Volterra Semigroups Have Invariant Subspaces

Turovskii, Yu. V.

doi:10.1006/jfan.1998.3368

Cited by 41 publications

(45 citation statements)

References 16 publications

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“…The other invariant subspace results [T2,ST1] are connected with the calculation of the joint spectral radius. It was established in [ST1] that for a precompact set M of compact operators on a Banach space X…”

Section: 32mentioning

confidence: 99%

Topological Radicals, V. From Algebra to Spectral Theory

Shulman

Turovskiı

2013

Operator Theory: Advances and Applications

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Abstract. We introduce and study procedures and constructions of the theory of general topological radicals that are related to the spectral theorythe centralization, primitivity and socle procedures, the scattered radical, the radicals related to the continuity of the usual, joint and tensor radii. Among other applications we find some sufficient conditions of continuity of the spectrum and spectral radii of different types, and in particular prove that in a GCR C*-algebra the joint spectral radius is continuous on precompact sets and coincides with the Berger-Wang radius.

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Section: 32mentioning

confidence: 99%

Topological Radicals, V. From Algebra to Spectral Theory

Shulman

Turovskiı

2013

Operator Theory: Advances and Applications

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“…Turovski's Theorem ( [4], [3], Theorem 8.1.11) states that a semigroup of quasinilpotent compact operators has a non-trivial invariant subspace, so S must contain operators that are not quasinilpotent.…”

Section: Proof Suppose There Is a K Such That R(ab) ≤ Kr(a)r(b)mentioning

confidence: 99%

On submultiplicativity of spectral radius and transitivity of semigroups

Radjavi

Rosenthal

2006

Proc. Amer. Math. Soc.

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Abstract. It is shown that a transitive, closed, homogeneous semigroup of linear transformations on a finite-dimensional space either has zero divisors or is simultaneously similar to a group consisting of scalar multiples of unitary transformations. The proof begins with the result that for each closed homogeneous semigroup with no zero divisors there is a k such that the spectral radius satisfies r(AB) ≤ kr(A)r(B) for all A and B in the semigroup.It is also shown that the spectral radius is not k-submultiplicative on any transitive semigroup of compact operators.

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“…cardinal) measure on T and θ : T → B(X ) has only the property (i) then ∞ (θ) and ∞ (θ, x) can also be defined by the above formulas. In this situation the algebraic spectral radius was used in [24], [20], [21], [10] etc. In the same situation, the local algebraic spectral radius was used in [11] and [21].…”

Section: Where the Series Is Norm Convergent In B(x ) (The First Termentioning

confidence: 99%

“…[24]; see also [20], [21]). The importance of that concept is also revealed by Theorem 4 of [10], which characterizes the existence of functional calculus with "holomorphic functions" on a polydisk corresponding to a basis in a nilpotent Lie algebra; this extends the natural calculus with polynomials in several non-commuting variables.…”

Section: Introductionmentioning

confidence: 99%

Analytic joint spectral radius in a solvable Lie algebra of operators

Beltiţă

2001

Studia Math.

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Abstract.We introduce the concept of analytic spectral radius for a family of operators indexed by some finite measure space. This spectral radius is compared with the algebraic and geometric spectral radii when the operators belong to some finite-dimensional solvable Lie algebra. We describe several situations when the three spectral radii coincide. These results extend well known facts concerning commuting n-tuples of operators.

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Volterra Semigroups Have Invariant Subspaces

Cited by 41 publications

References 16 publications

Topological Radicals, V. From Algebra to Spectral Theory

Topological Radicals, V. From Algebra to Spectral Theory

On submultiplicativity of spectral radius and transitivity of semigroups

Analytic joint spectral radius in a solvable Lie algebra of operators

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