2015
DOI: 10.4064/sm8218-12-2015
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Less than one implies zero

Abstract: In this paper we show that from the estimate $\sup_{t \geq 0}\|C(t) - \cos(at)I\| <1$ we can conclude that $C(t)$ equals $\cos(at) I$. Here $\left(C(t)\right)_{t \geq 0}$ is a strongly continuous cosine family on a Banach space.Comment: Corrected the previous version (in particular, a mistake in Lemma 2.1), streamlined and added recent references (see Introduction). 6 page

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Cited by 6 publications
(9 citation statements)
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“…Recently, Bobrowski and Chojnacki [5] established an analogue of this result for one-parameter cosine families: if a ∈ R and C = {C(t)} t∈R is a strongly continuous cosine family on a Banach space X such that sup t∈R C(t) − (cos at)I X < 1 2 , (1.1) then C(t) = (cos at)I X for each t ∈ R. Here, the cosine family {(cos at)I X } t∈R against which C is compared is an example of what is termed, following the nomenclature used in [5], a scalar cosine family. Schwenninger and Zwart [19] improved Bobrowski Chojnacki [6] in turn strengthened Schwenninger and Zwart's result by proving a general theorem that ensures that condition (1.2) alone, without C being necessarily strongly continuous, implies the coincidence of C and {(cos at)I X } t∈R . Schwenninger and Zwart [20] later showed that in the case a = 0, if C is strongly continuous, then the condition…”
Section: Introductionmentioning
confidence: 91%
“…Recently, Bobrowski and Chojnacki [5] established an analogue of this result for one-parameter cosine families: if a ∈ R and C = {C(t)} t∈R is a strongly continuous cosine family on a Banach space X such that sup t∈R C(t) − (cos at)I X < 1 2 , (1.1) then C(t) = (cos at)I X for each t ∈ R. Here, the cosine family {(cos at)I X } t∈R against which C is compared is an example of what is termed, following the nomenclature used in [5], a scalar cosine family. Schwenninger and Zwart [19] improved Bobrowski Chojnacki [6] in turn strengthened Schwenninger and Zwart's result by proving a general theorem that ensures that condition (1.2) alone, without C being necessarily strongly continuous, implies the coincidence of C and {(cos at)I X } t∈R . Schwenninger and Zwart [20] later showed that in the case a = 0, if C is strongly continuous, then the condition…”
Section: Introductionmentioning
confidence: 91%
“…Recently, Bobrowski and Chojnacki [1] established an analogue of this result for one-parameter cosine families: if a ∈ R and {C(t)} t∈R is a strongly continuous cosine family on a Banach space X such that sup t∈R C(t) − (cos at)I X < 1 2 , (1.1) [2] On cosine families close to scalar cosine families 167 then C(t) = (cos at)I X for each t ∈ R. This conclusion was further refined by Schwenninger and Zwart [7] who showed that condition (1.1) can be replaced by the condition sup t∈R C(t) − (cos at)I X < 1.…”
Section: Introductionmentioning
confidence: 94%
“…In this note we extend the first result of Schwenninger and Zwart (that is, the result of [7]) to cover the case of cosine families that are not necessarily indexed by real numbers, not necessarily operator-valued, and not necessarily continuous in any particular sense. A crucial step towards proving the relevant result will be the establishment of an analogue of the Cox-Nakamura-Yoshida-Hirschfeld-Wallen theorem for cosine sequences.…”
Section: Introductionmentioning
confidence: 95%
“…Bobrowski and Chojnacki proved in [4] that if a strongly continuous operator valued cosine function on a Banach space (C (t )) t ∈R satisfies sup t ≥0 C (t ) − c(t ) < 1/2 for some scalar bounded continuous cosine function c(t ) then C (t ) = c(t ) pour t ∈ R, and Zwart and F. Schwenninger showed in [18] that this result remains valid under the condition sup t ≥0 C (t ) − c(t ) < 1. The proofs were based on rather involved arguments from operator theory and semigroup theory.…”
Section: Introductionmentioning
confidence: 99%
“…Hence if u is divisible by 3, we have σ u 3 = 1.5 if u ∈ {3, 6,9,12,15,18,24, 30} , σ(u) > 1.5 otherwise. We then deduce from corollary 3.6 a complete description of the set Ω(1.5) = {a ∈ [0, π] | k(a) ≤ 1.5}.…”
mentioning
confidence: 99%