On the spectrum of tridiagonal operators in the context of orthogonal polynomials

Abstract

The basis for our studies is a large class of orthogonal polynomial sequences $(P_n)_{n\in {{\mathbb {N}}}_0}$, which is normalized by $P_n(x_0)=1$ for all $n\in {\mathbb {N}}_0$ where the coefficients in the three-term recurrence relation are bounded. The goal is to check if $x_0 \in {\mathbb {R}}$ is in the support of the orthogonalization measure $\mu$. For this purpose, we use, among other things, a result of G. H. Hardy concerning Cesàro operators on weighted $l^2$-spaces. These investigations generalize ideas from Lasser et al. (Arch Math 100:289–299, 2013).

1 Orthogonal polynomials on the real line and tridiagonal operators

Let $\mu$ be a probability measure on the real line. We denote the support of $\mu$ by $\mathcal {S}$ and assume its cardinality $\#\mathcal {S}=\infty$. Let $(p_n)_{n\in {\mathbb {N}}_0}$ denote the unique orthonormal polynomial sequence with respect to $\mu$, that is $\deg p_n =n$, $\int p_n p_m d\mu =\delta _{n,m}$, and $p_n$ has a positive leading coefficient for all $n,m\in {\mathbb {N}}_0$. The orthonormal polynomial sequence $(p_n)_{n\in {\mathbb {N}}_0}$ satisfies a recurrence relation

$$\begin{aligned} xp_n(x)=\lambda _n p_{n+1}(x)+\beta _np_n(x)+\lambda _{n-1} p_{n-1}(x) \end{aligned}$$

(1)

with $p_{-1}(x)=0$, $p_0(x)=1$, $\lambda _{-1}=0$, $\lambda _n>0$ and $\beta _n \in {\mathbb {R}}$ for all $n \in {\mathbb {N}}_0$.

Conversely, if $(p_n)_{n\in {\mathbb {N}}_0}$ is defined by (1), there is a probability measure $\mu$ such that $(p_n)_{n\in {\mathbb {N}}_0}$ is the orthonormal polynomial sequence with respect to $\mu$, see e.g. [2].

In the case $(\lambda _n)_{n\in {\mathbb {N}}_0}$ and $(\beta _n)_{n\in {\mathbb {N}}_0}$ are bounded $\mathcal {S}$ is compact and vice versa. The boundedness implies also that the orthogonalization measure $\mu$ is uniquely determined. The smallest interval containing $\mathcal {S}$ is called the true interval of orthogonality, see e.g. [2].

Now, let $x_0\in {\mathbb {R}}{\setminus } \mathcal {N}$, where $\mathcal {N}=\{x\in {\mathbb {C}}: \exists n\in {\mathbb {N}} \text{ with } p_n(x)=0\}$ is the set of zeros of all orthonormal polynomials. It is well known that $\mathcal {N}\subset {\mathbb {R}}$, see e.g. [2]. The normalized polynomials

$$\begin{aligned} P_n(x)={p_n(x)\over p_n(x_0)} \end{aligned}$$

(2)

form an orthogonal polynomial sequence $(P_n)_{n\in {\mathbb {N}}_0}$ with respect to $\mu$, that is

$$\begin{aligned} \int P_n P_m d\mu ={\delta _{n,m} \over h_n} \end{aligned}$$

(3)

with $h_n>0$. We call $x_0$ a normalizing point. The corresponding three-term recurrence relation is

$$\begin{aligned} xP_n(x)=\gamma _n P_{n+1}(x)+\beta _nP_n(x)+\alpha _n P_{n-1}(x) \end{aligned}$$

(4)

with $P_{-1}(x)=0$, $P_0(x)=1$,

$$\begin{aligned} \gamma _n= & {} {p_{n+1}(x_0) \over p_n(x_0)} \lambda _n,\end{aligned}$$

(5)

$$\begin{aligned} \alpha _{n}= & {} {p_{n-1}(x_0) \over p_n(x_0)} \lambda _{n-1}, \quad \text{ and }\end{aligned}$$

(6)

$$\begin{aligned} \alpha _n+\beta _n+\gamma _n= & {} x_0 \quad \text {for all}\quad n \in {\mathbb {N}}_0. \end{aligned}$$

(7)

Note that (6) implies $\alpha _0=0$. It is also important to emphasize that (7) applies if and only if $x_0$ is a normalization point and that our investigations heavily depend on Eq. (7).

Moreover, $\gamma _n\alpha _{n+1}=\lambda _n^2>0$. One easily shows

$$\begin{aligned} h_{n+1} \alpha _{n+1}=h_n\gamma _n \end{aligned}$$

(8)

which implies

$$\begin{aligned} h_n={ \gamma _0 \ldots \gamma _{n-1} \over \alpha _1 \ldots \alpha _n}=p_n^2(x_0)\quad \text{ for } \text{ all } n \in {\mathbb {N}}_0. \end{aligned}$$

(9)

Note that (9) also applies in the case $n=0$, where the nominator and denominator are empty products that means they are set equal 1 by default. Therefore, (3) as well as (9) yields $h_0=1$.

The so called Christoffel–Darboux formula is given by

$$\begin{aligned} \sum _{k=0}^n P_k(x)P_k(y)h_k=\gamma _n h_n {P_{n+1}(x)P_n(y)-P_n(x)P_{n+1}(y) \over x-y}, \end{aligned}$$

(10)

see [2]. Hence,

$$\begin{aligned} \sum _{k=0}^n P_k(x)^2 h_k=\gamma _n h_n (P_{n+1}'(x)P_n(x)-P_n'(x)P_{n+1}(x)), \end{aligned}$$

(11)

and in particular setting $x=x_0$ we get

$$\begin{aligned} P_{n+1}'(x_0)-P_n'(x_0) = {H_n \over \gamma _nh_n} \end{aligned}$$

(12)

with

$$\begin{aligned} H_n=\sum _{k=0}^n h_k\quad \text{ for } \text{ all }\quad n \in {\mathbb {N}}_0. \end{aligned}$$

(13)

Definition 1.1

If $\{P_{n+1}'(x_0)-P_n'(x_0): n\in {\mathbb {N}}_0\}$ is bounded, then we call $x_0$ a normalizing point with bounded growth of derivatives.

Note that further on speaking about $x_0$ as a normalizing point of bounded growth of derivatives is the same as to speak about the boundedness of $\{{H_n \over \gamma _nh_n}: n\in {\mathbb {N}}_0\}$.

Subsequently we deal with the case $\mathcal {S}=\text{ supp }\mu$ is compact which is equivalent with $(\gamma _n\alpha _{n+1})_{n\in {\mathbb {N}}_0}$ and $(\beta _n)_{n\in {\mathbb {N}}_0}$ are bounded sequences. Then the true interval of orthogonality is $[\min \mathcal {S},\max \mathcal {S}]$.

Lemma 1.1

In the case $x_0\ge \max \mathcal {S}$ we have $\alpha _{n+1}, \gamma _n >0$ for all $n\in {\mathbb {N}}_0$ and in the case $x_0\le \min \mathcal {S}$ we have $\alpha _{n+1}, \gamma _n <0$ for all $n\in {\mathbb {N}}_0$.

Proof

Since $\mathcal {N}\subset (\min \mathcal {S}, \max \mathcal {S})$ and the leading coefficient of all orthonormal polynomials is positive we have in the case $x_0\ge \max \mathcal {S}$ that $p_n(x_0)>0$ for all $n\in {\mathbb {N}}_0$. Whereas in the case $x_0\le \min \mathcal {S}$ the sign of $p_n(x_0)$ is alternating. $\square$

On the set of complex-valued sequences there acts a linear operator $T: {\mathbb {C}}^{{\mathbb {N}}_0}\rightarrow {\mathbb {C}}^{{\mathbb {N}}_0}$ determined by the recurrence relation (4). More precisely, for $\xi \in {\mathbb {C}}^{{\mathbb {N}}_0}$ put

$$\begin{aligned} (T\xi )_n=T\xi _n=\gamma _n\xi _{n+1}+\beta _n\xi _n+\alpha _n\xi _{n-1} \quad \text{ for } \text{ all }\quad n\in {\mathbb {N}}_0, \end{aligned}$$

(14)

where $\xi _{-1}=0$. Written as tridiagonal matrix the operator T has the form

$$\begin{aligned} T=\left( \begin{array}{ccccccc} \beta _0&{}\gamma _0&{}0&{}0&{}0&{}\cdots &{}\\ \alpha _1&{}\beta _1&{}\gamma _1&{}0&{}0&{}\cdots &{}\\ 0&{}\alpha _2&{}\beta _2&{}\gamma _2&{}0&{}\cdots &{}\\ 0&{}0&{}\alpha _3&{}\beta _3&{}\gamma _3&{}\cdots &{}\\ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\ddots &{}\ddots \end{array}\right) . \end{aligned}$$

(15)

Note that in our investigations T acts on the different spaces ${\mathbb {C}}^{{\mathbb {N}}_0}$, $l^1(h)$ and $l^2(h)$ which is clear from the respective context. First let us study T as an operator on

$$\begin{aligned} l^1(h)=\left\{ \xi \in {\mathbb {C}}^{{\mathbb {N}}_0}:\sum _{n=0}^\infty \mid \xi _n \mid h_n<\infty \right\} \end{aligned}$$

(16)

with norm $\Vert \xi \Vert _1=\sum _{n=0}^\infty \mid \xi _n\mid h_n$ for all $\xi \in l^1(h)$.

Proposition 1.1

In the case $\mid \alpha _n\mid$, $\mid \beta _n\mid$ and $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$ the operator T: $l^1(h)\rightarrow l^1(h)$ is well defined and continuous. Especially we have

$$\begin{aligned} \sum _{n=0}^\infty T\xi _nh_n=x_0\sum _{n=0}^\infty \xi _nh_n \quad \text{ and }\quad \Vert T\xi \Vert _1 \le C\Vert \xi \Vert _1 \end{aligned}$$

(17)

for all $\xi \in l^1(h)$, where $C=\min (3B,\mid x_0\mid +2B)$.

Proof

Set $\gamma _{-1}=\xi _{-1}=h_{-1}=0$.

Applying (8) and the assumed absolute convergence of the series we obtain

$$\begin{aligned} \sum _{n=0}^\infty T\xi _nh_n&=\sum _{n=0}^\infty (\gamma _n\xi _{n+1}+\beta _n\xi _n+\alpha _n\xi _{n-1})h_n\\&=\sum _{n=0}^\infty \alpha _{n+1}\xi _{n+1}h_{n+1}+\beta _n\xi _nh_n+\gamma _{n-1}\xi _{n-1}h_{n-1}\\&=\sum _{n=0}^\infty (\alpha _{n}+\beta _n+\gamma _{n})\xi _{n}h_{n} =x_0\sum _{n=0}^\infty \xi _nh_n.\\ \sum _{n=0}^\infty \mid T\xi _n\mid h_n&\le \sum _{n=0}^\infty (\mid \gamma _n\mid \mid \xi _{n+1}\mid +\mid \beta _n\mid \mid \xi _n\mid +\mid \alpha _n\mid \mid \xi _{n-1}\mid )h_n\\&=\sum _{n=0}^\infty \mid \alpha _{n+1}\mid \mid \xi _{n+1}\mid h_{n+1}+\mid \beta _n\mid \mid \xi _n\mid h_n\\&\quad +\mid \gamma _{n-1}\mid \mid \xi _{n-1}\mid h_{n-1}\\&=\sum _{n=0}^\infty (\mid \alpha _{n}\mid +\mid \beta _n\mid +\mid \gamma _{n}\mid )\mid \xi _{n}\mid h_{n} \le 3B \sum _{n=0}^\infty \mid \xi _{n}\mid h_{n}. \end{aligned}$$

At least two of the coefficients in $\mid \alpha _{n}\mid +\mid \beta _n\mid +\mid \gamma _{n}\mid$ do have the same sign. For instance, if $\text {sign}\alpha _n=\text {sign}\beta _n$, then $\mid \alpha _{n}\mid +\mid \beta _n\mid +\mid \gamma _{n}\mid =\mid \alpha _{n}+\beta _n+\gamma _n-\gamma _n\mid +\mid \gamma _{n}\mid \le \mid x_0\mid +2\mid \gamma _n\mid \le \mid x_0\mid +2B$. Proceeding the same way with all the other possibilities one gets alternatively $\sum _{n=0}^\infty \mid T\xi _n\mid h_n \le (\mid x_0\mid +2B)\sum _{n=0}^\infty \mid \xi _{n}\mid h_{n}$, which completes the proof. $\square$

We focus on the weighted Hilbert space

$$\begin{aligned} l^2(h)=\left\{ \xi \in {\mathbb {C}}^{{\mathbb {N}}_0}:\sum _{n=0}^\infty \mid \xi _n\mid ^2h_n<\infty \right\} \end{aligned}$$

(18)

with scalar product $\langle \xi ,\upsilon \rangle =\sum _{n=0}^\infty \xi _n\overline{\upsilon _n}h_n$ and norm $\Vert \xi \Vert _2=\sqrt{\langle \xi ,\xi \rangle }$ for all $\xi ,\upsilon \in l^2(h)$.

Proposition 1.2

In the case $\mid \alpha _n\mid$, $\mid \beta _n\mid$ and $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$ the operator T: $l^2(h)\rightarrow l^2(h)$ is a well defined, self-adjoint and continuous operator with

$$\begin{aligned} \Vert T\xi \Vert _2 \le C\Vert \xi \Vert _2, \end{aligned}$$

(19)

where $C=\min (3B,\mid x_0\mid +2B)$.

Proof

Set $\gamma _{-1}=\xi _{-1}=h_{-1}=\upsilon _{-1}=0$.

Now let $\xi \in l^2(h)$. Since

$$\begin{aligned} \mid T\xi _n\mid&\le \sqrt{\mid \gamma _n\mid }\sqrt{\mid \gamma _n\mid }\mid \xi _{n+1}\mid +\sqrt{\mid \beta _n\mid }\sqrt{\mid \beta _n\mid }\mid \xi _n\mid \\&\quad +\sqrt{\mid \alpha _n\mid }\sqrt{\mid \alpha _n\mid }\mid \xi _{n-1}\mid \end{aligned}$$

the Cauchy–Schwarz inequality implies

$$\begin{aligned} \mid T\xi _n\mid ^2 \le (\mid \gamma _{n}\mid +\mid \beta _n\mid +\mid \alpha _{n}\mid )(\mid \gamma _{n}\mid \mid \xi _{n+1}\mid ^2+\mid \beta _n\mid \mid \xi _n\mid ^2+\mid \alpha _{n}\mid \mid \xi _{n-1}\mid ^2). \end{aligned}$$

Therefore, proceeding like in the proof of Proposition 1.1

$$\begin{aligned} \sum _{n=0}^\infty \mid T\xi _n\mid ^2 h_n&\le C \sum _{n=0}^\infty (\mid \gamma _n\mid \mid \xi _{n+1}\mid ^2+\mid \beta _n\mid \mid \xi _n\mid ^2+\mid \alpha _n\mid \mid \xi _{n-1}\mid ^2)h_n\\&= C\sum _{n=0}^\infty \mid \alpha _{n+1}\mid \mid \xi _{n+1}\mid ^2 h_{n+1}+\mid \beta _n\mid \mid \xi _n\mid ^2 h_n\\&\quad +\mid \gamma _{n-1}\mid \mid \xi _{n-1}\mid ^2 h_{n-1}\\&= C\sum _{n=0}^\infty \; (\mid \alpha _{n}\mid +\mid \beta _n\mid +\mid \gamma _{n}\mid )\mid \xi _{n}\mid ^2 h_{n} \le C^2 \sum _{n=0}^\infty \mid \xi _{n}\mid ^2 h_{n}, \end{aligned}$$

which implies $\Vert T\xi \Vert _2 \le C \Vert \xi \Vert _2,$ where $C=\min (3B,\mid x_0\mid +2B)$.

For arbitrary $\xi ,\upsilon \in l^2(h)$ one gets due to the absolute convergence

$$\begin{aligned} \langle T\xi ,\upsilon \rangle= & {} \sum _{n=0}^\infty (\gamma _n\xi _{n+1}+\beta _n\xi _n+\alpha _n \xi _{n-1})\overline{\upsilon _n}h_n\\= & {} \sum _{n=0}^\infty \xi _n(\gamma _{n-1}\overline{\upsilon _{n-1}}h_{n-1}+\beta _n\overline{\upsilon _n}h_n+\alpha _{n+1}\overline{\upsilon _{n+1}}h_{n+1})\\= & {} \sum _{n=0}^\infty \xi _n(\alpha _n\overline{\upsilon _{n-1}}h_n+\beta _n\overline{\upsilon _n}h_n+\gamma _n\overline{\upsilon _{n+1}}h_n) = \langle \xi ,T\upsilon \rangle . \end{aligned}$$

$\square$

Corollary 1.1

In the case $\mid \alpha _n\mid$, $\mid \beta _n\mid$ and $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$ the spectrum $\sigma (T)$ is a subset of $[-C,C]$, where $C=\min (3B,\mid x_0\mid +2B)$.

The numerical range of T is the set

$$\begin{aligned} W(T)=\left\{ \langle T\xi ,\xi \rangle :\xi \in l^2(h),\Vert \xi \Vert _2=1\right\} . \end{aligned}$$

(20)

Since T is self-adjoint we have

$$\begin{aligned} \{ m(T),M(T)\}\subseteq \sigma (T) \subseteq \,\text{ co }(\sigma (T))\subseteq \overline{W(T)}=[m(T),M(T)], \end{aligned}$$

(21)

where $\text{ co }(\sigma (T))$ is the convex hull of $\sigma (T)$, $m(T)=\inf W(T)$ and $M(T)=\sup W(T)$, see [5, Intro]. Moreover, $\Vert T\Vert = \max (\mid m(T)\mid ,\mid M(T)\mid )$.

Proposition 1.3

In the case $\mid \alpha _n\mid$, $\mid \beta _n\mid$ and $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$ one gets

$$\begin{aligned} \langle (x_0\text{ id } -T)\xi ,\xi \rangle =\sum _{n=0}^\infty \gamma _n\mid \xi _n-\xi _{n+1}\mid ^2h_n \quad \text{ for } \text{ all }\quad \xi \in l^2(h). \end{aligned}$$

(22)

Proof

Set $\gamma _{-1}=\xi _{-1}=h_{-1}=0$. Using (8) and the absolute convergence of the series one gets for an arbitrary $\xi \in l^2(h)$ that

$$\begin{aligned} \sum _{n=0}^\infty (x_0 \xi _n-T\xi _n)\overline{\xi _n}h_n&=\sum _{n=0}^\infty ((\alpha _n+\beta _n+\gamma _n)\xi _n-\alpha _n\xi _{n-1}\\&\quad -\beta _n\xi _n-\gamma _n\xi _{n+1})\overline{\xi _n}h_n\\&=\sum _{n=0}^\infty (\gamma _n\xi _n\overline{\xi _n} -\gamma _n\xi _{n+1}\overline{\xi _n})h_n\\&\quad +(\gamma _{n-1}\xi _n\overline{\xi _n}-\gamma _{n-1}\xi _{n-1}\overline{\xi _n})h_{n-1}\\&=\sum _{n=0}^\infty \gamma _n(\xi _n\overline{\xi _n} -\xi _{n+1}\overline{\xi _n}+\xi _{n+1}\overline{\xi _{n+1}}-\xi _{n}\overline{\xi _{n+1}})h_{n}\\&=\sum _{n=0}^\infty \gamma _n\mid \xi _n-\xi _{n+1}\mid ^2h_n. \end{aligned}$$

$\square$

Lemma 1.2

The following statements apply.

• (i) $\sum _{n=0}^\infty \gamma _n\mid \xi _n-\xi _{n+1}\mid ^2h_n \ge 0$ for all $\xi \in l^2(h)$ with $\Vert \xi \Vert _2=1$ if and only if $\gamma _n>0$ for all $n\in {\mathbb {N}}_0$.

• (ii) $\sum _{n=0}^\infty \gamma _n\mid \xi _n-\xi _{n+1}\mid ^2h_n \le 0$ for all $\xi \in l^2(h)$ with $\Vert \xi \Vert _2=1$ if and only if $\gamma _n<0$ for all $n\in {\mathbb {N}}_0$.

Proof

If $\gamma _n>0$ for all $n\in {\mathbb {N}}_0$ then $\sum _{n=0}^\infty \gamma _n\mid \xi _n-\xi _{n+1}\mid ^2h_n \ge 0$ for all $\xi \in l^2(h)$. In the case we have not $\gamma _n>0$ for all $n\in {\mathbb {N}}_0$ there is an index $m\in {\mathbb {N}}_0$ such that $\gamma _m<0$ and $\gamma _n>0$ for all $n\in \{0,\ldots ,m-1\}$. Define $\zeta \in l^2(h)$ by $\zeta _n = (\sum _{k=0}^m h_k)^{-1/2}$ for all $n\in \{0,\ldots ,m\}$ and $\zeta _n = 0$ for all $n\in \{m+1,m+2\ldots \}$. Then $\Vert \zeta \Vert _2=1$ and $\sum _{n=0}^\infty \gamma _n\mid \zeta _n-\zeta _{n+1}\mid ^2h_n = \gamma _m \mid \zeta _m\mid ^2 h_m <0$.

The second statement is shown quite analogue. $\square$

Corollary 1.2

If $\mid \alpha _n\mid$, $\mid \beta _n\mid$, $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$ and $C=\min (3B, \mid x_0\mid +2B)$, then the following statements apply.

• (i) If $\gamma _n>0$ for all $n\in {\mathbb {N}}_0$, then $\overline{W(T)}\subseteq [-C,x_0]$. In particular, $\sigma (T)\subseteq [-C,x_0]$.

• (ii) If $\gamma _n<0$ for all $n\in {\mathbb {N}}_0$, then $\overline{W(T)}\subseteq [x_0,C]$. In particular, $\sigma (T)\subseteq [x_0,C]$.

• (iii) If there exist $k,l\in {\mathbb {N}}_0$ with $\gamma _k\gamma _l<0$, then $x_0 \in (\min \mathcal {S}, \max \mathcal {S})$.

Note that in the following $L^2({\mathbb {R}},\mu )$ is as usual a set of equivalence classes and a function used in this context represents an equivalence class. This is also expressed by using the formulation ’for $\mu$-almost all $x\in {\mathbb {R}}$’.

Define $\epsilon ^{(k)} \in l^2(h)$ by

$$\begin{aligned} \epsilon ^{(k)}_n=\frac{\delta _{k,n}}{h_k}\quad \text{ for } \text{ all }\quad n,k\in {\mathbb {N}}_0. \end{aligned}$$

(23)

Then obviously

$$\begin{aligned} \Vert \epsilon ^{(k)}\Vert ^2_2=\frac{1}{h_k}=\int P_k ^2d\mu \quad \text{ for } \text{ all }\quad k\in {\mathbb {N}}_0. \end{aligned}$$

(24)

Extending the map $\epsilon ^{(k)} \mapsto P_k$ linearly to the linear span of $\{\epsilon ^{(k)}:k\in {\mathbb {N}}_0\}$ and finally to the closure of the linear span we get the so-called Plancherel isomorphism

$$\begin{aligned} \mathcal {P}: l^2(h)\rightarrow L^2({\mathbb {R}},\mu ), \end{aligned}$$

which is an isometric isomorphism from $l^2(h)$ onto $L^2({\mathbb {R}},\mu )$. It is completely determined by

$$\begin{aligned} \mathcal {P}(\epsilon ^{(k)})=P_k\quad \text{ for } \text{ all }\quad k\in {\mathbb {N}}_0. \end{aligned}$$

Note that

$$\begin{aligned} T\epsilon ^{(k)}=\alpha _k\epsilon ^{(k-1)}+\beta _k\epsilon ^{(k)}+\gamma _k\epsilon ^{(k+1)}\quad \text{ for } \text{ all }\quad k\in {\mathbb {N}}_0, \end{aligned}$$

(25)

where $\epsilon ^{(-1)}_n=0$ for all $n\in {\mathbb {N}}_0$. Now we define an operator M on $L^2({\mathbb {R}},\mu )$ by

$$\begin{aligned} M(f)=\mathcal {P}\circ T\circ \mathcal {P}^{-1}(f)\quad \text{ for } \text{ all }\quad f \in L^2({\mathbb {R}},\mu ), \end{aligned}$$

(26)

where $\mathcal {P}^{-1}$ denotes the inverse operator of $\mathcal {P}$. Then $M\in B(L^2({\mathbb {R}},\mu ))$ with $\Vert M\Vert \,\le \min (3B,\mid x_0\mid +2B)$. Taking into account the three-term recurrence relation (4) we deduce that

$$\begin{aligned} M(P_k)(x)=\mathcal {P}(T\epsilon ^{(k)})(x)=\mathcal {P}(\alpha _k\epsilon ^{(k-1)}+\beta _k\epsilon ^{(k)}+\gamma _k\epsilon ^{(k+1)})(x)=xP_k(x) \end{aligned}$$

(27)

for $\mu$-almost all $x\in {\mathbb {R}}$ and for all $k\in {\mathbb {N}}_0$. If g is a function in the linear span of $\{P_k: k\in {\mathbb {N}}_0\}$, then the linearity of M yields

$$\begin{aligned} M(g)(x)=x g(x) \quad \text{ for }\quad \mu \text {-almost all}\; x\in {\mathbb {R}}. \end{aligned}$$

(28)

Since M is bounded and the closure of the linear span of $\{P_k: k\in {\mathbb {N}}_0\}$ is $L^2({\mathbb {R}},\mu )$ we get by standard arguments that

$$\begin{aligned} M(f)(x)=xf(x)\ \text{ for }\ \mu \text {-almost all}\; x \in {\mathbb {R}}\;\text{ and } \text{ for } \text{ all }\; f\in L^2({\mathbb {R}},\mu ).\qquad \end{aligned}$$

(29)

By [4, Definition 2.61 and Corollary 4.24] the spectrum $\sigma (M)$ is exactly the essential range

$$\begin{aligned} \mathcal {R}=\left\{ \lambda \in {\mathbb {R}}: \mu (\{x \in {\mathbb {R}}:\mid x-\lambda \mid <\epsilon \})>0 \text{ for } \text{ all } \epsilon >0\right\} , \end{aligned}$$

(30)

Obviously $\mathcal {R}=\text{ supp }\mu$ and $\sigma (M)=\sigma (T)$. Hence, we can add to Corollary 1.2 the following result.

Corollary 1.3

For orthogonal polynomials $(P_n)_{n\in {\mathbb {N}}_0}$ which are defined by (4) with $\mid \alpha _n\mid$, $\mid \beta _n\mid$ and $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$ we have

$$\begin{aligned} \mathcal {S}=\text{ supp }\mu =\sigma (T). \end{aligned}$$

(31)

2 A characterization of $x_0 \notin \mathcal {S}$

In the whole section we assume that $\mid \alpha _n\mid$, $\mid \beta _n\mid$ and $\mid \gamma _n\mid \le B$ for all $n\in {\mathbb {N}}_0$.

The main result of this paper will be a necessary and sufficient condition for $x_0 \in \mathcal {S}$. Moreover, in the case of $x_0 \notin \mathcal {S}$ we will present an explicit form of the inverse $(x_0\text{ id }-T)^{-1}$, which is based on a weighted Cesàro operator $C\in B(l^2(h))$.

Define $C\eta =((C\eta )_n)_{n\in {\mathbb {N}}_0}=(C\eta _n)_{n\in {\mathbb {N}}_0}$ by

$$\begin{aligned} C\eta _n=\frac{1}{H_n}\sum _{k=0}^n \eta _kh_k \quad \text{ for } \text{ all }\quad \eta \in l^2(h). \end{aligned}$$

(32)

Then C is a bounded linear operator on $l^2(h)$ with $\Vert C\Vert \le 2$, see [3, Theorem A]. It is straightforward to show that the adjoint operator $C^* \in B(l^2(h))$ is defined by

$$\begin{aligned} C^*\eta _n=\sum _{k=n}^\infty \eta _k\frac{h_k}{H_k}\quad \text{ for } \text{ all }\quad \eta \in l^2(h). \end{aligned}$$

(33)

Theorem 2.1

If $x_0 \notin \mathcal {S}= \sigma (T)$, then $x_0$ is a normalizing point with bounded growth of derivatives.

Proof

Given ${n \in {\mathbb {N}}}_0$ denote by $\chi ^{(n)}$ the sequence with $\chi ^{(n)}_k=1$ for $k\in \{0,\ldots ,n\}$ and $\chi ^{(n)}_k=0$ for $k\in \{n+1,n+2,\ldots \}$. An easy computation shows that

$$\begin{aligned} (x_0\text{ id }-T)\chi ^{(n)}_k= & {} 0\quad \text{ for } \text{ all }\quad k \in {\mathbb {N}}_0 \backslash \{n,n+1\},\\ (x_0\text{ id }-T)\chi ^{(n)}_n= & {} \gamma _n,\quad \text{ and }\\ (x_0\text{ id }-T)\chi ^{(n)}_{n+1}= & {} -\alpha _{n+1}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert (x_0\text{ id }-T)\chi ^{(n)}\Vert _2^2&=\gamma _n^2h_n+\alpha _{n+1}^2h_{n+1}=\gamma _n(\gamma _n+\alpha _{n+1})h_n\\&=\mid \gamma _n\mid \mid \gamma _n+\alpha _{n+1}\mid h_n. \end{aligned}$$

Since $x_0\notin \sigma (T)$, there exists $A = (x_0\text{ id }-T)^{-1}\in B(l^2(h))$. Then

$$\begin{aligned} \Vert A \circ (x_0\text{ id }-T)\chi ^{(n)}\Vert _2^2&= \Vert \chi ^{(n)}\Vert _2^2=\sum _{k=0}^nh_k=H_n,\quad \text{ and }\\ \Vert A \circ (x_0\text{ id }-T)\chi ^{(n)}\Vert _2^2&\le \Vert A\Vert ^2\Vert (x_0\text{ id }-T)\chi ^{(n)})\Vert _2^2\\&=\Vert A\Vert ^2\mid \gamma _n\mid h_n\mid \gamma _n+\alpha _{n+1}\mid \\&\le 2B\Vert A\Vert ^2\mid \gamma _n\mid h_n, \end{aligned}$$

which implies

$$\begin{aligned} H_n\le 2B\Vert A\Vert ^2\mid \gamma _n\mid h_n\quad \text{ for } \text{ all }\quad n\in {\mathbb {N}}_0. \end{aligned}$$

Therefore, $\left\{ \frac{H_n}{\gamma _nh_n}: n\in {\mathbb {N}}_0\right\}$ is bounded. $\square$

In order to prove the converse implication we start with determining a sequence $\omega =(\omega _n)_{n\in {\mathbb {N}}_0}$ such that $(x_0\text{ id }-T)(\omega )=\epsilon ^{(0)}$. Note that in the following lemma the operator T acts on ${\mathbb {C}}^{{\mathbb {N}}_0}$.

Lemma 2.1

A sequence $\omega =(\omega _n)_{n\in {\mathbb {N}}_0} \in {\mathbb {C}}^{{\mathbb {N}}_0}$ satisfies $(x_0 \text{ id }-T)(\omega )=\epsilon ^{(0)}$ if and only if

$$\begin{aligned} \omega _{n+1}=\omega _0-\sum _{k=0}^n \frac{1}{\gamma _kh_k} \quad \text{ for } \text{ all }\quad n\in {\mathbb {N}}_0. \end{aligned}$$

(34)

Proof

We have $((x_0 \text{ id }-T)\omega )_0=1/h_0=1$ if and only if $\omega _0-\omega _1=\frac{1}{\gamma _0}$. For $n\ge 1$ we see that $((x_0\text{ id }-T)\omega )_n=\omega _n-(\gamma _n\omega _{n+1}+\beta _n\omega _n+\alpha _n\omega _{n-1})=0$ if and only if $\gamma _n(\omega _{n+1}-\omega _n)=\alpha _n(\omega _n-\omega _{n-1})$. Now, by iteration we get

$$\begin{aligned} \omega _{n+1}-\omega _n=\frac{\alpha _n}{\gamma _n}(\omega _n-\omega _{n-1})=\frac{\alpha _n\alpha _{n-1}\cdots \alpha _1}{\gamma _n\gamma _{n-1}\cdots \gamma _1}\frac{-1}{\gamma _0}=\frac{-1}{\gamma _nh_n}. \end{aligned}$$

$\square$

Next we investigate under which assumptions a sequence $\omega =(\omega _n)_{n\in {\mathbb {N}}_0}$ of Lemma 2.1 is a member of $l^2(h)$.

Lemma 2.2

If $x_0$ is a normalizing point with bounded growth of derivatives, then

$$\begin{aligned} \sum _{k=0}^\infty \frac{1}{\mid \gamma _k\mid h_k}<\infty , \end{aligned}$$

(35)

and consequently the series $\sum _{k=0}^\infty \frac{1}{\gamma _kh_k}$ is convergent.

Proof

Due to the assumption there exists a $D>0$ with

$$\begin{aligned} \sum _{k=0}^n\left( \frac{1}{\gamma _kh_k}\right) ^2h_k\le D \sum _{k=0}^n\left( \frac{1}{H_k}\right) ^2h_k \quad \text{ for } \text{ all }\quad n \in {\mathbb {N}}_0. \end{aligned}$$

Since $C\epsilon ^{(0)}=\left( \frac{1}{H_n}\right) _{n\in {\mathbb {N}}_0} \in l^2(h)$, we have $\left( \frac{1}{\gamma _nh_n}\right) _{n\in {\mathbb {N}}_0}\in l^2(h)$, that is $\sum _{k=0}^\infty \frac{1}{\gamma _k^2h_k}<\infty .$ Finally, $\gamma _k^2 \le B\mid \gamma _k\mid$ yields $\sum _{k=0}^\infty \frac{1}{\mid \gamma _k\mid h_k}<\infty ,$ which implies the series $\sum _{k=0}^\infty \frac{1}{\gamma _k h_k}$ is convergent. $\square$

Now with respect to Lemma 2.1, if the series $\sum _{k=0}^\infty \frac{1}{\gamma _kh_k}$ is convergent, then the sequence $\omega =(\omega _n)_{n\in {\mathbb {N}}_0}$ is defined by

$$\begin{aligned} \omega _n=\sum _{k=n}^\infty \frac{1}{\gamma _kh_k}\quad \text{ for } \text{ all }\quad {n \in {\mathbb {N}}}_0. \end{aligned}$$

(36)

In order to prove that $\omega \in l^2(h)$ whenever $\left\{ \frac{H_n}{\gamma _nh_n}:n \in {\mathbb {N}}_0\right\}$ is bounded, we use the adjoint weighted Cesàro operator $C^*\in B(l^2(h))$. Define a sequence $\eta =(\eta _n)_{n\in {\mathbb {N}}_0}$ by

$$\begin{aligned} \eta _n=\frac{H_n}{\gamma _nh_n^2}\quad \text{ for } \text{ all }\quad n \in {\mathbb {N}}_0. \end{aligned}$$

(37)

Lemma 2.3

If $x_0$ is a normalizing point with bounded growth of derivatives, then $\eta \in l^2(h)$.

Proof

We have $\mid \eta _n\mid \le D \frac{1}{h_n}$ for all $n \in {\mathbb {N}}_0$. Hence, we have to show that $\left( \frac{1}{h_n}\right) _{n\in {\mathbb {N}}_0} \in l^2(h)$. According to Lemma 2.2 it follows

$$\begin{aligned} \sum _{k=0}^\infty \frac{1}{h_k^2} h_k\le B \sum _{k=0}^\infty \frac{1}{\mid \gamma _k\mid h_k}<\infty . \end{aligned}$$

$\square$

Since $C^*\eta =\omega ,$ Lemmas 2.3 and 2.1 yield the following proposition.

Proposition 2.1

If $x_0$ is a normalizing point with bounded growth of derivatives, then $\omega \in l^2(h)$ (defined by (36)) satisfies $(x_0\text{ id }-T)\omega =\epsilon ^{(0)}$.

Assuming that $x_0$ is a normalizing point with bounded growth of derivatives our next goal is to find sequences $\omega ^{(m)}\in l^2(h)$ with $(x_0\text{ id }-T)\omega ^{(m)}=\epsilon ^{(m)}$ for all $m \in {\mathbb {N}}.$

To that end, we introduce a sequence of operators $S_m \in B(l^2(h))$ by setting

$$\begin{aligned} S_{m+1}=\frac{1}{\gamma _m}\left( T\circ S_m-\beta _mS_m-\alpha _mS_{m-1}\right) \quad \text{ for } \text{ all }\quad m \in {\mathbb {N}}_0, \end{aligned}$$

(38)

where $S_{-1} = 0$ and $S_0 = \text{ id }$.

Proposition 2.2

The following two statements apply.

• (i)
  $$\begin{aligned} S_m\epsilon ^{(0)}=\epsilon ^{(m)}\quad \text{ for } \text{ all }\quad m \in {\mathbb {N}}_0. \end{aligned}$$

  (39)
• (ii)
  $$\begin{aligned} (S_m\omega )_k=\left\{ \begin{array}{lll} \omega _m &{}\quad \textrm{if}&{}\quad k=0,\ldots ,m, \\ \omega _k &{}\quad \textrm{if} &{}\quad k=m+1,m+2,\ldots \end{array}\right. \quad \text{ for } \text{ all }\; m \in {\mathbb {N}}_0. \end{aligned}$$

  (40)

Proof

In any case the proof is done by induction.

(i): By trivial means we have $S_0\epsilon ^{(0)}=\epsilon ^{(0)}$. Since $S_1=\frac{1}{\gamma _0}\left( T-\beta _0\text{ id }\right)$ we have $(S_1\epsilon ^{(0)})_0= \frac{1}{\gamma _0}(\beta _0-\beta _0)=0$, $(S_1\epsilon ^{(0)})_1= \frac{\alpha _1}{\gamma _0}\frac{1}{h_0}=\frac{1}{h_1}$ and $(S_1\epsilon ^{(0)})_k=0$ for all $k\ge 2$. Therefore, $S_1\epsilon ^{(0)}=\epsilon ^{(1)}$.

Assume that $S_m\epsilon ^{(0)}=\epsilon ^{(m)}$ and $S_{m-1}\epsilon ^{(0)}=\epsilon ^{(m-1)}$ for $m\in {\mathbb {N}}_0$ is already shown. Then

$$\begin{aligned} S_{m+1}\epsilon ^{(0)}= & {} \frac{1}{\gamma _m}\left( T\circ S_m-\beta _mS_m-\alpha _mS_{m-1}\right) \epsilon ^{(0)}\\= & {} \frac{1}{\gamma _m}\left( T\epsilon ^{(m)}-\beta _m\epsilon ^{(m)}-\alpha _m\epsilon ^{(m-1)}\right) \\= & {} \frac{1}{\gamma _m}\left( \alpha _m\epsilon ^{(m-1)} +\beta _m\epsilon ^{(m)}+\gamma _m\epsilon ^{(m+1)} -\beta _m\epsilon ^{(m)}-\alpha _m\epsilon ^{(m-1)}\right) \\= & {} \frac{1}{\gamma _m}\left( \gamma _m\epsilon ^{(m+1)}\right) =\epsilon ^{(m+1)}. \end{aligned}$$

(ii): By trivial means we have $(S_0\omega )_k=\omega _k$ for all $k \in {\mathbb {N}}_0$. Moreover, since $S_1={1\over \gamma _0}(T-\beta _0id)$, $T\omega =x_0id-\epsilon ^{(0)}$ and $x_0=\beta _0+\gamma _0$ we get

$$\begin{aligned} S_1\omega = \frac{1}{\gamma _0}T\omega -\frac{\beta _0}{\gamma _0}\omega =\frac{1}{\gamma _0}((\beta _0 + \gamma _0)\omega -\epsilon ^{(0)})-\frac{\beta _0}{\gamma _0}\omega = \omega -\frac{\epsilon ^{(0)}}{\gamma _0}. \end{aligned}$$

Hence, $(S_1\omega )_0=\omega _1$, $(S_1\omega )_1=\omega _1$, and $(S_1\omega )_k=\omega _k$ for all $k\ge 2$.

Assume again that the statement is already shown for $m \in {\mathbb {N}}$ and $m-1$. Then for $k=0,\ldots ,m-1$ we have

$$\begin{aligned} (S_{m+1}\omega )_k= & {} \frac{1}{\gamma _m}((T\circ S_m \omega )_k-\beta _m\omega _m-\alpha _m \omega _{m-1})\\= & {} \frac{1}{\gamma _m}(x_0\omega _m-\beta _m\omega _m-\alpha _m \omega _{m-1}) =\omega _m+\frac{\alpha _m}{\gamma _m}(\omega _m-\omega _{m-1})\\= & {} \omega _m-\frac{\alpha _m}{\gamma _m}\frac{1}{\gamma _{m-1}h_{m-1}} =\omega _m-\frac{1}{\gamma _m h_m}=\omega _{m+1}. \end{aligned}$$

For $k=m$ it follows

$$\begin{aligned} (S_{m+1}\omega )_k=\frac{1}{\gamma _m}(\alpha _m\omega _m+\beta _m\omega _m+\gamma _m\omega _{m+1}-\beta _m\omega _m-\alpha _m \omega _m)=\omega _{m+1}. \end{aligned}$$

Finally, for $k=m+1,m+2,\ldots$ we have

$$\begin{aligned} (S_{m+1}\omega )_k&=\frac{1}{\gamma _m}\left( \alpha _k\omega _{k-1}+\beta _k \omega _k+\gamma _k \omega _{k+1}-\beta _m\omega _k-\alpha _m\omega _k\right) \\&=\frac{1}{\gamma _m}\left( \gamma _k\left( \omega _k-\frac{1}{\gamma _kh_k}\right) +\beta _k\omega _k\right. \nonumber \\&\left. \quad +\alpha _k \left( \omega _k+\frac{1}{\gamma _{k-1}h_{k-1}}\right) -\beta _m\omega _k-\alpha _m\omega _k\right) \\&=\frac{1}{\gamma _m}\left( x_0 \omega _k-\frac{1}{h_k}+\frac{1}{h_k}-\beta _m\omega _k-\alpha _m\omega _k\right) =\omega _k. \end{aligned}$$

$\square$

Now our goal is met by setting $\omega ^{(m)} = S_m\omega$ for all $m \in {\mathbb {N}}_0$.

Proposition 2.3

If $x_0$ is a normalizing point with bounded growth of derivatives, then $S_m\omega \in l^2(h)$ satisfies $(x_0\text{ id }-T)S_m\omega =\epsilon ^{(m)}$ for all $m \in {\mathbb {N}}_0.$

Proof

Obviously $S_m$ commutes with $x_0\text{ id }-T$. Hence,

$$\begin{aligned} (x_0\text{ id }-T)S_m\omega =S_m(x_0\text{ id }-T)\omega =S_m\epsilon ^{(0)}=\epsilon ^{(m)}\quad \text{ for } \text{ all }\quad m \in {\mathbb {N}}_0. \end{aligned}$$

$\square$

For $m \in {\mathbb {N}}_0$ define the sequence $\eta ^{(m)}$ by

$$\begin{aligned} \eta ^{(m)}_k=\left\{ \begin{array}{lll} 0 &{}\quad \textrm{if}&{}\quad k=0,\ldots ,m-1, \\ \frac{H_k}{\gamma _k h_k^2}&{}\quad \textrm{if} &{}\quad k=m,m+1,\ldots \end{array}\right. . \end{aligned}$$

(41)

Note that $\eta ^{(0)}=\eta$. If $\left\{ \frac{H_n}{\gamma _nh_n}:n \in {\mathbb {N}}_0\right\}$ is bounded, then according to Lemma 2.3 we know that $\eta ^{(m)}\in l^2(h)$ for all $m \in {\mathbb {N}}_0$. Moreover,

$$\begin{aligned} C^*\eta ^{(m)}_n=\left\{ \begin{array}{lll} \sum _{k=m}^\infty \frac{\eta ^{(m)}_kh_k}{H_k} =\sum _{k=m}^\infty \frac{1}{\gamma _kh_k}=\omega _m &{}\quad \textrm{if}&{}\quad n\le m, \\ \sum _{k=n}^\infty \frac{1}{\gamma _kh_k}=\omega _n &{}\quad \textrm{if}&{}\quad n>m \end{array}\right. . \end{aligned}$$

(42)

By Proposition 2.2(ii) we have $C^*\eta ^{(m)} =S_m\omega$ for all $m \in {\mathbb {N}}_0$. Now we can combine the results above to determine the inverse operator of $x_0\text{ id }-T$. Define a sequence $\varphi =(\varphi _n)_{n \in {\mathbb {N}}_0}$ by

$$\begin{aligned} \varphi _n=\frac{H_n^2}{\gamma _nh_n^2}\quad \text{ for } \text{ all }\quad n \in {\mathbb {N}}_0. \end{aligned}$$

(43)

Note that $\frac{H_n}{\mid \gamma _n\mid h_n} \le D$ for all $n \in {\mathbb {N}}_0$ implies $\mid \varphi _n\mid \le D^2B$ for all $n \in {\mathbb {N}}_0$.

The multiplication with $\varphi \in l^\infty$ defines a bounded operator $M_\varphi$ on $l^2(h)$, where $M_\varphi (\xi )_n=\varphi _n\xi _n$ for all $\xi \in l^2(h)$, $n \in {\mathbb {N}}_0$.

Theorem 2.2

If $x_0$ is a normalizing point with bounded growth of derivatives, then $C^*\circ M_\varphi \circ C$ is the inverse of the operator $x_0\text{ id }-T$, where $\varphi$ is the sequence in (43).

Proof

Let $m \in {\mathbb {N}}_0$. We know that $C\epsilon ^{(m)}_k=0$ for all $k=0,\ldots ,m-1$ and $C\epsilon ^{(m)}_k=\frac{1}{H_k}$ for all $k=m,m+1,\ldots$. Hence, $M_\varphi \circ C\epsilon ^{(m)}=\eta ^{(m)}$ and $C^* \circ M_\varphi \circ C\epsilon ^{(m)}=S_m\omega$. In particular

$$\begin{aligned} (x_0\text{ id }-T)\circ (C^*\circ M_\varphi \circ C)\epsilon ^{(m)}=\epsilon ^{(m)}\quad \text{ for } \text{ all }\quad m \in {\mathbb {N}}_0. \end{aligned}$$

Furthermore, we obtain

$$\begin{aligned}&C^*\circ M_\varphi \circ C\circ (x_0\text{ id }-T)\epsilon ^{(m)}\\&\quad = (C^* \circ M_\varphi \circ C)(x_0 \epsilon ^{(m)}-(\gamma _m \epsilon ^{(m+1)}+\beta _m\epsilon ^{(m)}+\alpha _m\epsilon ^{(m-1)}))\\&\quad = x_0S_m\omega -(\gamma _m S_{m+1}\omega +\beta _m S_m\omega +\alpha _m S_{m-1}\omega )\\&\quad = x_0S_m\omega -T \circ S_m\omega =(x_0\text{ id }-T)S_m\omega =\epsilon ^{(m)}\quad \text{ for } \text{ all }\quad m \in {\mathbb {N}}_0. \end{aligned}$$

Therefore,

$$\begin{aligned} C^*\circ M_\varphi \circ C\circ (x_0\text{ id }-T)= \text{ id } =(x_0\text{ id }-T)\circ C^* \circ M_\varphi \circ C, \end{aligned}$$

i.e. $(x_0 \text{ id }-T)^{-1}=C^*\circ M_\varphi \circ C.$ $\square$

Summing up the results we gain the following theorem.

Theorem 2.3

$x_0 \notin \text{ supp }\mu = \sigma (T)$ if and only if $x_0$ is a normalizing point with bounded growth of derivatives.

Finally, we want to show the relationship of the results here with [1, Theorem 2.3]. For this we use the terms $\mathcal {A}$, A and $\mathcal {D}(A)$ with the same meaning as in [1]. Let

$$\begin{aligned} \mathcal {A}=\left( \begin{array}{ccccccc} \beta _0&{}\lambda _0&{}0&{}0&{}0&{}\cdots &{}\\ \lambda _0&{}\beta _1&{}\lambda _1&{}0&{}0&{}\cdots &{}\\ 0&{}\lambda _1&{}\beta _2&{}\lambda _2&{}0&{}\cdots &{}\\ 0&{}0&{}\lambda _2&{}\beta _3&{}\lambda _3&{}\cdots &{}\\ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\ddots &{}\ddots \end{array}\right) , \end{aligned}$$

(44)

where $(\beta _n)_{n\in {\mathbb {N}}_0}$ and $(\lambda _n)_{n\in {\mathbb {N}}_0}$ are the coefficients of (1). Then $\mathcal {A}$ can be regarded as a linear operator $\mathcal {A}: {\mathbb {C}}^{{\mathbb {N}}_0}\rightarrow {\mathbb {C}}^{{\mathbb {N}}_0}$, $\xi \mapsto \mathcal {A}\xi =(\mathcal {A}\xi _n)_{n\in {\mathbb {N}}_0}$, where

$$\begin{aligned} \mathcal {A}\xi _n=\lambda _{n-1}\xi _{n-1}+\beta _n\xi _n+\lambda _n\xi _{n+1}\quad \text{ for } \text{ all }\quad n\in {\mathbb {N}}_0. \end{aligned}$$

(45)

Note that $\lambda _{-1}=0$ and $\xi _{-1}$ can be chosen arbitrary.

Moreover, let $l^2=l^2(h)$ with $h_n=1$ for all $n\in {\mathbb {N}}_0$, and

$$\begin{aligned} c_{00}=\{\xi \in {\mathbb {C}}^{{\mathbb {N}}_0}: \#\{n\in {\mathbb {N}}_0: \xi _n\ne 0\}<\infty \}. \end{aligned}$$

(46)

Of course, $(c_{00},\Vert \Vert _2)$ is a subspace of the Hilbert space $(l^2,\Vert \Vert _2)$. As mentioned in [1] the linear operator

$$\begin{aligned} \mathcal {A}: c_{00}\rightarrow l^2, \quad \xi \mapsto \mathcal {A}\xi \end{aligned}$$

(47)

is closable and its closure is given by

$$\begin{aligned}{} & {} A: \mathcal {D}(A)\rightarrow l^2, \quad \xi \mapsto A \xi ,\quad \text{ where }\end{aligned}$$

(48)

$$\begin{aligned}{} & {} \mathcal {D}(A)=\left\{ \xi \in l^2: \exists (\xi ^{(k)})_{k\in {\mathbb {N}}_0} \subset c_{00}: \lim _{k\rightarrow \infty }\xi ^{(k)}=\xi \wedge \lim _{k\rightarrow \infty }\mathcal {A}\xi ^{(k)}\,\text{ exists }\right\} \nonumber \\ \end{aligned}$$

(49)

$$\begin{aligned}{} & {} \text{ and }\quad A \xi =\lim _{k\rightarrow \infty }\mathcal {A}\xi ^{(k)}. \end{aligned}$$

(50)

According to [1, Theorem 2.3.], the following statements hold true.

If $x_0\in \Omega (A)={\mathbb {R}}{\setminus }\sigma (A)$ then

$$\begin{aligned} \sup _{n\ge 0}{\sum _{k=0}^n p_k^2(x_0) \over \lambda _n^2 (p_n^2(x_0)+p_{n+1}^2(x_0))} < \infty . \end{aligned}$$

(51)

Provided A is bounded, also the converse is true.

Note that the assumptions made at the beginning of Sect. 2 imply the boundedness of $(\lambda _n)_{n\in {\mathbb {N}}_0}$ and $(\beta _n)_{n\in {\mathbb {N}}_0}$. One can show that the boundedness of $(\lambda _n)_{n\in {\mathbb {N}}_0}$ and $(\beta _n)_{n\in {\mathbb {N}}_0}$ imply that A is a bounded operator.

The relationship with our result can be derived from

$$\begin{aligned}{} & {} (\alpha _{n+1} +\gamma _n) {\sum _{k=0}^n p_k^2(x_0) \over \lambda _n^2 (p_n^2(x_0)+p_{n+1}^2(x_0))}\nonumber \\{} & {} \quad = \alpha _{n+1}{ p_n^2 (x_0)+p_{n+1}^2(x_0)\over p_n^2 (x_0)} {\sum _{k=0}^n p_k^2(x_0) \over \lambda _n^2 (p_n^2(x_0)+p_{n+1}^2(x_0))}\nonumber \\{} & {} \quad ={\sum _{k=0}^n p_k^2(x_0) \over {\lambda _n^2 \over \alpha _{n+1}}p_n^2(x_0)}= {\sum _{k=0}^n p_k^2(x_0) \over \gamma _n p_n^2(x_0) }={H_n \over \gamma _n h_n } \end{aligned}$$

(52)

and $\text {sign}\, \alpha _{n+1}= \text {sign}\,\gamma _n$.

In [1] there is no formula of the inverse as in Theorem 2.2 but there is no restriction $x_0\in {\mathbb {R}}{\setminus } \mathcal {N}$.