Muminov, M I

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EXT-2000-06801/10/1999Available at: http://www.ictp.trieste.it/~pub off IC/99/155United Nations Educational Scientic and Cultural OrganizationandInternational Atomic Energy AgencyTHE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICSON THE ESSENTIAL SPECTRUMOF UNBOUNDED NONSELF-ADJOINT FRIEDRICHS OPERATORM.I. Muminov1Uzbekistan Academy of Sciences (Samarkand Division),Temur Malik Street no. 3, 703000 Samarkand, UzbekistanandThe Abdus Salam International Centre for Theoretical Physics, Trieste, Italy.AbstractIn this paper, we consider the unbounded generalized Friedrichs operator H; i.e. the operatorof multiplication by the rational function u with the perturbation of integral operator with kernelK. We prove that if the kernel K satises some analyticity condition, then the essential spectrumof H coincides with the spectrum of the multiplication operator.MIRAMARE-TRIESTEOctober 19991E-mail: info@samuni.silk.org1 IntroductionIn L2(R) we consider the non-self-adjoint unbounded operator H dened by the formulaHf(x) = u(x)f(x) +ZRK(x; y)f(y)dy; f 2 Dui.e. H = H0+ V; where H0is the operator of multiplication by the rational function u and Vis integral operator with analytical kernel K(x; y) in W2= WWsatisfying the followingconditionjK(s+ iy; t+ iy)j Mjsj+ jtj+ 1;  > 1; (K)for all y 2 ( ; ); where W= fz 2 C : jImzj < g;  is a xed number, C the complexplane and Du= ff 2 L2(R) :RRju(x)f(x)j2dx <1g:The bounded generalized self-adjoint Friedrichs model is considered for the case of when u andK are analytic functions in [1]. In Lakaev S. N. [1], it was proved that the absolutely continuousspectrum of the operator H coincides with the spectrum of H0: Moreover, the singular spectrumof H is a nite set. In [2] the structure of essential spectrum for non-self-adjoint boundedFriedrichs operator is described.First, we remark that Du= D(H); where D(H)  is a domain of the adjoint operator H:Thus, the operator H is a closed operator (see theorem VIII.1 in [3]). Let (H) is a spectrumof H:Denition 1. A point  2 (H) is called discrete if  is isolated and the operatorP=12iIj j=(H   ) 1dis nite dimensional, where (H   ) 1is a resolvent of the operator H and the number  > 0such that f : j  j  g = fg: We denote by disc(H) a discrete spectrum of the operatorH.Denition 2. The essential spectrum of operator H is the setess(H) = (H) n disc(H):Denote by   the range of the function u: Let   be a set satisfying the following conditionC n   = D1[ ::: [Ds;where Diis an unbounded connected open set in C for i = 1; s: We observe that   = (H0);where (H0) is the spectrum of the operator H0:Theorem. The essential spectrum of operator H concides with the spectrum of the operatorH0; i.e.ess(H) = (H0):22 Auxiliary lemma and the proof of the main resultLet (z) be the Fredholms determinant of the operator I+V (H0 z) 1; where z 2 C n(H0), Iis a unit operator, (H0 z) 1is a resolvent of the operator H0: The function (z) is representedby the formula(z) = 1 +1Xn=1dn(z)n!; (1)dn(z) =ZR:::ZRDn(x1; :::; xn; z)dx1: : : dxn; (2)Dn(x1; :::; xn; z) =K(x1; x1) K(x1; x2)    K(x1; xn)K(x2; x1) K(x2; x2)    K(x2; xn)......   ...K(xn; x1) K(xn; x2)    K(xn; xn)1u(x1)  z: : :1u(xn)  z:Proposition 1. The series (1) converges for all z 2 C n (H0):Proof. We note that the function Dn(x1; :::; xn; z) is integrable, due to condition (K). LetN be a positive number. The function Dn(x1; :::; xn; z) is symmetric. Therefore, the integral (2)is represented by the following expression:dn(z) =ZR:::ZRKn(x1; :::; xn)dx1u(x1)  z: : :dxnu(xn)  z==Zjx1j>N:::Zjxnj>NKn(x1; :::; xn)dx1u(x1)  z: : :dxnu(xn)  z++ : : :+ CknZjx1j<N:::Zjxkj<NZjxn (k+1)j>N:::Zjxnj>NKn(x1; :::; xn)dx1u(x1)  z: : :dxnu(xn)  z++Zjx1j<N:::Zjxnj<NKn(x1; :::; xn)dx1u(x1)  z: : :dxnu(xn)  zwhereKn(x1; :::; xn) =K(x1; x1) K(x1; x2)    K(x1; xn)K(x2; x1) K(x2; x2)    K(x2; xn)......   ...K(xn; x1) K(xn; x2)    K(xn; xn):Consequently, we obtain:jdn(z)j 1(mz)nZjx1j>N:::Zjxnj>NjKn(x1; :::; xn)jdx1: : : dxn+ : : :++Ckn1(mz)nZjx1j<N:::Zjxkj<NZjxn (k+1)j>N:::Zjxnj>NjKn(x1; :::; xn)jdx1: : : dxn++1(mz)nZjx1j<N:::Zjxnj<NjKn(x1; :::; xn)jdx1: : : dxn3where mz= infx2Rju(x)  zj > 0 for z 2 C n (H0):We can choose a number N such that be following inequlityZjx1j>N:::Zjxnj>NjKn(x1; :::; xn)jdx1: : : dxnZjx1j<N:::Zjxkj<NZjxn (k+1)j>N:::Zjxnj>NjKn(x1; :::; xn)jdx1: : : dxnZjx1j<N:::Zjxnj<NjKn(x1; :::; xn)jdx1: : : dxn;holds.Then we havejdn(z)j 2n(mz)nZjx1j<N:::Zjxnj<NjKn(x1; :::; xn)jdx1: : : dxn:Thus, by using the condition (K) and applying Hadamard's theorem we obtain the followinginequalityjdn(z)j 2n(mz)nMn(2N)npnn:This proves proposition 1.Lemma 1. Let i be a xed positive integer. There exists an unbounded subset D0i Disuch that (z)! 1 as z !1 and z 2 D0i:Proof. We denote by a1; :::; akthe real poles of rational function u; mz= infx2Rju(x)  zj:a) Let z 2 Diand mz!1 as z !1: Then we havej1Xn=1dn(z)n!j 1Xn=11n!1(mz)nZR:::ZRjKn(x1; :::; xn)jdx1: : : dxn==1mz1Xn=11(mz)n 11n!ZR:::ZRjKn(x1; :::; xn)jdx1: : : dxn: (3)By using (3) we obtain limmz!1(z) = 1 in z 2 Di; since1Xn=01n!ZR:::ZRjKn(x1; :::; xn)jdx1: : : dxn<1:b) Now let mz! c = const as z !1 for z 2 Di: Then there exists an integer N, a positivenumber  <  and an unbounded connected set D0i= D0i(N; )  Disuch thatu 1(D0i)   = [ki=1fz : jz   aij < ; Imz > 0g [ fz = x+ iy : 0 < y < ; jxj > Ng;where N > jaij:Let z 2 D0iand  1(z); :::;  l(z) 2  be solutions of the equation u(x)   z = 0 and  N;acontour dened by N;= [i=1kfz : jz   aij = ; Imz  0g [ fz =  N + iy : 0  yg[4[fz = N + iy : 0  y  ; jxj  Ng [ fz = x+ i : jxj  Ng:Let K(t,s) be an analytic continuation in W2of the function K(x; y); x; y 2 R:The function d1(z) is represented asd1(z) =ZRK(x; x)dxu(x)  z=ZRn N;K(t; t)dtu(t)  z+Z N;K(t; t)dxu(t)  z== 2ilXj=1K( j(z);  j(z))u0( j(z))+Z N;K(x; x)dxu(x)  z:Since ju(t)  zj   > 0; for all z 2 D0i; there exists a number N > 0 such that the followinginequalityZ N;:::Z N;jDn(t1; :::; tn; z)jdt1: : : dtn 2ZRN;Z N;:::Z N;jDn(t1; :::; tn; z)jdt1: : : dtn 2nZRN;:::ZRN;jDn(t1; :::; tn)jdt1: : : dtnholds, whereRN;= ( N; a1  ) [ (a1+ ; a2  ) [ : : : [ (al 1+ ; al  ) [ (al+ ;N):We denote by Kza number dened byKz= maxfsup jK( j(z);  j(z))u0( j(z))j; supt2 N;jK(t;  j(z))u0( j(z))j; supt2 N;jK(t;  j(z))u0( j(z))j;supt2 N;jK(t;  j(z))u(t)  zj; supt2 N;jK( j(z)); tu(t)  zj; sups;t2RN;jK(t; s)u(t)  zjg: (4)Hence, we getjd1(z)j  2lKz+ 2ZRN;jK(t; t)jdtju(t)  zj 2lKz+ 2N;Kz:where N;is a length of the contour RN;:Proposition 2. limz!1Kz= 0 for z 2 D0i:Proof. Let u0(x) = const: Thenlimz!1 j(z) =1; for allj  l:If u0(x) 6= const: Thenlimz!1 j(z) =1; j  l:Thus, we havesup limz!1jK( j(z);  j(z))u0( j(z))j = 0:5Note that the contour RN;is bounded. So, we obtainlimz!1ju(t)  zj =1; t 2 RN;:Consequently,limz!1j sups;t2 N;K(t; s)u(t)  zj = 0:The relation limz!1Kz= 0 is proved analogously for all cases.Now, we considerdn(z) =ZR:::ZRK(x1; x1) K(x1; x2)    K(x1; xn)K(x2; x1) K(x2; x2)    K(x2; xn)......   ...K(xn; x1) K(xn; x2)    K(xn; xn)dx1u(x1)  z: : :dxnu(xn)  z=ZRn N;: : :ZRn N;K(t1; t1) K(t1; t2)    K(t1; tn)K(t2; t1) K(t2; t2)    K(t2; tn)......   ...K(tn; t1) K(tn; t2)    K(tn; tn)dt1u(t1)  z: : :dtnu(tn)  z+ : : :++CknZRn N;: : :ZRn N;Z N;dtn k 1: : :Z N;dtnK(t1; t1) K(t1; t2)    K(t1; tn)K(t2; t1) K(t2; t2)    K(t2; tn)......   ...K(tn; t1) K(tn; t2)    K(tn; tn)1u(t1)  z: : :1u(tn k  zdtn k 1u(tn k 1)  z: : :dtnu(tn)  z++Z N;: : :Z N;K(t1; t1) K(t1; t2)    K(t1; tn)K(t2; t1) K(t2; t2)    K(t2; tn)......   ...K(tn; t1) K(tn; t2)    K(tn; tn)dt1u(t1)  z: : :dtnu(tn)  z== (2i)nlXj1=1: : :lXjn=1K( j1;  j1) K( j1;  j2)    K( j1;  jn)K( j2;  j1) K( j2;  j2)    K( j2;  jn)......   ...K( jn;  j1) K( jn;  j2)    K( jn;  jn)1u0( j1(z)): : :1u0( jn(z))+6+(2i)n 1lXj1=1: : :lXjn 1=11u0( j1(z)): : :1u0( jn 1(z))Z N;K( j1;  j1) K( j1;  j2)    K( j1; xn)K( j2;  j1) K( j2;  j2)    K( j2; xn)......   ...K(xn;  j1) K(xn;  j2)    K(xn; xn)dxnu(xn)  z++ : : :+ Ckn(2i)n klXj1=1: : :lXjn k=11u0( j1(z)): : :1u0( jn k(z))Z N;: : :Z N;K( j1;  j1)    K( j1;  jn k) K( j1; tn k 1)    K( j1; tn)K( j2;  j1)    K( j2;  jn k) K( j2; tn k 1)    K( j2; tn)...   ......   ...K(tn;  j1)    K(tn;  jn kK(tn; tn k 1)    K(tn; tn)dtn k 1u(tn k 1)  zdtnu(tn)  z+Z N;: : :Z N;K(t1; t1) K(t1; t2)    K(t1; tn)K(t2; t1) K(t2; t2)    K(t2; tn)......   ...K(tn; t1) K(tn; t2)    K(tn; tn)dt1u(t1)  z: : :dtnu(tn)  z:Arguing as above (as for the case n=1) by using (4) and applying Hadamard's theorem we getthe following inequalityjdn(z)j  lnKnzpnn+ nln 12N0Knzpnn++Cknln k(2N;)kKnzpnn+ : : :++(2N;)nKnzpnn< 2n(2N;)nKnzpnn;for all z 2 D0iand n = 1; 2; : : : : As consequence we getj(z)  1j 1Xn=1jdn(z)jn!1Xn=12n(2N;)nKnzpnnn!= Kz1Xn=12n(2N;)nKn 1zpnnn!;for all z 2 D0i= D0i(N; ):Now, by proposition 2, we have the proof of lemma 1.Proof of theorem. The operator H   z is represented asH   z = (I + V (H0  z) 1)(H0  z)for z 2 Cn(H): Thus, (H z) 1exists if and only if (I+V (H0 z) 1)) 1exists. By proposition1, Fredholm's determinant (z) is dened for all z 2 C n (H0): V is a compact operator. Thefunction V (H0  z) 1is compact valued analytic in C n (H): By using the analytic Fredholmtheorem, we conclude that (I + V (H0  z) 1)) 1exists on C n (H0) except for discrete setD  C n (H0); since by lemma 1 (I + V (H0  z) 1)) 1exists for certain z 2 C n (H0): Soess(H)  (H0):7Now let 0be an arbitrary element of  (06= 1) and u(x0) = 0; u is a rational function.Then there exists positive N0> 0 such that function u is continuous in [x0:x0+1N0]: We setfn(x) =pn(n+ 1); as x 2 (x0+1n+1; x0+1n]0 as x =2 (x0+1n+1; x0+1n]for integer n > N0: Evidently, fn(x) 2 Duand ffng is an orthonormalized system. It is easy toshow thatlimn!1k (H   z0)fnk= 0:Thus, 02 (H): In other words   = (H0)  (H): Consequently, (H0)  ess(H):References1. Lakaev S. N. Some spectral properties of generalized Friedrichs model. Trudy seminara im.I. G. Petrovskogo, 1986. no 11, s 210-238 (in Russian).2. Lakaev S. N., Muminov M. I. On essential spectrum nonself-adjoint generalized Friedrichsmodel. Doklady Uzbek Academy of Sciences, no 4,(1997) c. 8-10 (in Russian).3. M. Reed, B. Simon. Methods of modern mathematical Physics, vol. 4. Analysis of operators,AMS (MOS) 1970.8

On the essential spectrum of unbounded nonself-adjoint Friedrichs operator

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