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Integral representations of linear operators

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Let $ ( X, \Lambda, \mu ) $ and $ ( Y, \Sigma, \nu ) $ be $ \sigma $- finite measure spaces (cf. Measure space) and let $ L _ {0} ( X, \mu ) $ and $ L _ {0} ( Y, \nu ) $ be the spaces of the complex-valued $ \mu $- measurable functions on $ X $ and the complex-valued $ \nu $- measurable functions on $ Y $, respectively. A linear subspace $ E = E ( X, \mu ) $ of $ L _ {0} ( X, \mu ) $ is called an ideal space, or a solid linear subspace, of $ L _ {0} $ if $ f \in L _ {0} $, $ g \in E $ and $ | f | \leq | g | $, $ \mu $- a.e., imply $ f \in E $. The classical $ L _ {p} $- spaces ( $ 1 \leq p \leq \infty $), the Orlicz spaces and, more generally, Banach function spaces (cf. also Orlicz space; Banach space) are typical examples of normed ideal spaces.

If $ E $, $ F $ are ideal spaces contained in $ L _ {0} ( Y, \nu ) $ and $ L _ {0} ( X, \mu ) $, respectively, then $ T \in {\mathcal L} ( E,F ) $, the linear space of all linear operators from $ E $ into $ F $, is called an integral operator, kernel operator, if there exists a $ ( \mu \times \nu ) $- measurable function $ T = T ( x,y ) $, $ ( x,y ) \in X \times Y $, such that for all $ f \in E $ and $ \mu $- a.e. with respect to $ x $, $ ( Tf ) ( x ) = \int _ {Y} {T ( x,y ) f ( y ) } {d \nu ( y ) } $.

Integral operators, also known as integral transforms, play an important role in analysis. It is a natural question to ask: Which $ T \in {\mathcal L} ( E,F ) $ are integral operators? J. von Neumann [a5] was the first to show that for the ideal spaces $ E = F = L _ {2} ( [ 0,1 ] ) $ the identity operator does not admit an integral representation. He proved, however, that a bounded self-adjoint linear operator $ T \in {\mathcal L} ( L _ {2} , L _ {2} ) $ is unitarily equivalent (cf. also Unitarily-equivalent operators) to an integral operator if and only if $ 0 $ is an element of the limit spectrum of $ T $.

$ T \in {\mathcal L} ( E,F ) $ is called a positive linear operator if for all $ 0 \leq f \in E $ one has $ T f \geq 0 $( $ \mu $- a.e.). An integral operator $ T $ with kernel $ T ( x,y ) $( $ ( x,y ) \in X \times Y $) is positive if and only if $ T ( x,y ) \geq 0 $, $ ( \mu \times \nu ) $- a.e.; $ T \in {\mathcal L} ( E,F ) $ is called regular if $ T $ maps order-bounded sets into order-bounded sets, i.e., for all $ f \in E $ there exists a $ g \in F $ such that for all $ h \in E $ satisfying $ | h | \leq | f | $, one has $ | {Th } | \leq g $; $ T \in {\mathcal L} ( E,F ) $ is ordered bounded if and only if $ T $ can be written as the difference of two positive linear operators if and only if its modulus $ | T | $, where for all $ 0 \leq f \in E $, $ | T | ( f ) = \sup \{ {| {Tg } | } : {| g | \leq f } \} $, is a positive linear operator mapping $ E $ into $ F $.

The following theorem holds: An integral operator $ T \in {\mathcal L} ( E,F ) $ is regular if and only if its modulus $ | T | $ is a positive linear operator mapping $ E $ into $ F $. In that case, the kernel of $ | T | $ is given by the modulus $ | {T ( x,y ) } | $( $ ( x,y ) \in X \times Y $) of the kernel of $ T $.

An integral transform need not be regular, as is shown, for instance, by the Fourier transform and the Hilbert transform.

Integral operators can be characterized via a continuity property: $ T \in {\mathcal L} ( E,F ) $ is a linear integral operator if and only if $ 0 \leq f _ {n} \leq f \in E $( $ n = 1,2, \dots $) and $ f _ {n} \rightarrow 0 $ in $ \nu $- measure as $ n \rightarrow \infty $ imply $ Tf _ {n} \rightarrow 0 $( $ \mu $- a.e.) as $ n \rightarrow \infty $.

An earlier version of this theorem for bilinear forms is due to H. Nakano [a4]. For regular linear operators defined on KB-spaces (cf. also $ K $- space), the result appeared in a slightly different form in a paper by G.Ya. Lozonovskii [a3]. The present version is due to A.V. Bukhvalov [a1]. A pure measure-theoretic proof and related results were given by A. Schep [a6]. For details and further results see [a2].

References

[a1] A.V. Bukhvalov, "A criterion for integral representability of linear operators" Funktsional. Anal. i Prilozhen. , 9 : 1 (1975) pp. 51 (In Russian)
[a2] "Vector lattices and integral operators" S.S. Kutateladze (ed.) , Mathematics and its Applications , 358 , Kluwer Acad. Publ. (1996)
[a3] G.Ya. Lozanovsky, "On almost integral operators in -spaces" Vestnik Leningrad Gos. Univ. , 7 (1966) pp. 35–44 (In Russian)
[a4] H. Nakano, "Product spaces of semi-ordered linear spaces" J. Fac. Sci. Hokkaidô Univ. Ser. I , 12 : 3 (1953) pp. 163–210
[a5] J. von Neumann, "Charakterisierung des Spektrums eines Integraloperators" , Actualités Sc. et Industr. , 229 , Hermann (1935)
[a6] A.R. Schep, "Kernel operators" Proc. Kon. Nederl. Akad. Wetensch. , A 82 (1979) pp. 39–53
How to Cite This Entry:
Integral representations of linear operators. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Integral_representations_of_linear_operators&oldid=52081
This article was adapted from an original article by W. Luxemburg (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article