The Detectable Subspace for the Friedrichs Model

This paper discusses how much information on a Friedrichs model operator can be detected from ‘measurements on the boundary’. We use the framework of boundary triples to introduce the generalised Titchmarsh–Weyl M-function and the detectable subspaces which are associated with the part of the operator which is ‘accessible from boundary measurements’. The Friedrichs model, a finite rank perturbation of the operator of multiplication by the independent variable, is a toy model that is used frequently in the study of perturbation problems. We view the Friedrichs model as a key example for the development of the theory of detectable subspaces, because it is sufficiently simple to allow a precise description of the structure of the detectable subspace in many cases, while still exhibiting a variety of behaviours. The results also demonstrate an interesting interplay between modern complex analysis, such as the theory of Hankel operators, and operator theory.


Introduction
In this paper, we determine detectable subspaces [4,6,7]-associated with the part of the operator which is 'accessible from boundary measurements'for the so-called Friedrichs model. The Friedrichs model is a toy model, first introduced in [9], and used frequently in the study of perturbation problems (see e.g. [18]). The particular form of the Friedrichs model we study here is a finite rank perturbation of the operator of multiplication by the independent variable acting on L 2 (R) and is given by the expression (Af )(x) = xf (x) + f, φ ψ(x), (1.1) and Dom ( A) = Dom (A * ) ∩ ker Γ 1 ∩ ker Γ 2 .
We are now ready to define one of the main concepts of the paper, the detectable subspaces, introduced in [4].
Fix Remark 2.1. In many cases of the Friedrichs model we will be considering, the spaces S B and T B coincide and are independent of B. This follows from [7,Proposition 2.9]. To avoid cumbersome notation, in many places we shall denote all these spaces by S. We will refer to S as the detectable subspace.
In [4,Lemma 3.4], it is shown that S is a regular invariant space of the resolvent of the operator A B : that is, (A B − μI) −1 S = S for all μ ∈ ρ(A B ).

The Friedrichs Model
In this section we introduce the Friedrichs model. We consider in L 2 (R) the operator A with domain f (x)dx exists and is zero , given by the expression (Af )(x) = xf (x) + f, φ ψ(x), (3.1) where φ, ψ are in L 2 (R). Observe that since the constant function 1 does not lie in L 2 (R) the domain of A is dense. We first collect some results from [4] where more details and proofs can be found: The adjoint of A is given on the domain Dom (A * ) = f ∈ L 2 (R) | ∃c f ∈ C : xf (x) − c f 1 ∈ L 2 (R) , (3.2) by the formula Note that Dom (A) ⊆ Dom (A * ) and that c f = 0 for f ∈ Dom (A). We introduce an operator A in which the roles of φ and ψ are exchanged: We immediately see that Dom ( A * ) = Dom (A * ) and that dx is uniquely determined, we can define trace operators Γ 1 and Γ 2 on Dom (A * ) as follows: which is the expression used in [4]. moreover, the following Green's formula holds We finish our review from [4] with the M -function and the resolvent: Here D is the function For the resolvent, we have that in which the coefficient c f is given by There is another approach to the Friedrichs model via the Fourier transform which may appear much more natural. It is easy to check that, denoting the Fourier transform by F and Ff =f , we get and where u(0 ± ) denotes the limit of u at zero from the left and right, respectively. Moreover, There are similar expressions for the adjoint operators and traces. In terms of extension theory it looks much more natural to use this Fourier representation compared to the standard form of the Friedrichs model (as a perturbed multiplication operator). However, despite the equivalence of both representations, for our later calculations the original model is more suitable, as it gives a simpler formula for the resolvent than working with the differential operator, and reduces many questions to more straightforward residue calculations.

Friedrichs Model: Reconstruction of M B (λ) from One Restricted Resolvent (A B − λ) −1 | S
In this section we show how to reconstruct M B (λ) explicitly from the restricted resolvent. The fact that even the bordered resolvent determines M B (λ) uniquely was proved in the abstract setting in [7], but of course methods of reconstruction depend on the concrete operators under discussion. We introduce the notation · for the Cauchy or Borel transform given by and P ± : L 2 (R) → H ± 2 (R) for the Riesz projections given by where the limit is to be understood in L 2 (R) (see [12] in the upper and lower complex half-plane, respectively. To simplify notation, we also sometimes write (f ) ± (k) = f (k ± i0) := 2πiP ± f (k). Proof. 1. Recovering the function ψ Take non-zero g ∈ S and λ ∈ C\(R ∪ σ(A B )). Observe that (3.12) may be rewritten in the form in which and D(λ) is given by (3.10). The left hand side of (4.3) is known as a function of λ, at least for g ∈ S. To determine ψ up to a scalar multiple it is therefore sufficient to find g and λ so that A(λ) is non-zero: in other words, find g such that the function A(·) is not identically zero. We proceed by contradiction. Suppose we have a non-trivial Friedrichs model (i.e. neither φ nor ψ is identically zero). If A(·) is identically zero then multiplying by M B (λ) −1 from (3.11) and using (3.13) we obtain For all non-real μ such that D(μ) is nonzero (this is true for a.e. non-real μ by analyticity), there exists g ∈ S in the range of the solution operator S μ,B . We know from (3.9) that such g have the form If we use the identity and use the notations from (4.1) then multiplying by (λ − μ), (4.7) becomes Performing the integral for the case in which λ · μ < 0, we obtain (4.10) Fix λ and let μ → i∞, so that D(μ) → 1 and φ(μ) → 0. This yields If, on the other hand, we consider λ · μ > 0 in (4.9) then the value of the integral is zero, and we obtain, upon letting μ → i∞, Equations (4.11,4.12) together imply that φ is identically zero, and hence so is φ. In this case the function ψ is irrelevant and so our Friedrichs model is trivial, a contradiction. Thus (4.3) determines ψ up to a constant multiple. We may choose this (non-zero) multiple arbitrarily, since φ can be rescaled if necessary to obtain the correct Friedrichs model.

Recovering the boundary condition parameterB
Returning to the parameter c f in (3.13) and using the notation (4.1), we have as λ → ∞, and uniformly in g. Now, for μ ∈ C\R with D(μ) = 0, we choose an element from S of the form exists such that g ∈ S, and indeed may be chosen as φ(μ)/D(μ), but we do not yet know φ and therefore do not claim that our particular choice of η is given by this formula. We fix some choice of η, so that g = g μ is determined and c f is known as a function of λ and μ. We have Assuming that λ · μ < 0, we know that Put λ = −μ and letting μ → ∞, we obtain For one choice of sign( λ) at least, iπsign( λ) − B = 0 and so we can recover B from the asymptotic behaviour of c f as λ → ∞.

Recovering φ(λ)/D(λ)
Once again we choose g = g μ of the form (4.13). Returning to (4.3) and indicating the μ-dependence of f by writing Since the left hand side of this equation is known and since ψ is known, this implies that is known. Substituting the known choice of g μ we discover that is known. Using identity (4.8) this means that is also known. We shall now fix λ and let μ → ∞, for which purpose we need to know how (λ − μ)c fµ (λ) will behave. From (3.13), we have as μ → ∞. Letting μ → ∞ in (4.14) therefore yields that is known. However, taking account of (3.11), the known quantity appearing in (4.16) is This means that α := M B (λ) φ(λ)D(λ) −1 is known, and simple algebra shows that which determines φ(λ) D(λ) and hence M B (λ) provided the factor 1 + α ψ(λ) is not identically zero; equivalently, provided iπsign( λ) − B is not zero.
We are therefore left to rule out just one pathological case: the case in which B = iπsign( (λ)) in one half-plane and φ ψ ≡ 0 in the same half-plane. This can only happen if M B (λ) −1 is zero in this half-plane, which means that every point in the half-plane is an eigenvalue of A B and the corresponding g λ given by belongs to L 2 (R) and also satisfies the conditions to lie in the domain of A B : (see (6.16) in [4]). This determines φ(λ)/D(λ), and the proof is complete.

Remark 4.3.
(Uniqueness of g μ ) An alternative approach can be found by examining the uniqueness of the function g μ in S defined in (4.13). If we know that the choice of η(μ) is unique then we can immediately determine φ(μ)/D(μ), which must be equal to η(μ). This is determined by g μ if g μ is unique with its required properties. We examine this now.
Definition 4.4. The non-uniqueness set is the set Equivalently, We also let Ω ± = C ± ∩Ω and call the sets C ± \Ω ± the uniqueness sets in the upper an lower half-planes. We can ignore the condition D(μ) = 0 since it can be removed by taking a closure. We can also assume that S = L 2 (R) since otherwise we know the whole resolvent (A B − λ) −1 , which means we know A B and hence M B . We consider two cases in C + (the situation in C − is similar): (I) C + \Ω + has measure 0 and (II) C + \Ω + has positive measure.
In case (II) the uniqueness set in C + , where we can recover φ(μ)/D(μ) immediately from g μ , has an accumulation point in C + and thus φ(μ)/D(μ) is uniquely determined in C + , by analyticity.
In case (I) we have that for almost all μ ∈ C + , the function x → (x−μ) −1 lies in S. However and so we have proved the following.

Determining S for the Friedrichs Model
This section is devoted to a detailed analysis of the space S for the Friedrichs model. We shall demonstrate how different aspects of complex analysis are brought into the problem of determining S and we compute the defect number def(S) = dim(S ⊥ ) for various different choices of the functions φ and ψ which determine the model.
We note that we analysed some cases of the Friedrichs model in [7]. In particular, it contains a comprehensive study of the case of disjointly supported φ and ψ.
Before proceeding, we introduce some notation. Let D(λ) be as in (3.10). Denote by D ± (λ) its restriction to C ± and (to shorten notation) by D ± := D ± (k ± i0), k ∈ R, the boundary values of these functions on R (which exist a.e., cf. [12,20]). In general, the functions D ± (λ) do not have a meromorphic continuation to the lower/upper half-plane. In cases when they do, we will continue to denote this extension by D ± (λ). Note that this extension will in general not coincide with D(μ) in the other half-plane.
We next give a characterisation of the space S, or, more precisely, its orthogonal complement from [7, Proposition 7.2]. The proof is based on the definition of S using (2.5) and on Lemma 3.2.
Proposition 5.1. Let P ± be the Riesz projections defined in (4.2) and D(λ) be as in (3.10).
if and only if   As an immediate consequence of (5.3), we get Furthermore, let η ∈ L ∞ (R) be a function such that η(k) = 0 a.e. and η[−P + (ψφ) + P + (φ)ψ], ηψφ, ηφ ∈ L ∞ (R). Define the operator L on L 2 (R) by Remark 5.4. Introducing a scaling parameter α ∈ C\{0} and replacing ψ by αψ, we denote the corresponding detectable subspace by S α . Then, under the conditions in the second part of Proposition 5.1, we get g ∈ S ⊥ α iff where the right hand side is the sum of a multiplication operator and the difference of two Hankel operators multiplied by φ. As in the theorem, we then get S ⊥ α = {0} iff 1/(2πiα) ∈ σ p (L) and S ⊥ α is given by the corresponding kernel. IEOT
• The rational function D + (μ), μ ∈ C + , has a meromorphic continuation to the lower half-plane and is given by D + (μ) = 1+2πi N j=1 c j φ(z j )(μ− z j ) −1 for μ ∈ C (note that this will not coincide with D(μ) in the lower half-plane and that for generic φ ∈ H + 2 the continuation of the function D − (μ) to C + will not even exist),

corresponds to a degenerated case and is given by
Remark 5.7. It is possible to choose rational φ and ψ in H + 2 (R) so that the defect number N − P of Theorem 5.6 takes any value between 0 and N − 1, while the corresponding defect numberÑ −P forS takes any value between 0 andÑ − 1, independently of the value of N − P . Therefore, any values can be realized for the defect numbers of S andS.
Proof. (outline) We use the fact that S = T where T is as defined in (2.5): the elements of T are found by solving (Ã * − μ)u = 0 and varying μ over the resolvent set of some appropriate operators A B . We therefore start by solving where φ, ψ ∈ H + 2 . Dividing by (x − μ) we find that u = (c u 1 − u, φ ψ)(x − μ) −1 . Taking the inner product with φ we get D(μ) u, φ − cu x−μ , φ = 0. There are two cases to consider.
There are two subcases to consider. ( There are some subcases to consider. In the case (2b) for any boundary condition B, by suitable choice of the two constants we see that μ belongs to the spectrum of A B . Therefore these functions are not included in the space S. In the case (2c) the function ψ x − μ should be included in S. There is only one B for which it is an eigenfunction (formally B = ∞), but even for this choice of B it can be approximated by elements from neighbouring kernels with D(μ) = 0. Note that this means that S is independent of B as expected. This proves the formula for T = S in the theorem.
We now obtain the expression for the dimension of S ⊥ , in the generic case M = 0 = M 0 , when ψ(x) = The first bracket gives D + and by Proposition 5.5 we know that g ∈ H − 2 and so, taking boundary values, (5.8) becomes in which D + (x) are the boundary values on the real line of the function x−μ dx, μ ∈ C + . Thus by the Residue Theorem, Therefore, by unique continuation of the meromorphic function to the lower half plane (see [12]) g is given by from which it is immediately clear that the space of all such g is at most Ndimensional. Note that the expression on the right hand side of the equality sign in (5.10) is not clearly an element of H − 2 ; to deal with this we substitute the particular ψ under consideration into the formula for D + and use residue calculations to obtain the following expression for its analytic continuation to C: If D + (μ) has no zeros in C − and if φ(z j ) = 0 for all j then we get , μ ∈ C − and the condition that lim μ→zj g(μ) = g(z j ) gives no additional restrictions, as can be confirmed by a simple explicit calculation. In this case, therefore, the defect of S is N . Now suppose D + has zeros in C − ; for simplicity we are assuming that they all lie strictly below the real axis. We let μ 1 , . . . , μ ν be the distinct poles of φ/D + , with orders p 1 , . . . , p ν and set P = ν j=1 p j . In order to ensure that g given by (5.10) lies in H + 2 we need that the conditions N j=1 c j (μ k − z j ) n g(z j ) = 0, n = 1, . . . , p k , k = 1, . . . , ν, (5.12) all hold -a total of P linear conditions on the numbers g(z 1 ), . . . , g(z N ). We now check that this is a full-rank system. Suppose for a contradiction that there is a non-trivial set of constants α i,k such that (μ k −z) n so that F has zeros at z 1 , . . . , z N . Observe that Q(z) := F (z) ν k=1 (μ k − z) p k is a polynomial of degree strictly less than P = ν k=1 p k , having N zeros. Now D + (μ) → 1 as (μ) → ∞, so D + has the same number of zeros as poles. In particular, D + has at least as many poles in C as it has zeros in C − , giving N ≥ P . Thus Q is a polynomial of degree < P ≤ N having N zeros. This means Q ≡ 0, so F ≡ 0, and the constants α i,k must all be zero. This contradiction shows that the set of linear constraints on the N values g(z j ) has full rank P , and so the set of allowable values for (g (z 1 ), . . . , g(z N )) has dimension N − P .
The degenerated case leading to non-zero M and M 0 can be analysed similarly by considering the local behaviour of φ/D + around zeroes of D + (x) on the real axis.
We conclude this part with an example. The details justifying the statements can be found below.
Moreover, we see that the bordered resolvent does not detect the singularities at the eigenvalues λ 0 ∈ C + or λ 0 ∈ C − : For λ ≈ λ 0 ∈ C + we have from (3.12) and (3.13) that with the Riesz projection P λ0 given by P λ0 = ·, u 1 u 2 , where Since u 1 ∈ S ⊥ , the singularity is cancelled by P S . u 2 is the eigenvector of A B − λ 0 (see [11]). For λ ≈ λ 0 ∈ C − we have again from (3.12) and (3.13) that is an eigenvector of A B for all (!) B and lies in S ⊥ , so P S cancels the singularity of the resolvent.
We note that this behaviour of the bordered resolvent is in accordance with Theorem 3.6 in [4].
Proof. (Statements in Example 5.8.) In this example, for λ ∈ C + we have by the residue theorem Clearly, this formula also gives the meromorphic continuation of D + to the lower half plane. We remark that this differs from D − which is given by We now calculate the numbers N, P, M, M 0 from Theorem 5.6. ψ has a simple pole at z 1 ∈ C − , hence N = 1. As φ has no zeroes, M 0 = 0. The function D + has one pole at z 1 ∈ C − , φ has a simple pole at w 1 ∈ C + . Thus all poles of φ/D + in C − stem from zeroes of D + . The only zero of this function is at λ 0 = z 1 + 2πiα(w 1 − z 1 ) −1 . Thus, if λ 0 ∈ C + then P = M = 0; if λ 0 ∈ C − then P = 1, M = 0; if λ 0 ∈ R then P = 0, M = 1. We next show the form of S ⊥ and S in the case λ 0 ∈ C + . Using φ ∈ H + 2 , from (5.1), we have g ∈ H − 2 and g = −2πiD −1 − φP − (ψg). Hence, Noting that g(z 1 ) is a free parameter, a short calculation shows that .
This is equivalent to (P + f )(w 1 ) = (P + f )(λ 0 ). Remark 5. 9. We note that in the case when φ, ψ ∈ H + 2 taking λ, μ ∈ C + , the M -function and the ranges of the solution operators S λ,B and S μ,B * do not depend on φ and ψ [see (3.11) and (3.9)]. In fact, we have In this highly degenerated case, only the boundary condition B can be obtained. Therefore, in this case a Borg-type theorem allowing recovery of the bordered resolvent from the M -function is not possible, even with knowledge of the ranges of the solution operators in the whole of the suitable half-planes. On the other hand, knowledge of the ranges of the solution operators in both half-planes, together with the M -function at one point allows reconstruction by [7].

Analysis for the Case φ, ψ ∈ H
If φ or ψ additionally lies in L ∞ , then this gives def(S B ) = +∞. However, we consider this choice of B as a degenerate case, since the hypotheses of [7, Proposition 2.9] are not satisfied.

The General Case ψ, φ ∈ L 2
We conclude this section by studying the general case. Without assumptions on the support, or the Hardy class of φ and ψ, the results are rather complicated. Therefore, in what follows we will not worry about imposing slightly stronger regularity conditions on φ and ψ. Thus we assume φ ∈ L 2 and ψ ∈ L 2 ∩ L ∞ or φ, ψ ∈ L 2 ∩ L 4 . (5.18) In some cases (which will be mentioned in the text), we will require the slightly stronger condition φ ∈ L 2 ∩ L 2+ε for some ε > 0 and ψ ∈ L 2 ∩ L ∞ or φ, ψ ∈ L 2 ∩ L 4 . 1 + 2πiα(P + (φψ) − ψ(P + (φ))) = 0 on E}. (5.20) Note that E 0 consists of those α such that the factor D + − 2πi(P + φ)ψ appearing in (5.3)-(5.5) vanishes on some non-null set E when ψ is replaced by αψ.
Remark 5.12. In many cases, such as when ψ is analytic on R, the set E 0 will be empty. However, it is possible to construct examples with non-empty E 0 .
We now give such an example. Take φ and ψ with disjoint supports. Then their product is 0 and the second term in formula (5.20) disappears. Choose the function φ additionally such that P + (φ), the multiple of ψ in the third term of (5.20), does not vanish on an interval, say [0, 1]. This is, for example, the case if φ has fixed sign and its support is an interval. One can then choose the function ψ on the interval [0, 1] such that, for some fixed non-zero value of the parameter α, the whole third term −2πiαψ(P + (φ)) in (5.20)  Proof. Let α ∈ E 0 \{0} and E be the set of positive measure on which 1 + 2πiα(P + (φψ) − ψ(P + (φ))) = 0. Set f = 2πi(P + (φψ) − ψ(P + (φ))). As 1 + αf | E = 0 then f | E = −1/α; this can only be true for a countable set of α. See, e.g. [7,Lemma 7.12].
Remark 5.15. When considering the corresponding S α note that the set E 0 := α : 1 + 2πiα(P + (ψφ) − φ(P + (ψ))) = 0 on a set of positive measure does not need to coincide with E 0 , so it is possible to have def S α = def S α even for α ∈ E 0 . For examples of this, see [7].
Proof. Without loss of generality, we assume α = 1. Let E be the set of positive measure from (5.20).
. By our assumptions on h and in (5.19), we have g ∈ L 2 . We next show that g satisfies the right hand side of (5.3) pointwise. Note that here and in several other places in this proof we use that P − P + f = 0. This is justified as our assumptions on h and in (5.19) guarantee that the functions f we apply this to are in appropriate function classes. We have Multiplying by 2πiψ and using that D + − D − = 2πiψφ on the real axis by the Sohocki-Plemelj Theorem (see [12]), gives We rewrite the D − -term as follows.
Inserting this in (5.21), and rearranging gives the identity Multiplying by φ and using that on E we have D + = 2πiψ(P + φ) this gives which, noting that (D + − 2πiψ(P + φ))χ E h = 0, is the equation on the right hand side of (5.3).
We now need to chose h ∈ L 2 (E) suitably to obtain an infinite dimensional subspace for the corresponding g. Choose E ⊂ E with |E | > 0 and sufficiently small such that Ω φ ⊆ E (as E has positive measure and φ is not identically zero this is always possible). Consider g = (P + φ)χ E h − φP − (χ E h). By the above arguments, g ∈ S ⊥ . Moreover, g| (E ) c = −χ((E ) c )φP − (χ E h). As χ((E ) c )φ ≡ 0 and P − (χ E h) are the boundary values of an analytic function and therefore non-zero a.e. on R, we have g ≡ 0 whenever P − (χ E h) ≡ 0 (see [12]), which gives an infinite dimensional set of such functions. Theorem 5.16. Let φ ∈ L 2 ∩ L ∞ and ψ ∈ L 2 ∩ C 0 (R), where C 0 (R) is the space of continuous functions vanishing at infinity, and assume α ∈ E 0 .
is a possibly unbounded multiplication operator and . Proof. (outline) The first part follows easily from (5.5) in Proposition 5.1 and standard results on compact operators. The compactness of the difference of Hankel operators follows from [19,Corollary 8.5].
For the second part, consider the analytic operator-valued function I + (M − μI) −1 K which is a compact perturbation of I. We need to know the values μ ∈ C for which this operator has non-trivial kernel. Each connected component of C\essran M either contains only discrete (countable) spectrum or else lies entirely in the spectrum. However for large μ, {0} = ker(I + (M − μ) −1 K), so by the Analytic Fredholm Theorem (see [21]), outside some bounded set there is no spectrum of M + K.
Although this theorem gives a description of S α for a rather general case of ψ and φ, for concrete examples as investigated in previous subsections it is useful to determine the space explicitly rather than just give the description in terms of operators K and M. However, this theorem shows the topological properties of the function def S α in the α-plane.
We wish to analyse the defect as a function of α. By Theorem 5.6, we need to determine the number of roots of the analytic continuation D + (λ) of D(λ) in C − . Now, . (5.22) After settingμ := 2πiα (z1−w1)(z2−w1) a short calculation shows that the roots of D + (λ) solve where In particular, forμ = 0 the roots are z 1 , z 2 ∈ C − . By continuity, for small |α|, by Theorem 5.6 we have def S α = 0. For a polynomial λ 2 + pλ + q = 0, an elementary calculation shows that it has a real root iff ( q) 2 = ( p) ( p q − q p) and 4 q ≤ |p| 2 . (5.24) We now analyse the defect in a few examples. (A) We first make the specific choice Then Allμ satisfying (5.25) satisfy the inequality in (5.24). This gives a parabola in the α-plane (or equivalently theμ-plane) with def S α = 0 inside or on the parabola and def S α = 1 outside. In the 1/α-plane this gives a curve whose interior is a petal-like shape with def S α = 0 for 1/α outside or on the curve and def S α = 1 for 1/α inside the curve. (B) We now return to the formula for D + in (5.22). Setting μ = (2πiα) −1 , we have . (5.26) Clearly for λ → ±∞, we have that μ = 0. We now choose c 1 , c 2 to get another real root at λ = 0. Consider and φ(x) = 1 x − i .
In the μ-plane this leads to one petal. As λ runs through R, this curve is covered twice (once for λ < 0 and once for λ > 0). We have def S α = 0 for μ outside the curve and def S α = 2 for μ inside the curve. On the curve we have def S α = 0. The double covering of the curve allows the jump of 2 in the defect when crossing the curve. (C) More generally, if ψ has N terms, then the problem of finding real roots of D + (λ) leads to studying the real zeroes of Generically ξ will not have real zeroes and we will only get one petal in the μ-plane. However, we can arrange it that ξ has N − 1 real zeroes which leads to N petals in the μ-plane. Assume a N = 0. Then to do this, we need to solve the linear system, where the matrix Z has jk-component given by z jk = (z k − λ j ) −1 . Z is invertible whenever all z k ∈ C − , λ j ∈ R are distinct.  Fig. 1, the defect in each of the components is given by 4 − ν − where ν − denotes the number of roots of D + in C − (by Theorem 5.6). At each curve precisely one of the roots crosses from the lower to the upper half-plane, thus increasing the defect by 1. On the curve itself, one root is on the real axis and by Theorem 5.6, the defect coincides with the smaller of the defects on the components on each side of the curve. By a similar reasoning at the three non-zero points of self-intersection of the curve the defect coincides with the smallest defect of the neighbouring components.
This example displays the analytical nature of finding the defect in terms of the location of roots of D + using Theorem 5.6. On the other hand, it also displays the topological nature of the same situation mentioned in Theorem 5.16. The complex 1/α-plane is separated into components in which the defect is constant everywhere (in this example the exceptional discrete set is empty). The curves are the range of 2πiM(t) on the real axis.
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