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In mathematics , the Stieltjes moment problem , named after Thomas Joannes Stieltjes , seeks necessary and sufficient conditions for a sequence (m 0 , m 1 , m 2 , ...) to be of the form
m
n
=
∫
0
∞
x
n
d
μ
(
x
)
{\displaystyle m_{n}=\int _{0}^{\infty }x^{n}\,d\mu (x)}
for some measure μ . If such a function μ exists, one asks whether it is unique.
The essential difference between this and other well-known moment problems is that this is on a half-line [0, ∞), whereas in the Hausdorff moment problem one considers a bounded interval [0, 1], and in the Hamburger moment problem one considers the whole line (−∞, ∞).
Existence
Let
Δ
n
=
[
m
0
m
1
m
2
⋯
m
n
m
1
m
2
m
3
⋯
m
n
+
1
m
2
m
3
m
4
⋯
m
n
+
2
⋮
⋮
⋮
⋱
⋮
m
n
m
n
+
1
m
n
+
2
⋯
m
2
n
]
{\displaystyle \Delta _{n}=\left[{\begin{matrix}m_{0}&m_{1}&m_{2}&\cdots &m_{n}\\m_{1}&m_{2}&m_{3}&\cdots &m_{n+1}\\m_{2}&m_{3}&m_{4}&\cdots &m_{n+2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\m_{n}&m_{n+1}&m_{n+2}&\cdots &m_{2n}\end{matrix}}\right]}
be a Hankel matrix , and
Δ
n
(
1
)
=
[
m
1
m
2
m
3
⋯
m
n
+
1
m
2
m
3
m
4
⋯
m
n
+
2
m
3
m
4
m
5
⋯
m
n
+
3
⋮
⋮
⋮
⋱
⋮
m
n
+
1
m
n
+
2
m
n
+
3
⋯
m
2
n
+
1
]
.
{\displaystyle \Delta _{n}^{(1)}=\left[{\begin{matrix}m_{1}&m_{2}&m_{3}&\cdots &m_{n+1}\\m_{2}&m_{3}&m_{4}&\cdots &m_{n+2}\\m_{3}&m_{4}&m_{5}&\cdots &m_{n+3}\\\vdots &\vdots &\vdots &\ddots &\vdots \\m_{n+1}&m_{n+2}&m_{n+3}&\cdots &m_{2n+1}\end{matrix}}\right].}
Then { m n n = 1, 2, 3, ... } is a moment sequence of some measure on
[
0
,
∞
)
{\displaystyle [0,\infty )}
n , both
det
(
Δ
n
)
>
0
a
n
d
det
(
Δ
n
(
1
)
)
>
0.
{\displaystyle \det(\Delta _{n})>0\ \mathrm {and} \ \det \left(\Delta _{n}^{(1)}\right)>0.}
{ m n n = 1, 2, 3, ... } is a moment sequence of some measure on
[
0
,
∞
)
{\displaystyle [0,\infty )}
m if and only if for all
n
≤
m
{\displaystyle n\leq m}
det
(
Δ
n
)
>
0
a
n
d
det
(
Δ
n
(
1
)
)
>
0
{\displaystyle \det(\Delta _{n})>0\ \mathrm {and} \ \det \left(\Delta _{n}^{(1)}\right)>0}
and for all larger
n
{\displaystyle n}
det
(
Δ
n
)
=
0
a
n
d
det
(
Δ
n
(
1
)
)
=
0.
{\displaystyle \det(\Delta _{n})=0\ \mathrm {and} \ \det \left(\Delta _{n}^{(1)}\right)=0.}
Uniqueness
There are several sufficient conditions for uniqueness, for example, Carleman's condition , which states that the solution is unique if
∑
n
≥
1
m
n
−
1
/
(
2
n
)
=
∞
.
{\displaystyle \sum _{n\geq 1}m_{n}^{-1/(2n)}=\infty ~.}
References
Reed, Michael; Simon, Barry (1975), Fourier Analysis, Self-Adjointness , Methods of modern mathematical physics, vol. 2, Academic Press, p. 341 (exercise 25), ISBN 0-12-585002-6 
![{\displaystyle \Delta _{n}=\left[{\begin{matrix}m_{0}&m_{1}&m_{2}&\cdots &m_{n}\\m_{1}&m_{2}&m_{3}&\cdots &m_{n+1}\\m_{2}&m_{3}&m_{4}&\cdots &m_{n+2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\m_{n}&m_{n+1}&m_{n+2}&\cdots &m_{2n}\end{matrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8206d416e2354bde00ecd3217b257609539b19c0)
![{\displaystyle \Delta _{n}^{(1)}=\left[{\begin{matrix}m_{1}&m_{2}&m_{3}&\cdots &m_{n+1}\\m_{2}&m_{3}&m_{4}&\cdots &m_{n+2}\\m_{3}&m_{4}&m_{5}&\cdots &m_{n+3}\\\vdots &\vdots &\vdots &\ddots &\vdots \\m_{n+1}&m_{n+2}&m_{n+3}&\cdots &m_{2n+1}\end{matrix}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/067bad63dc066fda432dc92c73a1724826746860)
