The Hamburger Moment Problem - Existence Theorem and Self-Adjoint Extensions

The Hamburger moment problem is a classical question in analysis that asks: given a sequence of real numbers $\{\gamma_n\}_{n=0}^{\infty}$, does there exist a measure $\mu$ on the real line such that $\gamma_n = \int_{-\infty}^{\infty} x^n \, d\mu(x)$ for all $n$? This problem has deep connections to operator theory, orthogonal polynomials, and functional analysis.

In this post, I’ll walk through one of the central results in the theory: the existence theorem. This theorem provides necessary and sufficient conditions for a solution to exist, and the proof beautifully intertwines the theory of sesquilinear forms with von Neumann’s theory of self-adjoint extensions of symmetric operators.

Theorem 1. A necessary and sufficient condition for there to exist a measure $d \rho$ with infinite support obeying $\gamma_n = \int_{-\infty}^{\infty} x^n \, d\rho(x)$ for all $n$ for a given sequence of real numbers $\{\gamma_n\}_{n=0}^{\infty}$ is that $\det( \mathcal{H}_N ) > 0$ for $N = 1, 2, ...$.

While working through Simon’s comprehensive paper (https://arxiv.org/pdf/math-ph/9906008), I found myself frequently jumping between sections and consulting additional references to follow the proof. To help myself (and hopefully others) understand the argument better, I’ve compiled the key steps into a single, linear narrative that presents the proof in a more accessible flow.

Setting Up: Sesquilinear Forms

Let $\{\gamma_n\}$ be a sequence of numbers. We define two sesquilinear forms, $H_N$ and $S_N$, on $\mathbb{C}^N$ for each $N$ by

\begin{equation} \label{eq:HN-form} H_N(\alpha, \beta) = \sum_{n,m=0}^{N-1} \bar{\alpha}_n \beta_m \gamma{n+m} \end{equation}

\begin{equation} \label{eq:SN-form} S_N(\alpha, \beta) = \sum_{n,m=0}^{N-1} \bar{\alpha}_n \beta_m \gamma{n+m+1} \end{equation}

and corresponding matrices $\mathcal{H}_N$ and $\mathcal{S}_N$ so that $H_N(\alpha, \beta) = \langle \alpha, \mathcal{H}_N \beta \rangle$ and $S_N(\alpha, \beta) = \langle \alpha, \mathcal{S}_N \beta \rangle$ in the usual Euclidean inner product. Our inner products are linear in the second factor and anti-linear in the first.

Proposition 1. $\{H_N\}_{N=1}^\infty$ are strictly positive definite forms if and only if $\mathcal{H}_N > 0$ for $N = 1, 2, \ldots$. Similarly, $\{S_N\}_{N=1}^\infty$ are strictly positive definite forms if and only if $\det( \mathcal{S}_N) > 0$ for $N = 1, 2, \ldots$.

Proof. This follows from Sylvester’s criterion. ∎

The Necessary Condition

Proposition 2. A necessary condition that $\gamma_i = \int_{\mathbb{R}} x^i \, d\mu$ holds for some measure $d\mu$ on $(-\infty, \infty)$ with infinite support is that each sesquilinear form $H_N$ is strictly positive definite.

Proof. Suppose there exists a measure $\mu$ on $(-\infty,\infty)$ with infinite support such that

\[\gamma_i = \int_{-\infty}^\infty x^i \, d\mu(x), \quad i = 0,1,2,\dots\]

and define the sesquilinear forms

\[H_N(\alpha,\beta) = \sum_{n,m=0}^{N-1} \bar{\alpha}_n \beta_m \gamma_{n+m} = \sum_{n,m=0}^{N-1} \bar{\alpha}_n \beta_m \int x^{n+m} \, d\mu(x),\] \[H_N(\alpha,\alpha) = \int \Big|\sum_{n=0}^{N-1} \alpha_n x^n \Big|^2 \, d\mu(x).\]

For any nonzero $\alpha \in \mathbb{C}^N$, let $p(x) = \sum_{n=0}^{N-1} \alpha_n x^n$.

First, note that $ \lvert p(x)\rvert^2 \ge 0 $ for all $ x $, so the integral $ H_N(\alpha,\alpha) $ can never be negative.

Second, since $p(x)$ is a nonzero polynomial, it has only finitely many zeros.

Because $\mu$ has infinite support, there are infinitely many points where $\lvert p(x) \rvert^2 > 0$ with positive measure, so

\[H_N(\alpha,\alpha) = \int |p(x)|^2 \, d\mu(x) > 0.\]

Thus $H_N$ is strictly positive definite for every $N$. ∎

Constructing the Hilbert Space

Suppose now that each $H_N$ is strictly positive definite. Let $\mathbb{C}[X]$ be the family of complex polynomials. Given $P(X) = \sum_{n=0}^{N-1} \alpha_n X^n$, $Q(X) = \sum_{n=0}^{N-1} \beta_n X^n$ (we suppose the upper limits in the sums are equal by using some zero $\alpha$’s or $\beta$’s if need be), define

\begin{equation} \label{eq:inner-product} \langle P, Q\rangle = H_N(\alpha, \beta). \end{equation}

This defines a positive definite inner product on $\mathbb{C}[X]$, and, in the usual way, we can complete $\mathbb{C}[X]$ to a Hilbert space $\mathcal{H}(\gamma)$ in which $\mathbb{C}[X]$ is dense. $\mathbb{C}[X]$ can be thought of as abstract polynomials or as infinite sequences $(\alpha_0, \alpha_1, \ldots, \alpha_{N-1}, 0, \ldots)$ which are eventually $0$ via $\alpha \sim \sum_{j=0}^{N-1} \alpha_j X^j$.

Define a densely defined operator $A$ on $\mathcal{H}(\gamma)$ with domain $\mathcal{D}(A) = \mathbb{C}[X]$ by

\begin{equation} \label{eq:operator} A[P(X)] = [XP(X)]. \end{equation}

In the sequence way of looking at things, $A$ is the right shift, that is,

\[A(\alpha_0, \alpha_1, \ldots, \alpha_N, 0, \ldots) = (0, \alpha_0, \alpha_1, \ldots, \alpha_N, 0, \ldots).\]

This means that

\begin{equation} \label{eq:inner-product-A} \langle P, A[Q]\rangle = S_N(\alpha, \beta) \end{equation}

and, in particular,

\begin{equation} \label{eq:moments-from-operator} \langle 1, A^n1\rangle = \gamma_n. \end{equation}

Since $H_N$ and $S_N$ are real-symmetric matrices, $A$ is a symmetric operator, that is, $\langle A[P], Q\rangle = \langle P, A[Q]\rangle$. Moreover, if we define a complex conjugation $C$ on $\mathbb{C}[X]$ by

\[C\left(\sum_{n=0}^{N-1} \alpha_n X^n\right) = \sum_{n=0}^{N-1} \bar{\alpha}_n X^n,\]

then $CA = AC$.

It follows by a theorem of von Neumann that $A$ has self-adjoint extensions. To understand the von Neumann theorem, let us review the von Neumann theory of self-adjoint extensions.

Von Neumann Theory of Self-Adjoint Extensions

Let $A$ be a symmetric operator on a Hilbert space $\mathcal{H}$. Define $K_{\pm} = \ker(A^* \mp i)$, that is, $K_+ = \{\varphi \in \mathcal{H} \mid A^*\varphi = i\varphi\}$. They are called the deficiency subspaces.

Now, we know the following identity: $\|(A \pm i)\phi\|^2 = \|A\phi\|^2 + \|\phi\|^2$, which follows from expanding $\langle (A + i)\phi, (A + i)\phi \rangle = \|A\phi\|^2 + i\langle A\phi, \phi\rangle - i\langle \phi, A\phi\rangle + \|\phi\|^2$ and noting that for a symmetric operator $A$, $\langle A\phi, \phi\rangle = \langle \phi, A\phi\rangle$, causing the cross terms to cancel. This identity implies $(A \pm i)$ are injective, since $(A \pm i)\phi = 0$ forces $\|\phi\| = 0$. If $A$ is closed, this identity further shows $(A \pm i)$ acts as an isometry from the domain $D(A)$ equipped with the graph norm to $\mathcal{H}$, ensuring $\operatorname{Ran}(A \pm i)$ is closed.

A fundamental relation between any densely defined operator $T$ and its adjoint is $\ker(T^*) = \operatorname{Ran}(T)^\perp$; applying this to $T = A \pm i$ (so $T^* = A^* \mp i$) gives $K_{\pm} = \ker(A^* \mp i) = \operatorname{Ran}(A \pm i)^\perp$. Since $\operatorname{Ran}(A \pm i)$ is closed, taking the orthogonal complement of both sides yields the final result:

\begin{equation} \label{eq:range-deficiency} \operatorname{Ran}(A \pm i) = K_{\pm}^\perp. \end{equation}

Place the graph norm on $\mathcal{D}(A^*)$, that is, $\|\varphi\|_{A^*}^2 = \langle \varphi, \varphi \rangle + \langle A^*\varphi, A^*\varphi \rangle$. This norm comes from an inner product, $\langle \varphi, \psi \rangle_{A^*} = \langle \varphi, \psi \rangle + \langle A^*\varphi, A^*\psi \rangle$.

Proposition 3. Let $A$ be a closed symmetric operator. Then

\[\mathcal{D}(A^*) = \mathcal{D}(A) \oplus K_+ \oplus K_-,\]

where $\oplus$ means orthogonal direct sum in the $\langle \cdot, \cdot \rangle_{A^*}$ inner product.

Proof. If $\varphi \in \mathcal{K}_+$ and $\psi \in \mathcal{K}_-$, then

\[\langle \varphi, \psi \rangle_{A^*} = \langle \varphi, \psi \rangle + \langle i\varphi, -i\psi \rangle = 0\]

so $\mathcal{K}_+ \perp_{A^*} \mathcal{K}_-$.

If $\varphi \in D(A)$ and $\psi \in \mathcal{K}_\pm$, then $\langle A^*\varphi, A^*\psi \rangle = \langle A\varphi, \pm i\psi \rangle = \langle \varphi, \pm iA^*\psi \rangle = -\langle \varphi, \psi \rangle,$

so $D(A) \perp_{A^*} \mathcal{K}_+ \oplus \mathcal{K}_-$. Let $\eta \in D(A^*)$. By equation \eqref{eq:range-deficiency}, $\operatorname{Ran}(A + i) + \mathcal{K}_+ = H$ so we can find $\varphi \in D(A)$ and $\psi \in \mathcal{K}_+$, so $(A^* + i)\eta = (A + i)\varphi + 2i\psi$. But then $(A^* + i)[\eta - \varphi - \psi] = 0$, so $\eta - \varphi - \psi \in \mathcal{K}_-$, that is, $\eta \in D(A) + \mathcal{K}_+ + \mathcal{K}_-$. ∎

Corollary 1.

Let $A$ be a closed symmetric operator. Then $A$ is self-adjoint if and only if $d_+ = d_- = 0$.
Let $A$ be a symmetric operator. Then $A$ is essentially self-adjoint if and only if $d_+ = d_- = 0$.

If $A \subset B$, then $B^* \subset A^*$, so if $B$ is symmetric, then $A \subset B \subset B^* \subset A^*$. Thus, to look for symmetric extensions of $A$, we need only look for operators $B$ with $A \subset B \subset A^*$, that is, for restrictions of $A^*$ to $D(B)$’s with $D(A) \subset D(B)$. By Proposition 3, every such $D(B)$ has the form $D(A) + S$ with $S \subset \mathcal{K}_+ + \mathcal{K}_-$.

On $D(A^*) \times D(A^*)$, define the sesquilinear form (sesquilinear means linear in the second factor and anti-linear in the first):

\begin{equation} \label{eq:Q-form} Q(\varphi, \psi) = \langle \varphi, A^\psi \rangle_H - \langle A^\varphi, \psi \rangle_H. \end{equation}

Proposition 4. Let $A$ be a closed symmetric operator. Then

The operators $B$ with $A \subset B \subset A^*$ are in one-one correspondence with subspaces $S$ of $\mathcal{K}_+ + \mathcal{K}_-$ under $D(B) = D(A) + S$.
$B$ is symmetric if and only if $Q \restriction_{D(B) \times D(B)} = 0$.
$B$ is symmetric if and only if $Q \restriction_{S \times S} = 0$.
$B$ is closed if and only if $S$ is closed in $\mathcal{K}_+ \oplus \mathcal{K}_-$ in $D(A^*)$ norm.
$\varphi \in D(A^*)$ lies in $D(B^*)$ if and only if $Q(\varphi, \psi) = 0$ for all $\psi \in D(B)$.
Let $J : \mathcal{K}_+ \oplus \mathcal{K}_- \to \mathcal{K}_+ \oplus \mathcal{K}_-$ by $J(\varphi, \psi) = (\varphi, -\psi)$. If $D(B) = D(A) + S$, then $D(B^*) = D(A) + J[S]^\perp,$ where $\perp$ is in $\mathcal{K}_+ \oplus \mathcal{K}_-$ in the $D(A^*)$ norm.
$\mathcal{K}_+(B) = \mathcal{K}_+ \cap S^\perp$, $\mathcal{K}_-(B) = \mathcal{K}_- \cap S^\perp$ (with $\perp$ in the $\langle \cdot, \cdot \rangle_{A^*}$ inner product).

Proof. We have already seen (i) holds and (ii) is obvious. (iii) holds since if $\varphi \in D(A)$ and $\psi \in D(A^*)$, then $Q(\varphi, \psi) = 0$ by definition of $A^*$ and symmetry of $A$. Thus, if $\varphi_1, \varphi_2 \in D(A)$ and $\psi_1, \psi_2 \in S$, then $Q(\varphi_1 + \psi_1, \varphi_2 + \psi_2) = Q(\psi_1, \psi_2)$.

(iv) is immediate if one notes that $\Gamma(B) = \Gamma(A) \oplus \{(\varphi, A^*\varphi) \mid \varphi \in S\}$ with $\oplus$ in $H \times H$ norm and that the $H \times H$ norm on $\{(\varphi, A^*\varphi) \mid \varphi \in S\}$ is just the $D(A^*)$ norm.

(v) follows from the definition of adjoint.

To prove (vi), let $\eta = (\eta_1, \eta_2)$, $\varphi = (\varphi_1, \varphi_2) \in \mathcal{K}_+ \oplus \mathcal{K}_-$. Then, direct calculations show that

\begin{equation} \label{eq:Q-calculation} Q(\eta, \varphi) = 2i[\langle \eta_1, \varphi_1 \rangle_H - \langle \eta_2, \varphi_2 \rangle_H] \end{equation}

\begin{equation} \label{eq:graph-inner-product} \langle \eta, \varphi \rangle_{A^*} = 2[\langle \eta_1, \varphi_1 \rangle_H + \langle \eta_2, \varphi_2 \rangle_H]. \end{equation}

Thus, $Q(\eta, \varphi) = 0$ if and only if $\eta \perp J\varphi$. (v) thus implies (vi).

To prove (vii), note that by (vi), $\mathcal{K}_\pm(B) = \mathcal{K}_\pm \cap D(B^*) = \mathcal{K}_\pm \cap J[S]^\perp = \mathcal{K}_\pm \cap S^\perp$ since $\varphi \in \mathcal{K}_\pm$ lies in $J[S]^\perp$ if and only if $J\varphi$ lies in $S^\perp$ if and only if $\varphi$ lies in $S^\perp$ by $J\varphi = \pm\varphi$ if $\varphi \in \mathcal{K}_\pm$. ∎

The Main Result

As a consequence, we arrive at the following proposition:

Proposition 5. If all $H_N$ are positive definite, then $A$ has self-adjoint extensions.

This completes the key step in the existence theorem for the Hamburger moment problem. The positive definiteness of the sesquilinear forms $H_N$ guarantees that we can construct a Hilbert space and a symmetric operator with self-adjoint extensions, which in turn provide the representing measures we seek. The full theorem, along with uniqueness conditions and further characterizations, builds on this foundation through the spectral theorem for self-adjoint operators.