Surjective Maps Preserving Local Spectral Radius

Let B(X) be the algebra of all bounded linear operators on a complex Banach space X. Let x0 is a nonzero fixed vector in X. We give the concrete form of every surjective map φ from B(X) into its self, such that the local spectral radius of φ(T )φ(S) + φ(R) at x0 equals the local spectral radius of TS + R at x0. We do not assume φ to be linear, or even additive. Mathematics Subject Classification: Primary 47A11; Secondary 47A10, 47B48


Introduction
Throughout this note, X and Y will denote a infinite-dimensional complex Banach spaces, and B(X) the algebra of all linear bounded operators on X and its unit will be denoted by I.The duality between the Banach space X and its dual, X * , will be denoted by ., . .For x ∈ X and f ∈ X * , we denote as usual by x ⊗ f the rank one operator on X given by z → z, f x.We say that a map φ : B(X) → B(X) is unital if φ(I) = I.The local resolvent of an operator T ∈ B(X) at some point x ∈ X, denoted by ρ T (x), is the union of all open subsets U ⊆ I C for which there exists an analytic function f : U → X such that (T − λ)f (λ) = x for all λ ∈ U.
The local spectrum of T at x is defined as the complement in I C of the local resolvent ρ T (x), i.e σ T (x) = I C \ ρ T (x).Let us observe that σ T (x) is a (possibly empty) closed subset of σ(T ), the usual spectrum of T ∈ B(X).The local spectral radius of T at x is defined by An operator T ∈ B(X) is said to have the single-valued extension property at λ 0 ∈ I C, abbreviated T has the SVEP at λ 0 , if for every neighbourhood U of λ 0 the only analytic function f : U → X which satisfies the equation is the constant function f ≡ 0. The operator T is said to have the SVEP if T has the SVEP at every λ ∈ I C. Evidently, T has SVEP at every λ ∈ I C \ int(σ p (T )), where int(σ p (T )) denotes the interior of the point spectrum of T .In particular, if σ p (T ) has empty interior, then T has SVEP.For example, every finite rank operator has SVEP.
In the finite dimensional case, we denote by M n the algebra of all complex n × n matrices.Every T belonging to M n has the single-valued extension property.We can see, for example, the books [1,15] for further information on the local spectral theory.
The problem of characterizing linear or additive maps on operator algebras preserving certain properties, subsets or relations has attracted the attention of many authors in the last decades (see for instance [2,4,5,9,10,13]).
In [5], Bourhim and Ransford studied additive maps on B(X) preserving the local spectrum at each point x ∈ X.They showed that the only additive map φ : B(X) → B(X) for which is the identity on B(X).
J. Bračič and V. Müller in [7] characterized the surjective linear map φ : B(X) → B(X) satisfies r T (x) = r φ(T ) (x) for all T ∈ B(X) and x is a nonzero vector fixed in X. Different versions of this result in the case when X is finitedimensional were considered in [4].In the infinite-dimensional case, the linear, surjective and continuous maps preserving different local spectral properties were characterized in [2].
in recent years a great activity has occurred in the question of relaxing the assumption of linearity or additivity (see, for example, [3,6,8,11,12]).
In [3], Bourhim and Mashreghi described surjective maps on B(X) which preserve the local spectral radius distance zero at points of X.They showed, in particular, that if φ is a surjective map on B(X) such that for every then there exists a nonzero scalar c and an operator While, Different versions of this result in the case where X is finite-dimensional and x is a nonzero fixed vector were considered in [6] and [11].
C. Costara in [11], showed the following result holds true without the linearity or additivity of the map.
then there exists a unimodular α ∈ I C and an invertible matrix A ∈ M n such that either Ax 0 = x 0 and or Ax 0 = x 0 and Where the matrix T is obtained from T by entrywise complex conjugation.
The purpose of this note is to study the same problem as in [3] but at nonzero vector fixed x 0 in X, with respect to the ring structure of the algebra.More precisely, we will show that if a surjective maps (not necessarily additive), if and only if there exists an invertible bounded linear or conjugate linear operator A : X −→ X such that φ(T ) = AT A −1 for all T ∈ B(X).We also give the version result in the finite dimensional case.

Preliminaries and Notations
In this section, we collect some lemmas that are needed for the proof of our main results.The first one plays a key role in the proof of Theorem 3.1.It is inspired [3,Theorem 3.1] in which the authors have obtained a local version of Zemánek's spectral characterization of the radical.Lemma 2.1 Let x 0 be a nonzero vector in X.Let T, S ∈ B(X) such that r T +R (x 0 ) = r S+R (x 0 ) for every R ∈ B(X) then T = S.
We shall make extensive use of the following result.
An operator P ∈ B(X) is called an idempotent operator if P 2 = P .We denote by P(X) = {P ∈ B(X) : P 2 = P } the set of all idempotent operators.The following lemma was obtained in [12], Theorem 1.1, in which the authors have given the form of every unital surjective map φ on B(X) such that T S is a nonzero idempotent if and only if φ(T )φ(S) is for all T, S ∈ B(X) when the dimension of X is at least 3.

Main Results
We begin this section with the following result which is the main theorem.Our arguments are influenced by the ones given in [14].
if and only if there exists an invertible bounded linear or conjugate-linear operator A : X −→ X such that Ax 0 = x 0 and φ(T ) = AT A −1 for all T ∈ B(X).
Proof.Let A : X −→ X be an invertible bounded linear or conjugate-linear operator such that Ax 0 = x 0 .By [2, Lemma 2.3], we have for every T, S, R ∈ B(X).This proves the "if" part.
Next, we state our main result for the finite dimensional case.
Where the matrix T is obtained from T by entrywise complex conjugation.
Proof.we will argue as in the proof of Theorem 3.1, So we get φ is a surjective map such that φ(0) = 0 satisfying the assumptions of Theorem 1.1, hence φ is either of the form (2), or of the form (3). Since φ(I) = I then α = 1 and consequently we get the desired result.