Dedicated to the 100th anniversary of Mark Krein.


 Ὀφιοῦχος Αβραμίδης
 8 ημέρες πριν
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1 Methods of Functional Analysis and Topology Vol 4 (28), no, pp 2 3 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM WHICH GENERATES PRESCRIBED FINITE POINT SPECTRUM F DIABA AND E CHEREMNIKH Dedicated to the th anniversary of Mark Krein Abstract The perturbations of Nevanlinna type functions which preserve the set of zeros of this function or add to this set new points are discussed Statement of the problem The point spectrum in the case of rank one perturbation of purely continuous spectrum may be very rich In general, this spectrum contains the eigenvalues as well the spectral singularities We will not give a review of the references (we only indicate [] and [2]) Let us consider the case where nonperturbed continuous spectrum coincides with the half line [, ) The eigenvalues of the perturbed operator is obtained as a set of zeros of the function () δ(ζ) = dτ, ζ [, ) τ ζ (called the denominator of the resolvent) Some properties of such function one can find in [4] Spectral singularities of the perturbed operator coincide with zeros of the functions (2) δ ± (σ) = lim δ(σ iε), σ > ε ± We discuss following questions: how to describe a) perturbations of the function such that the perturbed function δ(ζ) has new prescribed zeros without changing the other zeros; b) all perturbations of the function, which does not change the set of zeros of the function δ(ζ) The set of zeros of the function δ (σ) or δ (σ), σ >, are considered in the same way 2 The construction of the function δ(ζ) without zeros or with a prescribed finite set of zeros Let us consider, in the space L 2 (, ), the operator generated by the expression { ly = y (2), x > y() (y, η) L 2 (, ) = with a nonlocal boundary condition The function to study (the denominator of the resolvent) is γ(τ) (22) δ(ζ) = τdτ, ζ [, ), τ ζ 2 Mathematics Subject Classification 7A2, 7A3 Key words and phrases Finite point spectrum, perturbation of continuous spectrum, Hilbert transformation, set of zeros 2
2 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM 2 where πγ(τ) denotes the sinfourier transformation of the function η(x) Note that η L 2 (, ), γ L 2 ρ(, ), ρ(τ) = π τ and that the function (22) is of the form () Lemma 2 Let a i C, τ i >, i =,, n, be arbitrary numbers, then the function (23) δ(ζ) = ( ζ ia ) ( ζ ia n ) ( ζ i τ ) ( ζ i τ n ), Im ζ, admits the representation (22) Proof The expression (23) is a rational function of ζ, so, the decomposition into simple fractions gives δ(ζ) = A A n, ζ i τ ζ i τn A k = const Using the identity ( ) (24) τ ζ, τ τ L 2 ρ = i ζ i τ, Im ζ >, τ >, we obtain the representation (22) where Lemma is proved γ(τ) = A i A ni, τ > τ τ τ τ n If Im σ iε > then lim ε ± σ iε = ± σ So, due to (2) and (23) we have (25) δ ± (σ) = ( σ ± ia ) ( σ ± ia n ) ( σ ± i τ ) ( σ ± i τ n ), σ > The representation (23) (25) prove the following proposition Proposition 22 Let the representation (23) hold Then for k =,, n, ζ [, ) and σ >, we have the following: ) if a k > then δ(ζ), δ (σ), δ (σ), 2) if a k < then δ( a 2 k ) =, δ (σ), δ (σ), 3) if a k = iα k, α k > then δ(ζ), δ (α 2 k ) =, δ (σ) As a consequence one can give examples of the function γ(τ) such that the functions δ(ζ), δ (σ) and δ (σ) have a prescribed set of zeros In view of the condition τ i >, i =,, n, the function δ(ζ) is bounded, sup δ(ζ) < ζ [, ) (see (23)) We denote the relation () by δ(ζ) and the Hilbert transformation by Hh(σ) = V p dτ, σ > Theorem 23 Let δ (ζ) h (τ) and δ 2 (ζ) h 2 (τ) Suppose that h,2 L 2 (, ) C [, ) and that the function δ 2 (ζ) is bounded in the domain ζ [, ) Let δ(ζ) where (26) = h (τ) h 2 (τ) h (τ)hh 2 (τ) h 2 (τ)hh (τ) Then δ(ζ) = δ (ζ)δ 2 (ζ)
3 22 F DIABA AND E CHEREMNIKH Proof Denote δ(ζ) = δ (ζ)δ 2 (ζ), then we must prove that δ(ζ) = δ(ζ) According to the definition (22) we have h (τ) (27) δ(ζ) = δ2 (ζ) δ 2 (ζ) τ ζ dτ = h 2 (τ) τ ζ dτ δ h (τ) 2(ζ) τ ζ dτ Since h,2 L 2 h,2 (τ) (, ), the integrals dτ belong to a Hardy space in the domain τ ζ Imζ > and Imζ < The function δ(ζ) belongs to this Hardy space, too Therefore, there exists a function h L 2 (, ) such that δ(ζ) = τ ζ dτ Here, due to (27), h(σ) = ( δ (σ) 2πi δ (σ)) =, σ < In view of h,2 C [, ), the limitvalues (like (2)) exist as continuous functions, for example, So, for σ > we have (see (26) (27)) 2πi h(σ) = δ (σ) δ (σ) δ,± (σ) = ± πih (σ) Hh (σ), σ > = 2πih 2 (σ) ( πih 2 (σ) Hh 2 (σ))(πih (σ) Hh (σ)) ( πih 2 (σ) Hh 2 (σ))( πih (σ) Hh (σ)) = 2πih(σ), ie, h(σ) = h(σ), σ > Therefore, Theorem is proved δ(ζ) = τ ζ dτ = dτ = δ(ζ) τ ζ Corollary 24 To add a finite set of zeros to the function δ (ζ) h (τ) one can use the function δ 2 (ζ) h 2 (τ) with this set of zeros and replace the function h (τ) by the function (see (26)) 3 The perturbation which preserves the set of zeros We consider the perturbation γ(τ) of the function γ(τ) in the representation (3) δ(ζ) = γ(τ) τdτ, τ ζ ζ [, ) To simplify the calculations, we will study the function δ (σ), σ >, only (the functions δ (σ), σ > and δ(ζ), ζ [, ) are considered similarly) We are looking for a function γ(τ) such that replacing γ(τ) with γ(τ) γ(τ) in (3) we obtain the same set of zeros for the function δ (σ), σ > Let < σ < < σ n be the set of all zeros of the function δ (σ), σ > We suppose that γ C [, ] and that the multiplicities of the zeros are equal to We need the notations (32) R σ γ(τ) = γ(τ) γ(σ), Rγ(τ) = γ(τ) n j= A jr σj γ(τ), where the coefficients A k = const are defined below
4 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM 23 Lemma 3 Suppose that σ j >, j =,, n, are arbitrary numbers Let p(τ) = n j= ( j), q(τ) = p(τ)/( τ) n Then there exists the coefficients A k = const and polynomials f j (τ) of degree n, f j (σ j ) =, f j (σ k ) =, k j, k, j =,, n, such that following decomposition holds: (33) γ(τ) = γ(σ j )f j (τ) q(τ)rγ(τ) j= Proof We define the values A j as coefficients of the decomposition into simple fractions, γ(τ) q(τ) = j= n q(τ) = A j j γ(σ j ) j j= The decomposition (33) follows from the identity n [ A j γ(τ) q(τ) then f j (τ) = A j j Lemma is proved j= ] γ(τ) γ(σ j ) A j, j Note that the set of numbers γ(σ ),, γ(σ n ) and the function Γ(τ) = Rγ(τ) are independents composants of the decomposition (33) of the function γ(τ) Obviously, the function f j (τ) admits a decomposition in the form (34) f j (τ) = l= M lj (τ ) l, M lj = const The functions f (τ),, f n (τ) are linearly independents, so (35) det(m lj ) that In view of (3) and the relation γ(τ) = n j= γ(σ j)f j (τ)q(τ)γ(τ) we have for σ > (36) δ (σ) = where ( ) γ(σ i )l i (σ) q(τ) Γ(τ) τdτ, Γ = Rγ, i= (37) l i (σ) = ( Lemma 32 Let L = (l i (σ j)) Then det L = ) f i (τ) τdτ, σ > Proof Taking the derivative of the equality (24) in the case ζ σ with respect to τ and substituting τ = we obtain for l = 2, 3, ( ), l (τ ) l = P i α,l u(σ)α, u(σ) =, P σ i l,l In view of (34), (37), where π l i (σ) = (, f i ) α= = l= L i,α = ( ) M li, (τ ) l = n l=α M li P αl L i,α u(σ)α α=
5 24 F DIABA AND E CHEREMNIKH Obviously, the matrix L L n L n L nn = M i M 2i M ni P α,α P α,n P n,n is nondegenerate (see (35)) l (σ) So, the column is a nondegenerate stable (with respect to σ) transfor l n (σ) mation of the column because Lemma is proved u(σ) u(σ) n Since u(σ i ) u(σ j ), i j, we have det L u(σ ) u(σ n ) u(σ ) n u(σ n ) n We consider two functions δ(ζ) = γ(τ) τdτ, δ γ (τ) (ζ) = τdτ, τ ζ τ ζ Obviously, the set {σ j } is a set of zeros of the function δ (σ) iff < δ (σ), σ > ζ [, ) Lemma 33 Let {σ j } be a set of zeros of δ(σ) Then {σ j } is a set of zeros of δ (σ) iff (38) δ (σ) >, σ >, (39) δ (σ) δ (σ) =, j =,, n Note that due the identity (3) the condition inf (, ) δ (σ) δ (σ) = δ (σ) (, ) ( ) δ (σ) δ(σ) > follows from the condition inf δ (σ) > 2 inf (, ) (, ) δ(σ) if ( ) (3) sup δ (σ) δ (σ) < 2 inf (, ) δ (σ)
6 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM 25 Lemma 34 ) If γ(τ), γ (τ) are arbitrary functions, then δ (σ j ) δ(σ ( j ) = γ(σi ) γ (σ i ) ) l i (σ j) (32) i= R σj q(τ) ( Γ(τ) Γ (τ) ) τdτ, where Γ = Rγ, Γ = Rγ 2) If δ (σ j ) = δ(σ j ) =, then ( δ (σ) δ (σ) ) ( = γ(σi ) γ (σ i ) ) Rl i (σ) i= (33) R (R σ q(τ)) ( Γ(τ) Γ (τ) ) ( τdτ Γ(τ) Γ ) (τ) τdτ Proof ) The decomposition (36) for the functions δ (σ), δ(σ) follow from the equality (32) 2) The equality δ (σ i ) = signifies (see (36)) (34) γ(σ i )l i (σ j) R σj q(τ)γ(τ) τdτ =, i =,, n i= Like (33) we introduce the decompositions l i (σ) = = R σ q(τ) = n j= l i (σ j)f j (σ) Rl i (σ), f j (σ), j= R σj q(τ)f j (σ) R (R σ q(τ)) j= Subtracting from the equation (see (36)) δ (σ) = γ(σ i )l i (σ) R σ q(τ)γ(τ) ( τdτ i= the equations (34) which are multiplied by f j (σ) we obtain (35) δ n (σ) = γ(σ i )Rl i (σ) R (R σ q(τ)) Γ(τ) ( τdτ i= The same decomposition for the function δ (σ) gives (33) Lemma is proved Γ(τ) ) τ dτ Γ(τ) ) τ dτ Since det(l i (σ j)), the relation ( (36) γ(σi ) γ (σ i ) ) l i (σ j) R σj q(τ) ( Γ(τ) Γ (τ) ) τdτ = i= and (33) define uniquely a linear operator N such that ( (37) δ (σ) δ (σ) ) = (N (Γ Γ ))(σ)
7 26 F DIABA AND E CHEREMNIKH Let (( Γ = sup (( τ) Γ(τ) ) sup τ 3/2) ) Γ (τ) [; ) [; ) Theorem 35 Suppose that Γ < If f(τ) = Γ(τ) τ, then and Proof We use the estimates F (σ) V p for the following expressions ) Let σ > 4 and ( σ σ F (σ) = f(τ) dτ, τ, sup F (σ) C Γ, C = const [; ) sup τf(τ) Γ, sup τf (τ) 2 Γ [; ) [; ) σ σ σ σ Then σ σ a) F (σ) = f(τ) τ dτ τ() σ σ b) F 2 (σ) = V p f(τ) dτ σ = sup τf (τ) [2; ) σ σ σ σ σ σ ) f(τ) σ σ dτ F (σ) F 2 (σ) F 3 (σ) sup σ σ τf(τ) [; ) σ σ σ f(τ) f(σ) dτ σ, σ, σ here σ σ < c < σ σ c) F 3 (σ) = f(τ) dτ sup τf(τ) [; ) σ σ 2) Let < σ < 4 and ( 5 F (σ) = Then 5 σ σ dτ τ, σ σ dτ = ) f(τ) dτ G (σ) G 2 (σ) 5 τγ(τ) 5 a) G (σ) = V p dτ = Γ(τ) Γ(σ) τ dτ 5 τ Γ(σ)V p dτ = O(), σ and G (σ) C(sup Γ (τ) sup Γ(τ) ) C Γ b) G 2 (σ) = Theorem is proved [;5] 5 f(τ) dτ [;5] sup τf(τ) [5; ) Corollary 36 If Γ < then (see (37)) (N Γ)(σ), sup (N Γ)(σ) K Γ, [; ) 5 σ σ σ f (c)dτ dτ τ(), σ dτ τ() C Γ K = const σ, and
8 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM 27 Proof The function q(τ) is a linear combination of the functions and ( ) R σ (τ ) k = [ (τ ) k (σ ) k ] = (τ ) k, k =, 2 r(τ, σ) (τ ) k (σ ) k where r(τ, σ) is a polynomial in τ and σ of degree k The same form holds for R(R σ q(r)) Due to Theorem 35, the estimate and the presentation ( prove the Corollary 36 R σj q(τ)γ(τ) R σj q(τ) Γ τ Γ(τ) ) τ dτ = πiγ(σ) Γ(τ) τ σ V p dτ Corollary 37 If δ () then the value m(γ) inf δ (σ) [; ) is nonzero, ie, m(γ) > Proof The continuous function δ (σ), σ [; ), does not have zeros and, by Corollary 36, for Γ = γ and (35), δ (σ), σ Therefore, m(γ) > Let γ, γ be two functions, then we denote γ = γ γ, Γ = Γ Γ, Γ = Rγ, Γ = Rγ We denote by Ω(γ) the set of zeros of the function δ (σ), σ > Theorem 38 Let the function γ be such that m(γ ) > We define the function γ(τ) by its decomposition (38) γ(τ) = γ(σ i )f i (τ) q(τ) Γ(τ), i= Γ = Rγ, where Γ(τ) is an arbitrary function such that (see Corollary 36) (39) K Γ < 2 m(γ ) and the numbers γ(σ i ) are defined by Γ(τ) from the system (36) So, if γ = γ γ then Ω(γ) = Ω(γ ), ie, the sets of zeros of the functions δ (σ) and δ (σ) coincide Proof Let Ω(γ ) = {σ, σ n } Suppose that some function Γ satisfies the condition (39) The system (36) where Γ(τ) Γ (τ) = Γ(τ) defines the numbers γ(σ j ) γ (σ j ) = γ(σ j ) The relation (38) gives the function γ(τ), so, we obtain γ(τ) = γ (τ) γ(τ) Using the known function γ(τ) we define the function δ(ζ) by (35) The system (36) signifies that δ (σ j ) δ(σ j ) =, ie, δ (σ j ) = Therefore,
9 28 F DIABA AND E CHEREMNIKH Ω(γ) Ω(γ ) It remains to prove that δ (σ) if σ σ j, j =,, n We use the identity (3) According to Corollary 36 and (39), we have sup [; ) q(τ) (δ (σ) δ(σ)) = sup N ( Γ)(σ) K Γ < [; ) 2 m(γ ) = 2 inf [; ) δ (σ) Therefore, ie, Ω(γ) = Ω(γ ) Theorem is proved inf δ (σ) > 2 inf [; ) δ (σ) >, [; ) 4 Application to nonlocal SturmLiouville operator with trivial potential Let ly = y, we denote by B and A the operators in the space L 2 (, ) generated by the same differential expression ly and the corresponding boundary condition y() = and y() (y, η) L 2 (, ) = We need the Fourier transformation which diagonalizes the nonperturbed operator B, namely F: L 2 (, ) L 2 ρ(, ), ρ(τ) = π π, where and we need the relation Fy(τ) = y(x) sin x τ τ dx, τ >, (4) F(e i ζx )(τ) = τ ζ, τ >, Im ζ > We denote R ζ (B) = (B ζ), R ζ (A) = (A ζ) and (42) δ(ζ) = Fη(τ)ρ(τ) dτ, ρ(τ) = τ, ζ [, ), τ ζ π where the function η(x) defines the boundary condition for the perturbed operator A Theorem 4 Let ζ [, ), then ζ ρ(a) iff δ(ζ), in this case, (43) R ζ (A)f = R ζ (B)f δ(ζ) (R ζ(b)f, η) L 2 (, )e ζ where e ζ (x) = exp(i ζx), Im ζ > A value ζ [, ) is an eigenvalue of the operator A iff δ(ζ) = Proof Let e L 2 (, ) be an arbitrary function such that le L 2 (, ), e() =, and (e, η) L 2 (, ) = 2 Every element z L 2 (, ) admits the representation (44) z = y (y, η) L 2 (, )e where y = z (z, η) L 2 (, )e If z D(A) then z() (z, η) L 2 (, ) =, so y() =, ie, z D(B) Applying the operation l ζ to both sides of the equality (44) we obtain The equation (A ζ)z = f becomes (A ζ)z = (B ζ)y (y, η) L 2 (, )(l ζ)e (45) y (y, η) L 2 (, )R ζ (B)(l ζ)e = R ζ (B)f
10 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM 29 The identity Fle(τ) τfe(τ) e() = gives or F(l ζ)e(τ) Fe(τ) = τ ζ τ ζ FR ζ (B)(l ζ)e(τ) Fe(τ) = Fe ζ (τ), e ζ (x) = exp(i ζx) from which R ζ (B)(l ζ)e = e e ζ Substituting into (45) we obtain (46) y (y, η) L 2 (, )(e e ζ ) = R ζ (B)f The multiplication by η gives therefore, (y, η) L 2 (, )[ ( 2 (eζ, η) L 2 (, )) ] = (R ζ (B)f, η) L 2 (, ), (y, η) L 2 (, ) = δ(ζ) (R ζ(b)f, η) L 2 (, ) In view of (44), the equality (46) becomes (43) If δ(ζ) then the operator R ζ (A) is bounded, so ζ ρ(a) Other statements are simple to prove Theorem is proved We will consider the functions that are more general than (42), namely, (47) δ(ζ) = a τ ζ dτ, where a C, h L 2 (, ) We denote δ(ζ) (a, h) and assume that is a function such that the limit values δ ± ( ) of the function (47) are continuous on (, ) Let us introduce the normed space where the norm is U = {u = (a, h) : a C, h L 2 (, ), δ ± ( ) C[, )}, (48) u = sup δ ± (σ) δ δ L 2 (, ) σ> The definition (48) is correct, since the transformation (a, h) δ(ζ) is invertible, { a = limσ δ(σ), (49) = 2πi (δ (τ) δ (τ)) Lemma 42 Let some function δ(ζ) be holomorphic in the domain ζ [, ), have continuous on [, ) limitvalues δ ± ( ) and a finite value of the right side of (48) and a finite value of the limit a = lim σ δ(σ) Then the function δ(ζ) admits the representation (47), δ(ζ) (a, h), where (a, h) U Proof It is sufficient to consider the difference (see (49)) [ ] δ(ζ) a τ ζ dτ and use the principle of symmetry known as a property of analytic functions Lemma 43 The space U is closed Proof Let us rewrite the norm (48) in the form (4) u = sup δ ± (σ) 2π h L 2 (, ) σ> Let u n = (a n, h n ) U be a fundamental sequence, in view of (4) the sequence {h n } is fundamental in L 2 (, ) We denote h = lim h n Since δ n± (σ) δ m± (σ) u n u m, the sequence {δ n± (σ)} is fundamental too in the space of continuous function C[σ, σ 2 ] for every < σ < σ 2
11 3 F DIABA AND E CHEREMNIKH So, from the equality ( δ n (σ) δ m (σ) = a n a m ) h n (τ) h m (τ) dτ taking into account the boundedness of the Hilbert transform in the space L 2 (σ, σ 2 ) it follows that the sequence {a n } is fundamental Let a = lim n a n Obviously, (a, h) = lim n (a n, h n ) in the sense of the norm The functions δ ± ( ) corresponding to the pair (a, h) (see (47)) are continuous on [, ), being uniform limits of the sequence of the continuous functions {δ n± (σ)} So, (a, h) U, therefore the space U is closed Lemma is proved According to Lemma 43, the space U is a Banach space We introduce the product of the pairs u = (a, h ) and u 2 = (a 2, h 2 ) by the relation (see (47)) u u 2 δ (ζ)δ 2 (ζ) Lemma 44 If u,2 U, then u u 2 U and u u 2 u u 2 Proof The function δ(ζ) = δ (ζ)δ 2 (ζ) is holomorphic in the domain ζ [, ), has continuous on [, ) limitvalues δ ± (σ) and has the finite limit lim σ δ(σ) = a a 2, u u 2 = sup δ ± (σ)δ 2± (σ) δ δ 2 δ δ 2 L 2 (, ) σ> sup σ> sup σ> δ ± (σ) sup σ> δ (σ) δ 2 δ 2 L 2 (, ) u u 2 δ 2± (σ) sup δ 2 (σ) δ δ L 2 (, ) σ> According to Lemma 42, the function δ(ζ) admits the representation (47), δ (ζ)δ 2 (ζ) u, where u U Therefore, u u 2 U Lemma is proved According to Lemma 44, the Banach space U is a normed ring Let us come back to Theorem 23 If δ 2 (ζ), ζ [, ), then the relation (26) gives a sufficiency condition for such a transformation of the function h (τ) which preserves the roots of the function = δ (ζ) We will show that this condition is closed to a necessary condition Lemma 45 Let δ(ζ) be an arbitrary function in the form (47) and δ (ζ) be a function in the form (23) with the same roots as the function δ(ζ) Then δ(ζ) = δ (ζ)δ 2 (ζ) where the function (which has no roots) admits the representation (47) Proof results from Lemma 42 We will give some sufficient condition for the function δ(ζ) to not have roots (compare (42) and (4)) Theorem 46 Let (4) δ(ζ) = τ ζ dτ, ζ [, ) Suppose that the function and its Hilbert transform H(σ) = v p τ σ dτ are continuous on [, )
12 ON RANK ONE PERTURBATION OF CONTINUOUS SPECTRUM 3 Then, if sup ± πih(σ) Hh(σ) 2π h L 2 (, ) <, σ> then the function δ(ζ), ζ [, ), has no roots Proof results from the statement that an element u of the ring U is invertible if u < where denotes the pair (, ) Theorem 47 Let Ω = {ζ : dist(ζ, [τ, )) < ε} for some τ >, ε > Suppose that the function admits an analytic continuation, which belongs to the Hardy space H 2 (Ω) Suppose (42) sup ± πih(σ) Hh(σ) < σ> Then the function δ(ζ), ζ [, ), has no roots Proof We have the inequality τ ζ dτ <, ζ [, ) using PhragménLindelöf theorem and the estimate (42) Note that Theorem 46 and 47 complete the Theorem 38 and that one can easily rewrite all results in terms of the problem (2) 5 Conclusion The case where the function δ(ζ) has zeros with multiplicities > is considered in the same way The traditional inverse problem requires to find a unique operator such that its spectrum coincides with a given set in the complex plane But the problem to describe all the operators such that only a part of their spectrum coincides with a given set has a sense too The question of how to choose a perturbation as to obtain a given change is interesting One can compare such a problem with the construction of the wellknown transform of the potential which adds to the spectrum of the SturmLiouvill operator one new point only (see, eg [3]) The decomposition (33) is useful for the Friedrichs model (see [5]) References T Kato, Perturbation Theory for Linear Operators, SpringerVerlag, New York, B S Pavlov, S V Petras, The singular spectrum of a weakly perturbed multiplication operator, Funktsional Anal i Prilozhen 4 (97), no 2, 54 6 (Russian) 3 M Ablowitz, H Segur, Solitons and the Inverse Scattering Transform, SIAM, Philadelphia, 98 4 I S Kac, M G Krein, Rfunctions analytic functions mapping the upper half plane into itself, in: F V Atkinson, Discrete and Continuous Boundary Problems, Mir, Moscow, 968, pp (Appendix) (Russian) 5 E V Cheremnikh, Nonselfadjoint Friedrichs model and Weyl function, Reports Nation Acad Sci Ukraine (2), no 8, (Ukrainian) Laboratoire of Mathematics, Department of Mathematics, Faculty of Sciences, University Badji Mokhtar Annaba, Annaba, Algeria address: boumaraf Lviv Polytechnic National University, Lviv, Ukraine address: Received 4//27; Revised 27/8/27