A PROPERTY OF STRICTLY SINGULAR 1-1 OPERATORS

A PROPERTY OF STRICTLY SINGULAR - OPERATORS GEORGE ANDROULAKIS, PER ENFLO Abstract We prove that if T is a strictly singular - operator defined on an infinite dimensional Banach space X, then for every infinite dimensional subspace Y of X there exists an infinite dimensional subspace Z of X such that Z Y is infinite dimensional, Z contains orbits of T of every finite length and the restriction of T on Z is a compact operator.. Introduction An operator on an infinite dimensional Banach space is called strictly singular if it fails to be an isomorphism when it is restricted to any infinite dimensional subspace by operator we will always mean a continuous linear map. It is easy to see that an operator T on an infinite dimensional Banach space X is strictly singular if and only if for every infinite dimensional subspace Y of X there exists an infinite dimensional subspace Z of Y such that the restriction of T on Z, T Z : Z X, is a compact operator. Moreover, Z can be assumed to have a basis. Compact operators are special examples of strictly singular operators. If p < q then the inclusion map i p,q : l p l q is a strictly singular non-compact operator. A Hereditarily Indecomposable H.I. Banach space is an infinite dimensional space such that no subspace can be written as a topological sum of two infinite dimensional subspaces. W.T. Gowers and B. Maurey constructed the first example of an H.I. space [8]. It is also proved in [8] that every operator on a complex H.I. space can be written as a strictly singular perturbation of a multiple of the identity. If X is a complex H.I. space and T is a strictly singular operator on X then the spectrum of T resembles the spectrum of a compact operator on a complex Banach space: it is either the singleton {0} i.e. T is quasi-nilpotent, or a sequence {λ n : n =,,...} {0} where λ n is an eigenvalue of T with finite multiplicity for all n, and λ n n converges to 0, if it is an infinite sequence. It was asked whether there exists an H.I. space X which gives a positive solution to the Identity plus Compact problem, namely, every operator on X is a compact perturbation of a multiple of the identity. This question was answered in negative in [] for the H.I. space constructed in [8], for related results see [7], [9], and []. By [3], or the more general beautiful theorem of V. Lomonosov [0], if a Banach space gives a positive solution to the Identity plus Compact problem, it also gives a positive solution to the famous Invariant Subspace Problem I.S.P.. The I.S.P. asks whether there exists a separable infinite dimensional Banach space on which every operator has a non-trivial invariant subspace, by non-trivial we mean different than {0} and the whole space. It remains unknown whether l is a positive solution to the I.S.P.. Several negative solutions to the I.S.P. are known [4], [5], [], [], [3]. In particular, there exists a strictly singular operator with no non-trivial invariant subspace [4]. It is unknown whether every strictly singular operator on a super-reflexive Banach space has a non-trivial invariant subspace. Our main result Theorem. states 99 Mathematics Subject Classification. 47B07,46B03. The research was partially supported by NSF.

that if T is a strictly singular - operator on an infinite dimensional Banach space X, then for every infinite dimensional Banach space Y of X there exists an infinite dimensional Banach space Z of X such that Z Y is infinite dimensional, the restriction of T on Z, T Z : Z X, is compact, and Z contains orbits of T of every finite length i.e. for every n N there exists z n Z such that {z n, T z n, T z n,..., T n z n } Z. We raise the following Question. Let T be a quasi-nilpotent operator on a super-reflexive Banach space X, such that for every infinite dimensional subspace Y of X there exists an infinite dimensional subspace Z of X such that Z Y is infinite dimensional, T Z : Z X is compact and Z contains orbits of T of every finite length. Does T have a non-trivial invariant subspace? By our main result, an affirmative answer to the above question would give that every strictly singular, -, quasi-nilpotent operator on a super-reflexive Banach space has a nontrivial invariant subspace; in particular, we would obtain that every operator on the superreflexive H.I. space constructed by V. Ferenczi [6] has a non-trivial invariant subspace, and thus the I.S.P. would be answered in affirmative. Our main result is. The main result Theorem.. Let T be a strictly singular - operator on an infinite dimensional Banach space X. Then, for every infinite dimensional subspace Y of X there exists an infinite dimensional subspace Z of X, such that Z Y is infinite dimensional, Z contains orbits of T of every finite length, and the restriction of T on Z, T Z : Z X, is a compact operator. The proof of Theorem. is based on Theorem.3. We first need to define the biorthogonal constant of a finite set of normalized vectors of a Banach space. Definition.. Let X be a Banach space, n N, and x, x,..., x n be normalized elements of X. We define the biorthogonal constant of x,..., x n to be { } n bc{x,..., x n } := sup max α,..., α n : α i x i =. Notice that bc{x,..., x n } = inf { n β i x i : i= i= max β i = i n and that bc{x,..., x n } < if and only if x,..., x n are linearly independent. Before stating Theorem.3 recall that if T is a quasi-nilpotent operator on a Banach space X, then for every x X and η > 0 there exists an increasing sequence i n n= in N such that T in x η T in x. Theorem.3 asserts that if T is a strictly singular - operator on a Banach space X then for arbitrarily small η > 0 and k N there exists x X, x =, such that T i x η T i x for i =,,..., k +, and moreover, the biorthogonal constant of x, T x/ T x,..., T k x/ T k x does not exceed / η. Theorem.3. Let T be a strictly singular - operator on a Banach space X. Let Y be an infinite dimensional subspace of X, F be a finite codimensional subspace of X and k N. Then there exists η 0 0, such that for every 0 < η η 0 there exists x Y, x = satisfying },

a T i { x F and T i x } η T i x for i =,,..., k +, and T x b bc x,,..., T k x T x T k x η, where T 0 denotes the identity operator on X. We postpone the proof of Theorem.3. Proof of Theorem.. Let T be a strictly singular - operator on an infinite dimensional Banach space X, and Y be an infinite dimensional subspace of X. Inductively for n N we construct a normalized sequence z n n Y, an increasing sequence of finite families z j j Jn of normalized functionals on X i.e. J n n is an increasing sequence of finite index sets, and a sequence η n n 0,, as follows: For n = apply Theorem.3 for F = X set J =, k =, to obtain η < / 6 and z Y, z = such that T i z < η T i z for i =,, and bc{z, T z T z } <. η For the inductive step, assume that for n, z i n i= Y, z j j Ji i =,..., n, and η i n i= have been constructed. Let J n be a finite index set with J n J n and x j j Jn be a set of normalized functionals on X such that for every x span{t 3 i z j : j n, 0 i j} there exists j 0 J n such that x j 0 x x /. Apply Theorem.3 for F = j Jn kerx j, and k = n, to obtain η n < /n n+4 and z n Y, z n = such that 4 T i z n F and T i z n < η n T i z n for i =,,..., n +, and 5 bc{z n, T z n T z n,..., T n z n T n z n } <. ηn This finishes the induction. Let Z = span{t i z n : n N, 0 i n}, and for n N, let Z n = span{t i z n : 0 i n}. Let x Z with x = and write x = n= x n where x n Z n for all n N. We claim that 6 T x n < for all n N. n Indeed, write x = n n= i=0 a i,n T i z n T i z n and x n = n i=0 a i,n T i z n for n N. T i z n Fix n N and set x n = x + x + + x n. Let j 0 J n+ such that x n x j 0 x n by 3 for n replaced by n = x j 0 x since for n + m, J n+ J m thus by 4, x m kerx j 0 x j 0 x =. 3

Thus x n = x n x n x n + x n 4 where x 0 = 0. Hence, by and 5 we obtain that 7 a i,n 4bc{ T i z n T i z n Therefore T x n = n i=0 n i=0 : i = 0,..., n} 4 ηn for i = 0,..., n. a i,n T i+ z n T i z n n i=0 a i,n T i+ z n T i z n 4 ηn η n by, 4, and 7 = 4n η n < n by the choice of η n, which finishes the proof of 6. Let Z to be the closure of Z. Notice that Z Y {z n : n N} thus Z Y is infinite dimensional. We claim that T Z : Z X is a compact operator, which will finish the proof of Theorem.. Indeed, let y m m Z where for all m N we have y m =, and write y m = n= y m,n where y m,n Z n for all n N. It suffices to prove that T y m m has a Cauchy subsequence. Indeed, since Z n is finite dimensional for all n N, there exists ym m a subsequence of y m m such that T ym, m is Cauchy with the obvious notation that if ym = y p for some p, then ym,n = y p,n. Let ym m be a subsequence of ym m such that T ym, m is Cauchy with the obvious notation that if ym,n = y p for some p, then ym,n = y p,n. Continue similarly, and let ỹ m = ym m and ỹ m,n = ym,n m for all m, n N. Then for m N we have ỹ m = n= ỹm,n where ỹ m,n Z n for all n N. Also, for all n, m N with n m, ỹ t t m and ỹ t,n t m are subsequences of y m t t and y m t,n t respectively. Thus for all n N, T ỹ t,n t N is a Cauchy sequence. We claim that T ỹ m m is a Cauchy sequence. Indeed, for ε > 0 let m 0 N such that / m 0 < ε and let m N such that 8 T ỹ s,n T ỹ t,n < ε m 0 for all s, t m and n =,,... m 0. Thus for s, t m we have T ỹ s T ỹ t = T ỹ s,n T ỹ t,n n= m 0 T ỹ s,n T ỹ t,n + n= < m 0 ε m 0 + n=m 0 + n=m 0 + T ỹ s,n + by 6 and 8 n = ε + m 0 < ε by the choice of m 0, n=m 0 + T ỹ t,n which proves that T ỹ m m is a Cauchy sequence and finishes the proof of Theorem.. For the proof of Theorem.3 we need the next two results. 4

Lemma.4. Let T be a strictly singular - operator on an infinite dimensional Banach space X. Let k N and η > 0. Then for every infinite dimensional subspace Y of X there exists an infinite dimensional subspace Z of Y such that for all z Z and for all i =,..., k we have that T i z η T i z where T 0 denotes the identity operator on X. Proof. Let T be a strictly singular - operator on an infinite dimensional Banach space X, k N and η > 0. We first prove the following Claim: For every infinite dimensional linear submanifold not necessarily closed W of X there exists an infinite dimensional linear submanifold Z of W such that T z η z for all z Z. Indeed, since W is infinite dimensional there exists a normalized basic sequence z i i N in W having biorthogonal constant at most equal to, such that T z i η/ i+ for all i N. Let Z = span{z i : i N} be the linear span of the z i s. Then Z is an infinite dimensional linear submanifold of W. We now show that Z satisfies the conclusion of the Claim. Let z Z and write z in the form z = Σλ i z i for some scalars λ i such that at most finitely many λ i s are non-zero. Since the biorthogonal constant of z i i is at most equal to, we have that λ i 4 z for all i. Thus T z = λ i T z i i i λ i T z i i 4 z η = η z i+ which finishes the proof of the Claim. Let Y be an infinite dimensional subspace of X. Inductively for i = 0,,..., k, we define Z i, a linear submanifold of X, such that a Z 0 is an infinite dimensional linear submanifold of Y and Z i is an infinite dimensional linear submanifold of T Z i for i. b T z η z for all z Z i and for all i 0. Indeed, since Y is infinite dimensional, we obtain Z 0 by applying the above Claim for W = Y. Obviously a and b are satisfied for i = 0. Assume that for some i 0 {0,,..., k }, a linear submanifold Z i0 of X has been constructed satisfying a and b for i = i 0. Since T is - and Z i0 is infinite dimensional we have that T Z i0 is an infinite dimensional linear submanifold of X and we obtain Z i0 + by applying the above Claim for W = T Z i0. Obviously a and b are satisfied for i = i 0 +. This finishes the inductive construction of the Z i s. By a we obtain that Z k is an infinite dimensional linear submanifold of T k Y. Let W = T k Z k. Then W is an infinite dimensional linear submanifold of X. Since Z k T k Y and T is -, we have that W Y. By a we obtain that for i = 0,,..., k we have Z k T k i Z i, hence T i W = T i T k Z k = T k i Z k T k i T k i Z i = Z i since T is -. Thus by b we obtain that T i z η T i z for all z W and i =,,..., k. Obviously, if Z is the closure of W then Z satisfies the statement of the lemma. 5

Corollary.5. Let T be a strictly singular - operator on an infinite dimensional Banach space X. Let k N, η > 0 and F be a finite codimensional subspace of X. Then for every infinite dimensional subspace Y of X there exists an infinite dimensional subspace Z of Y such that for all z Z and for all i =,..., k + T i z F and T i z η T i z where T 0 denotes the identity operator on X. Proof. For any linear submanifold W of X and for any finite codimensional subspace F of X we have that 9 dimw/f W dimx/f <. Indeed for any n > dimx/f and for any x,..., x n linear independent vectors in W \F W we have that there exist scalars λ,..., λ n with λ,..., λ n 0,..., 0 and n λ i x i F since n > dimx/f. Thus n λ i x i F W which implies 9. i= Let RT denote the range of T. Apply 9 for W = RT to obtain 0 dimrt /RT F dimx/f <. Since T is - we have that dimx/t F dimrt /RT F. Indeed, for any n > dimrt /RT F and for any x,..., x n linear independent vectors of X\T F, we have that T x,..., T x n are linear independent vectors of RT \T T F = RT \F since T is -. Thus T x,..., T x n RT \RT F and since n > dimrt /RT n F, there exist scalars λ,..., λ n n RT F. Therefore T combining 0 and we obtain with λ,..., λ n 0,..., 0 such that λ i x i F, and hence n i= i= dimx/t F <. By we have that 3 dimx/t i F <, for i =,,..., k. i= i= λ i T x i λ i x i T F, which proves. By Thus dimx/w < where W = F T F T k F. Therefore if we apply 9 for W = Y and F = W we obtain 4 dimy/y W dimx/w <, and therefore Y W is infinite dimensional. Now use Lemma.4, replacing Y by Y W, to obtain an infinite dimensional subspace Z of Y W such that T i z η T i z for all z Z and i =,..., k +. Notice that for z Z and i =,..., k we have that z W thus T i z F. Now we are ready to give the 6

Proof of Theorem.3. We prove by induction on k that for every infinite dimensional subspace Y of X, finite codimensional subspace F of X, k N, function f : 0, 0, such that fη 0 as η 0, and for i 0 {0} N, there exists η 0 > 0 such that for every 0 < η η 0 there exists x Y, x = satisfying a T i { x F and T i x η T i } x for i =,,..., i 0 + k +. b T bc i 0 x, T i 0 + x,..., T i 0 +k x. T i 0 x T i 0 + x T i 0 +k x fη For k = let Y, F, f, and i 0 as above, and let η 0 0, satisfying 5 fη 0 < 6. Let 0 < η η 0. Apply Corollary.5 for k and η replaced by i 0 + and η/4 respectively, to obtain an infinite dimensional subspace Z of Y such that for all z Z and for i =,,..., i 0 + 6 T i z F and T i z η 4 T i z. Let x Z with x =. If bc{t i 0 x / T i 0 x, T i0+ x / T i0+ x } /fη then x satisfies a and b for k =, thus we may assume that { T i 0 } x 7 bc T i 0 x, T i0+ x > T i 0+ x fη. Let 8 0 < η η 4 min i i 0 T i0 x T i x min i 0 <i i 0 + T i x T i 0 x fη. Let z, z X, z = z =, z T i 0 x = T i 0 x and z T i 0+ x = T i 0+ x. Since ker z ker z is finite codimensional and T is -, by 3 we have that 9 dimx/t i 0 ker z ker z <. Apply Corollary.5 for F, k and η replaced by F T i 0 ker z ker z, i 0 + and η respectively, to obtain an infinite dimensional subspace Z of Y such that for all z Z and for all i =,,..., i 0 + 0 T i z F T i 0 ker z ker z and T i z η T i z. Let x X with x = x x = and let x Z ker x with T i 0 x = T i 0 and let x = x + x / x +. We will show that x satisfies a and b for k =. We first show that a is satisfied for k =. Since x, T x,..., T i 0+ x F by 6 and x, T x,..., T i 0+ x F by 0 we have that x, T x,..., T i 0+ x F. Before showing that the norm estimate of a is satisfied, we need some preliminary estimates: -3. If i < i 0 assuming that i 0 then 7

T i x = T i 0 T i 0 x x T i x T i 0 x η by 8 = T i 0 η by ηi 0 i T i η by applying 0 for z = x, i 0 i times T i since η by 8. Thus, by, for i < i 0 assuming that i 0 we have 3 T i x x + = T i x + T i T i x + T i 3 T i and 4 T i x x + = T i x + T i T i T i x T i. Also notice that 5 T i 0 x x + = T i 0 x + T i 0 T i 0 x + T i 0 = T i 0 x by, and 6 T i 0 x x + = T i 0 x + T i 0 z T i 0 x + T i 0 x = z T i 0 x = T i 0 x by 0 for z = x and i =. Also for i 0 < i i 0 + we have that by applying 0 for z = x, i i 0 times, we obtain T i η i i 0 T i 0 η T i 0 x by η < and T i 0 x = η T i x T i x 7 8 fη T i x by 8 T i x. Thus for i 0 < i i 0 + we have T i x x + = T i x + T i T i x + T i 9 3 T i x by 8. 8

Also for i 0 < i i 0 + we have T i x x + = T i x + T i T i x T i 30 T i x by 8. Later in the course of this proof we will also need that 3 T i 0+ x x + = T i 0+ x + T i 0+ T i 0+ x T i 0+ fη T i 0+ T i 0+ by 7 = fη T i0+ fη fη T i0+ since fη <. Finally we will show that for i i 0 + we have that T i x η T i x. Indeed if i = then T i x = x + T x + T x + T x + T η x + 4 x + η by 6 z = x, and 0 z = x η x + 4 x + η x + + x η = x + 4 + η x x + η by the choice of x η = x + 4 + η x x + x + η since x ker x 3 η 4 + η since x = η If < i < i 0 assuming that 3 i 0 we have that since η < η 4 by 8. 33 T i 3 x T i x x i by 3 and 4 < 3η by 0 < η by 8. 9

If i = i 0 > then 34 T i x T i x T i0 x T i 0 by 5 and 4 = 4 T i 0 T i 0 by < 4η by 0 for z = x and i = < η by 8. If i 0 < i i 0 + then 35 T i 3 x T i x x i x by 9 and 30 < η by 6 for z = x. Now 3, 33, 34 and 35 yield that for i i 0 + we have T i x η T i x, thus x satisfies a for k =. Before proving that x satisfies b for k = we need some preliminary estimates: 36-40. By 7 there exist scalars a 0, a with max a 0, a = and w < fη where T i 0 x 36 w = a 0 T i 0 x + a T i0+ x T i 0+ x. Therefore a 0 a = a T i 0 x 0 T i 0 x T i0+ x a T i 0+ x w < fη. Thus fη a 0, a and hence 37 a a 0 a 0 fη. Also by 36 we obtain that and thus 38 T i 0 x = Let T i 0 x = T i 0 x a 0 w T i 0 x a a 0 T i0+ x T i 0+ x T i 0 x w T i 0 x a T i0+ x x + a 0 a 0 T i 0+ x + T i 0 x. 39 w = T i 0 x + T i 0 x a T i0+ x x + a 0 T i 0+ x T i0 x x +. 0

Notice that 38 and 39 imply that w = T i 0 x / x + a 0 w and hence 40 w = T i 0 x x + a 0 w fη T i 0 x x + = fη z T i 0 x x + T i0 x fη using 37 and w < fη x + fη < by 5 fη since by the choice of z = fη z T i 0 x + T i 0 x by 0 for i = and z = x x + fη T i 0 x + x since z x + = = fη T i 0 x. Now we are ready to estimate the bc{t i 0 x/ T i 0 x, T i0+ x/ T i0+ x }. Let scalars A 0, A such that A T i 0 x T i0+ x 0 + A T i 0 x T i 0+ x =. We want to estimate the max A 0, A. By 39 we have = A 0 w T i 0 x a T i0+ x T i 0 x x + a 0 T i 0+ x + T i0 x T i0+ x + A x + T i 0+ x = A 0 T i 0 T i 0 x T i 0 x x + T i 0 x + A0 T i 0 x a A T i0+ x T i 0 + x + T i 0 x x + a 0 T i 0+ x x + T i 0+ x + A 0 A T i0+ x w + T i 0 x T i 0+ x x + A 0 T i 0 T i 0 x T i 0 x x + T i 0 x + A0 T i 0 x a A T i0+ x T i 0 + x + T i 0 x x + a 0 T i 0+ x x + T i 0+ x 4 A 0 fη A fη by the triangle inequality, 40 and 3. By 0 for i = we have that T i 0 x ker z and since zt i0+ x = T i0+ x it is easy to see that bc{t i 0 x / T i 0, T i0+ x / T i0+ x }. Thus 4 implies that 4 A 0 T i 0 x a A T i0+ x + T i 0 x x + a 0 T i 0+ x x + + 4fη A 0 + fη A and 43 Notice that 43 implies that A 0 T i 0 T i 0 x x + + 4fη A 0 + fη A. 44 A 0 4 + 8fη A 0 + 4fη A,

since T i 0 x x + T i 0 x by. Also by 4 we obtain A T i 0+ x T i 0+ x x + = T i 0 x + T i 0 T i 0 x T i 0 x + T i 0 T i 0 x = A 0 T i0 x a T i 0 x x + a 0 + 4fη A 0 + fη A. Thus 45 A 3 A 0 fη + 4fη A 0 + fη A by 9 for i = i 0 +, 37 and T i 0 x T i 0 x x + = Notice that 45 implies that T i0 x T i 0 x + T i 0 x T i0 x z T i 0 x + T i 0 x since z = = T i 0 x since x zt i T i 0 ker z 0 x by 0 for i = and z = x = by the choice of z. 46 A 6 + 8 5 A 0 since fη < /6 by 5. By substituting 46 into 44 we obtain A 0 4 + 8fη A 0 + 4fη 6 + 8 5 A 0 = 4 + 4fη + 5 fη A 0 5 + A 0 since fη < 5 4 Thus A 0 0. Hence 46 gives that A 6. Therefore { T i 0 } x T i0+ x bc, 6 T i 0 x T i 0+ x fη by 5. by 5. We now proceed to the inductive step. Assuming the inductive statement for some integer k, let a finite codimensional subspace F of X, f : 0, 0, with fη 0 as η 0 and i 0 N {0}. By the inductive statement for i 0, f and η replaced by i 0 +, f /4 and η/4 respectively, there exists η s.t. for 0 < η < η there exists x X, x = 47 T i x F and T i x η 4 T i x for i =,,..., i 0 + + k + and { T i 0 } + x 48 bc T i 0+ x, T i0+ x T i 0+ x,..., T i0++k x T i 0++k x fη /4.

Let η 0 satisfying 49 η 0 < η, fη 0 < 88, fη 0 <, 44k + let 0 < η < η 0 and let x X, x = satisfying 47 and 48. If { T i 0 } x bc T i 0 x, T i0+ x T i 0+ x,..., T i0+k+ x T i 0+k+ x fη then x satisfies the inductive step for k replaced by k +. Thus we may assume that { T i 0 } x 50 bc T i 0 x, T i0+ x T i 0+ x,..., T i0+k+ x > T i 0+k+ x fη. Let 5 0 < η < η 4 min i i 0 T i0 x T i x min i 0 <i i 0 +k+ T i x T i 0 x fη. Let J {, 3,...} be a finite index set and z, z j j J be norm functionals such that 5 z T i 0 x = T i 0 x, and 53 for every z span{t i0+ x,..., T i0+k+ x } there exists j 0 J with zj 0 z z. Since T is - we obtain by 3 that dimx/t i 0 ker zj <. Apply Corollary.5 j {} J for F, k, η replaced by F T i 0 ker zi, i 0 + k +, η respectively, to obtain an j {} J infinite dimensional subspace Z of Y such that for all z Z and for all i =,,..., i 0 +k + 54 T i z F T i 0 ker zj and T i z η T i z. j {} J Let x X, x = = x x and let x Z ker x with 55 T i 0 x = T i 0 and let x = x + x / x +. We will show that x satisfies the inductive statement for k replaced by k +. We first show that x satisfies a for k replaced by k +. The proof is identical to the verification of a for k =. The formulas 7, 8, 9, 30, and 35 are valid for i 0 < i i 0 + k +, and 3 is valid if i 0 + is replaced by any i {i 0 +,..., i 0 + k + }, and this will be assumed in the rest of the proof when we refer to these formulas. We now prove that b is satisfied for k replaced by k +. By 50 there exist scalars a 0, a,..., a k+ with max a 0, a,..., a k+ = and w < fη where k+ T i0+i x 56 w = a i T i 0+i x. i=0 3

We claim that 57 a 0 fη/4. Indeed, if a 0 < fη /4 / then max a,..., a k+ = and k+ T i0+ x a i T i 0+i x = w a T i 0 x 0 T i 0 x i= w + a 0 < fη + fη/4 < fη /4 since fη < /4 by 49 which contradicts 48. Thus 57 is proved. By 56 we obtain and thus 58 T i 0 x = Let 59 w = T i 0 x + T i 0 x = T i 0 x k+ a i w T i 0 x T i0+i x a 0 a i= 0 T i 0+i x T i 0 x k+ a i w T i 0 x T i0+i x x + a 0 a i= 0 T i 0+i x + T i 0 x. k+ i= a i a 0 T i0 x T i0+i x x + T i 0+i x T i0 x x +. Notice that 58 and 59 imply that w = T i 0 x / x + a 0 w and hence 60 w = T i 0 x x + a 0 w < = z T i 0 x x + fη3/4 by 5 T i0 x x + fη3/4 by w fη and 57 = z T i 0 x + T i 0 x fη 3/4 by 54 for i = and z = x x + T i 0 x + x fη 3/4 since z x + = = T i 0 x fη 3/4. Now we are ready to estimate the bc{t i0+i x / T i0+i x : i = 0,,..., k + }. Let scalars A 0, A,..., A k+ such that k+ T i0+i x A i T i 0+i x =. i=0 4

We want to estimate the max A 0, A,..., A k+. By 59 we have A 0 k+ a i T i0 x T i0+i x = w T i 0 x a i= 0 x + T i 0+i x + T i0 x k+ T i0+i x + A i x + T i 0+i x i= A 0 T i 0 T i 0 k+ x = T i 0 x x + T i 0 x + ai A 0 T i0 x a i= 0 T i 0 x x + + A i T i0+i x T i 0 +i x T i 0+i x x + T i 0+i x + A k+ 0 T i0+i x w + A T i i 0 x T i 0+i x x i= + A 0 T i 0 T i 0 k+ x T i 0 x x + T i 0 x + ai A 0 T i0 x T i 0 x x + + A i T i0+i x T i 0 +i x T i 0+i x x + T i 0+i x i= a 0 6 k+ A 0 fη 3/4 A i fη by 60 and 3; see the paragraph above 56. i= By 54 for i = and z = x we obtain that T i 0 x ker zj and by 53 and 48 it is easy to see that j J { T i 0 } x bc T i 0 x, T i0+i x T i 0+i x : i =,..., k + Since fη < 4 3 by 49, we have that 3 /fη /4, hence { T i 0 } x bc T i 0 x, T i0+i x T i 0+i x : i =,..., k + 3. fη /4 fη /4. Thus 6 implies that T i 0 6 A 0 T i 0 x x + k+ + fη 3/4 A fη /4 0 + A j fη, and for i =,..., k + 63 a i A 0 T i0 x T i 0 x x + + A i T i0+i x T i 0+i x x + k+ + fη 3 fη /4 4 A0 + A j fη. a 0 Since T i 0 x x + T i 0 x we have that 6 implies = T i 0 x + T i 0 T i 0 x j= T i 0 x + T i 0 T i 0 x 4 64 A 0 fη + k+ /4 8fη/ A 0 + 4 A j fη 3/4. 5 j= j= = by 55,

Notice also that 63 implies that for i =,..., k + T i0+i x A i T i 0+i x x + A 0 a i T i 0 x a 0 T i 0 x x + fη + 4fη k+ /4 A0 + A j fη 3 4. Thus 65 A i 3 A 0 fη /4 fη + 4fη k+ /4 A0 + A j fη 3 4 by 9 see the paragraph above 56, 57 and T i 0 x T i 0 x x + = T i0 x T i 0 x + T i 0 x For i =,..., k + rewrite 65 as A i 3 fη3/4 T i0 x z T i 0 x + T i 0 x j= j= since z = = T i 0 x since T i 0 x zt i ker z 0 by 54 for i = and z = x x = by 5. fη /4 + A i 3 fη /4 + 4fη / + Thus, since fη < 4/3 6 / 4 by 49, we obtain + fη /4 k+ A fη /4 0 + k+ A 0 + j= j i Hence, since /fη /4, we obtain that for i =,..., k + j= j i j= j i A j fη 3/4. 6 66 A i fη + 9 /4 fη A k+ 0 + 3 A /4 j fη 3/4. By substituting 64 in 66 we obtain that for i =,..., k +, 67 A i 6 36 + fη /4 fη + 7fη k+ / 4 A0 + 36 j= A j fη 3/4. k+ A j fη / + 3 A j fη 3/4. We claim that 64 and 67 imply that max{ A i : 0 i k + } /fη which finishes the proof. Indeed, if max{ A i : 0 i k + } = A 0 then 64 implies that A 0 4 fη + /4 8fη/ A 0 + 4k + A 0 fη 3/4 4 fη + /4 3 A 0 + 3 A 0 since fη < 6 4 j= j i k + 4 3 by 49

thus 68 A 0 fη /4 < fη since fη < 4/3 by49. Similarly, if there exists l {,..., k + } such that max{ A i : 0 i k + } = A l then 67 for i = l implies that 6 36 A l + fη /4 fη + 7fη / 4 Al + 36k + fη / A l + 3kfη 3/4 A l 4 fη + / 4 A l + 4 A l + 4 A l since /fη /4 /fη / and fη < 69 A l 68 fη / fη By 68 and 69 we have that max{ A i : proof. 88 4 44k+ by 49. Hence since fη < by 49 68. 0 i k + } /fη which finishes the References [] Androulakis, G.; Schlumprecht, Th. Strictly singular non-compact operators exist on the space of Gowers-Maurey, J. London Math. Soc. 64 00, no.3, 655-674. [] Androulakis, G.; Odell, E.; Schlumprecht, Th.; Tomczak-Jaegermann, N. On the structure of the spreading models of a Banach space, preprint. [3] Aronszajn, N.; Smith, K. T. Invariant subspaces of completely continuous operators, Ann. of Math. 60, 954. 345 350. [4] Enflo, P. On the invariant subspace problem in Banach spaces. Seminaire Maurey Schwartz 975 976 Espaces L p, applications radonifiantes et geometrie des espaces de Banach, Exp. Nos. 4-5, 7 pp. Centre Math., cole Polytech., Palaiseau, 976. [5] Enflo, P. On the invariant subspace problem for Banach spaces. Acta Math. 58 987, no. 3-4, 3 33. [6] Ferenczi, V. A uniformly convex hereditarily indecomposable Banach space. Israel J. Math. 0 997, 99 5. [7] Gasparis, I. Strictly singular non-compact operators on hereditarily indecomposable Banach spaces, preprint. [8] Gowers, W. T.; Maurey, B. The unconditional basic sequence problem. J. Amer. Math. Soc. 6 993, no. 4, 85 874. [9] Gowers, W.T. A remark about the scalar-plus-compact problem. Convex geometric analysis Berkeley, CA, 996, 5, Math. Sci. Res. Inst. Publ., 34, Cambridge Univ. Press, Cambridge, 999. [0] Lomonosov, V.I. Invariant subspaces of the family of operators that commute with a completely continuous operator. Russian Funkcional. Anal. i Priložen. 7 973, no. 3, 55 56. [] Read, C.J. A solution to the invariant subspace problem. Bull. London Math. Soc. 6 984, no. 4, 337 40. [] Read, C.J. A solution to the invariant subspace problem on the space l. Bull. London Math. Soc. 7 985, no. 4, 305 37. [3] Read, C.J. A short proof concerning the invariant subspace problem. J. London Math. Soc. 34 986, no., 335 348. [4] Read, C.J. Strictly singular operators and the invariant subspace problem. Studia Math. 3 999, no. 3, 03 6. 7

Department of Mathematics, University of South Carolina, Columbia, SC 908. giorgis@math.sc.edu Department of Mathematics, Kent State University, Kent, OH 4440. enflo@mcs.kent.edu 8