Orthogonal Frames in Krein Spaces ^†

Abstract

In this paper, we introduce the concept of orthogonal frames in Krein spaces, prove the independence of the choice of the fundamental symmetry, and from this, we obtain a number of interesting properties that they satisfy. We show that there is no distinction between orthogonal frames in a Krein space and orthogonal frames in its associated Hilbert. Furthermore, we characterize frames dual to a given frame, which is a useful tool for constructing examples.

1. Introduction

The theory of frames in Hilbert spaces was introduced in the year 1952 by Duffin and Schaeffer [1]; this means that this theory is relatively recent. Subsequently, in the year 1982, Daubechies, Grossman and Meyer [2] considered frames as overcomplete bases for finding the series expansions of functions in the Hilbert space of square integrable functions ${\mathbb{L}}_2\left(\mathbb{R}\right)$. In this sense, a frame is an overcomplete basis by virtue of the frame decomposition theorem (see [3]) which states that every vector of a Hilbert space can be written as a linear combination of the elements of the frame, and where the coefficients in the expansion are not necessarily unique, which is a highly applicable theory in signal processing, since even when one element of the frame is lost, it is possible to recover the information through the other elements. For more information see [3] (Section 8.5).

Krein spaces are a generalization of Hilbert spaces (see [4,5]); thus, it is natural to think about the extension of the frame theory in Hilbert spaces to Krein spaces. Such an extension was developed by K. Esmeral, O. Ferrer and E. Wagner in the year 2015 [6], who established the notion of discrete frames in Krein spaces, whose existence does not depend on the fundamental decomposition; thus. interesting properties were obtained, largely because the inner product is indefinite. Krein spaces find many applications in physics, since these had its appearance in quantum field theory. It was Dirac [7] who, in 1942, introduced the subject of indefinite inner products in quantum field theory; a year later, Pauli [8] took up Dirac’s ideas, and used these inner products for the quantization of fields, but the first mathematical treatment of a space with indefinite inner product was given by Pontrjagin [9], who was unaware of Dirac and Pauli’s investigations. Herein lies the importance of working in these spaces.

As for orthogonal frames in Hilbert spaces, these have been extensively developed by Bhatt [10,11,12], and the present paper is inspired by these works. Now, due to the fundamental decomposition of a Krein space, as an orthogonal direct sum of a positive definite space and a negative definite space—which turn out to be Hilbert spaces—given a pair of orthogonal frames in the first one and a pair of orthogonal frames in the second one, according to the definition given in [13], it is satisfied that the union of these pairs of frames are orthogonal in the Hilbert space associated to the Krein, as can be seen in Theorem 3, which motivated the study of these objects in Krein spaces.

This manuscript is divided into the following sections. Section 2 corresponds to the preliminaries, where the definitions and results necessary to develop the theory of orthogonal frames in Krein spaces are presented. In Section 3, the results obtained are presented, as well as some examples that complement the theory. Finally, Section 4 contains the conclusions of the work and possible future work and ways forward.

2. Preliminaries

The following definitions, propositions, theorems and comments are necessary to understand the results of this article. In the following, $\mathbb{K}$ denotes the field of real numbers or the field of complex numbers.

Definition 1

([4,5]). Let $\mathfrak{F}$ be a vector space over the field $\mathbb{K}$. An inner product in $\mathfrak{F}$ is a function

$$[·,·]:\mathfrak{F}\times \mathfrak{F}\to \mathbb{K}$$

which satisfies the following properties:

(i) 

$[\alpha x,y]=\alpha [x,y]$ for all $x,y\in \mathfrak{F}$ and every $\alpha \in \mathbb{K}$;

(ii) 

$[x+y,z]=[x,z]+[y,z]$ for all $x,y,z\in \mathfrak{F}$;

(iii) 

$[x,y]=\overline{[y,x]}$ for all $x,y\in \mathfrak{F}.$

The space $(\mathfrak{F},[·,·\left]\right)$ is called an inner product space. Similarly, the space $(\mathfrak{F},-[·,·\left]\right)$ is also an inner product space and is known as the anti-space of $(\mathfrak{F},[·,·\left]\right)$. For the condition $\left(iii\right)$ above, it is clear that $[x,x]\in \mathbb{R}$ for all $x\in \mathfrak{F}$, so, by the law of trichotomy of real numbers, it is possible to give the following definition.

Definition 2

([4,5]). Let $(\mathfrak{F},[·,·\left]\right)$ be a space with inner product and $x\in \mathfrak{F}$. We shall say that:

(i) 

x is a positive vector if $[x,x]>0;$

(ii) 

x is a negative vector if $[x,x]<0;$

(iii) 

x is a neutral vector if $[x,x]=0.$

Definition 3

([4,5]). Let $\mathfrak{V}$ be a vector subspace of $\mathfrak{F}$. If $\mathfrak{V}$ has only positive vectors and the null vector, $\mathfrak{V}$ is said to be defined positive. Similarly, if $\mathfrak{V}$ has only negative vectors and the null vector, $\mathfrak{V}$ is said to be defined negative. On the other hand, if it has both positive and negative elements, $\mathfrak{V}$ is said to be a space with indefinite inner product.

Definition 4

([4,5]). It is said that two vectors $x,y\in \mathfrak{F}$ are orthogonal $(x\perp y)$ if $[x,y]=0$ and that two sets $\mathfrak{V},\mathfrak{W}\subseteq \mathfrak{F}$ are orthogonal $(\mathfrak{V}\perp \mathfrak{W})$ if $x\perp y$ for all $x\in \mathfrak{V},y\in \mathfrak{W}$; in particular, if $\mathfrak{V}$ is reduced to a single vector $x$ is simply written $x\perp \mathfrak{W}$. In addition, if $\mathfrak{V}$ is a subset of $\mathfrak{F}$, the orthogonal complement of $\mathfrak{V}$ is given by ${\mathfrak{V}}^{\perp }:=\{x\in \mathfrak{F}:x\perp \mathfrak{V}\}$ in such a way that $\mathfrak{V}\subseteq {\mathfrak{V}}^{\perp \perp }$ and ${\mathfrak{V}}^{\perp }={\mathfrak{V}}^{\perp \perp \perp }$.

Definition 5

([4,5]). Let $\mathfrak{V},{\mathfrak{V}}^{\prime }$ be subspaces of $\mathfrak{F}$ such that $\mathfrak{V}\cap {\mathfrak{V}}^{\prime }=\left\{0\right\}$. The direct sum of $\mathfrak{V}$ and ${\mathfrak{V}}^{\prime }$ is denoted $\mathfrak{V}\left[\dot{+}\right]{\mathfrak{V}}^{\prime }$. In addition, if $\mathfrak{V}\perp {\mathfrak{V}}^{\prime }$ then it is called orthogonal direct sum and we write $\mathfrak{V}\oplus {\mathfrak{V}}^{\prime }$.

Definition 6

([4,5] Fundamental Decomposition). Let $(\mathfrak{F},[·,·\left]\right)$ be a space with inner product. We say that $\mathfrak{F}$ admits a fundamental decomposition if subspaces exist ${\mathfrak{F}}^0\subset \mathfrak{F}$, ${\mathfrak{F}}^{+}\subset \mathfrak{F}$ and ${\mathfrak{F}}^{-}\subset \mathfrak{F}$ such that $\mathfrak{F}={\mathfrak{F}}^0\oplus {\mathfrak{F}}^{+}\oplus {\mathfrak{F}}^{-}$, where $({\mathfrak{F}}^0,[·,·])$ is a neutral space, $({\mathfrak{F}}^{+},[·,·])$ is positive definite, and $({\mathfrak{F}}^{-},[·,·])$ is defined negative. In this case, we call $\mathfrak{F}={\mathfrak{F}}^0\oplus {\mathfrak{F}}^{+}\oplus {\mathfrak{F}}^{-}$ a fundamental decomposition.

The subspace ${\mathfrak{V}}^0:=\mathfrak{V}\cap {\mathfrak{V}}^{\perp }$ is called the isotropic part of $\mathfrak{V}$ and its non-zero elements are known as isotropic vectors. If ${\mathfrak{V}}^0=\left\{0\right\}$, it is said that $\mathfrak{V}$ is a non-degenerate subspace; otherwise, it is called a degenerate subspace.

Definition 7

([4,5]). A Krein space is a space with non-degenerate inner product $(\mathfrak{K},[·,·\left]\right)$ which admits a fundamental decomposition $\mathfrak{K}={\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}$ with $({\mathfrak{K}}^{+},[·,·])$ and $({\mathfrak{K}}^{-},-[·,·])$ Hilbert spaces.

Definition 8

([4,5]). Let $(\mathfrak{K},[·,·\left]\right)$ be a Krein space with fundamental decomposition $\mathfrak{K}={\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-},$ then we know that there are unique operators

$${\mathfrak{P}}^{+}:(\mathfrak{K},[·,·])⟶({\mathfrak{K}}^{+},[·,·]),{\mathfrak{P}}^{-}:(\mathfrak{K},[·,·])⟶({\mathfrak{K}}^{-},-[·,·]),$$

in the following way ${\mathfrak{P}}^{+}\left(k\right)=k^{+}$ and ${\mathfrak{P}}^{-}\left(k\right)=k^{-}$ for all $k\in \mathfrak{K}$ where $k^{+}\in {\mathfrak{K}}^{+}$, $k^{-}\in {\mathfrak{K}}^{-}$ and $k=k^{+}+k^{-}.$ To operators, ${\mathfrak{P}}^{+}$ and ${\mathfrak{P}}^{-}$ are known as fundamental projectors. The operator $\mathfrak{J}:(\mathfrak{K},[·,·\left]\right)⟶(\mathfrak{K},[·,·\left]\right)$ defined by $\mathfrak{J}:={\mathfrak{P}}^{+}-{\mathfrak{P}}^{-},$ that is, for all $k\in \mathfrak{K},$

$$\mathfrak{J}k={\mathfrak{P}}^{+}k-{\mathfrak{P}}^{-}k=k^{+}-k^{-},$$

is called the fundamental symmetry of Krein space $\mathfrak{K}$ associated to the fundamental decomposition. From now on, we will write $(\mathfrak{K},[·,·],\mathfrak{J})$ to denote Krein space $(\mathfrak{K},[·,·\left]\right)$ with fundamental symmetry $\mathfrak{J}$ associated with the fundamental decomposition $\mathfrak{K}={\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}.$

Proposition 1

([4,5]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space, then $\mathfrak{J}:(\mathfrak{K},[·,·\left]\right)⟶(\mathfrak{K},[·,·\left]\right)$ is a symmetric operator, $\mathfrak{J}$-isometric, self-adjoint and invertible with $\mathfrak{J}={\mathfrak{J}}^{-1}.$

Proposition 2

([4,5]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space. We define the function ${[·,·]}_{\mathfrak{J}}:\mathfrak{K}\times \mathfrak{K}⟶\mathbb{C}$ by means of the rule

$${[k_1,k_2]}_{\mathfrak{J}}=[\mathfrak{J}{k}_1,k_2],forall{k}_1,k_2\in \mathfrak{K},$$

then ${[·,·]}_{\mathfrak{J}}$ is a positive definite inner product, known as $\mathfrak{J}$-inner product.

Definition 9

([4,5]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space. The fundamental symmetry $\mathfrak{J}$ induces a norm in $\mathfrak{K}$ defined by

$${\parallel k\parallel }_{\mathfrak{J}}:={\sqrt{[k,k]}}_{\mathfrak{J}},forallk\in \mathfrak{K}.$$

This norm is known as the $\mathfrak{J}-$ norm of $\mathfrak{K}.$ In a more explicit form,

$${\parallel k\parallel }_{\mathfrak{J}}:={([k^{+},k^{+}]-[k^{-},k^{-}])}^{\frac{1}{2}},$$

for all $k\in \mathfrak{K}.$

Theorem 1

([4,5]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space. Then $(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ is a Hilbert space, known as the Hilbert space associated to the Krein space $(\mathfrak{K},[·,·],\mathfrak{J}).$

Example 1.

Let ${\ell }_2\left(\mathbb{N}\right)$ be the vector space of summable square sequences over $\mathbb{C}$. Then, the usual inner product which gives Hilbert space structure to ${\ell }_2\left(\mathbb{N}\right)$ is defined by

$${\langle ·,·\rangle }_{{\ell }_2}:{\ell }_2\left(\mathbb{N}\right)\times {\ell }_2\left(\mathbb{N}\right)\to \mathbb{C},{〈\{{\alpha }_n{{\}}_{n\in \mathbb{N}},\{{\beta }_n\}}_{n\in \mathbb{N}}〉}_{{\ell }_2}:=\sum _{n\in \mathbb{N}}{\alpha }_n\overline{{\beta }_n},$$

for all $\{{\alpha }_n{{\}}_{n\in \mathbb{N}},\{{\beta }_n\}}_{n\in \mathbb{N}}\in {\ell }_2\left(\mathbb{N}\right).$ However, to ${\ell }_2\left(\mathbb{N}\right)$, we can also see it as a Krein space with an inner product whose $\mathfrak{J}$-inner product coincides with the usual one. In this sense, we define the following mapping,

$${[·,·]}_{{\ell }_2}:{\ell }_2\left(\mathbb{N}\right)\times {\ell }_2\left(\mathbb{N}\right)\to \mathbb{C},{\left[\{{\alpha }_n{{\}}_{n\in \mathbb{N}},\{{\beta }_n\}}_{n\in \mathbb{N}}\right]}_{{\ell }_2}:=\sum _{n\in \mathbb{N}}{(-1)}^n{\alpha }_n\overline{{\beta }_n},$$

for all $\{{\alpha }_n{{\}}_{n\in \mathbb{N}},\{{\beta }_n\}}_{n\in \mathbb{N}}\in {\ell }_2\left(\mathbb{N}\right).$ Thus, if $\{e_n{\}}_{n\in \mathbb{N}}$ is the canonical orthonormal basis of ${\ell }_2\left(\mathbb{N}\right)$ then ${\ell }_2\left(\mathbb{N}\right)$ accepts the following fundamental decomposition:

$${\ell }_2\left(\mathbb{N}\right)={\ell }_2^{+}\left(\mathbb{N}\right)\oplus {\ell }_2^{-}\left(\mathbb{N}\right),$$

where ${\ell }_2^{+}\left(\mathbb{N}\right)=\overline{span}\{e_{2n}:n\in \mathbb{N}\}$ and ${\ell }_2^{-}\left(\mathbb{N}\right)=\overline{span}\{e_{2n+1}:n\in \mathbb{N}\}$ with associated fundamental symmetry:

$${\mathfrak{J}}_{{\ell }_2}:\left({\ell }_2\left(\mathbb{N}\right),{[·,·]}_{{\ell }_2}\right)\to \left({\ell }_2\left(\mathbb{N}\right),{[·,·]}_{{\ell }_2}\right)$$

given by ${\mathfrak{J}}_{{\ell }_2}{\left(\{{\alpha }_n\right\}}_{n\in \mathbb{N}}{)=\{{(-1)}^n{\alpha }_n\}}_{n\in \mathbb{N}}$ for all $\{{\alpha }_n{\}}_{n\in \mathbb{N}}\in {\ell }_2\left(\mathbb{N}\right).$ Therefore, ${[·,·]}_{{\mathfrak{J}}_{{\ell }_2}}={\langle ·,·\rangle }_{{\ell }_2}.$

From now on, whenever we see ${\ell }_2\left(\mathbb{N}\right)$ as a Krein space, we shall understand it to be endowed with a fundamental symmetry ${\mathfrak{J}}_{{\ell }_2}$ such that ${[·,·]}_{{\mathfrak{J}}_{{\ell }_2}}={\langle ·,·\rangle }_{{\ell }_2}$. An example of such symmetry is the one developed above, and a more trivial example is the symmetry given by the identity operator in ${\ell }_2\left(\mathbb{N}\right)$. Thus, we will write ${\Re }_2\left(\mathbb{N}\right)$ instead of ${\ell }_2\left(\mathbb{N}\right)$ when viewed as Krein space with such properties and the fundamental symmetry by ${\mathfrak{J}}_{{\Re }_2}$ to avoid confusion.

Definition 10

([4]). Let $({\mathfrak{K}}_1,{[·,·]}_1,{\mathfrak{J}}_1)$ and $({\mathfrak{K}}_2,{[·,·]}_2,{\mathfrak{J}}_2)$ be Krein spaces. The linear operator $T:({\mathfrak{K}}_1,{[·,·]}_1)⟶({\mathfrak{K}}_2,{[·,·]}_2)$ is said to be bounded if there is a real number $c>0$ such that for all $k\in {\mathfrak{K}}_1,$

$${\parallel Tk\parallel }_{{\mathfrak{J}}_2}\le c{\parallel k\parallel }_{{\mathfrak{J}}_1}.$$

In the following, $\mathcal{B}\left(\mathfrak{K}\right)$ denotes the space of linear and bounded operators in $(\mathfrak{K},[·,·\left]\right).$

Remark 1.

Given a bounded linear operator $T:({\mathfrak{K}}_1,{[·,·]}_1)⟶({\mathfrak{K}}_2,{[·,·]}_2),$ we can define the operator $T^{\circ }:({\mathfrak{K}}_1,{[·,·]}_{{\mathfrak{J}}_1})⟶({\mathfrak{K}}_2,{[·,·]}_{{\mathfrak{J}}_2})$, as $T^{\circ }\left(k_1\right):=T\left(k_1\right)$ for all $k_1$ in ${\mathfrak{K}}_1.$ Thus, note that $T^{\circ }$ is bounded linear and in essence different from the operator $T.$

Definition 11

([4]). Let $({\mathfrak{K}}_1,{[·,·]}_1,{\mathfrak{J}}_1)$ and $({\mathfrak{K}}_2,{[·,·]}_2,{\mathfrak{J}}_2)$ be Krein spaces. The adjoint of the bounded linear operator $T:({\mathfrak{K}}_1,{[·,·]}_1)⟶({\mathfrak{K}}_2,{[·,·]}_2)$ is the only bounded linear operator $T^{*}:(Dom\left(T^{*}\right)\subset {\mathfrak{K}}_2,{[·,·]}_2)⟶({\mathfrak{K}}_1,,{[·,·]}_1)$ such that for all $k_1\in {\mathfrak{K}}_1$ and $k_2\in Dom\left(T^{*}\right),$

$${[T{k}_1,k_2]}_2={[k_1,T^{*}{k}_2]}_1.$$

Remark 2.

Given a Krein space $(\mathfrak{K},[·,·],\mathfrak{J})$, it is of great importance for what follows, to denote with ${\mathfrak{J}}^{\mathfrak{J}}$ the linear and bounded mapping defined from Krein space $(\mathfrak{K},[·,·\left]\right)$ to the associated Hilbert space $(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ as ${\mathfrak{J}}^{\mathfrak{J}}\left(k\right):=\mathfrak{J}\left(k\right)$ for all $k\in \mathfrak{K}$. We also define the linear and bounded mappings: ${\mathfrak{id}}_{\mathfrak{J}}:(\mathfrak{K},{[·,·]}_{\mathfrak{J}})\to (\mathfrak{K},[·,·]),$${\mathfrak{id}}_{\mathfrak{J}}\left(k\right):={\mathfrak{id}}_{\mathfrak{K}}\left(k\right)=k,$ for all $k\in \mathcal{K},$${\mathfrak{id}}^{\mathfrak{J}}:(\mathfrak{K},[·,·])\to (\mathfrak{K},{[·,·]}_{\mathfrak{J}}),$${\mathfrak{id}}^{\mathfrak{J}}\left(k\right):={\mathfrak{id}}_{\mathfrak{K}}\left(k\right)=k,$ for all $k\in \mathfrak{K}$ and ${\mathfrak{J}}_{\mathfrak{J}}:(\mathfrak{K},{[·,·]}_{\mathfrak{J}})\to (\mathfrak{K},[·,·]),$${\mathfrak{J}}_{\mathfrak{J}}\left(k\right):=\mathfrak{J}\left(k\right),$ for all $k\in \mathfrak{K}.$ and adjoints are given by ${\left({\mathfrak{id}}_{\mathfrak{J}}\right)}^{*}={\mathfrak{J}}^{\mathfrak{J}}$ and ${\left({\mathfrak{id}}^{\mathfrak{J}}\right)}^{*}={\mathfrak{J}}_{\mathfrak{J}}.$

Proposition 3

([4,5]). Let $({\mathfrak{K}}_1,{[·,·]}_1,{\mathfrak{J}}_1)$, $({\mathfrak{K}}_2,{[·,·]}_2,{\mathfrak{J}}_2)$ be Krein spaces and consider $T:({\mathfrak{K}}_1,{[·,·]}_1)$$⟶({\mathfrak{K}}_2,{[·,·]}_2)$ a bounded linear operator, then $T^{*}={{\mathfrak{J}}_1}_{{\mathfrak{J}}_1}{\left(T^{\circ }\right)}^{*}{{\mathfrak{J}}_2}^{{\mathfrak{J}}_2}.$

Given that, by Theorem 1, $(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ is a Hilbert space, and since $\left({\mathfrak{K}}^{+},{[·,·]}_{\mathfrak{J}}\right)$ and $\left({\mathfrak{K}}^{-},{[·,·]}_{\mathfrak{J}}\right)$ are also, the following theorem is fully developed in the theory of frames in Hilbert spaces, and motivates the study of frames in Krein spaces. For a general overview of the theory of frames in Hilbert spaces, see [3,14,15,16,17].

Theorem 2

([18] (Proposition 3.7) Existence of Frames in Krein Spaces). Let $(\mathfrak{K}={\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-},[·,·],\mathfrak{J})$ be a Krein space. If ${\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}\subseteq {\mathfrak{K}}^{+}$ and ${\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}\subseteq {\mathfrak{K}}^{-}$ are frames for Hilbert spaces $\left({\mathfrak{K}}^{+},{[·,·]}_{\mathfrak{J}}\right)$ and $\left({\mathfrak{K}}^{-},{[·,·]}_{\mathfrak{J}}\right)$, respectively, then the sequence

$$\mathbb{X}:={\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}\cup {\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}\subseteq {\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-},$$

is a frame in Hilbert space $\left({\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}=\mathfrak{K},{[·,·]}_{\mathfrak{J}}\right)$.

Proof. 

As ${\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}\subseteq {\mathfrak{K}}^{+}$ and ${\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}\subseteq {\mathfrak{K}}^{-}$ are frames, then the pre-frame operators $T_{{\mathfrak{K}}^{+}}^{\circ }:\left({\ell }_2\left(\mathbb{N}\right),{〈·,·〉}_{{\ell }_2}\right)⟶\left({\mathfrak{K}}^{+},{[·,·]}_{\mathfrak{J}}\right)and{T}_{{\mathfrak{K}}^{-}}^{\circ }:\left({\ell }_2\left(\mathbb{N}\right),{〈·,·〉}_{{\ell }_2}\right)⟶\left({\mathfrak{K}}^{-},{[·,·]}_{\mathfrak{J}}\right)$, exist, are bounded and surjective. Furthermore, note that the pre-frame operator associated to the sequence $\mathbb{X}:={\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}\cup {\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}$ is given by

$$T_{\mathbb{X}}^{\circ }:\left({\ell }_2\left(\mathbb{N}\right),{〈·,·〉}_{{\ell }_2}\right)⟶\left({\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-},{[·,·]}_{\mathfrak{J}}\right),T_{\mathbb{X}}^{\circ }=T_{{\mathfrak{K}}^{+}}^{\circ }\oplus T_{{\mathfrak{K}}^{-}}^{\circ },$$

and therefore $T_{\mathbb{X}}^{\circ }$ is well-defined, bounded and surjective, since $T_{{\mathfrak{K}}^{+}}^{\circ }$ and $T_{{\mathfrak{K}}^{-}}^{\circ }$ have these properties. □

Definition 12

([6]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space. A sequence $\mathbb{X}={\left\{x_n\right\}}_{n\in \mathbb{N}}$ of elements of $\mathfrak{K}$ is a frame in $(\mathfrak{K},[·,·\left]\right)$ if there are constants $0<A\le B<+\infty$ such that

$$A{∥k∥}_{\mathfrak{J}}^2\le \sum _{n\in \mathbb{N}}{\left|[k,x_n]\right|}^2\le B{∥k∥}_{\mathfrak{J}}^2\forall k\in \mathfrak{K}.$$

(1)

The constants A and B are called bounds of the frame.

Definition 13

([6]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and the sequence $\mathbb{X}=\{x_n{\}}_{n\in \mathbb{N}}$ a frame in $(\mathfrak{K},[·,·\left]\right)$. It is said that $\mathbb{Y}=\{y_n{\}}_{n\in \mathbb{N}}$ is a dual frame to $\mathbb{X}$ if and only if

$$\sum _{n\in \mathbb{N}}[k,y_n]x_n=k,forallk\in \mathfrak{K}.$$

Proposition 4

([6,18]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space if $\mathbb{X}$ is a frame in $(\mathfrak{K},[·,·\left]\right)$. Then, the operator ${\mathcal{T}}_{\mathbb{X}}:({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{\Re })\to (\mathfrak{K},[·,·])$, ${\mathcal{T}}_{\mathbb{X}}\left({\left\{c_n\right\}}_{n\in \mathbb{N}}\right)={\sum }_{n\in \mathbb{N}}{c}_n{x}_n$ for all ${\left\{c_n\right\}}_{n\in \mathbb{N}}\in {\Re }_2\left(\mathbb{N}\right)$ is well defined and bounded. This operator is called pre-frame operator associated with $\mathbb{X}$.

Proof. 

It is sufficient to note that the following diagram commutes:

$$\begin{array}{ccc}({\Re }_2\left(\mathbb{N}\right),[·,·])& \stackrel{{\mathcal{T}}_{\mathbb{X}}}{\to }& (\mathfrak{K},[·,·\left]\right)\\ & & \\ {\mathfrak{id}}^{{\mathfrak{J}}_{{\Re }_2}}\downarrow & & \uparrow {\mathfrak{id}}_{\mathfrak{J}}\\ & & \\ ({\ell }_2\left(\mathbb{N}\right),{\langle ·,·\rangle }_{{\ell }_2})& \stackrel{{\mathcal{T}}_{\mathbb{X}}^{\circ }}{\to }& (\mathfrak{K},{[·,·]}_{\mathfrak{J}})\end{array}$$

where ${\mathcal{T}}_{\mathbb{X}}^{\circ }:({\ell }_2\left(\mathbb{N}\right),{\langle ·,·\rangle }_{{\ell }_2})\to (\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ is the pre-frame operator associated with $\mathbb{X}$ in $(\mathfrak{K},{[·,·]}_{\mathfrak{J}}).$ □

Proposition 5

([6,18]). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$ a frame in $(\mathfrak{K},[·,·\left]\right)$. Then, the adjoint operator of the pre-frame operator is given by

$${\mathcal{T}}_{\mathbb{X}}^{*}:(\mathfrak{K},[·,·])\to ({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{{\Re }_2}),{\mathcal{T}}_{\mathbb{X}}^{*}\left(k\right)={\mathfrak{J}}_{{\Re }_2}\left({\left\{[k,x_n]\right\}}_{n\in \mathbb{N}}\right),forallk\in \mathfrak{K},$$

and is known as the operator analysis associated with $\mathbb{X}.$

Definition 14

([18] (Definition 3.5)). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$ a frame in $(\mathfrak{K},[·,·\left]\right)$. The operator

$$\begin{array}{c}\hfill {\mathcal{S}}_{\mathbb{X}}:(\mathfrak{K},[·,·])\to (\mathfrak{K},[·,·]),\mathit{defined}\mathit{by}{\mathcal{S}}_{\mathbb{X}}={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*},\end{array}$$

is called frame operator associated with $\mathbb{X}.$

Remark 3.

Let k in $\mathfrak{K}$ any, then the frame operator is given by:

$$\begin{array}{ccc}\hfill {\mathcal{S}}_{\mathbb{X}}\left(k\right)& =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k\right)={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}\left({\mathfrak{J}}_{{\Re }_2}\left({\left\{[k,x_n]\right\}}_{n\in \mathbb{N}}\right)\right)\hfill \\ & =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}^2\left({\left\{[k,y_n]\right\}}_{n\in \mathbb{N}}\right)={\mathcal{T}}_{\mathbb{X}}\left({\left\{[k,x_n]\right\}}_{n\in \mathbb{N}}\right)\hfill \\ & =& {\displaystyle \sum _{n\in \mathbb{N}}[k,x_n]x_n.}\hfill \end{array}$$

In the following section, we present the main results of this work, extending the notion of orthogonal frames of Hilbert spaces to Krein spaces.

3. Orthogonal Frames in Krein Spaces

We then define an operator that will allow us to study orthogonal frames in Krein spaces.

Definition 15.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$, $\mathbb{Y}$ frames in $(\mathfrak{K},[·,·\left]\right)$. We define the operator

$${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}:(\mathfrak{K},[·,·])\to (\mathfrak{K},[·,·]),\mathit{by}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}:={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}.$$

The operator ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}$ can be seen as follows:

$$\begin{array}{cc}\hfill {\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\left(k\right)=& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}\left({\mathcal{T}}_{\mathbb{Y}}^{*}\left(k\right)\right)={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}\left({\mathfrak{J}}_{{\Re }_2}\left({\left\{[k,y_n]\right\}}_{n\in \mathbb{N}}\right)\right)\hfill \\ \hfill =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}^2\left({\left\{[k,y_n]\right\}}_{n\in \mathbb{N}}\right)={\mathcal{T}}_{\mathbb{X}}\left({\left\{[k,y_n]\right\}}_{n\in \mathbb{N}}\right)\hfill \\ \hfill =& {\displaystyle \sum _{n\in \mathbb{N}}[k,y_n]x_n,\forall k\in \mathfrak{K}.}\hfill \end{array}$$

Note that given two frames $\mathbb{X}$ and $\mathbb{Y}$ in a Krein space $\mathfrak{K}$, then it is clear that $\mathbb{Y}$ is dual to $\mathbb{X}$ if and only if the above operator satisfies that ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathfrak{id}}_{\mathfrak{K}}.$ In addition, note that if $\mathbb{X}=\mathbb{Y}$ then ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathcal{S}}_{\mathbb{X}},$ that is, the operator ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}$ turns out to be the frame operator associated with $\mathbb{X}.$

In a Hilbert space $\mathfrak{H}$, it is said that $\mathbb{X}$ is a frame orthogonal to $\mathbb{Y}$ if and only if ${\displaystyle ran\left({\left(T_{\mathbb{X}}^{\circ }\right)}^{*}\right)\perp ran\left({\left(T_{\mathbb{Y}}^{\circ }\right)}^{*}\right),}$ or equivalently, if and only if ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^0\equiv \mathbf{0}$, as can be seen in [10,11,12,13]. Now, the following theorem allows us to think about orthogonal frames in Krein spaces, attending to the above definition for the associated Hilbert space.

Theorem 3

(Existence of Orthogonal Frames in Krein Spaces). Let $\left(\mathfrak{K}={\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-},[·,·],\mathfrak{J}\right)$ be a Krein space. If ${\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}$ is a frame orthogonal to ${\left\{y_n^{+}\right\}}_{n\in \mathbb{N}}$ in Hilbert space $\left({\mathfrak{K}}^{+},{[·,·]}_{\mathfrak{J}}\right)$ and ${\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}$ is a frame orthogonal to ${\left\{y_n^{-}\right\}}_{n\in \mathbb{N}}$ in Hilbert space $\left({\mathfrak{K}}^{-},{[·,·]}_{\mathfrak{J}}\right)$, then the frame $\mathbb{X}:={\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}\cup {\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}$ is orthogonal to the frame $\mathbb{Y}:={\left\{y_n^{+}\right\}}_{n\in \mathbb{N}}\cup {\left\{y_n^{-}\right\}}_{n\in \mathbb{N}}$ in Hilbert space ${\displaystyle \left({\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}=\mathfrak{K},{[·,·]}_{\mathfrak{J}}\right)}$.

Proof. 

Let ${\mathbb{X}}^{+}:={\left\{x_n^{+}\right\}}_{n\in \mathbb{N}}$, ${\mathbb{Y}}^{+}:={\left\{y_n^{+}\right\}}_{n\in \mathbb{N}}$, ${\mathbb{X}}^{-}:={\left\{x_n^{-}\right\}}_{n\in \mathbb{N}}$ and ${\mathbb{Y}}^{-}:={\left\{y_n^{-}\right\}}_{n\in \mathbb{N}}$. Then, by Theorem 2, it follows that $\mathbb{X}={\mathbb{X}}^{+}\cup {\mathbb{X}}^{-}$ and $\mathbb{Y}={\mathbb{Y}}^{+}\cup {\mathbb{Y}}^{-}$ are frames in ${\displaystyle \left({\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}=\mathfrak{K},{[·,·]}_{\mathfrak{J}}\right)}$. In addition, by the hypothesis, it is satisfied that

$${\mathfrak{W}}_{{\mathbb{X}}^{+},{\mathbb{Y}}^{+}}^{\circ }:={\mathcal{T}}_{{\mathbb{X}}^{+}}^{\circ }{\left({\mathcal{T}}_{{\mathbb{Y}}^{+}}^{\circ }\right)}^{*}\equiv \mathbf{0},{\mathfrak{W}}_{{\mathbb{X}}^{-},{\mathbb{Y}}^{-}}^{\circ }:={\mathcal{T}}_{{\mathbb{X}}^{-}}^{\circ }{\left({\mathcal{T}}_{{\mathbb{Y}}^{-}}^{\circ }\right)}^{*}\equiv \mathbf{0}.$$

Thus, what we have to prove is ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }:={\mathcal{T}}_{\mathbb{X}}^{\circ }{\left(T_{\mathbb{Y}}^{\circ }\right)}^{*}\equiv \mathbf{0}.$ Indeed, note that, for any $k^{+}\oplus k^{-}\in {\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}$,

$$\begin{array}{ccc}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }\left(k^{+}\oplus k^{-}\right)\hfill & =\hfill & {\mathcal{T}}_{\mathbb{X}}^{\circ }{\left({\mathcal{T}}_{\mathbb{Y}}^{\circ }\right)}^{*}\left(k^{+}\oplus k^{-}\right)\hfill \\ \hfill & =\hfill & ({\mathcal{T}}_{{\mathbb{X}}^{+}}^{\circ }\oplus {\mathcal{T}}_{{\mathbb{X}}^{-}}^{\circ })\left({({\mathcal{T}}_{{\mathbb{Y}}^{+}}^{\circ })}^{*}\oplus {({\mathcal{T}}_{{\mathbb{Y}}^{-}}^{\circ })}^{*}\right)\left(k^{+}\oplus k^{-}\right)\hfill \\ \hfill & =\hfill & ({\mathcal{T}}_{{\mathbb{X}}^{+}}^{\circ }\oplus {\mathcal{T}}_{{\mathbb{X}}^{-}}^{\circ })\left({({\mathcal{T}}_{{\mathbb{Y}}^{+}}^{\circ })}^{*}\left(k^{+}\right)\oplus {({\mathcal{T}}_{{\mathbb{Y}}^{-}}^{\circ })}^{*}\left(k^{-}\right)\right)\hfill \\ \hfill & =\hfill & {\mathcal{T}}_{{\mathbb{X}}^{+}}^{\circ }\left({({\mathcal{T}}_{{\mathbb{Y}}^{+}}^{\circ })}^{*}\left(k^{+}\right)\right)+{\mathcal{T}}_{{\mathbb{X}}^{-}}^{\circ }\left({({\mathcal{T}}_{{\mathbb{Y}}^{-}}^{\circ })}^{*}\left(k^{-}\right)\right)\hfill \\ \hfill & =\hfill & {\mathfrak{W}}_{{\mathbb{X}}^{+},{\mathbb{Y}}^{+}}^{\circ }\left(k^{+}\right)+{\mathfrak{W}}_{{\mathbb{X}}^{-},{\mathbb{Y}}^{-}}^{\circ }\left(k^{-}\right)\hfill \\ & =\hfill & 0.\hfill \end{array}$$

Therefore, ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }\equiv \mathbf{0}$. Then, $\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in $({\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}=\mathfrak{K},{[·,·]}_{\mathfrak{J}})$. □

Definition 16

(Orthogonal Frames in Krein Spaces). Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$, $\mathbb{Y}$ frames in $(\mathfrak{K},[·,·\left]\right)$. We say that $\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in $(\mathfrak{K},[·,·\left]\right)$ if it is satisfied that $ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\right)\perp ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)$, where orthogonality is with respect to ${[·,·]}_{{\Re }_2}$.

Proposition 6.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and ${\mathfrak{J}}_{{\Re }_2}$, ${\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}$ fundamental symmetries of ${\ell }_2\left(\mathbb{N}\right)$ such that ${[·,·]}_{{\Re }_2}={〈·,·〉}_{{\ell }_2}={[·,·]}_{{\Re }_2^{{}^{\prime }}}$. Then, $\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in the Krein space $(\mathfrak{K},[·,·\left]\right)$ with respect to ${[·,·]}_{{\Re }_2}$ if and only if $\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in the Krein space $(\mathfrak{K},[·,·\left]\right)$ with respect to ${[·,·]}_{{\Re }_2^{{}^{\prime }}}$.

Proof. 

$(\Rightarrow )$ Suppose that $\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in $(\mathfrak{K},[·,·\left]\right)$ and with respect to ${[·,·]}_{{\Re }_2}$, that is,

$$\begin{array}{c}\hfill ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\right)\perp ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right).\end{array}$$

Let us see what $ran({\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}{\mathcal{T}}_{\mathbb{X}}^{*})\perp ran({\mathcal{T}}_{\mathbb{Y}}^{*})$; in effect, let ${\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}\in ran({\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}{\mathcal{T}}_{\mathbb{X}}^{*})$ and ${\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}\in ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)$ be any, then ${\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}$ and ${\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}$ are of the form

$$\begin{array}{c}\hfill {\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}={\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right)y{\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}={\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_2\right),k_1,k_2\in \mathfrak{K}.\end{array}$$

Then,

$$\begin{array}{cc}\hfill {[{\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}},{\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}]}_{{\Re }_2}=& {[{\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_2\right)]}_{{\Re }_2}={\left[{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_2\right)\right]}_{{\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}}\hfill \\ \hfill =& {\left[{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_2\right)\right]}_{{\mathfrak{J}}_{{\Re }_2}}={\left[{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_2\right)\right]}_{{\Re }_2}=0.\hfill \end{array}$$

Therefore, $ran({\mathfrak{J}}_{{\Re }_2^{{}^{\prime }}}^{{}^{\prime }}{\mathcal{T}}_{\mathbb{X}}^{*})\perp ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)$ and with this $\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in $(\mathfrak{K},[·,·\left]\right)$ with respect to ${[·,·]}_{{\Re }_2^{{}^{\prime }}}$.

$(\Leftarrow )$ Similarly to the previous test. □

Proposition 7.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X},\mathbb{Y}$ frames in $(\mathfrak{K},[·,·\left]\right),$ then the following statements are equivalent

(i) 

$\mathbb{X}$ is orthogonal to $\mathbb{Y}$ in $(\mathfrak{K},[·,·\left]\right)$;

(ii) 

$\mathbb{Y}$ is orthogonal to $\mathbb{X}$ in $(\mathfrak{K},[·,·\left]\right)$;

(iii) 

$W_{\mathbb{X},\mathbb{Y}}\equiv \mathbf{0}$;

(iv) 

$W_{\mathbb{Y},\mathbb{X}}\equiv \mathbf{0}$.

Proof. 

$\left(i\right)\Rightarrow \left(ii\right)$ Suppose that $\mathbb{X}$ is orthogonal to $\mathbb{Y}$, that is, $ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\right)\perp ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)$ with respect to ${[·,·]}_{{\Re }_2}$. Let us see what $\mathbb{Y}$ is orthogonal to $\mathbb{X}$. Let ${\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}\in ran\left({\mathcal{T}}_{\mathbb{X}}^{*}\right)$ and ${\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}\in ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\right)$ be any. Then, there are vectors $k_1,k_2\in \mathfrak{K}$ such that ${\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}={\mathcal{T}}_{\mathbb{X}}^{*}\left(k_2\right)$ and ${\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}={\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_1\right),$

$$\begin{array}{cc}\hfill {[{\left\{{\beta }_n\right\}}_{n\in \mathbb{N}},{\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}]}_{{\Re }_2}=& {[{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_1\right),{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_2\right)]}_{{\Re }_2}\hfill \\ \hfill =& {[{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_1\right),{\mathfrak{J}}_{{\Re }_2}^{*}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_2\right)]}_{{\Re }_2}\hfill \\ \hfill =& [\underset{\in ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)}{\underbrace{{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_1\right)}},\underset{\in ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\right)}{\underbrace{{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_2\right)}}{]}_{\Re 2}\hfill \\ \hfill =& 0\hfill \end{array}$$

Therefore, $ran\left({\mathfrak{J}}_{{\Re }_2}{T}_{\mathbb{Y}}^{*}\right)\perp ran\left({\mathcal{T}}_{\mathbb{X}}^{*}\right)$ with respect to ${[·,·]}_{{\Re }_2}$. We conclude that $\mathbb{Y}$ is orthogonal to $\mathbb{X}$.

$\left(ii\right)\Rightarrow \left(iii\right)$ Suppose that $ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\right)\perp ran\left({\mathcal{T}}_{\mathbb{X}}^{*}\right)$ of inner product ${[·,·]}_{{\Re }_2}$. Let us see what ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\equiv \mathbf{0}.$ Indeed, let $k_1,k_2\in \mathfrak{K}$ be any

$$\begin{array}{cc}\hfill [{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\left(k_1\right),k_2]=& [{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_1\right),k_2]\hfill \\ \hfill =& [\underset{\in ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\right)}{\underbrace{{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_1\right)}},\underset{\in ran\left({\mathcal{T}}_{\mathbb{X}}^{*}\right)}{\underbrace{{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_2\right)}}{]}_{\Re 2}\hfill \\ \hfill =& 0\hfill \end{array}$$

$\left(iii\right)\Rightarrow \left(iv\right)$ $\mathbf{0}\equiv {\mathbf{0}}^{*}={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{*}={\left({\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\right)}^{*}={\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)}^{*}{\mathfrak{J}}_{{\Re }_2}^{*}{\mathcal{T}}_{\mathbb{X}}^{*}={\mathcal{T}}_{\mathbb{Y}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}={\mathfrak{W}}_{\mathbb{Y},\mathbb{X}}.$

$\left(iv\right)\Rightarrow \left(i\right)$ Suppose that ${\mathfrak{W}}_{\mathbb{Y},\mathbb{X}}\equiv \mathbf{0}$. Let us see what $\mathbb{X}$ is orthogonal to $\mathbb{Y}$, that is, $ran\left({\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\right)\perp ran\left({\mathcal{T}}_{\mathbb{Y}}^{*}\right)$ with respect to ${[·,·]}_{{\Re }_2}$. Let ${\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}={\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),{\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}={\mathcal{T}}_{\mathbb{Y}}^{[*]}\left(k_2\right)$ for some $k_1,k_2\in \mathfrak{K}$. Let us note that

$$\begin{array}{ccc}\hfill {[{\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}},{\left\{{\beta }_n\right\}}_{n\in \mathbb{N}}]}_{{\Re }_2}& =& {[{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),{\mathcal{T}}_{\mathbb{Y}}^{*}\left(k_2\right)]}_{{\Re }_2}\hfill \\ & =& [{\mathcal{T}}_{\mathbb{Y}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(k_1\right),k_2]\hfill \\ & =& [{\mathfrak{W}}_{\mathbb{Y},\mathbb{X}}\left(k_1\right),k_2]\hfill \\ & =& [0,k_2]\hfill \\ & =& 0.\hfill \end{array}$$

Therefore, $\mathbb{X}$ is orthogonal to $\mathbb{Y}.$ □

Remark 4.

If $(\mathfrak{K},[·,·],\mathfrak{J})$ is a Krein space and $\mathbb{X}$, $\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$, we know that the operator ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }:(\mathfrak{K},{[·,·]}_{\mathfrak{J}})⟶(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ defined by ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }:={\mathcal{T}}_{\mathbb{X}}^{\circ }{\left({\mathcal{T}}_{\mathbb{Y}}^{\circ }\right)}^{*}$ turns out to be the operator $\mathbf{0}$. Thus, it is useful to ask about the relationship between ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }$ and ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}$ where ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}:(\mathfrak{K},[·,·])⟶(\mathfrak{K},[·,·])$ is given by ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}:={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}$, which is the same for the relationships between operators ${\mathfrak{W}}_{\mathbb{X},\mathfrak{J}\mathbb{Y}}$, ${\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathbb{Y}}$ and ${\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathfrak{J}\mathbb{Y}}$. The following result establishes some relationships between these operators.

Proposition 8.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$, $\mathbb{Y}$ frames in $(\mathfrak{K},[·,·\left]\right)$. Then, they are equivalent:

(i) 

$\mathbb{X}$, $\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$;

(ii) 

$\mathbb{X}$, $\mathfrak{J}\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$;

(iii) 

$\mathfrak{J}\mathbb{X}$, $\mathfrak{J}\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$;

(iv) 

$\mathfrak{J}\mathbb{X}$, $\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$.

Proof. 

Let $\mathbb{X}$, $\mathbb{Y}$ be frames in $(\mathfrak{K},[·,·\left]\right)$; then, it is clear that $\mathfrak{J}\mathbb{X}$, $\mathfrak{J}\mathbb{Y}$ are also frames in $(\mathfrak{K},[·,·\left]\right)$.

$\left(i\right)\Rightarrow \left(ii\right)$ Suppose that $\mathbb{X}$, $\mathbb{Y}$ are orthogonal in $(\mathfrak{K},[·,·\left]\right)$, i.e., ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\equiv \mathbf{0}$. However,

$$\begin{array}{ccc}\hfill {\mathfrak{W}}_{\mathbb{X},\mathfrak{J}\mathbb{Y}}& =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathfrak{J}\mathbb{Y}}^{*}={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\left(\mathfrak{J}{\mathcal{T}}_{\mathbb{Y}}\right)}^{*}={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{\mathfrak{J}}^{*}={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\mathfrak{J}={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\mathfrak{J}=\mathbf{0}\mathfrak{J}\equiv \mathbf{0}.\hfill \end{array}$$

$\left(ii\right)\Rightarrow \left(iii\right)$ Suppose that $\mathbb{X}$, $\mathfrak{J}\mathbb{Y}$ are orthogonal in $(\mathfrak{K},[·,·\left]\right)$. We want to prove that ${\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathfrak{J}\mathbb{Y}}\equiv \mathbf{0}$. In effect,

$$\begin{array}{c}\hfill {\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathfrak{J}\mathbb{Y}}={\mathcal{T}}_{\mathfrak{J}\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathfrak{J}\mathbb{Y}}^{*}=\mathfrak{J}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathfrak{J}\mathbb{Y}}^{*}=\mathfrak{J}{\mathfrak{W}}_{\mathbb{X},\mathfrak{J}\mathbb{Y}}=\mathfrak{J}\mathbf{0}\equiv \mathbf{0}.\end{array}$$

$\left(iii\right)\Rightarrow \left(iv\right)$ Suppose that ${\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathfrak{J}\mathbb{Y}}\equiv \mathbf{0}$. Let us see that ${\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathbb{Y}}\equiv \mathbf{0}$. In effect,

$$\begin{array}{c}\hfill {\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathbb{Y}}={\mathcal{T}}_{\mathfrak{J}\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}=\mathfrak{J}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}=\mathfrak{J}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{\mathfrak{id}}_{\mathfrak{K}}=\mathfrak{J}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\mathfrak{J}\mathfrak{J}={\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathfrak{J}\mathbb{Y}}\mathfrak{J}=\mathbf{0}\mathfrak{J}\equiv \mathbf{0}.\end{array}$$

$\left(iv\right)\Rightarrow \left(i\right)$ Suppose that ${\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathbb{Y}}\equiv \mathbf{0}$. Then, note that

$$\begin{array}{c}\hfill {\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{T}_{\mathbb{Y}}^{*}={\mathfrak{id}}_{\mathfrak{K}}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}=\mathfrak{J}\mathfrak{J}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}=\mathfrak{J}{\mathcal{T}}_{\mathfrak{J}\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}=\mathfrak{J}{\mathfrak{W}}_{\mathfrak{J}\mathbb{X},\mathbb{Y}}=\mathfrak{J}\mathbf{0}\equiv \mathbf{0}.\end{array}$$

□

Theorem 4.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$, $\mathbb{Y}$ frames in $(\mathfrak{K},[·,·\left]\right)$. Then, $\mathbb{X}$, $\mathbb{Y}$ are orthogonal in $(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ if and only if $\mathbb{X}$, $\mathbb{Y}$ are orthogonal in $(\mathfrak{K},[·,·\left]\right)$.

Proof. 

$(\Rightarrow )$ Suppose that $\mathbb{X}$, $\mathbb{Y}$ are orthogonal in $(\mathfrak{K},{[·,·]}_{\mathfrak{J}})$, i.e., ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }\equiv \mathbf{0}$. Now, note that the following diagram commutes:

$$\begin{array}{ccc}(\mathfrak{K},[·,·\left]\right)& \stackrel{{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}}{\to }& (\mathfrak{K},[·,·\left]\right)\\ & & \\ {\mathfrak{J}}^{\mathfrak{J}}\downarrow & & \uparrow {\mathfrak{id}}_{\mathfrak{J}}\\ & & \\ (\mathfrak{K},{[·,·]}_{\mathfrak{J}})& \stackrel{{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }}{\to }& (\mathfrak{K},{[·,·]}_{\mathfrak{J}})\end{array}$$

in effect, let $k\in \mathfrak{K}$ be anyone, then

$$\begin{array}{ccc}{\mathfrak{id}}_{\mathfrak{J}}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }{\mathfrak{J}}^{\mathfrak{J}}\left(k\right)\hfill & =\hfill & {\mathfrak{id}}_{\mathfrak{J}}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }\mathfrak{J}\left(k\right)={\mathfrak{id}}_{\mathfrak{J}}\left(\sum _{n\in \mathbb{N}}{[\mathfrak{J}\left(k\right),y_n]}_{\mathfrak{J}}{x}_n\right)\hfill \\ \hfill & =\hfill & {\mathfrak{id}}_{\mathfrak{K}}\left(\sum _{n\in \mathbb{N}}[{\mathfrak{J}}^2\left(k\right),y_n]x_n\right)=\sum _{n\in \mathbb{N}}[k,y_n]x_n={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}.\hfill \end{array}$$

Therefore, ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}=i{d}_{\mathfrak{J}}\mathbf{0}{\mathfrak{J}}^{\mathfrak{J}}\equiv \mathbf{0}$ and $\mathbb{X}$, $\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$.

$(\Leftarrow )$ Suppose that ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\equiv \mathbf{0}$. Let us see what ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }\equiv \mathbf{0}$. For this note that the following diagram commutes:

$$\begin{array}{ccc}(\mathfrak{K},{[·,·]}_{\mathfrak{J}})& \stackrel{{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }}{\to }& (\mathfrak{K},{[·,·]}_{\mathfrak{J}})\\ & & \\ {\mathfrak{J}}_{\mathfrak{J}}\downarrow & & \uparrow {\mathfrak{id}}^{\mathfrak{J}}\\ & & \\ (\mathfrak{K},[·,·\left]\right)& \stackrel{{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}}{\to }& (\mathfrak{K},[·,·\left]\right)\end{array}$$

Indeed, for any $k\in \mathfrak{K}$,

$${\mathfrak{id}}^{\mathfrak{J}}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}{\mathfrak{J}}_{\mathfrak{J}}\left(k\right)={\mathfrak{id}}^{\mathfrak{J}}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\mathfrak{J}\left(k\right)={\mathfrak{id}}^{\mathfrak{J}}\left(\sum _{n\in \mathbb{N}}[\mathfrak{J}\left(k\right),y_n]x_n\right)={\mathfrak{id}}_{\mathfrak{K}}\left(\sum _{n\in \mathbb{N}}{[k,y_n]}_{\mathfrak{J}}{x}_n\right)=\sum _{n\in \mathbb{N}}{[k,y_n]}_{\mathfrak{J}}{x}_n={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}$$

Thus, ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{\circ }={\mathfrak{id}}^{\mathfrak{J}}\mathbf{0}{\mathfrak{J}}_{\mathfrak{J}}\equiv \mathbf{0}$. This concludes the proof of the theorem. □

Remark 5.

Given a Krein space $(\mathfrak{K},[·,·],\mathfrak{J})$ and consider a bounded linear operator $\psi :(\mathfrak{K},[·,·\left]\right)\to (\mathfrak{K},[·,·\left]\right)$, then the following diagram commutes

$$\begin{array}{ccc}(\mathfrak{K},[·,·\left]\right)& \stackrel{\psi }{\to }& (\mathfrak{K},[·,·\left]\right)\\ & & \\ {\mathfrak{id}}^{\mathfrak{J}}\downarrow & & \uparrow {\mathfrak{id}}_{\mathfrak{J}}\\ & & \\ (\mathfrak{K},{[·,·]}_{\mathfrak{J}})& \stackrel{{\psi }^{\circ }}{\to }& (\mathfrak{K},{[·,·]}_{\mathfrak{J}})\end{array}$$

In fact, for any $k\in \mathfrak{K}$, we have that

$$\begin{array}{ccc}{\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }{\mathfrak{id}}^{\mathfrak{J}}\left(k\right)\hfill & =\hfill & {\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }{\mathfrak{id}}_{\mathfrak{K}}\left(k\right)={\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }\left(k\right)\hfill \\ \hfill & =\hfill & {\mathfrak{id}}_{\mathfrak{K}}\left(\psi \left(k\right)\right)=\psi \left(k\right).\hfill \end{array}$$

In addition, note that ψ is surjective if and only if ${\psi }^{\circ }$ es surjective.

Proposition 9.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\psi :(\mathfrak{K},[·,·\left]\right)\to (\mathfrak{K},[·,·\left]\right)$ be a bounded and surjective linear operator. Then, there exists ${\psi }^{†}:(\mathfrak{K},[·,·])\to (\mathfrak{K},[·,·])$ bounded linear such that $\psi {\psi }^{†}={\mathfrak{id}}_{\mathfrak{K}}$.

Proof. 

From the above observation, the operator ${\psi }^{\circ }:(\mathfrak{K},{[·,·]}_{\mathfrak{J}})\to (\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ is surjective and then there exists ${\left({\psi }^{\circ }\right)}^{†}:(\mathfrak{K},{[·,·]}_{\mathfrak{J}})\to (\mathfrak{K},{[·,·]}_{\mathfrak{J}})$ such that ${\psi }^{\circ }{\left({\psi }^{\circ }\right)}^{†}={\mathfrak{id}}_{\mathfrak{K}}^{\circ }$. Thus, if we consider ${\psi }^{†}:={\mathfrak{id}}_{\mathfrak{J}}{\left({\psi }^{\circ }\right)}^{†}{\mathfrak{id}}^{\mathfrak{J}}$, note that ${\psi }^{†}$ is a bounded linear operator of $(\mathfrak{K},[·,·\left]\right)$; also,

$$\begin{array}{ccc}\hfill \psi {\psi }^{†}& =& {\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }{\mathfrak{id}}^{\mathfrak{J}}{\mathfrak{id}}_{\mathfrak{J}}{\left({\psi }^{\circ }\right)}^{†}{\mathfrak{id}}^{\mathfrak{J}}={\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }{\mathfrak{id}}^{\mathfrak{J}}{\left({\mathfrak{id}}^{\mathfrak{J}}\right)}^{-1}{\left({\psi }^{\circ }\right)}^{†}{\mathfrak{id}}^{\mathfrak{J}}={\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }{\mathfrak{id}}_{\mathfrak{K}}^{\circ }{\left({\psi }^{\circ }\right)}^{†}{\mathfrak{id}}^{\mathfrak{J}}\hfill \\ & =& {\mathfrak{id}}_{\mathfrak{J}}{\psi }^{\circ }{\left({\psi }^{\circ }\right)}^{†}{\mathfrak{id}}^{\mathfrak{J}}={\mathfrak{id}}_{\mathfrak{J}}{\mathfrak{id}}_{\mathfrak{K}}^{\circ }{\mathfrak{id}}^{\mathfrak{J}}={\mathfrak{id}}_{\mathfrak{J}}{\mathfrak{id}}^{\mathfrak{J}}={\mathfrak{id}}_{\mathfrak{J}}{\left({\mathfrak{id}}_{\mathfrak{J}}\right)}^{-1}={\mathfrak{id}}_{\mathfrak{K}}.\hfill \end{array}$$

□

Proposition 10.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space, $\mathbb{X}$ a frame in $(\mathfrak{K},[·,·\left]\right)$ with bounds $B\ge A>0$ and $\psi \in \mathcal{B}\left(\mathfrak{K}\right)$. Then, $\psi \left(\mathbb{X}\right)$ is a frame in $(\mathfrak{K},[·,·\left]\right)$ if and only if ψ is surjective.

Proof. 

$(\Rightarrow )$ Suppose that $\psi \left(\mathbb{X}\right)$ is a frame in $(\mathfrak{K},[·,·\left]\right)$, then the frame operator ${\mathcal{S}}_{\psi \left(\mathbb{X}\right)}$ exists and is invertible. Furthermore,

$$\begin{array}{c}\hfill {\mathcal{S}}_{\psi \left(\mathbb{X}\right)}={\mathcal{T}}_{\psi \mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\psi \mathbb{X}}^{*}\end{array}$$

and for any ${\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}\in {\Re }_2\left(\mathbb{N}\right)$ is satisfied

$$\begin{array}{ccc}\hfill {\mathcal{T}}_{\psi \mathbb{X}}\left({\left\{{\alpha }_n\right\}}_{n\in \mathbb{N}}\right)& =& \sum _{n\in \mathbb{N}}{\alpha }_n\psi \left(x_n\right)=\underset{M\to \infty }{lim}\sum _{n=1}^M{\alpha }_n\psi \left(x_n\right)=\underset{M\to \infty }{lim}\psi \left(\sum _{n=1}^M{\alpha }_n{x}_n\right)\hfill \\ & =& \psi \left(\underset{M\to \infty }{lim}\sum _{n=1}^M{\alpha }_n{x}_n\right)=\psi \left(\sum _{n\in \mathbb{N}}{\alpha }_n{x}_n\right)=\psi {\mathcal{T}}_{\mathbb{X}}.\hfill \end{array}$$

Then,

$$\begin{array}{c}\hfill {\mathcal{S}}_{\psi \left(\mathbb{X}\right)}=\psi {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\left(\psi {\mathcal{T}}_{\mathbb{X}}\right)}^{*}=\psi {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}\left({\mathcal{T}}_{\mathbb{X}}^{*}{\psi }^{*}\right)=\psi \left({\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\right){\psi }^{*}=\psi {\mathcal{S}}_{\mathbb{X}}{\psi }^{*}\end{array}$$

and then $\psi$ is surjective.

$(\Leftarrow )$ Suppose that $\psi$ is surjective. Let us see that $\psi \left(\mathbb{X}\right)$ is frame in $(\mathfrak{K},[·,·\left]\right)$. Since $\psi$ is surjective then

$$\begin{array}{c}\hfill \psi {\psi }^{†}={\mathfrak{id}}_{\mathfrak{K}}\mathrm{o}{\left({\psi }^{†}\right)}^{*}{\psi }^{*}={\mathfrak{id}}_{\mathfrak{K}}.\end{array}$$

Thus, note that for any $k\in \mathfrak{K}$,${\psi }^{*}\left(k\right)\in \mathfrak{K}$,

$$\begin{array}{ccc}\hfill A\parallel {\psi }^{†}{\parallel }^{-2}{\parallel k\parallel }_{\mathfrak{J}}^2& =& A\parallel {\psi }^{†}{\parallel }^{-2}\parallel {\left({\psi }^{†}\right)}^{*}{\psi }^{*}\left(k\right){\parallel }_{\mathfrak{J}}^2\le A{\parallel {\psi }^{†}\parallel }^{-2}{\left(\parallel {\left({\psi }^{†}\right)}^{*}\parallel \parallel {\psi }^{*}\left(k\right){\parallel }_{\mathfrak{J}}^2\right)}^2\hfill \\ & =& A\parallel {\psi }^{†}{\parallel }^{-2}\parallel {\psi }^{†}{\parallel }^2\parallel {\psi }^{*}{\left(k\right)\parallel }_{\mathfrak{J}}^2=A{\parallel {\psi }^{*}\left(k\right)\parallel }_{\mathfrak{J}}^2\le \sum _{n\in \mathbb{N}}{\left|[{\psi }^{*}\left(k\right),x_n]\right|}^2\hfill \\ & =& \sum _{n\in \mathbb{N}}{\left|[k,\psi \left(x_n\right)]\right|}^2\le B\parallel {\psi }^{*}{\left(k\right)\parallel }_{\mathfrak{J}}^2\le B\parallel {\psi }^{*}{\parallel }^2{\parallel k\parallel }_{\mathfrak{J}}^2={B\parallel \psi \parallel }^2{\parallel k\parallel }_{\mathfrak{J}}^2.\hfill \end{array}$$

Therefore, ${\displaystyle A\parallel {\psi }^{†}{\parallel }^{-2}{\parallel k\parallel }_{\mathfrak{J}}^2\le \sum _{n\in \mathbb{N}}{\left|[k,\psi \left(x_n\right)]\right|}^2\le {B\parallel \psi \parallel }^2{\parallel k\parallel }_{\mathfrak{J}}^2}$ and then $\psi \left(\mathbb{X}\right)$ is a frame in $(\mathfrak{K},[·,·\left]\right)$ with bounds ${B\parallel \psi \parallel }^2\ge A{\parallel {\psi }^{†}\parallel }^{-2}>0$. □

Theorem 5.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space, $\mathbb{X}$, $\mathbb{Y}$ orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$ and ${\psi }_1$, ${\psi }_2\in \mathcal{B}\left(\mathfrak{K}\right)$ surjective. Then, ${\psi }_1\left(\mathbb{X}\right)$, ${\psi }_2\left(\mathbb{Y}\right)$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$.

Proof. 

Suppose that $\mathbb{X}$, $\mathbb{Y}$ are orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$ and ${\psi }_1$, ${\psi }_2\in \mathcal{B}\left(\mathfrak{K}\right)$ are surjective; then, by the above result ${\psi }_1\left(\mathbb{X}\right)$, ${\psi }_2\left(\mathbb{Y}\right)$ are frames in $(\mathfrak{K},[·,·\left]\right)$. It remains to prove that ${\mathfrak{W}}_{{\psi }_1\left(\mathbb{X}\right),{\psi }_2\left(\mathbb{Y}\right)}\equiv \mathbf{0}$. In effect,

$$\begin{array}{c}\hfill {\mathfrak{W}}_{{\psi }_1\left(\mathbb{X}\right),{\psi }_2\left(\mathbb{Y}\right)}={\mathcal{T}}_{{\psi }_1\left(\mathbb{X}\right)}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{{\psi }_1\left(\mathbb{Y}\right)}^{*}={\psi }_1{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{\psi }_2^{*}={\psi }_1{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}{\psi }_2^{*}={\psi }_1\mathbf{0}{\psi }_2^{*}\equiv \mathbf{0}.\end{array}$$

□

Theorem 6.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$, $\mathbb{Y}$ are frames in $(\mathfrak{K},[·,·\left]\right)$. Then, the following are equivalent

(i) 

${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathfrak{id}}_{\mathfrak{K}}$;

(ii) 

${\mathfrak{W}}_{\mathbb{Y},\mathbb{X}}={\mathfrak{id}}_{\mathfrak{K}}$;

(iii) 

${\mathfrak{W}}_{\mathfrak{J}\mathbb{Y},\mathfrak{J}\mathbb{X}}={\mathfrak{id}}_{\mathfrak{K}}$;

(iv) 

$[f,g]={\sum }_{n\in \mathbb{N}}[f,x_n][y_n,g]$, $\forall f,g\in \mathfrak{K}.$

Proof. 

$\left(i\right)\Rightarrow \left(ii\right)$

$${\mathfrak{W}}_{\mathbb{Y},\mathbb{X}}={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}^{*}={\left({\mathfrak{id}}_{\mathfrak{K}}\right)}^{*}={\mathfrak{id}}_{\mathfrak{K}}.$$

$\left(ii\right)\Rightarrow \left(iii\right)$

$$\begin{array}{ccc}\hfill {\mathfrak{W}}_{\mathfrak{J}\mathbb{Y},\mathfrak{J}\mathbb{X}}& =& {\mathcal{T}}_{\mathfrak{J}\mathbb{Y}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathfrak{J}\mathbb{X}}^{*}=\mathfrak{J}{\mathcal{T}}_{\mathbb{Y}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathfrak{J}}^{*}=\mathfrak{J}{\mathcal{T}}_{\mathbb{Y}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\mathfrak{J}\hfill \\ & =& \mathfrak{J}{\mathfrak{W}}_{\mathbb{Y},\mathbb{X}}\mathfrak{J}=\mathfrak{J}{\mathfrak{id}}_{\mathfrak{K}}\mathfrak{J}=\mathfrak{J}\mathfrak{J}={\mathfrak{J}}^2={\mathfrak{id}}_{\mathfrak{K}}.\hfill \end{array}$$

$\left(iii\right)\Rightarrow \left(iv\right)$ Let f, $g\in \mathfrak{K}$ be any,

$$\begin{array}{ccc}\hfill [f,g]& =& [\mathfrak{J}\left(f\right),\mathfrak{J}\left(g\right)]=\left[{\mathfrak{W}}_{\mathfrak{J}\mathbb{Y},\mathfrak{J}\mathbb{X}}\left(\mathfrak{J}\left(f\right)\right),\mathfrak{J}\left(g\right)\right]=\left[\sum _{n\in \mathbb{N}}[\mathfrak{J}\left(f\right),\mathfrak{J}\left(x_n\right)]\mathfrak{J}\left(y_n\right),\mathfrak{J}\left(g\right)\right]\hfill \\ & =& \sum _{n\in \mathbb{N}}[\mathfrak{J}\left(f\right),\mathfrak{J}\left(x_n\right)][\mathfrak{J}\left(y_n\right),\mathfrak{J}\left(g\right)]=\sum _{n\in \mathbb{N}}[f,x_n][y_n,g].\hfill \end{array}$$

$\left(iv\right)\Rightarrow \left(i\right)$ For any g, $f\in \mathfrak{K},$

$$\begin{array}{ccc}\hfill \left[g,f-\sum _{n\in \mathbb{N}}[f,y_n]x_n\right]& =& [g,f]-\left[g,\sum _{n\in \mathbb{N}}[f,y_n]x_n\right]=[g,f]-\sum _{n\in \mathbb{N}}\overline{[f,y_n]}[g,x_n]\hfill \\ & =& [g,f]-\sum _{n\in \mathbb{N}}[g,x_n][y_n,f]=[g,f]-[g,f]=0.\hfill \end{array}$$

Then, ${\sum }_{n\in \mathbb{N}}[f,y_n]x_n=f$, $\forall f\in \mathfrak{K}$. Then, ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathfrak{id}}_{\mathfrak{K}}$. □

Lemma 1.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space, $\mathbb{X}$ be a frame in $(\mathfrak{K},[·,·\left]\right)$ with bounds be $B\ge A>0$ and ${\left\{e_n\right\}}_{n\in \mathbb{N}}$ be the canonical basis of ${\ell }_2\left(\mathbb{N}\right)$. Then, the dual frames to $\mathbb{X}$ are the families of the form:

$$\mathbb{Y}={\left\{y_n\right\}}_{n\in \mathbb{N}}:={\left\{\psi \left(e_n\right)\right\}}_{n\in \mathbb{N}},$$

where $\psi :({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{{\Re }_2})⟶(\mathfrak{K},[·,·])$ is a bounded linear operator and left inverse of ${\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}$.

Proof. 

Let $\psi$ be a left inverse of ${\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}$, this is, ${\displaystyle \psi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}={\mathfrak{id}}_{\mathfrak{K}}}$. Therefore, $\mathbb{Y}$ is a frame in $(\mathfrak{K},[·,·\left]\right)$ with bounds ${B\parallel \psi \parallel }^2\ge A{\parallel {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\parallel }^{-2}>0$. Furthermore, for all $f\in \mathfrak{K},$

$$\begin{array}{ccc}\hfill {\mathfrak{id}}_{\mathfrak{K}}\left(f\right)& =& \psi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\left(f\right)=\psi {\mathfrak{J}}_{{\Re }_2}{\mathfrak{J}}_{{\Re }_2}\left({\left\{[f,x_n]\right\}}_{n\in \mathbb{N}}\right)=\psi \left({\left\{[f,x_n]\right\}}_{n\in \mathbb{N}}\right)\hfill \\ & =& \psi \left(\sum _{n\in \mathbb{N}}[f,x_n]e_n\right)=\sum _{n\in \mathbb{N}}[f,x_n]\psi \left(e_n\right)=\sum _{n\in \mathbb{N}}[f,x_n]y_n={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\left(f\right).\hfill \end{array}$$

Then, ${\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathfrak{id}}_{\mathfrak{K}}$ and thus $\mathbb{Y}$ is a dual frame a $\mathbb{X}$ in $(\mathfrak{K},[·,·\left]\right)$. On the other hand, if $\mathbb{Y}$ is a dual frame, a $\mathbb{X}$ in $(\mathfrak{K},[·,·\left]\right)$, then it is enough to take ${\displaystyle \psi :={\mathcal{T}}_{\mathbb{Y}}}$ and observe that

$$\begin{array}{c}\hfill {\mathcal{T}}_{\mathbb{Y}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}={\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}={\mathfrak{id}}_{\mathfrak{K}}.\end{array}$$

(2)

Additionally,

$$\begin{array}{c}\hfill \psi \left(e_n\right)={\mathcal{T}}_{\mathbb{Y}}\left(e_n\right)=\sum _{k\in \mathbb{N}}{\delta }_{n,k}{y}_k=y_n,\end{array}$$

where

$${\delta }_{n,k}:=\left\{\begin{array}{c}1,n=k,\hfill \\ 0,n\ne k.\hfill \end{array}\right.$$

□

Lemma 2.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$ be a frame in $(\mathfrak{K},[·,·\left]\right)$. Then, the bounded linear operators that are left as the inverse of ${\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}$ are of the form:

$$\begin{array}{c}\hfill {\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\phi \left({\mathfrak{id}}_{\mathfrak{K}}-{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\right),\end{array}$$

where $\phi :({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{{\Re }_2})⟶(\mathfrak{K},[·,·])$ is a bounded linear operator.

Proof. 

Let $\psi :={\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\phi \left({\mathfrak{id}}_{\mathfrak{K}}-{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\right)$ with $\phi :({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{{\Re }_2})⟶(\mathfrak{K},[·,·])$ be a bounded linear operator. Let us see that $\psi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}={\mathfrak{id}}_{\mathfrak{K}}$,

$$\begin{array}{ccc}\hfill \psi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}& =& \left({\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\phi \left({\mathfrak{id}}_{\mathfrak{K}}-{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\right)\right){\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\hfill \\ & =& {\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}+\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}-\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{T}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}\hfill \\ & =& {\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{S}}_{\mathbb{X}}+\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}-\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{S}}_{\mathbb{X}}\hfill \\ & =& {\mathfrak{id}}_{\mathfrak{K}}+\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}-\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}={\mathfrak{id}}_{\mathfrak{K}}.\hfill \end{array}$$

Thus, $\psi$ is a left inverse of ${\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}$. For the other implication, let us assume that $\psi$ is a left inverse of ${\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}$. Take $\phi :=\psi$. then,

$$\begin{array}{ccc}\hfill {\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\phi \left({\mathfrak{id}}_{\mathfrak{K}}-{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\right)& =& {\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\psi \left({\mathfrak{id}}_{\mathfrak{K}}-{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\right)={\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\psi -\psi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\hfill \\ & =& {\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\psi -{\mathfrak{id}}_{\mathfrak{K}}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}={\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}-{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}+\psi =\mathbf{0}+\psi =\psi .\hfill \end{array}$$

□

Theorem 7.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space and $\mathbb{X}$ a frame in $(\mathfrak{K},[·,·\left]\right)$. Then, the dual frames to $\mathbb{X}$ have the form:

$$\begin{array}{c}\hfill \mathbb{Y}={\left\{{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\left(e_n\right)+\phi \left(e_n\right)-\phi {\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}}\left(e_n\right)\right\}}_{n\in \mathbb{N}},\end{array}$$

where $\phi :({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{{\Re }_2})⟶(\mathfrak{K},[·,·])$ is a bounded linear operator and ${\left\{e_n\right\}}_{n\in \mathbb{N}}$, the canonical basis of ${\ell }_2\left(\mathbb{N}\right)$.

Proof. 

It is an immediate consequence of the Lemmas 1 and 2. □

Proposition 11.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space, $\mathbb{Y}$ be a dual frame $\mathbb{X}$ in $(\mathfrak{K},[·,·\left]\right)$ and $\phi :({\Re }_2\left(\mathbb{N}\right),{[·,·]}_{{\Re }_2})$$⟶(\mathfrak{K},[·,·\left]\right)$ be a bounded linear operator and surjective; then, ${\left({\phi }^{†}\right)}^{*}\left(\mathbb{Y}\right)$ is a dual frame a $\phi \left(\mathbb{X}\right)$.

Proof. 

Note that

$$\begin{array}{ccc}\hfill {\mathfrak{W}}_{\phi \left(\mathbb{X}\right),{\left({\phi }^{†}\right)}^{*}\left(\mathbb{Y}\right)}& =& {\mathcal{T}}_{\phi \left(\mathbb{X}\right)}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{{\left({\phi }^{†}\right)}^{*}\left(\mathbb{Y}\right)}^{*}=\phi {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{\left({\left({\phi }^{†}\right)}^{*}\right)}^{*}=\phi {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{\phi }^{†}\hfill \\ & =& \phi {\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}{\phi }^{†}=\phi {\mathfrak{id}}_{\mathfrak{K}}{\phi }^{†}=\phi {\phi }^{†}={\mathfrak{id}}_{\mathfrak{K}}.\hfill \end{array}$$

Thus, ${\left({\phi }^{†}\right)}^{*}\left(\mathbb{Y}\right)$ is a dual frame a $\phi \left(\mathbb{X}\right)$. □

Theorem 8.

Let $(\mathfrak{K},[·,·],\mathfrak{J})$ be a Krein space, $\mathcal{P}:(\mathfrak{K},[·,·\left]\right)⟶(\mathfrak{K},[·,·\left]\right)$ be an orthogonal projection that commutes with $\mathfrak{J}$ and $\mathbb{X},\mathbb{Y}$ orthogonal frames in $(\mathfrak{K},[·,·\left]\right)$. Then

(i) 

$\mathcal{P}\mathbb{X}$, $\mathcal{P}\mathbb{Y}$ define orthogonal frames in Krein space $(\mathcal{P}\mathfrak{K},[·,·\left]\right)$;

(ii) 

$({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbb{X}$, $({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbb{Y}$ define orthogonal frames in the Krein space given by $(({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathfrak{K},[·,·])$.

Proof. 

$\left(i\right)$${\mathfrak{W}}_{\mathcal{P}\mathbb{X},\mathcal{P}\mathbb{Y}}={\mathcal{T}}_{\mathcal{P}\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathcal{P}\mathbb{Y}}^{*}=\mathcal{P}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{\mathcal{P}}^{*}=\mathcal{P}{\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}\mathcal{P}=\mathcal{P}{\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}\mathcal{P}=\mathcal{P}\mathbf{0}\mathcal{P}\equiv \mathbf{0}.$

$\left(ii\right)$

$$\begin{array}{ccc}\hfill {\mathfrak{W}}_{({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbb{X},({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbb{Y}}& =& {\mathcal{T}}_{({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbb{Y}}^{*}=({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P}){\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}{({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})}^{*}\hfill \\ & =& ({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P}){\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{{\Re }_2}{\mathcal{T}}_{\mathbb{Y}}^{*}({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})=({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P}){\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\hfill \\ & =& ({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\mathbf{0}({\mathfrak{id}}_{\mathfrak{K}}-\mathcal{P})\equiv \mathbf{0}.\hfill \end{array}$$

□

Remark 6.

In ${\mathbb{C}}^n$, we have the following indefinite inner product ${\displaystyle {[·,·]}_{{\mathbb{C}}^n}:{\mathbb{C}}^n\times {\mathbb{C}}^n⟶\mathbb{C}}$, given by

$$\begin{array}{c}\hfill [\alpha ,\beta ]:=\sum _{i=1}^n{(-1)}^i{\alpha }_i\overline{{\beta }_i}\end{array}$$

for all $\alpha ={\sum }_{i=1}^n{\alpha }_i{e}_i$, $\beta ={\sum }_{i=1}^n{\beta }_i{e}_i\in {\mathbb{C}}^n$, where ${\left\{e_i\right\}}_{i=1}^n$ is the canonical orthonormal basis in $({\mathbb{C}}^n,{〈·,·〉}_{{\mathbb{C}}^n})$. Then, ${\mathbb{C}}^n$ admits the fundamental decomposition

$$\begin{array}{ccc}\hfill {\mathbb{C}}^n={\left({\mathbb{C}}^n\right)}^{+}\oplus {\left({\mathbb{C}}^n\right)}^{-},{\left({\mathbb{C}}^n\right)}^{+}& :=& \mathrm{span}\left\{e_{2i}:1\le 2i\le n,i\in \mathbb{N}\right\}\hfill \\ \hfill {\left({\mathbb{C}}^n\right)}^{-}& :=& \mathrm{span}\left\{e_{2i-1}:1\le 2i-1\le n,i\in \mathbb{N}\right\}\hfill \end{array}$$

with associated fundamental symmetry

$$\begin{array}{c}\hfill {\mathfrak{J}}_{{\mathbb{C}}^n}:({\mathbb{C}}^n,{\left[·,·\right]}_{{\mathbb{C}}^n})⟶({\mathbb{C}}^n,{\left[·,·\right]}_{{\mathbb{C}}^n})\end{array}$$

given by ${\mathfrak{J}}_{{\mathbb{C}}^n}\left(e_j\right)={(-1)}^j{e}_j$. Then, ${[·,·]}_{{\mathfrak{J}}_{{\mathbb{C}}^n}}={〈·,·〉}_{{\mathbb{C}}^n}$. When ${\mathbb{C}}^n$ is viewed as a Krein space with this fundamental symmetry ${\mathfrak{J}}_{{\mathbb{C}}^n}$, we will write $\left(\Re \left(n\right),{[·,·]}_{\Re \left(n\right)},{\mathfrak{J}}_{\Re \left(n\right)}\right).$

Example 2.

In $(\Re \left(2\right),{[·,·]}_{\Re \left(2\right)},{\mathfrak{J}}_{\Re \left(2\right)})$ we consider the sequences $\mathbb{X}=(e_1,e_1,e_1,e_2)$ and $\mathbb{Y}=\left\{-e_1-e_2,e_1,e_2,\mathbf{0}\right\}$ where ${\left\{e_i\right\}}_{i=1}^2$ is the canonical orthonormal basis in $({\mathbb{C}}^2,{〈·,·〉}_{{\mathbb{C}}^2})$. It is clear that both $\mathbb{X}$ and $\mathbb{Y}$ are frames in $(\Re \left(2\right),{[·,·]}_{\Re \left(2\right)})$ because the kernel of ${\mathcal{T}}_{\mathbb{X}}$ and ${\mathcal{T}}_{\mathbb{Y}}$ have dimension 2:

$$\begin{array}{c}\hfill ker\left({\mathcal{T}}_{\mathbb{X}}\right)=\mathrm{span}\left\{(-1,1,0,0),(1,0,-1,0)\right\}andker\left({\mathcal{T}}_{\mathbb{Y}}\right)=\mathrm{span}\left\{(1,1,1,0),(0,0,0,1)\right\}\end{array}$$

that is, ${\mathcal{T}}_{\mathbb{X}}$, ${\mathcal{T}}_{\mathbb{Y}}$ are both surjective linear transformations. Moreover, for any $({\beta }_1,{\beta }_2)\in \Re \left(2\right)$,

$$\begin{array}{ccc}\hfill {\mathfrak{W}}_{\mathbb{X},\mathbb{Y}}({\beta }_1,{\beta }_2)& =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{\Re \left(4\right)}{\mathcal{T}}_{\mathbb{Y}}^{*}({\beta }_1,{\beta }_2)\hfill \\ & =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{J}}_{\Re \left(4\right)}^2\left([({\beta }_1,{\beta }_2),-e_1-e_2],[({\beta }_1,{\beta }_2),e_1],[({\beta }_1,{\beta }_2),e_2],(0,0)\right)\hfill \\ & =& {\mathcal{T}}_{\mathbb{X}}{\mathfrak{id}}_{\Re \left(4\right)}\left([({\beta }_1,{\beta }_2),-e_1-e_2],[({\beta }_1,{\beta }_2),e_1],[({\beta }_1,{\beta }_2),e_2],(0,0)\right)\hfill \\ & =& [({\beta }_1,{\beta }_2),-e_1-e_2]e_1+[({\beta }_1,{\beta }_2),e_1]e_1+[({\beta }_1,{\beta }_2),e_2]e_1\hfill \\ & =& \left({\beta }_1-{\beta }_2-{\beta }_1+{\beta }_2\right)e_1\hfill \\ & =& 0e_1=0.\hfill \end{array}$$

Therefore, $\mathbb{X}$, $\mathbb{Y}$ are orthogonal frames in $(\Re \left(2\right),{[·,·]}_{\Re \left(2\right)})$.

Example 3.

Let $\mathfrak{K}={\mathbb{C}}^5$ be and consider the inner product ${[·,·]}_{\mathfrak{K}}:\mathfrak{K}\times \mathfrak{K}⟶\mathbb{C}$ given by

$$\begin{array}{c}\hfill [\alpha ,\beta ]:=-\sum _{i=1}^5{\alpha }_i\overline{{\beta }_{6-i}}for all\alpha =\sum _{i=1}^5{\alpha }_i{e}_i,\beta =\sum _{i=1}^5{\beta }_i{e}_i\end{array}$$

in $\mathfrak{K}$, where ${\left\{e_i\right\}}_{i=1}^5$ is the canonical orthonormal basis in $({\mathbb{C}}^5,{〈·,·〉}_{{\mathbb{C}}^5})$, which is an indefinite inner product. Note that we can define the bijective linear transformation

$$\begin{array}{c}\hfill \mathfrak{J}:(\mathfrak{K},[·,·])⟶(\mathfrak{K},[·,·]),\mathfrak{J}\left(e_i\right)=-e_{6-i},i=1,2,3,4,5.\end{array}$$

whose matrix associated with ${\left\{e_i\right\}}_{i=1}^5$ is

$$\left[\mathfrak{J}\right]=\left(\begin{array}{ccccc}0& 0& 0& 0& -1\\ 0& 0& 0& -1& 0\\ 0& 0& -1& 0& 0\\ 0& -1& 0& 0& 0\\ -1& 0& 0& 0& 0\end{array}\right)$$

and $\mathfrak{J}$ is the fundamental symmetry associated with the fundamental decomposition $\mathfrak{K}={\mathfrak{K}}^{+}\oplus {\mathfrak{K}}^{-}$, where

$$\begin{array}{ccc}\hfill {\mathfrak{K}}^{+}& :=& \mathrm{span}\left\{e_1-e_5,e_2-e_4\right\}\hfill \\ \hfill {\mathfrak{K}}^{-}& :=& \mathrm{span}\left\{e_1+e_5,e_2+e_4,e_3\right\}\hfill \end{array}$$

and also

$$\begin{array}{ccc}\hfill {[\alpha ,\beta ]}_{\mathfrak{J}}& =& {[\mathfrak{J}\left(\alpha \right),\beta ]}_{\mathfrak{K}}={\left[\mathfrak{J}\left(\sum _{i=1}^5{\alpha }_i{e}_i\right),\sum _{j=1}^5{\beta }_j{e}_j\right]}_{\mathfrak{K}}={\left[\sum _{i=1}^5{\alpha }_i\mathfrak{J}\left(e_i\right),\sum _{j=1}^5{\beta }_j{e}_j\right]}_{\mathfrak{K}}\hfill \\ & =& {\left[\sum _{i=1}^5{\alpha }_i(-e_{6-i}),\sum _{j=1}^5{\beta }_j{e}_j\right]}_{\mathfrak{K}}=\sum _{i=1}^5{\alpha }_i\sum _{j=1}^5\overline{{\beta }_j}[-e_{6-i},e_j]\hfill \\ & =& \sum _{i=1}^5{\alpha }_i\sum _{j=1}^5\overline{{\beta }_j}{〈e_i,e_j〉}_{{\mathbb{C}}^5}={〈\sum _{i=1}^5{\alpha }_i{e}_i,\sum _{j=1}^5{\beta }_j{e}_j〉}_{{\mathbb{C}}^5}={〈\alpha ,\beta 〉}_{{\mathbb{C}}^5}.\hfill \end{array}$$

Let us consider $\mathbb{X}=\left\{e_1+e_2,e_2+e_3,e_3+e_4,e_4+e_5,e_5\right\}$ in $(\mathfrak{K},{[·,·]}_{\mathfrak{K}})$. Note that

$$\begin{array}{ccc}\hfill {\mathcal{T}}_{\mathbb{X}}({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)& =& {\alpha }_1(e_1+e_2)+{\alpha }_2(e_2+e_3)+{\alpha }_3(e_3+e_4)+{\alpha }_4(e_4+e_5)+{\alpha }_5{e}_5\hfill \\ & =& \left({\alpha }_1,{\alpha }_1+{\alpha }_2,{\alpha }_2+{\alpha }_3,{\alpha }_3+{\alpha }_4,{\alpha }_4+{\alpha }_5\right)\hfill \end{array}$$

for all $\left({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5\right)\in \mathfrak{K}$ and $dim(ker\left({\mathcal{T}}_{\mathbb{X}}\right))=0$ which implies that $\mathbb{X}$ is a frame in $(\mathfrak{K},{[·,·]}_{\mathfrak{K}})$. Now, taking $\phi :(\Re \left(5\right),{[·,·]}_{\Re \left(5\right)})⟶(\mathfrak{K},{[·,·]}_{\mathfrak{K}})$, $\phi :={\mathfrak{id}}_{\Re \left(5\right)},$ and as ${\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_1\right)=-e_5$, ${\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_2\right)=-e_4+e_5$, ${\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_3\right)=-e_3+e_4-e_5$, ${\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_4\right)=-e_2+e_3-e_4+e_5$ and ${\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_5\right)=-e_1+e_2-e_3+e_4-e_5$, then

$$\begin{array}{ccc}\hfill \phi ({\mathfrak{id}}_{\Re \left(5\right)}-{\mathfrak{J}}_{\Re \left(5\right)}{\mathcal{T}}_{\mathbb{X}}^{*}{\mathcal{S}}_{\mathbb{X}}^{-1}{\mathcal{T}}_{\mathbb{X}})\left(e_i\right)& =& e_i-{\mathfrak{J}}_{\Re \left(5\right)}{\mathcal{T}}_{\mathbb{X}}^{*}{S}_{\mathbb{X}}^{-1}\left(x_i\right)\hfill \\ & =& e_i-\left({[{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_i\right),x_1]}_{\mathfrak{K}},{[{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_i\right),x_2]}_{\mathfrak{K}},\right.\hfill \\ & [{\mathcal{S}}_{\mathbb{X}}^{-1}& \left.\left(x_i\right),x_3{]}_{\mathfrak{K}},{[{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_i\right),x_4]}_{\mathfrak{K}},{[{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_i\right),x_5]}_{\mathfrak{K}}\right)\hfill \\ & =& \mathbf{0}\hfill \end{array}$$

for every $i,$ $1\le i\le 5.$ Therefore, $\mathbb{Y}=\left\{{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_1\right),{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_2\right),{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_3\right),{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_4\right),{\mathcal{S}}_{\mathbb{X}}^{-1}\left(x_5\right)\right\}$ is a dual frame a $\mathbb{X}$ in $(\mathfrak{K},{[·,·]}_{\mathfrak{K}})$.

4. Conclusions

In this paper, the definition of orthogonal frames in Krein spaces was introduced, it was shown that it does not depend on the fundamental symmetry, and that having orthogonal frames in Krein spaces is equivalent to having orthogonal frames in Hilbert spaces thanks to the Theorem 4. In addition, families of frames dual to a given one were characterized, and some properties that they satisfy were proven. In future work, we can think of giving conditions under which given a pair of frames and the families of frames dual to these, these families are orthogonal, while following the line of results obtained in the articles [10,11,12,13]. Likewise, we can think of introducing the definition of orthogonal frames in soft Hilbert spaces, since the theory of soft frames was already introduced in the article [19]. Similarly, following the ideas of this article, it is possible to introduce the notion of orthogonal continuous frames and extend the results obtained here. For continuous frames in Hilbert and Krein spaces, see [20,21], respectively. Therefore, much remains to be done; in the meantime, these results may have applications in signal processing theory, data analysis and mathematical physics.