Bessel Sequences, Riesz-Like Bases and Operators in Triplets of Hilbert Spaces

Abstract

Riesz-like bases for a triplet of Hilbert spaces are investigated, in connection with an analogous study for more general rigged Hilbert spaces performed in a previous paper. It is shown, in particular, that every $\omega$-independent, complete (total) Bessel sequence is a (strict) Riesz-like basis in a convenient triplet of Hilbert spaces. An application to non self-adjoint Schrödinger-type operators is considered. Moreover, some of the simplest operators we can define by them and their dual bases are studied.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

1 Introduction

A Riesz basis of a Hilbert space $\mathcal H$ is a sequence $\{\xi _n\}$ of elements of $\mathcal H$ that are transformed into an orthonormal basis of $\mathcal H$ by some bounded operator with bounded inverse. Riesz bases can also be viewed as frames [1–3]; i.e., there exist positive numbers c, C such that

$$\begin{aligned} c\Vert \xi \Vert ^2 \le \sum _{n=1}^\infty |\left\langle {\xi }\left| {\xi _n}\right. \right\rangle |^2 \le C \Vert \xi \Vert ^2, \quad \forall \xi \in \mathcal H. \end{aligned}$$

(1)

What distinguishes a frame from a Riesz basis is its minimality, i.e., once one of its elements is dropped out it ceases to be a frame.

In [4] it was studied a possible extension of the notion of Riesz basis of a Hilbert space to rigged Hilbert spaces, by introducing what we called Riesz-like bases, the main difference relying on the fact that the operator transforming $\{\xi _n\}$ into an orthonormal basis need not to be bounded. A motivation for this generalization stems from the following considerations.

Let us assume that $\{\xi _n\}$ is a sequence of vectors of $\mathcal H$ for which there exists an unbounded closed linear operator T, with dense domain D(T) and bounded inverse $T^{-1}$, such that $\xi _n \in D(T)$, for every $n \in \mathbb {N}$ and the sequence $\{T\xi _n\}$ is an orthonormal basis for $\mathcal H$. In this case we can endow $D(T)\subset \mathcal H$ with a new inner product

$$ \left\langle {\xi }\left| {\eta }\right. \right\rangle _{+1} :=\left\langle {T\xi }\left| {T\eta }\right. \right\rangle , \quad \xi , \eta \in D(T) $$

which makes it into a Hilbert space denoted by $\mathcal H_{+1}$. Since $T^{-1} \in {\mathcal B}(\mathcal H)$, the C*-algebra of linear bounded operators in $\mathcal H$, it follows that $c\Vert \xi \Vert \le \Vert \xi \Vert _{+1}$, for some $c>0$ and for every $\xi \in \mathcal H$, then $\mathcal H$ can be identified with a subspace of the conjugate dual $\mathcal H_{+1}^\times$ of $\mathcal H_{+1}$. This space, in turn, is isomorphic to the completion of $\mathcal H$ with respect to the norm induced by the inner product $\left\langle {\cdot }\left| {\cdot }\right. \right\rangle _{-1}$ defined by

$$ \left\langle {\xi }\left| {\eta }\right. \right\rangle _{-1} :=\left\langle {T^{*-1}\xi }\left| {T^{*-1}\eta }\right. \right\rangle , \quad \xi , \eta \in \mathcal H. $$

Thus, we put $\mathcal H_{-1}:=\mathcal H_{+1}^\times$. Hence, the sequence $\{\xi _n\}$ and the operator T (which is bounded from $\mathcal H_{+1}$ into $\mathcal H$, but unbounded in $\mathcal H$!) automatically generate a Gelfand triplet of Hilbert spaces $\mathcal H_{+1} \subseteq \mathcal H\subseteq \mathcal H_{-1}$, which is a typical example of a rigged Hilbert space and we will call $\{\xi _n\}$ a (strict) Riesz-like basis.

A second motivation comes from the so called Pseudo-Hermitian Quantum Mechanics. This recent development of Quantum Mechanics deals with non self-adjoint Hamiltonians that often can be made into self-adjoint operators by some (generalized) similarity transformation (see, e.g., [5–8]).

Assume, in fact, that $\mathsf{{H}}$ is a closed operator in Hilbert space whose dense domain $D(\mathsf{{H}})$ is regarded as a Hilbert space $\mathcal H_{\mathsf{{H}}}$ with the graph norm $\Vert \cdot \Vert _{\mathsf{{H}}}$. As we will see in Sect. 2.1, this automatically produces a rigged Hilbert space. Assume that $\mathsf{{H}}_{sa}$ is a self-adjoint operator in $\mathcal H$ with discrete spectrum and, for simplicity, that each eigenvalue $\lambda _k \in {\mathbb R}$ has multiplicity 1. Let $\psi _k$ be an eigenvector corresponding to $\lambda _k$. Then $\{\psi _k\}$ is an orthonormal basis for $\mathcal H$. Assume that there exists a bounded operator $T:\mathcal H_{\mathsf{{H}}}\rightarrow \mathcal H$, invertible and with bounded inverse $T^{-1}:\mathcal H\rightarrow \mathcal H_{\mathsf{{H}}}$ such that

$$\begin{aligned} \left\langle {\mathsf{{H}}\xi }\left| {T^\dag \eta }\right. \right\rangle = \left\langle {T\xi }\left| {\mathsf{{H}}_{sa}\eta }\right. \right\rangle , \quad \forall \xi \in \mathcal H_{\mathsf{{H}}},\, \eta \in D(\mathsf{{H}}_{sa}) \text{ s. } \text{ t. } T^\dag \eta \in \mathcal H. \end{aligned}$$

(2)

Put $\xi _k= T^{-1}\psi _k$, for every k. Then, the set $\{\xi _k\}$ is a Riesz-like basis for $\mathcal H_{\mathsf{{H}}}$ and an easy computation shows that, for every $\eta \in D(\mathsf{{H}}_{sa})$, such that $T^\dag \eta \in \mathcal H$

$$ \left\langle {\mathsf{{H}}\xi _n}\left| {T^\dag \eta }\right. \right\rangle =\lambda _n\left\langle {\xi _n}\left| {T^\dag \eta }\right. \right\rangle . $$

Thus, if $\{ \eta \in D(\mathsf{{H}}_{sa}): T^\dag \eta \in \mathcal H\}$ is dense in $\mathcal H$, we get $\mathsf{{H}}\xi _n =\lambda _n \xi _n$, for every n. Hence $\mathsf{{H}}$ has a family of eigenvectors that are mapped by T into the elements of an orthonormal basis of $\mathcal H$. It should be noticed that the operator T need not be bounded as an operator in $\mathcal H$. This situation is of interest because of the existence of physical models whose (non self-adjoint) Hamiltonian cannot be transformed into a self-adjoint one by similarity operators that are bounded, with bounded inverse [9].

The paper is organized as follows. In Sect. 2 we recall basic notions as that of rigged Hilbert space (RHS), of operators on a RHS, of Schauder basis and of (strict) Riesz-like basis and recall some results given in [4] about (strict) Riesz-like bases. In Sect. 3 we prove that every $\omega$-independent, complete (total) Bessel sequence is a strict Riesz-like basis in a convenient triplet of Hilbert spaces. Furthermore, we consider an example of application of this result to Schrödinger-type operators. In Sect. 4 we study some operators defined by strict Riesz-like bases and by their dual bases, proving their closedness, their self-adjointness and so on. We also prove that some of them are related by a weak intertwining relation and, moreover, we give a characterization of those which are quasi-Hermitian (this kind of operators are known also as pseudo-Hermitian operators in Pseudo-Hermitian Quantum Mechanics).

2 Preliminaries and Basic Aspects

2.1 Rigged Hilbert Spaces and Operators on Them

Let ${\mathcal D}$ be a dense subspace of $\mathcal H$. A locally convex topology t on ${\mathcal D}$ finer than the topology induced by the Hilbert norm defines, in standard fashion, a rigged Hilbert space (RHS)

$$\begin{aligned} {\mathcal D}[t] \hookrightarrow \mathcal H\hookrightarrow {\mathcal D}^\times [t^\times ], \end{aligned}$$

(3)

where ${\mathcal D}^\times$ is the vector space of all continuous conjugate linear functionals on ${\mathcal D}[t]$, i.e., the conjugate dual of ${\mathcal D}[t]$, endowed with the strong dual topology $t^\times = \beta ({\mathcal D}^\times ,{\mathcal D})$ and $\hookrightarrow$ denotes a continuous embedding. Since the Hilbert space $\mathcal H$ can be identified with a subspace of ${\mathcal D}^\times [t^\times ]$, we will systematically read (3) as a chain of topological inclusions: ${\mathcal D}[t] \subset \mathcal H\subset {\mathcal D}^\times [t^\times ]$. In this paper we will consider only the case where ${\mathcal D}$ itself is a Hilbert space, denoted by $\mathcal H_{+1}$, under a norm stronger than that of $\mathcal H$. Its conjugate dual is denoted by $\mathcal H_{-1}$.

As an example, let us be given a closed operator T with dense domain D(T) in Hilbert space $\mathcal H$. Then, a rigged Hilbert space, more precisely a triplet of Hilbert spaces , arises in a natural way. Indeed, the domain D(T) with the graph norm $\Vert \cdot \Vert _T$ defined by

$$ \Vert \xi \Vert _T= (\Vert \xi \Vert ^2 + \Vert T\xi \Vert ^2)^{1/2}= \Vert (I+T * T)^{1/2}\xi \Vert , \quad \xi \in D(T) $$

becomes a Hilbert space, namely $\mathcal H_T$. If $\mathcal H_T^\times$ denotes the Hilbert space conjugate dual of $\mathcal H_T$, then we get the triplet of Hilbert spaces

$$ \mathcal H_T \subset \mathcal H\subset \mathcal H_T^\times .$$

If $\mathcal H$ and ${\mathcal K}$ are two Hilbert spaces, we will indicate by ${\mathcal B}(\mathcal H,{\mathcal K})$ the Banach space of linear bounded operators in $\mathcal H$ into ${\mathcal K}$. If $\mathcal H={\mathcal K}$, then, to simplify the notation, we will put ${\mathcal B}(\mathcal H,\mathcal H)={\mathcal B}(\mathcal H)$.

Let $\mathcal H_{+1} \subset \mathcal H\subset \mathcal H_{-1}$ be a triplet of Hilbert spaces. An involution $X \mapsto X^{{\dag }}$ can be introduced in ${\mathcal B}(\mathcal H_{+1},\mathcal H_{-1})$ by the equality

$$ \left\langle {X^{{\dag }}\eta }\left| { \xi }\right. \right\rangle = \overline{\left\langle {X\xi }\left| {\eta }\right. \right\rangle }, \quad \forall \xi , \eta \in \mathcal H_{+1}. $$

Hence ${\mathcal B}(\mathcal H_{+1},\mathcal H_{-1})$ is a $^\dagger$-invariant vector space.

2.2 Schauder, Riesz-Like and Strict Riesz-Like Bases

Let ${\mathcal E}[t_{\mathcal E}]$ be a locally convex vector space and $\{\xi _n\}$ a sequence of vectors of ${\mathcal E}$. We adopt the following terminology:

• (i) the sequence $\{\xi _n\}$ is complete or total if the linear span of $\{\xi _n\}$ is dense in ${\mathcal E}[t_{\mathcal E}]$;

• (ii) the sequence $\{\xi _n\}$ is $\omega$ -independent if $\sum _{n=1}^\infty c_n \xi _n=0$ , implies $c_n=0$, for every $n \in {\mathbb N}$;

• (iii) the sequence $\{\xi _n\}$ is a topological basis for ${\mathcal E}$ if, for every $\phi \in {\mathcal E}$, there exists a unique sequence $\{c_n\}$ of complex numbers such that

  $$\begin{aligned} \phi = \sum _{n=1}^\infty c_n \xi _n, \end{aligned}$$

  (4)

  where the series on the right hand side converges in ${\mathcal E}[t_{\mathcal E}]$.

• (iv) a topological basis $\{\xi _n\}$ for ${\mathcal E}[t_{\mathcal E}]$ is a Schauder basis if the coefficient functionals $\{c_n=c_n(f)\}$, appearing in (4), are $t_{\mathcal E}$-continuous.

If $\{\xi _n\}$ is a topological basis for ${\mathcal E}$, then $\{\xi _n\}$ is $\omega$ -independent and therefore it consists of linearly independent vectors. Moreover, in Banach spaces, the two notions of topological basis and of Schauder basis do coincide.

In the remainder of the paper the Hilbert space $\mathcal H$ will always be meant as a separable one.

Consider a rigged Hilbert space ${\mathcal D}[t] \subset \mathcal H\subset {\mathcal D}^\times [t^\times ]$ and a Schauder basis $\{\xi _n\}$ for ${\mathcal D}[t]$. Every $f \in {\mathcal D}$ is the sum of a series $\sum _{n=1}^\infty c_n(f) \xi _n$, with uniquely determined, suitable coefficients $c_n(f)$. By the continuity of the linear functionals $c_n$ on ${\mathcal D}[t]$, it follows the existence and the uniqueness of a sequence $\{\zeta _n\} \subset {\mathcal D}^\times$ such that

$$ c_n(f) = \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }, \quad \forall n\in {\mathbb N}, f \in {\mathcal D}. $$

If we take $f= \xi _k$, then $c_n(\xi _k)=\overline{\left\langle {\zeta _n}\left| {\xi _k}\right. \right\rangle }= \delta _{n,k}$ i.e., the two sequences $\{\xi _n\}$ and $\{\zeta _n\}$ are biorthogonal .

The following statements on Schauder bases, given here only for triplet of Hilbert spaces $\mathcal H_{+1} \subset \mathcal H\subset \mathcal H_{-1}$, was proved in [4] for general rigged Hilbert spaces by adapting results given in [2, 10, 11].

Proposition 2.1

Let $\{\xi _n\}$ be a Schauder basis for $\mathcal H_{+1}$. Then there exists a unique sequence $\{\zeta _n\}$ of vectors of $\mathcal H_{-1}$ such that

• (i) the sequences $\{\xi _n\}$ and $\{\zeta _n\}$ are biorthogonal;

• (ii) for every $f\in \mathcal H_{+1}$,

  $$\begin{aligned} f= \sum _{n=1}^\infty \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\xi _n; \end{aligned}$$

  (5)

• (iii) The partial sum operator $S_n$, given by

  $$ S_n f = \sum _{k=1}^n \overline{\left\langle {\zeta _k}\left| {f}\right. \right\rangle }\xi _k, \quad f\in \mathcal H_{+1} , $$

  is continuous from $\mathcal H_{+1}$ into $\mathcal H_{+1}$ and has an adjoint $S_n^\dag$ everywhere defined in $\mathcal H_{-1}$ given by

  $$ S^\dag _n \varPsi = \sum _{k=1}^n {\left\langle {\varPsi }\left| {\xi _k}\right. \right\rangle }\zeta _k, \quad \varPsi \in \mathcal H_{-1}; $$

• (iv) the sequence $\{\zeta _n\}$ is a basis for $\mathcal H_{-1}$ with respect to the weak topology; i.e., if $\varPsi \in \mathcal H_{-1}$ one has

  $$\begin{aligned} \left\langle {\varPsi }\left| {f}\right. \right\rangle = \left\langle {\sum _{k=1}^\infty {\left\langle {\varPsi }\left| {\xi _k}\right. \right\rangle }\zeta _k}\left| {f}\right. \right\rangle = \sum _{k=1}^\infty {\left\langle {\varPsi }\left| {\xi _k}\right. \right\rangle }\left\langle {\zeta _k}\left| {f}\right. \right\rangle , \quad \forall f \in \mathcal H_{+1}.\end{aligned}$$

  (6)

Remark 2.2

Of course, (6) provides a weak expansion for every $h\in \mathcal H$; i.e., $h= \sum _{k=1}^\infty {\left\langle {h}\left| {\xi _k}\right. \right\rangle }\zeta _k$, weakly. In particular, for $f\in \mathcal H_{+1} \subset \mathcal H_{-1}$, (6) gives

$$ \Vert f\Vert ^2 = \sum _{k=1}^\infty {\left\langle {f}\left| {\xi _k}\right. \right\rangle }\left\langle {\zeta _k}\left| {f}\right. \right\rangle , \quad \forall f \in \mathcal H_{+1} $$

so that the series on the right hand side is convergent, for every $f \in \mathcal H_{+1}$.

Now we recall the notion of Riesz-like and strict Riesz-like bases we gave in [4] for a rigged Hilbert space ${\mathcal D}[t]\subset \mathcal H\subset {\mathcal D}^\times [t^\times ]$.

Definition 2.3

A Schauder basis $\{\xi _n\}$ for ${\mathcal D}[t]$ is called a Riesz-like basis for ${\mathcal D}[t]$ if there exists a one-to-one continuous operator $T:{\mathcal D}[t]\rightarrow \mathcal H$ such that $\{T\xi _n\}$ is an orthonormal basis for $\mathcal H$.

The range R(T) of T contains the orthonormal basis $\{e_k\}$ with $e_k:= T \xi _k$, $k\in {\mathbb N}$, hence R(T) is dense in $\mathcal H$.

If $\{\xi _n\}$ is a Riesz-like basis, we can find explicitly the sequence $\{\zeta _n\} \subset \mathcal H_{-1}$ of Proposition 2.1. The continuity of T and (5), in fact, imply

$$ Tf= \sum _{n=1}^\infty \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }T\xi _n = \sum _{n=1}^\infty \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }e_n, \quad \forall f \in \mathcal H_{+1}. $$

This, in turn, implies that $\overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }=\left\langle {Tf}\left| {e_n}\right. \right\rangle$, for every $f \in \mathcal H_{+1}.$ Hence $\zeta _n=T^\dagger e_n$, for every $n \in \mathbb {N }$.

Clearly, for every $n,k\in \mathbb {N}$,

$$ \left\langle {\zeta _k}\left| {\xi _n}\right. \right\rangle = \left\langle {T^\dagger e_k}\left| {\xi _n}\right. \right\rangle =\left\langle { e_k}\left| {T\xi _n}\right. \right\rangle =\left\langle {e_k}\left| {e_n}\right. \right\rangle = \delta _{k,n} $$

and $T^\dagger T\xi _n= \zeta _n$, for every $n \in {\mathbb N}$. This sequence is called the dual sequence .

Let $\{\xi _n\}$ be a Riesz-like basis for ${\mathcal D}[t]$. One can ask what happens if we strengthen the hypotheses on T, e.g. if we suppose that T is onto too and $T^{-1}$ is continuous from $\mathcal H$ into ${\mathcal D}[t]$. In other words, let us suppose that the operator T which makes of $\{T\xi _n\}$ an orthonormal basis for $\mathcal H$ has a continuous inverse $T^{-1}:\mathcal H[\Vert \cdot \Vert ]\rightarrow {\mathcal D}[t]$ (in particular, $T^{-1}$ is a bounded operator in $\mathcal H$). We say in this case that $\{\xi _n\}$ is a strict Riesz-like basis for ${\mathcal D}[t]$ . This assumption has important consequences on the involved topologies. Indeed, as shown in [4, Proposition 3.6]

Proposition 2.4

If the rigged Hilbert space ${\mathcal D}[t]\subset \mathcal H\subset {\mathcal D}^\times [t^\times ]$, with ${\mathcal D}[t]$ complete and reflexive, has a strict Riesz-like basis $\{\xi _n\}$ then it is (equivalent to) a triplet of Hilbert spaces $\mathcal H_{+1} \subset \mathcal H\subset \mathcal H_{-1}$. Moreover, $\{\xi _n\}$ is an orthonormal basis for $\mathcal H_{+1}$ and the dual sequence $\{\zeta _n\}$ is an orthonormal basis for $\mathcal H_{-1}$.

In other words the rigged Hilbert space is forced to be a triplet of Hilbert spaces. On the other hand, in a triplet of Hilbert spaces $\mathcal H_{+1} \subset \mathcal H\subset \mathcal H_{-1}$, if the operator T which makes of $\{T\xi _n\}$ an orthonormal basis for $\mathcal H$ is onto, then $T^{-1}$ is automatically continuous and so the basis $\{\xi _n\}$ is strict.

Remark 2.5

It is clear that, if $\{\xi _n\}$ is a strict Riesz-like basis, then it is an unconditional basis of $\mathcal H_{+1}$.

3 Bessel Sequences as Strict-Riesz Like Bases

Now, we will give an answer to the following natural questions: given a sequence $\{\xi _n\}\subset \mathcal H$, does there exist a rigged Hilbert space such that $\{\xi _n\}$ is a strict Riesz-like basis for it? Given a sequence $\{\xi _n\}\subset \mathcal H$, does there exist a triplet of Hilbert spaces $\mathcal H_{+1} \subseteq \mathcal H\subseteq \mathcal H_{-1}$ such that $\{\xi _n\}$ is an orthonormal basis for $\mathcal H_{+1}$?

Let $\{\xi _n\}$ be a Bessel sequence in $\mathcal H$, i.e., [11] there exists $C>0$ such that for every finite sequence of complex numbers $\{c_1,c_2,...c_n\}$, $n\in \mathbb {N},$

$$\begin{aligned} \left\| \sum _{k=1}^n c_k\xi _k\right\| ^2\le C \sum _{k=1}^n\left| c_k\right| ^2. \end{aligned}$$

(7)

Let $\{e_n\}$ be an orthonormal basis for $\mathcal H$ and define the operator

$$\begin{aligned} V:\sum _{k=1}^n c_k e_k\rightarrow \sum _{k=1}^n c_k\xi _k. \end{aligned}$$

(8)

It is clear that V is well-defined and bounded (by (7)) on ${\mathcal G}=\mathsf{span}\{e_n\}$, then it extends to a bounded operator, denoted again by V, to $\mathcal H$. Obviously, $Ve_n=\xi _n$, for every $n\in \mathbb {N}$.

We notice that $\{\xi _n\}$ is $\omega$-independent if and only if V, and $\{\xi _n\}$ is complete if and only if $V^*$ is injective. Now we give our main result.

Theorem 3.1

If $\{\xi _n\}$ is an $\omega$-independent complete Bessel sequence in $\mathcal H$, then, for every orthonormal basis $\{e_n\}$ of $\mathcal H$, there exists a triplet of Hilbert spaces ${\mathcal K}\subset \mathcal H\subset {\mathcal K}^\times$ which has $\{\xi _n\}$ as a strict Riesz-like basis. This triplet is unique up to unitary transformations.

Proof

We maintain the notations of the previous discussion. Let $\{e_n\}$ be an orthonormal basis of $\mathcal H$. If $\{\xi _n\}$ is a $\omega$-independent Bessel sequence in $\mathcal H$, the operator V defined in (8) is injective on $\mathcal H$. Indeed, since $\{e_n\}$ is an orthonormal basis for $\mathcal H$, for every $f\in \mathcal H$, $f=\lim _{N\rightarrow \infty }f_N$, where $f_N=\sum _{n=1}^N\left\langle {f}\left| {e_n}\right. \right\rangle e_n\in {\mathcal G}$. It follows that

$$Vf:=\lim _{N\rightarrow \infty }Vf_N=\lim _{N\rightarrow \infty }\sum _{n=1}^N\left\langle {f}\left| {e_n}\right. \right\rangle \xi _n=\sum _{n=1}^\infty \left\langle {f}\left| {e_n}\right. \right\rangle \xi _n.$$

If $Vf=0$, then $\left\langle {f}\left| {e_n}\right. \right\rangle =0$, for every $n\in \mathbb {N}$. Hence $f=0$.

Then V has an inverse $V^{-1}$ defined on the range $\mathsf{Ran}(V)$ of V and, since V is bounded, $V^{-1}$ is closed. Moreover, $\{\xi _n\}\subseteq \mathsf{Ran}(V)$, hence, by the completeness of $\{\xi _n\}$, the inverse of the operator V is densely defined. Now, we have

$$ \mathsf{Ran}(V)=\left\{ g\in \mathcal H: g= \sum _{n=1}^\infty c_n\xi _n \text{ with } \sum _{n=1}^\infty |c_n|^2<\infty \right\} . $$

The $\omega$-independence of $\{\xi _n\}$ guarantees the uniqueness of the expansion $g= \sum _{n=1}^\infty c_n\xi _n$ of every $g \in \mathsf{Ran}(V)$. Finally, we have

$$ V^{-1}g=V^{-1}\left( \sum _{n=1}^\infty c_n\xi _n\right) =\sum _{n=1}^\infty c_n e_n, \quad \forall g= \sum _{n=1}^\infty c_n\xi _n \in \mathsf{Ran}(V). $$

We put, for short, $T:=V^{-1}$ and $D(T)=\mathsf{Ran}(V)$. Then T is a closed densely defined operator such that $T\xi _n=e_n$ and has bounded inverse. Then, as we have already seen in Sect. 2.1, a triplet of Hilbert spaces arises in a natural way. More precisely, we get the triplet of Hilbert spaces

$$ \mathcal H_T \subset \mathcal H\subset \mathcal H_T^\times $$

where $\mathcal H_{T}=D(T)[\Vert \cdot \Vert _T]$ with

$$ \Vert \xi \Vert _T= (\Vert \xi \Vert ^2 + \Vert T\xi \Vert ^2)^{1/2}= \Vert (I+T^* T)^{1/2}\xi \Vert , \quad \xi \in D(T) $$

and the sequence $\{\xi _n\}$ is a strict Riesz-like basis for $\mathcal H_T$. Now, let us consider two different orthonormal bases $\{e_n\}$ and $\{e_n'\}$ of the Hilbert space $\mathcal H$. Then, as it is well-known, there exists a unitary operator $U:\mathcal H\rightarrow \mathcal H$ such that $Ue_n=e_n'$, therefore the two norms $\Vert \cdot \Vert _T$ and $\Vert \cdot \Vert _{UT}$ coincide and hence the two Hilbert spaces $\mathcal H_T$ and $\mathcal H_{UT}$ do.

Remark 3.2

If $\{\xi _n\}$ is an $\omega$-independent complete Bessel sequence in $\mathcal H$, then, for every orthonormal basis $\{e_n\}$ of $\mathcal H$, there exists a (unique) Hilbert space which has $\{\xi _n\}$ as an orthonormal basis since, once the triplet of Hilbert spaces $\mathcal H_T \subset \mathcal H\subset \mathcal H_T^\times$ is at hand (Theorem 3.1), then $\{\xi _n\}$ is an orthonormal basis for $\mathcal H_{+1} = D(T)[\Vert \cdot \Vert _{+1}]$ and, as a consequence of the uniqueness of $\mathcal H_T$, the Hilbert space $\mathcal H_{+1}$ (and the triplet), is unique too.

Remark 3.3

If $T=V^{-1}$ is also bounded, then $\{\xi _n\}$ is a Riesz basis for $\mathcal H$ and $\mathcal H_T$ coincides with $\mathcal H$ as a vector space but it carries a different albeit equivalent norm, as stated by the well-known theory of Riesz bases.

Remark 3.4

If $\{\xi _n\}$ is an $\omega$-independent complete Bessel sequence in $\mathcal H$, then Theorem 3.1 gives us full information on the possibility of expanding a vector $f \in \mathcal H$ in terms of $\{\xi _n\}$: indeed, $\{\xi _n\}$ determines a closed densely defined operator T and every vector f of the domain of T can be expanded uniquely as an unconditionally convergent series $f= \sum _{n=1}^\infty c_n \xi _n$, the convergence holds in the graph norm $\Vert \cdot \Vert _T$ of D(T), and then in the norm $\Vert \cdot \Vert$. Other vectors f of $\mathcal H$, by (6), can be obtained by a weakly convergent series $f= \sum _{k=1}^\infty {\left\langle {f}\left| {\xi _k}\right. \right\rangle }\zeta _k$, $\{\zeta _k\}$ being the dual sequence of $\{\xi _n\}$, in the sense that $\left\langle {f}\left| {\eta }\right. \right\rangle = \sum _{k=1}^\infty {\left\langle {f}\left| {\xi _k}\right. \right\rangle }\left\langle {\zeta _k}\left| {\eta }\right. \right\rangle , \forall \eta \in D(T)$.

If T is unbounded, then $0\in \sigma _c(T^{-1})$, the continuous spectrum of $T^{-1}$. Some more information on $\{\xi _n\}$ can be obtained just making some assumption on the spectral behaviour of $T^{-1}$. Assume, for instance, that $T^{-1}$ is compact, then the sequence $\{\xi _n\}$ converges to 0 in the norm of $\mathcal H$, being the image of an orthonormal basis through a compact operator. Of course one can go further and require that $T^{-1}$ belongs to some other well-known classes of operators, giving a more accurate description of how fast $\Vert \xi _n\Vert \rightarrow 0$. For a discussion on this subject see [12].

3.1 An Application

The importance of Theorem 3.1 is that, once we have at hand a non self-adjoint operator $\mathsf{H}$, with purely discrete real spectrum, it is possible to construct the Hilbert space of the system by finding out exactly the closed operator defining an inner product which makes the eigenvectors of $\mathsf{H}$ orthonormal. As expected, the inner product of the Hilbert space can be given in terms of the metric operator $Q=T^\dagger T$ which is unbounded as an operator in $\mathcal H$, whereas is bounded as an operator in $\mathcal H_{+1}$ into $\mathcal H_{-1}$ (see Proposition 4.4 in Sect. 4). This change of domain is not a deal by the physical point of view, because the observable of the system are in general unbounded linear operators defined on a dense set ${\mathcal D}$ of Hilbert space $\mathcal H$. As an example of this situation, let us consider the Hilbert space $\mathcal H=L^2(\mathbb {R})$ and the Hamiltonian

$$ H=-\frac{d^2}{dx^2}+\frac{x^2}{2}-\frac{4x}{1+x^2}\frac{d}{dx}-\frac{2}{1+x^2}=H_0+V $$

where $H_0$ is the Hamiltonian operator of the harmonic oscillator and $V=-\frac{4x}{1+x^2}\frac{d}{dx}-\frac{2}{1+x^2}$ (in spite of the notation, the operator V is not a physical potential, since it depends explicitly on the derivative operator). The set of its eigenvectors is $\{\xi _n=\frac{1}{\sqrt{2^n\,n!\sqrt{\pi }}} H_n(x)\frac{e^{-\frac{x^2}{2}}}{1+x^2},\,n\ge 0\}$, where $H_n(x)$ is the nth Hermite polynomial. The vectors $\xi _n$’s do not form a orthonormal basis for $\mathcal H$. However, they constitute an $\omega$-independent complete Bessel sequence in $\mathcal H$ as we will see in a while. Hence, by Theorem 3.1, there exists a triplet of Hilbert spaces which has $\{\xi _n\}$ as a strict Riesz-like basis and, even more important, there exists a Hilbert space $\mathcal H_{+1}$ such that $\{\xi _n\}$ is an orthonormal basis for $\mathcal H_{+1}$ and such that $H\in {\mathcal B}(\mathcal H_{+1})$ (H is closed and everywhere defined in $\mathcal H_{+1}$). Recall that, once we call $N_n=\frac{1}{\sqrt{2^n\,n!\sqrt{\pi }}}$, the set $\{e_n(x)=N_n H_n(x)e^{-\frac{x^2}{2}},\,n\ge 0\}$ is an orthonormal basis of $\mathcal H$. Hence the operator T which takes the sequence $\{\xi _n\}$ into $\{e_n\}$ is $T=1+x^2$. This is an unbounded continuous operator defined on the dense set $D(T)=\{f\in \mathcal H:\,(1+x^2)f\in \mathcal H\}$, with bounded inverse: $T^{-1}=\frac{1}{1+x^2}$. The Hamiltonian H is non self-adjoint and similar to $H_0$ by the intertwining operator T, $H=T^{-1}H_0T$, the eigenvectors of H are transformed into those of $H_0$ and H and $H_0$ have the same eigenvalues $\alpha _n=n+\frac{1}{2}$, for every $n\ge 0$ sorted n by n; (in particular, the ground state $\xi _0$ is transformed in that one of $H_0$). It remains to show that $\{\xi _n\}$ is an $\omega$-independent complete Bessel sequence in $\mathcal H$. Indeed, $\{\xi _n\}$ is a Bessel sequence since there exists $C=\Vert T^{-1}\Vert >0$ such that for every finite sequence of complex numbers $\{c_0,c_1,...c_n\}$, $n\in \mathbb {N},$

$$\begin{aligned} \left\| \sum _{k=0}^n c_k\xi _k\right\| ^2=\left\| \sum _{k=0}^n c_k T^{-1}e_k\right\| ^2\le C \sum _{k=0}^n\left| c_k\right| ^2. \end{aligned}$$

(9)

They are $\omega$-independent because if

$$\sum _{n=0}^\infty c_n \xi _n=0=\sum _{n=0}^\infty c_n T^{-1}e_n=T^{-1}\left( \sum _{n=0}^\infty c_n e_n\right) ,$$

then it implies $c_n=0$, for every $n \ge 0$, by the continuity and the injectivity of $T^{-1}$. Furthermore, they are a complete set because if $f\in \mathcal H$ is such that $\left\langle {f}\left| {\xi _n}\right. \right\rangle =0$ for every n, then

$$ 0=\left\langle {f}\left| {\xi _n}\right. \right\rangle =\left\langle {f}\left| {T^{-1}e_n}\right. \right\rangle =\left\langle {T^{-1}f}\left| {e_n}\right. \right\rangle =0 $$

which by the injectivity of $T^{-1}$ implies $f=0$. Notice that, albeit $\{\xi _n=T^{-1}e_n\}$ is an $\omega$-independent complete Bessel sequence in $\mathcal H$, it is not a Riesz basis for $\mathcal H$ because $T=(1+x^2)$ is an unbounded operator. Now, following what we saw before, the natural space where considering the previous operator H is $\mathcal H_{+1}=D(T)[\Vert \cdot \Vert _{+1}]$ with $\Vert \cdot \Vert _{+1}=\Vert (1+x^2)\cdot \Vert$.

4 Operators Defined by Strict Riesz-Like Bases

In this section some results in [13] are generalized to the case of operators defined in triplets of Hilbert spaces. Furthermore, we will prove some result about the similarity of operators introduced here, and a characterization of those which have real eigenvalues.

Let $\{\xi _n\}$ be a strict Riesz-like basis for the triplet $\mathcal H_{+1}\subset \mathcal H\subset \mathcal H_{-1}$ and $\{\zeta _n\}$ its dual basis. If ${{\varvec{\alpha }}}=\{\alpha _n\}$ is a sequence of complex numbers we can formally define, for $f\in \mathcal H_{+1}$,

$$\begin{aligned} A^{{\varvec{\alpha }}}f=\sum _{n=1}^\infty \alpha _n (\xi _n \otimes \overline{\zeta }_n) f=\sum _{n=1}^\infty \alpha _n\overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\xi _n \end{aligned}$$

(10)

$$\begin{aligned} B^{{\varvec{\alpha }}}f=\sum _{n=1}^\infty \alpha _n (\zeta _n \otimes \overline{\xi }_n) f=\sum _{n=1}^\infty \alpha _n{\left\langle {f}\left| {\xi _n}\right. \right\rangle }\zeta _n. \end{aligned}$$

(11)

$$\begin{aligned} R^{{\varvec{\alpha }}}f= \sum _{n=1}^\infty \alpha _n (\xi _n \otimes \overline{\xi }_n) f=\sum _{n=1}^\infty \alpha _n\left\langle {f}\left| {\xi _n}\right. \right\rangle \xi _n \end{aligned}$$

(12)

$$\begin{aligned} Q^{{\varvec{\alpha }}}f= \sum _{n=1}^\infty \alpha _n (\zeta _n \otimes \overline{\zeta }_n) f=\sum _{n=1}^\infty \alpha _n\overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\zeta _n \end{aligned}$$

(13)

Of course, these are the simplest operators that can be defined via $\{\xi _n\}$ and $\{\zeta _n\}$.

Remark 4.1

Before going further, a comment is in order. In [14] Balazs introduced the notion of Bessel multipliers (frame multipliers, Riesz multipliers) whose definition is apparently similar to those given above. To be more precise, if $\{\varphi _n\}$, $\{\psi _n\}$ are Bessel sequences respectively in two Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, fix ${{\varvec{m}}}=\{m_n\}$ a bounded sequence of complex numbers, the Bessel multiplier for the Bessel sequences above is an operator $M:\mathcal H_2\rightarrow \mathcal H_1$ defined by

$$ M= \sum _{n=1}^\infty m_n (\varphi _n \otimes \overline{\psi }_n). $$

The main differences with the operators in (10)–(11) is that the two sequences $\{\varphi _n\}$, $\{\psi _n\}$ are not necessarily biorthogonal (in particular, in [14, Corollary 7.5] a necessary and sufficient condition is given for $\{\varphi _n\}$, $\{\psi _n\}$ to be biorthogonal), and moreover, as we shall see in a while, we will also deal with possibly unbounded sequences. Thus, the two notions are not directly comparable.

Let $\mathcal H_{+1} \subset \mathcal H\subset \mathcal H_{-1}$ be a triplet of Hilbert spaces and $\{\xi _n\}$ a strict Riesz-like basis for $\mathcal H_{+1}$.

Clearly, the operator formally defined by (10) can take values in $\mathcal H_{+1}$ or in $\mathcal H$ or even in $\mathcal H_{-1}$, following the different topologies that make the series on the right hand side convergent. It is clear that, if $f \in \mathcal H_{+1}$, then

$$ \sum _{n=1}^\infty \alpha _n\overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\xi _n \text{ converges } \text{ in } \mathcal H_{-1} \Leftrightarrow \sum _{n=1}^\infty \left| \sum _{k=1}^\infty \alpha _k\overline{\left\langle {\zeta _k}\left| {f}\right. \right\rangle }\left\langle {\xi _k}\left| {\zeta _n}\right. \right\rangle _{-1} \right| ^2<\infty . $$

Since $\left\langle {\xi _k}\left| {\zeta _n}\right. \right\rangle _{-1}=\left\langle {\xi _k}\left| {\xi _n}\right. \right\rangle ,$ for every $k,n \in {\mathbb N}$, we can conclude that

$$ A^{{\varvec{\alpha }}}f \in \mathcal H_{-1} \Leftrightarrow \sum _{n=1}^\infty \left| \sum _{k=1}^\infty \alpha _k\overline{\left\langle {\zeta _k}\left| {f}\right. \right\rangle }G_{k,n} \right| ^2<\infty . $$

where $(G_{k,n})$ is the Gram matrix of the basis $\{\xi _k\}$; i.e., $G_{k,n}=\left\langle {\xi _k}\left| {\xi _n}\right. \right\rangle$, for $k,n \in {\mathbb N}$. Differently from the standard case, the Gram matrix of $\{\xi _k\}$ need not be bounded.

Similarly, since $\{e_n\}$ is an orthonormal basis in $\mathcal H$, we have

$$ A^{{\varvec{\alpha }}}f \in \mathcal H\Leftrightarrow \sum _{n=1}^\infty \left| \sum _{k=1}^\infty \alpha _k{\overline{\left\langle {\zeta _k}\left| {f}\right. \right\rangle }\left\langle {\xi _k}\left| {e_n}\right. \right\rangle } \right| ^2<\infty , $$

where, as before, $e_k=T\xi _k$, $k\in {\mathbb N}$.

Finally, as we shall see in Proposition 4.2,

$$ A^{{\varvec{\alpha }}}f \in \mathcal H_{+1}\Leftrightarrow \sum _{k=1}^\infty |\alpha _k|^2|\left\langle {\zeta _k}\left| {f}\right. \right\rangle |^2<\infty . $$

Of course, analogous considerations can be made for the operators defined in (11), (12) and (13). It is worth remarking that for the operators $B^{{\varvec{\alpha }}}$ and $R^{{\varvec{\alpha }}}$ the series on the right hand side of (11), (12) may converge also for some $f\in \mathcal H_{-1}$.

Now we examine more closely one of the cases listed above. In particular, we will suppose $A^{{\varvec{\alpha }}}f\in \mathcal H_{+1}$, for every $f\in \mathcal H_{+1}$. Under this assumption, let us define

$$ \left\{ \begin{array}{l} D(A^{{\varvec{\alpha }}})=\left\{ f \in \mathcal H_{+1}; \sum \limits _{n=1}^\infty \alpha _n \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\xi _n \text{ exists } \text{ in } \mathcal H_{+1}\right\} \\ { } \\ A^{{\varvec{\alpha }}}f= \sum \limits _{n=1}^\infty \alpha _n \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\xi _n, \; f \in D(A^{{\varvec{\alpha }}}) \end{array}\right. $$

$$ \left\{ \begin{array}{l} D(B^{{\varvec{\alpha }}})=\left\{ \varPsi \in \mathcal H_{-1}; \sum \limits _{n=1}^\infty \alpha _n \left\langle {\varPsi }\left| {\xi _n}\right. \right\rangle \zeta _n \text{ exists } \text{ in } \mathcal H_{-1} \right\} \\ { } \\ B^{{\varvec{\alpha }}}\varPsi = \sum \limits _{n=1}^\infty \alpha _n \left\langle {\varPsi }\left| {\xi _n}\right. \right\rangle \zeta _n, \; \varPsi \in D(B^{{\varvec{\alpha }}}) \end{array}\right. . $$

Then we have the following

$$\begin{aligned}&{\mathcal D}_\xi := \text{ span }\{\xi _n\} \subset D(A^{{\varvec{\alpha }}}); \nonumber \\&{\mathcal D}_\zeta := \text{ span }\{\zeta _n\} \subset D(B^{{\varvec{\alpha }}});\end{aligned}$$

(14)

$$\begin{aligned}&A^{{\varvec{\alpha }}}\xi _k =\alpha _k \xi _k, \; k=1,2, \ldots ; \nonumber \\&B^{{\varvec{\alpha }}}\zeta _k =\alpha _k \zeta _k, \; k=1,2, \ldots . \end{aligned}$$

(15)

Hence, $A^{{\varvec{\alpha }}}$ and $B^{{\varvec{\alpha }}}$ are densely defined and have the same eigenvalues. As we will see, if $\alpha _n\in \mathbb {R}$, $\forall n\in \mathbb {N}$, they are one the adjoint of the other.

It worths noting that the operators $(T^\dagger )^{-1}$ and $(T^{-1})^\dag$ do coincide [15, Remark 3.2].

Before continuing, we recall that if $X: D(X) \subseteq \mathcal H_{+1} \rightarrow \mathcal H_{+1}$ is a closed map and D(X) is dense in $\mathcal H_{+1}$, then there exists a closed densely defined map $X^\dag : D(X^\dag )\subseteq \mathcal H_{-1}\rightarrow \mathcal H_{-1}$ such that

$$ \left\langle {\varPhi }\left| {X\xi }\right. \right\rangle = \left\langle {X^\dag \varPhi }\left| {\xi }\right. \right\rangle , \quad \forall \xi \in \mathcal H_{+1}, \, \varPhi \in \mathcal H_{-1}.$$

If X is also closed as an operator in $\mathcal H$, then its Hilbert adjoint $X^*$ exists and $X^*= X^\dagger _{\upharpoonright D(X^*)}$ where $D(X^*)=\{\phi \in \mathcal H:\, X^\dag \phi \in \mathcal H\}$.

Proposition 4.2

The following statements hold.

• (i) $D(A^{{\varvec{\alpha }}})=\left\{ f \in \mathcal H_{+1}; \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {\zeta _n}\left| {f}\right. \right\rangle |^2 <\infty \right\}$,

  $D(B^{{\varvec{\alpha }}})=\left\{ \varPsi \in \mathcal H_{-1}; \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {\varPsi }\left| {\xi _n}\right. \right\rangle |^2 <\infty \right\}$.

• (ii) $A^{{\varvec{\alpha }}}$ and $B^{{\varvec{\alpha }}}$ are closed operators respectively in $\mathcal H_{+1} [\Vert \cdot \Vert _{+1}]$ and in $\mathcal H_{-1} [\Vert \cdot \Vert _{-1}].$

• (iii) $\left( A^{{\varvec{\alpha }}}\right) ^\dagger =B^{\overline{{\varvec{\alpha }}}}$, where ${\overline{{\varvec{\alpha }}}}=\{\overline{\alpha }_n\}$.

• (iv) $A^{{\varvec{\alpha }}}$ is bounded in $\mathcal H_{+1}$ if, and only if, $B^{{\varvec{\alpha }}}$ is bounded in $\mathcal H_{-1}$ and if, and only if, ${{\varvec{\alpha }}}$ is a bounded sequence. In particular $A^\mathbf{1 }=I_{\mathcal H_{+1}}$ and $B^\mathbf{1 }=I_{\mathcal H_{-1}}$, where $\mathbf{1 }$ is the sequence constantly equals to 1.

Proof

• (i): Since $\{\xi _n\}$ is an orthonormal basis for $\mathcal H_{+1}$, we have

  $$\begin{aligned} \left\| \sum _{k=n}^m \alpha _k \left\langle {\zeta _k}\left| {f}\right. \right\rangle \xi _k\right\| _{+1}^2=\sum _{k=n}^m |\alpha _k|^2 |\left\langle {\zeta _k}\left| {f}\right. \right\rangle |^2, \quad f \in \mathcal H_{+1} \end{aligned}$$

  (16)

  which shows that $f\in D(A^{{\varvec{\alpha }}})$ if and only if $\sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {\zeta _n}\left| {f}\right. \right\rangle |^2 <\infty {.}$

• (ii): The proof of this statement can be made by slight modifications of [13, Proposition 2.1 (2)].

• (iii): It is easy to show that $B^{\overline{{\varvec{\alpha }}}}=\sum _{n=1}^\infty \overline{\alpha }_n (\zeta _n \otimes \overline{\xi _n})\subseteq (A^{{\varvec{\alpha }}})^\dagger$. Conversely, let $\varPsi \in D((A^{{\varvec{\alpha }}})^\dagger )$; then there exists $\varPhi \in \mathcal H_{-1}$ such that

  $$ \left\langle {\varPsi }\left| {\sum _{n=1}^\infty {\alpha }_n \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\xi _n}\right. \right\rangle =\left\langle {\varPhi }\left| {f}\right. \right\rangle , \quad \forall f \in D(A^{{\varvec{\alpha }}}). $$

  By (14) and (15), ${\mathcal D}_\xi \subseteq D(A^{{\varvec{\alpha }}})$ and $A^{{\varvec{\alpha }}}\xi _k =\alpha _k \xi _k$, $k=1,2, \ldots$ . Thus, $\left\langle {\varPsi }\left| {\alpha _k \xi _k}\right. \right\rangle = \left\langle {\varPhi }\left| {\xi _k}\right. \right\rangle$, $k=1,2, \ldots$ . Hence

  $$ \sum _{k=1}^\infty |\alpha _k|^2 |\left\langle {\varPsi }\left| {\xi _k}\right. \right\rangle |^2=\sum _{k=1}^\infty |\left\langle {\varPhi }\left| {\xi _k}\right. \right\rangle |^2=\sum _{k=1}^\infty |\left\langle {(T^{-1})^\dagger \varPhi }\left| {e_k}\right. \right\rangle |^2=\Vert (T^{-1})^\dagger \varPhi \Vert ^2<\infty .$$

  This implies that $\varPsi \in D(B^{\overline{{\varvec{\alpha }}}})$.

• (iv): Let ${{\varvec{\alpha }}}$ be a bounded sequence, then there exists $M>0$ such that

  $$ \Vert A^{{\varvec{\alpha }}}f\Vert _{+1}=\left\| \sum _{k=1}^\infty \alpha _k \overline{\left\langle {\zeta _k}\left| {f}\right. \right\rangle }\xi _k\right\| _{+1}\le M\left\| \sum _{k=1}^\infty \overline{\left\langle {\zeta _k}\left| {f}\right. \right\rangle }\xi _k\right\| _{+1}, $$

  hence $A^{{\varvec{\alpha }}}$ is bounded in $\mathcal H_{+1}$.

In a very similar way one can prove (i), (ii) and (iv) for $B^{{\varvec{\alpha }}}$. This completes the proof.

Remark 4.3

In [15] Di Bella, Trapani and the author gave a definition of spectrum for continuous operators acting in a rigged Hilbert space ${\mathcal D}\subset \mathcal H\subset {\mathcal D}^\times$. We refer to that paper for precise definitions and results. So a natural question is: what is the spectrum (in that sense) of the operator $A^{{\varvec{\alpha }}}$ defined above? Let us assume that the sequence ${{\varvec{\alpha }}}$ is bounded, so that $A^{{\varvec{\alpha }}}$ is a bounded operator in $\mathcal H_{+1}$. The analysis is, in this case, particularly simple since, as usual, the set of eigenvalues consists exactly of the $\alpha _k$’s and, if $\lambda$ does not belong to the closure $\overline{\{\alpha _k; k \in {\mathbb N}\}}$ of the set of eigenvalues, then the inverse of $A^{{\varvec{\alpha }}}-\lambda I_{\mathcal H_{+1}}$ exists as a bounded operator in $\mathcal H_{+1}$. Hence, as expected, $\sigma (A^{{\varvec{\alpha }}})= \overline{\{\alpha _k; k \in {\mathbb N}\}}$. The situation for $B^{{\varvec{\alpha }}}$ is analogous.

Let us now consider the operators formally given by (12) and (13). They are, in fact, defined as follows:

$$ \left\{ \begin{array}{l} D(R^{{{\varvec{\alpha }}}})=\left\{ \varPsi \in \mathcal H_{-1}; \sum \limits _{n=1}^\infty \alpha _n \left\langle {\varPsi }\left| {\xi _n}\right. \right\rangle \xi _n \text{ exists } \text{ in } \mathcal H_{+1} \right\} \\ { } \\ R^{{{\varvec{\alpha }}}} \varPsi = \sum \limits _{n=1}^\infty \alpha _n \left\langle {\varPsi }\left| {\xi _n}\right. \right\rangle \xi _n, \; \varPsi \in D(R^{{{\varvec{\alpha }}}}) \end{array}\right. $$

$$ \left\{ \begin{array}{l} D(Q^{{{\varvec{\alpha }}}})=\left\{ f \in \mathcal H_{+1}; \sum \limits _{n=1}^\infty \alpha _n \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\zeta _n \text{ exists } \text{ in } \mathcal H_{-1} \right\} \\ { } \\ Q^{{{\varvec{\alpha }}}} f= \sum \limits _{n=1}^\infty \alpha _n \overline{\left\langle {\zeta _n}\left| {f}\right. \right\rangle }\zeta _n, \; f \in D(Q^{{{\varvec{\alpha }}}}) \end{array}\right. . $$

It is clear that

$$\begin{aligned}&{\mathcal D}_\zeta \subset D(R^{{{\varvec{\alpha }}}}) \text{ and } R^{{{\varvec{\alpha }}}} \zeta _k =\alpha _k \xi _k, \; k=1,2, \ldots ;\end{aligned}$$

(17)

$$\begin{aligned}&{\mathcal D}_\xi \subset D(Q^{{{\varvec{\alpha }}}}) \text{ and } Q^{{{\varvec{\alpha }}}} \xi _k =\alpha _k \zeta _k, \; k=1,2, \ldots \end{aligned}$$

(18)

Hence, $R^{{{\varvec{\alpha }}}}$ and $Q^{{{\varvec{\alpha }}}}$ are densely defined, and the following results can be established:

Proposition 4.4

The following statements hold.

• (1) $D(R^{{{\varvec{\alpha }}}})=\left\{ \varPsi \in \mathcal H_{-1}; \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {\varPsi }\left| {\xi _n}\right. \right\rangle |^2 <\infty \right\} =D(B^{{\varvec{\alpha }}})$,

  $D(Q^{{{\varvec{\alpha }}}})=\left\{ f \in \mathcal H_{+1}; \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {\zeta _n}\left| {f}\right. \right\rangle |^2 <\infty \right\} = D(A^{{\varvec{\alpha }}})$.

• (2) $R^{{{\varvec{\alpha }}}}$ and $Q^{{{\varvec{\alpha }}}}$ are closed.

• (3) $\left( R^{{{\varvec{\alpha }}}}\right) ^\dagger =R^{{\overline{{\varvec{\alpha }}}}}$ and $\left( Q^{{{\varvec{\alpha }}}}\right) ^\dagger =Q^{{\overline{{\varvec{\alpha }}}}}$, where ${\overline{{\varvec{\alpha }}}}=\{\overline{\alpha _n}\}$.

• (4) If $\{\alpha _n\} \subset {\mathbb R}$ (respectively, $\{\alpha _n\} \subset {\mathbb R}^+$) then $R^{{{\varvec{\alpha }}}}$ and $Q^{{{\varvec{\alpha }}}}$ are self-adjoint (respectively, positive self-adjoint). Furthermore, $R^{{{\varvec{\alpha }}}}$ is bounded from $\mathcal H_{-1}$ to $\mathcal H_{+1}$ if and only if $Q^{{{\varvec{\alpha }}}}$ is bounded from $\mathcal H_{+1}$ to $\mathcal H_{-1}$ and if, and only if, ${{\varvec{\alpha }}}$ is a bounded sequence.

• (5) If ${{\varvec{\alpha }}}={\varvec{1}}$, where, as before, ${\varvec{1}}$ denotes the sequence constantly equals to 1, then $R:=R^{\varvec{1}}$ and $Q:=Q^{\varvec{1}}$ are bounded positive self-adjoint operators respectively of ${\mathcal B}(\mathcal H_{-1},\mathcal H_{+1})$ and of ${\mathcal B}(\mathcal H_{+1},\mathcal H_{-1})$ and they are inverses of each other, that is $R= (Q)^{-1}$, and $R =T^{-1}(T^{-1})^\dag$, $Q = T^\dagger T$, where $T\in {\mathcal B}(\mathcal H_{+1},\mathcal H)$ is the operator such that $T\xi _n=e_n$, $\forall n\in \mathbb {N}$ and $\{e_n\}$ is an orthonormal basis for $\mathcal H$.

Proof

The proof is similar to that of Proposition 4.2 and we omit it.

Remark 4.5

From Proposition 4.4, we see that there exists a bounded invertible, positive self-adjoint operator Q from $\mathcal H_{+1}$ into $\mathcal H_{-1}$ that maps the strict Riesz-like basis $\{\xi _n\}$ into its dual basis $\{\zeta _n\}$.

Now, recall that $Q=Q^{\varvec{1}}$ and $R=R^{\varvec{1}}$, then we have the following

Proposition 4.6

Let ${{\varvec{\alpha }}}=\{\alpha _n\}$ be a sequence of complex numbers. The following equalities hold:

$$\begin{aligned} \begin{array}{l} QA^{{\varvec{\alpha }}}=B^{{\varvec{\alpha }}}Q=Q^{{\varvec{\alpha }}},\\ R B^{{\varvec{\alpha }}}= A^{{\varvec{\alpha }}}R=R^{{\varvec{\alpha }}}. \end{array} \end{aligned}$$

(19)

Proof

By Proposition 4.4 we have $D(A^{{\varvec{\alpha }}})=D(Q^{{\varvec{\alpha }}})$ and $D(B^{{\varvec{\alpha }}})=D(R^{{\varvec{\alpha }}})$. Moreover, from Proposition 4.2 and (18), if $f\in D(Q)$

$$\begin{aligned} Q f \in D(B^{{\varvec{\alpha }}})\Leftrightarrow & {} \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {Q f}\left| {\xi _n}\right. \right\rangle |^2<\infty \\\Leftrightarrow & {} \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {\zeta _n}\left| {f}\right. \right\rangle |^2<\infty \Leftrightarrow f \in D(Q^{{\varvec{\alpha }}}). \end{aligned}$$

Similarly one proves the equality $D(QA^{{\varvec{\alpha }}}) =D(Q^{{\varvec{\alpha }}})$. It is easily seen that $Q A^{{\varvec{\alpha }}}f= B^{{\varvec{\alpha }}}Q f = Q^{{\varvec{\alpha }}}f$, for every $f \in D(Q^{{\varvec{\alpha }}})$. The proof of the second equality in (19) is analogous.

Remark 4.7

Equations (19) show that the two operators $A^{{\varvec{\alpha }}}$ and $B^{{\varvec{\alpha }}}$ are similar, in the sense that Q and R act as intertwining operators, see e.g. [5, Definition 7.3.1]. The intertwining relations between operators have found some recent interest in Quantum Mechanics.

A simple consequence of previous results is the following corollary which generalizes the Theorem by Mostafazadeh¹ in [16] and thus gives a characterization of operators as $A^{{\varvec{\alpha }}}$ and $B^{{\varvec{\alpha }}}$ with real eigenvalues. Before continuing we recall the definition of (unbounded) quasi-Hermitian operator (see, e.g. [5, Definition 7.5.1]).

Definition 4.8

A closed operator A, with dense domain D(A) is called quasi-Hermitian if there exists a metric operator G, with dense domain D(G) in Hilbert space $\mathcal H$ such that $D(A)\subset D(G)$ and

$$\begin{aligned} \left\langle {A\xi }\left| {G\eta }\right. \right\rangle = \left\langle {G\xi }\left| {A\eta }\right. \right\rangle , \quad \xi , \eta \in D(A).\end{aligned}$$

(20)

If A is a quasi-Hermitian operator on $\mathcal H$, then by definition there exists an unbounded metric operator G such that

$$A^\dagger G=AG.$$

Corollary 4.9

Let T be the operator which transforms the strict Riesz-like basis $\{\xi _n\}$ into an orthonormal basis of Hilbert space $\mathcal H$. The following statements are equivalent.

• (i) The sequence ${{\varvec{\alpha }}}=\{\alpha _n\}$ consists of real numbers.

• (ii) $A^{{\varvec{\alpha }}}$ is quasi-Hermitian, with $G=Q=T^\dagger T$.

• (iii) $B^{{\varvec{\alpha }}}$ is quasi-Hermitian, with $G=R=T^{-1}\left( T^\dagger \right) ^{-1}$.

Proof

 

$(i)\Rightarrow (ii)$ :

Suppose first that $\{\alpha _n\} \subset {\mathbb R}$, then according to (iii) of Proposition 4.2 $\left( A^{{\varvec{\alpha }}}\right) ^\dagger =B^{{\varvec{\alpha }}}$. Then we can rewrite the first equality in (19) as

$$QA^{{\varvec{\alpha }}}=\left( A^{{\varvec{\alpha }}}\right) ^\dagger Q,$$

hence $A^{{\varvec{\alpha }}}$ is quasi-Hermitian, with $G=Q$.

$(ii)\Rightarrow (i)$ :

Let $A^{{\varvec{\alpha }}}$ be quasi-Hermitian, with $G=Q$ then

$$\begin{aligned} A^{{\varvec{\alpha }}}=Q^{-1}\left( A^{{\varvec{\alpha }}}\right) ^\dagger Q= T^{-1}(T^\dagger )^{-1}A^{{\varvec{\alpha }}}T^\dagger T.\end{aligned}$$

(21)

Put $H_0:=TA^{{\varvec{\alpha }}}T^{-1}$. It is an easy computation to prove that

$$ D\left( TA^{{\varvec{\alpha }}}T^{-1}\right) =D\left( (T^\dagger )^{-1}\left( A^{{\varvec{\alpha }}}\right) ^\dagger T^\dagger \right) =\left\{ f \in \mathcal H; \sum _{n=1}^\infty |\alpha _n|^2 |\left\langle {f}\left| {e_n}\right. \right\rangle |^2 <\infty \right\} ,$$

and from (21) we have

$$\begin{aligned} TA^{{\varvec{\alpha }}}T^{-1}=(T^\dagger )^{-1}\left( A^{{\varvec{\alpha }}}\right) ^\dagger T^\dagger .\end{aligned}$$

(22)

Since $D\left( (T^\dagger )^{-1}\left( A^{{\varvec{\alpha }}}\right) ^\dagger T^\dagger \right) \subseteq D\left( \left( TA^{{\varvec{\alpha }}}T^{-1}\right) ^\dagger \right)$ we can conclude that $H_0$ is symmetric and its eigenvalues are $\{\alpha _n\}\subset \mathbb {R}$.

$(i)\Leftrightarrow (iii)$ :

is analogous to $(i)\Leftrightarrow (ii)$.

 

Other operators defined by a strict Riesz-like basis and its dual basis, more precisely lowering and raising operators, have been considered in [18] to factorize, under opportune hypotheses, the operators $A^{{\varvec{\alpha }}}$ and $B^{{\varvec{\alpha }}}$.