Israel Journal of Mathematics, Volume 1, Issue 2 (1963)

ON C O M P A C T I F I C A T I O N OF METRIC SPACES* BY

M. REICHAW-REICHBACH

ABSTRACT

If f: X-+ X* is a homeomorphism of a metric separable space X into a compact metric space X* such that f(X) = X*, then the pair (f,X*) is called a metric compactification of X. An absolute G0-space (Fa-space) X is said to be of the first kind, if there exists a metric compactification (f,X*) of X such that ~o

f(X) = ~ G~, where Gi are sets open in X* and dim[Fr(Gi)] < dim X. i=1

(Fr(Gi) being the boundary of G~ and dim X - - t h e dimension of X). An absolute G0-space (Fa-space), which is not of the first kind, is said to be of the second kind. In the present paper spaces which are both absolute G~ and F,,-spaees of the second kind are constructed for any positive finite dimension, a problem related to one of A. Lelek in [11] is solved, axed a sufficient condition on X is given under which dim[X* - f ( X ) ] >__k, for any metric compactification (],X*) of X, where k =< dim X is a given number. Introduction. Let f : X - * X * be a h o m e o m o r p h i s m o f a separable metric space X into a c o m p a c t metric space X*, such t h a t f ( X ) = X*. The pair (f,X*) is then called a metric compactification o f X. I f X is an absolute G0-space (F~-space) (i.e. a G0-set (F~-set) in some c o m p a c t space), then X is said to be o f the first kind (cf. [6]) provided there exists a compactification ( f , X * ) o f X such that ~o G i, where Gi are sets open in X* and dim [Fr(Gi)] < dim X, f ( X ) = ¢3i=1 i = 1, 2 , . - . . (Fr(Gi) denotes the b o u n d a r y o f G i, and d i m X the dimension o f X in the sense o f Menger-Urysohn.) A n absolute G0-space (F~-space) which is n o t o f the first kind is said to be o f the second kind. The aim o f the present paper is: (i) to construct, for any positive finite dimension, spaces X which are b o t h absolute F~ and absolute G0-spaces o f the second kind; (ii) to solve a p r o b l e m related to one o f A . Lelek in [ 1 1 ] ; and (iii) to give a sufficient condition on X , such that, for a given k < d i m X , we have dim [ X * - f ( X ) ] > k for every c o m p a c tification (f,X*) o f X. The paper consists o f four parts. In Section I some k n o w n compactifications are mentioned; in Section II several problems concerning compactifications are

Received May 22, 1963 * This research has been sponsored by the U. S. Navy through the Office of Naval Research under contract No. 62558-3315. 61

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posed. Facts on coverings are quoted in Section III. Finally, Section IV contains a solution of the problems(1) posed in Section II. I. SOMECOMPACTIFICATIONSOF METRIC SPACES 1.1. Let X be a given topological space. Let X* = X o ( x * ) , where x*6X is an additional point, and let us define the toppology in X* by taking as open sets all sets open in X and all subsets U of X*, such that X * - U is a closed compact subset of X. Then the theorem of Alexandroff states: (1) The space X* is a compact topological space and X* is a Hausdorff space if and only if X is a locally compact Hausdorff space(2). The space X* is called the one-point compactification of the space X. A topological embedding is usually allowed rather than insist that X actually be a subset of X*. Thus by a compactification of a space X a pair (f, X*) is understood, such that f : X--, X* is a homeomorphism of X into a compact space X* and f ( X ) = X* (i.e. the image f ( X ) of X is dense in X*). In this sense the one-point compactification of a non compact space X is a pair (i, X*) where i: X ~ X* is the identity mapping and i(X) = X* = X u (x*). Another compactification of a topological space X is the Stone-(~ech compactification (e, fl(X))(3). This compactification is defined as follows: Let us take the set F(X) of all continuous functions f : X ~ d mapping X into the interval J = [0,1] and the product dVtX) with the Tychonoff topology. Let us define the mapping e: X ~ j~tx) by correlating with each point x e X the point e(x) whose f - t h coordinate is f(x), for each f e F(X). The mapping e(x) is a continuous mapping of X into jrtx), and in the case when X is a completely regular Tt-space it turns out to be a homeomorphism. In this case we define fl(X) by by fl(X) = e(X) and the pair (e, fl(X)) is called the Stone-t~ech compactification of X. Let us note that: (2) If (e, fl(X)) is the Stone-(2ech compactification of a completely regular Tl-space X and f : X ~ Y is a continuous mapping of X into a compact Hausdorff space Y, t h e n f [ e - ~ ( x ) ] has a continuous extension on fl(X) into y(4). Numerous other compactifications were constructed for various purposes. One of them, used in the dimension theory, is the Wallman compactification (~, w(X)). It turns out to be topologically equivalent to the Stone-C_.ech compactification provided w(X) is a Hausdorff space(5). (0 I learned recently that some problems considered in the present study have been solved by A. Lelek in an entirely different way (not published). (2) See [5], p. 150, also [3], p. 73. (3) See [5], p. 152. For properties of the Stonc-~ech compactification,seealso [2] and [13]. (4) See [5], p. 153. (s) Ibidem, p. 168. For properties of the Wallman compactification,[15].

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1.2. Considering the one-point compactification ( i , X * ) o f a metric space, we note that the space X* is generally not a metric space. For instance, if X is a metric space which is not locally compact, then by (1) X* cannot be a metric space (since every metric space is a Hausdorff space). Thus if we seek, for a given metric space X, a compactification (f, X*) where X* is also a metric space, we generally cannot achieve this by merely adding a single point, and should allow the set X* - f ( X ) to contain more than one point. In the present study we confine ourselves to metric compactifications (f,X*) o f metric separable spaces X only, i.e., we assume that X is a separable, metric space and X* a metric space. As already noted, the one-point compactification is generally not a metric compactification. Let us show that an analogous statement holds for the Stone-(3ech compactification (e,p(X)). THEOREM 1. If X is a non compact metric space and (e, fl(X)) the Stone-

Cech compactification of X, then fl(X) is not a metric space(6). Proof. Suppose, to the contrary, that fl(X) is a metric space. Let e(X) be the image of X in fl(X). Since X is not compact, there exists a sequence A = {a~}~= ~,2 ... of points a n ~ X which does not contain any convergent subsequence. Consider the points e(an) = b~. Since fl(X) is compact and metric, the sequence {bn}n = 1,z.... contains a convergent subsequence {b'~} c {b~}. Let b~--* b ~ fl(X) and consider the points a" = e-l(b'). By A' = { a ' } c A the sequence A' does not contain any convergent subsequence. Therefore A' is a closed subset of X. Let us define the real f u n c t i o n f : A' ~ J = [0, 1_] by

f(a'~) = { 01 forf°r nn== 2k2k_ 1 k = 1,2, .... Since A' does not contain any convergent subsequence, the function f : A ' ~ J is continuous and since A' is a closed subset of the metric space X, we can, using Tietze's extension theorem(7), extend this function, to a continuous function f : X ~ d (the extended function is denoted also by f ) . By (2), the f u n c t i o n f e - t has a continuous extension f t o the whole o f p(X). But since ,

f(bn) = f e

-1 ,

(bn)=f(a'n)=

{ 0 for n = 2 k 1 for n = 2 k - 1

and b" -* b, the function f c a n n o t be continuous at the point b. This contradiction shows that fl(X) is not a metric space. (6) This theorem seems to be well known. It was noted by A. Zabrodsky that the above proof may be applied to show that fl(X) can not even satisfy the first countability axiom. (7) See [8], p. 117.

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REMARK 1. Since the Wallman compactification (O,w(X)) is topologically equivalent to that of Stone-Cech, provided w(X) is a Hausdorff space it follows by Theorem I that if X is a non-compact metric space, then the space w(X) is not a metric space. II. PROBLEMS ON COMPACTIFICATIONS

II.1. The results of Section I indicate that metric compactifications of metric spaces are generally neither the Stone-Cech nor the one-point compactification. Now, since for metric compactifications the set X * - f ( X ) generally contains more than one point, there arises a problem of finding the structure of this set for some classes of metric spaces X. For example the following questions can be posed: (a) Is it always possible to find a compactification (f,X*) of X such that X * - f ( X ) would be countable? (b) Is it always possible to find a compactification (f,X*) such that dim [X* - f ( X ) ] < dimX? Regarding question (a) it is known that each space which does not contain a subset dense in itself, has a compactification (f,X*) such that X * - f ( X ) is countable(8). On the other hand, it is easily seen that for each compactification of the set X of rational numbers the set X* -f(X) is uncountable. Indeed, since f : X ~ X * is a homeomorphism, each point o f f ( X ) is a limit point and therefore X* is perfect. Hence X* is uncountable(a). Regarding (b) it is known(x°) that for each space X, there exists a compactification (f,X*) such that d i m X * - - d i m X and thus d i m [ X * - f ( X ) ] < dimX. Easy examples show that in many cases this weak inequality < can be replaced by the strong < . It suffices, for example to take any n-dimensional cube J"; n = 1,2, ... and any point p~J". The set X = J" -(p) can be compactified by adding this single point. We then have X* -- J "and dim[X* - f ( X ) ] = dim(p) = 0 < dimX, w h e r e f = i is the identity mapping. On the other hand, it is not always possible to achieve the strong inequality d i m ( X * - f ( X ) ) < d i m X . Indeed, for a 0dimensional space X, d i m ( X * - f ( X ) ) < d i m X = 0 means that X * - f ( X ) is empty and hence X is compact. It follows that for a 0-dimensional non compact space X this strong inequality is impossible. The problem of finding examples of n-dimensional spaces X, n > 0 of a simple topological structure for which dim [X* - f ( X ) ] < dim X does not hold for any compactification (f, X*) of X is more complicated. More precisely, this problem may be formulated as follows: (s) See [7], p. 194,IV. (9) See [3], p. 98. (lO) See [4], p. 65, Theorem V, 6. Also [9], p. 72.

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(c) Let X be a given n-dimensional space and k <= n an integer. Under what conditions on X shall we have dim [X* - f ( X ) ] > k for each compactification (f, X*) of X? II.2. B. Knaster discovered in [6] that there exist two kinds of absolute G~spaces (also called G~-spaces in compact spaces or topologically complete spaces). Their definition is(11): An absolute G6-space is said to be of the first kind, if there exists a compactification (f,X*) such that f ( X ) = r-)

G i and dim [Fr(Gi) ] < dim X,

where

G~, i = 1, 2, ..., are sets open in X* and Fr(G~) denotes the boundary of G~in X*. An absolute G~-space is said to be of the second kind if it is not of the first kind. It was shown by Lelek(12) that (3) An absolute G~-space of finite dimension is of the first kind, if and only if there exists a compactification (f, X*) of X such that dim [X* - f ( X ) ] < dim X. Now, it was shown in [6] that the Cartesian product N x J, where N is the set of irrational numbers in the interval J = [0,1], is an absolute G~-space of the second kind. It was further proved in [11] that if Z is any compact space with dim Z = n ->_0, then the space X = N x Z is an absolute G~-space of the second kind. These results provide a solution of problem (c) for n = k in the class of finite dimensional absolute G~-spaces. 1"he sequel will include a solution of the following problems: (al) Does there exist, for any positive finite dimension n = 1,2,..., a finite dimendsional space X, which is both an absolute F,, and G6-space of the second kind? (a2) Is it true that each absolute G,-space X of the second kind, of positive finite dimension n, contains a topological image of a set of the form N x Z, where N is the set of irrational numbers of the interval J = [0, 1] and dim Z = dim X? (a3) Problem (c). (a,) Construction of a weakly infinite dimensional absolute Fo and G~-space of the first kind such that for each compactification (f, X*) there is d i m ( X * - f ( X ) ) = 00(13). Before proceeding with a solution of problems (al)--(a4), we quote in the next section some facts on coverings.

(11) See: Introduction (22) See [11], p. 31, Theorem 1. (23) A space is called weakly infinite-dimensional if it is a union of a sequence of finite

dimensional spaces Xk, with dim Xk-+oo, for k~oo.

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III. COWRINGS

A covering of a space Y is a family ~ = {Gi} of sets G~ such that Y = [,.JtGt. If Gi are open (closed) sets, the covering is called open (closed). If the diameters

6(Gi) of all Gi are < e, (~ is called an e-covering, and if ~ is finite--a finite covering. d,(Y) denotes the infinum o f all numbers e > 0 such that there exists a finite open e-covering of Y satisfying (4) G i o n G ~ ' " C ~ G i . - - O , for any set of n + l indices i o < i l < ' " < i n (i.e., such that the intersection of any n + 1 different sets G~ is empty). It is known that for finite coverings of a space Y the existence of an open e-covering satisfying (4) is equivalent to the existence of a closed e-covering satisfying (4), and that for a compact space I7, dim Y __ 0 (the Lebesgue number of the family (F0,F1, ... Fro)) such that if there exists a point p e Z at distance =< 2 from all the sets Fko,Fkl, ..', Fk~, then f')~=oFk,-~ O. Proof.(15). Suppose the contrary. Then there exists a sequence of points p, e Z , n = 0,1,2,.-., and families S t = {Fk~o,...,FkJ.}, j = 0,1,2,..., of sets such that the point pj is at distance < ( l / ( j + 1)) from" all the sets Fk{ o f the family S i, but ~ o F k { = O. Since the number of different families S~, j = 0 , 1 , . . . constructed from a given finite family of sets {Fk)k=0.1 ..... is finite, some family - - s a y So--must appear in the sequence {So}j=o, t .... an infinite number of times. Thus there exists a subsequence {p~} ~ {p,} such that p,'is at distance < (1/(n + 1)) from all the sets Fk o, ..., FkOoof S o. Since Z is compact, the sequence {p',} contains a convergent subsequence to some point p ~ Z. Denoting this subsequence by {p'~}, we have p ' ~ p ~ Z . Now, since p(p'~,Fko) < (1/(n + 1)) for i = 0,1,..-, no, and n = 0,1, .--, and since p'~ ~ p, we have p(p, Fg,) = 0. Thus p e Fg~, i = 0,1,-.., no, which is incompatible with the fact (']7°__o Fk° = 0 (by the definition of S j). It follows by (5) that m F k, where (6) Let Y be a closed subset of a compact space Z and let Y c [.Jk=O F k are closed sets such that any different n + 1 of them have an empty intersection. Replacing each F k by its e-neighborhood/(16) Gk = S(Fk,e) (in Z), where 2e < 2, we obtain an open (in Z) covering ~ = {Gk} of Y, such that for the family {Gk} of closures of Gk, any n + 1 different sets G~ have an empty intersection(17).

(14) See [9], p. 60. (15) This is a standard proof and is given here for the sake of completeness only. (16) An e-neighborhood of a set F is by definition the union over all p e F of the sets

Sp

=

[z;q(p,z)< ~:zeZ] (17) For a proof of (6) see also [14], p. 414, Lcmma 2 and [10], p. 257.

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A n o t h e r consequence o f (5) is; (7) I f the closed sets F o, F1, "",Fro in a c o m p a c t space Z have an e m p t y intersection then there exists a n u m b e r e > 0 such that n o set o f diameter < e has a non empty intersection with each o f the sets Fo, F I , ' " , F,,. Indeed, it suffices to take e = 2 and to apply (5). We shall now give some properties o f coverings o f simplexes. Let tr S = (Po, "", Ps) be a closed s-dimensional simplex with vertices Po, P l , ' " , Ps in the Euclidean s-dimensional space E ~and let f : a s ~ Z be a h o m e o m o r p h i s m o f tr ~ into a space Z. Let tr s - l . , denote the (s - 1) dimensional closed face o f trS opposite to the vertex Pi ~ trS, i.e. a s- ~'~ = ( P o , ' " , Pi-1, Pi+ D ' " , Ps), i = O, 1,..., s, and let z ~ = f ( t r s) and z ~-1'i =f(tr~-l.i). Then z ~ is a curvilinear simplex with vertices qi =f(P~) and ( s - 1)-dimensional faces z ~-1'~, i = 0, 1,-.-,s. Since f is a h o m e o m o r p h i s m and [,-.si~=oa .~- 1 ~ = ~, we have that I~ 1i=o s ~s- ~,~ # ~. Thus applying (7) with m = s to the closed sets F i = z S- ~'~, we find that there exists a n u m b e r e > 0 such that no set with diameter < e intersects each o f the laces z s- i.~. (8) Let e > 0 be a n u m b e r such that no set with diameter < e intersects each face z S-l'i. Let further z S = /k.;k=O I,, F k, where F k are closed sets with diameters 6(F~) < e, k = 0, 1,-.-, m. Then some s + 1 sets Fko, " " , F k , have a non e m p t y intersection. Since tS(Fk) < e, no Fk containing a vertex qj o f z S intersects the face zs-l.~ opposite to qj. Since f is one-to-one, no s e t f - l ( F k ) containing a vertex p~ o f tr S intersects the face a s-~'j opposite to pj. N o w , the sets f - ~ ( F k ) , k = 0 , 1 , . . . m, cover the simplex a S a n d are closed, s i n c e f is continuous. Thus applying the same procedure as in the p r o o f of [2, 24] in (['1], p. 194) we obtain that some s + 1 s e t s f - ~ ( F ~ ) , j = 0 , 1 , . . . s , have a non e m p t y intersection. Hence also the sets Fk~, j = 0, 1,... s, have a n o n e m p t y intersection. IV. SOLUTION OF THE PROBLEMS FORMULATED IN II IV.1. A n n-dimensional absolute F~ and G~-space X a n d its properties. Let tr ~ = (Po, P l , ' " , P,,) be the n-dimensional closed simplex in the n-dimensional Euclidean space E" with vertices Po = (0, 0, ...,0) and p~ = (0, ... 0,1,0, ... 0), i = 1,2,-.., n. (i.e. Pi is the point in E "whose i-th coordinate is I and a l l / o t h e r coordinates are 0). Let A = {a j}, j = 1,2, ..., be the sequence o f points o f the f o r m a j = ( 1 / j ) , j = 1,2,..- on the real axis E l and let ao = 0 ~ E 1. D e n o t e by Fr(tr .) = U 7 = 0 tr,,-1,t the b o u n d a r y o f the simplex a *. Let (9)

X = (A x a ~) td [(ao) x F,'(tr~)]

Then X c E ' + l a n d the closure ~ o f X in E n÷ ~ is

.~' = (A x o'") k3 [(ao) x ~"] = I-A k.) (ao) ] x o" ". Since .~ is a c o m p a c t subset o f E ~+~ (as a p r o d u c t o f two c o m p a c t spaces /1 t.)(ao) and tr"), • is a c o m p a c t space, and since X can be written as a union

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[(ao) x Fr(tr')] U [-I,.Jj~=l(ai) x ~"] of a countable number of compact sets, it follows that X is an absolute Fo space• On the other hand the set 2 - X equals the interior of the simplex (ao) x tr ~. Since this interior is a union of compact sets, the set X - X is an F# set and therefore X is a G0-set in 37. It follows that (ba) The set X defined in (9) is both an absolute F , and G0-space. Evidently, d i m X = n. We shall now show that (b~) For each compactification (f, X*) of X we have dim IX* - f ( X ) ] > dim X = n. Indeed, suppose to the contrary that dim IX* - f ( X ) ] < n - 1 < dim X and consider the sets z nj = f [ ( a ~ ) x a"], j = 1,2,..., and z~n-- 1,i = f [ ( a j ) x 0""-- 1 , i ] , i = 0,1, .--, n, j = 0 , 1 , . - - . Since a j ~ a o for j---} oo, it follows that for every i = 0 , 1 , . . - , n , dist{[(a h x a " - l ' q , [(ao) x c r " - l " ] } ~ 0 for j--} ~ , where d i s t ( A , B ) = max[supx~ap(x,B), sup,,~p(A,x)] is the distance of the sets A and B in the sense of Hausdorff(xs). Since f : X - } X * is continuous and [A U (ao)] x a "- 1.i is compact it follows easily that (10)

• . - 1 , ~ ,T.o- 1 " ' ) ~ 0 f o r j ~ o o dlst(xj

and each i = 0 , 1 , . . . , n .

Now, the space X* being compact, there exists a subsequence {j'} of {j} such that the sequence of sets {~.,} converges to a continuum C = X*(~9). Writing j instead o f j ' , we have dist (~7, C) ~ 0 for j ~ oo. If there were C c~ [[,.ff= x z~] # ¢, then there would exist a point Yo and a sequence yj~ e xi~ of points, such that Yjk-~ Yo e ~o for k---} oo and some Jo. Then xjk = f - l ( y j k ) ~ f - 1 (Yo) = Xo, which is, because ofxj~ ~ (aj~) x a "and x o e (ajo) x a", incompatible with the openness of (ago) x a" in the union [,.J~=1(a j) x a". It follows that C n [ [..J~= 1 zT] = ¢, and since the set U~=oZg -1'I is an (n-1)-dimensional compact subset of C, it follows from the assumption dim IX* - f ( X ) ] < n - 1 and from Corollary 1 in ([4], p. 32), that dim C < n - 1. Thus, by the definition of d,(Y) (cf. section III), we obtain d,(C) = 0. Hence, by (6), there exists for every e > 0 an e-covering of C by sets Gk open in X*, k = 0 , 1 , . - . , m such that (11)

Gko n dk, n . . . n dk,, = ¢ for any set of subscripts/c o < kl < --- < k,.

• n T no- 1 •i = ¢ we may, according to (7), choose for this covering an e Since [")~=o so small that no Gk intersects each set z~ -~'~. Hence by (10) no set Gk intersects all the faces z~"- ~'~, i = 0,1,..., n, for sufficiently large j. Let G = I,Jk%0 Gk. Since C ~ G and dist (~, C) -~ 0 for j -~ ~ , there exists a Jo such that z~-= G for j ~_Jo. Fixing anyj----Jo, we find that the sets F k = z3 A ~k, k = O, 1, ...,m, satisfy the assumptions of (8) with s replaced by n and z by zj. Hence by (8) some n + 1 sets F k o , ' " , f k , , and therefore also the sets Gko,-.., Gk have a non empty intersection, which is incompatible with (11). Thus (b~) is proved.

Q8) See [8], p. 106 Qg) See [9], p. 110. Also [16], p. 11.

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F r o m (bl), (bl') and (3) we obtain THEOREM 2. The set X defined in (9) is both an absolute F~ and G#-space of the second kind and of dimension n. This theorem gives an answer to problem (al). IV. 2. On a problem of A. Lelek. The following problem P. 313 was formulated by Lelek in [111, p. 34). Does there exist, for each absolute G#-space X of the second kind with finite, positive dimension, a compact space Z with positive dimension, such that X contains a topological image of the set N x Z (N being the set of irrational numbers of the interval J = [0,11)? A negative answer to this question was given in [121. Now it is easily seen that a negative answer to problem (a2) posed in section 1I contains, as a special case, a negative answer to the problem of Lelek. (It suffices to take, in (a2) , n = dim X = 1.) We now proceed to prove that the answer to (a2) is negative. Indeed, let X be the space defined in (9). We shall show that there does not exist a space Z with dim Z = dim X = n such that N x Z has a topological image in X. Suppose, to the contrary, that such a space Z exists and let h : N x Z ~ X be a bomeomorphism of N x Z into X. Fix a point 4 6 N. Then the n-dimensional space (4) x Z has a topological image in X. Now X being a countable union of compact disjoint sets (a j) x trn and (ao) x Fr(~ ), j = 1,2, ..- and (4) x Z being n-dimensional,it follows(2°) that h [(4) x Z] has an n-dimensional intersection with some set (aj(¢))x trn. This intersection, as n-dimensional subset of tr ~, contains(21) an open subset of(aj(¢)) x tr~. Since h is one-to-one, the sets hi(4) x Z 1 and h [ ( ~ ' ) x Z 1 are disjoint for ~ ~ ~',4, 4 ' 6 N , and since N is uncountable, we get an uncountable family of disjoint open sets contained in X, which is impossible. IV. 3. Two theorems on eompaetifieation. We shall now prove two theorems which will enable us to provide an answer to problem (c) and to construct, for any n = 1,2, .-., No, a n-dimensional space X which is not locally compact at a single point and such that for each compactification ( f , X * ) of X we heva dim I-X* - f ( X ) l > 1. THnOREM 3. Suppose that the space X contains a sequence {Ci}i=l, 2.... of continua C i and a point p such that (cl) the sets Ci are closed and open in the union ~.J~°=1 Ci and Ci n Cj = ¢ for i ~ j ; (z0) This is a consequence of the Sum Theorem for Dimension n, Cf. [4], P. 30. (22) This follows easily from Theorem IV, 3 in [4], p. 44.

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(c2) there exists a number 6 > 0, such that 6(Ci) >__6 for each i = 1, 2,... ; oo

U, ol c , -

c , = (p).

Then X is not locally compact at the point p, and for each compactification (f,X*) of X we have dim[X* - f ( X ) ] > 1. Proof. Let Up be an arbitrary neighborhood containing the point p. We have to show that the closure Op is not compact. By (c0 and (ca) there exists a sequence oo of points p~ U~= t c~ such that p ~ p for i ~ oo and such that the sequence {P~}i=l,z .... has only a finite number of points in common with each C~. Thus we may assume that for each i = 1, 2,..-, we have p~ e Ct. Let S = S (p, r) be a spherical neighborhood of p with radius r < 6/2 contained in Up. Since pi ~ p, the sets C~ n S are not empty for i sufficiently large, and since C~ are connected, we obtain from (c2) that for these i; C~ n Fr(S) ~ ¢. Choose from each such set Ci n Fr(S) a point qi and consider the sequence {q,}. Since S = Op, we have {qi} = / ] p and since q~ e Fr(S), it follows that p(q~, p) = r > 0. Now, since qi e C~ for i sufficiently large, (cO and (ca) imply that any convergent subsequence of {q~} tends to p, which is impossible because P(qi, P) = r > 0. Thus Up is not compact. It remains to show that if (f,X*) is any compactification of X, then dim [X* - f ( X ) ] > 1. For this purpose let us consider the sets XI = U i ~ l c i u ( p ) and f ( X O . The closure f ( X O = X* = X* is a compactification of X~. Let y be any point of X* - f ( X 1 ) . Then the point y Cf(X). Indeed, if there would exist a point x e X such that y = f ( x ) , then we would have xCX1, since f is one-to-one. Now, y ef(X1) implies that there exists a sequence of points x, e Xx such that f ( x , ) -o y. By the continuity o f f - a we have x, -~ x e X - Xx. But by (ca) the set X1 is closed in X, and since x, e Xx it follows that x e X~. This contradiction shows that y Cf(X). Thus (12)

[X~' - f ( X 1 ) ] N f ( X ) = [f(X1) - f ( X x ) ] N f ( X ) = ¢.

Let us take further r < 6/2 and construct (in analogy with the first part of the proof) points pi-~ p, Pt e Ci and qi e C~, such that P(P, qi) = r > 0 for i sufficiently large. Since X* = f ( X 1 ) is compact andf(Cl) c X* we can choose a subsequence of the sequence {f(C~)} of continua converging(22) to some continuum C. Denoting the subscpripts of this subsequence by i we have therefore that dist [f(C~), C] ~ 0 for i ~ oo. Now, since Pi -~ P and pi e C~, it follows that C contains the pointf(p). If C would reduce to this point f(p), then ql e C~ would imply f(q~)~f(p), and since f - 1 is continuous we would also have q ~ p , in contradiction to P(P, qi) = r > 0. It follows that C contains at least two points, and since it is a continuum we have dim C > 1. Therefore dim [C - (f(p)] => 1. (22) Seo [9], p. 110.

1963]

ON COMPACTIFICATION OF M E T R I C SPACES

71

Now, by (cl) we have C C ) f ( C i ) = ~J for each i = 1,2, .... Therefore X * ~ X* and (12) imply that dim [X* - f ( X ) ] > 1. Theorem 3 is proved. EXAMPLE 1. Let X = (ao) t.) [[,.J~= 1 (a j) x J ] where ao = 0 and aj = 2-J+ 1 j = 1,2,..., are real numbers on the real axis and J = [0, 1] (Figure 1). This 1-dimensional space X is not locally compact at the single point ao = 0, and by Theorem 3 dim IX* - f ( X ) ] > 1 for any compactification (f,X*) of X. It is also easily seen that X is an absolute Fo and G~-space and thus, by (3) and d i m X = 1, we obtain that X is an absolute F , and G~-space of the second kind.

OO

04

03

OI

02

Figure 1

EXAMPLE 2. Let n = 2, 3,.--, No, and let X = (J~ - X1) L~ (O), where Xl={X; x=(xl,x2,...,xn), xl=0, O<xi 1 for any compactification (f,X*) of X (for n = 3, see Figure 2).

c

Bt

/ Figure 2 By Theorem 3 for each compactification ( f , X ,) of this full cube X excluding the full square OABC but including point O, dim IX* -- f(X)] --> 1.

72

M. REICHAW-REICHBACH

[June

Let us now show that (d) If the set X is a dosed subset of space Y and for each compactification (g,X*) of X we have dim I X * - g(X)] > k, then for each compactification (f, Y*) of Y we have dim iY* - f ( Y ) ] > k. Proof. The closure (in Y*)je(X)= X* of f(X) is a compactification (f,X*) of X and therefore by assumption, we have dim[X* - f ( X ) ] > k. Now, it is easily seen that~f(X) n f ( Y - X) = ¢. Indeed, otherwise we could find a point xo e Y - X and a sequence of points x n ~ X such that f(xn)...*f(xo). But sincef is homeomorphism on Y there would be x,,--, x o, which is incompatible with the closedness of X in Y. From f(X) n f ( Y - X ) = 0, we obtain

X* - f ( X ) c Y* - f ( Y ) , and therefore dim [Y* - f (Y) ] > k. As a consequence of (b~) and (d), we have the following answer to problem (c): THEOREM 4. If space Y contains topologically the set X defined in (9) and X is a closed subset of Y, then for each compactification (f, Y*) of Y* we have dimlY* - f ( Y ) ] > n. (The case n = 2 is illustrated in Figure 3).

B

o

A Figure 3 According to theorem 4, for each compactification (f, Y*) of this tu[1 cube Y excluding the interior of the square OABC (but including OA, AB, BC and CO) dim [Y*--f(Y)] _>--2.

IV. 4. A weakly infinite-dimensional absolute F~ and G~-space. As stated in (3), a finite dimensional absolute G~-space X is of the first kind if and only if there exists a compactification (f,X*) of X such that dim IX* - f ( X ) ] < dimX.

1963]

ON COMPACTIFICATION OF METRIC SPACES

73

We shall now show that the above condition is not necessary for infinite dimensional spaces. More precisely, we shall construct an absolute F , and G6-space of the first kind which is weakly infinite-dimensional and such that for each compactification (f,X*) of X we have d i m [ X * - f ( X ) ] = m. Let us take, for fixed n, the set A, of points x, ,, = 2-" + 2 - " , m = n + 1, n + 2, ..., on the real axis. Define X . = (A, x tr") U [(2-") × Fr(tr")]. where a" is an n-dimensional closed simplex with diameter 6(tr")= 2 - " . Let X = 0 , =~oi X ,. The set X can be considered as a subset of the Hilbert cube jSo, and its closure )? is ) ? = [.J,~=lx, w [ U ~ = I (2 - " ) x Int(tr")] u ( O ) where I n t a " = tr" - Fr(a") and 0 = (0, 0,..-) is the point all whose coordinates are zero. It is also easily seen that )? may be written in the form U ~ = a )~. u (0), where )~, = EAn u (2 -n)] X 0 "n. Since )? is a compact space and X is a countable union of compact sets, we find that X is an absolute F,-space. Further, we can write each set ( 2 - " ) × Int(a") as a union (_Jl =°°xF7 of compact sets FT, i = 1,2,-... Thus

n=l

is an

F,

set and

thus

X

is an

i=1

absolute

G~-space. Moreover, the sets

are open in 37, dim [Fr(Gs) ] < s and (']~= i G~ = X. Hence, X is an absolute F~ and G~-space of the first kind. By the defnition of X, it follows that X is a weakly infinite-dimensional space i.e. d i m X = oo(22). We shall now show that for each compactification ( f , X * ) of X we have d i m [ X * - f ( X ) ] = m. F o r this purpose, let us note that the set X . i~ homeomorphic with the space defined in (9), and hence by (b'l) we h a v e ' dim IX* - f ( X . ) ]

> dimX. = n

for each compactification ( f , X * ) of X,. Now it is easily seen that X . is a closed subset of X. Thus, applying (d) for X = X , and Y = X, we have dim [X* - f ( X ) ]

> n.

Since n is arbitrary, it follows that dim [X* - f ( X ) ]

= or.

Acknowledgment. The author is indebted to Mr. E. Goldberg for his kind help in editing this paper. (22) For weakly infinite-dimensional spaces X, dim X = a~ is sometimes written instead of dim X ----oo.

74

M.REICHAW-REICHBACH

[June

REFERENCES 1. Alexandroff, P. S., 1947, Kombinatornaya topologia, OGIZ. 2. C~,ch, E., 1937, On bicompact spaces, Ann. Math. (2), 38, 823-844. 3. Hocking, J. G. and Young, G. S., 1961, Topology, Addison-Wesley. 4. Hurewicz, W. and Wallman, H., 1941, Dimension Theory, Princeton Univ. Press. 5. Kelley, J. L., 1955, General Topology, Van Nostrand, New York. 6. Knaster, B., 1952, Un theoreme sur la compactification, Ann. See. Polon. Math., 23, 252-267. 7. Knaster, B., and Urbanik, K., 1953, Sur les espaces s6parables de dimension 0, Fund. Math., 40, 194-202. 8. Kuratowski, C., 1952, Topologie 1, Warszawa. 9. Kuratowski, C., 1952, Topologie II, Warszawa. 10. Lebesgue, H., Sur les correspondanees entre les points de deux espaces, Fund. Math., 2, 259-261. 11. Lelek, A., 1961, Sur deux genres d'espaces eomplets, Coll. Math. VII, 31-34. 12. Reichaw (Reichbach), M., A note on absolute G6-spaces, Prac. Amer. Math. Sac. (in press). 13. Stone, M. H., 1937, Applications of the theory of Boolean rings to general topology Trans. Amer. Math. Sac., 41, 375-481. 14. Urysohn, P. S., 1951, Trudy po topologii i drugim obtastyam matematiki, Vol. I. 15. Wallman, H., 1941, Lattices and topological spa~es, Ann. Math. (2), 42, 687-697. 16. Whyburn, G. T., 1958, Topological analysis, Princeton Univ. Press. TECHNION----ISRAEL INSTITUTE OF TECHNOLOGY,

HAIFA

ON A PROBLEM OF NACHBIN CONCERNING EXTENSION OF OPERATORS BY

JORAM

LINDENSTRAUSS

1

ABSTRACT

Problems I and II, stated below, are considered. It is shown that the answer to I may be negative even if X and Z are finite--dimensional and that the answer to II may be negative even if X and Z are separable and T compact. Concerning problem II some positive results are also obtained. For example, the answer to II is in the affirmative if Xis a conjugate space or an L1 space or if X = c or co and Z is separable. 1. Introduction. In the present note we are concerned with problem (6) of Nachbin [11]. We found it convenient to divide the problem into the following two parts.

I. Let Z, W and X be Banach spaces with Z D W and d i m Z / W = 2. Let T be an operator f r o m W into X . Suppose that f o r every Y with Z D Y D W and dim Y / W = 1 there is a norm preserving extension of T f r o m Y into X . Does T have a norm preserving extension f r o m Z into X? II. Let Z, W and X be Banach spaces with Z ~ W and dim Z / W = 00. Let T be a bounded linear operator f r o m W into X . Suppose that f o r every Y with Z ~ Y ~ W and dim Y / W < ov there is a norm preserving extension of T f r o m Y into X . Does T have a norm preserving extension f r o m Z into X ? In Section 2 we make some simple observations which show that the answer to both questions is, in general, in the negative. We shall see, however that in many situations the answer to problem II is in the affirmative. In Section 3 we show that the answer to II may be negative even if the spaces Z and X are separable and T is compact. The construction used in this section is similar to that used in [8] for giving a counterexample to a question closely related to I. I wish to express my thanks to Professor S. Kakutani for many valuable discussions concerning the subject of this note. NOTATIONS. All operators are assumed to be linear and bounded. All Banach spaces are assumed to be over the reals (this is only a matter of convenience, all the results proved here hold also in the complex case). The unit cell {x; l] x 11< 1} o f a Banach space X is denoted by Sx. Our notation for special spaces as LI, C(K), Received June 6, 1963 (1) Research supported in part by NSF Grant no. 25222. 75

76

JORAM LINDENSTRAUSS

[June

m and Co is standard. A Banaeh space X is called a ~3x space if from every Z D X there is a projection onto X of norm < ;t. For the basic facts concerning ~a spaces we refer to the book of Day [1, pp. 94-96]. A Banach space X is called an Ea space if from every Z D X with dim Z / X = 1 there is a projection onto X of norm < 2 (Grtinbaum [4]). The projection constant ~ ( X ) of X is defined by ~ ( X ) = inf {2; X is a ~

space}.

In a similar manner the expansion constant E(X) of X is defined. The projection constant (or expansion constant) is said to be exact if the inf appearing in its definition is attained. We say that the Banach space X has the metric approximation property if for every compact set K ~ X and every s > 0 there is an operator T from X into itself with a finite-dimensional range satisfying II TII = 1 and 11Zx- x II for x ~ K. This notion was introduced by Grothendieck [2]. It is not known whether there exists a Banach space which does not have this property. Let f be a functional defined on Y and let X c Y. The restriction o f f to X is denoted by.[ Ix. Similarly we denote restrictions of operators. 2. We begin with some positive results PROPOSITION 1. The answer to problem II is in the affirmative if X is a conjugate Banach space. Proof. This is a simple consequence of Tychonoff's theorem and the w* compactness of the unit cell of X. The details of the proof are identical with those given in the proof of (4) ~(9) in Theorem 2.2 of [5]. COROLLARY 1. The answer to problem II is in the affirmative if there is a conjugate space V ~ X and a projection P of norm l from V onto X. Proof. We extend T first in a norm preserving manner to an operator from Z into V and then apply P. REMARK. If X is an L 1 space then, as well known, there is a projection P with norm 1 from X** onto X, and hence the answer to problem II is in the affirmative for such X. It is easy to see that in general if X is a Banach space and if there is a conjugate space V D X from which there is a projection onto X with norm 2 then there is also a projection with norm < 2 from X** onto X. As we shall see in the next section the answer to problem II may be negative even for compact T. However, it follows easily from Proposition 1 that for compact T a slightly weaker version of I! has an affirmative answer. COROLLARY 2. Let the assumptions on Z, W, X and T be as in problem II. Suppose further that T is compact and that X has the metric approximation property. Then for every s > 0 there is an operator i" from Z into X with Tll=l]T[land~

T I w - TI[ --< s"

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ON A PROBLEM OF NACHBIN

77

Proof. Since T is a compact and X has the metric approximation property there is an operator To from X into itself having a finite-dimensional range B such that = 1 and [[ T o T ~. By Proposition 1 T o T h a s an extension ~ f r o m Z into B with ~ --
IITo II

zll---<

LEMMA 1. Let Z = W be Banach spaces with norm II II and unit cells S z and Sw respectively. Let 2 > 1. We define in Z a new norm Ill III by taking as its unit cell the closed convex hull of 2 - 1 S z u S w . Then

(a) xlllzlll____X (b) lllwlll=llwll

zll_>_lllzlll,

z ,

(c) Let Y satisfy Z ~ Y ~ W and let P be a projection f r o m Y onto W. T h e n

IllPll I =

1

if and

only

if fl P ll <__x(3).

LEMMA 2. Let W be a Banach space and let 2 > 1. There exists a Banach space X ~ W having the following property: Let Z be any Banach space containing IV. There is a projection of norm < 2 f r o m Z onto W if and only i f the identity operator f r o m W into X has a norm preserving extension f r o m Z into X .

Proof. Let V be a ~ 1 space containing a subspace isometric to ~, and let To be an isometry f r o m W into V. Let X = V ~ W where the n o r m is defined by Ii (v, w) ll -- max ( l] Let 7"1 be the identity operator of W and let T from W into X be defined by Tw = (To w, Tlw). T is an isometry. Let now Z be any space containing IV. Since V is a ~ t space To has a n o r m preserving extension from Z into V. 7"1 has an extension with n o r m ~/from Z into W if and only if there is a projection of n o r m ~/from Z onto W. Therefore T has a n o r m preserving extension from Z into X if and only if there is a projection of n o r m < 2 from Z onto W. This concludes the p r o o f of the lemma (we identify W with the subspace T W of X). Let W be a Banach space such that ~3(W) > E ( W ) . It m a y happen that from every Z ~ W with d i m Z / W = 2 there is a projection with n o r m < E ( W ) onto W (take for example W --- Co). However this seems to be an exceptional case. We do not know of general results in this direction but it is easy to give examples. F o r instance it is not difficult to construct a 4-dimensional space Z containing the 2-dimensional inner product space W such that there is no projection from Z

v][, Ilw U/x).

(2) Actually, since [[ToTI[may be smaller than IITI],we apply here the following version of Proposition 1. Let Z ~ W and X be Banach spaces with X being a conjugate space. Let T be an operator from W into X and let 2 _> IITIIbe given. If for every Y with Z ~ Y ~ W and dim Y/W < ~ there is an extension of T from Yinto Xwith norm =<2, then there is also an extension of T from Zinto Xwith norm __<2. The proof of this assertion is the same as that of Proposition 1. (3) Ill PHI = sublll ey III taken over a l l y with IIIy Iii = 1.

78

JORAM LINDENSTRAUSS

[June

onto W with norm < E(W) ( = 2/~/3, cf. Griinbaum [3], [4]).Put 2o = inf{ IIP II; P is a projection from Z onto W}. By Lemmas 1 and 2 (with 2 satisfying E(W) =< 2
1963]

ON A PROBLEM OF NACHBIN

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l f r o m Y into X whose restriction to W is the identity then the same is true f o r every Y ~ W (without any restriction on Y / W ) .

Proof. By Theorem 3 of [9] it follows that the assumptions in Proposition 2 (even if we consider only Y D W with dim Y / W = 1) imply that W is finite-dimensional and its unit cell is a polyhedron. Hence there is a finite-dimensional ~1 space Yo containing W. Let TObe an operator with norm 1 from ¥o into X whose restriction to W is the identity. Let Y be any space containing W. There is an operator Tt with norm 1 from Y into Yo whose restriction to W is the identity. T = ToTI is the required operator from Y into X. REMARKS. (a). Example (ii) above shows that Proposition 2 no longer holds if we require only that for every Y-~ W with dim Y / W = 1 there is an operator with norm 1 from Y to X whose restriction to W is the identity. (b) Lemma 2 and the result of Sobczyk mentioned above show that we cannot discard the requirement on Sx* in the statement of Proposition 2. 3. The counterexamples to problem II given in the previous section were based on the theorem of Sobczyk and therefore the space Z had to be non separable. We shall now construct an example in which all the spaces are separable. We introduce first some notations. As mentioned in the introduction we use a construction similar to that in [8]. The notations will be the same as in I-8] but the arguments used in the proofs and the purposes of the examples are quite different. Let /~ be the compact metric space of all the ordinals < 092 in the order topology(4). Let K,,, rn = 1, 2,... be the subset o f / ~ defined by (1)

Km = { ~ ; ( m - 1)co < ~ < mo~}.

Clearly K - { c o 2} = L)~=tK,,(5). Let N denote the set of positive integers. Let h(0t) be the function on /~ defined by S

1 if ot = moo + 2j - 1,

rn = 0,1, 2, ..., j = 1, 2, ..-

h(ct) i otherwise.

-

Further letf,, n ~ N be a sequence of continuous functions on J~ defined by -1 (3)

f,(0t) =

if~K2m_ t

m=l,2,..-,n

1 otherwise.

Let V be the space o f all the bounded real-valued functions on (the abstract set) /~ x N, with the usual vector operations and with the sup as norm. Let Xo be the (4) ~odenotes, as usual, the ordinal number of the well-ordered set of the integers. (5) {o~2}denotes the set consisting of the single point to2. We do not consider here 0 as an ordinal number.

80

JORAM LINDENSTRAUSS

[June

subspace of V consisting of all the functions v satisfying v(~, n) = v(~t, 1) for every E/~ and n s N, and v(~, 1) e C(/~). The mapping To from Xo onto C( /~) defined by (4)

Tox(~ ) = x(~, 1)

x e Xo,

~t ~ / ~

is clearly an isometry. Let Zo be the closed subspace of V spanned by Xo and the functions (5)

Zo(~, n) = f.(~)

~ E g,

n~N

and (6)

zk(~,n)=~k,,h(~ )

~eF., HEN, k=l,2,...(6)

With these notations we have the following LEMMA 3. (a) T h e r e is no projection f r o m Z o onto X o with n o r m < 5•4, (b) For every Y with Z o D Y ~ X o and d i m Y / X o < ~ , and for] every > 0 there is a projection o f norm < 1 + e f r o m Y onto X o.

Proof. (a) Let P be a projection from Z o onto X o with II P H = 4. Let g , , m = 1,2,... be the characteristic function of the set K2m x N . g m e X o and II2 g . - zo II = 1 for every m. Hence I12g.- Pzo II --<2 and thus Pzo(~t,n ) >=2 - 2 for 0t LJm = 1 K2m and n s N. By continuity of Pzo(~, 1) we obtain (7)

Pzo(092,1) > 2 - 4.

Let now gmj, m , j = 1,2,... denote the characteristic function of the set { m - 1 09 + j} x N. All the gmj belong to Xo and we have

[Izo+zl+z~+...+z.+2g~.+l,2~ll=2

m , j = 1,2,.-.

Hence ( P z o + z 1 + ... + Zm)(a,n) < - 2 + 24 for ~t = 2m09 + 2j,

and by continuity (8)

P ( z o + z 1 + ... + z,,)((2m + 2)09, I) -<_ - 2 + 24.

With the same g m j as above we have also tIZ1 + Z 2 + " "

+Zm--gk,2j+ll] = 1 ,

m,j,k=l,2,...

As above, we obtain from this that (9)

e ( z l + z2 + "" + z,,)(k09,1) > 1 - 2

(~) ~. k = 1 if n = k and 0 otherwise.

k , m = 1,2,...

1963]

ON A PROBLEM OF NACHBIN

81

By (8) and (9) Pzo((2m+l)og, l ) < - 3 + 32 for every m and hence Pzo(CO2,1) < - 3 + 32. This together with (7) implies that 2 ~ 5/4. We turn to the proof of (b). Assume first that Y is the span of Xo and k for some finite k. The operator Tkz(~) = z(~t, k + 1) maps Y into C(/~) and is clearly a norm preserving extension of To. Hence there is a projection of norm 1 from Y onto Xo. Let now Y be a general subspace of Zo containing Xo as a subspace of finite deficiency. That is Y = sp {X o, bl, "..bk} with k = dim Y[Xo and bi ~ Zo. There is an M < ~ such that k

z 12,1

k

Mltx ÷ •2,b,l[,

i=1

xsXo,2ireal.

i=l

Let b~, i = 1,..-, k be in the dense subspace of Zo spanned (linearly not topologically) by X and {zj}j°°_-o, such that Hb, - bi I[ < e/M. By what we have already shown there is a projection P of norm 1 from the subspace of Zo spanned by Xo and {~/}~=1 onto Xo. Define now P from Y onto Xo by

P(x + gi2ibi) = P(x + Ei2i[~i). We have(7)

IIP(x + ~iAib,)II =< IIx ÷

I1 :-<

< I x + ~,2,bifl + Z,12,1/M Z(1 ÷ )llx ÷ ~,2,b, II • Hence P is a projection of norm < 1 + e. REMARKS. 1. The question whether and when we can take 8 = 0 in (b) was treated in I-81. 2. It is clear that a construction similar to that done in Lemma 3 can be done for every compact metric K with (K')' # ¢(s). If we use for these K exactly the same construction (with the obvious modification obtained by replacing the characteristic functions of the sets Km by suitable Urysolm functions) we will get of course the same constant (that is 5/4) in (a). It seems likely that if we consider the spaces of ordinals < cok it is possible to construct similar examples with 5/4 replaced by a number Tk tending to ~ with k, and thus by taking direct sums we would get an example in which 5/4 can be replaced by ~ (i.e. in which there is no bounded projection at all from Zo into Xo). We have, however, not worked out the details of such constructions. Combining Lemma 3 (cf. also remark 2) with Lemma 1 we get (7) This inequality shows also that P is well defined, i.e. that no non trivial combination of the b~ belongs to X0. (s) K' denotes the set of limiting points of K.

82

JORAM LINDENSTRAUSS

[June

PROPOSITION 3. There is a 2 > 1 such that f o r every compact metric K with ( K ' ) ' ~ ¢ there exists a separable Z ~ C(K) satisfying (a) There is no projection of norm < 2 f r o m Z onto C(K). (b) F r o m every Y with Z ~ Y ~ C(K) and dim Y / C ( K ) < oo there is a projection of norm l f r o m Y onto C(K). Clearly, every ;t < 5/4 will do. Proposition 3 does not hold if ( K ' ) ' = ¢. Indeed, we have PROPOSITION 4. In problem II let X = C(K) with K compact metric. The answer to the problem is in the affirmative f o r every separable Z and W and f o r every 7" i f and only i f ( K ' ) ' = ¢. Proof. That the answer to II may be negative even for separable Z if ( K ' ) ' ~ ¢ follows from Proposition 3. That the answer is always in the affirmative if ( K ' ) ' = ¢ is is an easy consequence of LEMMA 4. Let Z D W be separable Banach spaces and let {f.}~=, be a w* convergent sequence in Sw*. Suppose that f o r every Y with Z ~ Y ~ W and f ~,, ~oo dim Y / W < oo there is a w* convergent sequence ty.s~=t in S t * such that , , oo Ynlw = f . f o r every n. Then there is a w* convergent sequence {z.}~= i in Sz* such that z.lw * = fn f o r every n. Proof. Let Z=sp(W,{z~}~°=l) and put Ym=sp(W,{zi}i~=~), m = l , 2 , . . . . Let {Ym,,}~=l . oo s Sr'., be a w* convergent sequence (to y*, say) such that Ym.~*I v/ = f ~ . By the diagonal method we choose a sequence mj such that exists for every i(9), and call this limit z*(z~). Next we choose an lim~ -,o~y.,j(zi) * oo increasing sequence {nk}kcO= 1 of integers and a subsequence {m k}koo= 1 o f { m j}j=l (for simplicity of notation we do not add another index) such that

I y*.k(z~) I

for i < k

z*(z~)l < k -

* Z i) - y~,~(z,)l * Y,.k,.( < k-

1

for n > n k,

i < k.

Let now z.* be any norm preserving extension of Ymk,n * to Z if nk < n < = nk+l, k = 1, " ' " " It is easily verified that the sequence ~z*/~ 1 has the required properties. L nJn = We conclude this paper by showing that the answer to problem II may be negative even for operators with a finite-dimensional range (and hence, in particular, for compact operators). PROPOSITION 5. There exist separable Banach spaces Z, W and X Z D W, X ~ W and dim W = 2 such that

(9) Ymj (zt) ISdefined if mj >= i.

with

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ON A PROBLEM OF NACHBIN

83

(a) T h e r e is no operator with norm 1 f r o m Z into X whose restriction to W is the identity. (b) For every finite-dimensional Y with Z ~ Y ~ W there is an operator with norm l f r o m Y into X whose restriction to W is the identity. Proof. Let X = C [0, n/2] and let Z = X be the space constructed in Proposition 3. Let W be the subspace o f X spanned by wl(t) = cos t and w2(t) = sint. We prove (b) first. Let Y be finite-dimensional with Z = Y = W. Let Yo be the subspace o f Z spanned by X and Y. By Proposition 3 (b) there is a projection o f n o r m 1 f r o m Y0 onto X. The restriction o f this projection to Y has the required properties. P r o o f o f (a). F o r t ~ [ 0 , n / 2 ] let tkt E X* be defined by (or(x)= x(t), x ~ X . F o r every t there is a w t ~ W such that w t ( t ) = l and [ w t ( s ) [ < l for s ¢ t . It follows that ~bt is the unique n o r m preserving extension o f ~bt I w into X. Suppose there were an operator T with n o r m 1 from Z into X whose restriction to W is the identity. Let z* = T*(o t, t ~ [0, 7r/2]. We have I1z* 11-<- 1 and z* I w = (T*q~t) lw = ~bt I w (since T I w is the identity). It follows that zt*I x is a n o r m preserving extension o f ~b, I w and hence, as observed above, z* I x = 4~t • Let n o w x ~ X. Then T x ( t ) = ~bt(Tx) = Tft(x) = z * ( x ) = zt*l x(X) = ~,(x) = x(t). Thus T x = x, in other words T is a projection and this contradicts Proposition 3 (a).

REFERENCES 1. Day, M. M., 1958, Normed linear spaces, Springer Verlag, Berlin. 2. Grothendieck, A., 1955, Produits tensoriels topologiques et espaces nucl6aires, Mere. Amer. Math. Soc. No. 16. 3. Gfiinbaum, B., 1959, On some covering and intersection properties in Minkowski spaces, Pacific J. Math., 9 487-494. 4. Grfinbaum, B., 1960, Some applications of expansion constants, Pacific J. Math., 10, 193-201. 5. Lindenstrauss, J., Extension of compact operators I, Technical note no. 28. Jerusalem 1962. Submittod for publication to the Trans. Amer. Math. Soc. 6. Lindenstrauss, J., Extension of compact operators III, Technical note no. 32, Jerusalem 1962. Submitted/or publication to Trans. Amer. Math. Soc. 7. Lindenstrauss, J., On the extension of operators with range in a C(K) space, Proc. Amer. Math. Soc. (to appear) 8. Lindenstrauss, J., On projections with norm 1 - - an example, Proc. Amer. Math. Soc. (to appear). 9. Lindenstrauss, J., 1963 Some results on the extension of operators, Bull. Amer. Math. Soc., 69, 584-588.

JORAM LINDENSTRAUSS

84

10. Lindenstrauss, J., On the extension of operators with a finite-dimensionalrange. Submitted for publication to Illinois J. Math. 11. Nachbin, L., 1961, Some problems in extending and lifting linear transformations, Proc. International Symposium on Linear Spaces, Jerusalem Academic Press, Jerusalem and Pergamon Press, Oxford, 340-350. 12. Sobczyk, A., 1941, Projection of the space m on its subspace co, Bull. Amer. Math. Soc. 47, 938-947.

YALE UNIVERSITY, NEW HAVEN, CONN.

REAL INVERSION A N D JUMP FORMULAE FOR THE LAPLACE TRANSFORM. PART I. BY

Z. DITZIAN 1 AND A. JAKIMOVSKI2 ABSTRACT

Generalizations of the Laplace asymptotic method are obtained and real inversion formulae of the Post-Widder type for the Laplace transform are generalized. §1. Introduction. In §2 of this paper we shall obtain generalizations of what are known as real inversion and jump formulae of the Post-Widder type for the Laplace transform. From our generalized inversion formula and certain heuristic considerations we shall obtain some new jump formulae. Generalizations of the Laplace asymptotic method necessary for proving the results of §2 are stated and proved in §3.

§2. Generalization of Post-Widder inversion and jump formula. Here the Laplace transformf(s) of the function q~(t) is defined by

(1.2)

J(s) = fo °° e-~' qS(t)dt- lim ~R e-S'qS(t) at R -* oo d O

where tk(t)~ L.(0, R) for each R > 0 and the right hand side is supposed convergent for some finite complex s. The Laplace-Stieltjes transform of the function ~(t) is defined by (2.2)

f(s)=fo

e-:td~(t) =- lira (R e - S'd~( t ) R~oo JO

where ct(t) is of bounded variation in the interval [0,R] for each R > 0, ct(0) = 0, ct(t) = 1/2(ct(t + ) + ct(t - ) ) and the right hand side of (2.2) is supposed convergent for some finite complex s. Received June 9, 1963 (1) This paper is to be a part of the first author's Ph.D. thesis written under the direction of the second author at The Hebrew University of Jerusalem. (2) The participation of the second author in this paper has been sponsored in part by the Air Force Office of Scientific Research OAR through the European Office,Aerospace Research, United States Air Force. 85

86

z. DITZIAN AND A. JAKIMOVSKI

[June

Here ~b(x _+) are defined by (3.2)

lim ~b(x -4-_h) - ~b(x _ )

h--*0

if these limits exist. tk(x _ 0) will denote the numbers for which (4.2)

f ~ [~b(x _+ y) - ~b(x ___0)] dy = o(h) as h +0

if such numbers exist. If ~b(x + 0) = ff(x - 0) then x is a Lebesgue point of ~b(u). Post [5] obtained a real inversion formula. Widder [6] and also Feller [2], Dubourdieu [1], and especially Pollard ([3] and [4])generalized Post's result. The final result obtained by Pollard is the following (see [5], Theorem 1.1, second part): THEOREM A. Suppose f(s) is a Laplace transform and both ~(t-t-O) exist; let {ak}, k > 1, be any sequence satisfying ak=O(kll2), k--* oo; then

0,]

+'

Pollard ([5], Theorem 1.1) stated that if in addition to the hypotheses of Theorem A we assume qS(t + 0) = tk(t - 0), then (5.2) is true for any sequence {ak} satisfying ak = o(k), k ~ oo. We could not prove this last result and shall show by an example that one of the main steps in his proof is incorrect. The step in question (see [5] top of page 449) is that, for 0 < 6 < 1 and Ok = o(k), k ~ 0% (5a.2)

['

]

d k [ ( k + Ok)

Pk(u)du <=fo -6 Pg(u)du

and

fk;(k+Ok)Pk(u)du< ff

where

( )d

Pk u u

-~-

+0

Pk(U) - (k(k+ --Ok)k+ e-(k+O~)"Uk-l(1 -- U) (1

k

u ~.

We shall show that these two inequalities are not true if, for example, x/ k = O(Ok), k ~ 0o. For k > ko (since 0k = o(k)) we have

k~ +Ok u l

O<(1-u)(1
2

< 6 , for 0 < _ u < _ l - 6 < l .

) =l-

UZ (1 + kkO-----L) u + k+Ok ,~

THE LAPLACE TRANSFORM. PART I

1963]

Hence, for k > ko,

fo 1-°ek(u)du <=6 (k-(k---i~i + Ok)k+l~o*-ae-(k+°~)"u*-i du The integrand in the last integral is increasing from u = 0 to u-

k-1 --(-}1 k + Ok

as k ~ m ) .

Therefore for k > k I > ko 1-~

~0

Pt(u)du < 6(1 - ,5)k (k (+k -Ok)1)! k+l e-(l-O)(k+O~)

(and by Stirling's formula) ,-, Aexp

{

-klog

~+(k+l)log(k+0t)-(1-a)(k+0k)

= Aexp

1 - k

log _----Z1 ~ - 5 + o(1)

+ -~-logk

(and since 5 < log 1/(1 - 3) for 0 < 5 < 1) - - * 0 a s k---* oo.

Hence 1 -(~

(5b.2)

Pt(u)du = 0

lim k -"~ co

Now

Hence

(5c.2)

fk t Ok)k+ 1 1" I d {uke_(k+Ok)u} du /(k+Ok) Pk(u)du _ ( k + .. -Jk/(k+O~,~l -- U) ~U kte -k (k + 0,) ~+* f ~ uke-(k+Ok), du k! Ok + k! JR/ck+Ok)

[1 + o(1)]Ok + 1£k+°~ vk e-~ dv. x/2=k

~.v

87

88

Z. D I T Z L A N A N D A. J A K I M O V S K I

[June

We have = < -~. o vk e-¢dv = 1.

v~e-Vdv If now x/k = o(0k)then I (5d.2)

lira J, k~o~

r'

as k - ,

and

P~(u)du

I=

I J kl(k+Ok)

O0.

Relations (5b.2) and (5d.2) are in contradiction with (5a.2). Similarly it can be shown that the second inequality used by Pollard is not true if ~/k = o(0k), k --* oo. Our first result is a generalization of Theorem A. In order to state our theorem we need the following notation. a) A sequence {ak}, k > 1, belongs to class A(~) for some real 2 if

b) A sequence {ak} belongs to class B if ak = o(k), k -.* m. c) A sequence {ak} belongs respectively to class B÷(B - ) if {ak} ~ B, I a~l k-1/2 "* m as k ~ co and for some k > ko, ak > 0(ak < 0). d) A sequence {ak} belongs to class A* if ak -- O ( ~ k ) as k ~ Qo. Denote by N(t) the normal distribution function, (6.2)

N(~)I--- - ~ f _ ~

e-"~/2du.

For k = 1, 2 ..... t > 0 and a given sequence {ak} the operator Lk,t,okIf(x)] is defined by (7.2,

L,.tak[f(s, ] -- ( k ~ ) ~ ( ~ - ~ ) ' + ~ f ( ~ ' ( ~ - ~ ) .

THEOREM 1.2. Suppose f(s) is the Laplace transform of c~(u). For fixed t > O, (i) I f {~k} ~ A(~) and both ~p(t +_O) exist then lira Lk,t,~ ~f(x)] = N(A)c~(t - 0) + (1 - N(2))c~(t + 0). k"* oO

(ii) I f {a~} ¢ B + and c~(t - ) exists then lira Lk.t.o~If(x)] = ~b(t - ). /¢~oo

(iii) I f {ak} ~ B - and dp(t + ) exists then lira Lk.t,o~I-f (x)] = ~b(t + ).

89

THE LAPLACE TRANSFORM. PART I

1963]

(iv) I f {ak} ~ B and c~(u) is continuous at a point t > 0 then lim Lk,,,a~[f(x)] = c~(t).

k"-~¢o

(v) / f {ak} eA* and both ck(t +_ O) exist and ?p(t + O) = c~(t - 0), then lim Lk,t,a~ If(x)] = ~b(t + 0).

k~co

We shall obtain the following two analogous results for the Laplace-Stieltjes transform. We begin by defining the operator Sk,t,,~: Given t > 0 and a sequence {ak}, k = 1,2, ..., we define Uk

Sk,t,a~[f(s)] =f(oo) + ( -- 1) k+l J(k+ak)l, f ~ f tk+t~ (u)-~. du

(8.2)

THEOREM 2.2. Suppose f(s) is the Laplace-Stieltjes transform of ~(t). Then

f N(2)~(t - ) + (1 - N(A))ct(t + ) if {ak) ~ A(2)

/

EB +

lim Sk,,.a~[f(s)] = I ct(t -- )

if {ak}

k-,~

if {ak} e B -

I ~(t + ) /

[ ~(t)

if ot(t - ) = ct(t + ) and {ak} e B.

THEOREM 3.2. If f (s) is the Laplace-Stieltjes transform of ~t(u), then f N(2)0t(t - ) + (1 -- N(2))~t(t + ) - ct(0+)if {ak} eA(2)

l o:(t ) o~(0+) lira f l Lk'"'ak[f(s)]du =

k-'~oo

/

~(t + ) - ~(0 + )

[ ~(t) -- ot(O + )

if {ak}

B+

if {ak} E B if ct(t -- ) = ~(t + ) and {ak} ~ B

If we choose in the last three theorems ~k = o(kl/2) k-~ oo we get Pollard's results [5]. Theorems 1.2, 2.2 and 3.2 will be proved in §4 by using improvements of the Laplace asymptotic method which are stated and proved in §3. These improvements are stated in a slightly more general form than is needed for proving the results of this paper. The more general form will be needed in a later paper. From Theorems 1.2, 2.2 and 3.2 we can deduce immediately the following trivial jump formulae. COROLLARY 1.2. Suppose f(s) is the Laplace transform of d~(u). I f f o r some t > 0 both ~(tl_+ 0) exist and (ak} ~A(2t) , (bk} ~ A(~,2) (21 :~ '~.2), then 1

N(;q)

-

lira {L~.t,b~If(s)] -- Lk,,.~ If(s)]) = ~(t + O) -- $(t -- 0). N(~2) k-~o

90

z. DITZIAN AND A. JAKIMOVSKI

[June

COROLLARY 2.2. Suppose f ( s ) is the Laplace transform of ¢(u). I f f o r some t > 0 both ¢(t + ) exist and {ak} ~ B +, {bk} e B - , then lim {Lk,,,b~[f(s)] -- Lk,, a~[f(s)]} = ¢(t + ) -- ¢(t -- ). k'-* oo

COROLLARY 3.2. Suppose f ( s ) is the Laplace-Stieltjes transform of a(u). For fixed t > 0 we have

(i) /f {ak} ~ A(,tl) and {b,} ~ A(22) (21 ~ ;t2), then 1

(a)

N(21) -- N(22) lim ( --

1) k+l ~{k+a.)/, U~

.

k ~ o~

.l (k + bk)/t

~ . . f (k+X)(u)du = a(t + ) -- a(t -- )

and r: lim J o {Lk'u'bk[f(x)] -- Lk .... k[f(x)]}du N(;q) - N(22) k-.co 1

(b)

= ~(t + ) - ~(t -

).

(ii) /f{ak}~B + and { b k } ~ B - , then (c)

lim {Sk,t,ok[f(x)] -- Sk,,,ak[f(x)]} = u(t + ) -- a(t -- ) &--*co

and (d)

lim (t{Lk,u,bk[f(x)] -- Lk,u,a~[f(x)]}du = ct(t + ) - a(t - ). k"* co d O

The following heuristic considerations lead to a new non-trivial jump formula. Choose in Corollary 1.2, b k - a k = (22 - 21) ~/E ¢(t + 0) - q~(t - 0) = lim N'2 )

1

.lim {Lk,t,b~ ' [f(x)] -- L,.t,ak[f(x)]}

(changing formally the order of the limits) =lim

lim

N(21)-N(;t2)

21-22

k!

_

~

(k)

} __

= l i m x/~-keX'/21im ~-.~o

"

1

al-.*= ( ' h - & ) x / f c

(-1)k k!

{ ( ~ + ~ _ ~ ) ' + ' ftk,

(~.~.)

1963] =lim k~

THE LAPLACE TRANSFORM. PART I

91

~r.~¢ '~ (-1)'+~ t, ,--v-,C ~+ ~I'~+'~,-T-,s(~'(~~--~ ~ ) -7

•

This suggests Theorem 4.2, which we shall prove formally below in ~4. THEOI~M 4.2. Suppose f(s) is the Laplace transform of d~(u) and that for some t > O, both q~(t ++_O) exist. Let {ak} e A(2), and Ok = O(X/k), k --* oo. Then

N/~k ( --~l1)k ~~

limk~oo

A - T1

{ ( - - --F - f -a-k) k k Jr Ok (k) ( ~ _ f ~ )

k_ _ a k

f(k+ 1)

Choosingin Theorem 4.2 2 = 0, a k

:

= q~(t + 0-- ~b(t- 0).

0 and Ok = 0 for k > 1we get

COROLLARY 4.2. Suppose f(s) is the Laplace transform of dp(u). If for some t > 0 both d?(t +_O) exist, then

= ~(t + 0) -

~b(t - 0 ) .

A formal computation similar to that used in obtaining Theorem 4.2 yields THEOREM 5.2. Suppose f(s) is the Laplace-Stieltjes transform of a(t) and let { ak} ~ A(2). Then lira ( - 1 ~-.~o

)~+i k! \ t = a(t + ) - - ~ ( t - - )

/

Choosing in Theorem 5.2 2 = 0, a~ = 0 for k = 1 we obtain the Widder jump formula (see [7] p. 298). In Theorems 4.2 and 5.2 {ak} e A()O, therefore a k = 0 ( x/ k) (k ~ oo). Hence

k

ak

e k~°g(l+("k/k)) ~ e"ke-1/2("~/k)

e"k-(~2/2)(k~ oo).

Therefore we may state the results of Theorem 4.2 and of Theorem 5.2 in, respectively, the following two equivalent forms:

= ~ ( t + 0 ) - q,(t - 0 )

lim e . + . ~ ( _ k-~oo

1)k f(k)(~-~)=~(t+)--¢t(t--).

92

z. DITZIAN AND A. JAKIMOVSKI

[June

§3. The Laplace asymptotic method. In this section we shall obtain and prove some results which are the basis for the p r o o f of our theorems in §2 and in part 2 of this paper. The results of this section are generalizations of the classic Laplace asymptotic method. TrmOREM 1.3. Let (i) a, b, rl, ~ and 2 be real numbers satisfying a < b, 6 > 0

andO= 1, is a sequence of real numbers satisfying g(k) ~ k, k ~ oo. Then 0.3)

lira

k-,® ~[ 2r(

e-g(k)h(")~/"~

Ja+mc.e(k)-1

e e(k)h(~)dx = 1 -- N(2~ x/--h"(a) )

Proof.

=

+ 27~

( ) dx - I k + J~ (say).

I. d a+ak . g ( k ) - I

+ql

By the arguments used in Widder 117]p. 278 we see that, for any 0 < r/t < t/, (2.3)

lira Jk = 0 k---~ oo

By (ii), for x e [-a - / 5 , a + th], we have h(x) - h(a) = [(x - a)2/2] h"(~(x)) where ~(x) e [x, a] if x < a or ~(x) ~ i'a, x] if x > a. Now

Ik =

~-kh"(a) 2•

f "+~l

J,+,k" g(k)-I exp

[

g(k) h"(~(x))

~ ]

dx.

For any fixed e satisfying - h"(a) > e > 0 we can define ~/1, 0 < rh < r/, and (51, 0 < 61 < 6 so that, for x~11a - 61,a + rh],

h"(a) - ~ < h"(x) < h"(a) + e.

(3.3) Denote, respectively, --

exp

2~ =

g(k)(h"(a) +_e) (x

,l a+ak" g(k) -1

~---k--~ h"(a) (N(r]lg(lc)t/2 ~ h"(a) +-------~e

x / - ( h " ( a ) ± e) )

-- N(akg(k) -1/2 x / - (h"(a) -T- e)))

dx

1963]

THE LAPLACE T R A N S F O R M . P A R T I

93

N(u) is a continuous function, and hence

(4.3)

k~limooIk~=

X/ h"(a)h"(a)4-e (1 - N(2 ~ / - (h"(a) 4- 8)).

By (2.3) and the definitions of Ik~and I k, Ik <=Ik <--I~. Hence by (4.3) and since e > 0 is arbitrary we get (5.3)

lim I k = 1 - N ( 2 ~ [ - h"(a) )

Our result follows from (2.3) and (5.3) combined.

Q.E.D.

COROLLARY 1.3. Let (i) a, b, ~l, t~ and 2 be real numbers satisfying a < b, 6 > 0 and O < ~l < b - a. (ii) h(x) ~ C2(a - ~1<- x <_ a + 6), h'(a) = 0, h"(a) < 0 and h(x) is nondecreasing in b <- x <_ a. (iii) {ak} and {g(k)} satisfy conditions (iii) and (iv) of Theorem 1.3. Then / - kh"(a)

(6.3)

lim

2~

k-*

I~a+ak'g(k)-t e-gtk)h(a)jb eg(k)h(x)dX = N(2 x/" - h"(a) )

ProoL The proof follows from Theorem 1.3 by substituting x = - z. THEOREM 2.3. Suppose (i) a, b, ~/, t~, 2, h(x), {ak} and {g(k)} satisfy conditions (i), (ii), (iii) and (iv) of Theorem 1.3. (ii) For the functions C~k(U),k)>=)l, defined on [a + akg(k)-l,b] and for some finite M we have ~(u du < M for k=>l +ak "glk)- t

(iii) lim k~oo~k(a + ak.g(k) -1 + O) exists and is equal to A. (iv) Denote ctk(x) -

[tkk(u) - q~k(a + a k g(k) -1 + 0)] du +a k " g ( k ) -

1

f o r k >=1 and x ~ [a + a k" g(k)- 2, b]. Suppose that for each el > 0 there exists a sufficiently small p(el), 0 < p(el) < (b - a)/2, such that for each k > 1 and all x satisfying 0 < x - (a + ak.g(k_6) < p(el) we have

I

<=

Ix -

(a + akg(k)-i)[.

Then

(7.3)

lira k ~

/

--

fb

kh"(a) e -g(k)h@) ~-~

dpk(x) e ~(k)h(~ dx

J~+~k" ~(k)- ~

= A(I - N ( 2 x ] ' - h"(a)).

94

z. DITZIAN AND A. JAKIMOVSKI

[June

Proof. In order to prove our theorem it is enough by (1.3) to show that (8.3)

lira ~/k

[4~k(x) - A] exp [g(k)(h(x) - h(a))] dx = 0.

k~oo

+ak "g(k)- 1

(1.3) yields by condition (iii) that (9.3)

lim ~/k k~oo

f;

[dA(a+ a k ' g ( k ) -~ + 0) -- A]"

+ak "g(k)- 1

" exp [g(k) (h(x) - h(a))] dx = O.

In order to prove (8.3) it is enough by (9.3) to show that Ik - ~fk

a+ak "g(k)- 1

[~k(X) -- q~k(a + ak" g ( k ) - 1 + 0)]" • exp [ g ( k ) ( h ( x ) - h(a))] dx = o(1) as k--, ~ .

Denote I k = lk,1 + Ik,2 where Ik, ~ and I,, 2 are respectively the integrals on the intervals [a + a k ' g ( k ) -1 ,a + ~h], [a + ~h,b]. Now by (ii) for each fixed ~h > 0 IIk.2 [ < X/-~ e x p [ g ( k ) ( h ( a + ~h) - h(a))]

- Ckk(a + a k ' g ( k ) -1+ 0)[ du = o(1) (k ~ oo).

Integration by parts yields Ik, I = ~fk exp [ g ( k ) ( h ( x ) - h(a))~k(x)]~+~'k, e(k)-' -.--

-

/~a+qt

~/k g(k)Ja+a k'e(~)-' ~ ( x ) h ' ( x ) e x p [g(k)(h(x) - h(a))] dx - Ikl 1 + Ik12

By (ii) and the fact Ctk(a + a k ' g ( k )- 1) = 0 we have Ikl 1 = o(1)~ k ~ ~ . For a given e I let p be that e~isting by (iv); let rh be the same as that in the proof of Theorem 1.3 and suppose also rh < p.

l I,,: [ <

kll2g(k)"

f

a+ffl

,,1 a+ak "g(k)- 1

I ~k(x) [ [ h'(x) I exp [g(k)(h(x)

- h(a))-l dx

(by (iv)) < ~1 kl/2" g(k)

fa+qt

(x - (a + a k g(k) - 1))1 h"(~(x))(x - a) I •

da+ak "g(k)- 1

•e x p [ g ( k ) h " ( , ( x ) )

(x-a)2-]dx

~a+rlt

<=~1 kI/2 " g ( k ) ( - h"(a) + e)

Ix - a ~ a+ak * g(k)- 1

II

- a - a k .g(k) -1 1"

"exp[g(k)(h~(a)+e)(x2a)2]dx

THE LAPLACE TRANSFORM. PART I

1963]

95

(and for Ml ~ -- h"(a) + 8) t~a+t/l

<--8tktla g(k) M tj .+o~.~,k)_, exp [ g ( k ) ( h " ( a ) + 8 ) ( x 2 a ) 2 ] , •

l ( x - a)2 + I x - a l l a k l g ( k ) - ~ ] d x

(and by the substitution u = x / - (g(k)/2)(h"(a) + e) ( x - a ) together with

lakl < ck'/2 <= c, g(k) ''2) < M,el

¢1)

e-U2u2du + M38t

ue-"2du _<-.exMa

By letting 81 ~0 we get Ikl~ = O(1) as k-* oo.

Q.E.D.

COROLLARY 2.3. Suppose (i) a, b, q, 6, 2, h(x), {ak} and {g(k)} satisfy conditions (i), (ii), (iii) and (iv) oJ Corollary 1.3. (ii) For the functions ~Pk(U) defined on [b,a + ak'g(k) -1] and for some finite M we have b"+"k'g~k'-~ 16~(x)ldx < M for k > 1. (iii) limk_.oo~ok(a + at.g(k) -1 - 0 ) exists and is equal to A. (iv) Denote

ek(x) = f x a +ak "glk)- 1 [q~k(U) -- Ckk(a + a~. g ( k ) - t _ 0)] dx for k > 1 and x e [ b , a + a,'g(k)-a]. Suppose that f o r each 81 > 0 there exists a sufficiently small P(el), 0 < p(el) < (a - b)/2 such that for each k >__1 and all x satisfying 0 < (a + ak.g(k) -1) -- x < p(el) we have

I k(x) I 8, Ix -

(a + a , . g ( k ) - ~ ) l "

Then (10.3)

lim k"* oo

~

2rr e-e(k)h(a)fo "+ak'~(il-' ~)k(X). egtk~ht~)dx - kh"(a) = AN(2\/-

h"(a)).

Proof. The same proof as that of Theorem 2.3 THEOREM 3.3. Suppose that (i) a, b, p, rl and 6 are real numbers satisD, ing p
96

z. DITZIAN AND A. JAKIMOVSKI

[June

(ii) h ( x ) e C 2 ( a - 6 < x < a + tl), h'(a) = O, h"(a) < O, h(x) is nondecreasing f o r x e [p, a] and nonincreasing for x e [a, b]. (iii) {Ok} ~ A*. (iv) {g(k)} satisfies condition (iv) of Theorem 1.3. (v) T h e f u n c t i o n s c~k(u)(k >= 1) are defined on p < x < b and f o r some finite M we have fpb I~k(u)]du < M

for

k >= 1.

(vi) T h e point a + a k ' g ( k ) - l is a Lebesgue limk_~(a + a k ' g ( k ) _ 6 ) exists and is equal to A. (vii) Deno te ak(X)==-

f;

point

of

q~k(U) and

[~k(U) -- ~k(a + a k • g(a)-l)] du

+ ak" glk)-

1

f o r k > 1 and p <_ x < b. Suppose that f o r each ~l > 0 there exists a sufficiently small P(~l), 0 < p(el) < Min ((a - p)/2, (b - a)/2) such that f o r k > 1 and all x satisfying] a + Ok" g(k) -1 -- x I < p(el) we have I k(x) 1 <= Ix - (a + ok. g(k)- * I Then

(! 1.3)

lira

J

- kh"(a)

-g(k)h(.)

f;

dPk(X)e ~(k)h(~) dx = A .

Proof. The same proof as that of Theorem 2.3 because I a k l <=C " k

1/2 .

TrmOREM 4.3. Suppose conditions (i), (ii), (iii) and (iv) of T h e o r e m 2.3 are satisfied. Then

(12.3)

lim - kh"(a) e-h'(a)A2]2 e -gt*)*t") k'-*oo

d?k(x). J a+ ak "g(k)-

1

• e g(*)htx)" (x -- a ) d x = A.

Proof. The argument used to prove Theorem 1.3 yields also (13.3)

fb lira k 3a k~oo

- 1 eh,,ta);t2/2 (X -- a) exp [g(k) (h(x) - h(a))] dx = h"(a---)+ a k " g(k) - 1

By (13.3), assumption (iii) of Theorem 2.3, and the arguments used in proving Theorem 2.3, it follows that it is enough to show that [~bk(X) -- ~k(a + ak" g(k) - 1 + O)](x - a)"

Ik ==-k + ak" g(k)- 1

• e x p [ g ( k ) ( h ( x ) - h ( a ) ) ] d x = o(1) as k-~ ~ .

1963]

97

THE LAPLACE TRANSFORM. PART I

Denote I k =-Ik~ + Ik~ where Ik, and Ik~ are respectively the integrals on [a + ak " g ( k ) - l , a + ~/~] and [a + r/~,b],ff, > 0. The argument used in the proof of Theorem 2.3 yields, since I C~k(X) -- ~k( a + ak" g ( k ) - ~ + 0) 11 a I is L integrable on [a + ak " g ( k ) - ~ , b], that Ik~ = O(1), k "-* oo. Integration by parts gives h(a))] ( x - a) a~+~' +ak "g(k)-

Iki = k a k ( x ) e x p [ g ( k ) ( h ( x ) -- k f ~ + ~

1

Ctk(x)exp[g(k)(h(x ) - h(a))]dx

.I a-l'ak " g ( k ) - 1

-- k g ( k ) f ~+~'

%(x)h'(x)(x - a)exp[g(k)(h(x) - h(a))]dx

J a+ak'g(k)-

1

~-- I k l l q - I k 1 2 + l k 1 3 ,

where Ctk(X) is defined by condition (iv) of Theorem 2.3. Clearly I k l l = O(1) as k --, oo. L e t 81 > O, P(el), e > 0, rh, 61 and ~(x) be the same as in the proof of Theorem 2.3, then for x ~ [a + a k • g ( k ) - 1, a + t/x], since h"(a) + 8 >= h"(~(x)), (14.3) e x p [ g ( k ) ( h ( x ) -

h(a))] = exp

oxp

g(k)h"(~(x)). ~

[

+

Let - L =- h"(a) + e. Then by (14.3) IIk12 [ < k81 f a+l/l

(x - a - a k • g(k) -1) E g(k)L(x-a)2/2 d x <=M181 .

da+ak "g(k)- 1

L e t h"(a) - ~ =-- - B, then, since - B < h"(~) < - L we have by (14.3)

[Ik13 [ < k g ( k ) 8 1 f a+~l

(x - a - ak " g ( k ) - l ) B ( x

--

a)2 e-g(~)'(~-°m2 dx

J a+ak • g(k)- 1

The substitution u = x/g(k)L/2 in the last integral together with the inequality I x - a - ak" g ( k ) - 11 ~ [ x - a I + I ak ]" g ( k ) - t and the fact lira k - ~ok - t g ( k ) = 1 gives

1I 131

fo uae-'2du+Maexla~lg(k) -112

tt 2 e - U2du

co

_-<M481. COROLLARY 4.3. S u p p o s e conditions (i), (ii), (iii) a n d (iv) o f C o r o l l a r y 2.3 are satisfied.

98

Z. DITZIAN AND A. JAKIMOVSKI

[June

Then

Proof. The same proof as that of Theorem 4.3. $4. Proof of the theorems of $2. In the proof of Theorem 1.2 we shall use the following result.

LEMMA 1.4. Suppose that the Laplace transform of 4 ( u ) exists and that a , = o(k) as k -t co. Then

kk+l +lim k!

S,+, e-kzz'+( m

+k

z -kr ) n,

~ Z = O

Proof. We shall prove (2,.4) only. The proof of (1.4) is similar. For real c and t > 0 let a(z) = a(z, c, t ) = f: e-'"4(utj d21. For any fixed t > 0 there exist constants c = c(t) and M = M(t) such that (3.4)

I a(z)

-

5 M M(t)

for z 2 0.

Define

Hence, by (3.4) (4.4)

Integration by parts in (2.4) yields

The maximum of u

. exp [ - ( k - c(k/( k + a,))) u ] is at u = (1 - (c/ ( k + a,)))- I .

THE LAPLACE TRANSFORM. PART I

1963]

99

Therefore, by (4.4), k

~i)~k k+,

for k > ko. The sequence { k / ( k + ak)} is bounded, and the argument used in proving Theorem 3a on page281 of Widder [7] (with t = 1, c - - 0 there) yields (2.4). Q.E.D. Proof of Theorem 1.2. Case (i). Let t > 0 and suppose that qS(t + 0) exist. (5.4)

1 Lk.,.ak[.[(x)] = ~

-

k ak - -

e

_(k+ak)u] t • u~/p(u)du

kk+l -~ {.(k+,,k)!k [.1+~ ~oo k, { f ' + Jl_~ +J(k+~)/k b J r ++, }

e-iZzk~ \ z K . e a k t / d z

=-- l_kl + lk 2 + Ik3 + lk4, {ak} ~ A(2), and so a~ = o(k), k --* oo. By Lemma 1.4 we have (6.4)

lim Ikl = lim I~4 = 0. k " * co

k~oo

In order to find the value of limk_.®Ika take in Theorem 2.3 h(z) = logz - z, g ( k ) = k , q S i ( z ) = d p ( z ( k / ( k + a k ) ) . t ) ( k > l ) , a = l and b = l + f . It can be verified that the functions ~bk(z)satisfy conditions (ii), (iii) and(iv) of Theorem 2.3. Hence, by Theorem 2.3 and Stirling's formula, we get (7.4)

lim Ika = (1 -- N(2))qb (t + 0). k--~ oo

In the same way, using Corollary 2.3 instead of Theorem 2.3, we get (8.4)

lim Ik2 = N(2) q5(t - 0) . k---~ co

Combining (7.4) and (8.4) we obtain the proof for case (i) of Theorem 1.2. Proof of Theorem 1.2. Case (v); The proof of this case is the same as that of case (i) but here we use Theorem 3.3 instead of Theorem 2.3. Proof of Theorem 1.2. Case (ii), Here {ak}6B +. Let t > O and suppose ~b(t - ) exists. For A > 0 and a sequence {bk} 6 A(A) we have, for k > ko,

kk+l fl(k+,,k)/k_k~zl, dz kk+t f(k+bk)/k = e - kZzkdz. 1 >= - - k! _~ e > k! " j ~_~ Therefore by (8.4), with the function ~b(t)~ l, we have

" . ak ke 1> limft~; -kZzkdz>N(2) k--* co

100

z. DITZIAN AND A. JAKIMOVSKI

[June

Now lim~too N(2) = 1, therefore (9.4)

lim f(k+ak)/ke--kzzkdz=. 1. k--~

,d I --di

For a given ~ > 0 there exists ~5-= tS(e), 0 < 6 < 1 and a constant kl - kl(el, {ak}) such that

tiP(

(10.4)

tz ) - ~ b ( t - ) < t for 1 - t S < z < ~

k

k-I'ak and k > k 1.

ak

=

Let Ikj(1 < i < 4) be the same as in (5.4). By (10.4) we have lim I Ik2

- )1 ____ lim ~

-

k~ao

dp I k + a k ] - dp(t - ) ] e-kZzkdz < t •

Letting e ~ 0 we get lira lk2 = O(t -- ). k~oo

(11.4)

The argument used in obtaining (4.4) shows that for fixed t > 0 there exist constants c = c(t) and MI (t) such that (12.4)

lilk(z)l------I k+okfxp--C -- -- kU

C~ - - ~ - -ak

du l < M

for k > ko • The only maximum of zkexp { -- k(1 - (c/(k + ak)))Z } is at z .~- (k -t- ak)" ( k + a k - c)- 1.

Also, limk-.~ ak = + OO, since {ak} ~ B +, and so (13.4)

for k > kl > ko.

k + ak ]£ + ak ~>k+ak--C

=

=

Integrating by parts the integral defining Ik3 in (5.4) we get

Ik3 = kk+t k! e - k O + ' ~ ' ( l + 3 ) k i l k ( l + 6)exp { k ~ a k C ( 1 + 6 ) } k!

|

ilk(Z) d z k exp

k

d(k +ak)/k

1

k + ak

The argument used in proving Lemma 1.4 yields now by (12.4) and (13.4)

(14.4)

lim ]k3 k~oo

=

0

.

1963]

THE LAPLACE TRANSFORM. PART I

l 01

By Lemma 1.4 we have lim k-,oo Ikl = lim k-~® Ik2 = 0. Combining the last result with (11.4) and (14.4) we get the proof of case (ii) of Theorem 1.2. Proof of Theorem. 1.2. Case (iii). q-he preof of this case is similar to the proof of case (ii) of Theorem 1.2. Proof of Theorem 1.2. Case (iv). It is known (and it follows from (9.4) too) that, for 0 < fi < 1,

k k+l I t ";~ lim ~ e-k"zkdz = 1. Since ¢(u) is continuous for u = t > 0, therefore for each e > 0 here is a t~ ~ 6(e), 0 < 6 < 1. such that

,z)

for 1

Let I~j (0 < j < 4) be defined as in (5.4). Then

li--~ Ilk2 + Ik3 -- ¢(t) l k"* oo

< lim e-kZz k k-'~ --~T" a~-~

¢

-~-------'tz + ak

--¢(t)

dz<e. --

Letting e~0 we get l i m k ~ o { I ~ 2 + I k a } = ¢ ( t ) . By Lemma 1.4 we have l i m k . ~ lkt = limk-.oolk4 = 0. This completes the proof of the case (iv) of Theorem 1.2. Proof of Theorem 2.2. By supposition ~(0) = 0. It is known (Widder [7]) that if f (x) is the Laplace-Stieltjes transform of a(t) then .](x)/x is the Laplace transform of a(t). If f (x) is the Laplace-Stieltjes transform of a(t) them (See [7], p. 294),

Substituting x = (k + ak)/t we get by (7.2) and (8.2)

The proof follows now from Theorem 1.2 since ~.(0)= 0, f ( x ) / x is the Laplace transform of a(t) and for each t > 0 ct(t +)exists, therefore ~(t + 0) exists and respectively ~(t ___0) = ~(t +). Q.E.D.

102

Z. DITZIAN AND A. JAKIMOVSKI

[June

Proof of Theorem 3.2. The argument used by Widder (see [7] p. 291-) yields for0
-

Y/

Lk,u.ak [jr(S)] du

k+ak(k~a----~k)

-k!

fJ° ~

e -(k+°")y/' yk-lot(y)dy

fo

e-,k+akl~lryk-

lct(y)dy =_ Jk,t -- J k,r

Now, Theorem 1.2 yields for a fixed t > 0 (because if {ok} belongs to any one of the classes A(2), B ± and B then {ak + 1} belongs to the same class and (k + a~)/k (k + ak)/k ~ 1 as k ---, oo) f N(2)ct(t - ) + (1 - N(2))0t(t + ) if {ak} eA(2) / (16.4)

lim JR., = 1. ct(t -- ) k-, ~o

[ ~t(t + )

Le(t) (17.4)

Jk,

if {ak} ~ B + if {ak} E B if ct(t + ) = ~t(t - ) and {ak} E B

k + a k C t ( 0 + ) k +=a k- ( ~- - - ~ ) k k! . y ~- 1 [ e ( y ) _ e ( 0 + )] dy

k

fo ~ e- (~+ok)y/,

k ---, oo,

N o w the argument used in [7] at the foot of page 291 and the top of 292 yields (18.4)

lim Jk, = at(0 + ) • r,o "

k+ak k

If {ak} belongs to any one of the classes A(2), B + and B then (k + a k ) / k ~ I. Hence by (18.4) lira lim Jk, = ~(0 + )

(19.4)

k-+oo r~O

'

The proof follows now by combining (15.4), (16.4) and (19.4). Proof of Theorem 4.2.

For a fixed t > 0 and {ak} ~ A(2), denote

~-. ( - 1)k+ , { ( k + a a Ik ~- e a2/2 X/'2nT-

By Theorem 1.2 Case (i) we get

lim~(~-~)l'f(k)( k-.~

k+ak ---7--)

=0.

1963]

THE LAPLACE TRANSFORM. PART I

103

Comparing the definition of Ik and the result of our theorem we see that in order to prove our theorem it is enough to show that lim Ik = ~(t + O) -- ~b(t -- 0).

(20.4)

k " * o0

We have (21.4)

_T__k yafo~{

l k = ~ / 2 X/k2rC--k( k +

kukq_ 1(

~ k + a k ) uk+l} .

kk+2 . e-~+,~.,4(uldu

=

e"~/~ , j ~ , o ~

k!

_ _ .

k+ ak

[ tkz e - k ~ z k ( z - 1)q~ i dz \ k + ak ]

Ik,1 q- Ik,2 q-lk,3 + lk,4. The arguments used in proving Lemma 1.4 yield here (22.4)

lira

Ikl =

k'-* oo

lim lk4 = 0. k~co

In order to estimate Ik, 3 we substitute in Theorem 4.3 a = 1, b = 1 + 6, h(z) = - z + logz, g(k) -- k and q~k(z) -- ~(tkz/( k + ak)) and get byTheorem 4.3 (23.4)

lim Ika = q~(t + 0). k---~ ou

In the same way, but using Corollary 4.3 instead of Theorem 4.3 we get lira lk2 = -- qb(t -- 0).

(24.4)

Combining (21.4), (22.4), (23.4)and (24.4) we get (20.4)and this completes the proof of our theorem. Proof of Theorem 5.2. For a fixed t > 0 and {ak} ~ A().) denote

k + a~ k ( - 1) k ~, ( k + a~] lk =-- eZ212~2-~ ( - - - t - - ) - - - k T f , t ] eZ2f2 x/2-~ ( k + ak]kf~e_tk~_,~.,ukda(t) =

k~

\----i--]jo

'~---- eZ2/2X/~k, \~---,(kq-ak) kdOf°°e -(k+akI"/t(kuk-l--(~)Uk)~t(u)du =

k!

kk+l

e-k~'z~'-l(1 -- z)~ \ k + a~]

"

The argument used in proving Theorem 4.2, but taking here (1/z)a(kzt/(k + ak) ) instead of d?(kzt/(k + ak)) there completes the proof.

104

Z. DITZIAN AND A. JAKIMOVSKI

[June

Added in proof. Theorem 2 of the paper by L. C. Hsu "Generalized StieltjesPost Inversion formula for Integral transforms involving a parameter," Amer. J. Math., 73 (1951), 199-210, is Pollard's Theorem 1.1 of [4]. As we have shown in §2 there is an incorrect step in Pollard's proof. There is also an incorrect step in Hsu's proof of his Lemma 2 which is used in proving his Theorem 2. Hsu proves on page 204 that if a sequence {x.} of Lebesgue's points of a function f(x) converges to a Lebesgue's point x of the same function then lim.-,oof(x,) = f(x) The following example shows that this is not true. Define f(x) in [ - 1 , ¼ ] by f(x)=O for - l < x < O , f ( x ) = l for 4 - " - 4 - " / 2 n < x < 4 - " (n>=l) and f ( x ) = 0 in all remaining points of [ - 1 , ¼ ] . The points x , = 4 - " - 4 - ( " + ~ ) / n and x = 0 for f(x). = 1, f(0) = 0 are Lebesgue's points of f(x). But x. --, 0 and f ( x , ) - - * 1 v~ 0 = f(O). BIBLIOGRAPHY 1. Dubourdieu, M. J., 1939, Sur un th6or~me de M. S. Bornstein relatifh la transformation de Laplace-Stioltjes, Compositio Matematica, 17, 96-111. 2. Foiler, W., 1939, Completely monotone function and sequences, Duke Mathematical $ournal, 5, 662--663. 3. Pollard, H., 1940, Note on the inversion of the Laplace integral, Duke Mathematical Journal, 6, 420-424 4. Pollard, H., 1940, Real inversion formulas for Laplace integrals, Duke Mathematical Journal, 7, 445-452. 5. Post, E. L., 1930, Generalized differentiation. Trans. Amer. Math. See., 32,723-781. 6. Widder, D.V., 1934, The inversion of the Laplace integral and related moment probioms, Trans. Amer. Math. Soc., 36, 107-200. 7. Widder, D. V., 1941, The Laplace transform, Princeton University Press. "raE HEnR~w UNIVERSITY OF JERUSALEM

A PROOF OF DILWORTH'S DECOMPOSITION THEOREM FOR PARTIALLY ORDERED SETS BY

MICHA A. PERLES ABSTRACT

A short proof of the following theorem is given: Let P be a finite partially ordered set. If the maximal number of elements in an independent subset of P is k, then P is the union of k chains. Let P be a partially ordered set. Two elements a and b of P are comparable if a < b or b < a. A subset C of P is a chain if every two distinct elements of C are comparable. A subset S of P is independent if no two elements of S are comparable. The following theorem is due to Dilworth I-3, Theorem 1.1]: THEOREM. If the maximal number of elements in an independent subset of P is k, then P is the union of k chains. This note contains a short proof of Dilworth's theorem for finite sets P. Proof. Denote by I PI the cardinal of P. The proof proceeds by induction on for all k simultaneously. If I PI = ~, there is nothing to prove. Assume, therefore, that the theorem holds for I PI < n, and let I PI = n. Denote by Pmax and Pmin the sets of all maximal, resp. minimal elements of P.

I P],

CASE 1. P contains an independent subset Po of k elements, different from both Pm,x and Pmin. Let Po = {Yl ..... YR} be such a set. Define

P+ = {xlxee,(Ey)1-yePo&y_-< x]}, P-

= ( x l x e P , ( E y ) [ y e P o & x < y]}.

It is easily verified that P + n P - = P o, P + n P - = P , P + ¢ P and P - C P (the first relation follows from the independence of Po, the second from the maximality of Po, the third ftom Po ¢ Pmin and the fourth from Po ¢ Pmax)" Now, I P + I < I P[' I P - I < I PI" By induction hypothesis, P + and P - decompose into k chains: k

k

P+= Uv,, P-= UL,. i=1

i=1

The elements of Po, being the minimal elements of P+ and the maximal elements Received June 10, 1963. 105

106

MICHA A. PERLES

[June

of P -, are the minimal elements of the chains U~ and the maximal elements of the chains L~. Assume, without loss of generality, that y~ is the minimal element of U i and the maximal element of L i (1 < i < k). Define Ci = Li U Ui. Cl is a chain, and we have P=P-L)P

+ = [,.JC~. /=1

CASE 2. Every independent subset of P containing k elements coincides with Pma~ or with Pmin" Take some a e Pmi,, and choose a b e Pma~' such that b > a (b may equal a). Define Ck = {a,b},ande'= P - {a,b}. CkiS a chain, IP'l
e = e ' U {a, b} = U c , .

Q.E.D.

i=1

REMARK. 1. Other proofs of Dilworth's theorem for finite sets may be found in [21, [31, I-41 and 1-5]. The original proof in 1-3] is direct, but somewhat complicated. The proof in 1-21 uses the duality theorem of linear programming. In 1-41, Dilworth's theorem is shown to be equivalent to a theorem of K6nig concerning bi-chromatic graphs ([8, p. 2321). In [51, it is obtained as a consequence of a theorem on the covering of a directed graph by a system of disjoint paths. REMARK 2. Dilworth's theorem for general sets P can be easily deduced from the finite case, applying the following result, which is a special case of a theorem of Rado ([9], [6], [1]). THEOREM. Let P be a set, K a finite set, and let ~ be the class of all finite subsets of P. For each F ~ ~ , let Or be a mapping o f F into K. Then there exists a mapping (a of P into K, having the following property. For every F e ~ there exists a G e ~ , such that G D_ F and q~(x) = ~bG(x)f o r all x e F. A very short proof of Rado's theorem, using Tychonoff's theorem, may be found in [9]. In [3], the infinite case of Dilworth's theorem is deduced from the finite case by another transfinite argument, using induction on k and Zorn's lemma.

BIBLIOGRAPHY 1. de Bruijn N. G. and ErdOs P. 1951, A colour problem for infinite graphs and, a problem in the theory of relations, Nederl. Akad. Wetensch. Proc. set. A, 54, 371-373. 2. Dantzig G. B. and Hoffman, A. J., 1956, Dilworth's theorem on partially orders sets, Linear inequalities and related systems, Annals o f Mathematics Studies No. 38, Princeton University Press, pp. 207-214. 3. Dilworth, R. P., 1950, A decomposition theorem for partially ordered sets, Ann. o f Math. 51, 161-166. 4. Fulkerson, D. R., 1956, Note on Dilworth's decomposition theorem for partially ordered sets, Proc. Amer. Math. Soc., 7, 701--702.

1963]

DILWORTH'S DECOMPOSITION THEOREM

107

5. Gallai, T. and Milgram, A. N., 1960, Verallgemeinerung eines graphentheoretischSatzes von R&tei, Acta Sci. Math. (Szeged), 21, 181-186. 6. Gottschalk, W. H., 1951, Choice functions and Tychonoff's theorem, Proc. Amer. Math. Soc., 2, 172. 7. Halmos, P. R., and Vaughan, H. E., 1950, The marriage problem, Amer. J. Math., 72, 214-215. 8. Ktinig, D., 1950, Theorie der endlichen und unendlichen Graphen, Chelsea Publishing Co., New York. 9. Rado, R., 1949, Axiomatic treatment of rank in infinite sets, Canad. J. Math., 1,337-343. THE HEBREWUNIVERSITYOl~ JERUSALEM

ON DILWORTH'S THEOREM IN THE IN FINITE CASE BY

MICHA A. PELES ABSTRACT

The followingtheorem is proved: Let c be an infinite cardinal. There exists a partially ordered set of cardinal c, which contains no infinite independent subset, and which is not decomposable into less than c chains. Let P be a partially ordered set, and k a natural number. Dilworth's theorem states that if the maximal number of mutually incomparable elements in P is k, then P is the union of k chains. (The terminology is explained in the preceding note [2].) P. Erd~Ss raised the question, whether Dilworth's theorem can be extended to the case where the cardinals of independent subsets of P are not bounded. The following theorem shows that such an extension is impossible. THEOREM. For every infinite cardinal N~ there exists a partially ordered set T~, such that (g) ]T~[ = N~; (B) T~ contains no infinite independent subset; (C) T~ is not decomposable into less than N~ chains. (I M] denotes the cardinal of the set M.) Proof. Let ¢0~ be the smallest ordinal power N~. Let T~ be the set of all ordered pairs (~, ~/) of ordinals ( < 0~, t / < ~0~. Define (~1, ~/1) -<-(~2, ~/2) iff ~1 < ~2 and t/1 < t/2. The relation =< is a partial ordering. If ~t < fl, then T~ c Tp, and the partial order relation of T# is an extension of the partial order of T~. Clearly IT I =

2

(~t,t/1) and (~2,]~2) a r e incomparable iff {1 < {2 and ~/2 < ~/i, or ~2 < ~1 and ~/1 < t/2. Therefore, if S is an independent subset of T~, then distinct elements of S have different first coordinates, and S may be well-ordered according to the magnitude of the first coordinates of its elements. If S were infinite, there would be an infinite sequence {(~i,~h) 10 __ t/s+ 1, for 0 < i < ¢0, which is impossible. This proves (B). In order to prove (C), consider three cases. CASE 0. c¢= 0. T 0 is not decomposable into less that No chains, since it contains, for every n < co, an independent subset S, = {(i, n - i) I 0 _< i _< n} having n + 1 elements. Received June 13, 1963. 108

1963]

DILWORTH'S THEOREM IN THE INFINITE CASE

109

CASE 1. t~ = fl + 1. Assume that c6 = {Cv Iv < cot} is a system of Ny totally ordered subsets of T~ (chains), where 7 < a, i.e. V < ft. Let C* be the set of all distinct second coordinates of elements of Cv (v < coy). Define i , -- { lv
D

U{cZlv

i1}.

Clearly, Ilw12={vlv r/for all r/e D, and ~/* < co~, since all elements of D are smaller than co~. Now, if v e 12, then I C~ I = N~, hence there exists an element ( 4 , r/,) e C, with r/~ > r/*. Choose such a pair (~,, r/v) for each v e 12, and let 4" = sup {4, [ v e 12} + 1. Since I I ___ and 4, < co~ for every v el2, we have also ~* < co~, and (~*,~/*) ~ T~. But (~*,r/*) cannot belong to any C, with v eI~, since ~?* is greater than any r/e C,. Also, (~*,~/*) cannot belong to any C, with v e 12, since (4*,r/*) and (~,, r/,) are incomparable (~* > ~, and ~* < r/0. Thus (¢*, ~/*) does not belong to any C ~ c 6 , and therefore U{c~[v < coy) ~ T~. CASE 2. ~ is a limit number, ct ¢ 0. Assume that cg = {Cv [ v < coy}is a system of Nr totally ordered subsets of T~, such that V < ~ (and therefore also ~, + 1 < ~), and 1,.J{C, Iv < coy} = T~. Let ~ ' = {C, ~ T~+I I v < coy}; then c~, is a system of at most Ny totally ordered subsets of Tr+l, and U { C , ~ Ty+, Iv < coy} = T~r'ITy+I = Ty+x which is impossible by case 1.

Q.E.D.

I~M~K. T~ is a distributive lattice. We may obtain a complete distributive lattice T~ satisfying the requirements of the theorem by defining T" = {(¢ r/) [ 4 < co~,r/< co,}. BIBLIOGRAPHY 1. Dilworth, R. P., 1950, A decomposition theorem for partially ordered sets, Ann. of Math., 51, 161-166. 2. Perles, M. A., 1963, A proof of Dilworth's decomposition theorem for partially ordered sets, lsraelJour. Math., 1, 105.

THE HEBREWUNIVERSITYOF JERUSALEM

CONTRIBUTION A DEUX PROBLEMES, CONCERNANT LES FONCTIONS DE LA CLASSE A PAR

J.-P. KAHANE ET Y. KATZNELSON* En souvenir de R. SALEM ABSTRACT

Construction explicite d'une fonction de la classe A =F(10 lipschitzienne d'ordre a > 0, ne satisfaisant pas la synth~se spectrale. Nouveaux exemples d'ensembles qui ne sont pas de r&solution au sens de Malliavin (en particulier, l'ensemble triadique de Cantor). Nouveaux examples d'ensembles E telles que les seules fonctions d'une variable r~elle op6rant dans l'alg~bre A(E) des restrictions a E des fonctions de la clesse A, soient les fonctions analytiques. Introduction. Par fonctions de la classe A nous entendons les fonctions sommes de s~ries de Fourier absolument convergentes

f(t) = E f(n)e'"' (sauf avis contraire, ~ = somme prise sur t o u s l e s entiers). Elles constituent une alg~bre de Banach, avec la norme

Ilfll, -- x I/(n)l. Les pseudomesures, c'est h dire les distributions ~t coefficients de Fourier born6s h(0 "-, E ~ ( n ) e ' ' d6finissent sur l'espace de Banach A des formes lin6aires

f ~ ~

f(t)h(t)dt = ~ f ( n ) ~ ( - n )

(sauf avis contraire, ~ -- int~grale sur le cercle), et forment un espace de Banach Am, avec la norme l] h IIo~ = suplh(n) l. P!

Nous appellerons pseudofonctions les pseudomesures dont les coefficients de Fourier tendent vers z6ro ~t l'infini. Nous nous occuperons de deux probl~mes, qui font d6j~t l'objet d'une abondante litt6rature. Probl~me de synth~se. On donne un ensemble ferm6 E, une fonction f ~ A, une pseudomesure h. Les hypotheses Received June 19, 1963 * The research reported in this document has been sponsored in part by the Air Force office of Scientific Research, OAR through the I European Office, Aerospace Research, United States Air Force". 110

1963]

FONCTIONS DE LA CLASSE A

(H)

f nulle sur E h port6e par E

(C)

fh(t)f(t)dt = 0?

111

entrainent-elles la conclusion

On a les 616ments de r6ponse suivants: Si .t" satisfait une condition de Lipschitz d'ordre ct > ½, (H) entraine (C) [14]. Si h est une mesure, (H) entraine (C). Si E a une fronti~re d6nombrable, ou si E est an ensemble triadique de Cantor, ou si E ne porte aucune pseudomesure qui ne soit une mesure, (H) entraine (C) [1, 2, 3, 6, 13]. Cependant, comme Malliavin l'a montr6 en 1959, la r6ponse au probl~me g6n6ral est n6gative [11] : il existe une f o n c t i o n f r6elle, ~ A, et une pseudomesure h port6e par l'ensemble des z6ros d e f (formellement, h = 6'(f), 6' 6tant la d6riv6e de la mesure de Dirac), telles que (C) n'ait pas lieu. La construction de f, faite par Malliavin, est assez difficile. On peut la remplacer par un th6or~me d'existence, en utilisant une m6thode probabiliste; on montre ainsi qu'il existe des f satisfaisant une condition de Lipschitz d'ordre positif, telles que (H) n'entraine pas (C) [4]. Dans le premier paragraphe, nous retrouverons ce r6sultat par une construction explicite, qui nous parait 8tre la d6monstration la plus simple du th6or~me de Malliavin. Si F est un ferm6 tel que les hypotheses (H) et E c F entrainent (C), on dit suivant Malliavin que F est un ensemble de r6solution. R6cemment [12], Malliavin a montr6 qu'un ensemble ferm6 "de multiplicit6", c'est g dire portant une pseudofonction T ¢ 0, n'est jamais un ensemble de r6solution; l'id6e est de d6finir formellement h = T. 6'(f), de sorte que le support de h soit ~ la fois contenu dans le support de T et dans celui de 6'(f). Nous donnerons un crit6re pour que F soit un ensemble de r6solution, montrant que non seulement tout ensemble de multiplicit6, mais certains ensembles "d'unicit6", comme l'ensemble de Cantor, ne sont pas ensembles de r6solution. Probl6me du caleul op6ratoire. Etant donn6 une classe (g de fonctions num6riques, on dit qu'une fonction F op6re dans ~ si, pour t o u t e f 6 (~ ~t valeurs dans l'ensemble de d6finition de F, la fonction compos6e F oJappartient/~ ~. Dans la suite, on se restreint/~ des fonctions F d~finies sur un ouvert de la droite r6elle. On salt depuis Wiener et L6vy que les fonctions analytiques op~rent dans A. On a montr~ en 1958 la r6ciproque, c'est ~ dire que seules les fonctions analytiques op~rent dans A [7]. On a cherch6 /t g6n6raliser ce r6sultat en rempla~ant A par A(E), l'alg~cre des restrictions d e s f e A / t un ensemble ferm6 E donn6. Comme A(E) a une structure d'alg~bre de Banach, les fonctions analytiques op~rent dans A(E). La r6ciproque

112

J.-P. KAHANE ET Y. KATZNELSON

[June

a lieu si E contient des progressions arithm6tiques arbitrairement riches [5], ou, plus g6n6ralement, des "mailles" arbitrairement riches (une "maille" de 2 k 61~merits 6tant un ensemble de 2k points !de la forme _+ u I _+ ... _ uk)[9]. Par contre, pour certains ensembles E ("ensembles de Helson") toutes les fonctions continues op~rent. Nous donnerons un crit~re g6n6ral pour la r6ciproque du th6or~me de L6vy, et nous montrerons qu'elle est satisfaite pour A(E) dans des cas assez 6tendus. Nous laissons ouverte la conjecture selon laquelle tout E ferm6 est soit ensemble de Helson (A(E) 6tant l'ensemble de toutes les fonctions continues sur E), soit ensemble "d'analyticit6" darts le sens que seules les fonctions analytiques op~rent dans A(E). P~r,n[~ r~Tm. Nous allons d'abord donner, par une construction explicite, une nouvelle d6monstration du th6or~me suivant, d6jh connu [4]. TH~OR~ME 1. II existe une f ~ A, satisfaisant une condition de Lipschitz d'ordre ~ > O, et une h ~ Aoo portde par l'ensemble des z~ros de f, telles que fh(t)f(t)dt ¢ O. On s'appuiera, comme toujours, sur la proposition suivante de Malliavin [11]. PROPOSITION 1.

Supposons f e A, f rdelle, et t ~ l u ) H e " : l t ~ d u < oo. 4-~o

Alors l'int~grale

f~_iue'U(Yfo-°)du, convergente dans Ao~, ddfinit pour tout a rdel une pseudomesure ho, portde par l'ensemble des z#ros de f(t) - a, et telle que, pour une infinitd de valeurs de a, f ha(t)(f(t - a)dt # O. I.¢ th~or~me 1 r~sultera done de la construction d ' u n e f ~ A, r~elle, lipschitzienne d'ordre • > O, telle que

(I)

~ e"l [l® =

O(u-~) (Ndonn6 > 2).

Nous construironsf sous la forme f ( t ) = £ anA(k.t) 1

oh A(t) est la fonction 2n-p6riodique, ~gale ~ tz/2 dans [ - n , n ] , {k.} une suite d'entiers croissants, et {a.} une suite positive sommable, qui seront d6termin6s plus tar& On utilisera les lemmes suivants.

] 963]

FONCTIONS DE LA CLASSE A

113

Posons

LEMME 1.

pg(u) = ~ feiuA(t)e-i~tdt

(2)

(u > 0).

Alors

(3)

2

Ip(u)l

D~monstration ~l~mentaire, ou par application du lemme de Van der Corput [15, vol I, p. 197]. LEMM~ 2. Soit dp une fonction gt variation bornde, ~k ~ L 2, k un entier positif, # un entier tel que [f~[ < k/2. Posons (4)

l f q~(t)~,(kt)e-'~'dt = ~---~f q~(t)e-"'dt ~--~f¢(t)dt + Y.

Alors

(5)

(C=~I~2~I)

IYt
zz\

D~monstration.

½)

i (n--½) 2

"

(4)s'6crit ~, ~ ( - n ) ~ ( n k + #) = ~b"(0)tk"(/~)+ Y

donc

lYI-5_(E n~:O Or

1

Var(~b) < 1 Var(~b) = 2re k(In 1 - ½ )

-<-2~ Ink + •1 lq~(nk+.)I2< [Var(~b)~ 22 ~ I & nk + .)I

Z .,o

l&.k ÷.)I~:. n~0

~-~ff-:

I

- (n -

(CVar(~)) 2 =

d'o/l r6sulte le lemme. LEMME 3. Soit {k,} une suite strictement croissante d" entiers positifs (n = 1,2, ...) avec kl = 1. Tout entier m s'dcrit (6)

m = 21k I + 22k2 + ...

(somme finie)

ou les 2j sont des entiers relatifs, tels que, pour tout j, (7)

]2~kl + ... + ,~jk:l <=½ki+ 1.

On posera dans la suite #i = ~1kl + "'" +'~jkj. D6monstration. S'il en est ainsi pour [m[ __
114

J.-P. K A H A N E ET Y. K A T Z N E L S O N

[June

Pour simplifier les calculs ~t venir, imposons d~s maintenant aux k. et aux a. les conditions suivantes: (8)

k. divise kn+l n- 1

(9)

E a j k j <= 1

ank ..

Posons (10)

f.(t) =

(11)

Xn,u(u)= ~ fei"$ne-lmdt.

~ ajA(kjt) 1

On 6crira y(t) pour Lo(0, et X~(u) pour Xoo.,(u). Pour chaque entier m, les 2j et #~ auront mSme sens que dans le lemme 3. Appliquons le lemme 2 dans les deux cas suivants:

(12)

r

k = k.+l

g = [an

dp(t) = e iuf"(t)

~l(t) = e iua'÷ t A ( t ) e - l a " + " k = k n+1

(13)

I t, = gn

~ t = etU$"(O

~O(kn+lt) = ei"(f-f")e -iC'-")'

(possible gr,~ce h (8)).

Tenant compte de n

Var(e i':n) = Var(ufn) < un z ~, aik j < 2urr 2 ank. (~ cause de (9)) 1

on obtient respectivement, dans les cas (12) et (13)

(14)

Ix.+,,..+,(u)l =< Ix.,,.(u)l Ip~.+,(a.+,u)l +2=ZCu~

(15)

Ix.,(.)l _-< Ix..,.(.)l +2nZCu k.+,"

an kn

Pour u donn6, nous nous servons de mani~re r6p6t6e de (14) pour n = s, s + 1,... q - 1, et de (15) pour n = q, les entiers s e t q restant ~t d6terminer en fonction de u; nous obtenons (les Pa 6tant tous de module < 1) q- 1

Ix=(~)l -<- Ix~,..I

J=~lP~,+,(a.,~)l -I

q

ajl.j

+ 2==c. E~ k.,

done, en tenant compte de I X~,,s 1< 1 et de (3),

Ix.(~)l

< =

(a~+l...aquq_~)-~ + 2rc2Cu ~, ajkj ~ kj+ 1"

1963]

FONCTIONS DE LA CLASSE A

115

C o m m e X,,(u) est le m-i6me coefficient de Fourier de e i"r, on a

( 2 ] q-~ -~ ~ ajkj \zt] (a~+~...aouq-~) " + C ' u ~ ~ (C'=2~zzC)

(16)

sous les conditions (8) et (9). Choissons maintenant (ce qui est compatible avec (8) et (9)) (17)

k, = 2 az"

a. = 2 - 2 -

(A entier > 1, n > 2).

Posons u = 2 :~ (x r6el), et choisissons, pour u assez grand, q = [tc]

q - s = 21

(1 entier > O)

Alors

as+ l ...aquq-s >: 2-2.2~+(q-s)2r >.~.//21-2

q aj kj - ~q 2 -(a+l)2J< 2 • 2 -(a+1)2~< 2U s

kj+ 1

-(A+1)2-2/-t

s

11 suffit donc de prendre, en fonction de N, l = N + 1 et A de sorte que (A + 1)2 -2t-1 > N + 1 pour avoir la condition (1) cherchde:

il e,.r

=

IJ'~ 00 cO

La fonetion f ( t ) = ~ a,A(k,t) satisfait d'ailleurs une condition de Lipschitz d'ordre 1/2A, puisque, lorsque k.Q 1 <= h < k~"l,

lf(t+h)-f(t)[<=nh j~,= t a j k j + 2

~, a j < C " a , , < C " h

TM

j=n+l

C" constante. Cela ach~ve la ddmonstration du Thdor~me 1. Par un choix diffdrent des a, et k., l'indgalitd (16) donne d'autres estimations intdressantes de Par exemple, la suite {a,} dtant donnde ddcroissante vers zdro, ddfinissons lafonction associFe ~ {a.} c o m m e (18)

A(u) = sup(aaa_,...a.u"). n

Pour tout u assez grand, ddfinissons q = q(u) par a~+~u < 1 < aqU. I1 est clair que A(u)= a I ...aqUL Soit maintenant B(u) une fonction de u > 0 telle que, pour tout n, B(u)/u" soit croissante et que B(u) < A(u) (par exemple, la fonction associde it une suite b . ~ 0 telle que b. < a,).Choisissons pour s = s(u) le plus grand indite inferieur t~ q(n) pour lequel (al ". a,u~)÷< B(u); alors (19)

(a,+ i"'" aquq-~) -½ < B(u)A-~'(u)

116

J.-P. K A H A N E E T Y. K A T Z N E L S O N

[June

et il est clair que s(u) tend vers l'infini quand u ~ oo. Imposons maintenant aux k,,, outre les conditions (8) et (9), les conditions suivantes (l prenant toutes les valeurs enti6res ~t partir d'un certain rang): q(t+ 1)

(20)

C'(l + 1)

Z s(t~

1.aJkj < A_~( l + 1). r~j + 1

Ces conditions sont comptabiles puisque chaque indice j n'y apparait qu'un nombre fini de fois. Compte tenu de (19) et (20), (16) donne II e"j' I]oo < A-~(u)( 1 + B(u)). Enongons le r6sultat. CO

Tm~OR~M~ 2. Soit {a.} une suite positive ddcroissante, telle que ~.1 a. < o~, A(u) la fonction associ~e ~ cette suite, B(u) une fonction de u > 0 croissant plus rite que tout polynome. On peut choisir les entiers k. de facon que la fonction oO

f ( t ) = ~, a. h(k.t) satisfasse 1

(21)

IIe'"s II~ = o(a-*(u)~(u)) (u ~ oo)

Remarques sur l'6nonc6 du th6or~me 2. 1. On peut remplacer A(t) par toute fonction D(t) telle que 11e'~° I1~ = O(u-*). 2. L'hypoth6se ~ a. < oo n'est essentieUe que pour assfirer f ~ A; sous les seules conditions (8), (9) et a.k. = o(k.. 1) (n ~ oo), on voit d'apr~s (15) que les e i.I,, ont une limite dans A~o , qu'on peut d6signer par e~";;quand la suite a. est d6croissante et tend vers z6ro, on peut encore choisir les k. de fagon que (21) ait lieu. Soit maintenant T une pseudofonction: T(t) ... ~, T(n)e ~" avec f ( n ) =

o(1). n~O0

Nous allons montrer que le th6or6me 2 vaut encore en remplagant II e"Wll~ par

II Te'"~ll~. Posons (22)

e,(T)=

sup

]~(n) l"

Inl>=k/2

Notons que, si Test une pseudofonction, limk~oo ek(T)= 0, et aussi bien, pour toute fonction q~eA, limk-.~oek(TqS)=0. NOUS utiliserons le lemme suivant. Lv.~v~n 4. Soit d~ une fonction de la classe A dont routes les fr~quences sont multiples de k (entier > 0): ~b(t) = ~,.q~(kn)e ik"t. Alors (23)

l[ T~

I1~ --- II Tll~ II~ I1~ + ~(T)11 ~ II~-

1963]

FONCTIONS DE LA CLASSE A

117

D ~monstration:

A Tq~(l) = ~T(l

-

kn)q~(kn)

et dans le second membre apparait une valeur de n e t une seule telle que - ( k / 2 ) < l - kn < k/2; ~ cette valeur correspond la majoration il zll~ II~ I1~, et ~t la somme des autres la majoration ek(T)[I ~b I1,. Conservons les notations (8), (9), (10). Supposons que, A(u) et B(u) 6tant donn6s comme dans le Th6or6me 2, on a choisi s = s(u) et q = q(u) comme plus haut, de fa~on que (19) et (20) aient lieu. Remarquons que dans (16) n'interviennent que des aj et des.kj tels q u e j > s; on peut donc remplacer dans le premier membre e ~: par e "~:-:~) d~s que (24)

p + 1 < s.

Nous d6finirons tout ~t l'heure p en fonction de u. Comme les fr6quences de

e i"(:-:") sont multiples des k~+ 1 (~t cause de (8)), le lemme 4 donne, compte tenu de (19) et (20),

II Te'": l]

<

cO ~

(25)

II Te'":'l[® IIe'~(:-:" I1~ +

--< II re'US" II~h-~(u)

ek(Teiuyp)

IIe'"(s-S"ll

1

(1 + n(u)) + ~k(Ze '~:" )e "~''~"

d6s que kp+ 1 > k. Choisissons maintenant, pour chaque u entier, p = p(u)comme te plus grand entier satisfaisant ~t la fois (24) et

(26)

I1Tlloo(1 +

( (1;)

Calu)... (1 + Capu) < B(u) C =

2 ~ -~

.

1

I1 est important de noter que p(u) augmente ind6finiment avec u. Le premier membre de (26) majore II quels que soient les k j, en vertu des estimations suivantes:

Te":'II~,

IITe'U:'ll~

<= IITllcOHe'~ll, ... IIe'~'^

II1

et, pour chaque ~b continfiment d6rivable par morceaux (ici, ~b = e i,,~^ )

II ¢ I11 = ~: I ~(,,)1 =< I~(o) 1 + c( ~: [ nqg(n)[z) tn¢O

<= II ,/' IIcO+ c 11,/" I1~. On aura donc, quels que soient les k~,

(27)

II Te'":" I[ ® <

B(u)

(p = p(u)).

Ayant choisi p = p(u), on choisit k = k(u, kl,...,kp) assez grand pour que (28)

ek(~,k,""k )(Te~:~') e"ET~" < A-1-(u).

118

J,-P. KAHANE ET Y. KATZNELSON

[Juna

Si l'on a

kp+ t >=k(u, k,,..., kp)

(29)

(p = p(u))

les in6galit4s (25), (27), (28) donnent

[1rei~I [1~o< A-~(u) (1 + B(u) + B2(u))

(u entier).

Or, pour p donn6, il n'y a qu'un nombre fini d'in6galit6s (29). I1 est done possible de choisir la suite ks. de fa~on A satisfaire/t la fois aux conditions (8), (9), (20) et (29). Si u est r4el non entier, soit l l'entier imm4diatement sup6rieur, et v = l - u. On peut 6crire [] Te*"I [l~ <=[[ TeUY 1[~ II e-iOI [], d'ott 1[Te '~ [[oo < CA-¢(u)(1 + B(u + 1) + B2(u + 1))

a-,ec C =

sup

}1e-'°' li,"

O__
Enon~ons le r~sultat, en remarquant qu'~ toute fonction B*(u) croissant plus rite que tout polynome on peut associer la fonction B(u) = x/B*(u - I).

TH~OREME3. Soit a, une suite positive ddcroissante telle que ~,;° a n < oo, A(u) la fonction associde ~ cette suite, B*(u) une fonction de u > 0 croissant plus rite que tout polynome, et T u n e pseudofonction. On peut choisir les entiers k, de fa~on que la fonction f(t) = ~ anA(k,t ) satisfasse (30)

l[ Te'"I [[oo= O(A-~(u)B*(u))

(u ~ oo).

Le th6or6me 3 peut se #n6raliser, en relachant l'hypoth~se que T e s t une pseudofonction. Etant donn6 2 > 1 et A > 0, nous noterons Sa,a toute partie de la droite r6elle qui est r6union infinie de segments de longueur ;t dont les distances mutuelles sont minor6es par A;t. Soit Tune pseudomesure iouissant de la propri&6 suivante: pour tout e > 0 et pour tout A > 0, il existe un 2 > 1 et un ensemble S du type Sa,a tels que (31)

[~(n) l__<e pour

n~S.

Nous dirons alors que T satisfait la condition (P). Evidemment, toute pseudofonction satisfait la condition (P). Autres exemples: les pseudomesures T telles que ~P(n) = 1-[~ cosJajn(ai >=0, ~,~ j a i < oo). Nous allons donner une autre forme it la condition (P). Lorsque (31) est r6alis6, convenons de dire que l'ensemble S porte T ?t e prks, ou encore, que c'est un e-porteur de T. Dire que Tsatisfait la condition (P), c'est encore dire (en choisissant

1963]

FONCTIONS DE LA CLASSE A

119

e = 1/p et A = 2 p ) q u ' i l existe une suite positive {2p} et des ensembles Sp du type Sa.,2p qui portent T ~t 1/p pros; on peuLd'ailleurs sans restriction supposer 2p croissant vers l'infini (p = 1,2, ...). Nous dirons qu'une suite positive {2"} croissant vers l'infini, et une suite d'ensembles {S*} sur la droite (p = 1,2, ...) sont associ6es ~t T si 1) pour chaque p, S* est du type Sa;,p 2) il existe une suite positive (cop} tendant vers l'infini et une suite {ep} tendant vers z6ro telles que, pour tout p, S* est somme directe d'un ep-porteur de T, soit S~, et de l'intervalle [-cop, cop]. I1 est clair que toute pseudomesure qui admet des suites ainsi associ6es satisfait la condition (P). Inversement, si T satisfait la condition (P), il suffit de poser 2 * = 32p et S * = Sp + [ - 2 ~ , 2 r ] pour avoir des suites assocides ~t 7:

LEMME 5. Si Tsatisfait la condition (P) et admet {2*} et {S*} comme suites assocides, il en est de m~me pour T~b, quelle que soit la fonction ~k de la classe A. D6monstration. Consid6rons les suites cot, 8p et S~,. intervenant dans la d6finition des suites associ6es. Soit ~p une suite d'entiers tendant vers l'infini, ~p < ½cop, et ~kp la somme de Fourier d'ordre ~p de ~k; on a II T~k - T~,l[o~ __<11TII~ I I ¢ ' - ¢'~11~ ="~. Soit Tp la restriction de T h S~; on a

et le support de Tp~bpest contenu dans la somme direete S~ + [ - ~p, cop]. En posant co; = ½to, et ~; = r/, + ¢' i1,, on voit que {2*} et {S*} sont bien associ6es ~t T~b. Le lemme 5 est d6montr$. LEMME 6. Si Tsatisfait la condition (P) et admet les suites A = {2*} et {S*} comme suites associ~es, et si ~k e A, on a (32) D6monstration.

(33)

lim II

p---~oO

= II zll II II "

Le coefficient de Fourier d'ordre l de T(t)~b(2t) est

~, T ( l - 2n)~(n) + [hi
~, ] ' ( l - 2n)~(n). }hi>q/2

Prenons 2 = 2* avec p > q. Dans la premiere somme, t o u s l e s termes, sauf au plus un, sont major6s en module par ep [I~bll~, le terme exccptionnel est major6 par r/TIl~ll~ll~ et, par un choix convenable de q et l, en est aussi voisin que l'on veut. La seconde somme est major6e par ]i T[I I (n)I" Quitte/t choisir d'abord q assez grand, puis p assez grand (fonction de q), IIz(t) ¢,~2"t)II est done aussi voisin qu'on veut de Ii rll ~ 11 II~. Le lemme 6 est d6montr6. Nous pouvons maintenant 6tablir,

120

J.-P. KAHANE ET Y. KATZNELSON

[June

TH~OR~ME 4. Ednonc~ du thdordme 3 reate valable, en supposant, non pas que T soit une pseudofonction, mais que T soit une pseudomesure satisfaisant la condition (P). D~monstration. I1 suffira, comme on l'a vu dans la d6monstration du th6or6me 3, d'avoir la majoration asymptotique (30) lorsque u est entier, quitte remplacer B*(u) par B ( u ) = B * ( u - 1). On conserve encore les notations et conventions (8), (9), (10). On d6finit maintenant s = s(u) comme le plus grand indice inf6rieur/t q(n) pour lequel (34)

11zll~(1

+ Calu )... (I + Ca~u)(a, ... asu~) ½<=B(u)

(C = (2 ] ~ 1/nl)i). De nouveau, s(u) augmente ind6finiment avec u, et (34) assure que, quels que soient les k~, (35)

II Tei"I" II~ (a, ... asu~)~ <=B(u).

Soit A (comme dans le lemme 6) une suite assoeide ~ T, donc (lemme 5)/t tout produit T~k, o~ ~ ~ A. On choisira les entiers kj dans la suite A de la mani~re suivante: kl, "", k. &ant choisis, on choisit (ce qui est possible d'apr& les lemmes 5 et 6) k,+, de fagon que, en posant T,.. = Te ~"I" et ¢.+1,. = ei"~"+'~, on ait (36)

IIr.+ ,,=I)~ =< I1r.,. I1~ II~.+,,. 11~ (1 + 2-")

pour toutes les valeurs de u entier telles que n >=s(u). I1 suit de (36) que (37)

II r e ' '

I1~ ~ 2 II Te""l]oo [I I!~s+,,, II~. $

Or, compte tenu du lemme 1, on a

(38)

H I[<+,,. Hoos (a~ ... a~u*)+*(a(u))-~

l[ Ye'"Yll~ < 2(A(u))-~ B(u) ce qui ach~ve la d6monstration du th~or~me 4. Comme consequence du tMor~me 4, on a: TI~OREME 5. Si l'ensemble fermd E porte une pseudomesure non nulle satisfaisant la condition (P), il n'est pas de r&olution. I1 suflit en effet d'appliquer la proposition suivante de Malliavin [12]: PROPOSITION 2. Supposons f~Alog, c'est a dire r&lle, TEA®, et

f ~ u I I1re'' II~du < oo.

ZILllog 11/f.[ <

oo, f

1963]

FONCTIONS DE LA CLASSE A

121

Alors l'integrale

/ ~°iuT( t) e ~U(t(')-")du, oo

convergente dans A~, ddfinit pour tout a rdel une pseudomesure h,, port~e par l' ensemble des zdros de f ( t ) - a et par le support de T, et telle que, pour une infinit~ de valeurs de a, f h,(t) (f(t) - a)dt v~ O. L'int6r~t du th6or~me 5 est qu'il s'applique ~t certains ensembles d'unicit6--c'est h dire ne portant aucune pseudofonction ~ 0 - - . I1 existe donc des ensembles d'unicit6 qui ne sont pas ensembles de synth~se. Nous allons retrouver ce r6sultat par une autre vole. Nous dirons qu'une pseudomesure T satisfait la condition (Q) si quels que soient les entiers positifs J et N, et e > 0, il existe un entier k = k(J, N, e) avec la propri6t6 suivante: quel que soit l'entier l

(39)

I~(nk+l+j)[<~

pour

IJl~J, [n]~N, nv~n,

nz 6tant un entier d6pendant de I. Toute pseudofonction satisfait la condition (Q). D'autre part, nous verrons tout ~ l'heure que la mesure naturelle construite sur l'ensemble de Cantor (qui n'est pas une pseudofonction, et ne satisfait pas non plus la condition (P)), satisfait la condition (Q). Nous aUons d6montrer, par une m&hode ind6pendante, un r6sultat analogue au th6or~me 5. THEOREME 6. Si l'ensemble ferm6 E porte une mesure p non nulle satisfaisant la condition (Q), il n'est pas de rdsolution.

D~monstration. 11 suffit, d'apr~s la proposition 2, de construire une r6elle, telle que ][P ei"s II = avec ~ > 2. Nous d6finironsf par

f~Alog

f ( t ) = ~ a.W(kd) 1

avec a. = 2 (40)

-2 n

, et qJ e A~og, qJ r6elle, telle que II e'"vll~o < u-4=

pour

u> 1

(on a mis en 6vidence de telles fonctions qs en d6montrant le th6or~me 1); les k, seront choisis dans la suite.

122

J.-P. K A H A N E E T Y. K A T Z N E L S O N _

")2m+l

Soit 22"` < u < _

[June

; alors

[[u ~ a~V(k.t)[!~ ~ 2HVI[, m+l

llexp(iu ?~ a.V(k~t)l]~ <~exp(2llVl[~) = C m+l

et par cons6quent

IIl~e'~f I1~ --- c II~e'~'ll

(41)

~

en posant fm(t) = ~Ta~F(kd). Supposons, pour simplifier les 6critures, que /~ soit de masse totale 1; alors

II~e"f"lloo < 1

(42)

quel que soit u.

Nous aUons d6montrer par r6currence qu'on peut choisir les entiers k. de fa~on que (43)

II/lei~Im lloo <

3 u -~

pour

22re
2 u -~

pour

2 2~÷~ ~ U < 2 2 " + :

Supposons donc qu'on ait pu choisir k l , ' . . , k s de fagon que (43) ait lieu; nous allons montrer qu'on peut choisir k = ks+ ~ de fagon ~t avoir (44)

[Ilae'UY"+lHoo <

~f3u-~

22'-÷1 < U < 2 2 m + 2

pour = 2u -~ pour 2 z"÷2 < u <

22'~' + 3

Posons ~bu = e ~j'm et d6montrons d'abord le lemme suivant

Soit N u n entier positif, Uo > O, e > O. II existe un entier posit(f k tel que, quel que soit l'entier l, LEMME 7.

(45)

]p(%(nk+l)[<e

pour

0
o,

[n[
n~ dtant un entier ddpendant de 1. En effet, on peut attacher/~ e et u o un entier d tel que ^

I qS,,(j)l < ~ pour 0 < u < Uo. l/l>J

Alors, en posant qS*(t)= ~.ljt<=jdp.(])e ijt, on a (46)

II~u~b.-/~q~,* II® < ~ pour 0 < u < no.

Or (47)

I~.(nk + 1) =

~. p(nk + I + j ) ~ . ( - j ) IJl
n~n~

1963]

FONCTIONS

DE LA CLASSE

123

A

et [q~ul est major6 par 1. Comme # satisfait la condition (Q), on peut choisir k = k(J,N,e/(4J + 2)) de fagon que (voir (39))

[f~(nk+l+j)[<4j+---~

pour [ j ] < J ,

[n[
n~n,

ce qui, joint & (46) et (47), donne (45); le lemme est ainsi d6montr6. Posons maintenant Su = ei. . . . . v. Soit u o = 2 2 m ÷ a et e ' = ~1 u o- - ~ t . Choisissons N de fa~on que (48)

•

[6,(n)[<e'

pour 0 < u < u

o.

Inl>N

Posons e = e ' / 2 N , et d6finissons kin+ x = k par le lemme 7. Le coefficient de Fourier d'ordre l de pe i":m÷' est

d, = E$,(-n)pdp,,(nk + l). Or, d'apr6s (42) et (48), on a

~,

.

/x.

[~O.(-n)#¢~(nk+l)]<

8~

pour O < u < u

o

In[>N

et d'apr~s (45), compte tenu de l[ ~O.II~ --- 1,

Z,

A

Inl-
] ~b.(- n) M,~(nk +

l) I < 2N8 pour 0 < u < Uo;

enfin ]~O~(-

n,) pc~,,(ntk +/)1

II

< II ~'~ ~ II,~.

II~

Le choix de a et a' donne donc, quel que soit l'entier l, (49)

[d,l<=uff ~ + I1~.1I~ II"~.11~ p°ur 0 < U < U o .

Consid6rons maintenant tour & tour les cas (i)

2 2m+x = < U < 2 2"~+2

(ii)

2 2"+2 = < U < 2 2 ~ + 3 = tt o.

Dans le cas (i), l'hypoth~se (43) dit que II ~,. I1~ <-- 1 et & (49), donne (50)

11,/,,,I1~o< (51)

ce qui, joint a

[dr [ < 3u-*.

Dans le cas (ii), l'hypoth~se (40) dit que

ce qui, joint

II~u[l~ -< 2u-~,

tl ~. II~ =<( u n t o + l )

(:")-'~

& [[p~b,, I!~ < 1 (voir (42)), donne [d, I =<2u-~'.

= u;,

-'~ , d'ofi

124

j.-P. KAHANE ET Y. KATZNELSON

[June

Moyennant (43), on a donc pu choisir km,l de fa~on que (44) ait lieu. Compte tenu de (41), cela entraine que la fonction

f(t) = ~ an~(kj) satisfait

1

--O(u

(u-, oo).

C o m m e f e Ato~, la d6monstration du th6or~me 6 est achev6e. Le th6or~me 6 v a n o u s permettre de monirer que l'ensemble triadique de Cantor n'est pas un ensemble de r6solution. On appelle g6n6ralement ensemble parfait symdtrique tout ensemble E de points de la forme

a o + a l +a2...+a~,+..ofi les aj sont des nombres donn6s (ao r6el, aj > 0 pour ] > 1 et ] ~ a j < 00), et o/l les 4- repr6sentent arbitrairement, et ind6pendamment les uns des autres, le signe + ou le signe - . La mesure "naturelle'" sur E (celle qui attribue des masses 6gales/t des portions 6gales) a pour transform6e de Fourier (si sa masse totale est 1) oo

/~(u) = ei'°° 1-I cos aju. 1

TH~OI~ME 7. Aucun ensemble parfait symdtrique n'est ensemble de rd-

solution. D6mo~stration. On va montrer que la mesure naturelle construite sur E satisfait la condition (Q). Sans inconv6nient, on peut supposer ao = 0, et, quitte /~ restreindre l'ensemble en fixant certains coefficients 4- (ce qui revient/t supprimer des 616ments de la suite {a j}, e t / t modifier ao), on pourra supposer sur les aj telle condition de d6croissance qui nous conviendra. Soit P~ l'ensemble des u r6els tels que ]/~(u) ] > e. Si # ne satisfait pas la condition (Q), il existe des entiers positifs J e t N, et un e > 0, tels que pour tout entier k positif, on puisse trouver un entier l, des entiers nl et nz de modules inf6rieurs/t N et des entiers Jt et J2 de modules inf6rieurs h J tels que

nlk + l + jl e P~, n2k + l + j2 ~ P~; il s'ensuit que

(hi - n2)lc~P~- P~ + ] - 2J,2J] of/le second membre repr6sente une somme directe. Doric, pour que # satifasse la condition (Q), il suffit qu'elle satisfasse la condition (Q'): quel que soit le choix de 2, N, e, il existe un entier/c tel que

nk~P~-P,+[-2J,2d]

pour n entier,

Inl _<_2N, n ~ O .

1963]

FONCTIONS DE LA CLASSE A

125

Notons P,,~ l'ensemble des u r6els tels que 1-[~° Icos j l > riv6e de 1-I~cosaju ne d6passe pas p~ = ~ f a j,

P~.~ + [ - J , d ]

C o m m e la d6-

c Pei2,x

pourvu que Jp~ < e/2. Etant donn6 J et e, nous choisirons x p o u r qu'il en soit ainsi. I1 s'agit donc de montrer que, 6tant donn6 N, ~, r, il existe un entier k tel que (52)

nk~P,/,_,,,~ - P~/2,,~ p o u r n = 1 , 2 , . . . , 2 N .

Fixons une lois p o u r route un g entre 0 et n/I2. Soit v u n entier tel que (53)

8

cos'~t < ~.

Si u~P~/z,,~, il existe au plus v valeurs de j > x telles que autrement dit, I c o s a j u ] > cosg

et

Icosaju I cos ,

I sin a~u [ < sin

pour tous les entiers j > x sauf au plus v valeurs. Si u et v appartiennent P~/z,~, on a donc

I sin aju [ < sin at et I sin a i r I < sin d ' o ~ lsinaj(u- v) I < 2 s i n ~ p o u r tous les j > x sauf au plus 2v valeurs. Airtsi, quels que soient les entiers to, > r et ;t > 2v, si l'on pose tcn+Z

Sn(w) =

I-[ sinajw ICrt

on a (54)

] S,,(w) I < (2 sin ~)a - 2,

si

w e P~I2 ,,~ - Pe/2, ~.

Si les aj sont assez rapidement d6croissants (il suffit que toutes les sommes finies de la forme _ a~ _ a2 ". + am soient distinctes), la moyenne quadratique de S,(w) est 2 -a/z. Supposons 2 choisi de sorte que

(55)

(2 sin ct)a- 2, < ~ 2-~/2.

I1 existe alors un T = T. tel que, sur tout intervalle I n de longueur T., la moyenne quadratique de Sn(w ) d6passe 2(2sin~)~-2";/l fortiori, il existe sur un tel intervalle un w tel que S.(w) > 2(2sin~)~-2'et, c o m m e la d6riv6e de Sn(w ) ne d6passe pas PK., In contient un intervalle 0 n de longueur [Onl = p~l (2 sin~) ~-2" tel que

(56)

] S.(w) ] > (2 sin ~)~-2" si w E 0n.

126

J.-P. KAHANE ET Y. KATZNELSON

[June

Montrons maintenant que, dtant donnd, N, e., r, il existe un entier k tel qu'on ait (52). On ddfinit v e t 2 de fa~on ~ avoir (53) et (55). Ii suffit, d'apr~s (54), de ddfinir des entiers k, xl, r2, "'"/~2N tels que (57)

[S,(nk)] > (2sinct) ~'-2" pour n = 1,2,...,2N.

Pour cela, on choisit x 1 de fa~on que [ 81 [ > 1 (ainsi chaque intervalle 81 contient au moins un entier), puis ~c2 de fa¢on que

1

18 1> 2T1

(ainsi chaque intervalle 82/2, homothdtique de 02 par rapport/t 0 dans le rapport ½, contient un intervalle 11, donc un intervalle 81), puis ~3 de fa~on que

1 183I > ~2Tz (ainsi chaque intervalle 8a/3 contient un intervalle 82/2)et ainsi de suite. Les x, dtant ainsi choisis, on ddfinit/t partir de n = 2N des intervalles 8n sur lesquels on a (56), chaque 8n/n contenant 8n-1/(n-1). L'intervalle 01, commun h tous les 8,In, contient un entier k en lequel on a (57), et cela ach6ve la ddmonstration du thdor~me 7. DEUXIEME PARTIE E dtant une partie compacte du cercle, on ddsigne par A(E) l'ensemble des restrictions /t E des fonctions de la classe A. A(E) est munie d'une structure d'alg~bre de Banach, comme quotient de l'alg~bre A par l'iddal des fonctions de la classe A s'annulant sur E. Dans tout ce qui suit, F est une fonction, ddfinie sur l'intervalle ] - 1 , 1 [ , qui op~re dans A(E), et N(R) la fonction de croissance exponentielle imaginaire dans A(E), ddfinie par (58)

N(R) =

sup IIe'I Ilfll_-
lIE

les normes &ant prises darts A(E). L'intdrSt de considdrer cette fonction apparait dans le thdor~me suivant. THE'ORI~ME 8. Si, pour un a > 0, N(R) > e Qn (R > O) toute fonction F qui opdre dans A(E) est analytique. La ddmonstration se fonde sur les lemmes suivants. LEMME 8. Soit x o un point d'accumulation de E a4 K > 0; il existe des nombres positifs e et rl tels que, si f e A(E) est une fonction rdelle satisfaiaant

19631

(59)

FONCTIONS DE LA CLASSE A

Support f c

127

V~ = E n i x o - t/, x o + t/[

on ait IIF(f)[]~ < K.

(60) D6monstration du lemme.

On peut supposer Xo = 0 et (quitte ~ remplacer

F(x) par F(x)-F(O)) F ( 0 ) = 0 . Montrons d'abord l'existence de e et r/tels que (59) implique (60) pour toute f qui s'annule au voisinage de Xo. Supposons, par l'absurde, que e et ~/ n'existent pas. On peut alors d6finir par r6currence une suite de nombres r6els r/, et une suite tk, e A(E) tels que (61) (62)

II .l[ < 2-", Support ~b.= Vn., 4 . = 0

11F( .)

> K

sur [ - 2 q , + l , 2 q . + ~ ] .

Pour tout 4 > 0 d6signons par ~¢ la fonction de la classe A, paire, dgale ~t 1 entre 0 et 4, lin6aire entre 4 et 24, nulle sur [24, oo[. On a II ZZT¢I11 < 3, et il est bien connu que, s i f ~ A , f ( O ) = 0, on a lim [[fza7~Ill = 0. ¢--,0

IlfzTe 1[~

Par cons6quent, si f E A(E) et f(0) = 0, on a lim¢_~o = 0 (la norme 6tant prise dans A(E)) et, sous la seule hypoth6se f ~ A(E), on a (63)

limsup [Ifz:zze[[~ < 3If(0) I. g--*0

Posons maintenant tk = ~t~b, et ~ = V(tk). D'apr~s (61), dpEA(E), done • ~A(E), et de plus ~ ( 0 ) = 0. D'apr~s (63), l i m , . ~ [[fzz7~.[[r = 0. Or, d'apr~s (62), F(~bn) = (c7~. - z ~ . . ,)O, done lim,_,~o ]]F(~b,)lie = 0, contrairement h (61). Pour tout K > 0, on peut done trouver e = eK > 0 et q = r/K > 0 tels que (59) implique (60) pour toute f nulle au voisinage de x o = 0. E 6tant un ensemble infini, la fonction F est continue. Choisissons maintenant > 0 de sorte que 1 K e < ~erl 5 et Ixl<~sup Iv(x)[ < --15 et r/= qK. Consid6rons unefeA(E) satisfaisant (59). On a

F(f) = F(f)(1 - z:7{) + F(f)zzye F(f)(1 - c7~) = F(f(1 - z:7¢/2))(1 - z:y¢) puisque F(f) et F ( f ( 1 - A¢/2)) ne different que dans l'intervalle [ - 4 , 4 ] off 1 - zz7¢ est nulle. Comme f(1 - z::7¢/2) s'annule au voisinage de 0 et que sa norme ne d6passe pas 4 llf < 48 < e,,5, on a K

IIF ( f ( 1 -

< 3-

128

J.-P. KAHANE ET Y. KATZNELSON

[June

d'o~ (quel que soit ~.) 4K 5 D'autre part, d'aprbs (63), on peut choisir ~ assez petit pour que

IIF ( f ) ( 1 - ~ ) 11~<

K

IIF ( f ) ~ 11~< 3 ] Fff(0))l < 3-" I1 en r~sulte bien

IIf(f)lie < K,

et le lemme est d6montr6.

LEMME 7. E dtant donn~, il existe deux nombres positifs ~ et K tels que, si f E A(E), f r~elle, tlftlE < ~, aZors l]F(y)11~< g. C'est un corollaire du lemme 8 au moyen d'un r6sultat connu sur la continuit6 de l'op~ration f--+ F ( f ) [8, p. 88, lemme 3.1]. Remarquons que les lemmes 8 et 9 restent valables si l'on remplace A(E) par une alg~bre de Banach r~guli~re sans idgaux primaires eontenant, pour chaque point x de son spectre E, et pour chaque voisinage V de x, une fonction ~b de norme inf6rieure ~t un nombre dgpendant de x, mais non de V, qb ayant son support dans Vet 6tant ggale ~t 1 dans un voisinage de x. Pour d6montrer le thgor~me 8, il suffit de prouver que F est analytique h l'origine. D'apr~s le lemme 7, si l'on choisit 0~> 0 assez petit, la fonction Fl(x)= F(~sinx), qui op~re, elle aussi, dans A(E), reste bornge (en norme) darts le cylindre (64)

{f + t, Ill I1~ < 1, I t l --<-~).

Les coet~cients de Fourier de F 1 satisfont (65)

1

"~

Pl(n)e i*: = ~-~f_~ Fl(f + t)e-~'~dt

pour t o u t e f ~ A(E) telle que F~(f + t) soit int6grable en tant que fonction de t ~t valeurs vectorielles. En particulier, (65) a lieu s i f est une combinaison lin6aire finie d'idempotents de A(E). Si E est de mesure positive, on sait d6j~ que seules les fonetions analytiques op~rent [8, p. 118, th6or~me I14]. Sinon, on sait que les idempotents engendrentAE [10]. Etant donn6 n entier et e > 0, il existe done une fonctionfeA(E), r&lle, combinaisons lin6aire finie d'idempotents, telle que (66)

Ilfll~ < 1 et I1e": II~ --- eol~l-

~-1.

Appliquant (65) et (66), on obtient

I :~o) 1 --
e)

K Otant la borne sup6rieure de ~F(g)II~ quand g appartient an cylindre (64).

1963]

FONCTIONS DE LA CLASSEA

129

Cela implique l'analyticit6 de Fa, done l'analyticit6 ~t l'origine de F, qu'il fallait dfmontrer. Cela ach~ve la d6monstration du th6or~me 8. Comme application du th~or~me 8, nous allons donner un crit~re plus explicite pour que toute fonction F qui op&e dans A(E) soit analytique. Nous dirons que E satisfait la condition (R) si, quel que soit N, il existe une mesure T port~e par E et un entier positif 2 tels que (67)

sup

]~

p

Iml _-
I~(p-m2)l
K ne d6pendant que de E. Nous v6rifierons plus loin que E satisfait la condition (R) d~s que l'une des conditions suivantes est r6alis6e: a) E contient des progressions arithm6tiques arbitrairement riches b) E contient des mailles arbitrairement riches (voir introduction) c) E est "ensemble de multiplicit6 au sens strict", c'est h dire qu'il porte une mesure pseudofonction non nulle d) plus g6n6ralement, E porte des mesures T telles que le rapport limsuPl,l_.oo [ 2f'(n)[/suP, l~(n) l soit arbitrairement petit. TH~OR~ME 9. Si E satisfait la condition (R), toute fonction F, ddfinie sur ] - 1 , 1 [ , opdrant dans A(E), est analytique. La d6monstration se fait en trois 6tapes. LEMME 8. La condition (R) entraine la suivante: (R1) pour toute ~ A , il existe une mesure T portde par E et un entier positif 2 tels que (68)

1, 11T(t)¢('zt)ll~o < 3KII¢II~.

T(0)=

En effet, le p-i6me coefficient de Fourier de T(t)q~ (2 0 est

E,~(m)~(p-m,l) = m

E

+

Iml>N

E Iml
et l'on a

I Z l ~ llrll~ Z I~(m>l Iml>/V

Iml>N

Iml~_N

Iml___
On peut choisir N de sorte que

z I~(m) 1 <__gll¢llo ~

Im[>N

puis T et ;t de fagon h avoir (67), et enfin, quitte ~t multiplier T(t) par une exponentielle convenable, on a I ~(0) 1_-_(2/3) 11TII~ et l'on peut, sans restriction, prendre ] ' ( 0 ) = 1.

130

J.-P. K A H A N E ET Y. KATZNELSON

[June

LEMME 9. Soit a < (2/Tr2)log (zr/2). Pour tout R assez grand, il existe une y e A rdelle, de norme R, telle que IIe'~ I1~---e-°~" D6monstration. Soit R > n2/2, q = [2R/n 2] et u = 2R/zr2q ; alors 1 < u < 2. Choisissons les ky (j = 1,2,-..,q) de teUe sorte que q

1 k--~,~ < C'6tant la constante absolue figurant dans l'in6galit6 (16), et posons q

f ( t ) = u Z A(kjt) 1

I1 r6sulte de (16) que [[elf[[~o = 2

=< zcexp -

-~--log

d'o/l te r6sultat. LEMME 10. Sous la condition (RI), N(R) > e al~dds que a < (2/zt 2)log (re/2), et que R est assez grand. C'est 1/l le point essentiel. Choisissons f comme dans lemme 9. Pour taote mesure T port6e par E et tout 2 entier, on a (dans la notation de l'introduction)

It(0)l = ~

T(t)e'f(me-'f(mdt

<

llz(t)df'~')ll~lle-'f~')ll~

Choisissons ~b = e ~¢, puis T et ;t de fa¢on ~ satisfaire (68). On obtient

IIe-"'~')I1~ >-- ~ IIe" IIZ 1

Ilf(~t)

Comme I1~ <-- R et II e is I1~ <= e -:.R avec a < a' < (2/~ ~)log (hi2) quand R est assez grand, on a 1 a'R > eaR N(R) > ]---ge

quand R est assez grand, et le lemme est &abli. Du th6or6me 8 r6sulte imm6diatement le th6or~me 9. Remarquons que le condition (R) est satisfaite d6s que la suivante est r6alis6e: (R'): quels que soient N entier et e > 0, il existe une mesure T port6e par E et un entier positif 2 tels que [[ T[[~ -- 1 et que, pour tout p, on ait

pour m = O, + 1,... + N sauf peut-~tre pour une valeur de m.

1963]

FONCTIONS DE LA CLASSE A

131

A son tour, ( R ' ) rfsulte de (R"): quels que soient N entier et e > 0, il existe une mesure T p o r t f e par E et un entier positif ;t tels que, en dfsignant par E~(T)l'ensemble des entiers n tels que ] ~(n) I > e, aucun des m2 (m = -t- 1,..., _ 2N) n ' a p p a r t i e n n e ~t l'ensemble E~(T) - E~(T).

La fin de la d f m o n s t r a t i o n du thfor~me 7 (~t partir de (52)) m o n t r e que, si E contient des mailles arbitrairement riches, la condition ( R ' ) est satisfaite, d o n c E satisfait la condition (R), c o m m e annonc6 plus haut. D ' a u t r e part, si, p o u r chaque e > 0, E porte une mesure T telle que = a et < e , il satisfait encore la condition (R'), puisque p o u r la mesure T considfrfe, E~(T) est bornf.

IIzll

limsupt,,v.® I¢(n) ]

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