Bounds for the spectral abscissa of an element in a Banach algebra |
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Authors: | K L Olifirov |
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Institution: | 1. Leningrad State University, USSR
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Abstract: | For an arbitrary element x with spectrum sp(x) in a Banach algebra with identity e ≠ 0 we define the upper (lower) spectral abscissa \(\mathop {\sigma + (x)}\limits_{( - )} = \mathop {\max }\limits_{(\min )} \operatorname{Re} \lambda ,\lambda \in sp(x)\) . With the aid of the spectral radius \(\rho (x) = \mathop {\max }\limits_{\lambda \in sp(x)} \left| \lambda \right| = \mathop {\lim }\limits_{n \to + \infty } \parallel x^n {{1 - } \mathord{\left/ {\vphantom {{1 - } n}} \right. \kern-0em} n}\) we prove the following bounds: γ?(x)?σ?(x)?Γ?(x)?+(x)?σ+(x)?γ+(x), Γ(±)(x)=(2δ(±))?1 (ρ δ 2 )(±)?δ (±) 2 ?ρ 0 2 )(δ(±)≠0), γ(±)(x)= (±)ρδ(±)?δ(±), δ+?0, δ??0 ρ (±) δ = ρ(x+eδ(±)). We mention a case where equality is achieved, some corollaries,and discuss the sharpness of the bounds: for every ? > 0 there is a δ: ¦δ¦ ≥ρ 0 2 /2?, such that Δ: = ¦γ(±) x?Γ(±) x¦?ε and conversely, if the bounds are computed for some δ ≠ 0, then △ ≤ρ 0 2 /2 ¦δ¦. An example is considered. |
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