1993
DOI: 10.1007/bf01200295
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A near-factorization theorem for integrable functions

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Cited by 4 publications
(5 citation statements)
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“…Assume that x ∈ H is such that x(ω) 2 ≥ h(ω) almost everywhere, and g ∈ L 1 (µ) satisfies an inequality of the form |g(ω)| ≤ kh(ω) almost everywhere, with k a constant. Then the function y ∈ L 2 (µ, F) defined by y(ω) = g(ω)x(ω)/ x(ω) 2 if x(ω) = 0, y(ω) = 0 if x(ω) = 0, satisfies the equality x • y = g. This observation immediately implies the following factorization result.…”
Section: Factorization Resultsmentioning
confidence: 64%
See 1 more Smart Citation
“…Assume that x ∈ H is such that x(ω) 2 ≥ h(ω) almost everywhere, and g ∈ L 1 (µ) satisfies an inequality of the form |g(ω)| ≤ kh(ω) almost everywhere, with k a constant. Then the function y ∈ L 2 (µ, F) defined by y(ω) = g(ω)x(ω)/ x(ω) 2 if x(ω) = 0, y(ω) = 0 if x(ω) = 0, satisfies the equality x • y = g. This observation immediately implies the following factorization result.…”
Section: Factorization Resultsmentioning
confidence: 64%
“…The main result of this section is that the function x in an approximate factorization of h can be chosen so that x(ω) 2 ≥ h(ω) almost everywhere.…”
Section: Factorization Resultsmentioning
confidence: 99%
“…without, however, separability assumptions on the measure space. That condition was removed in [8] provided that the measure space is separable.…”
Section: Dual Spaces Of Operators and Factorizationsmentioning
confidence: 99%
“…On the same line, it was subsequently proved that every such operator is reflexive (see [16]). The essentially boundedness condition appearing in [5] was removed in [8], under the condition that H is separable. Now, even if one is not able to find the localizing vectors sitting in arbitrary finite codimensional subspaces one still can get approximating vectors, as the following result from [11] shows: Theorem 2.4 (cf.…”
Section: Prunaru Ieotmentioning
confidence: 99%
“…This is equivalent to saying that, the space L 2 (µ) is separable.) A closer look at the proof of this theorem given in [11] reveals the fact that the separability assumption on the measure space {Ω, B, µ} is used only to reduce the problem, via the above mentioned theorem from [8], to the case when the bounded functions are dense in H. Therefore we can drop the separability condition, assuming instead the denseness of bounded functions and we get the following result which will be useful in cases when the underlying measure space may not be separable:…”
Section: Prunaru Ieotmentioning
confidence: 99%