Almost-invariant and essentially-invariant halfspaces
aa r X i v : . [ m a t h . F A ] O c t ALMOST-INVARIANT AND ESSENTIALLY-INVARIANTHALFSPACES
GLEB SIROTKIN AND BEN WALLIS
Abstract.
In this paper we study sufficient conditions for an operator to havean almost-invariant half-space. As a consequence, we show that if X is an infinite-dimensional complex Banach space then every operator T ∈ L ( X ) admits anessentially-invariant half-space. We also show that whenever a closed algebra ofoperators possesses a common AIHS, then it has a common invariant half-spaceas well. Introduction and the first result
The invariant subspace problem in its full generality was famously closed in thenegative by Enflo in a series of papers beginning with [En76] and ending with [En87],where he constructed an operator acting on a separable Banach space which fails toadmit any nontrivial invariant subspace, that is, invariant subspace which is nonzero,proper, and closed. Independently Charles Read produced an outstanding series ofexamples [Re84, Re85, Re86, Re91, Re97], in particular, an operator acting on ℓ [Re85] which fails to admit any nontrivial invariant subspace. Thus, the answer isstill negative even for “nice” Banach spaces. Current work on the invariant subspaceproblem now focuses on the separable Hilbert space case, which remains unsolvedfor the time being.A related but independent problem was posed in [APTT09]. Let us say that asubspace Y of a Banach space X is almost-invariant under T whenever T Y ⊆ Y + E for some finite-dimensional error subspace E . In this case, the smallestpossible dimension of E we call the defect . To make things nontrivial, we restrictour attention to the case where Y is a halfspace , that is, a closed subspace withboth infinite dimension and infinite codimension in X . Let us use the abbreviations AIHS for “almost-invariant halfspace” and
IHS for “invariant halfspace.” Thenwe can ask, does every operator on an infinite-dimensional Banach space admit anAIHS? Let us call this the
AIHS problem .It turns out that for closed algebras of operators existence of a common AIHSand IHS is equivalent. In [MPR13, Theorem 2.3], this was proved for the case whenAIHS is complemented, but the proof can be adapted to the general case as well. Letus demonstrate this. Note also that, although our other results are validated onlyfor complex Banach spaces, this next theorem works even for real Banach spaces.
Theorem 1.1.
Let X be a (real or complex) Banach space, and let A be a norm-closed algebra of operators in L ( X ) . If there exists a halfspace Y of X which isalmost-invariant under every A ∈ A , then there exists a halfspace Z which is in-variant under every A ∈ A . The authors thank William B. Johnson for his helpful comments.Mathematics Subject Classification: 15A03, 15A18, 15A60, 47L10, 47A10, 47A11, 47A15Keywords: functional analysis, Banach spaces, surjectivity spectrum, point spectrum, invariantsubspaces.
Proof.
Let us assume, without loss of generality, that A is a unital algebra satisfyingthe hypothesis of the theorem. By a result of Popov ([Po10, Theorem 2.7]), there isnumber M ∈ N such that for every A ∈ A we have AY ⊆ Y + F A with dim( F A ) ≤ M .For each A ∈ A , define J A ∈ L ( Y, X/Y ) by J A := qA | Y , where q : X → X/Y isthe canonical quotient map. Let us also assume that there are vectors y , . . . , y M ∈ Y and that there is T ∈ A such that for W := J T ([ y i ] Mi =1 ) we have dim( W ) = M . Itwas proved in [Po10, Lemma 3.4] that if Y is an AIHS under T and V := Y ∩ T − ( Y )then the defect of Y under T is precisely dim( Y /V ). Thus, dim(
Y /V ) = M . Noticethat V = N ( J T ) and Y = V ⊕ [ y i ] Mi =1 .Fix any A ∈ A and any vector v ∈ V and let us show that J A v ∈ W holds.Indeed, if not, let δ > k P Mi =1 α i J T y i + α M +1 J A v k over theset of all ( α i ) M +1 i =1 from the unit sphere of K M +1 . The minimum is attained due tothe compactness of the sphere and, thus, strictly positive due to the linear indepen-dence of the vectors J T y , . . . , J T y M , J A v . Again by compactness, the maximum of k P Mi =1 α i J A y i k over the unit sphere in K M +1 exists. Hence, we could find λ > λ k P Mi =1 α i J A y i k < δ . Then, using the fact that J T v = 0, weobtain M + 1 linearly independent vectors( λJ A + J T ) y , · · · , ( λJ A + J T ) y M , ( λJ A + J T ) (cid:18) λ v (cid:19) . We can see this due to the fact that if ( α i ) M +1 i =1 is in the unit sphere of K M +1 thenwe would have (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) M X i =1 α i ( λJ A + J T ) y i + α M +1 ( λJ A + J T ) (cid:18) λ v (cid:19)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≥ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) M X i =1 α i J T y i + α M +1 J A v (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) − λ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) M X i =1 α i J A y i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) > . Since λJ A + J T = J λA + T we obtain a contradiction to the fact that the defect of λA + T ∈ A must be no larger than M .Thus, we have shown that there is an M -dimensional subspace F T ⊂ X such thatfor every A ∈ A we have AV ⊆ Y + F T . Now we have the following inclusion Z = span { Av : A ∈ A , v ∈ V } ⊂ Y + F T . Moreover, since V is finite-codimensional in a halfspace Y , it is a halfspace itself.Therefore, Z is infinite-dimensional as A is unital. We conclude by observing that Z is a half-space which is A -invariant due to A being an algebra.The connections between the existence of an invariant subspace and the existenceof an AIHS for an individual operator are different. If T ∈ L ( X ), X an infinite-dimensional complex Banach space, fails to admit an invariant subspace, then T hasno eigenvalues and hence, by [PT13, Theorem 2.3], admits an AIHS of defect ≤ T admits no AIHS of defect ≤ ≤
1, andthen in [SW14] the authors showed that if X is an infinite-dimensional complexBanach space and T ∈ L ( X ) is quasinilpotent or weakly compact then it admits anAIHS of defect ≤
1. Together with [MPR12, Remark 2.9] and [PT13, Theorem 2.3], it follows that if T has only countably many eigenvalues (for example, if T is strictlysingular) then it admits an AIHS of defect ≤
1. In section 2, we will show that if T commutes with a strictly singular operator with infinite-dimensional range thenit admits an AIHS of defect ≤
1. Note that results of this kind draw heavily uponthe tools of spectral theory, and so do not obviously carry over to the case where X is a real Banach space.Suppose the answer to the AIHS problem is negative in the complex case. Insection 3, we describe the structure of a hypothetical operator acting on an infinite-dimensional complex Banach space, but which fails to admit an AIHS of defect ≤ Y isalmost-invariant under T ∈ L ( X ) with defect ≤ d if and only if Y is invariant under T + F for some F ∈ F ( X ) of rank ≤ d , where F denotes the class of finite-rankoperators. Let us say that a Y is essentially-invariant under T ∈ L ( X ) just incase it is invariant under T + K for some K ∈ K ( X ), where K denotes the classof compact operators. We can then ask, does every operator acting on an infinite-dimensional Banach space admit an essentially-invariant halfspace ( EIHS )? Notethat every AIHS is an EIHS, and making this a formally weaker question.To the best of our knowledge, EIHS’s made their first appearance in the literaturein 1971, in a paper by Brown and Pearcy. There, they showed that every operatoron a complex infinite-dimensional Hilbert space admits an EIHS ([BP71, Corollary3.2]). Later, in [Ho78], Hoffman summarized the results of a few earlier papers,namely [FSW72], [Vo76], and [Ar77]. By that time he was able to observe thatevery operator on a separable, infinite-dimensional complex Hilbert space is essen-tially reducing, that is, admits a halfspace Y such that both Y and its orthogonalcomplement are essentially-invariant under the operator. In section 4, we will showfurther that every operator acting on an infinite-dimensional complex Banach spaceadmits an EIHS. Using the same techniques, we will also prove that for any infinite-dimensional complex Banach space X , the set of operators admitting an AIHS ofdefect ≤ L ( X ).All notation in this paper is standard, such as seen in [LT77], [AA02], and [AK06].Let us just mention three kinds of spectrum of an operator T : X → X that weconstantly use in this paper: the spectrum σ ( T ) = { λ ∈ C : λ − T is not invertible } ,the point spectrum σ p ( T ) = { λ ∈ C : λ − T is not injective } , and the surjectivityspectrum σ su ( T ) = { λ ∈ C : λ − T is not surjective } .2. An analog for Lomonosov’s Theorem
Let us formally state and prove an important result which was alluded to but notexplicitly given in [SW14].
Theorem 2.1 ([SW14]) . Let X be an infinite-dimensional complex Banach space,and let T ∈ L ( X ) . If T has no more than countably many eigenvalues then it admitsan AIHS of defect ≤ .Proof. In [MPR12, Remark 2.9] it was observed that if T fails to admit an IHSthen there exists a finite-codimensional T -invariant subspace W such that σ ( T | W ) GLEB SIROTKIN AND BEN WALLIS is a connected component of σ ( T ). Then, it was shown in [SW14, Theorem 3.11]that every quasinilpotent operator acting on an infinite-dimensional complex Banachspace admits an AIHS of defect ≤
1. Hence, if σ ( T | W ) is a singleton then we aredone. Otherwise, since σ ( T | W ) is a connected and compact subset of C which is nota singleton, ∂σ ( T | W ) must be uncountable. It was shown in [PT13, Theorem 2.3]that if the boundary of the spectrum contains a non-eigenvalue then the operatoradmits an AIHS of defect ≤
1. Since σ p ( T ) and hence σ p ( T | W ) is countable, such anon-eigenvalue must lie in the uncountable set ∂σ ( T | W ). This gives us an AIHS ofdefect ≤ T | W , and hence also for T .So, for example, any strictly singular operator acting on a complex infinite-dimensional Banach space admits an AIHS of defect ≤
1. This is due to the fact thatsuch operators always have countable spectrum (cf., e.g., [AA02, Theorem 7.11]).However, we can say something considerably stronger.
Theorem 2.2.
Let X be an infinite-dimensional complex Banach space and T ∈L ( X ) . If T commutes with a strictly singular (or compact) operator with infinite-dimensional range, then T admits an AIHS of defect ≤ . This can be viewed as an analog to Lomonosov’s Theorem, which states that any op-erator acting on an infinite-dimensional Banach space and commuting with a nonzerocompact operator admits a nontrivial hyper-invariant closed subspace ([Lo73]). Notethat unless the AIHS problem has an affirmative answer in the complex case, sim-ply being nonzero is insufficient. To see this, let T ∈ L ( X ) be an operator actingon a complex Banach space X which fails to admit an AIHS of defect ≤
1. Then T ⊕ ∈ L ( X ⊕ C ) also fails to admit an AIHS of defect ≤
1, even though it commuteswith the nonzero compact operator 0 ⊕ I C , where I C is the identity acting on C .On the other hand, Lomonosov’s Theorem cannot be strengthened to use strictlysingular operators instead of compact ones, since Read has famously constructeda strictly singular operator acting on a Banach space with no nontrivial invariantclosed subspace ([Re91]).First, let us prove the following lemma. Lemma 2.3.
Let X be an infinite-dimensional complex Banach space, and suppose S ∈ L ( X ) and T ∈ L ( X ) satisfy the following properties. (i) σ p ( S ) is countable; (ii) S is not a multiple of the identity, that is, S = µI for any µ ∈ C ; (iii) σ p ( T ∗ ) = (cid:13) ; and (iv) S commutes with T , that is, ST = T S .Then T admits an AIHS of defect ≤ .Proof. We may assume σ p ( T ) is uncountable by Theorem 2.1, say σ p ( T ) = ( λ α ) α ∈ A for some uncountable index set A . Notice that each eigenspace N ( λ α − T ), α ∈ A ,is S -invariant, since given u ∈ N ( λ α − T ) we get T Su = ST u = λ α Su so that Su is either zero or a λ α -eigenvector under T . Furthermore, we can assumethat each such eigenspace is finite-dimensional since otherwise, due to the fact thatevery infinite-dimensional closed subspace contains a halfspace ([MPR12, Lemma This Remark appears in the preprint [MPR12], but not the published version [MPR13]. Itsremoval has to do with organizational purposes and the near-simultaneous publication of [PT13]rather than any logical problem. The same thing happened for [MPR12, Lemma 2.1], which weuse in the proof of Lemma 2.3, but which was also stricken from [MPR13]. α ∈ A we can find an S -eigenvector v α ∈ N ( λ α − T ). Thisgives us an uncountable set ( v α ) α ∈ A of linearly independent S -eigenvectors, whichare also T -eigenvectors. Let ( µ α ) α ∈ A be the corresponding set of eigenvalues under S . Since σ p ( S ) is countable, it must be that ( µ α ) α ∈ A has only countably manydistinct values. Then we can find a sequence ( α n ) such that ( v α n ) all share the same S -eigenvalue, say µ . So, it cannot be that [ v α n ] = X , else we would have S = µI ,contradicting the fact that S is not a multiple of the identity. However, given thateach v α n is also a T -eigenvector, [ v α n ] is an infinite-dimensional T -invariant closedsubspace. Recall that if σ p ( T ∗ ) = (cid:13) then every infinite-dimensional T -invariantsubspace is either the whole space or a halfspace ([SW14, Proposition 3.9]). Inparticular, [ v α n ] is an IHS under T .Thus, we have the following immediate consequence. Proposition 2.4.
Let X be an infinite-dimensional complex Banach space and T ∈L ( X ) satisfies the following conditions. (i) σ p ( T ∗ ) = (cid:13) ; and (ii) T commutes with a nonzero strictly singular (or compact) operator.Then T admits an AIHS of defect ≤ .Proof. Let S ∈ L ( X ) be a nonzero strictly singular operator commuting with T .Clearly, S is not a multiple of the identity, and by the Spectral Theorem for strictlysingular operators (cf., e.g., [AA02, Theorem 7.11]), σ ( S ) and hence σ p ( S ) are count-able subsets of C . Thus we can apply Lemma 2.3.Now, we are ready to prove Theorem 2.2. Proof of Theorem 2.2.
Consider a strictly singular operator S ∈ L ( X ) which com-mutes with T and has infinite-dimensional range. According to [SW14, Proposition3.5], there exists a finite-codimensional and T -invariant closed subspace W of X such that σ p ( T | ∗ W ) = (cid:13) . In fact, if we look at the proof to [SW14, Proposition 3.5],we see that W has the form W = p ( T ) X for some polynomial p . Note that SW ⊆ SW = Sp ( T ) X = Sp ( T ) X = p ( T ) SX ⊆ p ( T ) X = W. so that W is S -invariant. Thus, for any w ∈ W we have S | W T | W w = ST w = T Sw = T | W S | W w so that S | W commutes with T | W . Since S is strictly singular, so is S | W , and it isnonzero since S has infinite-dimensional range. Thus we can apply Proposition 2.4to obtain an AIHS of defect ≤ T | W , and hence also for T . Remark 2.5.
More generally, to obtain an AIHS under T ∈ L ( X ) of defect ≤ ,where X is an infinite-dimensional complex Banach space, it is enough to find acommuting operator S which has infinite-dimensional range, is not a multiple of theidentity on any finite-codimensional subspace, and such that σ p ( S ) is countable. Structure of an operator without an AIHS
In this section, we demonstrate a two-side shift-like structure for those hypothet-ical operators on an infinite-dimensional complex Banach space which fail to admitan AIHS of defect ≤
1. First, we shall describe a right-shift-like structure.
GLEB SIROTKIN AND BEN WALLIS
Theorem 3.1.
Let X be an infinite-dimensional complex Banach space, and let T ∈ L ( X ) . If T fails to admit an AIHS of defect ≤ then there exists e ∈ X suchthat [ T n e ] ∞ n =0 is finite-codimensional in X .Proof. Suppose T fails to have an AIHS of defect ≤
1. By [MPR12, Remark 2.9],we can find a finite-codimensional T -invariant subspace W such that σ ( T | W ) isconnected. Let U := λ − T | W ∈ L ( W ) for some λ ∈ ∂σ ( T ). By [SW14, Proposition2.12], the null spaces N ( U n ) are strictly increasing and finite-dimensional. Thuswe can apply [SW14, Proposition 2.9] to obtain e ∈ W such that V := [ U n e ] ∞ n =0 isinfinite-dimensional. Since T fails to admit an AIHS of defect ≤
1, it must be that V is finite-codimensional, which finishes the proof as V = [( λ − T | W ) n e ] ∞ n =0 = [( T | W ) n e ] ∞ n =0 = [ T n e ] ∞ n =0 . To describe a left-shift-like structure in the Theorem 3.6 we will need a couple ofadditional steps.
Proposition 3.2.
Let X be a complex Banach space and T ∈ L ( X ) be a continuousoperator. Suppose that the null space N ( T ) is complemented in X and let W be aclosed subspace such that X = W ⊕ N ( T ) . Let P W : X → W denote the continuouslinear projection onto W along N ( T ) . Then P W T | W ∈ L ( W ) , and σ su ( P W T | W ) ⊆ σ su ( T ) ⊆ σ su ( P W T | W ) ∪ { } . If furthermore σ su ( T ) is connected then σ su ( T ) = σ su ( P W T | W ) .Proof. Write S := P W T | W , and let P N ( T ) : X → N ( T ) denote the continuous linearprojection onto N ( T ) along W .For the first inclusion, let λ ∈ ρ su ( T ) and w ∈ W . Then there is x ∈ X suchthat ( λ − T ) x = w . Then T P W x = T ( P W x + P N ( T ) x ) = T x = λx − w = λ ( P W x + P N ( T ) x ) − w and hence SP W x = P W T P W x = λP W x − w . So w = ( λ − S ) P W x , which givesus λ ∈ ρ su ( S ).For the second inclusion, let λ ∈ ρ su ( S ) \ { } = ( σ su ( S ) ∪ { } ) c and x ∈ X . Thenthere is w ∈ W such that ( λ − S ) w = P W x , and hence x = P W x + P N ( T ) x = ( λ − S ) w + P N ( T ) x = ( λ − T ) w + P N ( T ) T w + P N ( T ) x = ( λ − T ) w + λP N ( T ) λ − ( T w + x ) = ( λ − T )( w + P N ( T ) λ − ( T w + x ))so that λ ∈ ρ su ( T ).To prove the last part of the Proposition, recall that the surjectivity spectrum isalways a compact subset of C (cf., e.g., Theorem 2.42 in [Ai04]). Thus, if σ su ( T ) isconnected, then we must have σ su ( T ) = σ su ( P W T | W ).As an aside, we might also ask whether the above Proposition can be improved toexclude the union with { } . The following example shows that this is not possiblein general. Example 3.3.
For any < p < ∞ , there exists a continuous linear operator T ∈L ( ℓ p ) and a closed subspace W such that ℓ p = W ⊕ N ( T ) with ∈ σ su ( T ) ∩ σ p ( T ) but / ∈ σ ( P W T | W ) , where P W : ℓ p → W denotes the continuous linear projectiononto W along N ( T ) . Proof.
Let ( e n ) ∞ n =1 denote the canonical basis of ℓ p . Set R : ℓ p → ℓ p as the unweightedright-shift operator, that is, the 1-bounded linear operator defined by Re n = e n +1 forall n ∈ Z + . Consider subspace W := [ e n ] ∞ n =2 with basis projection P W and operator T ∈ L ( W ) defined by T = (2 − R ) P W .Notice that operator P W T | W = T | W , as an operator on W ∼ = ℓ p , acts exactly as2 − R . Hence, the spectrum σ ( P W T | W ) equals σ (2 − R ) = { λ : | λ − | ≤ } (cf., e.g.,[AA02, Example 6.21]) which implies that zero is not in σ ( P W T | W ). In particular,we have N ( T ) = [ e ] and, thus, P W is a projection onto W along N ( T ). It is clear,that T is not onto, which concludes the example.Next, let us consider a sequence of null spaces N ( T n ). These subspaces satisfythe inclusion N ( T n ) ⊆ N ( T n +1 ) for every n ∈ N . In addition, if, for some n , wehave a decomposition X = W n ⊕ N ( T n ) and P W n is the respective projection onto W n along N ( T n ), then P W n [ N ( T m )] ⊆ N ( T m ) and hence(1) P W n [ N ( T m )] = N ( T m ) ∩ W n for any m ≥ n. We can see this by applying T m to both sides of the decomposition x = x n + P W n x where x is any vector from N ( T m ) and x n is in N ( T n ). Let us use these observationsto extend Proposition 3.2. Lemma 3.4.
Let X be an infinite-dimensional complex Banach space and T ∈L ( X ) . Then at least one of the following must be true. (i) T admits an IHS. (ii) There exists a nonincreasing sequence X = W ⊇ W ⊇ W ⊇ · · · of complements of N ( T n ) , and matching continuous linear projections P W n : X → W n onto W n along N ( T n ) , such that for all n ∈ N we have σ su ( P W n T | W n ) ⊆ σ su ( T ) ⊆ σ su ( P W n T | W n ) ∪ { } . If furthermore ∈ ∂σ ( T ) and σ ( T ) is connected, then, for all n ∈ N , σ su ( P W n T | W n ) = σ su ( T ) . Proof.
Let us assume (i) is false. Then, by [SW14, Lemma 2.11], every N ( T n ) isfinite-dimensional and, hence, complemented in X . We set W = X and buildthe chain of W n ’s by induction. As W n +1 we select any complement of the finitedimensional subspace P W n [ N ( T n +1 )] = N ( T n +1 ) ∩ W n in W n . Due to N ( T n +1 ) ⊆ P W n (cid:2) N ( T n +1 ) (cid:3) ⊕ N ( T n ) ⊆ N ( T n +1 ) ,W n +1 is also a complement of N ( T n +1 ) in X . This, in particular, implies that theoperator P W n +1 | W n is the projection from W n onto W n +1 along P W n [ N ( T n +1 )].Next, we claim that P W n [ N ( T n +1 )] = N ( P W n T | W n ). Since T [ N ( T n +1 )] ⊂ N ( T n )is annihilated by P W n , and P W n [ N ( T n +1 )] = N ( T n +1 ) ∩ W n by (1) above, we con-clude that P W n [ N ( T n +1 )] ⊆ N ( P W n T | W n ).For the reverse inclusion, suppose x ∈ N ( P W n T | W n ). Then we can write T n +1 x = T n ( P W n T x + P N ( T n ) T x ) = 0 + T n P N ( T n ) T x = 0 , and the claim is proved. Thus, P W n +1 can be viewed as the projection from W n onto W n +1 along N ( P W n T | W n ). GLEB SIROTKIN AND BEN WALLIS
Now, we are ready to prove the inclusion of spectra using Proposition 3.2. Thecase n = 0 is clear and, then we have σ su ( P W n +1 T | W n +1 ) = σ su ( P W n +1 ( P W n T | W n ) | W n +1 ) ⊆ σ su ( P W n T | W n ) ⊆ σ su ( T ) ⊆ σ su ( P W n T | W n ) ∪ { } ⊆ σ su ( P W n +1 T | W n +1 ) ∪ { } . This proves all but the last statement of (ii).Now let us consider the case when σ ( T ) is connected. If equality did not hold, thenwe would have σ su ( P W n T | W n ) = σ su ( T ) \ { } with 0 ∈ ∂σ ( T ) ⊂ σ su ( T ). However,this is impossible since the surjectivity spectrum is a nonempty compact subset of C (cf., e.g., [Ai04, Theorem 2.42]).Notice that in the proof of the last lemma, if T has no IHS, we get(2) N ( P W n T | W n ) = N ( T n +1 ) ∩ W n . Therefore, if N ( T n ) = N ( T n +1 ) happens for any n , then N ( P W n T | W n ) = { } . Thismeans 0 is not an eigenvalue for operator P W n T | W n . If, in addition, zero is a limitpoint of ∂σ su ( T ), then by Lemma 3.4 together with compactness of the surjectivityspectrum (cf., e.g., [Ai04, Theorem 2.42]), we would get 0 ∈ ∂σ su ( P W n T | W n ). Inthat case, by [SW14, Proposition 2.8], P W n T | W n and hence T would admit an AIHSof defect ≤
1. This proves the following lemma.
Lemma 3.5.
Let X be an infinite-dimensional complex Banach space and T ∈ L ( X ) be such that ∈ ∂σ su ( T ) is a limit point of ∂σ su ( T ) , but T fails to admit an AIHSof defect ≤ . Then dim N ( T n +1 ) / N ( T n ) > for each n ∈ N and, consequently, S ∞ n =0 N ( T n ) has infinite dimension and finite codimension. Now we are ready to prove this section’s main result.
Theorem 3.6.
Let X be an infinite-dimensional complex Banach space and T ∈L ( X ) . Then either T admits an AIHS of defect ≤ , or else there is a finite-codimensional closed subspace V , an eigenvalue λ ∈ σ p ( T ) ∩ σ p ( P V T | V ) , and alinearly independent sequence ( v n ) ∞ n =1 ⊆ V such that [ v n ] ∞ n =1 = V , and satisfying Sv n +1 = v n for all n ∈ Z + , and Sv = 0 , where P V ∈ L ( X ) is a projection onto V , and S := P V ( λ − T ) | V = λ − P V T | V ∈ L ( V ) . Proof.
Assume T fails to admit an AIHS of defect ≤
1. By Theorem 2.1 we mayassume that σ ( T ) and hence ∂σ ( T ) are infinite. Recall from [Ai04, Theorem 2.42]that the boundary of the spectrum is always contained in the boundary of thesurjectivity spectrum, and that by [SW14, Proposition 2.8], ∂σ su ( T ) ⊆ σ p ( T ). Inparticular, σ p ( T ) ∩ ∂σ ( T ) should be infinite, and so, by shifting for convenience, wemay assume that zero is a limit point of ∂σ ( T ). Moreover, our target statement isnow about operator T instead of λ − T .Set K j := N ( T j ) and observe that since b T : K j +2 /K j +1 → K j +1 /K j definedby b T ( x + K j +1 ) := T x + K j is a linear injective map we have dim( K j +2 /K j +1 ) ≤ dim( K j +1 /K j ) for all j ∈ Z + . Moreover, by Lemma 3.5, lim j →∞ dim( K j +1 /K j ) mustbe a strictly positive number. Hence, there exist integers N and r > K N + n +1 /K N + n ) = r for every n > N = 1. Next,we use Lemma 3.4 to select a sequence of spaces ( W n ) ∞ n =1 complementing K n ’s in X . Recall from (2) that N ( P W n T | W n ) = K n +1 ∩ W n , so that K n +1 ∩ W n mustcontain a nonzero vector, on pain of having σ p ( P W n T | W n ) = (cid:13) . However, the latter is impossible by Theorem 2.1, since we have assumed T and hence P W n T | W n fails toadmit an AIHS of defect ≤ v n ∈ ( K n +1 ∩ W ) \ K n inductively. We could guarantee theclaim of the theorem if for each n we have T v n +1 = v n + z n with z n ∈ K . Indeed,the last identity guarantees that for V = [ v n ] ∞ n =1 we have T ( V ) ⊂ V + K . Thefact that v n ∈ K n +1 \ K n makes sure that V is infinite-dimensional. Therefore, V isfinite-codimensional as T does not have an IHS but T ( V + K ) ⊆ V + K and dim K < ∞ . Finally, since V ⊂ W , there is a projection P V onto V which annihilates K and, hence, every z n . That is, P V T v n +1 = v n for every n and P V T v = 0.At last, let us select vectors v n . Set v to be any vector in ( K ∩ W ) \ K . Bythe paragraph above, we will be done if we prove that for every n > x ∈ ( K n +1 ∩ W ) \ K n there are vectors v ∈ ( K n +2 ∩ W ) \ K n +1 and z ∈ K suchthat T v = x + z . To see this, consider linear map e T : K n +2 ∩ W → K n +1 /K naturally defined by e T x := T x + K . Notice that N ( e T ) = K ∩ W and, by (1), K = K ⊕ ( K ∩ W ). So, we conclude that dim N ( e T ) = dim K /K = r , andhence, due to the fact that dim K n +1 /K = nr , obtaindim range e T = dim K n +2 ∩ W − dim N ( e T ) = ( n + 1) r − r = dim K n +1 /K , which implies that e T is a surjection. Therefore, for each x ∈ ( K n +1 ∩ W ) \ K n thereis v ∈ K n +2 ∩ W such that e T v = x + K . The latter means T v = x + z for some z ∈ K ⊂ K n . Notice that v / ∈ K n +1 as, otherwise, T v would have been in K n andsame would have been true for T v − z = x , whereas x / ∈ K n .4. The essentially-invariant halfspace problem
Does every operator acting on an infinite-dimensional Banach space admit anEIHS? Let us call this the
EIHS problem . In this section, we give an affirmativeanswer in the complex case. As a corollary to the techniques we use, we will alsoobserve that if X is an infinite-dimensional complex Banach space then the set ofall operators in L ( X ) which admit an AIHS of defect ≤ L ( X ).To get the job done, we will need a straightforward fact, whose proof we includefor completeness. Proposition 4.1.
Let X be a Banach space and let ( y n ) ∞ n =1 be a seminormalizedbasic sequence in X with basis constant B ≥ . Suppose K : [ y n ] ∞ n =1 → X is acontinuous linear operator satisfying P ∞ n =1 k Ky n k < ∞ . Then K can be extended toa compact operator e K ∈ K ( X ) satisfying k e K k ≤ B ∞ X n =1 k Ky n k . Proof.
Let y ∗ n ∈ [ y n ] ∗ be coordinate functionals for the basis ( y n ) ∞ n =1 . Denote theirnorm-preserving extensions to the whole X by the same y ∗ n . Since k y ∗ n k ≤ B foreach n and due to the absolute convergence of the series P Ky n , we may define acontinuous operator by the following: e Kx = ∞ X n =1 y ∗ n ( x ) Ky n for all x ∈ X. Since we are using coordinate functionals e K is an extension of K to X and it is ofthe desired norm. To show that e K is compact, it is sufficient to observe that it isthe norm-limit of its finite-rank partial sums. Theorem 4.2.
Let X be an infinite-dimensional complex Banach space, and let T ∈ L ( X ) . Then at least one of the following are true. (i) T admits an AIHS of defect ≤ ; or (ii) There exists µ ∈ C such that for any ǫ > there exists a halfspace Y ⊆ X and a compact operator e K ∈ K ( X ) with k e K k < ǫ , such that ( µ − T ) y = e Ky for all y ∈ Y .Proof. Let us assume (i) is false. In [SW14, Theorem 3.5], it was shown that when-ever an operator T acting on a complex infinite-dimensional Banach space fails to ad-mit an AIHS of defect ≤
1, there exists a finite-codimensional subspace W for which σ p ( T | ∗ W ) = (cid:13) . Since the surjectivity spectrum is always a nonempty set (cf., e.g.,[Ai04, Theorem 2.42]), we can find µ ∈ σ su ( T | W ), and define S := µ − T | W ∈ L ( W ).Notice that this means σ p ( S ∗ ) = (cid:13) and 0 ∈ σ su ( S ).We claim that for any finite-codimensional closed subspace U ⊆ W , the restric-tion S | U ∈ L ( U, W ) cannot be bounded below. For suppose otherwise, towards acontradiction, that is, suppose S | U is bounded below for some finite-codimensionalclosed subspace U . Recall that bounded below operators have closed range (cf., e.g.,[AA02, Theorem 2.5]). In particular, S | U has closed range, and since U is finite-codimensional, so does S . Recall also that λ / ∈ σ p ( S ∗ ) just in case λ − S has denserange (cf., e.g. [AA02, Theorem 6.19]). Since σ p ( S ∗ ) = (cid:13) this means S must havedense range. Along with its closed range, this gives us SW = W , which contradictsthe fact that 0 ∈ σ su ( S ). Thus the claim is proved, and S | U is not bounded belowfor any finite-codimensional closed subspace U ⊆ W .The last paragraph, by [LT77, Proposition 2.c.4], implies that for every ǫ > V ⊂ W such that S | V is a compactoperator and k S | V k < ǫ . Of course, we may assume that V = [ v n ] ∞ n =1 where ( v n ) ∞ n =1 is a normalized basic sequence. By taking appropriate subsequences, if needed, wemay assume that V is a half-space and that Sv n converges. It follows, that forseminormalized block basic sequence y n = v n − v n − the space Y = [ y n ] is a half-space, operator S | Y is compact, and k S | Y k < ǫ . Moreover, we have Sy n → Corollary 4.3.
Every operator acting on an infinite-dimensional complex Banachspace admits an EIHS.
Corollary 4.4.
Let X be an infinite-dimensional complex Banach space. Then theset { T ∈ L ( X ) : T admits an AIHS of defect ≤ } is norm-dense in L ( X ) . References []AA02 Yuri A. Abramovich and C.D. Aliprantis.
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Department of Mathematical Sciences, Northern Illinois University, DeKalb, IL60115
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