Positive Solutions of a Class of Operator Equations
Positive solutions of a class of matrix equations were studied by Bhatia, et al., Bull. London Math. Soc., 32, 214 (2000), SIAM J. Matrix Anal. Appl., 14, 132 (1993) and 27, 103–114 (2005), by Kwong, Linear Algebra Appl., 108, 177–197 (1988), and by Cvetkovi? and Milovanovi?, [Linear Algebra Appl....
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| author | Cvetković, A. S. Milovanović, G. V. Stanić, M. P. Цветковіч, А. С. Міловановіч, Г. В. Станіч, М. П. |
| author_facet | Cvetković, A. S. Milovanović, G. V. Stanić, M. P. Цветковіч, А. С. Міловановіч, Г. В. Станіч, М. П. |
| author_sort | Cvetković, A. S. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:47:54Z |
| description |
Positive solutions of a class of matrix equations were studied by Bhatia, et al., Bull. London Math. Soc., 32, 214 (2000), SIAM J. Matrix Anal. Appl., 14, 132 (1993) and 27, 103–114 (2005), by Kwong, Linear Algebra Appl., 108, 177–197 (1988), and by Cvetkovi? and Milovanovi?, [Linear Algebra Appl., 429, 2401–2414 (2008)]. Following the idea used in the last paper, we study a class of operator equations in infinite-dimensional spaces and prove that the positivity of solutions can be established for this class of equations under the condition that a certain rational function is positive semidefinite. |
| first_indexed | 2026-03-24T02:16:20Z |
| format | Article |
| fulltext |
UDC 517.9
A. S. Cvetković (Univ. Belgrade, Serbia),
G. V. Milovanović (Serb. Acad. Sci. and Arts, Belgrade and State Univ. Novi Pazar, Serbia),
M. P. Stanić (Univ. Kragujevac, Serbia)
POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS*
ДОДАТНI РОЗВ’ЯЗКИ ДЕЯКОГО КЛАСУ ОПЕРАТОРНИХ РIВНЯНЬ
The positive solutions of certain class of matrix equations have been recently considered by Bhatia et al. [Bull. London
Math. Soc. – 2000. – 32. – P. 214 – 228], [SIAM J. Matrix Anal. and Appl. – 1993. – 14. – P. 132 – 136; 2005. – 27. –
P. 103 – 114], Kwong [Linear Algebra and Appl. – 1988. – 108. – P. 177 – 197] and Cvetković and Milovanović [Linear
Algebra and Appl. – 2008. – 429. – P. 2401 – 2414]. Following the idea used in the last paper, we study a class of operator
equations in infinite-dimensional spaces for which we prove that the positivity of a solution can be established provided
that a certain rational function is positive semidefinite.
Додатнi розв’язки деякого класу матричних рiвнянь було нещодавно вивчено в роботах Бхатiа та iн. [Bull. London
Math. Soc. – 2000. – 32. – P. 214 – 228], [SIAM J. Matrix Anal. and Appl. – 1993. – 14. – P. 132 – 136; 2005. – 27. –
P. 103 – 114], Квонга [Linear Algebra and Appl. – 1988. – 108. – P. 177 – 197] та Цветковича та Мiловановича [Linear
Algebra and Appl. – 2008. – 429. – P. 2401 – 2414]. З використанням iдеї, запропонованої в останнiй роботi, вивчено
клас операторних рiвнянь в нескiнченновимiрних просторах, для якого доведено, що додатнiсть розв’язку можна
встановити за умови, що деяка рацiональна функцiя є позитивно напiввизначеною.
1. Introduction and preliminaries. Matrix equations appear in several fields of mathematics, e.g.,
linear algebra, differential equations, numerical analysis, optimization theory, etc. (cf. [1, 9, 11]).
Also, these equations play important roles in many applications in system theory, e.g., stability
analysis and optimal control (cf. [10, 20]), observer design [8], as well as in other computational
sciences and engineering.
In a survey paper, Lancaster [18] reviewed the existence and uniqueness results, as well as the
methods for obtaining explicit representations for a solution X for matrix equations of the form
p∑
k=1
AkXCk = B, (1.1)
with Ak, Ck, and B being known matrices not necessarily square. A special case of (1.1)
AX +XC = B (1.2)
is known as the Sylvester equation (cf. [10], Chapter 9). A further important special case is obtained
by putting C = A∗, where A∗ is the conjugate transpose of A (or C = AT in the real case). Such an
equation
AX +XA∗ = B (1.3)
is the well-known Lyapunov equation, which has been studied extensively. The equation (1.3) has a
great deal with the analysis of the stability of motion (cf. [10, 20]).
In a recent paper [3], Bhatia and Drisi have considered the following matrix equations:
* The authors were supported in part by the Serbian Ministry of Education, Science and Technological Development
(grant numbers #174015 and #44006).
c© A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ, 2015
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2 245
246 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
AX +XA = B,
A2X +XA2 + tAXA = B,
A3X +XA3 + t(A2XA+AXA2) = B,
A4X +XA4 + t(A3XA+AXA3) + 6A2XA2 = B,
A4X +XA4 + 4(A3XA+AXA3) + tA2XA2 = B,
(1.4)
where A is a given positive definite matrix and matrix B is positive semidefinite. The first equation
in (1.4) has the form (1.2), with C = A. The second equation has been studied by Kwong in [17],
where he gave proof of the existence of the positive semidefinite solution. In [3] (see also [2, 4])
necessary and sufficient conditions for the parameter t were given in order that the previous equations
have positive semidefinite solutions, provided that B is positive semidefinite matrix. There is also
a strong connection between the question of positive semidefinite solutions of these equations and
various inequalities involving unitarily equivalent matrix norms (see [2, 12, 13, 15, 16]).
Cvetković and Milovanović [7] have considered the existence of positive semidefinite solutions
of a general matrix equation of the following form:
m∑
ν=0
aνA
m−νXAν = B, (1.5)
where A is a positive definite matrix, B is a positive semidefinite matrix, aν = am−ν ∈ R, ν =
= 0, 1, . . . ,m, and a0 = am > 0.
For problems connected with differential equations where it is necessary to consider the operator
equations similar to those of the form (1.1), Daleckii and Krein in [9] (§ 3) considered general
equations of the form
n∑
j,k=0
cj,kA
jXBk = Y,
where B is a bounded linear operator on a certain Banach space B1, A is a bounded linear operator
on a certain Banach space B2, and operator Y as well as the unknown operator X are bounded linear
operators from space B1 to B2. They gave the conditions under which there exists a unique solution
of such an operator equation (see [9], Theorem 3.2).
In this paper we continue with ideas presented in [7] and study a certain class of operator equations
on infinite dimensional spaces.
Let V be a separable Hilbert space (for example `2) and let B(V ) be the space of bounded linear
operators on V. We consider the following operator equation:
p∑
k=0
akA
kXAp−k = B, a0, a1, . . . , an ∈ R, A,B ∈ B(V ), (1.6)
with symmetry ak = ap−k, k = 0, 1, . . . , p, and a0 = ap > 0. Our aim is to give conditions which
ensure that there exists a unique positive solution of the equation (1.6).
To express results easier, we introduce the polynomial
qp(x, y) =
p∑
k=0
akx
kyp−k.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 247
We note that qp is a homogeneous polynomial of degree p, for which we require that the coefficients
ak, k = 0, 1, . . . , p, be such that qp(x, y) > 0, x, y > 0.
We want to use an information from the spectrum of the operator A to solve this equation. We
assume that A and B are symmetric, and that A is strictly positive ((Ax, x) > 0, x 6= 0) and compact
operator. In this settings B has to be compact, if we are looking for the continuous solution X,
otherwise it need not. We assume that spectral resolutions of the linear operators A and B, provided
B is compact, are given by
A =
∑
k∈N
λAk P
A
k , B =
∑
k∈N
λBk P
B
k ,
where PAk and PBk are orthogonal projections onto the eigenspaces corresponding to the eigenvalue
λAk of the operator A and to the eigenvalue λBk of the operator B, respectively.
For the brevity we introduce the notation qpk,` = qp(λAk , λ
A
` ).
Also, we denote the identity operator simply by 1, which will not lead to confusion since it will
be clear from the context when 1 denotes the identity operator and when it denotes the number.
Solution of the equation (1.6) need not be compact, even when both of operators A and B are
compact. For example,
p∑
k=0
AkXAp−k = (p+ 1)Ap
has the solution X = 1 which is not compact.
The paper is organized as follows. In Section 2 we present some auxiliary results. The main
results on the positive solution of the operator equation (1.6), as well as two examples are given in
Section 3.
2. Auxiliary results. The following result is well-known, but we present the proof for the sake
of completeness.
Lemma 2.1. Let Pk, k ∈ N, be orthogonal projections with the properties
PkP` = 0, k 6= `, k, ` ∈ N and
(∑
k
Pk
)
x→ x, x ∈ V.
If X is continuous, then for every x ∈ V and every ε > 0 there exists n0, such that for every
n,m > n0 ∥∥∥∥∥
(
X −
m∑
k=1
n∑
`=1
P`XPk
)
x
∥∥∥∥∥ < ε.
In other words,
∑
k,`
P`XPk converges to X strongly.
Proof. We obtain result easily. If X = 0 the statement is trivial. So, we assume that X 6= 0.
First, due to property PkP` = 0, k 6= ` k, ` ∈ N, we know that
∑n
k=1
Pk is an orthogonal projection.
Fix ε > 0, then there exists n1 ∈ N such that for n > n1 we have∥∥∥∥∥
(
n∑
k=1
Pk
)
x− x
∥∥∥∥∥ < ε
6‖X‖
,
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
248 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
and there exists n2 ∈ N such that for n > n2 we obtain∥∥∥∥∥
(
n∑
k=1
Pk
)
Xx−Xx
∥∥∥∥∥ < ε
2
.
Let n0 = max {n1, n2}. Then for n,m > n0 we get∥∥∥∥∥Xx−
m∑
k=1
n∑
`=1
PkXP`x
∥∥∥∥∥ =
∥∥∥∥∥Xx−
n∑
`=1
XP`x+
n∑
`=1
XP`x−
m∑
k=1
n∑
`=1
PkXP`x
∥∥∥∥∥ ≤
≤
∥∥∥∥∥X
(
1−
n∑
`=1
P`
)
x
∥∥∥∥∥+
+
∥∥∥∥∥
(
1−
m∑
k=1
Pk
)
X
n∑
`=1
P`x−
(
1−
m∑
k=1
Pk
)
Xx+
(
1−
m∑
k=1
Pk
)
Xx
∥∥∥∥∥ ≤
≤ ‖X‖
∥∥∥∥∥
(
1−
n∑
`=1
P`
)
x
∥∥∥∥∥+
∥∥∥∥∥
(
1−
m∑
k=1
Pk
)
X
(
1−
n∑
`=1
P`
)
x
∥∥∥∥∥+
∥∥∥∥∥
(
1−
m∑
k=1
Pk
)
Xx
∥∥∥∥∥ ≤
≤ ‖X‖
∥∥∥∥∥
(
1−
n∑
`=1
P`
)
x
∥∥∥∥∥+
(
‖1‖+
∥∥∥∥∥
m∑
k=1
Pk
∥∥∥∥∥
)
‖X‖
∥∥∥∥∥
(
1−
n∑
`=1
P`
)
x
∥∥∥∥∥+
+
∥∥∥∥∥
(
1−
m∑
k=1
Pk
)
Xx
∥∥∥∥∥ ≤
≤ 3‖X‖
∥∥∥∥∥
(
1−
n∑
`=1
P`
)
x
∥∥∥∥∥+
∥∥∥∥∥
(
1−
m∑
k=1
Pk
)
Xx
∥∥∥∥∥ < ε.
This clearly proves the statement. We used the fact that
∥∥∥∑n
k=1
Pk
∥∥∥ = 1, since it is orthogonal
projection.
Lemma 2.2. Assume X is the continuous solution of (1.6), then
PA` XP
A
k = (1/qp`,k)P
A
` BP
A
k , k, ` ∈ N.
If B = 0, then X = 0 is the unique continuous solution of (1.6). If (1.6) has a continuous
solution, then it is unique solution in the set of continuous solutions.
Proof. Since PAk A = APAk = λAk P
A
k , k ∈ N, we have
PA` BP
A
k = PA`
p∑
ν=0
aνA
νXAp−νPAk =
p∑
ν=0
aν(λA` )ν(λAk )p−νPA` XP
A
k =
= qp(λA` , λ
A
k )PA` XP
A
k .
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 249
Conclusion holds, since the operator A is strictly positive.
It is obvious that X = 0 is a solution of the homogeneous equation
p∑
k=0
akA
kXAp−k = 0.
But, for any k, ` ∈ N, we get PA` 0PAk = 0, hence PA` XP
A
k = 0, for any continuous solution X. For
any x ∈ V, we obtain Xx =
∑
k,`∈N
0x = 0, hence, X = 0 is the unique solution.
If X1 and X2 are two continuous solutions of (1.6), then X1 − X2 is the solution of the
homogeneous equation. Hence, X1 −X2 = 0 which proves our statement.
Theorem 2.1. Let B be compact. The series∑
k,`∈N
1
qp`,k
PA` BP
A
k
converges strongly if and only if equation (1.6) has continuous solution.
Proof. According to [19, p. 166], if the sequence of linear operators converges, then it strongly
converges to a bounded linear operator. Hence, strongly convergent series
∑
k,`∈N
1
qp`,k
PA` BP
A
k
converges to some continuous X. To prove that X is a solution of (1.6), we note
p∑
ν=0
aνA
νXAp−ν =
p∑
ν=0
aνA
ν
∑
k,`∈N
1
qpk,`
PA` BP
A
k
Ap−ν =
=
∑
k,`∈N
1
qpk,`
p∑
ν=0
aνA
νPA` BP
A
k A
p−ν =
=
∑
k,`∈N
1
qpk,`
(
p∑
ν=0
aν(λA` )ν(λA` )p−ν
)
PA` BP
A
k =
∑
k,`∈N
PA` BP
A
k = B.
If X is a continuous solution of (1.6), then the series∑
k,`∈N
PA` XP
A
k =
∑
k,`∈N
1
qp`,k
PA` BP
A
k ,
converges strongly to X.
Example 2.1. Let us illustrate the previous discussion using an example. Consider the case
PAk = PBk = (·, ek)ek, k ∈ N, p = 1, q1(x, y) = x+ y and λAk = (λBk )2/2 = 1/(2k2), k ∈ N, where
{ek}k∈N is the Hilbert basis of V. Then
1
λA` + λAk
PA` BP
A
k =
δk,`
λBk
PBk = δ`,kkP
B
k , `, k ∈ N. (2.1)
Consequently, for x =
∑
m∈N
(1/m)em, we have
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
250 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
1
λA` + λAk
PA` BP
A
k x = δ`,kk(x, ek)ek = δ`,kek.
Hence, the series
n∑
k=1
m∑
`=1
1
λAk + λA`
PA` BP
A
k x =
min{m,n}∑
k=1
ek
is not convergent. This means that even if B is strictly positive and compact, a continuous solution
need not exist (Theorem 2.1).
There is a special case in which we can claim that a continuous solution exists.
Theorem 2.2. Let B = CAp, C ∈ B(V ), and AC = CA. Then the equation (1.6) has the
solution X = 1/qp(1, 1)C.
Proof. For x ∈ V, we have
n∑
k=1
m∑
`=1
1
qpk,`
PAk CA
pPA` x = C
n∑
k=1
m∑
`=1
(λAk )p
qp(λAk , λ
A
` )
PAk P
A
` x =
= C
min{m,n}∑
k=1
(λAk )p
qp(λAk , λ
A
k )
PAk x =
1
qp(1, 1)
C
min{m,n}∑
k=1
PAk x→
1
qp(1, 1)
Cx,
where we use the fact that AC = CA implies PAC = CPA, where PA is any spectral projection of
A (see [5, p. 150]).
In the sequel we are dealing with the unbounded solutions of (1.6). We denote
V A,n
0 =
n⊕
k=1
PAk V, n ∈ N, V A
0 =
⋃
n∈N
V A,n
0 .
Note that V A
0 is a linear space. Every x ∈ V A
0 is a linear combination of eigenvectors of the
operator A. In what follows we assume that B is a bounded operator, not necessarily compact. As a
consequence, we are searching for unbounded solutions of (1.6).
Lemma 2.3. Suppose x ∈ V A
0 , then the series
∑
k,`∈N
1
qp`,k
PA` BP
A
k x converges.
Proof. Since x ∈ V A
0 , by definition of V A
0 , there exists some h ∈ N such that x ∈ V A,h
0 . Hence,(∑h
k=1
PAk
)
x = x and PAk x = 0, k > h. For m > h we have
n∑
`=1
m∑
k=1
1
qp`,k
PA` BP
A
k x =
h∑
k=1
(
n∑
`=1
1
qp`,k
PA`
)
(BPAk x)→
h∑
k=1
(qp(λAk , A))−1BPAk x,
as n,m→ +∞, where we used the fact that (qp(λAk , ·))−1 :σ(A)→ R, k = 1, . . . , h, is bounded on
the spectrum of A.
Lemma 2.4. The closure of V A
0 coincides with V.
Proof. Take any x ∈ V, then V A
0 3
∑n
k=1
PAk x→ x, as n→ +∞.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 251
Lemma 2.5. Let X0 :V A
0 → V be the operator defined by
X0x =
∑
k,`∈N
1
qp`,k
PA` BP
A
k x.
Then X0 is a linear operator densely defined on V. Operator X0 is symmetric and closable.
For any x ∈ V A
0 , we have
p∑
ν=0
aνA
νX0A
p−νx = Bx.
Proof. According to Lemma 2.3, we see that X0 is well defined. If x, y ∈ V A
0 and if α, β are
scalars, then αx+ βy ∈ V A
0 , and according to
X0(αx+ βy) =
∑
k,`∈N
1
qp`,k
PA` BP
A
k (αx+ βy) =
= α
∑
k,`∈N
1
qp`,k
PA` BP
A
k x+ β
∑
k,`∈N
1
qp`,k
PA` BP
A
k y = αX0x+ βX0y,
we infer that X0 is a linear operator. Hence, according to Lemma 2.4, X0 is a densely defined linear
operator on V.
We prove that X0 is symmetric. Let x, y ∈ V A
0 , then
(X0x, y) = lim
n→+∞
n∑
k,`=1
1
qpk,`
PAk BP
A
` x, y
=
= lim
n→+∞
x, n∑
k,`=1
1
qpk,`
PAk BP
A
` y
= (x,X0y).
We prove that X0 is closable. According to [5, p. 66], we have to prove that if xk ∈ D(X0),
limxk = 0 and limX0xk = y, then y = 0.
Let z ∈ V A
0 be arbitrary. Then we get
(y, z) = lim
n→+∞
(X0xn, z) = lim
n→+∞
lim
m→+∞
m∑
k,`=1
1
qpk,`
PAk BP
A
` xn, z
=
= lim
n→+∞
xn, lim
m→+∞
m∑
k,`=1
1
qpk,`
PAk BP
A
` z
=
= lim
n→+∞
(xn, X0z) = (0, X0z) = 0.
Since V A
0 is dense in V, we get y = 0. Thus, we conclude that X0 is closable.
Choose now an arbitrary x ∈ V A,n
0 . Then Aνx ∈ V A,n
0 , ν = 0, 1, . . . , p. We conclude that the
left-hand side of (1.6) is well defined and we have
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
252 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
p∑
ν=0
aνA
νX0A
p−νx =
p∑
ν=0
aνA
ν
∑
k,`∈N
1
qpk,`
PA` BP
A
k
Ap−νx =
=
∑
k,`∈N
PA` BP
A
k x = Bx.
Definition 2.1. Operator X is the minimal closed extension of X0. We call X the solution of the
equation (1.6).
Trivially X is symmetric, as being closure of a symmetric operator X0. We call X the solution,
despite the fact, that we can claim that equation (1.6) is valid only on V A
0 .
There is a special case in which we can give some stronger results.
Lemma 2.6. Let B(V A
0 ) ⊂ V A
0 . The solution X is self-adjoint.
Proof. If B satisfies the mentioned condition, then clearly solution of the equation (1.6), on the
set V A
0 , can be given by
Xx =
n∑
k,`=1
1
qpk,`
PAk BP
A
` x,
where n = max {m1,m2} and m1 is such that x ∈ V A,m1
0 and m2 is such that BPA` x ∈ V
A,m2
0 ,
` = 1, . . . ,m1. Since X is closure of X0, we know that X �V A
0
= X0 and (X − λ) �V A
0
= X0 − λ,
where A �V A
0
denotes restriction of operator A to V A
0 .
Let λ ∈ C\R be arbitrary and choose x ∈ ker (X0 − λ). Then, we have λ(x, x) = (X0x, x) =
= (x,X0x) = λ(x, x). It follows that x = 0 and ker (X0 − λ) = {0} for every λ ∈ C\R. Let us fix
λ ∈ C \R and let x ∈ rang (X0 − λ)⊥ ∩ V A
0 . For y ∈ V A
0 arbitrary, we have 0 = ((X0 − λ)y, x) =
= (y, (X0 − λ)x). We conclude x ∈ ker (X0 − λ) = {0}. Hence, it must be rang (X0 − λ) = V A
0
for any λ ∈ C\R.
Let X∗ denote the adjoint of X. Since X is densely defined and closed, we know that there exists
X∗, which is closed and densely defined. Also (see [5, p. 70])
rang (X − λ)
⊕
ker (X∗ − λ) = V.
Since X is symmetric on D(X), we get X ⊂ X∗ (see [5, p. 97]). For λ ∈ C\R, we conclude that
ker (X∗ − λ) = (rang (X − λ))⊥ ⊂ (rang (X0 − λ))⊥ = (V A
0 )⊥ = {0}.
According to von Neumann’s formulae (see [5, p. 106]), we know that
D(X∗) = D(X)+̇ ker (X∗ − λ)+̇ ker (X∗ − λ),
where λ ∈ C\R is arbitrary. We conclude directly that D(X∗) = D(X), which gives X = X∗ and
X is self-adjoint.
According to the previous lemmas we are ready to formulate the following statement.
Theorem 2.3. Let A and B be symmetric, and let A be strictly positive and compact and B
bounded. There exists symmetric and closed X such that the equation (1.6) is valid on V A
0 . Moreover,
if B(V A
0 ) ⊂ V A
0 , then X is self-adjoint.
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POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 253
It is interesting to give an interpretation of the example given in (2.1). It can be easily seen that
X =
∑
k∈N
k(·, ek)ek, is actually, spectral resolution of the self-adjoint operator X.
Another obvious interpretation of the result is for the case B = 2. Then, X is the solution of
the equation AX + XA = 2. Due to symmetry of A and ker (A) = {0}, we know that A does not
have residual spectrum. Consequently, the range of A has to be dense in V and there must exist the
self-adjoint inverse of A.
3. Positive solutions. We denote by 2N and 2N− 1 the sets of even and odd positive integers.
Consider now the linear operators Ik :PAk V → Crang (PA
k ), k ∈ N, defined on some orthonormal
basis Hk = {ek,1, . . . , ek,rang (PA
k )}, k ∈ N, by Ikek,` = f`, ` = 1, . . . , rang (PAk ), k ∈ N, where
{f1, . . . , frang (PA
k )} is the natural basis of Crang (PA
k ), and respective direct sum In0 = ⊕nk=1Ik. It is
trivial fact that In0 :V A,n
0 → Cdim(V A,n
0 ) is an isometrical isomorphism.
In what follows we adopt the following definitions:
An operator A is nonnegative, positive, strictly positive if and only if for all x ∈ D(A) we have
(Ax, x) ≥ 0, (Ax, x) ≥ 0 and A 6= 0, (Ax, x) > 0, respectively.
A matrix A is positive definite, positive semidefinite if and only if (Ax, x) > 0, (Ax, x) ≥ 0,
respectively.
A function f :R → C is positive definite if and only if for all n ∈ N and any given points xk,
k = 1, . . . , n, the matrix ‖f(xk − x`)‖nk,`=1, is positive semidefinite.
Lemma 3.1. If the function x 7→ ϕp(x), given by
1
ϕp(x)
=
p/2−1∑
ν=0
aν cosh
(p
2
− ν
)
x+
1
2
ap/2, p ∈ 2N,
(p−1)/2∑
ν=0
aν cosh
(p
2
− ν
)
x, p ∈ 2N− 1,
is positive definite, then the linear operator Cn :V A,n
0 → V A,n
0 , defined by
Cnx =
n∑
k,`=1
1
qp`,k
(PAk x, (I
n
0 )−11v)P
A
` (In0 )−11v, (3.1)
where 1v = (1, 1, 1, . . . , 1) ∈ Cdim(V A,n
0 ), is positive and the matrix In0Cn(In0 )−1 is positive semide-
finite.
Proof. It is easy to prove that Cn is a symmetric linear operator. For x ∈ V A,n
0 , we have
(Cnx, x) =
n∑
k,`=1
1
qp`,k
(PAk x, (I
n
0 )−11v)P
A
` (In0 )−11v, x
=
=
n∑
k,`=1
1
qp`,k
(PAk x, (I
n
0 )−11v)(P
A
` (In0 )−11v, x) =
=
n∑
k,`=1
1
qp`,k
(PAk x, (I
n
0 )−11v)((I
n
0 )−11v, P
A
` x) =
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254 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
=
n∑
k,`=1
1
qp`,k
(PAk x, (I
n
0 )−11v)(PA` x, (I
n
0 )−11v).
Accordingly, we note that if matrix Dn = ‖1/qp`,k‖
n
`,k=1
is positive semidefinite, the operator Cn is
positive. To prove semidefiniteness of Dn we use the same arguments as in [7]. Since λk, k ∈ N, is
a positive sequence, we can represent it as λk = exk , xk ∈ R, k ∈ N. Then we conclude that
∥∥∥∥ 1
qp(λA` , λ
A
k )
∥∥∥∥ = diag (epx`/2)
∥∥∥∥∥∥ 1∑p
ν=0
aνe
(p/2−ν)(x`−xk)
∥∥∥∥∥∥diag (epxk/2) =
= Z‖ϕp(x` − xk)‖Z∗.
We recognize that the matrix Dn is congruent with the matrix
En = ||ϕp(x` − xk)||,
hence, positive semidefiniteness of Dn and En are equivalent. The matrix Z is simply diagonal
matrix with the positive entries 1/
√
2epxk/2, k ∈ N. According to the condition of this lemma, the
matrix En is positive semidefinite, hence, matrix Dn is positive semidefinite, and Cn is positive.
Positive semidefiniteness of the matrix In0Cn(In0 )−1 is a consequence of a positivity of the operator
Cn, since, for x ∈ Cdim(V A,n
0 ), we get
(In0Cn(In0 )−1x, x) = (Cn(In0 )−1x, (In0 )−1x) = (Cn((In0 )−1x), (In0 )−1x).
Lemma 3.2. Let B be positive operator and
Bn =
n∑
`=1
m∑
k=1
PA` BP
A
k , n ∈ N.
Then {Bn} is a sequence of the nonnegative linear operators, B = limBn, and there exists n0 ∈ N
such that for all n ≥ n0 the operator Bn is positive.
Proof. According to Lemma 2.1 and continuity of B, a sequence of the linear operators
Bn =
n∑
`=1
n∑
k=1
PA` BP
A
k =
(
n∑
`=1
PA`
)
B
(
n∑
k=1
PAk
)
, n ∈ N,
converges to B. Even more, we see that Bn, n ∈ N, is the sequence of nonnegative operators, since
for every x ∈ V A,n
0 , we have
(Bnx, x) =
((
n∑
`=1
PA`
)
B
(
n∑
k=1
PAk
)
x, x
)
=
(
B
(
n∑
k=1
PAk
)
x,
(
n∑
k=1
PAk
)
x
)
≥ 0.
Since 0 6= B = limBn it follows that there exists n0 ∈ N such that Bn 6= 0 for all n ≥ n0.
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POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 255
Theorem 3.1. Let us assume that operator B is positive and that the function x 7→ ϕp(x), given
by
1
ϕp(x)
=
p/2−1∑
ν=0
aν cosh
(p
2
− ν
)
x+
1
2
ap/2, p ∈ 2N,
(p−1)/2∑
ν=0
aν cosh
(p
2
− ν
)
x, p ∈ 2N− 1,
is positive definite. The solution of the operator equation (1.6), described in Theorem 2.3, is positive.
Proof. Let X be a solution of (1.6). Then we have strong convergence of the sequence
Xn =
∑n
k=1
∑n
`=1
PA` XP
A
k , as n→ +∞, on V A
0 .
Let Cn be an operator defined in (3.1). The linear operators In0Bn(In0 )−1, In0Cn(In0 )−1 and
In0Xn(In0 )−1 on CdimV A,n
0 , can be represented using matrix multiplication as matrices. Even more,
the matrix In0Xn(In0 )−1 is a Schur product of the matrices In0Cn(In0 )−1 and In0Bn(In0 )−1. However,
for n ≥ n0 the matrix In0Bn(In0 )−1 is positive semidefinite, due to positivity of Bn, and
(In0Bn(In0 )−1a, a) = (Bn(In0 )−1a, (In0 )−1a) ≥ 0, a ∈ Cdim(V A,n
0 ).
The matrix In0Cn(In0 )−1 is positive semidefinite according to Lemma 3.1. Accordingly, for all n ≥ n0
the operator Xn is positive, since
0 ≤ (In0Xn(In0 )−1a, a) = (Xn(In0 )−1a, (In0 )−1a), a ∈ Cdim(V A,n
0 ).
Using this observations, we simply derive that for every x ∈ V A
0 we have
(Xx, x) = lim
n→+∞
(Xnx, x) = lim
n→+∞
n∑
k,`=1
1
qp`,k
(P`XPkx, x) =
= lim
n→+∞
n∑
k,`=1
1
qp`,k
(XPkx, P`x) = lim
n→+∞
n∑
k,`=1
1
qp`,k
(XPkπnx, P`πnx) =
= lim
n→+∞
n∑
k,`=1
1
qp`,k
(P`XPkπnx, πnx) =
= lim
n→+∞
n∑
k,`=1
1
qp`,k
P`XPkπnx, πnx
= lim
n→+∞
(Xnπnx, πnx) ≥ 0,
and X 6= 0, where we used the fact that Pkπn = Pk, k = 1, . . . , n, for πn =
∑n
k=1
Pk, which is
orthogonal projection onto V A,n
0 .
For every x ∈ D(X) there exists a sequence xn ∈ V A
0 , such that xn → x, Xxn → Xx as
n→ +∞. Therefore, since (Xxn, xn) ≥ 0, we have (Xx, x) = lim(Xxn, xn) ≥ 0.
As in [7] we can define a characteristic polynomial for the equation (1.6).
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256 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
Definition 3.1. For even p we define the characteristic polynomial Qp for the equation (1.6) to
be Qp(cosh t) = 1/ϕp(t), and for odd p we define the corresponding characteristic polynomial to be
Qp(cosh t) = 1/(cosh (t/2)ϕp(t)).
Now, we can use this characteristic polynomial to give the following statement.
Theorem 3.2. Suppose we are given the equation (1.6), with a strictly positive and compact
operator A, with the characteristic polynomial Qp which has k1 real zeros contained in the interval
[−1, 1) and k2 zeros smaller than −1, with k1 ≥ k2, for even p, and k1 + 1 ≥ k2, for odd p, where
k1 + k2 = [p/2]. Then the corresponding function ϕp is positive definite, i.e., the equation (1.6) has
a positive symmetric and closed solution, provided B is positive.
Proof. It is proved in [7] that under this condition function ϕp is positive definite. According to
Theorem 3.1, in this case we have a symmetric and closed solution.
We give now an example with integral operators acting on the space L2(0, 1). Denote by C2[0, 1]
the space of twice continuously-differentiable functions on [0, 1] and by H2[0, 1] the corresponding
space of twice differentiable functions with the second derivative being an element of L2(0, 1). In
addition, let C2
0 [0, 1] and H2
0 [0, 1] be their subspaces, with the additional conditions
f ′(0) + f ′(1) = 0 and f ′(1) = f(0) + f(1), (3.2)
respectively.
We need also C4[0, 1] as the space of four times continuously-differentiable functions on [0, 1] and
H4[0, 1] as the space of four times differentiable functions with fourth derivative being an element of
L2(0, 1). With C4
0 [0, 1] and H2
0 [0, 1] we denote their subspaces, with the additional conditions (3.2)
and
f ′′′(0) + f ′′′(1) = 0 and f ′′′(1) = f ′′(0) + f ′′(1), (3.3)
respectively.
Lemma 3.3. Let an integral operator C : L2(0, 1)→ L2(0, 1) be defined by
(Cf)(x) =
1∫
0
|x− t|f(t) dt, x ∈ [0, 1].
Then ker (C) = {0}, rang (C) = H2
0 [0, 1], rang (C) = L2(0, 1), C is compact and self-adjoint and
D2C = CD2 = 2, where D2 :C2
0 [0, 1]→ L2(0, 1) is the second derivative and C2
0 [0, 1] = L2(0, 1).
Operator A = C2 is strictly positive, self-adjoint, with ker (A) = {0}, rang (A) = H4
0 [0, 1],
rang (A) = L2(0, 1) and D4A = AD4 = 4, where D4 :C4
0 [0, 1] → L2(0, 1) is the forth derivative
and C4
0 [0, 1] = L2(0, 1).
Proof. For every x ∈ [0, 1] we have |x − t| ∈ L2(0, 1), so that C is defined everywhere on
L2(0, 1). We know that C is compact and self-adjoint since its kernel |x − t| is continuous and
symmetric (see [14, 21]). Let εn be a sequence of real numbers converging to zero. For a fixed
x ∈ [0, 1], consider the sequence of functions
gn(t) =
|x+ εn − t| − |x− t|
εn
, n ∈ N.
We have an integrable and uniform bound
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 257∣∣∣∣ |x+ εn − t| − |x− t|
εn
∣∣∣∣ ≤ |x+ εn − t− x+ t|
|εn|
= 1 ∈ L2(0, 1),
as well as the pointwise convergence
lim
n→+∞
gn(t) =
{
1, x > t,
−1, x < t,
= g(t) ∈ L2(0, 1).
Using the Lebesgue theorem on dominated convergence (see [6]), we get
lim
n→+∞
1∫
0
gn(t)f(t) dt =
1∫
0
lim
n→+∞
gn(t)f(t) dt =
=
1∫
0
sgn (x− t)f(t) dt = (Cf)′(x).
Let εn, n ∈ N, be again a sequence of real numbers converging to zero. Then
lim
n→+∞
1
εn
1∫
0
(sgn (x+ εn − t)− sgn (x− t)) f(t) dt =
= 2 lim
n→+∞
1
εn
∫
[x,x+εn]
f(t) dt = 2f(x),
for a.e. x ∈ [0, 1], according to the Lebesgue differentiation theorem (see [6]). Since
(Cf)′(0) = −
1∫
0
f(t) dt, (Cf)′(1) =
1∫
0
f(t) dt,
(Cf)(0) + (Cf)(1) =
1∫
0
f(t) dt,
we see that Cf satisfies the conditions (3.2). Hence, for every f ∈ L2(0, 1) we have Cf ∈ H2
0 [0, 1]
and 2f(x) = (Cf)′′(x), for a.e. x ∈ [0, 1]. Therefore, D2C = 2.
On the other hand, using an integration by parts, for f ∈ C2
0 [0, 1] we get
(CD2f)(x) =
1∫
0
|x− t|f ′′(t) dt = 2f(x) + x(f ′(0)− f ′(1)) + f ′(1)− (f(0) + f(1)),
and, due to the conditions (3.2), we find CD2 = 2.
If f ∈ ker (C), we have f(x) = 1/2(Cf)′′(x) = 1/2(0)′′(x) = 0, for a. e. x ∈ [0, 1]. We
conclude that ker (C) = {0}. Finally, it is a trivial fact that C2
0 [0, 1] = H2
0 [0, 1] = L2(0, 1), where
the closure is taken in L2-norm.
The statement for the operator A can be obtained in the same fashion.
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258 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
Lemma 3.4. The eigenvalues of the operator A given in Lemma 3.3 are given by λk = λ̃2k,
k ∈ N0, where λ̃0 = 2/α2
0, λ̃k = −2/ν2k , k ∈ N, α0 is the unique solution of the equation
4 + e−α(2 + α) + eα(2− α) = 0 and νk, k ∈ N, are positive solutions of the equation 2 + 2 cos ν +
+ ν sin ν = 0. The corresponding eigenvectors are
f0(x) =
1 + e−α0
1 + eα0
eα0x + e−α0x, fk(x) =
1 + cos νk
sin νk
cos νkx+ sin νkx, k ∈ N.
Proof. We first find eigenvalues of the operator C. Starting with the equation (Cf)(x) = λf(x),
by differentiating we get the following differential equation:
λf ′′(x)− 2f(x) = 0, (3.4)
with the boundary conditions (3.2). It is obvious that λ = 0 is not an eigenvalue.
For λ > 0, the solution of the differential equation (3.4) is given by f(x) = C1e
αx + C2e
−αx,
C1, C2 ∈ R, where α =
√
2/λ. Using the boundary conditions (3.2), for C1 and C2, we get the
system of linear equations
C1 − C2 + C1e
α − C2e
−α = 0,
C1 + C2 + C1e
α + C2e
−α = αC1e
α − αC2e
−α,
which determinant is given by
∆ = 4 cosh
α
2
(
2 cosh
α
2
− α sinh
α
2
)
.
The equation ∆ = 0 has the unique solution α = α0 > 0. Thus, the operator C has one eigenvalue
λ̃0 = 2/α2
0 greater than zero, and the corresponding eigenvector is
f0(x) =
1 + e−α0
1 + eα0
eα0x + e−α0x.
For λ < 0, the solution of differential equation (3.4) is given by f(x) = C1 cos νx + C2 sin νx,
C1, C2 ∈ R, where ν =
√
−2/λ. Using the boundary conditions (3.2) we get the following system
of linear equations for C1, C2:
C2 − C1 sin ν + C2 cos ν = 0,
C1 + C1 cos ν + C2 sin ν = −νC1 sin ν + νC2 cos ν.
Therefore, C1 = C2(1 + cos ν)/sin ν and, since C2 6= 0 (because we are looking for nontrivial
solutions), we get 2 cos ν/2(cos ν/2 + ν sin ν/2) = 0. It is easy to see that if cos ν/2 = 0, then
C1 = C2 = 0, and the values for ν are not eigenvalues of the operator C. Let us denote by νk,
k ∈ N, the positive solutions of the previous equations (one solution in each of intervals of the form
[kπ, (k + 1)π), k ∈ N0). Then, λ̃k = −2/ν2k , k ∈ N, are eigenvalues of the operator C, and
fk(x) =
1 + cos νk
sin νk
cos νkx+ sin νkx, k ∈ N,
are the corresponding eigenvectors.
It is easy to see that the eigenvalues of the operator A = C2 are given by λk = λ̃2k, k ∈ N0, and
that fk, k ∈ N0, are the corresponding eigenvectors (λkfk = λ̃kCfk = C(λ̃kfk) = C2fk).
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POSITIVE SOLUTION OF A CERTAIN CLASS OF OPERATOR EQUATIONS 259
Example 3.1. Consider the equation
p∑
k=0
AkXpA
p−k = 1, p = 1, (3.5)
where A is an operator given in Lemma 3.3. Since V A
0 = 1(V A
0 ), according to Lemma 2.6, the
solution X1 of the equation (3.5) is self-adjoint.
Let us denote by {ek}k∈N0 the orthonormal set of eigenvectors of the operator A. Then A =
=
∑∞
k=0
λkPλk , where Pλk , k ∈ N0, is a projection onto the eigenspace which correspond to the
eigenvector λk. Since Pλk = (·, ek)ek, k ∈ N0, we get
Af =
∞∑
k=0
λk(f, ek)ek, f ∈ L2(0, 1).
The solution of the equation (3.5) can be easily found as X1 = 1/8D4.
For p = 1 we get Q1(t) = 1, and the solution X1 is positive, according to Theorem 3.2. For
f ∈ C4
0 [0, 1], a direct computation gives
(D4f, f) =
1∫
0
(D4f)(t)f(t) dt =
= (D3f)(t)f(t)
∣∣∣1
0
− (D2f)(t)(Df)(t)
∣∣∣1
0
+
1∫
0
((D2f)(t))2 dt.
Using the boundary conditions (3.2) and (3.3) we get
(D3f)(1)f(1)− (D3f)(0)f(0)− (D2f)(1)(Df)(1) + (D2f)(0)(Df)(0) =
= (D3f)(1)((Df)(1)− f(0))− (D3f)(0)f(0)−
−(D2f)(1)(Df)(1)− (D2f)(0)(Df)(1) =
= (D3f)(1)(Df)(1)− f(0)((D3f)(1) + (D3f)(0))−
−(Df)(1)((D2f)(1) + (D2f)(0)) = 0.
Therefore,
(D4f, f) =
1∫
0
((D2f)(t))2 dt ≥ 0.
Example 3.2. Consider the equation
p∑
k=0
AkXpA
p−k = 1, p = 2, (3.6)
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260 A. S. CVETKOVIĆ, G. V. MILOVANOVIĆ, M. P. STANIĆ
where A is an operator given in Lemma 3.3. In the same way as in Example 3.1, we conclude that the
solutionX2 = 1/48D8 of the equation (3.6) is self-adjoint. Since for p = 2 we have Q2(t) = t+1/2,
the solution X2 is positive, according to Theorem 3.2.
Similarly as in Example 3.1, for f ∈ C8
0 [0, 1] = {f ∈ C8[0, 1] | f (5)(1) = f (4)(0) + f (4)(1),
f (5)(0) + f (5)(1) = 0, f (7)(1) = f (6)(0) + f (6)(1), f (7)(0) + f (7)(1) = 0} we get
(D8f, f) =
1∫
0
(D8f)(t)f(t) dt =
= (D7f)(t)f(t)
∣∣∣1
0
− (D6f)(t)(Df)(t)
∣∣∣1
0
+
1∫
0
((D4f)(t))2 dt =
1∫
0
((D4f)(t))2 dt ≥ 0.
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Received 01.12.12
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
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| id | umjimathkievua-article-1977 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:16:20Z |
| publishDate | 2015 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/b2/017b0d78fbdf95a42780558eb4aeacb2.pdf |
| spelling | umjimathkievua-article-19772019-12-05T09:47:54Z Positive Solutions of a Class of Operator Equations Додатні розв'язки деякого класу операторних рівнянь Cvetković, A. S. Milovanović, G. V. Stanić, M. P. Цветковіч, А. С. Міловановіч, Г. В. Станіч, М. П. Positive solutions of a class of matrix equations were studied by Bhatia, et al., Bull. London Math. Soc., 32, 214 (2000), SIAM J. Matrix Anal. Appl., 14, 132 (1993) and 27, 103–114 (2005), by Kwong, Linear Algebra Appl., 108, 177–197 (1988), and by Cvetkovi? and Milovanovi?, [Linear Algebra Appl., 429, 2401–2414 (2008)]. Following the idea used in the last paper, we study a class of operator equations in infinite-dimensional spaces and prove that the positivity of solutions can be established for this class of equations under the condition that a certain rational function is positive semidefinite. Додатні розв'язки деякого класу матричних рівнянь було нещодавно вивчено в роботах Бхатіа та ін. [Bull. London Math. Soc. - 2000. - 32. - P. 214-228], [SIAM J. Matrix Anal. and Appl. - 1993. - 14. - P. 132-136; 2005. - 27. -P. 103 -114], Квонга [Linear Algebra and Appl. - 1988. - 108. -P. 177-197] та Цветковича та Міловановича [Linear Algebra and Appl. - 2008. - 429. - P. 2401 -2414]. З використанням ідеї, запропонованої в останній роботі, вивчено клас операторних рівнянь в нескінченновимірних просторах, для якого доведено, що додатність розв'язку можна встановити за умови, що деяка раціональна функція є позитивно напіввизначеною. Institute of Mathematics, NAS of Ukraine 2015-02-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1977 Ukrains’kyi Matematychnyi Zhurnal; Vol. 67 No. 2 (2015); 245-260 Український математичний журнал; Том 67 № 2 (2015); 245-260 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1977/975 https://umj.imath.kiev.ua/index.php/umj/article/view/1977/976 Copyright (c) 2015 Cvetković A. S.; Milovanović G. V.; Stanić M. P. |
| spellingShingle | Cvetković, A. S. Milovanović, G. V. Stanić, M. P. Цветковіч, А. С. Міловановіч, Г. В. Станіч, М. П. Positive Solutions of a Class of Operator Equations |
| title | Positive Solutions of a Class of Operator Equations |
| title_alt | Додатні розв'язки деякого класу операторних рівнянь |
| title_full | Positive Solutions of a Class of Operator Equations |
| title_fullStr | Positive Solutions of a Class of Operator Equations |
| title_full_unstemmed | Positive Solutions of a Class of Operator Equations |
| title_short | Positive Solutions of a Class of Operator Equations |
| title_sort | positive solutions of a class of operator equations |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1977 |
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