COMPACT INTERPOLATION ON AX = Y IN ALGL†

Journal of Applied Mathematics & Informatics.
2014.
Sep,
32(3_4):
441-446

- Received : November 07, 2013
- Accepted : December 24, 2013
- Published : September 28, 2014

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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
H
. Then there exists a compact operator A in Alg
L
such that AX = Y if and only if sup
and
Y
is compact. Moreover, if the necessary condition holds, then we may choose an operator
A
such that
AX
=
Y
and ∥
A
∥ =
K
.
AMS Mathematics Subject Classification : 47L35.
C
be a collection of operators acting on a Hilbert space
H
. An interpo-lation question for
C
asks for which operators X and Y on
H
when is there a bounded linear operator
A
(usually satisfying some other conditions) such that
AX
=
Y
? The "other conditions" that have been of interest to us involve re-stricting A to lie in the algebra associated with a subspace lattice. 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 Douglas
[2]
. Another interpolation question for a given subal-gebra ⌋ of
B
(
H
) asks for which vectors
x
and
y
in
H
is there a bounded operator
A
∈
C
that maps
x
to
y
. Lance
[6]
initiated the discussion by considering a nest
N
and asking what conditions on
x
and
y
will guarantee the existence of an operator
A
in Alg
N
such that
Ax
=
y
. Hopenwasser
[3]
extended Lance's result to the case where the nest
N
is replaced by an arbitrary commutative subspace lattice
L
. Munch
[7]
considered the problem of finding a Hilbert-Schmidt oper-ator
A
in Alg
N
that maps
x
to
y
, whereupon Hopenwasser
[4]
again extended to Alg
L
. In
[1]
, authors studied the problem of finding
A
so that
Ax
=
y
and
A
is required to lie in certain ideals contained in Alg
L
(for a nest
L
).
Roughly speaking, when an operator maps one thing to another, we think of the operator as the interpolating operator and the equation representing the mapping as the interpolation equation. The equations
Ax
=
y
and
AX
=
Y
are indistinguishable if spoken aloud, but we mean the change to capital letters to indicate that we intend to look at fixed operators
X
and
Y
, and ask under what conditions there will exist an operator
A
satisfying the equation
AX
=
Y
.
Let
x
and
y
be vectors in a Hilbert space. Then ⟨
x
,
y
⟩ means the inner product of vectors
x
and
y
. Note that the "vector interpolation" problem is a special case of the "operator interpolation" problem. Indeed, if we denote by
x
⊗
u
the rank-one operator defined by the equation
x
⊗
u
(
w
) = ⟨
w
,
u
⟩
x
, and if we set
X
=
x
⊗
u
, and
Y
=
y
⊗
u
, then the equations
AX
=
Y
and
Ax
=
y
represent the same restriction on
A
.
Let
H
be a Hilbert space and ℕ = {1, 2, ...}. A bounded operator
A
on
H
has
finite-rank
if rangeA is finite dimensional. A bounded operator
A
on
H
is called
compact
if
A
(ball
H
) has compact closure in
H
, where ball
H
= {
h
∈
H
: ∥
h
∥ ≤ 1}. We denote
B
_{0}
(
H
) the set of all compact operators on
H
.
We will study finite-rank operator interpolation problems on
L
and find a compact operator
A
in Alg
L
such that
AX
=
Y
for given
X
and
Y
in
B
(
H
) as convergence of finite-rank operators. Also, we will study this problem for given countable operators
X
_{1}
,
X
_{2}
,... and
Y
_{1}
,
Y
_{2}
,...
Theorem 1.1
(
[2]
).
Let X and Y be bounded operators acting on a Hilbert space H. Then the following statements are equivalent:
Moreover, if (1), (2), and (3) are valid, then there exists a unique operator A so that
Lemma 1.2.
Let A and X be bounded operators acting on a Hilbert space H.
Proof
. Let
ƒ
∈
H
. Then (a)
AXƒ
=
A
(
y
⊗
x
)
ƒ
=
A
⟨
ƒ, x
⟩
y
= ⟨
ƒ, x
⟩
Ay
. So
AX
is a rank-one operator.
Theorem 1.3.
Let X and Y be bounded operators acting on a Hilbert space H. Then the following are equivalent:
Proof
. If there exists a rank-one operator
A
=
y
⊗
x
such that
AX
=
Y
. Then
rangeY
^{∗}
⊆
rangeX
^{∗}
. Since
rangeX
⊄ {
x
}
^{⊥}
, there exists a vector
h
in
H
such that ⟨
Xh, x
⟩ ≠ 0. For any
ƒ
in
H
,
Y ƒ
= (
y
⊗
x
)
Xƒ
= ⟨
Xƒ
,
x
⟩
y
. So
Y
is a rank-one operator.
Conversely, suppose
rangeY
^{∗}
⊆
rangeX
^{∗}
and
Y
=
y
_{1}
⊗
x
_{1}
. Then there exists an operator
B
in
B
(
H
) such that
BX
=
Y
. Then for each
h
∈
H
,
So
B
(
rangeX
) =
rangeY
=
sp
{
y
_{1}
}. Define
by
Ah
=
Bh
if
h
∈
rangeX
and
Ah
= 0 if
h
∈
B
(
H
). Since
rangeA
⊂
sp
{
y
_{1}
}. Since
rangeA
is a linear subspace containing
y
_{1}
,
Hence
A
is a rank-one operator and
AX
=
Y
.
Theorem 1.4
.
Let X and Y be bounded operators acting on a Hilbert space H. If rangeY
^{∗}
⊆
rangeX. and Y is a finite-rank operator, then there exists a finite-rank operator A such that AX
=
Y
.
Proof
. If
Y
is a finite-rank operator, then
where
y
_{1}
, ...,
y_{n}
are linearly independent. Since
rangeY
^{∗}
⊆
rangeX
^{∗}
, there exists an operator
B
in
B
(
H
) such that
BX
=
Y
. For each
h
∈
H
,
So
B
(
rangeX
) =
sp
{
y
_{1}
, ...,
y_{n}
} Define
by
Ah
=
Bh
if
h
∈
rangeX
and
Ah
= 0 if
Then
A
∈
B
(
H
). And
Therefore
dim
(
rangeA
) =
n
and
AX
=
Y
.
Theorem 2.1.
Let L be a subspace lattice on H and let X and Y be operators in B
(
H
).
Assume that the range of X is dense in H. Then the following are equivalent:
Moreover, if condition (2) holds, we may choose an operator A such that
∥
A
∥ =
K
.
Proof
. Assume that sup
Then there exists an operator
A
in Alg
L
such that
AX
=
Y
by Theorem 3.1
[5]
. Since
Y
is compact, there is a sequence {
Y_{n}
} of fInite-rank operators that converges to
Y
in the norm topology on
B
(
H
). From the construction of
Y_{n}
, since
for each
n
∈ ℕ,
for each
n
∈ ℕ. By Theorems 1.4, there is a FInite-rank operator
A_{n}
such that
A_{n}X
=
Y_{n}
for each
n
∈ ℕ. Since
Y_{n}
→
Y
in the norm topology on
B
(
H
), ∥
A_{n}
−
A
∥ → 0. Hence
A
is compact. The proof of the converse is obvious.
Theorem 2.2.
Let L be a subspace lattice on H and let
X
_{1}
, ...,
X_{n}
and Y
_{1}
, ...,
Y_{n}
be bounded operators acting on H. Let k be a fixed natural number in
{1, 2, ...,
n
}
and assume that X_{k} has dense range. Then the following are equivalent:
Moreover, if condition (2) holds, we may choose an operator A such that
∥
A
∥ =
K
.
Proof
. If sup
and for given
k
in {1, 2, ,,,
n
},
Y_{k}
is compact, then by Theorem 3.2
[5]
, there exists an operator
A
in Alg
L
such that
AX_{i}
=
Y_{i}
for
i
= 1, 2, ,,,
n
. Since
Y_{k}
is compact, there is a sequence {
Y_{km}
}of finite-rank operators that converges to
Y_{k}
in the norm topology on
B
(
H
). From the construction of
Y_{km}
, we know that
for each
m
∈ ℕ. Therefore
for each
m
∈ ℕ. By Theorem 2.1, for each
m
∈ ℕ, there is a finite-rank operator
A_{km}
such that
A_{km}
X_{k}
=
Y_{km}
. Since
Y_{km}
→
Y_{k}
in the norm topology on
B
(
H
), ∥
A_{km}
−
A
∥ → 0. Hence
A
is compact. We omit the proof of the converse since it can be proved easily.
Theorem 2.3
(
[5]
).
Let L be a subspace lattice on H and let
X
_{1}
, ...,
X_{n}
and
Y
_{1}
, ...,
Y_{n}
be bounded operators acting on H. Let k be a fixed natural number in
{1, 2, ...,
n
}
and assume that X_{k} has dense range and Re
⟨
E
^{⊥}
X_{i}f
,
E
^{⊥}
X_{j}g
⟩ ≥ 0
for each E in L, i
<
j and all f, g in H. Then the following are equivalent:
By the Theorem 2.3, we can get the following Theorem.
Theorem 2.4.
Let L be a subspace lattice on H and let X
_{1}
, ...,
X_{n} and Y
_{1}
, ...,
Y_{n} be bounded operators acting on H. Let k be a fixed natural number in
{1, 2, ...,
n
}
and assume that X_{k} has dense range. If for i
<
j in
{1, 2, ...,
n
}
and all f, g in H, Re
⟨
E
^{⊥}
X_{i}f
,
E
^{⊥}
X_{j}g
⟩ ≥ 0,
then the following are equivalent:
If we observe the proof of the above theorems, we can generalize Theorem 2.2 to the countable case easily.
Theorem 2.5.
Let L be a subspace lattice on H and let X_{i} and Y_{i} be bounded operators acting on H for all i
= 1, 2, ...,
n. Let k be a fixed natural number in
{1, 2, ...,
n
}
and assume that X_{k} has dense range. Then the following are equivalent:
Moreover, if condition (2) holds, we may choose an operator an operator A such that
∥
A
∥ =
K
.
Theorem 2.6.
Let L be a subspace lattice on H and let X_{i} and Y_{i} be bounded operators acting on H for all i
= 1, 2, ...,
n
.
Let k be a fixed natural number in
{1, 2, ...,
n
}
and assume that X_{k} has dense range and Re
⟨
E
^{⊥}
Y_{i}f
,
E
^{⊥}
Y_{j}g
⟩ ≤
Re
⟨
E
^{⊥}
X_{i}f
,
E
^{⊥}
X_{j}g
⟩
for each E in L, i
<
j and allf, g in H, then the following are equivalent:
Joo Ho Kang received her Ph.D. at the University of Alabama under the 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, Kyungpook, Korea.
e-mail: jhkang@daegu.ac.kr

1. Introduction

Let
- (1) rangeY∗⊆range X∗
- (2) Y∗Y≤λ2X∗Xfor some λ≥ 0
- (3) there exists a bounded operator A on H so that AX = Y.

- (a)∥A∥2= inf{μ:Y∗Y≤μX∗X}
- (b) kerY∗=kerA∗and
- (c)

- (a) If X=y⊗x is a rank-one operator and Ay≠ 0,then AX is a rank-one operator.
- (b) If X= (y1⊗x1)+(y2⊗x2)is a rank-two operator and Ay1and Ay2are linearly independent, then AX is a rank-two operator.
- (c) Ifis a rank-n operator and Ay1, ...,Aynare linearly independent, then AX is a rank-n operator.
- (d) If X is a rank-n operator, then AX is a rank-m operator for m≤n.

- (b)AXƒ=A((y1⊗x1)+(y2⊗x2))ƒ=A⟨ƒ,x1⟩y1+A⟨ƒ,x2⟩y2= ⟨ƒ,x1⟩Ay1+⟨ƒ,x2⟩Ay2for allƒ∈H.
- (c)for allƒ∈H.
- (d) can be easily obtained by using (a), (b) and (c).

- (1) There is a rank-one operator A=y⊗x such that AX=Y and range X⊄ {x}⊥.
- (2) rangeY∗⊆rangeX∗and Y is a rank-one operator.

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2. Results

- (1) There exists a compact operator A in AlgL such that AX=Y.
- (2)supand Y is compact.

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- (1) There exists a compact operator A in AlgL such that AXi=Yifor each i= 1, 2, ,,,n.
- (2)supand Ykis compact.

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- (1)for each E in L and i= 1, 2, ...,n.
- (2) There exists an operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.
- (3)sup

- (1)for each E in L and i= 1, 2, ...,n, and Ykis compact.
- (2) There exists a compact operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.

- (1) There exists a compact operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.
- (2)supand Ykis com-pact.

- (1) There exists M≥ 0such thatsupfor each i∈ ℕand Ykis compact.
- (2) There is a compact operator A in AlgL such that AXi=Yifor i= 1, 2, ...,n.

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Citing 'COMPACT INTERPOLATION ON AX = Y IN ALGL†
'

@article{ E1MCA9_2014_v32n3_4_441}
,title={COMPACT INTERPOLATION ON AX = Y IN ALGL†}
,volume={3_4}
, url={http://dx.doi.org/10.14317/jami.2014.441}, DOI={10.14317/jami.2014.441}
, number= {3_4}
, journal={Journal of Applied Mathematics & Informatics}
, publisher={Korean Society of Computational and Applied Mathematics}
, author={KANG, JOO HO}
, year={2014}
, month={Sep}