A Hausdorff-like moment problem and the inversion of the Laplace transform

Math. Nachr. 279, No. 11, 1147 – 1158 (2006) / DOI 10.1002/mana.200510414

A Hausdorff-like moment problem and the inversion of the Laplace trans-form

Nguyen Dung∗1, Nguyen Vu Huy∗∗2, Pham Hoang Quan∗∗∗2, and Dang Duc Trong†2

1 Institute of Applied Mechanics, Vietnamese Academy of Science and Technology, 291 Dien Bien Phu Str., Dist. 3, Hochim-inh City, Vietnam

2 Hochiminh City National University, 227 Nguyen Van Cu, Q5, Hochiminh City, Vietnam

Received 30 August 2005, accepted 28 November 2005Published online 7 July 2006

Key words Ill-posed problem, inversion of the Laplace transform, Hausdorff moment problem, Muntz poly-nomial, polynomial approximation

MSC (2000) Primary: 65R30; Secondary: 41A10

Dedicated to Professor Phan Dinh Dieu

We consider the problem of finding u ∈ L2(I), I = (0, 1), satisfying

ZI

u(x)xαk dx = µk ,

where k = 0, 1, 2, . . . , (αk) is a sequence of distinct real numbers greater than −1/2, and µ = (µkl) is agiven bounded sequence of real numbers. This is an ill-posed problem. We shall regularize the problem byfinite moments and then, apply the result to reconstruct a function on (0, +∞) from a sequence of values of itsLaplace transforms. Error estimates are given.

c© 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

This paper deals with two problems. The first one is that of finding u in L2(I), I = (0, 1), satisfying∫I

u(x)xαk dx = µk (1.1)

where k = 0, 1, 2, . . . , (αk) is a sequence of distinct real numbers such that

αk > −12

for all k = 0 , 1 , 2 , . . . ,

and (µk) is a given bounded sequence of real numbers.The second problem is that of approximating u0 in L2(0, +∞) such that∫ ∞

0

u0(x)e−βkx dx = µ0k , k = 0 , 1 , 2 , . . . , (1.2)

∗ Corresponding author: e-mail: [email protected], Phone: (84.8) 9300944, Fax: (84.8) 9308300∗∗ e-mail: [email protected], Phone: (84.8) 8350098, Fax: (84.8) 8350096∗∗∗ e-mail: [email protected], Phone: (84.8) 8350098, Fax: (84.8) 8350096† e-mail: [email protected], Phone: (84.8) 8350098, Fax: (84.8) 8350096

c© 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1148 Nguyen Dung, Nguyen Vu Huy, Pham Hoang Quan, and Dang Duc Trong: A Hausdorff-like moment

where (βk) is a sequence of distinct real numbers. In fact, put s = e−x and w0(s) = u0(− ln s). It follows from(1.2) that ∫

I

w0(s)sαk ds = µ0k , k = 0 , 1 , 2 , . . . , (1.3)

where αk = βk − 1, k = 0, 1, 2, . . . , i.e., we shall get the first problem.As known, problem (1.1) is ill-posed, i.e., solutions do not always exist, and in the case of existence, these

do not depend continuously on the given data (that are represented by the right-hand side of (1.1)). In [5], theauthors considered a particular case, in which (αi) is a sequence of positive integers

αi = i , i = 0 , 1 , 2 , . . .

In [9], a two-dimensional version of the problem was presented. However, concrete estimates were not givenin the paper. Now, in the present paper, we shall give some sharp error estimates. As in [9], we shall approx-imate the unknown function u by its orthogonal projection on the space generated by an orthonormal systemL0,L1, . . . ,Ln where Li are the Muntz polynomials (see, e.g., [7] or the next section). We recall that thesepolynomials are obtained through an orthonormalization of the linear space spanxα0 , . . . , xαn.

Consider now the problem of the inversion of the Laplace transform using discrete data. Consider the equation

Au0 =∫ ∞

0

e−stu0(t) dt = g(s) , u0 ∈ L2(0,∞) , (1.4)

where A denotes the Laplace transform operator. It is noted that there are no universal methods of inversion ofthe Laplace transform, and, moreover, the right-hand side is known only on (0,∞) or a subset of (0,∞) and theuse of the Bromwich inversion formula ([16, p. 67]) is therefore not feasible. We can use the Tikhonov methodto regularize the problem. In fact, in this method, we can approximate u0 by functions uβ satisfying

βuβ + A∗Auβ = A ∗ g , β > 0 .

Since A is self-adjoint (cf. [6]), the latter equation can be written as

βuβ +∫ ∞

0

uβ(s)s + t

ds =∫ ∞

0

e−stg(s) ds .

The latter problem is well-posed. Another method is developed by Saitoh and his group ([2, 8, 12, 13]), wherethe function u0 is approximated by integrals having the form

uN(t) =∫ ∞

0

g(s)e−stPN (st) ds , N = 1 , 2 , . . . ,

and PN is known (see [8]). Using the Saitoh formula, we can get directly error estimates.In this paper, we consider the case that a Laplace transform is known only on a subset of the right half plane

Ω+. Note that the Laplace transform Af of an L2-function f is analytic on Ω+. If it is known on an opensubset of Ω+, then it is known on the whole of Ω+. More generally, if Af is known on a countable subset of Ω+

accumulating at a point then, in view of the analyticity of Au0 in Ω+, it is known on the whole Ω+. Thus we shallconsider the particular case (1.2). It is a moment problem. For a construction of an approximate solution of (1.2),we note that the sequence of functions

(e−βnt

)is (algebraically) linear independent and moreover the vector

space generated by the latter sequence is dense in L2(0,∞). The method of truncated expansion as presented in([5, Section 2.1]) is applicable and we refer the reader to this reference for full details. The case Au0 is knownon positive integers, i.e., Au0(n) = µn, n = 1, 2, . . . , is treated in details in [5] with an estimate of the errorin the solution for a given estimate of the error in the right-hand side (µn). In the present paper, using Muntzpolynomials, we shall regularize the general problem (1.2) and give some sharp error estimates.

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Math. Nachr. 279, No. 11 (2006) 1149

The remainder of this paper consists of two sections. Section 2 contains the main results for the first problem,namely, using solutions of the finite moment problem∫

I

u(x)xαk dx = µk , k = 0, n − 1 , (1.5)

we shall give a regularization of problem (1.1). Section 3 deals with a moment problem associated with theLaplace transform.

2 Regularized approximation by finite moments

Let Lm be the polynomial

Lm(x) =m∑

j=0

Cmjxαj (2.1)

where

Cmj =√

2αm + 1

(∏m−1r=0 (αj + αr + 1)∏mr=0,r =j(αj − αr)

). (2.2)

Recall that the αi’s are distinct numbers. If αk = k, k = 0, 1, 2, . . . , we have the Legendre polynomials

Lm(x) =m∑

j=0

Cmjxj (2.3)

where

Cmj =√

2m + 1 (−1)m−j (m + j)!(j!)2(m − j)!

. (2.4)

For each µ = (µk), we define the sequence

λ = λ(µ) = (λk) (2.5)

as follows

λk = λk(µ) =k∑

j=0

Ckjµj

and we put

pn = pn(µ) =n−1∑k=0

λk(µ)Lk . (2.6)

Then the problem (1.5) is equivalent to that of finding u ∈ L2(I) satisfying∫I

u(x)Lk(x) dx = λk , k = 0 , . . . , n − 1 . (2.7)

In what follows, we shall denote by Pm and Pn the orthogonal projections onto the spaces

span1, x, . . . , xm−1

and span

xα0 , xα1 , . . . , xαn−1

,

respectively. We have

Theorem 2.1 Let µ = (µk) be a given sequence of real numbers and let the following assumption hold:

∞∑i=0

2αi + 1(2αi + 1)2 + 1

= ∞ . (2.8)

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Then problem (1.1) has at most one solution. A solution exists if and only if

∞∑k=0

λ2k =

∞∑k=0

⎛⎝ ∞∑

j=0

Ckjµj

⎞⎠

2

< ∞ (2.9)

where Ckj is as in (2.2) if j ≤ k and Ckj = 0 if k < j.If u is the solution of (1.1) then

pn(µ) −→ u in L2(I) as n −→ ∞ .

Moreover, if the solution u ∈ H1(I) and

limk→∞

αk = α < 0 , 0 < δ < 2αk + 1 for all k (δ is a constant) , (2.10)

then, for each n > 1/δ, we have pn(µ) = Pn u and

‖pn(µ) − u‖ ≤(√

n

3r− 1

)−1(F (u)

)1/2 + C ‖u‖ q√

n3r −1

where r and q are arbitrary real numbers satisfying

r >ln

`3 + 2

√2

´δ

, q =`3 + 2

√2

´e−rδ (< 1) , and F (u) =

ZI

x(1 − x) |u′(x)|2 dx , (2.11)

and ‖·‖ is the L2(I)-norm.

Remark 2.2 If limk→∞ αk = α ≥ 0, we put

w(x) = u(x)xα+1/4 , βk = αk − α − 1/4 .

In this case, problem (1.1) is equivalent to one of finding w such that∫I

w(x)xβk dx = µk , k = 0 , 1 , 2 , . . . ,

where limk→∞ βk = −1/4 < 0.

In Theorem 2.1 above, it is assumed that a solution exists. In the case of non-exact data, Theorem 2.3 belowwill be useful. First, put

An = max

⎧⎪⎨⎪⎩n;

n−1∑k=0

⎛⎝ k∑

j=0

|Ckj |⎞⎠

2⎫⎪⎬⎪⎭ ,

and let f be a strictly increasing function such that

f(t) = (An+1 − An)t + (n + 1)An − nAn+1 for all t ∈ [n, n + 1] . (2.12)

Then, we have

Theorem 2.3 Let u0 ∈ L2(I) be the unique solution of (1.1) corresponding to the exact data µ0 = (µ0k) in

the right-hand side of (1.1). For each ε > 0, put

n(ε) =[f−1

(ε−1

)],

where [x] is the largest integer not greater than x.


Math. Nachr. 279, No. 11 (2006) 1151

Then, there exists a function η(ε) such that η(ε) → 0 as ε → 0 and that for all sequences µ satisfying∥∥µ − µ0∥∥∞ ≡ sup

k,l

∣∣µkl − µ0kl

∣∣ < ε ,

we have

‖pn(ε)(µ) − u0‖ ≤ η(ε) .

Moreover, if u0 ∈ H1(I) and (2.10) holds, then

‖pn(ε)(µ) − u0‖ ≤ ε1/2 +

(√n(ε)3r

− 1

)−1 (F (u0)

)1/2 + C ‖u0‖ q

qn(ε)3r −1 (2.13)

with F , q and r as in (2.11).

P r o o f o f T h e o r e m 2.1. We first have

Lemma 2.4 (a) The set Ln is orthonormal in L2[0, 1].(b) For every n, j = 1, 2, . . . , one has

‖xj − Pn xj‖ =1√

2j + 1

n−1∏k=0

∣∣∣∣1 − 2αk + 1j + αk + 1

∣∣∣∣ . (2.14)

Moreover, Ln is complete in L2[0, 1] if and only if

∞∑i=0

2αi + 1(2αi + 1)2 + 1

= ∞ . (2.15)

P r o o f. Cf. [7] for a proof.

By the completeness and the orthonormality of Lk, if (1.1) has a solution u then

pn(µ) = Pn u −→ u in L2(I) and∞∑

k=0

λ2k = ‖u‖2 < ∞ .

Here, ‖.‖ is the L2(I)-norm. Conversely, if (2.9) is satisfied then, by completeness of Lk, the sum u of theseries

∑∞k=0 λkLk, which is convergent, is the solution of (1.1).

Now, we consider the case that the solution u ∈ H1(I) (with the assumption (2.10)) and we shall establish anestimate for the quantity ‖u − Pn u‖. To this end, we divide the proof into two steps. In the first step, we shallestimate ‖Lm − Pn Lm‖ for m ∈ N. In the second step we shall give the desired estimate.

Step 1. Estimate of ‖Lm − Pn Lm‖. For convenience, we denote by xj (j = 0, 1, 2, . . . ) the polynomial

x −→ xj on I = (0, 1) .

Since limk→∞ αk = α < 0, we can assume that

for all k , αk < 0 , which implies2αk + 1

j + αk + 1< 1 .

From (2.14) and the known inequality 0 < 1 + t < et for each t ≥ −1, we get

‖xj − Pn xj‖ ≤ 1√2j + 1

exp

(−

n−1∑k=0

2αk + 1j + αk + 1

)

≤ 1√2j + 1

exp

(− 1

2j + 1

n−1∑k=0

(2αk + 1)

)≤ 1√

2j + 1e− nδ

2j+1 .

(2.16)

We have



Lemma 2.5 Let Cmj be coefficients of the Legendre polynomials defined in (2.4). Then

m∑j=0

|Cmj | ≤ 2

√3 + 2

√2

π√

2

(3 + 2

√2)m

. (2.17)

A proof of Lemma 2.5 will be given later. Combining (21) with Lemma 2.5 above, we have∥∥∥∥∥∥Lm(x) −m∑

j=0

CmjPnxj

∥∥∥∥∥∥2

=

∥∥∥∥∥∥m∑

j=0

Cmj

(xj − Pnxj

)∥∥∥∥∥∥2

≤⎧⎨⎩

m∑j=0

|Cmj | 1√2j + 1

e− nδ

2j+1

⎫⎬⎭

2

≤ max0≤j≤m

12j + 1

e− 2nδ

2j+1

⎛⎝ m∑

j=0

|Cmj |⎞⎠

2

≤ 4

(3 + 2

√2

π√

2

)(3 +

√2)2m max

0≤j≤m

12j + 1

e− 2nδ

2j+1 .

(2.18)

Note that

d

dt

(1t

e− 2nδt

)=

1t2

e− 2nδt

(2nδ

t− 1

)> 0

for each t < 2nδ, which means the function t → t−1e− 2nδt is increasing on (0, 2nδ]. So, for 2m + 1 < 2nδ, we

get, in view of (2.18) that∥∥∥∥∥∥Lm(x) −m∑

j=0

CmjPnxj

∥∥∥∥∥∥ < 2

√3 + 2

√2

π√

2

(3 + 2

√2)m 1√

2m + 1e− nδ

2m+1

which implies

‖Lm − PnLm‖ ≤∥∥∥∥∥∥Lm −

m∑j=0

CmjPnxj

∥∥∥∥∥∥ ≤ 2

√3 + 2

√2

π√

2

(3 + 2

√2)m 1√

2m + 1e− nδ

2m+1 (2.19)

for 2m + 1 ≤ 2nδ.

Step 2. Estimate of ‖u − Pnu‖. Now, let

m−1∑j=0

ajLj = Pmu .

Recall that Pm is the orthogonal projection of u onto span1, x, . . . , xm−1

. We have

˛˛˛Pmu −

m−1Xj=0

ajPnLj

˛˛˛2

=

˛˛˛m−1Xj=0

aj

`Lj − PnLj

´˛˛˛2

≤ (2.20)

≤m−1Xj=0

|aj |2.m−1Xj=0

`Lj − PnLj

´2(2.21)

≤ ‖u‖2m−1Xj=0

`Lj − PnLj

´2, (2.22)


Math. Nachr. 279, No. 11 (2006) 1153

which combines with (2.19) to imply that

∫I

∣∣∣∣∣∣Pmu −m−1∑j=0

ajPnLj

∣∣∣∣∣∣2

dx ≤ ‖u‖2m−1∑j=0

∥∥Lj − PnLj

∥∥2(2.23)

≤ ‖u‖2 4(3 + 2

√2)

π√

2· m

(3 +

√2)2m

2m + 1e− 2nδ

2m+1 . (2.24)

We have

Lemma 2.6

‖u − Pmu‖ ≤ 1m

(F (u))1/2, with F (u) as in (2.11) . (2.25)

A proof of Lemma 2.6 will be given later.By Lemma 2.6 we get∥∥∥∥∥∥u −

m−1∑j=0

ajPnLj

∥∥∥∥∥∥≤ ‖u − Pmu‖ +

∥∥∥∥∥∥Pmu −m−1∑j=0

ajPnLj

∥∥∥∥∥∥≤ 1

m(F (u))1/2 + 2 ‖u‖

√(3 + 2

√2)

π√

2

√m

2m + 1(3 +

√2)m

e−nδ

2m+1

≤ 1m

(F (u))1/2 + 2 ‖u‖√(

3 + 2√

2)

π√

2

(3 +

√2)m

e−rδm

<1m

(F (u))1/2 + C ‖u‖qm

(2.26)

if 2m + 1 < 2nδ and rm(2m + 1) ≤ n. Here, r and q are defined as in (2.11), C is a constant independent of n

and m.Finally, we choose an integer m such that

2m + 1 < 2nδ and rm(2m + 1) < 2r(m + 1)2 ≤ n ,

for example, m =√

n3r , the largest integer not greater than

√n3r . Then using (2.26) and noting that

n−1∑j=0

ajPnLj ∈ spanxα1 , . . . , xαn ,

we derive from the minimality of ‖u − pn(µ)‖

‖u − pn(µ)‖ = ‖u − Pnu‖ ≤∥∥∥∥∥∥u −

m−1∑j=0

ajPnLj

∥∥∥∥∥∥ (2.27)

≤(√

n

3r− 1

)−1

(F (u))1/2 + C ‖u‖ q√

n3r −1. (2.28)

We are now ready for a proof of Theorem 2.3.



P r o o f o f T h e o r e m 2.3. We have

‖pn(µ) − u0‖ ≤ ‖pn(µ) − pn(µ0)‖ + ‖pn(µ0) − u0‖ . (2.29)

From the completeness and the orthonormality of the Lk’s in L2(I), we have

‖pn(µ0) − u0‖2 =∞∑

k≥n

|λk(µ)|2 =∑

k≥n(ε)

⎛⎝ k∑

j=0

Ckjµ0j

⎞⎠

2

. (2.30)

From (2.6), we have

pn(µ) − pn(µ0) =n−1∑k=0

(λk(µ) − λk(µ0)

)Lk =n−1∑k=0

k∑j=0

Ckj(µk − µ0k)Lk .

Hence, in view of (2.12)

‖pn(µ) − pn(µ0)‖2

=n−1∑k=0

( k∑j=0

Ckj(µk − µ0k))2

≤ ε2n−1∑k=0

( k∑j=0

|Ckj |)2

≤ ε2An = ε2f(n) .(2.31)

So we have

‖pn(ε)(µ) − pn(ε)(µ0)‖2 ≤ ε2f(n(ε)

) ≤ ε2f(f−1

(ε−1

))= ε . (2.32)

If we put

η(ε) =√

ε +

⎧⎪⎨⎪⎩

∑k≥n(ε)

⎛⎝ k∑

j=0

Ckjµ0j

⎞⎠

2⎫⎪⎬⎪⎭

1/2

then by (2.29)–(2.32), it follows that

‖pn(ε)(µ) − u0‖ ≤ η(ε) .

As ε → 0, we have f−1(ε−1

) → ∞ and n(ε) → ∞. By (2.9)

∑k≥n(ε)

⎛⎝ k∑

j=0

Ckjµ0j

⎞⎠

2

−→ 0 as ε −→ 0 .

Hence η(ε) → 0 for ε → 0.Now, if u0 ∈ H1(I) and (2.10) holds, we get (2.13) by using (2.29), (2.32) and Theorem 2.1.

We now give a proof Lemma 2.5.

P r o o f o f L e m m a 2.5. Put

Pm(t) =m∑

j=0

(m + j)!(j!)2(m − j)!2j

(t − 1)j .

Then

Lm(x) =m∑

j=0

Cmjxj = (−1)m

√2m + 1Pm(1 − 2x) ,


Math. Nachr. 279, No. 11 (2006) 1155

which implies

m∑j=0

|Cmj | = (−1)mLm(−1) =√

2m + 1Pm(3) . (2.33)

Moreover, one has (cf. [10, p. 48], or [1, Eq. (22.10.10)])

Pm(3) =1π

∫ π

0

(3 + 2

√2 cosφ

)mdφ (m ∈ N0) .

We claim that

Pm(3) ≤ 2

√3 + 2

√2

π√

2· 1√

2m + 1

(3 + 2

√2)m (m ∈ N0) . (2.34)

Indeed, one has

Pm(3) =1π

∫ π

0

(3 + 2

√2 cosφ

)mdφ (2.35)

=1π

(∫ π2

0

+∫ π

π2

)(3 + 2

√2 cosφ

)mdφ (2.36)

=1π

∫ π2

0

[(3 + 2

√2 cosφ

)m +(3 − 2

√2 cosφ

)m]dφ , (2.37)

from which it follows with t = cosφ that

Pm(3) ≤ 2π

∫ π2

0

(3 + 2

√2 cosφ

)mdφ ≤ 2

π

∫ 1

0

(3 + 2t

√2)m

√1 − t2

dt ≤ 2π

∫ 1

0

(3 + 2t

√2)m

√1 − t

dt .

Putting s =√

1 − t and ζ = s√

2√

23+2

√2

we get

Pm(3) ≤ 4π

(3 + 2

√2)m

√3 + 2

√2

2√

2

∫ 1

0

(1 − ζ2

)dζ

after some computations.Letting ζ = cosφ, we have

Pm(3) ≤ 4π

(3 + 2

√2)m

√3 + 2

√2

2√

2

∫ π2

0

sin2m+1 θ dθ . (2.38)

Putting In =∫ π/2

0 sinn θ dθ we have (cf. [15, p. 681])

I2k =1.2 . . . (2k − 1)

2.4 . . . (2k)· π

2and I2k−1 =

2.4 . . . (2k − 2)1.3 . . . (2k − 1)

.

It follows that

InIn−1 =π

2n(n ∈ N) . (2.39)

But

In−1 =∫ π

2

0

sinn−1 θ dθ >

∫ π2

0

sinn θ dθ = In .



Hence (2.39) gives

I2n ≤ InIn−1 =

π

2n. (2.40)

Substituting (2.40) into (2.38), we get (2.34).Now, using (2.33) and (2.34) one obtains

m∑j=0

|Cmj | ≤ 2

√3 + 2

√2

π√

2

(3 + 2

√2)m (m ∈ N0) ,

which completes the proof of Lemma 2.5.

P r o o f o f L e m m a 2.6. Put

lk =∫

I

u(x)Lk(x) dx for all k = 0 , 1 , 2 , . . .

Then, by completeness and the orthonormality property of the Lm’s in L2(I), one has

Pmu =m−1∑k=0

lkLk , u =∞∑

k=0

lkLk and ‖u‖2 =∞∑

k=0

|lk|2 .

Hence

‖Pmu − u‖2 =∞∑

k=m

|lk|2 . (2.41)

Moreover, we have the identity (cf. [14])

F (u) =∫

I

x(1 − x) |u′(x)|2 dx =∞∑

k=0

k(k + 1)l2k for u ∈ H1(I) .

So∞∑

k=m

m2l2k ≤∞∑

k=m

k2l2k ≤ F (u) .

Combining this with (2.41), we obtain

‖Pm u − u‖2 ≤ 1m2

F (u) ,

which completes the proof of Lemma 2.6.

3 A moment problem from Laplace transform

As discussed, we shall consider the second problem of approximating u0 ∈ L2(0,∞), such that∫ ∞

0

u0(x)e−βkx dx = µ0k , k = 0 , 1 , 2 , . . . , (3.1)

where (βk) is a sequence of distinct real numbers. Put s = e−x and w0(s) = u0(− ln s). It follows from (3.1)that ∫

I

w0(s)sαk ds = µ0k , k = 0 , 1 , 2 , . . . , (3.2)

where αk = βk − 1, k = 0, 1, 2, . . .


Math. Nachr. 279, No. 11 (2006) 1157

Note that ∫I

|w0(s)|2 ds =∫ ∞

0

|u0(x)|2 e−x dx .

Then we have w0 ∈ L2(I), since u0 ∈ L2(0,∞).

Theorem 3.1 Let u0 ∈ L2(0,∞) be the solution of (3.1) corresponding to µ0 = (µ0k) ∈ 2 in the right-hand

side of (3.1). Suppose the sequence (βk) satisfies

βk >12

and∞∑

k=0

2βk − 1(2βk − 1)2 + 1

= ∞ for all k ,

which means the sequence (αk), αk = βk − 1, satifies (2.15) in Lemma 2.4. With n(ε) as in Theorem 2.3 andpn(µ) as in (2.6), put

n(ε)(µ)(x) = pn(ε)(µ)(e−x) .

Then there exists a function η(ε) such that η(ε) → 0 as ε → 0 and that for all sequences µ satisfying

‖µ − µ0‖∞ ≡ supk

|µk − µ0k| < ε ,

we have

‖n(ε)(µ) − u0‖ρ ≤ η(ε) , (3.3)

where the norm ‖·‖ρ is defined by ‖h‖2ρ =

∫∞0 |h(x)|2e−x dx.

Moreover, if u0 ∈ H1(0,∞) and the sequence (αk), αk = βk − 1, satisfies (2.10) in Theorem 2.1, then

‖n(ε)(µ) − u0‖ρ ≤ √ε +

(√n(ε)3r

− 1

)−1(F(u0))1/2 + C ‖u0‖ρ q

qn(ε)3r −1 (3.4)

where F(u0) =∫∞0

(1 − e−x) |u′0(x)|2 dx.

P r o o f. The results of this theorem follow from Theorem 2.3 by noting that

s = e−x , w0(s) = u0(− ln s) = u0(x) , (3.5)

∥∥pn(ε)(µ) − w0

∥∥2=

∫I

∣∣∣pn(ε)(µ)(s) − w0(s)∣∣∣2 ds

=∫

Q

∣∣∣pn(ε)(µ)(e−x) − u0(x)∣∣∣2 e−xdx = ‖n(ε)(µ) − u0‖2

ρ ,(3.6)

F (w0) =∫

I

s(1 − s)|w′0(s)|2ds =

∫ ∞

0

(1 − e−x)|u′0(x)|2dx = F(u0) , (3.7)

and we complete the proof of Theorem 3.1.

Acknowledgements This work was supported by the Council for Natural Sciences of Vietnam.

References

[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions (Dover, New York, 1972).[2] K. Amano, S. Saitoh, and M. Yamamoto, Error estimates of the real inversion formulas of the Laplace transform, Integral

Transforms Spec. Funct. 10, 165–178 (2000).



[3] D. D. Ang and N. Dung, Inversion of the Laplace Transform, in: Proceedings of Workshop “Engineering MechanicsToday”, to appear.

[4] D. D. Ang, R. Gorenflo, and D. D. Trong, A multidimensional Hausdorff moment problem: regularization by finitemoments, Z. Anal. Anwendungen 18, No. 1, 13–25 (1999).

[5] D. D. Ang, R. Gorenflo, L. K. Vy, and D. D. Trong, Moment Theory and Some Inverse Problems in Potential Theoryand Heat Conduction, Lecture Notes in Mathematics Vol. 1792 (Springer-Verlag, Berlin – Heidelberg, 2002).

[6] D. D. Ang, J. Lund, and F. Stenger, Complex variables and regularization method of inversion of the Laplace transform,Math. Comp. 54, No. 188, 589–608 (1989).

[7] P. Borwein and T. Erdelyi, Polynomials and Polynomial Inequalities, Graduate Texts in Mathematics Vol. 161 (Springer-Verlag, Berlin – Heidelberg, 1995).

[8] D.-W. Byun and S. Saitoh, A real inversion formula for the Laplace transform, Z. Anal. Anwendungen 12, 597–603(1993).

[9] Nguyen Vu Huy and Dang Duc Trong, A Hausdorff moment problems with non-integral powers: Approximation byfinite moments, Vietnam J. Math. 32:4, 371–377 (2004).

[10] N. N. Lebedev, Special Functions and their Applications (Dover Publications Inc., New York, 1972).[11] A. L. Rabenstein, Introduction to Ordinary Differential Equations (Acad. Press, New York et al., 1972).[12] S. Saitoh, Integral Transforms, Reproducing Kernels and their Applications, Pitman Research Notes in Mathematics

Series Vol. 369 (Addison Wesley Longman Ltd., U.K., 1997).[13] S. Saitoh, Vu Kim Tuan, and M. Yamamoto, Conditional stability of a real inverse formula for the Laplace transform,

Z. Anal. Anwendungen 20, 193–202 (2001).[14] G. Talenti, Recovering a function from a finite number of moments, Inverse Problems 3, 501–517 (1987).[15] A. Taylor, Advanced Calculus (Blaisdell Publ. Comp., New York et al., 1965).[16] D. V. Widder, The Laplace Transform (Princeton University Press, Princeton, 1946).


A Hausdorff-like moment problem and the inversion of the Laplace transform

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