Stieltjes Transformation on Ordered Vector Space of Generalized Functions and Abelian and Tauberian Theorems

Int. J. Contemp. Math. Sciences, Vol. 3, 2008, no. 33, 1619 - 1628

K. V. Geetha and N. R. Mangalambal

Department of Mathematics, St. Joseph’s College Irinjalakuda, Kerala, India [email protected] [email protected]

Abstract. Stieltjes transformation of the form ˆ ∞ f(t) f(x)= 0 (xm+tm)ρ dt, m, ρ > 0 has been studied extensively. We apply this form of the transform to an ordered vector space of generalized functions to which the topology of bounded convergence is assigned. Some of the order properties of the transform and its inverse are studied. The asymptotic behaviour of the transform and its inverse are also studied.

Mathematics Subject Classiﬁcation: Primary 46F10, Secondary 46A04, 46A08

Keywords: Stieltjes transformation, tempered distributions, Strongasymp- totic behaviour

1. Introduction In an earlier paper [1] we had associated the notion of order to multinormed spaces and their duals. The topology of bounded convergence was assigned to the dual spaces. Some properties of these spaces were also studied. As illustrative examples the spaces D, E, La,b, L(w, z) and their duals were also studied. m In the present paper, an ordered testing fuction space Ma,b is deﬁned and the topology of bounded convergence is assigned to its dual. Some properties m m of the cone in Ma,b and in (Ma,b) are studied in section 2. Stieltjes transformation and its properties have been studied by John J. K. [2]. In section 3 we apply some of these results to the new context when the topology of bounded convergence is assigned to the ordered vector space m (Ma,b) . It follows that the Stieltjes transform and its inverse are order bounded. 1620 K. V. Geetha and N. R. Mangalambal Order relation is deﬁned on the space ζ of test functions of rapid descent and its dual ζ, the space of tempered distributions. The topology of bounded convergence is assigned to ζ. The notion of strongasymptotic behaviour is m introduced and the strongasymtotic behaviour of Sρ (f) with reference to the strongasymtotic behaviour of f and vice versa are studied in section 4.

m 2. The testing function space Ma,b and its dual m Let Ma,b denote the linear space of all complex-valued smooth functions defined on (0, ∞). Let (Km) be a sequence of compact subsets of R+ such that K1 ⊆ K2 ⊆ ... and such that each compact subset of (0, ∞) is contained in one Kj, j =1, 2,.... On each Kj define m m(1−a+k) m a−b | 1−m d k | μa,b,Kj,k(φ) = sup t (1 + t ) (t ) φ(t) . t∈Kj dt a, b ∈ R, k =0, 1, 2,..., m ∈ (0, ∞). { m }∞ k k m μa,b,Kj,k k=0 is a multinorm on Ma,b,Kj where Ma,b,Kj is a subspace of Ma,b consisting of functions with support contained in Kj. The above multinorm m m m generates the topology τa,b,Kj on Ma,b,Kj . The inductive limit topology τa,b as m Kj varies over all compact sets K1,K2,... is assigned to Ma,b. With respect m to this topology Ma,b is complete. m On each Ma,b,Kj an equivalent multinorm is given by m m (1) μ¯a,b,Kj ,k(φ) = sup μa,b,Kj,k (φ) 0≤k≤k m m Definition 2.1. The positive cone C of Ma,b when Ma,b is restricted to real- m valued functions is the set of all non-negative functions in Ma,b. m Then C + iC is the positive cone in Ma,b which is also denoted as C. m Theorem 2.1. The cone C of Ma,b is not normal. m Proof. Suppose that Ma,b is restricted to real-valued functions. Let j be a fixed ∩ m +ve integer and let (φi) be a sequence of functions in C (Ma,b,Kj ) such that λi = sup{φi(t),t ∈ Kj} converges to 0 but (φi) does not converge to 0 for m Ma,b,Kj . λi,t∈ Kj+1 Define ψi = . 0,t∈ Kj+1 Let ξi be the regularization of ψi defined by ∞ ξi(t)= θα(t − t )ψi(t )dt 0 ∈ m Then ξ Ma,b,Kj+2 for all i. m Also, 0 ≤ φi ≤ ξi and (ξi) converges to 0 for τa,b. In Proposition 1.3, chapter 1, [3], it has been proved that the positive cone in an ordered topological vector space E(τ) is normal if and only if for any two nets {xβ : β ∈ I} and Stieltjes transformation 1621

{yβ : β ∈ I} in E(τ)if0≤ xβ ≤ yβ for all β ∈ I and if {yβ : β ∈ I} converges to 0 for τ then {xβ : β ∈ I} converges to 0 for τ. So we conclude that C is not normal for the Schwartz topology. It follows that C + iC is not normal for m Ma,b m Theorem 2.2. The cone C is a strict b-cone in Ma,b. m Proof. Let Ma,b be restricted to real-valued functions. Let B be the saturated m m class of all bounded circled subsets of Ma,b for τa,b. m ∪ Bc Bc { ∩ − ∩ ∈ B} Then Ma,b = B∈ where = (B C) (B C):B is a fundamental system for B and C is a strict b-cone, since B is a collection of all τ-bounded m sets in Ma,b. m m Let B be a bounded circled subset of Ma,b for τa,b. Then all functions in B have their support in some compact set Kj0 and there exists a constant M>0 such that |φ(t)|≤M, for all φ ∈ B, t ∈ Kj0 . Let M, t ∈ Kj +1 ψ(t)= 0 0,t∈ Kj0+1. Then the regularization ξ of ψ is deﬁned by ∞ ξ(t)= θα(t − t )ψ(t )dt 0 and ξ has its support in Kj0+2. Also, B ⊂ (B + ξ) ∩{ξ} ⊂ (B + ξ) ∩ C − (B + ξ) ∩ C It follows that C is a strict b-cone. We conclude that C + iC is a strict b- cone.

m m Order and topology on the dual of Ma,b. Let (Ma,b) denote the linear m space of all linear functionals defined on Ma,b. An order relation is defined on m m (Ma,b) by identifying the positive cone of (Ma,b) to be the dual cone C of C in m 0 m m Ma,b. The class of all B , the polars of B as B varies over all σ(Ma,b, (Ma,b) )- m m bounded subsets of Ma,b is a neighbourhood basis of 0 in (Ma,b) for the locally m m m convex topology β((Ma,b) ,Ma,b). When (Ma,b) is ordered by the dual con C m m and is equipped with the topology β((Ma,b) ,Ma,b) it follows that C is a normal m m m m cone since C is a strict b-cone in Ma,b for τa,b. Note that β((Ma,b) ,Ma,b)is the G-topology corresponding to the class G of all compete, bounded, convex, m m m circled subsets of (Ma,b) for the topology σ(Ma,b, (Ma,b) ). Anthony L. Perissini [3] has observed that a subset L(E,F), the vector space of all continuous linear mappings of E into F is bounded for the topology of pointwise convergence if and only if it is bounded for the G-topology corresponding to the class of all complete, bounded, convex, circled subsets of E(τ). 1622 K. V. Geetha and N. R. Mangalambal m + Theorem 2.3. The order dual (Ma,b) is the same as the topological dual m m (Ma,b) when (Ma,b) is equipped with the topology of bounded convergence. Proof. Every positive linear functional is continuous for the Schwartz topology m m τa,b on Ma,b. m + m m (Ma,b) = C(Ma,b,R) − C(Ma,b,R) m where C(Ma,b,R) is the linear subspace consisting of all non-negative order m m bounded linear functionals on Ma,b of L(Ma,b,R), the linear space of all order m m + m bounded linear functionals on Ma,b. It follows that (Ma,b) ⊆ (Ma,b) . m On the other hand, the space (Ma,b) equipped with the topology of bounded m m convergence β((Ma,b) ,Ma,b) and ordered by the dual cone C of the cone C m in Ma,b is a reflexive space ordered by a closed a normal cone. Hence if D is m a directed (≤) set of elements that is either majorized in (Ma,b) or contains m m m a β((Ma,b) ,Ma,b)-bounded section, then sup D exists in (Ma,b) and the filter m m F(D) of sections of D converges to sup D for β((Ma,b) ,Ma,b). m + m Hence (Ma,b) =(Ma,b) . m m m We conclude that (Ma,b) with respect to the topology β((Ma,b) ,Ma,b)is both topologically complete and order complete

3. The Stieltjes transformation In this section we study some properties of the Stieltjes transformation. The inverse of the transformation is also discussed here. m For φ ∈ Mα,β the Stieltjes transform is defined as ∞ m ˆ φ(t) Sρ (φ)=φ(x)= m m ρ dt 0 (x + t ) for a fixed m>0, ρ ≥ 1. With suitable integrability properties the double integral ∞ ∞ f(x)φ(t) m m ρ dtdx 0 0 (x + t ) can be evaluated in two different ways so that f,φˆ = f,φˆ . This relation is the basis for the application of the method of adjoints to the Stieltjes transformation. We have the following theorem due to John J. K. [2]. − 1 − 1 Theorem 3.1. Let α>1 m , β<ρ+1 m . Then the Stieltjes transform m m maps C ∩ Mα,β continuously into C ∩ Ma,b if ≤ ≤ 1 − 1 a 1, a m + α + ρ and a<1 if α = ρ +1 m ≥ − ≥ 1 − − − 1 b 1 ρ, b m + β ρ and b>1 ρ if β =1 m . Stieltjes transformation 1623 m m m m Now let f ∈ (Ma,b) . For each φ ∈ Mα,β we have Sρ (φ) ∈ Ma,b. Then the adjoint mapping m m Sρ (f),φ = f,Sρ (φ) m m defines the Stieltjes transform Sρ (f) ∈ (Mα,β) of f. Theorem 3.2. The Stieltjes transform is strictly positive and order bounded. m Proof. (Ma,b) is an ordered vector space. We define an order relation on the field of complex numbers by identifying the positive cone to be the set of m m complex numbers α + iβ, α>0, β>0. If f>0, Sρ (f) > 0. So Sρ is strictly positive and hence is order bounded. 1 For a non-negative integer n, a differential operator Ln for ρ + n> m is defined by

Ln = Ln,x(φ(x)) (−1)nm1−2nΓ(ρ) d d d = (x1−m )n−1x2mn+mρ−m(x1−m )nφ(x) 1 − 1 Γ(n + m )Γ(ρ + n m ) dx dx dx The Stieltjes transform can be inverted by the application of this diﬀerential operator. The formal adjoint of this operator is itself. We make the following conclusions from the results proved by John J. K. [2]. 1 ∞ m m −ρ Result 3.1. If ρ + n> m , 0 Ln,x(x + t ) dt =1. m m Result 3.2. Ln,t maps C ∩ Ma,b continuously into C ∩ Mα,β, provided − 1 − 1 1 − 1 − α>1 m , β<ρ+1 m , a = m + α ρ, b = m + β ρ.

Result 3.3. If Ln is the differential operator and S is the Stieltjes transform 1−mρ mρ−1 1−mρ mρ−1 operator then either x Ln and Sx or Lnx and x S commute m on Ma,b where 1 1 a = + α − ρ, b = + β − ρ. m m That is mρ−1 m mρ−1 ∧ x Sρ [Ln,t(φ)] = x [Ln,t(φ)] ∞ m m −ρ mρ−1 = Ln,x (x + t ) t φ(t)dt 0 m mρ−1 m = Ln,xSρ (t φ),φ∈ Ma,b. − 1 − 1 { ˆ } Result 3.4. If α>1 m , β<ρ+1 m , the sequence Ln,xφ(x) converges m in C ∩ Mα,β to φ(x). 1 − 1 − ∧ Result 3.5. Let a = m + α ρ, b = m + β ρ. Then (Lnφ) converges to φ m in Ma,b as n →∞. 1624 K. V. Geetha and N. R. Mangalambal m Result 3.6. Let f ∈ (Ma,b) . Then f ∈ C if and only if for every non-negative m m integer n, LnSρ (f) ∈ C where C is the positive cone in (Ma,b) . It follows that Ln is strictly positive and hence is orderbounded. We have the following results on inversion of the Stieltjes transform. m m For f ∈ (Mα,β) , φ ∈ Mα,β m m Sρ (Lnf),φ = f,Ln(Sρ φ) → f,φ as n →∞. m m For f ∈ (Ma,b) , φ ∈ Ma,b m m Ln(Sρ (f)),φ = f,Sρ (Lnφ) → f,φ . 4. Abelian and Tauberian Theorems We have made a study of the spaces D, E and their duals with order relation defined on these spaces and with the topology of bounded convergence assigned to the dual spaces [1]. Here we consider the space ζ of test functions of rapid descent with an order relation defined on it and its dual ζ, the space of tempered distribution. An order relation is defined on ζ and the topology of bounded convergence is assigned to it. We introduce the notion of stron- m gasymtotic behaviour of f and study the strongasymtotic behaviour of Sρ (f) with reference to the strongasymtotic behaviour of f and vice versa in the present context. We have the relations D⊆ζ and ζ ⊆D [4]. Definition 4.1. Restricting ζ to the real-valued functions an order relation is defined on ζ by identifying the positive cone to be the set of all non-negative functions in ζ. The cone C + iC is the positive cone of ζ which is also denoted as C. Result 4.1. The restriction of C to D is the positive cone of D. Definition 4.2. An order relation is defined on ζ by identifying the positive cone of ζ to be the dual cone C of C in ζ. Result 4.2. The topology of ζ is the topology of bounded convergence which is the same as the subspace topology on ζ when D is assigned the topology of bounded convergence. ◦ ◦ ◦ Proof. We prove that Bζ = BD ∩ ζ where Bζ is the polar of Bζ = {ψ ∈ ζ : |f(ψ)| 0, for all t>s, t, s ∈ R and ◦ BD is the polar of BD, the corresponding set of elements on D. We know that the elements of ζ are precisely those members of D that have continuous ◦ extensions to ζ .Soiff1 ∈ Bζ , f1 ∈ ζ , |f1(ψ)| < 1 for all ψ ∈ Bζ .Ifψ ∈ BD, |f(ψ)| 0 the limit ν(ak) lim = ar exists. k→∞ ν(k) Definition 4.4. A distribution f in ζ+ has strongasymtotic behaviour at ∞ with respect to the regularly varying function ν(k) if the limit f(kt) lim = F (t) k→∞ ν(k) exists, in the sense of convergence in ζ with respect to the topology of bounded convergence, provided F =0. F is a homogeneous function of order r and ∈ ⊆ ∞ f(kt) hence F ζ and support of F (0, ). By saying that limk→∞ ν(k) = F (t) exists in ζ with respect to the topology of bounded convergence what we mean is that if F (t) ∈ B◦, where B◦ is a basis element for the topology of bounded f(kt) ∈ ◦ ≥ convergence, ν(k) B for k k0. i.e., f(kt) | ,ψ(t) | < 1, ∀ψ ∈ B, k ≥ k0 ν(k) whenever | F (t),ψ(t) | < 1, for all ψ ∈ B where B = {ψ ∈ ζ : |f(ψ)| < t for some f ∈ ζ} for some >0, for all t>s, t, s ∈ R. 1 − 1 − { m Lemma 4.1. Let a = m + α ρ, b = m + β ρ. Then the set A = Sρ (φ): m m φ ∈ Mα,β} is dense in Ma,b..

m m Proof. Let φ ∈ Ma,b. Then φn = Sρ (Ln,tφ) are elements of A and φn → φ in m Ma,b

m Theorem 4.1. Let f ∈ (Ma,b) and ν(k) be a regularly varying function of order r>(−ma). The following statements are equivalent m (i) f has in C ∩ (Ma,b) strongasymtotic behaviour at ∞ with respect to ν(k). m m (ii) Sρ (f) has in C ∩ (Mα,β) strongasymtotic behaviour at ∞ with respect to 1−mρ 1 m k ν(k) and k1−mρν(k) Ln,xSρ (f)(kx), for k>0 is uniformly continuous in m the topology of bounded convergence in C ∩ (Ma,b) for all values of n. Proof. (i) ⇒ (ii) ∈ m ∈ m f(kt) Let φ Ma,b, φ1 Mα,β. Also let limk→∞ ν(k) = g(t). Then we have 1 lim f(kt),φ(t) = g(t),φ(t) k→∞ ν(k) 1626 K. V. Geetha and N. R. Mangalambal m m m If φ1 ∈ Mα,β, Sρ (φ1) ∈ Ma,b. Hence m 1 m g(t),Sρ (φ1) = lim f(kt),Sρ (φ1)(t) k→∞ ν(k) 1 m = lim Sρ (f)(kx),φ1(x) . k→∞ k1−mρν(k) m m But g(t),Sρ (φ1) = Sρ (g)(x),φ1(x) . So we conclude that 1 m m lim Sρ (f)(kx)=Sρ (g)(x). k→∞ k1−mρν(k) m m This means that Sρ (f) has in C ∩ (Mα,β) strongasymptotic behaviour at inﬁnity with respect to k1−mρν(k). 1 m f(kt) m LnSρ (f)(kx),φ(x) = ,Sρ Ln(φ)(t) . k1−mρν(k) ν(k) f(kt) f(kt) ∈ ≥ Since limk→∞ ν(k) = F (t)inζ , ν(k) ,ψ < 1, for all ψ B, k k0 whenever | F (t),ψ | < 1 for all ψ ∈ B where B = {ψ ∈ ζ : |f(ψ)| 0, for all t>s, t, s R. Since ν(k) is a continuous linear functional, there exists a positive constant C1 and integers j, q such that 1 m m m LnSρ (f)(kx),φ(x) ≤ C1μ¯a,b,K ,q(Sρ Ln(φ)) k1−mρν(k) j m m ≤ m m − m μ¯a,b,Kj,q(Sρ Ln(φ)) μ¯a,b,Kj ,q(Sρ Ln(φ) φ)+¯μa,b,Kj ,q(φ). Using results 3.3, 3.4 and 3.5, m m − m 1−mρ m mρ−1 − μ¯a,b,Kj,q(Sρ Ln(φ) φ)=¯μa,b,Kj ,q(x (LnSρ (t φ(t)) φ(x))) ≤ m m − C2μ¯α,β,Kj,q(LnSρ (φ0) φ0) mρ−1 ∈ m where φ0(t)=t φ(t) Mα,β,Kj ≤ m nC2μ¯α,β,Kj,q+1(φ0) ≤ m nC3μ¯a,b,Kj ,q+1(φ). m m ≤ m Consequently,μ ¯a,b,Kj ,q(Sρ Ln(φ)) C4μ¯a,b,Kj ,q+1(φ). Hence 1 m m LnSρ (f)(kx),φ ≤ C5μ¯a,b,K ,q+1(φ) k1−mρν(k) j for all n, uniformly for k ≥ k0 where C5 is a constant depending only on f,j and q. (ii) ⇒ (i) Let m Sρ (f)(kt) m lim = Sρ (g)(t), say. k→∞ k1−mρν(k) Stieltjes transformation 1627 Then 1 m lim Sρ (f)(kx),φ1(x) k→∞ k1−mρν(k) m = Sρ (g)(x),φ1(x) m = g(t),Sρ (φ1)(t) But 1 m lim Sρ (f)(kx),φ1(x) k→∞ k1−mρν(k) 1 m = lim f(kt),Sρ (φ1)(t) k→∞ ν(k) f(kt) It follows that limk→∞ ν(k) = g(t) on a dense set of elements of the space m { f(kt) ≥ } m Ma,b. Now we prove that the set ν(k) : k k0 is bounded in (Ma,b) .By m the theorem of uniform convergence it follows that f has in (Ma,b) stron- m gasymtotic behaviour at ∞ with respect to ν(k). Let φ ∈ Ma,b. Since 1 m ≥ m k1−mρν(k) Ln,xSρ (f)(kx), k k0 is uniformly continuous in (Ma,b) there exists a constant C1 > 0 and positive integers j, q such that 1 m m LnSρ (f)(kx),φ1(x) ≤ C1μ¯a,b,K ,q(φ) k1−mρν(k) j for k ≥ k0, for all n ∈ Z+. Since 1 m 1 m LnSρ (f)(kx),φ(x) = f(kt),Sρ (Lnφ)(t) k1−mρν(k) ν(k) it follows that 1 ≤ m f(kt),φ(t) C2μ¯a,b,Kj ,q+1(φ). ν(k) Hence the theorem. m Corollary 4.2. If (fα)α∈J is a monotone net in C ∩ (Ma,b) having strongasymtotic behaviour at ∞ with respect to a regularly varying function ν(k) m m then (Sρ (fα))α∈J converges to a function f ∈ C ∩ (Ma,b) having strongasymtotic behaviour with respect to k1−mρν(k). 1 m ≥ Also, k1−mρν(k) Ln,xSρ (f)(kx) for k k0 is uniformly continuous with respect m to the topology of bounded convergence in C ∩ (Ma,b) for all values of n. m Proof. Follows from Theorem 4.1 since (Mα,β) is order complete. m Theorem 4.3. Let f ∈ (Ma,b) and ν(k) be a regularly varying function of m order γ>(−ma).IfSρ (f) has strongasymtotic behaviour at ∞ with respect to 1−mρ ∩ m 1 m ≥ k v(k) in C (Ma,b) then k1−mρν(k) Ln,xSρ (f)(kx) for k k0 is uniformly m continuous with respect to the topology of bounded convergence in C ∩ (Ma,b) for all values of n. 1628 K. V. Geetha and N. R. Mangalambal m m Corollary 4.4. Let (fα)α∈J be a monotone net in C ∩(Ma,b) such that (Sρ (fα))α∈J 1−mρ m has strongasymtotic behaviour at ∞ with respect to k v(k) in C ∩ (Ma,b) m where ν(k) is regularly varying function of order r>(−ma). Then (Sρ (fα))α∈J ∈ ∩ m 1 m ≥ converges to a function f C (Ma,b) and k1−mρν(k) Ln,xSρ (f)(kx) for k k0 is uniformly continuous with respect to the topology of bounded convergence in m C ∩ (Ma,b) for all values of n. m Proof. Follows from Theorem 4.3 since (Ma,b) is order complete. Acknowledgment. The authors gratefully acknowledge the guidance of Dr. T. Thrivikraman in the preparation of this paper. References

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Received: May 11, 2008