1660-5446/17/030001-8 published onlineMay 26, 2017
c Springer International Publishing 2017
Bounded Local Resolvents for Operators with Single-Valued Extension Property
S. E. Benharkakia, A. Daoui and E. H. Zerouali
Abstract.
Let
Tbe a bounded operator with (SVEP) on its localizable spectrum
σloc(
T). We show that for every open subset
Uof
σloc(
T), there exists a unit vector
xwhose local spectrum coincides with the closure of
U, and such that its local resolvent function is bounded. This result answers positively to an open question stated by several authors, and extends the both cases of operators with trivial divisible subspace and operators whose point spectrum has empty interior.
Mathematics Subject Classification.
Primary 47A 11.
Keywords.
(SVEP), bounded local resolvent function, localizable spec- trum.
1. Introduction
Let X be a complex Banach space and let B ( X ) be the algebra of all bounded linear operators on X . For T ∈ B ( X ), we denote by ρ ( T ) its resolvent set, by σ ( T ) := C\ρ ( T ) its spectrum and by R
T( . ) : λ → R
T( λ ) = ( T −λ )
−1∈ B ( X ) its resolvent map defined on ρ ( T ).
For an element x ∈ X , we will say that a complex number λ belongs to the local resolvent set of T at x , denoted by ρ
T( x ), if there exists an analytic function f : U → X , defined on a neighborhood U of λ , which satisfies
( T − μ ) f ( μ ) = x for every μ ∈ U. (1.1) The local spectrum of T at x is defined as the set σ
T( x ) := C\ρ
T( x ). Clearly, σ
T( x ) is a closed subset of σ ( T ) and from Eq. (1.1), it follows easily that σ
T( x ) = σ
T( αx ) for every non-zero number α . In particular σ
T( x ) = σ
T(
xx).
It is also easy to see that
σ
T( x ) ∪ σ
T( y )\σ
T( x ) ∩ σ
T( y ) ⊂ σ
T( x + y ) ⊂ σ
T( x ) ∪ σ
T( y )
for every x, y ∈ X . In particular, if σ
T( x ) ∩ σ
T( y ) = ∅, we get σ
T( x + y ) = σ
T( x ) ∪ σ
T( y ). For further information and results on local spectra, see [8].
The third author was supported by the Project URAC 03 of the National Center of Re- search and by Hassan II Academy of Sciences. We are indebted to the anonymous referee, for his valuable comments that improved this paper.
An operator T ∈ B ( X ) is said to have the single-valued extension prop- erty [(SVEP), for short] if, for every open set U ⊂ C, the only analytic solution f : U → X of the equation
( T − μ ) f ( μ ) = 0 ( for every μ ∈ U ) (1.2) is the function f ≡ 0.
The class of operators with (SVEP) is very large. In finite dimensional space, every linear operator has (SVEP). More generally, any operator with empty interior point spectrum satisfies (SVEP).
If T ∈ B ( X ) has (SVEP), then for every x ∈ X , there exists a unique analytic function R
T( ., x ) defined on ρ
T( x ), such that ( T − μ ) R
T( μ, x ) = x for all μ ∈ ρ
T( x ). This function is called the local resolvent function of T at x , and coincides with ( T − μ )
−1x on ρ ( T ). It is a known fact that the local spectrum σ
T( x ) may be empty even for x = 0. We derive, see Finch [6], that T has (SVEP) if and only if σ
T( x ) = ∅ for every x = 0.
For a bounded operator, it is not difficult to show that R
T( λ ) ≥ 1
dist ( λ, σ ( T )) for every λ ∈ ρ ( T ), where dist ( λ, σ ( T )) =: inf
μ∈σ(T)|λ − μ|
is the distance of λ to σ ( T ) [4, Corollary 1 . 5]. It follows that the resolvent map is never bounded. In contract with resolvent maps, Goz´ alez observed in [7], that the behavior of the local resolvent function may be quite different.
More precisely it is shown in [7, Theorem 2], that a normal operator admits a non-zero bounded local resolvent function if and only if its spectrum has non-empty interior.
In [11, Proposition 2.1] and [12], M. Neumann provided the same char- acterization for multiplication operators on the space of continuous functions on a Hausdorff space. M. Neumann extended this result to non-separable Ba- nach spaces. He also showed that the multiplication operator by a continuous function, on the Banach algebra C (Ω) of continuous functions on a compact Hausdorff space Ω, has non-trivial bounded local resolvent if and only if its spectrum has non-empty interior. The general question of the existence of bounded local resolvent for convolution operators is stated in [11] and the situation where such operators avoid Wiener–Pitt phenomena is solved in [10].
We will say that an operator T has non-trivial bounded local resolvent functions if there exists a non-zero vector x ∈ X with bounded local resolvent function.
The notion of localizable spectrum, denoted by σ
loc( T ), plays a crucial role in this paper and is defined next.
Let T ∈ B ( X ) have (SVEP), the localizable spectrum of T is the set of
all complex number λ with the following property : for each open neighbor-
hood U of λ there exists a non-zero vector x ∈ X such that σ
T( x ) ⊂ U . For
an arbitrary operator T ∈ B ( X ), the localizable spectrum σ
loc( T ) is always
contained in the approximate point spectrum σ
ap( T ). For a large class of op-
erators including normal operators, the localizable spectrum coincides with
σ ( T ). In contrast with the previous observation, a pure subnormal bounded
operator has empty localizable spectrum. See [5,8] for more information and further results.
In Braˇ ciˇ c and M¨ uller [2], proved that for every operator T on a Banach space X , such that both its point spectrum and its localizable spectrum have non-empty interior, there is a vector x such that the local resolvent function at x , R
T( ., x ) is bounded on ρ
T( x ). In the same paper, a decomposable oper- ator whose spectrum has empty interior and with non-trivial bounded local resolvent function is given.
A qualitative version of the previous problem is the following:
Let U be a prescribed open subset of σ ( T ), can we find x ∈ X with bounded local resolvent function and such that σ
T( x ) = U ?
In this direction, M¨ uller with Braˇ ciˇ c in [2, Theorem 4] and with Neu- mann [10, Theorem 4] obtained the next results,
Theorem 1.1. Let T ∈ B ( X ) be an operator such that either σ
p( T ) has empty interior or E
T(∅) :=
λ∈C
( T − λ )( X ) = {0} and such that σ
loc( T ) has non- empty interior. Let U be a non-empty open subset of σ
loc( T ) and let u ∈ X be a vector with σ
T( u ) ⊂ U . Then for every > 0; there exists some x ∈ X such that u − x ≤ ε , σ
T( x ) = U and such that R
T( ., x ) is bounded.
Since any of the conditions intσ
p( T ) = ∅ and E
T( ∅ ) = { 0 } implies (SVEP), the question of relaxing the assumption on T to (SVEP) has been stated as an open problem in [2,10]. Using a slight modification of the proofs in [2,10], we answer positively to this question. Our main result is
Theorem 1.2. Let T ∈ B ( X ) be with (SVEP) and such that σ
loc( T ) has non- empty interior. Let U be a non-empty open subset of σ
loc( T ) and let u ∈ X be a vector with σ
T( u ) ⊂ U . Then for every > 0; there exists some x ∈ X such that u − x ≤ ε , σ
T( x ) = U , and such R
T( ., x ) is bounded.
2. Proof of the Main Results
The next useful observation is not difficult to obtain. It provides a weak infinite form of injectivity to be used in our proof.
Lemma 2.1. Let T ∈ B ( X ) have (SVEP), x ∈ X and λ ∈ C . Then λ ∈ ρ
T( x ) if and only if there is a unique family of vectors ( x
i)
i≥0∈ X , such that x
0= x and ( T − λ ) x
i+1= x
i( i ≥ 0) and such that sup
i≥1x
ii
< + ∞ . In particular, for λ ∈ ρ
T( x ) we have x ∈
n≥0
( T − λ )
nX . For convienence, we will write ( x
i)
i= (( T − λ )
−ix )
ieven if T − λ is not one to one.
We start our proof by extending [2, lemma 3] that is the key point of the proof. It encloses the main modifications of the previous works.
Lemma 2.2. Let T ∈ B ( X ) have (SVEP), λ ∈ σ
loc( T ) and V a neighborhood
of λ . Then there exists some vector x ∈ X such that λ ∈ σ
T( x ) ⊂ V .
Proof. For λ ∈ σ
loc( T ) and V
0a given neighborhood of λ , we first choose
x
0∈ X such that σ
T( x
0) ⊂ V
0. If λ ∈ σ
T( x
0), we take x = x
0and the proof
is complete. If λ / ∈ σ
T( x
0), using the definition of the localizable spectrum, there exists a sequence of elements x
n∈ X and a sequence of decreasing open neighborhoods of λ such that σ
T( x
n) ⊂ V
n\V
n+1. If λ ∈ σ
T( x
n) for some n , the lemma is proved by taking x = x
n. Otherwise, we let V
nsatisfying
n≥0
V
n= {λ} and we associate with x
n, a sequence of non-negative numbers α
ndecreasing enough to zero such that,
(1) The vector x =:
∞n=0
α
nx
nconverges.
(2) The function f ( z ) =:
n≥1
α
nR
T( z, x
n) for z ∈ V converges.
To get the second assertion on C\V , we use dist ( V, σ
T( x
n)) > 0 for every n . Indeed, that M
n= sup
C\VR
T( z, x
n) < +∞ and it suffices to assume that
n≥1
α
nM
n< ∞. From the expression ( T − z ) f ( z ) =
n≥1
α
nR
T( z, x
n) =
∞ n=0α
nx
n= x for z ∈ V , we get σ
T( x ) ⊂ V . It remains to show that λ ∈ σ
T( x ).
For each n ≥ 1, we have ( T − z )
i≥n+1
α
iR
T( z, x
i) =
∞n+1
α
ix
i, and hence
σ
T⎛
⎝
i≥n+1
α
ix
i)
⎞
⎠ ⊂ V
n+1.
Also, since the local spectra of α
1x
1, . . . , α
nx
nand
i≥n+1
α
ix
iare mutually disjoint, we obtain
σ
T( x ) =
n i=1
σ
T( x
i) ∪ σ
T⎛
⎝
i≥n+1
α
ix
i)
⎞
⎠ .
We take λ
narbitrary in σ
T( x
n). It will follow that λ
n∈ σ
T( x ) for every n ≥ 1 and we derive that lim
n→∞λ
n= λ ∈ σ
T( x ).
Proof of Theorem 1.2. We first fix a dense sequence ( λ
n)
n≥1in U such that λ
i= λ
jfor i = j and u ∈ X such that σ
T( u ) ⊂ U . We use Lemma 2.2, to provide a family ( x
n)
n≥1of unit vectors such that λ
1∈ σ
T( x
1) ⊂ U and λ
n∈ σ
T( x
n) ⊂ U \{λ
1, . . . , λ
n−1}.
Let ε > 0 be arbitrary small, we will construct inductively a sequence ( α
n)
n≥1of non-negative numbers and a subset M ⊂ N, for each n ∈ M a non-negative integer a
n, and, for every n ∈ N\M and k ∈ N, a positive integer m ( n, k ) such that the following conditions are fulfilled:
(1) λ
n∈ σ
T( u +
ni=1
α
ix
i), (2) α
nx
n≤ ε
2
n,
(3) sup
z∈UR
T( z, α
nx
n) ≤ 1 2
n, (4) α
nx
n∈ 1
2
n( T − λ
j)
aj+1B
Xfor j < n, j ∈ M (here B
Xis the unit ball of X ),
(5) ( T − λ
j)
−m(j,k)α
nx
n≤ k
m(j,k)2
nfor j < n, j / ∈ M, k ∈ N
∗,
(6) n ∈ M ⇐⇒ u +
ni=1
α
ix
i∈ /
k≥1
( T − λ
n)
k, and if n ∈ M , then a
n= max
k ; u +
ni=1
α
ix
i∈ ( T − λ
n)
k, (7) ( T − λ
n)
−m(n,k)( u +
ni=1
α
ix
i) ≥ k
m(n,k)for n / ∈ M, k ∈ N . The first step n = 1 is trivial. Let n > 1 and suppose that α
i, . . . , α
n−1, the set M ∩ {1 , . . . , n − 1}, the numbers a
jfor ( j < n − 1 , j ∈ M ) and m ( j, k ) for ( j < n − 1 , j ∈ N\M, k ∈ N) satisfying (1)–(7) have been already constructed. To prove the n th-step, we distinguish two cases:
(i) If λ
n∈ σ
Tu +
n−1 i=1α
ix
i, we take α
n= 0 and then (1)–(5) are trivially satisfied.
(ii) If not, since λ
n∈ σ
T( x
n) ⊂ U\{λ
1, . . . , λ
n−1} , we deduce that (1) is satisfied for each positive number α
n> 0 and x
n∈
k≥1
( T − λ
j)
kfor every j < n . Thus, (1) to (4) are satisfied for all α
n> 0 small enough.
To check (5), we see that for each j < n such that j ∈ M , we have λ
j∈ σ
T( x
n). Consequently, x
n∈
k≥1
( T − λ
j)
kand sup
m≥1( T − λ
j)
−mx
nm1
< +∞. Thus, for α
n> 0 small enough, we will have ( T − λ
j)
−mα
nx
n≤ k
m2
nfor all k ∈ N
∗. Finally (5) is satisfied for all α
n> 0 which are small
enough.
We turn now to prove the alternative cases (6) and (7). If u +
ni=1
α
ix
i∈ /
k≥1
( T − λ
n)
k, we put n ∈ M and a
n=: max {k ; u +
ni=1
α
ix
i∈ ( T − λ
n)
k.
Suppose first that n ∈ M . It follows that u +
ni=1
α
ix
i∈
k≥1
( T − λ
n)
kand since λ
n∈ σ
T( u +
ni=1
α
ix
i), we have sup
m≥1( T − λ
n)
−m( u +
ni=1
α
ix
i)
m1
= +∞, (otherwise, we can construct f an- alytic solution of (1 . 1) for x = u +
ni=1
α
ix
i). Therefore, for each k ∈ N there exists m ( n, k ) ∈ N such that
( T − λ
n)
−m(n,k)u +
n i=1α
ix
i≥ k
m(n,k). Set x = u +
n≥1
α
nx
n, where ( α
n)
n≥1is as above. We have x−u ≤ ε and R
T( z, x ) = R
T( z, u ) +
i≥1
R
T( z, α
ix
i). It follows, sup
z∈U
R
T( z, x ) ≤ sup
z∈U
R
T( z, u ) +
i≥1
sup
z∈U
R
T( z, α
ix
i)
≤ sup
z∈U
R
T( z, u ) +
i≥1
1 2
i< +∞.
Hence, σ
T( x ) ⊂ U and the local resolvent function R
T( ., x ) is bounded on
C\U .
To show that U ⊂ σ
T( x ), we only need to see that λ
n∈ σ
T( x ) for every n ≥ 0. To this aim, we distinguish two cases:
(a) If n ∈ M , then u +
ni=1
α
ix
i∈ ( T − λ
n)
an+1and by (4), we get
+∞
i=n+1
α
ix
i∈
+∞
i=n+1
1
2
i( T −λ
n)
an+1B
X⊂ ( T −λ
n)
an+1B
X⊂ ( T − λ
n)
an+1. Consequently, x ∈ ( T − λ
n)
an+1, and so λ
n∈ σ
T( x ).
(b) If n ∈ M ,
m≥1
sup
( T − λ
n)
−mx 1
m ≥ sup
k≥1
( T − λ
n)
−m(n,k)x 1 m ( n, k )
≥ sup
k≥1
( T − λ
n)
−m(n,k)u +
ni=1
α
ix
i−
( T − λ
n)
−m(n,k)+∞
i=n+1
α
ix
i1 m ( n, k )
≥ sup
k≥1
k
m(n,k)−
+∞
i=n+1
k
m(n,k)2
i1 m ( n, k )
= sup
k≥1
1 − 1
2
nk
m(n,k)1
m ( n, k )
≥ sup
k≥1
1 2 k
m(n,k)1
m ( n, k ) = +∞.
Hence, λ
n∈ σ
T( x ). Finally λ
n∈ σ
T( x ) for every n ≥ 1, and thus σ
T( x ) = U .
It is easy to see that for an operators with the decomposition property ( δ ), the localizable spectrum coincides with the spectrum. In particular it is the case of decomposable operators or, more generally, the quotient of the decomposable operators by a closed invariant subspace. For further details, see [9]. An immediate consequence of the previous theorem is the following corollary
Corollary 2.3. Let T be a bounded decomposable operator on Banach space X such that σ ( T ) has non-empty interior. Then there exists a non-zero vector x in X for which R
T( ., x ) is bounded on ρ
T( x ).
Example 2.4. Let X be the Banach space of all bounded complex-valued functions on D := {z ∈ C; |z| < 1} equipped with the supremum norm and let T ∈ B ( X ) be the operator, multiplication by the independent variable z in D . It is easy to see that T is decomposable, and hence T has (SVEP).
Since σ
p( T ) = D has non-empty interior, [8, p 15], there exists a non-zero
vector x in X , such that R
T( ., x ) is bounded on ρ
T( x ).
On the other hands, in Braˇ ciˇ c and M¨ uller [3], constructed an example of a bounded linear operator T with (SVEP) and such that σ
p( T ) has non- empty interior. Moreover, T has the decomposition property ( δ ), as a quotient of a decomposable operator by a closed invariant subspace. It will follow that σ
loc( T ) = σ ( T ), and hence that the localizable spectrum has non-empty interior. We apply theorem 1.2, to provide a non-zero vector x in X , such that R
T( ., x ) is bounded on ρ
T( x ).
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