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THE LAPLACE TRANSFORM
PRINCETON MATHEMATICAL SERIES Editors: Marston Morse, H. P. Robertson, A. W. Tucker
1. THE CLASSICAL GROUPS: THEIR INVARIANTS AND REPRESENTATIONS By Hermann Weyl
2. TOPOLOGICAL GROUPS By L. Pontrjagin. Translated from the Russian by Emma Lehmei
3. AN INTRODUCTION TO DIFFERENTIAL GEOMETRY WITH USE OF THE TENSOR CALCULUS By Luther Pfahler Eisenhart
4. DIMENSION THEORY By Witold Hurewicz and Henry Wallman
5. THE ANALYTICAL FOUNDATIONS OF CELESTIAL MECHANICS By Aurel Wintner
6. THE LAPLACE TRANSFORM By David Vernon Widder
THE LAPLACE TRANSFORM B y DAVID VERNON WIDDER
PRINCETON PRINCETON UNIVERSITY PRESS LONDON I HUMPHREY MILFORD : OXFORD UNIVERSITY PRESS
1941
Copyright, 1941, by Princeton University Press
Printed in the United States of America
PREFACE The material of this book is the outgrowth of a course of lectures which I have given from time to time at Harvard University on the subject of Dirichlet series and Laplace integrals. It is designed for students who have such a knowledge of analysis as might be obtained by reading the more fundamental parts of the familiar text of E. C. Titchmarsh on the theory of functions. I have taken pains to include proofs of results which such a student might not know even though they might be available elsewhere. For example, the first chapter is devoted largely to the study of the Riemann-Stieltjes integral. Al though this material is in constant use by analysts it seems not to have been collected in convenient form. There are only a few instances where I have had to depart from this aim of having the book complete in itself. If he desires, the student may omit such parts without losing the fundamental ideas. I wish to express here my gratitude to Professor G. D. Birkhoff for suggesting the writing of this book. I also wish to acknowledge my indebtedness to Professor G. H. Hardy. For it was his course of lec tures delivered in Princeton University in 1928 that first aroused my interest in the type of analysis here presented. My thanks are also due to Professor Marston Morse and his committee for requesting the inclusion of the book in this series. I gratefully acknowledge the aid provided by the Milton Fund of Harvard University for the preparation of manuscripts. Finally, I wish to extend my thanks to the John Simon Guggenheim Memorial Foun dation. It was during the year in which I held a Guggenheim Fellow ship that I worked out much of the material appearing in the last chapter. D. Y. W.
ν
CONTENTS CHAPTER I THE STIELTJES INTEGRAL SECTION
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
PAGE
Introduction Stieltjes integrals Functions of bounded variation Existence of Stieltjes integrals Properties of Stieltjes integrals The Stieltjes integral as a series or a Lebesgue integral Further properties of Stieltjes integrals Normalization Improper Stieltjes integrals Laws of the mean. . . . Change of variable Variation of the indefinite integral Stieltjes integrals as infinite series; second method Further conditions of integrability Iterated integrals The selection principle Weak compactness
3 3 6 7 8 10 12 13 15 16 19 20 22 24 25 26 33
CHAPTER II FUNDAMENTAL FORMULAS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
Region of convergence Abscissa of convergence Absolute convergence Uniform convergence Analytic character of the generating function Uniqueness of determining function Complex inversion formula Integrals of the determining function Summability of divergent integrals Inversion when the determining function belongs to L2 Stieltjes resultant Classical resultant Order on vertical lines Generating function analytic at infinity Periodic determining function Relation to factorial series
:
35 38 46 50 57 59 63 70 75 80 83 91 92 93 96 97
CONTENTS
viii
CHAPTER III THE MOMENT PROBLEM SECTION
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
PAGE
Statement of the problem Moment sequence An inversion operator Completely monotonic sequences Function of L p .... Bounded functions .... Hausdorff summability Statementoffurthermomentproblems.. .... . The moment operator TheHamburgermomentproblem Positive definite sequences Determinant criteria The Stieltjes moment problem Moments of functions of bounded variation A sufficient condition for the solubility of the Stieltjes problem . ... Indeterminacy of solution
100 101 107 108 109 Ill 113 125 126 129 132 134 136 138 140 142
CHAPTER IV ABSOLUTELY AND COMPLETELY MONOTONIC FUNCTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
Introduction Elementary properties of absolutely monotonic functions Analyticity of absolutely monotonic functions Bernstein's second definition Existence of one-sided derivatives Higher differences of absolutely monotonic functions Equivalence of Bernstein's two definitions Bernstein polynomials ... . . Definition of Gruss Equivalence of Bernstein and Gruss definitions ' Additionalproperties of absolutely monotonic functions. .. .... Bernstein's theorem Alternative proof of Bernstein's theorem Interpolation by completely monotonic functions Absolutely monotonic functions with prescribed derivatives at a point . Hankel determinants whose elements are the derivatives of an absolutely monotonic function · Laguerre polynomials A linear functional Bernstein's theorem Completely convex functions
144 144 146 147 149 150 151 152 154 155 156 160 162 163 165 167 168 171 175 177
CHAPTER Y TAUBERIAN THEOREMS 1. Abelian theorems for the Laplace transform 2. Abelian theorems for the Stieltjes transform
*
180 183
CONTENTS SECTION
3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
ix PAGE
Tauberian theorems Karamata's theorem Tauberian theorems for the Stieltjes transform Fourier transforms Fourier transforms of functions of L The quotient of Fourier transforms A special Tauberian theorem Pitt's form of Wiener's theorem Wiener's general Tauberian theorem Tauberian theorem for the Stieltjes integral One-sided Tauberian condition Application of Wiener's theorem to the Laplace transform Another application ; The prime-number theorem Ikehara's theorem
185 189 198 202 204 207 209 210 212 213 215 221 222 224 233
CHAPTER VI THE BILATERAL LAPLACE TRANSFORM 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.
Introduction Region of convergence Integration by parts Abscissae of convergence Inversion formulas Uniqueness Summability Determining function belonging to L2 The Mellin transform Stieltjes resultant Stieltjes resultant at infinity Stieltjes resultant completely defined Preliminary results ' The product of Fourier-Stieltjes transforms Stieltjes resultant of indefinite integrals Product of bilateral Laplace integrals . Resultants in a special case Iterates of the Stieltjes kernel Representation of functions Kernels of positive type Necessary and sufficient conditions for representation
237 238 239 240 241 243 244 245 246 248 249 250 251 252 256 257 259 262 265 270 272
CHAPTER VII INVERSION AND REPRESENTATION PROBLEMS FOR THE LAPLACE TRANSFORM 1. 2. 3. 4.
Introduction Laplace's asymptotic evaluation of an integral Application of the Laplace method Uniform convergence
276 277 280 283
Χ
CONTENTS
SECTION
5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.
PAGE
Uniform convergence; continuation The inversion operator for the Laplace-Lebesgue integral The inversion operator for the Laplace-Stieltjes integral Laplace method for a new integral The jump operator The variation of the determining function A general representation theorem Determining function of bounded variation Modified conditions for determining functions of bounded variation. ... Determining function non-decreasing The class L p , ρ > 1 Determining function the integral of a bounded function The class L , The general Laplace-Stieltjes integral
285 288 290 296 298 299 302 306 308 310 312 315 317 320
CHAPTER VIII THE STIELTJES TRANSFORM 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27.
Introduction Elementary properties of the transform Asymptotic properties of Stieltjes transforms Relation to the Laplace transform Uniqueness The Stieltjes transform singular at the origin Complex inversion formula A singular integral The inversion operator for the Stieltjes transform with a(t) an integral.. The inversion operator for the Stieltjes transform in the general case .. The jump operator The variation of a(t) ... A general representation theorem Order conditions ... Generalrepresentationtheorems.. The function a(t) of bounded variation The function a(t) non-decreasing and bounded The function a(t) non-decreasing and unbounded The class L p , ρ > 1.. . The function 0, J— 00
contrary to hypothesis. Hence a(t) has but a finite number of points of increase. By virtue of Theorem 10 and of the fact that every posi tive sequence is either definite or semidefinite the proof is com plete. Theorem 12a a gives us another proof of Corollary 11. For, a com pletely monotonic sequence has, by Theorem 4a, the representation (1) with a(t) non-decreasing in (0, 1) and constant elsewhere. Another result of the same nature, but less useful since it has no counterpart for the positive semidefinite case is contained in: THEOREM 126. A necessary and sufficient condition that equations (1) should have a non-decreasing solution a{t) with infinitely many points of increase is that*
(2)
μο > 0,
μο
μι
βΐ
β2
> ο,
μο μι
μι
μι Μ2
μζ > 0
β2
μ4
Μ3
This follows at once from Theorem 11 and a familiar necessary and sufficient condition from algebra that a quadratic form should be posi tive definite.f It is clear further that if equations (1) have a non-decreasing solution ait) with a finite number of points of increase then the determinants (2) are positive or zero. For this is true if the forms §11 (1) are all positive (definite or semidefinite). But it is not true conversely that equations (1) have a non-negative solution whenever determinants (2) are nonnegative. For example, if (η = 0, 1, 2, 3) βη
(η = 4,5,
* These determinants are called Hankel determinants, having been introduced by H. Hankel [1861] in his thesis. t For a simple proof of the theorem in question see L. M. Blumenthal [1928],
136
THE MOMENT
PROBLEM
[CH. ILL
we have
with all successive determinants zero.
But the quadratic form
is neither positive definite nor positive semidefinite. — 1 when
For it reduces to
Hence by Theorems 10 and 11 equations (1) can have no non-decreasing solution a(t). 13. The Stieltjes Moment Problem The Stieltjes problem may be treated as a special case of the Hamburger problem. THEOREM 13a. A necessary and sufficient condition that there should exist a non-decreasing function a(t) such that (1)
(n = 0, 1, 2,
...),
the integrals all converging, is that the sequences {nn}o and \ii„}i should be positive, or that the quadratic forms (2)
(» = 0, 1, 2, • • •)
(3)
(n = 0, 1, 2, . . . )
should be positive (definite or semidefinite*). The equivalence of the two forms of the condition is apparent by Theorem 11. We prove the result in the latter form involving quadratic * See the footnote to Theorem 11.
§13]
STIELTJES MOMENT PROBLEM
137
forms. For the necessity, let the sequence have the form (1). By Theorems 10 and 11 the forms (2) are positive, since we may regard a(t) as constant in the interval (— oo, 0). Furthermore,
Since /?( 1. Since f(x) is completely monotonic for it is analytic there. Consequently the integral (4) must converge for x > 0 by Theorem 5b of Chapter II. Finally, we see that is bounded, for otherwise This would imply, since is monotonic, that contrary to hypothesis. This completes the proof. 20. Completely Convex Functions By analogy with the completely monotonic functions of S. Bernstein we introduce the class of completely convex functions according to the following definition. A function f(x) is completely convex in an interval (a, b) if it has derivatives of all orders there and if DEFINITION 2 0 .
in that interval. For example the functions sin x and cos x are completely convex in the intervals
and
respectively.
We shall show that any
such function is necessarily entire.* We need two preliminary results. The first is a familiar result of J. Hadamard LEMMA 20a. If f(x) is of class in an interval (a, b) and if
then (1) For, by Taylor's theorem we have
Subtracting these two equations gives
* See D . V . W i d d e r [1940]. Boas.
T h e proof here presented was given later b y R . P.
t See T . Carleman [1926] p. 11.
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ABSOLUTELY MONOTONIC FUNCTIONS
[CH. IV
From this, inequality (1) is immediate if we choose for x a value at which f'(x) attains its maximum LEMMA 206. If f(x) is of class and if f(x) and — f"(x) are nonnegative in (a, b), then
(2) For, suppose that
Then by comparing the area under the curve y = f(x) with the area of the triangle with vertices (a, 0), (b, 0), (c, /(c)) we obtain (2). We are now in a position to prove the result stated above. THEOREM 20. Iff(x) is completely convex in an interval (a, b), it is entire. By consideration of the function one sees that there is no restriction in replacing the given interval by the interval (0, 1). Integration by parts gives
so that
Since the function gives
is also completely convex the same result
Repeating the process we obtain
Now let
be any positive number less than
Then
§20]
COMPLETELY CONVEX FUNCTIONS
179
and by Lemma 20b
By Lemma 20a
That is
uniformly in By use of Taylor's formula with remainder we recognize at once that the power series expansion of f(x) must converge for all x and converge to f(x) in (0, 1). This proves* the theorem. * We have in f a c t p r o v e d a little m o r e : that f(x) is at m o s t of order u n i t y , type
CHAPTER V TAUBERIAN THEOREMS 1. Abelian Theorems for the Laplace Transform The theorems which we shall treat first are called Abelian because they are generalizations of a familiar result of Abel. This states that if f(x) is defined by the series
the series converging for x numerically less than one, then
whenever the series on the right converges. An integral analogue of this result is that if /(s) is defined for real s > 0 by the convergent integral
then
provided the integral on the right converges. result would be that if*
Another form of the
then
whenever the limit on the right exists. It is this and similar results which we wish to prove in the present section. We shall assume, unless otherwise stated, that s is a real variable. * W e remind the reader that in Laplace-Stieltjes integrals the determining f u n c t i o n a(t) is assumed t o be normalized (see Section 8 of Chapter I ) . 180
§1] T H E O R E M 1.
ABELIAN THEOREMS
181
If
(1) then for any
and any constant A
(2) (3) Since the integral (1) converges for s > 0, we have by Theorem 2.3a of Chapter II
If T is any positive number, then
(4)
For any positive
we can find* a constant M such that
whence
The right-hand side of this inequality approaches zero as s becomes infinite, so that we have from (4)
* See T h e o r e m 2.2a of C h a p t e r I I .
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TAUBERIAN THEOREMS
[Ch. V
The left-hand side being independent of T we allow T to approach zero and obtain
We have thus established (3).
In a similar way we have
from which (2) follows by allowing T to become infinite. COROLLARY la. If for some non-negative number y
then
Note that if A = 0 we must write
In particular we note that the existence of a( °°) implies Also /(«>) = a ( 0 + ) . COROLLARY lb. If ( 1 ) converges for s > 0, then
The proof of this requires only slight modifications of the argument above and is left to the reader.
2. Abelian Theorems for the Stieltjes Transform We consider here the equation (1) It will be shown* in a later chapter that if (1) converges for a single positive value of s it converges uniformly in any closed interval of the positive s axis not including the origin and that then (2) In order to study the asymptotic behavior of /(s) at s = 0 and at s = » as determined by the properties of a{t) we need the following result. LEMMA 2. If the integral ( 1 ) converges for s > 0, and if T > 0, then (3)
It is clear from (2) and the convergence of (1) that the integral on the left-hand side of (3) converges. Since (4)
it follows from the result stated above that (4) converges for s + T > 0 and hence uniformly in a neighborhood of s = 0, from which (3) follows. THEOREM 2a. If ( 1 ) converges for s > 0 and if 0 < 7 ^ P. then for any constant A
* Theorem 2c and Corollary 3a3 of Chapter V I I I .
184
T A U B E R I A N THEOREMS
[CH.
V
v>here
For, (s > 0)
This follows from the familiar formula
Hence if T > 0,
Making use of Lemma 2 we have
Since T was arbitrary it follows that
Also,
But
Hence
This concludes the proof of the theorem.
§3] COROLLARY 2a.
185
T A U B E R I A N THEOREMS If
for
then
The case in which 7 = 0 requires special attention. vergence of (1) implies that THEOREM 2b. If ( 1 ) converges for s > 0 and if p >
The mere con0,
then
If we set
then
Hence
Since / 3 ( » ) = 0 the theorem is established.
3. Tauberian Theorems The converse of Abel's theorem, stated in Section 1, is not true. That is, the series (1)
186
T A U B E R I A N THEOREMS
[CH. V
may converge for and fix) may tend to a limit as x tends to 1 without having the series
convergent. The example fix) = (x + l ) - 1 shows this. Another example, showing the failure of the converse of Abel's theorem in its integral form, is
Here
Yet
diverges. It was A. Tauber [1897] who first gave a conditional converse of Abel's theorem. He proved that if (1) converges for if and if / ( I —) = A, then
We prove the integral analogue of this result. THEOREM 3a. If ait) belongs to L in ( 0 , II) for every R and if the integral
converges for s > 0, then the conditions
(1) imply
To prove this we note first that under the hypotheses of the theorem we have
187
T A U B E R I A N THEOREMS
[Ch.
V
Now consider the difference
Then
Hence
It follows that
This proves the theorem. Note that in the example given above a(t) was cos t which fails to satisfy the condition a(t) = o(i _1 ). A more general result is contained in: THEOREM 3 b .
Let
converge for s > 0, and let (2) Then
(3)
if and only if (4)
Note that if
188
T A U B E R I A N THEOREMS
then
implies that
[CH. V
so that Theorem 3a is included in Theorem 3 b. To prove the theorem we assume without loss of generality that A = 0. First suppose that (3) holds and prove (4). This is obvious from the equation assume (4) and prove (3). Now (5)Conversely,
By Corollary la
By (2)
Consequently (5) shows that
(6) But by (4) we may apply Theorem 3a to the integral (6) and obtain
Integrating by parts, we have
which is what we wished to prove. If we choose a(t) a step-function with jumps at the positive integers this becomes a familiar result of A. Tauber [1897], It states that a series
which is summable in the sense of Abel to the number A
converges to A if and only if
§4]
KARAMATA'S THEOREM
189
4. Karamata's Theorem It was Littlewood [ 1 9 1 0 ] who first showed that the condition in Tauber's theorem could be replaced by the condition We consider the integral analogue of the theorem and give an ingenious proof of J. Karamata [ 1 9 3 1 ] . 4.1. We shall prove first the following preliminary result. LEMMA 4 . 1 . If g(x) is continuous almost everywhere in ( 0 , 1 ) and is bounded there, and if y > 0, then for every e > 0 there exist polynomials p(x) and P(x) such that
e.
First suppose that
Let JJ be an arbitrary positive number. Clearly there exists a continuous function h(x), which may coincide with g(x) except in the neighborhood of the points a and /3, such that
By the Weierstrass approximation theorem there exists a polynomial Q(x) such that
Set then
190
T A U B E R I A N THEOREMS
[CH. V
In a similar way we can determine the polynomial p(x) < g(x) such that
Since the right-hand sides of these last two inequalities can be made arbitrarily small, our lemma is true for the step-function g(x) just considered. It is consequently true for any step-function with a finite number of jumps, for any such function is a linear combination of functions of the type just treated. We turn now to the general case. With a positive e we can determine positive numbers S and R so that (1)
Since g(e~t)e~tty~1 is Riemann-integrable in (8, R) we see from the definition of an integral that there exist two step-functions gi(x) and gi(x), each with a finite number of jumps such that
(2) (3) If we complete the definition of gi(x) and gz{x) by the equations
inequalities (2) hold for all positive t, and by the first part of the proof there exist polynomials p(x) and P(x) such that
(4) (5) From (1) and (3) we have
§4]
KARAMATA'S THEOREM
191
Combining this with (4) and (5) there results the required inequality
4.2. We turn now to the proof of the following theorem due to Karamata. THEOREM 4 . 2 . Let a(t) be non-decreasing and such that the integral
converges for s > 0 ; and for some positive number y let
(1) Let g(x) be of bounded variation in (0, 1).
Then
(2) s varying through the set of points for which the integral on the left exists. Let a(t) be continuous except perhaps in the set of points xa, xx, x2, • • • and let g(e~') be continuous except perhaps in the points y0, yi, 2/2, • • • • Denote by E the set of points ?/>/Xj (i, j = 0, 1, 2, • • •). If s is not in E the integral on the left of (1) exists since a(t) and g(e~st) will have no common discontinuities. Since E is a countable set its complement is dense in (0, oo) so that s may approach zero or become infinite while remaining in the set of points for which the integral on the left of (2) exists. Let e be an arbitrary positive number. Determine polynomials (Lemma 4.1) p(x) and P(x) so that (3)
Since a(t) is non-decreasing we have (4)
(5)
192
T A U B E R I A N THEOREMS
[Ch. V
If we replace s by (n + l)s in (1) we obtain
for any positive integer n, whence
for any polynomial P(x).
From (4) we have
(6).
s approaching zero or becoming infinite in the set complementary to E. From (3), (5) and (6) we see that
the desired result. 4.3. From this result by specializing g(x) we may obtain the following Tauberian theorem. THEOREM 4.3. If a(t) is non-decreasing and such that the integral
converges for s > 0, and if for some non-negative number y
then (1) It is no restriction to take A = 1. First suppose 7 = 0. Since a(t) is non-decreasing it either approaches a limit or becomes infinite as From Corollary 1 c the latter case would imply
contradicting our hypothesis. The finite limit which a(t) must approach is then certainly unity. A similar argument applies to the case
§4]
KARAMATA'S THEOREM
193
If y > 0 we choose the function gf(e~!) of Theorem 4.2 as follows:
Then the conclusion of Theorem 4.2 is precisely (1) provided t approaches its limit through the set of points where a(t) is continuous. Obviously this restriction may be removed since a(t) is monotonia 4.4. We prove now a very useful theorem due to Hardy and Littlewood [1912] in its original form. We employ the slightly more general form of Landau* THEOREM 4 . 4 . If f(x) has a second derivative in the interval 0 < x < -X and if for some real value of a (1) (2) then (3)
Let 5 be an arbitrary positive number less than unity and let x be positive. By Taylor's remainder theorem (4) By (2) there exists a constant M such that (5)
By (1), (4), and (5) we obtain
* E. Landau [1929] p. 58.
194
T A U B E R I A N THEOREMS
[CH. V
Hence
Since 5 is arbitrary we have in either case
which is equation (3). It is obvious that in condition (2) the direction of the inequality may be reversed.
COROLLARY
4.4a.
If condition
(1)
is replaced by
then (3) becomes
This follows at once from the theorem by consideration of the function Of course if a = 0 our conclusion becomes
COROLLARY 4 . 4 6 . If fix) is of class C' in non-negative number a
then
0 SI X
0, if
and if there exists a constant K such that the function (1) is a non-decreasing function of t in (0, » ) , then (2) if We note first that Theorem 3a is included in the present result.
For,
then (3)
at least for large values of t. Evidently the behavior of a(l) for small values t is unimportant * See of E. Landau [1906] p. 218.in Theorem 3a provided only that f(s) is
196
T A U B E R I A N THEOREMS
[CH. V
defined for s > 0. Hence we may suppose that (3) holds for all t, so that j3 (t) is non-decreasing for K = 1. To prove the theorem we have
Since /3( 0, if
§4]
KARAMATA'S THEOREM
197
and if a constant K exists such that (4)
then
Otherwise expressed this result states that a series whose terms satisfy (4) cannot be summable in the sense of Abel unless it converges. It follows from the theorem since (4) implies (1) if a(t) is a suitable step-function. 4.6. By way of generalizing Theorem 4.3 we add the following result. THEOREM 4.6. Let the integral
converge for s > 0 and let constants K, A, and y > 0 exist such that the function
is non-decreasing in 0 ^ t
0 , if the integral
converges, and if
then the relation
implies
Write
Since we wish to show that I\ tends to A as s becomes infinite we need only prove that / 2 and I 3 tend to zero. This results from the following relations
Allowing first s and then X to become infinite, we have
* Hardy and Littlewood [1930] p. 34.
199
TAUBERIAN-THEOREMS
[CH.
V
Treating Js(s) in a similar way we have
This completes the proof of the lemma. By its use we establish the following result also due to Hardy and Littlewood.* THEOREM 5a. Let 4>it) belong to L in ( 0 , R) for every R > 0 , and let the integral
converge. If constants K and A exist such that (1) (2) then
Consider the function
It is clear from (2) and (1) that
and that
* Hardy and Littlewood [1930] p. 34.
By use of Theorem 4.4 we obtain at once that
That is
200
T A U B E R I A N THEOREMS
[CH. V
From this we see that
But the integral on the left can be expressed as the product of two Laplace transforms, (3)
One sees this by interchanging the order of integration, a step which is valid by virtue of (1) and Fubini's theorem. By (1) we have
Hence we may apply Theorem 4.6 to the integral on the right-hand side of (3) and obtain (4)
The second derivative of this function is
By (1) we have This with (4) shows that
so that we may again apply Theorem 4.4 to the function (4) and obtain
201
T A U B E R I A N THEOREMS
[Ch.
V
Again, using (1) and Theorem 4.6 we have
Now applying Lemma 5 we obtain the desired result
Another result of this type which we shall need in a later chapter is THEOREM 5b. Let ) if it is defined there and if
The new form of the result is contained in THEOREM 1 0 6 . If a(x) belongs to S*B on ( 0 ,
if g(x) belongs to L
212
T A U B E R I A N THEOREMS
[CH. V
and is such that
then
implies a(cc)
= A.
11. Wiener's General Tauberian Theorem We now obtain Wiener's general Tauberian theorem. THEOREM 11a. If gi(x) belongs to W, g2(x) to L, and p(t) to B in , and if
(1) then (2) For, we have only to apply Theorem 10a taking g(x) = gi(x) and
It is no restriction to assume that A = 0. Clearly a(x) is bounded. Our result will be established if we can show that a(x) belongs to S and that §10 (1) holds. But
The right-hand side approaches zero with y — x since* gz{x) belongs to L. Finally,
* See, for example, N. Wiener [1933] p. 14.
§12]
T H E O R E M FOR STIELTJES I N T E G R A L S
213
By (1) the inner integral approaches zero boundedly as x becomes infinite. Hence §10 (1) is established. The conclusion of Theorem 10a gives (2). Another form of this result is obtained by an exponential change of variable. THEOREM 1 1 6 . If g\(x) and g2(x) belong to L, p(x) to B in ( 0 , oo), if
and if
then
It should be observed that the condition g\ e W in Theorem 11a is a necessary one. Suppose that it were not satisfied, that
for some x0. Then
Hence A = 0.
Choose gf2(0 so that it belongs to W and choose p(t) =
But
This function clearly approaches no limit as x becomes infinite. That is, the conclusion of Theorem 11a does not apply if g\(x) is not assumed to belong to W.
12. Tauberian Theorem for the Stieltjes Integral Another result of Wiener is also contained in Theorem 10a. To state it we must first introduce Wiener's class M. DEFINITION 1 2 . The function f(x) belongs to the class M if it is continuous in the interval — a> < x < °° and if
It is obvious that a function of M belongs to L.
214
TAUBERIAN THEOREMS
[CH. V
THEOREM 12. If g^x) belongs to MW, if g2(x) belongs to M, if a{x) is of bounded variation in every finite interval and is such that the function
belongs to B, and if
then
To prove this set
This integral exists for all x since
Moreover The series on the right converges uniformly for y in some neighborhood so that a{x) is a slowly oscillating function. Now of y = 0. Each term of the series approaches zero with y so that
§13]
ONE-SIDED TAUBERIAN CONDITION
215
The change in the order of integration is justified since
Again changing the order of integration we have
By hypothesis
and the integral on the left has a bound independent of x. Lebesgue's convergence theorem
Hence, by
All conditions of Theorem 10a are satisfied so that
and our result is established.
13. One-sided Tauberian Condition For the applications of Theorem 11a, it is frequently convenient to suppose that p(t) is bounded only on one side. That the theorem is no longer true without further hypotheses is seen by the following example.
Then p(x) is bounded on one side but not on both, gi(x) belongs to W, gi(x) to L, and
216
T A U B E R I A N THEOREMS
[CH. V
The functions h\(x) and hi{x) do not approach a common limit as x becomes infinite. The case when p{t) is bounded on one side can be treated by the following theorem due to H. R. Pitt [19386], THEOREM 1 3 a .
Let
(1) (2) (3)
except that gt{x) may be discontinuous at a set of points of measure zero, and let the function (4)
exist for all x and belong to B.
Then
(5)
implies
We note first that it is no restriction to take A — 0. Our second remark is that wre may assume gi(x) continuous. For, the function
satisfies conditions (1) and is also continuous. It is clearly positive and integrable. To styow that it belongs to W we have
§13]
ONE-SIDED T A U B E R I A N C O N D I T I O N
217
To show that (5) is satisfied when g^x) is replaced by g(x) we have
Since hi(x) is bounded this last integral exists. By use of (2) and Fubini's theorem the interchange of the order of integration is justified. Since hi(x) is bounded we may take the limit as x becomes infinite under the sign of the last integral and obtain the limit zero as desired. We now replace g(x) by gi{x) and assume it to be continuous. Finally we observe that we may assume that
for this can always be brought about by multiplying gi(t) and gr2(0 by suitable constants. Now consider the function
where 8 is an arbitrary positive constant. Since S > 0 we have
Hence
(6) But we can show that (7)
For,
so that (7) is evident. We are now in a position to apply Theorem 11a to the function S$(x). Choosing the function g2(x) of that theorem as a suitable step-function we have for every D > 0
Hence inequalities (6) show that
§13]
ONE-SIDED T A U B E R I A N C O N D I T I O N
219
In particular
Since
it follows from (3) that
exists for all x. Let e be an arbitrary positive number. N = 2V(e) such that
By (3) we can determine
Then
(8) Since gzix) is Riemann integrable we can determine step-functions qi(x) and 0 and if
then
We see by Corollary la that
Obviously
222
T A U B E R I A N THEOREMS
[CH. V
Hence it is no restriction to suppose that p(t) is identically zero in (0, 1). In Theorem 1-36 choose
Conditions (10), (12) and (13) of Section 13 are obviously satisfied. Condition (11) becomes holds since Condition (14) Finally (15) is satisfied since
is clearly continuous in the interval (0 < x < °°) and approaches finite limits when x approaches zero or becomes infinite. It approaches A as x becomes infinite, and since
it is clear that hi(x) approaches zero with x. All conditions of the theorem are satisfied and when we apply it we obtain the desired result. One could easily obtain more general results by altering the choice of gi(x) and g?.(x). Since no new principles are involved and since we have already obtained the results by Karamata's method, we content ourselves here with the above special case. It should be observed that Karamata's method is much simpler than Wiener's for the particular Laplace kernel e~"x. The great importance of Wiener's methods and results lies in their extreme generality.
15. Another Application As another example of the use of Wiener's theorem let us prove the following special case of Theorem 4.4. Here again the complete theorem could be obtained by Wiener's method, but since we have already obtained the result by a simpler method we content ourselves here with the special case by way of illustration.
§14]
A P P L I C A T I O N OF W I E N E R ' S T H E O R E M
THEOREM 1 5 .
223
If f(x) has a continuous second derivative for -positive x
such that (1) and if
(2) then (3)
We note first that (1) implies the existence of the integral
so that / ' ( o o ) exists.
It must be zero for otherwise we should have
obtained by integration by parts, making use of (3) and (4). rem 116 choose
contrary and (4)Now from consider to(1)(2). the Hence equation
In Theo-
224
T A U B E R I A N THEOREMS
[CH. V
Clearly gi(t) and g2(t) belong to L on (0, I.
The proof of this result follows in an obvious way from Theorem 2.4a of Chapter II and is omitted. It is clear that when the integral (1) converges in a proper strip, then it converges uniformly in any closed bounded region inside the strip and not touching the boundary of the strip. Moreover, there is also uniform convergence in a Stolz region corresponding to a point of the boundary of the strip at which (1) converges. Hence it is clear that the integral (1) represents an analytic function f(s) at interior points of the strip, that /w(s) = (-1)* f e~ s t t h da{t)
(k = 0, 1, 2, · · ·)
§5]
INVERSION
FORMULAS
241
at such points, and that /(«) is continuous at those boundary points of the strip at which (1) converges. We say that the integral (1) converges absolutely if the integral
converges. Here the function u(t) is the variation of a(x) in the interval 0 ^ x ^ t if t is positive, is zero if t is zero and is the negative of the variation of a(x) in the interval t ^ z ^ 0 if t is negative. Clearly u(t) is a non-decreasing function in ( — which vanishes at the origin. We now define abscissas of absolute convergence (x) is really independent of a. This follows at once by Cauchy's integral theorem, using the analyticity of f(s) and (2). * Compare H. Hamburger [1921], p . 416.
266
BILATERAL LAPLACE TRANSFORM
[CH. VI
Let be an arbitrary point of the strip o-i and c — p, but by Theorem 56 of Chapter II, it must also converge for a > a. Then we conclude our proof by analytic extension. Conversely, it is evident from Theorem 13a, Chapter III, that if (7) holds then f(s) is analytic for a > a and the sequences (6) are positive for every c greater than a. COROLLARY 196. If the sequences and are positive, then
For under the present hypotheses / ( — s) satisfies the conditions of Theorem 196 in the interval ( —/3, —a). T H E O R E M 19C. If f(s) is analytic in the strip and if for some real c in (a < a < /3) the sequence is positive, then
where a{t) is a non-decreasing function. By Theorem 10 of Chapter III there exists a non-decreasing function /3 (t) such that
As in the previous proof we have (8).
Then
Also (10) by the inequality of Schwarz and (10)
§19]
R E P R E S E N T A T I O N OF F U N C T I O N S
2691
That is, (ID and in like manner (12) B y Lemma 196 the inequalities (11) and (12) enable us to conclude
respectively.
Adding and setting
(13) we have
As in the previous proof this equation must hold throughout the strip of analyticity. Again it is clear that the converse of the theorem is trivial. COROLLARY 19C. If is a positive sequence, the result holds. For the sequences and are both positive if one is, as is easily verified by an examination of the defining quadratic forms. Following the same order of ideas we prove: THEOREM 19D. If f(s) is analytic at a real point s — c where the sequence (14)
is completely monotonic, then for all complex s (15) where a(t) is non-decreasing and bounded in For, by Theorem 4a of Chapter III there exists a non-decreasing bounded function j3(t) such that
270
BILATERAL LAPLACE TRANSFORM
[C H. VI
Since/(s) is analytic at s = c Λ») - Σ/"ω Η =0
(emit)
-Σ
TL I
W=0
for some neighborhood, [ s — c | < this gives
p,
of c.
JQ
By Theorem M of Chapter I
where a(t) is defined by (13) in (0 ^ t ^ 1). But this integral defines an entire function, so that we can complete our proof by analytic extension. It is obvious conversely that if /(s) has the representation (15) the sequence (14) is completely monotonic for every real c.
20. Kernels of Positive Type In order to obtain necessary and sufficient conditions for the repre sentation of a function as a bilateral Laplace integral we need a pre liminary discussion of kernels of positive type.* These are the con tinuous analogues of positive or semidefinite quadratic forms. DEFINITION 20. A real function k(x, y) which is continuous in the square {a ^ χ ^ b, a ^ y ^ b) is of positive type there if for every real function φ(χ) continuous in (a ^ χ ^ b) b
(1)
pb
/ k(x, y) (x) {y)dxdy ^ 0. / ι "a
As an example take k ( x , y ) = g ( x ) g { y ) where g ( x ) is any function con tinuous in (a ^ χ b). Note that the integral (1) may vanish without having φ(χ) identically zero. Thus in our example we have only to c h o o s e φ( χ ) o r t h o g o n a l t o g ( x ) o n ( a , b ) . A kernel is said to be positive definite if it is of positive type and if integral (1) can vanish for no real continuous function φ(χ) except φ(χ) = 0. As an example take a = 0, b = π and QO
" cos nx cos ny.
k(x,y) = The integral (1) becomes
Σe OO
-Ti
a2n
T
φ ( χ ) cos n x d x * See J. Mercer [1909], p. 242.
( n = 0, 1, 2, · · ·).
§20]
K E R N E L S OF P O S I T I V E
TYPE
271
But (2) cannot be zero unless all the an are zero. But by the completeness of the cosine set on this implies that is identically zero. We now prove an important result of J . Mercer [ 1 9 0 9 ] which brings out the connection between kernels and quadratic forms. THEOREM 2 0 . A continuous kernel k(x, y) is of positive type if and, only if for every finite sequence of distinct numbers of the quadratic form (3)
is positive (definite or semidefinite). Suppose first that the quadratic forms (3) are positive. Let us prove that k{x, y) is of positive type. Choose an arbitrary continuous function (x). The integral ./() can be expressed as the limit of a sum
Choosing
in (3) we see that J () is the limit of a non-negative function of n, is therefore itself non-negative. Let us turn next to the converse. Suppose that we could find some form (3) which is not positive, with Then we can choose the £» so that has a negative value —A. N o w define an auxiliary function Let c be a point of Let e and 17 be so small that and are points of the same interval. Now is defined as zero in and as unity in In the rest of the interval it is to be linear and such that it is continuous in (a, b). Set
where e and rj have been chosen so small that no two intervals overlap and such that and Set
272
BILATERAL LAPLACE TRANSFORM
Then we may easily compute
[CH. VI
J(6):
where
Here qt] is the region between the square and the square Since
for (x, y) in g,,- we have easily
where M is the maximum of in the whole square. Since F„(0, 0) = —A we can find i? so small that
Then
and for all e sufficiently small. We can choose e so small that the righthand side of this inequality is negative. For this continuous function d(x) we see that J(6) is negative, contradicting (1). Hence there can be no Qn , with the x, interior points of (a, b), which can have a negative value. Neither can there be a Q„ having a negative value even if x0 or xn is allowed to be an end point. This is clear by the continuity of k(x, y). The proof of the theorem is complete. A kernel which is continuous in an open square is said to be of positive type there if it is of positive type in every closed square21. interior to theand open square. Conditions for Representation Necessary Sufficient By way of making the results of the present section plausible let us make an analogy with the moment problem. W e have already com-
§21]
C O N D I T I O N S FOR
REPRESENTATION
273
pared the Hausdorff moment problem with the representation problem for the unilateral Laplace integral. Here we shall then expect to compare the Stieltjes moment problem with the bilateral representation problem. We recall that the system of equations (1) has a non-decreasing solution fi(t) if and only if the quadratic forms (2) are all positive. If in (1) the integer n is replaced by a continuous variable x and if we obtain
where
Clearly a(u) is non-decreasing when fi(t) is. When one changes from the discrete to the continuous one would expect the quadratic forms (2) to coalesce into a single double integral
which would be required to be non-negative for all continuous functions i-(x). That is, the kernel /j (x + y) would be of positive type. W e are thus led to the following* result. THEOREM 21. A necessary and sufficient condition that the function f(x) can be represented in the form (3)
where a(t) is non-decreasing and the integral converges for a < x < b, is that f(x) should be analytic there and that the kernel f(x + y) should be of positive type in the square (a < 2x < b, a < 2y < b). First suppose that (3) holds. Then if a < a < /3 < 6 we will show that f(x + y) is of positive type in the square * Compare S. Bochner [1932] p. 76, M. Mathias [1923], and D . V. Widdcr [19346],
274
BILATERAL LAPLACE TRANSFORM
Clearly f(x) is analytic in the interval a < x < b. function 4>{x) we have
[CH. VI
For any continuous
T o justify the interchange in the order of integration we have only to observe that the integral (3) converges uniformly in (a ^ x /?). This completes the proof of the necessity of the condition. Conversely, if f(x + y) is of positive type in the square (a ^ 2x j3, a i£ 2y ^ /3), then by Theorem 20, choosing x{ = (c/2) + iS for some number c of the quadratic forms (4) are positive, provided that S is chosen for each n so that
In particular, if we set ( 5 ) recall that and
we have
(6) If the value of similar way to (7)
from (5) is substituted in (6) the latter reduces in a
§21]
C O N D I T I O N S FOR
REPRESENTATION
Since this form was obtained from (4) it is itself positive. by in (7) and let S approach zero. We obtain
275
Now replace
Now by Corollary 19c
for some non-decreasing function a(t). This completes the proof of the theorem. W e observe that it would be sufficient to assume that f(x) is continuous in (a < x < b). For, it could be shown that this with the fact that f(x + y) is of positive type would insure the analyticity* of/(s). * See R. P. Boas, Jr. and D. V. Widder [19406] and R. P. Boas, Jr. [1941],
CHAPTER VII INVERSION AND REPRESENTATION PROBLEMS FOR THE LAPLACE TRANSFORM 1. Introduction We have seen that the Laplace transform may be regarded as a generalization of Taylor's series. Thus the series (1) and the integral
(2) may be regarded as the discrete and continuous aspects, respectively, of the same Stieltjes integral (3)
Making a change of variable we obtain the Laplace-Stieltjes integral (4)
One familiar determination of the coefficients of (1) (5)
involves a knowledge of the derivatives of F(z) at 2 = 0. By analogy we should expect the existence of an inversion formula for (4) which depends on a knowledge of the derivatives of f(s) in a neighborhood of s = oo; say along the positive real axis, since such a neighborhood is the transform in the z-plane under the transformation z = e~s of a righthanded real neighborhood of the origin in the z-plane. Such an inversion formula was discovered by E. L. Post [1930] for the case when (4) becomes
276
§2]
A S Y M P T O T I C M E T H O D OF L A P L A C E
277
The general case (4) was first treated by the author [1934a], formula is
Post's
(6) Since a limit process is involved it is clear that one needs to know derivatives of f(s) only for large real values of s. In this sense (6) is analogous to (5). The present chapter will be devoted to an elaboration of formula (6) and to the representation results which grow out of the formula. 2. Laplace's Asymptotic Evaluation of an Integral In our development of an inversion formula it will be convenient to apply an asymptotic method of Laplace [1820]. For the reader's convenience we prove here the results which we shall use. LEMMA 2. If a < b, 0 < y, then
For, simple changes of variable give
so that
THEOREM 2 a .
If
1. 2. h{x) is non-increasing
then
Let e be an arbitrary number such that may choose S less than 77 and so small that (1) Consider the integral
Then we
278
INVERSION
OF
LAPLACE
TRANSFORM
where l'k and l ' l correspond to the intervals respectively. Then
Since h"(a) < 0 it is clear that to zero with 1/k for any positive 8.
[CH. V I I
and
Hence l'l tends By Taylor's formula with remainder
By (1)
By use of Lemma 2
Since e is arbitrary we have
and our theorem is proved. T H E O R E M 2b. If in addition to the hypotheses theorem we have
1.
and
2.
of the previous
3.
(2) then
Set
By Theorem 2a we need only show that Ik tends to zero when k becomes infinite. Define e as in the proof of Theorem 2a. Choose 5 so that (1) holds and so that
This is possible by hypothesis 3.
As in the proof of Theorem 2a,
§2]
A S Y M P T O T I C M E T H O D OF L A P L A C E
break the integral intervals
279
into two parts and / " corresponding to the and , respectively. Then
Integrating the integral l'k by parts we obtain
But
so that by use of (1) we have
Making the change of variable we obtain Hence it is clear that
or, since e is arbitrary, that
and this establishes our result. COROLLARY 2 6 . 1 . If the conditions of the theorem hold except that then
COROLLARY 2 6 . 2 .
1. 2. is non-decreasing in
If
280
I N V E R S I O N OF L A P L A C E T R A N S F O R M
[CH. VII
3.
then
COHOLLAKY 2F>.3.
Equation
(2)
of Theorem 2b may be replaced by
This is clear since the continuity of (x) on the right at x = a implies (2). It is in the form of Corollary 2b. 3 that the theorem is ordinarily stated, but we shall have need for the more general result. 3. Applications of the Laplace Method We proceed at once to apply the results of the previous section to an integral which will lead directly to the inversion formula desired. THEOREM 3 a .
If
1. 2.
for a fixed t and every larger R; i dx converges for a fixed positive c;
3.
then
Choose any positive number S and set
Then by hypothesis 2. there exists a constant M such that
Integration by parts gives
if k > tc.
In this case the above limit is zero.
The function
§3]
A P P L I C A T I O N S OF L A P L A C E M E T H O D
considered as a function of u has a maximum number less than for k sufficiently large, say for
But
Since
we have
We apply Theorem 2b to the integral
taking
We have
so that the desired hypotheses are satisfied.
and by Stirling's formula
which is what we wished to prove.
Hence
281 a Hence
282
I N V E R S I O N OF L A P L A C E T R A N S F O R M
[CH. VII
A result of a similar nature is contained in: THEOREM 3 b .
1. 2.
If
for a fixed-t and every smaller positive e; dx converges for a fixed real constant r;
3.
then
Choose a positive 5 less than t and set
Then maximum that so Integration tThe Using — By that 5 function Ik use by the for tends hypothesis of ksame atby Corollary sufficiently in to parts device brackets zero2. gives 26.2 with there as large, in for we considered l /exists the kkand say have >since proof for hence ra constant kas of>isathe k non-decreasing function . previous M Hence suchofthat theorem u in has its weonly see
§4]
UNIFORM CONVERGENCE
283
This completes the proof of the theorem. THEOREM 3 C . I f
1.
for every R > 1; converges for a fixed c > 0;
2. 3.
converges for a fixed r;
4.
then
The proof is obtained in an obvious way using Theorem 3a and Theorem 36. COROLLARY 3c. 1. If hypothesis 4 . is replaced by 4'.
the conclusion holds. Of course hypothesis 4 / implies the existence of limits on the right and on the left of at u = t. COROLLARY 3c.2. If hypothesis 4. is omitted the conclusion holds for almost all positive t. For, it is known that condition 1. implies 4. for the points t of the Lebesgue set for the function 4. Uniform Convergence In this section we extend the Laplace method of asymptotic evaluation of an integral to the case in which one function of the integrand involves a parameter. THEOREM 4 a .
If
1. h(x) is non-increasing in 2.
then (1) uniformly in If a
and
where b' < b. then
and the
284
I N V E R S I O N OF L A P L A C E T R A N S F O R M
[CH. VII
integral (1) exists by virtue of hypothesis 2. Let e be arbitrary except that Choose 5 < n] and so small that
(2) when The second inequality follows by the uniform continuity of (x). For, if
then
so that tx is in the region of continuity of Since for all x in it is clear that S may be chosen so small that (2) holds. Set
and write Ik as the sum of two integrals /,' and / " corresponding to the intervals and respectively. Then
where A is independent of t i we have
n
T
o
show its existence
(3) The right-hand side is clearly a continuous function of t in a tk t ^ if a > 0. If a = 0 the same is true, for then