 1, and H (z0 ) = 21 . Taking R(t) = z(t) ? z(t0 ) ? z0 (t0)(t ? t0 ) then, Z
t
k R(t) kk t (z0 (s) ? z0 (0))ds k jt ? t0 j1+ H (z0 ): 0
Now we have
k z(t0 + 1) ? z(t0 ) k=k z0 (t0 ) + R(t0 + 1) k 21 :
This shows that H (z) 21 and therefore,
k z0 kC 0 2H (z) H (z0 )1? : Let us suppose that k df kC 0 = 1. For every small there exists x0 2 U , h 2 E with k h k= 1 such that 1 ? k df (x0)h k 1. We de ne z(t) = f ( x0;h(1 ? e?t)).Then k df kC 0 ? k z0 kC 0 2H (z) H (z0 )1? M k f kC r k f k1C?t 29
for every . Thus we obtain (17) k f kC s M k f kC r k f k1C?t When n > 1 we can deduce the inequality (17) by induction. To obtain the complete result we only have to apply the previous lemma. Using the inequality (16) and arguing by induction as in the nite dimensional case, considered in [3], we obtain: Lemma A.4 Assume that U F , V G are regular subsets. Let t 1, f 2 C t(U; F ), g 2 C t(V; G) with f (U ) V . If k f kC 1 K there exist a constant Mt , depending on K such that k g f kC t Mt (k g kC t + k g kC 1 k f kC t )
A.4 Di erentiability properties of composition
We will systematically study di erentiability properties of comp : C r (U; F ) C s(V; G) ! C t(U; G) First we will study continuity and di erentiability with respect to each of the components and then, join continuity and di erentiability. In order to obtain g f we need f (U ) V . We assume this condition in proposition (A.5). To consider this composition as a function of the rst factor it is necessary that d(f (U ); V c) > 0 to make it possible in a complete neighborhood U of f . We assume this condition from lemma A.6 on. The following is a direct consequence of lemma (A.2) Proposition A.5 Let r; s 0 and ( if 0 s < 1; 0 r < 1 t rs min(r; s) if 1 max(r; s) Let f 2 C r (U; F ). Then the map f : C s(V; G) ?! C t (U; G) g ?! g f is linear and continuous . 30
Now we consider the variation of the rst factor.
Proposition A.6 Let r; s; t 0, U , V balanced subsets. and g 2 C s(V; G) Let U C r (U; F ) be such that for every f 2 U we have d(f (U; v c) > > 0. The map g : U C r (U; F ) ?! C t (U; G) f ?! g f is continuous in the following cases i) for t = 0 when r 0; s > 0: ii) for 0 < t < 1? when s > t; r t; rs > t: iii) for t = k + with k 1, 2 [0; 1) when s > t; r t: iv) for t = k + 1? with k 0 when s > k + 1, r t
Proof: i) It is similar to (i) of proposition (A.2). Thus if 0 s 1?
then
k g(f1 ) ? g(f2) kC 0 k g kC s k f1 ? f2 ksC 0 ii.1) First we consider that 0 < s 1?. We can also assume 0 < t < r < 1.
Using lemma (A.1) and (i1) of proposition (A.2) we have
k g (f1) ? g(f2) kC t =k g f1 ? g f2 kC t rs?t
t
2 k g f1 ? g f2 kCrs0 k g f1 ? g f2 kCrsrs
2 k g kC s [2+ k f1 ksC r + k f2 ksC r ] rst k f1 ? f2 kCs?0 r (18) ii.2) We assume that s = 1 + with 0 < < 1 and 0 t r < 1. Let f1 2 U and = d(f1(U ); V c) > 0. We take x; y 2 U and make z1 = f1 (x); w1 = f1(y). For > 0 xed there is a path 1 joining z1 to w1 such that Z 1 g(z1) ? g(w1) = dg( 1(t)) _1 (t)dt 0 and Z 1 k _ 1(t) k dt (kV + ) k z1 ? w1 k 0
31
t
Let f2 2 U with k f2 ? f1 kr < . If z2 = f2 (x); w2 = f2(y) then
2(t) = 1(t) + (1 ? t)(z2 ? z1) + t(w2 ? w1) is a path contained in U joining z2 to w2 and
g(z2) ? g(w2) = We take
S1 =k
Z
0
1
1
Z
0
dg( 2(t)) _ 2 (t)dt
[dg( 1(t)) ? dg( 2(t))] _ 1(t)dt k
k g kC 1+ (k z1 ? z2 k + k w1 ? w2 k) (kV + ) k z1 ? w1 k 1 S2 =k dg( 2(t))( _ 1(t) ? _ 2(t))dt kk g k1k z2 ? z1 + w1 ? w2 k 0 Z
thus
k g(f1(x)) ? g(f2(x)) ? g(f1(y)) + g(f2(y)) k S1 + S2 2 k g kC 1+ k f1 ? f2 k C 0 kV k f1 kC r k x ? y kr + k g kC 1 k f1 ? f2 kC r k x ? y kr This shows that
k g f1 ? g f2 kC r 2kV k g kC 1+ (k f1 kC r k f1 ? f2 k C 0 + k f1 ? f2 kC r ) hence there is a constant M such that (19) k g f1 ? g f2 kC r M k g kC 1+ k f1 ? f2 k C r iii) Let us assume that t = r = k + ; s = k + with 0 < < 1. The
case = 0 is included in [5] , so we consider > 0. Let > 0 and f1 ; f2 2 U with k fi kr < i = 1; 2. For k = 1 we have
k g(f1 ) ? g (f2) kC 0 k g kC 1 kV k f1 ? f2 kC 0 Moreover d(g f ) 2 C r?1(U; L(E; G)) and k d(g f1) ? d(g f2) kC =k dg f1 d(f1 ? f2 ) ? (dg f2 ? dg f1 ) df2 kC 2 k dg f1 kC k d(f1 ? f2) kC +2 k dg f2 ? dg f1 kC k df2 kC (20) 32
On the other hand
s?r
k dgf2 ?dgf1 kC 2 k dgf2?dgf1 kC 0 (k dgf2 kC + k dgf1 kC ) 2 k g kC 1+ (k f1 k C 1 kU + k f2 k C 1 kU + 2) k f1 ? f2 kCs?0r (21) Substituting (21) in (20) we nd a constant M such that
k g f1 ? g f2 kC t M k g kC 1+ k f1 ? f2 kCs?rr
(22)
The veri cation of the inequality (22) for every n is a routine induction argument. iv) Is a consequence of the arguments used in (ii) and (iii).
De nition A.4 Let IR3+ be the set of indices (r; s; t) which satisfy some of the conditions stated in proposition A.6. We say that is the set of the exponents of continuity
Proposition A.7 Let (r; s; t) 2 IR3 ? and > 0. There are Banach spaces E , F and g 2 C s(B2 F; IR) such that g : U C r (B E; F ) ?! C t(B ; IR) is discontinuous.
The continuity in (i) requires the uniform continuity of g. In the remaining cases if (r; s; t) 62 it satis es some of the conditions considered below. If 0 < s 1?; 0 < r 1? we take ' to be a C 1 cut-o function which takes the value 1 on [? 21 ; 21 ] and the value zero outside of [?1; 1]. We set
g(x) =
1
X
i=6
jx ? 1i js'(2i(x ? 1i ))(?1)i:
Since each of the terms in the sum has disjoint support, the sum converges. It is clear that it de nes a C s function out of zero. At zero g(0) = 0 and since 2?s(i?1) sup jx ? 1 js'(2i(x ? 1 )) 2si; i i x 33
2?s(i?1)
sup
x2[ 1 ?2?i ; 1 +2?i ] i
i
jg(x) ? g(0)j=jxjs
jg(x)j 2si
It follows that is bounded and hence g is C s: We take fn(x) = jxjr + n1 and f0(x) = jxjr . Clearly, k fn ? f0 kC r ! 0. Let n > m be integers, m; n of di erent parity. Then jg fn ? g fmj(x) = 2jxjrs ?n?1 ?n?1 when x 2 [?2 r ; 2 r ] so that k g fn ? g fm kC rs 2. Hence, it is impossible that g fn converges in C rs. If we consider in (ii) or (iii) the option s = t, r 1 the continuity also fails. This fact is well known for integer exponents and we will sharpen it below. If s = t = k + with k 2 IN; 0 < 1?, it suces to take g = xk jxj ; fn = x + n1 . Clearly k fn ? Id kC r ! 0 for any r. (g fn)(k)(x) = jx + n1 j (k + ) ( + 1) Hence, j(g fn ? g Id)(k)( ?n1 ) ? (g fn ? g Id)(k) (0)j = 2 n1 (k + ) ( + 1) so that k g fn ? g Id kC t K > 0. We deal the other cases in in nite-dimensional Banach spaces. For p 1 we introduce the usual space of summable sequences
lp with the norm
1
X = fx = (xn n=1 2 IRIN = jxnjp n=1
)1
< 1g
1
k x k= ( jxnjp)1=p X
n=1
which converts it into a Banach space. We denote by en the sequence with all the components 0 except the n-th one which takes the value 1. First, we take 0 < r < 1 and p = 1=r. Let us consider the real functions g0;n(x) = n(x ? 1) [1;1+1=n] + [1+1=n;1] R and in general gk;n(x) = 0x gk?1;n(s)ds. We de ne gk : B3 lp ?! P IR (xj )j2IN ?! 1 j =1 gk;j (xj ) 34
Given x = (xj )j2IN 2 B3(x) there exist indices j1 ; : : : ; j6 such that if y 2 B1=2 (x) then jyj j < 1 for j 2 IN ? fj1 ; : : : ; j6 g. If we denote
x : lp ! IR6; x ! (xj1 ; : : : ; xj6 ) and
gk;x : IR6 ! IR; (x1 ; : : : ; x6 ) !
6
X
i=1
gk;ji (xi)
then gk (y) = gk;x x(y), for every y 2 B1=2 (x). Hence gk 2 C k (B3 ; IR). Let ' be a cut-o function which takes the value 1 on [ ?31 ; 13 ] but zero outside of [ ?32 ; 32 ] and set h(x) = (1 + jx ? 1jr )'(x ? 1). We de ne
f : B3=2 l1 ?! lp (xj )j2IN ?! (xj )j2IN
Then
1
k f (x) ? f (y) k= y( jh(xj ) ? h(yj )jp)1=p k h kC r k x ? y kr X
j =1
and fn = f + en=2n satis es this same inequality for every n 2 IN. It is obvious that f ; fn 2 C r (B3=2 l1 ; lp) and k fn ? f kC r ! 0. However for n big enough if x 2 [1; 1 + (1=2n)?r ] then
jg1 fn (xen) ? g1 f (xen )j = jx ? 1jr =2 + 1=8n so that k g1 fn ? g1 f kC r K > 0. In this example there is not continuity
for s = 1 and t = r. This completes (ii). Finally if we consider r = 1 and h(x) = x'(x ? 1) in the above relations then k gk+1 fn ? gk+1 f kC k+1? K > 0. In this example there is not continuity for s = k + 1, t = k + 1?. Remark If F is a nite dimensional space then f (U ) is a compact subset of V . Every continuous function on V is uniformly continuous on f (U ). For this reason under the additional hypothesis that F is nite dimensional, we can supplement the cases in proposition (A.6) with v) 0 < t = r 1?, s = 1. vi) t = s = k, r k. 35
vii) s = k + 1, t = k + 1?, r k + 1?. We will denote the corresponding set of exponents of continuity by 0 .
Proposition A.8 Let r; s; t 0 and n 2 IN such that (r; s ? n; t) 2 . If g 2 C s(V; G) then g is n times di erentiable. Proof: In these conditions (dng) : U ?! C t(U; Lk (F; G)) is continuous. Moreover if f0 2 U there are positive constants M; ; such that
(23) k (dng)(f1) ? (dng)(f2) kC t M k g kC s k f1 ? f2 kC r n when f1 ; f2 2 U = ff 2 U = k f2 ? f1 kC r < . In consequence d (g) 2 C (U ; C t(U; Lk (F; G)).
We will show that g is n times continuously di erentiable and (24) k dk (g)(f1) ? dk (g)(f2 ) kC r;t M 0 k g kC s k f1 ? f2 kC r for an adequate constant M 0 . Let us consider n = 1. We will verify that d(g) : U ?! L(C r (U; F ); C t(U; G)) exists is de ned by d(g)(f )[h] = dg f h and is continuous. For k h kC r 1 and small enough we have k g (f + h) ? g(f ) ? dg f h kC t = Z 1 g ( f + h ) ? g f ? dg f h t k kC =k [dg (f + sh) ? dg f ] hds kC t 0
1
Z
2 k (dg)(f + sh) ? (dg)(f ) kC t k h kC t ds 2M k g kC s Moreover l : C t(U; L(F; G)) ?! L(C r (U; G); C t(U; G)) u ?! l(u)[] = [x ! u(x) (x)] is a continuous linear map and d(g) = l (dg). From proposition (A.3) and proposition(A.4) we deduce that d(g) 2 C (U; L(C r (U; G); C t(U; G)) and obtain the inequality (24). Now the induction carries on immediately. Again proposition (A.8) contains the best conditions on di erentiability. 0
36
Example A.5 If (r; s ? 1; t) 62 then there is g 2 C s((?2; 2); IR) such that g : C r ((?1; 1); IR) ! C t ((?1; 1); IR) is not di erentiable. We only deal with the cases (ii) and (iii) of lemma A.6. Let us consider g = xn+1jxj with n 2 IN; 0 < < 1. If n 1 , the map g : C n+1((?1; 1); IR) ! C n+ ((?1; 1); IR) is continuous but not di erentiable. In e ect taking f = Id and h = 1, then for every x 2 (?1; 1) g(f + h)(x) ? g(f )(x) = lim g(x + ) ? g(x) = g0(x) (25) lim !0 !0 However 1 [g(n)(x + ) ? g(n) (x) ? g(n+1)(x)] j0 = x= ?2 (n + 1 + )(n + ) (2 + )[1 + 2 ] This shows that the convergence fails in C n+ - norm. On the other hand if n = 0; 0 < < 1 and we take r = , t = then g : C r ((?1; 1); IR) ! C t ((?1; 1); IR) is not di erentiable . Choosing f = jxj ; h = 1 we nd for every x 2 (?1; 1) g(f + h)(x) ? g(f )(x) = lim g(jxjr + ) ? g(jxjr ) = g0(jxjr ) (26) lim !0 !0 However 1 [g(jxjr ? ) ? g(jxjr ) + g0(jxjr )]jxr ==2 = ?[1 + ] 0 ? 2 this shows that the convergence fails in C t - norm. Now we treat the di erentiability of the composition operator in the space product. We transfer its analysis to the partial derivatives
Lemma A.5 Let r; s; t 0 and k 2 IN such that (r; s ? k; t) 2 . Then i) The map
R : U C r (U; F ) ?! L(C s(V; G); C t(U; G)) f ?! f is k times continuously di erentiable.
37
Proof: i) We assume k = 0. Let f1 ; f2 2 U with k f1 ? f2 kr small
enough in order that inequality (23) applies, then
k R(f1) ? R(f2 ) kC s;r = sup k g f1 ? g f2 kC t M k f1 ? f2 kC r kgkCs 1
When k 1 we complete the proof by induction. We will show that R is continuously di erentiable and
dR : U C r (U; F ) ?! L(C r (U; F ); L(C s(V; G); C t(U; G))) f ?! dR(f )(h) = [g ! dg f h] Indeed if k h kC r 1 and is a small parameter then k R(f + h) ? R(f ) ? dR(f )(h) kC s;t = sup 1 k g (f + h) ? g f ? dg f h kC t M kgkCs 1 We consider R1 : U C r (U; F ) ?! L(C s?1(V; L(F; G)); C t(U; L(F; G)) f ?! f
l : L(C s?1(V; L(F; G)); C t(U; L(F; G)) ?! L(C r (U; F ); L(C s(V; G); C t(U; G))) u ?! l(u)(h) = [g ! u(dg) h] We know that R1 is k ? 1 times di erentiable, l is a continuous linear map and dR = l R1. As a consequence of lemma [ A.3] we otain that R is k times continuously di erentiable. We summarize our conclusions
Theorem A.1 Let r; s; t be positive numbers and K 2 IN such that (r; s ? k; t) 2 . The map comp : U C r (U; F )] C s(V ; G) ?! C t (U; G) (f; g) ?! g f is continuous and k times continuously di erentiable.
38
Proof: The continuity is a consequence of lemma A.5 and lemma A.6. Let k 1 and consider L : [U C r (U; F )] C s(V; G) ?! L(C r (U; F ); C t(U; G)) (f; g) ?! L(f; g) = [h ! dg f h] Note that dcomp is the sum of L and the map R de ned above. Moreover L is the compositions of the maps (f; g) ?! (f; dg) ?! dg f ?! [h ! dg f h] The rst and the third maps are linear. As a consequence of the induction processes we deduce that the intermediate map is k ? 1 times continuously di erentiable. Thus we obtain that the composition operator is k times continuously di erentiable.
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