On the classical moment problems

We introduce to the classical moment problem, a short history. We refer to Akhiezer’s book:

N. I, Akhiezer, The Classical Moment Problem and Some Related Questions in
Analysis, Oliver & Boyd, Edinburgh/London, 1965.

and Christiansen’s notes: Moment (on Steven Miller’s page ).

The moment problem is a classical problem in analysis. This problem occurs for the first time in the work of Chebychev in 1873. After that, T.Stieltjes (1894-1895) and A.Markov consider more general case. Chebychev and A.Markov took the moment problem in the relationship with probability theory. The first solution and discussion of extended moment problem is due to Hamburger, he studied Classical moment problem (one-dimensional).
Classical moment problem (one-dimensional) Given an infinite sequence of real numbers ${\{s_n\}_n}$ (${s_0 = 1}$). Does there exist a positive Borel measure ${\mu}$ such that:

$\displaystyle s_n = \int_{\mathbb{R}}x^nd\mu(x).$

In general, we have Classical moment problem (multidimensional)
Given a function ${s : \mathbb{N}^k \rightarrow \mathbb{R}}$. Does there exist a positive Borel measure ${\mu}$ such that:

$\displaystyle s(n) = \int_{\mathbb{R}^k}x_1^{n_1}\dots x_n^{n_k}d\mu(x_1,\dots,x_k)<\infty \quad (*).$

In the case one-dimensional moments, the sequence ${\{s_n\}}$ is a function and we have ${s_n = s(n), n \in \mathbb{N}}$.
Two measure ${\mu}$ and ${\nu}$ are called equivalent if they satisfy:

$\displaystyle s_n = \int_{\mathbb{R}^k}x^n d\mu(x) = \int_{\mathbb{R}^k}x^n d\nu(x).$

In other words, we say they have same moments.
The measure ${\mu}$ is called determinate if there only exists ${\mu}$ such that ${s_n = \int_{\mathbb{R}^k}x^n d\mu(x)}$ and indeterminate otherwise.
The aims of the multidimensional moment problem are:

1. To find necessary and sufficient conditions for existence of measure ${\mu}$ satisfying (*).
2. To be able to decide determinacy.
3. In the indeterminate case to give a complete description of all measures satisfying (*).

On the Spectral Theorem (from Linear algebra)

The form we prefers says that every bounded self-adjoint operator is a multiplication operator. This means that given a bounded self-adjoint operator on a Hilbert space ${\mathcal{H}}$, we can always find a measure ${\mu}$ on a measure space ${M}$ and a unitary operator ${U: \mathcal{H} \rightarrow L^2(M, d\mu)}$ so that

$\displaystyle (UAU^{-1}f)(x) = F(x)f(x),$

for some bounded real-valued measurable function ${F}$ on ${M}$.

In the case of real, the Hilbert space $\mathcal{H}$ becomes a Euclidean space with an inner product. So an operator in $\mathcal{H}$ corresponds with an orthogonal matrix $A$. And an orthogonal matrix can be diagonalized, i.e. there exists $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ and $S^{-1} = S^T$ such that

$\displaystyle S^TAS = diag(\lambda_1, \dots, \lambda_n)$.

(cont.)

The tangent plane of a surface and the tangent space of a manifold (at one point)

The tangent plane of a surface
By a tangent vector to $S$ at a point $p \in S$, we mean the tangent vector $\alpha'(0)$ of a differentiable parametrized curve $\alpha: (-\epsilon, \epsilon) \to S$ with $\alpha(0) = p$.

What is the tangent plane of a surface? That is a plane which containes all of the tangent vectors of this surface at point $p \in S$.

Proposition
Let $\varphi: U \subset \mathbb{R}^2 \to S$ be a parametrization of a regular surface $S$ and let $q \in U$. The vector subspace of dim $2$,

$d\varphi_q(\mathbb{R}^2) \subset \mathbb{R}^3,$

coincides with the set of tangent vectors to $S$ at $\varphi(q)$

There is a similar situation in here. In the case of manifold, the tangent space is also built from the set of all of tangent vectors of a manifold.

Tangent spaces of a manifold

In Milnor’s book (\cite{Milnor}), the tangent space $TM_x$ at $x$ for arbitrary smooth manifold $M \subset \mathbb{R}^k$ is defined:

Choose a parametrization $g : U \to M \subset \mathbb{R}^k$ of a neighborhood $g(U)$ of $x$ in $M$, with $g(u) = x$. We have $dg_u: \mathbb{R}^m \to \mathbb{R}^k$. So the image $dg_u(\mathbb{R}^m)$ of $dg_u$ is equal to $TM_x$.

References

J. W. Milnor, Topology from differentiable viewpoint, 1965.

M. do Carmo, Differential geometry, curves and surfaces, 1976.