1995
DOI: 10.14492/hokmj/1380892538
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Perfectness and semiperfectness of abelian $*$-semigroups without zero

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Cited by 4 publications
(6 citation statements)
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“…A complex-valued function on a * -semigroup S is Stieltjes singular if it vanishes on S + S. For s ∈ S define G s = {σ ∈ S * : σ(s) = 0} and G If S is a perfect semigroup then so isS [25]. If S is a * -semigroup having a zero 0 then σ(0) = 1 for all σ ∈ S * , as is readily seen.…”
Section: The Main Resultsmentioning
confidence: 97%
See 1 more Smart Citation
“…A complex-valued function on a * -semigroup S is Stieltjes singular if it vanishes on S + S. For s ∈ S define G s = {σ ∈ S * : σ(s) = 0} and G If S is a perfect semigroup then so isS [25]. If S is a * -semigroup having a zero 0 then σ(0) = 1 for all σ ∈ S * , as is readily seen.…”
Section: The Main Resultsmentioning
confidence: 97%
“…. Since the paper is already quite long, we merely refer to the corresponding result for perfectness instead of Stieltjes perfectness [25].…”
Section: But This Follows From (3) This Proves (6)mentioning
confidence: 99%
“…Existence: Let S = H ∪ {0} be the semigroup with zero obtained by adjoining to H a zero external to H. From the fact that H is quasi-perfect it follows that S is perfect (see [6]). The following facts are taken from [8], proof of Theorem 3.2 (where the measures were assumed to be Radon measures but the proof carries over). For y ∈ H + H there is a unique complex measure µ y on A(S * ) such that…”
Section: Resultsmentioning
confidence: 99%
“…for all x ∈ S. (Given y ∈ H + H, choose h, k ∈ H such that y = h + k * and let µ y be the measure denoted by µ h,k in [8] where ϕ s (y) = µ y 1 {0} and ϕ m (y) = G σ(y) dµ(σ). In general, however, the set 1 {0} need not be measurable.…”
Section: Resultsmentioning
confidence: 99%
“…If X is perfect then H is quasi-perfect by Theorem 2.4 since H is an ideal of X. Conversely, if H is quasi-perfect then the perfectness of X follows just as in the proof of [6], Theorem 3.2, that perfectness of H implies perfectness of H ∪ {0}, only the argument with the Cauchy-Schwarz inequality has to be applied twice, first to get from the integral representation on H + H + H + H (⊂ H + H + H ) to the integral representation on H + H and a second time to get it on all of H . Corollary 2.6.…”
Section: Corollary 25 a * -Semigroup H Is Quasi-perfect If And Onlymentioning
confidence: 85%