SOLVING OPERATOR EQUATIONS AX = Y AND Ax = y IN ALGL†

Journal of Applied Mathematics & Informatics.
2015.
May,
33(3_4):
417-424

- Received : December 20, 2014
- Accepted : March 09, 2015
- Published : May 30, 2015

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Abstract. In this paper the following is proved: Let
L
be a subspace lattice on a Hilbert space
H
and
X
and
Y
be operators acting on a Hilbert space
H
. If
XE
=
EX
for each
E
∈
L
, then there exists an operator
A
in Alg
L
such that
AX
=
Y
if and only if sup
and
YE
=
EYE
.
Let
x
and
y
be non-zero vectors in
H
. Let
P_{x}
be the orthogonal pro-jection on
sp
(
x
). If
EP_{x}
=
P_{x}E
for each
E
∈
L
, then the following are equivalent.
(1) There exists an operator
A
in Alg
L
such that
Ax
=
y
.
(2) <
f
,
Ey
>
y
=<
f
,
Ey
>
Ey
for each
E
∈
L
and
f
∈
H
.
AMS Mathematics Subject Classification : 47L35.
A
satisfying
AX
=
Y
for two operators
X
and
Y
acting on a Hilbert space
H
in 1966
[1
,
2
,
3
,
4
,
5
,
6]
. Douglas used the range inclusion property of operators to show necessary and sufficient conditions for the existence of an operator
A
such that
AX
=
Y
.
A
condition for the operator
A
to be a member of
A
which is a specified subalgebra of
B
(
H
) can be given. In this paper, authors investigated to find sufficient and necessary conditions that there exists an operator
A
in Alg
L
satisfying
AX
=
Y
for operators
X
and
Y
acting on a Hilbert space
H
and there exists an operator
B
in Alg
L
satisfying
Bx
=
y
for two vectors
x
and
y
in
H
. And authors investigated the above interpolation problems for finitely or countably many operators and vectors.
The simplest case of the operator interpolation problem relaxes all restrictions on
A
, requiring it simply to be a bounded operator. In this case, the existence of
A
is nicely characterized by the well-known factorization theorem of Douglas.
Theorem 1.1
(R.G. Douglas
[1]
).
Let X and Y be bounded operators acting on a Hilbert space H. Then the following statements are equivalent:
(1) rangeY^{∗}
⊆
range X^{∗}
(2) Y^{∗}Y
≤
λ
^{2}
X^{∗}X for some
λ
≥ 0
(3) there exists a bounded operator A on H so that AX = Y .
Moreover, if (1), (2) and (3) are valid, then there exists a unique operator A so that
(a)
║
A
║
^{2}
= inf{
μ
:
Y
^{∗}
Y
≤
μX
^{∗}
X
}
(b) kerY^{∗} = kerA^{∗} and
(c) rangeA^{∗}
⊆
rangeX^{−}.
We need to look at the proof of Theorem A carefully. Then we know that the image of
A
on
is 0 from the proof of (3) by (2).
H
be a Hilbert space. A
subspace lattice L
is a strongly closed lattice of orthogonal projections on
H
containing the trivial projections 0 and I. The symbol Alg
L
denotes the algebra of bounded operators on
H
that leave invariant every projection in
L
; Alg
L
is a weakly closed subalgebra of
B
(
H
). A lattice
L
is a
commutative subspace lattice
, or CSL, if the projections in
L
commute; in this case, Alg
L
is called a
CSL algebra
. Let
x
_{1}
,⋯,
x_{n}
be vectors of
H
. Then sp({
x
_{1}
,⋯,
x_{n}
}) = {
α
_{1}
x
_{1}
+
α
_{2}
x
_{2}
+ ⋯ +
α_{n}x_{n}
:
α
_{1}
;
α
_{2}
, ⋯
α_{n}
∈ ℂ }. Let
M
be a subset of
H
. Then
means the closure of
M
and
the orthogonal complement of
. Let ℕ be the set of natural numbers and ℂ be the set of complex numbers.
Let
L
be a subspace lattice and
A
,
X
and
Y
be operators acting on a Hilbert space
H
such that
AX
=
Y
. If
XE
=
EX
, then ║
YEf
║ = ║
AXEf
║ = ║
A_{E} Xf
║ ≤ ║
A
║║
XEf
║ for all
E
∈
L
and for all f in
H
. If we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then the inequality above may be stated in the form
Theorem 2.1.
Let L be a subspace lattice on a Hilbert space H and X and Y be operators acting on the Hilbert space H. If XE = EX for each E in L, then the following are equivalent.
(1) There exists an operator A in AlgL such that AX = Y.
(2) sup
and YE = EY E for each E in L .
Proof
. Assume that sup
and
Y E
=
EY E
for each
E
in
L
. Then for each
E
in
L
, there exists an operator
A_{E}
in
B
(
H
) such that
A_{E}
(
XE
) =
Y E
=
EY E
by Theorem A. In particular, if
E
=
I
, then we have an operator
A_{I}
in
B
(
H
) such that
A_{I}X
=
Y
. So
A_{E}
(
XE
) =
A_{I}XE
=
EA_{I}XE
for each
E
in
L
. Since
EX
=
XE
for each
E
∈
L
,
A_{I}XE
=
EA_{I}EX
. Hence
A_{I}E
=
EA_{I}E
on
. Let
h
be in
. Since
EX
=
XE
for each
E
in
L
, <
Eh
,
X f
>=<
h
,
EXf
>=<
h
,
XEf
>= 0. So
Eh
∈
. By the definition of
A_{I}
, (
A_{I}E
)
h
= 0 = (
EA_{I}E
)
h
. Hence
A_{I}E
=
EA_{I}E
on
. So
A_{I}
is an operator in Alg
L
. □
Assume that
X
_{1}
,⋯,
X_{n}
and
Y
_{1}
,⋯,
Y_{n}
are operators in
B
(
H
) and
A
is an operator in Alg
L
such that
AX_{i}
=
Y_{i}
for each
i
= 1,⋯,
n
. Then
Y_{i}Ef_{i}
=
AX_{i}Ef_{i}
for each
i
= 1,⋯,
n
,
E
∈
L
and each
f_{i}
in
H
. Hence
for all
E
∈
L
and all
f_{i}
in
H
. If, for convenience, we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then the inequality above may be stated in the form
Theorem 2.2.
Let X_{1},⋯,X_{n} and Y_{1},⋯,Y_{n} be bounded operators acting on H. If X_{i}E = EX_{i} for each E in L and i
{1, 2,⋯,
n
},
then the following are equivalent
.
(1) There exists an operator A in AlgL such that AX_{i} = Y_{i} for i = 1; 2,⋯,n
.
(2)
sup
and Y_{i}E = EY_{i}E for each i
= 1,⋯,
n and E in L
.
Proof
. Assume that sup
and
Y_{i}E
=
EY_{i}E
for each
i
= 1,⋯,
n
and
E
∈
L
. Let
E
be in
L
and
Define
A_{E}
:
M_{E}
→
H
by
. Then
A_{E}
is welldefined and bounded linear. Extend
A_{E}
on
continuously. Define
A_{E}f
= 0 for each
f
∈
. Then
A_{E}
:
H
→
H
is a bounded linear and
A_{E}
EX_{i}
=
Y_{i}E
for
i
= 1,⋯,
n
. If
E
=
I
, then
A_{I}X_{i}
=
Y_{i}
for
i
= 1,⋯,
n
. Since
EX_{i}
=
X_{i}E
and
Y_{i}E
=
EY_{i}E
for each
i
= 1,⋯,
n
,
A_{E}X_{i}E
=
A_{I}X_{i}E
=
A_{I}
EX_{i}
and
A_{E}X_{i}E
=
EA_{I}X_{i}E
=
EA_{I}EX_{i}
. Hence
A_{I}E
=
EA_{I}E
on
. Let
h
be in
. Then since
EX_{i}
=
X_{i}E
for each
i
= 1,⋯,
n
, <
Eh
,
X_{i}f
>=<
h
,
EX_{i}f
>=<
h
,
X_{i}Ef
>= 0 for each
f
∈
H
. So
By the definition of
A_{I}
,
A_{I}Eh
= 0 =
EA_{I}Eh
for each
E
∈
L
. Hence
A_{I}E
=
EA_{I}E
on
. So
A_{I}
is an operator in Alg
L
□
We can generalize the above Theorem to the countable case easily.
Theorem 2.3.
Let X_{i} and Y_{i} be bounded operators acting on H for all i
= 1, 2,⋯.
If X_{i}E = EX_{i} for each E in L and i in
ℕ,
then the following are equivalent
.
(1) There exists an operator A in AlgL such that AX_{i} = Y_{i} for i
= 1, 2,⋯.
(2)
sup
and Y_{i}E = EY_{i}E for each i
= 1,⋯
and E
∈
L
.
Proof
. Assume that sup
and
Y_{i}E
=
EY_{i}E
for each
i
= 1,⋯. Let
E
be in
L
and
Define
A_{E}
:
N_{E}
→
H
by
. Then
A_{E}
is welldefined and bounded linear. Extend
A_{E}
on
continuously. Define
A_{E}f
= 0 for each
f
∈
. Then
A_{E}
:
H
→
H
is a bounded linear and
A_{E}
EX_{i}
=
Y_{i}E
for
i
= 1,⋯. If
E
=
I
, then
A_{I}X_{i}
=
Y_{i}
for
i
= 1,⋯. Since
EX_{i}
=
X_{i}E
and
Y_{i}E
=
EY_{i}E
for each
i
= 1,⋯ ,
A_{E}X_{i}E
=
A_{I}X_{i}E
=
A_{I}
EX_{i}
and
A_{E}X_{i}E
=
EA_{I}X_{i}E
=
EA_{I}EX_{i}
. Hence
A_{I}E
=
EA_{I}E
on
. Let
h
be in
. Then since
EX_{i}
=
X_{i}E
for each
i
= 1,⋯ , <
Eh
,
X_{i}f
>=<
h
,
EX_{i}f
>=<
h
,
X_{i}Ef
>= 0 for each
f
∈
H
. So
for each
n
∈ ℕ. By the definition of
A_{I}
,
A_{I}Eh
= 0 =
EA_{I}Eh
for each
E
∈
L
. Hence
A_{I}E
=
EA_{I}E
on
. So
A_{I}
is an operator in Alg
L
. □
x
and
y
be non-zero vectors in a Hilbert space
H
. Let
X
=
x
⊗
y
and
Y
=
y
⊗
y
. Then for f in
H
and
E
∈
L
,
If for convenience, we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then for f in
H
and
E
∈
L
,
is
Hence sup
,
Y Ef
=<
f
,
Ey
>
y
and
EY Ef
=<
f
,
Ey
>
Ey
for each f in
H
and each
E
∈
L
.
We can obtain the following theorem by Theorem 2.1.
Theorem 3.1.
Let L be a subspace lattice on H and let x and y be non-zero vectors in H. Let P_{x} be the orthogonal projection on sp
(
x
).
If EP_{x} = P_{x}E for each E
∈
L, then the following are equivalent
.
(1) There exists an operator A in AlgL such that Ax = y.
(2)
<
f
,
Ey
>
y
=<
f
,
Ey
>
Ey for each E
∈
L and f
∈
H
.
Let
x_{i}
,
y_{i}
(
i
= 1,⋯,
n
) be non-zero vectors in
H
. Let
X_{i}
=
x_{i}
⊗
y_{i}
and
Y_{i}
=
y_{i}
⊗
y_{i}
. Then for
f_{i}
in
H
and
E
∈
L
Hence
and
Y_{i}Ef
=<
f
,
Ey_{i}
>
y_{i}
and <
EY_{i}Ef
>=<
f
,
Ey_{i}
>
Ey_{i}
for each
E
∈
L
,
f
∈
H
and
i
= 1,⋯,
n
.
We can obtain the following theorem by Theorem 2.2.
Theorem 3.2.
Let L be a subspace lattice on H and let x
_{1}
,⋯,
x_{n} and y
_{1}
,⋯,
y_{n} be vectors in H. Let Px_{i} be the orthogonal projection on sp
(
x_{i}
).
If EP_{xi} = Px_{i}E for each E
∈
L and i
= 1,⋯,
n, then the following are equivalent
.
(1) There exists an operator A in AlgL such that Ax_{i} = y_{i} for i
= 1; 2,⋯,
n
.
(2) sup
and < f , Ey_{i} > y_{i} =< f , Ey_{i} > Ey_{i} for each E
∈
L, f
∈
H and i
= 1,⋯,
n
.
We can extend Theorem 3.2 to countably infinite vectors and get the following theorem from Theorem 2.3.
Theorem 3.3.
Let L be a subspace lattice on H and let
{
x_{i}
}
and
{
y_{i}
}
be vectors in H for i
∈ ℕ.
Let Px_{i} be the orthogonal projection on sp
(
x_{i}
).
If EP_{xi} = Px_{i}E for each E
∈
L and
= 1, 2,⋯ ,
then the following are equivalent
.
(1) There exists an operator A in AlgL such that Ax_{i} = y_{i} for i
= 1; 2,⋯.
(2)
sup
and
<
f
,
Ey_{i}
>
y_{i}
=<
f
,
Ey_{i}
>
Ey_{i} for each E ∈ L, f
∈
H and i
= 1; 2,⋯.
Theorem 3.4.
Let L be a subspace lattice on a Hilbert space H and x and y be vectors in H. Let Px be the orthogonal projection on sp
(
x
).
If EP_{x} = P_{x}E for each E
∈
L and
sup
,
then there exists an operator A in AlgL such that Ax = y
.
Proof
. Assume that sup
. Then for each
E
in
L
, there exists an operator
A_{E}
in
B
(
H
) such that
A_{E}Ex
=
Ey
by Theorem 1.1. In particular, if
E
=
I
, then we have an operator
A_{I}
in
B
(
H
) such that
A_{I}x
=
y
. Let’s put
A_{I}
=
A
. So
A_{E}Ex
=
Ey
=
EAx
for each
E
∈
L
. Hence
A_{E}E
=
EA
on
sp
(
x
). Let
h
be in
sp
(
x
)
^{⊥}
. Since
EP_{x}
=
P_{x}E
for each
E
∈
L
, <
Eh
,
Ex
>=<
h
,
Ex
>=<
h
,
EP_{x}x
>=<
h
,
P_{x}Ex
>= 0. Hence
Eh
∈
sp
(
Ex
)
^{⊥}
. By the definition of
A_{E}
and
A
,
A_{E}Eh
= 0 =
EAh
for each
E
in
L
. Hence
A_{E}E
=
EA
on
H
for each
E
in
L
. So
A
=
EA
. Therefore
A
is in Alg
L
. □
Theorem 3.5.
Let L be a subspace lattice on a Hilbert space H and x
_{1}
,⋯,
x_{n} and y
_{1}
,⋯,
y_{n} be vectors in H. Let Px_{i} be the orthogonal projection on sp
(
x_{i}
).
If EP_{xi} = P_{xi}E for each E
∈
L and
= 1,⋯,
n and
then there exists an operator A in AlgL such that Ax_{i} = y_{i} for i
= 1,⋯,
n
.
Proof
. Assume that sup
. Let
E
be in
L
. Define
A_{E}
:
sp
({
Ex
_{1}
,⋯,
Ex_{n}
}) →
H
by
. Then
A_{E}
is well-defined and bounded linear. Define
A_{E}f
= 0 for each
f
∈
sp
({
Ex
_{1}
,⋯,
Ex_{n}
})
^{⊥}
. Then
A_{E}
:
H
→
H
is bounded linear and
A_{E}Ex_{i}
=
Ey_{i}
for
i
= 1,⋯,
n
. If
E
=
I
, then
A_{I}x_{i}
=
y_{i}
for
i
= 1,⋯,
n
. Let’s put
A_{I}
=
A
. So
A_{E}Ex_{i}
=
Ey_{i}
=
EAx_{i}
for each
E
∈
L
. Hence
A_{E}E
= EA on
sp
({
x
_{1}
,⋯,
x_{n}
}). Let
h
be in
sp
({
x
_{1}
,⋯,
x_{n}
})
^{⊥}
. Since <
Eh
,
Ex_{i}
>=<
h
,
Ex_{i}
>= 0, <
Eh
,
. So
Eh
∈
sp
({
Ex
_{1}
,⋯,
Ex_{n}
})
^{⊥}
. By the definition of
A_{E}
and
A
,
A_{E}Eh
= 0 =
EAh
for each
E
in
L
. Hence
A_{E}E
=
EA
on
H
for each
E
in
L
. So
A
=
EA
. Therefore
A
is in Alg
L
. □
We ca generalize the above theorem for countable case.
Theorem 3.6.
Let L be a subspace lattice on a Hilbert space H and
{
x_{i}
}
and
{
y_{i}
}
be vectors in H. Let P_{xi} be the orthogonal projection on sp
(
x_{i}
)
for each i
= 1; 2,⋯.
If EP_{xi} = P_{xi}E for each E
∈
L and i
= 1; 2,⋯
and
then there exists an operator A in AlgL such that Ax_{i} = y_{i} for i
= 1, 2,⋯.
Sang Ki Lee received his Ph.D at Keimyung University under direction of Chul Soon Han. He has been a professor of Daegu University since 1982. His research interest is topology and mathematics education.
Department of Mathematics Education, Daegu University, Daegu, Korea.
e-mail: sangklee@daegu.ac.kr
Joo Ho Kang received her Ph.D at the University of Alabama under direction of Tavan T. Trent. She has been a professor of Daegu University since 1977. Her research interest is an operator theory.
Department of Mathematics, Daegu University, Daegu, Korea.
e-mail: jhkang@daegu.ac.kr

1. Introduction

Interpolation problems have been developed by many mathematicians since Douglas considered a problem to find a bounded operator
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2. The EquationAX=Yin AlgL

Let
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3. The EquationAx=yin AlgL

Let
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BIO

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(1966)
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17
413 -
415
** DOI : 10.1090/S0002-9939-1966-0203464-1**

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,
Kang J.H.
(2003)
Interpolation problems in CSL-algebras AlgL
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(2005)
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,
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3269 -
** DOI : 10.1090/S0002-9939-02-06610-8**

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(2014)
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441 -
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** DOI : 10.14317/jami.2014.441**

Lance E.C.
(1969)
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68
** DOI : 10.1112/plms/s3-19.1.45**

Citing 'SOLVING OPERATOR EQUATIONS AX = Y AND Ax = y IN ALGL†
'

@article{ E1MCA9_2015_v33n3_4_417}
,title={SOLVING OPERATOR EQUATIONS AX = Y AND Ax = y IN ALGL†}
,volume={3_4}
, url={http://dx.doi.org/10.14317/jami.2015.417}, DOI={10.14317/jami.2015.417}
, number= {3_4}
, journal={Journal of Applied Mathematics & Informatics}
, publisher={Korean Society of Computational and Applied Mathematics}
, author={LEE, SANG KI
and
KANG, JOO HO}
, year={2015}
, month={May}