FINITELY STRICTLY SINGULAR OPERATORS BETWEEN JAMES SPACES

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Niveau: Supérieur, Doctorat, Bac+8
FINITELY STRICTLY SINGULAR OPERATORS BETWEEN JAMES SPACES ISABELLE CHALENDAR, EMMANUEL FRICAIN, ALEXEY I. POPOV, DAN TIMOTIN, AND VLADIMIR G. TROITSKY Abstract. An operator T : X ? Y between Banach spaces is said to be finitely strictly singular if for every ? > 0 there exists n such that every subspace E ? X with dimE > n contains a vector x such that ?Tx? < ??x?. We show that, for 1 6 p < q < ∞, the formal inclusion operator from Jp to Jq is finitely strictly singular. As a consequence, we obtain that the strictly singular operator with no invariant subspaces constructed by C. Read is actually finitely strictly singular. These results are deduced from the following fact: if k 6 n then every k-dimensional subspace of Rn contains a vector x with ?x??∞ = 1 such that xmi = (?1)i for some m1 < · · · < mk. 1. Introduction Recall that an operator T : X ? Y between Banach spaces is said to be strictly singular if for every ? > 0 and every infinite dimensional subspace E ? X there is a vector x in the unit sphere of E such that ?Tx? < ?. Furthermore, T is said to be finitely strictly singular if for every ? > 0 there exists n ? N such that for every subspace E ? X with dimE > n there exists a vector x in the unit sphere of E such that ?Tx? < ?.

• finitely strictly


• containing u¯1


• ?u¯1 ?u¯1


• inclusion operator


• every ?


• singular


• jp ?


• operator has

Sujets

• Singular

Informations

• Publié par: mijec
• Nombre de lectures: 19
• Langue: English

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FINITELY STRICTLY SINGULAR OPERATORS
BETWEEN JAMES SPACES
ISABELLE CHALENDAR, EMMANUEL FRICAIN, ALEXEY I. POPOV, DAN TIMOTIN,
AND VLADIMIR G. TROITSKY
Abstract. An operator T: X → Y between Banach spaces is said to be finitely
strictly singular if for every ε > 0 there exists n such that every subspace E ⊆ X
with dimE > n contains a vector x such that kTxk < εkxk. We show that, for
16p<q <∞,theformalinclusionoperatorfromJ toJ isfinitelystrictlysingular.p q
As a consequence, we obtain that the strictly singular operator with no invariant
subspaces constructed by C. Read is actually finitely strictly singular. These results
narededucedfromthefollowingfact: ifk6ntheneveryk-dimensionalsubspaceofR
icontains a vector x with kxk = 1 such that x = (−1) for some m <···<m .ℓ m 1 k∞ i
1. Introduction
Recall that an operator T : X → Y between Banach spaces is said to be strictly
singular if for every ε > 0 and every infinite dimensional subspace E ⊆ X there is
a vector x in the unit sphere of E such that kTxk < ε. Furthermore, T is said to be
finitely strictly singular if for every ε > 0 there exists n ∈N such that for every
subspace E ⊆X with dimE>n there exists a vector x in the unit sphere of E such
that kTxk < ε. Finitely strictly singular operators are also known in literature as
superstrictly singular. Note that
compact ⇒ finitely strictly singular ⇒ strictly singular,
and that each of these three properties defines a closed subspace inL(X,Y ). Actually,
each property defines an operator ideal. We refer the reader to [2, 7, 9, 10, 11, 14] for
more information about strictly and finitely strictly singular operators. All the Banach
spaces in this paper are assumed to be over real scalars.
We say that a subspace E ⊆ X is invariant under an operator T : X → X if
{0} =E =X andT(E)⊆E. Every compact operator has invariant subspaces by [1].
On the other hand, Read constructed in [12] an example of a strictly singular operator
without nontrivial closed invariant subspaces (this answered a question of Pe lczyn´ski).
Read’s operator acts on an infinite direct sum which involves James spaces. Recall
Date: September 4, 2008.
1
662 I. CHALENDAR, E. FRICAIN, A. I. POPOV, D. TIMOTIN, AND V. G. TROITSKY
∞that James’ p-space J is a sequence space consisting of all sequences x = (x ) inp n n=1
c satisfyingkxk <∞ where0 Jp
n−1 1 n oX
ppkxk = sup |x −x | : 16k <···<k , n∈NJ k k 1 np i+1 i
i=1
is the norm inJ . For more information on James’ spaces we refer the reader to [3, 6,p
7, 8, 13].
It was an open question whether every finitely strictly singular operator has invariant
subspaces. Some partial results in this direction were obtained in [2, 11]. We answer
this question in the negative by showing that the operator in [12] is, in fact, finitely
strictly singular. As an intermediate result, we prove that the formal inclusion operator
from J toJ with 16p<q<∞ is finitely strictly singular. The latter statement inp q
a certain sense refines the result of Milman [9] that the formal inclusion operator from
ℓ to ℓ with 16p<q<∞ is finitely strictly singular.p q
nMilman’s proof is based on the fact that every k-dimensional subspace E of R
contains a vector “with a flat”, namely, a vector x with sup-norm one with (at least)
k coordinates equal in modulus to 1. For such a vector, one has kxk ≪kxk . Theℓ ℓq p
proofs of our results are based on the following refinement of this observation. We
will show thatx can be chosen so that thesek coordinates have alternating signs. For
such a “highly oscillating” vectorx one haskxk ≪kxk . More precisely, a finite orJ Jq p
infinite sequence of real numbers in [−1, 1] will be called a zigzag of order k if it has
a subsequence of the form (−1, 1,−1, 1,... ) of length k. Our results will be based on
the following theorem; two different proofs of it will be presented in Sections 2 and 3.
nTheorem 1. For every k6n, every k-dimensional subspace ofR contains a zigzag
of order k.
Corollary 2. Let k∈N, then every k-dimensional subspace of c contains a zigzag of0
order k.
nProof. LetF be a subspace ofc with dimF =k. For everyn∈N, defineP : c →R0 n 0
∞ nvia P : (x ) → (x ) . Let n be such that dimP (F ) = k. There exists n suchn i i 1 n 2i=1 i=1 1
that every vector in F attains its norm on the first n coordinates. Indeed, define2

g : F \{0}→N via g(x) = max i : |x| =kxk . Then g is upper semi-continuous,i ∞

hence bounded on the unit sphere ofF , so that we putn = max g(x) : x∈F,kxk =2

1 .FINITELY STRICTLY SINGULAR OPERATORS BETWEEN JAMES SPACES 3
nPutn = max{n ,n }. SinceP (F ) is ak-dimensional subspace ofR , by Theorem 11 2 n
there exists x∈F such that P x is a zigzag of order k. It follows from our definitionn
of n that x is a zigzag of order k in F .
Suppose that 1 6 p < q. Since kxk is defined as the supremum of ℓ -norms ofJ pp
certain sequences, k· 6k· implies k· 6k· . It follows that J ⊆J and theℓ ℓ J J p qq p q p
formal inclusion operator i : J → J has norm 1. We show next that it is finitelyp,q p q
strictly singular. The proof is analogous to that of Proposition 3.3 in [14]. The main
difference, though, is that we use Corollary 2 instead of the simpler lemma from [9, 14].
Theorem 3. If 1 ≤ p < q < ∞ then the formal inclusion operator i : J → J isp,q p q
finitely strictly singular.
q−pq pProof. Given any x ∈ J , then |x −x | 6 2kxk |x −x | for every i,j ∈ N,p i j ∞ i j
pp1− qqso that kxk 6 2kxk kxk . Fix an arbitrary ε > 0. Let k ∈ N be suchJ ∞q Jp
1 1− 1p qthat (k − 1) > . Suppose that E is a subspace of J with dimE = k. Bypε
Corollary 2, there is a zigzag z ∈ E of order k. By the definition of norm in J , wep
1
phavekzk > 2(k− 1) .Jp
1
z 1 −
pPut y = . Then y∈E withkyk = 1. Obviously,kyk 6 (k− 1) , so thatJ ∞pkzk 2Jp
p
1 1
− q
q pki (y)k =kyk 6 (k− 1) kyk <ε.p,q J Jq q Jp
Hence, i is finitely strictly singular. p,q
We will now use Theorem 3 to show that the strictly singular operatorT constructed
by Read in [12] is finitely strictly singular. Let us briefly outline those properties ofT
that will be relevant for our investigation. The underlying spaceX for this operator is
defined as theℓ -direct sum ofℓ andY ,X = (ℓ ⊕Y ) , whereY itself is theℓ -direct2 2 2 ℓ 22
L
∞sum of an infinite sequence of J -spaces Y = J , with (p ) a certain strictlyp p iii=1 ℓ2
increasing sequence in (2, +∞). The operatorT is a compact perturbation of 0⊕W ,1
where W : Y →Y acts as a weighted right shift, that is,1
W (x ,x ,x ,... ) = (0,β x ,β x ,β x ,... ), x ∈J1 1 2 3 1 1 2 2 3 3 i pi
with β → 0. Note that one should rather write βi x instead of βx . Clearly, iti i p ,p i i ii i+1
suffices to show that W is finitely strictly singular.1
For n∈N, define V : Y →Y vian
V (x ,x ,x ,... ) = (0,β x ,...,β x , 0, 0... ), x ∈J .n 1 2 3 1 1 n n i pi4 I. CHALENDAR, E. FRICAIN, A. I. POPOV, D. TIMOTIN, AND V. G. TROITSKY
It follows from β → 0 that kV −W k→ 0. Since finitely strictly singular operatorsi n 1
from Y to Y form a closed subspace of L(Y ), it suffices to show that V is finitelyn
strictly singular for every n. Given n∈N, one can write
nX
V = βj i P,n i i+1 p ,p ii i+1
i=1
where P : Y →J is the canonical projection and j : J →Y is the canonical inclu-i p i pi i
sion. Thus, V is finitely strictly singular because finitely strictly singular operatorsn
form an operator ideal. This yields the following result.
Theorem 4. Read’s operator T is finitely strictly singular.
In the remaining two sections, we present two different proofs of Theorem 1, one
based on combinatorial properties of polytopes and the other based on the geometry
of the set of all zigzags and algebraic topology.
2. Proof of Theorem 1 via combinatorial properties of polytopes
kBy a polytope inR we mean a convex set which is the convex hull of a finite set. A
set is a polytope iff it is bounded and can be constructed as the intersection of finitely
many closed half-spaces. A facet of P is a face of (affine) dimension k− 1. We refer
the reader to [5, 15] for more details on properties of polytopes.
A polytope P is centrally symmetric iff it can be represented as the absolutely
convex hull of its vertices, that is, P = conv{±u¯ ,...,±u¯ } where ±u¯ ,...,±u¯ are1 n 1 n
the vertices of P . Clearly, P is centrally symmet