A unified construction yielding precisely Hilbert and James sequences spaces
aa r X i v : . [ m a t h . GN ] A p r A UNIFIED CONSTRUCTION YIELDING PRECISELYHILBERT AND JAMES SEQUENCES SPACES
Duˇsan Repovˇs and Pavel V. Semenov
Abstract.
Following James’ approach, we shall define the Banach space J ( e ) foreach vector e = ( e , e , ..., e d ) ∈ R d with e = 0. The construction immediatelyimplies that J (1) coincides with the Hilbert space i and that J (1; −
1) coincideswith the celebrated quasireflexive James space J . The results of this paper showthat, up to an isomorphism, there are only the following two possibilities: (i) either J ( e ) is isomorphic to l ,if e + e + ... + e d = 0 (ii) or J ( e ) is isomorphic to J . Sucha dichotomy also holds for every separable Orlicz sequence space l M .
0. Introduction
In infinite-dimensional analysis and topology – in Banach space theory, two se-quences spaces – the Hilbert space l and the James space J – are certainly presentedas a two principally opposite objects. In fact, the Hilbert space is the ”simplest”Banach space with a maximally nice analytical, geometrical and topological prop-erties. On the contrary, the properties of the James space are so unusual andunexpected that J is often called a ”space of counterexamples” (see [3,5]).Let us list some of the James space properties: (a) J has the Schauder basis, butadmits no isomorphic embedding into a space with unconditional Schauder basis[1,3,4]; (b) J and its second conjugate J ∗∗ are separable, but dim( J ∗∗ /χ ( J )) = 1,where χ : J → J ∗∗ is the canonical embedding (see [1]); (c) in spite of (b), thespaces J and J ∗∗ are isometric with respect to an equivalent norm (see [2]); (d) J and J ⊕ J are non-isomorphic and moreover, J and B ⊕ B are non-isomorphic foran arbitrary weakly complete B (see [3, 4]); (e) on J there exists a C -functionwith bounded support, but there are no C -functions with bounded support (see[7]); (f) there exists an infinite-dimensional manifold modelled on J which cannotbe homeomorphically embedded into J (see [4,7]); and (g) the group GL ( J ) ofall invertible continuous operators of J onto itself is homotopically non-trivial withrespect to the topology generated by operator’s norm (see [8]), but it is contractiblein pointwise convergency operator topology (see [10] and the book [3] for morereferences).In this paper we shall define the Banach space J ( e ) for each vector e = ( e , e , ..., e d ) ∈ R d with e = 0. The construction immediately implies that J (1) = l and Mathematics Subject Classification . Primary: 54C60, 54C65, 41A65; Secondary: 54C55,54C20.
Key words and phrases.
Hilbert space;Banach space; James sequence space; Invertible contin-uous operator. Typeset by
AMS -TEX DUˇSAN REPOVˇS AND PAVEL V. SEMENOV J (1; −
1) = J . Surprisingly, there are only these two possibilities, up to an isomor-phism. It appears that J ( e ) is isomorphic to l , if e + e + ... + e d = 0 (see Theorem5) and J ( e ) is isomorphic to J otherwise (see Theorem 6).Such a dichotomy holds not only for the space l (which is clearly defined byusing the numerical function M ( t ) = t , t ≥ l M defined by an arbitrary Orlicz function M : [0; + ∞ ) → [0; + ∞ )with the so-called ∆ -condition. Then there are also exactly two possibilities for J ( e ): either J ( e ) is isomorphic to l M , or J ( e ) is isomorphic to the James-Orliczspace J M (see [9]). For simplicity, we restrict ourselves below for M ( t ) = t , t ≥
1. Preliminaries
Let d be a natural number and e = ( e , e , ..., e d ) ∈ R d a d -vector with e = 0.Having in our formulae many brackets we shall choose the special notation a ∗ b forthe usual scalar product of two elements a ∈ R d and b ∈ R d . A d -subset ω of N isdefined by setting ω = { n (1) < n (2) < ... < n ( d ) < n ( d + 1) < ... < n ( kd − < n ( kd ) } ⊂ N for some natural k and then the subsets ω (1) = { n (1) < n (2) < ... < n ( d ) } ,ω (2) = { n ( d +1) < n ( d +2) < ... < n (2 d ) } , ......, ω ( k ) = { n (( k − d +1) < ... < n ( kd ) } are called the d -components of the set d -set ω .For each d -set ω and each infinite sequence of reals x = ( x ( m )) m ∈ N ∈ R N wedenote x ( ω ) = ( x ( m )) m ∈ ω and x ( ω ; i ) = ( x ( m )) m ∈ ω ( i ) . Definition 1.
For each d ∈ N , e ∈ R d , x ∈ R N and d -set ω = { n (1) < ... < n ( kd ) } the ( e, ω ) -variation of x is defined by the equality ( e, ω ) = vuut k X i =1 ( e ∗ x ( ω ; i )) = p ( e x ( n (1)) + ... + e d x ( n ( d ))) + ... + ( e x ( n (( k − d + 1)) + ... + e d x ( n ( kd ))) Definition 2.
For each d ∈ N , e ∈ R d , x ∈ R N the e -variation of x is defined bythe equality || x || e = sup { e ( x, ω ) : ω are d -subset of N } . Definition 3.
The set of all infinite sequences of reals tending to zero with finite e -variation is denoted by J ( e ) . We omit the routine verification of the following proposition.
Proposition 4. ( J ( e ); || · || e ) is a Banach space for each e = ( e , e , ..., e d ) ∈ R d with e = 0 . (cid:3) Note that a restriction e = 0 is purely technical. It avoids the case e = 0 ∈ R d and guarantees that k (1; 0; 0; ... ) k e >
0. Below we fix such a hypotesis.
UNIFIED CONSTRUCTION 3
Theorem 5. If e + e + ... + e d = 0 , then J ( e ) and l are isomorphic. Theorem 6. If e + e + ... + e d = 0 , then J ( e ) and J are isomorphic. Theorem 5 is proved in Section 2 as the corollary of Lemmas 7-10. We believethat Lemma 9 is of interest independently of Theorem 5 and its proof. Theorem6 is proved in Section 3 as a corollary of Lemmas 11-13. Lemma 11 really stressesthe importance of equality e + e + ... + e d = 0.Lemma 13 is the most difficult to prove. In the last case some special combi-natorial Sublemma 14 is needed. Roughly speaking, it states that each 2-subsetof naturals admits a representation as a union of at most N = [0 , d ] + 2 of its2-subsets which consist of d separated pairs. It seems that this statement is newand possibly interesting for geometric combinatorics. For example one can try tofind an analog of Sublemma 14 for finite planar subsets.One more open question concerns analogs of Theorems 5 and 6 for spaces offunctions over the segment [0; 1]. The main obstruction here is that the Jamesfunctional space JF has a non-separable dual space [4]. Also, we believe thatTheorems 5 and 6 are true for a generalizations of J in the spirit of results of [6].
2. Proof of Theorem 5Lemma 7.
The inclusion operator id : l → J ( e ) is well-defined and continuous. Proof.
Let k · k be the standard Euclidean norm. Fix any x = ( x , x , x , ... ) ∈ l and pick any d -set ω = ω (1) ∪ ω (2) ∪ ... ∪ ω ( k ) with d -components ω (1) , ω (2) , ...ω ( k ).Then ( e ∗ x ( ω ; i )) ≤ k e k · k x ( ω ; i ) k due to the Cauchy inequality. Hence,( e ( x ; ω )) = k X i =1 ( e ∗ x ( ω ; i )) ≤ k e k · ( k X i =1 k x ( ω ; i ) k ) ≤ ( k e k k x k ) and therefore k x k e = sup { e ( x ; ω ) : ω } ≤ k e k k x k = C k x k . (cid:3) Lemma 8. det e e . . . e d e e . . . e d · · · · e e . . . e d = ( − d ( d Y i =1 e i )( d Y i =1 e i ) . (cid:3) Lemma 9.
For each d ∈ N , e ∈ R d with e = 0 and e + e + ... + e d = 0 thereexists a constant C = C e > such that for every sequence of reals x (1) , x (2) , ..., x ( d ) , x ( d + 1) the inequality | d X i =1 e i x ( n ( i )) | ≥ C | x (1) | holds for some d-set ≤ n (1) < ... < n ( d ) ≤ d + 1 . Proof.
The assertion is obvious for x (1) = 0. So let x (1) = 0 and consider thecase when all numbers e , e , ..., e d are non-zero. Denote by L the linear mapping of DUˇSAN REPOVˇS AND PAVEL V. SEMENOV R d +1 into itself defined by the matrix from Lemma 5. By this lemma, L : R d +1 → R d +1 is an isomorphism. Consider R d +1 with the max - norm k ( x (1) , x (2) , ..., x ( d ) , x ( d + 1)) k = max {| x ( j ) | : 1 ≤ J ≤ d + 1 } , i.e. as the Banach space l d +1 ∞ of dimension d + 1. Define the constant C as thedistance between the origin and the L image of the set of all elements with the firstcoordinate equal to ± C = dist(0; { L ( y (1) , y (2) , ..., y ( d ) , y ( d + 1) : y (1) = ± } ) > . Next, pick x = ( x (1) , x (2) , ..., x ( d ) , x ( d + 1)) ∈ l d +1 ∞ with x (1) = 0 and set y ( i ) = x ( i ) · ( x (1)) − , i = 1 , , ..., d, d + 1 . Then y (1) = 0 and k L ( y (1) , y (2) , ..., y ( d ) , y ( d +1)) k ∞ ≥ C, k L ( y (1) , y (2) , ..., y ( d ) , y ( d +1)) k ∞ ≥ C | x (1) | . By definition of the max norm and by the definition of the isomorphism L we seethat | P di =1 e i x ( n ( i )) | ≥ C | x (1) | , for some indices 1 ≤ n (1) < ... < n ( d ) ≤ d + 1.It is easy to check that for an arbitrary e ∈ R d with e = 0 and e + e + ... + e d =0 the constant C e , works properly, where the vector e , consists of all non-zerocoordinates of the vector e . (cid:3) Lemma 10.
The inclusion operator Id : l → J ( e ) is a surjection. Proof.
Suppose to the contrary that k x k e < ∞ but k x k = ∞ for some x =( x ( m )) m ∈ N ∈ R N . Due to the equality ∞ X m =1 x ( m ) = d +1 X i =1 ( ∞ X k =1 x ( k ( d + 1) + i ))we see that for some 1 ≤ i ≤ d + 1 the series P ∞ k =1 x ( k ( d + 1) + i ) is divergent.So let C be the constant from Lemma 9. Applying this lemma for each natural k to the reals x ( k ( d + 1) + i ) , x ( k ( d + 1) + i + 1) , x ( k ( d + 1) + i + 2) , ..., x ( k ( d + 1) + i + d )we find some d -set, say ω ( k ), such that | e ∗ x ( ω ( k )) | ≥ C | x ( k ( d + 1) + i ) | . Hence, ∞ X k =1 ( e ∗ x ( ω ( k ))) ≥ C ∞ X k =1 x ( k ( d + 1) + i ) = ∞ and this is why k x k e = ∞ . (cid:3) Note that Theorem 5 implies that for e + e + ... + e d = 0 it suffices to define J ( e )as the set of all sequences with a finite e -variation. In this situation the convergenceof coordinates to zero is a corollary of finiteness of the e -variation. UNIFIED CONSTRUCTION 5
3. Proof of Theorem 6
As it was mentioned above we first explain the reason for the appearance of therestriction e + e + ... + e d = 0. Lemma 11.
The inclusion operator Id : J (1; − → J ( e ) is well-defined andcontinuous. Proof.
For arbitrary reals t , t , ..., t d we see that | e t + e t + ... + e d t d | = | e ( t − t ) + ( e + e ) t + ... + e d t d | == | e ( t − t ) + ( e + e )( t − t ) + ( e + e + e ) t + ... + e d t d | == | d − X i =1 ( e + e + ... + e i )( t i − t i +1 ) | ≤ C d − X i =1 | t i − t i +1 | =and ( e t + e t + ... + e d t d ) ≤ ( C d − X i =1 | t i − t i +1 | ) ≤ C ( d − d − X i =1 ( t i − t i +1 ) where C = max {| e + e + ... + e i | : 1 ≤ i ≤ d − } . Now pick any d -set ω = ω (1) ∪ ω (2) ∪ ... ∪ ω ( k ) with d -components ω (1) , ω (2) , ..., ω ( k ).Making the estimates above we see that( e ( x ; ω )) = k X j =1 ( e x ( n (( j − d + 1)) + e x ( n (( j − d + 2)) + ... + e d x ( n ( jd ))) ≤≤ C ( d − k X j =1 d − X j =1 ( x ( n (( j − d + i )) − x ( n (( j − d + i +1))) ≤ C ( d − k x k J (1; − according to the definition of one of equivalent norms in the James space J = J (1; − k x k J ( e ) ≤ C √ d − k x k J (1; − . (cid:3) The following lemma gives a chance to pass from an arbitrary vector e =( e , e , ..., e d ) ∈ R d to the special( d + 1)-vector u d = (1 , − , , , ... ∈ R d +1 . Lemma 12.
The inclusion operator Id : J ( e ) → J ( u d ) is well-defined and contin-uous. Proof.
Fix x ∈ J ( e ) and pick any ( d + 1)-set ω = ω (1) ∪ ω (2) ∪ ... ∪ ω ( k ) with( d + 1)-components ω (1) , ω (2) , ..., ω ( k ). For each component ω ( j ) = { n (( j − d + 1) + 1) < n (( j − d + 1) + 2) < ... < n ( j ( d + 1)) } let ω , ( j ) = ω ( j ) \ { n (( j − d + 1) + 1) } and ω , ( j ) = ω ( J ) \ { n (( j − d + 1) + 2) } . Then ω , = ω , (1) ∪ ω , (2) ∪ ... ∪ ω , ( k ) and ω ,, = ω ,, (1) ∪ ω ,, (2) ∪ ... ∪ ω ,, ( k ) are two d -sets with d -components ω , (1) , ..., ω , ( k ) and with d -components ω ,, (1) , ω ,, (2) , ..., ω ,, ( k ).Consider for simplicity the case j = 1. Then e ( x ( n (2))) − x ( n (1))) = ( e x ( n (2))) + e x ( n (3))) + ... + e d x ( d + 1)))) − DUˇSAN REPOVˇS AND PAVEL V. SEMENOV − ( e x ( n (1))) + e x ( n (3))) + ... + e d x ( d + 1)))) = e ∗ x ( ω , ; 1) − e ∗ x ( ω ,, ; 1)and ( x ( n (2))) − x ( n (1))) ≤ e (( e ∗ x ( ω , ; 1)) + ( e ∗ x ( ω ,, ; 1)) ) . Having such an estimate for each j = 2 , , ..., k and summarizing all inequalities wesee that ( u d ( x ; ω )) ≤ e ( e ( x ; ω , )) + ( e ( x ; ω ,, )) ) ≤ e k x k e = C k x k e . Passing to the supremum over all ( d + 1)-sets, we finally obtain k x k u d ≤ C k x k e . (cid:3) So our final lemma shows that dependence on d ∈ N can in fact be eliminatedand we can return to the original vector (1; −
1) = u . Together with Lemmas 11and 12 it completes the proof of the theorem. Lemma 13.
The inclusion operator Id : J ( u d ) → J ( u ) is well-defined and con-tinuous. Proof.
First, we need the following purely combinatorial sublemma. We will tem-porarily say that a 2-set∆ = { d (1) < d (2) < . . . < d (2 s − < d (2 s ) } ⊂ N is d − dispersed if s = 1 , or if s > d (2 j +1) ≥ d (2 j )+ d for all j = 1 , , . . . , d − . (cid:3) Sublemma 14.
Every 2-set ω = { n < n < ... < n k − < n k } can be decom-posed into a union of at most [0,5d]+2 pairwise disjoint, d − dispersed − subsets . Proof of sublemma.
Induction on k . The initial step k = 1 is trivial. So let ω , = ( n ; n ) ∪ ( n ; n ) ∪ ... ∪ ( n k − ; n k ) ∪ ( n k +1 ; n k +2 ) = ω ∪ ( n k +1 ; n k +2 ) . By induction hypothesis we have that ω = ∆ ∪ ... ∪ ∆ m , m ≤ [0 , d ] + 1for some [0 , d ] + 2 pairwise disjoint, d -dispersed 2-subsets ∆ , ..., ∆ m . There areexactly two possibilities:a) Inequality max ∆ i ≤ n k +1 − d − ≤ i ≤ m . Then one can sim-ply add the pair ( n k +1 ; n k +2 ) to ∆ i . Clearly the 2-sets ∆ i = ∆ i ∪ ( n k +1 ; n k +2 )is also d -dispersed and ω , = ∆ ∪ ... ∪ ∆ i − ∪ ∆ i ∪ ∆ i +1 ∪ ... ∪ ∆ m , m ≤ [0 , d ] + 1 . Hence, in this case the number of items in the decomposition of ω , into d -dispersed2-subsets is the same as for ω .b) Inequalities max ∆ i ≥ n k +1 − d are true for all 1 ≤ i ≤ m . This means that ineach 2-subset ∆ i its maximal pair intersects with the segment [ n k +1 − d ; n k +1 − i consist of a pairwise disjoint, linearly ordered pairs. Therefore on this UNIFIED CONSTRUCTION 7 segment of the fixed length d can in general, be placed either at most [0 , d ] pairs,or at most [0 , d − i .Hence in this case m ≤ max { [0 , d ] , , d − } ≤ [0 , d ] + 1 . This implies that one can simply consider the pair ( n k +1 ; n k +2 ) as an additional,separate item in the decomposition of ω , into union of d -dispersed 2-subsets. (cid:3) Let us return to the proof of Lemma 13. The main advantage of a d -dispersed 2-set ∆ is that one can ”extend” it up to a ( d + 1)- set ∇ by adding the ( d −
1) naturalnumbers which immediately follow d (2 j ) to each 2-component { d (2 j − d (2 j ) } of∆. Namely,∆(1) = { d (1); d (2) } ⇒ ∇ (1) = { d (1); d (2); d (2) + 1; d (2) + 2; ... ; d (2) + d − } ∆(2) = { d (3); d (4) } ⇒ ∇ (2) = { d (3); d (4); d (4) + 1; d (4) + 2; ... ; d (4) + d − } ∆( s ) = { d (2 s − d (2 s ) } ⇒ ∇ ( s ) = { d (2 s − d (2 s ); d (2 s ) + 1; ... ; d (2 s ) + d − } . Clearly max ∇ (1) < min ∇ (2) < max ∇ (2) < min ∇ (3) < ...max ∇ ( s − < min ∇ ( s )and that is why the sets ∇ (1) , ∇ (2) , ..., ∇ ( s ) really are ( d + 1) − components of theirunion ∇ . So for each x = ( x ( m )) m ∈ N ∈ R N we have( u ( x, ∇ )) = s X j =1 ( x ( d (2 j )) − x ( d (2 j − == s X j =1 ( x ( d (2 j )) − x ( d (2 j −
1) + 0 · x ( d (2 j ) + 1) + ... + 0 · x ( d (2 j ) + d − == ( u d ( x, ∇ )) and finally for an arbitrary 2-set ω = { n < n < ... < n k − < n k } we obtain( u ( x, ω )) = k X i =1 ( x ( n (2 i )) − x ( n (2 i − = m X j =1 ( X i ∈ ∆ j ( x ( n (2 i )) − x ( n (2 i − ) == m X j =1 ( u ( x, ∆ j )) = m X j =1 ( u d ( x, ∇ j )) ≤ m k x k J ( u d ) . Hence the inclusion operator id : J ( u d ) → J ( u ) is a well-defined mapping and itsnorm does not exceed the constant p [0 , d ] + 2 . (cid:3) Acknowledgements
The first author was supported by the Slovenian Research Agency grants No.P1-0292-0101-04 and Bl-RU/05-07/7. The second author was supported by theRFBR grant No. 05-01-00993.
DUˇSAN REPOVˇS AND PAVEL V. SEMENOV
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