Exercise 23, § 5

We take up again the notation of the preceding exercise; in particular, we consider a sequence $\boldsymbol{\mu}= (\mu_n)_{n\in\mathbf{N}}$ of real numbers satisfying conditions (24) and such that the sesquilinear form $b_{\boldsymbol{\mu}}$ is separating. We identify $\mathbf{C}[X]$ with a dense subspace of $E_{\boldsymbol{\mu}}$.

a) There exists a unique sequence $(p_n)_{n\in\mathbf{N}}$ of polynomials in $\mathbf{C}[X]$ such that the following conditions are satisfied ("orthogonal polynomials of the first species associated with $\boldsymbol{\mu}$"):

(i) The family $(p_n)$ is an orthonormal basis of $E_{\boldsymbol{\mu}}$;

(ii) The polynomial $p_n$ is of degree $n$ for every $n\in \mathbf{N}$;

(iii) The coefficient of $X^n$ in $p_n$ belongs to $\mathbf{R}_+^*$.

Moreover, one has $p_n\in \mathbf{R}[X]$ for every $n\in \mathbf{N}$ and $p_0= 1/\overline{\mu_0}$. We also set $p_{-1}= 0$.

b) There exist sequences $(a_n)_{n\geqslant-1}$ in $\mathbf{R}^*_+$ and $(b_n)_{n\in\mathbf{N}}$ in $\mathbf{R}$ such that $a_{-1}= 1$ and

for every $n\in \mathbf{N}$. These sequences are unique.

c) For every $n\geqslant 1$, the coefficient of $X^n$ in the orthogonal polynomial $p_n$ is equal to $(a_0\cdots a_{n-1})^{-1}$.

d) Let $\mathscr{F}$ be the complex vector space of sequences $(z_n)_{n\geqslant-1}$ of complex numbers such that $z_{-1}= 0$. We denote by $\widetilde{v}_{\boldsymbol{\mu}}$ the linear mapping of $\mathscr{F}$ into itself such that $\widetilde{v}_{\boldsymbol{\mu}}(z_n) = (w_n)_{n\in\mathbf{N}}$ where $w_{-1}= 0$ and

for every $n\in \mathbf{N}$.

For every $s\in \mathbf{C}$, the sequence $z_s= (p_n(s))_{n\geqslant-1}$ is the unique sequence in $\mathscr{F}$ such that $(z_s)_0=p_0(z)$ and $\widetilde{v}_{\boldsymbol{\mu}}(z_s) =sz_s$.

e) Let $z$ and $w$ be elements of $\mathscr{F}$. We define the sequence $W(z, w) =$ $(W_n(z, w))_{n\in\mathbf{N}}$ by setting

for $n\in \mathbf{N}$. One has

for every $n\in \mathbf{N}$.

f) Let $P\subset \ell^2(\mathbf{N})$ be the space of sequences with finite support. For every $z\in P$, the sequence $\widetilde{v}_{\boldsymbol{\mu}}(z)$ belongs to $\ell^2(\mathbf{N})$.

We denote by $v_{\boldsymbol{\mu}}$ the partial operator on $\ell^2(\mathbf{N})$ with domain P defined by $z\mapsto \widetilde{v}_{\boldsymbol{\mu}}(z)$. It is symmetric.

g) The adjoint of $v_{\boldsymbol{\mu}}$ is the partial operator with domain

dom($v^*_{\boldsymbol{\mu}}$) $=\{z\in \ell^2(\mathbf{N})|\widetilde{v}_{\boldsymbol{\mu}}(z)\in \ell^2(\mathbf{N})\}$

such that $v^*_{\boldsymbol{\mu}}(z) =\widetilde{v}_{\boldsymbol{\mu}}(z)$ for $z\in$ dom($v_{\boldsymbol{\mu}}^*$).

h) For all $z$ and $w$ in dom($v_{\boldsymbol{\mu}}^*$), the limit of $W_n(z, w)$ as $n\rightarrow +\infty$ exists and is equal to $\langle w|v^*_{\boldsymbol{\mu}}(z)\rangle - \langle v_{\boldsymbol{\mu}}^*(w)|z\rangle$.

i) Let $s\in \mathbf{C}$. The kernel of $v_{\boldsymbol{\mu}}^*-s\cdot 1_{\ell^2(\mathbf{N})}$ is of dimension $\leqslant 1$; it is of dimension 1 if and only if $z_s\in \ell^2(\mathbf{N})$, and is then generated by $z_s$.

j) The closure of the partial operator $v_{\boldsymbol{\mu}}$ (resp. of $u_{\boldsymbol{\mu}}$) has deficiency index $(0,0)$ or $(1,1)$; the former case holds if and only if there exists $s\in \mathbf{C}-\mathbf{R}$ such that $z_s\notin\ell^2(\mathbf{N})$. (For the case of $u_{\boldsymbol{\mu}}$, note that the mapping $X^n\mapsto e_n$, where $(e_n)$ is the canonical basis of $\ell^2(\mathbf{N})$, induces an isometric isomorphism of $E_{\boldsymbol{\mu}}$ into $\ell^2(\mathbf{N})$ which permits one to identify $u_{\boldsymbol{\mu}}$ with $v_{\boldsymbol{\mu}}.$)

k) Let $s\in \mathbf{C}$ and $f\in$ Ker($u^*_{\boldsymbol{\mu}}-s\cdot 1_{E_{\boldsymbol{\mu}}}$). If $\langle f|1\rangle = 0$, then $f= 0$.

l) Let $u_1$ and $u_2$ be self-adjoint extensions of $u_{\boldsymbol{\mu}}$ on $E_{\boldsymbol{\mu}}$. One has $u_1=u_2$ if and only if there exists $s\in \mathbf{C}-\mathbf{R}$ such that $\langle R(u_1, s)1|1\rangle =\langle R(u_2, s)1|1\rangle$. (Suppose that such an equality holds; set $f_1= R(u_1, s)1$, and now prove that $f_1$ does not belong to the domain of $\overline{u}_{\boldsymbol{\mu}}$; deduce that $u_1$ coincides with the closure of the restriction of $u_1$ to dom($\overline{u}_{\boldsymbol{\mu}}$)$\oplus \mathbf{C}f_1$. Argue analogously with $u_2$ and $f_2= R(u_2, s)1$, then prove that $f_1=f_2$ to conclude that $u_1=u_2.$) m) Suppose that $u_{\boldsymbol{\mu}}$ is not essentially self-adjoint. There then exist at least two measures $\nu \in \mathscr{M}_m(\mathbf{R})$ satisfying $\nu (x^n) =\mu_n$ for every $n\in \mathbf{N}$. n) Suppose in the sequel that $u_{\boldsymbol{\mu}}$ is essentially self-adjoint. For every $s\in$ C-R, there exists a sequence $(r_{s,m})_{m\in\mathbf{N}}$ of polynomials such that $(x-s)r_{s,m}$ converges in $E_\mu$ to the constant function 1 as $m\rightarrow +\infty$.

o) Let $\nu \in \mathscr{M}_m(\mathbf{R})$ be a measure such that $\nu (x^n) =\mu_n$ for every $n\in \mathbf{N}$. For every $m\in \mathbf{N}$, the inequality

is valid.

p) The measure $\nu$ is unique.