Classical moment problems (Hamburger, Stieltjes, trigonometric)

Let $\left\{ m_n : n \geq 0\right\}$ be a sequence of real numbers and show that the following statements are equivalent.

1. There is a positive regular Borel measure $\mu$ on $[0, \infty )$ such that $m_n = \int t^n \, d\mu (t)$ for all $n \geq 0$.

2. If $\alpha_0, \ldots , \alpha_n \in \mathbb {C}$, then $\sum_{j,k=0}^{n} m_{j+k} \alpha_j \overline{\alpha_k} \geq 0$ and $\sum_{j,k=0}^{n} m_{j+k+1} \alpha_j \overline{\alpha_k} \geq 0$.

3. There is a self-adjoint operator $A$ with $\sigma (A) \subseteq [0, \infty )$ and a vector $e$ in $\operatorname {dom} A^n$ for all $n \geq 0$ such that $m_n = \left\langle A^n e, e \right\rangle$ for $n \geq 0$.

Let $\left\{ m_n : n \in \mathbb {Z}\right\} \subseteq \mathbb {C}$ and show that the following statements are equivalent.

1. There is a positive measure $\mu$ on $\partial \mathbb {D}$ such that $m_n = \int z^n \, d\mu (z)$ for all $n$ in $\mathbb {Z}$.

2. If $\alpha_{-n}, \ldots , \alpha_{-1}, \alpha_0, \alpha_1, \ldots , \alpha_n \in \mathbb {C}$, then $\sum_{j,k=-n}^{n} m_{j-k} \alpha_j \overline{\alpha_k} \geq 0$.

3. There is a unitary operator $U$ and a vector $e$ such that $m_n = \left\langle U^n e, e \right\rangle$ for all $n$.

Recall the proof that (7.1b) implies (7.1c) in the Stieltjes moment problem (Theorem 7.1): given a sequence $\left\{ m_n : n \geq 0\right\}$ of real numbers satisfying $\sum_{j,k=0}^{n} m_{j+k}\alpha_j\overline{\alpha_k} \geq 0$ and $\sum_{j,k=0}^{n} m_{j+k+1}\alpha_j\overline{\alpha_k} \geq 0$ for all $\alpha_0, \ldots , \alpha_n \in \mathbb {C}$, one lets $\mathcal{H}_0$ be the vector space of all finitely nonzero sequences $\left\{ \alpha_n : n \geq 0\right\}$ of complex numbers, with the semi-inner product $[x, y] = \sum_{j,k=0}^{\infty } m_{j+k} \alpha_j \overline{\beta_k}$ for $x = \left\{ \alpha_n\right\}$, $y = \left\{ \beta_n\right\}$ in $\mathcal{H}_0$; lets $T_0 x = \left\{ 0, \alpha_0, \alpha_1, \ldots \right\}$ for $x = \left\{ \alpha_n\right\}$ in $\mathcal{H}_0$, so that $[T_0 x, y] = [x, T_0 y]$; lets $\mathcal{K}_0 = \left\{ x \in \mathcal{H}_0 : [x,x] = 0\right\}$, shows $T_0 \mathcal{K}_0 \subseteq \mathcal{K}_0$, and lets $T$ be the densely defined symmetric operator induced by $T_0$ on the Hilbert space $\mathcal{H}$ obtained by completing $\mathcal{H}_1 = \mathcal{H}_0/\mathcal{K}_0$ with respect to the inner product induced by $[\cdot ,\cdot ]$; uses the conjugation $J_0(\left\{ \alpha_n\right\} ) = \left\{ \overline{\alpha_n}\right\}$ on $\mathcal{H}_0$ (which satisfies $J_0 \mathcal{K}_0 \subseteq \mathcal{K}_0$, so induces a conjugation $J$ on $\mathcal{H}$ satisfying the hypotheses of Proposition 7.2) to produce a self-adjoint extension $A$ of $T$; and takes $e_0 = \left\{ 1, 0, 0, \ldots \right\} \in \mathcal{H}_0$ and $e = e_0 + \mathcal{K}_0 \in \mathcal{H}_1$, so that $e \in \operatorname {dom} T^n \subseteq \operatorname {dom} A^n$ for all $n \geq 0$ and $\left\langle A^n e, e \right\rangle = [T_0^n e_0, e_0] = m_n$ for all $n \geq 0$.

Show that the operator $A$ constructed in this way is cyclic.