DOI: 10.24193/subbmath.2019.4.10
Ascent, descent and additive preserving problems
Mourad Oudghiri and Khalid Souilah
Abstract. Given an integern≥1, we provide a complete description of all ad- ditive surjective maps, on the algebra of all bounded linear operators acting on a complex separable infinite-dimensional Hilbert space, preserving in both di- rections the set of all bounded linear operators with ascent (resp. descent) non- greater thann. In the context of Banach spaces, we consider the additive pre- serving problem for semi-Fredholm operators with ascent or descent non-greater thann.
Mathematics Subject Classification (2010):47B49, 47L99, 47A55, 47B37.
Keywords:Linear preserver problems, ascent, descent, semi-Fredholm operators.
1. Introduction
LetX be an infinite-dimensional Banach space over the real or complex fieldK, and letB(X) be the algebra of all bounded linear operators onX.
For a subset Λ ⊂ B(X), we say that a map Φ on B(X) preserves Λ in both directions (or, equivalently, that Φ is apreserver ofΛ in both directions) if for every T ∈ B(X),
T ∈Λ if and only if Φ(T)∈Λ.
For an operatorT ∈ B(X), write ker(T) for its kernel, ran(T) for its range and T∗for its adjoint on the topological dual spaceX∗. Theascenta(T) anddescentd(T) ofT ∈ B(X) are defined by
a(T) = inf{k≥0 : ker(Tk) = ker(Tk+1)}
and
d(T) = inf{k≥0 : ran(Tk) = ran(Tk+1)},
where the infimum over the empty set is taken to be infinite (see [15, 19]). Clearly, a bounded linear operator is injective (resp. surjective) if and only if its ascent (resp.
descent) is zero.
Over the last years, there has been a considerable interest in the so-calledlinear preserver problemsthat concern the question of determining the form of all linear, or
additive, maps onB(X) that leave invariant certain subsets. The most linear preserver problems were solved in the finite-dimensional context, and extended later to the infinite-dimensional one. For excellent expositions on linear preserver problems, the reader is referred to [7, 11, 12, 13, 16] and the references therein.
One of the most famous problems in this direction is Kaplansky’s problem [8], asking whether bijective unital linear maps Φ, between semi-simple Banach algebras, preserving in both directions invertibility, are Jordan isomorphisms (i.e. Φ(a2) = Φ(a)2 for alla). This problem was first solved in the finite-dimensional case [10], and it was later extended to von Neumann algebras [1]. In the case of the algebraB(X), A. A. Jafarian and A. R. Sourour established in [7] that every unital surjective linear map Φ onB(X), preserving in both directions invertibility, has one of the following two forms
T 7→AT A−1 or T 7→AT∗A−1, (1.1)
whereAis a bounded linear operator between suitable spaces. Later, it was shown in [6] that every unital surjective additive preserver of injective operators or of surjective operators in both directions takes one of the two forms (1.1).
Since injective and surjective operators are precisely those operators with zero ascent and descent respectively, the following question arises: What can we say about surjective linear maps onB(X) preserving in both directions operators of finite ascent and descent, respectively?
LetH be a separable complex infinite-dimensional Hilbert space, and denote by A(H) (resp.D(H)) the set of all operators inB(H) of finite ascent (resp. descent). In [11], the authors showed that a surjective additive continuous map Φ :B(H)→ B(H) preservesA(H) orD(H) in both directions if and only if
Φ(T) =cAT A−1 for all T ∈ B(H), (1.2) where c is a non-zero scalar and A : H → H is an invertible bounded linear, or conjugate linear, operator. An analog result was proved for A(H)∪ D(H) by the same authors, see [12]. It should be noted that the question of removing the continuity condition or extending these results to the context of Banach spaces is still open.
The above results motivated us to continue the study of additive preservers involving the ascent and descent. This study may be considered as a key step towards a deeper understanding of operators with finite ascent or descent and their topological properties. In this paper, we will show that if we limit the variation of the ascent and the descent, then we obtain the same conclusion as in [11] without considering continuous preservers.
For each integern≥1 let us introduce the following subsets ofB(H):
1. An(H) the set of all operatorsT ∈ B(H) with a(T)≤n;
2. Dn(H) the set of all operatorsT ∈ B(H) with d(T)≤n.
Now, we summarize the first main result in the following theorem:
Theorem 1.1. LetΦ :B(H)→ B(H)be an additive surjective map. Then the following assertions are equivalent:
1. Φpreserves An(H)in both directions;
2. Φpreserves Dn(H)in both directions;
3. Φpreserves An(H)∪ Dn(H)in both directions;
4. there exist a non-zero scalar c and a bounded invertible linear, or conjugate linear, operator A:H →H such that
Φ(T) =cAT A−1 for all T ∈ B(H).
Unfortunately, the approach used here does not allow us to obtain an analogue result in the context of Banach spaces. More precisely, one of the most important steps in the proof of the previous theorem consists in determining the topological interior of An(H), Dn(H), andAn(H)∪ Dn(H) using that of A(H)∪ D(H), which is known only in the context of separable Hilbert spaces, see [12].
Recall that an operatorT ∈ B(X) is calledupper(resp.lower)semi-Fredholmif ran(T) is closed and dim ker(T) (resp. codim ran(T)) is finite. The following properties will be used tacitly throughout the paper (see [15, Section 16]):
1. If the codimension of the range ran(T) of an operatorT ∈ B(X) is finite, then ran(T) is automatically closed;
2. The composition of two upper (resp. lower) semi-Fredholm operators is an upper (resp. lower) semi-Fredholm operator;
3. IfSTis an upper (resp. lower) semi-Fredholm operator, thenT(resp.S) is upper (resp. lower) semi-Fredholm.
In [14], the authors studied all linear maps Φ onB(H) preserving in both direc- tions semi-Fredholm operators. It has been shown that such maps Φ preserve in both directions the ideal of compact operators, and that the induced maps on the Calkin algebra are Jordan automorphisms. The problem of determining the structure of such maps on the whole spaceB(H) has remained open, and hence they conjectured that Φ is of the formT 7→AT B+ Ψ(T) whereA, B∈ B(H) are Fredholm operators and Ψ is a linear map onB(H) whose range is contained in the ideal of compact operators.
In this paper, we prove that if we limit the variation of the ascent (resp. descent) of upper (resp. lower) semi-Fredholm operators, then we obtain the complete descrip- tion of all additive preservers of such operators in the context of Banach spaces. More precisely, we consider additive preservers of the following subsets ofB(X):
1. Fn+(X) the set of all upper semi-Fredholm operatorsT ∈ B(X) with a(T)≤n;
2. Fn−(X) the set of all lower semi-Fredholm operatorsT ∈ B(X) with d(T)≤n;
3. Fn±(X) =Fn+(X)∪ Fn−(X).
The second main result of the present paper is stated as follows:
Theorem 1.2. Let Φ : B(X) → B(X) be an additive surjective map preserving any one of the subsets Fn+(X),Fn−(X) or Fn±(X) in both directions. Then there exist a non-zero scalarc, and either a bounded invertible linear, or conjugate linear, operator A:X→X such that
Φ(T) =cAT A−1 for all T∈ B(X),
or, a bounded invertible linear, or conjugate linear, operatorB :X∗ →X such that Φ(T) =cBT∗B−1 for all T ∈ B(X).
As an application of Theorem 1.2, we derive the following corollary:
Corollary 1.3. LetΦ :B(H)→ B(H)be an additive surjective map. Then the following assertions are equivalent:
1. Φpreserves Fn+(H)in both directions;
2. Φpreserves Fn−(H) in both directions;
3. Φpreserves Fn±(H) in both directions;
4. there exist a non-zero scalar c and a bounded invertible linear, or conjugate linear, operator A:H →H such that
Φ(T) =cAT A−1 for all T ∈ B(H).
The paper is organized as follows. In the second section, we give the topological interior of each of the subsetsAn(H),Dn(H), andAn(H)∪Dn(H). The third section is devoted to establish some useful results on rank-one perturbations of these topological interiors. These results are needed for proving our theorems in the last section.
2. Topological interior of A
n(H), D
n(H), and A
n(H) ∪ D
n(H)
Recall that the hyper-kernel and the hyper-rangeof an operator T ∈ B(X) are respectively the subspacesN∞(T) = [
k≥0
ker(Tk) andR∞(T) = \
k≥0
ran(Tk).
Let us introduce the following subsets ofB(X):
1. B+n(X) ={T ∈ B(X) : ran(T) is closed and dimN∞(T)≤n};
2. B−n(X) ={T ∈ B(X) : codimR∞(T)≤n};
3. B±n(X) =Bn+(X)∪ Bn−(X).
One of the most important steps in the proof of our main theorems is to show that the maps we are dealing with preserve the subsetsBn+(X),Bn−(X) andBn±(X) in both directions. In order to prove this implication, we establish that the topological interior of An(H), Dn(H) and An(H)∪ Dn(H) is respectively Bn+(H), B−n(H) and B±n(H). Similar results are given forFn+(X),Fn−(X) andFn±(X).
It should be noted that the ascent and the hyper-kernel of an operatorT ∈ B(X) are related by the following inequality (see [17])
a(T)≤dimN∞(T). (2.1)
Similarly, the descent is related to the hyper-range by
d(T)≤codimR∞(T). (2.2)
Remark 2.1. ForT ∈ B(X), it follows easily from the definition of the ascent and of the descent that:
1. dim ker(Tn+1)≤nif and only if dimN∞(T)≤n;
2. codim ran(Tn+1)≤nif and only if codimR∞(T)≤n.
Proposition 2.2. Bn+(X),Bn−(X)andB±n(X)are open subsets ofFn+(X),Fn−(X)and Fn±(X), respectively.
Proof. It follows from (2.1) and (2.2) thatB+n(X) andB−n(X) are subsets ofFn+(X) and Fn−(X) respectively, and so B±n(X) is a subset of Fn±(X). Let S ∈ Bn+(X). In particular, we have dim ker(Sn) = dim ker(Sn+1) ≤ n and Sn+1 is an upper semi- Fredholm operator. Hence, it follows by [15, Theorem 16.11] that there existsη >0 such that forT ∈ B(X) withkT−Sn+1k< η, we have thatT is upper semi-Fredholm and
dim ker(T)≤dim ker(Sn+1)≤n. (2.3) On the other hand, since the functionT 7→Tn+1 is continuous onB(X), there exists ε >0 such that
kTn+1−Sn+1k< η for allT ∈ B(X) with kT−S k< ε. (2.4) Combining (2.4) and (2.3) we obtain thatTn+1 is upper semi-Fredholm and
dim ker(Tn+1)≤dim ker(Sn+1)≤n,
and so T ∈ Bn+(X) for allT ∈ B(X) with k T−S k< ε. This shows that B+n(X) is open.
Similarly, we prove thatB−n(X) is open, and henceBn±(X) is also open.
From [5, Lemma 1.1], given a non-negative integerd, we have
a(T)≤d⇔ker(Tm)∩ran(Td) ={0} for somem≥1. (2.5) Remark 2.3. LetT ∈ B(X). Then the following assertions hold:
1. IfT has finite ascent and descent then a(T) = d(T) andX= ker(Tk)⊕ran(Tk), wherek= a(T) and the direct sum is topological (see [15, Corollary 20.5]).
2. IfT =T1⊕T2with respect to any decomposition ofX, then it follows from [18, Theorem 6.1] that
a(T) = max{a(T1),a(T2)} and d(T) = max{d(T1),d(T2)}.
The following example shows thatB+n(X),B−n(X) andBn±(X) are proper subsets ofFn+(X),Fn−(X) andFn±(X), respectively, and that there exist operators with finite ascent and descent which are not semi-Fredholm.
Example 2.4. LetY ⊂Xbe a closed subspace of dimensionn+1, and writeX=Y⊕Z where Z is a closed subspace of X. With respect to this decomposition, consider the operator T = 0⊕I. According to the previous remark, one can easily see that a(T) = d(T) = 1. SinceN∞(T) = ker(T) = Y and R∞(T) = ran(T) = Z, then T belongs toFn+(X)∩ Fn−(X) and not toB±n(X).
Similarly, forS=I−T, we have a(S) = d(S) = 1, ker(S) =Z and ran(S) =Y. Thus,S is not a semi-Fredholm operator.
Recall that an operatorT ∈ B(X) is calledupper(resp.lower)semi-Browderif it is upper (resp. lower) semi-Fredholm of finite ascent (resp. descent). Clearly, every operator inFn+(X) (resp. Fn−(X)) is upper (resp. lower) semi-Browder.
Theorem 2.5. LetT ∈ B(X)be non-zero. The following assertions are equivalent:
1. T ∈ B±n(X)(resp.B+n(X),Bn−(X));
2. for everyS ∈ B(X)there existsε0>0such thatT+εS∈ Fn±(X)(resp.Fn+(X), Fn−(X)), for all numbers (equivalently, rational numbers)|ε|< ε0.
Proof. (1)⇒(2) follows immediately from the previous proposition.
(2) ⇒ (1). Suppose that for every S ∈ B(X) there exists ε0 > 0 such that T +εS ∈ Fn±(X) for all numbers |ε| < ε0. In particular, we have T ∈ Fn±(X), and so T is either upper semi-Browder or lower semi-Browder. It follows from [15, Theorem 20.10] that there exist two closed T-invariant subspaces X1 and X2 such that X =X1⊕X2, dimX1 <∞,T1 =T|X1 is nilpotent andT|X2 is either bounded below or onto, respectively. We claim that dimX1 ≤n. Let {ei : 0 ≤ i ≤ p} be a basis of X1 such that T e0 = 0 and T ei =εiei−1 for 1 ≤i ≤ pwhere εi ∈ {0,1}.
With respect to the decomposition ofX, consider the operator S ∈ B(X) given by S=S1⊕0 where S1e0= 0 andS1ei=ei−1 for 1≤i≤p. Clearly, for ε /∈ {−1,0}we have
(T1+εS1)e0= 0 and (T1+εS1)ei= (εi+ε)ei−1 for 1≤i≤p.
Hence (T1+εS1)pep =λe06= 0 whereλ= (εp+ε). . .(ε1+ε).
Thereforee0∈ker(T1+εS1)∩ran(T1+εS1)p, and consequently a(T1+εS1) = d(T1+εS1)≥p+ 1 by (2.5). But, we have also
a(T1+εS1)≤a(T+εS) and d(T1+εS1)≤d(T+εS).
SinceT+εS∈ Fn±(X), then a(T+εS)≤nor d(T+εS)≤n. Thus dimX1≤n.
Now, ifT ∈ Fn+(X) (resp.Fn−(X)) thenT is upper (resp. lower) semi-Browder, and so the space X1 (resp. X2) is uniquely determined and X1 =N∞(T) (resp. X2 = R∞(T)) (see [15, Theorem 20.10]). This proves thatT ∈ B+n(X) (resp. B−n(X)).
For a subset Γ⊆ B(X), we write Int(Γ) for its interior. As a consequence of Theorem 2.5, we derive the following corollary.
Corollary 2.6. We have Int(Fn+(X)) = Bn+(X), Int(Fn−(X)) = B−n(X) and Int(Fn±(X)) =B±n(X).
Proof. Let us show that Int(Fn+(X)) = Bn+(X). Note that B+n(X) ⊆ Int(Fn+(X)) becauseB+n(X) is open. LetT /∈ Bn+(X), then Theorem 2.5 ensures the existence of an operatorS∈ B(X) and a sequence (εk) converging to zero such thatT+εkS /∈ Fn+(X) for allk≥0. Consequently,T /∈Int(Fn+(X)).
Similarly, we prove that Int(Fn−(X)) =Bn−(X) and Int(Fn±(X)) =B±n(X).
Theorem 2.7. LetH be a separable complex infinite-dimensional Hilbert space and let T ∈ B(H). Then the following assertions are equivalent:
1. T ∈ B±n(H)(resp.Bn+(H),Bn−(H));
2. for everyS∈ B(H)there existsε0>0such thatT+εS ∈ An(H)∪Dn(H)(resp.
An(H),Dn(H)), for all numbers (equivalently, rational numbers)|ε|< ε0.
Proof. (1)⇒(2) follows immediately from Proposition 2.2.
(2) ⇒ (1). Suppose that for every S ∈ B(H) there exists ε0 > 0 such that T +εS ∈ An(H)∪ Dn(H) for all |ε| < ε0. Then, using [12, Proposition 2.5], we get that T is a semi-Browder operator. The rest of the proof is similar to the proof of
Theorem 2.5.
Using a similar proof of Corollary 2.6, we get the following result.
Corollary 2.8. We have Int(An(H)∪ Dn(H)) =B±n(H),Int(An(H)) =Bn+(H) and Int(Dn(H)) =Bn−(H).
3. B
+n(X), B
n−(X) and B
n±(X) under rank-one perturbations
Letz∈Xand letf ∈X∗be non-zero. We denote byz⊗f the rank-one operator defined by (z⊗f)(x) = f(x)z for all x∈X. Note that every rank-one operator in B(X) can be written in this form.
In [13], the authors proved that for a rank-one operator F ∈ B(X) and forT ∈ B(X) with dim ker(T)≤n, we have either dim ker(T+F)≤nor dim ker(T−F)≤n.
In the following, we extend this result to the setting of the hyper-kernel subspace.
Proposition 3.1. Let T ∈ B(X)be such thatdimN∞(T)≤n, and letF ∈ B(X)be a rank-one operator. Then eitherdimN∞(T+F)≤nordimN∞(T−F)≤n.
Before giving the proof of this proposition, we need to establish some lemmas.
ForT, F ∈ B(X), let
M(T, F) ={x∈ N∞(T) :F Tix= 0 for alli≥0}.
Clearly, M(T, F) is aT-invariant subspace ofN∞(T)∩ker(F). Furthermore, ifT has a finite ascent, then M(T, F) is closed.
Lemma 3.2. Let T ∈ B(X) be non-zero, and let F =z⊗f be a rank-one operator such that ker(T)∩ker(F) = {0}. Assume that there exist an integer m ≥ 0 and a vectorx∈ker(T+F)m+1\ker(T+F)m such that x /∈M(T, F). Thenxis a linear combination of linearly independent vectorsxi,0≤i≤m, such that
(T+F)x0= 0, (T+F)xi=xi−1 for1≤i≤m, andf(xi) =δi0 for0≤i≤m.
Proof. Letui = (T+F)m−ixfor 0≤i≤m. It follows thatui, 0≤i≤m, are linearly independent vectors, (T +F)u0 = 0 and (T +F)ui = ui−1 for 1 ≤ i ≤ m. Since ker(T)∩ker(F) ={0}, we infer that f(u0) 6= 0. Without loss of generality we may assume thatf(u0) = 1. Consider the scalarsc0, c1, . . . , cm−1 defined inductively by
c0=−f(u1)
c1=−c0f(u1)−f(u2)
c2=−c1f(u1)−c0f(u2)−f(u3) ...
cm−1=−cm−2f(u1)− · · · −c0f(um−1)−f(um).
This means that we have f(ui) +
i
X
k=1
ci−kf(uk−1) = 0 for 1≤i≤m. (3.1)
Let x0 = u0 and xi = ui +
i
X
k=1
ci−kuk−1 for 1 ≤ i ≤ m. Clearly, the vectors xi, 0≤i≤m, are linearly independent. Moreover, it follows from (3.1) thatf(xi) =δi0
for 0≤i≤m. Furthermore, we have (T+F)x0= (T+F)u0= 0 and
(T+F)xi = (T+F)ui+
i
X
k=1
ci−k(T+F)uk−1=ui−1+
i
X
k=2
ci−kuk−2=xi−1 for 1≤i≤m. Finally, we have
x=um∈Span{ui: 0≤i≤m}= Span{xi: 0≤i≤m}.
This completes the proof.
The following lemma is a special case of Proposition 3.1, and it will be required for proving that proposition.
Lemma 3.3. Let T ∈ B(X) be such that dimN∞(T) ≤ n, and let F ∈ B(X) be a rank-one operator such thatker(T)∩ker(F) ={0}. Then eitherdimN∞(T+F)≤n ordimN∞(T−F)≤n.
Proof. Write F = z⊗f where z ∈ X and f ∈ X∗ are non-zero. Clearly, if either ker(T+F)n+1or ker(T−F)n+1is contained in M(T, F), then either dimN∞(T+F)≤ nor dimN∞(T−F)≤nrespectively. Hence, we may assume that ker(T+F)n+1* M(T, F) and ker(T−F)n+1*M(T, F). Let 0≤m, p≤nbe the biggest integers for which there existx∈ker(T+F)m+1\ker(T+F)mandy∈ker(T−F)p+1\ker(T−F)p such that x, y /∈M(T, F). Without loss of generality we can assume thatm≤p. We will show that dimN∞(T+F)≤n. Using the previous lemma, we infer that y is a linear combination of linearly independent vectorsyi, 0≤i≤p, such that
(T −F)y0= 0, (T−F)yi=yi−1for 1≤i≤p, andf(yi) =δi0for 0≤i≤p.
From this, one can easily see that (T +F)y0 = 2z and (T+F)yi =T yi =yi−1 for 1≤i≤p, and so (T +F)kyi=yi−k for 0≤k≤i≤p. Thus, we get easily that
I+
p
X
i=0
yi⊗f(T+F)i =
p
Y
i=0
I+yi⊗f(T+F)i .
Furthermore, since f((T +F)iyi) = f(y0) = 1 for 0 ≤ i ≤ p, the above equation defines an invertible operator denoted byS.
Let u∈ker(T+F)n+1 be an arbitrary non-zero vector, and let 0≤r ≤nbe such thatu∈ker(T+F)r+1\ker(T +F)r. Ifu∈M(T, F), thenf(Tiu) = 0, and so (T+F)iu=Tiufor everyi≥0. Hence,Su=u∈M(T, F)⊆ N∞(T). Consider the
case when u /∈M(T, F). Then, Lemma 3.2 asserts that uis a linear combination of linearly independent vectorsxi, 0≤i≤r, satisfying
(T+F)x0= 0, (T+F)xi=xi−1 for 1≤i≤r, andf(xi) =δi0 for 0≤i≤r.
It follows that (T +F)kxi =xi−k for k ≥0 and 0 ≤i ≤r, where we set formally xj = 0 for j < 0. Now, by the definition of m, we have r ≤ m ≤ p. This allows us to obtain easily that Sxi = xi +yi for 0 ≤ i ≤ r. It follows that TiSxi = x0+y0 ∈ker(T), and henceSxi∈ N∞(T) for 0≤i≤r. Consequently, we get that Su∈ N∞(T). The vector uwas arbitrary, therefore Sker(T+F)n+1 ⊆ N∞(T). So that dim ker(T+F)n+1≤n. According to Remark 2.1, this completes the proof.
ForT, F ∈ B(X), we denote respectively by ˜T and ˜F the operators induced by T andF onX/M(T, F). Note that the hyper-kernels of ˜T+cF˜ andT+cF are related by the following relation (see [17, Lemma 2.9])
N∞( ˜T+cF˜) =N∞(T+cF)/M(T, F) for allc∈K. (3.2) Proof of Proposition3.1. Firstly, if ˜F = 0, then it follows from (3.2) that
N∞( ˜T+ ˜F) =N∞(T+F)/M(T, F) =N∞( ˜T) =N∞(T)/M(T, F).
So that dimN∞(T+F) = dimN∞(T)≤n.
Now, consider the case ˜F 6= 0. Thenz /∈M(T, F), and for everyx∈X, we have x+ M(T, F)∈ker( ˜T)∩ker( ˜F) ⇔ T x∈M(T, F) andF x=f(x)z∈M(T, F)
⇔ T x∈M(T, F) andf(x) = 0
⇔ x∈M(T, F).
This implies that ker( ˜T)∩ker( ˜F) ={0}.
Since dimN∞( ˜T)≤n−q whereq= dim M(T, F), the previous lemma ensures that either dimN∞( ˜T+ ˜F)≤n−qor dimN∞( ˜T−F˜)≤n−q. Thus, we get that either dimN∞(T +F)≤nor dimN∞(T −F)≤n. This completes the proof.
Throughout the sequel, Λ will denote any of the subsets B+n(X), Bn−(X) or B±n(X). Also, the subsetBn(X) =Bn+(X)∩ B−n(X), introduced and studied in [17], will be used in the rest of this paper.
Recall that for a semi-Fredholm operatorT ∈ B(X), theindex is defined by ind(T) = dim ker(T)−codim ran(T),
and if the index is finite,T is said to beFredholm. It should be noted that if ind(T) = 0 then a(T) = d(T) (see [12, Lemma 2.3]). Moreover, in this case
T ∈Λ ⇔ T ∈ Bn(X) ⇔ dimN∞(T)≤n.
Proposition 3.4. Let T ∈Λ and let F ∈ B(X) be a rank-one operator. Then either T+F ∈Λ orT−F ∈Λ.
Before proving this proposition, a duality relation betweenBn+(X) andBn−(X) should be established first. For a subsetM ⊆X, we denote byM⊥={f ∈X∗:M ⊆ker(f)}
its annihilator.
Lemma 3.5. Let T be a bounded operator onX. Then :
T ∈ B+n(X) (resp.B−n(X)) ⇔ T∗∈ B−n(X∗) (resp.B+n(X∗)).
Proof. Suppose thatT ∈ B+n(X). In particular, T is a semi-Fredholm operator, and so ran(Tk) is closed for everyk ≥0. Since a(T) ≤n, it follows from [15, Corollary A.1.17] that
ker(Tn+1)⊥= ker(Tn)⊥= ran((T∗)n+1) = ran((T∗)n).
Thus, d(T∗)≤n. Using [15, Theorem A.1.20] we get that
codimR∞(T∗) = codim ran((T∗)n) = dim ker(Tn) = dimN∞(T)≤n.
So that T∗ ∈ B−n(X∗). The proofs of the converse and of the statement for Bn−(X)
are similar.
Proof of Proposition3.4. Let T ∈ Λ, and let F ∈ B(X) be a rank-one operator.
It follows from [15, Theorem 16.16] that T +F and T −F are semi-Fredholm. If T ∈ B+n(X) then Proposition 3.1 implies that eitherT+F ∈ B+n(X) orT−F ∈ Bn+(X).
The case whenT ∈ Bn−(X) follows from the first one by duality.
The following theorem, will play a crucial role in proving the main results.
Theorem 3.6. Let F ∈ B(X) be a non-zero operator. Then the following assertions hold:
1. There exists an invertible operator T ∈ B(X) such thatT+F /∈Λ.
2. If dim ran(F)≥2, then there exists an invertible operator T ∈ B(X)such that T+F /∈Λ andT−F /∈Λ.
Proof. Suppose first that ran(F) has an infinite dimension. Then codim ker(F) =∞, and hence there exist linearly independent vectors xi, 0≤i≤2n+ 1, that generate a subspace having trivial intersection with ker(F). It follows that the vectors F xi, 0≤i≤2n+ 1, are linearly independent. Write
X = Span{xi: 0≤i≤2n+ 1} ⊕Y = Span{F xi: 0≤i≤2n+ 1} ⊕Z, whereY, Z are two closed subspaces andY =F−1Z. Then there exists an invertible operatorT ∈ B(X) such thatT Y =Z, andT xi = (−1)iF xi for 0≤i≤2n+ 1.
Clearly,x2i+1∈ker(T+F) andx2i∈ker(T−F) for 0≤i≤n, and hence dim ker(T±F)> n.
But, we have also
ran(T+F)⊆Span{F x2i: 0≤i≤n} ⊕Z, and
ran(T−F)⊆Span{F x2i+1: 0≤i≤n} ⊕Z.
Then codim ran(T ±F)> n, and soT ±F /∈Λ. This establishes the assertions (1) and (2).
Assume now that F is finite-rank, and let p= min{dim ran(F),2}. It follows from [17, Proposition 2.12] that there exists an invertible operator T ∈ B(X) such that T+F /∈ Bn(X) and T−(−1)pF /∈ Bn(X). But, T +F and T −(−1)pF are
Fredholm operators of index zero, thenT+F /∈Λ andT−(−1)pF /∈Λ. This completes
the proof.
4. Proofs of the main results
As a consequence of Theorem 3.6 and Proposition 3.4, we have the following result.
Lemma 4.1. Let Φ :B(X)→ B(X)be an additive surjective map. IfΦpreservesΛ in both directions, then Φ is injective and it preserves the set of rank-one operators in both directions.
Proof. Suppose on the contrary that there exists F 6= 0 such that Φ(F) = 0. Then, by Theorem 3.6, there exists an invertible operator T ∈ B(X) satisfyingT+F /∈Λ.
But, Φ(T +F) = Φ(T)∈Λ. This contradiction proves that Φ is injective.
LetF ∈ B(X) with dim ran(F)≥2. Then it follows again by Theorem 3.6 that there exists an invertible operatorT ∈ B(X) such thatT+F andT−F do not belong to Λ, and hence Φ(T+F) and Φ(T−F) do neither. Therefore, by Proposition 3.4, we obtain that dim ran(Φ(F)) ≥ 2. Since Φ is bijective and Φ−1 satisfies the same properties as Φ, we obtain that Φ preserves the set of rank-one operators in both
directions.
Recall that an operatorT ∈ B(X) is said to bealgebraicif there exists a non-zero complex polynomialP for whichP(T) = 0. Such an operatorT has finite ascent and descent (see [3, Theorem 2.7] and [4, Theorem 1.5]). Moreover, we have
T ∈Λ ⇔ T ∈ Bn(X) ⇔ dimN∞(T)≤n.
Lemma 4.2. LetΦ :B(X)→ B(X)be an additive surjective map preservingΛin both directions. ThenΦ(I) =cI wherec is a non-zero scalar.
Proof. We claim first thatS = Φ(I) is an algebraic operator. Letx∈X be non-zero.
If the set{Six: 0≤i≤2n+1}is linearly independent, then there exists a linear form f ∈X∗such thatf(Six) =−δi,2n+1for 0≤i≤2n+ 1. LetT =S+Sn+1x⊗f Sn+1. It follows that
T(Six) =Si+1x, for 0≤i≤n−1, and T(Snx) = 0.
Hence a(T)≥n+ 1. On the other hand, we have
T∗(f Si) =f Si+1, for 0≤i≤n−1, andT∗(f Sn) = 0.
Then a(T∗) ≥n+ 1, and so d(T)≥ n+ 1. Thus T /∈ Λ. This contradiction shows that {Six: 0≤i≤2n+ 1} is a linearly dependent set. The vectorxwas arbitrary, therefore it follows from [2, Theorem 4.2.7] thatS is algebraic.
Now assume, on the contrary, that S is not a scalar multiple of the identity.
Then there existsy1∈X such that the vectorsy1 andSy1 are linearly independent.
Since S ∈ Λ, the subspace ran(S) has an infinite dimension, and hence there exists
yi ∈ X, 2 ≤ i ≤ n, such that {y1, Syi : 1 ≤ i ≤ n} is a linearly independent set.
Consider linear formsgi∈X∗ such that
gi(y1) = 0 andgi(Syj) =−δij for 1≤i, j≤n.
If we letF =Pn
i=1S2yi⊗gi, we obtain easily thatSyj∈ker(S+F), for 1≤j≤n, and (S+F)y1 = Sy1 ∈ ker(S +F). Consequently, dimN∞(S+F) ≥ n+ 1. But, we have also that S+F is an algebraic operator (see [4, Proposition 3.6]), therefore S+F /∈Λ. By Lemma 4.1, Φ is bijective and preserves rank-one operators in both directions. Hence, we obtain that K = Φ−1(F) is of rank non-greater than n and I+K /∈ Λ. However, I+K is algebraic and ker((I +K)n+1) ⊆ ran(K), and so
I+K∈Λ. This contradiction completes the proof.
Letτ be a field automorphism ofK. An additive mapA:X →Y between two Banach spaces is calledτ-semi linearifA(λx) =τ(λ)Axholds for allx∈Xandλ∈K. Moreover, we say simply thatAisconjugate linearwhenτis the complex conjugation.
Notice that ifAis non-zero and bounded, then τ is continuous, and consequently, τ is either the identity or the complex conjugation (see [9, Theorem 14.4.2 and Lemma 14.5.1]). Moreover, in this case, the adjoint operator A∗ : Y∗ → X∗, defined by A∗(g) =τ−1◦g◦A for allg∈Y∗, is againτ-semi linear.
Lemma 4.3. LetΦ :B(X)→ B(X)be an additive surjective map preservingΛin both directions. Then there exists a non-zero scalar c, and either
1. there exists an invertible bounded linear, or conjugate linear, operator A:X → X such thatΦ(F) =cAF A−1 for all finite-rank operators F ∈ B(X), or 2. there exists an invertible bounded linear, or conjugate linear, operatorB:X∗→
X such thatΦ(F) =cBF∗B−1 for all finite-rank operatorsF ∈ B(X). In this case, X is reflexive.
Proof. The existence of a non-zero scalarcsuch that Φ(I) =cIis ensured by Lemma 4.2. Clearly, we can suppose without loss of generality that Φ(I) = I. Since Φ is bijective and preserves the set of rank-one operators in both directions, then by [16, Theorems 3.1 and 3.3], there exist a ring automorphism τ :K→ Kand either two bijectiveτ-semi linear mappingsA:X →X andC:X∗→X∗ such that
Φ(x⊗f) =Ax⊗Cf for allx∈X andf ∈X∗, (4.1) or two bijectiveτ-semi linear mappingsB:X∗→X andD:X →X∗such that
Φ(x⊗f) =Bf ⊗Dx for allx∈X andf ∈X∗. (4.2) Suppose that Φ satisfies (4.1), and let us show that
C(f)(Ax) =τ(f(x)) for allx∈X andf ∈X∗. (4.3) Clearly, it suffices to establish that for allx∈X andf ∈X∗,f(x) =−1 if and only if C(f)(Ax) =−1. Letx∈X andf ∈X∗. We can choose linearly independent vectors
z1, . . . , zn in ker(f)∩ker(C(f)A). Then, it follows from [17, Lemma 3.8] that f(x) =−1 ⇔ ∃{gi}ni=1 ⊆X∗:I+x⊗f+
n
X
i=1
zi⊗gi∈ B/ n(X)
⇔ ∃{gi}ni=1 ⊆X∗:I+x⊗f+
n
X
i=1
zi⊗gi∈/ Λ
⇔ ∃{gi}ni=1 ⊆X∗:I+Ax⊗Cf+
n
X
i=1
Azi⊗Cgi ∈/ Λ
⇔ ∃{gi}ni=1 ⊆X∗:I+Ax⊗Cf+
n
X
i=1
Azi⊗Cgi ∈ B/ n(X)
⇔ C(f)(Ax) =−1.
Thus, relation (4.3) holds, and arguing as in [16], we get thatτ,A,Care continuous, τ is the identity or the complex conjugation, and C = (A−1)∗. Therefore,τ−1 = τ and, for everyu∈X, we have
Φ(x⊗f)u=τ(f A−1u)Ax=A(f(A−1u)x) =A(x⊗f)A−1u.
Thus, Φ(x⊗f) = A(x⊗f)A−1 for allx∈X and f ∈X∗; that is, Φ(F) = AF A−1 for all finite-rank operators F∈ B(X).
Now suppose that Φ satisfies (4.2), and let us show that
D(x)(Bf) =τ(f(x)) for allx∈X andf ∈X∗. (4.4) Let x∈ X and f ∈X∗. Choose linearly independent linear forms h1, . . . , hn ∈X∗ such that hi(x) = 0 and D(x)(Bhi) = 0 for 1 ≤ i ≤ n. Then, it follows from the surjectivity ofD and from [17, Lemma 3.8] that
D(x)(Bf) =−1 ⇔ ∃{ui}ni=1⊆X :I+Bf ⊗Dx+
n
X
i=1
Bhi⊗Dui ∈ B/ n(X)
⇔ ∃{ui}ni=1⊆X :I+Bf ⊗Dx+
n
X
i=1
Bhi⊗Dui ∈/ Λ
⇔ ∃{ui}ni=1⊆X :I+x⊗f+
n
X
i=1
ui⊗hi∈/Λ
⇔ ∃{ui}ni=1⊆X :I+x⊗f+
n
X
i=1
ui⊗hi∈ B/ n(X)
⇔ ∃{ui}ni=1⊆X :I+f⊗Jx+
n
X
i=1
hi⊗Jui∈ B/ n(X∗)
⇔ f(x) =−1,
where J :X →X∗∗ is the natural embedding. Thus, relation (4.4) holds, and arguing as in [16], we get that τ, B, D are continuous, τ is the identity or the complex conjugation, andD= (B−1)∗J. But, the operatorsDand (B−1)∗, and therefore also
J are bijections, which implies the reflexivity of X. Furthermore, τ−1 =τ and, for every u∈X, we have
Φ(x⊗f)u = (Bf ⊗(B−1)∗J(x))u= (B−1)∗J(x)(u)·Bf
= τ(J(x)(B−1u))·Bf =B(J(x)(B−1u)f)
= B(f ⊗J(x))B−1u=B(x⊗f)∗B−1u.
Thus, Φ(x⊗f) =B(x⊗f)∗B−1 for allx∈X andf ∈X∗. Hence, Φ(F) =BF∗B−1 for all finite-rank operatorF∈ B(X). This completes the proof.
Theorem 4.4. Let Φ :B(X) → B(X) be an additive surjective map preserving Λ in both directions. Then there exists a non-zero scalarc, and either
1. there exists an invertible bounded linear, or conjugate linear, operator A:X → X such thatΦ(T) =cAT A−1 for allT ∈ B(X), or
2. there exists an invertible bounded linear, or conjugate linear, operatorB:X∗→ X such thatΦ(T) =cBT∗B−1 for all T ∈ B(X).
Proof. Since Φ preserves Λ in both directions, it follows that Φ takes one of the two forms in Lemma 4.3.
Suppose that Φ(F) =cAF A−1 for all finite-rank operatorsF ∈ B(X). Let Ψ(T) =c−1A−1Φ(T)A for allT ∈ B(X).
Clearly, Ψ satisfies the same properties as Φ. Furthermore, Ψ(I) =I and Ψ(F) =F for all finite-rank operatorsF ∈ B(X). LetT ∈ B(X) and choose an arbitrary rational numberλsuch thatT−λand Ψ(T)−λare invertible. LetF∈ B(X) be a finite-rank operator. SinceT −λ+F and Ψ(T)−λ+F are Fredholm of index zero, then
T−λ+F ∈ Bn(X) ⇔ T−λ+F ∈Λ ⇔ Ψ(T)−λ+F ∈Λ
⇔ Ψ(T)−λ+F ∈ Bn(X).
Hence, we get by [17, Proposition 2.17] that Ψ(T) =T. This shows that Φ(T) =cAT A−1for allT ∈ B(X).
Now suppose that Φ(F) = cBF∗B−1 for all finite-rank operators F ∈ B(X). Then Lemma 4.3 ensures thatX is reflexive. By considering
Γ(T) =c−1J−1(B−1Φ(T)B)∗J for allT ∈ B(X),
we get in a similar way that Γ(T) =T for allT ∈ B(X). Thus, Φ(T) =cBT∗B−1 for allT ∈ B(X), as desired. This finishes the proof.
With these results at hand, we are ready to prove our main results.
Proof of Theorem 1.1. (1)⇒(4). Suppose that Φ preservesAn(H) in both directions.
Using the fact that Φ is surjective, it follows by Theorem 2.7 that, for everyT ∈ B(H), T ∈ B+n(H)⇔ ∀S∈ B(H),∃ε0>0 :{T+εS:ε∈Qand|ε|< ε0} ⊆ An(H)
⇔ ∀S∈ B(H),∃ε0>0 :{Φ(T) +εΦ(S) :ε∈Qand|ε|< ε0} ⊆ An(H)
⇔ ∀R∈ B(H),∃ε0>0 :{Φ(T) +εR:ε∈Qand|ε|< ε0} ⊆ An(H)
⇔ Φ(T)∈ Bn+(H).
Thus Φ preserves B+n(H) in both directions. It follows that Φ takes one of the two forms in Theorem 4.4. Let us show that Φ cannot take the form
Φ(T) =cBT∗B−1 for all T ∈ B(H). (4.5) Suppose on the contrary that Φ takes the form (4.5). Let{en:n≥0}be an arbitrary orthonormal basis of H. Consider the weighted unilateral shift operator U ∈ B(H) given by
U en= (n+ 1)−1en+1 for every n≥0. (4.6) Clearly,U is an injective quasi-nilpotent operator.
Thus, a(U∗) = d(U∗) =∞,U ∈ Bn+(H) andU∗∈ B/ ±n(H).
So that Φ(U) =cBU∗B−1∈ B/ ±n(H), a contradiction.
(2)⇒(4). Now, suppose that Φ preserves Dn(H) in both directions. As above, using Theorem 2.7 we infer that Φ preservesBn−(H) in both directions, and so Φ takes one of the two forms in Theorem 4.4. Consider the unilateral shift operatorS∈ B(H) given by
Se0= 0 and Sen=en−1forn≥1.
Clearly,S is surjective and a(S) =∞.
Thus, d(S∗) = ∞, S ∈ B−n(H) and S∗ ∈ B/ n−(H). This contradiction shows that Φ cannot take the form (4.5).
(3)⇒(4) is similar to the first implication with the same example (4.6).
(4)⇒(1), (2) and (3) are obvious.
Proof of Theorem 1.2. Follows from Theorems 2.5 and 4.4.
Proof of Corollary1.3. The proof is similar to the proof of Theorem 1.1.
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Mourad Oudghiri
Universit´e Mohammed Premier
D´epartement Math, Labo LAGA, Facult´e des Sciences d’Oujda 60000 Oujda, Maroc
e-mail:[email protected] Khalid Souilah
Universit´e Mohammed Premier
D´epartement Math, Labo LAGA, Facult´e des Sciences d’Oujda 60000 Oujda, Maroc
e-mail:[email protected]
