1 Introduction

Let X and Y be two inner product spaces, a map \(T:X \rightarrow Y\) is called orthogonality preserving if Tx and Ty are orthogonal for any orthogonal vectors x and y in X; see [6, 18].

For a given \(\varepsilon \in [0,1)\), two vectors \(x,y\in X\) are said to be approximately orthogonal or \(\varepsilon \)-orthogonal, denoted by \(x \perp ^\varepsilon y\), if \(|\langle x, y\rangle |\le \varepsilon \Vert x\Vert \Vert y\Vert \), where \(\langle \cdot , \cdot \rangle \) denotes the inner product in X and \(\Vert \cdot \Vert \) denotes the norm in X induced by \(\langle \cdot , \cdot \rangle \). A map \(T:X \rightarrow Y\) is said to be approximately orthogonality preserving if \(x \perp y \Rightarrow Tx \perp ^\varepsilon Ty\) for \(x,y\in X\).

Approximately orthogonality preserving maps have been studied in several settings; see [1, 5, 8,9,10,11,12,13, 17].

Chmieliński in [9, Theorem 1] proved that a linear map \(T:X \rightarrow Y\) is orthogonality preserving if and only if there exists \(\gamma >0\) such that \(\langle T(x), T(y)\rangle =\gamma ^2\langle x, y\rangle ,\) for \( x, y\in X\). Furthermore, he showed that for a nonzero linear \(\varepsilon \)-orthogonality preserving map \(T:X \rightarrow Y\) for some \(\varepsilon \in [0, 1)\) there is \(\gamma >0\) such that

$$\begin{aligned} |\langle T(x), T(y)\rangle -\gamma ^2\langle x, y\rangle |\le \delta \min \{\gamma ^2 \Vert x\Vert \Vert y\Vert , \Vert T(x)\Vert \Vert T(y)\Vert \}, \text {for}\,\, x,y\in X, \end{aligned}$$

with \(\delta =4\varepsilon \big (\frac{1}{1-\varepsilon } +\sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\big ).\)

In the present paper, we generalize these results to the setting of Krein spaces. We first recall some basic facts on these structures; more details can be found, for example, in [3, 4, 7, 14,15,16].

Definition 1.1

Suppose that \({\mathcal {K}}\) is a linear vector space equipped with a map \([\cdot ,\cdot ]:{\mathcal {K}} \times {\mathcal {K}}\rightarrow {\mathbb {C}}\) such that

$$\begin{aligned}{}[x, y]&=\overline{ [y, x]},\\ [\alpha x +\beta y , z]&= \alpha [x , z] +\beta [y, z] \end{aligned}$$

for each \(x,y,z\in {\mathcal {K}}\) and \(\alpha ,\beta \in {\mathbb {C}}\). Then \(({\mathcal {K}},\,\, [\cdot ,\cdot ])\) is called an indefinite inner product space.

Recall that an indefinite inner product has all the properties of the usual inner product except positive definiteness.

Vector spaces equipped with indefinite inner products were used for the first time in the quantum field theory in physics and mechanics.

We denote the sets of all positive, negative, and neutral elements of \({\mathcal {K}}\), by \({\mathcal {K}}^{++}\equiv \{x\in {\mathcal {K}} :[x, x]>0\},\,\,\,{\mathcal {K}}^{--} \equiv \{x\in {\mathcal {K}} :[x, x]<0\},\) and \({\mathcal {K}}^{0} \,\,\,\,\equiv \{x\in {\mathcal {K}} :[x, x]=0\}\), respectively.

Let \(x,y\in {\mathcal {K}}\). We say that x is orthogonal to y, denoted by \(x[\perp ]y\), when \([x, y]=0\). If \({\mathcal {L}}\) is a subspace of \({\mathcal {K}}\), then the subspace \({\mathcal {L}}^{[\perp ]} \equiv \{x\in {\mathcal {K}} :[x, y]=0\), for all \(y\in {\mathcal {L}}\}\), is called the orthogonal complement of \({\mathcal {L}}\) with respect to \([\ , \ ]\), and \(x_0\in {\mathcal {L}}\) is an isotropic element of \({\mathcal {L}}\) if \(x_0\ne 0\) and \(x_0[\perp ]{\mathcal {L}}\). Also \({\mathcal {L}}^{0}\equiv {\mathcal {L}}\bigcap {\mathcal {L}}^{[\perp ]}\) is the isotropic part of \({\mathcal {L}}\). If \({\mathcal {L}}^{0}=\{0\}\), then \({\mathcal {L}}\) is called nondegenerate.

Azizov in [4, Theorem 1.24], proved that if \({\mathcal {L}}\) is a subspace of \({\mathcal {K}}\) such that \({\mathcal {L}}\) admits a decomposition \( {\mathcal {L}}={\mathcal {L}}^{+}\oplus {\mathcal {L}}^{-}\), where \({\mathcal {L}}^+\subseteq {\mathcal {K}}^{++}\cup \{0\}\) and \({\mathcal {L}}^-\subseteq {\mathcal {K}}^{--}\cup \{0\}\) into the direct sum of the subspaces, then \({\mathcal {L}}\) is nondegenerate. In addition, if \({\mathcal {L}}^{+}[\perp ] {\mathcal {L}}^{-}\), then we write

$$\begin{aligned} {\mathcal {L}}={\mathcal {L}}^{+}[\oplus ] {\mathcal {L}}^{-}. \end{aligned}$$

This decomposition is called a canonical decomposition of the subspace \( {\mathcal {L}}\).

Definition 1.2

An indefinite inner product space (\({\mathcal {K}},[\cdot ,\cdot ]\)) is called a Krein space, if the vector space \({\mathcal {K}}\) admits a canonical decomposition \({\mathcal {K}}={\mathcal {K}}^+[\oplus ]{\mathcal {K}}^-\) such that \(({\mathcal {K}}^+, [\cdot , \cdot ])\) and \(({\mathcal {K}}^-,- [\cdot , \cdot ])\) are Hilbert spaces relative to the norms \(\Vert x\Vert =[x,x]^{\frac{1}{2}} \ (x\in {\mathcal {K}}^+)\) and \(\Vert x\Vert =(-[x,x]^{\frac{1}{2}}) \ (x\in {\mathcal {K}}^-)\).

Suppose that \(({\mathcal {K}}, [\cdot , \cdot ])\) is a Krein space with a canonical decomposition \({\mathcal {K}}={\mathcal {K}}^+[\oplus ]{\mathcal {K}}^-\) such that \(P^+:{\mathcal {K}}\rightarrow {\mathcal {K}}^+\) and \(P^-:{\mathcal {K}}\rightarrow {\mathcal {K}}^-\) are two mutually orthogonal projection operators generated by the canonical decomposition above so that \(P^++P^-=I\), where I is the identity operator on \({\mathcal {K}}\). Thus for any \(x\in {\mathcal {K}}\), we have \(x=x^++x^-\), where \(x^+=P^+x\) and \(x^-=P^-x\). The linear operator \(J: {\mathcal {K}}\rightarrow {\mathcal {K}}\), defined by \(J=P^+-P^-\), is called the canonical symmetry operator of the Krein space \({\mathcal {K}}\). Thus J is a bounded self-adjoint operator such that \(J^2=I\) and \(J^{-1}=J^*=J\).

By using the canonical decomposition \({\mathcal {K}}={\mathcal {K}}^+[\oplus ]{\mathcal {K}}^-\) on the Krein space \(({\mathcal {K}}, [\cdot , \cdot ])\), we can define an inner product as follows:

$$\begin{aligned} \langle x, y\rangle =[x^+, y^+]-[x^-, y^-], \end{aligned}$$
(1.1)

where \(x=x^++x^-\) and \(y=y^++y^-\) are elements in \({\mathcal {K}}\).

Then \(({\mathcal {K}}, \langle \cdot , \cdot \rangle )\) is a Hilbert space with respect to the norm \(\Vert x\Vert ^2=\langle x, x\rangle \). In fact, \(\langle x, y\rangle =[x, y]\), for vectors \(x, y\in {\mathcal {K}}^+\), and \(\langle x, y\rangle =-[x, y]\) for \(x, y \in {\mathcal {K}}^-\).

Relation (1.1) between the indefinite inner product \([\cdot ,\cdot ]\) and the definite inner product \(\langle \cdot , \cdot \rangle \) on \({\mathcal {K}}\) implies that

$$\begin{aligned} \langle x, y\rangle =[Jx, y],\quad \quad [x, y]= \langle Jx, y\rangle \end{aligned}$$

for each \(x, y\in {\mathcal {K}}\), where J is the canonical symmetry operator and \(J( x^++x^-)=J(x^+)+J(x^-)=x^+-x^-\). In addition, \(\Vert x\Vert ^2=\langle x, x\rangle =[Jx, x]\). A straightforward computation shows that

$$\begin{aligned}&[x, y]=[x^+, y^+]+[ x^-, y^-],\quad [x, y]=\langle x^+, y^+\rangle -\langle x^-, y^-\rangle , \\&[x, x] =\Vert x^+\Vert ^2-\Vert x^-\Vert ^2. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert y\Vert ^2= & {} [Jy,y] =[J(y^++y^-) , y^++y^-]=[y^+-y^-, y^++y^-]\\= & {} [y^+, y^+]+[-y^-, y^-]. \end{aligned}$$

By the definition of J, we have \(J(y^+)=y^+\), \(J(y^-)=-y^-\). Therefore

$$\begin{aligned} \Vert y\Vert ^2=[Jy^+, y^+]+[Jy^-, y^-]=\Vert y^+\Vert ^2 +\Vert y^-\Vert ^2. \end{aligned}$$

2 Approximate orthogonality in Krein spaces

In this section, we define approximate orthogonality in the framework of Krein spaces. We start our work with the following lemmata and prove them by similar strategies to those of [9].

Lemma 2.1

Let \(({\mathcal {K}}, [\cdot , \cdot ], J)\) be a Krein space, let \(x,y\in {\mathcal {K}}\), and let \(\varepsilon \ge 0\). If

$$\begin{aligned} |[Jx+y,x-Jy]|\le \varepsilon \Vert Jx +y\Vert \Vert x-Jy\Vert , \end{aligned}$$

then

$$\begin{aligned} |\Vert x\Vert ^2-\Vert y\Vert ^2|\le \varepsilon (\Vert x\Vert ^2+\Vert y\Vert ^2). \end{aligned}$$

Proof

By a straightforward computation, we get

$$\begin{aligned} |[Jx+y,x-Jy]|^2&= [Jx+y,x-Jy][x-Jy,Jx+y]\\&=(\Vert x\Vert ^2-[x,y]+[y,x]-\Vert y\Vert ^2)(\Vert x\Vert ^2+[x,y]-[y,x]-\Vert y\Vert ^2)\\&=\Vert x\Vert ^4+2i\Vert x\Vert ^2 \mathrm{Im}\ [x,y]-\Vert x\Vert ^2\Vert y\Vert ^2-2i \Vert x\Vert ^2\mathrm{Im}[x,y]\\&\quad +\,4(\mathrm{Im}[x,y])^2+2i\mathrm{Im}[x,y]\Vert y\Vert ^2-\Vert x\Vert ^2\Vert y\Vert ^2\\&\quad -\,2i\mathrm{Im}[x,y]\Vert y\Vert ^2+\Vert y\Vert ^4\\&=\Vert x\Vert ^4+\Vert y\Vert ^4-2\Vert x\Vert ^2\Vert y\Vert ^2 +4(\mathrm{Im} [x,y])^2\\&= (\Vert x\Vert ^2-\Vert y\Vert ^2)^2+4(\mathrm{Im}[x,y])^2. \end{aligned}$$

On the other hand,

$$\begin{aligned} \Vert Jx+y\Vert ^2\Vert x-Jy\Vert ^2&= ([x+Jy,Jx+y])([Jx-y,x-Jy])\\&=(\Vert x\Vert ^2+[x,y]+[y,x] \\&\quad +\Vert y\Vert ^2)(\Vert x\Vert ^2-[x,y]-[y,x]+\Vert y\Vert ^2)\\&=\Vert x\Vert ^4+\Vert y\Vert ^4+2\Vert x\Vert ^2\Vert y\Vert ^2-4(\mathrm{Re} [x,y])^2\\&= (\Vert x\Vert ^2+\Vert y\Vert ^2)^2-4(\mathrm{Re} [x,y])^2. \end{aligned}$$

The first inequality of the assumption implies that

$$\begin{aligned} (\Vert x\Vert ^2-\Vert y\Vert ^2 )^2+4(\mathrm{Im}[x,y])^2&\le \varepsilon ^2( (\Vert x\Vert ^2-\Vert y\Vert ^2 )^2 -4(\mathrm{Re} [x,y])^2) \\&\le \varepsilon ^2 (\Vert x\Vert ^2 +\Vert y\Vert ^2)^2. \end{aligned}$$

Hence \( \big |\Vert x\Vert ^2-\Vert y\Vert ^2\big | \le \varepsilon (\Vert x\Vert ^2 +\Vert y\Vert ^2).\)\(\square \)

Lemma 2.2

Suppose that \(({\mathcal {K}}_1, [\cdot , \cdot ], J_1)\) and \(({\mathcal {K}}_2, [\cdot , \cdot ], J_2)\) are two Krein spaces and that \(T: {\mathcal {K}}_1\rightarrow {\mathcal {K}}_2\) is a map such that for each \(x\in {\mathcal {K}}_1\), \(T( x^{\pm })={ (T x)}^{\pm }\) and for some \(\delta \ge 0\) and \(\gamma >0\), the functional inequality \( \Big | [T(x), T(y)]-\gamma ^2[x,y]\Big | \le \delta \gamma ^2 \Vert x\Vert \Vert y\Vert \) holds. Then

$$\begin{aligned} \Big | [J_2T(x), T(y)]-\gamma ^2[J_1x,y]\Big | \le 2 \delta \gamma ^2 \Vert x\Vert \Vert y\Vert \quad \quad x, y\in {\mathcal {K}}_1. \end{aligned}$$

Proof

From the properties of canonical symmetry operators, and from \(\Vert x^+\Vert \le \Vert x\Vert \) and \(\Vert x^-\Vert \le \Vert x\Vert \), we get

$$\begin{aligned} \Big | [J_2T(x), T(y)]-\gamma ^2[J_1x,y]\Big |&=\Big | [{T(x)}^+-{ T(x)}^-,T(y)]-\gamma ^2 ([x^++x^-, y])\Big |\\&=\Big | [T(x^+), T(y)]-[T(x^-),T(y)]\\&\quad -\,\gamma ^2 ([x^+, y] -[x^-, y])\Big |\\&\le \Big | [T(x^+), T(y)]\\&\quad -\,\gamma ^2[x^+, y]\Big |+\Big | [T(x^-), T(y)]-\,\gamma ^2[x^-, y]\Big |\\&\le \delta \gamma ^2 \Vert x^+\Vert \Vert y\Vert \\&\quad +\,\delta \gamma ^2 \Vert x^-\Vert \Vert y\Vert \le 2\delta \gamma ^2 \Vert x\Vert \Vert y\Vert . \end{aligned}$$

\(\square \)

Lemma 2.3

Under the assumptions of Lemma 2.2, T is a quasi-linear map, that is, T is quasiadditive,

$$\begin{aligned} \Vert T(x+y)-T(x)-T(y)\Vert \le \sqrt{8}\sqrt{\delta }\gamma (\Vert x\Vert +\Vert y\Vert ),\quad (x,y \in {\mathcal {K}}_1 ) \end{aligned}$$

and quasihomogeneous

$$\begin{aligned} \Vert T(\lambda x)-\lambda T(x)\Vert \le \sqrt{8}\sqrt{\delta } \gamma |\lambda | \Vert x\Vert .\quad ( x\in {\mathcal {K}}_1,\lambda \in {\mathbb {C}}). \end{aligned}$$

Proof

Let \(x,y\in {\mathcal {K}}_1\) be arbitrary. Then

$$\begin{aligned} {\Vert T(x+y)-T(x)-T(y)\Vert }^2 =&[J_2(T(x+y)-T(x)-T(y)),T(x+y)-T(x)-T(y)]\\ \le&\Big | [J_2T(x+y), T(x+y)]-\gamma ^2[J_1(x+y),x+y]\Big |\\&+\Big |-[J_2T(x+y),T(x)]+\gamma ^2[J_1(x+y),x]\Big |\\&+\Big |-[J_2T(x+y),T(y)]+\gamma ^2[J_1(x+y),y]\Big |\\&+\Big |-[J_2T(x),T(x+y)]+\gamma ^2[J_1(x),x+y]\Big |\\&+\Big |[J_2T(x),T(x)]-\gamma ^2[J_1(x),x]\Big |\\&+\Big |[J_2T(x),T(y)]-\gamma ^2[J_1(x),y]\Big |\\&+\Big |-[J_2T(y),T(x+y)]+\gamma ^2[J_1(y),x+y]\Big |\\&+\Big |[J_2T(y),T(x)]-\gamma ^2[J_1(y),x]\Big |\\&+\Big |[J_2T(y),T(y)]-\gamma ^2[J_1(y),y]\Big |. \end{aligned}$$

By using the above lemma and the assumption, we obtain

$$\begin{aligned} {\Vert T(x+y)-T(x)-T(y)\Vert }^2 \le&2\delta \gamma ^2({ \Vert x+y\Vert }^2+2 \Vert x\Vert \Vert x+y\Vert \\&+2 \Vert y\Vert \Vert x+y\Vert +2 \Vert x\Vert \Vert y\Vert + {\Vert x\Vert }^2+{\Vert y\Vert }^2)\\ \le&8\delta \gamma ^2{( \Vert x\Vert +\Vert y\Vert )}^2. \end{aligned}$$

In a similar way, for \(\lambda \in {\mathbb {C}}\), we have

$$\begin{aligned} {\Vert T(\lambda x)-\lambda T(x)\Vert }^2 \le&\Big | [J_2T(\lambda x), T(\lambda x)]-\gamma ^2[J_1(\lambda x),\lambda x]\Big |\\&+\Big |-{\overline{\lambda }} [J_2T(\lambda x), T(x)]+\gamma ^2{\overline{\lambda }}[J_1(\lambda x),x]\Big |\\&+ \Big |-\lambda [J_2T(x), T(\lambda x)]+\gamma ^2\lambda [J_1 x,\lambda x]\Big |\\&+ \Big |{|\lambda |}^2 [J_2T(x), T(x)]+\gamma ^2{|\lambda |}^2[J_1 x, x]\Big |. \end{aligned}$$

Hence

$$\begin{aligned} {\Vert T(\lambda x)-\lambda T(x)\Vert }^2&\le 2\delta \gamma ^2({ \Vert \lambda x\Vert }^2+2|\lambda |\Vert x\Vert \Vert \lambda x\Vert +{|\lambda |}^2 {\Vert x\Vert }^2)\\&=8\delta \gamma ^2{ \Vert \lambda x\Vert }^2. \end{aligned}$$

\(\square \)

Lemma 2.4

[4, Lemma 7.7] Let \(({\mathcal {K}},[\cdot , \cdot ])\) be a Krein space and let \({\mathcal {L}}\) be a subspace of \({\mathcal {K}}\). Then \(\overline{{\mathcal {L}}+ {\mathcal {L}}^{[\perp ]}}={\mathcal {K}}\) if and only if \({\mathcal {L}}\) is nondegenerate.

Recall that \(\langle y\rangle \) is the closure of the linear span \(\{y\}\). The following lemma states the condition under which \(\langle y\rangle \) is nondegenerate.

Proposition 2.5

Let \(({\mathcal {K}},[\cdot , \cdot ], J)\) be a Krein space and let y be a nonzero element of \({\mathcal {K}}\). Then \(\langle y\rangle \) is nondegenerate if and only if y is not a neutral vector.

Proof

Let \(\langle y\rangle \) be nondegenerate. On the contrary, if y is neutral, then

$$\begin{aligned}{}[y,y]=0\Rightarrow ([y,\lambda y]=0 \ \hbox { for some}\ \lambda \in {\mathbb {C}})\Rightarrow y\in \langle y\rangle ^{[\perp ]}. \end{aligned}$$
(2.1)

Therefore \(y\in \langle y\rangle ^{[\perp ]} \cap \langle y\rangle \). Then \(\langle y\rangle ^{[\perp ]} \cap \langle y\rangle \ne \{0\}\), that is a contradiction, since \(\langle y\rangle \) is nondegenerate.

Conversely, let y not be neutral. On the contrary, if \(\langle y\rangle ^{[\perp ]} \cap \langle y\rangle \ne \{0\}\), then there exists a nonzero scalar number \(\lambda \in {\mathbb {C}}\) such that \(\lambda y \in \langle y\rangle ^{[\perp ]} \cap \langle y\rangle \). Then for all \(\mu \in {\mathbb {C}}\), we reach \(0=[\lambda y, \mu y]=\lambda {\bar{\mu }}[y,y]\). Thus \([y,y]=0\), that is, y is a neutral vector which is a contradiction. \(\square \)

Corollary 2.6

Let \(({\mathcal {K}},[\cdot , \cdot ])\) be a Krein space. If y is not a neutral vector in \({\mathcal {K}}\), then \(\langle y\rangle +\langle y\rangle ^{[\perp ]}={\mathcal {K}}\).

Proof

Let y not be a neutral vector. Then Proposition 2.5 ensures that \(\langle y\rangle \) is nondegenerate and Lemma 2.4 yields \(\overline{\langle y\rangle +\langle y\rangle ^{[\perp ]}}={\mathcal {K}}\). On the other hand, since \(\langle y\rangle \) is a finite-dimensional subspace and \(\langle y\rangle ^{[\perp ]}\) is closed, so \(\langle y\rangle +\langle y\rangle ^{[\perp ]}\) is closed and we can write \(\langle y\rangle +\langle y\rangle ^{[\perp ]}={\mathcal {K}}\). \(\square \)

In Corollary 2.6, the condition that y is not a neutral vector is a necessary condition. The following example shows that if y is a neutral vector, then it may happen that \({\mathcal {K}}\ne \langle y\rangle +\langle y\rangle ^{[\perp ]}\).

Example 2.7

Consider the Krein space \(({\mathbb {C}}_2,[\cdot , \cdot ])\) with the sesquilinear form \([x,y]=x_1y_1-4x_2y_2\). Take \(y=(4,2)\). Then \([(4,2), (4,2)]=0\), and so y is a neutral vector and \(y^{[\perp ]}=\{k(2,1);\ k\in {\mathbb {R}}\}\). If \(x=(5,6)\), then there are no \(\lambda _1,\lambda _2\in {\mathbb {C}}\) such that \(x=\lambda _1 y_1+\lambda _2 y_2\) for some \(y_1\in \langle y\rangle \) and \(y_2 \in \langle y\rangle ^{[\perp ]}\).

The main theorem is based on the property of Hilbert spaces that states that, if \(x_1\perp x_2\) and \(x=x_1+x_2\), then \(\Vert x\Vert ^2=\Vert x_1\Vert ^2+\Vert x_2\Vert ^2\), but we do not have this property in Krein spaces [2, 11]. This means that we may have \(x=x_1+x_2\) and \(x_1[\perp ]x_2\) but \(\Vert x\Vert ^2\ne \Vert x_1\Vert ^2+\Vert x_2\Vert ^2\). In the following lemma, we investigate some conditions under which we can write \(\Vert x\Vert ^2=\Vert x_1\Vert ^2+\Vert x_2\Vert ^2\) for an element x in a Krein space.

Lemma 2.8

Let \(({\mathcal {K}},[\cdot , \cdot ], J)\) be a Krein space. If \(x,y\in {\mathcal {K}}\) are linearly independent so that \(y\in {\mathcal {K}}^+\) (or \(y\in {\mathcal {K}}^-\)) is not neutral, then we can write

$$\begin{aligned} x=x_1+x_2, \qquad \Vert x\Vert ^2=\Vert x_1\Vert ^2+\Vert x_2\Vert ^2, \end{aligned}$$

where \(x_1\in \langle y\rangle \) and \(x_2\in \langle y\rangle ^{[\perp ]}\).

Proof

Let \(y\in {\mathcal {K}}^+\) and y is not neutral. By Corollary 2.6, we can write \(\langle y\rangle +\langle y\rangle ^{[\perp ]}={\mathcal {K}}\). Note that in this case \(\langle y\rangle \subseteq {\mathcal {K}}^+\). Thus \(x=x_1+x_2\), where \(x_1\in \langle y\rangle \) and \(x_1[\perp ] x_2\). Then

$$\begin{aligned} \Vert x\Vert ^2&= \Vert x_1+x_2\Vert ^2=[Jx_1+Jx_2, x_1+x_2]\\&=[Jx_1, x_1]+[Jx_1, x_2]+[Jx_2, x_1]+[Jx_2, x_2]\\&= \Vert x_1\Vert ^2 +\Vert x_2\Vert ^2 +2\mathrm{Re} [Jx_1, x_2]\\&=\Vert x_1\Vert ^2 +\Vert x_2\Vert ^2 +2\mathrm{Re} [x_1, x_2]. \end{aligned}$$

Thus \(\Vert x\Vert ^2=\Vert x_1\Vert ^2+\Vert x_2\Vert ^2\). Similarly, it is true, when \(y\in {\mathcal {K}}^-\). \(\square \)

The next theorem shows that an approximately orthogonality preserving linear map T between two Krein spaces such that \(T({\mathcal {K}}_1^{\pm })\subseteq {\mathcal {K}}_2^{\pm }\) is injective, continuous, and satisfies (2.2). To achieve the next result we adopt some arguments from [9, Theorem 3.1]. We also need the following facts about the polarization formula for [x, y] in the Krein space. A straightforward computation shows that

$$\begin{aligned}{}[x, y]&=\frac{1}{4} (\Vert x+y\Vert ^2 -\Vert x-y\Vert ^2 )-\frac{1}{2} (\Vert x^-+y^-\Vert ^2 -\Vert x^- -y^-\Vert ^2)\\&\quad + \frac{i}{4} (\Vert x+iy\Vert ^2 -\Vert x-iy \Vert ^2)-\frac{i}{2} (\Vert x^- +i y^-\Vert ^2 -\Vert x^- -i y^- \Vert ^2 ). \end{aligned}$$

Theorem 2.9

Let \(({\mathcal {K}}_1,[\cdot , \cdot ]_1, J_1)\) and \(({\mathcal {K}}_2,[\cdot , \cdot ]_2, J_2)\) be two Krein spaces, let \(T: {\mathcal {K}}_1 \rightarrow {\mathcal {K}}_2\) be a nonzero linear map such that \(T({\mathcal {K}}_1^{\pm })\subseteq {\mathcal {K}}_2^{\pm }\), and let T be an approximately orthogonality preserving map for some \(\varepsilon \in [0, 1)\). Then T is injective and continuous, and there exists \(\gamma >0\) such that for \(x,y\in {\mathcal {K}}_1\),

$$\begin{aligned} |[T(x),T(y)]-\gamma ^2 [x,y]|\le \delta \min \{\gamma ^2 \Vert x\Vert \Vert y\Vert , \Vert T(x)\Vert \Vert T(y)\Vert \} \end{aligned}$$
(2.2)

with

$$\begin{aligned} \delta =12\varepsilon \left( \frac{1}{1-\varepsilon } +\sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\right) . \end{aligned}$$

Conversely, if \(T: {\mathcal {K}}_1 \rightarrow {\mathcal {K}}_2\) satisfies (2.2) with some \(\delta \ge 0\) and \(\gamma >0\) and \(T(x^{\pm })=(T x )^{\pm }\), then T is a quasilinear approximately orthogonality preserving map and

$$\begin{aligned} x [\perp ] y \Rightarrow T(x) [\perp ] ^\delta T(y) \quad \text {and}\quad T(x)[\perp ] T(y)\Rightarrow x [\perp ]^\delta y \end{aligned}$$

for x, y in \({\mathcal {K}}_1\).

Proof

If \(\dim {\mathcal {K}}_1=1\), then the assertion trivially holds. We assume that \(\dim {\mathcal {K}}_1 \ge 2\).

Let x and y be two nonzero elements in \({\mathcal {K}}_1\), and we want to show that

$$\begin{aligned} \frac{1}{\delta _1} \lambda (y) \le \lambda (x) \le \delta _1\lambda (y), \end{aligned}$$
(2.3)

where \(\lambda (x)=\frac{\Vert T(x)\Vert }{\Vert x\Vert }\) with \(\delta _1=\sqrt{\frac{1+\varepsilon }{1-\varepsilon }+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}}\ge 1\).

Let \(y=y^++y^-\) be a decomposition of y, and put

$$\begin{aligned} y'={\left\{ \begin{array}{ll} y^+, &{} \lambda (y^+)<\lambda (y^-),\\ y^-, &{} \lambda (y^-)< \lambda (y^+); \end{array}\right. } \end{aligned}$$

therefore \(y'\) is not neutral and \(\Vert y\Vert ^2=[Jy, y]=[Jy^+, y^+]+[Jy^-, y^-]=\Vert y^+\Vert ^2+\Vert y^-\Vert ^2\) (recall that \(Jy^+=y^+\) and \(Jy^-=-y^-\)). Also, we have \(\Vert Ty\Vert ^2=\Vert Ty^+\Vert ^2+\Vert Ty^-\Vert ^2.\)

There are three cases for x and \(y'\).

(i) Suppose that x and \(y'\) are linearly dependent, that is, \(x= \mu y'\) for some \(\mu \in {\mathbb {C}}\). Then

$$\begin{aligned} \lambda (y')&=\lambda (x). \end{aligned}$$
(2.4)

If \(\lambda (y^+)<\lambda (y^-)\), then \(\frac{\Vert T(y^+)\Vert }{\Vert y^+\Vert }<\frac{\Vert T(y^-)\Vert }{\Vert y^-\Vert }\).

Similarly, if \(\lambda (y^-)<\lambda (y^+)\), then \(\frac{\Vert T(y^-)\Vert }{\Vert y^-\Vert }<\frac{\Vert T(y^+)\Vert }{\Vert y^+\Vert }\).Thus

$$\begin{aligned} \frac{\Vert T(y^+)\Vert ^2}{\Vert y^+\Vert ^2} \le \frac{\Vert T(y^+)\Vert ^2 +\Vert T(y^-)\Vert ^2}{\Vert y^+\Vert ^2+\Vert y^-\Vert ^2}\le \frac{\Vert T(y^-)\Vert ^2}{\Vert y^-\Vert ^2} \end{aligned}$$

or

$$\begin{aligned} \frac{\Vert T(y^-)\Vert ^2}{\Vert y^-\Vert ^2} \le \frac{\Vert T(y^+)\Vert ^2 +\Vert T(y^-)\Vert ^2}{\Vert y^+\Vert ^2+\Vert y^-\Vert ^2}\le \frac{\Vert T(y^+\Vert ^2}{\Vert y^+\Vert ^2}. \end{aligned}$$

That means \(\lambda (y')^2 \le \lambda (y)^2\), so equivalently from (2.4), \(\lambda (x) \le \lambda (y)\). Put \(\delta _1=1\).

(ii) Suppose that x and \(y'\) are linearly independent and \(x[\perp ] y'\).

Define \(u=\frac{x}{\Vert x\Vert },\quad v=\frac{y'}{\Vert y'\Vert }\). Then \(\Vert u\Vert =\Vert v\Vert =1,\quad \Vert T(u)\Vert =\lambda (x)\) and \(\Vert T(v)\Vert =\lambda (y')\); also \([u,v]=0\) and \(J_1u+v[\perp ] u-J_1v\). By the assumption

$$\begin{aligned} J_2 T(u)+T(v)[\perp ] ^\varepsilon T(u)-J_2T(v). \end{aligned}$$

Thus \(|[J_2 T(u)+T(v), T(u)-J_2T(v)]|\le \varepsilon \Vert J_2 T(u)+T(v)\Vert \Vert T(u)-J_2T(v)\Vert \). Lemma 2.1 implies that

$$\begin{aligned} |\Vert T(u)\Vert ^2-\Vert T(v)\Vert ^2|\le \varepsilon (\Vert T(u)\Vert ^2 +\Vert T(v)\Vert ^2), \end{aligned}$$

that is, \(|\lambda (x)^2-\lambda (y')^2| \le \varepsilon (\lambda (x)^2+\lambda (y')^2)\) and we reach \(\lambda (x)\le \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\lambda (y')\). So by (2.4), we obtain

$$\begin{aligned} \lambda (x) \le \sqrt{\frac{1+\varepsilon }{1-\varepsilon }} \lambda (y') \le \sqrt{\frac{1+\varepsilon }{1-\varepsilon }} \lambda (y). \end{aligned}$$
(2.5)

Then \(\lambda (x)\le \sqrt{\frac{1+\varepsilon }{1-\varepsilon }} \lambda (y)\). Put \(\delta _1=\sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\).

(iii) Now, we assume that x and \(y'\) are linearly independent but not orthogonal. By Lemma 2.8, we can choose two nonzero elements \(x_1\) and \(x_2\) in \({\mathcal {K}}_1\) such that \(x=x_1+x_2,\,\,\,x_1\in \langle y'\rangle \), and \(x_2[\perp ]x_1\). \(\Vert x\Vert ^2 =\Vert x_1\Vert ^2 +\Vert x_2\Vert ^2\) and therefore

$$\begin{aligned} \Vert x_1\Vert \le \Vert x\Vert ,\quad \Vert x_2\Vert \le \Vert x\Vert ,\quad \Vert x_1\Vert \Vert x_2\Vert \le \Vert x\Vert ^2. \end{aligned}$$

Since \(x_1=\mu y'\) for some \(\mu \in {\mathbb {C}}\) by (i), \(\lambda (x_1)=\lambda (y')\) and since \(x_2[\perp ]x_1\) by (ii), we have

$$\begin{aligned} \sqrt{\frac{1-\varepsilon }{1+\varepsilon }}\lambda (y')\le \lambda (x_2)\le \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\lambda (y'). \end{aligned}$$

We prove that \(| [J_2T(x_1), T(x_2)]|=|[T(x_1), T(x_2)]|\)

$$\begin{aligned}&T(x_1)\in {\mathcal {K}}_2^-\Rightarrow [J_2T(x_1), T(x_2)]=[-T(x_1), T(x_2)]=-[T(x_1), T(x_2)] \\&T(x_1)\in {\mathcal {K}}_2^+\Rightarrow [J_2T(x_1), T(x_2)]=[T(x_1), T(x_2)]. \end{aligned}$$

Therefore

$$\begin{aligned}{}[J_2T(x_1), T(x_2)]=\pm [T(x_1), T(x_2)]\Rightarrow | [J_2T(x_1), T(x_2)]|=|[T(x_1), T(x_2)]| \end{aligned}$$

and also since \(T(x_1)=T(\lambda y')\) that is \(T(x_1)\in {\mathcal {K}}_2^+\) or \(T(x_1)\in {\mathcal {K}}_2^-\), we get \(|[J_2T(x_1),T(x_2)]|=|[T(x_1),T(x_2)]|\)

$$\begin{aligned} \mathrm{Re} [J_2T(x_1), T(x_2) ]&\le |[J_2T(x_1),T(x_2)]|=|[T(x_1),T(x_2)]|\le \varepsilon \Vert T(x_1)\Vert \Vert T(x_2)\Vert \\&=\varepsilon \lambda (y')\Vert x_1\Vert \lambda (x_2)\Vert x_2\Vert \le \varepsilon \sqrt{\frac{1-\varepsilon }{1+\varepsilon }} \lambda (y')^2 \Vert x\Vert ^2. \end{aligned}$$

Therefore

$$\begin{aligned} \Vert T(x)\Vert ^2&= \Vert T(x_1)+T(x_2)\Vert ^2\\&=\big [J_2\big (T(x_1)\big )+J_2\big (T(x_2)\big ),T(x_1)+T(x_2)\big ]\\&= \Vert T(x_1)\Vert ^2 +\Vert T(x_2)\Vert ^2 +2 \mathrm{Re} [J_2T(x_1), T(x_2)]\\&\le \lambda (y')^2\Vert x_1\Vert ^2 +\frac{1+\varepsilon }{1-\varepsilon }\lambda (y')^2\Vert x_2\Vert ^2+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\lambda (y')^2\Vert x\Vert ^2\\&=\lambda (y')^2 \left( \Vert x_1\Vert ^2 +\Vert x_2\Vert ^2 + \big ( \frac{1+\varepsilon }{1-\varepsilon }-1\big ) \Vert x_2\Vert ^2 +2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\Vert x\Vert ^2\right) \\&\le \lambda (y')^2 \Big ( \Vert x\Vert ^2 +\big (\frac{1+\varepsilon }{1-\varepsilon }-1\big ) \Vert x\Vert ^2+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\Vert x\Vert ^2\Big )\\&=\lambda (y')^2 \Big (\frac{1+\varepsilon }{1-\varepsilon }+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\Big ) \Vert x\Vert ^2, \end{aligned}$$

and we get

$$\begin{aligned} \lambda (x)&\le \lambda (y') \sqrt{\frac{1+\varepsilon }{1-\varepsilon }+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}} \\&\le \lambda (y) \sqrt{\frac{1+\varepsilon }{1-\varepsilon }+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}} . \end{aligned}$$
(2.6)

Put \(\delta _1=\sqrt{\Big (\frac{1+\varepsilon }{1-\varepsilon }+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\Big )}\) and hence \(\lambda (x)\le \delta _1\lambda (y)\).

If x and y are arbitrary, then \(\delta _1=\max \{1,\sqrt{\frac{1+\varepsilon }{1-\varepsilon }+2\varepsilon \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}} , \sqrt{\frac{1+\varepsilon }{1-\varepsilon }}\}\) and we reach (2.3).

Let \(\ker T \ne \{0\}\). Then there exists a nonzero element \(y\in \ker T \) such that \(T(y)=0\), so \(\lambda (y)=0\). By (2.3), \(\lambda (x)=0\) for all \(x\ne 0\), that is \(T\equiv 0\), which is contrary to the assumptions. Therefore \(\ker T=\{0\}\) and T is injective.

Now, we show that T is continuous. Fix a nonzero element \(y_0\) of \({\mathcal {K}}_1\), and take \(\gamma =\lambda (y_0)>0\). From (2.3), we have

$$\begin{aligned} \frac{1}{\delta _1}\gamma \Vert x\Vert \le \Vert T(x)\Vert \le \delta _1 \gamma \Vert x\Vert , \quad x\in {\mathcal {K}}_1. \end{aligned}$$
(2.7)

This inequality gives the continuity of T. Moreover, (2.7) gives that

$$\begin{aligned} \big |\Vert T(x)\Vert -\gamma \Vert x\Vert \big |&\le (\delta _1-1) \gamma \Vert x\Vert , \quad x\in {\mathcal {K}}_1, \end{aligned}$$
(2.8)
$$\begin{aligned} \big |\Vert T(x)\Vert ^2-\gamma ^2\Vert x\Vert ^2\big |&\le (\delta _1^2-1) \gamma ^2\Vert x\Vert ^2, \quad x\in {\mathcal {K}}_1, \end{aligned}$$
(2.9)

and

$$\begin{aligned} \frac{1}{\delta _1}\Vert T(x)\Vert \le \gamma \Vert x\Vert \le \delta _1 \Vert T(x)\Vert , \quad x\in {\mathcal {K}}_1. \end{aligned}$$

In a similar fashion, from inequalities (2.8) and (2.9), we get

$$\begin{aligned} \big |\Vert T(x)\Vert -\gamma \Vert x\Vert \big |&\le (\delta _1-1) \gamma \Vert T(x)\Vert , \quad x\in {\mathcal {K}}_1, \end{aligned}$$
(2.10)
$$\begin{aligned} \big |\Vert T(x)\Vert ^2-\gamma ^2\Vert x\Vert ^2\big |&\le (\delta _1^2-1) \gamma ^2\Vert T(x)\Vert ^2, \quad x\in {\mathcal {K}}_1. \end{aligned}$$
(2.11)

Let \(x,y\in {\mathcal {K}}_1\) be arbitrary. Then

$$\begin{aligned} \Big |[T(x), T(y)]-\gamma ^2 [x,y]\Big |= & {} \Big | \frac{1}{4} (\Vert T(x)+T(y)\Vert ^2 -\Vert T(x)-T(y)\Vert ^2)\\&\quad -\frac{1}{2} (\Vert T(x^-)+T(y^-)\Vert ^2 -\Vert T(x^-)-T(y^-)\Vert ^2)\\&\quad +\frac{i}{4} (\Vert T(x)+iT(y)\Vert ^2-\Vert T(x)-iT(y)\Vert ^2)\\&\quad -\frac{i}{2}(\Vert (Tx)^-+i(Ty)^-\Vert ^2-\Vert (Tx)^--i(Ty)^-\Vert ^2)\\&\quad -\gamma ^2 \Big (\frac{1}{4} (\Vert x+y\Vert ^2 -\Vert x-y\Vert ^2) \\&\quad -\frac{1}{2} (\Vert x^-+y^-\Vert ^2 -\Vert x^--y^-\Vert ^2)\\&\quad +\frac{i}{4} (\Vert x+iy\Vert ^2-\Vert x-iy\Vert ^2)\\&\quad -\frac{i}{2}(\Vert x^-+iy^-\Vert ^2-\Vert x^--iy^-\Vert ^2)\Big )\Big |\\&\le \frac{1}{4} \Big |\Vert T(x)+T(y)\Vert ^2 -\gamma ^2 \Vert x+y\Vert ^2\Big |\\&+\frac{1}{4}\Big | \Vert T(x)-T(y)\Vert ^2 -\gamma ^2 \Vert x-y\Vert ^2\Big |\\&+\frac{1}{2}\Big | \Vert T(x^-)+T(y^-)\Vert ^2 -\gamma ^2 \Vert x^-+y^-\Vert ^2\Big |\\&+\frac{1}{2}\Big | \Vert T(x^-)-T(y^-)\Vert ^2 -\gamma ^2 \Vert x^--y^-\Vert ^2\Big |\\&+\Big |\frac{i}{4} \Big (\Vert T(x)+iT(y)\Vert ^2 -\gamma ^2 \Vert x+iy\Vert ^2\Big )\Big |\\&+\Big | -\frac{i}{4}\Big ( \Vert T(x)-iT(y)\Vert ^2 -\gamma ^2 \Vert x-iy\Vert ^2\Big )\Big |\\&+\Big |-\frac{i}{2}\Big ( \Vert T(x^-)+iT(y^-)\Vert ^2 -\gamma ^2 \Vert x^-+iy^-\Vert ^2\Big )\Big |\\&+\Big | \frac{i}{2}\Big ( \Vert T(x^-)-iT(y^-)\Vert ^2 -\gamma ^2 \Vert x^--iy^-\Vert ^2\Big )\Big |. \end{aligned}$$

From inequality (2.9), we get

$$\begin{aligned} \Big |[T(x), T(y)]-\gamma ^2 [x,y]\Big |&\le \frac{1}{4} \gamma ^2(\delta _1^2 -1)(\Vert x+y\Vert ^2 +\Vert x-y\Vert ^2)\\&\quad +\frac{1}{2} \gamma ^2(\delta _1^2 -1)(\Vert x^-+y^-\Vert ^2+\Vert x^--y^-\Vert ^2)\\&\quad +\frac{1}{4} \gamma ^2(\delta _1^2 -1)(\Vert x-iy\Vert ^2+\Vert x+iy\Vert ^2)\\&\quad +\frac{1}{2} \gamma ^2(\delta _1^2 -1)(\Vert x^--iy^-\Vert ^2+\Vert x^-+iy^-\Vert ^2)\\&\le \gamma ^2 (\delta _1^2 -1)\frac{1}{4} \Big ( 4\Vert x\Vert ^2+4\Vert y\Vert ^2\Big )+\frac{1}{2} \Big ( 4\Vert x^-\Vert ^2+4\Vert y^-\Vert ^2\Big )\\&=\gamma ^2 (\delta _1^2 -1) (\Vert x\Vert ^2 +\Vert y\Vert ^2+2\Vert x^-\Vert ^2 +2\Vert y^-\Vert ^2\\&\le \gamma ^2 (\delta _1^2 -1) (\Vert x\Vert ^2+\Vert y\Vert ^2+2\Vert x\Vert ^2+2\Vert y\Vert ^2)\\&\le 3 \gamma ^2 (\delta _1^2 -1) (\Vert x\Vert ^2+\Vert y\Vert ^2). \end{aligned}$$
(2.12)

Similarly, from inequality (2.11), we obtain

$$\begin{aligned} \Big |[T(x), T(y)]-\gamma ^2 [x,y]\Big |&\le 3(\delta _1^2-1) \Big (\Vert T(x)\Vert ^2+\Vert T(y)\Vert ^2\Big ). \end{aligned}$$

Thus

$$\begin{aligned} \Big |[T(x), T(y)]-\gamma ^2 [x,y]\Big |\le & {} 3(\delta _1^2 -1)\min \{\gamma ^2 \Vert x\Vert ^2 \nonumber \\&+\gamma ^2 \Vert y\Vert ^2, \ \Vert T(x)\Vert ^2 +\Vert T(y)\Vert ^2 \}. \end{aligned}$$
(2.13)

Now, suppose that x and y are two nonzero elements in \( {\mathcal {K}}_1\). By applying (2.12) to vectors \(\frac{x}{\Vert x\Vert }\) and \(\frac{y}{\Vert y\Vert }\), we get

$$\begin{aligned} \Big |[\frac{T(x)}{\Vert x\Vert },\frac{ T(y)}{\Vert y\Vert }]-\gamma ^2 \Big [\frac{x}{\Vert x\Vert },\frac{y}{\Vert y\Vert }\Big ]\Big |&\le 3(\delta _1^2 -1) \Big ( \gamma ^2 \big \Vert \frac{x}{\Vert x\Vert }\big \Vert ^2 +\gamma ^2 \frac{y}{\Vert y\Vert }\big \Vert ^2 \Big )\\&\le 6\gamma ^2 (\delta _1^2-1). \end{aligned}$$

Hence

$$\begin{aligned} \Big |[T(x), T(y)]-\gamma ^2 [x,y]\Big | \le 6(\delta _1^2 -1)\gamma ^2 \Vert x\Vert \Vert y\Vert . \end{aligned}$$

Applying (2.13) to vectors \(\frac{x}{\Vert T(x)\Vert }\) and \(\frac{y}{\Vert T(y)\Vert }\), we arrive at

$$\begin{aligned} \Big |[T(x), T(y)]-\gamma ^2 [x,y]\Big | \le 6(\delta _1^2 -1) \Vert T(x)\Vert \Vert T(y)\Vert . \end{aligned}$$
(2.14)

Since \(6(\delta _1^2-1)=\delta \), equation (2.2) follows for all \(x,y\in {\mathcal {K}}_1\).

Conversely, if \(T: {\mathcal {K}}_1 \rightarrow {\mathcal {K}}_2\) satisfies (2.2) with some \(\delta \ge 0\) and \(\gamma >0\), then by Lemma 2.3, T is quasilinear.

If \(x[\perp ]y\), then \([x,y]=0\), and by (2.2), we reach \(|[Tx, Ty]|\le \delta \Vert Tx \Vert \Vert Ty\Vert \). This means that

\(Tx[\perp ] ^\delta Ty\). Thus T is an approximately orthogonality preserving map. \(\square \)