Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 22, 1997, 123–164

ONE-DIMENSIONAL GRAPH PERTURBATIONS OF SELFADJOINT RELATIONS

Seppo Hassi and Henk de Snoo

University of Helsinki, Department of Statistics P.O. Box 54, FIN-00014 Helsinki, Finland; hassi@cc.helsinki.fi University of Groningen, Department of Mathematics Postbus 800, NL-9700 AV Groningen, Nederland; desnoo@math.rug.nl

Abstract. Let A be a selfadjoint operator (or a selfadjoint relation) in a Hilbert space ℌ, let Z be a one-dimensional subspace of ℌ^{2} such that A ∩ Z = {0, 0} and define S = A ∩ Z^{∗}. Then S is a closed, symmetric operator (or relation) with defect numbers (1, 1) and, conversely, each such S and a selfadjoint extension A are related in this way. This allows us to interpret the selfadjoint extensions of S in ℌ as one-dimensional graph perturbations of A. If Z = span {φ, ψ}, then the function Q(l) = l[φ, φ] + [(A − l)^{−1}(lφ − ψ), l̄φ − ψ], generated by A and the pair {φ, ψ}, is a Q-function of S = A∩Z^{∗} and A. It belongs to the class N of Nevanlinna functions and essentially determines S and A. Calculation of the corresponding resolvent operators in the perturbation formula leads to Kreĭn’s description of (the resolvents of) the selfadjoint extensions of S. The class N of Nevanlinna functions has subclasses N_{1} ⊃ N_{0} ⊃ N_{−1} ⊃ N_{−2} , each defined in terms of function-theoretic properties. We characterize the Q-functions belonging to each of these classes in terms of the pair {φ, ψ}. If Q(l) belongs to the subclass N_{k}, k = 1, 0, −1, −2, then all but one of the selfadjoint extensions of S have a Q-function in the same class, while the exceptional extension has a Q-function in N \ N_{1} . In particular, if S is semi-bounded, the exceptional selfadjoint extension is precisely the Friedrichs extension. We consider our perturbation formula in the case where the Q-function Q(l) belongs to the subclass N_{k}, k = 1, 0, −1, −2, or if it is an exceptional function associated with this subclass. The resulting perturbation formulas are made explicit for the case that A or its orthogonal operator part is the multiplication operator in a Hilbert space L^{2}(dρ).

0. Introduction

Let S be a closed symmetric relation in a Hilbert space ℌ, whose defect numbers are (1, 1). Then S has canonical selfadjoint extensions, i.e., selfadjoint extensions within the space ℌ. In this paper we fix a canonical selfadjoint extension A of S. The other canonical selfadjoint extensions of S are now described as one-dimensional graph perturbations of A. Our description of the graph perturbations is in terms of a one-dimensional subspace Z of ℌ^{2} , which connects S and A by S = A ∩ Z^{∗} (here Z^{∗} denotes the adjoint of the subspace Z in the graph sense), and which satisfies A ∩ Z = {0, 0}. Note that S is obtained from A by a restriction on the graph of A. Special choices of the subspace Z provide restrictions on the domain and restrictions on the range of A, respectively.

1991 Mathematics Subject Classification: Primary 47B25, 47B15; Secondary 47A55, 47A57.

The use of finite-dimensional subspaces to restrict a symmetric relation goes back to [C2], [CD1], [CD2], and was continued in [DSS]. In these papers the underlying idea was to broaden the classes of boundary value problems. In the present context this method seems appropriate as it provides us with coordinates to describe perturbations.

It is the purpose of this paper to give various forms of the perturbation formula depending on the behaviour of the so-called Q-function Q(l) of A and S = A ∩ Z^{∗}, and to relate them to known results in terms of operator models. The Q-function belongs to the class N of Nevanlinna functions, i.e., it is holomorphic on C \ R, satisfies Q(l)^{∗} = Q(l̄) and Im Q(l)/ Im l ≥ 0. Conversely, for each Nevanlinna function Q(l) there exist a symmetric relation (which may be taken to be completely nonselfadjoint and hence is an operator) and a canonical selfadjoint extension, which generate Q(l) as the corresponding Q-function, cf. [HLS], [LT]. The Q-function Q(l) is an ingredient in Kreĭn’s formula, which describes the resolvent operators of the canonical selfadjoint extensions of S in terms of onedimensional range perturbations of the resolvent operator of A. Calculation of the resolvent operators in terms of our perturbation formula gives Kreĭn’s formula.

The class of Nevanlinna functions has subclasses N_{1} and N_{0} ⊂ N_{1} of functions Q(l) in N with the property that

∫_{1}^{∞} \frac{Im Q(iy)}{y} dy < ∞,

and that

sup_{y>0} y Im Q(iy) < ∞,

respectively. The theory of the class N_{1} goes back to Kac, cf. [Ka]; see also [HLS] and [KK]. The class N_{0} in its turn has subclasses N_{−1} and N_{−2} ⊂ N_{−1} of functions Q(l) in N_{0} with the property that

∫_{1}^{∞}( sup_{y>0} y Im Q(iy) − y Im Q(iy)) dy < ∞,

and

sup_{y>0} y^{2}( sup_{y>0} y Im Q(iy) − y Im Q(iy)) < ∞,

respectively, cf. [HS1].

If there is a canonical selfadjoint extension with a Q-function belonging to N_{1} , N_{0} , N_{−1} , or N_{−2} , then, with the exception of one, all the other canonical selfadjoint extensions have a Q-function belonging to N_{1} , N_{0} , N_{−1} , or N_{−2} , respectively. In each case the exceptional canonical selfadjoint extension has a Q-function which belongs to N \ N_{1} .

We will give some examples of exceptional extensions. The first example occurs when S is a densely defined, semibounded operator: then all but one of its canonical selfadjoint extensions have a Q-function in N_{1} and the exceptional selfadjoint extension is the Friedrichs extension, cf. [HLS] and [C1], [CS], [K1], [K2], [KY]. This case has also been considered from the point of view of rank one perturbations: see [GS], [KS], and [S]. Related results concerning nonnegative relations in a Kreĭn space setting were obtained in [JL1] and [JL2]. Another example occurs when S is a nondensely defined operator: then all but one of its selfadjoint extensions are operators and the exceptional selfadjoint extension is a proper relation. It coincides with the Friedrichs extension when S in addition is semibounded, see [CS]. An exposition of these facts can be found in [HLS]. In this paper a model space was constructed for a Nevanlinna function Q(l) belonging to N_{1} in which all the canonical selfadjoint extensions, including the exceptional one, were exhibited. The present graph perturbation formula covers this case.

Let A be a selfadjoint relation and let Z = span {φ, ψ} satisfy A∩Z = {0, 0}. Without loss of generality we assume that S = A ∩ Z^{∗} is a closed symmetric operator and that A is a selfadjoint relation whose multivalued part is at most one-dimensional. We characterize when the Q-function Q(l) of A and S = A∩Z^{∗} belongs to the above subclasses of Nevanlinna functions by means of {φ, ψ}. The Q-function Q(l) of A and S = A ∩ Z^{∗} belongs to N_{1} or N_{0} if and only if A is an operator and φ ∈ dom |A|^{1/2} or φ ∈ dom A, respectively. In addition, the Q-function Q(l) of A and S = A ∩ Z^{∗} belongs to N_{−1} or N_{−2} , if and only if A is an operator, φ ∈ dom A, and Aφ − ψ belongs to dom |A|^{1/2} or to dom A, respectively.

We will present the perturbation formula according to the behaviour of the Q-function of A and S = A ∩ Z^{∗}, including a description of the exceptional extension.

The first case is where A is an operator. If Q(l) belongs to N_{1} some refinements of our perturbation formula can be given. In particular, if Q(l) belongs to N_{0} the general graph perturbation can be reduced to a one-dimensional perturbation of the range of A: A+τR, τ ∈ R, where R is the orthogonal projection onto a one-dimensional subspace. The exceptional selfadjoint extension corresponds to the parameter τ = ∞; it is the only canonical selfadjoint extension which is not an operator and its domain is equal to the domain of S. See [D, Chapter 6] for the case of a selfadjoint operator in a finite-dimensional Hilbert space. The Qfunction Q(l) belongs to N_{−1} or N_{−2} if and only if the range of R belongs to dom |A|^{1/2} or dom A, respectively. The case when Q(l) belongs to N_{1} \ N_{0} is similar to the case Q(l) ∈ N_{0} , but not so explicit. In this case a description in terms of space triplets (see the last section of this paper) is more suitable. Further details are given in [HKS], [HS2].

The second case is when A is a selfadjoint relation whose multivalued part is one-dimensional. Then the Q-function Q(l) is an exceptional function corresponding to the class N_{0} . We can obtain an appropriate version of the perturbation result. Refinements in this case can be given depending on whether Q(l) is an exceptional function corresponding to N_{−1} or N_{−2} , see also [HS1].

The results described above are specialized for the selfadjoint operator of multiplication by the independent variable in a Hilbert space L^{2}(dρ) with ρ a nondecreasing function on R, to which is possibly orthogonally added a one-dimensional multivalued part. We will give a description of all canonical selfadjoint extensions; in the case where the Q-function belongs to N_{1} or to N_{0} the description reduces by means of an explicit isometric isomorphism to the results in [HLS].

In Section 1 of this paper we will present some preliminary results concerning Q-functions associated with symmetric and selfadjoint relations. In Section 2 we then give the general perturbation formula and in Section 3 show the connection with Kreĭn’s formula. We will reformulate some facts concerning Q-functions of A and S = A ∩ Z^{∗} in terms of Z in Section 4. In Sections 5 and 6 we study the perturbation formula in the cases, where A is an operator or that A has a onedimensional multivalued part and S is an operator, respectively. Finally, Section 7 contains further refinements of the perturbation formula when A or its operator part is a multiplication operator in some L^{2}(dρ)-space, where ρ is a nondecreasing function on R.

1. Preliminaries

In this section we will give a short introduction and present some notation and terminology, concerning relations in Hilbert spaces and Q-functions.

Consider a Hilbert space ℌ with inner product [ · , ·]. A linear relation S in ℌ is just a linear subspace of the orthogonal sum ℌ ⊕ ℌ of ℌ with itself. For instance, a linear operator is a linear relation when we identify the operator and its graph. The relation is said to be closed if it is closed as a subspace of ℌ ⊕ ℌ. The null space ker S is given by ker S = {f ∈ ℌ : {f, 0} ∈ S} and the multivalued part mul S is given by mul S = {g ∈ ℌ : {0, g} ∈ S}. A linear relation S is (the graph of) an operator if and only if mul S = {0}. Its domain and range are given by dom S = {f : {f, g} ∈ S} and ran S = {g : {f, g} ∈ S}, respectively. By S^{−1} we mean the linear relation S^{−1} = {{g, f} : {f, g} ∈ S}. Note that dom S^{−1} = ran S. Clearly, ker S^{−1} = mul S. We define the linear relation S_{∞} by S_{∞} = {0} ⊕ mul S. When S is closed, the operator part S_{s} of S is defined by

S_{s} = {{f, g} ∈ S : g ⊥ mul S}.

In this sense a relation S = S_{s} ⊕ S_{∞} is a multivalued linear operator.

For each l ∈ C we define the linear relation S −l = {{f, g −lf} : {f, g} ∈ S}. Clearly, ker (S − l)^{−1} = mul S. When S is closed, the resolvent set ρ(S) is the set of all l ∈ C for which (S − l)^{−1} = {{g − lf, f} : {f, g} ∈ S} is (the graph of) a bounded linear operator defined on all of ℌ. The resolvent set ρ(S) is open and the resolvent operator (S − l)^{−1} , l ∈ ρ(S), satisfies

(S − l)^{−1} − (S − λ)^{−1} = (l − λ)(S − l)^{−1}(S − λ)^{−1}, l, λ ∈ ρ(S),

which is the usual resolvent identity.

For any subset S in ℌ^{2} we define the adjoint S^{∗} by

S^{∗} = {{h, k} ∈ ℌ^{2} : ⟨{h, k}, {f, g}⟩ = 0, for all {f, g} ∈ S},

where the form ⟨{f, g}, {h, k}⟩ is defined by [g, h] − [f, k]. Then S^{∗} is automatically a closed linear relation and (S^{∗})^{∗} is the closure of the linear subspace spanned by S. A linear relation S is called symmetric if S ⊂ S^{∗} and selfadjoint if S = S^{∗}. For a closed symmetric relation S we have von Neumann’s formula, which expresses S^{∗} as a direct sum:

(1.1)  S^{∗} = S ˙+Mμ̄(S^{∗}) ˙+M_{μ}(S^{∗}),  μ ∈ C \ R,

where M_{λ}(S^{∗}) = {{f, g} ∈ S^{∗} : g = λf}, λ ∈ C; note that ker (S^{∗} − λ) = dom M_{λ}(S^{∗}). The symbol ˙+ denotes the componentwise sum in ℌ ⊕ ℌ. The defect numbers (m, n) of S are the dimensions of the defect subspaces ker (S^{∗}−μ̄) and of ker (S^{∗} − μ), respectively, if μ ∈ C^{+} . In general, the symmetric relation S has selfadjoint extensions if we extend the original Hilbert space. The operator S has canonical selfadjoint extensions, i.e., selfadjoint extensions in the original Hilbert space if and only if the defect numbers of S are equal. Finally we observe that

(1.2)  ker (S^{∗} − l) ∩ dom S = {0},  l ∈ C \ R.

To see this, let {χ(l), lχ(l)} ∈ S^{∗} be nontrivial and assume that χ(l) ∈ dom S. Then there exists an element k such that {χ(l), k} ∈ S, so that

[k, χ(l)] − l̄[χ(l), χ(l)] = ⟨{χ(l), k}, {χ(l), lχ(l)}⟩ = 0,

which contradicts the fact that [k, χ(l)] is real.

For any closed symmetric linear relation S the relation S_{∞} is selfadjoint in mul S and S_{s} is a closed symmetric operator in ℌ ⊖ mul S with the same defect numbers as S. A symmetric relation is called completely nonselfadjoint if it is not the nontrivial orthogonal sum of a symmetric and a selfadjoint relation. By the above reasoning a completely nonselfadjoint symmetric relation is automatically an operator. A closed symmetric relation S is completely nonselfadjoint if and only if

ℌ = \overline{span} \{ker (S^{∗} − l) : l ∈ C \ R\}.

This characterization goes back to M.G. Kreĭn, cf. [LT, Proposition 1.1].

Let S be a closed symmetric relation in a Hilbert space ℌ with defect numbers (1, 1), and let A be a canonical selfadjoint extension of S. For some μ ∈ C\R we choose a nontrivial element χ(μ) ∈ ker (S^{∗} − μ) and we define the vector function χ(l), l ∈ C \ R, by

(1.3)  χ(l) = (I + (l − μ)(A − l)^{−1})χ(μ).

Then ker (S^{∗} − l) is spanned by χ(l) and furthermore

(1.4)  S = {{f, g} ∈ A : [g − lf, χ(l̄)] = 0},

as it is easily verified that the righthand side of (1.4) defines a closed symmetric extension of S with defect numbers (1, 1). A function Q(l) is a Q-function of A and S if it is a solution of:

(1.5)  \frac{Q(l) − Q(λ)^{∗}}{l − λ̄} = [χ(l), χ(λ)],  l, λ ∈ C \ R.

Up to a real constant the function Q(l) is uniquely defined and

(1.6)  Q(l) = Q(μ)^{∗} + (l − μ̄)[(I + (l − μ)(A − l)^{−1})χ(μ), χ(μ)].

It is clear from the definition of the Q-function of S and A that we may always take out selfadjoint parts of S. In particular, we may assume that S is completely nonselfadjoint, see also [LT]. Conversely, a Q-function essentially determines a pair (S, A) in a model space, where S is completely nonselfadjoint. These model spaces can be constructed in various ways, see [LT] and, for instance, [HLS].

We denote the orthogonal projection onto ℌ ⊖ mul A by P and the spectral family of the operator part A_{s} of A (in ℌ ⊖ mul A) by E(t). We recall [HLS, Proposition 2.1]:

Proposition 1.1. Let S be a closed symmetric relation with defect numbers (1, 1) and let A be a selfadjoint extension of S. Let Q(l) be the Q-function of S and A with the representation (1.6). Then Q(l) admits an integral representation

(1.7)  Q(l) = α + βl + ∫_{R}(\frac{1}{t − l} − \frac{t}{t^{2} + 1}) dσ(t),

where

(i) α = Q(i)^{∗} + i[χ(i), χ(i)] = Re Q(i), (ii) β = [(I − P)χ(i), (I − P)χ(i)] (= lim_{y→∞} Im Q(iy)/y), (iii) dσ(t)/(t^{2} + 1) = d([E(t)P χ(i), P χ(i)]), and where the function σ(t) is nondecreasing on R and satisfies

(1.8)  ∫_{R} \frac{dσ(t)}{t^{2} + 1} < ∞.

The singular part S_{∞} is selfadjoint and we may reduce S to the Hilbert space ℌ ⊖ mul S. In that case the multivalued part of any selfadjoint extension of S is at most one-dimensional as stated in the following proposition.

Proposition 1.2. Let S be a closed symmetric relation with defect numbers (1, 1) and let A be a selfadjoint extension of S. Then the multivalued part mul S of S is given by

(1.9)  mul S = {g ∈ mul A : [g, (I − P)χ(μ̄)] = 0}.

Hence

(1.10)  dim (mul A ⊖ mul S) ≤ 1.

In particular, if S is an operator then

mul A = span {(I − P)χ(μ̄)},

and mul A is at most one-dimensional.

Proof. The inclusion mul S ⊂ mul A and the identity

mul S = {g ∈ mul A : [g, χ(μ̄)] = 0}

follow immediately from (1.4). For g ∈ mul A we may write

[g, χ(μ̄)] = [(I − P)g, χ(μ̄)] = [g, (I − P)χ(μ̄)],

so that (1.9) follows. Clearly, (I − P)χ(μ̄) ∈ mul A and (1.10) follows. Now assume that S is an operator. If A is an operator then the identity mul A = span {(I − P)χ(μ̄)} is trivially true. Next suppose that A is not an operator. As S is an operator it follows from (1.9) that (I − P)χ(μ̄) ≠ 0 and that mul A is one-dimensional. This completes the proof.

The following result is a slight extension of [HLS, Proposition 2.2] where it was assumed that S is completely nonselfadjoint. Moreover, we include some information about the classes N_{−1} and N_{−2} .

Proposition 1.3. Let Q(l) be a Q-function belonging to S and A. Then (i) lim_{y→∞} Im Q(iy)/y = 0 if and only if χ(l) ∈ \overline{dom} A for some (and, hence,

for all) l ∈ C \ R. Assume that S is a closed operator. Then (ii) lim_{y→∞} Im Q(iy)/y = 0 if and only if A is an operator. (iii) Q(l) ∈ N_{1} if and only if A is an operator and χ(l) ∈ dom |A|^{1/2} for some

(and, hence, for all) l ∈ C \ R. (iv) Q(l) ∈ N_{0} if and only if A is an operator and χ(l) ∈ dom A for some (and,

hence, for all) l ∈ C \ R.

(v) Q(l) ∈ N_{−1} if and only if A is an operator and χ(l) ∈ dom |A|^{3/2} for some (and, hence, for all) l ∈ C \ R.

(vi) Q(l) ∈ N_{−2} if and only if A is an operator and χ(l) ∈ dom A^{2} for some (and, hence, for all) l ∈ C \ R.

As for the statements (i)–(iv), it suffices to discuss (ii). First assume that A is an operator. Then P = I and hence β = 0 in (1.7). Conversely, assume that β = 0. According to (i) χ(l) ∈ \overline{dom} A. From (1.1) it then follows that dom S^{∗} ⊂ (mul A)^{⊥}. However, by assumption, (dom S^{∗})^{⊥} = mul S^{∗∗} = mul S = {0}, which implies that S^{∗} is densely defined. Therefore, it follows that mul A = {0}, and A is an operator. The items (v) and (vi) follow from Proposition 1.1, in the same way as (iii) and (iv), cf. [HLS, Proposition 2.1] and [HS1, Proposition 1.2].

The next result is similar to (iv) and (vi) of Proposition 1.3, but stated in terms of the symmetric operator S. For this purpose, suppose that Q(l) belongs to N_{0} . Let A be the corresponding selfadjoint operator extension of S. Then χ(l) ∈ dom A and it follows from von Neumann’s formula (1.1) that dom S^{∗} ⊂ dom A, so that

dom S^{∗} = dom A.

Proposition 1.4. Assume that S is a closed operator.

(i) There is a Q-function Q(l) ∈ N_{0} of S if and only if S is not densely defined. (ii) There is a Q-function Q(l) ∈ N_{−2} of S if and only if S is not densely

defined and mul S^{∗} ⊂ dom S^{∗}, or equivalently, if and only if dom S^{∗} = dom S + mul S^{∗}.

Proof. We first prove (i). Assume that there is a Q-function Q(l) ∈ N_{0} . Then the previous proposition shows that the corresponding selfadjoint extension A is an operator and that χ(l) ∈ dom A. Hence it follows from (1.4) that

(1.11)  dom S = {f ∈ dom A : [f, (A − μ)χ(μ)] = 0},  μ ∈ C \ R,

which implies that S is not densely defined. Conversely, assume that S is not densely defined, in which case there is a nontrivial γ ∈ mul S^{∗}. We choose a selfadjoint operator extension A of S. Then it follows from the definition that (A−l)^{−1}γ ∈ ker (S^{∗}−l), so that the defect subspaces of S are contained in dom A. Hence the Q-function Q(l) of A belongs to N_{0} , by the previous proposition.

Next we prove (ii). Assume that there is a Q-function Q(l) ∈ N_{−2} . In particular Q(l) belongs to N_{0} , so that the corresponding selfadjoint extension A is an operator and (1.11) holds, which implies that (A−μ)χ(μ) ∈ mul S^{∗}. Moreover, the previous proposition shows that χ(l) ∈ dom A^{2} . Therefore (A − μ)χ(μ) ∈ dom A and, by dimension arguments, we conclude that mul S^{∗} = span {(A − μ)χ(μ)} ⊂ dom A and that dom A = dom S + mul S^{∗}, which is the same as dom S^{∗} = dom S + mul S^{∗}. Conversely, assume that dom S^{∗} = dom S + mul S^{∗}. Then S is nondensely defined, since otherwise dom S^{∗} = dom S, which is impossible, see (1.1) and (1.2). By (i) there is a selfadjoint operator extension A of S whose Q-function Q(l) belongs to N_{0} , and hence χ(μ) ∈ dom A. Again, it follows from (1.11) that mul S^{∗} = span {(A − μ)χ(μ)}, and consequently

(A − μ)χ(μ) ∈ dom S^{∗} = dom A,

which implies χ(μ) ∈ dom A^{2} . Therefore Q(l) ∈ N_{−2} .

The equivalence of the last two statements in (ii) is obvious by the above reasoning. This completes the proof.

A corresponding statement about the classes N_{1} or N_{−1} would involve the theory of triplets of Hilbert spaces, see [HKS] and Section 7 of the present paper.

2. One-dimensional perturbations

In this section we consider a closed symmetric relation with defect numbers (1, 1). Its canonical selfadjoint extensions are interpreted as perturbations of the graph of one fixed canonical selfadjoint extension.

We associate with a selfadjoint relation A and a one-dimensional subspace Z of ℌ^{2} the relation S = A ∩ Z^{∗}. Clearly, S is a closed symmetric relation and if Z = span {φ, ψ}, then

(2.1)  S = {{f, g} ∈ A : ⟨{f, g}, {φ, ψ}⟩ = 0}.

Moreover, the adjoint S^{∗} is given by the componentwise sum in ℌ^{2} :

(2.2)  S^{∗} = A ˙+Z.

To prove this identity we note that

S^{∗} = (A ∩ Z^{∗})^{∗} = \overline{span} (A ˙+\overline{span} Z) = \overline{span} (A ˙+Z) = A ˙+Z,

where the second equality holds since A is selfadjoint, and hence closed, and the last two equalities hold since Z is one-dimensional and hence closed.

Lemma 2.1. The closed symmetric relation S = A ∩ Z^{∗} is selfadjoint and equal to A if and only if Z ⊂ A.

Proof. Suppose that S is selfadjoint, then S = A = S^{∗} and it follows from (2.2) that Z ⊂ A. Conversely, let Z ⊂ A. Then (2.2) shows that S^{∗} = A, and hence S = A. This completes the proof.

Since Z is one-dimensional, it is clear that Z is not contained in A if and only if the componentwise sum A ˙+Z is direct, i.e.,

(2.3)  A ∩ Z = {0, 0}.

Unless otherwise stated, this condition will be assumed in the rest of this paper. A more general version of the first part of the following lemma can be found in [DSS, Lemma 5.1].

Lemma 2.2. Let the selfadjoint relation A and the one-dimensional subspace Z satisfy (2.3). Then S = A ∩ Z^{∗} is a closed symmetric relation with defect numbers (1, 1). Conversely, if S is a closed symmetric relation with defect numbers (1, 1) and A is a canonical selfadjoint extension of S, then there exists a onedimensional subspace Z, which satisfies (2.3), such that S = A ∩ Z^{∗}.

Proof. Let Z = span {φ, ψ}. Choose l ∈ C \ R, then since A is selfadjoint, there is an element {φ̃, ψ̃} ∈ A for which ψ̃−lφ̃ = ψ−lφ. Define χ̃ = φ̃−φ. Then {χ̃, lχ̃} = {φ̃, ψ̃} − {φ, ψ} ∈ S^{∗} by (2.2). Moreover, χ̃ ≠ 0 by (2.3). This shows that ker (S^{∗} − l) ≥ 1. Now let {χ, lχ} ∈ S^{∗}. Then {χ, lχ} = {f, g} − c{φ, ψ} for some {f, g} ∈ A and c ∈ C by (2.2). Hence, {χ, lχ}−c{χ̃, lχ̃} = {f, g}−c{φ̃, ψ̃} ∈ A, and since A has no eigenvalue at l ∈ C \ R, this shows that χ = cχ̃. We conclude that dim ker (S^{∗} − l) = 1, l ∈ C \ R, i.e., the symmetric relation has defect numbers (1, 1).

Next we prove the converse. We have that S ⊂ A ⊂ S^{∗} and dim S^{∗}/S = 2. Therefore, since A is selfadjoint, we have that dim S^{∗}/A = 1. Hence, we can choose a one-dimensional subspace Z ⊂ S^{∗} such that S^{∗} = A ˙+Z and A ∩ Z = {0, 0}. Then clearly S = A ∩ Z^{∗}. This completes the proof.

Let A be a selfadjoint relation in a Hilbert space ℌ. We call two pairs {φ, ψ} and {φ̃, ψ̃} equivalent (with respect to A) if {φ − φ̃, ψ − ψ̃} ∈ A. Two subspaces Z and Z̃ are called equivalent if they are spanned by equivalent pairs. If Z and Z̃ are equivalent and if Z satisfies (2.3), then also Z̃ satisfies (2.3). In particular, if Z = span {φ, ψ} and Z̃ = span {φ, P ψ}, where P is the orthogonal projection onto ℌ⊖mul A, then the subspaces Z and Z̃ are equivalent, as {φ, ψ}−{φ, P ψ} = {0, (I − P)ψ} ∈ A. Observe that in the definition of S = A ∩ Z^{∗} we can always replace Z by an equivalent subspace Z̃.

Lemma 2.3. Let the selfadjoint relation A and the one-dimensional subspace Z satisfy (2.3). Then there is a one-dimensional symmetric subspace Z̃ which is equivalent to Z.

Proof. Let Z = span {φ, ψ}. Since A ∩ Z = {0, 0}, there is an element {h, k} ∈ A for which ⟨{h, k}, {φ, ψ}⟩ = 1. Then

\{φ̃, ψ̃\} = \{φ, ψ\} − \frac{1}{2}⟨\{φ, ψ\}, \{φ, ψ\}⟩\{h, k\},

spans a one-dimensional symmetric subspace which is equivalent to Z. This completes the proof.

The following theorem gives a description of all canonical selfadjoint extensions of S = A ∩ Z^{∗} as perturbations (of the graph) of A, a fixed canonical selfadjoint extension of S. Clearly, Z is symmetric if and only if S ˙+Z is a selfadjoint extension of S. Note that only in the case that the subspace Z is symmetric, there is a value τ ∈ R ∪ {∞} for which 1/τ + [ψ, φ] = 0.

Theorem 2.4. Let A be a selfadjoint relation and let Z be a one-dimensional subspace for which (2.3) holds. The canonical selfadjoint extensions of S = A∩Z^{∗} are in one-to-one correspondence with τ ∈ R ∪ {∞}. These canonical selfadjoint extensions A(τ) are given by A(0) = A and for τ ≠ 0 and 1/τ + [ψ, φ] ≠ 0 by

(2.4)  A(τ) = \{\{f, g\} − \frac{⟨\{f, g\}, \{φ, ψ\}⟩}{1/τ + [ψ, φ]}\{φ, ψ\} : \{f, g\} ∈ A\},

while for 1/τ + [ψ, φ] = 0 the canonical selfadjoint extension A(τ) is given by

(2.5)  A(τ) = S ˙+Z.

Proof. By definition the relation A is selfadjoint. The relation A(τ) in (2.4) when τ ≠ 0, 1/τ +[ψ, φ] ≠ 0, and the relation A(τ) in (2.5) when 1/τ +[ψ, φ] = 0, are clearly true symmetric extensions of S and therefore they are selfadjoint. Observe that in this last case Z is symmetric.

Now we will show that each selfadjoint extension of S has the indicated form. Let H be a canonical selfadjoint extension of S. The case H = A corresponds to τ = 0.

Next assume that H ≠ A. Assume that {h, k} is a fixed element in H, and write {h, k} = {f, g} + c{φ, ψ}, according to the decomposition (2.2) with {f, g} ∈ A and c ∈ C. Assume that {h, k} does not belong to S. This implies that c ≠ 0, as otherwise {h, k} would belong to H and A at the same time, which is impossible in the present case H ≠ A. Since H is selfadjoint, it follows that ⟨{h, k}, {h, k}⟩ = 0, which leads to

(2.6)  \begin{aligned}0 & = ⟨\{f, g\} + \{φ, ψ\}c, \{f, g\} + \{φ, ψ\}c⟩ \\ & = ⟨\{f, g\}, \{f, g\}⟩ + ⟨\{f, g\}, \{φ, ψ\}⟩c̄ \\ & + ⟨\{φ, ψ\}, \{f, g\}⟩c + ⟨\{φ, ψ\}, \{φ, ψ\}⟩|c|^{2}. \end{aligned}

Clearly, ⟨{f, g}, {f, g}⟩ = 0 since {f, g} ∈ A. Since c ≠ 0, we rewrite (2.6) as

(2.7)  \frac{1}{c}⟨\{f, g\}, \{φ, ψ\}⟩ + \frac{1}{c̄}⟨\{φ, ψ\}, \{f, g\}⟩ + ⟨\{φ, ψ\}, \{φ, ψ\}⟩ = 0,

or equivalently

(2.8)  \frac{1}{c}⟨\{f, g\}, \{φ, ψ\}⟩ + [ψ, φ] = −\frac{1}{c̄}⟨\{φ, ψ\}, \{f, g\}⟩ + [φ, ψ].

As the righthand side of (2.8) is the complex conjugate of the lefthand side, we conclude that the lefthand side must be real, say −1/τ for some τ ∈ R ∪ {∞}, τ ≠ 0:

(2.9)  \frac{1}{c}⟨\{f, g\}, \{φ, ψ\}⟩ = −(1/τ + [ψ, φ]).

First assume that ⟨{f, g}, {φ, ψ}⟩ = 0. According to (2.9) this assumption is equivalent to 1/τ + [ψ, φ] = 0. It follows from (2.7) that Z is symmetric. Our assumption means that {f, g} ∈ S and thus, {h, k} ∈ S ˙+Z. Hence, as the element {h, k} belongs at the same time to the canonical selfadjoint extensions H and S ˙+Z and as it does not belong to S, we conclude that H = S ˙+Z. Therefore the selfadjoint extension H is equal to A(τ) for 1/τ + [ψ, φ] = 0, where A(τ) is of the form (2.5).

Now assume that ⟨{f, g}, {φ, ψ}⟩ ≠ 0. Solve the equation (2.9) for c:

c = −\frac{⟨\{f, g\}, \{φ, ψ\}⟩}{1/τ + [ψ, φ]}.

We observe that

\{h, k\} = \{f, g\} + c\{φ, ψ\} = \{f, g\} − \frac{⟨\{f, g\}, \{φ, ψ\}⟩}{1/τ + [ψ, φ]}\{φ, ψ\}.

Hence, as the element {h, k} belongs at the same time to the canonical selfadjoint extensions H and A(τ) and does not belong to S, we conclude that H = A(τ), where A(τ) is of the form (2.4).

Thus we have shown that any canonical selfadjoint extension of S, different from A, is of the form (2.4) or (2.5) for some τ ∈ R∪{∞}, τ ≠ 0. This completes the proof.

3. Q-functions and Kreĭn’s formula

In this section we express the Q-function of S = A∩Z^{∗} and A, Z being a onedimensional subspace satisfying (2.3), in terms of Z = span {φ, ψ}. We calculate the resolvent operators in Theorem 2.4 in terms of the resolvent operator for A. In this way we obtain a new proof of Kreĭn’s description of all canonical selfadjoint extensions of a closed symmetric relation with defect numbers (1, 1), cf. [HLS].

In terms of {φ, ψ} we introduce the vector function χ(l), l ∈ C \ R, by

(3.1)  χ(l) = (A − l)^{−1}(lφ − ψ) + φ.

Using the resolvent identity, it can easily be checked that χ(l) satisfies (1.3). Moreover, S = A ∩ Z^{∗} can also be written in the form (1.4), since

(3.2)  ⟨{f, g}, {φ, ψ}⟩ = [g − lf, χ(l̄)] = ⟨{f, g}, {χ(l̄), l̄χ(l̄)}⟩.

In particular, ker (S^{∗} − l) is spanned by χ(l). We define the function Q(l) by

(3.3)  Q(l) = [χ(l), l̄φ − ψ] + [φ, ψ],

or equivalently

(3.4)  Q(l) = l[φ, φ] + [(A − l)^{−1}(lφ − ψ), l̄φ − ψ].

Note that if the condition (2.3) is not satisfied, so that {φ, ψ} ∈ A, then (3.1) shows that χ(l) = 0. Moreover (3.3) shows that Q(l) = [φ, ψ], a real constant, since {φ, ψ} ∈ A.

Suppose we replace {φ, ψ} in (3.1) by the equivalent pair of the form {φ̃, ψ̃} = {φ, ψ} + {h, k}, with {h, k} ∈ A. Since (A − l)^{−1}(k − lh) = h, the function χ(l) remains the same. Note also the converse: if for {φ, ψ}, {φ̃, ψ̃} ∈ ℌ^{2}

(A − l)^{−1}(lφ̃ − ψ̃) + φ̃ = (A − l)^{−1}(lφ − ψ) + φ,

then (A − l)^{−1}(l(φ̃ − φ) − (ψ̃ − ψ)) = φ − φ̃, so that {φ̃ − φ, ψ̃ − ψ} ∈ A.

However, the definition of the function Q(l) in (3.3) depends on the choice of the pair {φ, ψ}. When {φ, ψ} is replaced by {φ, ψ} + {h, k}, {h, k} ∈ A, then the function in (3.3) is now given by

(3.5)  Q(l) + 2 Re[h, ψ] + [h, k],

which differs from the original function Q(l) in (3.3) by a real constant. Of course, when {φ, ψ} is replaced by {φ, P ψ}, it follows from (3.5) that we obtain precisely the same function.

The element χ(l) spans ker (S^{∗} − l). Clearly,

(3.6) {χ(l), lχ(l)} = {(A − l)^{−1}(lφ − ψ), (I + l(A − l)^{−1})(lφ − ψ)} + {φ, ψ}.

This decomposition corresponds to the direct sum (2.2). In particular, for each l ∈ C \ R the subspace span {χ(l), lχ(l)} is equivalent to Z = span {φ, ψ}. Note that for each l ∈ C \ R the subspace span {χ(l), lχ(l)} is nonsymmetric:

Im[lχ(l), χ(l)] = (Im l)[χ(l), χ(l)].

This remark provides a nonsymmetric version of Lemma 2.3. Moreover, with the pair {χ(l̄), l̄χ(l̄)} the definition (2.1) takes the form (1.4).

In case A is a selfadjoint operator and φ ∈ dom A, the formulas (3.1) and (3.4) for χ(l) and Q(l) reduce to

(3.7)  χ(l) = (A − l)^{−1}(Aφ − ψ),

and

(3.8)  Q(l) = [ψ, φ] + [φ, ψ] − [Aφ, φ] + [(A − l)^{−1}(Aφ − ψ), (Aφ − ψ)].

In fact, then the pair {φ, ψ} is equivalent to the pair {0, ψ − Aφ}, as

{φ, ψ} = {0, ψ − Aφ} + {φ, Aφ}.

Moreover, the symmetric relation S = A ∩ Z^{∗} is an operator, and the identity (2.1) reduces to

(3.9)  S = {{f, g} ∈ A : [f, Aφ − ψ] = 0}.

Proposition 3.1. The function Q(l) satisfies the equation (1.5). Hence Q(l) is a Q-function of A and S = A ∩ Z^{∗}.

Proof. Write out the inner product

[χ(l), χ(λ)] = [(A − l)^{−1}(lφ − ψ) + φ, (A − λ)^{−1}(λφ − ψ) + φ].

This gives the following terms

\begin{aligned}[(A − l)^{−1}(lφ − ψ), (A − λ)^{−1}(λφ − ψ)] & + [(A − l)^{−1}(lφ − ψ), φ] \\ & + [φ, (A − λ)^{−1}(λφ − ψ)] + [φ, φ]. \end{aligned}

Application of the resolvent identity to the first term leads to the following decomposition:

(l − λ̄)[χ(l), χ(λ)] = [(A − l)^{−1}(lφ − ψ), λφ − ψ]
− [(lφ − ψ), (A − λ)^{−1}(λφ − ψ)] + (l − λ̄)[(A − l)^{−1}(lφ − ψ), φ]
+ (l − λ̄)[φ, (A − λ)^{−1}(λφ − ψ)] + (l − λ̄)[φ, φ].

Combining the first and third term, and the second and the fourth term on the righthand side, it is easily seen, by means of (3.4), that the righthand side is equal to Q(l) − Q(λ)^{∗}. This completes the proof.

We will now calculate the resolvent operators of the selfadjoint extensions in (2.4) and (2.5). It turns out that we obtain a formula due to Kreĭn describing abstractly all selfadjoint extensions of a symmetric operator. The connection between Kreĭn’s formula and von Neumann’s formula was made explicit in [HLS].

Theorem 3.2. The resolvents of the canonical selfadjoint extensions A(τ) of S = A ∩ Z^{∗} in (2.4) and (2.5) are given by (A(0) − l)^{−1} = (A − l)^{−1} and for τ ∈ R ∪ {∞}, τ ≠ 0, by

(3.10)  (A(τ) − l)^{−1} = (A − l)^{−1} − χ(l)\frac{1}{Q(l) + 1/τ} [ · , χ(l̄)].

Proof. Assume that τ ≠ 0 and let h ∈ ℌ. Define k = (A(τ) − l)^{−1}h, so that {k, h + lk} ∈ A(τ). According to Theorem 2.4 the element {k, h + lk} has the decomposition

(3.11)  {k, h + lk} = {f, g} − c{φ, ψ}.

If 1/τ + [ψ, φ] ≠ 0 the element {f, g} in (3.11) belongs to A, and c ∈ C satisfies

(3.12)  c(1/τ + [ψ, φ]) = ⟨{f, g}, {φ, ψ}⟩.

If 1/τ + [ψ, φ] = 0 the element {f, g} in (3.11) belongs to S ⊂ A, and c ∈ C is uniquely determined by {k, h + lk} because of (2.3). Note that in this case (3.12) is automatically satisfied as the righthand side of (3.12) is now 0. It follows from (3.11) that

{f, g} = {k, h + lk} + c{φ, ψ} ∈ A,

so that

{g − lf, f} = {h, k} + c{ψ − lφ, φ} ∈ (A − l)^{−1},

or equivalently

(A − l)^{−1}h = k + c(φ + (A − l)^{−1}(lφ − ψ)).

Due to the definition (3.1) and the definition k = (A(τ) − l)^{−1}h, this gives

(3.13)  (A(τ) − l)^{−1}h = (A − l)^{−1}h − cχ(l).

It also follows from (3.11) that h = g − lf + c(lφ − ψ), which yields

(3.14)  [h, χ(l̄)] = [g − lf, χ(l̄)] + c[lφ − ψ, χ(l̄)].

Due to the definition (3.3), the inner product in the second term on the righthand side of (3.14) is equal to

[lφ − ψ, χ(l̄)] = Q(l) − [ψ, φ],

recall that (1.5) implies that Q(l)^{∗} = Q(l̄). Therefore it follows from (3.2) that

(3.15)  [h, χ(l̄)] = ⟨{f, g}, {φ, ψ}⟩ + c(Q(l) − [ψ, φ]).

First consider the case 1/τ + [ψ, φ] ≠ 0. Substitution of ⟨{f, g}, {φ, ψ}⟩ from (3.12) into (3.15) gives

(3.16)  [h, χ(l̄)] = c(Q(l) + 1/τ).

Next consider the case 1/τ + [ψ, φ] = 0. The first term on the righthand side of (3.15) is zero as {f, g} ∈ S. Hence (3.15) reduces to (3.16). Thus, in both cases 1/τ + [ψ, φ] ≠ 0 and 1/τ + [ψ, φ] = 0, we can solve the equation (3.16) for c and substitute the result in (3.13). This gives the desired result. Hence, in each case the resolvent of A(τ) is given by (3.10). This completes the proof.

We recall the following proposition concerning Q-functions, which are normalized as indicated, cf. [HLS, Proposition 3.2].

Proposition 3.3. Let Q(l) be the Q-function of S and A, normalized by Re Q(μ) = 0. For each τ ∈ R ∪ {∞} the Q-function Q_{τ} (l) of the canonical selfadjoint extension A(τ) and S, normalized by Re Q_{τ} (μ) = 0, is given by

(3.17)  Q_{τ} (l) = \frac{Q(l) − τ(Im Q(μ))^{2}}{τQ(l) + 1} ,  τ ∈ R ∪ \{∞\}.

We emphasize that a Q-function of S and A is determined up to a real constant; the only condition on a Q-function is that it satisfies (1.5).

With any Nevanlinna function Q(l) and a parameter τ ∈ R ∪ {∞} we may associate the function Q_{τ} (l), defined by the linear fractional transform (3.17), see [HLS]. In [HLS] and [HS1] the following facts were proved using a functiontheoretic method. If Q(l) belongs to N_{1} , N_{0} , N_{−1} or N_{−2} , then for all but one τ ∈ R ∪ {∞}, the function Q_{τ} (l) belongs to N_{1} , N_{0} , N_{−1} or N_{−2} , respectively, whereas for the exceptional value of τ, given by 1/τ + γ = 0, the exceptional function Q_{τ} (l) does not belong to N_{1} . Moreover, the exceptional function was characterized in a function-theoretic manner.

4. Characterization of subclasses of Q-functions

Let A be a selfadjoint relation and let Z = span {φ, ψ} be a one-dimensional subspace which satisfies (2.3). Associated with S and its selfadjoint extension A is a Q-function Q(l) given by (3.3), see Proposition 3.1. In this section we will give necessary and sufficient conditions in terms of Z for Q(l) to belong to the subclasses N_{1} , N_{0} , N_{−1} , or N_{−2} . If Q(l) belongs to N_{1} we can evaluate lim_{y→∞} Q(iy) and if Q(l) belongs to N_{0} , N_{−1} , or N_{−2} we can evaluate the corresponding moments in terms of A and Z.

In the following we assume that S = A∩Z^{∗} and A are restricted to the Hilbert space ℌ ⊖ mul S. According to Proposition 1.2 we may restrict ourselves, without loss of generality, to the situation that S = A ∩ Z^{∗} is a closed symmetric operator with defect numbers (1, 1) and where A is an operator or A is a multivalued operator whose multivalued part mul A is one-dimensional. In this case mul A = span {(I − P)φ}, cf. Proposition 1.2. Here P denotes the orthogonal projection onto ℌ ⊖ mul A.

Proposition 4.1. Let A be a selfadjoint relation in a Hilbert space ℌ and let Z = span {φ, ψ} be a one-dimensional subspace which satisfies (2.3). Assume that S = A ∩ Z^{∗} is an operator, so that mul A is at most one-dimensional.

(i) If mul A = {0}, then S is not densely defined if and only if φ ∈ dom A. In this case mul S^{∗} = span {Aφ − ψ}.

(ii) If mul A is one-dimensional, then S is not densely defined and dom S = dom A and mul S^{∗} = span {(I − P)φ}.

Proof. Suppose that mul A = {0}, i.e., that A is an operator. We first assume that S is not densely defined. Then there exists a nontrivial element {0, g} ∈ mul S^{∗} and it follows that {0, g} = {h, Ah}+c{φ, ψ} for some h ∈ dom A and c ∈ C. Clearly, c ≠ 0 and thus φ ∈ dom A follows. Suppose conversely that φ ∈ dom A. Then the definition (2.1) of S can be written as

S = {{f, g} ∈ A : [f, Aφ − ψ] = 0}.

Note that Aφ − ψ ≠ 0, due to (2.3). This shows that S is not densely defined and that mul S^{∗} has the indicated form.

To prove the second statement note that, if mul A is one-dimensional then A is not an operator, so that A and thus S are not densely defined. Moreover, (I − P)φ ≠ 0. Now let {f, g} ∈ A and denote c = ⟨{f, g}, {φ, ψ}⟩/[(I − P)φ, φ]. Then with g̃ = g − c(I − P)φ we have {f, g̃} ∈ A and it is easy to check that ⟨{f, g̃}, {φ, ψ}⟩ = 0, i.e., {f, g̃} ∈ S. This shows that dom A ⊂ dom S and completes the proof.

Proposition 4.2. Assume that the conditions of Proposition 4.1 hold.

(i) If φ ∈ dom S, then S ˙+Z is a selfadjoint extension of S which is not an operator.

(ii) Conversely, if S ˙+Z is not an operator, then φ ∈ dom S.

Proof. If φ ∈ dom S, then {φ, ψ′} ∈ S for some ψ′ ∈ ℌ. Since ⟨{φ, ψ′}, {φ, ψ}⟩ = 0, it follows that [φ, ψ] = [ψ′, φ] ∈ R. Hence, Z is symmetric and S ˙+Z is selfadjoint. Moreover {0, ψ′ − ψ} ∈ S ˙+Z and ψ′ ≠ ψ on account of (2.3). The converse statement is obvious.

Let Q(l) be the Q-function of S = A ∩ Z^{∗} and A as defined by (3.3). From the general theory of Nevanlinna functions we know that

lim_{y→∞} \frac{Re Q(iy)}{y} = 0,  lim_{y→∞} \frac{Im Q(iy)}{y} ≥ 0.

It follows from (3.1) that (I − P)χ(l) = (I − P)φ. Hence, the last limit can be expressed in terms of Z: if we apply (ii) of Proposition 1.1, we obtain

(4.1)  lim_{y→∞} \frac{Im Q(iy)}{y} = [(I − P)φ, (I − P)φ].

We state Proposition 1.3 in terms of the subspace Z.

Proposition 4.3. Let Q(l) be the Q-function of S = A ∩ Z^{∗} and A. Then (i) lim_{y→∞} Im Q(iy)/y = 0 if and only if φ ∈ \overline{dom} A. Assume that S is an operator. Then (ii) lim_{y→∞} Im Q(iy)/y = 0 if and only if A is an operator.

(iii) Q(l) ∈ N_{1} if and only if A is an operator and φ ∈ dom |A|^{1/2} . (iv) Q(l) ∈ N_{0} if and only if A is an operator and φ ∈ dom A.

(v) Q(l) ∈ N_{−1} if and only if A is an operator, φ ∈ dom A, and Aφ − ψ ∈ dom |A|^{1/2} .

(vi) Q(l) ∈ N_{−2} if and only if A is an operator, φ ∈ dom A, and Aφ−ψ ∈ dom A.

Proof. The first statement follows immediately from (4.1). Note that the first element on the righthand side of (3.1) belongs to dom A and that dom A = dom |A| ⊂ dom |A|^{1/2} . Therefore, φ belongs to dom |A|^{1/2} or to dom A if and only if χ(l) belongs to dom |A|^{1/2} or to dom A, respectively. If A is an operator and φ ∈ dom A, it follows from (3.7) that χ(l) belongs to dom |A|^{3/2} or dom |A|^{2} , if and only if Aφ − ψ belongs to dom |A|^{1/2} or dom |A| = dom A, respectively. Now apply Proposition 1.3. The proof is completed.

Note that if {φ̃, ψ̃} is equivalent to {φ, ψ}, then φ̃ ∈ dom |A|^{1/2} if and only if φ ∈ dom |A|^{1/2} , and φ̃ ∈ dom A if and only if φ ∈ dom A, in which case Aφ − ψ = Aφ̃ − ψ̃ spans mul S^{∗}, see Proposition 4.1.

A function Q(l) in N belongs to N_{1} if and only if the integral representation (1.7) of Q(l) reduces to

(4.2)  Q(l) = γ + ∫_{R} \frac{1}{t − l} dσ(t),

where γ ∈ R and ∫_{R} dσ(t)/(|t| + 1) < ∞, cf. [Ka] and [KK]. It follows from this representation that

(4.3)  γ = lim_{y→∞} Q(iy).

In particular, the function Q(l) belongs to N_{0} if and only if the function σ(t) satisfies ∫_{R}dσ(t) < ∞. Moreover, the function Q(l) belongs to N_{−1} or N_{−2} if and only ∫_{R}(|t| + 1) dσ(t) < ∞ or ∫_{R}(|t|^{2} + 1) dσ(t) < ∞, respectively, see [HS1]. Hence, in these cases the moments

(4.4)  m_{k}(σ) = ∫_{R} t^{k} dσ(t),

associated with the spectral measure of the function Q(l), are well-defined for k = 0, k = 0, 1, or k = 0, 1, 2, respectively. If Q(l) is the Q-function of S = A∩Z^{∗} and A, we will express the limit γ and the moments m_{0}(σ), m_{1}(σ), and m_{2}(σ), respectively, in terms of the operator A and the subspace Z, cf. [A], [K3], [KL]. For this we need the polar decomposition of A = U|A|. Here the operators U (= U^{∗} = U^{−1} ) and |A| are given in terms of the spectral decomposition A = ∫_{R} t dE(t) of A as follows

(4.5)  |A| = ∫_{R} |t| dE(t),  U = ∫_{R} \frac{t}{|t|} dE(t).

The functions σ(t) in (1.7) and E(t) are related by (iii) of Proposition 1.1. Hence if Q(l) belongs to N_{0} , it follows from (3.7) that

(4.6)  dσ(t) = d([E(t)(Aφ − ψ), Aφ − ψ]).

Proposition 4.4. Let Q(l) be the Q-function of S = A ∩ Z^{∗} and A. If Q(l) ∈ N_{1} , then

(4.7)  γ = lim_{y→∞} Q(iy) = [ψ, φ] + [φ, ψ] − [U|A|^{1/2}φ, |A|^{1/2}φ].

Moreover, if Q(l) belongs to N_{0} , then γ = [ψ, φ] + [φ, ψ] − [Aφ, φ] and Q(l) has the operator representation

(4.8)  Q(l) = γ + [(A − l)^{−1}ψ̃, ψ̃],

where ψ̃ = Aφ − ψ.

(i) If Q(l) ∈ N_{0} , then

m_{0}(σ) = lim_{y→∞} −iy(Q(iy) − γ) = [ψ̃, ψ̃].

(ii) If Q(l) ∈ N_{−1} , then

m_{1}(σ) = lim_{y→∞} −(iy)^{2}(Q(iy) − γ + \frac{m_{0}(σ)}{iy}) = [U|A|^{1/2}ψ̃, |A|^{1/2}ψ̃].

(iii) If Q(l) ∈ N_{−2} , then m_{1}(σ) = [Aψ̃, ψ̃] and

m_{2}(σ) = lim_{y→∞} −(iy)^{3}(Q(iy) − γ + \frac{m_{0}(σ)}{iy} + \frac{m_{1}(σ)}{(iy)^{2}}) = [Aψ̃, Aψ̃].

Proof. Observe that Q(l) in (3.4) can be written as

(4.9) \begin{aligned}Q(l) = l[φ, φ] & + [(A − l)^{−1}ψ, ψ] \\ & − l[(A − l)^{−1}ψ, φ] − l[(A − l)^{−1}φ, ψ] + l^{2}[(A − l)^{−1}φ, φ]. \end{aligned}

Consider the behaviour of this function for l = iy, y → ∞. The second term [(A − l)^{−1}ψ, ψ] does not give a contribution to the limit. The terms

−l[(A − l)^{−1}ψ, φ]  and  − l[(A − l)^{−1}φ, ψ]

give contributions [ψ, φ] and [φ, ψ], respectively. The two remaining terms l[φ, φ] and l^{2}[(A − l)^{−1}φ), φ] are combined to

(4.10) l[(I + l(A − l)^{−1})φ, φ] = l[A(A − l)^{−1}φ, φ] = ∫_{R} \frac{lt}{t − l} d([E(t)φ, φ]),

where we have used

(4.11)  l(A − l)^{−1} = −I + A(A − l)^{−1}.

The assumption φ ∈ dom |A|^{1/2} is equivalent to ∫_{R} |t| d([E(t)φ, φ]) < ∞. Note that

|\frac{iyt}{t − iy} | = \frac{y|t|}{\sqrt{t^{2} + y^{2}}} ≤ |t|.

Application of the dominated convergence theorem to (4.10) with l = iy, y → ∞, gives the limit − ∫_{R} t d([E(t)φ, φ]). Next observe that

[U|A|^{1/2}φ, |A|^{1/2}φ] = ∫_{R} \frac{t}{|t|}|t|^{1/2} d([E(t)φ, |A|^{1/2}φ]) = ∫_{R} t d([E(t)φ, φ]),

which proves (4.7).

Clearly, if Q(l) ∈ N_{0} , then γ in (4.7) is given by γ = [ψ, φ] + [φ, ψ] − [Aφ, φ]. Comparing this with the operator representation (3.8) we obtain (4.8). The identity (4.11) and the representation (4.8) imply that

(4.12)  −l(Q(l) − γ) = [ψ̃, ψ̃] − [A(A − l)^{−1}ψ̃, ψ̃].

The statement in (i) follows from (4.12) and the integral representation (4.2). If Q(l) ∈ N_{−1} , then ψ̃ ∈ dom |A|^{1/2} and

[A(A − l)^{−1}ψ̃, ψ̃] = [(A − l)^{−1}U|A|^{1/2}ψ̃, |A|^{1/2}ψ̃].

Hence (4.12), (i), and (4.11) give (4.13)

−l^{2}(Q(l) − γ + \frac{m_{0}(σ)}{l}) = [U|A|^{1/2}ψ̃, |A|^{1/2}ψ̃] − [A(A − l)^{−1}U|A|^{1/2}ψ̃, |A|^{1/2}ψ̃].

The statement in (ii) follows from (4.13) and the integral representation (4.2). If Q(l) ∈ N_{−2} , then ψ̃ ∈ dom A by Proposition 4.3, and

[A(A − l)^{−1}U|A|^{1/2}ψ̃, |A|^{1/2}ψ̃] = [(A − l)^{−1}Aψ̃, Aψ̃],

and m_{1}(σ) = [Aψ̃, ψ̃]. Hence (4.13), (ii), and (4.11) give

(4.14)  −l^{3}(Q(l) − γ + \frac{m_{0}(σ)}{l} + \frac{m_{1}(σ)}{l^{2}}) = −l[(A − l)^{−1}Aψ̃, Aψ̃].

The statement in (iii) follows from (4.14) and the integral representation (4.2). This completes the proof.

5. The operator case

Let A be a selfadjoint operator in the Hilbert space ℌ and let Z = span {φ, ψ} be a one-dimensional subspace which satisfies (2.3). We will now consider Theorem 2.4 in greater detail.

First we present a suitable formulation of the definition of S and its adjoint S^{∗}.

Lemma 5.1. The closed symmetric operator S = A ∩ Z^{∗} is given by

(5.1)  S = {{f, Af} : f ∈ dom A, [Af, φ] − [f, ψ] = 0}.

Its adjoint S^{∗} is given by

(5.2)  S^{∗} = {{h, A(h + cφ) − cψ} : h + cφ ∈ dom A, c ∈ C}.

Proof. Since A is an operator, (5.1) is just a restatement of (2.1). Similarly, consider an arbitrary element of (2.2), i.e., an element of the form

(5.3)  {f, g} − c{φ, ψ},  {f, g} ∈ A, c ∈ C.

With h = f −cφ, it follows that f = h+cφ ∈ dom A and that g−cψ = Af −cψ = A(h + cφ) − cψ, so that the element (5.3) is equal to

(5.4)  {h, A(h + cφ) − cψ},  h + cφ ∈ dom A, c ∈ C,

and belongs to the righthand side of (5.2).

Conversely, consider an arbitrary element in the righthand side of (5.2), i.e., an element of the form (5.4). With f = h + cφ ∈ dom A, c ∈ C, it follows that the element (5.4) is equal to (5.3) and belongs to (2.2). This completes the proof.

The form (5.2) in which the adjoint S^{∗} is written is now used for the description of all canonical selfadjoint extensions of S. We recall from Proposition 4.1 that S is not densely defined if and only if φ ∈ dom A, in which case mul S^{∗} = span {Aφ − ψ}.

Theorem 5.2. The canonical selfadjoint extensions A(τ), τ ∈ R ∪ {∞}, of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0, 1/τ + [ψ, φ] ≠ 0, by

(5.5)  \begin{aligned}A(τ) = \{\{h, A(h + cφ) − cψ\} : & h + cφ ∈ dom A, c ∈ C, \\ & c = \frac{[A(h + cφ), φ] − [h + cφ, ψ]}{1/τ + [ψ, φ]}\}, \end{aligned}

while for 1/τ + [ψ, φ] = 0 the canonical selfadjoint extension A(τ) is given by

(5.6)  \begin{aligned}A(τ) = \{\{h, A(h + cφ) − cψ\} : & h + cφ ∈ dom A, c ∈ C, \\ & [A(h + cφ), φ] − [h + cφ, ψ] = 0\}. \end{aligned}

Proof. Consider an arbitrary element in A(τ) as given by (2.4) for τ ≠ 0, 1/τ + [ψ, φ] ≠ 0, or by (2.5) for 1/τ + [ψ, φ] = 0. That is, consider an element of the form

(5.7)  {f, Af} − c{φ, ψ}.

If 1/τ + [ψ, φ] ≠ 0, assume that c ∈ C satisfies

(5.8)  c(1/τ + [ψ, φ]) = [Af, φ] − [f, ψ].

If 1/τ + [ψ, φ] = 0, then assume that {f, Af} ∈ S and note that (5.8) is still satisfied, cf. (5.1). Define h = f − cφ, then f = h + cφ ∈ dom A and g − cψ = A(h + cφ) − cψ. Therefore (5.7) is equal to

(5.9)  {h, A(h + cφ) − cψ},

while, in addition, c ∈ C must satisfy

(5.10)  c(1/τ + [ψ, φ]) = [A(h + cφ), φ] − [h + cφ, ψ].

Hence A(τ) is contained in the righthand side of (5.5) or of (5.6), respectively.

Conversely, consider an arbitrary element in the righthand side of (5.5) with τ ≠ 0, 1/τ + [ψ, φ] ≠ 0, or of (5.6) with 1/τ + [ψ, φ] = 0. That is, consider an element of the form (5.9) where c satisfies (5.10). Define f = h + cφ, then f ∈ dom A and the element (5.9) has the form (5.7), while c satisfies (5.8). Note that if 1/τ + [ψ, φ] = 0, then it follows from (5.8) that {f, Af} ∈ S, cf. (5.1). Therefore, the righthand side of (5.5) or of (5.6) is contained in A(τ) as given by (2.4) or by (2.5), respectively. This completes the proof.

In general it is impossible to solve the constant c in terms of A from the equation (5.5). However, since φ ∈ dom S^{∗}, the condition h+cφ ∈ dom A implies that h ∈ dom S^{∗}. This observation leads to an interpretation of (5.5) in terms of triplet spaces, see [HS2]. If the Q-function Q(l) of A and S is an exceptional function corresponding to the class N_{1} , then it follows from Proposition 3.3 that all canonical selfadjoint extensions A(τ), τ ≠ 0, have Q-functions belonging to N_{1} , cf. [HLS]. It is in the case where the Q-function Q(l) belongs to N_{1} (or equivalently φ ∈ dom |A|^{1/2} ), that the perturbation formula in Theorem 5.2 can be simplified. For then the limit γ = lim_{y→∞} Q(iy) exists as a real number, cf. Proposition 4.4, and in the description of all canonical selfadjoint extensions A(τ), τ ∈ R ∪ {∞}, of S = A ∩ Z^{∗}, there exists precisely one exceptional value of the parameter τ, given by 1/τ + γ = 0, for which the Q-function corresponding to A(τ) and S = A ∩ Z^{∗} belongs to N \ N_{1} . In the case that Q(l) belongs to N_{1} , it is also possible to explicitly “solve” c ∈ C in terms of A from (5.5). Moreover, we will characterize the canonical selfadjoint extension A(τ), 1/τ + γ = 0, in terms of a “boundary condition”.

Proposition 5.3. Assume that the Q-function Q(l) of S = A ∩ Z^{∗} and A belongs to N_{1} . Then dom S^{∗} ⊂ dom |A|^{1/2} . All canonical selfadjoint extensions A(τ) of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0, 1/τ + γ ≠ 0, by

(5.11)  \begin{aligned}A(τ) = \{\{h, A(h + cφ) − cψ\} : & h + cφ ∈ dom A, c ∈ C, \\ & c = \frac{[U|A|^{1/2}h, |A|^{1/2}φ] − [h, ψ]}{1/τ + γ}\}, \end{aligned}

while for 1/τ +γ = 0 the exceptional canonical selfadjoint extension A(τ) is given by

(5.12)  \begin{aligned}A(τ) = \{\{h, A(h + cφ) − cψ\} : & h + cφ ∈ dom A, c ∈ C, \\ & [U|A|^{1/2}h, |A|^{1/2}φ] − [h, ψ] = 0\}. \end{aligned}

Proof. Since φ ∈ dom |A|^{1/2} , it follows from h + cφ ∈ dom A, that h ∈ dom |A|^{1/2} , as dom A ⊂ dom |A|^{1/2} . Hence it follows from (5.2) that dom S^{∗} ⊂ dom |A|^{1/2} .

Consider an arbitrary element in A(τ) in the righthand side of (5.5) with τ ≠ 0, 1/τ + [ψ, φ] ≠ 0, or in the righthand side of (5.6) with 1/τ + [ψ, φ] = 0. That is, consider an element of the form

(5.13)  {h, A(h + cφ) − cψ},

where h + cφ ∈ dom A and c ∈ C satisfies (5.10). As φ ∈ dom |A|^{1/2} implies that h ∈ dom |A|^{1/2} , the equation (5.10) may be written as

(5.14) c(1/τ +[ψ, φ]) = c[U|A|^{1/2}φ, |A|^{1/2}φ]−c[φ, ψ]+[U|A|^{1/2}h, |A|^{1/2}φ]−[h, ψ].

By Proposition 4.4 this leads to

(5.15)  c(1/τ + γ) = [U|A|^{1/2}h, |A|^{1/2}φ] − [h, ψ].

Thus we have shown that A(τ) as given by (5.5) or (5.6), is contained in the righthand side of (5.11) or of (5.12), respectively.

Conversely, consider an arbitrary element in the righthand side of (5.11) with τ ≠ 0, 1/τ + γ ≠ 0, or of (5.12) with 1/τ + γ ≠ 0. That is, consider an element of the form (5.13), where c ∈ C satisfies (5.15). Insert the expression for γ from Proposition 4.4 in (5.15). This shows that c ∈ C then satisfies (5.14). As h + cφ ∈ dom A, the identity (5.14) is equivalent to (5.10). Hence this shows that the righthand side of (5.11) or (5.12) is contained in A(τ) as given by (5.5) or by (5.6), respectively. This completes the proof.

Another way to express the exceptional selfadjoint extension A(τ), 1/τ +γ = 0, is via

A(τ) = {{h, k} ∈ S^{∗} : [U|A|^{1/2}h, |A|^{1/2}φ] − [h, ψ] = 0},

i.e., with a “boundary condition” of the form [U|A|^{1/2}h, |A|^{1/2}φ]−[h, ψ] = 0. This is similar to S = {{h, k} ∈ A : [U|A|^{1/2}h, |A|^{1/2}φ] − [h, ψ] = 0}.

Next we will consider the case that the Q-function Q(l) of S = A ∩ Z^{∗} and A belongs to the class N_{0} (or equivalently that φ ∈ dom A). In this case S is given by (3.9): S = {{f, Af} : f ∈ dom A, [f, Aφ − ψ] = 0}. Note that S is a nondensely defined operator, due to the domain restriction. It follows from (5.2) that dom S^{∗} ⊂ dom A. Furthermore, as φ ∈ dom A implies that φ ∈ dom |A|^{1/2} , the following result is obtained as a special case of Proposition 5.3.

Proposition 5.4. Assume that the Q-function Q(l) of S = A ∩ Z^{∗} and A belongs to N_{0} . Then dom S^{∗} = dom A. All canonical selfadjoint extensions A(τ) of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0, 1/τ + γ ≠ 0, by

(5.16)  A(τ) = \{\{h, Ah + \frac{[h, Aφ − ψ]}{1/τ + γ}(Aφ − ψ)\} : h ∈ dom A\},

while for 1/τ +γ = 0 the exceptional canonical selfadjoint extension A(τ) is given by

(5.17) A(τ) = {{h, Ah + c(Aφ − ψ)} : h ∈ dom A, c ∈ C, [h, Aφ − ψ] = 0}.

We explicitly state what happens when Z is spanned by an element of the form {0, ψ}, ψ ≠ 0. Clearly, Z is symmetric and S = {{f, g} ∈ A : [f, ψ] = 0}. The identities (3.7) of χ(l) and (3.8) of Q(l) reduce to

χ(l) = −(A − l)^{−1}ψ,  Q(l) = [(A − l)^{−1}ψ, ψ]  (∈ N_{0}),

so that γ = 0 and τ = ∞ is the exceptional value. Recall that the Q-function Q(l) changes if the pair {φ, ψ} is replaced by an equivalent pair, while the function χ(l) stays the same. For τ ∈ R the canonical selfadjoint extensions A(τ) of S are densely defined operators and when ∥ψ∥ = 1, they are one-dimensional perturbations of A of the form A(τ) = A + τR, where R denotes the orthogonal projection onto mul S^{∗} = span {ψ}, see Proposition 4.1. For τ = ∞ we obtain

A(∞) = S ˙+Z = S ˙+({0} ⊕ span {ψ}),

which is the unique selfadjoint extension of S which is not an operator. For finite dimensional Hilbert spaces the connection between the one-dimensional range perturbations and Kreĭn’s formula is derived in [D, Chapter 6].

From the explicit expression for the canonical selfadjoint extensions A(τ) in (5.5) and (5.6) the following result is immediately clear.

Corollary 5.5. Assume that the Q-function Q(l) of S = A ∩ Z^{∗} and A belongs to N \ N_{0} . Let τ ∈ R ∪ {∞} and τ ≠ 0. Then dom A(τ) ∩ dom A = dom S.

The next corollary shows a completely different behaviour of the domains of the canonical selfadjoint extensions when Q(l) belongs to N_{0} .

Corollary 5.6. Assume that the Q-function Q(l) of S = A ∩ Z^{∗} and A belongs to N_{0} . Let τ ∈ R ∪ {∞}, then

dom A(τ) = dom A if 1/τ + γ ≠ 0 and dom A(τ) = dom S if 1/τ + γ = 0.

It follows from Corollary 5.6 that A(τ), 1/τ + γ = 0, is the only selfadjoint extension of S, which is not an operator. In fact then we have that

A(τ) = S ˙+({0} ⊕ span {Aφ − ψ}),

or, equivalently,

A(τ) = {{h, k} ∈ S^{∗} : [h, Aφ − ψ] = 0},

which is similar to

S = {{h, k} ∈ A : [h, Aφ − ψ] = 0}.

Corollaries 5.5 and 5.6 show the different behaviours in the cases Q(l) ∈ N\N_{0} and Q(l) ∈ N_{0} . If Q(l) ∈ N_{1} \ N_{0} the application of space triplets gives a description which is similar to Corollary 5.6. This is already apparent in the operator models for the cases Q(l) ∈ N_{1} \ N_{0} and Q(l) ∈ N_{0} , see [HLS] and Section 7 of the present paper. For further details we refer to [HKS] and [HS2].

If the Q-function Q(l) of S = A ∩ Z^{∗} and A belongs to N_{0} , then the Q-function Q_{τ} (l) of A(τ) and S, 1/τ + γ ≠ 0, also belongs to N_{0} , while for 1/τ + γ = 0 the Q-function H(l) of A(τ) in (5.17) has the property that lim_{y→∞} Im H(iy)/y > 0. Likewise, if Q(l) belongs to N_{−1} or N_{−2} , then Q_{τ} (l), 1/τ + γ ≠ 0, belongs to the same subclass, while for 1/τ + γ = 0 the function H(l) satisfies

H(l) − ( lim_{y→∞} \frac{Im H(iy)}{y})l ∈ N_{1} or N_{0},

respectively, cf. [HS1]. We relate the moments for Q_{τ} (l), 1/τ + γ ≠ 0, to the moments for Q(l). We use the notation q = (Im Q(μ))^{2} .

Proposition 5.7. Let the Q-function Q(l) of S = A ∩ Z^{∗} and A belong to N_{0} . Then for 1/τ + γ ≠ 0 the moments m_{0}(σ_{τ}) are given by

(5.18)  m_{0}(σ_{τ}) = \frac{(τ^{2}q + 1)m_{0}(σ)}{(τγ + 1)^{2}}.

If Q(l) ∈ N_{−1} , then the moments m_{1}(σ_{τ}), 1/τ + γ ≠ 0, are given by

(5.19)  m_{1}(σ_{τ}) = \frac{(τ^{2}q + 1) [(τγ + 1)m_{1}(σ) + τm_{0}(σ)^{2}]}{(τγ + 1)^{3}}.

Furthermore, if Q(l) ∈ N_{−2} , then the moments m_{2}(σ_{τ}), 1/τ + γ ≠ 0, are given by (5.20)

m_{2}(σ_{τ}) = \frac{(τ^{2}q + 1)[(τγ + 1)^{2}m_{2}(σ) + 2(τγ + 1)τm_{1}(σ)m_{0}(σ) + τ^{2}m_{0}(σ)^{3}]}{(τγ + 1)^{4}}.

Proof. Assume that Q(l) ∈ N_{0} . It follows from (3.17) that for τγ + 1 ≠ 0 the limit γ_{τ} = lim_{y→∞} Q_{τ} (iy) exists as a real number and is given by

γ_{τ} = \frac{γ − τq}{τγ + 1}.

Then a simple calculation shows that

−iy(Q_{τ} (iy) − γ_{τ} ) = \frac{−iy(Q(iy) − γ)(τ^{2}q + 1)}{(τγ + 1)(τQ(iy) + 1)}.

Taking limits and using (i) of Proposition 4.4 gives (5.18). Next assume that Q(l) ∈ N_{−1} and use the identity:

\begin{matrix}− (iy)^{2}(Q_{τ} (iy) − γ_{τ} + \frac{m_{0}(σ_{τ})}{iy}) \\ = \frac{(−(iy)^{2}[Q(iy) − γ + m_{0}(σ)/iy](τγ + 1) − (iy)[Q(iy) − γ]τm_{0}(σ))(τ^{2}q + 1)}{(τγ + 1)^{2}(τQ(iy) + 1)}. \end{matrix}

Hence, taking limits and using (i) and (ii) of Proposition 4.4 gives (5.19). Now assume that Q(l) ∈ N_{−2} . A straightforward calculation shows that

−(iy)^{3}(Q_{τ} (iy) − γ_{τ} + \frac{m_{0}(σ_{τ})}{iy} + \frac{m_{1}(σ_{τ})}{(iy)^{2}})

= −\frac{(iy)^{3}[Q(iy) − γ + m_{0}(σ)/iy + m_{1}(σ)/(iy)^{2}](τ^{2}q + 1)(τγ + 1)^{2}}{(τγ + 1)^{3}(τQ(iy) + 1)}

− \frac{(iy)^{2}[Q(iy) − γ + m_{0}(σ)/iy](τ^{2}q + 1)(τγ + 1)τm_{0}(σ)}{(τγ + 1)^{3}(τQ(iy) + 1)}

− \frac{(iy)[Q(iy) − γ](τ^{2}q + 1)[(τγ + 1)τm_{1}(σ) + τ^{2}m_{0}(σ)^{2}]}{(τγ + 1)^{3}(τQ(iy) + 1)}.

Again by taking limits and using (i), (ii), and (iii) of Proposition 4.4 yields (5.20). This completes the proof.

6. The multivalued case

Let A be a selfadjoint relation in the Hilbert space ℌ and let Z = span {φ, ψ} be a one-dimensional subspace which satisfies (2.3). We may assume, without loss of generality, that mul A is one-dimensional and that S is a closed symmetric operator. For this case we will consider Theorem 2.4 in greater detail.

Let A = A_{s} ⊕ A_{∞} be a selfadjoint relation whose multivalued part mul A is one-dimensional, so that

ℌ = (ℌ ⊖ mul A) ⊕ mul A.

Recall that dom A = dom A_{s} is dense in ℌ ⊖ mul A. Choose a unit vector in mul A and identify ℌ = (ℌ ⊖ mul A) ⊕ mul A with ℌ = (ℌ ⊖ mul A) ⊕ C in the obvious way. We use vector notation for the elements in this last Hilbert space. Hence we may write

(6.1)  A = \{\{( \begin{matrix}h \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}h \\ u \end{matrix} )\} : h ∈ dom A_{s}, u ∈ C\} .

Now let Z = span {φ, ψ} be any one-dimensional subspace in ℌ^{2} ; it is equivalent to the subspace span {φ, P ψ}. Hence, without loss of generality, we may assume that

(6.2)  Z = span \{( \begin{matrix}φ_{0} \\ δ \end{matrix} ) , ( \begin{matrix}ψ_{0} \\ 0 \end{matrix} )\} ,

with elements φ_{0}, ψ_{0} ∈ ℌ ⊖ mul A and δ ∈ C.

If A is a selfadjoint relation whose multivalued part is one-dimensional, then S = A ∩ Z^{∗} is a closed symmetric relation and mul S is at most one-dimensional. We now consider a suitable formulation of the definition of S and its adjoint S^{∗}.

Lemma 6.1. Let A be a selfadjoint relation as in (6.1) and let Z be a onedimensional subspace as in (6.2). Then the closed symmetric relation S = A ∩ Z^{∗} is given by

(6.3) S = \{\{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}f \\ u \end{matrix} )\} : f ∈ dom A_{s}, uδ̄ = [f, ψ_{0}] − [A_{s}f, φ_{0}], u ∈ C\} .

Its adjoint S^{∗} is given by (6.4)

S^{∗} = \{\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}A_{s}(h + cφ_{0}) − cψ_{0} \\ v \end{matrix} )\} : h + cφ_{0} ∈ dom A_{s}, c, v ∈ C\} .

Proof. The identity (6.3) follows from the definition S = A ∩ Z^{∗} and the special form (6.1) and (6.2) of A and Z, respectively. In order to prove (6.4), let

\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}k \\ v \end{matrix} )\} , h, k ∈ ℌ ⊖ mul A, c, v ∈ C,

be an element of S^{∗}. Then for all f ∈ dom A_{s} we have

⟨\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}k \\ v \end{matrix} )\} , \{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}f \\ u \end{matrix} )\}⟩ = 0,

with uδ̄ = [f, ψ_{0}] − [A_{s}f, φ_{0}]. This is equivalent to

[k + cψ_{0}, f] − [h + cφ_{0}, A_{s}f] = 0, f ∈ dom A_{s}.

Hence h + cφ_{0} ∈ dom A_{s} and A_{s}(h + cφ_{0}) = k + cψ_{0} . Therefore, S^{∗} is contained in the righthand side of (6.4).

Conversely, assume that h+cφ_{0} ∈ dom A_{s}. Then for every f ∈ dom A_{s} with uδ̄ = [f, ψ_{0}] − [A_{s}f, φ_{0}] we have

\begin{matrix}⟨\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}A_{s}(h + cφ_{0}) − cψ_{0} \\ v \end{matrix} )\} , \{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}f \\ u \end{matrix} )\}⟩ \\ = [A_{s}(h + cφ_{0}), f] − [h + cφ_{0}, A_{s}f] = 0. \end{matrix}

Hence the righthand side of (6.4) is contained in S^{∗}. This completes the proof.

Note that

(6.5)  [( \begin{matrix}ψ_{0} \\ 0 \end{matrix} ) , ( \begin{matrix}φ_{0} \\ δ \end{matrix} )] = [ψ_{0}, φ_{0}].

From (6.1) and (6.2) it follows that Z ⊂ A if and only if δ = 0 and {φ_{0}, ψ_{0}} ∈ A_{s}. Hence the condition (2.3) is equivalent to

(6.6)  δ ≠ 0  or  {φ_{0}, ψ_{0}} ∉ A_{s}.

Clearly, if δ = 0 and {φ_{0}, ψ_{0}} ∉ A_{s} it follows from (6.3) that mul S is onedimensional, and reduction to ℌ ⊖ mul S gives us the situation considered in the previous section. (Note that this case occurs if φ ∈ dom A.) On the other hand, it follows from (6.3) that if δ ≠ 0 then S is an operator. Therefore we will assume in the rest of this section that

(6.7)  δ ≠ 0,

which is equivalent to (2.3) and S being an operator. In a similar manner as before, the form (6.4) in which the adjoint S^{∗} is written is now used for the description of all canonical selfadjoint extensions of S.

Theorem 6.2. Let A be a selfadjoint relation given by (6.1), let the onedimensional subspace Z be given by (6.2) and assume (6.7). Then the Q-function Q(l) of S = A ∩ Z^{∗} and A satisfies

(6.8)  lim_{y→∞} \frac{Im Q(iy)}{y} = |δ|^{2}.

The canonical selfadjoint extensions A(τ), τ ∈ R ∪ {∞}, of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0 by (6.9)

\begin{aligned}A(τ) & = \{\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}A_{s}(h + cφ_{0}) − cψ_{0} \\ u \end{matrix} )\} : h + cφ_{0} ∈ dom A_{s}, \\ & c, u ∈ C, uδ̄ = c(1/τ + [ψ_{0}, φ_{0}]) − [A_{s}(h + cφ_{0}), φ_{0}] + [h + cφ_{0}, ψ_{0}]\}. \end{aligned}

Proof. It follows from (4.1) that Q(l) satisfies (6.8). In order to prove (6.9) we apply Theorem 2.4. Consider an arbitrary element in A(τ) as given by (2.4) for τ ≠ 0, 1/τ + [ψ_{0}, φ_{0}] ≠ 0, or by (2.5) for 1/τ + [ψ_{0}, φ_{0}] = 0, cf. (6.5). That is, consider an element of the form

(6.10)  \{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}f \\ u \end{matrix} )\} − c \{( \begin{matrix}φ_{0} \\ δ \end{matrix} ) , ( \begin{matrix}ψ_{0} \\ 0 \end{matrix} )\} .

If 1/τ + [ψ_{0}, φ_{0}] ≠ 0, assume that c ∈ C satisfies

(6.11)  \begin{aligned}c(1/τ + [ψ_{0}, φ_{0}]) & = ⟨\{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}f \\ u \end{matrix} )\} , \{( \begin{matrix}φ_{0} \\ δ \end{matrix} ) , ( \begin{matrix}ψ_{0} \\ 0 \end{matrix} )\}⟩ \\ & = [A_{s}f, φ_{0}] − [f, ψ_{0}] + uδ̄. \end{aligned}

If 1/τ + [ψ_{0}, φ_{0}] = 0, then assume that

( \begin{matrix}f \\ 0 \end{matrix} ) ∈ dom S.

Note that then (6.11) is still satisfied due to (6.3). Define h = f − cφ_{0} , then h + cφ_{0} = f ∈ dom A_{s} and A_{s}f − cψ_{0} = A_{s}(h + cφ_{0}) − cψ_{0} . Therefore the element (6.10) has the form

(6.12)  \{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}A_{s}(h + cφ_{0}) − cψ_{0} \\ u \end{matrix} )\} ,

where c must satisfy

(6.13)  c(1/τ + [ψ_{0}, φ_{0}]) = [A_{s}(h + cφ_{0}), φ_{0}] − [h + cφ_{0}, ψ_{0}] + uδ̄.

Hence A(τ) is contained in the righthand side of (6.9).

Conversely, consider an arbitrary element in the righthand side of (6.9) with τ ≠ 0. That is, consider an element of the form (6.12) where c ∈ C satisfies (6.13). Define f = h + cφ_{0} , then f ∈ dom A_{s} and the element (6.12) has the form (6.10). Moreover (6.11) is satisfied. Note that if 1/τ + [ψ_{0}, φ_{0}] = 0, then it follows from (6.11) and (6.3), that

\{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}A_{s}f \\ u \end{matrix} )\} ∈ S.

Therefore, the righthand side of (6.9) is contained in the righthand side of (2.4) or of (2.5), respectively. This completes the proof.

Note that in Theorem 6.2 the relation A(0) = A is the only canonical selfadjoint extension of S = A ∩ Z^{∗}, which is not an operator. All the other extensions A(τ), τ ∈ R ∪ {∞}, τ ≠ 0, are (densely defined) selfadjoint operators, which can also be immediately seen from (6.9). Moreover, it can be seen from (6.3) that dom A = dom S, cf. Proposition 4.1; comparing (6.4) and (6.9) we see that dom S^{∗} = dom A(τ) for all τ ≠ 0, cf. Proposition 5.4. The Q-function Q(l) of S and A is an exceptional function corresponding to the class N_{0} ; all selfadjoint extensions A(τ), τ ≠ 0, in (6.9) have a Q-function in N_{0} . Recall that if we normalize Q(l) and Q_{τ} (l) by Re Q(μ) = 0 and Re Q_{τ} (μ) = 0 for some μ ∈ C \ R, then the functions Q(l) and Q_{τ} (l) are related by (3.17).

We can use this model to interpret some results concerning N_{−1} and N_{−2} functions, cf. [HS1]. Now the operator part A_{s} plays an important role in the descriptions. Moreover, we will present some transformation formulas for the corresponding moments in terms of the exceptional extension.

Using the special form (6.1) and (6.2) of A and Z, respectively, and rearranging terms in (3.3), we obtain

(6.14)  Q(l) − |δ|^{2}l = l[φ_{0}, φ_{0}] + [(A_{s} − l)^{−1}(lφ_{0} − ψ_{0}), l̄φ_{0} − ψ_{0}].

In the Hilbert space (ℌ ⊖ mul A)^{2} we introduce the, at most one-dimensional, subspace Z_{0} by

(6.15)  Z_{0} = span {φ_{0}, ψ_{0}}.

We define the symmetric operator S_{0} in the Hilbert space ℌ ⊖ mul A by S_{0} = A_{s} ∩ Z_{0}^{∗}, i.e.,

(6.16)  S_{0} = {{f, A_{s}f} : f ∈ dom A_{s}, [A_{s}f, φ_{0}] − [f, ψ_{0}] = 0}.

Note that

S_{0} = S ∩ (ℌ ⊖ mul A)^{2}.

If Z_{0} ⊂ A_{s}, then we have the degenerate case where S_{0} = A_{s}, and it follows from (6.14) that

(6.17)  Q(l) = |δ|^{2}l + [φ_{0}, ψ_{0}].

Now we assume that Z_{0} = span {φ_{0}, ψ_{0}} is one-dimensional and that

(6.18)  A_{s} ∩ span {φ_{0}, ψ_{0}} = {0, 0}.

Then S_{0} is a closed symmetric operator in ℌ ⊖ mul A with defect numbers (1, 1). By comparing the righthand side of (6.14) with (3.4), we obtain the following result.

Lemma 6.3. The function Q(l) − |δ|^{2}l is the Q-function of S_{0} and A_{s}. It belongs to N_{1} or to N_{0} if and only if φ_{0} ∈ dom |A_{s}|^{1/2} or φ_{0} ∈ dom A_{s}, respectively.

The following result is obtained from [HS1, Propositions 3.2 and 4.2].

Corollary 6.4. The Q-function Q(l) of S = A∩Z^{∗} and A is an exceptional function corresponding to N_{−1} or N_{−2} , if and only if φ_{0} ∈ dom |A_{s}|^{1/2} or φ_{0} ∈ dom A_{s}, respectively. In these cases the Q-functions of the selfadjoint extensions A(τ), τ ≠ 0, in (6.9) belong to N_{−1} or N_{−2} , respectively.

The expression for A(τ), τ ≠ 0, in (6.9) can be simplified in each of these cases, cf. Propositions 5.3 and 5.4. We will not pursue this, but instead, we will express the corresponding moments in terms of Q(l). We observe that if Q(l) − |δ|^{2}l ∈ N_{1} , then there exists a real limit

(6.19)  η = lim_{y→∞}(Q(iy) − |δ|^{2}iy),

and that if Q(l) − |δ|^{2}l ∈ N_{0} , then

(6.20)

m_{0} = sup_{y>0} y Im(Q(iy) − |δ|^{2}iy) = lim_{y→∞} −iy(Q(iy) − |δ|^{2}iy − η)
= ∫_{R} dϱ(t) < ∞,

and is positive, since we have assumed that Q(l) − |δ|^{2}l does not reduce to a real constant. Denote the spectral function of Q_{τ} (l) (cf. the integral representation (4.2)) by σ_{τ} (t). Again we use the notation q = (Im Q(μ))^{2} .

Proposition 6.5. Let τ ∈ R ∪ {∞}, τ ≠ 0. The moments m_{0}(σ_{τ}) are given by

(6.21)  m_{0}(σ_{τ}) = \frac{τ^{2}q + 1}{τ^{2}|δ|^{2}}.

If Q(l) − |δ|^{2}l belongs to N_{1} , then the moments m_{1}(σ_{τ}) are given by

(6.22)  m_{1}(σ_{τ}) = −\frac{(τ^{2}q + 1)(τη + 1)}{τ^{3}|δ|^{4}}.

Moreover, if Q(l) − |δ|^{2}l belongs to N_{0} , then the moments m_{2}(σ_{τ}) are given by

(6.23)  m_{2}(σ_{τ}) = \frac{(τ^{2}q + 1)[(τη + 1)^{2} + τ^{2}|δ|^{2}m_{0}]}{τ^{4}|δ|^{6}}.

Proof. Clearly, lim_{y→∞} Q_{τ} (iy) = 1/τ and hence

iy(Q_{τ} (iy) − 1/τ) = −\frac{τ^{2}q + 1}{1/iy(τ^{2}Q(iy) + τ)}.

Taking limits and using (6.8) and (i) of Proposition 4.4 gives (6.21). Now assume that Q(l) − |δ|^{2}l belongs to N_{−1} and use the identity:

(iy)^{2}(Q_{τ} (iy) − \frac{1}{τ} + \frac{τ^{2}q + 1}{iyτ^{2}|δ|^{2}} ) = iy(\frac{(τ^{2}q + 1)(τ(Q(iy) − |δ|^{2}iy) + 1)}{τ^{2}|δ|^{2}(τQ(iy) + 1)}).

Hence, taking limits and using (6.8), (6.19), and (ii) of Proposition 4.4 gives (6.22). Now assume that Q(l) − |δ|^{2}l belongs to N_{0} and use the identity:

(iy)^{3}(Q(iy) − \frac{1}{τ} + \frac{τ^{2}q + 1}{iyτ^{2}|δ|^{2}} − \frac{(τ^{2}q + 1)(τη + 1)}{(iy)^{2}τ^{3}|δ|^{4}})

= iy(\frac{(τ^{2}q + 1)iy(Q(iy) − |δ|^{2}iy − η)τ^{2}|δ|^{2}}{(τQ(iy) + 1)τ^{3}|δ|^{4}} )

− iy(\frac{(τ^{2}q + 1)(τη + 1)(τ(Q(iy) − |δ|^{2}iy) + 1)}{(τQ(iy) + 1)τ^{3}|δ|^{4}}).

Again by taking limits and using (6.8), (6.19), (6.20), and (iii) of Proposition 4.4 yields (6.23). This completes the proof.

Finally, we will consider the linear fractional transform of H(l) = Q(l)−|δ|^{2}l. It is the Q-function of S_{0} and A_{s}. Assume that it is normalized by Re H(μ) = 0. Then for each τ ∈ R ∪ {∞} the Nevanlinna function H_{τ} (l) is defined by

(6.24)  H_{τ} (l) = \frac{H(l) − τ(Im H(μ))^{2}}{τH(l) + 1} ,  τ ∈ R ∪ \{∞\}.

It is the Q-function of S_{0} and the canonical selfadjoint extension A_{0}(τ) of S_{0} in ℌ ⊖ mul A, defined by

(6.25)  \begin{matrix}A_{0}(τ) = & \{\{h, A_{s}(h + cφ_{0}) − cψ_{0}\} : h + cφ_{0} ∈ dom A_{s}, c ∈ C, \\ c = \frac{[A_{s}(h + cφ_{0}), φ_{0}] − [h + cφ_{0}, ψ_{0}]}{1/τ + [ψ_{0}, φ_{0}]}\}, \end{matrix}

while for 1/τ + [ψ_{0}, φ_{0}] = 0 the canonical selfadjoint extension A_{0}(τ) is given by

(6.26)  \begin{matrix}A_{0}(τ) = & \{\{h, A_{s}(h + cφ_{0}) − cψ_{0}\} : h + cφ_{0} ∈ dom A_{s}, c ∈ C, \\ [A_{s}(h + cφ_{0}), φ_{0}] − [h + cφ_{0}, ψ_{0}] = 0\}. \end{matrix}

Note that A(τ) and A_{0}(τ) are related by

A_{0}(τ) = {{P f, g} : {f, g} ∈ A(τ), g ∈ ℌ ⊖ mul A},

and that A(τ) ∩ (ℌ ⊖ mul A)^{2} = S_{0}, τ ≠ 0.

7. Multiplication operators

In this section we consider a real-valued nondecreasing function ρ on R and associate with this function the Hilbert space L^{2}(dρ). Multiplication by the independent variable:

ℳ_{ρ} = {{f, g} ∈ (L^{2}(dρ))^{2} : g = tf},

is a densely defined selfadjoint operator, cf. [AG]. We will consider either this selfadjoint operator ℳ_{ρ} in the Hilbert space L^{2}(dρ), or the selfadjoint relation

ℳ_{ρ} ⊕ ({0} ⊕ C),

in the Hilbert space L^{2}(dρ)⊕C. In each of these cases we will consider a symmetric restriction with defect numbers (1, 1) and describe its selfadjoint extensions.

The operator case. Let ℌ = L^{2}(dρ) and A = ℳ_{ρ}. Let Z = span {φ, ψ} be a one-dimensional subspace of ℌ^{2} which satisfies (2.3). Define the function ω by ω(t) = tφ(t) − ψ(t). The function ω does not vanish on R because of (2.3). Note that the function ω remains the same if the pair {φ, ψ} is replaced by an equivalent pair. Moreover, pairs producing the same ω are equivalent. The function ω does not necessarily belong to L^{2}(dρ). In fact, we have

(7.1)

φ ∈ dom |A|^{1/2}  if and only if  φω̄ ∈ L^{1}(dρ)

if and only if ∫_{R} \frac{|ω(t)|^{2}}{|t| + 1} dρ(t) < ∞,

and

(7.2)  φ ∈ dom A  if and only if  ∫_{R} |ω(t)|^{2} dρ(t) < ∞.

Clearly, the conditions φ ∈ dom A and Aφ − ψ ∈ dom |A|^{1/2} are equivalent to \sqrt{|t| + 1} ω(t) ∈ L^{2}(dρ) and the conditions φ ∈ dom A and Aφ − ψ ∈ dom A are equivalent to (|t| + 1) ω(t) ∈ L^{2}(dρ).

For each f ∈ dom A the function fω̄ is in L^{1}(dρ) and we denote the integral ∫_{R} f\overline{ω} dρ by [f, ω], abusing the notation for the inner product in L^{2}(dρ). Observe that for each f ∈ dom A

(7.3)  ⟨{f, Af}, {φ, ψ}⟩ = [f, ω].

Let h ∈ L^{2}(dρ) and c ∈ C, then clearly

(7.4)  th + cω ∈ L^{2}(dρ)  if and only if  h + cφ ∈ dom A.

Hence, in this case (h + cφ)\overline{ω} belongs to L^{1}(dρ) and the integral ∫_{R}(h + cφ)\overline{ω} dρ is thus denoted by [h + cφ, ω].

We now express S = A ∩ Z^{∗}, S^{∗} and the corresponding null spaces of S^{∗} − l in terms of the function ω. The following lemma is a restatement of (5.1), (5.2) and (3.1).

Lemma 7.1. The symmetric operator S is given by

(7.5)  S = {{f, tf} : f ∈ dom ℳ_{ρ}, [f, ω] = 0}.

The adjoint relation S^{∗} is given by

(7.6)  S^{∗} = {{h, th + cω} : c ∈ C, h, th + cω ∈ L^{2}(dρ)}.

In particular, S is not densely defined if and only if ω ∈ L^{2}(dρ), in which case mul S^{∗} = span {ω}. The null space ker (S^{∗} − l) is spanned by

(7.7)  χ(l) = \frac{ω}{t − l} ,  l ∈ C \ R.

We describe all canonical selfadjoint extensions of S by means of Theorem 5.2, applied to the case that ℌ = L^{2}(dρ) and A = ℳ_{ρ}. Our description is in terms of restrictions of the adjoint S^{∗} in (7.6).

Proposition 7.2. All canonical selfadjoint extensions A(τ) of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0 and 1/τ + [ψ, φ] ≠ 0, by

(7.8) A(τ) = \{\{h, th + cω\} : c ∈ C, h, th + cω ∈ L^{2}(dρ), c = \frac{[h + cφ, ω]}{1/τ + [ψ, φ]}\}.

If 1/τ + [ψ, φ] = 0, then A(τ) is given by

(7.9) A(τ) = {{h, th + cω} : c ∈ C, h, th + cω ∈ L^{2}(dρ), [h + cφ, ω] = 0}.

We now assume that Q(l) ∈ N_{1} and formulate the counterpart of Proposition 5.3 for the multiplication operator on L^{2}(dρ). Recall from Proposition 4.4 that the exceptional value τ corresponding to Q(l) is given by 1/τ + γ = 0. For h ∈ L^{2}(dρ) for which hω̄ ∈ L^{1}(dρ), we denote the integral ∫_{R} h\overline{ω} dρ by [h, ω], abusing the notation for the inner product in L^{2}(dρ). The case Q(l) ∈ N_{0} gives the same formal results.

Proposition 7.3. Assume that Q(l) belongs to N_{1} . Then all canonical selfadjoint extensions A(τ) of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0 and 1/τ + γ ≠ 0, by

(7.10) \begin{aligned}A(τ) & = \{\{h, th + \frac{1}{1/τ + γ} [h, ω] ω\} : \\ & h ∈ L^{2}(dρ), hω̄ ∈ L^{1}(dρ), th + \frac{1}{1/τ + γ} [h, ω] ω ∈ L^{2}(dρ)\}. \end{aligned}

These extensions are all operators. If 1/τ + γ = 0, then the exceptional canonical selfadjoint extension A(τ) is given by

(7.11) \begin{aligned}A(τ) & = \{\{h, th + c ω\} : \\ & h ∈ L^{2}(dρ), hω̄ ∈ L^{1}(dρ), [h, ω] = 0, th + c ω ∈ L^{2}(dρ)\}. \end{aligned}

The extension A(τ), 1/τ + γ = 0, is not an operator if and only if ω ∈ L^{2}(dρ), in which case mul S^{∗} = span {ω}.

The Q-function Q(l) of S and A in (3.3) has an integral representation in terms of the measure dρ. It is given by

(7.12)  Q(l) = [ψ, φ] + [φ, ψ] + ∫_{R}(\frac{|ω(t)|^{2}}{t − l} − t|φ(t)|^{2}) dρ(t).

Observe that the function t|φ|^{2} acts as a regularizing factor, which ensures the integrability of the integrand in (7.12). To see (7.12), observe that it follows from the definition of ω that

\frac{|ω|^{2}}{t − l} − t|φ|^{2} = l|φ|^{2} − φ\overline{ψ} − ψ\overline{φ} + \frac{(lφ − ψ)(l\overline{φ} − \overline{ψ})}{t − l}.

Since φ and ψ belong to L^{2}(dρ) and (t − l)^{−1} is bounded for t ∈ R it is clear that the righthand side belongs to L^{1}(dρ). Hence the integral in (7.12) is welldefined and the identity in (7.12) for Q(l) follows from rewriting (3.4). We can characterize the Q-function Q(l) in terms of the function ω(t) as follows:

(i) Q(l) belongs to N_{1} if and only if ω(t)/\sqrt{|t| + 1} belongs to L^{2}(dρ), (ii) Q(l) belongs to N_{0} if and only if ω(t) belongs to L^{2}(dρ), (iii) Q(l) belongs to N_{−1} if and only if \sqrt{|t| + 1} ω(t) belongs to L^{2}(dρ), (iv) Q(l) belongs to N_{−2} if and only if (|t| + 1) ω(t) belongs to L^{2}(dρ). In these cases (7.12) can be rewritten as

(7.13)  Q(l) = γ + ∫_{R} \frac{|ω(t)|^{2}}{t − l} dρ(t).

In Proposition 1.4 we have given necessary and sufficient conditions for the existence of a Q-function in N_{0} or N_{−2} , in terms of the symmetric operator S. In the present case of a multiplication operator on L^{2}(dρ) we obtain

(a) the function Q(l) belongs to N_{0} if and only if dom S is not dense in L^{2}(dρ),

(b) the function Q(l) belongs to N_{−2} if and only if dom A = dom S + span {ω}. It is now possible to obtain similar results for the classes N_{1} and N_{−1} . For this purpose, we define the Hilbert space L_{+}^{2}(dρ) ⊂ L^{2}(dρ) as follows

L_{+}^{2}(dρ) = L^{2}((|t| + 1)dρ).

Then dom A ⊂ L_{+}^{2}(dρ) and dom A is dense in this Hilbert space. In addition we define the Hilbert space L_{+}^{2}(S) as the closure of dom S in the space L_{+}^{2}(dρ).

Proposition 7.4. We have

(i) the function Q(l) belongs to N_{1} if and only if dom S is not dense in L_{+}^{2}(dρ), (ii) the function Q(l) belongs to N_{−1} if and only if L_{+}^{2}(dρ) = L_{+}^{2}(S) + span {ω}.

Proof. We prove (i) by showing that dom S is not dense in L_{+}^{2}(dρ) if and only if ω/\sqrt{|t| + 1} ∈ L^{2}(dρ). If ω/\sqrt{|t| + 1} ∈ L^{2}(dρ), then h ∈ dom S means that

0 = ∫_{R} h(t)\overline{ω(t)} dρ(t) = ∫_{R} h(t)\frac{\overline{ω(t)}}{|t| + 1}(|t| + 1) dρ(t).

Since ω/(|t| + 1) ∈ L_{+}^{2}(dρ) we conclude that dom S is not dense in L_{+}^{2}(dρ).

For the converse statement we assume that there exists a nontrivial g_{0} ∈ L_{+}^{2}(dρ), such that for all h ∈ dom S

∫_{R} h(t)\overline{g_{0}(t)}(|t| + 1) dρ(t) = 0.

Since dom A is dense in L_{+}^{2}(dρ) there exists an element f_{0} ∈ dom A for which

c_{0} = ∫_{R} f_{0}(t)\overline{g_{0}(t)}(|t| + 1) dρ(t) ≠ 0.

Clearly f_{0} does not belong to dom S. We define c_{1} and ω_{0}(t) by

c_{1} = ∫_{R} f_{0}(t)\overline{ω(t)} dρ(t), ω_{0}(t) = ω(t)/(|t| + 1).

It follows that for every f ∈ dom A

(7.14)  ∫_{R} f(t)(c_{1}\overline{g_{0}(t)} − c_{0}\overline{ω_{0}(t)} )(|t| + 1) dρ(t) = 0,

as this is clear for f = f_{0} and for f ∈ dom S, and dom A is the span of f_{0} and dom S. Let Δ be a compact subinterval of R. Let f be any function in L_{Δ}^{2}(dρ) and extend f to all of R by setting it equal to zero outside Δ. Then f ∈ dom A and, hence, it follows from (7.14) that

∫_{Δ} f(t)(c_{1}\overline{g_{0}(t)} − c_{0}\overline{ω_{0}(t)} )(|t| + 1) dρ(t) = 0.

Therefore we conclude that c_{1}\overline{g_{0}(t)} − c_{0}\overline{ω_{0}(t)} = 0 in Δ almost everywhere with respect to the measure (|t|+1)dρ(t). Note that here ω_{0}(t) = (tφ(t)−ψ(t))/(|t|+1), restricted to Δ, belongs to L_{Δ}^{2}((|t|+1)dρ). Since Δ is arbitrary we conclude that c_{1}\overline{g_{0}(t)}−c_{0}\overline{ω_{0}(t)} = 0 almost everywhere with respect to (|t|+1)dρ(t). Now c_{0} ≠ 0 forces ω_{0} ∈ L_{+}^{2}(dρ), which is equivalent to ω/\sqrt{|t| + 1} ∈ L^{2}(dρ).

We prove (ii) by showing that L_{+}^{2}(dρ) = L_{+}^{2}(S) + span {ω} if and only if \sqrt{|t| + 1} ω ∈ L^{2}(dρ). If \sqrt{|t| + 1} ω ∈ L^{2}(dρ), then ω ∈ L_{+}^{2}(dρ) and certainly ω/(|t| + 1) ∈ L_{+}^{2}(dρ). Moreover, in the sense of L_{+}^{2}(dρ) we have the orthogonal decomposition

dom A = dom S [+] span \{\frac{ω}{|t| + 1}\}.

This implies that

L_{+}^{2}(dρ) = L_{+}^{2}(S) [+] span \{\frac{ω}{|t| + 1}\}.

Now ω ∈ L_{+}^{2}(dρ) and ω ∉ L_{+}^{2}(S), imply L_{+}^{2}(dρ) = L_{+}^{2}(S) + span {ω}. The converse statement is obvious. This completes the proof.

We make some comments about the usual Riesz–Herglotz integral representation. For simplicity we assume that ω does not vanish on a set of positive dρ-measure, and we associate with the measure dρ and the function ω a measure dσ by dσ = |ω|^{2} dρ. Since the function χ(i) belongs to L^{2}(dρ) (cf. Lemma 7.1), it follows that the measure dσ satisfies the integrability condition

(7.15)  ∫_{R} \frac{dσ(t)}{t^{2} + 1} < ∞.

In terms of the measure dσ, the function Q(l) in (7.12) has the integral representation

Q(l) = α + ∫_{R}(\frac{1}{t − l} − \frac{t}{t^{2} + 1}) dσ(t),

where α ∈ R is given by

\begin{aligned}α & = ∫_{R} \frac{t}{t^{2} + 1}(|ψ(t)|^{2} − |φ(t)|^{2}) dρ(t) \\ & + ∫_{R} \frac{1}{t^{2} + 1}φ(t)\overline{ψ(t)} dρ(t) + ∫_{R} \frac{1}{t^{2} + 1}ψ(t)\overline{φ(t)} dρ(t). \end{aligned}

It follows from (7.12) that the function Q(l) belongs to N_{1} , N_{0} , N_{−1} , N_{−2} if and only if

\begin{matrix}∫_{R} \frac{dσ(t)}{|t| + 1} < ∞,  ∫_{R} dσ(t) < ∞, \\ ∫_{R}(|t| + 1) dσ(t) < ∞,  or  ∫_{R}(t^{2} + 1) dσ(t) < ∞, \end{matrix}

respectively, cf. [HS1]. In [HLS] an operator model was constructed involving the Hilbert space L^{2}(dσ) and multiplication by the independent variable ℳ_{σ} in that space. Let ι be the mapping that assigns to each function f on R a function of the form f/ω. Note that ιω = 1. Then ι provides an isometric isomorphism between L^{2}(dρ) and L^{2}(dσ); in addition the selfadjoint operators ℳ_{ρ} and ℳ_{σ} are isometrically isomorphic under ι. The symmetric operator S in L^{2}(dρ) given by (7.5) is isometrically isomorphic to the symmetric operator {{f, g} ∈ ℳ_{σ} : [f, 1] = 0}. If ω(t) = 1, then the two measures coincide: dσ = dρ and ι is the identity mapping. If we start with a measure dρ which satisfies

∫_{R} \frac{dρ(t)}{t^{2} + 1} < ∞,

then we can choose φ and ψ in L^{2}(dρ) such that ω = 1. Take any ψ ∈ L^{2}(dρ), which behaves like −1 + O(t) near 0. Then the function φ defined by φ(t) = (ψ(t) + 1)/t belongs to L^{2}(dρ). As we mentioned before any other pair giving the same ω = 1 is equivalent to this choice.

The above descriptions are all stated in terms of the function ω(t), which does not necessarily belong to the Hilbert space L^{2}(dρ). However, if for instance Q(l) belongs to N_{1} , then ω(t) belongs to the Hilbert space L_{−}^{2}(dρ), where L_{−}^{2}(dρ) is the L^{2} space with measure dρ(t)/(|t| + 1), which contains the original Hilbert space L^{2}(dρ) as a proper dense subset. Its dual is the Hilbert space L_{+}^{2}(dρ), which is contained in the original Hilbert space L^{2}(dρ). The inner product of the Hilbert space L^{2}(dρ) can be extended to serve as a duality between these spaces. The results in the present section act as models for the interpretation of the previous results by means of specific triplets of Hilbert spaces. In general, this program is carried out for the case that Q(l) belongs to N_{1} in [HKS] and for the case that Q(l) satisfies less restrictive conditions in [HS2].

The multivalued case. Let ℌ = L^{2}(dρ) ⊕ C and define A = ℳ_{ρ} ⊕ ({0} ⊕ C). Let Z ⊂ ℌ^{2} be defined by

Z = span \{( \begin{matrix}φ_{0} \\ δ \end{matrix} ) , ( \begin{matrix}ψ_{0} \\ 0 \end{matrix} )\} ,

with elements φ_{0}, ψ_{0} ∈ L^{2}(dρ) and δ ∈ C. We assume that (6.7) is satisfied, which is equivalent to (2.3) and S being an operator. Define the function ω_{0} by ω_{0}(t) = tφ_{0}(t) − ψ_{0}(t). The following lemma is just a translation of (6.3), (6.4) and (3.1).

Lemma 7.5. The closed symmetric operator S = A ∩ Z^{∗} is given by

(7.16)  S = \{\{( \begin{matrix}f \\ 0 \end{matrix} ) , ( \begin{matrix}tf \\ u \end{matrix} )\} : f ∈ dom ℳ_{ρ}, uδ̄ + [f, ω_{0}] = 0\} .

Its adjoint S^{∗} is given by

(7.17)  S^{∗} = \{\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}th + cω_{0} \\ v \end{matrix} )\} : th + cω_{0} ∈ L^{2}(dρ), c, v ∈ C\} .

The null space ker (S^{∗} − l) is spanned by

(7.18)  χ(l) = ( \begin{matrix}ω_{0}/(t − l) \\ δ \end{matrix} ) ,  l ∈ C \ R.

As before, the form (7.17) is now used for the description of all canonical selfadjoint extensions of S, cf. Theorem 6.2.

Proposition 7.6. The canonical selfadjoint extensions A(τ), τ ∈ R ∪ {∞}, of S = A ∩ Z^{∗} are given by A(0) = A and for τ ≠ 0 by

(7.19)  \begin{aligned}A(τ) & = \{\{( \begin{matrix}h \\ −cδ \end{matrix} ) , ( \begin{matrix}th + cω_{0} \\ u \end{matrix} )\} : th + cω_{0} ∈ L^{2}(dρ), c, u ∈ C, \\ & uδ̄ = c(1/τ + [ψ_{0}, φ_{0}]) − [h + cφ_{0}, ω_{0}] \}. \end{aligned}

We give an integral representation of the Q-function Q(l) in terms of the measure dρ and the constant δ in (6.2). It follows from (6.8) and (6.14) that the Q-function Q(l) in (3.3) of S = A ∩ Z^{∗} and A is given by

(7.20)  Q(l) = [ψ_{0}, φ_{0}] + [φ_{0}, ψ_{0}] + |δ|^{2}l + ∫_{R}(\frac{|ω_{0}(t)|^{2}}{t − l} − t|φ_{0}(t)|^{2}) dρ(t).

The function H(l) = Q(l) − |δ|^{2}l can be further characterized in terms of the function ω_{0} by means of (6.14):

(i) H(l) belongs to N_{1} if and only if ω_{0}(t)/\sqrt{|t| + 1} belongs to L^{2}(dρ), (ii) H(l) belongs to N_{0} if and only if ω_{0}(t) belongs to L^{2}(dρ) If H(l) ∈ N_{1} , then

(7.21)  H(l) = γ_{0} + ∫_{R} \frac{|ω_{0}(t)|^{2}}{t − l} dρ(t),

where γ_{0} = [ψ_{0}, φ_{0}] + [φ_{0}, ψ_{0}] − [tφ_{0}, φ_{0}]. In this case the conditions in (7.19) of Proposition 7.6:

th + cω_{0} ∈ L^{2}(dρ), uδ̄ = c(1/τ + [ψ_{0}, φ_{0}]) − [h + cφ_{0}, ω_{0}],

are equivalent to

hω̄_{0} ∈ L^{1}(dρ), uδ̄ = c(1/τ + γ_{0}) − [h, ω_{0}].

Note that it follows from (7.17) that \binom{0}{1} spans mul S^{∗}. Moreover, we see directly from (7.17) that ω_{0} ∈ L^{2}(dρ) if and only if \binom{0}{1} ∈ dom S^{∗}. Hence, we have proved H(l) ∈ N_{0} if and only if mul S^{∗} ⊂ dom S^{∗}. We have mentioned in Corollary 6.4, that H(l) ∈ N_{0} if and only if Q_{τ} (l) ∈ N_{−2} for τ ≠ 0, cf. Proposition 1.4. Similar comments are valid when H(l) ∈ N_{1} , in which case again space triplets are needed.

Also in this case it is possible to reduce the integral representation to the usual Riesz–Herglotz representation. Assume for simplicity that ω_{0} does not vanish on a set of positive dρ-measure. Then we associate with the measure dρ and the function ω_{0} the measure dσ_{0} , dσ_{0} = |ω_{0}|^{2} dρ. The measure dσ_{0} satisfies the integrability condition

∫_{R} \frac{dσ_{0}(t)}{t^{2} + 1} < ∞,

cf. (7.18), and the Q-function Q(l) in (7.20) has the integral representation

(7.22)  Q(l) = α + βl + ∫_{R}(\frac{1}{t − l} − \frac{t}{t^{2} + 1}) dσ(t),

One-dimensional graph perturbations of selfadjoint relations 163

where β = |δ|^{2} and α ∈ R is given by

\begin{aligned}α & = ∫_{R} \frac{t}{t^{2} + 1}(|ψ_{0}(t)|^{2} − |φ_{0}(t)|^{2}) dρ(t) \\ & + ∫_{R} \frac{1}{t^{2} + 1}φ_{0}(t)\overline{ψ_{0}(t)} dρ(t) + ∫_{R} \frac{1}{t^{2} + 1}ψ_{0}(t)\overline{φ_{0}(t)} dρ(t). \end{aligned}

Again, an isometric isomorphism between the various spaces and operators can be made explicit.

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Received 5 October 1995