CC BY
6120 Probl. Anal. Issues Anal. Vol. 7 (25), No. 2, 2018, pp. 20-38
DOI: 10.15393/j3.art.2018.4630
UDC 517.983
V. M. BRUK
GENERALIZED RESOLVENTS OF OPERATORS GENERATED BY INTEGRAL EQUATIONS
Abstract. We define a minimal operator L0 generated by an integral equation with an operator measure and give a description of the adjoint operator We prove that every generalized resolvent of L0 is an integral operator and give a description of boundary value problems associated to generalized resolvents.
Key words: integral equation, Hilbert space, symmetric operator, generalized resolvent, boundary value problem
2010 Mathematical Subject Classification: 46G12, 45N05, 47A10
1. Introduction. In [13], A.V. Straus described generalized resolvents of a symmetric operator generated by formally selfadjoint differential expression in the scalar case. In [4] these results were extended to the operator case. Further, the generalized resolvents of differential operators were studied in many works (a detailed bibliography is available, for example, in [10], [12]).
In this paper, we consider the integral equation
t t y(t) = xo - iJ dp(s)y(s) — iJ f (s)ds, (1)
where y is an unknown function; f G L2(H; a,b), J is an operator in a separable Hilbert space H, J = J *, J2 = E (E is the identical operator); p is an operator-valued measure defined on Borel sets A C [a, b] and taking values in the set of linear bounded operators acting in H; f stands for /[tot) if t0 < t, for — Jjtot) if t0 > t, and for 0 if t0 = t. We assume that
©Petrozavodsk State University, 2018
the measure p is self-adjoint, and p has a bounded variation, and the set Sp of single-point atoms of measure p can be arranged in the form of an increasing sequence.
We define the minimal operator L0 generated by equation (1) and give a description of the adjoint operator L£. We prove that every generalized resolvent of L0 is an integral operator. Unlike differential operators, the domain and the range of the characteristic function of a generalized resolvent are spaces of sequences. Moreover, we give a description of generalized resolvents in terms of boundary value problems.
2. Preliminary assertions. Let H be a separable Hilbert space with scalar product (•, •) and norm ||-||. We consider a function A ^ P(A) defined on Borel sets A C [a, b] and taking values in the set of bounded linear operators acting in H. The function P is called an operator measure on [a, b] (see, e. g., [3, ch. 5]) if it is zero on the empty set and the equality P (UAn) = P(An) holds for disjoint Borel sets An, where the series converges weakly. Further, we extend any measure P on [a,b] to a segment [a, bo] (bo >b) letting P(A) = 0 for all Borel sets A C (b, bo].
By VA(P) we denote VA(P) = p(A) = sup£j ||P(Aj)||, where the supremum is taken over finite sums of disjoint Borel sets Aj C A. The number Va(P) is called variation of the measure P on the Borel set A. Suppose that the measure P has the bounded variation on [a,b]. Then for p-almost all £ G [a,b] there exists an operator function £ ^ ^P(£) such that possesses the values in the set of bounded linear operators acting in H, ||^P(£)|| = 1, and the equality P(A) = fA ^P(£)dp holds for each Borel set A C [a, b] ([3, ch. 5]). A function h is integrable with respect to the measure P on a set A if there exists the Bochner integral /Atfp(t)h(t)dp = JA(dP)h(i). Then the function y(t) =//o(dP)h(s) is continuous from the left.
Denote by SP a set of single-point atoms of the measure P (i.e., a set t G [a, b] such that P({t}) = 0). The set SP is at most countable.
In following Lemma 1, p1, p2, q are operator measures having bounded variations and taking values in the set of linear bounded operators acting in H. Suppose that the measure q is self-adjoint, i.e., (q(A))* = q(A) for each Borel set A C [a, b]. We assume that these measures are extended to the segment [a, b0] D [a, b0) D [a, b] in the manner described above.
Lemma 1. [8] Let f, g be functions integrable on [a,b0] with respect to the measure q. Then any functions
y(t)= yo — iJJ dPi(s)y(s) — iJJ dq(s)f(s),
to to
t t
z(t) = z0 — iJ Jdp2(s)z(s) — iJ Jdq(s)g(s) (a ^t0 < b0, t0 ^t ^ b0)
to to
satisfy the following formula (analogous to the Lagrange one):
C2 C2
J (dq(t)f (t),z (t)) — J(y(t),dq(t)g(t)) = (iJy(c*),z((*)) —
ci ci
C2 C2
— (iJy(cl),z(cl)) +J C(y(t),dp2(t)z(t)) — J(dPi(t)y(t),z(t)) —
c1c1
— J (iJp1({t})y(t), P2({t})z(t)) —
tesPl nSp2 n[ci ,C2)
— J] (iJq({t})f (t), p2({t})z(t)) —
teSqnSp2 n[ci,c2)
— JJ (iJpl({t})y(t), q({t})g(t)) —
teSpl nsqn[ci,c2)
— J (iJq({t})f (t), q({t})g(t)) , to ^ d <c2 ^ bo. (2)
teSqn[ci,c2)
Let a segment [h,l2] C [a,b0]. We consider a set of Borel measurable functions, ranging in H, bounded on [l\,l2], continuous from the left, and
constant on [h,l2] If (b,bo\. We introduce the norm ||u|L ¡] = sup ||u(t)||
[l, 21 te[h,h]
on this set and obtain a Banach space denoted by C[l\, l2].
Theorem 1. [7] For any function g G C[a,b0] there exists a unique solution of the equation
t
y(t) = Jdp(0y(0 + g(t), a ^ to ^ bo, (3)
to
belonging to the space C[to — 5,bo], where a ^ to < bo, 5 = 5(to) > 0 is small enough if to > a and 5 = 0 if to = a, the measure p has the bounded variation on [a,b].
t
t
Corollary 1. Suppose t0 = a. Then for any function g G C[a,b0] there exists a unique solution of equation (3) belonging to the space C[a,b0].
Remark 1. In general, a solution of (3) can be non-extendable to the left (see [7]). However, if the measure p in (3) is continuous, then a solution can be extended to the left up to the point a and this extension is unique.
Suppose further that p is a self-adjoint measure with the bounded variation. We consider the equation
t t t y(t) = x0 — iJj dp(s)y(s) — iJ\j y(s)dp(s) — iJJ f (s)dp(s), (4)
a a a
where À G C; p is the usual Lebesque measure on [a, b] (p([a, ft)) = ft — a for all a, ft G [a,b],a<ft ) extended to [a,b0] by the equality p(A) = 0 for each Borel set A C (b,b0]; x0 G H ; f G L2(H ; a,b) and f = 0 on (b,b0].
We construct the continuous measure po (i.e., a measure without single-point atoms) from the measure p in the following way. We set p0((tfc}) = 0 for tk G Sp and we set p0(A) = p(A) for all Borel sets such that A nSp = 0. The measure p0 is self-adjoint. Replace p by p0 in (4) to obtain the equation
t t t y(t) = X0 — ijj dp0 (s)y(s) — UàJ y(s)dp(s) — IJ j f (s)dp(s). (5)
a a a
By Corollary 1, it follows that equations (4), (5) have unique solutions. Denote by W the operator solution of the equation
t t W (t, À)x0 = x0 — iJ dp0(s)W (s, À)x0 — iJÀ W (s, À)x0 dp(s),
where x0 G H. In Lemma 1 we take pi = p0 + Àp, p2 = po + Àp, q = p, f = g = 0, y(t) = W (t, À)x0, z(t) = W (t, À)z0, x0,z0 G H. Since the measure p0 is self-adjoint and the equality Sp0 = 0 holds, we obtain
(iJW (c2,À)x0,W (C2,À)Z0) — (iJW (ci,À)x0,W (ci ,À)Z0) = 0
for all c1, c2 (a ^ c1 ^ c2 ^ b0). In this equality we take c2 = t, c1 = a. Then we get
W *(t,À)JW (t,À) = J. (6)
The functions t ^ W(t,X) and t ^ W-1(t, 26 X) = JW*(t,X)J are continuous with respect to the uniform operator topology. Consequently, there exist constants a > 0, 3 > 0 such that the inequality
a ||x||2 ^ \\W(t,X)x\\2 ^ 3 ||x||2 (7)
holds for all x E H, t E [a,b0], X E C C C (C is a compact set). The function X ^ W(t, X) is holomorphic for any fixed t.
Lemma 2. [7,8] The function y is a solution of the equation (5) if and only if y has the form
t
y(t) = W(t, X)xo - W(t, X)iJ JW*(s,X)f (s)dv(s),
a
where x0 E H, a ^ t ^ b0.
3. Linear operators generated by the integral equation. In
this section, we introduce a minimal operator L0 generated by equations (4), (1) and give a description of the adjoint operator L*. Further the following notation is used: D(A) is the domain of an operator A, R(A) is the range of A. Since all considered operators are linear, we shall often omit the word «linear».
Let L2(H, a, b0) be the space of ^-measurable functions y with values in H such that Jb° ||y(t)||2 d^(t) < x>. This space coincides with the space H = L2(H; a, b) since ^(A) = 0 for each Borel set A C (b, bo].
Let us define the minimal operator L0 in the following way. The domain D(L0) consists of functions y E H for each of which there exists a function f E H such that (4) holds with X = 0 and y satisfies the conditions
y(a)= y(bo) = y(tk) = 0, tk ESP. (8)
Then we set L0y = f. By Lemma 1, the operator L0 is symmetric. If equalities (4), (8) hold, then y eD(L0 - XE) and (L0 - XE)y = f (X E C).
We claim that if y E D(L0) then y(t) = 0 for all t E [b,b0]. Indeed, limt^b+0y(t) = 0 since y(b0) = 0. If b E Sp, then y(b) = 0. If b E Sp, then equality (8) implies y(b) = 0. Since ^(A) = 0 for each Borel set A C (b,b0], we obtain the desired assertion.
It follows from (8), that y(a) = 0. In this case y is independent of the condition a E Sp. Thus the operator L0 does not change if the
measure p is replaced by a measure pi such that pi ({a}) = Pi ({b}) = 0 and p^A) = p(A) for all Borel sets A C [a,b] \ {a,b}. Therefore, without loss of generality, it can be assumed that b0 = b, and p({a}) = p({b}) = 0 (i.e., a,b E Sp), and p is the usual Lebesque measure on [a,b]. Further we write ds instead of dp(s).
Remark 2. It is possible that D(L0) = {0}. An example is available in [7]. In this case L0 = H x H, i. e., L0 is a linear relation. (The terminology on linear relations can be found, for example, in [2]).
Lemma 3. [8] The operator L0 is closed. The function y belongs to the domain D(L0 — XE) if and only if the equalities
t
y(t) = W (t,\)ijjw *(s,X)f (s)ds,
a
Sk
y(sk) = W(sk, \)ijjW*(s,X)f (s)ds = 0
a
hold, where sk E Sp U {b}, f = (L0 — XE)y.
Corollary 2. The function f E H belongs to the range R(L0 — XE) if and only if f satisfies the condition
Sk
jw *(s,X)f (s)ds = 0 (9)
a
for all sk E Sp U {b}.
Remark 3. Condition (9) is equivalent to the following:
Sk
J W*(s,X)f(s)ds = 0, sk ESp U {a, b}. (10)
Sk-1
Further, suppose that the set Sp of single-point atoms {tk} can be arranged in the ascending order t1 <t2 < ... < tk < ... and the limit point is b. By xb denote the characteristic function of a set B.
Lemma 4. The domain D(L0) of the operator L0 is dense in H.
Proof. Suppose that there exists a function h E H such that the equality (h,z)H = 0 holds for all z E D(L0). By y denote a solution of equation (5), in which X = 0 and the function f is replaced by h. Suppose that
z E D(L0) and denote zk(t) = X[tk-i;tk]z (t0 = a, tk E Sp, k E N, N is the natural number set). It follows from Lemma 3 that zk E D(L0). We apply Lagrange's formula (2) to the functions y, h and zk, L0zk for c1 = tk-1, C2 = tk, Pi = P2 = Po, q = Then we obtain (y,Lozk)h = (h,zk)h = 0. Hence,
for each function z E D(L0). By (7), it follows that a set of functions t ^ W(t, 0)ck is closed in the space L2(H; [tk-1,tk]), where ck E H. Using corollary 2 and equality (10), we obtain that there exists ck E H such that y(t) = W(t, 0)ck (tk-1 ^ t ^ tk). Lemma 2 implies h(t) = 0 for t E [tk-1,tk]. Since k is arbitrary (k E N), we get h = 0. □
We denote Wk(t,X) = xt-1;tk)(t)W(t,X)W-1(tk-i,X), to = a, k E N. Let Wn(t, X) = (w1 (t, X),... ,wn(t, X)) be the operator one-row matrix. For fixed t, X, the operator Wn(t, X) maps Hn to H continuously; here Hn is the Cartesian product of n copies of H. It is convenient to treat elements from Hn as one-column matrices, and to assume that
where we denote £n = col(^1,..., £n) E Hn, £k E H.
Let kerk(X) be a linear space of functions t ^ wk(t,X)£k, £k E H. By (7), it follows that kerk (X) is closed in H. The spaces kerk(X) and kerj (X) are orthogonal for k = j. We denote Kn(X) = ker1(X) © ... © kern(X). Obviously, Kn(X) C Km(X) for n < m.
Lemma 5. The set UnKn(X) is dense in ker(L* — XE).
Proof. It follows from Corollary 2 and (10) that the range R(L0 — XE) consists of functions f E H orthogonal to functions of the form wk(■, X)£k, where £k E H, k E N. The equality ker(L*0 — XE) © R(L0 — XE) = H implies the desired assertion. The Lemma is proved. □
Denote the operator ^ Wn(■, X)£n (£n E Hn) by Wn(X). The operator Wn(X) maps Hn continuously and one-to-one onto Kn(X) C H. Consequently, the adjoint operator W*n(X) maps H onto Hn continuously. We find the form of the operator W* (X). For all £n E Hn, f E H, we have
n
Wn(t,X)£n = wk (t,X)£k,
b
(f, Wn(X) Uh = i(f (s),Wn(s,X)Cn)ds
b
i(W:(s,X)f (s),l)ds =(W*n(X)fJn).
Therefore,
b
Wn(X)f = W:(s,X)f (s)ds. (11)
a
So we obtain the following statement:
Lemma 6. The operator Wn(X) maps Hn continuously and one-to-one onto Kn(X). The adjoint operator W*n(X) maps H continuously onto Hn and acts by (11). Moreover, W*n(X) maps Kn(X) one-to-one onto Hn.
Lemma 7. There exist a,P > 0 such that the inequalities
nn
a £ Ak \\Tk II2 ^ \\Wn(X)Tn\\H ^ P £ Ak \\Tk\\2, T = (Ti,..,Tn) E Hn, k=1 k=1
(12)
nn
A-1\\wk\\2^ \\Wn(X)Tn\\l ^P£ A-1\\wk\\2 (13)
k=i k=1
hold for all n G N, where
tk
Ak = tk — tk-1, Wk = Jwl(s, X)wk(s, X)Tkds.
tk-1
Proof. Using (7), we get
tk
aAk \\Tk\\2 ^ \\wk(s,X)Tk\\2 ds ^ PAk \\Tk\\2 , a,P> 0.
tk-1
Therefore,
tk
a Ak 2
n n n
Ak \\Tk\\2 ^^ / \\wk (s,X)Tk\\2 ds ^ Ak \\Tk \
k=i k=i+ k=i
tk-1
This implies (12). To prove (13), use (7) to obtain
tk
ai^k \\Tk|| ^
J w*k(s, X]Wk(s, X)Tkds
tk-1
^ PA \\Tk \
where ai,Pi > 0. Hence, aiAk \\Tk\\2 ^ A-1 \\<fk\\2 ^ Pi Ak \\Tk\\2. Thus,
n n n
Ak \\Tk\\2 A--1 \\<pk\\2 ^ Pi Ak \\Tk\\2. k=i k=i k=i
Now, using (12), get (13). The Lemma is proved. □
Let H_, H+, H0 = l2(H) be linear spaces of sequences, respectively, f = {Tk}, cp = [<pk}, £ = [£k} such that the series ^^ Ak \\Tk\\2,
Er=i A _ \Wk\\ , \£k\\ converge, where Tk, , £k s H. These
spaces become Hilbert spaces if we introduce scalar products as k=1 k=1
(f,f)o = (f,p) = ), t,t sHo.
k=1
In these spaces, the norms are defined by the equalities
<x <x <x
2 J2Ak \\Tk\\2, \\f\\2+ = ^2A_i\\pk||2, \\f\\0 = Y1 \\£k\\2. k=i k=i k=i
The spaces H+, H_ can be treated as spaces with positive and negative norms with respect to H0 (see [3, ch.1], [9, ch.2]). So, H+ C H0 C H_ and a \\f\\_ ^ \\p\\0 ^ P \\f\\ + , where if s H+, a,P > 0, i.e., the space H0 is equipped with the spaces H+, H_. The "scalar product" (f,f) = (f, i)0 is defined for if s H+, f s H_. If f sH0, then (if, f)0 coincides with the scalar product in H0.
Let T C H_ be a set of sequences vanishing starting from a certain number (its own for each sequence). The set T is dense in the space H_. The operator Wn(A) is the restriction of Wn+i(A) to Hn. By W'(A) denote an operator in T, such that W(A)t = Wn(A)fn for all n s N,
k
where f = (rn, 0,...). It follows from (12) that the operator W'(A) admits an extension by continuity to the space H-. By W(A) denote the extended operator. Moreover, we denote W(t,A)r = (W(A)r)(t), where r = [rk} E H-. For a fixed t, the operator W(t,A) maps H- into H. Lemmas 5, 6 imply the following assertion.
Lemma 8. The operator W(A) maps H- continuously and one-to-one onto ker(L0 — AE). A function u belongs to ker(L* — AE) if and only if there exists r = {rk} E H- such that u(t) = (W(A)F)(t) = W(t, A)r.
The adjoint operator W*(A) maps H continuously onto H+. Let us find the form of W*(A). Suppose f E H, f E T, f = {ln) 0,...}. Then
b b (f, W*(A)f) = (W(A)f f )h = J(W(t, A)f f (t))dt =j(I W*(t, A)f (t))dt.
a a
Since W*(A)f E H+ and the set T is dense in H-, we obtain
b
W *(A)f = J W *(t,A)f (t)dt. (14)
a
Thus we obtain the following statement.
Lemma 9. The operator W*(A) maps H continuously onto H+ and acts by formula (14). Moreover, W*(A) maps ker(L* — AE) one-to-one onto H+ and kerW*(A) = R(L0 — AE).
Lemma 10. Suppose that f E H and functions Fan, Fbn are defined as
t
Fan (t) = —2-lWn(t,A)iJn J W*(sA)f (s)ds, (15)
a b
Jfm(t) = 2-1Wn(t,A)iJn JW*(s,A)f (s)ds.
t
Then Fan, F^n E D(L**) for all n E N. If the function f vanishes on [tn, b], then L**(Fan) — AFan = L0(Fbn) — AF'hn = 2-1 f. Here Jn is an operator in Hn acting by the formula JnFn = (J^1,... ,J£n).
Proof. Using (15), we get
t
Fan(t) = J2 Fk(t), Fk(t) = -2-lwk(t,A)iJ I'wk(s,A)f (s)ds.
k=\ , ^
tk-1
The function Fk is continuous on the interval [tk-i,tk) and vanishes outside this interval. The function Fk does not change in the space H if changed at one point. Therefore, without loss of generality, the function Fk can be assumed to be continuous from the left at the point tk. Then, taking into account Lemma 2, we obtain that Fk is a solution of equation (5) (in which a = tk-i and f is replaced by 2-1 f) on the segment [tk-l,tk]. In [8] it is proved that every function y E D(L0) is a solution of equality (5) in which f is replaced by g = L0y. Therefore, we can apply Lagrange's formula (2) to the functions y E D(L0), Fk for cl = tk-l, c2 = tk, q = p, pl = p2 = po. Since the measure po is continuous and the equality y(tk-l) = y(tk) = 0 holds, we obtain
tk tk J (2-lf (s) + XFk (s),y(s))ds = J (Fk(s),g(s))ds.
tk-1 tk-1
This implies that Fan E V(L*0) and L*0(Fan) - XFan = 2-lf if f (t) = 0 for t > tn. We denote
b
tin = 2-llJn [W*(t,X)f(t)dt = 2-liJnW*n(\)f; Un(t) = Wn(t,\)dn.
By Lemma 6, it follows that un E kern(A). Now the equality Fbn(t) = Un(t) + Fan(t) implies Fbn E D(L0) and LOFbn - AFbn = 2—f if f (t) = 0 for t > tn. The Lemma is proved. □
Theorem 2. A function y E H belongs to D(L0) if and only if there exists a function f E H such that
IX 1
y(t) = W(t,A)F - ^ wk (t,A)lJ wk (s,A)f(s)ds, F = [r^ EH-; (16) k=l {
in this case L0y — Ay = f. The series in (16) converges in H.
Proof. First we prove that if y has form (16), then y e D(L0). It follows from Lemma 8 that W(X)reker(L0 — \E). The function
t t
zk(t) = wk(t, X)iJ jw*k(s, X)f (s)ds = wk(t, X)iJ jw*k(s, X)f (s)ds
a tk-i
vanishes outside the interval [tk-1,tk). We denote fk(t) = X[tk-i;tk)(t)f(t). By (7), it follows that
tk
\\Zk (t)\\ ^ 3 j \\f (s)\\ ds ^ f3Af ||x[tk_i; tk )f IIh
tk-1
Therefore,
tk
\\Zk\\H = j \\Zk (t)\\2 dt ^ 3 2Ak Ixitk-i; tk)f IIH . (17)
tk-1
We denote un(t) = Y^k=1 zk(t) and claim that the sequence {un} converges in H. Indeed, using (17), we get
nn ||u l|2 IU l|2 ^ A ||„. . fll2 s- Q2,u „\Uf\\2
\\zk\\H ^ 3^ 11X[tk-i;tk)f NH ^ 32(b — a) k=1 k=1
Therefore the sequence {un} converges to some function u e H and
t
<X r.
i(t) = — ^ Wk (t, X)iJ wk(s,J)f (s)ds, \\u\\h ^ 3i \\f \\« , 3i > 0
k=l
By Lemma 10, it follows that un = 2Fan E D(L0) and L0un — \un = = Y^k=l X[tk-i;tk)f • Since the operator L0 is closed, we see that u E D(L0) and L0 u — \u = f •
Now suppose that a function y E D(L0) and L0y — \y = f. If the function y has the form (16), then the function y — y E ker(L0 — \E). According to Lemma 8, there exists £ E H- such that y — y = W(X)F. Therefore, y has form (16). The Theorem is proved. □
By standard transformations, equality (16) is reduced to the form
X 1
1-1 % - 1 • • wi
y(t) = W(t, X)j - Wk(t, X)iJ w*k(s, X)f (s)ds+
x b
+ 2-1 ^ Wk (t,X)iJ wt(s,X)f (s)ds, (18)
k=1 t tk
where j= {(k }eH-, (k = Tk - 2-1iJ j w*k (s,X)f (s)ds.
tk-1
Let J denote an operator in H- acting by the formula J{£k} = {J£k}. Taking into account the convergence of the series in (18), we write equality (18) in the form
t
y(t) = W(t,X)J - 2-1W(t,X)ijjW*(s,X)f (s)ds+
a
b
+ 2-1 W (t,X)ijj W *(s,X)f (s)ds, (19) t
where j eH-, f = L0y - Xy.
4. The description of generalized resolvents. Let A be a symmetric operator acting in a Hilbert space H and A be a selfadjoint extension of A to H, where H is a Hilbert space, H D H, and scalar products coincide in H and H. By P denote an orthogonal projection of H onto H. The function X ^ R\ defined as R\ = P(A - XE)-1 |H, ImX = 0, is called a generalized resolvent of the operator A (see, e. g., [1, ch.9])
Theorem 3. Every generalized resolvent R\ (ImX = 0) of the operator L0 is the integral operator
b
Rxf = f K (t,s,X)f (s)ds.
a
The kernel K(t, s, X) has the form
K(t, s, X) = W(t, X)(M(X) + 2-1sgn(s - t)iJ)W*(s,X), where M(X):H+ is the bounded operator such that M(X) = M*(X),
(ImX)-1Im(M(X)x,x) ^ 0 (20)
for every X (ImA = 0) and for every x e U+. The function X ^ M(X)x is holomorphic for every X e U+ in half-planes ImA = 0.
Proof. Suppose y = R\f. Then y has form (19). In this equality, X e U-is uniquely determined by f and X, ImA = 0, i.e., X = X(f,X). Indeed, if f = 0, then W(t,X)<X = R\0 = 0. It follows from^Lemma 8 that X = 0. Moreover, X depends on f linearly. Consequently X = S(X)f, where S(X): H is a linear operator for fixed X. We claim that the operator
S(X) is bounded. Indeed, if a sequence {fn} converges to zero in the space H as n ^ <, then the sequence {yn}={R\fn} converges to zero in H. Hence, the sequence {W(X)Xn} (where Xn = S(X)fn) converges to zero in H. By Lemma 8, it follows that the sequence {S(X)fn} converges to zero in the space U-. Therefore S(X) is a bounded operator.
Now we prove that X(f,X) is uniquely determined by the element W*(X)f eU+. Suppose W*(X)f = 0. Consider a function equal to the sum of the last two summands in (19). This function belongs to D(Lk — XE). Therefore, W(X)X(f,X) belongs to the range R(Rj) of the operator Rx. Hence, X(f,X) = 0. Thus, S(X)f = M(X)W*(X)f, where M(X): U+ ^U-is an everywhere defined operator. Let Wk (X) be a restriction of W* (X) to ker(L0 — XE). By Lemma 9, it follows that M(X) = S(X)(W*(X))-1. Hence M(X) is the bounded operator.
Let us prove that the function X ^ M(X)x is holomorphic for every X e U+ (ImX = 0). It follows from (19) and the holomorphicity of the function X^R\ that the function X^W(X)S(X)f is holomorphic. Using (6), we obtain that the function X ^ S(X)f is holomorphic. Now the holomorphicity of the function X ^ M(X) follows from Lemma 11. This Lemma is formulated after the proof of the Theorem. In Lemma 11 it should be taken that Bx = H, B2 = U+, B3 = U-, T^X) = W*(X), T2(X) = M(X), T3(X) = S (X). _
Note that the equality R*x = Rj implies M(X) = M*(X).
Let us prove that (20) holds. It follows from Lemma 9 that there exists a function f e H such that X = W*(X)f. Let pk : U- ^ H be the operator defined by the formula pkX = £k, where X = {£k} e U-. We denote Mk(X) = pkM(X) and
<x
z(t) = W (t,X)(M (X)X — 2-1Jx) = ^ Wk (t,X)(Mk (X)X — 2-1Jxk),
k=1
where X = W*(X)f, xk = pkX. We shall apply formula (2) to the functions
y = Ra/, z on the interval [tk-1,tk]. Using the argument from the proof of Lemma 10, we can assume that the function wk(t, A) is continuous from the left at the point tk. We note that wk(tk-1, A) = E. Hence,
y(tk) = z(ifc) = w(tk)(Mfc(A)x - 2-1iJxfc),
y(tfc_i) = w(tk)(Mk(A)£ + 2-1iJxk), z(tk_i) = w(tk)(Mk(A)£ - 2-1iJxk). Using (2), we get
¿k ¿k ¿k
(A - A)_^ y"(RA/,/ )dt -J (/, Ra/)dt) -J (Ra/,Ra/)dt+
ik-1 ik-1 ¿k-1 tk
+ J ||z(t)||2 dt = (ImA)-1Im(Mk (A)x,x).
tk-1
Therefore,
(ImA)-1Im(RA/, /)« - (Ra/, Ra/)h + ||z||J = (ImA)-1Im(M(A)x,x).
Since (ImA)-1Im(RA/,/)h - (Ra/, Ra/)h ^ 0, we see that (20) holds. □
The function A^M(A) is called characteristic function (see [13]).
Lemma 11. [6] Let B1, B2, B3 be Banach spaces. Let bounded operators Ta(A) : B1 ^ B3, 71 (A) : B ^ B2, Ti(A) : B2 ^ B3 satisfy the equality 73(A) = T2(A)T1(A) for every fixed A belonging to some neighborhood of a point Ao and suppose the range of operator T1 (Ao) coincides with B2. If the functions T1(A), T3(A) are strongly differentiable at the point A0, then the function T2(A) is strongly differentiable at A0.
5. Boundary value problems connected with generalized resolvents. To shorten the notation, we shall denote W(t, 0) = W(t), w(t, 0) = w(t), W(0)= W. We put A = 0 in formula (19). By Theorem 2, it follows that y E D(Lq) and L0y = / if and only if y has the form
t b y(t) = W(t)C - 2-1W(t)i^W*(s)/(s)ds + 2-1 W(t)^W*(s)/(s)ds,
a t
where X e U-. Each function y e D(L0) represented by (21) is associated with a pair of boundary values {Y,Y'} e U- x U+, where
b
Y = T1y = X Y' = T2y = W *f = j W *(s)f (s)ds.
a
Let r denote the operator that takes each y eD(L*) to the ordered pair {Y, Y'}, i.e., Ty = {T1y, T2y}.
Theorem 4. The range R(T) of the operator r coincides with U- x U+ and "the Green formula"
(L*y, z)h — (y, L*z)h = (Y', Z) — (Y, Z') (22)
holds, where y,z e V(L*k), Ty = {Y, Y'}, Tz = {Z, Z'}.
Proof. The equality R(r) = U- x U+ follows from Lemmas 8, 9. Let us prove (22). Suppose that the function y has form (21) and
t b z(t) = W (t)X — 2-1W (t)ijj W* (s)g(s)ds + 2-1W (t)uj W * (s)g(s)ds,
at
(23)
where x e U-, g = L*z. Then
(f, WX) = (W* f, X) = (Y', Z); (Wl g) = ((,W*g) = (Y, Z'). (24) In (21), we denote
t
T* /
Xa(t) = —2-1W (t)iX W* (s)f (s)ds = —J2 2-1Wk (t)iJ w*k (s)f (s)ds,
a k=1 a
b _ b
<
Fb(t) = 2-1W(t)U W *(s)f (s)ds = ^ 2-1Wk(t)iJ w*k(s)f (s)ds.
t k=1 t
We introduce the similar notation Ga, Gb for equality (23) by changing f to g. We define functions Fk, Gk by formulas
t
Fk (t) = —2-1wk (t)ujwk (s)f (s)ds,
tk-1
t
Gk (t) = -2_V (t)Jw*(s)g(s)ds.
ifc-i
Also, as in the proof of Lemma 10, it can be assumed, without loss of generality, that the functions Fk, Gk are continuous from the left at the point tk. Arguing as in proof of Lemma 10, we apply Lagrange's formula (2) to the functions Fk, 2-1/ and Gk, 2_1g on the segment [tk-i, tk]. Taking into account (6), we obtain
tk tk
j (2-1/(s),Gk (s))ds -J (Fk (s), 2_1g(s))ds =
tk-i tk-1
tk tk
4_1( iJW(tk)iJ IW*(s)/(s)ds, W(tk)iJ IW*(s)g(s)ds
Therefore,
tk -1 tk-1 tk tk
= W *(s)/(s)ds^ W *(s)g(s)ds
tk-1 tk -1
(2_7,Go)« - (Fo, 2-1g)H = 4_1(iJW/, W*g). (25)
We denote u(t) = 2-1W(t)iJW*/, v(t) = 2_1W(t)iJW*g. By Lemma 8, it follows that u, v Gker(L0) and Fb(t) = u(t) + F0(t), <P6(t) = v(t) + G0(t). Using (25), we get
(2-1/, g6)h - (Fft, 2-1g)H = (2-1/, g«)H - (Fo, 2_^)« + (2-1/, v)a-
- (u, 2-1g)H = 4_1(iJW/, W*g) - 4_1(iJWW*g)-
- 4_1(iJW*/, W*g) = -4_1(iJW*/, W*g). (26) From (24), (25), (26), we obtain (22). The Theorem is proved. □
We introduce operators : H_ ^ H0, : H+ ^ Ho by the formulas = (A^Vk}, i+p = (A_1/2^k}, where P = (rk} G H_, p = (pk} G H+. The operator (#+) maps continuously and one-to-one H_ onto H0 (H+ onto H0, respectively). Suppose that y G D(L0). We put Y = Y1y = Y' = Y2y = and Yy = {71y,72y}. Then
R(Y) = Ho x Ho. Using (22), we get
(L0y,z)h - (y,L0z)H = (Y',Z) - (Y,Z'), (27)
t
where y,z E V(L*), yv = {Y, Y'}, Yz = {Z, Z'}.
It follows from (27) that the ordered triple (H0,Yi,Y2) is the space of boundary values (a boundary triplet in another terminology) for the operator L0 in the sense of papers [11], [5] (see also [9], [12]). We consider the boundary value problem
L0y = Xy + h, (K(X) - E)Y' - i(K(X) + E)Y = 0, (28)
where {Y,Y'} = YV; h E H; X^K(X) is a holomorphic operator function in H0 such that \\K(X)|| ^ 1; lmX > 0.
From [5] and (27) we obtain the following statement.
Theorem 5. There exists a one-to-one mapping between boundary problems (28) and generalized resolvents of the operator L0. For any solution y of problem (28), a function R\ defined by the equality y = R\h is a generalized resolvent and, conversely, for any generalized resolvent R\ there exists a function K(X) such that y = R\h is the solution of (28).
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Received April 04, 2018.
In revised form, August 08, 2018.
Accepted August 11, 2018.
Published online August 31, 2018.
Saratov State Technical University
77, Politehnicheskaja str., Saratov 410054, Russia
E-mail: [email protected]
