Séminaire d’analyse fonctionnelle École Polytechnique R. I. OVSEPIAN A. P EŁCZY ŃSKI The existence in every separable Banach space of a fundamental total and bounded biorthogonal sequence and related constructions of uniformly bounded orthonormal systems in L2 Séminaire d’analyse fonctionnelle (Polytechnique) (1973-1974), exp. no 20, p. 1-15 <http://www.numdam.org/item?id=SAF_1973-1974____A22_0> © Séminaire analyse fonctionnelle (dit "Maurey-Schwartz") (École Polytechnique), 1973-1974, tous droits réservés. L’accès aux archives du séminaire d’analyse fonctionnelle implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/legal.php). Toute utilisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright. Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques http://www.numdam.org/ THE EXISTENCE iN EVERY SEPARABLE BANACH SPACE OF A FUNDAMENTAL TOTAL AND BOUNDED BIORTHOGONAL SEQUENCE AND RELATED CONSTRUCTIONS BOUNDED ORTHONORMAL SYSTEMS IN OF UNIFORMLY L2 by R.I. OVSEPIAN (Erevan) and A. PE0142CZY0144SKI (Warszawa) * Institute of Mathematics of the Armenian Academy of Sciences. #0 Institute of Mathematics, Polish Academy of Sciences. Abstract. 1) * n n (x,) Tn every separable Banach space X biorthogonal sequence a is constructed such that linear combinations of the xl "" dense in X, for every x in X if x (x) s are n 4~ 0 for all n then x = 0 and n CJ , n 11 11 2) Linear subspacps of which admit basis consisting of uniformly bounded functions are an orthonormal characteri-zed. XX.1 The present paper consists In the first and 4) we using one of three sections. trick invented a Olevskii (~9~ by Lemmas 3 prove Theorem 1 : In every Banach space X there exists separable a fundamental * and total biorthogonal (xn ,xn) n sequence such that * Recall that a sequence n n of Banach space X and bounded linear functionals a of’ X - the dual of X is said to be X, i.e. elements on biorthogonal if x*(x ) n ~ n,m = 1,2,... A biorthogonal combinations of the v:’s are of elements of pairs consisting = m amn for sequence (xn ,x n) dense in X, and is total if the condition is fundamental if linear -.L -M- x*n(x) = 0 for Theorem 1 1,2, ... implies that n = answers question a weaker result has been previously ([1J, p.238). of Banach obtained by A slightly Davis and Johnson [4]. The main result of the second section is Theorem 2 : L2(03BC) a where 03BC Let E be is separable a probability e, set S. Then E admits an bounded functions if and Moreover if E n orthonormal basis can linear measure subspace on a orthonormal basis a si g me field of subsets of of consisting separable subspace be constructed space which is dense in the Hilbert space a uniformly if only is of norm 11 - 11 so .0 then the of that it spans in E n L-(I,) . a linear sub- XX.2 As corollary we obtain that every subspace codimension admits an orthonormal basis consisting a infinitely of H. many times differentiable functions. In the third section an we isometric that there exists j(X) ded a as Using a question a answers consider the class of such Banach spaces X embedding, scalar-valued continuous functions ball of uniformly bounded of This of finite [14J~ Shapiro which admit L ~ (0~1) oi a is not on a j, into a space a L2(~), space X has the above (ii) satisfies the condition [13] result of Rosenthal i.e. j(X) 11 of all S such show that we regar- of Theorem 2. property if and only if it contains to the space of all S such that the unit on probability measure V totally bounded subset of L2( ) C(S) compact Hausdorff space Borel subspace of recent profound subspace isomorphic say a Banach closed linear a absolutely convergent series of scalars. 1. Proof of Theorem 1 . O1 evski i’ s tion of Lemma 3 of [A] Banach space X, then rated by Lemma 1 : We [9]. a a lemma which is If A is a non-empty subset of xo, x1,...,x 2n-l 2n-1 subspace Banach space and let of n be elements 0 f X and let ill, elements of X exists and then a such that unitary real a subspace of X geneX generated by A. be a - Let a denotes the closed linear A and lin A - the linear Let X be modifica- with begin X" 0 positive integer. -lk X*1 0 x1, I I * ..., ...7 x 2n_1 x2n-1 bebe. XX.3 Proof : The conditions 2x2 - matrix. The (3) and (4) are specific unitary satisfied for every unitary matrix for which (1) and (2) hold is defined to be the matrix which transform the unit vector basis 2 -dimensional of the 12-n Hilbert space onto the Haar basis of this space. We put We have Clearly (5) implies (1) Proposition ---- -- - - .. --- -- -. 1 : Let -. (x aAl and (2). x 1* ) be a fundamental and total n I sequence in a Banach space X such that there exists - nite sequences (nk) such that an biorthogonal increasing inf i- 1, I Then there exists XX.4 a fundamental and total biorthogonal sequence (e ,e ) n n in X such that iix )) Proof : one may assume that Without loss of n generality f or all n. sequence Next Pick (m r) a permutation p(.) of the indices so of the indices and that an = 1 increasing if put we m akr. k, j verify where are defined as in Lemma 1 that such defined sequence for n = m r. Using * (e n ,e n) Lemma 1 has the desired we easily properties. XX.5 We shall Proof of Theorem 1 : of X implies assume that there exist sequences 3~ and that dim X F1 C F2 C .M, of subspaces of X E1 c E2 x = 0. *M* In view of 11 Proposition 1 it is (x) = enough ~. C .., such that dim is dense in X and if f then = Then the of E~ _ separability subspaces of X dim F~ ; i for 0 for all to construct a biortho- XX.6 inductive hypothesis implies that and to we a inf er that S 3. We define Thus linear functional on x. Remark 1 : Using in the C and since dim ] H H ~F. to be any n 3° Day’s technique (cf. [3]) case the Borsuk antipodal mapping theorem one can choose (both g extension of prezerving norm which bases in the case on of * complex scalars) real and of for s Now the 1,2,... = and x3s x- 3s inspection so of the that proof separable Banach space for every e> 0 and bounded biorthogonal sequence (e such n n in every en) for all ced by n. 1+ However, c. 1 : Corollary norm We do not know whether for every III ~ 111 In every was separable ,e*) with n by n and sup lle*ll n C. there exists yields a that lie n nil lien (1+V2)2 + n . 0 this bound Bessaga we be repla- can have equivalent biorthogonal an fundamental and total 1. n biorthogonal sequence in X such thatile n II co .. n Remark 2 : A similar argument Theorem 1 allows to prove the that fundamental a Banach space X there exists = n is any fundamental and total for all observed such that there exists in X (e n sequence as c > of Theorem 1 to that which is used in the following proof of = 1 c XX.7 let T : Theorem 11 : Let X and Y be Banach spaces and one bounded linear operator. compact, then * quences (x n’ ,x n) in X is not If X is separable, T(X) X -- - Ybe one-to- is dense in Y and T there exists fundamental and total biorthogonal se- * and (y’ ny) n in Y such that and for all 2. n. Constructions of uniformly bounded orthonormal sequences. We (= a employ the following notation. If ~ negative normalized measure) on a sigma non . for any p-absolutely and L2(J1) is probability a field of subsets of ~ _ as usually a /~ square summable scalar valued functions denote measure x the Banach spaces of those and y on S. x respectively. The proof Instead of ([9J, Lemma of Theorem 2 is similar to the Proposition 1, a set S. product apply the following of Theorem 1. result due to Olevskil 4). Proposition 2 : of we proof Let . , . Let p be aprobability measure on a sigma field of subsets be an infinite orthonormal (with respect to the inner (xn) n >) sequence of functions in Then there exists an orthonormal sequence Loo(03BC) (e n) such that 1 im inf such that n oo. XX.8 and Proposition 2 can be obtained by a non essential modification of the proofs of Lemma 1 and Proposition 1. Actually 01evskil stated Proposition 2 f or the Lebesgue measure on [0,I]. The of proof To prove Theorem 2 it is convenient to Let Lemma 2 : gn L4(p)) converges to dimensional no > k (g n) and a subspace be a use normalized sequence in the following simple L 2(p) which fact. weaklyY (in .. zero Then for every finite and let of say function h in the F, and for k>0 orthogonal complement there exist index an of F such that and Let Proof : 20132013201320132013 p = dim F. Let l 2 p any orthonormal basis for F. ’ Pick e>0so that Since (gI ) conver g es Clearly h belongs weakl y to 0 in L 2(g), there exists I to the orthogonal complement of an index n° and > k XX.9 We have and (i) It follows from Proof of Theorem 2 : that there exists in E creasing sequence of finite dimensional subspaces 00 U dim F = p and ~ is dense in E. F p 00 subset of Loo(p) Loo(03BC) is dense in E n there exists in E In view of can one sequence a Proposition 2, in (h ) mal sequence (Fp) choose the sequence in the L°(p) it L’(p) (gn) is P E norm. is so that the p a in- such that separable union U the to define that for s= (ii) yields that assumption of Lemma 2. inductively an orthonor- 1,2, .., where I We define h1 as F1 with the functions some the conditions I,et any element of us consider (7) and separately ~ Suppose that for have been defined to (8) and two so that cases. F The condition satisfying enough so if Clearly F112 C:F2 c ..,. an satisfy XX.10 complete the induction and the proof of the sufficiency conditions (i) and (ii). The necessity is trivial. This A similar Remark 1 : Banach space X into a and T is not sup n jjx j) and Remark 2 : and E by x2m(t) F.G. by = 0 Arutunian T(x),x n denotes the inner >H It is 2m-1 If E in X such that (x ) sequence n is a an orthonormal m=2 enough where and nor E H, then (xn) biortho g onal decomposition E1 for x E X and for H admit of L 2 0, 1] (x ) x2 can sequen- onto uniformly bounded and to define n x. > of product fundamental and total such that neither such that the functions and a n n * orthonormal bases. 2 T(X). operator from is separable # is defined There exists E1 bounded linear orthonormal basis f or E. Moreover if X is n ., E2 = an * E X n=1,2..., where . , . be chosen so that (x subspaces is one Let E = Hilbert space H. a (T(x ) ) * and to one a then there exists compact oo separable argument gives Let T : X -- -~H be Theorem 2’ : of the is any orthonormal basis for are unbounded, (m= 19 2y (unpublished) we have However x 2m-1 0 as was for 0t2 observed earlier XX.11 Corollary 2 : If E i s a linear subspace of a separable where p is a non purely atomic probability measure and complement of E is finite orthonormal baxis. can a Moreover if E --, be chosen from elements of E Q Proof : It is to show that enough space the L-(J.1) orthogonal uniformly bounded in E then the basis is L~(~.) . rE] (ii) of Theorem 2 To check (i) first observe that the density r regarded integer as L2(~) of subspace a p and for every in L2( ) implies linearly independent " 00 there exist (i) satisfies the conditions of that for every f1,f9,...,f 2 1 .1 p+1 and L Ð (~) positive in such that the matrix is invertible. Let for i = 1, 2, ... , p+1. thogonal complement existence of to any basis of the or- The above observation applied of E and any element f of y in an be the inverse matrix and let i,k L 00 (J.1) non such that zero y~f> ~ 1 and [E] yields the ~r, g > ~ 0 for all g orthogonal complement of E. The last condition means that y E Hence there is no f/0 in [E] which is orthogonal to all y E Lco(g) is dense in [EJ- Hence (E] satisfies (i). equivalently in the The"moreover" part of the Corollary follows from the observation that if [E] satisfies (ii) than E also satisfies An immediate consequence of Corollary 3 : (ii). Corollary 2 is Let f be an y unbounded function in Then the ortho- uniformly bounded orthonormal basis consisting of trigonometrical polynomials. This basis has no extension to any uniformly bounded orthonormal basis for gonal complement Corollary of f admits a 3 answers a question of Shapiro [l4] XX.12 Fat 3. Definition : space S. to a A closed linear a spaces. subspace Loop) - L2(~) I fat with : or Z of regarded as measure C(S) a Hausdorff compact a on is said to be fat with respect L2( ) subset of the Hilbert space denote the natural respect to p iff operator Borel probability bounded set. totally Let rem be Let 03BC C(S) if the unit ball of Z is not Z is of subspaces equivalently injection. A It is clear that the rest.riction of I if E T I to Z is not p satisfies the condition (ii) (Z) compact a of Theo- 2n Our next result characterizes Banach spaces which admit fat isometric embeddings in terms of C(S) spaces. Some of the equivalent conditions are stated 2-absolutely summing operators, i.e. such bounded linear ope- into rators which admit C12’ measure (cf. factorization a and through a equivalent : (a) there exists such that injection follwing conditions uniformly bounded sequence (x n) of elements subsequence of (x n) is a weak Cauchy sequence, no there exists a 2-absolutely summing operator from (d) there exists a 2-absolutely summing y non X onto 1~ y compact operator from X 12, y for every for there exists is fat with Proof : to of X 11, (c) subspace isomorphic a some are a X contains (e) for p e (b) into I [8]). For every Banach space X the Proposition 3 : natural some a probability Let T ba embedding j of X into Borel measure 03BC on a C(S) space j(X) S such that to p. respect (a) ~ (b). This is (b) ~ (c). isometric profound a a recent result of Rosenthal bounded linear operator from 11 [13]. onto 12 XX.13 (cf. ~2~ for the existence of such ~7] (cf. Grothendieck T admits extension to an a T is Then by a result of 2-absolutely summing. Hence, by 2-absolutely summing operator from I) (c) = (d). Obvious. (d) ~ (e). Let T : operator and let S be [11], and Pietsch exists a Borel image j(X) (e) =~ (a). (x n) A : L 2(p) __ .12. of j(X) is under (x n) does Cauchy (e) with n uniformly a 1 for Cauchy sequences strong Cauchy sequences. 3 was subset bounded n/m (n,m=1, 2,...) not contain weak Proposition the respect to . 2 m some because recently independently by Weis [16J. discovered Our last result is related to result of Sidon Corollary 4 : tends a that there exists sequences into our is not Ip A l 03BC j for is not compact, totally bounded Since T subspace of C(S) fat a A similar result to Let result of Persson a S such that T= on 2 in X such that takes weak compact 2-absolutely summing non 2 It follows from Thus the sequence of S. a for every isometric embedding j : X -- -~ C( S) there operator Thus sequence be compact Housdorff space. By a of the unit ball of -- ~12 X probability measure p bounded linear -vjea.’,,-.1y to infinite of a [15]. I,et p ( gn) Gaposhkin’s [6] generalization be a zero be a probability uniformly in L2(p) measure on a sigme bounded sequence in Loo(J.1) and lim sup 0. field of subsets such that subsequence and c > 0 such that nk for every finite sequence of scalars (gn) Then there exists n an X 12. onto I f.1 F8]) also operator). an c ,c ,...,c (p~l,2,...). XX.14 Without loss of Proof : Then (g) weakly does not have conver g es in converges to zero. generality Cauchy (in L2(p) Thus g~ ~ regarded L ( ~) ---L 2(p) inj ect ion L 2(p)) but zero as that a because subsequences subse q uence no of (g ) (g ) strongly L’(p) sequence of elements of sequences because the natural takes weak in stron g Cauchy sequences assume may Lm(p)) does not contain weak into to we L 2(p). 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