NON-COMPACT PERTURBATIONS OF THE SPECTRUM OF MULTIPLIERS GIVEN WITH FUNCTIONS Текст научной статьи по специальности «Математика»

Non-compact perturbations of the spectrum of multipliers given with functions

R. R. Kucharov1, R. R. Khamraeva1'2

1 National University of Uzbekistan, 100174, Tashkent, Uzbekistan 2Westminster International University in Tashkent, 100010, 12, Istiqbol str., Tashkent, Uzbekistan [email protected], [email protected]

DOI 10.17586/2220-8054-2021-12-2-135-141

The change in the spectrum of the multipliers H0 f (x, y) = xa + y& f (x, y) and H0f (x, y) = xay& f (x, y) for perturbation with partial integral operators in the spaces L2 [0, 1]2 is studied. Precise description of the essential spectrum and the existence of simple eigenvalue is received. We prove that the number of eigenvalues located below the lower edge of the essential spectrum in the model is finite.

Keywords: essential spectrum, discrete spectrum, lower bound of the essential spectrum, partial integral operator.

Received: 25 January 2021

Revised: 10 March 2021

Final revision: 12 March 2021

1. Introduction

The first results on the finiteness of the discrete spectrum of N - particle Hamiltonians with N > 2 were obtained by Uchiyama in 1969 [1-3]. He found sufficient conditions for the finiteness of the number of discrete eigenvalues for energy operators in the space L2 (R6) for the system of two identically charged particles in the field of a fixed center with or without an external electromagnetic field. In 1971, Zhislin proved the finiteness of the discrete spectrum for energy operators in symmetry spaces of negative atomic ions with nuclei of any mass and of molecules with infinitely heavy nuclei under the assumption that the total charge of the system is less than -1 [4].

Let and Q2 be closed bounded sets in RV1 and RV2, respectively. In the space Lp(Q1 x Q2), p > 1 partially integral operator (PIO) T of the Fredholm type in general is given by the equality [5]:

T = To + T1 + T2 + K, (1)

where the operators T0,T1,T2, K have the following view:

Tof (x,y) = ko(x,y)f (x,y), T1f (x,y) = J k>1(x,s,y)f (s,y)d^(s),

n1

T2f (x,y) = J k2(x,t,y)f (x,t)d^2 (t), Kf (x,y)= J J k(x,y; s,t)f (s,t)d^1(s)d^2(t).

Here, the functions k0, k1,k2, and k are given measurable functions in the concept of Lebesque on Q1 x Q2, ^ x Q2, ^ x Q2 and (Q1 x Q2)2, respectively, and integration of functions is understood in the concept of Lebesgue, where ^k(•) - Lebesque measure on Qk, k = 1,2.

In the Hilbert space L2(Q x Q), where Q = [a, b]v, consider the following model operator:

H = Ho - (jT1 + №), Y> 0, 0. (2)

Here, the actions of the operators H0, T1 and T2 are determined by formulas:

Hof (x, y) = ko(x, y)f (x, y),

T1f (x,y)= J V1(x)¥1(s)f (s,y)ds, T2f (x,y)= J ^2(y)^2(t)f (x,t)dt, n n

where ko(x, y) is a nonnegative continuous function on Q x Q, (•) is a continuous function on Q and

J v2j(t)dt = 1, j = 1,2.

n

Via p(), &(•), ffess(:) and adjsc(•) denote, respectively, the resolvent set, spectrum, essential spectrum and discrete spectrum self-adjoint operators [6].

In [7], sufficient conditions for finiteness and infinity were obtained in the discrete spectrum for aess(H) = a(H0). In work [8] proved the existence of the Efimov effect in model (2) for given k0(x,y). In [9], the essential spectrum and the number eigenvalues below the lower bound of the essential spectrum in model (2), when the function

k0(x, y) has the form: k0(x, y) = u(x)u(y), where u(x) is a nonnegative continuous function on Q = Q1 = Q2 and

/dx

—— < to. In [10] studied the existence of an infinite number of eigenvalues (the existence of Efimov's effect) for

u(x)

Q

a selfadjoint partial integral operators.

2. The lower boundary of the essential spectrum of V

Consider the multiplier:

= (xa +yP

Vof (x, y) = (xa + y3)f (x, y), a> 0, 0.

Let us define a partially integral operator (PIO) V:

V = Vo - y(Ti + T2), Y> 0,

where:

(3)

1 1 Tif (x,y) = J f (s,y)ds, T2f (x,y)= J f (x,t)dt, f e L2[0,1]2. oo In the space L2 [0,1] we define the operators H1 and H2 in Friedrichs models:

k1 k1 Hif(x) = xa<p(x) - y J y(s)ds, H2^(y) = y3 ^(y) - Y J ^(t)dt.

oo

Lemma 1. [11] The number A e R\[0,1] is the eigenvalue of the operator H1 (of the operator H2) if and only if

Ai(A) = 0 (A2(A) = 0), where:

A1 (A) = 1 - Yjj^, A2(A) = 1 - y J

ds

3A

Lemma 2.

A)

B)

lim A1(A)

A^ 0 —

lim A2(A)

1

1 — a

,

1

1 - fi

,

if 0 < a < 1; if a > 1,

if 0 <fi< 1; if fi > 1.

Proof. First, we prove the statement A.

a) Let 0 < a < 1. Consider an arbitrary increasing sequence {An}neN of negative numbers approaching to zero, i.e An < An+1 < 0 and lim An = 0. Then:

and

0 < -< -1-, n e N

sa - An~ sa - An+1

-< — for almost all s e [0,1].

sa - An < sa L J

The function h0(s) = — is integrable by [0,1] in the concept of Lebesgue and:

sa

ho(s)ds

1a

Hence, due to Lebesgue theorem on limited transition under the sign of the Lebesgue integral it follows that:

1

1

Y

Y

1

1

k1

ds

lim

A^-o- y sa - A 1 - a

o

Thus, we have:

lim A1(A) = 1--—.

A^o- 1 - a

b) Let a > 1. Suppose that a =1. Then:

1

lim A1(A) = 1 - y lim f-- = 1 - y lim ln ( 1--- | = -to.

A^o- A^o- J s - A A^o- y -A J

o

If a > 1 , then we have:

Hence:

i.e.

ds ds

lim -- > lim -- = +to,

A^o- / sa — A A^o- / s - A

lim A1(A) = -to. Ao

Proposition 1.1) Let 0 < a < 1 (0 < fi < 1). Then:

a) if a + y < 1 (fi + Y < 1), then the operator H1 (operator H2) outside the essential spectrum has the only eigenvalue of the operator;

b) if a + y > 1 (fi + Y > 1), then the operator H1 (operator H2) outside the essential spectrum has the only eigenvalue of the operator ^ (the eigenvalue value £2), while £k is a simple proper value Hk £k < 0, k = 1, 2.

II) Let a > 1 (fi > 1). Then the operator H1 (operator H2) outside the essential spectrum has a unique eigenvalue ^ (eigenvalue £2), for this £k is a simple eigenvalue of the operator Hk and £k < 0, k = 1,2.

Proof. It is easy to note that the function A1(A) by (-to, 0) is strictly decreasing and A1(A) > 0 to (1, to), thus, the operator H1 on the set (1, to) has no eigenvalue.

Let 0 < a < 1. By Lemma 2 and monotonicity, the function A1(A) to (-to, 0) states a and b, since:

Ran(A1) = ( 1--—, 1

\ 1 - a

Let a > 1. Then from Lemma 2 we obtain: Ran(A1) = (-to, 1). Due to monotonicity functions A1(A) (-to, 0) equation A1(A) = 0 (-to, 0) has a unique solution ^ < 0 ^ is a simple eigenvalue operator H1. Proposition 1 for the operator H2 is proved similarly. Theorem 1. Let 0 < a < 1 and 0 < fi < 1. Then:

a) if a + y < 1 and a + fi < 1, then:

&(V) = aeSS(V) = a(Vo) = [0, 2];

b) if a + y > 1 and a + fi < 1, then

&(V) = aeSS(V) = a(Vo) U [£1,1 + £1],

where £1 - negative eigenvalue of operator H1;

c) if a + y < 1 and a + fi > 1, then:

& ess (V )= a(Vo) U [£2, 1+ £2],

where £2 is a negative eigenvalue of operator H2;

d) if a + Y > 1 and a + fi > 1 , then:

& ess (V) = a(Vo) U [£1,1 + £1 ] U [£2, 1 + £2] and &disc(V) = {wo},

where wo = £1 + £2 and wo is a simple eigenvalue of operator V.

Proof. It is easy to note that the operator V will be unitarily equivalent to the operator H1 < E+E< H2 (see [12]). Then &(V) = &(H1) + &(H2) and for of multiplicity nV(w) eigenvalue w e &(V) \ &ess(V) of the operator V the following equality takes place:

1

s

1

1

nV M = nHt (p) • nn2 (q),

p+q=u

(P'q)ea(Hi)xa(H2)

where nHl (p) and nH2 (q) - multiplicity of the eigenvalues p and q of the operators H1 and H2, respectively. This and Proposition 1 imply the proof the theorem. Theorem 2. Let a > 1 and 0 < 3 < 1. Then:

a) if / + y < 1, then:

°(V ) = & ess (V ) = a(Vo) U [&, 1 + &]■,

b) if 3 + Y > 1, then:

&ess(V) = a(Vo) U [&, 1 + £1 ] U [£2, 1 + £2] and adisc(V) = j^o},

where uo = ^ + £2 and uo is a simple eigenvalue of the operator V. Theorem 3. Let a > 1 and 3 > 1. Then:

&ess(V) = a(Vo) U [£1,1 + £1 ] U [£2, 1 + £2] and &disc(V) = j^o},

where uo = ^ + £2 e uo - is a simple eigenvalue of the operator V. Corollary 1. Let 0 < a < 1 and 0 < 3 < 1. Then:

0, if a + y < 1 and 3 + 7 < 1,

£1, if a + y > 1 and 3 + Y < 1,

£2, if a + y < 1 and 3 + Y > 1,

minj£1, £2}, if a + y > 1 and 3 + Y > 1.

Emin(V ) =

3. Discrete spectrum of partial integral operators Let's define the multiplier Ho:

Hof (x, y) = xayP f (x, y), a > 0, 3> 0,

and the operators T1 , T2 :

Tif (x,y) = J f (s,y)ds, T2f (x,y) = J f (x,t)dt.

0

Let us define a self-conjugate PIO H:

H = Ho - y(T1 + T2), Y > 0. We have a(Ho) = [0,1]. For each A G R\ [0,1] define the function A^y; A) on [0,1] (A2(x; A) on [0,1]) by formula:

1 1 ds ds

Ai(y;X) = 1 -^S^oyhx, A2(x;X) = 1 -y/:

sayß - X' 2y ' J 'J xasß - X'

0 0

In the space L2 [0,1] we define the family {H1(t)}te[01] of the self-adjoint operators in the Friedrichs' model:

1

H1(t)y(x) = tßxay(x) - y J y(s)ds.

0

Similarly, in the space L2[0,1] we define the family jH2(t)}te[o1]:

1

H2(t)^(y)= tayP 4(y) - yJ Hs)ds.

o

Lemma 3. Function:

n(t) = mf (Hj(t)y, y), t e [0,1] (j = 1, 2) (3)

\m\ = i

is non-positive, continuous and increasing on [0,1].

Proof. In work [9], there is a proof of the continuity and non-positivity of the function nj (t) on [0,1]. We will show the monotonicity of the function nj (t) on [0,1]. We define the family of the {Ho(t)}ie[o1] multipliers:

Ho(t)p(x) = xat3p(x), <p e L2[0,1]. Then it follows from t1 < t2, t1,t2 e [0,1] that:

Ho(t1) < Ho(t2).

Therefore, we have:

n1(t1)= inf (H1(t1)^,^)= inf [(Ho(t1)^,^) - y(Khp,v)\ <

11^11=1

inf (Hi(t2)if, if) = inf [(H0(t2)^,^) — Y(Kf,f)} = n1(t2),

11^11=1

Where:

i

Kif(x) = j i(s)ds.

o

This means that the function n1(t) is increasing on the set [0,1].

Obviously, for each y e [0,1] the function A1(A) = A1(y; A) is strictly decreasing on (-to, 0). Therefore, for

each y e [0,1] there exists finite or infinite limit lim A1(y; A). Moreover, there is:

A^o-

Lemma 4. a) if 0 < a < 1, then for each y e (0,1]:

Y1

lim A1(y; A) = 1 - —L---3;

A^o- 1 - a y3

b) if a > 1, then for each y e (0, 1]:

lim A1(y; A) = -to.

A^o-

Proof. a) Let 0 < a < 1. Then, for y e (0,1] we get

11

soyf-x < ho(s,y) = lay?

and for any ascending sequence {An} negative numbers decreasing to zero we have:

—^ ^ < —-, n e N.

say3 - An ~ say3 - An+1

On the other hand, for each y e (0,1] there exists a Lebesgue integral from function h1 (s, y) on s e [0,1]:

k1

J ho(s,y)ds =

1

1 1

l0(s,V)ds = --—5.

1 — a yß

0

Then, by Lebesgue's theorem on the limited transition under the sign of the Lebesgue integral, we obtain:

Y 1

lim Ai(y; A) = 1 - ---j, y e (0,1];

A^0- 1 — a yß

b) Let a > 1 and assume that a = 1. It is obvious that for y = 0 we have: lim A1(y; A) = —to. For each y e (0,1] we have:

1 1

f ds f ds 1 , ( y3\

1 ' In 1 - V .

J sayß — A J syß — A yß \ A

00

Therefore for y g (0,1} we get:

lim A1(y; A) = 1 — Y lim ln f 1 — — ) = —to;

Suppose that a > 1. Then from inequality:

i i

/ds f ds ,

-у--W syd-x, y € [0,1]

0 0

we get that

lim f-dps—- = y G [0,1]

x^o-J sayp - A 1 1

o

and accordingly, lim A1 (y;A) = -to.

x^o-

Obviously, the function:

Y 1

hi(y)=\im Ai(y;-) = 1 -

1 - ay"

Y

increases by (0,1] from -то to h1ax = hi(1) = 1 - -——.

1 — a

We put:

nmax = max п (t), j = 1, 2.

Then nmax = (1).

Lemma 5. Let 0 < a < 1 (0 < в < 1). Then:

a) if y + a < 1(y + в < 1), then п^ = 0 (nf^ = 0);

b) if y + a > 1 (y + в> 1), then nfax < 0 (п1^ < 0). Proof. Let 0 < a < 1. a) Assume that: y + a = 1. We have:

hfax = hi(1) = lim Ai(1; -) = 1 - = 0.

1 - a

Hence, taking into account the monotonicity of the function Ai (1; -) by - < 0 we get that Ai (1; -) > 0 for any - < 0. Then, according to Proposition 1, the operator Hi(1) has no eigenvalue below the bottom edge the essential spectrum of the operator Hi(1). By the minmax principle and from equality (3) we obtain that nfax = ni(1) = Emin (Hi(1))=0.

If y + a < 1. Then hf = hi(1) > 0. On the other hand Ai(1;-) > hfax. Then according to the proposition 1, the operator Hi(1) has no negative eigenvalue value. It follows that nfax = 0. b) Let y + a > 1. Then:

hi(y) < hfax = 1 0.

1-a

Therefore, for for each y € (0,1] we have hi(y) = lim Ai(y; -) < 0. Hence, since the function Ai(y; -) is

monotonic with respect to - < 0 implies the existence of a unique number -0 = -0 (y) < 0 (for each y € (0,1]) such that Ai(y; -0(y)) = 0. For y = 0 we have Ai(0; -y) = 0, i.e -0 = -0(0) = -y is a solution to the equation Ai(0; -) = 0. Due to minmax principle [13] solution of - 0(y) equation Ai(y; -) = 0 is defined using continuous function ni(t), i.e. - 0(t) = ni(t), t € [0,1]. However -0(y) < 0, y € [0,1]. Therefore ni(1) = nf* < 0. Lemma 6. Let a > 1 (в > 1). Then nfax < 0 (nfax < 0). Proof. For y = 1 we get:

i

ds

Ai(-) = Ai(1;-) = 1 -J -o—^.

0

We have:

lim Ai(-) = 1 and lim Ai(-) = -то.

Then, due to the monotonicity of the function A1( ) for < 0 we obtain the existence of unique number 0 < 0, such as Ai(-0) = 0. Therefore, -0 = nfax < 0.

By the theorem 3.3 [14] and Lemma 3,4 or the essential spectrum operator H we obtain the following statement. Theorem 4. For the essential spectrum of the operator H there is a place for equality:

aess(H) = [-Y,Y0] U [0,1],

where y0 = max{n1ax, nf*}.

From the positivity of the operators H0, Ti and T2 for the operator H we have:

a(H) С (-то, 1],

l

i.e. above the upper edge of the essential spectrum &ess(H) of the operator H eigenvalues are missing. Then, by the theorem 4 the discrete spectrum of the operator H lies in the set of negative numbers. We put:

£ = 1

£o (1 + a)(1 + fi).

Theorem 5. If y > £o, then the operator H has the negative eigenvalue, lying to the left of the bottom edge of the essential spectrum.

Proof. Assume the conditions y > £o. Put fo(x, y) = 1. Then ||fo|| = 1 and

(Hfo, fo) = (Hofo, fo) - Y((Tfo, fo) + (T2fo, fo)) = £o - 2y. Then, by the theorem 4 we have Emin(H) = - y and from y > £o we get that

Ao = (Hfo, fo) <- Y = Emin(H).

Hereof and according to the minmax principle we get that Ao e &disc(H), i.e. Ao = £o - 2y - is the eigenvalue of the operator H.

Corollary 2. Number of eigenvalues of the operator H is at most one and for y > £o the discrete spectrum of the operator H is not empty.

References

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