Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 1, 13–25

A Multi-Dimensional Hausdorff Moment Problem: Regularization by Finite Moments

D. D. Ang, R. Gorenflo and D. D. Trong

Dedicated to Prof. Dr. G. Anger on occasion of his 70th birthday

Abstract. We consider the multi-dimensional Hausdorff moment problem over the unit cube: to reconstruct an unknown function from the (inaccurately) given values of the integrals of the unknown function multiplied by all power-products of the independent variables. We describe a regularization scheme using orthogonalization by the tensor product of (shifted) Legendre polynomials and ”approximation” of the unknown function by a finite sum, the dimension of the space of approximation playing the role of the regularization parameter. For the case of square integrability of the unknown function we present an estimate of the regularization error that implies convergence if the data error tends to zero.

Keywords: Ill-posed problems, Hausdorff moment problem, polynomial approximation

AMS subject classification: Primary 65 R 30, secondary 41 A 10

1. Introduction

We propose to present a regularization of a Hausdorff moment problem, with error estimates given. The literature on the moment problem is impressive, we shall list only a few of them in the bibliography (cf. [2 - 8, 11, 13]). Of particular relevance to the present work are the papers by Ang, Vy and Gorenflo [3] and by Talenti [11] (see also Inglese [7]). In [11], the author considered the one-dimensional case and proved the stability of the finite moment approximation with stability estimates given. But the relationship of the finite moment approximation to the original moment problem was not exhibited as transparently as we feel desirable. In the present paper, we shall consider the multi-dimensional case, specifically the following problem.

Problem (MP). Find a function u in L^{2}(I) satisfying the sequence of equations

∫_{I} u(x_{1}, …, x_{d})x_{1}^{k_{1}} ⋯ x_{d}^{k_{d}} dx_{1} ⋯ dx_{d} = μ_{k_{1}···k_{d}}  (k_{i} ∈ ℕ_{0}; i = 1, …, d) (1)

where d ∈ ℕ, I = [0, 1]^{d} ⊂ ℝ^{d} and μ = (μ_{k_{1}···k_{d}}) is a given bounded sequence of real numbers.

D. D. Ang: Nat. Univ., Dept. Math., 227 Nguyen Van Cu, Q5, Hochiminh City, Vietnam R. Gorenflo: Freie Universität Berlin, FB Math. und Inf., Arnimallee 2–6, D-14195 Berlin D. D. Trong: Nat. Univ., Dept. Math., 227 Nguyen Van Cu, Q5, Hochiminh City, Vietnam

ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

It can be proved that (MP) is an ill-posed problem. Note also that (MP) does not always have a solution for an arbitrary bounded sequence (μ_{k_{1}...k_{d}}) in the right-hand side of (1). Hence it is realistic to deal also with cases where problem (MP) does not have a solution. We refer to the paper [3] of which the present work can be seen as a continuation. However while [3] deals with a general moment problem, the Hausdorff moment problem has it own peculiarities that we have taken into account, in departure from the general case, especially in the crucial choice of the function f(t), [f^{−1}(ε^{\frac{1}{2}} )] being the dimension of our finite approximation.

In our construction of finite-dimensional approximations, we obtain an orthonormalization of the basis functions {x_{1}^{k_{1}} ⋯ x_{d}^{k_{d}}} (k_{i} ∈ ℕ_{0}; i = 1, …, d) not through the Gram-Schmidt process but by using the polynomials

L_{k_{1}...k_{d}}(x_{1}, …, x_{d}) = L_{k_{1}}(x_{1}) ⋯ L_{k_{d}}(x_{d}),  (2)

where

L_{m}(x) = ∑_{j=0}^{m} C_{mj}x^{j}  (m ∈ ℕ_{0})  (3)

with

C_{mj} = (2m + 1)^{\frac{1}{2}} (−1)^{j} \frac{(m + j)!}{(j!)^{2}(m − j)!}  (j = 0, 1, …, m).  (4)

The sequence (L_{k_{1}} ⋯ L_{k_{d}}) (k_{i} ∈ ℕ_{0}; i = 1, …, d) forms a complete orthonormal set in L^{2}(I). In fact,

L_{m}(x) = (2m + 1)^{\frac{1}{2}} P_{m}(1 − 2x),  (5)

where P_{m} is the Legendre polynomial:

P_{m}(t) = ∑_{j=0}^{m} \frac{(m + j)!}{(j!)^{2}(m − j)!2^{j}} (t − 1)^{j}, ∥P_{m}∥_{L^{2}(−1,1)}^{2} = \frac{2}{2m + 1}

(cf. [10: page 227]). Thus (L_{m}) is a complete orthonormal basis in L^{2}(0, 1) and hence (L_{k_{1}} ⋯ L_{k_{d}}) is a complete orthonormal set in L^{2}(I). From (3), we have

L_{k_{1}···k_{d}}(x_{1}, …, x_{d}) = ∑_{p_{1}=0}^{k_{1}} ⋯ ∑_{p_{d}=0}^{k_{d}} C_{k_{1}p_{1}} ⋯ C_{k_{d}p_{d}}x_{1}^{p_{1}} ⋯ x_{d}^{p_{d}}.  (6)

If μ = (μ_{k_{1}···k_{d}}) is a real sequence, we define the sequence λ = λ(μ) = (λ_{k_{1}···k_{d}}) as follows:

λ_{k_{1}···k_{d}} = λ_{k_{1}···k_{d}}(μ) = ∑_{p_{1}=0}^{k_{1}} ⋯ ∑_{p_{d}=0}^{k_{d}} C_{k_{1}p_{1}} ⋯ C_{k_{d}p_{d}}μ_{p_{1}···p_{d}}.

Now put

p^{n} = p^{n}(μ) = ∑_{k_{1},...,k_{d}=0}^{n} λ_{k_{1}···k_{d}}(μ)L_{k_{1}···k_{d}}.  (7)

2. Results

We now give our results in form of three theorems that we shall prove in Section 3.

Theorem 1. Let μ = (μ_{k_{1}···k_{d}}) (k_{i} ∈ ℕ_{0}; i = 1, …, d) be a given sequence of real numbers. If u is the (unique) solution of (1) in L^{2}(I), then

∑_{k_{1},...k_{d}=0}^{∞} ( ∑_{p_{1},...,p_{d}=0}^{∞} C_{k_{1}p_{1}} ⋯ C_{k_{d}p_{d}}μ_{p_{1}···p_{d}})^{2} < ∞  (8)

and

p^{n}(μ) → u  in L^{2}(I) as n → ∞,  (9)

where C_{ij} is defined by (4) if j ≤ i and C_{ij} = 0 if j > i. Moreover, if the solution u is in H^{1}(I), then

∥p^{n}(μ) − u∥ ≤ \frac{1}{n + 1}(F (u))^{\frac{1}{2}}  (n ∈ ℕ)  (10)

where ∥ · ∥ is the L^{2}(I)-norm and

F (u) = ∑_{i=1}^{d} ∫_{I} x_{i}(1 − x_{i}) |\frac{∂u}{∂x_{i}} |^{2} dx_{1} ⋯ dx_{d}.

In Theorem 1 above, it is assumed that a solution exists. In the general case that a solution may not exist, Theorem 2 below will be useful.

Theorem 2. Let u_{0} ∈ L^{2}(I) be the unique solution of (1) corresponding to μ^{0} = (μ_{k_{1}···k_{d}}^{0}) (k_{i} ∈ ℕ_{0}; i = 1, …, d) in the right-hand side of (1). Let

f(t) = \frac{1}{\sqrt{(2π)^{d}}} (3 + 2\sqrt{2})^{dt+d}  (11)

and for 0 < ε < 1, put

n(ε) = [f^{−1}(ε^{− \frac{1}{2}} )]  (12)

where [x] is the largest integer n ≤ x. Then there exists a function η(ε) (0 < ε < 1) such that η(ε) → 0 as ε → 0 and that for all sequences μ satisfying

∥μ − μ^{0}∥_{∞} := sup_{k_{1},...,k_{d}∈ℕ_{0}} |μ_{k_{1}···k_{d}} − μ_{k_{1}···k_{d}}^{0} | ≤ ε

we have ∥p^{n(ε)}(μ) − u_{0}∥ ≤ η(ε). Moreover, if u_{0} ∈ H^{1}(I) and 0 < ε < (\frac{\sqrt{2π}}{3+2\sqrt{2}})^{2d}, then

∥p^{n(ε)}(μ) − u_{0}∥ ≤ ε^{\frac{1}{2}} + \frac{|F (u_{0})|^{\frac{1}{2}}}{C_{1} + C_{2} ln (\frac{1}{ε})},  (13)

where

C_{1} = \frac{ln (\sqrt{2π}(3 − 2\sqrt{2}))}{ln(3 + 2\sqrt{2})}  and  C_{2} = \frac{1}{2d ln(3 + 2\sqrt{2})}.  (14)

Remark. In the first version of this paper (see also [2]), we had derived (13) with C_{1} and C_{2} replaced by

C̃_{1} = \frac{2 ln (\frac{8\sqrt{5}}{27\sqrt{2}})}{2 ln 3 + \frac{1}{2} ln 2}  and  C̃_{2} = \frac{1}{d(2 ln 3 + \frac{1}{2} ln 2)},

respectively.

Estimate (13) with C_{1} and C_{2} given by (14) was derived using formula (31) as suggested by one of the referees. Clearly, the constants C_{1} and C_{2} give a better approximation than do C̃_{1} and C̃_{2}, respectively. While it is not known to us whether C_{1} and C_{2} are the best constants, we shall show below that the estimate of the order of magnitude of the error for ε → 0 as given by (13) is optimal in the sense of the following theorem.

Theorem 3. Assume that there exist two functions m : (0, 1] → ℕ and α : (0, 1] → (0, ∞) such that lim_{ε↓0} m(ε) = +∞ and lim_{ε↓0} α(ε) = 0, and

∥p^{m(ε)}(μ) − u_{0}∥ ≤ (1 + (F (u_{0}))^{\frac{1}{2}} ) α(ε)  (15)

for all u_{0} ∈ H^{1}(I), I = [0, 1]^{d}, and all μ = (μ_{k_{1}···k_{d}}) satisfying

∥μ − μ_{0}∥_{∞} = sup_{k_{1},...,k_{d}≥0} |μ_{k_{1}···k_{d}} − μ_{k_{1}···k_{d}}^{0} | ≤ ε,

where

μ_{k_{1}...k_{d}}^{0} = ∫_{I} u_{0}(x_{1}, …, x_{d})x_{1}^{k_{1}} ⋯ x_{d}^{k_{d}} dx_{1} ⋯ dx_{d}.

Then we have

lim inf_{ε↓0} α(ε) ln (\frac{1}{ε}) ≥ \frac{d}{2} ln(3 + 2\sqrt{2}).  (16)

3. Proofs

For definiteness, we shall give the proofs of Theorems 1 - 3 in the case d = 3 only. The proofs carry almost verbatim to the general case. For d = 3, we put (x_{1}, x_{2}, x_{3}) = (x, y, z), (k_{1}, k_{2}, k_{3}) = (i, j, k), μ_{k_{1}k_{2}k_{3}} = μ_{ijk}, λ_{k_{1}k_{2}k_{3}} = λ_{ijk} and L_{k_{1}k_{2}k_{3}}(x, y, z) = L_{i}(x)L_{j}(y)L_{k}(z) = L_{ijk}(x, y, z).

Proof of Theorem 1. We easily get the first result of Theorem 1 using the orthonormality property of the L_{ijk}′s. In fact, one has in view of (3)

∫_{I} uL_{ijk} dxdydz = ∫_{I} u(x, y, z) ∑_{p=0}^{i} ∑_{q=0}^{j} ∑_{r=0}^{k} C_{ip}C_{jq}C_{kr} x^{p}y^{q}z^{r} dxdydz

= ∑_{p=0}^{i} ∑_{q=0}^{j} ∑_{r=0}^{k} C_{ip}C_{jq}C_{kr} (∫_{I} u(x, y, z) x^{p}y^{q}z^{r} dxdydz)

= ∑_{p=0}^{i} ∑_{q=0}^{j} ∑_{r=0}^{k} C_{ip}C_{jq}C_{kr}μ_{ijk}

= λ_{ijk}.

A Multi-Dimensional Hausdorff Moment Problem 17

Hence, by completeness and the orthonormality property of the L_{ijk}′s in L^{2}(I) one has

u = ∑_{i,j,k=0}^{∞} λ_{ijk}(μ)L_{ijk} and ∥u∥^{2} = ∑_{i,j,k=0}^{∞} |λ_{ijk}(μ)|^{2} < ∞, (17)

i.e. (8) holds. To prove (9), we subtract the first equation in (17) from (7) to get

p^{n}(μ) − u = − ∑_{max(i,j,k)>n} λ_{ijk}(μ)L_{ijk}.

Hence

∥p^{n}(μ) − u∥^{2} = ∑_{max(i,j,k)>n} |λ_{ijk}(μ)|^{2}.  (18)

Combining (8) and (18) gives (9).

For the proof of (10) we rely, with α_{k} = ∫_{0}^{1} v(x)L_{k}(x) dx, on the identity (cf. [11])

∫_{0}^{1} x(1 − x)|v′(x)|^{2} dx = ∑_{k=0}^{∞} k(k + 1)α_{k}^{2}  ∀ v ∈ H^{1}(0, 1).  (19)

In view of (17) and (19), we have

∑_{i,j,k=1}^{∞} i^{2}λ_{ijk}^{2} ≤ ∫_{I} x(1 − x)|\frac{∂u}{∂x}(x, y, z)|^{2}dxdydz  (20)
∑_{i,j,k=1}^{∞} j^{2}λ_{ijk}^{2} ≤ ∫_{I} y(1 − y)|\frac{∂u}{∂y}(x, y, z)|^{2}dxdydz  (21)
∑_{i,j,k=1}^{∞} k^{2}λ_{ijk}^{2} ≤ ∫_{I} z(1 − z)|\frac{∂u}{∂z}(x, y, z)|^{2}dxdydz.  (22)

Adding (20) - (22) together, we get

∑_{i,j,k=1}^{∞} (i^{2} + j^{2} + k^{2})λ_{ijk}^{2} ≤ F (u),  (23)

where, we recall,

\begin{aligned}F (u) & = ∫_{I} (x(1 − x)|\frac{∂u}{∂x}(x, y, z)|^{2} + y(1 − y)|\frac{∂u}{∂y}(x, y, z)|^{2} \\ & + z(1 − z)|\frac{∂u}{∂z}(x, y, z)|^{2})dxdydz. \end{aligned}

Note that

i^{2} + j^{2} + k^{2} ≥ (n + 1)^{2} for max(i, j, k) > n. (24)

We get in view of (18), (23) and (24) that

(n + 1)^{2}∥p^{n}(μ) − u∥^{2} ≤ ∑_{max(i,j,k)>n} (i^{2} + j^{2} + k^{2})λ_{ijk}^{2}(μ) ≤ F (u),

which gives the desired inequality. The proof of Theorem 1 is completed

18 D. D. Ang, R. Gorenflo and D. D. Trong

Proof of Theorem 2. We have

∥p^{n}(μ) − u_{0}∥ ≤ ∥p^{n}(μ) − p^{n}(μ^{0})∥ + ∥p^{n}(μ^{0}) − u_{0}∥ (n ∈ ℕ). (25)

We shall estimate the two terms in the right-hand side of (25). Using (18) gives

∥p^{n}(μ^{0}) − u_{0}∥^{2} = ∑_{max(i,j,k)>n} ( ∑_{p,q,r=0}^{∞} C_{ip}C_{jq}C_{kr}μ_{pqr}^{0})^{2}.  (26)

For the first term, we get in view of (7)

p^{n}(μ) − p^{n}(μ^{0}) = ∑_{i,j,k=0}^{n} ∑_{p,q,r=0}^{∞} C_{ip}C_{jq}C_{kr}(μ_{pqr} − μ_{pqr}^{0})L_{ijk}.  (27)

Hence

\begin{aligned}∥p^{n}(μ) − p^{n}(μ^{0})∥^{2} & = ∑_{i,j,k=0}^{n} ( ∑_{p,q,r=0}^{∞} C_{ip}C_{jq}C_{kr}(μ_{pqr} − μ_{pqr}^{0}))^{2} \\ & ≤ ∥μ − μ^{0}∥_{∞}^{2} ∑_{i,j,k=0}^{n} ( ∑_{p,q,r=0}^{∞} |C_{ip}C_{jq}C_{kr}|)^{2} . \end{aligned}  (28)

Noting that C_{mj} = 0 for j > m, we have

∑_{p,q,r=0}^{∞} |C_{ip}C_{jq}C_{kr}| = ∑_{p=0}^{i} |C_{ip}| ∑_{q=0}^{j} |C_{jq}| ∑_{r=0}^{k} |C_{kr}|.  (29)

Substituting (29) into (28) gives

∥p^{n}(μ) − p^{n}(μ^{0})∥^{2} ≤ ∥μ − μ^{0}∥_{∞}^{2} (∑_{i=0}^{n} (∑_{p=0}^{i} |C_{ip}|)^{2})^{3} .  (30)

From (3) - (5), one gets

∑_{p=0}^{i} |C_{ip}| = L_{i}(−1) = (2i + 1)^{\frac{1}{2}} P_{i}(3).  (31)

On the other hand, one has (cf. [9: page 48], or [1: Equation (22.10.10)])

P_{i}(3) = \frac{1}{π} ∫_{0}^{π} (3 + 2\sqrt{2} cos ϕ)^{i} dϕ (i ∈ ℕ_{0}).

We claim that

P_{i}(3) ≤ 2\sqrt{\frac{3 + 2\sqrt{2}}{π\sqrt{2}}} · \frac{1}{\sqrt{2i + 1}} (3 + 2\sqrt{2})^{i}  (i ∈ ℕ_{0})  (32)

the proof of which will be given later. Using (31) and (32) one gets after some rearrangements

∑_{p=0}^{i} |C_{ip}| ≤ 2\sqrt{\frac{3 + 2\sqrt{2}}{π\sqrt{2}}} (3 + 2\sqrt{2})^{i} (i ∈ ℕ_{0})

and

∑_{i=0}^{n} (∑_{p=0}^{i} |C_{ip}|)^{2} ≤ \frac{4(3 + 2\sqrt{2})}{π\sqrt{2}} ∑_{i=0}^{n} (3 + 2\sqrt{2})^{2i} ≤ \frac{1}{2π} (3 + 2\sqrt{2})^{2n+2}.

Substituting the latter inequality into (30) gives

∥p^{n}(μ) − p^{n}(μ^{0})∥ ≤ ∥μ − μ^{0}∥_{∞} \frac{1}{\sqrt{8π^{3}}} (3 + 2\sqrt{2})^{3n+3}.  (33)

Put

f(t) = \frac{1}{\sqrt{8π^{3}}} (3 + 2\sqrt{2})^{3t+3}  and  n(ε) = [f^{−1}(ε^{− \frac{1}{2}} )].

Then we have in view of (33)

∥p^{n}(μ) − p^{n}(μ^{0})∥ ≤ ε^{\frac{1}{2}} .  (34)

If we let

η(ε) = ε^{\frac{1}{2}} + ∑_{max(i,j,k)>n(ε)} ( ∑_{p,q,r=0}^{∞} C_{ip}C_{jq}C_{kr}μ_{pqr}^{0})^{2} ,

then, in view of (25), (26) and (34), we have ∥p^{n(ε)}(μ) − u_{0}∥ ≤ η(ε). As ε → 0, we have f^{−1}(ε^{− \frac{1}{2}} ) → ∞ and n(ε) → ∞. By (8)

∑_{max(i,j,k)>n(ε)} ( ∑_{p,q,r=0}^{∞} C_{ip}C_{jq}C_{kr}μ_{pqr}^{0})^{2} → 0 as n(ε) → 0.

Hence η(ε) → 0 as ε → 0.

Now, let u_{0} ∈ H^{1}(I). Then by Theorem 1 one has

∥p^{n(ε)}(μ^{0}) − u_{0}∥ ≤ \frac{(F (u_{0}))^{\frac{1}{2}}}{n(ε) + 1} .  (35)

The definition of n(ε) implies n(ε) + 1 > f^{−1}(ε^{− \frac{1}{2}} ), and f being a monotone increasing function, f(n(ε) + 1) > ε^{− \frac{1}{2}} or, equivalently,

\frac{1}{\sqrt{8π^{3}}} (3 + 2\sqrt{2})^{3(n(ε)+1)+3} > ε^{− \frac{1}{2}} .

It follows that

(3 + 2\sqrt{2})^{n(ε)+2} > \sqrt{2π}ε^{− \frac{1}{6}} ,

20 D. D. Ang, R. Gorenflo and D. D. Trong

i.e.,

n(ε) + 2 > \frac{1}{ln (3 + 2\sqrt{2})} ln (\frac{\sqrt{2π}}{ε^{1/6}} ).

Thus we get after some calculations

n(ε) + 1 > \frac{ln (\sqrt{2π}(3 − 2\sqrt{2}))}{ln(3 + 2\sqrt{2})} + \frac{1}{6 ln(3 + 2\sqrt{2})} ln (\frac{1}{ε}).

It can be checked directly that the right-hand side of the latter inequality is positive if and only if

0 < ε < (\frac{\sqrt{2π}}{3 + 2\sqrt{2}})^{6}.

Let ε satisfy this inequality. By the preceding arguments, (35) gives

∥p^{n(ε)}(μ^{0}) − u_{0}∥ ≤ \frac{(F (u_{0}))^{\frac{1}{2}}}{C_{1} + C_{2} ln (\frac{1}{ε})}  (36)

with

C_{1} = \frac{ln (\sqrt{2π}(3 − 2\sqrt{2}))}{ln(3 + 2\sqrt{2})} and C_{2} = \frac{1}{6 ln(3 + 2\sqrt{2})}.

Combining (34) and (36) gives

∥p^{n(ε)}(μ) − u_{0}∥ ≤ ε^{\frac{1}{2}} + \frac{(F (u_{0}))^{\frac{1}{2}}}{C_{1} + C_{2} ln (\frac{1}{ε})}.

Now the proof of Theorem 2 will be completed once (32) is proved. In fact, one has

P_{i}(3) = \frac{1}{π} ∫_{0}^{π} (3 + 2\sqrt{2} cos ϕ)^{i}dϕ

= \frac{1}{π} (∫_{0}^{π/2} + ∫_{π/2}^{π}) (3 + 2\sqrt{2} cos ϕ)^{i} dϕ

= \frac{1}{π} ∫_{0}^{π/2} ((3 + 2\sqrt{2} cos ϕ)^{i} + (3 − 2\sqrt{2} cos ϕ)^{i})dϕ.

(37)

From (37), with t = cos ϕ, it follows that

P_{i}(3) ≤ \frac{2}{π} ∫_{0}^{π/2}(3 + 2\sqrt{2} cos ϕ)^{i}dϕ ≤ \frac{2}{π} ∫_{0}^{1} \frac{(3 + 2\sqrt{2}t)^{i}}{\sqrt{1 − t^{2}}} dt ≤ \frac{2}{π} ∫_{0}^{1} \frac{(3 + 2\sqrt{2}t)^{i}}{\sqrt{1 − t}} dt.

Putting s = \sqrt{1 − t} and ζ = s\sqrt{\frac{2\sqrt{2}}{3+2\sqrt{2}}} we get, after some computations,

P_{i}(3) ≤ \frac{4}{π} (3 + 2\sqrt{2})^{i}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} ∫_{0}^{1}(1 − ζ^{2})^{i}dζ.

Letting ζ = cos θ, we have

P_{i}(3) ≤ \frac{4}{π} (3 + 2\sqrt{2})^{i}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} ∫_{0}^{π/2}sin^{2i+1} θ dθ.  (38)

Putting I_{n} = ∫_{0}^{π/2} sin^{n} θ dθ we have (cf. [12: p. 681])

I_{2k} = \frac{1 · 3 ⋯ (2k − 1)}{2 · 4 ⋯ (2k)} · \frac{π}{2}  and  I_{2k−1} = \frac{2 · 4 ⋯ (2k − 2)}{1 · 3 ⋯ (2k − 1)}.

It follows that

I_{m}I_{m−1} = \frac{π}{2m}  (m ∈ ℕ).  (39)

But

I_{m−1} = ∫_{0}^{π/2}sin^{m−1} θ dθ > ∫_{0}^{π/2}sin^{m} θ dθ = I_{m}.

Hence (39) gives

I_{m}^{2} ≤ I_{m}I_{m−1} = \frac{π}{2m}.  (40)

Substituting (40) into (38), we get

P_{i}(3) ≤ \frac{4}{π} (3 + 2\sqrt{2})^{i}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} · \sqrt{\frac{π}{2(2i + 1)}}.

This establishes (32) and completes the proof of Theorem 2

Proof of Theorem 3. We choose

u_{0}(x, y, z) = \frac{1}{m(ε)}L_{m(ε)+1}(x) = \frac{1}{m(ε)}L_{m(ε)+1,0,0}(x, y, z),

μ_{pqr}^{0} = ∫_{I} u_{0}(x, y, z)x^{p}y^{q}z^{r} dxdydz.

Since ∫_{0}^{1} L_{i}(x)x^{j} dx = 0 (0 ≤ j < i) one has

μ_{pqr}^{0} = 0  for p ≤ m(ε),  (41)

i.e., u_{0} is orthogonal for all the L_{ijk}’s, 0 ≤ i ≤ m(ε). Put μ^{0} = (μ_{pqr}^{0}) and μ = (μ_{pqr}) with

μ_{pqr} = ε(−1)^{p+q+r} + μ_{pqr}^{0}.  (42)

It is clear that ∥μ − μ^{0}∥_{∞} = sup_{p,q,r} |μ_{pqr} − μ_{pqr}^{0}| = ε. Hence, from assumption (15) of Theorem 3, one has

∥p^{m(ε)}(μ) − u_{0}∥ ≤ α(ε)(1 + (F (u_{0}))^{\frac{1}{2}} ).  (43)

22 D. D. Ang, R. Gorenflo and D. D. Trong

We estimate the left-hand side of (43). The orthonormality of the sequence (L_{ijk}) implies

∥p^{m(ε)}(μ) − u_{0}∥^{2} = ∥p^{m(ε)}(μ)∥^{2} + ∥u_{0}∥^{2} = ∥p^{m(ε)}(μ)∥^{2} + \frac{1}{m^{2}(ε)}.  (44)

On the other hand, one has in view of (42), (29) and (31)

∥p^{m(ε)}(μ)∥^{2} = ε^{2} ( ∑_{i=0}^{m(ε)} (∑_{p=0}^{i} |C_{ip}|)^{2})^{3} = ε^{2} ( ∑_{i=0}^{m(ε)}(2i + 1)|P_{i}(3)|^{2})^{3} . (45)

From (37) we conclude

P_{i}(3) ≥ \frac{1}{π} ∫_{0}^{π/2}(3 + 2\sqrt{2})^{i}dϕ.  (46)

We shall make use of the inequality

(2i + 1)|P_{i}(3)|^{2} ≥ \frac{(3 + 2\sqrt{2})^{2i}}{2π^{2}} ,  (47)

the proof of which will be given later. Combining (47) with (45) gives

\begin{aligned}∥p^{m(ε)}(μ)∥^{2} & ≥ ε^{2} ( ∑_{i=0}^{m(ε)} \frac{(3 + 2\sqrt{2})^{2i}}{2π^{2}} )^{3} \\ & = \frac{ε^{2}}{8π^{6}} (\frac{(3 + 2\sqrt{2})^{2(m(ε)+1)} − 1}{(3 + 2\sqrt{2})^{2} − 1})^{3} \\ & ≥ \frac{ε^{2}}{64π^{6}} (3 + 2\sqrt{2})^{6m(ε)}. \end{aligned}  (48)

From (43), (45) and (48) we get

(1 + (F (u_{0}))^{\frac{1}{2}} )^{2}α^{2}(ε) ≥ \frac{ε^{2}}{64π^{6}} (3 + 2\sqrt{2})^{6m(ε)} + \frac{1}{m^{2}(ε)} ≥ min_{t>0} g(t),  (49)

where

g(t) = \frac{ε^{2}}{64π^{6}} (3 + 2\sqrt{2})^{6t} + \frac{1}{t^{2}} .

It is clear that g attains its minimum at t_{ε} > 0 satisfying

g′(t_{ε}) = \frac{3ε^{2}}{32π^{6}} (3 + 2\sqrt{2})^{6t_{ε}} ln(3 + 2\sqrt{2}) − \frac{2}{t_{ε}^{3}} = 0.  (50)

From (49) one has

(1 + (F (u_{0}))^{\frac{1}{2}} )^{2}α^{2}(ε) ≥ g(t_{ε}) ≥ t_{ε}^{−2},

A Multi-Dimensional Hausdorff Moment Problem 23

which implies

α(ε) ln (\frac{1}{ε}) ≥ \frac{1}{1 + (F (u_{0}))^{\frac{1}{2}}} · \frac{ln (\frac{1}{ε})}{t_{ε}}.  (51)

From (50) one has

t_{ε}^{3}(3 + 2\sqrt{2})^{6t_{ε}} = \frac{64π^{6}}{3ε^{2} ln(3 + 2\sqrt{2})}.

Hence

3 ln t_{ε} + 6t_{ε} ln(3 + 2\sqrt{2}) = ln (\frac{64π^{6}}{3 ln(3 + 2\sqrt{2})}) + 2 ln (\frac{1}{ε}) .

Multiplying this equality by t_{ε}^{−1}, letting ε → 0 and noting that t_{ε} → +∞ as ε → 0, we get

lim_{ε↓0} \frac{ln (\frac{1}{ε})}{t_{ε}} = 3 ln(3 + 2\sqrt{2}).  (52)

On the other hand, in view of (19) and the definitions of F (u_{0}) and u_{0} we have

\begin{aligned}F (u_{0}) & = ∫_{I} x(1 − x)|\frac{∂u_{0}}{∂x} (x, y, z)|^{2}dxdydz \\ & + ∫_{I} y(1 − y)|\frac{∂u_{0}}{∂y} (x, y, z)|^{2}dxdydz \\ & + ∫_{I} z(1 − z)|\frac{∂u_{0}}{∂z} (x, y, z)|^{2}dxdydz \\ & = \frac{1}{m^{2}(ε)}(m(ε) + 1)(m(ε) + 2). \end{aligned}  (53)

Hence, in view of (52) and (53), inequality (51) implies the desired inequality (16).

To complete the proof of Theorem 3, we shall prove inequality (47). Putting t = cos ϕ and s = \sqrt{1 − t}, and then letting ζ = s\sqrt{\frac{2\sqrt{2}}{3+2\sqrt{2}}}, we get the inequality

P_{i}(3) ≥ \frac{1}{π} (3 + 2\sqrt{2})^{i}\sqrt{\frac{3 + 2\sqrt{2}}{\sqrt{2}}}J,  (54)

where

J = ∫_{0}^{√\overline{\frac{2\sqrt{2}}{3+2\sqrt{2}}}} (1 − ζ^{2})^{i}dζ.

We shall estimate J. One has

J = I_{2i+1} − K_{i},

where

I_{2i+1} = ∫_{0}^{π/2} sin^{2i+1} θ dθ  and  K_{i} = ∫_{√\overline{\frac{2\sqrt{2}}{3+2\sqrt{2}}}}^{1} (1 − ζ^{2})^{i}dζ.

Using (39) one has I_{2i+1}I_{2i+2} = \frac{π}{2(2i+2)}. Hence

I_{2i+1}^{2} ≥ I_{2i+1}I_{2i+2} = \frac{π}{2(2i + 2)}.  (55)

On the other hand,

\begin{aligned}K_{i} & = ∫ 1_{\frac{2\sqrt{2}}{3+2\sqrt{2}}} (1 − ζ)^{i} \frac{dζ}{2\sqrt{ζ}} \\ & ≤ \frac{1}{2}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} ∫ 1_{\frac{2\sqrt{2}}{3+2\sqrt{2}}} (1 − ζ)^{i}dζ \\ & = \frac{1}{2(i + 1)}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} (\frac{3}{3 + 2\sqrt{2}})^{i+1} . \end{aligned}  (56)

From (55) and (56) we deduce

J ≥ \sqrt{\frac{π}{2(2i + 2)}} − \frac{1}{2(i + 1)}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} (\frac{3}{3 + 2\sqrt{2}})^{i+1}

= \frac{1}{\sqrt{i + 1}} (\sqrt{\frac{π}{4}} − \frac{1}{2\sqrt{i + 1}}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} (\frac{3}{3 + 2\sqrt{2}})^{i+1})

≥ \frac{1}{\sqrt{i + 1}} (\sqrt{\frac{π}{4}} − \frac{1}{2}\sqrt{\frac{3 + 2\sqrt{2}}{2\sqrt{2}}} · \frac{3}{3 + 2\sqrt{2}})

> \frac{1}{2\sqrt{i + 1}}.

In view of the latter inequality, (54) gives

(2i + 1)|P_{i}(3)|^{2} ≥ \frac{2i + 1}{4(i + 1)π^{2}\sqrt{2}}(3 + 2\sqrt{2})^{2i+1} ≥ \frac{(3 + 2\sqrt{2})^{2i}}{2π^{2}} .

This establishes (47) and completes the proof of Theorem 3

Acknowledgement. The work of the first author was completed with a financial support from the National Program of Basic Research of Vietnam (project ”Nondestructive Evaluation”). The second author appreciates financial support from the Research Commission of Free University of Berlin (project ”Convolutions”) and travel grants from DFG (Deutsche Forschungsgemeinschaft) in Bonn. All three authors would like to thank the referees for their valuable comments and constructive criticisms. Our special thanks go to one of the referees for his (or her) suggestions leading to a sharpening of our estimates in Theorem 2.

A Multi-Dimensional Hausdorff Moment Problem 25

References

[1] Abramowitz, M. and I. A. Stegun: Handbook of Mathematical Functions. New York: Dover 1972.

[2] Ang, D. D., Gorenflo, R. and D. D. Trong: A multidimensional Hausdorff moment problem. Preprint. Berlin: Freie Universität, Preprint Mathematics A-23/97.

[3] Ang, D. D., Vy, L. K. and R. Gorenflo: A regularization method for the moment problem. In: Inverse Problems: Principles and Applications in Geophysics, Technology and Medicine (Math. Research: Vol. 74). Berlin: Akademie Verlag 1993, pp. 37 – 45.

[4] Ang, D. D., Gorenflo, R. and L. K. Vy: Backus-Gilbert regularization of a moment problem. Preprint. Berlin: Freie Universität, Preprint Mathematics A-7/93.

[5] Askey, R., Schoenberg, I. J. and A. Sharma: Hausdorff’s moment problem and expansions in Legendre polynomials. J. Math. Anal. Appl. 86 (1982), 237 – 245.

[6] Bertero, M., De Mol, C. and E. Pike: Linear inverse problems with discrete data. Part I: General formulation and singular system analysis. Inverse Problems 1 (1985), 301 – 330.

[7] Inglese, G.: Recent results in the study of the moment problem. In: Theory and Practice of Geophysical Data Inversion (eds.: A. Vogel et al.). Braunschweig und Wiesbaden: Vieweg-Verlag 1992, pp. 73 – 84.

[8] Kirsch, A., Schomburg, B. and G. Berendt: The Backus-Gilbert method. Inverse Problems 4 (1988), 771 – 783.

[9] Lebedev, N. N.: Special Functions and Their Applications. New York: Dover Publications Inc. 1972.

[10] Rabenstein, A. L.: Introduction to Ordinary Differential Equations. New York et al.: Acad. Press 1972.

[11] Talenti, G.: Recovering a function from a finite number of moments. Inverse Problems 3 (1987), 501 – 517.

[12] Taylor, A.: Advanced Calculus. New York et al.: Blaisdell Publ. Comp. 1965.

[13] Shohat, J. A. and J. D. Tamarkin: The Problem of Moments (AMS Mathematical Surveys). Providence (R.I.): Amer. Math. Soc. 1943.

Received 23.12.1997; in revised form 14.12.1998