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2004

Asymptotic Laplace transformsClaudiu MihaiLouisiana State University and Agricultural and Mechanical College, mihai2@math.lsu.edu

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Recommended CitationMihai, Claudiu, "Asymptotic Laplace transforms" (2004). LSU Doctoral Dissertations. 3941.https://digitalcommons.lsu.edu/gradschool_dissertations/3941

ASYMPTOTIC LAPLACE TRANSFORMS

A Dissertation

Submitted to the Graduate Faculty of theLouisiana State University and

Agricultural and Mechanical Collegein partial fulfillment of the

requirements for the degree ofDoctor of Philosophy

in

The Department of Mathematics

byClaudiu Mihai

B.S. in Math., Bucharest University, 1993M.S., Louisiana State University, 2001

December 2004

Acknowledgments

This dissertation would not be possible without several contributions. I would like

to express my deepest gratitude to my adviser Dr. Frank Neubrander for his help

and support during the last five years. Special thank you notes to my committee

members Dr. William Adkins, Dr. Donald Kraft, Dr. Jimmie Lawson, Dr. Gestur

Olafsson and Dr. Michael Tom for their time and help throughout my graduate

studies. This dissertation is dedicated to my children Alexandru and Ana, for their

love, support and encouragement.

ii

Table of Contents

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii

Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1. The Classical Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2. Asymptotic Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Appendix. Remarks on Complex Version of Phragmen-MikusinskiInversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Vita . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

iii

Abstract

In this work we discuss certain aspects of the classical Laplace theory that are rel-

evant for an entirely analytic approach to justify Heaviside’s operational calculus

methods. The approach explored here suggests an interpretation of the Heaviside

operator · based on the ”Asymptotic Laplace Transform.” The asymptotic ap-

proach presented here is based on recent work by G. Lumer and F. Neubrander

on the subject. In particular, we investigate the two competing definitions of the

asymptotic Laplace transform used in their works, and add a third one which we

suggest is more natural and convenient than the earlier ones given. We compute

the asymptotic Laplace transforms of the functions t 7→ etn for n ∈ N and we show

that elements in the same asymptotic class have the same asymptotic expansion

at ∞. In particular, we present a version of Watson’s Lemma for the asymptotic

Laplace transforms.

iv

Introduction

The Laplace transform theory has a rich history. It carries the name of a French

mathematician, Pierre-Simon Marquis de Laplace, who used it in his monograph

”Theory Analytique des Probabilites” from 1812. The Laplace transform of a func-

tion f : [0,∞) → R is the function f : λ→∫ ∞

0e−λtf(t) dt, where f is defined for all

λ ∈ C with Reλ > a and a depends on the growth of the function f or, more pre-

cisely, on the exponential growth of its antiderivative F (t) :=∫ t

0f(s) ds. The most

significant property of the Laplace transform is that, essentially, integration and

differentiation become multiplication and division; i.e., if λ→ f(λ) is the Laplace

transform of f, then it follows from∫ ∞

0e−λtf ′(t) dt = −f(0)+λ

∫ ∞0e−λtf(t) dt and∫ ∞

0e−λtF (t) dt = 1

λ

∫ ∞0e−λtf(t) dt that

f ′(λ) = λf(λ)− f(0), and

F (λ) =1

λf(λ).

With these properties, the Laplace transform changes linear integral and differen-

tial equations into algebraic equations which, at least theoretically, are easier to

solve. In the 1900’s, an English physicist, mathematician and electrical engineer,

Oliver Heaviside (1850-1925), had set the ground for an ”Operational Calculus”

that allowed him to solve ordinary differential equations by transforming them into

algebraic problems. In his monograph on ”Electro-Magnetic Theory” ([14], Vol II,

p 32) he says:

”We have now to consider a number of problems which can be solved

at once without going to the elaborate theory of Fourier series and

integrals. In doing this, we shall have, preliminarily, to work by instinct,

1

not by rigorous rules. We have to find out first how things go in the

mathematics as well in the physics. When we have learnt the go of it

we may be able to see our way to an understanding of the meaning

of the processes, and bring them into alignment with other processes.

And I must write here a caution. I may have to point out sometimes

that my method leads to solutions much more simply than Fourier’s

method. I may, therefore, appear to be disparaging and endeavoring to

supersede his work. But it is nothing of the sort. In a complete treatise

on diffusion, Fourier’s and other methods would come in side by side

- not as antagonists, but as mutual friends helping one another. The

limitations of the space forbid this, and I must necessarily keep Fourier

series and integrals rather in the background. But this is not to be

misunderstood in the sense suggested. No one admires Fourier more

than I do. It is the only entertaining mathematical work I ever saw. Its

lucidity has always been admired. But it was more than lucid. It was

luminous.”

Let us have a look how Heaviside’s method leads to solutions of ordinary differen-

tial equations ”much more simply than Fourier’s method.” Essentially, Heaviside

postulated that there exists an one-to-one operator · from the world of functions

defined on [0,∞) into a world of symbols containing the complex numbers, where

neither world is made precise, with the following properties. The operator · is

assumed (postulated) to be an algebra homomorphism; i.e., for all functions f, g

and all c ∈ C,

cf + g = cf+ g, (1)

f ∗ g = fg, (2)

2

where the multiplication ∗ is defined by the finite convolution product (f ∗ g)(t) =∫ t

0f(t− s)g(s)ds. Heaviside assumes that in the world of symbols the usual prop-

erties of addition and multiplication will hold; i.e., the world of symbols is a field.

Inspired by the Fourier transform which maps differentiation into multiplication,

Heaviside postulates further that

f ′ = Df := λf − f(0), (3)

where λ is an unspecified, fixed symbol and where the number f(0) is considered

to be an element in the world of symbols. The postulates (1) and (3) imply that

0 = 1′ = λ1 − 1, or

1 =1

λ. (4)

By (2), (4), and the fact that (1∗1)(t) =∫ t

01 ds = t, it follows that 1 t = 1∗1 =

1 · 1 = 1λ2 . Continuing in this fashion and using that (1 ∗ t)(t) =

∫ t

0s ds = 1

2t2,

we get t2

2 = t ∗ 1 = 1 ∗ 1 ∗ 1 = 1

λ3 . By induction, for all n ∈ N0,

tn

n! = 1 ∗ 1 ∗ . . . ∗ 1 =

1

λn+1. (5)

In order to be able to assign a symbol to transcendental functions, Heaviside pos-

tulates further that in the world of symbols infinite addition poses no problem. As

a consequence,

e−t = 1− t+t2

2. . . = 1 − t+ t

2

2+ . . . =

1

λ− 1

λ2+ . . . =

1

λ+ 1,

or, more general,

ect =1

λ− c. (6)

With these operational rules at hand, Heaviside has a simple, yet powerful frame-

work to solve linear ordinary differential equation. As an example of the Heaviside

1For notational simplicity we write tn instead of en, where en(t) := tn.

3

method consider the linear, first order, inhomogeneous initial value problem

f ′(t) + f(t) = (2t+ 1)et2 , f(0) = 1. (7)

Using the postulated operator · defined above and its properties (1) − (6), we

see that

f ′(t) + f(t) = (2t+ 1)et2 and f(0) = 1

⇔ f ′(t)+ f(t) = (2t+ 1)et2 and f(0) = 1

⇔ λf − 1 + f = (2t+ 1)et2

⇔ (λ+ 1)f = 1 + (2t+ 1)et2

⇔ f = 1λ+1

+ 1λ+1

(2t+ 1)et2

⇔ f = e−t+ e−t(2t+ 1)et2

⇔ f = e−t + e−t ∗ (2t+ 1)et2

= e−t +∫ t

0e−(t−s)(2s+ 1)es2

ds

= e−t + e−t∫ t

0(2s+ 1)es2+s ds

= e−t + e−t[et2+t − 1]

⇔ f = et2

⇔ f(t) = et2 .

It is important to observe that by using Heaviside’s ”Operational Calculus”, one

can solve a differential equation with basic algebra and first semester calculus

alone and, more important, that the procedure yields existence and uniqueness at

the same time. It is worthwhile mentioning that Heaviside used this method for

all differential equations appearing in his treatise on electro-magnetic theory and,

even more remarkable, that his method always provides the right result. Since for

mathematicians the result does not always justify the means, the mathematical

community was not too impressed by what it saw. An anonymous Fellow of the

4

Royal Society in a letter to Sir Edmund T. Whittaker writes: (See, J.L.B. Cooper

[7]).

”There was a sort of tradition that a Fellow of the Royal Society could

print almost anything he liked in the Proceedings untroubled by ref-

erees: but when Heaviside had published two papers on his symbolic

methods, we felt the line had to be drawn somewhere, so we put a stop

to it”.

Looking once again at Heaviside postulates, natural questions that arises are:

What is the world of functions, what is the operator ·, and what

is the world of symbols?

As a first response to these questions, mathematicians like D.V. Widder and

G. Doetsch developed the nowadays classical Laplace transform theory in the first

half of the twentieth century (see [10], [26]). Laplace transform theory quickly re-

placed Heaviside’s operational calculus as the engineers preferred method to solve

linear differential equations. In Laplace transform theory, the world of functions is

the set of all functions f ∈ L1loc[0,∞) with exponentially bounded antiderivative

F , the operator · is the Laplace transform f 7→ f , and the world of symbols

is the set of analytic functions r in a right half plane that are representable as

Laplace transforms; i.e. r = f for some f ∈ L1loc. Although Laplace transform

theory is a powerful tool with which one can treat most of the important linear

engineering problems, it can only partially explain the success of Heaviside’s oper-

ational methods. First of all, since Heaviside’s world of functions contains rapidly

growing functions like t 7→ et2 , it is definitely larger than the set of all Laplace

transformable functions. Second, Heaviside’s world of symbols is a field, whereas

the set of analytic functions that are representable as Laplace transform is not even

5

a ring 2. Third, since Laplace transform theory always requires artificial growth

conditions, it cannot easily produce uniqueness results. In the 1950’s, the Polish

mathematician Jan Mikusinski developed a full mathematical foundation of Heav-

iside’s operational calculus based on the fact that continuous functions defined

on [0,∞) form an integral domain with respect to addition and convolution. In

the Mikusinski algebraic foundation of Heaviside’s operational calculus, functions

and symbols are interpreted as elements of the quotient field constructed from the

integral domain (C[0,∞),+, ∗) (see, for example, [19], [20]).

In this dissertation we discuss certain aspects of an alternative, entirely analytic

approach to justify Heaviside’s methods employing an interpretation of the Heav-

iside operator · based on the ”Asymptotic Laplace Transform.” This concept

was developed around 1939 by the Argentinian mathematician J.C. Vignaux [27]

and investigated among others by L. Berg [6], M. Cotlar [28], M. Deakin [8], W.A.

Ditkin [9] and Yu. I. Lyubich [18]. The asymptotic approach presented here is

based on more recent papers by G. Lumer and F. Neubrander on the subject ([16],

[17]). In particular, we investigate the two competing definitions of the asymp-

totic Laplace transform used in these papers, and add a third one. Before studying

the asymptotic Laplace transform in Chapter 2, we collect in Chapter 1 some of

the background materials from classical Laplace transform theory, the convolution

transform, and the recent convolution approach to generalized functions taken by

B. Baumer, G. Lumer, and F. Neubrander (see [3], [16], [17]). Our own results in

Chapter 1 are concerned with the characterization of uniqueness sequences (see

Proposition 1.9 and Theorem 1.10) and a more detailed investigation of the con-

nection between the abscissa of convergence and the exponential growth bounds of

2We note that if r(λ) =∫∞0 e−λtf(t) dt and u(λ) =

∫∞0 e−λtg(t) dt for some f, g ∈ L1

loc([0,∞), X) and some

λ ∈ C such that∫∞0 e−λt |f(t)| dt and

∫∞0 e−λt |g(t)| dt do not exist for λ, then (f ∗ g)(t) =

∫ t0 f(t − s)g(s) ds

might not be Laplace transformable at λ; i.e.,∫∞0 e−λt(f ∗ g)(t) dt might not exists.(see [1]).

6

functions of bounded variation (see Proposition 1.2). Unfortunately, our attempt to

generalize Baumer’s Phragmen-Mikusinski type inversion formula for the Laplace

transform from real uniqueness sequences to complex uniqueness sequences was

not successful, and we can only present a slightly more transparent proof of this

crucial and far-reaching result. Since we still feel that a generalization to complex

uniqueness sequences should be possible we collected our investigations of this

topic in the Appendix.

In Chapter 2, our main contributions to the still developing theory of the asymp-

totic Laplace transform are as follows. We prove that all operational properties of

the Laplace transform extend to the asymptotic settings whenever one uses the

second definition given by G. Lumer and F. Neubrander in [17] or the new defini-

tion we add. With the new definition of the asymptotic Laplace transform given in

this dissertation, all operational properties remain valid but the equivalence classes

considered in our definition of the asymptotic Laplace transform are large enough

to contain easily computable transforms; i.e., asymptotic Laplace transforms are

more readily computable with our definition than with the definition given in [17].

We remark that with the first definition given by G. Lumer and F. Neubrander in

[16] asymptotic Laplace transforms were defined by even larger equivalent classes

(and therefore even more readily computable), but unfortunately, not all (only

almost all) operational properties hold for the huge equivalent classes considered

there. To demonstrate the usefulness of our definition, we compute the asymp-

totic Laplace transform of the functions t 7→ etn , n ∈ N. As indicated above,

the asymptotic Laplace transform f of a function f consists of an equivalence

class of analytic functions. We show that if one element in the asymptotic Laplace

transform f of f ∈ L1loc([0,∞), X) has an asymptotic expansion in terms of 1

λas

λ→∞, then any other element in the asymptotic class has the same asymptotic

7

expansion in terms of 1λ

as λ→∞ (see Proposition 2.15). In this context, we also

generalize Watson’s Lemma for the asymptotic Laplace transform; i.e., we prove

that if f ∈ L1loc([0,∞), X) has an asymptotic expansion in terms of tnn∈N as

t→ 0+, f(t) ∼∑∞

n=0 cntn, then for any u ∈ f we have that

u(λ) ∼∞∑

n=0

cnn!

λn+1as λ→∞

(see Proposition 2.16). It is remarkable that Watson’s Lemma holds no matter

which definition of the asymptotic Laplace transform is chosen. We conclude the

chapter with a brief description of the asymptotic Laplace transform of generalized

functions and indicate a few applications.

8

1. The Classical Laplace Transform

In this chapter we will review, discuss, and expand results from the classical theory

of the Laplace transform which are needed later. We are considering functions

f : R+ → X, where X is a complex Banach space and denote by L1loc(R+, X) the

space of all f : R+ → X which are Bochner integrable on [0, T ] for all T > 0.

Clearly, the existence of the Laplace integral

f(λ) :=

∫ ∞

0

e−λtf(t) dt := limT→∞

∫ T

0

e−λtf(t) dt (1.1)

for f ∈ L1loc(R+, X) and λ ∈ C depends on the growth of the function f. To discuss

the existence of f and for the discussion of many other results in Laplace transform

theory it is advantageous to use a generalization of of the Laplace integral; i.e, the

Laplace-Stieltjes integral

dF (λ) :=

∫ ∞

0

e−λt dF (t) := limT→∞

∫ T

0

e−λtdF (t), (1.2)

where F : [0,∞) → X is a function of bounded semivariation (see below for a

short summary of their basic properties and [1] for a complete summary of the

properties of the Riemann-Stieltjes integral). If F ′ = f almost everywhere for

some f ∈ L1loc([0,∞), X), then (1.2) reduces to the Laplace transform (1.1). By

the fundamental theorem of calculus, this holds if F is absolutely continuous and

if either X is finite dimensional (X = C or X = Cn), or X is reflexive, or X has

the Radon-Nikodym property (see [1], Chapter 1, for a discussion of the Radon-

Nicodym property). If1 F (t) =∑

i≥0 aiχ(ti,∞)(t) with ai ∈ X and 0 ≤ t0 < t1 <

t2 . . . , then (1.2) represents a Dirichlet series∑i≥0

aie−λti . (1.3)

1clearly, χI denotes the characteristic function of I ⊂ R.

9

In general, (1.2) is a true generalization of (1.1) and (1.3); i.e., there are functions

r : C ⊃ Ω → X which have a representation (1.2) but which cannot be expressed

either in form (1.1) or (1.3).

Recall that a function F : [a, b] → X is of bounded semivariation if there exists

M ≥ 0 such that ‖∑

i(F (ti)− F (si))‖ ≤ M for every choice of a finite number of

non overlapping intervals (si, ti) in [a, b]. A function F is of bounded variation on

[a, b] if there exists M ≥ 0 such that∑

i ‖(F (ti)− F (ti−1)‖ ≤ M for every finite

partition a = t0 < t1 < . . . < tn = b of [a, b]. The set of functions F : [a, b] → X of

bounded semivariation is denoted by BSV ([a, b], X). A function F : R+ → X is

in BSVloc(R+, X) if it is of bounded semivariation on any compact subinterval of

R+. Clearly, any function of bounded variation is of bounded semivariation, and in

finite dimensions the two concepts are equivalent: i.e. every function F : [a, b] → C

which is of bounded semivariation is also of bounded variation (see [1], Section

1.9). However, in infinite dimensions there are functions of bounded semivariation

which are not of bounded variation. Take, for instance, F : [0, 1] → L∞[0, 1] de-

fined by F (t) := χ[0,t]. Then it is easy to check that F is not of bounded variation,

but it is of bounded semivariation.

To discuss the domain of existence of f we introduce the abscissa of convergence

of f by

abs(f) := infReλ : f(λ) exists

. (1.4)

It will be shown in Proposition 1.1 that the set of λ ∈ C for which the Laplace in-

tegral converges is either empty (i.e., abs(f) = +∞), or a right half plane bounded

to the left by abs(f) < ∞, or the entire complex plane (i.e., abs(f) = −∞). To

estimate abs(f) in terms of f : R+ → X one considers the exponential growth

10

bound

ω(f) := inf

ω ∈ R : sup

t≥0

∥∥e−ωtf(t)∥∥ <∞

. (1.5)

Clearly, abs(f) ≤ ω(f). However, in general, abs(f) < ω(f). In fact, take f(t) =

eteetsin(eet

). Then, clearly, ω(f) = +∞. However, since F (t) =∫ t

0f(s) ds =

− cos(eet) + cos(e) is bounded, integration by parts shows that

limT→∞

∫ T

0

e−λtf(t) dt = limT→∞

∫ T

0

e−λtF ′(t) dt

= limT→∞

[−e−λT cos(eeT

)− cos e− λ

∫ T

0

e−λtF (t) dt]

exists for Reλ > 0. This shows that, in general, abs(f) < ω(f). Yet, the fundamen-

tal result about the existence of the Laplace integral states that abs(f) is always

determined by the exponential growth of the normalized antiderivative

F0(t) := F (t)− F (∞) (t ≥ 0),

where F (t) :=∫ t

0f(s) ds, F (∞) := limt→∞ F (t) =

∫ ∞0

dF (t) if the limit exists,

and F (∞) := 0 otherwise: i.e.,

F0(t) =

−∫ ∞

tf(s) ds if the integral exists,∫ t

0f(s) ds otherwise.

(1.6)

We will show in Proposition 1.2 below that abs(f) = ω(F0). To do so, we define

abs(F ) := infReλ : dF (λ) exists

. (1.7)

From ∫ T

0

e−λt dF (t) =

∫ T

0

e−λt dF0(t) =

∫ T

0

e−λtf(t) dt,

by letting T →∞, it follows that

abs(f) = abs(F0) = abs(F ), (1.8)

where F, F0 are defined by (1.6).

11

Proposition 1.1. Let F ∈ BSVloc(R+, X). Then dF (λ) converges if Reλ >

abs(F ) and diverges if Reλ < abs(F ).

Proof. It follows from the definition of abs(F ) that dF (λ) does not exist if Reλ <

abs(F ). For λ0 ∈ C define G0(t) =∫ t

0e−λ0sdF (s) (t ≥ 0). Then∫ t

0

e−λsdF (s) =

∫ t

0

e−(λ−λ0)s dG0(s) (λ ∈ C, t ≥ 0).

Integration by parts yields∫ t

0

e−λs dF (s) = e−(λ−λ0)tG0(t) + (λ− λ0)

∫ t

0

e−(λ−λ0)sG0(s) ds. (1.9)

If dF (λ0) exists, then G0 is bounded. Therefore, dF (λ) exists if Reλ > Reλ0 and

dF (λ) = (λ− λ0)

∫ ∞

0

e−(λ−λ0)sG0(s)ds (Reλ > Reλ0). (1.10)

This shows that dF (λ) exists if Reλ > abs(F ).

Also, we see from∫ T

0e−λt dF (t) = F (T )e−λT − F (0) + λ

∫ T

0e−λtF (t) dt that∫ ∞

0e−λt dF (t) converges when ‖F (t)‖ ≤Meωt and Reλ > ω and then∫ ∞

0

e−λt dF (t) = λ

∫ ∞

0

e−λtF (t) dt− F (0) (Reλ > ω). (1.11)

Thus, for any function F of bounded semivariation

abs(F ) ≤ ω(F ).

Since abs(f) = abs(F ), to prove abs(f) = ω(F ) it remains to be shown that

ω(F ) ≤ abs(F ) for F ∈ BSVloc(R+, X). For X = C, the following proposition is

due to D.V. Widder [26].

Proposition 1.2. Let F ∈ BSVloc(R+, X).

(a) If∫ ∞

0e−λt dF (t) exists for λ = λ0 = γ + iδ , γ > 0, then F (t) = o(eγt) as

t→∞.

12

(b) If∫ ∞

0e−λt dF (t) converges for λ = λ0 = γ + iδ , γ < 0, then F (∞) exists and

F0(t) = F (t)− F (∞) = o(eγt) as t→∞.

(c) ω(F0) = abs(F ).

(d) In particular, if F (t) =∫ t

0f(s) ds for some f ∈ L1

loc([0,∞), X), then

abs(f) <∞ if and only if ω(F ) <∞.

Proof. (a) First define G(t) :=∫ t

0e−λ0sdF (s). Then we have that∫ t

0

eλ0sdG(s) =

∫ t

0

e−λ0seλ0sdF (s) =

∫ t

0

dF (s) = F (t)− F (0).

On the other hand, integrating by parts we obtain∫ t

0

eλ0s dG(s) = eλ0tG(t)−∫ t

0

G(s) d(eλ0s) = eλ0tG(t)− λ0

∫ t

0

G(s)eλ0s ds.

So we have that F (t)− F (0) = eλ0tG(t)− λ0

∫ t

0G(s)eλ0s ds. It follows that

[F (t)− F (0)]e−λ0t = G(t)− λ0e−λ0t

∫ t

0

G(s)eλ0s ds.

Finally, taking the limit as t→∞, we get

limt→∞

[F (t)− F (0)]e−λ0t = limt→∞

[G(t)− λ0e−λ0t

∫ t

0

G(s)eλ0sds].

To finish the proof we have to show that

limt→∞

[G(t)− λ0e−λ0t

∫ t

0

G(s)eλ0sds] = 0.

For this let G(∞) := limt→∞∫ t

0e−λ0sdF (s) =

∫ ∞0e−λ0sdF (s) which exists from the

hypothesis. Then we have

limt→∞

[G(t)− λ0e−λ0t

∫ t

0

G(s)eλ0s ds] = G(∞)− limt→∞

λ0e−λ0t

∫ t

0

G(s)eλ0s ds

= limt→∞

λ0e−λ0t

∫ t

0

(G(∞)−G(s))eλ0sds.

13

At last, let ε > 0. From the definition of G(∞) it follows that there exists a T0

such that ‖G(∞)−G(t)‖ < ε for s > T0. Then

limt→∞

λ0e−λ0t

∫ t

0

(G(∞)−G(s))eλ0s ds = limt→∞

λ0e−λ0t[

∫ T0

0

(G(∞)−G(s))eλ0s ds

+

∫ t

T0

(G(∞)−G(s))eλ0s ds].

The first term in the limit will go to zero because it does not depend on t and

the second term is bounded by ε because∥∥∥λ0e

−λ0t∫ t

T0(G(∞)−G(s))eλ0sds

∥∥∥ ≤

λ0e−λ0t

∫ t

T0ε eλ0sds = ε e−λ0t(eλ0t − eλ0T0). Since ε can be chosen arbitrarily small

we have that

limt→∞

[F (t)− F (0)]e−λ0t = 0

which implies that F (t) = o(eγt) (as t→∞).

(b) Since γ < 0 it follows from Proposition 1.1 that

limT→∞

∫ T

0

dF (t) = F (∞)− F (0)

exists. So, the existence of F (∞) is assured. Next consider

F (∞)− F (t) =

∫ ∞

t

eλ0sdG(s),

where G(t) is defined as in (a). Integration by parts gives

F (∞)− F (t) = e−λ0tG(t)− λ0

∫ ∞

t

eλ0sG(s) ds.

Multiplying with e−λ0t and taking the limit we get

limt→∞

[F (∞)− F (t)]e−λ0t = −G(∞)− limt→∞

λ0e−λ0t

∫ ∞

t

eλ0sG(s) ds

and the same argument as in (a) finishes the proof.

(c) From (a) it follows that, if abs(F ) > 0 then ω(F0) = ω(F ) ≤ abs(F ). From

(b) it follows that, if abs(F ) < 0 then ω(F0) ≤ abs(F ). To finish the argument

14

assume abs(F ) = 0. There are two cases to consider. Case 1: If∫ ∞

0dF (t) does

not exists, then F (∞) = 0 and F0(t) = F (t). Since abs(F ) = 0, F exists for any

λ := ε + iδ with ε > 0. Then by (a) it follows that F (t) = o(eεt) which implies

ω(F0) = ω(F ) < ε. Taking the limit as ε→ 0 it follows that ω(F0) = ω(F ) ≤ 0 =

abs(F ).

Case 2. If∫ ∞

0dF (t) exists, then F0(t) = F (t)− F (∞). As above

‖F0(t)‖ ≤ ‖F (t)‖+ ‖F (∞)‖ = o(eεt) + ‖F (∞)‖ .

for all ε > 0. Thus, ω(F0) ≤ 0 = abs(F ).

(d) Statement (d) follows now from the fact that F ′(t) = f(t) almost everywhere

and therefore we have that dF (λ) =∫ ∞

0e−λt dF (t) =

∫ ∞0e−λtf(t) dt = f(λ).

Hence, dF (λ) exists if and only if f(λ) exists, and abs(f) = abs(F ) = ω(F0) (by

(c)). Now, since F0(t) = F (t) − F (∞) it follows that ω(F ) < ∞ if and only if

ω(F0) <∞. Hence ω(F ) <∞ if and only if ω(F0) = abs(F ) = abs(f) <∞.

The previous lemma will not hold for the case when λ0 ∈ iR as it can be seen

in the next example.

Example 1.3. Consider the Heaviside function F (0) := 0 and F (t) := 1 for t > 0.

Then∫ ∞

0e−λt dF (t) converges for any choice of λ > 0 but F (t) 6= o(1) as t → ∞.

It is still to be noticed that F (t) = o(eεt) for any choice of ε > 0. Also, if F (t) := 2

for t ∈ [0, 1] and F (t) := 2√t for t > 1 then, at λ = ir (r 6= 0), we have∫ ∞

0

e−λt dF (t) =

∫ ∞

1

e−irt

√tdt.

Using integration by parts we have that∫ ∞

1

e−irt

√tdt =

1

ir− 1

2ir

∫ ∞

1

e−irt

√t3dt.

So, F (ir) converges for all r 6= 0, yet F (∞) does not exists.

15

One of the main aims of this chapter is to describe and analyze uniqueness se-

quences for the Laplace transform. To do so we recall first the Riesz-Representation

theorem for Lipschitz-continuous functions. We say that F ∈ Lip0([0,∞), X) if

F (0) = 0 and ‖F‖Lip := supt,s≥0‖F (t)−F (s)‖

|t−s| < ∞. In order to see how the Laplace

transform relates to the Laplace-Stieltjes transform, let f ∈ L∞([0,∞), X). Then

F (t) :=∫ t

0f(s) ds is in Lip0([0,∞), X) and the Laplace-Stieltjes transform re-

duces to the Laplace transform: i.e.,∫ ∞

0e−λt dF (t) =

∫ ∞0e−λtf(t) dt. The key fact

for the Laplace transform theory is the Riesz-Stieltjes representation of bounded

linear operators from L1(0,∞) into X. For a proof of this theorem see [1], Section

2.1.

Theorem 1.4. (Riesz-Stieltjes Representation). There exists an isometric iso-

morphism RS : Lip0([0,∞), X) → L(L1(0,∞), X) given by RS(F ) := T where

Tg :=∫ ∞

0g(t) dF (t) for all continuous functions g ∈ L1(0,∞) and Tχ[0,t] = F (t)

for all t ≥ 0.

The Riesz-Stieltjes Representation Theorem is important because when considering

the Laplace-Stieltjes transform LS it allows us to see how properties of F and its

transform dF relate to each-other. Since

dF (λ) = TF e−λt and F (t) = TFχ[0,t]

for all λ, t > 0 it follows that the function F determines the operator TF on the

set of characteristic functions, which is total in L1(0,∞). Therefore TF , and in

particular TF e−λt = dF (λ), is completely determined. On the other hand, any

information on dF (λ) for λ > 0 translates into information on TF on the set

of exponential functions which is also total2 in L1(0,∞) (see Proposition 1.8).

Therefore, dF determines the properties of TF and, in particular, of TFχ[0,t] =

2A set D ⊂ X is called total in X if its linear span is dense in X.

16

F (t), (t ≥ 0). The following results form complex analysis will be used, their

proofs can be found in [22], pages 300-311.

Proposition 1.5. Suppose that h is a bounded analytic function on the unit disc

not identically zero, and α1, α2, . . . are the zeros of h. Then

∞∑n=1

(1− |αn|) <∞

Proposition 1.6. Suppose that rn is analytic on the region Ω for each n ∈ N, that

no rn is identically zero there, and that

∞∑n=1

|1− rn(λ)| (1.12)

converges uniformly on compact subsets of Ω. Then the product

r(λ) =∞∏

n=1

rn(λ) (1.13)

converges uniformly on compact subsets of Ω. Hence r is analytic on Ω. Moreover

we have that

m(r, λ) =∞∑

n=1

m(rn, λ), λ ∈ Ω, (1.14)

where m(r, λ) is defined to be the multiplicity of the zero of r at λ. (If r(λ) 6= 0,

then m(r, λ) = 0).

Motivated by Proposition 1.5 and the fact that the map λ 7→ λ−1λ+1

maps the open

right half plane on the unit circle, we make the following definition.

Definition 1.7. A sequence of distinct complex numbers (λn)n∈N with no accumu-

lation point and with the property that Reλn ≥ γ > 0 for some γ > 0, is called a

uniqueness sequence ifN∑

n=1

1− |λn − 1||λn + 1|

→ ∞.

17

Proposition 1.8. Let (λn)n∈N be a uniqueness sequence. Then the exponential

functions t 7→ e−λnt are total in L1(0,∞).

Proof. a) Let n0 ∈ N be such that n0 > γ. The family of monomials

M = 1

n0

tn

n0 : n ∈ N0

contains the set 1n0, t, t2 . . .. So, it is total in C[0, 1] and thus in L1(0, 1). Define

Φ : L1(0, 1) → L1(0,∞) by Φ(f)(t) := n0e−n0tf(e−n0t).

Then Φ is an isometric isomorphism and

Φ(1

n0

tn

n0 ) = n0e−n0t 1

n0

(e−n0t)n

n0 = e−(n0+n)t.

Therefore, the family of monomials M is mapped onto the set of exponential func-

tions

E = e−n· : n ≥ n0.

Since M is total in L1(0, 1) it follows that E is total in L1(0,∞).

b) To finish the proof we show that H, which is defined to be the closure of the

linear span of the functions e−λn·n∈N, contains the set E . If H ⊇ E , then the

conclusion follows from a). Suppose that H does not contain the set E . Then there

exists m ≥ n0 such that t 7→ e−mt /∈ H. By the Hahn-Banach theorem there exists

φ ∈ L∞(0,∞) = L1(0,∞)∗ such that φ(H) ≡ 0 and 〈e−m·, φ〉 6= 0. Let 0 < β < γ.

The function

λ→ Ψ(λ) :=⟨e−λ·, φ

⟩=

∫ ∞

0

e−λtφ(t) dt

is analytic and bounded forReλ ≥ β and the function ρ : z → 1+z1−z

+β is a conformal

map between the unit disk and the half plane Reλ > β. Define h(z) := Ψ(ρ(z))

and µn := ρ−1(λn) = λn−β−1λn−β+1

. Then, h is analytic and bounded in the unit disk and

18

h(µn) = Ψ(λn) = φ(e−λn·) = 0 for all n ∈ N. Let an := 1− |µn| = |λn−β+1|−|λn−β−1||λn−β+1|

and bn := 1 − |λn−1||λn+1| = |λn+1|−|λn−1|

|λn+1| . Then, since |λ+ 1|2 − |λ− 1|2 = 4Reλ, it

follows that

an

bn=

|λn + 1| (|λn − β + 1| − |λn − β − 1|)(|λn + 1| − |λn − 1|) |λn − β + 1|

=|λn + 1|

|λn − β + 1|· |λn − β + 1|2 − |λn − β − 1|2

|λn + 1|2 − |λn − 1|2· |λn + 1|+ |λn − 1||λn − β + 1|+ |λn − β − 1|

=|λn + 1|

|λn − β + 1|· Reλn − β

Reλn

· |λn + 1|+ |λn − 1||λn − β + 1|+ |λn − β − 1|

.

Since for 0 < β < γ and Reλn ≥ γ it follows that the distance of λn to −1 is larger

than the distance of λn−β to −1 and to 1; i.e., |λn + 1| ≥ |λn − β ± 1| . Moreover,

it follows from |λn + 1| ≤ |λn − 1|+ 2 that |λn − 1| ≥ |λn + 1| − 2. Thus,

an

bn≥ Reλn − β

Reλn

.|λn + 1|+ |λn − 1|

|λn − β + 1|+ |λn − β − 1|

≥ Reλn − β

Reλn

.2 |λn + 1| − 2

2 |λn + 1|.

≥ (1− β

γ)(1− 1

1 + γ) > 0.

Since∑N

n=1 bn → ∞ as N → ∞, we obtain that∑N

n=1 an =∑N

n=1 1 − |µn| → ∞

as N → ∞. Using Proposition 1.5 it follows that h(µ) = 0 for |µ| < 1. But this

implies that Ψ(λ) = 0 for Reλ > β, contradicting Ψ(m) 6= 0. Thus H = L1(0,∞).

Some examples of sequences satisfying the condition in the previous proposition

are the equidistant sequences λn = a + nb (a, b > 0). Examples of sequences not

satisfying the condition are given by λn = nα for α > 1 and λn = 1 + in. Next

we will show that there are sequences satisfying the condition in the previous

proposition, on any vertical line x = γ > 0, and on any line y = αx for any

α ∈ R. Let λ = (x, y) ∈ C be such that Reλ > 0. Also let A = (−1, 0) and

19

B = (1, 0). Obviously, the distance from λ to A is larger than the distance from λ

to B, therefore

0 ≤ |λ− 1||λ+ 1|

= ε < 1.

Now fix 0 < ε < 1. We are looking for the points λ = (x, y) in the right half plane

such that

0 ≤ |λ− 1||λ+ 1|

= ε

Therefore, we have

√(x−1)2+y2√(x+1)2+y2

= ε, so√

(x− 1)2 + y2 = ε√

(x+ 1)2 + y2. Squaring

both sides we obtain that

x2(1− ε2)− 2x(1 + ε2) + y2(1− ε2) + (1− ε2) = 0.

Dividing by (1− ε2) and completing the squares we obtain that

(x− 1 + ε2

1− ε2)2

+ y2 =(1 + ε2

1− ε2)2 − 1 =

4ε2

(ε2 − 1)2

which is a equation of a circle centered on the real axis at (1+ε2

1−ε2, 0) with radius

2ε1−ε2

. Therefore, the set of points λ = (x, y) in the right half plane for which

|λ− 1||λ+ 1|

= ε

is a circle centered at (1+ε2

1−ε2, 0) with radius 2ε

1−ε2. We should note that as ε → 1−

both the radius and the x-coordinate of the center are going to infinity. Define

εn = n−1n. Then there is a circle Cn with center at (2n2−2n+1

2n−1, 0) and radius 2n2−2n

2n−1

such that for any choice of λn ∈ Cn we have that

N∑n=1

1− |λn − 1||λn + 1|

=N∑

n=1

1− εn =N∑

n=1

1

n→∞

as N →∞. At this point we can show now that there are uniqueness sequences on

any vertical line x = γ with γ > 0 and on any line y = αx (x > 0), where α is any

20

real number. For this , let γ > 0. Then Cn intersects the line x = γ at the points

(γ, yn) where

yn = ±√

(2n− 1− γ)(γ − 1

2n− 1).

Therefore, for any γ > 0 there exists an nγ ∈ N such that the previous equation

has two real roots for any n ≥ nγ. The same type of argument is used to show that

Cn intersects any line y = αx for any n ≥ nα and, therefore there are uniqueness

sequences on any line y = αx.

We say that sequence of complex numbers (λn)n∈N with Reλn ≥ γ > 0 is

called a Laplace uniqueness sequence if given a Laplace transformable function

f ∈ L1loc([0,∞), X) with the property that f(λn) = 0 for each n, implies f ≡ 0.

In the following we will see how the concept of Laplace uniqueness sequence and

uniqueness sequences as defined in Definition 1.7 are related. We need the follow-

ing preliminary result for Lipschitz continuous functions, which is an extension of

Theorem 1.7.3 in [1], which was first mentioned by Pastor in 1919 and is a special

case of a result of Yu-Cheng Shen, see [23].

Proposition 1.9 (Uniqueness). Let (λn)n∈N be a sequence of complex numbers

with no accumulation point, Reλn ≥ γ > 0, and |arg(λn)| ≤ θ for some θ < π2

and

all n ∈ N. The following are equivalent.

(i) (λn)n∈N is a uniqueness sequence.

(ii) For all F ∈ Lip0([0,∞), X) we have that dF (λn) = 0 for all n ∈ N if and

only if F = 0.

Proof. To prove that (i) implies (ii) we have to show that if dF (λn) = 0 for any n ∈

N then F ≡ 0. To accomplish this task we combine Proposition 1.8 with the Riesz-

Stieltjes Representation Theorem 1.4. By Proposition 1.8, if (λn)n∈N is a sequence

21

such that∑N

n=1 1 − |λn−1||λn+1| → ∞ as N → ∞, then the family of exponentials

e−λn· is dense in L1(0,∞). Thus, the operator TF := TFg :=∫ ∞

0g(t) dF (t) has

the property that

TF (e−λn·) =

∫ ∞

0

e−λnt dF (t) = dF (λn) = 0.

Since the induced operator is zero on a total set it follows that it is identically

zero. Therefore the function F is identically zero which implies that (λn)n∈N is a

Laplace uniqueness sequence.

To prove that (ii) implies (i) we have to show that if (λn)n∈N is a Laplace uniqueness

sequence then∑N

n=1 1− |λn−1||λn+1| →∞. We will prove this by contradiction. Suppose∑∞

n=1 1− |λn−1||λn+1| <∞. For λn = dne

iαn we have that

|λn + 1| − |λn − 1|cos(αn)

=4√

1 + 2 cos(αn)dn

+ 1d2

n+

√1− 2 cos(αn)

dn+ 1

d2n

→ 2 as n→∞.

Thus, |λn+1|−|λn−1|cos(αn)

≥ 1 for sufficiently large n. In particular, |λn + 1| − |λn − 1| ≥

cos(αn) ≥ cos(θ) for sufficiently large n. Thus,

1− |λn − 1||λn + 1|

=|λn + 1| − |λn − 1|

|λn + 1|>

cos(θ)

|λn + 1|.

Therefore,∑∞

n=11

|λn+1| < ∞ and for any ε > 0 we have that∑∞

n=11

|λn+ε+1| <∑∞n=1

1|λn+1| <∞. Moreover,

∞∑n=1

1− |λn + ε− 1||λn + ε+ 1|

=∞∑

n=1

|λn + ε+ 1| − |λn + ε− 1||λn + ε+ 1|

≤∞∑

n=1

2

|λn + ε+ 1|.

Hence,∑∞

n=1 1− |λn+ε−1||λn+ε+1| <∞.

For each n ∈ N define µn = λn + ε and rn(λ) := µn−λ2+µn+λ

. Then rn is analytic

on the region Reλ > 0 and it has a zero at λ = µn. Now we will check that the

22

conditions in the Proposition 1.6 are satisfied. First we show that

∞∑n=1

|1− rn(λ)| =∞∑

n=1

∣∣∣∣ 2λ+ 2

2 + µn + λ

∣∣∣∣ <∞. (1.15)

Let U be a compact subset of the region Reλ > 0 and let

an :=

∣∣∣∣ 2λ+ 2

2 + µn + λ

∣∣∣∣ and bn :=1

|µn + 1|. (1.16)

Then

limn→∞

an

bn= lim

n→∞

|µn + 1| |2λ+ 2||1 + µn + 1 + λ|

= limn→∞

2 |λ+ 1|∣∣∣1 + 1+λµn+1

∣∣∣ , (1.17)

which exists for any λ ∈ U. Therefore∑∞

n=1 |1− rn(λ)| converges uniformly on

U. By Proposition 1.6 it follows that the product r(λ) =∏∞

n=1 rn(λ) converges

uniformly on compact subsets of the open right half plane Reλ > 0. Hence r(λ) is

analytic on Reλ > 0. Moreover, since only a finite number of factors rn(λ) have

|rn(λ)| > 1 it follows that 0 ≤ |r(λ)| < 1. By Proposition 1.6, r(λ) = 0 if and only

if λ = µn for some n ∈ N. Now let q(λ) := r(λ)λ. Then |λq(λ)| = |r(λ)| ≤ 1 for

Reλ > 0. By [1], Theorem 2.5.1, it follows that there exists f ∈ C0(R+, X) such

that

supt>0

∥∥∥∥f(t)

t

∥∥∥∥ <∞ and q(λ) = λf(λ)

for Reλ > 0. Thus,

f(λ) =q(λ)

λ=

1

λ

∞∏n=1

µn − λ

2 + µn + λ.

Since supt>0

∥∥∥f(t)t

∥∥∥ < ∞ it follows that ‖f(t)e−εt‖ ≤ M for any ε > 0. Therefore,

h(t) := f(t)e−εt has the Laplace transform h(λ) =∫ ∞

0e−(λ+ε)tf(t) dt = f(λ + ε).

Since h is continuous and bounded it follows that H(t) :=∫ t

0h(s) ds belongs to

Lip0([0,∞), X) and we have that

dH(λ) = h(λ) = f(λ+ ε).

23

Thus, dH(λn) = f(λn + ε) = f(µn) = 0, for all n ∈ N, which is a contradiction.

This finishes the proof of the proposition.

Theorem 1.10. Let (λn)n∈N be a sequence of complex numbers with Reλn ≥ γ > 0

and |arg(λn)| ≤ θ < π2

for all n ∈ N. The following are equivalent.

(i) (λn)n∈N is a uniqueness sequence.

(ii) For all f ∈ L1loc([0,∞), X)with abs(f) < γ we have that f(λn) = 0 for all

n ∈ N if and only if f = 0.

Proof. The proof of this statement is a consequence of the previous proposition

and the fact that there is a isometric isomorphism between Lip0([0,∞), X) and

Lipω([0,∞), X) given by

Iω(F )(t) :=

∫ t

0

e−ωsdF (s),

where Lipω([0,∞), X) := F : R+ → X : F (0) = 0, ‖F‖Lipω< ∞ and where

‖F‖Lipω:= supt>s≥0

‖F (t)−F (s)‖∫ ts eωrdr

. For a proof of this fact see [1], Section 2.4. Now, by

Proposition 1.2, if f ∈ L1loc(0,∞) is Laplace transformable, then F (t) :=

∫ t

0f(s)ds

is absolutely continuous and exponentially bounded. Thus, G(t) :=∫ t

0F (s) ds ∈

Lipω[0,∞). Since

f(λ) =

∫ ∞

0

e−λtf(t) dt = λ

∫ ∞

0

e−λtF (t) dt = λ2

∫ ∞

0

e−λtG(t) dt

it follows that (λn)n∈N is a Laplace uniqueness sequence for f if and only if is a

uniqueness sequence for G. Therefore G(t) = 0 implies f(t) = 0, which completes

the proof.

Next we will present an inversion formula for the Laplace transform which is

due to B. Baumer (see [2] and [3]). Recall that a sequence (βn) ⊂ R+ is a Muntz

24

sequence provided that, for all n ∈ N,

βn+1 − βn ≥ 1 and∞∑

n=1

1

βn

= ∞. (1.18)

We should note that Muntz sequences are uniqueness sequences since

∞∑n=1

1− |βn − 1||βn + 1|

=∞∑

n=1

1− βn − 1

βn + 1=

∞∑n=1

2

βn + 1= ∞

because∑∞

n=11

βndiverges. With this inversion formula one can prove many im-

portant properties of the Laplace and convolution transform (see below). We have

tried to generalize this theorem for real Muntz sequences (βn)n∈N to any complex

uniqueness sequence, unfortunately without much success. See the Appendix for

some partial results. For technical reasons, we state the inversion formula first for

the finite Laplace transform q(λ) =∫ T

0e−λtf(t) dt. Later on, in Corollary 1.15, we

will state it for the Laplace Transform f(λ) =∫ ∞

0e−λtf(t) dt.

Theorem 1.11 (Phragmen-Mikusinski Inversion). Let f ∈ C0([0, T ], X) for

some T > 0 and q(λ) :=∫ T

0e−λtf(t) dt. Let (βn)n∈N be a Muntz sequence. Then

f(t) = limn→∞

Nn∑i=1

αn,iq(βni)eβnit, (1.19)

where Nn is chosen such that∑Nn

j=11

βnj≥ T and αn,i are defined by

αn,i := βnie−βni

∑Nnj=1

1βnj

Nn∏j=1,j 6=i

βnj

βnj − βni

. (1.20)

Moreover, the limit in (1.19) is uniform on [0, s] for all 0 < s < T.

25

Proof. First we show that for any T > 0 and any n ∈ N there exists an Nn such

that∑Nn

j=11

βnj≥ T. Since 1

βnj> 1

βnj+ifor any 1 ≤ i ≤ n we have that

n∞∑

j=1

1

βnj

=∞∑

j=1

1

βnj

+ . . .+∞∑

j=1

1

βnj

(n times)

≥∞∑

j=1

1

βnj+1

+ . . .+∞∑

j=1

1

βnj+n

=∞∑

k=n+1

1

βk

= ∞.

Since n is fixed we obtain that∑∞

j=11

βnj= ∞ which assures the existence ofNn. Let

cn :=∑Nn

j=11

βnj. The main idea of this proof is to construct a sequence of functions

φn which will ”converge” to the δ-function. This will be done by showing that the

antiderivatives Φn := 1 ∗ φn converge to the Heaviside function H : t→ χ(0,∞)(t),

where the convergence will be pointwise for any t 6= 0 and uniform for |t| > ε. We

define φn by

φn(t) :=

βne−βn(·) ∗ βn2e

−β2n(·) ∗ . . . ∗ βnNne−βnNn(·)(t+ cn) for t ≥ −cn,

0 otherwise.

Consider the shifted antiderivative of φn given by

ψn := 1 ∗ βne−βn(·) ∗ βn2e

−β2n(·) ∗ . . . ∗ βnNne−βnNn(·).

It’s Laplace transform is∫ ∞

0

e−λtψn(t) dt =1

λ

βn

λ+ βn

. . .βnNn

λ+ βnNn

= γn,01

λ+ γn,1

βn

λ+ βn

+ . . .+ γnNn

βnNn

λ+ βnNn

,

where the coefficients γn,i are found using the partial fraction decomposition and

are given by

γn,i = −Nn∏

j=1,j 6=i

βnj

βnj − βni

.

26

Here we have used the operational property of the Laplace transform of mapping

convolution into multiplication. Finally taking the inverse Laplace transform, and

knowing that the inverse Laplace transform of 1λ+βni

is e−βnit, we have that

ψn(t) = 1 +Nn∑i=1

γn,ie−βnit (1.21)

for t ≥ 0. Since cn =∑Nn

j=11

βnj, the antiderivative of φn, denoted by Φn, is given by

Φn(t) = ψn(t+ cn) = 1 +Nn∑i=1

γn,ie−βni(t+cn)

= 1−Nn∑i=1

Nn∏j=1,j 6=i

βnj

βnj − βni

e−βnicne−βnit

= 1−Nn∑i=1

Nn∏j=1,j 6=i

βnj

βnj − βni

e−βni

∑Nnj=1

1βnj e−βnit.

By definition, αn,i = βnie−βni

∑Nnj=1

1βnj

∏Nn

j=1,j 6=iβnj

βnj−βni. Thus,

Φn(t) = 1−Nn∑i=1

αn,i

βni

e−βnit

for all t ≥ −cn. To show that Φn converges to the Heaviside function we need some

estimates for |αn,i|. For this consider

ln

∣∣∣∣αn,i

βni

∣∣∣∣ = ln

∣∣∣∣∣e−βni∑Nn

j=11

βnj

Nn∏j=1,j 6=i

βnj

βnj − βni

∣∣∣∣∣= −βni

Nn∑j=1

1

βnj

+Nn∑

j=1,j 6=i

ln

∣∣∣∣ βnj

βnj − βni

∣∣∣∣= −βni

Nn∑j=1

1

βnj

+i−1∑j=1

lnβnj

βni − βnj

+Nn∑

j=i+1

lnβnj

βnj − βni

= S1 + S2 + S3.

We estimate first S1+S3. For this consider the function g(t) := −βni

t+ln t

t−βni. Since

limt→∞ g(t) = 0 and g′(t) =−β2

ni

t2(t−βni)it follows that g is a positive and decreasing

function on the interval (βni,∞). Using this and the fact that βni + n(j − i) ≤ βnj

27

implies g(βni +n(j− i)) > g(βnj); i.e., − βni

βnj+ln

βnj

βnj−βni< − βni

βni+n(j−i)+ln βni+n(j−i)

n(j−i)

for any j > i we obtain that

S1 + S3 = −βni

Nn∑j=1

1

βnj

+Nn∑

j=i+1

lnβnj

βnj − βni

≤Nn∑

j=i+1

−βni

βnj

+ lnβnj

βnj − βni

≤Nn∑

j=i+1

(− βni

βnj + n(j − i)+ ln

βni + n(j − i)

n(j − i)

)

≤∞∑

j=i+1

(− βni

βni + n(j − i)+ ln

βni + n(j − i)

n(j − i)

)

=∞∑

j=1

− βni

βni + nj+ ln

βni + nj

nj

<

∫ ∞

0

(− βni

βni + nt+ ln

βni + nt

nt

)dt.

Finally, making the substitution t = βniun

we obtain that

S1 + S3 ≤βni

n

∫ ∞

0

(− 1

1 + u+ ln

1 + u

u

)du =

βni

n

(u ln

1 + u

u

) ∣∣∣∣∞0

.

Since(u ln 1+u

u

) ∣∣∞0

= 1 it follows that S1 + S3 ≤ βni

n. It remains to estimate S2.

For this we observe that βnj ≤ βni − n(i − j) for j < i and that the function

h(t) := tβni−t

is increasing on (0, βni). Therefore,

βnj

βni − βnj

≤ βni − n(i− j)

βni − (βni − n(i− j))=βni − n(i− j)

n(i− j).

Now it follows that

S2 ≤i−1∑j=1

lnβni − n(i− j)

n(i− j)=

i−1∑j=1

lnβni − nj

nj

≤∫ i−1

0

lnβni − nt

ntdt =

βni

n

∫ n(i−1)/βni

0

ln(1

t− 1) dt.

Since ln(1t− 1) < 0 if t ∈ (1

2, 1) it follows that

S2 ≤βni

n

∫ 1/2

0

ln(1

t− 1) dt =

βni

n

∫ 1/2

0

1

1− tdt =

βni ln 2

n.

28

Finally, we have that

S1 + S2 + S3 ≤βni ln 2

n+βni

n=βni(1 + ln 2)

n≤ 2βni

n.

Thus,

|αni| ≤ βnie2n

βni . (1.22)

Now we are able to show that Φn(t) → 1 for all t > 0. Let t > 0. Then

|Φn(t)− 1| ≤Nn∑i=1

∣∣∣∣αn,i

βni

∣∣∣∣ e−βnit ≤∞∑i=1

e2βni

n e−βnit

≤∞∑i=1

e−βnit

2 ≤∞∑i=1

e−nit

2 =e−nt/2

1− e−nt/2.

where the inequalities hold for n > 4t. Therefore, Φn(t) → 1 as n → ∞ uniformly

for t > ε > 0. Next we have to show that Φn(t) → 0 for t < 0. We will show this

in two steps. First we show that∫ ∞

0

e−tΦn(t)dt→ 1.

This follows from the fact that the convolution of two positive functions is positive.

So φn is positive and Φn = 1 ∗ φn is positive and monotonically increasing. The

second step is to show that ∫ ∞

−∞e−tΦn(t)dt→ 1

which will imply again from the positivity and monotonicity that Φn(t) → 0 for all

t < 0. Using the well known inequalities 1+ t < et (t > 0) and et < 11−t

(0 < t < 1)

we obtain that

βni + 1

βni

= 1 +1

βni

< e1

βni <1

1− 1/βni

=βni

βni − 1.

Multiplying by βni

βni+1we have that

1 <βni

βni + 1e

1βni <

β2ni

β2ni − 1

< 1 +1

(ni)2 − 1< e

1(ni)2−1 .

29

Since Φn = 0 for t < −cn it follows that∫ ∞

−∞e−tΦn(t)dt =

∫ ∞

−cn

e−tΦn(t)dt = ecn

∫ ∞

0

e−tΦn(t− cn)dt

= e∑Nn

i=1 1/βni

Nn∏i=1

βni

βni + 1=

Nn∏i=1

βni

βni + 1e1/βni .

Therefore,

1 ≤∫ ∞

−∞e−tΦn(t)dt =

Nn∏i=1

βni

βni + 1≤

∞∏i=1

e1/((ni)2−1) = e∑∞

i=11

(ni)2−1 .

Since∑∞

i=11

(ni)2−1= 1

n2

∑∞i=1

1(i)2−1/n2 it follows that

∫ ∞−∞ e−tΦn(t)dt → 1 which

completes the argument. The only thing left to complete the proof is the converges

of the inversion formula: i.e.∥∥∥f(t)−∑

αn,iq(βni)eβnit

∥∥∥ → 0 as n→∞.

Using that q(βni) =∫ T

0e−βnisf(s)ds we obtain that∥∥∥f(t)−

∑αn,iq(βni)e

βnit∥∥∥ =

∥∥∥∥f(t)−∫ T

0

∑αn,ie

βnite−βnisf(s)ds

∥∥∥∥=

∥∥∥∥f(t)−∫ T

0

φn(s− t)f(s)ds

∥∥∥∥ .Let ε > 0. Then there exists a δ > 0 such that ‖f(t)− f(s)‖ < ε for |t− s| < 2δ.

Also choose a n0 such that Φn(−δ) + 1 − Φn(δ) < ε for all n > n0. Then for

t ∈ [δ, T − δ] we have that∥∥∥∥f(t)−∫ T

0

φn(s− t)f(s)ds

∥∥∥∥ ≤ ∫ t−δ

0

φn(s−t) ‖f(s)‖ ds+∫ T

t+δ

φn(s−t) ‖f(s)‖ ds

+

∥∥∥∥f(t)−∫ t+δ

t−δ

φn(s− t)f(t)ds

∥∥∥∥ +

∫ t+δ

t−δ

φn(s− t) ‖f(t)− f(s)‖ ds

≤ ‖f‖∫ t−δ

0

φn(s− t)ds+ ‖f‖∫ T

t+δ

φn(s− t)ds

+ ‖f‖∣∣∣∣1− ∫ t+δ

t−δ

φn(s− t)ds

∣∣∣∣ + ε

∫ t+δ

t−δ

φn(s− t)ds

≤ ‖f‖ (Φn(−δ)− Φn(−t)) + ‖f‖ (Φn(T − t)− Φn(δ))

+ ‖f‖ |1− Φn(δ) + Φn(−δ)|+ ε(Φn(t+ δ)− Φn(t− δ)) ≤ ε(3 ‖f‖+ 1).

30

Finally, using that f(0) = 0 we obtain for t ∈ [0, δ] that

∥∥∥∥f(t)−∫ T

0

φn(s− t)f(s)ds

∥∥∥∥≤ ‖f(t)‖+

∫ t+δ

0

φn(s− t) ‖f(s)‖ ds+

∫ T

t+δ

φn(s− t) ‖f(s)‖ ds

≤ ε+ ε(Φn(δ)− Φn(t)) + ‖f‖ (Φn(T − t)− φn(δ)) ≤ ε(2 + ‖f‖).

Therefore,∑Nn

i=1 αn,iq(βni)eβnit converges uniformly to f(t) on [0, s].

The Phragmen-Mikusinski inversion theorem has some remarkable properties and

corollaries. To state them we have to introduce the notion of functions of exponen-

tial decay T > 0.

Definition 1.12. Let T > 0. We say that r : (ω,∞) → X is of exponential decay

T, if lim supλ→∞1λ

ln ‖r(λ)‖ ≤ −T. To simplify notation we will write r ≈T 0

whenever r is of exponential decay T.

Observe that a function is of exponential decay T if for all ε > 0 there exists

λ0 > ω such that ‖r(λ)‖ ≤ Meλ(−T+ε) for any λ > λ0. Examples of functions

which are of exponential decay T for all T > 0 are λ 7→ e−λ ln λ and λ 7→ e−λα

for α > 1. We show next that if f ∈ L1loc([0,∞), X) is Laplace transformable

then∫ ∞

Te−λtf(t) dt is of exponential decay T. In particular, for any Laplace trans-

formable f ∈ L1loc([0,∞), X) with supp(f) ⊂ [T,∞) it follows that f(λ) is of

exponential decay T.

Lemma 1.13. Let f ∈ L1loc[0,∞) such that f(λ) exists for λ > ω > 0 and let

T > 0. Then a(λ) :=∫ ∞

Te−λtf(t)dt is of exponential decay T ; i.e., a(λ) ≈T 0.

Proof. Let F (t) :=∫ t

0f(s)ds. From Proposition 1.2 it follows that f exists for

Reλ > ω > 0 if and only if there exist M > 0 such that ‖F (t)‖ ≤ Meωt. Let

31

T ≥ 0, using integration by parts we obtain∫ ∞

T

e−λtf(t)dt = −e−λTF (T ) + λ

∫ ∞

T

e−λtF (t) dt. (1.23)

Therefore, there exists a constant CT > 0 such that

‖∫ ∞

T

e−λtf(t) dt‖ = ‖ − e−λTF (T ) + λ

∫ ∞

T

e−λtF (t) dt‖

≤ e−λT‖F (T )‖+Mλ

λ− ωe−(λ−ω)T

≤ CT e−λT

for all λ > 2ω. This shows that a(λ) =∫ ∞

Te−λtf(t) dt ≈T 0 for all T > 0.

The next corollary states that the Phragmen-Mikusinski inversion formula is

invariant to perturbation of exponential decay; i.e., if for t ∈ [0, T ),

f(t) = limn→∞

Nn∑i=1

αn,iq(βni)eβnit

for some function q then

f(t) = limn→∞

Nn∑i=1

αn,iq(βni)eβnit

for all perturbed functions q = q + r, where r is a perturbation of exponential

decay T.

Corollary 1.14 (Perturbation). Let r : (ω,∞) → X and let (βn)n∈N be a Muntz

sequence. If r is of exponential decay T > 0, then

limn→∞

Nn∑i=1

αn,ieβnitr(βni) = 0

for all 0 ≤ t < T and the limit is uniform for all t ∈ [0, s] and 0 < s < T.

Proof. Let t ∈ [0, T ). Then T > 2T+t3. Since r is of exponential decay T it follows

that there exists a n0 such that e−T−t

3βn < 1

βnand ‖r(βn)‖ ≤ M

βne−

2T+t3

βn and such

32

that 2n0< T−t

3. From 1.22 we know that |αn,i| < βnie

2n

βni < βnieT−t

3βni . Therefore∥∥∥∥∥

Nn∑i=1

αn,ieβnitr(βni)

∥∥∥∥∥ ≤Nn∑i=1

βnieT−t

3βnieβnit

M

βni

e−2T+t

3βni

≤ M

∞∑i=1

eβni(T−t

3+t− 2T+t

3) ≤M

∞∑i=1

e−T−t

3βni

≤ M

∞∑i=1

e−T−t

3ni = M

e−T−t

3n

1− e−T−t

3n→ 0

as n→∞ uniformly for all t ∈ [0, s] and for all 0 < s < T.

Corollary 1.15 (Phragmen-Mikusinski Inversion, Part II).

Let f ∈ C0([0,∞), X) be Laplace transformable and let f(λ) =∫ ∞

0e−λtf(t) dt.

Let (βn)n∈N be a Muntz sequence and let αn,i and Nn be as in Theorem 1.11. Then,

for all t ≥ 0,

f = limn→∞

Nn∑i=1

αn,if(βni)eβni(·),

where the limit is uniform on compact sets.

Proof. The proof is an immediate consequence of the Theorem 1.11, Lemma 1.13

and Corollary 1.14.

The following theorem is another consequence and it will characterize the maximal

interval [0, T ] on which a function can vanish in terms of the growth of the Laplace

transform evaluated at Muntz points.

Theorem 1.16 (Support Theorem). Let 0 ≤ T and let f ∈ L1loc([0,∞), X) be

Laplace transformable. Then the following are equivalent:

(i) Every Muntz sequence (βn)n∈N satisfies lim supn→∞1

βnln

∥∥∥f(βn)∥∥∥ = −T.

(ii) For every Muntz sequence (βn)n∈N there exists a Muntz subsequence (βnk)k∈N

satisfying

limk→∞

1

βnk

ln∥∥∥f(βnk

)∥∥∥ = −T.

33

(iii) There exists a Muntz sequence (βn)n∈N satisfying

lim supn→∞

1

βn

ln∥∥∥f(βn)

∥∥∥ = −T.

(iv) f(t) = 0 almost everywhere on [0, T ] and T ∈ supp(f).

(v) lim supλ→∞1λ

ln∥∥∥f(λ)

∥∥∥ = −T.

Proof. We show first that (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i), and then that (iv) is

equivalent with (v).

Suppose (i) holds, and let (βn)n∈N be a Muntz sequence. For each ε = 1k, k ∈ N we

denote by (βεn)n∈N a subsequence of (βn)n∈N obtained by dropping all the elements

of (βn)n∈N for which∥∥∥f(βn)

∥∥∥ ≤ e−(T+ε)βn . The dropped sequence denoted by αεn has

the property that lim supn→∞1

αεn

ln∥∥∥f(αε

n)∥∥∥ ≤ −(T + ε), and by (i) it cannot be a

Muntz sequence, so it converges. Hence the remaining sequence will still satisfy the

Muntz condition. Next, for ε = 1 we pick the first k1 elements of the (β1n)n∈N such

that∑k1

i=11β1

i≥ 1. For ε = 1

2we pick k2 elements such that

∑k1

i=11β1

i+

∑k2

i=11

β1/2i

≥ 2.

Continuing this process in the same fashion we obtain a Muntz sequence which

has the property that limk→∞1

βnkln

∥∥∥f(βnk)∥∥∥ = −T. Clearly, (ii) → (iii).

Suppose that (iii) holds. Since f ∈ L1loc([0,∞), X) is Laplace transformable it

follows that F (t) :=∫ t

0f(s) ds is in C0([0, T ], X) and by 1.23 for T = 0, F (λ) =

1λf(λ). Let t ∈ [0, T ), let FT (λ) :=

∫ T

0e−λtF (t) dt and a(λ) :=

∫ ∞Te−λtF (t) dt. .

Then, by the Phragmen-Mikusinski inversion theorem

F (t) = limn→∞

Nn∑i=1

αn,iFT (βni)eβnit = lim

n→∞

Nn∑i=1

αn,i(F (βni)− r(βni))eβnit

= limn→∞

Nn∑i=1

αn,iF (βni)eβnit − lim

n→∞

Nn∑i=1

αn,ir(βni)eβnit,

34

where the second term goes to zero by Lemma 1.13 and Corollary 1.14. It remains

to be shown that the fist term goes to zero. This follows from∥∥∥∥∥Nn∑i=1

αn,iF (βni)eβnit

∥∥∥∥∥ ≤Nn∑i=1

∥∥∥αn,iF (βni)eβnit

∥∥∥≤

Nn∑i=1

βnieT−t

3βnieβnit

1

βni

∥∥∥f(βni)∥∥∥

≤∞∑i=1

eT−t

3βnieβnite−

2T+t3

βni

≤∞∑i=1

e−T−t

3ni =

e−T−t

3n

1− e−T−t

3n→ 0.

Hence F (t) = 0 for all t ∈ [0, T ). Thus, f = 0 on [0, T ). To show that T ∈ supp(f),

suppose that f = 0 almost everywhere on [0, T + ε). Then it follows that f(λ) =∫ ∞T+ε

e−λtf(t) dt which is of exponential decay T + ε contradicting (iii). Thus, (iv)

holds.

Finally, suppose (iv) holds. Then, from Lemma 1.13 we have that

lim supλ→∞1λ

ln∥∥∥f(λ)

∥∥∥ ≤ −T. Let (βn)n∈N be a Muntz sequence such that

lim supλ→∞

1

βn

ln∥∥∥f(βn)

∥∥∥ ≤ −T − ε.

Then we will have again that F (t) =∫ t

0f(s) ds = 0 for t ∈ [0, T +ε), contradicting

T ∈ supp(f), and therefore (i) holds. Moreover, suppose that (v) does not hold;

i.e.,

lim supλ→∞

1

λln

∥∥∥f(λ)∥∥∥ < −T.

This contradicts (i) so, (v) must be true. Finally, assuming that (v) holds, it follows

that (iii) holds. Thus, (iv) holds and the proof is finished.

A crucial corollary of the Support Theorem 1.16 is the following result due to

E. C. Titchmarsh (see [25]).

35

Corollary 1.17. Let 0 < T , and let k ∈ L1loc[0,∞) and f ∈ L1

loc([0,∞), X) be

Laplace transformable. Then k ∗ f = 0 on [0, T ] implies that there exists x1, x2 > 0

with x1 + x2 ≥ T such that k = 0 almost everywhere on [0, x1] and f = 0 almost

everywhere on [0, x2].

Proof. Let x1 := − lim supλ→∞1λ

ln∣∣∣k(λ)

∣∣∣ and x2 := − lim supλ→∞1λ

ln∥∥∥f(λ)

∥∥∥ .Then, by the Support Theorem 1.16, k and f are zero on [0, x1] and [0, x2] respec-

tively. Furthermore, by the same theorem there exists a Muntz subsequence such

that x1 := − limn→∞1

βnln

∣∣∣k(βn)∣∣∣ and x2 := − limn→∞

1βn

ln∥∥∥f(βn)

∥∥∥ . Since

limn→∞

1

βn

ln∥∥∥k(βn)f(βn)

∥∥∥ = limn→∞

1

βn

ln∣∣∣k(βn)

∣∣∣ + limn→∞

1

βn

ln∥∥∥f(βn)

∥∥∥ = −x1 − x2

we obtain that x1 + x2 ≥ T.

Another remarkable corollary is the celebrated Titchmarsh-Foias Theorem. For

numerical functions (i.e., X = C), the equivalence of (i) and (ii) was proved by

E.C. Titchmarsh in 1926 (see [25]) and the equivalence of (i) and (iii) by C. Foias

in 1961 (see [12]). For Banach space valued functions, the following result is due

to B. Baumer, G. Lumer and F. Neubrander (see [2], [3] and [5]).

Corollary 1.18 (Titchmarsh-Foias). Let k ∈ L1[0, T ] and consider the con-

volution operator Tk : f 7→ k ∗ f from C([0, T ], X) into C0([0, T ], X). Then the

following are equivalent:

(i) 0 ∈ supp(k);

(ii) Tk is one-to-one;

(iii) The range of Tk is dense.

Proof. The fact that (i) is equivalent with (ii) follows essentially from the Corollary

1.17. Suppose that 0 ∈ supp(k); then k is not zero on any interval [0, ε], for ε > 0.

36

Then, by Corollary 1.17 it follows that f = 0 on the interval [0, x1] with x1 ≥ T

whenever k ∗ f = 0 on [0, T ]. This shows that Tk is one-to-one. For the converse

suppose that Tk is one-to-one, but 0 /∈ supp(k); i.e., k(t) = 0 for t ∈ [0, ε]. Then,∫ t

0k(t− s)f(s) ds =

∫ t−ε

0k(t− s)f(s) ds. Now choose f with supp(f) ⊂ [1− ε, 1].

Then k ∗ f = 0 but f 6= 0 contradicting that Tk is one-to-one. It remains to be

shown that (i) is equivalent with (iii). We will show first that (i) implies (iii). Let

βn be a Muntz sequence which satisfies the condition limn→∞1

βnln

∥∥∥k(βn)∥∥∥ = 0

and let εn := max 2√n,− 2

βnln

∥∥∥ k(βn)βn

∥∥∥ and define

gn(t) :=Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn)

where Nn and αn,i are defined in Theorem 1.11. Clearly gn ∈ C([0, T ], X) and we

will show that k ∗ gn → f uniformly on compacts intervals.

(k ∗ gn)(t) =Nn∑i=1

αn,if(βni)

k(βni)

∫ t

0

eβni(t−εn−s)k(s)ds

=Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn)

∫ t

0

e−βnisk(s)ds

=Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn)

[k(βni)−

∫ T

t

e−βnisk(s)ds

]

=Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn)k(βni)

−Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn)

∫ T

t

e−βnisk(s)ds

By Theorem 1.11 the first sum converges to f , therefore to finish the argument we

have to show that the second sum goes to zero. Using the estimates for αn,i from

the Theorem 1.11 and the definition of εn we obtain that∥∥∥∥∥Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn)

∫ T

t

e−βnisk(s)ds

∥∥∥∥∥ ≤37

≤∞∑i=1

βnie2βni/n C

βnie−βniεn/2eβni(t−εn)e−βnit

= C

∞∑i=1

eβni(2/n−εn/2)

≤ Ce2−

√n

1− e2−√

n→ 0

To complete the proof we finally have to show that (iii) implies (ii). Let f ∈

C([0, T ], X). Suppose k ∗ f = 0. We will show that f = 0 which will complete the

proof. Since the range of Tk is dense it follows that there exists a sequence gn such

that k∗gn → Id. Then it follows that 0 = k∗f = lim k∗f∗gn = lim k∗gn∗f = Id∗f.

Therefore∫ t

0(t− s)f(s) ds = 0 for all t ≥ 0 and thus f = 0.

Before continuing, we would like to destile an important fact from the proof

above.

Corollary 1.19 (Inversion of the Convolution Transform). Let k ∈ L1[0, T ]

with 0 ∈ supp(k), and let (βn)n∈N be a Muntz sequence such that

limn→∞

1

βn

ln∣∣∣k(βn)

∣∣∣ = 0.

Then, for all f ∈ C([0, T ], X) we have that

k ∗ gn → f

uniformly on [0, T ], where

gn(t) :=Nn∑i=1

αn,if(βni)

k(βni)eβni(t−εn),

and where the constants Nn, αn,i are defined as in Phragmen-Mikusinski Inversion

Theorem 1.11 and εn := max 2√n,− 2

βnln

∥∥∥ k(βn)βn

∥∥∥.Let f ∈ C0([0, T ], X) and k ∈ L1[0, T ] with 0 ∈ supp(k). To find a solution g of

the equation

k ∗ g = f

38

based on the corollary above, we have to find a topology on C([0, T ], X) for which

the functions gn converge towards a function g. This will be done in the reminder of

this chapter where we define the field of generalized functions via linear extensions

of the spaces C([0, a], X).

A linear extension of a Banach space X is, by definition, a completion of X with

respect to a new norm. In our set-up, new norms are generated through one-to-one

operators T ∈ L(X,Y ), where L(X,Y ) denotes the Banach space of all bounded

linear operators from X into a Banach space Y. The following lemma is essential

for our construction.

Proposition 1.20. Let X, Y be Banach spaces and let T ∈ L(X, Y ) be one-to-one.

On X define a new norm by ‖x‖T := ‖Tx‖ . Then ‖ ‖T is a new norm and if XT

denotes the completion of X with respect to ‖ ‖T we have that

XT T−−−→ Im(T )x xX

T−−−→ Im(T )

i.e. X is continuously embedded in XT , and the operator T extend to an isometric

isomorphism T between XT and Im(T ). In particular (X, ‖ ‖T ) is already a Banach

space if and only if Im(T ) is closed in Y.

Proof. Clearly, ‖ ‖T is a new norm. The identity map X → XT is continuous

since ‖x‖T = ‖Tx‖ ≤ ‖T‖ ‖x‖ . Next we need to show that T is an isometric

isomorphism. Let x ∈ XT . Then there exists a sequence xn ∈ X such that xn → x

in XT and therefore xn is Cauchy in XT and Txn is a Cauchy sequence in Y.

Hence limn→∞ Txn exists and equals y. Now, suppose that vn → x as well. Then,

the same argument shows that limn→∞ Tvn = w. Finally,

‖y − w‖T = limn→∞

‖Txn − Tvn‖ = limn→∞

‖xn − vn‖T = ‖z − z‖T = 0.

39

Therefore T (x) := limn→∞ Txn is a well defined extension of T mapping XT into

Im(T ) which is linear and∥∥∥T x∥∥∥ = limn→∞

‖Txn‖ = limn→∞

‖xn‖T = ‖x‖T

for all x ∈ XK . That shows that T is an isometry and is one-to-one. It remains to

be shown that T is onto. Consider y ∈ Im(T ) and the sequence yn ∈ Im(T ) with

yn → y in Y. Let xn ∈ X be the corresponding sequence with Txn = yn. Then

(xn)n∈N is a Cauchy sequence in XT . Since XT is complete let x = limn→∞ xn.

Then

T (x) = limn→∞

Txn = limn→∞

yn = y.

Therefore we have shown that T is an isometric isomorphism between XT and

Im(T ). To finish the proof we need to show thatX is complete under the norm ‖ ‖T

if and only if Im(T ) is closed in Y. Suppose (X, ‖ ‖T ) is complete. Let y ∈ Im(T ).

Then there exists a convergent sequence (yn)n∈N such that yn ∈ Y and yn → y.

Therefore there exists a sequence xn ∈ X for any n ∈ N such that Txn = yn. so,

we have

limn→∞

‖yn − y‖ = limn→∞

‖Txn − y‖ = 0.

Since Txn converges it follows that (xn)n∈N is Cauchy in (X, ‖ ‖T ). By complete-

ness, it follows that there exists an x such that xn → x and Tx = y. So, y ∈ Im(T ).

Conversely, suppose that Im(T ) is closed. We need to show that any Cauchy se-

quence with respect to ‖ ‖T converges. Let (xn)n∈N be a Cauchy sequence. Then

Txn is a Cauchy sequence in Y and because of the closeness of Im(T ) it follows

that Txn → y for some y ∈ Im(T ). Hence, there exists an x ∈ X such that Tx = y

and xn → x. Therefore (X, ‖ ‖T ) is a Banach space and the proof is complete.

Now, if k ∈ L1[0, a] and 0 ∈ supp(k) then it follows from Corollary 1.18 that the

convolution operator Tk : f 7→ k∗f is a one-to-one operator defined on C([0, a], X)

40

with a dense range in C0([0, a], X). Therefore, by Proposition 1.20, we obtain the

following diagram

C([0, a], X)TkTk−−−→ C0([0, a], X) = Im(Tk)x x

C([0, a], X)Tk−−−→ C0([0, a], X),

where Tk is an isometric isomorphism between the completion of C([0, a], X) with

the norm ‖f‖Tk:= ‖Tkf‖ and C0([0, a], X). For example, if k(t) = 1, then the

antiderivative operator Tk : f 7→ 1∗f defines an isometric isomorphism Tk between

C([0, a], X)Tk and C0([0, a], X). In particular, for any f ∈ C0([0, a], X) there exists

a unique generalized derivative f ′ := Tk

−1f ∈ C([0, a], X)Tk . More general, one

can consider the convolution operators induced by kα : t 7→ tα−1

Γ(α)for α > 0.

Since kα ∈ L1[0, 1] we denote by Tα the operator f 7→ kα ∗ f. Then Tα(Tβ(f)) =

kα ∗ (kβ ∗ f) = (kα ∗ kβ) ∗ f = Tα+β. For a detailed exposition of these facts see

[2], Sections II.2 and II.4. As an immediate consequence of this construction we

obtain the following consequence from Corollary 1.19.

Corollary 1.21. Let k ∈ L1[0, T ] with 0 ∈ supp(k). Then, for all f ∈ C0([0, a], X)

there exists a unique g ∈ C([0, T ], X)Tk such that

k ∗ g := Tkg = f.

Moreover g = Tk − lim gn, where the sequence gn ∈ C([0, a], X) is defined as in

Corollary 1.19.

When one looks at arbitrary generalized function spaces; i.e., C[0, a]Tki with

i = 1, 2 induced by k1, k2 ∈ L1[0, 1] the following shed some light on the structure

of these spaces. We identify elements f ∈ C[0, a]Tk1 and f ∈ C[0, a]Tk2 if they have

41

the same embedding in C[0, a]Tk1∗k2 ; i.e., f = f if

k2 ∗ (k1 ∗ f) = k1 ∗ (k2 ∗ f). (1.24)

By doing so, one is able to define the convolution between the two generalized

functions f ∈ C[0, a]Tk1 and g ∈ C[0, a]Tk2 by

f ∗ g := T−1k1∗k2

((k1 ∗ f) ∗ (k2 ∗ g)) ∈ C[0, 1]Tk1∗k2 .

Clearly, the definition also works if one of the functions is X− valued. Since (f ∗

k1) ∗ (g ∗ k2) = (g ∗ k2) ∗ (f ∗ k1) it follows that f ∗ g = g ∗ f. Suppose that f can

be identified with f ∈ C[0, a]Tk3 ; i.e., k3 ∗ (k1 ∗ f) = k1 ∗ (k3 ∗ f). Then f ∗ g can

be identified with f ∗ g. To see this, observe that f ∗ g ∈ T−1k1∗k2

((k1 ∗ f) ∗ (k2 ∗ g))

can be identified with f ∗ g ∈ T−1k3∗k2

((k3 ∗ f) ∗ (k2 ∗ g)) if and only if

(k3 ∗ k2) ∗ (k1 ∗ k2) ∗ (f ∗ g) = (k1 ∗ k2) ∗ (k3 ∗ k2) ∗ (f ∗ g)

⇔ (k3 ∗ k2) ∗ (k1 ∗ f) ∗ (k2 ∗ g) = (k1 ∗ k2) ∗ (k3 ∗ f) ∗ (k2 ∗ g)

⇔ k2 ∗ [k3 ∗ (k1 ∗ f)] ∗ (k2 ∗ g) = k2 ∗ [k1 ∗ (k3 ∗ f)] ∗ (k2 ∗ g).

Moreover, for all f ∈ C[0, a]Tk2 we have that k1 ∗ f ∈ C[0, a]Tk2 and, for h :=

T−1k1∗k1∗1(1 ∗ k1) we have that

h ∗ (k1 ∗ f) = T−1k1∗k1∗1∗k2

(k1 ∗ k1 ∗ 1 ∗ h ∗ k2 ∗ (k1 ∗ f))

= T−1k1∗k1∗1∗k2

(1 ∗ k1 ∗ k2 ∗ (k1 ∗ f))

= f.

This shows that that the convolution g = k1 ∗ f has a convolution inverse h (i.e.,

h ∗ g = f) if 0 ∈ supp(k1). In order to obtain a convolution inverse for functions k1

with 0 /∈ supp(k1), let us consider the convolution with a function k ∈ L1loc[0,∞)

42

with k = 0 on some interval [0, a] and a ∈ supp(k). Then we have that

(k ∗ f)(t) =

∫ t

0

f(t− s)k(s) ds =

∫ t

a

f(t− s)k(s) ds

=

∫ t−a

0

f(t− a− s)k(s+ a) ds = (k−a ∗ f)(t− a),

where k−a(t) := k(t+a) is the shift of k to the left by a. This means that convoluting

with a function which is zero on [0, a] is the same as convoluting with the left-shifted

function k−a and then right-shifting the result. Since the shift operations are not

invertible in intervals or half lines we will adjust our functions by identifying them

with their zero continuations onto (−∞,∞). In considering functions on (−∞,∞)

the shift operations become invertible. Next, we adapt the the definition of the

convolution operator for functions on half lines. For k ∈ L1loc[a,∞) with a ∈ supp(k)

and f ∈ C[b,∞) with b ∈ supp(f) we define

k ∗ f : t 7→∫ ∞

−∞k(t− s)f(s) ds,

where for notational simplicity both k and f are identified with their zero contin-

uations on (−∞,∞). We remark here that, for a = b = 0 this convolution is the

convolution we have discussed so far. Now for k, f as above we have that

(k ∗ f)(t) =

∫ ∞

−∞k(t− s)f(s) ds =

∫ t−a

b

k(t− s)f(s) ds

=

∫ t−a−b

0

k(t− b− s)f(s+ b) ds

=

∫ t−a−b

0

k(t− a− b− s+ a)f(s+ b) ds

=

∫ t−a−b

0

k−a(t− a− b− s)f−b(s) ds

= (k−a ∗ f−b)(t− a− b) = (k−a ∗ f−b)a+b(t).

First observe that k∗f ≡ 0 on (−∞, a+b). Second, since k−a and f−b are both zero

on (−∞, 0) and the convolution transform is one-to-one on C[0,∞) we obtain that

43

the convolution on the space of continuous functions with supp(f) ⊂ [a,∞) for

some a ∈ R has no zero divisors. In particular, if k1 ∈ L1loc[a,∞) and f ∈ C[b,∞)

with a ∈ supp(k1) and b ∈ supp(f), then the convolution operator is given by

Tkf = Sa+b Tk−a S−bf,

where Sa+b denotes the right shift operator f 7→ f(· − a − b), S−b is the left shift

operator f 7→ f(·+ b) and where f, k are identified with their zero continuation to

(−∞,∞). At this point, we also extend spaces like C[a,∞) to spaces C[a,∞)Tk

via the operators Tk : C[a,∞) → C[a,∞) with

Tkf : f 7→∫ t

a

k(t− s)f(s) ds

for k ∈ C[0,∞) by taking the completion of the space C[a,∞) with respect

to the seminorms ‖f‖α,Tk:= supt∈[a,α] ‖Tkf(t)‖ . For any generalized function

f ∈ C0[b,∞)Tk there is a natural embedding in C0[a,∞)Tk for any a ≤ b just

by identifying k ∗ f with its zero extension to the left and then taking T−1k . In

a similar manner with the finite interval case we identify f ∈ C0[a,∞)Tk1 with

g ∈ C0[b,∞)Tk2 if k2 ∗ k1 ∗ f = k1 ∗ k2 ∗ g, where, again the functions k2 ∗ k1 ∗ f and

k1 ∗ k2 ∗ g are considered as functions in C(−∞,∞), extending them with zero to

the left.

Theorem 1.22 (The field of generalized scalar functions). Let

F := f ∈ C0[a,∞)Tk : a ∈ R, k ∈ C[0,∞) with 0 ∈ supp(k),

and define

f ∗ g := T−1k1∗k2

(k1 ∗ f ∗ k2 ∗ g)

for f ∈ C0[a,∞)Tk1 and g ∈ C0[a,∞)Tk2 . Then F is a field with respect to addition

and convolution.

44

Proof. Clearly, F is an additive group. The convolution is defined for all f, g ∈ F

by the embedding of the functions with different domains in C(−∞,∞) extending

them with zero to the left. If f ∗g = 0 then we have that k1∗f ∗k2∗g = 0. Therefore,

either k1 ∗ f = 0 or k2 ∗ g = 0 which will imply that f = 0 or g = 0. Next we

show that there exists an identity element with respect to the convolution. For

this, let f ∈ C0[a,∞)Tk , f 6= 0 for some k ∈ C[0,∞) with 0 ∈ supp(k ∗ f). Define

h := (f ∗k)−a. Then h ∈ C[0,∞) with 0 ∈ supp(h). Define f−1 := T−1h (k−a). Then

for any g ∈ F , g ∈ C0[a,∞)Tk1 we have that

f ∗ f−1 ∗ g = T−1k∗h∗k1

(k ∗ f ∗ h ∗ T−1h (k−a) ∗ k1 ∗ g) = T−1

k∗h∗k1(h−a ∗ k−a ∗ k1 ∗ g) = g.

Therefore f ∗f−1 is the identity with respect to the convolution. Since, f ∗(g+h) =

f ∗ g + f ∗ h it finally follows that F is a field.

In this construction we have only considered scalar valued functions. The vec-

tor valued generalized functions cannot form a field, since the convolution of two

X−valued functions is not defined if the space X has no algebra structure. How-

ever, we can consider spaces C0([a,∞), X)Tk as above and we can define the con-

volution of a scalar valued generalized function with a vector valued generalized

function. In this way, the vector valued generalized functions form a vector space

over the field of scalar valued generalized functions.

Corollary 1.23 (The vector space of generalized functions). Let X be a

Banach space and let F be the field of generalized scalar functions. Then

V := f ∈ C([a,∞), X)Tk : a ∈ R, k ∈ C[0,∞) with 0 ∈ supp(k)

is a vector space over F where scalar multiplication of a vector f ∈ V with some

h ∈ F is defined by

h ∗ f := T−1k1∗k2

(k1 ∗ h ∗ k2 ∗ f), whereh ∈ C[a,∞)Tk1 , f ∈ C[b,∞)Tk2 .

45

2. Asymptotic Laplace Transforms

The purpose of this chapter is to present asymptotic versions of the Laplace trans-

form which have the benefit that all f ∈ L1loc([0,∞);X) and all generalized func-

tions f ∈ C([0,∞), X)Tk become transformable while, depending on the definition,

all or almost all operational properties of the classical Laplace transform remain

valid. The concept of asymptotic Laplace transforms has its roots in the theory of

asymptotic power series of analytic functions; see, e.g., R. Remmert [21]. Asymp-

totic Laplace transforms were considered first by J.C. Vignaux [27] and further

investigated by J.C Vignaux and M. Cotlar [28], W.A. Ditkin [9], L. Berg [6], Yu. I.

Lyubich [18] and M. Deakin [8]. Since slightly less transparent definitions were used,

the method and the scope of its applicability remained largely unnoticed. In 1999,

G. Lumer and F. Neubrander [16] revisited the asymptotic Laplace transform and

applied the concept to verify general theorems for distributions and hyperfunctions

semigroups. Although the Lumer-Neubrander definition of the asymptotic Laplace

transform was superior to earlier versions, it still had the disadvantage that the

operational property f ′(λ) = (−tf(t)) did not extend to the asymptotic setting. To

remedy this defect, G. Lumer and F. Neubrander gave yet another definition of the

Asymptotic Laplace transform in [17]. In this chapter we will prove the operational

properties of the asymptotic Laplace Transform as defined by G. Lumer and F.

Neubrander in [17], a task that was omitted there. We will also propose a third def-

inition of the asymptotic Laplace transform. Our definition is located somewhere

“in-between” the two definitions proposed by G. Lumer and F. Neubrander, and

enjoys the good properties of both: i.e., all operational Laplace transform prop-

erties are valid and Laplace transforms are ”easily computed”. As we have seen

46

in Proposition 1.1, a function f ∈ L1loc([0,∞);X) is Laplace transformable if and

only if its antiderivative F : t 7→∫ t

0f(s)ds is exponentially bounded. In this case

the Laplace transform f(λ) =∫ ∞

0e−λtf(t)dt = λ

∫ ∞0e−λtF (t)dt is analytic and

bounded in a right half plane.

To state the first definition of the asymptotic Laplace transform given by G. Lumer

and F. Neubrander in [16] we need the following notations. We denote by Σ a post-

sectorial region in C; i.e., an open subset of the right half plane such that for all

angles 0 < φ0 <π2

there exists r0 > 0 such that λ = reiφ ∈ Σ for all r > r0 and

|φ| < φ0.

We also denote by A(Σ, X) the set of all analytic, X−valued functions defined

on some postsector Σ. Recall from Definition 1.12 that a ≈T 0 if a is of exponential

decay T ; i.e.,

lim supλ→∞

1

λln ‖a(λ‖ ≤ −T.

Definition 2.1. Let 0 < T < ∞. The T-asymptotic Laplace transform (of type

one) of f ∈ L1loc([0,∞), X) is given by

fT1 := r ∈ A(Σ, X) : r(λ)−

∫ T

0

e−λtf(t) dt ≈T 0

=

∫ T

0

e−λtf(t) dt+ 0T1 ,

where 0T1 = a ∈ A(Σ, X) : such that a ≈T 0. The asymptotic Laplace trans-

form (of type one) is defined as

f1 :=⋂T>0

fT1

= r ∈ A(Σ, X) : r(λ)−∫ T

0

e−λtf(t) dt ≈T 0 for all T > 0.

Observe that, if r ∈ f1 then, for all T > 0, there exists aT such that

r(λ)−∫ T

0

e−λtf(t) dt =: aT (λ) ≈T 0.

47

Clearly, since λ 7→∫ T

0e−λtf(t) dt is entire and r ∈ A(Σ, X), it follows that aT ∈

A(Σ, X). Since this definition does not include any restrictions on the growth of the

elements in the asymptotic class outside of the real line, the equivalence classes are

huge as soon as they are nonempty (which they always are, see the Existence and

Uniqueness Theorem 2.11 below). The lack of growth restriction on r ∈ A(Σ, X)

allows us to easily compute the asymptotic Laplace transform (of type one) of the

functions t 7→ et2 and t 7→ etn (n ∈ N).

Example 2.2. Let Γ be the directed rectangle consisting of the paths Γ1 := t ∈

[0, T ],Γ2(N) := t = T + ir; r ∈ [0, N ],Γ3(N) := t = r + iN ; r ∈ [0, T ], and

Γ4(N) := t = ir; r ∈ [0, N ]. Since∫

Γ3(N)e−λtet2 dt =

∫ T

0e−λ(r+iN)er2−N2+2irN dr

→ 0 as N →∞, it follows that∫ T

0

e−λtet2 dt = limN→∞

(−

∫Γ2(N)

+

∫Γ4(N)

)e−λtet2 dt.

Define

r(λ) := limN→∞

∫Γ4(N)

e−λtet2 dt = i

∫ ∞

0

e−iλre−r2

dr.

Then r(λ) is entire and r(λ)−∫ T

0e−λtet2 dt = a(λ), where the remainder λ→ a(λ)

is given by a(λ) := limN→∞∫

Γ2(N)e−λtet2 dt = i

∫ ∞0e−λ(T+ir)eT 2−r2+2iT r dr. For

λ > 0, the estimate ‖a(λ)‖ ≤ e−λT eT 2 ∫ ∞0e−r2

dr implies that r2 ∈ et21; that is

et21 = i

∫ ∞

0

e−iλre−r2

dr + 01, (2.1)

where

01 = a ∈ A(Σ, X) : such that a ≈T 0 for all T > 0. (2.2)

Following almost the same argument, we can also compute the asymptotic Laplace

transform of the function t 7→ etn as follows. Let Γ be the directed path consisting

of the paths Γ1 := t ∈ [0, T ], Γ2 := t = T + ir; r ∈ [0, T sin(π/n)], Γ3 := t =

48

rei cos(π/n); r ∈ [0, Tcos(π/n)

]. Then, by Cauchy’s theorem, we have that∫Γ1

e−λt+tn dt+

∫Γ2

e−λt+tn dt+

∫Γ3

e−λt+tn dt = 0.

Define

r(λ) := limT→∞

−∫

Γ3

e−λt+tn dt = −ei πn

∫ ∞

0

e−λrei πn erneiπ

dr

= −ei πn

∫ ∞

0

e−λr(cos( πn

)+i sin( πn

))e−rn

dr.

Then r is entire and

r(λ)−∫ T

0

e−λtetn dt = a(λ),

where

a(λ) :=

∫Γ2

e−λt+tn dt+ ei πn

∫ ∞

T/ cos(π/n)

e−λr(cos(π/n)+i sin(π/n))−rn

dr.

To show that r ∈ et21 we need to show that a ≈T 0. Note that

|a(λ)| =

∣∣∣∣∫Γ2

e−λt+tn dt+ ei πn

∫ ∞

T/ cos(π/n)

e−λr(cos(π/n)+i sin(π/n))−rn

dr

∣∣∣∣≤

∣∣∣∣∣i∫ T sin( π

n)

0

e−λ(T+ir)e(T+ir)n

dr

∣∣∣∣∣ +

∣∣∣∣∣ei πn

∫ ∞

Tcos( π

n )

e−λr(cos( πn

)+i sin( πn

))−rn

dr

∣∣∣∣∣= I1 + I2

For I1 we have that

I1 ≤ e−λT

∫ T sin(π/n)

0

e(T+ir)n

dr = MT e−λT .

For I2 we also have that

I2 =

∫ ∞

Tcos( π

n )

e−λr cos( πn

)−rn

dr ≤∫ ∞

Tcos( π

n )

e−λr cos( πn

) dr =e−λT

λ cos(πn).

Thus, |a(λ)| ≤ MT e−λT + e−λT

λ cos( πn

)≤ e−λT (MT + 1

λ cos( πn

)). Therefore a ≈T 0 and

r ∈ etn1. In particular

etn = −ei πn

∫ ∞

0

e−λr(cos( πn

)+i sin( πn

))e−rn

dr + 01, (2.3)

where 01 is defined by (2.2).

49

As we will see later in this chapter, with this definition all but one of the op-

erational properties of the Laplace transform extend to the asymptotic Laplace

transform. The one operational property of the Laplace transform that does not

extend is f ′(λ) 6= (−tf(t)).

Example 2.3. Let X = C, 1 < α < 2, and r(λ) := e−λαsin(eλ2

). Clearly, we have

that r ≈T 0 for any T > 0. Thus, r ∈ 01. But

lim supλ→∞

1

λln |r′(λ)| = lim sup

λ→∞

1

λln

∣∣∣−e−λα

λα−1 sin(eλ2

) + 2λeλ2−λα

cos(eλ2

)∣∣∣

= +∞.

Therefore, r′ /∈ 01 = −t01. Hence, in general f′1 6= −tf1.

To repair this defect, in the next definition of the asymptotic Laplace transform

given by G. Lumer and F. Neubrander, a smaller class of analytic functions appears;

namely, functions of minimal exponential type on some postsector Σ.

We say that an analytic function u : C ⊃ Σ → X is of minimal exponential type

on a postsector Σ if for every sector

Σφ =λ : |arg(λ)| ≤ φ <

π

2, λ 6= 0

and all ω > 0 there exist M > 0 such that ‖u(λ)‖ ≤ Meω|λ| in Σ ∩ Σφ. We

denote by OΣ(X) the set of X − valued analytic functions of minimal exponential

type on a given postsector Σ. Also, we denote by O(Σ, X) :=⋃

ΣOΣ(X) the set

of analytic functions that are of minimal exponential type on some postsector Σ.

Clearly, O(Σ, X) and OΣ(X) are closed under addition, multiplication and scalar

multiplication; i.e., if u1 ∈ OΣ1(X) and u2 ∈ OΣ2(X) then u3 = u1 + u2 ∈ OΣ3(X)

and u3 = u1 · u2 ∈ OΣ3(X) where Σ3 ⊂ Σ1 ∩ Σ2. With these notations at hand we

can state the second definition of the asymptotic Laplace transform given in [17].

50

Definition 2.4. For 0 < T < ∞, the T-asymptotic Laplace transform (of type

two) of f ∈ L1loc([0,∞), X) is given by the set of analytic functions r defined on

some post-sectorial region Σ with values in X which are of the minimal exponential

type and are asymptotically equal to the finite Laplace transform of f ; i.e.,

fT2 := r ∈ O(Σ, X) : r(λ)−

∫ T

0

e−λtf(t) dt ≈T 0

=

∫ T

0

e−λtf(t) dt+ 0T2 ,

where 0T2 = a ∈ O(Σ, X) such that a ≈T 0. The asymptotic Laplace transform

(of type two) is defined as the intersection of the sets fT2 ; i.e.,

f2 := ∩T>0fT2

= r ∈ O(Σ, X) : r(λ)−∫ T

0

e−λtf(t) dt ≈T 0 for all T > 0.

Observe again that if r ∈ f2, then r ∈ O(S,X). Let aT (λ) := r(λ) −∫ T

0e−λtf(t) dt ≈T 0. Since λ 7→

∫ T

0e−λtf(t) dt is entire and bounded for Reλ > 0, it

follows that aT ∈ O(Σ, X). That is, as in Definition 2.1, the remainder aT inherits

the regularity of r.

As we will see in Proposition 2.13 below, the fact that the elements in the asymp-

totic equivalence class of∫ T

0e−λtf(t) dt are assumed to be of minimal exponential

type is strong enough to ensure the validity of the operational property f′ =

−tf. As a first indication why this might be true consider the function a(λ) =

e−λαsin(eλ2

) discussed in the (counter) example above. Then a is not of minimal

exponential type on any cut sector. For λ = reiθ we have that

a(λ) = e−rαeiθ eir2ei2θ − e−ir2ei2θ

2

= e−rα(cos(αθ)+i sin(αθ)) eir2(cos(2θ)+i sin(2θ)) − e−ir2(cos(2θ)+i sin(2θ))

2

= e−rα(cos(αθ)+i sin(αθ)) eir2 cos(2θ)−r2 sin(2θ)) − e−ir2 cos(2θ)+r2 sin(2θ))

2

51

Using the equality |z1 − z2|2 = |z1|2 + |z2|2 − 2Re(z1 · z2) we obtain that

|a(λ)| = e−rα cos(αθ)√

(e−r2 sin 2θ)2 + (er2 sin 2θ)2 − 2 cos2(r2 cos 2θ))

Therefore, we have

|a(λ)|eωr

= e−rα cos(αθ)−ωr√

(e−r2 sin 2θ)2 + (er2 sin 2θ)2 − 2 cos2(r2 cos 2θ))

= e−rα cos(αθ)−ωr+r2 sin 2θ

√(e−r2 sin 2θ)2

(er2 sin 2θ)2+ 1− 2 cos2(r2 cos 2θ))

(er2 sin 2θ)2

Assuming π2> θ > 0 we obtain that

limr→∞

|a(λ)|eωr

= ∞.

Thus, a is not of minimal exponential type on any cut sector and, therefore, Exam-

ple 2.3 is not an example disproving f′2 = −tf2. In fact, as we will see below,

this operation property holds for ·2 and for ·3.

Before we start proving the operational properties of the asymptotic Laplace trans-

form f1 and f2, let us add yet another definition of the asymptotic Laplace

transform. This definition is based on our observation that the use of postsectors

by G. Lumer and F. Neubrander is not necessary in order for these operational

properties to be valid. To state the third definition we need the following notation.

Denote by Sφ an open cut sector on the right half plane; i.e.,

Sφ := λ : λ = reiθ, r > r0, |θ| < φ <π

2.

We say that an analytic function u : Sφ → X is of minimal exponential type on

Sφ if for all ω > 0 there exists M > 0 such that ‖u(λ)‖ ≤ Meω|λ| in Sφ. We

denote by OSφ(X) the set of X−valued analytic functions of minimal exponential

type on a given open cut sector Sφ. Also, we denote by O(S,X) :=⋃

SφOSφ

(X)

the set of analytic functions that are of minimal exponential type on some open

52

cut sector Sφ. Again, O(S,X) and OSφ(X) are closed under addition and scalar

multiplication; i.e., if u1 ∈ OS1(X) and u2 ∈ OS2(X) then u3 = u1 + u2 ∈ OS3(X)

and u3 = u1 · u2 ∈ OS3(X) where S3 = S1 ∩ S2.

Definition 2.5. For 0 < T < ∞, the T-asymptotic Laplace transform (of type

three) of f ∈ L1loc([0,∞), X) is given by the set of analytic functions r defined on

some sectorial region S with values in X which are of the minimal exponential type

and are asymptotically equal to the finite Laplace transform of f ; i.e.,

fT3 := r ∈ O(S,X) : r(λ)−

∫ T

0

e−λtf(t) dt ≈T 0

=

∫ T

0

e−λtf(t) dt+ 0T3 ,

where 0T3 = a ∈ O(S,X) such that a ≈T 0. The asymptotic Laplace transform

(of type three) is defined as the intersection of the sets fT3 ; i.e.,

f3 := ∩T>0fT3

= r ∈ O(S,X) : r(λ)−∫ T

0

e−λtf(t) dt ≈T 0 for all T > 0.

We remark here that with these definitions we have the following inclusions

f2 ⊂ f3 and f2 ⊂ f1. (2.4)

Since the functions of minimal exponential type play an important role in what

follows, we give some examples of such functions.

Example 2.6. (a) The functions

u(λ) = λnecλγ

are of minimal exponential type in any postsector contained in the region λ :

|λ| > 1 ∩ λ : Reλ > 0 for any n ∈ R , γ ∈ [−1, 1) and c ∈ R. Notice that u is

also of minimal exponential type for γ = 1 provided that c ≤ 0.

53

To see this, it is enough to show that the two factors are of minimal exponential

type. Clearly, for n ≤ 0, λn is of minimal exponential type since it is bounded

for |λ| ≥ ε > 0. For n > 0, consider the positive real function g(x) = xn

eωx (x ≥ 0)

for some ω > 0. Then g(0) = 0 = limx→∞ g(x) = 0 and it follows from g′(x) =

[nxn−1 − ωxn]e−ωx that g attains its maximum at x = nω. Since n

en ≤ e−1 for all n

it follows that xn ≤ nωen e

ωx ≤ 1ωe−1eωx. Therefore,

|λ|n ≤ 1

ωe−1eω|λ| for |λ| ≥ ε > 0.

Next, the function u(λ) = ecλγis of minimal exponential type for γ ∈ [−1, 1). We

have that |u(λ)| = ec|λ|γ cos(γ arg(λ)) ≤ ec|λ|γ . Since ec|λ|γ

eω|λ| = e|λ|[c|λ|γ−1−ω] and

lim|λ|→∞ |λ| [c |λ|γ−1 − ω] = −∞ it follows that for |λ| > 1 there exists a M > 0

such that e|λ|[c|λ|γ−1−ω] ≤ eM . Thus |u(λ)| ≤ eMeω|λ| for |λ| > 1.

(b) The function

u(λ) := ln(λ) := ln |λ|+ iarg(λ)

with −π < arg(λ) < π (the principal branch of the logarithm) is of minimal

exponential type in any postsector contained in the region λ : |λ| > 1 ∩ λ :

Reλ > 0. To see this observe that |ln(λ)| ≤ ln(|λ|) + |arg(λ)| ≤ |λ| + π for

|λ| ≥ 1. Thus, for all ω > 0 there exits M > 0 such that |lnλ| ≤ Meω|λ| for all

|λ| ≥ 1.

(c) The function

u(λ) := e−λ ln λ

is of minimal exponential type in any postsector contained in the region λ : |λ| >

1 ∩ λ : Reλ > 0. To see this,let ω > 0 and λ = reiα. Then

Re(−λ lnλ) = Re[(−r cosα− ir sinα)(ln r + iα)] = −r cosα ln r + rα sinα

54

and therefore

|u(λ)|eω|λ| = e−r cos α ln r+rα sin α−ωr.

The real function g(α) := −r cosα ln r+rα sinα−ωr is an even function so, it will

be enough to study g for α ∈ [0, θ] where 0 < θ < π2. Since g′(α) = r sinα ln r +

r sinα + rα cosα > 0, it follows that g(α) is increasing for α ∈ [0, θ]. Because g

is even it follows that max|α|≤θ g(α) = g(θ). Finally, since limr→∞(−r cos θ ln r +

rθ sin θ − ωr) = −∞, it follows that there exists a M > 0 such that

e−r cos α ln r+rα sin α−ωr ≤ eg(θ) ≤ eM

for all α ∈ [−θ, θ].

(d) The function

u(λ) := e−λ2

is of minimal exponential type on the sector Sπ4

:= λ : λ = reiθ, |θ| < π4, but

is not of minimal exponential type on any postsector Σ. To see this observe that

|u(λ)| = e−r2 cos 2α. It follows that u is of minimal exponential type for |α| < π4. If

α > π4, then u(λ) is not of minimal exponential type.

We should note here that, by definition, all examples of analytic functions of

minimal exponential type on some postsector are of minimal exponential type on

all open cut sectors with a sufficiently large cut. Next we will prove a useful lemma

which describes the behavior of functions of minimal exponential type which are

also of exponential decay T.

Lemma 2.7. Let a ∈ O(Sφ, X). The following are equivalent

(a) a ≈T 0; i.e., lim supλ→∞1rln ‖a(λ)‖ ≤ −T,

(b) lim supr→∞1rln ‖a(reiθ)‖ ≤ −T cos θ for all −φ < θ < φ.

55

Proof. Clearly, (b) implies (a). To show that (a) implies (b) we define the

Phragmen-Lindelof function h by

h(φ) := lim supr→∞

1

rln ‖a(reiφ)‖ <∞. (2.5)

To complete this proof we need the following lemma taken form E.C. Titchmarch

[24], Chapter 5, Section 5.7.

Lemma 2.8. Let α < φ1 < φ2 < β, and φ2−φ1 < π, and let h(φ1) ≤ h1, h(φ2) ≤

h2. Let H(φ) be the function of the form a cosφ + b sinφ which takes the values

h1, h2 at φ1, φ2. i.e.,

H(φ) =h1 sinφ2 − h2 sinφ1

sin(φ2 − φ1))cosφ+

h1 cosφ2 − h2 cosφ1

sin(φ1 − φ2))sinφ,

which takes the values h1, h2 at φ1, φ2. Then

h(φ) ≤ H(φ) (for all φ such that φ1 ≤ φ ≤ φ2).

Using this lemma with φ1 = 0, φ2 = π2− µ, where 0 < µ < π

2, we have that

h(φ1) = h(0) ≤ −T ; i.e., h1 = −T and h(φ2) = lim supr→∞1rln ‖a(reiφ2)‖ ≤ ω, for

any ω > 0; i.e., h2 = 0. Therefore, a = H(0) = −T and 0 = H(π2− µ) which gives

b =T cos(π

2−µ)

sin(π2−µ)

. Hence,

h(φ) ≤ H(φ) = −T cosφ+T cos(π

2− µ)

sin(π2− µ)

sinφ,

for all 0 < φ < π/2− µ. Thus, taking the limit as µ→ 0 we have that

h(φ) ≤ −Tcosφ for |φ| < π/2.

This finishes the proof of the Lemma 2.7.

Lemma 2.9. Let f ∈ L1loc([0,∞), X) such that f(λ) exists for λ > ω > 0. Then

f ∈ fi for i = 1, 2, 3. Moreover

56

(a) f is of minimal exponential type on the half plane Reλ > ω+ε, for any ε > 0

and

(b) a(λ) :=∫ ∞

Te−λtf(t)dt is of minimal exponential type on the half plane Reλ >

ω + ε and a ≈T 0 for any T > 0.

Proof. (a) Let F (t) :=∫ t

0f(s)ds. Since f(λ) exists for Reλ > ω > 0 if and only if

there exist M > 0 such that ‖F (t)‖ ≤Meωt, integration by parts yields∫ ∞

0

e−λtf(t) dt = +λ

∫ ∞

0

e−λtF (t) dt. (2.6)

Therefore,∥∥∥f(λ)

∥∥∥ ≤ |λ|∫ ∞

0e−ReλtMeωtdt ≤ M |λ|

Reλ−ω. This shows that f is of

minimal exponential type on the half plane λ : Reλ > ω + ε. Therefore, f ∈

f2 ⊂ f3 and f ∈ f1.

To prove (b) observe that a(λ) =∫ ∞

0e−λtf(t)χ(T,∞)(t) dt. Therefore, a is of minimal

exponential type on the half plane Reλ > ω + ε. By Lemma 1.13 it follows that

a ≈T 0.

Remark 2.10. Let f ∈ L1loc([0,∞), X). Then the finite Laplace transform

fT (λ) :=

∫ ∞

0

e−λtf(t)χ[0,T ](t)dt =

∫ T

0

e−λtf(t) dt

is bounded on the half plane λ : Reλ > 0. In particular fT (λ) ∈ fTi for

i = 1, 2, 3.

In the following we will prove that all operational properties of the Laplace

transform extend to the asymptotic Laplace transform. Clearly, f → fi is linear:

i.e.,(f + g)i = fi + gi and cfi = cfi for f, g ∈ L1loc([0,∞), X), c ∈ C

and i = 1, 2, 3. Also notice that ≈T defines an equivalence relation on O(Σ, X).

Furthermore, if r ∈ O(Σ, X) and q ∈ O(Σ, C) with r ≈T 0 and q ≈S 0, then

qr ≈T+S 0. The same holds for O(S,X) and A(Σ, X).

57

Theorem 2.11 (Existence and Uniqueness). 1 Let f, g ∈ L1loc([0,∞);X) and

0 < T ≤ ∞. Then for i = 1, 2, 3

(a) ∅ 6= fi ⊂ fTi ;

(b) if f exists, then f ∈ fi;

(c) fTi ∩ gT

i 6= ∅ if and only if f = g a.e. on [0, T ].

Proof. (a) By definition fi ⊂ fTi . We remark here that in order to prove

statements (a) and (b) it is sufficient to prove them for the smaller equivalence

class; i.e., if (a) and (b) hold for f2 then they hold for f3 and f1. For

f ∈ L1loc([0,∞), X) and 0 < T <∞ define fT := f · χ[0,T ]. Then the finite Laplace

transform fT (λ) =∫ T

0e−λtf(t) dt =

∫ ∞0fT (t) dt is an entire function of minimal

exponential type (even bounded on Reλ > 0) and

fT2 = fT + 0T

2 6= ∅, (2.7)

where 0T2 := a ∈ O(Σ;X) : a ≈T 0. Let 0 < T ′ < T < ∞, F (t) :=

∫ t

0f(s) ds

(t ≥ 0) and h(λ) :=∫ T

T ′ e−λtf(t) dt = e−λTF (T ) − e−λT ′

F (T ′) + λ∫ T

T ′ e−λtF (t) dt.

Since F is bounded on [T ′, T ], it follows that h ≈T ′ 0. Thus

fT2 = fT ′ + h+ 0T

2 ⊂ fT ′ + h+ 0T ′

2 = fT ′ + 0T ′

2 = fT ′

2 .

This shows that

f2 = ∩T>0fT2 ⊂ fT

2 ⊂ fT ′

2 if 0 < T ′ < T <∞.

1This theorem is essentially due to Vignaux [27] who extended Ritt’s Theorem about asymptotic series toasymptotic integrals; see, e.g., R. Remmert [21], 9.6.4 and also J.C. Vignaux and M. Cotlar [28], L. Berg [6]. Inthe present form, the result was proved for 1 by G. Lumer and F. Neubrander in [16].

58

To see that the asymptotic Laplace transform f2 of f is well defined (i.e.,

f 6= ∅ for all f ∈ L1loc((0,∞), X)), define

r(λ) := λ

∫ ∞

0

e−λt(1− e−d(λ)G(t) )F (t) dt, (2.8)

where G(t) := max‖F (t)‖, 1, and λ→ d(λ) is an analytic function to be chosen

below such that r ∈ fT for all T > 0. Since |1− e−z| = |z∫ 1

0e−zt dt| ≤ |z| for all

z ∈ C with Re(z) ≥ 0, it follows that ‖e−λt(1 − e−d(λ)G(t) )F (t)‖ ≤ e−Re(λ)t|d(λ)| for

all t ≥ 0 and λ ∈ Ω := λ ∈ C : Re(λ) > 0 and Re(d(λ)) ≥ 0. If we assume that

Ω contains a post-sectorial region and that d(λ) is of minimal exponential type it

follows that r ∈ O(Σ, X) from the growth estimate

‖r(λ)‖ ≤ 1

Re(λ)|λd(λ)|.

Consider a(λ) = r(λ) −∫ T

0e−λtf(t) dt = r(λ) + λe−λTF (T ) − λ

∫ T

0e−λtF (t)dt

= λ∫ ∞

Te−λt(1 − e−

d(λ)G(t) )F (t) dt − λ

∫ T

0e−λte−

d(λ)G(t)F (t) dt + λe−λTF (T ) = a1(λ) −

a2(λ) + a3(λ). Then r ∈ f if and only if a ≈T 0 for all T > 0. Clearly, a3(λ) =

λe−λTF (T ) satisfies a3 ≈T 0 for all T > 0. Consider a2(λ) = λ∫ T

0e−λte−

d(λ)G(t)F (t) dt.

If λ ∈ Ω∩R+, then ‖a2(λ)‖ ≤MTλe−Re(d(λ))/M , where M = sup0≤t≤T ‖G(t)‖, and

therefore lim supλ→∞1λ

ln ‖a2(λ)‖ ≤ − lim infλ→∞1λRe(d(λ)). Thus, to ensure that

a2 ≈T 0 for all T > 0 we assume that 1λRe(d(λ)) → ∞ as λ → ∞. Finally,

consider a3(λ) = λ∫ ∞

Te−λt(1 − e−

d(λ)G(t) )F (t) dt. Then ‖a3(λ)‖ ≤ |d(λ)|e−λT for all

λ ∈ Ω ∩ R+. This shows that a3 ≈T 0 for all T > 0 if lim supλ→∞1λ

ln(|d(λ)|) ≤ 0.

The considerations above show that r as defined in 2.8 is an element of f if

the damping function d is analytic on a post-sectorial subregion Σ of the open

right half-plane, Re(d(λ)) ≥ 0 for all λ ∈ Σ, 1λRe(d(λ)) → ∞ as λ → ∞, and

lim supλ→∞1λ

ln(|d(λ)|) ≤ 0.

An example of a damping function with these properties is given by

d(λ) = λ ln(λ).

59

From the Example 2.6 it follows that d(λ) is of minimal exponential type. Now

for λ = reiθ we have that Re(d(λ)) = r ln r cos θ − rθ sin θ which implies that

Re(d(λ)) ≥ 0 for all λ ∈ Σ, 1λRe(d(λ)) → ∞ as λ → ∞ and that Σ = λ ∈

C : Re(λ) > 1 ∩ λ = reiα : r > eθ tan(θ), −π/2 < θ < π/2. Since |d(λ)| =√r2(ln r)2 + r2θ2 it follows that lim supλ→∞

1λ

ln(|d(λ)|) = 0, so d(λ) = λ lnλ

satisfies all the required conditions.

(b) Follows from Lemma 2.9.

(c) Clearly, the approximate Laplace transform fTi =

∫ T

0e−λtf(t) dt + 0T

i of

any f ∈ L1loc([0,∞), X) comprises a “large” set of analytic functions. Nevertheless,

the sets are disjoint for different f . If (c) holds for f1 then it holds for f2. To

see this let 0 < T ≤ ∞, and r ∈ fT1 ∩gT

1 . Then there exist a, b ≈T 0 such that

r(λ) =

∫ T

0

e−λtf(t) dt+ a(λ) =

∫ T

0

e−λtg(t) dt+ b(λ)

for all λ > ω. Thus, hT (λ) =∫ T

0e−λth(t) dt = c(λ), where h := f−g, c := b−a, and

c ≈T 0. Now statement (c) follows from the Proposition 1.16; see also B. Baumer

and F. Neubrander [4]. The proof for f3 is identical with the one above.

The usefulness of the Laplace transform in applications to differential and inte-

gral equations is due to the fact that it maps differentiation, integration, and more

generally, convolution onto multiplication. Next it will be shown that these crucial

operational properties extend to the asymptotic Laplace transform as well. Recall

that if f ∈ L1loc([0,∞), X) and , g ∈ L1

loc([0,∞),R) then

f ∗ g : t→∫ t

0

f(t− s)g(s) ds (t ≥ 0)

denotes the convolution of f and g. For all scalar, locally Lebesgue integrable

functions functions g, h and locally Bochner integrable functions f the convolution

60

f ∗g is in L1loc[0,∞), f ∗g = g ∗f , and (g ∗h)∗f = g ∗ (h∗f).2 To simplify notation

we define f∞i := f.

Proposition 2.12. Let 0 < T ≤ ∞, i ∈ 1, 2, 3, and f ∈ L1loc([0,∞);X). If f is

absolutely continuous and f ′ exists a.e., then

f ′Ti = λfT

i − f(0).

If g ∈ L1loc([0,∞); C) , then

f ∗ gTi = fT

i · gTi .

Proof. Let 0 < T <∞. If q ∈ f ′Ti , then there exist ω > 0 and a ≈T 0 such that

q(λ) =∫ T

0e−λtf ′(t) dt+ a(λ) = λ

[∫ T

0e−λtf(t) dt+ 1

λe−λTf(T ) + 1

λa(λ)

]− f(0) for

all λ > ω. Thus q ∈ λfTi − f(0). Conversely, if r ∈ λfT

i − f(0) then there

exist ω > 0 and a ≈T 0 such that r(λ) = λ[∫ T

0e−λtf(t) dt+ a(λ)

]− f(0) =∫ T

0e−λtf ′(t) dt− e−λTf(T )+λa(λ) for all λ > ω. Thus, r ∈ f ′T

i . This shows that

f ′Ti = λfT

i − f(0) for all 0 < T <∞ and thus also for T = ∞.

To prove the other statement, let 0 < T < ∞, r(λ) =∫ T

0e−λtf(t) dt + a(λ), and

q(λ) =∫ T

0e−λsg(s) ds+b(λ), where a, b ≈T 0. Define w(λ) := a(λ)

∫ T

0e−λsg(s) ds+

b(λ)∫ T

0e−λtf(t) dt+ a(λ)b(λ). Then w ≈T 0 and

r(λ)q(λ) =

∫ T

0

∫ T

0

e−λ(t+s)f(t)g(s) dtds+ w(λ)

=

∫ T

0

∫ T+s

s

e−λtf(t− s)g(s) dtds+ w(λ)

=

∫ T

0

∫ t

0

e−λtf(t− s)g(s) dsdt+

∫ 2T

T

∫ T

t−T

e−λtf(t− s)g(s) dsdt+ w(λ)

=

∫ T

0

e−λt(f ∗ g)(t) dt+ e−λT

∫ T

0

e−λt

∫ T

t

f(t+ T − s)g(s) dsdt+ w(λ).

Thus, fTi ·gT

i ⊂ f ∗gTi . If r ∈ f ∗gT

i , then r(λ) =∫ T

0e−λt(f ∗g)(t) dt+a(λ)

for some a ≈T 0. Since∫ T

0e−λt(f ∗ g)(t) dt =

(∫ T

0e−λtf(t) dt

) (∫ T

0e−λsg(s) ds

)−

2For a proof of the scalar case, see N. Dunford and J. Schwartz [11], VIII.1.24, 633-636.

61

e−λT∫ T

0e−λt

∫ T

tf(t + T − s)g(s) dsdt it follows that r ∈ fT

i gTi + 0T

i . This

shows that f ∗ gTi ⊂ fT

i gTi + 0T

i ⊂ f ∗ gTi + 0T

i = f ∗ gTi . Thus,

f ∗ gTi = fT

i · gTi + 0T

i , or fTi · gT

i = f ∗ gTi + 0T

i = f ∗ gTi .

Proposition 2.13. Let f ∈ L1loc([0,∞), X) and i ∈ 2, 3. Then

−tfi = f′i.

Proof. Let u ∈ fi be analytic and of minimal exponential type on a postsector

Σ or an open cut sector S. Let T > 0. Consider a := u− fT . Then a is of minimal

exponential type since it is a difference of two functions of minimal exponential

type and a′(λ) = u′(λ) − f ′T (λ) = u′(λ) − (−tf)T (λ). To show that u′ ∈ −tf2

it will be enough to show that a′ is of minimal exponential type on some adjusted

postsector Σ′ and that a′ ≈T 0. Take a small r > 0 and define Σ′ := λ ∈ Σ

such that the disc Ωr(λ) := ξ : |ξ − λ| ≤ r ⊂ Σ. Then Σ′ is a postsector and

for λ ∈ Σ′ and ξ ∈ δΩr we have that ξ = reiarg(ξ) + λ and ‖a(ξ)‖ ≤ Mωeω|ξ|, for

sufficiently large ξ. By Cauchy’s Theorem,

‖a′(λ)‖ =

∥∥∥∥ 1

2πi

∫δΩr

a(ξ)

(ξ − λ)2dξ

∥∥∥∥ ≤ 1

rmaxξ∈δΩr

‖a(ξ)‖ ≤ Mωeω|ξ|

r=Mωe

ωr

reω|λ|.

This shows that a′ is of minimal exponential type. It remains to be shown that

a′ ≈T 0. Using that a ≈T 0, ξ = |ξ| eiθλ , we obtain that for any ε > 0

‖a′(λ)‖ ≤ 1

rmaxξ∈δΩr

‖a(ξ)‖

≤ 1

re(−T+ε)|ξ| cos θξ

≤ 1

re(−T+ε)(λ+r) cos θξ .

Thus,

1

λln ‖a′(λ)‖ ≤ 1

λln(

1

re(−T+ε) cos θξ(λ+r)) = (−T + ε) cos θξ(1 +

r

λ)− 1

λln r.

62

Now taking the limit as λ → ∞ it follows that lim supλ→∞1λ

ln ‖a′(λ)‖ ≤ −T + ε

since θξ → 0 as λ→∞. Since ε was arbitrary, we conclude that a′ ≈T 0 and thus

u ∈ −tf2. One should note that the previous argument shows that if a function

u is of minimal exponential type on some postsector Σ then the same holds for u′

on a smaller postsector Σ′ and that a ≈T 0 implies a′ ≈T 0.

For the inclusion −tf2 ⊂ f′2, let T > 0 and u ∈ −tf. Then u(λ) =∫ T

0−tf(t)e−λtdt + a(λ), where a ≈T 0. It follows that u(λ) = (

∫ T

0f(t)e−λtdt)′ +

a(λ). It remains to be shown that there exits a A(λ) of minimal exponential

type and A ≈T 0 such that A′(λ) = a(λ). Consider A(λ) :=∫ ∞

λa(µ)dµ =

limλ1→∞∫ λ1

λa(µ)dµ where we integrate along the ray Γ := µ = seiα : α = arg λ.

Then A(λ) is analytic and, by Lemma 2.7, for all ε > 0 there exists r0 > 0 such

that ‖a(µ)‖ ≤ e(−T+ε)s for all s > r0. Thus,

‖A(λ)‖ ≤ limλ1→∞

∫ λ1

λ

‖a(µ)‖ dµ

≤ limr1→∞

∫ r1

r

e−Ts cos α+εs ds

≤ limr1→∞

e−Ts cos α+εs

−T cosα+ ε

∣∣∣∣r1

r

=e−Tr cos α+εr

T cosα− ε

Therefore A(λ) is of minimal exponential type. Moreover, on rays λ = reiα we have

that

lim supr→∞

1

rln

∥∥A(reiα)∥∥ = lim sup

r→∞

1

r(−Tr cosα− εr − ln(T cosα− ε)) ≤ −T cosα.

Now using Lemma 2.7 it follows that A ≈T 0. To complete the proof we have

to show that A′(λ) = a(λ). Consider Γ := Γ1 ∪ Γ2 ∪ Γ3 ∪ Γ4 where, Γ1 := ξ :

ξ = reiα, |λ| ≤ r ≤ r1, Γ2 := ξ : ξ = (1 − t)(Reλ1 + iReλ1 tanα) + t(Reλ1 +

iReλ1 tan β), β = arg(λ + h), t ∈ [0, 1], Γ3 := ξ : ξ = reiβ, Reλ1 cos β ≥ r ≥

63

|λ+ h|,Γ4 := ξ := rλ+(1−r)(λ+h) : r ∈ [0, 1]. Then by Cauchy’s Theorem we

have that∫

Γa(µ)dµ = 0. Since,

∣∣∣∫Γ2a(µ)dµ

∣∣∣ ≤ML ≤ e−TReλ1Reλ1(tan β − tanα)

where M := maxµ∈Γ2 |a(µ)| and L is the length of the curve Γ2 it follows that∫Γ2a(µ)dµ→ 0 as λ1 →∞. Finally,

limh→0

∥∥∥∥A(λ+ h)− A(λ)

h− a(λ)

∥∥∥∥ = limh→0

∥∥∥∥1

h[

∫ ∞

λ+h

a(µ) dµ−∫ ∞

λ

a(µ) dµ]− a(λ)

∥∥∥∥≤ lim

h→0

∥∥∥∥1

h

∫ λ+h

λ

a(µ)dµ− a(λ)

∥∥∥∥≤ lim

h→0

∥∥∥∥1

h

∫ λ+h

λ

a(µ)− a(λ)dµ

∥∥∥∥≤ 1

h‖a(µ)− a(λ)‖h

≤ ε,

by the continuity of a. Therefore A′(λ) = a(λ). Replacing Σ by S and ·2 by ·3

all arguments remain valid. This finishes the proof of the proposition.

In Example 2.2 we have seen that

et21 = i

∫ ∞

0

e−iλre−r2

dr + 01.

Unfortunately, it is difficult to see whether r0 : λ 7→ i∫ ∞

0e−iλre−r2

dr is of minimal

exponential type on a certain open cut sector Sθ. We suspect not, but in absence of

a strong argument for or against the statement r0 ∈ et22 or r0 ∈ et23 we go a

third road and discuss other functions in et2. As shown in the proof of Theorem

2.11, the function

r1(λ) := λ

∫ ∞

0

e−λt(1− e−λ ln(λ)G(t) )F (t) dt

is an asymptotic Laplace transform of f (i.e., r1 ∈ f), where F (t) =∫ t

0es2

ds

and G(t) := max‖F (t)‖, 1. Since r1 is not a classical function of analysis one

might want to find more familiar functions in f. We present one further method,

64

motivated by a method used by L. Berg [6], of constructing elements of et2 which

depend more on the special nature of the function f(t) = et2 . Consider the function

r2(λ) :=

∫ λ/2

0

e−λt+t2 dt.

Since the function λ 7→ e−λt+t2 is analytic in C it follows that r2(λ) is analytic in

C and that the integral is independent of the path chosen [15]. Moreover, r(λ) is of

minimal exponential type on the open cut sector Sπ4, since for the parametrization

t = sλ2

we have that r(λ) = λ2

∫ 1

0e−

λ2

2s+λ2

4s2ds and therefore

‖r2(λ)‖ ≤∣∣∣∣λ2

∣∣∣∣ ∫ 1

0

∣∣∣e−λ2

2s+λ2

4s2

∣∣∣ ds.Now for λ = a+ ib we have that

∣∣∣e−λ2

2s+λ2

4s2

∣∣∣ = e−a2−b2

2s+a2−b2

4s2

which is a positive

decreasing function on [0, 1] with minimum at s = 1, provided that a2 − b2 is

positive. So for λ in the sector Sπ4

:= λ = reiθ, −π4≤ θ ≤ π

4 we have that

‖r2(λ)‖ ≤∣∣λ2

∣∣ which implies the minimal exponential type. Now for λ a real positive

number the convex function t → e−λt+t2 attains its minimum at t = λ2. It follows

that r2(λ) −∫ T

0e−λtet2 dt =

∫ λ/2

Te−λt+t2 dt ≤ (λ

2− T )e−λT+T 2

for all sufficiently

large λ > 0. Therefore, r2 ∈ fT3 for all T > 0 and thus r2 ∈ et23; i.e.,

et23 =

∫ λ/2

0

e−λt+t2 dt+ 03. (2.9)

This example can be generalized for any function of the type t 7→ etn for n > 1

as follows. Define

r(λ) :=

∫ ( λn

)1

n−1

0

e−λt+tn dt.

Since λ 7→ e−λt+tn is analytic in C it follows that r is analytic in C and the integral

is independent of the path chosen. For the parametrization t = s(λn)

1n−1 we obtain

that

r(λ) = (λ

n)

1n−1

∫ 1

0

e−λ( λn

)1

n−1 s+( λn

)n

n−1 sn

ds.

65

Therefore, ‖r(λ)‖ ≤∣∣∣(λ

n)

1n−1

∣∣∣ ∫ 1

0

∣∣∣∣e−λ( λn

)1

n−1 s+( λn

)n

n−1 sn

∣∣∣∣ ds. Since

∣∣∣∣e−λ( λn

)1

n−1 s+( λn

)n

n−1 sn

∣∣∣∣ =

∣∣∣∣∣e−λn

n−1

n1

n−1

(s− 1n

sn)

∣∣∣∣∣ = e−Reλ

nn−1

n1

n−1

(s− 1n

sn)

and the fact that the function f(s) = e−c(s− 1n

sn) is a positive decreasing function on

[0, 1] for c > 0 it follows that ‖r(λ‖ ≤∣∣∣(λ

n)

1n−1

∣∣∣ . Hence r is of minimal exponential

type. It remains to be shown that r ∈ etn3. Again, for λ a real positive number,

the convex function t 7→ e−λt+tn attains its minimum at t = (λn)

1n−1 , hence

r(λ)−∫ T

0

e−λtetn dt =

∫ ( λn

)1

n−1

T

e−λt+tn dt ≤ ((λ

n)

1n−1 − T )e−λT+T n

for all sufficiently large λ > 0. Therefore, r ∈ etnT3 for any T > 0 and thus

r(λ) ∈ etn3. We should notice here that the examples 2.2 and 2.9 and are valid

only when we consider the asymptotic Laplace transform as being the class of

analytic functions of minimal exponential type on a sector. It appears that working

with analytic functions on sectors will simplify the procedure of finding examples,

and it is one of the main ideas of this dissertation to adopt this definition instead of

the ones given earlier by G. Lumer and F. Neubrander. We turn now to the question

about the characteristic properties of an equivalence class fi.We will prove below

that if one of the members of the asymptotic class f has an asymptotic expansion

in terms of 1λ

as λ → ∞ then all members of the asymptotic Laplace transform

have the same asymptotic expansion. The investigation of this topic is what follows

next.

Recall that f(x) = O(g(x)) as x→∞ if there exists a > 0 such that ‖f(x)‖ ≤

M ‖g(x)‖ for x ∈ [a,∞). Also, f(x) = o(g(x)) as x → ∞ if for each ε > 0 there

exists aε such that ‖f(x)‖ ≤ ε ‖g(x)‖ for all x ∈ [aε,∞). The following properties

of the order symbols hold.

66

1 If c1, c2 are constants, then c1O(g) + c2O(g) = O(g).

2 If c1, c2 are constants, then c1o(g) + c2o(g) = o(g).

3 O(O(g)) = O(g).

4 O(o(g)) = o(O(g)) = o(o(g)) = o(g).

5 O(f)O(g) = O(fg).

6 O(f)o(g) = o(fg).

A finite or infinite sequence of functions φn(x)n∈N defined on an interval (a,∞)

is called an asymptotic sequence as x → ∞ if the following two conditions are

satisfied:

(1) φn(x) 6= 0, for n = 1, 2 . . .

(2) φn+1(x) = o(φn(x)), as x→∞.

Obvious examples of asymptotic sequences are given by φn(x) = 1x

or φn(x) = e−nx.

We say that a functionf(x) has an asymptotic development toN terms with respect

to the asymptotic sequence φn(x)n∈N if there exists constants c1, c2, . . . , cN such

that

f(x) = c1φ1(x) + . . .+ cNφN(x) + o(φN(x)) as x→∞.

In the case that f(x) has an asymptotic development to N terms for any N ∈ N,

we say that f(x) has an asymptotic expansion in terms of the sequence φn(x)n∈N

and write

f(x) ∼∞∑

n=1

cnφn(x) as x→∞.

The following theorem is essentially due to G. Lumer (unpublished).

67

Theorem 2.14. Let f : [0,∞) → X extend to an entire analytic function f(z) =

a0+a1z+a2z2+. . .. Then each r ∈ fi admits the uniquely determined asymptotic

expansion at ∞ in 1λn

f(λ) ∼ a0

λ+a1

λ2+

2!a2

λ3+ . . . (2.10)

Proof. We must show that there exists a sequence of vectors (an)n∈N, an ∈ X, such

that for any u ∈ fi and any n ≥ 0,

u(λ) =a0

λ+ . . .+

n!an

λn+1+ o(

1

λn+1) (2.11)

as λ→∞. We do this in several steps.

a) Let T > 0, and n ∈ N be fixed. Then fT (λ) ≈T u(λ), where

fT (λ) =

∫ T

0

e−λtf(t)dt = I1 + I2 (2.12)

with

I1 :=

∫ T

0

e−λt(a0 + a1t+ a2t2 + . . .+ ant

n)dt

I2 :=

∫ T

0

e−λt(an+1tn+1 + . . .)dt

Since f(z) is entire we have that j√‖aj‖ → 0; i.e for any ε > 0 there exists nε > n

such that

‖aj‖ ≤ (ε)j (2.13)

for j ≥ nε. Let 0 < α < 1 and choose ε > 0 such that Tε ≤ α. Write I2 = I∗∗2 + I∗2 ,

where

I∗2 :=

∫ T

0

e−λt∑

m≥nε

amtmdt

and

I∗∗2 :=

∫ T

0

e−λt

nε−1∑m=n+1

amtmdt.

68

Let supt∈[0,T ]

∥∥∑nε−1m=n+1 amt

m−n−1∥∥ := M(nε, T ). Then

‖I∗∗2 ‖ ≤M(nε, T )

∫ T

0

e−λttn+1dt ≤ M(nε, T )(n+ 1)!

λn+2. (2.14)

In particular, ‖I∗∗2 ‖ = o( 1λn+2 ) as λ→∞. For I∗2 we have∥∥∥∥∥ ∑

m≥nε

amtm

∥∥∥∥∥ ≤ tnε(εnε + εnε+1t+ εnε+2t2 + . . .)

≤ αnε

T nεtnε(1 + α+ α2 + . . .)

=αnε

T nε

1

1− αtnε .

Thus,

‖I∗2‖ ≤∫ T

0

e−λt

∥∥∥∥∥ ∑m≥nε

amtm

∥∥∥∥∥ dt≤ αnε

T nε

1

1− α

(nε)!

λnε+1

≤ M∗

λn+2

if λ > 1 and M := αnε

T nε

(nε)!1−α

. In particular,

‖I2‖ = O(1

λn+2) = o(

1

λn+1) (2.15)

as λ→∞.

b) To estimate I1 =∫ T

0e−λt(a0 + a1t+ a2t

2 + . . .+ antn)dt define

E1(λ) : =

∫ ∞

0

e−λt(a0 + . . .+ antn)dt− I1

=

∫ ∞

T

e−λt(a0 + a1t+ . . .+ antn)dt.

Let M1 := supt∈[T,∞)

∥∥a0

tn+ . . .+ an

∥∥ . Then

‖E1(λ)‖ ≤∫ ∞

T

e−λttn∥∥∥a0

tn+

a1

tn−1+ . . .+ an

∥∥∥ dt≤ M1

∫ ∞

T

e−λttndt

≤ M1e−λT (

T n

λ+nT n−1

λ2+ . . .+

n!

λn+1)

≤ M2(n, T )e−λT

69

where M2(n, T ) = supλ≥1

∣∣∣T n

λ+ nT n−1

λ2 + . . .+ n!λn+1

∣∣∣ = T n + nT n−1 + . . .+ n!

To finish the proof, let a(λ) := u(λ) − fT (λ) ≈t 0. In particular, for all µ > 0

there exists Mµ > 0 such that ‖a(λ)‖ ≤ Mµe−λ(T+µ). This implies that ‖a(λ)‖ =

O(e−λ(T+µ)) for all µ > 0. Therefore,∥∥∥∥u(λ)− (a0

λ+a1

λ2+ . . .+

n!an

λn+1)

∥∥∥∥≤

∥∥∥u(λ)− fT (λ)∥∥∥ +

∥∥∥∥I1(λ) + I2(λ)− (a0

λ+a1

λ2+ . . .+

n!an

λn+1)

∥∥∥∥≤

∥∥∥u(λ)− fT (λ)∥∥∥ + ‖I2(λ)‖+ ‖E1(λ)‖

≤ O(e−λ(T+µ)) + o(1

λn+1) +O(e−λT )

= o(1

λn+1)

for any µ > 0. Hence 2.11 holds so the theorem is proved.

This theorem has inspired us to prove the following result, which in essence

states that if one of the functions in the elements of the equivalence class of f

has an asymptotic expansion in terms of 1λ

as λ → ∞, then any other element of

the equivalence class has the same asymptotic expansion.

Proposition 2.15. Let r1, r2 ∈ fi for some f ∈ L1loc([0,∞), X) be such that

r1 has an asymptotic expansion in terms of 1λ

as λ → ∞. Then r2 has the same

asymptotic expansion in terms of 1λ

as λ→∞.

Proof. Suppose that r1, r2 ∈ fi and

r1(λ) = a0 +a1

λ+ . . .+

an

λn+ o(

1

λn+1).

from the definition of the asymptotic Laplace transform we have that for any T > 0

ri(λ)−∫ T

0

e−λtf(t)dt ≈T 0 (2.16)

70

for i = 1, 2. We need to show that

r2(λ)− a0 −a1

λ− . . .− an

λn= o(

1

λn+1)

as λ→∞. Since ∥∥∥r2(λ)− a0 −a1

λ− . . .− an

λn

∥∥∥ =

=

∥∥∥∥r2(λ)−∫ T

0

e−λtf(t)dt+

∫ T

0

e−λtf(t)dt− r1(λ) + r1(λ)− a0 − . . .− an

λn

∥∥∥∥≤

∥∥∥∥r2(λ)−∫ T

0

e−λtf(t)dt

∥∥∥∥ +

∥∥∥∥∫ T

0

e−λtf(t)dt− r1(λ)

∥∥∥∥+

∥∥∥r1(λ)− a0 − . . .− an

λn

∥∥∥≤ e−λT + e−λT + o(

1

λn+1)

= o(1

λn+1)

we have that

r2(λ) = a0 +a1

λ+ . . .+

an

λn+ o(

1

λn+1)

and the proof is complete.

Recall that f(x) = O(g(x)) as x→ 0+ if there exists a > 0 such that ‖f(x)‖ ≤

M ‖g(x)‖ for x ∈ (0, a). Also, f(x) = o(g(x)) as x→ 0 if for each ε > 0 there exists

aε such that ‖f(x)‖ ≤ ε ‖g(x)‖ for all x ∈ (0, aε) Next we prove a generalization

of Watson’s Lemma for the asymptotic Laplace transform.

Theorem 2.16 (Watson’s Lemma). Suppose that f ∈ L1loc([0,∞), X) has an

asymptotic expansion in terms of tnn∈N as t→ 0+; i.e., suppose that

f(t) ∼∞∑

n=0

cntn.

Then, for any u ∈ fi we have that

u(λ) ∼∞∑

n=0

cnn!

λn+1as λ→∞ (2.17)

71

Proof. We need to show that

∥∥∥∥∥u(λ)−N∑

n=0

cnn!

λn+1

∥∥∥∥∥ = o(1

λN+1) as λ→∞

for any N > 0. Let T > 0. Since∑N

n=0 cnn!

λn+1 =∫ ∞

0

∑Nn=0 e

−λtcntn dt we have that

∥∥∥∥∥u(λ)−N∑

n=0

cnn!

λn+1

∥∥∥∥∥ =

∥∥∥∥∥u(λ)−N∑

n=0

∫ ∞

0

e−λtcntn dt

∥∥∥∥∥=

∥∥∥∥∥u(λ)−∫ T

0

N∑n=0

cntne−λtdt−

∫ ∞

T

N∑n=0

cntne−λtdt

∥∥∥∥∥≤

∥∥∥∥∥u(λ)−∫ T

0

N∑n=0

cntne−λtdt

∥∥∥∥∥ +

∥∥∥∥∥∫ ∞

T

N∑n=0

cntne−λtdt

∥∥∥∥∥= I1 + I2

Now

I1 =

∥∥∥∥∥u(λ)−∫ T

0

e−λtf(t) dt+

∫ T

0

e−λtf(t) dt−∫ T

0

N∑n=0

cntne−λt dt

∥∥∥∥∥≤

∥∥∥∥u(λ)−∫ T

0

e−λtf(t) dt

∥∥∥∥ +

∥∥∥∥∥∫ T

0

e−λt(f(t)−N∑

n=0

cntn) dt

∥∥∥∥∥≤ ‖a(λ)‖+

∥∥∥∥∥∫ T

0

e−λt(f(t)−N∑

n=0

cntn) dt

∥∥∥∥∥From the definition of the asymptotic Laplace transform it follows that for any

ε > 0 ‖a(λ)‖ ≤ e−λ(T−ε) for sufficiently large la and from the fact that f(t) ∼∑∞n=0 cnt

n and by using Proposition 1.16 for the function (f(t)−∑N

n=0 cntn)χ[r,T ]

it follows that ∥∥∥∥∥∫ T

0

e−λt(f(t)−N∑

n=0

cntn) dt

∥∥∥∥∥72

≤∫ r

0

e−λt

∥∥∥∥∥f(t)−N∑

n=0

cntn

∥∥∥∥∥ dt+

∥∥∥∥∥∫ T

r

e−λt(f(t)−N∑

n=0

cntn) dt

∥∥∥∥∥≤

∫ r

0

e−λtcN tN dt+

∥∥∥∥∥∫ T

r

e−λt(f(t)−N∑

n=0

cntn)χ[r,T ] dt

∥∥∥∥∥≤

∫ ∞

0

e−λtcN tN dt+Me−λr dt

= cN(N)!

λN+1+Me−λr

= o(1

λN+2).

It remains to be shown that I2 = o( 1λn+1 ).

I2 =

∥∥∥∥∥∫ ∞

T

N∑n=0

cntne−λtdt

∥∥∥∥∥≤ M1

∫ ∞

T

e−λttndt

≤ M1e−λT (

T n

λ+nT n−1

λ2+ . . .+

n!

λn+1)

≤ M2(n, T )e−λT ,

where M2(n, T ) = supλ≥1

∣∣∣T n

λ+ nT n−1

λ2 + . . .+ n!λn+1

∣∣∣ = T n + nT n−1 + . . . + n! In

particular I2 = O(e−λT ) and therefore I2 = o( 1λn+1 ). Thus∥∥∥∥∥u(λ)−

N∑n=0

cnn!

λn+1

∥∥∥∥∥ ≤ o(1

λN+1) +O(e−λT )

= o(1

λN+2)

for any N > 0 and the proof is complete.

In the next part of the dissertation we will briefly describe how to take asymp-

totic Laplace transforms of generalized functions f ∈ C([a′∞), X)Tk . To simplify

the notation, in this section we will drop the index used for different definitions

of the asymptotic Laplace transform and remark that, if used consistently any of

these definitions can be used.

73

Definition 2.17. Let k ∈ C[0,∞) with 0 ∈ supp(k) and f ∈ C0([0,∞), X)Tk .

Then k ∗ f ∈ C0[0,∞), X) and we define the asymptotic Laplace transform of f to

be

f :=k ∗ fk

.

The functions r = pq

with p ∈ k ∗ f and q ∈ k in the asymptotic class

are going to be meromorphic functions in a postsectorial (sectorial region) with

possible poles at the zeros of q ∈ k. Since 0 ∈ supp(k) it follows that k 6= 0.

Thus, the zeros of q ∈ k cannot form a uniqueness sequence.

We need to show that this extension of the asymptotic Laplace transform is well

defined and that the operational properties of the asymptotic Laplace transform

extend to the generalized function case. Recall from Chapter 1 that we identify

two generalized functions f ∈ C0([0,∞), X)Tk1 and g ∈ C0([0,∞), X)Tk2 if

k2 ∗ k1 ∗ f = k1 ∗ k2 ∗ g,

considered as functions on C(−∞,∞) by continuing them with zero to the left.

Suppose now that f, g as above can be identified. Then

f =k1 ∗ fk1

=(k1 ∗ f) ∗ k2k1 ∗ k2

=(k1 ∗ f ∗ k2)k1 ∗ k2

=k1 ∗ (g ∗ k2)k1 ∗ k2

=(g ∗ k2)k2

= g.

where we have used the operational properties of the asymptotic Laplace transform

which are valid for any functions in C0([0,∞), X).

We will show next that the asymptotic Laplace transform is linear. Suppose that

f ∈ C0([0,∞), X)Tk1 and g ∈ C0([0,∞), X)Tk2 and k1, k2 ∈ C[0,∞) with 0 ∈

74

supp(k1) ∩ supp(k2). Then f + g = T−1k1∗k2

(k2 ∗ k1 ∗ f + k1 ∗ k2 ∗ g) and thus

f + g =(k2 ∗ k1 ∗ f + k1 ∗ k2 ∗ g)

k1 ∗ k2

=(k2 ∗ k1 ∗ f)+ (k1 ∗ k2 ∗ g)

k1 ∗ k2

=k2(k1 ∗ f)+ k1(k2 ∗ g)

k1k2

=(k1 ∗ f)k1

+(k2 ∗ g)k2

= f+ g

Recall that one cannot define the convolution of two vector valued generalized

function. However, in the case that one of them is scalar valued the following

operational property holds. Let f ∈ C0([0,∞), X)Tk2 and h ∈ C0[0,∞)Tk1 . Then

the convolution h ∗ f := T−1k1∗k2

(k1 ∗ h ∗ k2 ∗ f). Therefore,

h ∗ f =k1 ∗ h ∗ k2 ∗ f

k1 ∗ k2=k1 ∗ hk1

k2 ∗ fk2

= hf.

In Chapter 1 we have shown that the scalar valued generalized functions

F := f ∈ C0[a,∞)Tk : a ∈ R, k ∈ C[0,∞) with 0 ∈ supp(k),

form a field and that the vector valued generalized functions form a vector space

over that field.

V := f ∈ C([a,∞), X)Tk : a ∈ R, k ∈ C[0,∞) with 0 ∈ supp(k)

is a vector space over F where scalar multiplication of a vector f ∈ V with some

h ∈ F is defined by

h ∗ f := T−1k1∗k2

(k1 ∗ h ∗ k2 ∗ f),

In order to extend the definition of the asymptotic Laplace transform for a gener-

alized function f ∈ C0([a,∞), X)Tk1 we first observe that for f ∈ L1loc([0,∞), X)

75

we have that the asymptotic Laplace transform of the shifted function

fa : t 7→

0 if 0 ≤ t ≤ a

f(t− a) otherwise,

satisfies fa = e−λaf since

faT =

∫ T

0

e−λtfa(t) dt+ 0T

=

∫ T

a

e−λtf(t− a) dt+ 0T

= e−λa

∫ T−a

0

e−λsf(s) ds+ 0T

= e−λa[

∫ T−a

0

e−λsf(s) ds+ eλa0T ]

= e−λafT−a.

We should note that eλa0T = 0T−a since for a ≈T 0 we have that

lim supλ→∞

1

λln

∥∥eλaa(λ)∥∥ ≤ a+ lim sup

λ→∞

1

λln ‖a(λ‖ = a− T = −(T − a).

With this property at hand we can define the asymptotic Laplace transform of a

generalized function f ∈ C0([a,∞), X)Tk for some k ∈ C[0,∞) with 0 ∈ supp(k),

as

f := e−λak ∗ f−ak

.

The asymptotic class is now formed by meromorphic functions defined on a post-

sector (or on a open cut sector), except at the poles. We denote the space of mero-

morphic functions with M(Σ, X) and respectively M(S,X). We also remark that

M(Σ, X) is a vector space over C as well as over M(Σ,C). First, we need to show

that the asymptotic Laplace transform is well defined for generalized functions.

Recall that we identify f ∈ C([a,∞), X)Tk1 with g ∈ C([b,∞), X)Tk2 if

k2 ∗ k1 ∗ f = k1 ∗ k2 ∗ g,

76

considered as functions on C(−∞,∞)) by continuing them with zero to the left.

Then we have that

f = e−λa(k1 ∗ f)−ak1

= e−λa(k1 ∗ f)−a ∗ k2k1 ∗ k2

= e−λ(a+b)(k1 ∗ f ∗ k2)−(a+b)k1 ∗ k2

= e−λbk1 ∗ (g ∗ k2)−bk1 ∗ k2

= e−λb(g ∗ k2)−bk2

= g.

Proposition 2.18. The map · : (V ,+, ∗) →M(Σ, X) has the following proper-

ties

(a) f + g = f+ g.

(b) h ∗ g = hg,

for any f ∈ C0([0,∞), X)Tk1 , g ∈ C0([b,∞), X)Tk2 , h ∈ C0[a,∞)Tk1 and k1, k2 ∈

C[0,∞) with 0 ∈ supp(k1) ∩ supp(k2).

Proof. Without loss of generality we can assume that a ≤ b. By definition, f +g =

T−1k1∗k2

(k2 ∗ k1 ∗ f + k1 ∗ k2 ∗ g). Therefore, using the properties of the asymptotic

Laplace transform for continuous functions we obtain statement (a) as follows

f + g = e−λa(k2 ∗ k1 ∗ f + k1 ∗ k2 ∗ g)k1 ∗ k2

= e−λak2(k1 ∗ f)a+ k1(k2 ∗ g)ak1 ∗ k2

= e−λak2(k1 ∗ f)ak1k2

+ e−λak1(k2 ∗ g)ak1k2

= e−λa(k1 ∗ f)ak1

+ e−λae−λ(b−a)(k2 ∗ g)bk2

= f+ g.

To prove statement (b) recall that for h ∈ C0[a,∞)Tk1 and g ∈ C0([b,∞), X)Tk2

the scalar multiplication in the vector space V is defined as

h ∗ g : T−1k1∗k2

(k1 ∗ h ∗ k2 ∗ g).

77

Thus,

h ∗ g = e−λ(a+b)(k2 ∗ k1 ∗ h ∗ k1 ∗ k2 ∗ g)a+bk2 ∗ k1

= e−λak2(k1 ∗ h)ak2k1

e−λbk1(k2 ∗ g)bk2k1

= hg.

Since integrating is the same as convoluting with the Heaviside function 1 =

χ[0,∞) it follows that we can define the antiderivative F of a generalized function

f ∈ C0([a,∞), X)Tk as F := 1 ∗ f. It follows from Proposition 2.18 that

F := 1 ∗ f = 1f =1

λf.

Thus, f = λF and we have that f ′ = λf. In other words in the space

of generalized functions multiplying with λ corresponds to differentiation while

dividing by λ corresponds to integrating the generalized function.

We conclude this dissertation with some applications of the asymptotic Laplace

transform.

Example 2.19. Consider once again, the linear, first order, inhomogeneous initial

value problem

f ′(t) + f(t) = (2t+ 1)et2 , f(0) = 1. (2.18)

Taking now the asymptotic Laplace transform in both sides of the equation and

using its properties we obtain

78

f ′(t) + f(t) = (2t+ 1)et2 and f(0) = 1

⇔ f ′(t)+ f(t) = (2t+ 1)et2 and f(0) = 1

⇔ λf − 1 + f = (2t+ 1)et2

⇔ (λ+ 1)f = 1 + (2t+ 1)et2

⇔ f = 1λ+1

+ 1λ+1

(2t+ 1)et2

⇔ f = e−t+ e−t(2t+ 1)et2

⇔ f = e−t + e−t ∗ (2t+ 1)et2

= e−t +∫ t

0e−(t−s)(2s+ 1)es2

ds

= e−t + e−t∫ t

0(2s+ 1)es2+s ds

= e−t + e−t[et2+t − 1]

⇔ f = et2

⇔ f(t) = et2 .

Hence, taking the asymptotic Laplace transform yields existence and uniqueness

of the solution of the equation. We remark here that using any of the definition of

the asymptotic Laplace transform will yield the same solution since the property

f′ = −tf is not required to solve this equation. The next example deals with

a more general class of ordinary differential equations.

Example 2.20. Consider the Laplace equation

(a2t+ b2)y′′(t) + (a1t+ b1)y

′(t) + (a0t+ b0)y(t) = 0, (2.19)

where ai, bi are given complex or real numbers with a2 6= 0, t ≥ 0. We may assume

that b2 = 0 by taking (t− b2a2

) as our new variable. Therefore we discuss the equation

(a2t)y′′(t) + (a1t+ b1)y

′(t) + (a0t+ b0)y(t) = 0, (2.20)

with a2 6= 0 and initial conditions y′(0) = y(0) = 0.

79

First observe that

ty′′(t) = [ty(t)]′′ − 2y′(t) and ty′(t) = [ty(t)]′ − y(t).

It follows from f ′ = λf − f(0) and from f′ = −tf that

ty′′(t) = [ty(t)]′′ − 2y′(t) = λ[ty(t)]′ − 2λy(t) = −λ2y(t)′ − 2λy(t)

ty′(t) = [ty(t)]′ − y(t) = λty(t) − y(t) = −λy(t)′ − y(t)

y′′(t) = λ2y(t)

y′(t) = λy(t)

Thus, taking the asymptotic Laplace transform on both sides of the equation we

have

(a2t)y′′(t) + (a1t+ b1)y

′(t) + (a0t+ b0)y(t) = 0.

Using the formulas derived above we obtain

a2[−λ2y′ − 2λy] + a1[−λy′ − y] + b1λy − a0y′ + b0y = 0.

Thus,

y′(−a2λ2 − a1λ− a0) + y((b1 − 2a2)λ+ b0 − a1) = 0. Therefore,

y′

y=

(b1 − 2a2)λ+ b0 − a1

a2λ2 + a1λ+ a0

=γ1

λ− λ1

+γ2

λ− λ2

where, λ1, λ2 are the distinct roots (in the case they exists) of the equation a2λ2 +

a1λ+ a0 = 0, and γ1, γ2 are obtain by using the partial fraction decomposition

γ1 =λ1(b1 − 2a2) + b0 − a1

λ1 − λ2

and

γ2 =λ2(b1 − 2a2) + b0 − a1

λ2 − λ1

.

80

Hence,

ln(y(t)) =

∫γ1

λ− λ1

+γ2

λ− λ2

dλ

= γ1 ln(λ− λ1) + γ2 ln(λ− λ2) + c.

It follows that

y(t) 3 eγ1 ln(λ−λ1)+γ2 ln(λ−λ2)+c = c(λ− λ1)γ1(λ− λ2)

γ2 .

It follows from Example 2.6 that the function r : λ 7→ (λ − λ1)γ1(λ − λ2)

γ2 is a

function of minimal exponential type on a certain open cut sector or postsector.

Now, let q1(λ) := (λ − λ1)γ1 and q2(λ) := (λ − λ2)

γ2 . If γi < 0 then qi(λ) = hi(λ)

where, hi(t) = 1Γ(γi)

t−γi−1eλit.

If γi ≥ 0, then

1

(λ− λi)γi+2qi(λ) =

1

(λ− λi)2∈ teλit.

Let ki(t) := tγi+1

Γ(γi+2)eλit. Then

ki(λ) =1

(λ− λi)γi+2

and ki(λ)qi(λ) ∈ teλit. Let hi := T−1ki

(teλit) ∈ C[0,∞)Tki . Then ki ∗ hi = teλit.

Thus,

ki(λ)qi(λ) ∈ ki ∗ hi

and therefore qi(λ) ∈ ki∗hiki

= hi ⊂ ki∗hiki = hi. Thus,

y(t) = ch1h2 and therefore, y = c(h1 ∗ h2).

81

References

[1] W. Arendt, C.J.K. Batty, M. Hieber and F. Neubrander, Vector-ValuedLaplace Transforms and Cauchy Problems, Monographs in Mathematics,Birkhauser Verlag, 2001.

[2] B. Baumer, A Vector Valued Operational Calculus and Abstract Cauchy Prob-lems, Dissertation, Louisiana State University, 1997.

[3] B. Baumer, On the inversion of the convolution and Laplace transform, Trans-actions of The American Mathematical Society, Vol. 355, 3, 1201-1212.

[4] B. Baumer and F. Neubrander, Laplace Transform methods for evolutionequations, Conference del Seminario di Mathematica dell’ Universita di Bari,259, 1994, 27-60.

[5] B. Baumer, G. Lumer and F. Neubrander, Convolution kernels and generalizedfunctions, Generalized Functions, Operator Theory, and Dynamical Systems.C.R.C. Res. Notes Math., Chapman Hall, 1998.

[6] L. Berg, Asymptotiche Auffassung der Operatorenrechnung, Studia Math. 21,1962, 215-229.

[7] J.L.B. Cooper, Heaviside and the operational calculus, Math. Gaz., 36, 1952,5-19.

[8] M. Deakin, Laplace transform for superexponential functions, J. Austral.Math. Soc. Ser., B 36, 1994, 17-25.

[9] M. Ditkin, On the theory of operational calculus, Dokl. Akad. Nauk SSSR,123, 1958, 395-396.

[10] G. Doetsch, Theorie und Anwendung der Laplace-Transformation, Dover Pub-lications, New York, 1943.

[11] N. Dunford and J. Schwartz, Linear Operators, Vol. I, Interscience, 1958.

[12] C. Foias, Approximation des operateurs de J. Mikusinski par des fonctionscontinues, Studia Mathematica, 21, 1961, 73-74.

[13] M. Gonzales, Classical Complex Analysis, Monographs and textbooks in pureand applied mathematics, Marcel Dekker Inc., 1992.

[14] O. Heaviside, Electro-Magnetic Theory, Vol I-III, D. Van Nostrand Company,London, 1893-1899.

[15] E. Hille, Analytic Function Theory, Vol. I, Ginn and Company, 1959.

82

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[18] Y.I. Lyubich, The classical and local Laplace transformation in an abstractCauchy problem. Usepi Math. Nauk, 21, 1966, 3-51.

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83

Appendix. Remarks on Complex Versionof Phragmen-Mikusinski Inversion

In this Appendix we have collected some partial results we obtained trying to

extend the Phragmen-Mikusinski inversion for sequences in the complex plane.

Although we were not successful, we strongly believe that an inversion formula

based on uniqueness sequences in the complex plane should be valid. Let f ∈

C0([0, T ], X), and let q(λ) :=∫ ∞

0e−λtf(t)dt. Let (λn)n∈N be a uniqueness sequence

such that Reλn is a Muntz sequence. Let Nn be such that cn :=∑Nn

k=11

Reλnk≥ T.

Since Reλn is a Muntz sequence the same argument will hold to show that such

an Nn exists. We proceed in the same manner defining the functions φn by:

φn(t) =

λne−λn(·) ∗ λn2e

−λ2n(·) ∗ . . . ∗ λnNne−λnNn(·)(t+ cn) for t ≥ −cn,

0 otherwise.

Following the same idea we want to show that φn converges to the δ− function:

i.e. The antiderivative Φn := 1 ∗ φn converges to the Heaviside function H(t) =

χ(0,∞)(t). Let ψn := 1∗λne−λn(·)∗λn2e

−λ2n(·)∗. . .∗λnNne−λnNn(·). Taking the Laplace

transform of it we get

ψn(λ) =

∫ ∞

0

e−λtψn(t)dt =1

λ

λn

λ+ λn

. . .λnNn

λ+ λnNn

=1

λ+

Nn∑k=1

γn,k1

λ+ λnk

where the γn,k are obtained using the partial fractions method and are given by:

γn,k = −Nn∏

j=1,j 6=k

λnj

λnj − λnk

Since the inverse Laplace transform of 1λ+λnk

is equal to e−λnkt we obtain that

ψn(t) = 1 +Nn∑k=1

γn,ke−λnkt for all t ≥ 0.

84

Therefore

Φn(t) = ψn(t+ cn) = 1 +Nn∑k=1

γn,ke−λnk(t+cn)

= 1−Nn∑k=1

αn,k

λnk

e−λnkt

Where the αn,k are defined to be

αn,k := λnke−λnk

∑Nnj=1

1Reλnj

Nn∏j=1,j 6=k

λnj

λnj − λnk

= −λnkγn,ke−λnk

∑Nnj=1

1Reλnj

Next we will show that ∣∣∣∣αn,k

λnk

∣∣∣∣ ≤ e1+ln 2

nReλnk

For the argument to work we will need an extra assumption on the sequence

(λn)n∈N: i.e. all λn should lie on a line passing through the origin.

ln

∣∣∣∣αn,k

λnk

∣∣∣∣ = ln

∣∣∣∣γn,ke−λnk

∑Nnj=1

1Reλnj

∣∣∣∣= −Reλnk

Nn∑j=1

1

Reλnj

+k−1∑j=1

ln

∣∣∣∣ λnj

λnj − λnk

∣∣∣∣ +Nn∑

j=i+1

ln

∣∣∣∣ λnj

λnj − λnk

∣∣∣∣≤ −Reλnk

Nn∑j=1

1

Reλnj

+k−1∑j=1

ln|λnj|

|λnk| − |λnj|+

Nn∑j=i+1

ln|λnj|

|λnj| − |λnk|

= −Reλnk

Nn∑j=1

1

Reλnj

+k−1∑j=1

ln1

Reλnk

Reλnj− 1

+Nn∑

j=k+1

ln1

1− Reλnk

Reλnj

= −Reλnk

Nn∑j=1

1

Reλnj

+k−1∑j=1

lnReλnj

Reλnk −Reλnj

+Nn∑

j=k+1

lnReλnj

Reλnj −Reλnk

Since Reλn is a Muntz sequence it follows that

ln

∣∣∣∣αn,k

λnk

∣∣∣∣ ≤ Reλnk(1 + ln 2)

n

85

Next we show that Φn → 1 for all t > 0.

|Φn(t)− 1| =

∣∣∣∣∣Nn∑j=1

αn,j

λnj

e−λnjt

∣∣∣∣∣≤

Nn∑j=1

∣∣∣∣αn,j

λnj

∣∣∣∣ e−Reλnjt

≤∞∑

j=1

eReλnj2

n e−Reλnjt

≤∞∑

j=1

e−Reλnjt

2

≤∞∑

j=1

e−njt2 =

e−nt/2

1− e−nt/2

Therefore Φn(t) → 1 as n→∞.

86

Vita

Claudiu Mihai was born on October 25, 1967, in Gheraesti Village, Neamt, Roma-

nia. He finished his undergraduate studies at Bucharest University, June 1993. In

August 1999 he came to Louisiana State University to pursue graduate studies in

mathematics. He earned a master of science degree in mathematics from Louisiana

State University in May 2001. He is currently a candidate for the degree of Doctor

of Philosophy in mathematics, which will be awarded in December 2004.

87