Introduction to Transform Theory

4. The Prime Number Theorem

72

Lemma 2.3.

(2.7)

X

+ ( - W [ P1P2 . . .

,j.

Here the bracket indicates “ largest integer 5 and the denominators in the (k + 1)th row run through the (pk) products of the first p primes taken k at a time, For example, ”

We prove this well-known formula by induction on x. It is valid for 0 < x < 1 when each side of equation (2.7) is zero. Assume it true in the interval 0 < x < N , where N is an integer not divisible by any of the first p primes. It remains true in 0 < x < N + 1 for, as x moves into the larger interval only the first line of formula (2.7) changes; it increases by unity on both sides. Next make the same assumption when N is divisible by just m of the first p primes. Then the (k 1)th row ( k 5 m) adds (- l)k (t) as x moves into the larger interval. The remaining rows (k > m) do not change. That is, a total of

+

5

(-I)k(T)

= (1 -

1)“ = 0

k=O

is added, so that (2.7) again remains true in the larger interval. The proof is complete. Theorem 2.3. .(X)

= o(x),

x + co.

4.2. The Function m ( x )

73

Inequality (2.6) gives a relation between x(x) and np(x). Now observe that if the brackets are omitted in formula (2.7) the right-hand side becomes

Since dropping a bracket alters an individual term by at most unity and terms in the ( k + 1)th row the result is an increase of since there are at most

(0

f ();

= (1

k=l

+

1)P

= 2p.

It is true that in alternate rows the dropping of brackets decreases the sum of the row, but our statement remains true a fortiori. Hence

n(x) < p

+ +x 2p

k=t

Allowing x to become infinite we obtain

The left-hand side of (2.8) must be zero since it is 20 and is independent of p, whereas the right-hand side tends to zero as p -+ 00. Hence lim n(x)/x = 0, as we wished to prove. As observed above Theorem 2.3 must be considered a corollary of Theorem 2.2, since it essentially expresses the same result, in terms of the function n(x) inverse to en.It is rare that a theorem and its corollary describe opposite trends. But this s e e m to be the situation here, in the following sense. Theorem 2.2 describes the distribution of primes as “ rather thick ”-not so “ sparse ” for example as the square numbers ( x k - 2 < 00). On the contrary Theorem 2.3 describes the distribution as “rather thin”. Thus, the probability of picking a prime from the first n positive integers is x(n)/n, so that by Theorem 2.3 the probability of picking a grime from all the integers is zero (in a precise sense). By use of these two conflicting trends one might conjecture the prime number theorem. It can be shown that the functions

are essentially inverse functions (see Exercise 1 of this chapter).

74

4. The Prime Number Theorem

But ifp, behaved likef(n) for large n and n(x) like g(x) for large x both the above theorems would be satisfied. We shall later prove the prime number theorem, which states that

-

-

X

n log n or n(x) log x as the variable becomes infinite. pn

3. The Function 9(x) To make further progress in the study of primes we define a new function, more natural to the study than n(x) since it involvesproducts of primes rather than sums.

Definition 3.

(3-1) The notation means that ifp, ,p 2 , . ..,pmare all the primes

5 x then

(3.2) The Stieltjes integral representation (3.1) follows from the fact that ~ ( x is) a step-function with unit jumps at the primes, For example, 9(10) = 5.347. A simple numerical observation about the binomial coefficient (") now gives us an upper estimate for g(x).

Theorem 3.1.

9(x)

= O(x),

x-+ 00.

For,

(3 m!m!

(3.3)

=

(2in)! (2rn)(2rn - 1) . . . ( r n 1 . 2 . . . . in

+ 1)

The numerator on the right of (3.3) is clearly divisible by all the primes between m 1 and 2m, so that the same must be true of the integer on the left:

+

(3.4)

4.3. The Function 8 ( x )

75

Here the vertical line i s to be read "divides" or "is a factor of". But (':) < 22m since it is only one of the (positive) terms in the binomial expansion (1 -iIf m is a power of 2, m = 2", (3.4) gives

(3.5)

2"-

n

1
ks

pk

< 22".

2"

Multiplying together n inequalities (3.5), corresponding to the first n positive integers, gives

From (3.2) and (3.6)

9(2") < 2" log 4. Since 9(x) E 7 , then for 2"-' < x < 2"

$(x) ~

X

2" log4 <2 "-'< log 16.

Thus 9(x)/x is bounded above, and our theorem is proved. This result now gives an improved estimate for x(x).

Theorem 3.2.

For, from (3.1)

9(x) = n(x) log x log x

(3.7)

. ( X ) y

=

- Jlx

t

dt,

35) + -x1 Jl n(t) - dt. x t

By theorem 2.3 the integrand of (3.7), and hence the last term of (3.7), is bounded above. The same is true of 9(x)/x by Theorem 3.1, so that the result is established. A second corollary of Theorem 3.1 is that p,/(n log n) is bounded away from zero.

Theorem 3.3. p n > Bn log n,

some B > 0, n = 1, 2 , 3,. . . .

76

4. The Prime Number Theorem

For, by Theorem 3.2 there exists B > 0 such that 1 x

.(X)

Setting x

= pn, this

< --

Blogx’

x > 1.

becomes

n<--<-1 P n B log p n

Pn B log n ’

4. The Function $ ( x )

In order to obtain a lower bound for the rate of growth of n(x) we introduce a new function $(x), somewhat larger than $(x). The fact that it is larger enables us to use the same device, involving a binomial coefficient, as was used for 8(x). As we shall see, $(x) is not so much bigger than $(x) as to spoil its usefulness in estimating the rate of growth of ~ ( x ) .

Definition 4.1.

This notation means that we are to add l o g p , for each integer 5 x which is a power of ph. For example, $(lo)

=

3 log 2

+ 2 log 3 + log 5 + log 7 = log (23.32*5.7).

Note that in this example $(lo) is the log of the LCM of the first 10 positive integers. Set U(n) = LCM of 1,2, 3,. . . ,n = p l h 1 p 2 k z - * * p k m ,

where kj is the largest integer for which p ; ~5 n and where no integer I n has a larger prime factor than p , . We thus see that $(a> = log U(n>,

as in the special case n

=

10.

Let us prove next two results from elementary number theory.

4.4. The Function $ ( x )

77

Lemma 4.1. 1. 0 < x,

m

1,2,3,.

=

Set [x/m] = n. By definition of the bracket symbol we have .A

ng-
mnSx<mn+m,

I!$[

n s -Cxl
=n.

Lemma 4.2. 1. o < x

[2x] - 2[x]

=$-

For, x

<[XI

s 1.

+ 1,

[2x] 5 2x < 2[x] + 2,

[2x] - 2[x] < 2.

But an integer less than 2 cannot be greater than 1.

Lemma 4.3. 1. m = l , 2 , 3,...,

pisaprime

The highest power of p which divides m ! is

=3

m

There are m, = [m/p] multiples of p which are 5 m and their product is pi”’ml!. Repeat this statement when m is replaced by m, , thus defining m2 = [m,/p]. By Lemma 4.1 m2 = [m/p2].Repeat the statement for m 2 , defining n ~ 3and , so on. The required power will be m, + m2 m3 + . . . , as stated. All terms of the series after a certain point will be zero since p k > m for all large k. Hence we may write (4.1) as a finite series as follows:

+

H(p, m ! ) =

1 k = l

I“

Pk

7

78

4. The Prime Number Theorem

where the integer M is determined by the inequalities

We now obtain an estimate from below of the rate of growth of $(x).

Theorem 4.3.

$(x) > Bx,

some B > 0, 2 5 x < co.

We show first that the binomial coefficient (",) divides the least common multiple of all positive integers <= 2m:

(4.3)

(2m)! m!m!

Let p be any prime divisor of

(2).By Lemma 4.3

(4.4) N

I

(4.5)

C

1=N,

k=l

where

(4.6) Here we have used Lemma 4.2 and (4.2). It is to be noted that the second bracket on the right of (4.2) will be zero when k is near N , but this fact does not alter the validity of (4.5). By (4.6) p N is one of the integers 5 2m so that p N divides U(2m).Thus (4.3) is established, and hence

$(2rn) 2 m log 2. Now if 2 m - 2 < x < 2 m

$(x) $(2m - 2)2m - 2 1 -> > - log 2, X

2m-2

2m

2

m > 2.

We thus have a positive lower bound for $(x)/x valid for all x > 2, as desired.

4.4. The Function $ ( x )

79

Since the inequality pk" x used in defining $(x) is equivalent to pk 5 xl/",it is easy to relate $(x) to 9(x):

+ 9 (x"2) + S(x"3) + . . .

$(x) =$(x)

*

This series has all terms zero when

$(x) =$(x) +

(4.7)

xlik

becomes less than 2, so that

N

c

2N 5 X < 2N+1.

S(Xllk),

k=2

This relation shows that $(x) and $(x) have the same order of magnitude in the following sense.

Theorem 4.4.

$(x)

= $(x)

+ o(x),

x -+ 03.

For, from Theorem 3.1 we have (4.8)

N

N

k= 2

k=2

1 g ( x l l k ) s A C x ' i k s AN&,

some A > 0.

But by (4.7) N 5 (log x)/log 2, from which we see that the right-hand side of (4.8)= o(x) as x + co. This proves the result.

Theorem 4.5.

$(x) > Ax,

some A > 0, 2 < x < 00.

For, by Theorem 4.3 some B > 0. The term o(1) > - B/2, for example, for large x, so that the proof is easily completed.

Theorem 4.6.

$(x) = O(x),

x -+ co.

The proof is similar to the previous one, using Theorems 3.1 and 4.4. We can now apply our estimates on the growth of $(x) and $(x) to obtain corresponding ones for ~ ( x and ) pn.

80

4. The Prime Number Theorem

Theorem 4.7. .(X)

log x

= $(x)

+ o(x),

x + m.

This follows from equation (3.1) and from Theorem 2.3. By the latter we conclude that for each E > 0 t,here is a number c such that ~ ( t <) et when t > c. Hence

This implies that the integral on the left of (4.9) is o(x) as x + co,and the proof is complete.

Theorem 4.8. (4.10)

X

.(X)

> A -,

2<x<m.

someA>O,

log x

For, from Theorems 4.5 and 4.7

n(x) log x > B x

+ o(x) > B x -

3x 2

Bx 2

- = -,

some B > 0

for all large x . The conclusion is trivial for small x > 2.

Theorem 4.9. n

pn = O(n log n),

--f

00.

Replacing x by pn in (4.10) gives

(4.1 1)

n

'A P " / h PI1

log n > log A

7

+ logp, - log logp,,

log A log n 1>>l+-log P n 1% P"

log log p , IOgP" .

Allowing n to become infinite gives (4.12)

(4.13)

log n lim -= 1, n + m log P, log p , < 2 log n,

n > m, some m.

4.4. The Function + ( x )

81

From (4.11) and (4.13) 2 p n < - n log n, A

n > m.

The desired conclusion is now immediate. The conclusions of $3 and 44 may now be summarized in the following theorem of Tchebychev.

Theorem 4.10. There exist positive numbers A and B such that Bx < 8(x) < A x , Bx < +(x) < A x , BX Ax < n(x)< log x

log x ’

Bn log n < p n < An log n. While this result gives good estimates for the rate of growth of the four functions involved, it falls short of the prime number theorem, which asserts, for example, that pn n log n as n + co.The following theorem shows that there are equivalent statements involving the other three functions. N

Theorem 4.11. The existence of any one of the following limits implies that of the other three: (4.14)

lirn

n log n

= 1, lim

n ( x ) log x

=

X

1, lim

-= 1,

$(x) X

Y

In view of Theorems 4.4 and 4.7 it will be sufficient to prove the equivalence of the first two limits. Assume the second and let x + 03 through the primes. Then (4.15)

* log P. 1, lim R’m

Pn

82

4. The Prime Number Theorem

But this implies the first limit (4.14) by (4.12). Conversely, assume the first limit (4.14) and let x lie between two primes p , and p n + * ,so that (4.16)

n log Pn

-< Pn+ 1

-

.(X)

log x

Pn+ 1 5 (n + 1) log . ~

Pn

X

.

But from (4.12) and the assumed limit it is clear that the two extreme terms of (4.16) tend to unity as n 00. This gives the desired conclusion. -+

5. Five Lemmas

In the proof of the prime number theorem we shall need several preliminary results from analysis. Some of these are already proved in an earlier chapter, but for clarity we restate them here in appropriate context. Lemma 5.1.

Since t - 1 < [ t ] 5 t, we obtain upper and lower bounds for the integral (5,1), after integration by parts, as follows: 1 x - -- log 2~ < 2

When divided by x the two extremes of this continued inequality as x -+ 03, and the lemma is proved.

-+

1

The next result is the familiar Riemann-Lebesgue theorem. We assume it known for a finite range and prove it for an infinite one. Lemma 5.2. 1. f ( X ) € L ( - W ,

O3)

4.5. Five Lemmas

83

For, if R > 0

R

m

eixtf(t)

eix'f(t)dt 151,

dt-j -R

~.f(t> ldt. rl>R

Since the left side is a number independent of R and since the right side is o(1) as R -+ co,the proof is complete. The next result is a special case of Theorem 9.2, Chapter 2.

Lemma 5.3.

This is equation (9.10) of Chapter 2 when p = 2, A, o = log x. It becomes (9.11) of Chapter 2 for f ( s ) = [(s). We shall need some properties of the function [(s) essentially proved in Chapter 3.

= log

k,

+ (['(s)/{(s)),

Lemma 5.4.

*

A. f ( s ) is analytic for

CT

2 1,

B. f ( a + iY) = 0 ( b 9 IYlL uniformly in 1

s

CT

(YI

-+

5 2.

These were essentially contained in $8 of Chapter 3. As we shall see in more detail in Chapter 8 a Tauberian theorem is one which draws a conclusion about the limit of a function from a hypothesis about the limit of its average. It is just such a result which we now state.

84

4. The Prime Number Theorem

Lemma 5.5. 1.

1sx<

Xf(X)€t,

2. j x f ( t ) dt

N

ax,

x -,

1

*

x+

f(x)-a,

00;

+ co

+ co.

Let 6 be an arbitrary positive number. By hypothesis 2, (5.2)

s : f ( t ) dt

= ax

+ o(x),

x + co,

+ adx - ax + o(x) = a6x + o(x), By hypothesis 1 we have for 1 < x 2 t 5 x(l + S), xf(x) 5 t f ( t ) 5 x(1 + S)f(X + ax), X X $(x) 5 f(t ) 5 - (1 + S)f(x + Sx). t dt

= ax

But

1

x

-5-51, 1-tS-t

-

so that

s f ( t ) 5 (1

1+6-

--

+ S)f(X + 6x),

f ( t ) dt 5 Sx(1

+ 6)f(x + ax).

Now using (5.2) we obtain

li;;; f(x) X+

m

lim f(x x+ m

s a(1 + S), + 6 x ) = !& X+

m

a

f ( x )2 -

l+S'

x -, co.

4.6. Proof of the Prime Number Theorem

85

Allowing 6 to approach zero, we see that a 5 h”f(x)

s E f ( x ) 5 a,

and the lemma is proved.

6. Background and Proof of the Prime Number Theorem Before proving the main theorem let us sketch the motivating idea. By use of Euler’s product one immediately obtains a double series expansion for log {(s), 61

log [(s) = -

2 log (1 - p i S ) ,

CT

>1

k= 1

which converges, by the very way in which it was obtained, for c i > 1. Rearranging this double series as a simple series expresses log [(s), and hence its derivative, as a Dirichlet series,

where

A(k) = log p , = 0,

k

= p“

k #pn.

That is, A(k) is zero when k is not the power of a prime, is log p when k is any power of the primep. Observing that $(x) is a step-function with jump A(k) at the integer k = p m , we also see that

86

4. The Prime Number Theorem

From Theorem 9.1 of Chapter 2, x not an integer,

(6.3) If one had sufficient information s5out [(s) in the critical strip, one might hope to change the constant a in (6.3) to b < 1, thus introducing the residue x of the integrand at s = 1 : $(x) = x

--

ds.

If this new integral could be shown to be o(x) as x co,we would have $(x) x, as desired. In the original proofs by Hadamard and de la Vallee Poussin, the procedure outlined above was generally followed. Due to the paucity of information about [(s) in the critical strip the line d = b in (6.4) had to be replaced, as path of integration, by a curve having the line o = 1 as asymptote. In the modified proof (due to E. Landau, G . H. Hardy, A. E. Ingham, et al) which we shall give, there will be essentially three changes in the above procedure : A. Replace - ['I( of (1) by f(s) of Lemma 5.4, B. Replace s in the denominator of (6.3) by s2, C. Replace b by 1 in (6.4). The last alteration is possible since f ( s ) is analytic on 0 = 1. Cauchy's residue theory is replaced, as a tool, by the Riemann-Lebesgue theorem. By Theorem 4.11 the prime number theorem is equivalent to the following theorem.

-

--f

Theorem 6. $(x)-,x,

In view of equation (6.2)

By Lemma 5.3

x + co.

4.7. Further Developments

By Lemma 5.4,for 1 that

87

a 5 2 there exists a positive constant M such

By Lebesgue’s limit theorem or by uniform convergence the last integral

(6.5) approaches

as a -,1-, and the latter integral is o(1) as x -, + oc, by Lemma 5.2. Now an appeal to Lemma 5.1 and equation (6.5) shows that

(6.7)

Ji)2 log

dlC/(t) =

IX 112

t

dt = x

+ a(x),

x + 03.

Since $(t) E t, the hypotheses of Lemma 5.5 are satisfied for f ( t ) = $ ( t ) / t , and we conclude that $(x) x, as was to be proved.

-

7. Further Developments

The nonvanishing of [( 1 + iy) was an essential feature in the proof of Theorem 6 . Just how essential may be judged by the following theorem.

Theorem 7. 1. $ ( x ) ~ x ,

x+oc,=>~(l+iy)#O,

-oo
To obtain a contradiction assume that i(s) has a root of order m at so = 1 + iyo ,yo # 0. Since the logarithmic residue at so is then equal to m we have

From equation (6.2) we have, after integration by parts, that

88

4. The Prime Number Theorem

In particular, take s = a + iyo , s - so = u - 1. For an arbitrary E > 0 there exists by hypothesis 1 an R such that I $(t) - t 1 < E t when t > R . Hence

This gives a contradiction if E is chosen <m/(l + yo2)1/2; it shows in fact that the limit in question is zero rather than m. This theorem is of interest in view of a recent “ elementary ” proof of the prime number theorem (by A. Selberg and P. Erdos) which does not involve complex variables. For an exposition of the elementary proof see N. Levinson [1969, p. 2251. The logarithmic integral li(x)=lim

(~~-81$+Jl+610gt) x dt = 1 . 0 4 . * * + s 2log, dt

8+0

being asymptotically equivalent to x/log x,

log x lim .. li x = 1 , may also be used as an approximation to n(x). In fact, C . F. Gauss proposed this approximation as a result of his examination of tables of primes. It is interesting to compare these two approximations with n(x) itself for two large values of x for which n(x) is known exactly.

107 lo9

664,579 50,847,478

620,421 48,254,940

li (x)

n(4 li ( x )

664,918 50,849,235

0.999490 0.999965

Observe that n(x) < li (x) for these two values of x. Indeed this inequality holds for all x for which n ( ~ )has been computed. Yet Littlewood proved in 1914 that the reverse inequality holds for infinitely many

4.8. Summary

89

values of x. S. Skewes proved in 1937 that this reverse inequality holds for some value of x less than xg (see Skewes [1955]), xg

=

1o’O

1010~

The order of magnitude of the difference n(x) - li (x)has been studied. Littlewood proved that n(x) = li (x)

+~

( exp[ x -a(log x log log x ) ” ~ ] ) ,

x -,00

for some positive number a. If the Riemann hypothesis is assumed a better result can be proved: n(x> = li (x)

+ o(& log x),

x + co.

We conclude this section with the statement of an interesting result, conjectured by L. Euler and proved by von Mangoldt. It involves the Mobius function p(n): p(1) = 1,

p(n) = (- l)q,

p(n) = 0,

n = plpz . . .pq, ( p j are distinct primes),

otherwise.

It is an elementary result that for cr > I

It is a much deeper result that this series still converges for s = 1 and has the value zero there. It is of interest to us here because E. Landau has shown that the convergence of (7.1) at s = 1 is “equivalent to the prime number theorem. ”

8. Summary

By elementary means we have found the correct order of magnitude of n(x), the number of primes 5 x: BX < .(x)

log x

Ax
2gx
90

4. The Prime Number Theorem

where A and B are positive constants. The essential device for obtaining the upper bound (8.1) resulted from the observation that the product of all the primes between an integer rn and its double divides the binomial coefficient (im).This same binomial coefficient, in turn, is a divisor of the least common multiple of all positive integers 5 x , and this remark provided the lower bound (8.1). By relating n(x) to the zeta-function of Riemann we obtained the exact asymptotic behavior of n(x),in the sense that

-X

.(X)

X+

log x ’

co.

The essential property of (‘(s) used was that (‘(1 + iy) # 0 ,

< y
--oo

A clear and brief summary about the distribution of primes is to be found in G. H. Hardy [1929], his 1928 Gibbs lecture. EXERCISES

-

-

1. Assuming n(x) x/log x , prove log p n - p n / n . By taking logarithms of the latter relation (permissible?) show that logp, log n. Thus prove, without use of Theorem 4.10 that n(x)

--

X

log x

*p”

-

n log n.

2. Prove the converse conclusion of Lemma 5.5 without hypothesis 1 :

j: -

f ( m )= u *

f ( t )dt

UX,

x + 00.

Of course,f(x) E L in (1, R ) for every R > 1.

+

3. By consideration off(x) = 1 cos x show that the first hypothesis of Lemma 5.5 cannot be omitted. 4. Bertand’s conjecture asserts that there is a prime between n and 2n, n = 2, 3, 4,. . . . Prove it if 0.8 x log x

1.2 x log x

-< T(X) < ~,

x 2 64.

Exercises

5.

lim [U(n)]""

=

?,

U ( n ) = LCM (1, 2,

. ..,n).

n-+ m m

6.

Find ca for

U(k)e - k s . k= 1

7. Prove

c" Pk1 -log logn, -

n + 00.

k= 1

[Hint: (1

E)k log k < p k < (1

+ c)k log k,

and

may be estimated by the analogous integral.]

8. Prove

9. Prove Theorem 4.6. 10. Prove Theorem 6 by use of the integral

large k

91

This Page Intentionally Left Blank

5 The Laplace Transform

1. Introduction

We turn now to a particular transform of the type described as case A in the first sentence of Chapter 2. The kernel G(s, t ) is here CS', so that the transform is .m

and is the continuous analog of a Dirichlet series. We shall derive a set of theorems and formulas for (1.1) quite analogous to the corresponding ones in Chapter 2. However, to see the correspondence more clearly it is convenient to replace (1.1) by another,

where the integral is now a Stieltjes integral. For the student unfamiliar with such integrals it is suggested that the basic facts may be quickly learned by reading the first few sections of Chapter V, Widder [1961]. Indispensable are the existence theorem, integration by parts, and the 93

94

5. The Laplace Transform

special cases when the integrator function is a step-function or is absolutely continuous. If a(t) in (1.2) is a step-function, that transform reduces to a Dirichlet series; if

4 0 = j'+(Y)

dY,

0

(1.2) reduces to (1.1). The greater flexibility of the " Laplace-Stieltjes " integral (1.2) is thus apparent. As we shall see, it also reduces under suitable conditions to type (1.1) after integration by parts: f ( s ) = s,

Jb"

e-"a(t) dt.

2. Definitions and Examples

We begin with the following definition. Definition 2.1. The integral

I

R

m

(2.1)

f(s)

=

e-S' da(t) = lim R-m

0

e-S' da(t), 0

where a(t) is of bounded variation in 0 2 t 5 R for every positive R, is called a Laplace-Stieltjes integral. We automatically assume that ~ ( t is) a complex-valued function which satisfies the condition of the definition and that a(0) = 0 when the integral (2.1) is written. In particular, if .(t) is absolutely continuous,

then (2.1) takes the classical form f ( s )=

lom e-"4(t) dt.

We shall refer to either equation (2.1) or (2.3), as a Laplace transform. If a(t) is a suitable step-function, (2.1) reduces to a Dirichlet series.

5.2. Definitions and Examples

95

In $2 of Chapter 1 we gave a list of examples of Laplace transforms. A useful Laplace-Stieltjes integral is

where

U(0)= 0,

t > 0.

U ( t ) = 1,

Definition 2.2. The integral m

(2.4)

f(s)

=

j- e-S' dcr(t)

=

m

s

m

edS' dor(t)

0

+

m

est d [ - @ ( - t ) ] 0

is called a bilateral Laplace-Stieltjes integral. Again, if m(t) is defined by (2.2), then (2.4) takes the form

If we set t

= e-x,

(2.5) becomes a Mellin transform: f ( s )=

jOwxS-14(e-X)dx.

Examples are : m

dt =

-m

B.

.n -=j ns

c.

xs-l

-dx,

x2

0

ns 71

=

xS-le-' dx,

0

m

2 sin 2

1

m

e-S' exp( -e-')

A. T(s) =

m

+1

0<0<2.

joxS-I xn dx, X

-1 < a < l .

2 cos 2

E.

y JOmxsP1sinx

r(s)sin - =

dx,

-1 < CJ < 1.

CJ

> 0.

5. The Laplace Transform

96

71s

F. T(s) cos - =

Jorn

G. exp s2c =

e-"k(t, c) d t ,

2

0 < a < 1.

xS-' cos x d x ,

> 0, -m < a c co,

c

-m

where

See ErdClyi [1954, p. 3441, for example.

3. Convergence As might be expected from experience with Dirichlet series the region of convergence of a Laplace integral is a half-plane. Let us prove first the following result.

Lemma 3.

For, if we set B ( t ) equal to the integral (3.1) we have

= B(R) expC - (s

- s0)N

R

+ (s - so)

exp[-(s - s,)t]

B(t)

dt.

b

Since B(R) is bounded by hypothesis lim P(R) exp [ -(s - sO)R]= 0, R-+m

and

1

a > a.

,

m

(3.2)

Ibrne-"da(t) = (s - so)

b

exp[-(s - s o ) t ] P ( t ) dt

5.3. Convergence

97

if either integral converges. But jbmexP[-(s - so)tlB(t) dt

4 jbmexp[-(o

-o,)tl~ dt =

M exp[ - b(o - o,)] , c - 60

o>u(J.

The conclusion of the lemma is now immediate. Note that integration by parts of the integral on the left of (3.2), which may be only conditionally convergent, has replaced it by an absolutely convergent one. Theorem 3. (3.3)

1.

jom exp( - so t ) dct(t)

converges

Since inequality (3.1) with b = 0 is an immediate consequence of (3.3), the proof is evident. This tlleorem enables one to define and show the existence of an abscissa of convergence o,, just as for Dirichlet series. Since Theorem 3 is trivially true if convergence is replaced by absolute convergence throughout, we may also infer the existence of o, , the abscissa of absolute convergence. The two abscissas need not be equal, as the examples of Chapter 2 show. Another example of classical form is

som

e-s'e'sin e' dt =

11*

dx,

for which a, = 0, a, = 1. Corollary 3. If (3.4)

converges for s = oo + iz, and for s = o1 + iz, with converges for all s in the strip oo < o < 6,.

6, < cl,then

it

5. The Laplace Transform

98

For the bilateral Laplace integral one defines two abscissas of convergence oc’ oC”. In examples A, B, C of $2 they are (0, a), (0, 2), and (- 1, l), respectively. 4. Uniform Convergence

We show next that a Laplace integral converges uniformly in certain Stolz regions.

Theorem 4.1. If

jam e-’‘ dct(t) converges at so, it converges uniformly in St(so). Given

E

> 0, we must show that there exists R,, independent of s

in

I arg(s - so)\

a

71

< - ;o > o o 2

such that when R > Ro

We choose R, so that when t > R > R,

This is possible by Cauchy’s criterion for the existence of a limit. Clearly R, has no dependence on s in St(so). Then for o > nowe have by Lemma 3 (4.3) Since

1 s - so I /(o - go) = sec[arg(s - so)] 5 sec a in St(so), (4.2) follows from (4.3).

5.5. Formulas for uCand U ,

99

By use of Weierstrass's theorem we have at once the following theorem. Theorem 4.2. If m

f ( s >=

j-

e-s'

rc'< (T < cCN,

du(t),

W

thenf(s) is analytic there, and for p = 1,2, 3, . . .

1-

00

f'"(s) =

e-"'( - t)" dor(t),

(T'

< (T < oc".

co

Of course, this theorem applies to the unilateral Laplace integral as a special case. 5. Formulas for

uc and U.

As in the corresponding derivation of rc for Dirichlet series we prove two preliminary results.

Lemma 5.1. 1. a(t) = O(ea'),

For,

lo

t

R

KS' da(t) = a(R)e-sR

Since for

(T

--f

co

R

+s

e-''a(t) dt. 0

> CI u(R)e-'R

= O ( e - u R + a R= ) o(l),

R

+ a,

we have m

(5.1)

Jo e-'' du(t) = s

j

00

e-"u(t) dt 0

whenever either integral converges. But m

jome-''a(t) dt 4

e-"'Me"'dt, 0

some M .

100

5. The Laplace Transform

The dominant integral converges for CT > a, and the proof is complete. Note that the integration by parts used in (5.1) replaces a (conditionally) convergent integral by an absolutely convergent one. It is to be observed that condition 2 of Lemma 6.1, Chapter 2, is not needed here.

Lemma 5.2.

*

a(?) = U(e"'),

t -+00.

Set

I' e-'"

P(t) =

0 5 t < co.

da(u),

0

Then

j' da(u) = I'eaudp(u)

a(?) =

0

0

= P(t)e"'

-a

If

P(u)e"" du.

0

Since P ( t ) = 0(1),t -+ co,there exists M > 0 such that a(t)l

5 Me"' + a M j' e""du 5 2Me"'. 0

This completes the proof. Note that this lemma is false for a < 0. For, if

the integral

lo m

e-S' da(t) =

converges for

CT

> - 1. But

I

00

e-s'e-' d t

0

a(t) #

O(eat) for any negative a since

ct(co) = 1.

We can now obtain easily a formula for

CT~.

5.5. Formulas for ucand u,,

101

Theorem 5.1. 1.

I

r-tm

2. a > o

S,

m

(5.3)

=>

0, for

e-S'

da(t) is a

We prove only the case a < 00. If

E

(or +a>.

> 0, then hypothesis 1 gives t-+ 03.

a(t) = O(e(a+e)t),

By Lemma 5.1 the integral (5.3) converges for 6 > CI + E , every E > 0, and hence CT > a. Suppose next that (5.3) converged for some point to the left of the line D = a. There would then exist a point s = /3 > 0, p c a, where (5.3) would also converge, and by Lemma 5.2 a(?)= O(epr),

t

-+ co.

That is la(t) 1

s Meat,

sume M , 0 5 t < co,

The contradiction shows that (5.3) must diverge for 6 < a, as stated. The same argument, as that used in $6 of Chapter 2, shows that if 6, is known in advance to be positive, then it is given by the limit superior (5.2). If 4(t) E L in (0, R) for every R > 0, then 6, for the integral

lom dt e-"'+(t)

is given by (5.4)

6, =

1

lim - log r-tm

t

1 J+)

du

[

if either side is positive. For 6, replace 4(u) by I4(u)l in (5.4) and 1 a(?)1 by u(t) in (5.2), where u(t) is the total variation of a(x) in 0 5 x 5 t . We give without proof the formulas for 6,' and 6," of the bilateral transform (2.4):

102

5. The Laplace Transform

valid when 0 # a,.'' > a,' # 0. A consequence of Lemma 5.2 of frequent use follows.

Theorem 5.2.

the integral ( 5 . 5 ) converging absolutely

For, let so be arbitrary with oo > 0 and a. > 6,. By Lemma 5.2 (5.6) Hence

a(t) = U(exp(ao t/2))

t -+

co.

a(t) exp(- so t ) = O(exp(- a. t/2)) = o(l),

t + co,

and integration by parts gives m

f ( s o )= so

S, exp(-

so t>a(t) dt.

This integral converges absolutely by virtue of (5.6), and the proof is complete.

6. Behavior on Vertical Lines

A function represented by a Laplace integral behaves on vertical lines in much the same way as one which is the sum of a Dirichlet series.

uniformly in

crc

+ 6 5 o < 03.

5.6. Behavior on Vertical Lines

103

+

Choose so = uc (6/2). Since (6.1) converges at so there exists a constant M such that

and for any R > 0

Hence by Lemma 3 we have for u > no and any R > 0

The right-hand side of (6.2) is independent of u, so that the uniformity will follow automatically if it can be made arbitrarily small by choice ofRand (z(.But

so that the result follows immediately. This theorem may be applied in an obvious way to the bilateral Laplace transform. It is of interest to verify it for Example F of $2. By Stirling's formula (Titchmarsh [1939; p. 2591)

1 r(u+ iZ) I

-

(2x1'12

1 Z1 (2~-1)12e-~lrl12,

Hence

r(a + i z ) cos n-2 (u + ir) = O( I z I(2a-1)/2 1, But (20 - 1)/2 < 1 in 0 < u < I, as required.

IZI

+a.

104

5. The Laplace Transform

7. Inversion

The inversion formula for the Laplace-Stieltjes integral is quite analogous to the one developed in Chapter 2 for Dirichlet series. We could prove it in much the same way as before, using Theorem 6 to treat that part of the integral which is extended over R 5 t < 00, a neighborhood of infinity. To treat the remaining part, over 0 _I t 5 R, we employ some elementary Fourier analysis. Since this may be used equally well for both parts we abandon the original approach. Before recalling the Fourier theorem needed we summarize a few standard facts from analysis.

j l T d t <M,

A.

C.

f(t)

--oo

< x,

E L (- co, a)* lim R-oo

y<

1-

00,

some M .

m

f(t) sin R t df = 0.

m

D. a(t) E V (bounded variation) a 5 t 5 b ; a(t) real; .(a) = 0 u(t) = total variation of a(x) on a S x 5 t 3

= a10) - cr,(t>, 4)= alto + (m,

for some nondecreasing functions u1 and a2 both vanishing at a. The first of these facts is an immediate consequence of the convergence of the integral

Statement B is Bonnet's form of the second law of the mean; statement C is the Riemann-Lebesgue theorem (see Lemma 5.2 of Chapter 4). In the special case in which a(t) is absolutely continuous, statement D follows immediately from the decomposition

5.7. Inversion

105

By use of the above facts we can now prove two lemmas.

Lemma 7.1. 05t 5A

1. a(t) E V (bounded variation),

(7.2)

1 A sin Rt dt lim - Jo a(t)

t

*

a(O+)

=-

2

R+w

,

By virtue of (7.1) it is no restriction to assume that a(0 + ) = 0. On account of the linearity of the relation (7.2) we may also suppose first that a(t) is real and then (by D) that a(t) E By B

r.

= a(6)

s6 se+ 5 dt

5

t

A

a(t)

sinRt

dt

Oj4;26,

S

and by A

I Z(R) I 5 Ma(6) +

I

JsA

sin Rt dt

1.

Since a(t)/t E L on (6, A ) we may apply C to get

i'k II(R)I S M a ( 6 ) = 0 ( 1 ) ,

6+0+.

R-rm

Hence lim I ( R ) = 0, R+CX

as we wished to prove. The bounded variation condition on a(t) was needed only locally, as the following extension of the previous result shows.

106

5. The Laplace Transform

Lemma 7.2.

(-a< t < co);

1. M(t)EL,

2. a(t) (7.3)

1 lim R-rrnz

1-

E

V in a neighborhood of t = xo

a(t)

m

+

sin R ( x , - t) a(xO+) a(xO-) dt = xo - t 2

By hypothesis 2 a(xO+ t) integral (7.3) is equal to

E

V is - A

5 t 5 A for some A. The

(7.4) sin R t

dt

1 sin Rt dt. + -1 a(xO + t)IfILA t

The first of these integrals tends to [a(xo+)+ a(xO- ) ] / 2 by Lemma 7.1 ; the second approaches zero by C since u(xO+ t)/t E L on ( A , 00) and on (- co, - A). This completes the proof. Condition 1 is practical for our purposes but is stronger than necessary. It could clearly be replaced by a(t)/( I t I + 1) E L on (- co, GO). For, the Riemann-Lebesgue theorem could still be applied to the second integral (7.4). Note that Fourier's double integral formula (7.5)

l R lim exp(-ix, t ) dt

R-rn271

-R

I

m

exp(iyt)a(y) dy

-m

=

4xo+)

+ a(xo-) 2

is an immediate consequence of this lemma. We now prove the principal inversion formula for the LaplaceStieltjes integral.

Theorem 7.1.

5.7. Inversion

(7.6)

*

lim R-oo

1 a+iR esf a(t+) + a(t-) f(s) - ds = , 27ci "(a-iR s 2

707

t>O

= 0,

t

< 0.

By Theorem 5.2 (7.7) the integral converging absolutely on the line cr = a and uniformly (by Theorem 4.1) on any finite segment of it. Hence the integration with respect to s described in (7.6) may be performed under the integral sign (7.7) to produce

Defining a ( y ) to be 0 for y < 0, a(y)e-"YELon (- 00, co) and Lemma 7.2 is applicable to the integral (7.8), giving the desired result. A uniqueness theorem for the Laplace-Stieltjes integral is now immediate. Of course, the determining function can be altered at many points without changing the generating function; so some sort of normalization is needed. We say that a function of bounded variation a(t) is normalized, a E V*, on 0 5 t < R if and only if cc(t+) a ( t - ) a(0) = 0, a(t) = , O
+

Theorem 7.2.

(7.9)

*

0 5 t < R, every R > 0;

1.

a(t) E V*,

2.

jOm e-S' dor(t) = 0, a(t)

= 0,

cr > go, some 0 5 t < 00.

go

108

5. The Laplace Transform

For, we have only to choose the constant a of Theorem 7.1 > O and >go to see that the limit (7.6) is zero for t > 0. But the right-hand , the proof is complete. side of (7.6) is ~ ( t ) and

Corollary 7.2. 0 5 t < R, every R > 0;

1.

# ( t ) E L,

2.

lom dt = 0,

(7.10)

-

e "'#(t)

*

q5(t) = 0

0

> go, some oo

for almost all t > 0.

For, equations (7.9) and (7.10) are the same if (7.11) The present a(t) is continuous and hence normalized, and by the theorem is identically zero. Hence its derivative, which is almost everywhere equal to 9, is identically zero. This corollary is of course in constant use when problems are solved by use of tables of Laplace transforms. See Theorem 1 of Chapter 1. Let us invert next the Laplace-Lebesgue transform. Here the bilateral case can be treated as simply as the unilateral one. Note that formal differentiation of equation (7.6) would cancel the factor (l/s) from the integrand and would give # ( t ) in (7.11). The result is true under limited circumstances. Theorem 7.3.

=l

m

(7.12) 1. f ( s )

e-"q5(t) dt

converges absolutely for

0

= a;

-m

2. # ( t )

E

V in a neighborhood of to #(to+)

+ #(to-) 2

For, the integral (7.12) converges uniformly on the line segment s = a + iy, R 5 y 5 R. Hence integration under the integral sign over that range is permitted :

5.7. Inversion

f ( s ) exp s t , ds =

2rcl

109

jmCp(u) du ftiRexp s ( t o - u) ds -m

a-iR

sin R(to - u) to

-#

du.

But $(u)e-"" E L (- co < u < a)by hypothesis 1 ; so that we are in a position to apply Lemma 7.2. This gives the stated result. Contrast Theorems 7.1 and 7.3. In the former the constant a must be >O, not in the latter; in the former absolute convergence is not in the hypothesis but is there in the latter. Corollary 7.3a. 1. f ( s ) =

lom X ' - ' ~ ( X )dx converges absolutely for

2. $(x)

V in a neighborhood of xo

E

0

= a;

This is just a restatement of the theorem in the Mellin form. Corollary 7.3b. 1.

$@)EL,

(-03

< 2 < 00);

2. $(t) E I/ in a neighborhood of to; "00

(7.13)

This is the special case a = 0 of the theorem in which we have set g(y) = f ( i y ) . It gives an inversion of the Fourier transform (7.13).

Assuming the validity of transform G of $2 let us invert it as an example : t2 (7.14) exp s2c = e-S' ex+ dt, - w < < < < , c>O.

1-

m

Sl

(4nc)"2

110

5. The Laplace Transform

By Theorem 7.3 with u = 0

(7.15)

(4Ttc)"2

=2Tt

1

exp( - y2c)eiY' d y .

-00

Note that there is no need to use the Cauchy principle value of the integral since it clearly converges. But we can check (7.15) in this special case by (7.14) itself by setting s = - it/(2c) therein (renaming the variable of integration). In particular, if c = +we obtain the classic formula

k) 2

exp( -

2

1

=

p / ~ ~ i Y ' e x-p (f)d y ,

-00

< t < 03,

which states that exp( - tz/2) is its own Fourier transform except for a constant factor. 8. Convolutions

1

When two power series, ak zh and xbhZk, are multiplied together the result is a new power series C c k z k with n

A similar result holds for Laurent series, but now m

cn=

1 akbn-h,

n=O, *1,+2,

....

k=-m

Following our usual analogy with power series we would expect that the product of two Laplace integrals, f eCs'a(t) dt, e-"'b(t) dt is the Laplace integral f C S ' c ( t )dt, where

c(x) =

1; a(t)b(x- t ) dt, / a(t)b(x- t ) dt,

05x

< co,

m

c(x) =

- 00 < x < co,

--03

in the unilateral and bilateral cases, respectively. We show here that this is the case under suitable restrictions on the functions involved. We introduce the following definition and notation.

5.8. Convolutions

Definition 8. The convolution of

4 ( x ) with $(x)

17 1

is

when this integral converges. A simple change of variable shows that 4 * t,b = $ * 4. In particular, if 4 and $ vanish for t < 0, (8.1) reduces to

so that the unilateral case is a special case of the bilateral. It will result from the following theorem that o ( x )is defined almost everywhere when 4 and $ E L (-co, co).

Theorem 8. m

e-"4(t) dt converges absolutely for cr = a;

1. f ( s ) = -m m

2.

g(s) =

1

e-"$(c) dt

converges absolutely for

0=a

-m m

3

A. w(x)

=j

4(t)$(x

- t ) dt converges absolutely almost

-m

(8.3) B. f(s)g(s)=

everywhere ;

jme - s ' o ( t ) dt

converges absolutely for

0 = a.

-m

For,

since the inner integral is equal to

e-"'

Sm

e - a y \$(y)I dy

-m

and this function of t when multiplied by 4(t)E L by hypotheses 1 and 2. Hence by Fubini's theorem we may invert the order of integration in (8.4) to obtain (8.5)

jm e-ay d y jm 14(t)+(y - t ) l dt < a. -m

--m

5. The Laplace Transform

112

As a consequence, the integral defining w(x) converges absolutely almost everywhere and the absolute convergence of the integral (8.3) is proved. Finally, equation (8.3) is established by inverting the order of integration in /-mmg5(t)

dt

Im -m

e-"+(y - t ) d y

= f(s)g(s),

This concludes the proof. If a = 0, then 4 and that w(x) E Lalso.

(T

= a.

+ E L and (8.5) shows

Corollary 8. 1. f ( s ) =

IOm YW1q5(x) dx converges absolutely for

2. g(s) =

Jo

(T

= a;

(T

=a

a3

*

x"-'+(x) dx converges absolutely for x

A. o ( x ) = Jm 0

d(t)+(;)

dt

converges absolutely, almost all

x>o; m

B. .f(s)y(s) =

x S - l w ( x )dx

converges absolutely for (T = a.

0

This is the Mellin form of the "product theorem." It follows by an easy change of variable. Note the factor (lit). As an application of the product theorem let us reprove Lemma 7.2 of Chapter 3. From equation (2.1) of that chapter

Hence

I [f] 1

((s)r(s) =

xS-'

d x Iomxs-lxe-xd x .

0

By Corollary 8

1

m

c(s)r(s)=

0

X'-'o(x) dx,

(T

> 1,

5.9. Fractional Integrals

113

where

Replacing the lower limit 1 in the second integral (8.6) by 3J (leaving its value unchanged), we obtain

9. Fractional Integrals

We shall seek to obtain here the analog of Theorem 9.2 of Chapter 2. The number p of that theorem was a positive integer. Here we shall permit it to be any positive number. First we introduce the notion of fractional integrals. Recall that the nth iterated integral of a function ~ ( t may ) be put in the form

The nth derivative with respect to x of this function is ~ ( x )With . this in mind the following definition is a natural one. Definition 9. The fractional integral of E(X)of order p is (9.1)

a,(x) = D-Pa(x) = - (X

- t)"'Ct(t)

dt,

p > 0.

The fractional derivative of a(x) of order p is

(9.2)

&')(x) = DPa(x) = D[Pl+'(DP-[Pl-' a(x>>,

P < 0.

The first is called the fractional integral of order p ; the second is the fractional derivative of order p . [ p ] is the greatest integer S p .

114

5. The Laplace Transform

Note that p - [ p ] - 1 < 0 so that the function in parenthesis (9.2) is defined by (9.1). For example,

and if p

= n,

a positive integer, this reduces to xr

(r

+n

+ n)(r + n - 1) -

. ( r + I)

'

As another example

Note that if a(0) = 0, then

To obtain the desired inversion formula we need a preliminary result.

Lemma 9. 1. a(x) E

v,

0 s x < 00;

2. p > o a,(xj

E

v .c,

0 5 x < 00.

Since the operation D - p is clearly a linear one it is sufficient to assume that a(t) is real, a(0) = 0, a(t) E 7 , and to prove that a,(x) E t . C. For 6 > 0, x 2 0,

r(P + 1",(x

+ 6) - a,(x)l X+d = J" ( x + 6 - t ) P dor(t) + j"[(X + 6 - t),

- ( x - t)P] da(t).

0

The right-hand side of this equation is 2 0 since the two integrands are 2 0 and a(t) E t. Also both integrands are decreasing if p > 1 though the second is increasing if 0 < p < 1. Hence

5.9. Fractional Integrals

xt6

5

i,

da(t)

5 6P

s,

da( t ) f 6 P I’d.(

+ [(x +

6)P

s’

- xP]

da(r),

0

7 15

p>1

xtd

X+d

=6P

j

t)

0

da(t),

0
0

s 1.

In either case the right-hand side tends to zero with 6 so that a,(x +) = a,(x). In like manner we have for 0 < 6 < x

X-6

5 6”

da(t) j1-6

5 6”

da(t),

p > 1,

0

da(t) jxx-6

= 6”

+ [x” - (x - 6)P-J j +

r-’da(t) 0

po),

0 < p 5 1.

This shows that uP(x -) = a,(x) for x > 0 so that ap(x) is continuous, as stated. We can now prove the desired result. Theorem 9. d

2. a > 0 ,

a>a,,

> a,;

p >O O
By Theorem 5.2 (9.5)

116

5. The Laplace Transform

the integral converging absolutely. Now use Theorem 8.2, as applied to unilateral integrals, to multiply the integral (9.5) by the integral

a > 0.

dt,

Since ap(t) is precisely the convolution obtained from that theorem we obtain

:g

= ~ome-srap(t) dt,

a > 0,

a > a,

We may now invert this integral by use of Theorem 7.3, noting that the hypotheses thereof are here satisfied by virtue of Lemma 9. Note that a Dirichlet series,

1ak exp( - Ak s) = 1 e-S‘ dcl(t), m

f(s) =

k= 1

m

0

is included in the theorem. In this case

We thus recapture Theorem 9.2,Chapter 2, if p is integral. In particular, for the Riemann zeta-function we have

These are two formulas used in Chapter 4 but there obtained from the theory of Dirichlet series. No Cauchy value is needed for these complex integrals. They converge since [(8.2), Chapter 31

10. Analytic Behavior of Generating Functions

We prove here a result which generalizes Theorem 11.1 of Chapter 2. A generating function need have no singularity on the axis of convergence. However, if the determining function is nondecreasing the following theorem shows that there is surely a singularity on the axis.

5.10. Analytic Behavior of Generating Functions

117

Theorem 10.1. 1. f ( s ) = JOm

2. a(t) E

e-S'

t,

da(t),

u > u, > - co;

0 < t < 00

f ( s ) # A at s = u,.

3

As in Theorem 11.1, Chapter 2, we may assume u, = 0. We shall deduce a contradiction from the assumption f ( s ) E A at s = 0. This would imply f ( s ) E A in some neighborhood of s = 0 and hence that the Taylor expansion

(10.1) would converge for some real and negative value of s, s = -6. But by Theorem 4.2 equation (10.1) becomes for s = -6

And now since all factors are positive and a(t) E 7 we may interchange the symbols C, J (using the series analog of Fubini's theorem) to obtain

But this contradicts the hypotheses cr, = 0. As an example, consider formula 6 of $2, Chapter 1, 2s --" cosh t d t . Here u(t) = sinh t E f , 0, = 1, and the left-hand side is indeed singular at s = 1, as predicted. Let us prove here a simple result, frequently useful. It is that the asymptotic behavior of the generating function at co is determined by the behavior of the determining function near the origin. Theorem 10.2. .m

(10.2)

*

f ( + 00)

=

lim f(a) = a(O +).

a++m

118

5. The L aplace Transform

By Theorem 5.2

Since

1

m

ct(0 +) = o

e-"'cc(O +) dt,

0

we may assume, without loss of generality, that cc(O+) = 0. Then for any 6 > 0 and any positive go > ac f(o) = oJoe-"'a(r)dt

+o

Jd

exp[-(o - oO)t]exp(-a,t)cr(t) d t ,

(0

If(o)l 5 sup l a ( t ) l c J e - " ' d t O$rjd

0

1 exp(-a,t)(cr(t)( m

+ oexpC-(o-

a,)6]

dt,

d

lim u+ m

If(o)l 5 sup

Icr(t)l.

Ojrs6

But by our assumption the right-hand side tends to zero with 6. Since the left-hand side is independent of 6 it must be zero. One useful consequence is that a Laplace-Lebesgue transform always approaches zero as o -+ + CL, : lim Jome-.V(t) dt = 0. U+

m

Note also that if a(m) exists in (10.2) thenf(O+) = cr(c0). This is an immediate consequence of Theorem 4.1.

11. Representation

We shall obtain here two sets of sufficient conditions that are highly practical for the representation of functions as bilateral and as unilateral Laplace integrals. Unless one restricts the determining function to lie in some special class (t, B, LP, etc.) it is impractical to ask for

5.11. Representation

119

necessary and sufficient conditions. In $12 we shall prove such a theorem and in Chapter 6 another, Theorem 7. A result for the bilateral transform is usually attributed to H. Hamburger [1921, p. 2431.

Theorem 11.1. 1.

f(S)EA,

a
2. f(ao + iy) E L

3.

(--GO

< y < m) for each oo,

lim f ( o + i y ) = 0 uniformly

M

M

< a < /3

for some function +(t), the integral converging absolutely in M < a < p. By Fourier's integral formula, ( 7 3 , applicable by virtue of hypothesis 2, we have

1

R

f(go

(11.2)

+ i y ) = Rdao2n lim !-ediy*d t: j -R 1 " exp[-(o, 2lr -"

=-

j

x x :/p[(o,

e"'f(oo m

+ ir) d r

+ iy)t] d t

+ i r ) t ] f ( a o + ir) d r ,

provided that the outer integral (1 1.2) converges. We shall show that it does. Set (11.3)

We note first that this integral is indeed independent of o,as the notation suggests. By Cauchy's theorem, valid in the presence of hypothesis 1, this will be true if

120

5. The Laplace Transform

for any pair of numbers a1 and a2 in tl < CT < p. But this clearly follows from hypothesis 3. From (11.3) and hypothesis 2

If in this estimate we choose a near a we can use it to show that

converges absolutely for a > a ; if we choose a near /3 we see that -0

converges absolutely for a < p. Thus the absolute convergence of (1 1.1) is established, and we were correct in dispensing with the Cauchy value in (11.2). As an illustration consider the transform C of 92. The generating function sec (ns/2)is analytic in the strip - 1 < CT < 1 and

Hence hypotheses 2 and 3 are satisfied in any vertical strip. Thus the region of validity of formula C should be la1 < 1, as indicated in 92. The corresponding result for the unilateral transform is contained in the following theorem.

Theorem 11.2. 1. f ( S ) € A ,

a
2. f(ao + iy) E L 3.

(-00

< y < co) for each no,

c1 < g o

< 00;

lim f ( s ) = 0 Isl-+m

for some function 4(t), the integral converging absolutely in tl < CT < co. Since the inequality

5.7 1. Representation

121

+

for I IZ iy I > R and cr > a certainly implies the same inequality for 1 y > R and IZ > u, it is clear that hypothesis 3 of the present theorem implies hypothesis 3 of Theorem 11.1 for any fi > a. Hence the conclusion of the latter theorem must hold:

I

e-"+(t) dt,

f(s) --m

the integral converging absolutely for a < t~ < co. [$(t) cannot depend on fi since it is uniquely determined by formula (11.3)]. It remains only to show that $(t) vanishes for negative t. By Cauchy's theorem the integral of f(s)est over the line segment IZ = c > CI, - R 5 y 5 R , is equal to the integral of that function over n 7L the semicircumference s = c +Reie, - - 5 8 5 - * 2 2' (11.4)

e(c+iy)ff(c+ i y ) d y

i -R

exp[t(c

+ Re")]f(c + Reie)eiedo.

I

The integral on the right is in absolute value at most equal to (11.5)

Re"

sup -n/2sesa/2

1 f ( c + Re")\

J n i 2 etRcose do. -n/2

But for t < 0

By hypothesis 3 it is thus clear that the function (11.5) tends to zero when R -+ co. The same must be true of the first integral (1 1.4). Hence by formula (11.3) + ( t ) = 0 for t < 0.

122

5. The Laplace Transform

As examples consider the two function I/(? + 1) and l/(s2- 1). Both satisfy condition 3 for any CY and both belong to L on any vertical line not passing through a pole. Hence 6, in these cases must be governed by hypothesis 1, so that nc= 0 for the first and nc= 1 for the second: m 1 n > 0, = e-"sin t d t ,

lo

1

m

=

joKS*sinht d t ,

n > 1.

12. Generating Functions Analytic a t Infinity

In this section we shall restrict the determining function to be entire. In this way we shall be able to arrive at a necessary and sufficient condition for representation. Let us first recall certain facts about entire functions and introduce a notation. The order of an entire function is a real number which describes the rate of growth of its modulus as the variable approaches 00, as compared with the same rate for exponential functions. Thus the order is p forf(z) if f ( z ) = O(exp IzIP+'), I z [ + 00 for every E > 0 and for no E < 0. For example exp(z") is entire of order n, n = 0, 1, 2, . . .; cos is entire of order 1/2. For exp{ez},

&

p = 03.

If 0 < p < co the type off(z) is y if (12.1)

f ( z ) = 0 (exP(Y

+ 41ZlP)>

IZI

-+

03

for every E > 0 and for no E < 0. For example z2ez2 is of order 1 and of type 2; cos UJ; is of order 1/2 and of type la1 . Let us now introduce the notation f ( z ) E { p , y } by the following definition. (See, for example, R. P. Boas [1954, p. 181.) Definition 12. The entire function f ( z ) E { p , y ] , or has growth { P ? Y> A. f ( z ) is of order < p ,

or

0

B. f ( z ) is of order p and of type

sy.

5.12. Generating Functions Analytic at Infinity

123

Part of the usefulness of this notation stems from the fact that the class {pl, yl} is included in { p 2 , y 2 } if only p1 < p z . Thus sin 3z E (1, 3}, but also sin 32 E (2, y} for any y. If a power series expansion for f ( z ) is available there is a convenient formula for determining its growth (R. P. Boas [1954, p. 111).

Lemma 12.

These two formulas are seen to be equivalent by use of Stirling's formula. A convenient way to remember them is to recall the expansion

Since for n > 1 many powers of z are missing in the expansion lim k ( b k ( " l k= lim n k ] b , k ( ' / k k+m

k-i m

We can now obtain the desired representation criterion.

Theorem 12. (1 2.2) 0

f ( s ) E A for 1 s 1 > y and f(m) = 0.

Assume first the representation (1 2.2). Since d4t) = 0

(exp(y + 40,

t + a,

by (1 2. I), the integral (1 2.2) converges absolutely for > 0, and hence for CJ > y. If

E

IJ

> y + E , every

124

5. The Laplace Transform

then by Lemma 12 (12.3) But (12.4) provided that (12.5) Here we have used the series analog of Fubini's theorem to integrate term by term in (12.4). Inequality (12.5) follows for (i > y from (12.3) by the familiar formula for the radius of convergence of a power series. Hence equation (12.4) is valid in the half-plane (i > y. But since series (12.4) converges for Is( > y it serves to extendf(s) analytically into that region and the desired conclusion follows. It is clear from series (12.4) that f ( w )= 0. Conversely, if the series (12.4) converges for 1s I > y then (12.3) holds and the argument is reversible, to obtain the representation (12.2). Two examples are S

(12.6)

S

m

=

joeFS' cosh t d t ,

=

joe-S' cos t d t ,

o > 1,

m

(i

> 0.

In both cases the determining function has growth (1, l} and the generating function is analytic for I s 1 > 1, as predicted by the theorem. We noted above that the integral (12.2) must always converge absolutely for (i > y, but example (12.6) shows that the integral may at times converge in a larger half-plane. The Laplace integral (12.2) may thus sometimes serve as an analytic continuation for the series (12.4), just as series (12.4) always does so for the integral (12.2) under the conditions of the theorem.

5.13. The Stieltjes Transform

125

13. The Stieltjes Transform

A transform studied by T. J. Stieltjes [I8941 in connection with his investigations of infinite continued fractions is defined by

(13.1)

f(s) =

lrnfldt. 0

s+t

It is of special interest here because it arises naturally from the iteration of the Laplace transform:

Under suitable conditions on 4(t) the interchange in the order of integration may be justified, but we prefer to study the transform (13.1) directly. As always, we assume that 4 ( t ) E L on (0, R) for every R > 0 without further statement. We prove the following result.

Theorem 13. 1.

*

lrn dt, 0

2. f ( s )

some so # 0;

=smfl

df

0

*

converges,

so+t

s+t

f ( s ) is well defined and is analytic in the complex s-plane

cut along the negative real axis. Let D be an arbitrary compact region having no point in common with the negative real axis, - 00 < 5 0, z = 0. Denote by 6 the minimum distance between that segment and D and by A the maximum value for Is[,s E D. Set

Then (13.2)

a(00) = f ( s o ) , and

we have by integration by parts that

126

5. The Laplace Transform

for any s in D.In fact we show that the integral (13.2) represents an analytic function at interior points of D.If R > A

so that the integral (13.2) is dominated for, S E D, by a convergent integral independent of s. It converges uniformly in D and consequently represents an analytic function, as stated, by a familiar theorem of Weierstrass. The proof is thus complete. As an example, consider dt

n

(13.3)

3

= J-0 ( s

+ t)$'

This is easily proved for (r > 0 by iterating the familiar Laplace transform of l/$. But we see by Theorem 13 and by analytic continuation of n/& that (13.3) is valid throughout the cut plane of the theorem. The example also shows that some sort of cut as that used in the theorem was essential.

14. Inversion of the Stieltjes Transform

We give here the classical inversion of the integral (13.1), originally due to Stieltjes.

Theorem 14.1. (14.1)

1. f ( s ) =

a,

0

2. $(t) (14.2)

E

$(to)

C =

fl dt s+t at t lim

e+O+

converges for some s # 0 ; =

to > 0

f( - to - i E ) - f( - to 2ni

+ iE)

5.14. Inversion of the Stieltjes Transform

127

By Theorem 13 the quotient (14.2) is well defined, and (14.3)

=-I ( t -4(t)d t

f( - to - i z ) - f( - to + i ~ ) 2n i

1

Edt

O3

0

(t

- to)2

+

toy + E 2 '

71 0

Observe first that the theorem is true if (14.4)

E

4(t) = 1. For, then

=cot-'(+J)

-in,

&-+O+.

E2

Hence without loss of generality we may assume +(to)= 0. Now choose R > to and express the integral (14.3) as the sum of two others I1(&) and 12(&) corresponding to the two intervals (0, R) and ( R , a),respectively. Then if 0 < to - 6 < to + 6 < R,

If-iol > d

By (14.4) we have

-

lim II,(E)I5 E-0

sup IcP(t)t. It--rol>6

Consequently, I'(0-t) = 0, by hypothesis 2. The integral 12(&) need not converge absolutely. Accordingly we introduce the function

Integration by parts gives - EU(R) I 2 ( E ) = n[(R - to)2 4-E z ]

2E

+-I n:

R

4 0 ( t - to) [ ( t - to)2

+

dt,

&2]2

and 12(0+) = 0 also, provided only that the integral on the right of (14.5) converges. By Theorem 14 the integral (14.1) converges at s = 1, for example, so that the function

128

5. The Laplace Transform

is bounded on (0, co). Integration by parts gives

a(t) = P(t)(t + 1) -

f P(t) dt

= O(t),

t -+ 03.

0

Hence the convergence of the integral on the right of (14.5) is immediate, and the proof is complete. In formula (14.2) the variable s of the functionf(s) approaches -to from above and below along the vertical line CT = - t o . A more general approach is possible, as we now show. Let o(1) be a real function of E which approaches zero with E . Then it may be shown that

+

+

f(--to 4 1 ) - k)- f(-to o(1) 4-ie) lim = &(to). 2ni + The previous proof goes through, mutatis mutandis. We omit details but observe that (14.4) becomes Edt -to o(1) =cot-( & )-n, E-+O+.

(14.6)

&*O

+

Accordingly we must now prove that

4(t>dt JOrn(t- to

+ o(1))2 + E 2 - 0 ,

E j O + ,

and this is accomplished much as before. By use of (14.6) we can now prove that the approach of s to -to may be along a circular arc, a result we shall need in Chapter 9.

Corollary 14.1.

For, if we relate the variables E and q of theorem and corollary, respectively, by the equation E = to sin q, then they tend to zero together and the function o(1) of equation (14.6) becomes o(1) = to(l - cos q). This function does indeed tend to zero with E and q, and (14.6) becomes (14.7), as desired.

5.75. Summary

129

As an example let us use Corollary 14.1 to invert the transform (13.3). Equation (14.7) becomes

and we have recovered the determining function. In Chapter 9 we shall be considering in some detail the potential transform

Since it is equivalent to the Stieltjes transform

we can already obtain an inversion of it by use of Corollary 14.1. Theorem 14.2. 1. f ( s ) = -

j

71 0

2.

4(t) E C

t’(t) dt s2

converges for some s # 0;

+ t2

at t

= to

>0

As an example, consider the pair equation (14.8) becomes COSE

=

lit, f = l/s. In this case

1

lim -= E*O+ to to’ and the inversion is verified. 15. Summary

The summary at the end of Chapter 2 would serve equally well here. The main additional material involves the matter of representation.

730

5. The L aplace Transform

Let us list the basic inversion formulas and the Hamburger conditions for representation. If

then 1 27ti

(15.1)

a(t) = -

c+im

j

f(s)es' -ds.

c-im

s

If f ( s )=

jom eTSrq5(t)dt,

then 1

ctim

27Li j

(15.2)

+(t) = -

f ( s ) e s tds.

c-im

The integrals (15.1) and (15.2) may diverge, but the Cauchy principle value is always effective. Note that (15.2) follows formally from (15.1) by differentiation under the sign, as one would expect since + ( t ) is equal to the derivative of a(t) almost everywhere. The Hamburger sufficient conditions for the validity of f ( s )=

jme-"$(t)

dt

-m

with the integral converging absolutely in the strip a < 0 < p are that (1) f ( s ) E A there, (2) f ( s ) E L on each vertical line of the strip, and (3) thatf(o + iy) = o(1) uniformly in a < < p as l y l -+ co. If is replaced by + co in conditions (1) and (2), and if (3) is replaced by (3') f ( ~=) o( 1) as I s 1 + co in the half-plane 0 > a, the conditions become sufficient for the validity of

a

f ( s )=

jom e-s'+(t> dt,

0

> a.

Finally it is useful to recall that any function analytic at infinity and vanishing there is always a Laplace transform of an entire function of predictable maximum growth.

Exercises

137

EXERCISES 1. Let z and s be complex constants with positive real parts, w a complex variable. Prove by Cauchy’s theorem that the integral of e-wwz has the same value over the two paths 0 5 t < 03, (a) w = st, 0 5 t < 03. (b) w = t , Thus prove that

Re z > 0, R e s > O .

(16.1)

2. Show that the integral (16.1) converges at s = i if 0 < Re z < 1. Obtain its value from (16.1) and Theorem 4.1. Thus establish formulas E and F of $2, at least for 0 < CT < 1 . Complete the proof of formula E by analytic continuation, using Theorem 4.2. In particular check E at s = 0, s = - 1/2. 3. Prove formula B of $2 by expressing 1/(x2+ 1) as a Laplace integral and reversing the order of integration in the resulting iterated integral. Justify by Fubini’s theorem, at least for 1 < 0 < 2. Complete the proof by analytic continuation.

4. Prove formula C of $2 by change of variable in B. 5. Prove formula G of $2 by completing the square in the exponents of the integrand. 6. Use Corollary 8 to express T(s/2)T(1 - (42)) as a Mellin transform and thus get a new proof of formula B, $2. 7. Show that the capital 0 of Lemma 5.2 may be replaced by small 0 . 8. Find a Mellin transform representation for T(s) valid in -1 < a < O ; in - 2 < a < -1; ... in - n - 1 < C T -n. < [Hint: T(s) = T(s + l)/s and

9. If x, 6, p are positive constants, find

Max [(x + 6 - t)” - (x - t)”], os t jx

thus verifying inequalities (9.3).

132

5. The Laplace Transform

10. If a < 0, < 0, show Dp(D"f(x)) = D"'pf(x). What are you assuming aboutf(x)? Discuss DUD-"and D-"D".

11. From equation (2.1) of Chapter 3 show that 1

C(s)

= s Jo x s - l [ l / x ] dx,

0

> 1.

By Corollary 8 express r(s)C(s) as a Mellin transform thus producing a new proof of Lemma 7.2 of Chapter 3. o ( x ) will be o(x) = x Jlme-"[y] dy =

1

-

ex- 1 '

12. Show that f ( s ) = l/(l - sz) satisfies the hypotheses of Theorem 11.1 in - 1 < 0 < 1 and find the corresponding determining function. e - It1 Ans. d ( t ) = -. 2

13. Show that any function analytic a t co is a Laplace-Stieltjes transform. 14. Show thatf(s) = 1/(1 - s 2 ) satisfies the conditions of Theorem 12 with y = 1 and find the corresponding determining function.

Ans.

4(t)= - sinh t .

15. Find by two methods the determining function corresponding to the generating function l/[s(sz l)].

+

16. Find +(x) so that m

S

s

m=

Jo

xs-'(b(x) dx,

I 01 < 2. 4

Ans. -

n(x2

x2

+ 1)Z'

6

ReaZ Inversion Theory

1. tntroduction

In 97 of Chapter 5 we gave the classical inversion of the Laplace transform and in $14 we gave the analogous inversion of the Stieltjes transform. Both involve the determining functions for complex values of the variable. There are two more recent inversions, studied extensively by the author, [I9341 and [1938], which involve only real values of the variable. In Chapter 7 we shall develop these real inversions as special cases of a much more general theory, but in the present chapter we treat them more expeditiously, using the specific properties of the kernels involved. The chief analytic equipment needed for this direct approach is the Laplace asymptotic method, and we shall begin with a rapid development thereof. It involves the estimation of integrals of the form -6

for large values of the parameter n. It is important that the function be positive, for the positiveness enables one to alleviate the conditions to be imposed on Q(t). In our application to the inversion

g(t) should

133

134

6. Real Inversion Theory

of the Laplace transform we are thus able to relax the restrictive local hypothesis on the determining function, such as hypothesis 2 of Theorem 7.3, Chapter 5. 2. Laplace's Asymptotic Method

A very useful tool, indispensable for the work of this chapter, is Laplace's method of estimating the behavior of certain integrals for large values of the parameter. In most cases these integrals cannot be explicitly evaluated. We shall begin by giving a heuristic approach, illustrating and checking the method by use of integrals which can be evaluated. Consider an integral in which the positive parameter R (which need not be an integer) occurs as a power of one factor of the integrand: b

I, =

1enh@)4(t)dt. b

[g(t)]"$(t) a

dt =

a

It is required to obtain the asymptotic behavior of I,, as n -+ we seek a function A , such that

00.

That is,

In = 1. A, Since large values of a power get larger as the power increases and small values get smaller it would be natural to expect that the maximum value of g or of h in (a,6) would play an important r61e in the solution. This is in fact the case. If there are several points of (a, b) where the maximum is taken on we can break the integral into several parts, in each of which the maximum is at an end of the interval of integration. Let us consider a variety of such cases. In the present section let us assume that h(t) and $(t) have sufficient properties for the Taylor expansions used. Later we shall make all conditions precise. Case I h(t) E in (a, b); h'(a) < 0; +(a) # 0. Then

lim

n-rm

6.2. Laplace3 Asymptotic Method

135

To see why this is a reasonable result, note first that the first factor in the integrand of I,,is largest at t = a. Consequently we replace each factor of the integrand by the leading term in its Taylor development:

I,,

~

Ja*@hl.)

+ h ‘ @ ) V - ~ ) l + ( ~d ) t

= +(a)enh(a)

This integral can be evaluated, but its size is not greatly changed if the upper limit of integration is replaced by co. It then becomes a Laplace integral whose value is I/[-nh’(a)]. We could now establish (2.1) rigorously following the method just sketched. We shall not do so in this simple case, since there will be no application of it made later. Let us rather illustrate it by the example .1

(2.2) I,, = J (1 - t)”(2 0

+7 2 + 3t) d t = ( n +2nl)(n + 2) iyn - +

00.

Using (2.1) we have a = 0, h(0) = 0, h’(0) = - 1, +(O) and (2.2) agree.

= 2,

so that (2.1)

Case I1 h(t) E J in (a, b ) ; h’(a) < 0 ; +(a) = 0 ; +’(a) # 0. Then

+’(a)enh(a) [nh‘(a)I2 ’

n + co.

Proceeding as in Case I, we are led to the estimate

As an example take 2

(2.4)

In=

2“+2

J (2 - t)”t dt = ( n + l)(n + 2 ) 0

Here h(a) = h(0) = log 2, h’(0) = (2.3) and (2.4) agree.

-+,

2n+2 N-

n2 ’

n+m.

+(O) = 0 , +’(O) = 1, so that

136

6. Real Inversion Theory

Case I11 h(t) E

5. in (a, b ) ;

Then

- Ji

h’(a) = 0; h”(a)< 0 ; 4 ( a ) # 0.

enhca)4(a)

[-2nh”(a)]’/2’

n + co.

This is the case which we shall find most useful in subsequent work. This time we need one more term in Taylor’s expansion of h(t) and thus arrive at the estimate

and this is equal to the right-hand side of (2.4) by use of the probability integral 00 1 Jo exp( - t 2 ) dt = 2

fi.

For our illustration here we choose

4“n!n! By use of Stirling’s formula

To apply formula (2.5) we have h(t) = log (1 - t2), &(t) = 1, h(0) = 0, h’(0) = 0 , h”(0)= - 2 . Hence it gives the same result. We can now impose a set of sufficient conditions for the validity of the resuIt conjectured in Case TI1 above.

Theorem 2.1. 1. 4 ( t ) E L , a 5 t 2 b, & I + exists; ) 2. h ( t ) E c2. 1, aI t 5 b, h’(a) = 0, h”(a)< o

(2.6)

=>

Z,

4(t)dt

= Jab e”h(t)

-

~4(~+)e~~(”’[-2nh”(a)]-~’~,

n-m.

6.2. LaplaceS Asymptotic Method

137

Since we have not postulated here that $(a+) +O, (2.6) must be understood to mean that

when 4(u+) = 0. By use of a linear transformation we may assume that a = 0. Again, by multiplying (2.6) by enh(")we see that we may take h(u) = 0 without loss of generality. Let us first treat the special case $ ( t ) = 1. We must show

Ji I,, = d j Z [-2h"(0)1-'/~.

lim n-tm

For an arbitrary positive when 0 5 t 5 6

E

- 2q2 = h"(0) - E

< -h"(O) there corresponds a 6 such that h"(t) 5 h"(o) + E = - 2p2.

Here we have introduced positive numbers p and q as abbreviations. They both approach [ - h"(0)/2]* I 2 as E + 0. By Taylor's formula t2 h(t) = h"(4) -, 2

0 < 5 < t,

so that

(2-7)

5 h(t) 5 - p V ,

-932

ogtjs.

Since h(t)E 1

Ji

b

j,

enfi(')

dt

5 J S e n h ( a ) ( b - 6) = o(l),

n -+ co.

Here we have used the fact that h(6) < 0 by virtue of (2.7). Hence

f i /~exp(-nq2t2)dt + o(1) 2 ,/.I, Tiqd,xp( - t 2 ) d t 4

0

5 ,/.

/ exp(-np2t2) dt + o(1) d

0

1

Jnp6

P

O

+ o(1) 5 J n I , 2 -

exp( - t 2 ) d t

+ o(l),

138

6. Real Inversion Theory

Allowing n to become infinite we see, by use of the probability integral, that

The two inner terms of this continued inequality do not depend on while the two outer ones approach the same value as E + 0. Hence

lim

(2.8)

n+

Ji I, = ,L

E,

(-2h"(0))-"~,

00

as desired. Now in the general case, when 4(t)is not constant, it will be enough to show that lim ,/n n+m

jbenh("[4(~)- 4(0+)]

dt = 0,

0

or to assume that 4(0+) = 0 and to show that

But for any 6 between 0 and b

Hence

lim Ji

n+m

5 sup

lr$(Z)l

c,

O i t j 6

where c is the constant (2.8). Now as 6 approaches zero the right side does also, and the proof is complete. We now make several immediate extensions of the previous theorem. If the maximum value of h(t) occurs at b rather than at a, then a simple change of variable shows that

1

b

(2.9)

enh(')4(t)dt

a

-

n 4 co. J714(b-))enh(b)[-2~h"(b)]-112,

Or, if the maximum is at an intermediate point c, then

j en*(")4(t)dt - 2 J 7 1 ~ ( ~ ) e ~ ~ ( ~ ) [ - 2 n h " ( c ) ]n- " ~ , b

(2.10)

-+ 03.

a

6.2. LaplaceS Asymptotic Method

139

In many applications of the Laplace method the integral I,, is extended over an infinite range. To treat such cases we state and prove the following theorem.

Theorem 2.2. 1. Hypotheses 1 and 2 of Theorem 2.1 hold for every b > a ;

2. (2.11)

JGm

*

emh(t)1 $(t) 1 dt < co, Jbmenh(t)$(f) dt

-

some m

&$(a+)

e"h(")[-2nh"(u)]-1/2,

Again reducing to the case a = 0, $(O+) lim &I,,

=

n-t a,

lim n'm

fi

n -+ co.

we must show that

= 0,

m

e""(')$(t) dt = 0. 0

Butforn>mandb>a

Since h(b) < 0 the right side

Jb

-,0 as n

b

& I,, = n/,

enh(')4(t)dt

But the first term on the right also completes the proof.

-+

--f

co. Hence

+ o(l),

n --f co.

0 as n -+ co by Theorem 2.1. This

As an application let us prove Stirling's formula: (2.12)

n!

-

nne-n

JG,

n

--f

00.

By a familiar Laplace transform

n! = n"+' /om(e-9ydt. Apply (2.9) for the interval (0, 1) and (2.1 1) for interval (1, a).In both cases h(t) = - t + log t , $ ( t ) = 1. We may choose m = 1 since

1

m

1

e - ' t d t < co.

140

6. Real Inversion Theory

Hence h(1) = - 1, h'(1)

= 0,

h"(1) = - 1, and (2.10) becomes

(2.13)

so that (2.12) is proved.

3. Real Inversion of the Laplace Transform

There are two classical inversions of a power series, one the Cauchy method of contour integration, the other the Taylor method using derivatives. We may expect analogous methods for the Laplace transform. In the previous chapter we have already found for the Laplace integral an inversion by contour integration. We now obtain one by use of derivatives. In the Taylor method for inverting the series,

all derivatives of F at x = 0 are used. These could all be computed from the values of F(x) in a neighborhood (0,s) of the origin. To convert (3.1) to a Dirichlet series, the prototype for a Laplace transform, we replace x by e P x thus converting the neighborhood (0,s) into (log 1/6, a).Thus we would look for an operator which would utilize the values of the generating function in a neighborhood of +a and which presumably would involve all of its derivatives. We now define such an operator. Definition 3. For integers k > 0

Note that this operator will be defined for any positive y and all large k if f ( x ) E C" in a neighborhood (however small) of + 00. We shall see that as k + co it does generally invert the Laplace-Lebesgue integral.

6.3. Real Inversion of the Laplace Transform

For example, Lk. Y

[$]

=

141

k+lY7 k

(3.3) Iff(x) has an abscissa of absolute convergence o,, we shall show that the operator of Definition 3 does indeed approach the determining function as k + oo at all points of continuity thereof. Theorem 3. (3.4)

(3.5)

*

We know from Chapter 5 that differentiation under the integral sign is valid in the region of convergence. Hence

when k / y > o,. After change of variable this equation becomes

and we are in a position to apply Laplace's asymptotic method. We may choose the constant m of Theorem 2.2 as any number greater than yo,; for, then

J~'(e -

tt)m

I 4(ty>I dt c oo .

But now the computations are the same as those used in $2 for the proof of Stirling's formula. Consequently the integral (3.6) (without the exterior factor) is asymptotic to c$(y)J%e-k/Jk. This is equation (2.13). Now taking into account the factor k k + ' / k ! we , have the desired result.

142

6. Real Inversion Theory

As examples, consider l W 2 = joe - 9 d t ,

1

O~ =

0,

w e-xf

~ r= S,; J n t d t ,

o,, = 0.

It is easy to compute the limits of the functions (3.2) and (3.3) to obtain y and l/(~y)’’~,respectively, for all y > 0, as predicted by the theorem.

The results of the previous section can be greatly extended. We have postulated the absolute convergence of the given transform and the continuity of the determining function in order to keep proofs simple. However, it is known that if (3.4) converges somewhere and if +(t)E L in (0, R) for every R > 0, then (3.5) holds almost everywhere. The present inversion should be contrasted with Theorem 7.3 of Chapter 5 where the strong local condition 4(t)E V was required. That was needed on account of the intervention of the second mean-value theorem, which can be avoided here since the kernel (e-?)‘ is positive. 4. The Stieltjes Transform

As a second application of the Laplace method let us seek an inversion of the Stieltjes transform,

We shall assume that 4(t) E L in 0 5 t 5 R for every R so that the transform is defined when the limit (4.1) exists. We note first that if it exists for one positive value x = xo it does so for all positive values. See Theorem 13 of Chapter 5. Similarly, if (4.1) converges absolutely at xo > 0, it does so for all x > 0. If (4.1) converges absolutely for x > 0, then by Fubini’s theorem it is a Laplace transform of the Laplace transform of 4, W

f(x)

= Jo

e - x t dt JomC-ly4(y) d y .

We thus see from Chapter 5 that (4.1) can be differentiated as often as desired under the integral sign.

6.4. The Stieltjes Transform

143

We now introduce a Linear differential operator, somewhat analogous to that of Definition 3.

Definition 4. For integers k > 0

We shall show that this operator plays the same r61e for the Stieltjes transform as that of Definition 3 did for the Laplace. For example,

M,,,[~I= -

1 c Y 5 p

+ 1oY4p4)+ 2oY3j(3q.

This is seen to be an Euler differential operator; that is, a linear one in which each coefficient is a power of y of degree equal to the order of the corresponding derivative. This is true for all k . For arbitrary k it could be characterized as the linear differential operator of order 2k + 1 with fundamental solutions x " , n = O , + I , _+2,..., + k

and with leading coefficient (- l ) k + ' y 2 k + ' / ( k ! ) 2 .

As an example, (4.3)

M k ,Y

[?]

+

=

2(2k 1)!(2k)! 1 42k+ l(k!)4

3'

+

As another example let us apply the operator to the function ( x t ) - ' , the kernel of the Stieltjes transform. To facilitate the computation we use a simple lemma about homogeneous functions of two variables.

Lemma 4.1. 1. f ( x , t ) E

c2,

x, t > 0;

2. J(dx, A t ) = A-'f(Ax, At),

a

f(x, ax

a t t ) = - -- f ( x , t),

at x

all d > o

x,t > 0.

144

6. Reallnversion Theory

For, tf(x, t ) is homogeneous of degree zero. By Euler's theorem

and this is equivalent to the stated result.

Lemma 4.2. For x > 0 , I > 0

For,

Multiply by y 2 k + 1 / tto k obtain a homogeneous function of order - 1 and apply Lemma 4.1 k times. At each stage the factor t / y must be inserted, so that (-t)"k + I)! ak y k + l Mk3)"G] = k !k ! s ( y + t)k+2' 1

Performing the indicated differentiation now yields the stated result. We can now invert the Stieltjes transform.

Theorem 4. 1

f(x)=

jmfl dt, x+t

converges absolutely x > 0;

0

2. $(t) E C at t = y > 0

As we have noted above, differentiation under the integral sign is valid, so that Lemma 4.2 yields

(4.4)

6.5. The Hausdorff Moment Problem; Uniqueness

145

We now apply Laplace's asymptotic method, using (2.9) for the interval (0, 1) and (2.11) for (1, 00). In both cases h(t) = log t - 2 log ( t l), h(1) = - log 4, h'(1) = 0, h"(1)= - f. Thus the integral (4.4) (without the external numerical factor) is asymptotic to

+

Now using Stirling's formula we have

so that the theorem is proved. Note that the constant m of Theorem 2.2 may here be chosen as zero since

As an example take

- 1= - J 1 Ji

"

7.t

0

dt

J&+ t ) '

This may be verified by equation (4.2) with 4 ( y ) = y-'". But now it is easy to compute directly the limit of the right-hand side of (4.3) as k -+ 03 to obtain

"I

limM,,, - =k+m

J.

1

XJL7

as predicted by the theorem. 5. The Hausdorff Moment Problem; Uniqueness

In his study of methods for summing divergent series F. Hausdorff [1921] was led to inquire what sequences p, can be written as

(5.i)

1

P, =Jo t"dcc(t),

n = 0, 1,2, .. .

6. Real Inversion Theory

146

with ~ ( tE) t. The sequence it,, = 1 /(n + 1) has the property (a(r) = t ) , whereas any unbounded sequence does not. Hausdorff found a neat characterization of those that do. We present it here because of its usefulness in obtaining a representation theorem for the Laplace integral. We show first that a sequence cannot have more than one representation (5.1) by a normalized function a(t) of bounded variation. Recall the definition of V* from 47 of Chapter 5: (5.2)

a(r) =

a(0) = 0 ,

a(t+)

+ a@-) , 2

O
The uniqueness will follow from the next theorem.

Theorem 5.1. 1.

E(t)EV*,

OStZl;

2. Jol t" da(t) = 0,

*

n

= 0, I ,

u(r) = 0,

2, ..

I

0 5 t 5 1.

Two integrations by parts show that hypothesis 2 implies

(5.3)

lo1

t"p(r) dr = 0 ,

n

= 0 , 1,2,

...

where

By (5.4) P(r) is continuous together with its conjugate P(r). (We are here assuming, for increased generality, that a(r) is complex.) By Weierstrass's theorem of approximation there corresponds to every E > 0 a polynomial P(t) such that [ p ( t ) - P(r)[ < E for 0 S t 5 1 . By equations (5.3)

The left side, being independent of 6 , can only be zero if (5.5) is to hold for all E . That is, p ( t ) and P'(t) are identically zero. But B'(t) = a(r) at

6.5. The Hausdorff Moment Problem: Uniqueness

147

the dense set of points of continuity of a(t). Then for 0 < 1 < 1, cc(t+) = a(t -) = 0, because these one-sided limits are known to exist for a(t) E V and hence may be computed by use of a restricted approach through the above mentioned dense set where a(?)is already known to be zero. By (5.2) a(t) = 0 for 0 5 t < 1, and a(1) = 0 by hypothesis 2 (n = 0).

Corollary 5.1. 1.

4(t)EL,

2.

I 4(t)t”dt

OltSl;

1

= 0,

Iz = 0,1,2,

...

0

*

4(t)= 0 almost everywhere,

05t

1.

If we set

4 0 = J0k

Y ) dY,

then the continuous function a(t) vanishes identically by Theorem 5.1. But its derivative, also zero, is equal to +(t)almost everywhere. We may use Theorem 5.1 to improve Theorem 7.2 of Chapter 5. There we showed that the identical vanishing of a generating function implied the identical vanishing of the determining function. We can now draw the same conclusion if the generating function vanishes at a certain set of points in arithmetic progression. The result is usually called Lerch‘s theorem, named after M. Lerch [1903].

148

6. Real Inversion Theory

and since hypothesis 2 is really unaltered if a finite number of the zeros off at the beginning of the progression are omitted from consideration, we may assume that g o > 0, q,> gC.Then

jOm e-"'P(t) dt = 0,

(5.7)

n = 0, 1,2, . . .

But (5.7) implies

sd

(

3

t"P log-

= 0,

dt

n = 0 , 1,2,

. . .,

and by Corollary 5.1 P ( t ) is zero almost everywhere. The same is true of a(t), and from (5.6) we have f(s) = 0, as desired. Even though Theorem 5.2 is valid if the progression starts at an arbitrary point of the region of convergence of the transform (however far to the right) it must not be supposed that Theorem 5.1 is valid if the sequence of integers n of hypothesis 2 is truncated at its beginning. Thus if a(t) = U(t), the unit function, all moments pn are zero except p o . But U ( t )f 0. In this connection we call attention to a generalization due to C . Muntz [1914]: Theorem 5.2 remains true if the sequence of exponents n of hypothesis 2 is replaced by another sequence p n (not necessarily integers) such that (5.8)

po = 0,

limp,

=

co,

n+ m

1 pn =

00.

n=l

Also the Weierstrass approximation theorem can be similarly improved as follows: Any f ( x ) E C on a 5 x 5 b can be uniformly approximated there arbitrarily closely by a finite linear combination of the functions xPn,equations (5.8) holding. 6. Hausdorff's Moment Theorem

We now prove Hausdorff's [1921] main result. Define the difference operator A as

AK

= P n +I - A >

and introduce the following terminology.

6.6. Hausdorff's Moment Theorem

149

Definition 6.1. A sequence {p,): is completely monotonic (c.m.)

k , n = 0, 1,2,

(- 1)"Akpn2 0,

0

...

Examples of such sequences are 1,1, 1, ... 1,0,0,

...

1,a,a2, . . .

(6.1)

osag1.

1 1 1 , - -,... 2' 3 1 2'

# c.m. For, Azpo = - - Also

[-(n] +1 l)! " # c.m. 0

Hausdorffs theorem now reads as follows. Theorem 6. 1

(6.3)

n = 0, 1,2,

t"dm(t),

1. pn =

...;

0

2. (6.4)

0

a(t)E

t,

05 t 5 1

{p,,}; ~ c . m .

For example, the sequences (6.1) and (6.2) are 1

a" = Jo t" dU(t - a),

1 =

n+l

lot" dt, 1

and in both cases the integrator function is nondecreasing. The necessity of condition (6.4) is immediate. For, from (6.3) we have (- A)"pi,=

J

1

0

(1 - t ) kt" dct(t) >= 0.

150

6, Real Inversion Theory

To prove the sufficiency we establish first a series of lemmas, involving a " moment operator" M which we now define. It depends on a given sequence of constants { p n } z . Definition 6.2. For any polynomial

c n

P(x) =

QkXk,

k=O

n

MIP(x)l =

akpk

*

k=O

Note that M is defined so as to operate on polynomials only and that it is attached to a given sequence pn. Observe also that if it were known that the sequence had the representation (6.3) then

1 P(t) 1

M [ P ]=

dct(t).

0

Thus, after the theorem is proved, it will follow that M is a positive operator, carrying positive polynomials into positive numbers. It is clearly linear :

MCc,P,(x)

+ C,P,(Xll

= c,M[P,(xll

+ czMCpz(xll.

Lemma 6.1. 1. N[P(x)] is defined as in Definition 6.2 for M except that pk is replaced by p k + .

*

M[xP(x)] = N[P(x)].

The proof is immediate.

Lemma 6.2. M [ ( 1 - X ) ~ X " ]= (- A)'"Pn (6.5) For, if E is the translation operator, = Pn+l?

then - A = 1 - E and

6.6. HausdorffS Moment Theorem

751

Both sides of (6.5) are seen to be

Lemma 6.3.

For, by Lemma 6.2 this sum is

M [k = O

(p(1

- x)”-k]

= M{l] =Po.

For example, if n = 2, (6.6) becomes

- 2Pl

(Po

+ PLZ) + 2(Pl - P J + Pz = Po

We now define a special type of polynomial. It is the Bernstein polynomial of degree n for the function x m . It is known that this polynomial tends uniformly on 0 5 x 5 1 to x”’ as the degree becomes infinite. This property provides the motivation of our proof of Theorem 6, but no use of the property will be made therein.

Definition 6.3. Fm,,(x) = k

=

(!)m(;)xk(l ~

n

-x

y - k .

Lemma 6.4. For m,n = 1,2,3,. ..

If we note that

then Fm,,(x)- x” may be written as

n-1 k=O

k=O

152

6. Real Inversion Theory

This first sum is Fm-l Jx) - x”, so we need only check that the second is Fm-l,n-.l(x) except for the factor (1 - x) (n - l)m-l/nm-l. But this is immediate from Definition 6.3. For example, if n = 2 and m = 3, the lemma becomes (332x(1 - x)

+ x2 =

(2’ -

2x(1- x)

+ x2 -

G)‘

- (1

- x)x.

Lemma 6.5.

This lemma contains the essence of the proof. For note that if we assumed the representation (6.3), equation (6.8) would be 1

lim

1

Fm,.(t)dcr(t) =

n+ m JO

tm

da(t),

0

and this would follow from the known uniform convergence of the Bernstein polynomial Fm,, to its defining function tm. We prove the lemma by induction on m. For m = 0 the sequence on the left of (6.8) is p o for aZZ n by Lemma 6.3. Now assume (6.8) true for a fixed integer m - 1 but for every moment operator M (that is, corresponding to every sequence p,). We now prove (6.8) for m by applying the operator M to equation (6.7). On the right we obtain by Lemma 6.1.

and by the induction assumption this approaches Pm-1

- Pm-1

+Pm,

as n + 00. Here we have used the fact that the sequence corresponding to N is derived from that for M by increasing each subscript by unity. This completes the proof.

6.6. HausdorffS Moment Theorem

153

Lemma 6.6. n = 0 , 1 , ...) 0 5 x 2 1 ;

1. an(x)Ef,

2.

=-

Ict,(x)I S A ,

n = 0,1,

...,

There exists integers no < n, < n, < 05x

some A and a(x) E f,

5 1, such that lim a,,(x)

0 S x 5 1.

= a(x),

k+m

This is E. Helly's theorem and will be assumed. See D. V. Widder [1946, p. 271. We turn now to the proof of Theorem 6. By Lemma 6.2 equation (6.8) becomes

and this sum may be written as the Stieltjes integral

where a,(t) is a step-function with jumps defined as follows a,(O+) = a,(O) = 0,

an(1) -an(l-)

By Lemma 6.3 a,(l) = p o and by (6.4) a,(t)

lim

E

f. By Helly's theorem

a,,(t) = a(r)

k+ m

for a suitable a(t) E

7. Hence (6.8) becomes

If the limit and integral signs may be interchanged this becomes (6.9)

Jbol t mda(t)

as was to be proved.

= pm ,

m

= 0, 1,2,

.. .,

154

6. Real Inversion Theory

That the interchange is valid becomes evident if one integrates by parts and applies Lebesgue’s bounded convergence theorem. Note that (6.9) is trivially true for m = 0 since a,(]) = a(1) = p o . This completes the proof.

7. Bernstein’s Theorem

We now make use of Hausdorff’s result, Theorem 6 , to obtain a necessary and sufficient condition for the representation of a function as a Laplace-Stieltjes integral. The condition to be used is that the function should be completely monotonic, a property of functions analogous to that defined in 96 for sequences. Definition 7.

A function f(x) is completely monotonic (f(x) E c.m.) a < x < co o

A. f(x)ECm, B.

a<x
(- l)kf(k’(~) 2 0,

k

= 0,

If, in addition,f(a+) < 00, thenf(x)

E

1,2,

. . .,

a < x < 03.

c.m. in a 6 x < co.

As examples, the functions

are all completely monotonic in 0 < x < 03 but not in 0 5 x < co. On the other hand 1

x 1’

+

w

e-kx

c,,

k = l

are completely monotonic in 0 5 x < co. But (1 + x2)-l $ c.m. on (a, co) for any a. This will be apparent from Theorem 7. S. Bernstein’s theorem [1928, p. 561 follows.

6.7. BernsteinS Theorem

155

Theorem 7. 1. f(x)Ec.m.,

(7.1) o

~ ( x=)

a s x < co

jow e-X' dar(t),

a 5 x < co,

where a(t) E t . B on 0 2 t < co. Since f ( x + a) E c.m. on 0 5 x < co under hypothesis 1 and since j.;e-ay

d4Y) E t

when ~ ( tE)1, we may assume without loss of generality that a = 0. If equation (7.1) holds then

Jb

m

(- ~ ) ~ ( k )= (x>

f(O+)

=

j

e-xttk

da(t) 2 0 ,

o < x < co,

W

dor(t) < 00.

0

We have here used Theorems 4.1 and 4.2 of Chapter 5. Thus the necessity of hypothesis 1 is proved. To prove the sufficiency we use the following lemma.

Lemma 7. 1. f ( x ) E c.m.,

2.

0 5 x < 00;

6>0

*

{f(nS)}," E c.m.

Set g(x) = f ( d x ) . Then g ( x ) E c.m., 0 5 x < 03. Now note that - A g ( x ) = g(x) - g(x

+ 1) E c.m.,

0 5 x < co.

For, (-I)k[g'k)(x)-g'k'(x

+ I)]

= (-I)k+lg(k+l)(~),~ome~,x <x <

x+~,

by the mean-value theorem. The right-hand side is 20. Applying the result successively k times we see that (- A)kg(x) is completely monotonic and hence 20 on 0 5 x < 00 for k = 1,2,3, . . . . In particular, (- I)kAkg(n)= (- l)k A"fnc5)>= 0, k,n = 0, 1,2, . . ., as we wished to prove.

156

6. Real Inversion Theory

Now to prove the theorem choose 6 of the lemma equal to Ilm, the reciprocal of a positive integer. Hence

E

c.m. and by Haus-

dorffs theorem

(3s,'

f-

=

n = 0, 1,2, . ..

t"dcc,(t),

for some normalized nondecreasing and bounded function am(f). That is,

1

=

j0t" dcrl(t),

(rn = 1).

By Theorem 5.1 am(tl'm)= al(t). Thus equation (7.2) becomes

1 1

1

f(;)

= j o t n dcrl(t") =

t"~'"da,(t),

n = 0, 1,2, . .

0

Equation (7.3) shows that F(x) andf(x) coincide for all positive rational values of x. Since both are continuous on 0 5 x < cx), they coincide there and f ( x ) has the desired representation. It is of course evident that a(t), coming as it does from Hausdorffs representation, is bounded. Corollary 7. f ( x ) E c.m.,

where a(t) E t on 0

t < co.

a <x

< co

6.8. Bounded Determining Function

Again taking a = 0, we note that for every 6 > O.f(x 0 S x < 03. By Bernstein’s theorem co

f(x

where

a,(t) E

+ 6) = J”0

7 . B on 0

e-X‘

ciag(t),

157

+ 6) E c.m. on

o s x < co,

t < co. That is,

By Theorem 5.2, the uniqueness of a Laplace representation, a(t) must be independent of 6 in spite of appearances. Since 6 was arbitrary (7.4) must hold for x > 0. For the above examples, the function a(t) corresponding to the completely monotonic function l/x is the unbounded function t ; that corresponding to I/(x + 1) is the bounded function 1 - e-I. Since

and since 1 - cos t is not monotonic, it follows that l/(x* be completely monotonic in any neighborhood of co.

+

+ 1) cannot

8. Bounded Determining Function

An application of Bernstein’s theorem enables us to obtain necessary and sufficient conditions for the representation of a function as a Laplace-Lebesgue integral with bounded determining function.

Theorem 8.

(8.1)

o

f ( x ) = jome-x‘4(t)dt,

I4(t)I

5 M.

758

6. Real Inversion Theory

Assuming first the representation (8.1), we have for k

= 0, 1,2,

.. .

Thus the necessity of conditions 1 is proved. Conversely, from conditions 1 we see at once that

This is equivalent to saying that ( M / x ) - f ( x ) and .f(x) + ( M / x ) are completely monotonic in 0 < x < 00. By Corollary 7 there must exist nondecreasing functions P ( t ) and y ( t ) such that

By the uniqueness theorem

where a(t) = Mt - P(t) = y ( t )

(8.3)

- Mt.

Here we have considered M / x as the Laplace-Stieltjes transform of Mt. Equations (8.3) show that u ( t ) is of bounded variation and hence has a derivative 4(t) almost everywhere. But (8.3) also shows that for any 6 > -t (8.4)

-

Hence we may use Lebesgue's limit theorem to show that

6.8. Bounded Determining Function

159

A simple transformation enables us to express the left side of (8.5) as

But (8.4) shows that a ( t ) C, ~ 0 S t < 00, so that (8.6) tends to a(?) - a(0) = a([) as 6 + 0. Thus a(?) is absolutely continuous, a(t) =

i'4(Y> 0

dY

and (8.2) takes the form (8.1). The boundedness of from (8.4), and the proof is complete.

4 follows at once

Corollary 8.

m

o

f(x)=

joe - x ' $ ( t ) dt,

I4(t)l

S Me"'.

This result follows from the theorem after a translation in the variable x . As an example, take f ( x ) = e-"/x. Then

Thusf(x) must have the representation (8.1) with M e-r

-= JlmFXt dt, X

=

1. Clearly

0 < x < co,

so that $ ( t ) = 0 on (0, 1 ) and + ( t ) = 1, on (1, co) in accord with the theorem. As a second example, consider the equation

160

6. Real Inversion Theory

Since U ( t ) is not absolutely continuous the hypotheses 1 of Theorem 8 or of Corollary 8 must be violated. They clearly are, for k = 0, whatever the choice of M . 9. An Application of Bernstein's Thoerem

As an amusing consequence of Bernstein's theorem let us prove the following result of Y. Tagamlitzki [1946, p. 9401. Theorem 9.

1.

*

If(k)(~)I

Se-x,

f ( x ) = Ae-"

0Sx<00,

for some A ,

k=0,1,2, IAl

..,

5 1.

The hypothesis is equivalent to

- (- l)k(e-")(k)5 (- I)kf(k)(~) 5 (- I)k(e-x)(k), so that the functions e-x + f ( x ) are completely monotonic on 0 < x < 00. Hence by Bernstein's theorem

e-x + j ( x ) =

J

m

e-xr dy(t),

0

y(t)E

t

(0,~).

That is,

S, e-x' m

j(x) =

d [ ~ (t 1) - ~ ( t ) ]=

J

m

e-x' d [ y ( t ) - ~ ( -t I)].

0

By the uniqueness theorem U(t - 1) - P ( t ) = y ( t ) - U(t - I),

(9.1)

2 A U ( t - 1) = A/?(r) + Ay(t),

+

where A is the difference operator, Af(t)= f ( t 6) - f ( t ) . Since each term on the right is >= 0 when 6 > 0, it follows that each must be zero when the left is zero. Since U(t - 1) is constant except for a single jump at t = 1, the functions B(t) and y ( t ) must have the same property.

6.10. Completely Convex Functions

161

Thus f ( x ) = where A is the amount of the jump of U(t - 1) - P ( t ) at t = 1. That [ A ]5 1 now follows from hypothesis 1 with k = 0. This completes the proof. 10. Completely Convex Functions

The hypotheses of Bernstein's theorem involve only the signs of the successive derivatives of a function. We now prove another result of the same kind.

Definition 10. f ( x ) is completely convex on a < x < b

A. f ( x ) ~ C " , a < x < b ;

0

B.

(- l)kf'2"(~)2 0,

a < x < b.

For example, sin x is completely convex on (0, n), cos x on (- n/2, n/2). We shall prove that any such function is entire, no matter how small the defining interval (a, b) may be. More precisely, f ( x ) can be extended analytically into the complex plane so as to be entire. We prove this by use of a series of lemmas. In all of these we assume without further statement that f ( x ) E C" in a 5 x 5 b, though less continuity would obviously be needed in some of them.

Lemma 10.1. 1.

I f ( k ) ( ~ ) I j M , a s x j b , someM, k = 0 , 1 , 2, . . .

*

f ( x ) is entire.

+

For, if c = (a b)/2 we have by Taylor's series with remainder, after applying hypothesis 1 to be the remainder, that

Hence

7 62

6. Real In version Theory

But the series on the right converges for all x by hypothesis 1 and hence provides the desired analytic continuation.

Lemma 10.2. 1.

M k =rnaxIf(x)I,

k=0,1,2

asxsb

2Mo + M 2 MI5-((b b-a 2

- a).

Let c be a point where If’(x)l attains its maximum. By Taylor’s theorem (see Exercise 15 of this chapter) f(b)-f(c) f(a)-f(c)

+ f f ” ( X ) ( b - c)’, = (a - c)j-’(c) + if”( Y)(a - c)2, = (b - c)~’(c)

c

< X < b,

a

< Y < c.

Subtracting these equations, we have

( b - a)M,

s 2Mo + -21 M z [ ( b - c)’ + ( a - c)’]

1 2

2 2Mo + - M z ( b - a)’.

We have here used the fact that the bracket, considered as a function of c, takes its maximum at an end point of (a, b). This result is due to J. Hadamard. For a reference and the above proof see T. Carleman [1926, p.111.

Figure 1

6.10. Completely Convex Functions

163

Lemma 10.3. 1. f ( x ) Z O ,

agxsbb;

2. f"(x) 5 0,

a5x 5b

It is obvious graphically, from the convexity of the curve y = f ( x ) , that the triangle of the figure lies below the curve. By a comparison of areas we have the desired result. Theorem 10. 1. f ( x ) is completely convex in a < x < b f ( x ) is entire.

=>

It is no restriction to take a = 0, b = 71. Then Jonf(x)sin x d x = f ( n ) + f ( 0 ) - r f " ( x ) sin x d x , 0

- Jo f " ( x ) sin x d x 5

(10.1)

Jof( x ) sin x d x .

Since (- l ) k f ' 2 k ' ( ~is) completely convex we may apply (10.1) to obtain (- l)kj n j ( " ) ( x ) sin x d x

5 (- l)k-l j ( 2 k - 2 ) ( xsin ) x dx

0

Let 6 be a small positive number and apply Lemma 10.3 with a = 6, b=n-6, n-d

0 5 (-

f ( Z k ) (sin ~ )x dx 5 A/sin 6, d

164

6. Real Inversion Theory

That is, the even derivatives of f ( x ) satisfy the hypothesis of Lemma 10.1. That the odd ones do also follows from Lemma 10.2, and the proof is complete. T h s result was originally due to the author [1940], but the present proof is due to R. P. Boas. 11. Summary

The principal results of the present chapter involve sereval definitions and analytic formulas. We summarize them as follows:

A. The Laplace method. b Ja

enh(x)4(x) dx

-

J~4(c)enh(c)(-nh”(c))-’/2,

It

-,

co7

if h(x) has a single maximum a t c, a < c c b, where h”(c)< 0. B. The Laplace inversion operator.

C. The Stieltjes inversion operator.

D. Hausdorffs theorem (completely monotonic sequences).

lo t“ dor(t), 1

(- l)kAkpn 2 O e p n=

a(t) E t.

E. Bernstein’s theorem (completely monotonic functions).

F. Completely convex functions (- l)kf(2k)(x) >= 0 =.f(x) is entire.

Exercises

165

EXERCISES 1.

Check by expressing the integral in terms of the Gamma function and using Stirling’s formula. Ans. $(z/n)’”. 1

(3 - e‘)” dt

N

n -+

?

03.

JO

Ans. 2”“ln.

s,^ (sin

t)zn

Ans. 4.

J:(l

cos 2t dt- ?

n -+ co.

- (n/n)’”.

-sint)”dt-

?

n+m.

Ans. 2/n. 5. If h”(x)< 0 in a x b, h’(c) = 0 for a < c < b, +(x) E C in a 5 x S b except at c where +(c+) and +(c-) exist, apply the Laplace asymptotic method to

/obe”h(x)+(x) dx. -

Ans. J n

[+(c+)

+

+(c-)]e“h(C)

J - 2nh“(c)

6 . If [XI means “greatest integer S x , ”prove

kk+‘ 2 5 lim - e-k‘[t + 2]tk d t = -. k! 2

k-+m

7.

jo

If a(t) E V* in 0 S t 5 R for every R and if f(x)=

dor(t),

x > 0,

166

6. Reallnversion Theory

prove

1;

8.

en(sin x - x )

x4 d x - ?

Ans.

n+m.

6*I3r(5/3) 3,,5/3

'

9. Show that the sequence l / ( n + l)!, n = 0, 1,2, . .. is not completely monotonic.

10. Show that l/( x3

+ x) is completely monotonic on 0 < x < co.

11. Prove

12. If

compute by use of formula (2.10)

13. For a > 0 and k a positive integer prove that

ae 14. For a > 0 prove that

aet 5 eat,

0 < t < co.

Use this result and Exercise 13 to provide an illustration for Corollary 8. 15.

In the proof of Lemma 10.2 we tacitly assumed a < c < b. If that assumption was not justified, complete the proof of the lemma.

Exercises

167

16. If for I > 0

0 < x < 00, k = 0, 1,2,

I f ‘ k ’ ( ~ )5( Ike-IX, prove f(x)

= Ae-Ax,where

. . .,

I A I 5 1.

17. If for a > 0 , b > 0 If(”’(x)]S

+ b2ke-2X,

prove thatf(x) = Ae-”

+ Be-’”,

0 < x < 00, k where [ AI

= 0,

1,2,.

5 a, IBI

..,

b. 18. If g(x) is the completely monotonic sum of a Dirichlet series convergent for x > 0 and if

If‘k’(~)I

sg‘k’(X),

O < X < ~ O , k = 0 , 1 , 2 , ...,

prove that f(x) is also the sum of a Dirichlet series.

This Page Intentionally Left Blank

7

The!Convolution Transform

-

1 Introduction

The Laplace and Stieltjes transforms both take the following form, as we shall see, after exponential changes of variable,

m

This is called the convolution transform. Since we were able to invert both of those special cases of (1.1) by linear differential operators it is natural to seek an analogous inversion of (1.1) generally. In each special case it was possible to choose the operator in such a way as to produce a modified kernel G, which took the form of a power g", so that the Laplace asymptotic method was available. For more general kernels G we abandon this special tool but retain many of the features of the method. For a large class of kernels G we shall be able to discover a linear differential operator of order ri, with constant coefficients, which when applied to Swill produce a kernel G, which is everywhere positive and which tends to the Dirac 6-function as n becomes infinite, just as the powers g" did in the two special cases. We turn now to the details. 169

170

7. The Convolution Transform

2. Definitions and Examples

We introduce the transform to be studied by the following definition. Definition 2. The functionf(x) is a convolution transform of 4 ( t ) if = G(x)

(2.1)

m

*

=

j-

m

G(x - 0#(4 4

the integral converging for some x. The function G(x) is called the kernel of the transform, 4 is the determining, and f the generating function. We illustrate the definition by a number of examples. Example A. The unilateral Laplace transform ~ ( x= )

lo

e-X'

a@

dt.

This takes the form (2.1) if f ( x ) = exF(ex), G(x) = ex exp( - ex),

4(t) = cD(e-'). Example B. The Stieltjes transform

Here

f ( x ) = F(ex),

1 G(x) = ex+ I '

4(r) = cD(e').

Example C.

lX m

f ( x ) = ex

e - ' 4 ( t ) dr.

This transform already has the form (2.1) without any change of variable if G ( t ) = e' (- co,0), G(t) = 0 (0, 03).

7.3. Operational Calculus

171

Example D. F(x) =

-5

2 " 71 0

t ~

x2

+ t2 @(t) dt.

This might be called the potential transform, since it is clearly related to the Poisson integral representation of a function which is harmonic in a half-plane. We again need an exponential change of variable to attain the convolution form : f ( x ) = F(e-"),

2 G(x) = -

1 ~

nI

+ e-2x'

4(t) = a?(e-').

3. Operational Calculus To predict an inversion formula for the convolution transform it is useful to apply the notions of operational calculus. In it we treat some symbol, such as D for differentiation, as if it were a number throughout a series of calculations and then at the end we give it its original meaning. The justification of such a procedure lies in the algebra of the operator used. We shall not be concerned with the justification of the calculus but use it heuristically only. Definition 3.1. eaDf(x)= f ( x

+ a).

If D were a number we should have eaD=

C akDk k!. k=O

If D kis now interpreted as a derivative, we have m

e a D f ( x )=

k (k)

Ca - f(x + a) k=o k!

by Maclaurin's series, at least if f ( x ) is analytic. But note that the definition is taken to apply to any function.

172

7. The Convolution Transform

We can use Definition 3.1 to obtain the interpretation as an operator of any function of D which has a Laplace integral representation. For example, if

then a suitable interpretation of l/E(D) should be

G ( t ) 4 ( x - t ) dr.

G(t)[edtD4(x)]dt =

= J

--m

J-a3

That is,

We are thus led to a symbolic inversion of the convolution transform. On account of equation (3.1) we make the following definition.

Definition 3.2. The reciprocal of the bilateral Laplace transform of the kernel of a convolution transform is called the inversion function of the convolution transform.

Let us illustrate by Example C above. Here

so that (3.1) becomes m

(3.3)

(1 - 0 ) e . J

e-'4(r) dt = 4(x).

X

But this may be checked directly by differentiation if +(x) E C, for example. Indeed if $(x) E L in 0 5 x 6 R for every R > 0 and if the integral (3.3) converges for x > 0, then equation (3.3) holds for almost all positive x. This conclusion is typical of the general case, though we shall not here attempt proofs for such generality.

7.4. The Laguerre-Pblya Class

173

4. The Laguerre-Ptjlya Class The conclusion suggested by the heuristic developments of the previous section can be proved rigorously when the inversion function E(s) belongs to a large class of entire functions studied earlier by E. Laguerre [1882, p. 8281 and G. P6lya [1913, p. 2791 in another connection. We shall call the class E.

Definition 4.1. An entire function E(s) belongs to class E

n m

(4.1) o E(s) = A exp(bs - cs2)

S”

k= 1

where A , b, c, a, are real, c 2 0, n is a nonnegative integer, and (4.2) We shall assume known the elements of the Weierstrass theory of primary factors. The convergence of the series (4.2) guarantees that the product (4.1) should converge and represent an entire function. Laguerre proved, and Polya added a refinement, that a sequence of polynomials, having real roots only, which converge uniformly in every compact set of the complex $-plane, approaches a function of class E. Conversely, every function of class E is the uniform limit of such a sequence. For example,

and the polynomials (1 - ~ ~ / / k have ~ ) ~ *real roots only. This example helps one see why the constant c of the definition must be 20. In the definition we do not assume that the a, are distinct,, as indeed they were not in the above example. Moreover, we wish to include the case in which the product is a finite one. We may assume, without loss of generality, that the roots ak are arranged in an order of increasing absolute value,

7. The Convolution Transform

174

Examples of functions of E are

(4.3)

sin s coss, -

1, (1 - s), exp(-s*),

s

1

1

- ’ r(i - s)’ r(s)‘

We also introduce a notation E, for a subset of E as follows. Definition 4.2.

(4.4)

-

E(s) E Eo m

where ‘b and a, are real and finite, and “

1

We have thus omitted the cases c > 0 and those for which E(s) has only a finite number of roots. Moreover, now E(0) = 1. Thus the first three functions (4.3) and the last are not members of E,; the others are. Even though the whole class E is composed of functions that can be inversion functions of convolution transforms we restrict attention here to the smaller class E, for brevity of exposition. It is not the property of the class, originally used by Laguerre to define it, that we shall need, but other properties which we shall now develop. 5. Some Statistical Terms

Although we shall make no use of any facts from the theory of statistics it is advantageous to employ a few terms therefrom. A frequency function p ( x ) is any function 2 0 on - 00 < x < co for which

1- p(x) dx m

=

m

The mean of a frequency functionp(x) is

when the number is finite.

1.

7.6. Properties of Laguerre-P6lya Kernels

175

The variance of a frequency function p ( x ) of mean M, is

v, = J

m

(x - M,)2 p ( x ) dx.

-m

For example, if p ( x ) = k(x, t ) =

exp( - x2/4t) (4741’2

then p ( x ) is a frequency function for each positive value of the parameter t with mean 0 and variance 2. These facts are easily derived from formula G, $2 of Chapter 5 , (4.5)

exp(ts2) =

S

m

e-”/c(y, t ) dy,

t > 0.

-m

It is sometimes helpful to think of p(x) as defining the manner in which a unit mass is spread over the x axis. Then Mp is the center of gravity of this linear bar and V, is its moment of inertia about the center of gravity. The more the mass is concentrated near this “fulcrum” the smaller will be the bar’s moment of inertia. Thus the variance measures the tendency of the mass to lie away from or to vary from the mean. 6. Properties of the Laguerre-P6lya Kernels

We shall call any function G(x) which has a bilateral Laplace transform whose reciprocal lies in E a Laguerre-Polya kernel. We now develop some of the properties of such kernels. Theorem 6.1.

E(s) E El3 l E ( u + i y ) l z IE(a)l,

For, if E(s) is defined by (4.4), then

-cocou
176

7. The Convolution Transform

Theorem 6.2. 1. E ( s ) E E o ;

2. R > O , (6.1)

1

=-

uniformly in

p>O

E(a

+ iy)

I CT I 5 R.

That is, l/E(s) is uniformly small in any vertical strip of the complex s-plane, smaller than any power of l/lyl. This theorem would not be true for functions of the whole class E, as one sees by observing the first two functions (4.3). To prove it choose N > p and so large that I I >= R when k > N . This is possible since I ukI + co as k + co. Set

By Theorem 6.1

Since ebeI E,(D) I has no roots and is continuous in I a1 S R it has a positive lower bound there. Hence for a suitable constant M

This proves the result.

Theorem 6.3. 1. E(s) E E,

for-somefunction G(t). Here a, is a root of E(s) nearest the origin. That is, the reciprocal of every function E(s) of E,, is a bilateral Laplace transform which converges in a strip symmetric about the imaginary axis and extending out to the nearest zero of E(s). To prove

7.6. Properties of Laguerre-P6lya Kernels

177

this we have only to apply Theorem 11.1, Chapter 5. By Theorem 6.2 withp = 2weseethat 1/E(o iy) E L ,-a < y < co foreachoin 1 0 1 < la, 1 and that 1/E(a iy) tends uniformly to zero in that strip. It is clearly analytic there, so that all conditions of Hamburger’s theorem are satisfied. We shall now prove that the function G(t) of Theorem 6.3 is a frequency function and compute its mean and variance.

+

+

Theorem 6.4.

=>

A. G ( t ) is a frequency function; B. M G = b ;

In defining G(t) by equation (6.3) we have inverted the integral (6.2) by Theorem 7.3 of Chapter 5 . It is clear from Theorem 6.2 that the integral (6.3) converges absolutely for all t. Indeed, it defines G(t) as a function of class C“ by virtue of the relations (6.1) .To prove conclusion B we differentiate equation (6.2) with respect to s and set s = 0 ,

E’(o) - E’(0) = --

E*(O)

I’

a

tG(t) d t .

-m

But from the definition of E(s) we have by logarithmic differentiation

The validity of this step and the convergence of the series (6.4) derive from the general Weierstrass theory which we are assuming. Setting s = 0 in (6.4) gives B. Set F(s) = e“/E(s) and compute F”(0)from the integral (6.2),

1

m

F”(0)=

-03

( t - b)’G(t) dt.

178

7. The Convolution Transform

On the other hand 1

ak

again by logarithmic differentiation. In particular, F'(0) = 0. A second differentiation gives FF" - (F')' FZ

=f-k = 1 (s

1

- ak)'.

For s = 0

and conclusion C is established. Finally, to prove A set s = 0 in (6.2),obtaining 1=

j

m

G(t) dr. -m

It remains only to show that G(t) 2 0 , and we shall do this by a limit process. We set

and show first that

where g,(r) is a frequency function. If g,,(t) + G(t) when n --t co, as it is reasonable to expect since P,(s) -P E(s), then G(t)will also be a frequency function. Clearly

7.7. Inversion

179

By use of Theorem 8 of Chapter 5 we can now compute the function g,(t) of equation (6.5). Set

do = la1 Ig(a1t - a1b - 0, g2(t) = I a2 I&, t - 1) * Sl(t), .

.

. .

. .

.

.

.

gn(t)= IanIg(ant-1) * g n - l ( t > *

Then equation (6.5) holds with this function g,(t), and since g(t) 2 0 it is clear that gn(t)2 0. The integral converges absolutely for I <

CI

lull.

Inverting the integral (6.5) gives

This integral converges absolutely when n

>= 2 since

Inequality (6.7) also enables us to apply Lebesgue’s limit theorem to take the limit under the integral sign in (6.6), since

Thus 1

lim g,(t) n-+m

=-

2ni

im

,sr

__ ds = G ( t ) ,

E(s)

-im

and G(t)2 0, as we wished to prove. The basic fact used in the latter part of this proof is that the convolution of two frequency functions is again a frequency function. 7 . Inversion

We are now in a position to invert a n y convolution transform whose kernel has a bilateral Laplace transform whose reciprocal is in E , . We define the inversion operator E ( D ) in the following natural way.

180

7. The Convolution Transform

Definition 7. If E(s) E E, , m

k= 1

then

E ( D ) f ( x ) = lim n+m

fi (1

+b +

-")J(x

k= 1

k=l

ak

L).

a,

The operator E ( D ) may be described accordingly as a linear differential operator with constant coefficients, of infinite order, plus certain translations. I n particular the translations can be accumulated into a single one through theamount b + x y a;' in the special case for which that series converges.

Theorem 7.1.

n m

1. E ( s ) E E , ,

E(s)=eb"

k= 1

1 2. G ( f ) = - J

2x1

3. 4(t)EB.C,

im

est

-ds,

-co
E(s)

-im

-a

4. f ( x ) = G * 4 =

1

< t < 00;

m

G(x - t ) $ ( t ) dt

-03

*

E ( D ) f ( x ) = $(x),

- co

< x < co.

As in $6 we define

so that E(s) = P,(s)E,(s). We begin by applying the operator P,(D) to the kernel G(x). Set (7.1)

The functions G,(t) are Laguerre-P6lya kernels and have all the properties described in Theorem 6.4 since the functions En($)clearly are

7.7. Inversion

181

members of E,, . The integrals (7.1) converge absolutely for - co < t < co by Theorem 6.2. Since formal differentiation gives 1

DkG(t)= 2ni

1

im

estSk

-ds

-im

E(s)

and since

I 2ni 1

eoDG(f)= G(t

+ a) = -

im

-im

esteUs

-ds,

E(s)

we have the formal result

P f l ( W ( r )= G,(f). To justify equation (7.2) we have only to apply the classical theorem on differentiation under an integral sign by showing that the resulting integral (7.2) converges uniformly on every compact set of the t axis. Actually it does so over the entire axis since

The dominant integral converges by Theorem 6.2. We next apply the operator P,(D) to f ( x ) : (7.3) P,(D)f(x) = P,(D)

J

m

G(x - t)4(t)d t

=J

m

-m

--a0

G,(x

- t)+(t) d t

provided again that the resulting integral is uniformly convergent on compact sets of the x axis, say on 1x1 5 A for arbitrary A . If I $(t) I < M , -00 < t < 00, then

1

J:IG,(x

- t)6(t)d t

G,(t) d t

5M

\

*

m

-A+R

G,(t) dt, 1x1

IA.

The dominant integral is independent of x and is small for large R,so that the integral -m

182

7. The Convolution Transform

does converge uniformly on 1x1 5 A . A similar proof applies to the integral extended over (0, a).Thus equation (7.3) is valid. Finally we must show that P,(D)f(x) + 4(x) as i 7 -+ co,or that (7.4)

lim J m G,(x - t)[4(t)- 4(x)] d t n-+m

-m

Here we have used the fact that C,(x) is itself a frequency function and has all the properties in the conclusion of Theorem 6.4. For an arbitrary 6 > 0 we break the integral (7.4) into two parts corresponding to the ranges of integration 1 t 1 < 6 and 1 t 1 2 6, denoting the parts by I,, and J, ,respectively. Again by Theorem 6.4 we have for a fixed x

1

6

(7.5)

IInI

5 m a x ~4(x - t ) - ~ ( x ) I If166

Gn(t) d t

-6

Since It I 2 6 * 1 S t 2 / S 2 we have

Here we have used the fact that the mean of Cn(t)is zero and have computed its variance by Theorem 6.4. From (7.5) and (7.6) it follows that (7.7)

-

lim I Z , + J , ( S m a x l + ( x - t ) - 4 ( x ) l .

n-+m

11166

Since this limit superior is independent of 6 and since the right-hand side of (7.7) 4 0 as 6 -+ 0, the left side must also be zero and equation (7.4) is proved. It is apparent from the proof that the continuity of 4(x) is used only at the point x where the inversion of the transform is to be accomplished. In fact, the theorem can be improved drastically by removing all local

7.8. The Laplace Transform as a Convolution

183

conditions on 4 ( x ) and requiring only that the integral G * 4 should converge. The conclusion E(D)f(x)= 4(x) is still true for almost all x . The situation is in marked contrast to that involved in Theorem 7.3, Chapter 5, for example, where some strong local condition like bounded variation is required. It is the positiveness of the kernels G,(t) that makes the difference. See I. I. Hirschman and D. V. Widder [1955, p. 1391. In one of our applications we shall need a slightly more general result than that contained in Theorem 7.1. If E(D) is replaced by limn+a,exp(E,D)P,,(D), where E, + O as n -P 00, the new operator still acomplishes the inversion.

-

Theorem 7.2.

1.

En

= o(l),

n-PO3;

2. All conditions of Theorem 7.1

*

lim exp(e,D) P,(D)f(x) = 4(x), n+

- 03

Since exp(e,D) means a translation through that .m

(7.8)

< x < 00.

W

lim n+m

J

C,(t)$(x - t -m

+

E,)

E,

we now must prove

d t = 4(x).

As before, we break the integral (7.8) into two parts corresponding to 1 t 1 < 6 and I t ( >= 6. Inequality (7.5) now becomes

I 1, I 5

max I4(x - t + E,) - 4(x) I. I f 1 58

By the continuity of $(t) we have

lim n+ a,

11,,] max

It156

I~(x

- t ) - $(x)l,

and the rest of the proof follows as before. 8. The Laplace Transform as a Convolution

We now apply the general inversion formula of $7 to the special case of Example A of $2, thus recapturing the inversion of the Laplace transform given in Chapter 6. We need a preliminary result involving linear differential operators.

184

7. The Convolution Transform

Theorem 8.1. 1.

F(X)E

C",

0 < x < co

Hence if we operate successively, in the order k = 1,2, 3,. .. we obtain the result at once. Since the coefficients of the differential operator are constants the order of applying the factors is not involved, is only important for quick computation. For B, simple calculation gives for k = r t , n - 1, ..., 1,

This time it is convenient to apply the separate factors of B in the opposite order, k = 12, n - 1, . . . ,1. Conclusion B results at once. We now show that the Laplace transform, Example A of $2, may be inverted as a special case of Theorem 7.2. As noted in 52, the transform (2.2) is equivalent to (2.1) with kernel G(x) = e" exp( -ex).

We show first that this is a Laguerre-P6lya kernel:

1

W

-a,

00

e-"e'exp(-e')

e - Y s dt = r(l - s).

dt = 0

The familiar product expansion of T(s) is

7.8. The Laplace Transform as a Convolution

185

From the identity

we obtain

Here y is Euler's constant, y=lim n-+m

[I " -1- l o g n ] . k=lk

Clearly l/r(l- s) E E,,. The constants of Definition 4.2 are b ak = k,

We apply Theorem 7.2 with

Then

E,

4 0 as n -+ co by (8.1) and exp(E,D)P,(D) = no

fi (1 -;).

k= 1

Of course nDf(ex) = eD'og n f(e") =f(nex).

Using the notation of Example A of $2, Theorem 7.2 gives

exF(ex)=
k=l

= -7,

7. The Convolution Transform

186

By Theorem 8.1, conclusion A, we have ."t 1

(8.2)

lim exp(&,D)P,(D)e"F(e") fl-

=

m

lim ( - 1)" -ecn+')"F(")(nex) n- m n!

= cD(e-").

Setting e-"

=y

equation (8.2) becomes

We have thus proved

Theorem 8.2. 1. @(t)eB.C, 2. F(x)

O
00;

=I e-x'@(t) dt n

Except for an unimportant difference in hypothesis this is Theorem 3 of Chapter 6 . 9. The Stieltjes Transform as a Convolution We turn next to Example B of $2. If we assume only that @(t)E B.C, as in Theorem 7.1, the Stieltjes transform need not converge. For example if @ ( t ) = 1, F(x) is undefined. However, to apply Theorem 7.1 we have only to make a preliminary differentiation. Thus we have

and this assumes the form (2.1) if we set f ( x ) = -exF'(ex), G(x) = ex(ex + l)-',

+ ( t ) = @(e').

For the inversion function E(s) we have the equation

1 E(s) -=

m

e-St

ei

J-a,(et + 1)'

m

dt=J

x-s ~

n

(X

+

71.5

dx=-. sin rcs 1)'

7.9. The Stieltjes Transform as a Convolution

187

Hence the symbolic statement of Theorem 7.1 becomes - sin 71 D e"F'(e") = @(ex). 7 1 0

This symbolic statement, interpreted as usual by the infinite product expansion of the inversion function, becomes

But by conclusion A of Theorem 8.1

In order to apply conclusion B of Theorem 8.1 we temporarily introduce a new function H by the equation

Then

Now replace ex by a new variable y and eliminate H between equations (9.3) and (9.4). We obtain

where Dy stands for differentiation with respect t o y . In the notation of Definition 4 of Chapter 6 equation (9.2) becomes lim Mn,y[F(~)l= @(Y),

n-t

03

and we are led to the following theorem. Theorem 9.1. 1. @(t)€B.C,

O
188

7. The Convolution Transform

Note that we have now postulated the convergence of the given Stieltjes transform. As noted above, hypothesis 1 alone does not produce it. The only part of the theorem left unproved is that equation (9.1) follows from the hypotheses 1 and 2. But for 0 < 6 =< x < co

so that the integral (9.1) converges uniformly in 6 5 x < co, and consequently differentiation under the integral sign (9.5) is permissible. and t by we obtain If in Example D of $2 we replace x by

&

fi

so that the potential transform is another form of the Stieltjes transform. We are thus led to the following result.

Theorem 9.2.

1. @(t)€B.C, 2. F ( x )

=-I 2 n

0

O
x2

+ t Zdt

converges,

0 < x < 00,

An example is provided by the pair @ = l/t, F = l/x. In this case the differentiations may be completed explicitly, and equation (9.6) becomes lim n+m

(2n

+ 1)!(2n)!

(n!)424n+1

1

Jr;=&'

1 o<x<0O,

and this may be verified by use of Stirling's formula. In general the calculations involved in formula (9.6) are not feasible and other methods of inversion are more useful. See Chapter 9 and Theorem 14.2 of Chapter 5.

Exercises

189

10. Summary

The inversion function E(s) for the convolution transform (10.1)

f ( x ) = Jm C(x - t)4(t)dt -m

is the reciprocal of the bilateral Laplace transform of the kernel G, (10.2)

1

-=

E(s)

J’

a,

eCsfG(t)dt.

-m

That is,

E ( D ) f ( x )= 4 ( x ) , (10.3) where D stands for differentiation with respect to x. The symbolic equation (10.3) can be realized practically for a large class of kernels G. These derive by equation (10.2) from functions E(s) which are in the Laguerre-P6lya class, that is, those functions which are uniform limits of polynomials with real roots only. We have here treated a subclass, those functions E(s) which have an expansion a,

m

1 l/a,2 < co. k= 1

Then equation (10.3) is interpreted to mean

where eaDf(x)= f ( x + a). The Laplace transform has inversion function E(s) = l / r ( l - s) and the Stieltjes transform, E(s) = (sin m)/(m).In each case equation (10.4) recaptures the inversion of Chapter 6.

EXERCISES 1. If g ( x ) = ex(x < 0),= 0 (x 2 0), show that g * g = I X I ex ( x < 0), = O ( x 2 0). If 4 E B.C on ( - co, co)show that -co < x < a). (1 - 0 ) 2 g * g * 6 = 4,

190

7. The Convolution Transform

2. Generalize the previous exercise.

3. If h(x) = e-1"1/2 and

4 E B.C on (- co, 00)

(1- D Z ) h * $ = f $ ,

show that

-m<x
4. If h(x) is the function of Exercise 3, compute h * h and prove that (1 - D2)2 h * h * 5.

4 = 4,

- co < x < 00.

Using Definition 3.2 compute the inversion functions for the kernels of Exercises 1 and 2, designating the regions of convergence of the Laplace integrals involved.

6. Same problems for Exercises 3 and 4. 7. If g(x) and h(x) are the functions of Exercises 1 and 3 compute g * h and the corresponding inversion function.

Ans. e-lxl(lxl - x

+ 1)/4.

8. By differentiating equation (4.5) (or otherwise) show that

1;

m

y 2 k ( y , t ) d y = 24

:1

y4k(y, t ) d y = 12t2,

t > 0,

00

k(x - y , t)y4 d y = x4

+ 12x2t + 12t2,

t > 0.

9. If for any real t we define

(11.1)

exp(tP)f(x) =

1 ktk. f ( Z k ) ( ~ ) ,

k=o

show that exp(t~')x" = k(x, t ) * x",

t > 0.

10. Using Definition 3.2 show that the inversion function for the kernel k(x, to), to.> 0, is exp( - tos2).Using (1 1.1) prove exp( - tDz)(k(x, t ) * x")

= x",

as the symbolic equation (3.1) would predict.

Exercises

11. Generalize exercises 8, 9, 10:

exp( - t ~ ~ ) ( k t( )x*,x") = x". 12. Using (10.1) show that for any polynomial P(x)

exp(tP)P(x) = k(x, t ) * P(x),

t > 0.

[Hint: Expand P(x - y ) in Maclaurin's series.] 13.

If 4(t)E B.C on (0,a),p is a positive integer, and

show that

Check for the case $(t) = 1.

79 1

This Page Intentionally Left Blank

8

Tauberian Theorems

1. Introduction

We recall that a series

is sermmable (A), in the sense of Abel, to the number A if the power series m

f(x)=

akXk

k= 0

converges for 1x1 < 1 and iff(1 -) = A . The classical theorem of Abel is a regularity or consistency result to the effect that the convergence of (1.1) to A implies its Abel summability to the same value. The example ak = ( - l)k shows that the converse statement is not true. It is this very fact that makes summability of divergent series a useful concept. A " corrected " converse is, however, true. For example, summability (A) implies convergence if one of the following conditions holds : 193

194

8. Tauberian Theorems

The first of these results is trivial as a consequence of the following inequalities : n

xakXksf(X),

O<X
n = O , 1 , 2 , ...,

k=O

Since the partial sums s, E t, they must approach a limit if f(1 -) is finite; hence summability implies convergence. The case B was first proved by A. Tauber [1897], case C by J. E. Littlewood [1910]. The latter result and closely related ones might be said to form the underlying theme of the present chapter. We recall further that series (1.1) is summable (C), in the sense of Cesaro, to the number A if lim

SO

n-t OCI

+ s1 + ..*+ s, n+l

= A.

This method of summability is also regular. (See Exercise 1 in this chapter.) The same example used above shows that summability (C) does not imply convergence. Again a “corrected” converse can be proved. For example, G . H. Hardy [1910, p. 3011 proved that condition C above is a suitable correction. We shall prove this later. Note that f(x)=

so + S I X

+ s2x2 + ...

1 S X +x2

+

...

so thatf(x) may be regarded as a weighted average, with weights xk, of all sk . Abel summability results in the limiting case when all weights are unity. Similarly, equation (1.3) asserts, in a sense, that the arithmetic average of “all” s k is A. These two examples are typical of

8.1. Introduction

195

summability theory. In general we say that a theorem is Abeliun if it asserts something about an average of a sequence from a hypothesis about its ordinary limit; it is Tuuberiun if conversely the implication goes from average to limit. Usually the latter results must have in the hypothesis some supplementary condition, like A, B, or C above. In this chapter we shall be dealing chiefly with Tauberian theorems involving integrals, which are the continuous analogs of series. After such results are proved it is then a fairly simple matter to obtain, as special cases, the classical ones about series. See $7. We conclude this section with the proof of a typical Abelian result, which will be useful later. Theorem 2.1.

Throughout this chapter we shall assume without further statement that u(t) E Lin (0, R) for every R > 0. Since

it is no restriction to assume that A = 0, u(t) = o(tY).Then to an arbitrary E > 0 there corresponds an R such that lu(t)l

Hence

< EtY,

R 5 t < co.

196

8. Tauberian Theorems

It follows that f ( x ) = o(x-y-'),

x -+ Of,

as we wished to prove. 2. Integral Analogs

In this chapter we shall be primarily concerned with Tauberian theorems involving Laplace integrals, and these may be regarded as continuous analogs of Tauber's original series theorem, or generalizations thereof. We say that the integral

is summable (A) to the value A if the Laplace integral f(x)=

(2.2)

I

m

e-%(t) dt

0

converges for x > 0 and iff(l-) = A . It is summable (C) if (2.3)

A(x) = Joxa(f) d t ,

lim

Our first result is the regularity theorem.

Theorem 2.1. m

1. f ( x ) =

e-%(t) dt,

0 < x < co;

0

2. =>

JOm

a(t) dt = A

f(U+)

= A.

That is, the convergence of (2.1) implies its Abel summability to the same value. This result is an immediate consequence of Theorem 4.1 of Chapter 5. A converse of this, corresponding to Condition A of $1 follows.

8.2. Integral Analogs

197

Theorem 2.2. m

1. f ( x ) =

J'0

0 < x < 00;

e-x'a(t) dt,

2. f ( O + ) = A ; 0 < t < co

3. a(t) 2 0 ,

a(t) dt

JOm

= A.

Analogous to the corresponding proof for series given in $1, this result follows easily from the inequality

A(R) = lim x+o+

loR

e "'a(t) dt -

5 l b f(x). x-o+

The example a(t) = cos t shows that hypothesis 3 cannot be omitted. For, in this case f ( x ) = x/(x' 1), f ( O + ) = 0, but the integral (2.1) diverges. Our next result is considered Abelian though it is a comparison of two kinds of averages. It states that Ceshro summability implies Abel summability.

+

Theorem 2.3. 1. f ( x ) = (2.4)

lorn0 < x < C x t a ( t )dt,

2. /:o(t)dt - A x ,

(2.5)

A

f(x)--,

X

If a(t) is an integral

then it is the integral

x+

03

x+o+

00;

198

8. Tauberian Theorems

that is summable ( C ) by hypothesis 2 and which is summable (A) by (2.5). However, the theorem as stated, in terms of a(t) rather than b(t), suggests a different interpretation. The integral (2.1) diverges to + 00 ( A > 0), but the partial integrals A(x) become infinite in the prescribed way (2.4). The conclusion is that f(x) also becomes infinite, but at a resultant rate (2.5), as x + 0 . This result is essentially contained in Theorem 1. For, an integration by parts gives

+

f(x) = x

-

low e-XtA(t) dt,

x > 0,

and (2.4) becomes A(x) Ax, X + co. The conclusion of Theorem 1, y = 1, applied to the integralf(x)/x now yields the desired result. An example to show that the theorem is not reversible is a(t) = 1

(2.6)

+ sin t + t cos t , f(x)

1 x

=-

-

A(t) = t

+ t sin t

+ (x' 2x2 +

1)2'

Here f(x) l/x, x + 0 + , but A (t)/t approaches no limit as t + co. Accordingly, to obtain a corrected converse we must expect to impose some additional condition on a(t). It turns out that boundedness is a suitable one. Note that a(t) as defined by (2.6) is not bounded.

Theorem 2.4. W

1. f(x> =

J0

e-xta(t)

dt,

2. a(t) E B (bounded), 3. f(x)--, (2.7)

=r

1:

A X

a(t) dt

o < x < co; 0 < t < 00;

x+o+

-

Ax,

x + 00.

We shall obtain this result in the next section as a special case of a much more general theorem. But let us illustrate it here by the example

8.3. A Basic Theorem

199

The Laplace transform of a(t) is (Exercise 14) (2.8)

1

1

f(x) = - tan-’ X x

7

I

-

x+o+.

2x’

Since a(t) E B we may apply the theorem to obtain (2.7) with A But this may be verified directly since Joxa(f)dt = r ( x

sin y

.x

sin y

Y

O

Y

- y ) -d y = x J -dy

0

a(t) dt = x-rm X

I

c0

O

sin y

7I

Y

2

= 42.

+ cos x - 1,

-d y = -.

Exercise 3 shows that hypothesis 2 is essential here. 3. A Basic Tauberian Theorem

We observe that hypothesis 3 of Theorem 2.4 may be written as

if g(t) = e - t and that the conclusion (2.7) takes the same form if g(t) is replaced by h(t) = 1 on (0, l), h(t) = 0 on (1, 00). Thus Theorem 2.4 states that, under the restriction of boundedness on a(& if (3.1) holds for one “ kernel ’’ g(x) it also holds for at least one other kernel h(t). Our basic theorem will show that the conclusion will hold for many kernels h(t) even if g(t) is not e-t. The essential property of e - * for our purpose is contained in the familiar uniqueness property of the Laplace transform: the vanishing of the generating function implies that of the determining function. We dignify this property in a kernel g(t) by a notation as follows. Definition 3.1. A function g(x) E U (has the uniqueness property) on (0, co)if g(x) E L on (0, co) and if the equation Jomg(;)a(t)

dt

E

0,

0 < x < co

for a(t) E B. C implies a(t) = 0, 0 < t < co.

8. Tauberian Theorems

200

Clearly an exponential change of variables converts the integral (3.1) into a convolution, the range of integration becoming the whole real axis. Our uniqueness property takes the following form for the interval (-a, a). (-

Definition 3.2. A function g(x) E U on (- co,00) if g(x) E L on co) and if the equation

00,

rm

g(x - t) a(t) dt

= 0,

0 <x < m

for a(t) E B. C implies a(t) = 0, - co < t < co. Note that if g(x) E U on (0, 03) then g(e")e" E U on (- 00, m). For example e-" E U on (0, 00) and exp(-e")e" E U on (- co, 00). On the other hand a function g(x) which is constant over a finite interval and is zero elsewhere 4 U on (- 00, 00). This can be seen'by choosing a(t) as a suitable periodic function. (See Exercise 12 of this chapter.) We now state our basic Tauberian Theorem as follows.

Theorem 3.1. 1. g(x)E

u,

2. a(x) E B ,

(-a, a); (-a, co);

3. h(x) EL,

(-a, m);

11m

4.

Jm

g(x - t)a(t) d t --t A

g(t) dt,

x+m

m

-m

m

- t)a(t)dt

3

+A

h(t) d t ,

x + m.

m

By considering the function a(t) - A we see that there is no loss of generality in assuming A = 0. Set G=g*a, H=h*a. From the assumption G(m) = 0 we wish to prove H(m) = 0. Assuming the contrary, there must exist 6 > 0 and a sequence x,,tending to co with n such that H(xJ I > 6. Now observe that G * h = H * g = g * h *a. (3.2)

I

+

This follows by Fubini's theorem in the presence of hypotheses 1, 2, 3.

8.3. A Basic Theorem

201

Now consider the sequence = H(x

s,(x)

+ x,).

We shall show that it is bounded and equicontinuous on - co < x < co. For, if M is an upper bound for I a(t) 1, then

-co < x < co,

IH(x)I 5 M I m Ih(t)l dt, -m

m

I s,(x) - s,(y) I 5 M J”- m 1 h(x + X, - t ) - h(y + X, - t ) [

dt

=~(l),

y-x40.

The latter result follows from a classical result of Lebesgue theory ; see N. Wiener [1933, p. 141. By Ascoli’s lemma (see below) we now select from the functions sn(x) a subset s,(x) which tends to a limit s(x), continuous on (- co, co). By Lebesgue’s limit theorem, applicable since H E B, g EL, we have

1- s,,(x m

lirn k-ao

m

- t ) g ( t ) dt

= lim k+m

m

m

(3.3)

= lirn k-m

H(x m

J”-ao G(x + xnk- t)h(t) dt =

1

+ x,,

- t)g(t) dt

W

s(x - t ) g ( t ) dt = g

* s.

-m

Here we have used equation (3.2). Since G(m) = 0, another application of the Lebesgue limit theorem shows that the limit (3.3) is zero. Since s(x) E B.C we infer from hypothesis 1 and Definition 3.2 that s(x) -= 0. But Is(0)l = lirn Is,,(O) I = lim IH(x,,,)l 2 6 > 0. k+m

k-m

The contradiction shows that H(co) = 0, as we wished to prove. For the reader’s convenience we prove the form of Ascoli’s lemma which we have used.

Lemma 3.1. 1. {s,(x)}~ is equicontinuous on - co < x < co;

2.

Is,(x)l < M ,

+ A.

lim s,(x)

n

=

= s(x),

I , 2, ..., -co < x <

some M

some integers nk , some s(x);

k+ m

B. s ( x ) E C ,

00,

-w<x<<.

202

8. Tauberian Theorems

Arrange all real rational numbers in a sequence xl, x 2 ,x3, . . . . From the bounded double sequence s,(x,,) we may pick, by the familiar diagonal process, a subsequence snk(xm)such that (3.4)

lim snk(x,)= s(x,),

m = 1,2, 3,

. . ..

k- m

See, for example, D. V. Widder [1946, p. 261. The function s(x) is defined by the limit (3.4) at all rational points. Now let y be irrational and let E > 0 be arbitrary. By hypothesis 1, if x , is sufficiently near to y ,

I snk(y) - S n k ( x m ) I < E , IS,,(Y)

< E,

- t9,(x,)l

1,2937 . . ., j = 1,2,3, ....

k

=

Such an x , exists since y is a limit point of rational points. By Cauchy's limit criterion equation (3.4) implies the existence of a number K such that k >K, j > KISnk(Xrn) - snl(Xrn)l < E , By the triangle inequality

I S,,(Y)

- s,,(Y>

I <3 ~ ,

k >K, j >K,

from which we may infer the existence of lim

(3.5)

k+m

S",(Y> = 4 Y > .

It remains to show only that s(x), now defined for all x, is continuous. By the equicontinuity of the sequence snk(x)we have for a given E the existence of a number 6, independent of nk, such that

1

snk(x)

- snk(y)

Is(x) - 4 Y ) I

I < 8,

s

E,

1 x - Y I < 6, Ix - Yl < 8.

Thus the lemma is proved. We point out that Ascoli's result is usually stated for compact regions, in which case the limit (3.5) is uniform. That the approach need not be uniform on (- 00, co)may be seen by considering sn(x) = sin (x/n). Here s(x) = 0, but any inequality of the form Is,(x)I < E < 1 must fail (at x = ns1/2, for example). The previous theorem, restated for the interval (0, co) becomes the following.

8.4. Hardy and Littlewood Theorems

203

Theorem 3.2.

u,

(0, co);

2. u(x) E B,

(0, co);

3. h(x) E L ,

(0, co);

1. g ( x ) E

We may specialize this result by choosing g(x) = e - x x a - l and h(x) = xP-' on (0, l), h(x) = 0 on (1, co). The hypotheses 1 and 3 are evidently satisfied if a > 0, > 0.

Theorem 3.3

2. a(x) E B, 3. f ( x )

(0, a);

- ma) XU

'

x -+

o+

In particular if a = /? = 1, this result reduces to Theorem 2.4, the proof of which we had deferred. 4. Hardy's and Littlewood's Integral Tauberian Theorems

As we noted in $1, Hardy proved first that if ak = O(l/k), k 3 co, then the series (1.1) cannot be summable in the sense of CesAro unless it converges; Littlewood (1912, p. 434) proved the same result for Abel summability. In this section we shall prove the integral analogs of these results, deriving therefrom i n 97 the classical series result. We need first a new definition.

8. Tauberian Theorems

204

Definition 4. A function f ( x ) E SO (slowly oscillating) on (0, co) if lim M

Y ) -f(x)l

=0

when y and x + 00 in such a way that y / x + 1. For example, if f ( x ) f ( x ) E SO there. For,

I f ( Y ) - f ( x >I 5

and x l f ’ ( x ) l < M on (0, co) then

E C’

lpv) I s

log($

= o(l),

L 1 . X

Thus sin (log x ) E SO on (0, 00). On the other hand, sin x $ SO on (0, co). (See Exercise 6 of this chapter.)

Theorem 4.1. 1. a(x) E so, -x

2. Jo a(t) dt

*

(0, co);

-

AX,

co

X +

a(co) = A .

As usual, we may assume A = 0. From hypothesis 1 we see that to an arbitrary E > 0 there correspond numbers K and S such that y > K, x > K,

< E,

la(y) - a(x)I

That is, for x - 6x < y < x

:1

- 11 < 6.

+ 6x

a(x) - E

< a ( y ) < a(x) + E .

Integration gives

1

x(li6)

(4.1)

[a(x) - ~ 1 2 < 6 ~

Now let x Thus

+ 00.

a(y) dy

x(1-a)

lim

< [a(x) + E ] 2Sx.

The integral (4.1) is o(x), x [a(x)

- E l 2s 2 0,

l&ij

[a($

-,co, by

+ E l 2s 2 0,

X+ 00

-&

5 lim a(x) 5 ihi a(x) 5 E .

The conclusion is now immediate.

hypothesis 2.

8.4. Hardy and Littlewood Theorems

205

Hardy's integral Tauberian theorem is an immediate consequence of this result.

Theorem 4.2. (0, GO);

1. xa(x) E B,

2. A ( x ) =

lx dt ; a(?)

0

3. (4.2)

Jx 0

A ( t ) d ?- A x ,

J,

~ ( tdt)

x-+GO

= A.

Hypothesis 1 corresponds to condition C of $1 and hypothesis 3 states that the integral (4.2) is summable (C) to the value A . We apply Theorem 4.1 to the function A(x). It belongs to SO since

Here M is a bound for 1 xa(x) 1. The conclusion of Theorem 4.1 is A(m) = A , equivalent to (4.2). The proof of Littlewood's integral Tauberian theorem is slightly more complicated. Theorem 4.3.

m

(4.3) +-

jo a(t) = A .

By use of Theorem 2.4 we show first that, in the presence of hypothesis 1, the Abel summability of the integral (4.3) implies its Ceshro

206

8. Tauberian Theorems

summability, and we can then apply Theorem 4.2. An obvious integration by parts enables us to write hypothesis 2 as

lom e-"'A(t) dt - A x ,

x +0

+.

To prove that A ( x ) E B, consider the difference a(t)[1 - e-'IX]

dt -

j

m

e-'lxu(t)

dt.

X

By hypothesis I

5M

i;"

+M

dt

2M.

Since f ( x ) E B on (0, co) by hypothesis 2, the same must be true of A(x). Hence if we apply Theorem 2.4 to A(x) we find that

j;A(t) dt - A x ,

x -+ co.

That is, the integral (4.3) is summable ( C ) to A , and the convergence of (4.3) now follows from Hardy's theorem. As an illustration take a(t) = (sin t)/t. Its Laplace transform is tan-'(llx) which tends to 742 as x + O + . Obviously ta(t) E B, so that

somSF dt

= 71 -

2'

5. One-sided Tauberian Conditions

In Hardy's and Littlewood's theorems the condition u(x)x EB can be replaced by a one-sided condition in which a(x)x is bounded below (or above) only. With a view to proving this we first establish several results of J. Karamata [1931, p. 271.

8.5. One-sided Conditions

207

Theorem 5.1. 1. g ( x ) = O , O S x < c < l ;

2. E > 0 ,

* (5.1) (5.2)

g(x)EC,

c5xg1;

y>o

There exist polynomials p ( x ) , P ( x ) such that p(x) < g(x) < P(x), Jol

[log(;)]y-

05x5 17

b ( x ) - p ( x ) ] d x < E.

For an arbitrary positive number q we can determine h(x) E C on 0 5 x 5 1 such that g(x)

5 h(x),

0 5 x 2 1.

For example, if g ( c + ) > 0 (the only case needed later) we may take h(x) = g(x) except in a small interval (c - 6, c) where h(x) = g(c+)(x - c + 6)/6. The value of the integral (5.3) is less than

Hence inequality (5.3) may be satisfied by choice of 6. By the Weierstrass approximation theorem we determine a polynomial Q(x) such that Ih(x) - Q(x)I < q on 0 S x 5 1, and we define P(x) = Q(x) q. Then g(x) 5 h(x) < P(x) and

+

x [(P - Q) +

I Q - h I + ( h - g)l dx <

+ Y ~ ( Y )+ V .

Similarly, there is a polynomial p(x) such that p(x) < g(x) on (0, 1) and such that

208

8. Tauberian Theorems

+

Adding inequalities (5.4) and (5.5) and choosing g = &/(4r(y) 2), we have (5.2). Hypothesis 1 could be very considerably relaxed if desired; see D. V. Widder [1941, p. 1891. Theorem 5.2.

0 < x < co;

2. a(x) 2 0 ,

3. g(x)=O, 4. f ( x )

osx
A

y>o,

N -

xy '

g(x)EC, c l x S 1 ;

x+o+.

x+o+.

We begin by determining the polynomials p and P of Theorem 5.1, corresponding to a given E . Since a(x) 2 0 we have (5.7)

/

e-"'p(e-"')u(t) dt

m

1

W

e-"'g(e-"')a(t) dt

5

0

e-"'P(e-"')a(t) dt,

0

If we replace x by (n + 1)x in hypothesis 4 we obtain

Thus the conclusion (5.6) is valid if g(x) is replaced by X" and hence by any polynomial. From (5.7) and (5.9) we have m

e-'p(e-')tY-' dt

5 l b xT(y) x-o+

e-X' g(e-"')a(t) dt 0

00

S fi;;; x T ( y ) / x+o+

0

e-"'g(e-"')a(t)

dt

5A

/

m

0

e-'P(e-')ty-'

dt.

8.6. One-sided Littlewood Theorem

209

By (5.2) the two extremes of this inequality differ by AE.Hence the two terms in the middle must be equal to each other and to the middle term of (5.8). That is, (5.6) is established. We now specialize the function g(x) to obtain the following Tauberian theorem. Theorem 5.3.

3. f ( x )

-

A

-

xy

'

y>o,

x-bo+

Except for the present one-sided Tauberian condition 2 this result would be included in Theorem 3.3. To prove it, choose g(x) = l / x on (l/e, l ) , g(x) = 0 on (0, l/e) in Theorem 5.2. Then (5.6) becomes

which is equivalent to (5.10).

6. One-sided Version of Littlewood's Integral Theorem

To obtain a one-sided version of Theorem 4.3 we need several preliminary results, of interest in themselves. We introduce a definition. Definition 6. A function f ( x ) E SD (slowly decreasing) on (0,co) if (6.1)

lim M Y ) - f W l 2 0,

when y and x become infinite in such a way that y > x and y/x + 1 .

8. Tauberian Theorems

210

For example, any increasing function belongs to this class. Or,

f(x) E SD if xf’(x) > - M for some constant M . For,

(;),

f(Y) - A x ) > -Mlog

Y > x,

from which (6.1) is immediate. We now prove a one-sided version of Theorem 4.1.

Theorem 6.1. (0, 00);

1. a(x) E SD,

2.

/ox

a(t) dt

Ax,

x + co

a(w) = A.

=>

Assume A

-

= 0.

For an arbitrary

> 0 we have from (6.1) that x < y < (1 3. 6)x. E

a(x) - E < a(y), (6.2) Integrating with respect to y and using hypothesis 2, we obtain

7

lim a(x)

(6.3)

5 E.

x-t m

Rewrite (6.2) as

and integrate with respect to x :

This with (6.3) shows that a(co) = 0, as we wished to prove. Corollary 6.1. 1.

xa(x) > - M ,

2.

JOm

a(x) dx

someM, O < x < co;

is summable (C)to A

6.8. One-sided Littlewood Theorem

211

This is a one-sided version of Theorem 4.2. It follows at once by observing that A@) now satisfies the conditions of Theorem 6.1. We prove next a result of E. Landau [1929,p. 581.

Theorem 6.2.

1.

osx
f(X)EC2,

2. f ( 0 ) = A ; -M 3. f”(x) > x2

’

someM, O < x < c o

(3

*

f’(x)=o - ,

x-+o+.

From Taylor’s theorem we have for 0 < 6 c I

Now let x obtain

-+

0

+ , noting that f ( x + ax) - f ( x )

-+

0 by hypothesis 2.We

-M6 M6 < lim x f ‘ ( x ) 5 G xf’(x) 5 -. 2(1 - S ) 2 =xz+ x + o + 2 Since 6 is arbitrary we obtain at once the desired conclusion. We derive next a consequence of Theorem 4.3, using a modification of the first hypothesis thereof.

Theorem 6.3. 1. B(x) =

jxta(t) dt = o(x),

x

-+

co ;

0

2. f ( x ) = Jorn e-x‘a(t) dt -+ A , (6.4)

=>

jo* a(t) dt = A .

x

-+

o+

212

8. Tauberian Theorems

This states that hypothesis 1 is an alternative condition to Littiewood's for the Abel summability of the integral (6.4) to imply its convergence. Assume A = 0. Integration by parts gives

3 dt = B(l)e-" - x Jme-"' 9 dt + 1e -%(t) t2 t m

(6.5) Jme-xf 1

1

dt.

1

The determining function B(t)/t2of this first integral is U ( l / t ) by hypothesis 1. Hence we may apply Theorem 4.3. Its limit as x 0 + , from the right-hand side of (6.5) is --f

1

1

B(l) -

a(t) dt.

0

For, lim x

f,*e - X fB(t) dt = 0

x+o+

by Theorem 2.1, and

lim fpe-.i2ct, dt

=

- lim

1

Jol

e-*'a(t) dt

=

- l 0 u ( t )dt

by hypothesis 2. Thus the conclusion of Theorem 4.3 yields

jlm $?dt = B ( l ) - j a(t) dt, 1

0

=B(l)

(6.7)

+

m

a(t) dt. 1

To obtain (6.7) we have integrated integral (6.6) by parts. Equation (6.4), with A = 0, is now immediate by comparing (6.6) with (6.7). We can now obtain the desired one-sided version of Littlewood's theorem very easily.

Theorem 6.4.

2. f ( x ) =

loebxta(t)dt JOm

a(t) dt

+A ,

= A.

x + O+

8.7. Classical Series Results

2 13

From hypothesis 1 we have

By Theorem 6.2f'(x) = o(l/x), x +O+. Thus Jbme-"'[M

+ ta(t)] dt = M-f ' ( x ) X

-

M

-, X

x -+

O+

Since M + ta(t) > 0 we may apply Theorem 5.3 to this integral to obtain

Jb'

[ M + tu(t)] dt

N

Mx,

x + 00.

This implies that hypothesis 1 of Theorem 6.3 is valid. The conclusion of that theorem is the desired result here. 7. Classical Series Results

We have hitherto proved the integral analogs of the Tauberian theorems about series mention in $1. It is now possible to obtain these discrete results, not by analogy, but as direct consequence of the continuous cases. We do so in a few instances, using the more general onesided versions. First, the discrete analog of Theorem 2.4 or of Theorem 5.3 (y = 1): Theorem 7.1. m

(7.1)

1.

xake-kx-tA,

x+O+;

k=O n

2.

-M<s,=Cak,

someM, n = 0 , 1 , 2 ,...

k= 0

=>

That is, if the sequence of partial sums of a series is bounded on one side it cannot be summable in the sense of Abel unless it is also summable in the sense of Ceshro. It is obvious that for y = 1 in Theorem 5.3 the lower bound zero of hypothesis 2 may be replaced by any number -M. (See Exercise 15 of this chapter.)

214

8. Tauberian Theorems

Define a function A(x) so that A(0) = 0 and

c a, > - M ,

[XI

A(x) =

x > 0.

k=O

If the sum of the series (7.1) isf(x), we have

1 e-”‘A(t) dt, 03

W

f ( x ) = Jo e-”‘dA(t) = x

x

> 0.

0

Thus by hypothesis 1 j o r n e - X t A ( t ) d t -A- ,

x-bo+.

X

The conclusion of Theorem 5.3, with x replaced by n

+ 1, is

This is the desired result. We turn next to the series form of Hardy’s theorem.

Theorem 7.2.

n

n

Observe k s t that hypothesis 2 implies that sn= o(n) and hence that an = o(n), n -+ 00. Define A(x) as in the previous theorem and apply Theorem 6.1 to it. Set m = [ y ] and n = [x], y > x. Then

8.7. Classical Series Results

215

From these inequalities it is clear that A(x) E SD on (0, co). Further, if n 1 S x < n + 2

+

Then by hypothesis 2

j; A ( t ) dt

N

Ax,

x

+

co.

The conclusion of Theorem 6.1 is m

A(0)=

Uk k= 0

= A,

as we wished to prove. The series version of Littlewood’s theorem follows. Theorem 7.3. 1.

ak

-M

someM, k = 0 , 1,2, ...;

>k + 1’

W

We apply Theorem 6.4 defining a(x) by U(x)=Uk,

ksx
k = 0 , 1 , 2 ,....

From hypothesis 1 it is clear that a(x) > - M / x for 1 4 x < co. The conclusion of Theorem 6.4 is

as we wished to prove.

216

8. Tauberian Theorems

8. Summary

A typical Abelian theorem is the consistency or regularity one which states that a series or integral which converges is also summable, say in Abel‘s sense. A typical Tauberian theorem is a modified converse, say Littlewood’s, to the effect that the integral a(t) dt, where a(t) = O(l/t), t + 00, cannot be summable, in Abel’s sense, unless it converges. In proving this and similar results we used two approaches. In the first, a modification of Wiener’s method, we used integrals with very general kernels but imposed Tauberian conditions that were “ twosided ’”;in the second we used the Laplace kernel e-X’ only but imposed the less restrictive “ one-sided ” conditions. The results thus overlap in generality. It is noteworthy that no appeal to the theory of Fourier integrals was needed, nothing more recondite having been used than Ascoli’s selection theorem in the first method and Weierstrass’s approximation theorem for the second.

:5

EXERCISES 1. Show that Cesliro summability, as defined by (1.3) for series and

by (2.3) for integrals, is regular. 2. Illustrate Theorem 2.3 by the example a(t) = 1 - cos t , f ( x )

= 1/(x3

+ x).

+

3. Use the example a(t) = 1 t sin t to show that Theorem 2.4 would be false without hypothesis 2. 4. Prove the power series analog of Theorem 2.4 by use of that theorem :

1. l a k l < M , 2.

OD

CakX--

k=O

someM, k = 0 , 1 , 2 ,....

A , 1- x

x+l-.

n

Cak-An, k=O

n4oo.

Exercises

5.

2 ?7

Use Theorem 2.1 to prove that

1

00

a(co) = A

e-xta(t) dt

0

-

A

-

x+o+.

X'

6. Prove that sin x # SO on (0, 00). [Hint: sin(x + n) - sin x = -2 sin x.]

7. Prove thatf(x)

E

SD on (0, 00) iff(x)

E B,

x f ( x ) E t.

+ b cos x E SD on (0, co). if a > 0 Show thatf(x) = x + cos x + cos(1og x ) E SD on (0,

8. Show that ax 9.

10. Show that f ( x ) in Exercise 9 is such that x f ( x ) [Hint: (xf(x))' = x(2

- sin x ) + cos x + a s i n

00).

1on (0, 03).

(: - log -

x

1.1

11. Usef(x) ofExercise 9 to determine whether Lemma 5.5 of Chapter 4 includes the present Theorem 6.1. 12. If h(t) = 1 on (1, e), h(t) = 0 elsewhere, show h(t) # U on (0, co). d Hint: Take a(t) = - sin(2 n log t), for example. dt

1

[

13. By integrating formula 3, $2 of Chapter 1, prove rigorously that

tan-'

(L) som

e-Xi

=

sin t

-dt, t

x > 0.

14. Prove that

1

- tan-'($ X

lo m

=

e-Xi dt

jo siny -d y , Y

x

> 0.

[Hint: Integrate (9.1) by parts.]

15. Show that Theorem 5.3 remains true if y 2 1 and hypothesis 2 is replaced by a(x) 2 - M on (0, a).

218

8. Tauberian Theorems

16. Show that if M is replaced by 0 in hypothesis 1 of Corollary 6.1 the conclusion follows trivially. dt

2 A(R)

17. Admitting equation (6.7) and Theorem 4.10 of Chapter 4, complete the proof of the prime number theorem by use of Exercise 7 and Theorem 6.1 of the present chapter.

9

Inoersion by Series

1. Introduction

In this chapter we shall treat two special transforms, both of which may be written as

with suitable kernels K(t). By exponential change of variable, (1.1) becomes the convolution transform

S_

4 ,

f(e-x) =

K(e-x+')#(e-') dt, m

whose inversion function E(s) is defined by

For both kernels K(x) which we have in mind E(s) will belong to the Laguerre-Pblya class. See §4 of Chapter 7. Consequently,the inversion of (1.1) can then be accomplished by use of the operator E(D), as in 219

220

9. Inversion by Series

Theorem 7.1 of that chapter. However, for these special kernels there is an interesting alternate method of inversion when the generating function f ( x ) has a power series expansion

We shall show that then the determining function 4(x) also has a power series expansion, which can be obtained from (1.3) by the introduction of a sequence of multipliers: E( -k)Xk. 4 ( ~=) k2 =O 2. The Potential Transform

The first of the kernels to be treated is 2 K(x) = -

1 + 1'

nx2

The transform (1.1) becomes

The integral is clearly related to the Poisson representation of a harmonic function u(x, y) in the half-plane y > 0. We shall make this more explicit later, but on account of this relationship we propose to call (2.1) the potential trunsform. We can obtain the inversion function E(s) at once from equation (1.2): 2 w x s - l d x - 1m,

1 -=-I E(s) n

=-.(1;, n

0

x2+1

1 2 )

ff

y(s-22)/2 0

=-. sin

dY l+y 1

(7)

9.3. A Brief Table 221

Thus E(s) = sin(ns/2) is an entire function with real roots in arithmetic progression. It fails to belong to the class E, of $4, Chapter 7, only because E(0) is not equal to unity. This causes no real difficulty S / ~ ) belong to E, . because E ( S ) / ( T C does The inversion operational equation TCD sin -f ( e U x ) = 4(e-x) 2

(2.2) must be interpreted as

sinE) (2.3)

if we wish to apply Theorem 7.1 of Chapter 7. We shall make this remark more explicit in $5. We observe also that (2.1) can be reduced to the Stieltjes transform :

3. A Brief Table Let us list here a few pairs (f,

4) which satisfy equation (2.1).

A.

1 x+l

t tz+ 1

B.

e-x

sin t

e-x

cos t -

C.

-

t

X

D.

tan- x X

(0'

O
222

9. Inversion by Series

O
,{: log x

F.

n -- 1 2 t2+1

x2 - 1 x log x x2 - 1

G.

t log t t2 1

+

From the integral (2.1) it is clear that the generating functionf(x) is always an even function. In the above table it is to be understood that x > 0, so that absolute value signs can be omitted. Let us indicate briefly how this table could be derived. Formula A follows by direct integration of the identity 1

t2

Formula 3 of 52, Chapter 1, was 03

U G~ = jo e-X'sin at dt.

If we regard this as a sine-transform, the inverse is sin at da, from which Formula B above follows by setting t = 1. Formula C can be obtained in a similar way from the Laplace transform of cos at. Formulas D and E are familiar ones in the integral calculus. To obtain Formulas F and G we introduce a parameter s and integrate the following identity with respect to t from 0 to 03, tS ---=

t2

+1

tS

x2

tS(X2

+ t2

(t2

- 1)

+ 1)(2 +

t2)'

From the familiar formula

'jm tSd t = n x2 + t 2 71s' xs-

0

1

cos 2

-1 < s < 1,

9.4. The Inversion Algorithm

223

we have

The Mellin integral clearly converges for - 1 c s c 3 and accordingly represents an analytic function of s in that interval. Hence the identity (3.1) must hold there, taking account of the removable singularity at s = 1, where the analytic function on the right has the value (2/n)(log x)/(x2 - 1). This result is equivalent to Formula F. Note that we get a new proof of Formula A by setting s = 2 in (3.1). Finally t o obtain Formula G differentiate (3.1) with respect to s and set s = 2. The differentiation under the integral sign is valid by Theorem 4.2 of Chapter 5. We obtain

’s

a,

71 0

t210gt dt (t2 + l)(x2 t 2 )

+

xlogx x2 - 1’

=-

as indicated in Formula G. See Exercise 6-10 of this chapter for alternate derivations of some of these transform pairs.

4. The Inversion Algorithm In this section let us experiment with an inversion procedure made plausible by the operational equation (2.2). It is applicable, as described below, when the generating function has an expansion for x > 0 in a convergent series of powers. Let us describe the conjectured algorithm in three steps: I. Expand the generating function in powers of x . 11. Cancel the even powers. 111. Change signs of alternate terms in the resulting series. The sum of the final series should be the determining function Let us try the process in a few examples.

4(x).

224

9. Inversion b y Series

Exampk A .

1 x+l

I. -- 1 - x + x2 - x3 +

111. x - x3

+ x5 - x7 +

*.*

. a * .

X =x2 1'

+

Note that the expansion could have been in negative powers of x. 1.

1 -1 x+l x

1

1 + T x2 x - . - - .

1 1 1 111. - - - + - x3

x

x5

X ... = -

x2 f 1'

Example B.

x3

111. x - - + 3!

x5 - . . * = sin x. 5!

Example C. e-x 1 1. -=--I+--x

111.

x

1 x x3 ---+--... x

2!

4!

x

x2

2!

31

+ * * * *

=-.cos x X

In Example D there is an expansion in positive powers of x , but they are all even powers, so that the algorithm leads to 4 ( x ) = 0. This is correct on the interval (0, 1) at least. For Example E the expansion must be in negative even powers of x, and again $(x) E 0. This too is correct on the interval (1, a) at least. Examples F and G have no Maclaurin expansions; and some modification of the algorithm is needed. We shall treat these examples later. The conjectured inversion procedure seems to have encouraging possibilities. Let us obtain some precise results involving it.

9.5. The Inversion Operator

225

5. The Inversion Operator

Let us alter the symbolic equation (2.2) so as to avoid the exponential change of variable. Since the derivative off(e-x) is -e-xf'(e-X) we would do well to define a new operator 8 as follows:

"(41

= - xf'(x).

In terms of this new operator equation (2.2) becomes

ne f ( x ) = -n 8 *

4(x) = sin-

2

2

n

k=l

As in Chapter 7 we now truncate the infinite product to define a linear differential operator of order 2n 1 as follows.

+

Definition 5.

LxCfl = lim L , x [ f l . n+ 03

Note that this operator, when applied to a monomial xa reproduces it except for a constant factor. For €'[x"] = -ax".

Accordingly, if we define

we have or

Since &(a) vanishes at the even integers between -2n and 2n it is clear is xZk,k = 0, that a set of fundamental solutions for the operator Ln,x f 1, & 2 , . . . , & n. Hence sin (n8/2) may be described as a linear

226

9. Inversion by Series

differential operator which annuls all even powers of x . We illustrate by exhibiting L, ,~explicitly :

+

A

Ll,” = - [ x 3 D 3 3x2D2 - ~ x D ] . 8

Clearly it annuls the function x - ~ ,1, x 2 and is an Euler direrential operator (with the order of each derivative equal to the power of its monomial coefficient), as we might have predicted from the elementary theory of differential equations.

Theorem 5.1. 1. q5(t)~C.B, (5.2) 2. f(x) =

-s A

(5.3)

*

O< t<

00;

tq5(t) d t ,

0

x2

+ t2

L,[fl = q5(x),

convergesfor O < x < a, 0 < x < m.

Note that hypothesis 1 alone is not sufficient to guarantee the convergence of the integral (5.2). If 4 ( t ) is constant, for example, the integral clearly diverges. This is further corroboration that Theorem 7.1 of Chapter 7 is not directly applicable to the present transform. It is equation (2.3) rather than (2.2) which we must use. It becomes, in terms of operator 8,

This integral is obtained by applying 6’ to the integral (5.1), a step which may be justified by showing that the integral (5.4) converges uniformly on (6, m) for an arbitrary positive 6. This is immediate from the inequality

If now we apply Theorem 7.1 to the integral (5.4)the conclusion (5.5) follows. As in that theorem we need not assume 4(t)E C ; (5.3) will still follow almost everywhere or at any points of continuity.

9.6. Series Inversion

227

We point out here that this theorem can be proved easily without any appeal to the general theory of the convolution transform. It may be shown by induction that

See D. V. Widder 11966, p. 3571. Here the integrand is essentially the (2n + 2)th power of a function so that Laplace’s asymptotic method is applicable (Theorem 2.2 of Chapter 6). 6. Series Inversion

Let us now suppose that in addition to its representation by a potential transform f ( x ) also has a power series expansion convergent for some positive x. Theorem 6.1. 1. f(x)

=-

2. f(x)

=

sm +

-

x t24 ( t )t 2 d t ,

7r 0

o<x
m

(6.1)

ukxk-O,

0 < x < p,

some a ;

k=O

(6.2)

=>

4(t) =

m

7c

uk sin - ( a - k ) t k - a ,

2

k=O

at least at points t where

0 < t < 6, some 6,

4(t)E C.

Observe first that the function &(a), defined by (5.1) satisfies the inequality

Now apply the operator Ln,,to the series (6.1), a step surely valid for any convergent power series since nothing worse than differentiation is involved:

c m

(6.3)

L,,,[f] =

k=O

Uk

&(a

- k ) +-a,

0 < t