I nternat. J. Math. Math. Sei. 441 Vol. 2 # (1979) 441-458 THE STIELTJES TRANSFORM OF DISTRIBUTIONS U. N. TIWARI Department of Mathematics John Abbott College Montreal, Canada J. N. PANDEY Department of Mathematics Carleton University Ottawa, Canada (Received September 28, 1978 and in revised form April 3, 1979) ABSTRACT. In the present work, two complex inversion formulas of Byrne and Love for generalized Stieltjes transformation are shown to be valid for a class of distributions. This is accomplished by transfering the complex inversi( formulas on the testing function space of a class of distributions and then showing that the limiting process in the resulting formula converges in the topology of the testing function space. KEY WORDS AND PHRASES. Generalized Functions, Stieltj Transformation. AMS (MOS) SUBJECT CLASSIFICATION (1970) CODES. i. P 46A40, Secondary 42A25. INTRODUCTION. Let p be any complex number except zero and the negative integers. Then 442 U.N. TIWARI and J. N. PANDEY for all s in the "cut plane", that is all complex numbers except those which are negative real or zero, the StleltJes Transform in its general form is defined by: (,+)P The follorl inversion theorems for particular values of p and s are well known. THEKEEM A (Widder). If f(t) belongs to L(O,R) for every positive R and is such that he integral o :+x converges for x > O, then F(s) exists for complex s in the cut plane and llm f (-f -i) -F(-+i) +)+f ( -) 2 2i for any positive at which THBOREM B (Smer). f(+) If p and f(-) > 0, f(t) both exist. is locally integrable in [0,=], the improper Lebesgue integral. o (+t) converges (for a certain value of s in the cut plane and so for all), t . the limits f(t_+0) exist, then: t [(,:+o)+(,:-o)] li,. j’ 0+ o d ’ c x (z)P’lF’(z) > 0 and 443 STIELTJES TRANSFORM OF DISTRIBUTIONS where C x is a contour in the cut plane from THEOREM 1.3 (Byrne and Love). If Re p -x-i > I, m I(x+0)+ (x-0)} -x+i. f is locally integrable in [0,=], > O; then, improper Lebesgue integral (I.I)converges, and x for which the Lebesgue limits to for each positive f(x_+O) exists, 2i O+ f (x+t)P’2{F(t’i)-F(t+i)}dt" [2, p. 349] -x THEOREM I 4 (Byrne and Love). If Re p f(t) > 1, 2- E L(0 ,=) and the improper l+t Lebesgue integral o (+t) p converges, then for each positive x for which the Lebesgue limit I 2 [f(x+O)+f(x-O)} llm -K)+ THEOREM i.i and Pandey [14]. f-x (x+t) p-2 |F (t-i)-F (t+i)}dr f(m0_ exists, [2, p. 352] has been extended to distributions by Pandey and Zemanlan 13] Theorem 1.2 was extended to distribution by Pathak [15]. Our object is to extend theorems 1.3 and 1.4 of Byrne and Love to generalized functions (dlstrlbutlons). 2. THE TESTING FUNCTION SPACE, S(1) AND ITS DUAL. An infinitely dlfferentlable complex valued function (x) defined over I (0,) belongs to the testing function spaces S(1) if, sup 0<x< .l+x_ (x) < (R) 444 U.N. TIWARI and J. N. PANDEY for all k 0, I, 2, Clearly, S(I) is is a fixed real number. where The zero element a vector space with respect to the field of complex numbers. of the vector space zero. S(1) is the function defined over I which is identically The topology over S(1) is generated by the collection of semlnorms }= [24; p. 8]. We say that a sequence | converges in S(1) to (x) if for each fixed k, tends to space. . yk( tends to zero as The space B(1) is a vector subspace of S(1) and the topology of D(I) of any member of S(1) to D(1) is in D(1) by S(I) and as such the restriction D’(1), where S(1) and D’(I) denote the dual spaces of S(I) and D(I) respectively. non-negative integer k as and - We say that a sequence belongs to S(I) is a Cauchy sequence in S(I) if 3. I) The space S(I) is a locally convex Hausdorff tologlcal vector is stronger than the topology induced on other. belongs to where yk( 1 where ) goes to zero for any both tend to infinity independently of each It can be readily seen that S(1) is sequentlally complete. THE D.I.STRIB,UIONAL S,TIELT,J,E,S .TNSFORMA.TION For a complex s not negative or zero. 1 (s+x)P belongs to S’ where a < Rep. Therefore, the distributional Stieltes transformation F(s) of an arbitrary element F(s) f =n S’, a < (x), < Re p, is defined by I (s+x)P > (3.) where s belongs to the complex plane cut along the negative real axis including the origin. THEOREM 3.1. (x) If m and k both assume non-negatlve integral values and is a STIELTJES TRANSFORM OF DISTRIBUTIONS 445 compact set of the complex plane not meeting the negative real axis, then for fixed non-negative integers m and k, there exists a constant B satisfying of the complex plane not meeting uniformly for all s lying in the compact set the negative real axis on the origin. Using the compactness of the set PROOF. < and the fact that I S’; (s+x)e+k Re p, the theorem is immediate. THEOREM 3.2. For an arbitrary the equation (3.1). d (-s) where (-I) m (p)m PROOF. Then, for (’) m m < (x), (+x) p(p+l)(p+2) f S’ and a < Re p let F(s) be defined by I, 2, (P)m > p+m (.) (p+m-1). p If p is such that s does not have a branch cut in the complex plane, the proof can be given in a way similar to that given in [3, Lemma 2a]. If p is p such that s has a branch cut (along the negative real axis for the sake of deflnlteness), then choose the contour of integration as shown in the complexplane cut along the negative real axis below. 446 U. N. TIWARI and J. N. PANDEY S/AS Y X Here, C I and C 2 are arcs of two concentric circles with centre at origin and C is The radii of C the contour of integration as shown, 1 and C 2 and paths L I and L2 are so chosen that the point s is contained in the region bounded by the contour. Let d inf l-sl and choose F(s+as)-F(s). < (t), < f(t), "P 2i As IAsl d < NOW > z-s-s (z+t)p z-s (z.s)2j dz > where C is the contour shown in the diagram < f(t), 8As > where eAs As 2hi f C 1 (z+t )p We now wish to show that Using 1 (z-s )P(z-s-As) 8As 0 in dz S (l) as c AsO. Theorem 3.1, we have ,yk(SAs ) <_ Bc IAsl 2 L d 3 (3.4) 447 STIELTJES TRANSFORM OF DISTRIBUTIONS where L is the length of the contour C and B (p)k(l+t)a’tk As-, 0 in > 0. (3.3) and using (3.4), we get -p < e(t), ’() is the uniform botmd of for all z lying on the closed contour Camd t (z+t) +k Letting c (+t) > + Now; the theorem follows from induction on he order of the derivative of F(s). THEOREM 3.3. transform of The function F for real x where F(s) is the 0[x as x 0[x as x 0[ x "k’Re if cz as x 0 Re p Re p if cz o P] < <_ Re if The proof is Insnedlate from the boundedness 4. StleltJes S’, satisfies the following relation: f E(m)(x) (m)(x) p roperty of dlstributios 9, p. 18]. COI,LE INVERSION OREMS We are now ready to prove our first inversion theorem. THEOREM 4.1. be the im StleltJes < ! 2i <1 For a fixed transform of and Re p > I, let f f(t) as defined by (3.1). J (x+t)P -z[ F(t-i)-F(t+i)]dt, Ten, (x)> -x < f, > ur 11 PIOOF. 1 D(1) First onslder 1 (x+t)P’2F(t-i)dt f -x -x (d:) p-2 1 < f(y), (y+ -) > S’(1) and let F(s) dt U.N. TIWARI and J. N. PANDEY 448 For fixed x and t, Since, in view of Theorem 3.2, F(s) is analytic in the cut plane, the left-hand side integral in (4.1) is meaningful. > 0, can be shown that for -x By using the technique of Riemann sums it (x+t)P "2F (t’i) dt (x+t) < f(y), -x+ f(Y)’ p-2 (y+t-i) dt p [ (y+A.i)P-l ] cP’I (x+x)P_ "I I. I > p---" (x-y+i) (y-x+-i [by Lenna 5,1, p. 333] (say). 0+, One can "easily check that as and for fixed %, and x. I (x+t)P’2F(t’f) dt cp -i (-x+ -i)P -I Therefore, letting . (x+t)P’2F(t-f)dt >I > 0, we get: (p-l) (y-x-i) (y+% -i) In view of Lemma 3.5* [7, p. 12], it follows that: Therefore, letting k 0 in S (I) for Re p < f(y), -x for fixed x and )p_i as p’I > (4.2) % (4.3) in (4.2), we obtain < f(Y)’ p--- y’-x-i > -x * The proof was provided by Professor E.R. Love. (4.4) 449 STIELTJES TRANSFORM OF DISTRIBUTIONS Using a similar argument, we can show that f (x+t)P -2F (t+i)dt < f(y), p.-- y-x+i > -x Combining Equations (4.4) and (4.5), we get J 2i (x+t)P -x < f(y), [ (y-x)2 2] (4.6) J> Now using the technique of Riemann sums, we obtain < p-I 2 J (x+t )p-2 [(-+/-)-(+i)]a, (x) > -X b < f(Y)’ (x) dx __7 2a [y-x)2] (x) where the support of > D(1) Is contained in (a,b), b > a > 0. same techniques as followed in proving Theorem 2 of e Usi (3) one can show that b a (y -x ) (y) 2. in the topology of S (4.8) (I) as 0+. Therefore, letting 0+ in (4.7), we have llm <p-I (x+t 2i )p-2 [(:-i)-(+t)]dt, (x) > < ,> This completes the proof of the theorem. To prove our other inversion theorems, we require a couple of Lemmas. EMM 4.2. Let t, s, > 0. Then, b lim ->+ (l+x)c uniformly for all J" (t_x)2+T2 a x > O. for finite b > a > 0 and < I, U.N. TIWARI and J. N. PANDEY 450 PROOF. Let b a l(l+x) Since sup x>N>b a<t<b positive > N and b (t-X) <q<1 0 > 0, is bounded, for (t.x)24 2 there exists a such hat (4.9) (0,q) uniformly for all and x > N. Now assume that 6 is a positive number _< rain (1, a ) and for 6 <_ x <_N let us write. I a+x-6 (I+x)C . b a+x+6 + Denote e ree eresslons and respectively. + I on e t’xl dt rlghtd side of Eqn. (4.10) by I1, 12 Now a+x+5 a+x-s 2s Now choose (t’x)2 + s() a ( 5 such that 5(I) < Chs way. and fix Therefore (4.) uniformly for all x [6,N] and [0,q]. x b It-xl dt STIELTJES TRANSFORM OF DISTRIBUTIONS There fore (l+x) (a+6)22 Therefore, I 3 0 if 8 <x< b O+ as (4. 2) [5,N]. x uniformly for all , 451 Next, a+x -5 (c-x)2 2 a Now for <_ 5 x <_ It-xlat a (a’S)2.l.q.l (a-x 0 2 (4.3) as uniformly for all x lying in [5,a]. For a < x < N, 3’ I -2 0 /,n (a_:)2.. + 3.2 n k 2 O+ uniformly for all x as [a,N] (l+x) (4.14) CombininE results (4. II) through (4.14) we have lim _< < x 0 Now, for III <_ As I11 x uniformly for all <_ 8, (l+) n _> 0 (4.5) we can see that [(b’x)2+2] 0 (a-x) O+ uniformly for all x [0,]. (4.16) U.N. TIWARI and J. N. PANDEY 452 Combining results (4.15) and (4.16), we get lim III <-- _+ x uniformly for all > 0. Thus, the proof of the lemma is complete. LEMMA 4.3. numbers and (t) D(1). . >I > Let Re p Assume that t, x, I and are all positive Then, I S where the support 0+ in the topology of as contained in PROOF. 0 as (a,b); > b a of (t) E D(1) is > 0. We have to show that for each m 0, I, 2, (l+x)LxmDl(x,) 0 uniformly. It can be easily shewn that uniformly for all (l+x)xmDxm(x,) uniformly for all 0 satisfying 0 as x < I. < (R) Therefore, for (0,i). So, > 0 there exists N > 0 such that (4. I) uniformly for all x >N, and Now consider the case 0 and complete the proof for m m= 0. < 0 < x <_ . < I. I, 2, 3, We will first give the proof for m by using the result for the case 0 453 STIELTJES TRANSFORM OF DISTRIBUTIONS I. O. Write For m (l+x) (].+x)=: (x,]) 2i -1 l+ (x_t)2+T2 kx+X +i First, consider ]: (x,,Q) I l(x,). (say) ]:2 (x,) Since, I 1 (x+x+) ’i (x+ (p-l) -)’ - ++ z "pdz x+x (xq.x) <_ 21p.llT Therefore "l (t)dt err X’m Re P Pl/ A Re p we can find a constant B(p) independent of x and such that b Z(x,)l <_ B (+x) j. a (x-t)2+ In view of Lemma 4.2, the rlght-hand side converges to 0 uniformly for x > 0 as 0+. 12(x,). We now consider (1) For Re p x++i all _> 2, t 1 < For 0 < < b, using Lemma 3.3 (7, p. 9) we get x <_ Ip-lle t<x 2TT (4. 8) U.N. TIWARI and J. N. PANDEY 454 (ll) For Re p _> 2, t>x ++)+ ) t+x+i Re p -2 x+X + i 2 2+T P’,[ e2 m P]’ Re p-1 X+l)Re p-2 (b+ (4.19) (iii) For Re p < 2, t <_ x and ARe For <_ re p 2, E [a,b], b > a > 0, p-2 !P’ll’e2wllm Pl(a+A)Re Re(p-2’)2 -I (iv) t t > x, p-1 t [l+x’i] [a,b], b > a > 0, 0 < I. < (! t-x1+) ( i+ 0 < T] b)Re p-2 (4.20) X" < i -1 (4.) Therefore, from inequalities (4.19) through (4.21), it is evident that for Re p >I and 0 < < I, there exists a positive constant K(X,p) independent of t and x satisfying -i < K ,p) (t-x) + ] t { (a,b) Similarly, under the same set of conditions -1 <_ K(,p) (t-x) + ], t (a,b) In view of these inequalities we, therefore, have established that STIELTJES TRANSFORM OF DISTRIBUTIONS 455 a<t<b That is a (t’x)2+2 a (t’x)2+2 Since (z+), as b 2 j, dr. (t-) + ’ o 0+ uniformly for all x follows that 12(x,) 0 The case m > 0. Therefore, O+ uniformly for all as in view of Lemma 5.2, it x > O. 1, 2, 3, A careful computation along with integration by parts will show that b + (Z+x) x (x+i) Using the technique of induction, we obtain: (xq -i) -p] (t+A)p-2 (t)dt U.N. TIWARI and J. N. PANDEY 456 "F" + [’(x++i) " (x+)p-2 (m-) (x+-i) "p] a (t)p-2 (m-2) (t)dt (4.22) Denote the integrals on the right-hand side of (5.22) by in that order. 0, In view of case m Jl0 as To show that other integrals converge to 0 as consider the most general integral Jm+l" (x+)’’m+l- (x--i)’P’ml -)+ Jl’ 32’ Jm+l uniformly for all x E (0,N). -+ uniformly for x E (0,N), we As before, (-p-re+l) x+ z’’mdzl f x+ -i --’eVl z’m Pl p+m <_ (x+X)Re Therefore, we can find a positive constant C independent of x and (l+x) CZxm 0 as such that 0+ (x uniformly for all x e (0,N) and each fixed m i, 2, Thus, we have proved that (l+x)xmDml(x,) m + 0 as uniformly for all 0, l, 2, 3, Combining this fact with inequality (4.17), we have lira l(l+x)zxmDmx(x,l)l < x (0,N) and each fixed STIELTJES TRANSFORM OF DISTRIBUTIONS uniformly for all x > 0 and each fixed m 457 O, I, 2, since is arbitrary our claim is established. THEOREM 4.4. Let Re p >I > and S’. f(t) > transform of f(t) defined by (3.1) then for im <P2i f(y-i)-F(y+i)] (y+t)P-2dy, 0 If F(s) is the StieltJes and each (t) > 0 X PROOF. By using the same technique as used in proving Theorem 4.1, it can be shown that where the support of (t) is contained in (a,b), b > a > 0, < f(), a > (t)dt. Letting 0+, the result follows in view of Lemma 4.3. THEOREM 4.5. the For a fixed StleltJes transform llm (By using Riemann’s sum technique) <p-I 2i < I < Re p, let f(t) of f(t) defined by (3.1). J (x+t)P’2[F(t’i)’F(t+i)]dt’ S’(I) and let F(s) be Then, {(x) >= <f,{> U.N. TIWARI and J. N. PANDEY 458 for all 6 D(1) and A > 0. The result follows quite easily in view of Theorems 4.1 and 4.4. PROOF. The authors are grateful to Professor E.R. Love, for his ACKNOWLEDGEMENTS. valuable suggestions, in particular for providing the proof of Lemma 4.3. The work of the second author was supported by a National Research Council Grant Number A52 98. REFERENCES. i. 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