Abstract

This paper is concerned with the theory for 𝐽-Hermitian subspaces. The defect index of a 𝐽-Hermitian subspace is defined, and a formula for the defect index is established; the result that every 𝐽-Hermitian subspace has a 𝐽-self-adjoint subspace extension is obtained; all the 𝐽-self-adjoint subspace extensions of a 𝐽-Hermitian subspace are characterized. This theory will provide a fundamental basis for characterizations of 𝐽-self-adjoint extensions for linear nonsymmetric expressions on general time scales in terms of boundary conditions, including both differential and difference cases.

1. Introduction

The spectral theory for differential and difference has been investigated extensively. In general, under certain definiteness conditions, a formally symmetric differential expression can generate a minimal operator which is symmetric, that is, a densely defined Hermitian operator, in a related Hilbert space and its adjoint, is the corresponding maximal operator (see, e.g., [13]). There are many results on self-adjoint extensions of the minimal operators since self-adjoint extension problems are fundamental in the study of spectral theory for differential expressions [26]. However, for some formally symmetric differential expressions, their minimal operators may be nondensely defined, or their maximal operators may be multivalued (e.g., [7, Example 2.2]). Further, for a formally symmetric difference expression, even a second-order one, its minimal operator is nondensely defined, and its maximal operator is multivalued in the related Hilbert space in general [8]. Therefore, the classical von Neumann self-adjoint extension theory and the Glazman-Krein-Naimark (GKN) theory for symmetric operators are not applicable in these cases.

The appropriate framework is linear subspaces (linear relations in the terminology of [7, 9, 10]) in a Hilbert space to study the linear differential or difference expressions for which the corresponding operators are nondensely defined or multivalued. Lesch and Malamud studied formally symmetric Hamiltonian systems in the framework of linear subspaces [7]. Coddington studied self-adjoint extensions of Hermitian subspaces in a product space [11]. He had extended the von Neumann self-adjoint extension theory for symmetric operators to Hermitian subspaces. By applying the relevant results in [11], Shi established the GKN theory for Hermitian subspaces [12]. For more results about nondensely defined Hermitian operators or Hermitian subspaces, we refer to [1315].

The study of spectral problems involving linear differential and difference expressions with complex-valued coefficients is becoming a well-established area of analysis, and many results have been obtained [1, 1622]. Such expressions are not formally symmetric in general, and hence, the spectral theory of self-adjoint subspaces is not applicable. To study such problems, Glazman introduced a concept of 𝐽-symmetric operators in [23], where 𝐽 is a conjugation operator given in Section 2. The minimal operators generated by certain differential and difference expressions with complex-valued coefficients are 𝐽-symmetric operators in the related Hilbert spaces (e.g., [18, 24, 25]). 𝐽-self-adjoint extension problems are also fundamental in the spectral theory for such expressions. Many results have been obtained on 𝐽-self-adjoint extensions [2427]. Knowles gave a complete solution to the problem of describing all the 𝐽-self-adjoint extensions of any given 𝐽-symmetric operator provided the regularity field of this operator is nonempty [26]. Given a differential or difference expression, it is in practice difficult however to determine whether the appropriate 𝐽-symmetric operator has empty or nonempty regularity field. Therefore, Race established the theory for 𝐽-self-adjoint extensions of 𝐽-symmetric operators without the restrictions on the regularity fields [24]. However, the appropriate framework is also linear subspaces in a Hilbert space to study the linear nonsymmetric differential or difference expressions for which the corresponding minimal operators are nondensely defined, or the corresponding maximal operators are multivalued. So, the 𝐽-self-adjoint extension theory mentioned the above needs to be extended to linear subspace when we consider the nonsymmetric Hamiltonian systems which induce the nondensely defined or multivalued operators.

In this present paper, the concept of the defect indices of 𝐽-Hermitian subspaces is given and a formula for the defect indices is obtained. Further, the result that every 𝐽-Hermitian subspace has a 𝐽-self-adjoint subspace extension is given, and the characterizations for all the 𝐽-self-adjoint subspace extensions of a 𝐽-Hermitian subspace are established, which can be regarded as the GKN theorem for 𝐽-Hermitian subspaces.

Remark 1.1. We will apply the results obtained in the present paper to characterizations of 𝐽-self-adjoint extensions for linear Hamiltonian difference systems in terms of boundary conditions in the near future.

The rest of this present paper is organized as follows. In Section 2, some basic concepts and fundamental results about linear subspaces are introduced. In Section 3, the defect index of a 𝐽-Hermitian subspace is defined, and a formula for the defect index is given. Section 4 pays attention to the existence of 𝐽-self-adjoint subspace extensions and the GKN theorem for 𝐽-Hermitian subspace.

2. Preliminaries

In this section, we introduce some basic concepts and give some fundamental results about linear subspaces in a product space.

Let 𝑋 be a complex Hilbert space with the inner product ,. The norm is defined by 𝑓=𝑓,𝑓1/2 for 𝑓𝑋. Let 𝑋2 be the product space 𝑋×𝑋. By definition, the elements of 𝑋2 consist of all possible ordered pairs (𝑥,𝑓) with 𝑥𝑋 and 𝑓𝑋, and for arbitrary two elements (𝑥,𝑓), (𝑦,𝑔)𝑋2 and 𝛼𝐶,𝛼(𝑥,𝑓)=(𝛼𝑥,𝛼𝑓),(𝑥,𝑓)+(𝑦,𝑔)=(𝑥+𝑦,𝑓+𝑔).(2.1) The null element of 𝑋2 is (0,0). The inner product in 𝑋2 is defined by(𝑥,𝑓),(𝑦,𝑔)=𝑥,𝑦+𝑓,𝑔,(𝑥,𝑓),(𝑦,𝑔)𝑋2,(2.2) and denotes the induced norm.

Let 𝑇 be a linear subspace in 𝑋2 which is called to be a linear relation in [7, 9, 10]. For brevity, a linear subspace is only called a subspace. For subspace 𝑇 in 𝑋2, we shall use the following definitions and notations:𝐷(𝑇)={𝑥𝑋(𝑥,𝑓)𝑇forsome𝑅𝑓𝑋},(𝑇)={𝑓𝑋(𝑥,𝑓)𝑇forsome𝑇𝑥𝑋},𝑇(𝑥)={𝑓𝑋(𝑥,𝑓)𝑇},ker(𝑇)={𝑥𝑋(𝑥,0)𝑇},1={(𝑓,𝑥)(𝑥,𝑓)𝑇},𝑇𝜆={(𝑥,𝑓𝜆𝑥)(𝑥,𝑓)𝑇}.(2.3) Clearly, 𝑇(0)={0} if and only if 𝑇 can determine a unique linear operator from 𝐷(𝑇) into 𝑋 whose graph is 𝑇, and 𝑇1 is closed if and only if 𝑇 is closed. Since the graph of a linear operator in 𝑋 is a subspace in 𝑋2 and a linear operator is identified with its graph, the concept of subspaces in 𝑋2 generalizes that of linear operators in 𝑋.

Definition 2.1 (see [11]). Let 𝑇 be a subspace in 𝑋2.(1)Its adjoint, 𝑇, is defined by 𝑇=(𝑦,𝑔)𝑋2.𝑓,𝑦=𝑥,𝑔(𝑥,𝑓)𝑇(2.4)(2)𝑇 is said to be a Hermitian subspace if 𝑇𝑇.(3)𝑇 is said to be a self-adjoint subspace if 𝑇=𝑇.

Lemma 2.2 (see [11]). Let 𝑇 be a subspace in 𝑋2, then 𝑇 is a closed subspace in 𝑋2, 𝑇=(𝑇), and 𝑇=𝑇, where 𝑇 is the closure of 𝑇.

Definition 2.3. An operator 𝐽 defined on 𝑋 is said to be a conjugation operator if for all 𝑥,𝑦𝑋, 𝐽𝑥,𝐽𝑦=𝑦,𝑥,𝐽2𝑥=𝑥.(2.5)

It can be verified that 𝐽 is a conjugate linear, that is, 𝐽(𝑥+𝑦)=𝐽𝑥+𝐽𝑦 and 𝐽(𝜆𝑥)=𝜆𝐽𝑥 for 𝑥,𝑦𝑋 and 𝜆𝐶, and norm-preserving bijection on 𝑋 satisfying𝐽𝑥,𝑦=𝐽𝑦,𝑥𝑥,𝑦𝑋.(2.6) For example, the complex conjugation 𝑥𝑥 in any 𝐿2 space is a conjugation operator on 𝐿2.

Definition 2.4. Let 𝑇 be a subspace in 𝑋2, and let 𝐽 be a conjugation operator.(1)Its 𝐽-adjoint, 𝑇𝐽, is defined by 𝑇𝐽=(𝑦,𝑔)𝑋2.𝑓,𝐽𝑦=𝑥,𝐽𝑔(𝑥,𝑓)𝑇(2.7)(2)𝑇 is said to be a 𝐽-Hermitian subspace if 𝑇𝑇𝐽.(3)𝑇 is said to be a 𝐽-self-adjoint subspace if 𝑇=𝑇𝐽.

Remark 2.5. (1) It can be easily verified that 𝑇𝐽 is a closed subspace. Consequently, a 𝐽-self-adjoint subspace 𝑇 is a closed subspace since 𝑇=𝑇𝐽. In addition, 𝑆𝐽𝑇𝐽 if 𝑇𝑆.
(2) From the definition, we have that 𝑓,𝐽𝑦=𝑥,𝐽𝑔 for all (𝑥,𝑓)𝑇 and (𝑦,𝑔)𝑇𝐽, and that 𝑇 is a 𝐽-Hermitian subspace if and only if 𝑓,𝐽𝑦=𝑥,𝐽𝑔(𝑥,𝑓),(𝑦,𝑔)𝑇.(2.8)
(3) The concepts of 𝐽-Hermitian and 𝐽-self-adjoint subspaces generalize those of 𝐽-symmetric and 𝐽-self-adjoint operators, respectively (see, e.g., [1, 24] for the concepts of 𝐽-symmetric and 𝐽-self-adjoint operators).

Lemma 2.6. Let 𝑇 be a subspace in 𝑋2, then(1)𝑇={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)𝑇𝐽},(2)𝑇𝐽={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)𝑇}.

Proof. Result (2) follows from result (1) and the second relation of (2.5). So, one needs only to prove result (1). Set 𝐷1={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)𝑇𝐽}. Let (𝑦,𝑔)𝑇, then 𝑓,𝑦=𝑥,𝑔 for all (𝑥,𝑓)𝑇. So, the second relation of (2.5) yields that 𝑓,𝐽(𝐽𝑦)=𝑥,𝐽(𝐽𝑔) for all (𝑥,𝑓)𝑇. Hence, (𝐽𝑦,𝐽𝑔)𝑇𝐽. Then (𝑦,𝑔)=(𝐽2𝑦,𝐽2𝑔)𝐷1, which implies that 𝑇𝐷1. Conversely, let (𝑦,𝑔)𝐷1, then there exists (̃𝑦,̃𝑔)𝑇𝐽 such that (𝑦,𝑔)=(𝐽̃𝑦,𝐽̃𝑔). It follows from (̃𝑦,̃𝑔)𝑇𝐽 that 𝑓,𝐽̃𝑦=𝑥,𝐽̃𝑔 for all (𝑥,𝑓)𝑇, which implies that (𝐽̃𝑦,𝐽̃𝑔)𝑇, that is, (𝑦,𝑔)𝑇. So, 𝐷1𝑇. Consequently, 𝑇=𝐷1, and result (1) holds.

Remark 2.7. Let 𝑇 be a subspace in 𝑋2, then from Lemmas 2.2 and 2.6, and the closedness of 𝑇𝐽, one has that 𝑇𝐽=(𝑇)𝐽, and 𝑇 is 𝐽-Hermitian if 𝑇 is 𝐽-Hermitian.

Lemma 2.8. Let 𝑇 be a closed 𝐽-Hermitian subspace, then (𝑦,𝑔)𝑇 if and only if (𝑦,𝑔)𝑇𝐽 and 𝑓,𝐽𝑦=𝑥,𝐽𝑔 for all (𝑥,𝑓)𝑇𝐽.

Proof. Let 𝑇 be a closed 𝐽-Hermitian subspace. Clearly, the necessity holds by (2) of Remark 2.5. Now, consider the sufficiency. Suppose that (𝑦,𝑔)𝑇𝐽 and 𝑓,𝐽𝑦=𝑥,𝐽𝑔 for all (𝑥,𝑓)𝑇𝐽, then we get from (2.6) that 𝑦,𝐽𝑓=𝑔,𝐽𝑥 for all (𝑥,𝑓)𝑇𝐽. This, together with (1) of Lemma 2.6, implies that 𝑦,𝑓=𝑔,̃𝑥 for all (̃𝑥,𝑓)𝑇. So, (𝑦,𝑔)𝑇, and hence, (𝑦,𝑔)𝑇 by Lemma 2.2. So, the sufficiency holds.

Lemma 2.9. Let 𝑇 be a closed 𝐽-Hermitian subspace in 𝑋2, then(1)(𝑇𝐽)={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)𝑇},(2)𝑇={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)(𝑇𝐽)}.

Proof. Since result (1) and the second relation of (2.5) imply that result (2) holds, it suffices to prove result (1). Set 𝐷={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)𝑇}. Let (𝑦,𝑔)𝐷, then there exists (̃𝑦,̃𝑔)𝑇 such that (𝑦,𝑔)=(𝐽̃𝑦,𝐽̃𝑔). By Lemma 2.8, (̃𝑦,̃𝑔)𝑇 implies that 𝑓,𝐽̃𝑦=𝑥,𝐽̃𝑔 for all (𝑥,𝑓)𝑇𝐽. Then (𝐽̃𝑦,𝐽̃𝑔)(𝑇𝐽), that is, (𝑦,𝑔)(𝑇𝐽). So, 𝐷(𝑇𝐽). Conversely, let (𝑦,𝑔)(𝑇𝐽), then 𝑓,𝑦=𝑥,𝑔 for all (𝑥,𝑓)𝑇𝐽, that is, 𝑓,𝐽(𝐽𝑦)=𝑥,𝐽(𝐽𝑔)(𝑥,𝑓)𝑇𝐽.(2.9) Clearly, (2.9) holds for all (𝑥,𝑓)𝑇 since 𝑇𝑇𝐽. So, (𝐽𝑦,𝐽𝑔)𝑇𝐽. This, together with (2.9) and Lemma 2.8, implies that (𝐽𝑦,𝐽𝑔)𝑇. So, (𝐽(𝐽𝑦),𝐽(𝐽𝑔))𝐷, that is, (𝑦,𝑔)𝐷. Then, (𝑇𝐽)𝐷 and then (𝑇𝐽)=𝐷.

Remark 2.10. Since 𝑇𝐽=(𝑇)𝐽 by Remark 2.7, Lemma 2.9 yields that (𝑇𝐽)={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)𝑇} and 𝑇={(𝐽𝑦,𝐽𝑔)(𝑦,𝑔)(𝑇𝐽)} for a 𝐽-Hermitian subspace 𝑇 which may not be closed.

3. Defect Index of a 𝐽-Hermitian Subspace

In this section, the definition of the defect index of a 𝐽-Hermitian subspace is introduced, and a formula for the defect index is obtained.

Let 𝑇 be a closed 𝐽-Hermitian subspace. It has been known that 𝑇𝐽 is a closed subspace by (1) of Remark 2.5. Then the closedness of 𝑇 and 𝑇𝐽 and 𝑇𝑇𝐽 gives that𝑇𝐽=𝑇𝒯,(3.1) where 𝒯 denotes the orthogonal complement of 𝑇 in 𝑇𝐽, that is, 𝒯=𝑇𝐽𝑇. Now, let 𝑆 be a closed 𝐽-Hermitian subspace extension of 𝑇, that is, 𝑇𝑆 and 𝑆 is 𝐽-Hermitian. Then, it follows from the closedness of 𝑆 and 𝑇 that there exists a unique subspace 𝐾𝑆,𝑇=𝑆𝑇 such that𝑆=𝑇𝐾𝑆,𝑇.(3.2) Clearly, 𝑆𝑇𝐽 since 𝑆𝐽𝑇𝐽 and 𝑆𝑆𝐽. Then by (3.1) and (3.2), 𝐾𝑆,𝑇 can be expressed as𝐾𝑆,𝑇=(𝑦,𝑔)𝒯thereexist𝑦1,𝑔1𝑦𝑇,2,𝑔2𝑆,suchthat𝑦2,𝑔2=𝑦1,𝑔1.+(𝑦,𝑔)(3.3) Further, we have the following result.

Theorem 3.1. Let 𝑇 be a closed 𝐽-Hermitian subspace, and let 𝑆 be a 𝐽-self-adjoint subspace extension (briefly, 𝐽-SSE) of 𝑇, that is, 𝑇𝑆 and 𝑆 is 𝐽-self-adjoint, then 𝑇dim𝐽𝑆𝑆=dim𝑇.(3.4)

Proof. If 𝑇 is a 𝐽-self-adjoint subspace, then 𝑇=𝑇𝐽 and 𝑇 is the only 𝐽-SSE of itself. So, (3.4) holds. Now, suppose that 𝑇 is 𝐽-Hermitian but not 𝐽-self-adjoint, that is, 𝑇𝑇𝐽 and 𝑇𝑇𝐽. It follows that (3.2) holds with 𝐾𝑆,𝑇{0}. Let dim𝑆/𝑇=𝑚, then (3.2) yields that dim𝐾𝑆,𝑇=𝑚. In the case of 𝑚<, let {(𝑥𝑗,𝑓𝑗)}𝑚𝑗=1 be a basis of 𝐾𝑆,𝑇, then 𝑥𝑆=𝑇span1,𝑓1,𝑥2,𝑓2𝑥,,𝑚,𝑓𝑚.(3.5) Define 𝑇𝑗𝑥=𝑇span1,𝑓1,𝑥2,𝑓2𝑥,,𝑗,𝑓𝑗,𝑗=1,2,,𝑚.(3.6) Clearly, 𝑇𝑇𝑗𝑇𝑗+1 for 𝑗=1,2,,𝑚1 and 𝑆=𝑆𝐽=𝑇𝑚𝐽𝑇𝑚1𝐽𝑇2𝐽𝑇1𝐽𝑇𝐽,(3.7) since 𝑇𝑇1𝑇2𝑇𝑚=𝑆 and 𝑆 is 𝐽-self-adjoint. Further, 𝑇𝑗 (𝑗=1,2,,𝑚) is a closed subspace since 𝑇 is closed. It holds that 𝑇𝑇𝑗𝑇𝑗+1,𝑗=1,2,,𝑚1.(3.8) Otherwise, for example, suppose that 𝑇1=𝑇2, then by Lemma 2.2, we have 𝑇1=𝑇1=𝑇2=𝑇2 since 𝑇1 and 𝑇2 are closed. It contradicts 𝑇1𝑇2. So, (3.8) holds. It follows from (3.8) and (2) of Lemma 2.6 that 𝑇𝐽𝑇𝑗𝐽𝑇𝑗+1𝐽,𝑗=1,2,,𝑚1.(3.9) We get from (3.7) and (3.9) that dim𝑇𝐽/𝑆𝑚=dim𝑆/𝑇.
In the case of 𝑚=+, we have the linear span of an infinite set in (3.5). So we can construct an infinite sequence of subspaces of the form (3.6) which satisfies the relations like those in (3.7) and (3.9). So, we have dim𝑇𝐽/𝑆=+=dim𝑆/𝑇.
Next, we prove that dim𝑆/𝑇dim𝑇𝐽/𝑆. Since 𝑇𝐽 and 𝑆 are closed subspaces, there exists uniquely a closed subspace 𝐾𝑇𝐽,𝑆=𝑇𝐽𝑆 such that 𝑇𝐽=𝑆𝐾𝑇𝐽,𝑆. Set dim𝑇𝐽/𝑆=𝑚. Then dim𝐾𝑇𝐽,𝑆=𝑚. If 𝑚<, let {(𝑦𝑗,𝑔𝑗)}𝑚𝑗=1 be a basis of 𝐾𝑇𝐽,𝑆, then𝑇𝐽𝑦=𝑆span1,𝑔1,𝑦2,𝑔2𝑦,,𝑚,𝑔𝑚.(3.10)Define 𝑆𝑗𝑦=𝑆span1,𝑔1,𝑦2,𝑔2𝑦,,𝑗,𝑔𝑗,𝑗=1,2,,𝑚.(3.11) Clearly, it holds that 𝑆𝐽=𝑆𝑆𝑗𝑆𝑗+1 for 𝑗=1,2,,𝑚1 and 𝑇𝐽=𝑆𝑚𝑆𝑚1𝑆2𝑆1𝑆=𝑆𝐽,(3.12) since 𝑆𝑆1𝑆2𝑆𝑚=𝑇𝐽 and 𝑆 is 𝐽-self-adjoint. Further, 𝑆𝑗 (𝑗=1,2,,𝑚) is a closed subspace since 𝑆 is closed. Similarly, it holds that 𝑆𝐽𝑆𝑗𝑆𝑗+1,𝑗=1,2,,𝑚1.(3.13) We get from (3.12) and (3.13) that dim(𝑆𝐽)/(𝑇𝐽)𝑚. It can be verified that 𝑆dim𝐽𝑇𝐽𝑆=dim𝑇(3.14) by Lemma 2.9. So, dim𝑆/𝑇𝑚=dim𝑇𝐽/𝑆. Further, it can be verified that dim𝑆/𝑇dim𝑇𝐽/𝑆 also holds for 𝑚=+. Based on the above discussions, (3.4) holds.

Remark 3.2. (1) From Theorem 3.1 and its proof, one has that if one of the two dimensions in (3.4) is finite, so is the other and they are equal, and if one of the two dimensions is infinite, so is the other. Here, there is no distinction between degrees of infinity.
(2) The case for 𝐽-symmetric operators was established in [24, Theorem 3.1].

Remark 3.3. Note that 𝑇𝐽=(𝑇)𝐽. We get from Theorem 3.1 that if 𝑆 is a 𝐽-SSE of 𝑇, which may not be closed, then it holds that 𝑇dim𝐽𝑇𝑆=2dim𝑇.(3.15)
Now, we give the concept of the defect index of a 𝐽-Hermitian subspace. The concept of the defect index of a 𝐽-symmetric operator in 𝑋 was given by [24, Definition 3.2].

Definition 3.4. Let 𝑇 be a 𝐽-Hermitian subspace, then 𝑑(𝑇)=(1/2)dim𝑇𝐽/𝑇 is called to be the defect index of 𝑇.

Remark 3.5. (1) It will be proved that every 𝐽-Hermitian subspace has a 𝐽-SSE in Theorem 4.3 in Section 4. So, by (3.15) we have that the defect index of every 𝐽-Hermitian subspace is a nonnegative integer.
(2) Since 𝑇𝐽=(𝑇)𝐽 by Remark 2.7 and every 𝐽-SSE is closed, we have that a 𝐽-symmetric subspace 𝑇 and its closure 𝑇 have the same defect index and the same 𝐽-SSEs.

Definition 3.6 (see [12]). Let 𝑇 be a subspace in 𝑋2. The set Γ(𝑇)={𝜆𝐂thereexists𝑐(𝜆)>0suchthat𝑓𝜆𝑥𝑐(𝜆)𝑥(𝑥,𝑓)𝑇}(3.16) is called to be the regularity field of 𝑇.
It is evident that Γ(𝑇)=Γ(𝑇) for a subspace 𝑇.

Lemma 3.7. Let 𝑇 be a subspace in 𝑋, then(1)𝑅(𝑇𝜆)=ker(𝑇𝜆) for each 𝜆𝐂,(2)for each 𝜆Γ(𝑇),𝑋=𝑅𝑇𝑇𝜆ker𝜆(orthogonalsumin𝑋),(3.17)(3)𝑋=𝑅(𝑇𝐽𝜆) for each 𝜆Γ(𝑇).

Proof. (1) Let 𝜆𝐂. It is clear that 𝑇ker𝜆=𝑥𝑋𝑥,𝜆𝑥𝑇.(3.18) For every 𝑥ker(𝑇𝜆), we have from (3.18) that (𝑥,𝜆𝑥)𝑇. So, 𝑔,𝑥=𝑦,𝜆𝑥 for all (𝑦,𝑔)𝑇, which implies that 𝑔𝜆𝑦,𝑥=0. Therefore, 𝑥𝑅(𝑇𝜆), and then ker(𝑇𝜆)𝑅(𝑇𝜆). Conversely, for every 𝑥𝑅(𝑇𝜆), we have that 𝑔𝜆𝑦,𝑥=0 for all (𝑦,𝑔)𝑇. It follows that 𝑔,𝑥=𝑦,𝜆𝑥 for all (𝑦,𝑔)𝑇. So, (𝑥,𝜆𝑥)𝑇, and hence 𝑥ker(𝑇𝜆) by (3.18). So, 𝑅(𝑇𝜆)ker(𝑇𝜆). Consequently, 𝑅(𝑇𝜆)=ker(𝑇𝜆), and result (1) is proved.
(2) By result (1) and Lemma 2.2, we have that 𝑅(𝑇𝜆)=ker((𝑇)𝜆)=ker(𝑇𝜆). So, by the projection theorem, in order to prove (3.17), it suffices to show that 𝑅(𝑇𝜆) with 𝜆Γ(𝑇) is closed in 𝑋. It is evident that 𝑇𝜆1=(𝑓𝜆𝑥,𝑥)(𝑥,𝑓)𝑇.(3.19) Let 𝜆Γ(𝑇). Since Γ(𝑇)=Γ(𝑇), one has that 𝜆Γ(𝑇), that is, there exists a constant 𝑐(𝜆)>0 such that (𝑓𝜆𝑥𝑐𝜆)𝑥(𝑥,𝑓)𝑇.(3.20) Then we get from (3.19) and (3.20) that (𝑇𝜆)1(0)={0}. So, (𝑇𝜆)1 determines a linear operator from 𝑅(𝑇𝜆) to 𝑋. Further, the closedness of 𝑇𝜆 and (3.20) imply that this operator is a closed and bounded operator. So, its domain 𝑅(𝑇𝜆) is closed in 𝑋. Therefore, (3.17) holds, and result (2) is proved.
(3) Let 𝜆Γ(𝑇). We first show that there exists a constant 𝑀>0 such that 𝑥𝑀𝑓𝜆𝑥(𝑥,𝑓)𝑇𝐽.(3.21) Assume the contrary, then there exists a sequence {𝑓𝑘𝜆𝑥𝑘}𝑘=1𝑅(𝑇𝐽𝜆) with 𝑓𝑘𝜆𝑥𝑘=1 (𝑘=1,2,) such that 𝑥𝑘>𝑘,𝑘=1,2,.(3.22) Clearly, (𝑥𝑘,𝑓𝑘)𝑇𝐽 and (𝑇)𝐽=𝑇𝐽 imply that (𝑥𝑘,𝑓𝑘)(𝑇)𝐽. So, 𝑔,𝐽𝑥𝑘=𝑦,𝐽𝑓𝑘 for all (𝑦,𝑔)𝑇 by (1) of Remark 2.5, which, together with 𝜆𝑦,𝐽𝑥𝑘=𝑦,𝐽(𝜆𝑥𝑘), implies that for 𝑘=1,2,, 𝑔𝜆𝑦,𝐽𝑥𝑘𝑓=𝑦,𝐽𝑘𝜆𝑥𝑘(𝑦,𝑔)𝑇.(3.23) Define 𝜙𝑘(𝑔𝜆𝑦)=𝑔𝜆𝑦,𝐽𝑥𝑘 for (𝑦,𝑔)𝑇. Then 𝜙𝑘, 𝑘=1,2, are linear functionals in 𝑅(𝑇𝜆). Since 𝜙𝑘(𝑔𝜆𝑦)=𝑦,𝐽(𝑓𝑘𝜆𝑥𝑘) by (3.23) and 𝐽(𝑓𝑘𝜆𝑥𝑘)=𝑓𝑘𝜆𝑥𝑘=1, we have that {𝜙𝑘(𝑔𝜆𝑦)}𝑘=1 is bounded for any given 𝑔𝜆𝑦𝑅(𝑇𝜆). Note that 𝑅(𝑇𝜆) with 𝜆Γ(𝑇) is closed by the proof of result (2), and hence it is a Hilbert space with the inner product ,. So, by the resonance theorem, {𝜙𝑘}𝑘=1 is bounded, that is, {𝐽𝑥𝑘}𝑘=1 is bounded. Since 𝐽𝑥𝑘=𝑥𝑘, 𝑘=1,2,, we have a contradiction with (3.22). So, (3.21) holds.
Inserting (3.21) into 𝑥+𝑓𝑓𝜆𝑥+(1+|𝜆|)𝑥, we get that ||𝜆||𝑀𝑥+𝑓1+1+𝑓𝜆𝑥.(3.24) It can be easily verified from the closedness of 𝑇𝐽 and (3.24) that 𝑅(𝑇𝐽𝜆) is closed in 𝑋. So, result (1) implies that 𝑇𝑋=𝑅𝐽𝑇𝜆ker𝐽𝜆.(3.25) By Remark 2.10, ker((𝑇𝐽)𝜆)={𝐽𝑦𝑦ker(𝑇𝜆)}, while it can be verified that ker(𝑇𝜆)={0} for 𝜆Γ(𝑇)=Γ(𝑇). Therefore, (3.25) yields that result (3) holds.

If Γ(𝑇), we have the following results which give a formula for the defect index of a 𝐽-Hermitian space.

Theorem 3.8. Assume that 𝑇 is a 𝐽-Hermitian subspace with Γ(𝑇). Let 𝜆Γ(𝑇), then 𝑇𝐽=𝑇̇+𝒩,(3.26) where 𝒩=(𝑦,𝑔)𝑇𝐽𝑇𝑔𝜆𝑦ker𝜆,(3.27)𝑑(𝑇)=dim𝑅(𝑇𝜆)𝑇=dimker𝜆.(3.28)

Proof. We first prove (3.26). Since 𝑇 is 𝐽-Hermitian, one has that 𝑇 is 𝐽-Hermitian, and hence 𝑇(𝑇)𝐽=𝑇𝐽. Clearly, 𝒩𝑇𝐽. So, 𝑇+𝒩𝑇𝐽. On the other hand, let (𝑥,𝑓)𝑇𝐽. It follows from (3.17) and 𝜆Γ(𝑇) that there exist (𝑦,𝑔)𝑇 and 𝑤ker(𝑇𝜆) such that 𝑓𝜆𝑥=𝑔𝜆𝑦+𝑤, that is, (𝑓𝑔)𝜆(𝑥𝑦)=𝑤. Let (̃𝑥,𝑓)=(𝑥𝑦,𝑓𝑔), then (𝑥,𝑓)=(𝑦,𝑔)+(̃𝑥,𝑓) and (̃𝑥,𝑓)𝒩. So, 𝑇𝐽𝑇+𝒩, and consequently, 𝑇𝐽=𝑇+𝒩.
Now, let (𝑢,)𝑇𝒩, then 𝜆𝑢𝑅𝑇𝑇𝜆,𝜆𝑢ker𝜆.(3.29) Consequently, 𝜆𝑢=0 by 𝑅(𝑇𝜆)=ker(𝑇𝜆), which can be obtained from (3.17). Since 𝜆Γ(𝑇)=Γ(𝑇), one has by Definition 3.6 that there exists a constant 𝑐(𝜆)>0 such that 0=𝜆𝑢𝑐(𝜆)𝑢. It follows that 𝑢=0, which, together with 𝜆𝑢=0, implies that =0. Then, (𝑢,)=(0,0) if (𝑢,)𝑇𝒩. So, (3.26) holds.
Next, we prove (3.28). Let 𝜆Γ(𝑇). Set 𝑈1={(𝑥,𝜆𝑥)𝑇𝐽}, then 𝑈1 is closed since 𝑇𝐽 is closed. Let 𝑈2=𝒩/𝑈1. We will show that dim𝑈2𝑇=dimker𝜆.(3.30) Let {𝜓𝑗}𝑚𝑗=1ker(𝑇𝜆) be linearly independent, then, by (3) of Lemma 3.7, there exists (𝑢𝑗,𝑗)𝑇𝐽 such that 𝜓𝑗=𝑗𝜆𝑢𝑗, 1𝑗𝑚. It follows from 𝜓𝑗ker(𝑇𝜆) that (𝑢𝑗,𝑗)𝑈2 for 1𝑗𝑚. In addition, if 𝑚𝑗=1𝑐𝑗𝑢𝑗,𝑗𝑈1,𝑐𝑗𝐂,(3.31) then 𝑚𝑗=1𝑐𝑗(𝑗𝜆𝑢𝑗)=0, that is, 𝑚𝑗=1𝑐𝑗𝜓𝑗=0. So, 𝑐𝑗=0 for 1𝑗𝑚, and hence {(𝑢𝑗,𝑗)}𝑚𝑗=1𝑈2 is linearly independent (mod 𝑈1). Conversely, let {(𝑢𝑗,𝑗)}𝑚𝑗=1𝑈2 be linearly independent (mod 𝑈1), and let 𝜓𝑗=𝑗𝜆𝑢𝑗, then 𝜓𝑗ker(𝑇𝜆). If 𝑚𝑗=1𝑐𝑗𝜓𝑗=0, then 𝑚𝑗=1𝑐𝑗(𝑢𝑗,𝑗)𝑈1. So, 𝑐𝑗=0 (1𝑗𝑚), and hence the set {𝜓𝑗}𝑚𝑗=1 is linearly independent. Based on the above discussions, (3.30) holds. On the other hand, it is evident that dim𝑥,𝜆𝑥𝑇𝑇=dimker𝜆.(3.32) Further, by Lemma 2.6, we have that dim𝑈1=dim𝑥,𝜆𝑥𝑇.(3.33) It follows from (3.30)–(3.33) that dim𝑈1=dim𝑈2, and hence (3.26) implies that 1𝑑(𝑇)=2𝑇dim𝒩=dimker𝜆.(3.34) So, (3.28) holds by (1) of Lemma 3.7.

Remark 3.9. From Theorem 3.8, one has the following result: for a 𝐽-Hermitian subspace 𝑇, dim𝑅(𝑇𝜆) and dimker(𝑇𝜆) are constants in Γ(𝑇) which are equal to the defect index of 𝑇. This result extends [24, Theorem 5.7] for 𝐽-symmetric operators to 𝐽-Hermitian subspaces. Similarly, there is no distinction between degrees of infinity.

4. 𝐽-Self-Adjoint Subspace Extensions of a 𝐽-Hermitian Subspace

In this section, we consider the existence of 𝐽-SSEs of a 𝐽-Hermitian space and the characterizations of all the 𝐽-SSEs.

Define the form [] as[](𝑥,𝑓)(𝑦,𝑔)=𝑓,𝐽𝑦𝑥,𝐽𝑔,(𝑥,𝑓),(𝑦,𝑔)𝑇𝐽.(4.1) Then, 𝑓,𝐽𝑦=𝑥,𝐽𝑔 if and only if [(𝑥,𝑓)(𝑦,𝑔)]=0. Further, for all 𝑌𝑗=(𝑥𝑗,𝑓𝑗)𝑇𝐽(𝑗=1,2,3) and 𝜇𝐂,𝑌3𝑌1+𝑌2=𝑌3𝑌1+𝑌3𝑌2,𝑌1+𝑌2𝑌3=𝑌1𝑌3+𝑌2𝑌3,𝜇𝑌1𝑌2=𝑌1𝜇𝑌2𝑌=𝜇1𝑌2,𝑌1𝑌2𝑌=2𝑌1.(4.2) Since the closure 𝑇 of a 𝐽-Hermitian subspace 𝑇 is also a 𝐽-Hermitian subspace by Remark 2.7, and 𝑇 and 𝑇 have the same defect indices and the same 𝐽-SSEs by (2) of Remark 3.5, we shall assume that 𝑇 is closed in the rest of this section. Let 𝑇 be a closed subspace in 𝑋2, and let 𝐾𝒯 be a subspace, where 𝒯 is given in (3.1). Let 𝐾𝐽|𝒯 be a restriction of 𝐾𝐽 to 𝒯, that is,𝐾𝐽||𝒯[]={(𝑦,𝑔)𝒯(𝑥,𝑓)(𝑦,𝑔)=0(𝑥,𝑓)𝐾},(4.3)

then 𝐾 is called to be 𝐽-Hermitian in 𝒯 if 𝐾𝐾𝐽|𝒯, and 𝐾 is called to be 𝐽-self-adjoint in 𝒯 if 𝐾=𝐾𝐽|𝒯.

Remark 4.1. From the definition, we have that [(𝑥,𝑓)(𝑦,𝑔)]=0 for all (𝑥,𝑓)𝐾 and (𝑦,𝑔)𝐾𝐽|𝒯, and 𝐾 is 𝐽-Hermitian in 𝒯 if and only if [(𝑥,𝑓)(𝑦,𝑔)]=0 for all (𝑥,𝑓), (𝑦,𝑔)𝐾.

Lemma 4.2. Let 𝑇 be a closed 𝐽-Hermitian subspace, and let 𝐾𝒯 be a subspace. Assume that 𝑆=𝑇𝐾, then(1)𝑆 is 𝐽-Hermitian if and only if 𝐾 is 𝐽-Hermitian in 𝒯,(2)𝑆 is 𝐽-self-adjoint if and only if 𝐾 is 𝐽-self-adjoint in 𝒯.

Proof. (1) Suppose that 𝑆 is 𝐽-Hermitian. It can be easily verified that 𝐾 is 𝐽-Hermitian in 𝒯 by (2) of Remark 2.5, 𝐾𝑆, and Remark 4.1. So, the necessity holds. We now prove the sufficiency. Suppose that 𝐾 is 𝐽-Hermitian in 𝒯. For all (𝑥,𝑓), (𝑦,𝑔)𝑆, we get from 𝑆=𝑇𝐾 that 𝑥(𝑥,𝑓)=1,𝑓1+𝑥2,𝑓2,𝑥1,𝑓1𝑥𝑇,2,𝑓2(𝑦𝐾,𝑦,𝑔)=1,𝑔1+𝑦2,𝑔2,𝑦1,𝑔1𝑦𝑇,2,𝑔2𝐾.(4.4) Since 𝑇 is 𝐽-Hermitian and 𝐾𝐾𝐽|𝒯𝑇𝐽, it can be obtained from (2) of Remark 2.5 and Remark 4.1 that []=𝑥(𝑥,𝑓)(𝑦,𝑔)1,𝑓1𝑦1,𝑔1+𝑥1,𝑓1𝑦2,𝑔2+𝑥2,𝑓2𝑦1,𝑔1+𝑥2,𝑓2𝑦2,𝑔2=0.(4.5) So, 𝑆 is 𝐽-Hermitian. The sufficiency holds, and result (1) is proved.
(2) First, consider the necessity. Suppose that 𝑆 is 𝐽-self-adjoint, then 𝐾 is 𝐽-Hermitian in 𝒯, that is, 𝐾𝐾𝐽|𝒯, by result (1). It suffices to show that 𝐾𝐽|𝒯𝐾. For any given (𝑥,𝑓)𝑆, there exist (𝑥1,𝑓1)𝑇 and (𝑥2,𝑓2)𝐾 such that the first relation of (4.4) holds. Let (𝑦,𝑔)𝐾𝐽|𝒯, then [(𝑥2,𝑓2)(𝑦,𝑔)]=0 by Remark 4.1. Note that (𝑦,𝑔)𝑇𝐽 by 𝐾𝐽|𝒯𝑇𝐽. Then [(𝑥1,𝑓1)(𝑦,𝑔)]=0 by (2) of Remark 2.5. Therefore, the first relation of (4.4) yields that []=𝑥(𝑥,𝑓)(𝑦,𝑔)1,𝑓1+𝑥(𝑦,𝑔)2,𝑓2(𝑦,𝑔)=0(𝑥,𝑓)𝑆.(4.6) So, (𝑦,𝑔)𝑆𝐽. Therefore, 𝑆=𝑆𝐽 yields that (𝑦,𝑔)𝑆, which, together with (𝑦,𝑔)𝐾𝐽|𝒯, 𝐾𝐽|𝒯𝑇={0}, and 𝑆=𝑇𝐾, implies that (𝑦,𝑔)𝐾. Hence, 𝐾𝐽|𝒯𝐾, and hence 𝐾=𝐾𝐽|𝒯. The necessity holds.
Next, consider the sufficiency. Suppose that 𝐾 is 𝐽-self-adjoint in 𝒯. By result (1), one has that 𝑆𝑆𝐽. It suffices to show that 𝑆𝐽𝑆. Let (𝑦,𝑔)𝑆𝐽, then (𝑦,𝑔)𝑇𝐽 since 𝑆𝐽𝑇𝐽 by 𝑇𝑆. It follows from (3.1) that there exist (𝑦1,𝑔1)𝑇 and (𝑦2,𝑔2)𝒯 such that 𝑦(𝑦,𝑔)=1,𝑔1+𝑦2,𝑔2.(4.7) We claim that (𝑦2,𝑔2)𝐾. In fact, since (𝑦,𝑔)𝑆𝐽, we have [](𝑥,𝑓)(𝑦,𝑔)=0(𝑥,𝑓)𝑆.(4.8) Inserting the first relation of (4.4) and (4.7) into (4.8) and using (2) of Remark 2.5 and Remark 4.1, we get that [(𝑥2,𝑓2)(𝑦2,𝑔2)]=0 for all (𝑥2,𝑓2)𝐾. Then, (𝑦2,𝑔2)𝐾𝐽|𝒯=𝐾. It follows from (𝑦2,𝑔2)𝐾, 𝑆=𝑇𝐾, and (4.7) that (𝑦,𝑔)𝑆. So, 𝑆𝐽𝑆, and hence 𝑆=𝑆𝐽. The sufficiency holds.

Now, we give the following result on the existence of 𝐽-SSEs.

Theorem 4.3. Every closed 𝐽-Hermitian subspace has a 𝐽-SSE.

Proof. Let 𝑇 be a closed 𝐽-Hermitian subspace. If 𝑇 is 𝐽-self-adjoint, then this conclusion holds. So, we assume that 𝑇𝑇𝐽. To prove that 𝑇 has a 𝐽-SSE, it suffices to prove that there exists a 𝐽-self-adjoint subspace 𝐾 in 𝒯 by Lemma 4.2. The proof uses Zorn’s lemma. Since 𝑇𝑇𝐽, one has that 𝒯{0}. Choose 0(𝑥0,𝑓0)𝒯 and set 𝐾0=span{(𝑥0,𝑓0)}. Then 𝐾0 is 𝐽-Hermitian in 𝒯 since [(𝑥0,𝑓0)(𝑥0,𝑓0)]=0. Let 𝒦 be the set of all the 𝐽-Hermitian subspaces in 𝒯 which contain 𝐾0, then 𝒦 is not empty since 𝐾0𝒦. Further, let 𝒦 be ordered by extension, that is, 𝐴<𝐵 if and only if 𝐴𝐵, and let ={𝐾𝛼} be an arbitrary totally ordered subset of 𝒦. Set 𝐾=𝛼𝐾𝛼. Then, it can be verified that 𝐾 is 𝐽-Hermitian in 𝒯 by Remark 4.1 and the fact that all the elements of are 𝐽-Hermitian in 𝒯. So, 𝐾 is an upper bound of in 𝒦. Therefore, 𝒦 has a maximal element by Zorn’s lemma. This means that 𝐾0 has a maximal 𝐽-Hermitian subspace extension, denoted by 𝐾, in 𝒯. We now prove 𝐾=𝐾𝐽|𝒯. Suppose that 𝐾𝐾𝐽|𝒯 on the contrary. Note that 𝐾𝐾𝐽|𝒯. Choose (̃𝑥0,𝑓0)𝐾𝐽|𝒯 satisfying (̃𝑥0,𝑓0)𝐾, and set ̇𝐾=𝐾+span{(̃𝑥0,𝑓0)}. It can be verified by Remark 4.1 and the fact that 𝐾 is 𝐽-Hermitian in 𝒯 that [(𝑥,𝑓)(𝑦,𝑔)]=0 holds for all 𝐾(𝑥,𝑓),(𝑦,𝑔). Note that 𝐾𝒯. Then, 𝐾 is 𝐽-Hermitian in 𝒯, which contradicts the maximality of 𝐾. Hence, 𝐾=𝐾𝐽|𝒯.

Remark 4.4. Since 𝑇 and its closure have the same 𝐽-SSEs, we have that every 𝐽-Hermitian subspace has a 𝐽-SSE. In addition, Theorem 4.3 extends the relevant result, for example, [1, Chapter III, Theorem 5.8], for 𝐽-symmetric operators to 𝐽-Hermitian subspaces.
The following result will give a characterization of all the 𝐽-SSEs.

Theorem 4.5. Let 𝑇 be a closed 𝐽-Hermitian subspace. Assume that 𝑑(𝑇)=𝑑<+, then a subspace 𝑆 is a 𝐽-SSE of 𝑇 if and only if 𝑇𝑆𝑇𝐽, and there exists {(𝑥𝑗,𝑓𝑗)}𝑑𝑗=1𝑇𝐽 such that(1)(𝑥1,𝑓1),(𝑥2,𝑓2),,(𝑥𝑑,𝑓𝑑) are linearly independent (mod𝑇),(2)[(𝑥𝑠,𝑓𝑠)(𝑥𝑗,𝑓𝑗)]=0 for 𝑠,𝑗=1,2,,𝑑,(3)𝑆={(𝑦,𝑔)𝑇𝐽[(𝑦,𝑔)(𝑥𝑗,𝑓𝑗)]=0,𝑗=1,2,,𝑑}.

Proof. First, consider the necessity. Suppose that 𝑆 is a 𝐽-SSE of 𝑇, then it holds that 𝑇𝑆𝑇𝐽 since 𝑆𝐽𝑇𝐽 and 𝑆=𝑆𝐽. We also have that (3.2) holds and 𝐾𝑆,𝑇 in (3.2) is 𝐽-self-adjoint in 𝒯 by Lemma 4.2. Note that dim𝑆/𝑇=𝑑 by Theorem 3.1. Then dim𝐾𝑆,𝑇=𝑑, and let {(𝑥𝑗,𝑓𝑗)}𝑑𝑗=1 be a basis of 𝐾𝑆,𝑇, then we get from (3.2) that (1) holds. In addition, since 𝐾𝑆,𝑇 is 𝐽-self-adjoint in 𝒯, one has that (2) holds by Remark 4.1. For convenience, set𝐷=(𝑦,𝑔)𝑇𝐽𝑥(𝑦,𝑔)𝑗,𝑓𝑗.=0,𝑗=1,2,,𝑑(4.9) Now, we prove 𝑇𝐾𝑆,𝑇=𝐷, that is, 𝑆=𝐷. Let (𝑦,𝑔)𝑇𝐾𝑆,𝑇, then (𝑦,𝑔)𝑇𝐽 by 𝑆𝑇𝐽, and there exist (̃𝑦,̃𝑔)𝑇 and 𝑐𝑗𝐶 such that (𝑦,𝑔)=(̃𝑦,̃𝑔)+𝑑𝑗=1𝑐𝑗𝑥𝑗,𝑓𝑗.(4.10) Inserting (4.10) into [(𝑦,𝑔)(𝑥𝑠,𝑓𝑠)] and using (2) of Remark 2.5 and (2) of this theorem, we get that [(𝑦,𝑔)(𝑥𝑠,𝑓𝑠)]=0 for 𝑠=1,2,,𝑑. So, (𝑦,𝑔)𝐷, and hence 𝑇𝐾𝑆,𝑇𝐷. Conversely, suppose that (𝑦,𝑔)𝐷, then by (3.1), there exist (𝑦1,𝑔1)𝑇 and (𝑦2,𝑔2)𝒯 such that (4.7) holds. The definition of 𝐷, (4.7), and (2) of Remark 2.5 implies that for 𝑠=1,2,,𝑑, 𝑦2,𝑔2𝑥𝑠,𝑓𝑠=𝑥(𝑦,𝑔)𝑠,𝑓𝑠𝑦1,𝑔1𝑥𝑠,𝑓𝑠=0.(4.11) So, (𝑦2,𝑔2)(𝐾𝑆,𝑇)𝐽|𝒯, which implies that (𝑦2,𝑔2)𝐾𝑆,𝑇 since 𝐾𝑆,𝑇 is 𝐽-self-adjoint in 𝒯. One has from (4.7) that (𝑦,𝑔)𝑇𝐾𝑆,𝑇, and consequently, 𝐷𝑇𝐾𝑆,𝑇. Hence, 𝑇𝐾𝑆,𝑇=𝐷, that is, 𝑆=𝐷. The necessity holds.
Next, consider the sufficiency. Suppose that there exists {(𝑥𝑗,𝑓𝑗)}𝑑𝑗=1𝑇𝐽 such that conditions (1) and (2) hold and 𝑆 is given in condition (3). From (3.1), we have 𝑥𝑗,𝑓𝑗=𝑥𝑗0,𝑓𝑗0+̃𝑥𝑗,𝑓𝑗,𝑥𝑗0,𝑓𝑗0𝑇,̃𝑥𝑗,𝑓𝑗𝒯,𝑗=1,2,,𝑑.(4.12) It can be easily verified that the set {(̃𝑥𝑗,𝑓𝑗)}𝑑𝑗=1 satisfies conditions (1) and (2). Let 𝐾=span{(̃𝑥1,𝑓1),(̃𝑥2,𝑓2),,(̃𝑥𝑑,𝑓𝑑)}, then 𝐾 is 𝐽-Hermitian in 𝒯 since {(̃𝑥𝑗,𝑓𝑗)}𝑑𝑗=1 satisfies condition (2) and 𝐾𝒯. By the proof of Theorem 4.3, there exists 𝐾1𝒯 such that 𝐾1 is 𝐽-self-adjoint in 𝒯 and 𝐾𝐾1. Then, by Lemma 4.2, 𝑇𝐾1 is a 𝐽-SSE of 𝑇, which, together with Theorem 3.1, yields that 𝑑=dim𝑇𝐾1𝑇=dim𝐾1𝑑.(4.13) Therefore, 𝐾𝐾1 and dim 𝐾=𝑑 imply that 𝐾=𝐾1, and hence 𝐾 is 𝐽-self-adjoint in 𝒯. With a similar argument to the proof of 𝑇𝐾𝑆,𝑇=𝐷, we have 𝑇𝐾=𝐷,(4.14) where 𝐷=(𝑦,𝑔)𝑇𝐽(𝑦,𝑔)̃𝑥𝑗,𝑓𝑗.=0,𝑗=1,2,,𝑑(4.15) On the other hand, it can be easily verified that 𝐷=𝑆. So, 𝐾𝑆=𝑇, and hence 𝑆 is 𝐽-self-adjoint by (2) of Lemma 4.2. The sufficiency holds.

Remark 4.6. The case for 𝐽-symmetric operators is given by [24, 27]. Theorem 4.5 can be regarded as the GKN theorem for 𝐽-Hermitian subspaces, which will be used for characterizations of 𝐽-self-adjoint extensions for linear Hamiltonian difference systems in terms of boundary conditions.

Acknowledgments

This research was supported by the NNSFs of China (Grants 11101241 and 11071143), the NNSF of Shandong Province (Grant ZR2011AQ002), and the independent innovation fund of Shandong University (Grant 2011ZRYQ003).