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Stieltjes moment problem - Wikipedia

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In mathematics, the Stieltjes moment problem, named after Thomas Joannes Stieltjes, seeks necessary and sufficient conditions for a sequence (m₀, m₁, m₂, ...) 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 (−∞, ∞).

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 )} with infinite support if and only if for all 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 )} with finite support of size m if and only if for all n ≤ m {\displaystyle n\leq m} , 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}

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.}

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 ~.}

• 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

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