Equations $AX=Y$ and $Ax=y$ in Alg$mathcal L$

EQUATIONS AX = Y AND Ax = y IN ALGL

Young Soo Jo, Joo Ho Kang, and Dongwan Park

Abstract. Let L be a subspace lattice on a Hilbert space H and X and Y be operators acting on a Hilbert space H. Let P be the projection onto R(X), where RX is the range of X. If P E = EP for each E ∈ L, then there exists an operator A in AlgL such that AX = Y if and only if

sup{kE

Y f k/kE

Xf k : f ∈ H, E ∈ L} = K < ∞.

Moreover, if the necessary condition holds, then we may choose an operator A such that AX = Y and kAk = K.

Let x and y be vectors in H and let P

be the projection onto the singlely generated space by x. If P

E = EP

for each E ∈ L, then the assertion that there exists an operator A in AlgL such that Ax = y is equivalent to the condition

K

:= sup{kE

yk/kE

xk : E ∈ L} < ∞.

Moreover, we may choose an operator A such that kAk = K

whose norm is K

under this case.

1. Introduction

Let H be a Hilbert space and B(H) be the class of all bounded opera- tors acting on H. A subspace is a closed linear manifold and we identify the subspace with the orthogonal projection whose range it is. If L is a collection of projections that is closed under the operations of meet and join, then it is a lattice. A subspace lattice L is a strongly closed lattice of orthogonal projections on H containing the trivial projections 0 and I. The symbol AlgL denotes the algebra of bounded operators on

Received January 31, 2005. Revised June 28, 2005.

2000 Mathematics Subject Classification: 47L35.

Key words and phrases: interpolation problem, subspace lattice, algL, CSL-algL.

H that leave invariant all projections in L; AlgL is a weakly closed sub- algebra of B(H). A lattice L is a commutative subspace lattice, or CSL, if the projections in L all commute; in this case, AlgL is called a CSL algebra. Let x

, . . . , x n be vectors of H. Then the generating space by {x

, . . . , x _n } is the set {α

x

+α

x

+· · ·+α _n x _n : α

, α

, . . . , α _n ∈ C }.

Let M be a subset of H. Then M means the closure of M and M ^⊥ the orthogonal complement of M . Let N be the set of natural numbers and C be the set of complex numbers.

Let X and Y be operators acting on H and A be a subalgebra of B(H). An interpolation problem for A is that given two operators X and Y , under what conditions can we be sure that there is an operator A in A such that AX = Y ? And we consider the above problem for finite and countable operators X _i and Y _i (i = 1, 2, . . . ).

Given two vectors x and y in H, when is there an operator A in A that maps x to y? In [3], the problem is studied under the conditions that L is a commutative subspace lattice on H and X is an operator of rank 0 or 1 acting on H. Let R(X) be the range of X. In [5] and [6], authors considered it when R(X) is dense in H and have investigated the interpolation problem for AlgL with strong desire for finding weaker conditions than the dense range of X. In this paper, we obtain a con- dition that we wanted to find. We introduce a history of interpolation problems to know how it is developed.

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 fac- torization theorem.

Theorem A. (R. G. Douglas [2]) Let X and Y be bounded oper- ators acting on a Hilbert space H. Then the following statements are equivalent:

(1) R(Y ^∗ ) ⊆ R(X ^∗ );

(2) Y ^∗ Y ≤ λ

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 oper- ator A so that

(a) kAk

= inf{µ : Y ^∗ Y ≤ µX ^∗ X}

(b) ker Y ^∗ = ker A ^∗ and (c) R(A ^∗ ) ⊆ R(X ⁻ ).

We need to look at the proof of Theorem A carefully. Then we know

that the image of A on R(X) ^⊥ is 0 from the proof of (3) by (2).

In [8], Katsoulis, Moore and Trent found a necessary and sufficient condition of the existence of an interpolation operator A for a nest al- gebra. That is, the following:

(∗) sup{kE ^⊥ Y f k/kE ^⊥ Xf k : f ∈ H and E ∈ N } = K < ∞, where we use the convention

= 0, when necessary.

In [3], Hopenwasser found a result that, if X is rank-one or zero, then the condition (∗) is also sufficient for a commutative subspace lattice on H. In [10], Moore and Trent showed that the condition (∗) is a necessary and sufficient condition that there exists an operator A in AlgL such that AX = Y where L is a commutative subspace lattice. In [5], Jo and Kang obtained a necessary and sufficient condition for the interpolation problem in AlgL .

Theorem B. (Jo, Kang [5]) Let L be a subspace lattice on H. Let X and Y be operators in B(H). Assume that R(X) is dense in H. Then the following statements are equivalent:

(1) There exists an operator A in AlgL such that AX = Y and every E in L reduces A.

(2) sup{k X n i=1

E _i Y f _i k/k X n

i=1

E _i Xf _i k : n ∈ N, E ∈ L and f _i ∈ H} < ∞ _. In [6], the authors showed the condition (∗) is a necessary and suf- ficient condition for the interpolation problem in AlgL when R(X) is dense in H and L is a subspace lattice on H.

2. The equation AX = Y in AlgL

We use the convention

= 0, when necessary.

Theorem 2.1. Let L be a subspace lattice on H and let X and Y be operators acting on H. Let P be the projection onto the R(X). If P E = EP for each E ∈ L, then the following are equivalent:

(1) There exists an operator A in AlgL such that AX = Y . (2) sup{kE ^⊥ Y f k/kE ^⊥ Xf k : f ∈ H, E ∈ L} = K < ∞.

Proof. Assume that sup{kE ^⊥ Y fk/kE ^⊥ Xfk : f ∈ H, E ∈ L} = K < ∞.

Then for each E in L, there exists an operator A _E in B(H) such that

A _E (E ^⊥ X) = E ^⊥ Y and kA _E k ≤ K by Theorem A. In particular, if

E = 0, then we have an operator A

in B(H) such that A

X = Y and kA

k ≤ K. So A _E (E ^⊥ X) = E ^⊥ Y = E ^⊥ A

X. Since P E = EP for each E ∈ L, A _E E ^⊥ = E ^⊥ A

on H. For, since A _E (E ^⊥ X) = E ^⊥ Y = E ^⊥ (A

X), A _E E ^⊥ = E ^⊥ A

on R(X) for each E in L. Let E be in L and f ∈ R(X) ^⊥ . Then for any g ∈ H, ­

f, E ^⊥ Xg ®

= ­

f, E ^⊥ P Xg ®

­ =

f, P E ^⊥ Xg ®

= 0. Also by the definition of A

, E ^⊥ A

f = 0. Hence A _E E ^⊥ h = 0 = E ^⊥ A

h for h ∈ (R(X) ^⊥ ∩ E ^⊥ ) + (R(X) ^⊥ ∩ E). Since P E = EP , P ^⊥ E = EP ^⊥ , P ^⊥ E and P ^⊥ E ^⊥ are projections onto R(X) ^⊥

∩E and R(X) ^⊥ ∩ E ^⊥ , respectively. Hence

(R(X) ^⊥ ∩ E ^⊥ ) + (R(X) ^⊥ ∩ E) = P ^⊥ E ^⊥ + P ^⊥ E

= P ^⊥ E ^⊥ ⊕ P ^⊥ E

= P ^⊥ (E ^⊥ ⊕ E) = P ^⊥ . Therefore A _E E ^⊥ = E ^⊥ A

on H.

For each E in L,

E ^⊥ A

E ^⊥ = A E E ^⊥ E ^⊥ = A E E ^⊥ = E ^⊥ A

So A

is an operator in AlgL.

If (1) is assumed, then AX = Y and E ^⊥ AE ^⊥ = E ^⊥ A for all E ∈ L.

So for any vector f , since E ^⊥ AE ^⊥ X = E ^⊥ Y , E ^⊥ AE ^⊥ Xf = E ^⊥ Y f . Hence

kE ^⊥ Y f k ≤ kE ^⊥ AkkE ^⊥ Xf k ≤ kAkkE ^⊥ Xf k for all E ∈ L and all f ∈ H. So

sup{kE ^⊥ Y f k/kE ^⊥ Xf k : f ∈ H, E ∈ L} < ∞. ¤ In [10], authors induced the second condition of Theorem 2.1 for the interpolation problem in AlgL under the condition that R(X) is dense in H and L is a commutative subspace lattice. We proved it without the dense range condition and commutative condition of L in Theorem 2.1.

We give two examples to compare a difference between the Theorem in [10] and Theorem 2.1.

Let P be the projection onto the R(X).

Example 1. Let H be a Hilbert space with an orthonormal basis

{e

, e

, e

, . . . }. Let L _∞ be the subspace lattice generated by {[e

],

[e

, e

, e

] : k = 1, 2, . . . }. Then AlgL ∞ is the collection of all matrices of the following form



 

 

 

 

 

 

 



∗ ∗

∗

∗ ∗ ∗

∗

∗ ∗ ∗

∗

∗ ∗ ∗

∗ . ..



 

 

 

 

 

 

 

 ,

where non-starred entries are zero.

AlgL _∞ is called a tridiagonal algebra. If R(X) is dense, then P = I.

But if X is an operator and P E = EP for every E in L , then all of off-diagonal entries are zero.

Example 2. Let H = [e

, e

, e

, . . . ] and let E

= 0 and E _n = [e

, e

, . . . , e n ]. Let L be the subspace lattice generated by {E n | n = 0, 1, 2, . . . }. Then A ∈ AlgL iff A has the form



 

 

 

 

 

 



∗ ∗ ∗ ∗ ∗ ∗

∗ ∗ ∗ ∗ ∗ ∗

∗ ∗ ∗ ∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗ . ..



 

 

 

 

 

 

 ,

where non-starred entries are zero.

If R(X) is dense, then P = I. And if P E = EP for every E in L,

then P has the following form



 

 

 

 

 

 



∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗ . ..



 

 

 

 

 

 

 ,

where non-starred entries are zero.

Theorem 2.2. Let X

, . . . , X _n and Y

, . . . , Y _n be bounded opera- tors acting on H. Let P _j be the projection onto R(X _j ) for all j = 1, 2, . . . , n. If P _k E = EP _k for some k in {1, 2, . . . , n}, then the follow- ing are equivalent:

(1) There exists an operator A in AlgL such that AX _i = Y _i for i = 1, 2, . . . , n.

(2) sup{kE ^⊥ ( X n i=1

Y i f i )k/kE ^⊥ ( X n

i=1

X i f i )k : f i ∈ H, E ∈ L} = K < ∞.

Theorem 2.3. Let X i and Y i be bounded operators acting on H for all i = 1, 2, . . . . Let P _j be the projection onto R(X _j ) for all j = 1, 2, . . . . If P k E = EP k for some k 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{kE ^⊥ ( X m i=1

Y i f i )k/kE ^⊥ ( X m i=1

X i f i )k : f i ∈ H, E ∈ L, m ∈ N} = K < ∞.

Proof. If we assume that there is an operator A in AlgL such that AX i = Y i for i = 1, 2, . . . , then for f i ∈ H, E ∈ L and m ∈ N,

kE ^⊥ ( X m i=1

Y _i f _i )k = kE ^⊥ ( X m i=1

AX _i f _i )k = kE ^⊥ A(

X m i=1

X _i f _i )k

≤ kE ^⊥ AE ^⊥ ( X m i=1

X _i f _i )k

≤ kE ^⊥ AkkE ^⊥ ( X m i=1

X _i f _i )k

≤ kAkkE ^⊥ ( X m i=1

X _i f _i )k.

If kE ^⊥ ( P _m

i=1 X i f i )k is not zero, then kE ^⊥ (

X m i=1

Y _i f _i )k/kE ^⊥ ( X m

i=1

X _i f _i )k < kAk.

Otherwise, we use the convention

= 0. So sup{kE ^⊥ (

X m i=1

Y _i f _i )k/kE ^⊥ ( X m

i=1

X _i f _i )k : f _i ∈ H, E ∈ L, m ∈ N} < ∞.

Conversely, we assume that sup{kE ^⊥ (

X m i=1

Y i f i )k/kE ^⊥ ( X m i=1

X i f i )k : f i ∈ H, E ∈ L, m ∈ N}

= K < ∞.

Then for each E in L, there exists an operator A _E in B(H) such that A E (E ^⊥ X i ) = E ^⊥ Y i for i = 1, 2, . . . and kA E k ≤ K by Theorem A. In particular, if E = 0, then we have an operator A

in B(H) such that A

X _i = Y _i for i = 1, 2, . . . and kA

k ≤ K. So A _E (E ^⊥ X _i ) = E ^⊥ Y _i = E ^⊥ A

X i . Since P k E = EP k for some k ∈ {1, 2, . . . }, for each E ∈ L, A _E E ^⊥ = E ^⊥ A

on H. For, since A _E (E ^⊥ X _i ) = E ^⊥ Y _i = E ^⊥ (A

X _i ), A E E ^⊥ = E ^⊥ A

on R(X i ) for each E in L and i = 1, 2, . . . . We as- sumed that P _k E = EP _k for some k ∈ {1, 2, . . . }. Let E be in L and f ∈ R(X _k ) ^⊥ . Then for any g ∈ H, ­

f, E ^⊥ X _k g ®

= ­

f, E ^⊥ P _k X _k g ®

­ =

f, P k E ^⊥ X k g ®

= 0. Also by the definition of A

, E ^⊥ A

f = 0. Hence A _E E ^⊥ f = 0 = E ^⊥ A

f for f ∈ (R(X _k ) ^⊥ ∩ E ^⊥ ) + (R(X _k ) ^⊥ ∩ E).

Since P _k E = EP _k , P _k ^⊥ E = EP _k ^⊥ , P _k ^⊥ E and P _k ^⊥ E ^⊥ are projections onto R(X k ) ^⊥ ∩ E and R(X k ) ^⊥ ∩ E ^⊥ , respectively. Hence

(R(X k ) ^⊥ ∩ E ^⊥ ) + (R(X k ) ^⊥ ∩ E) = P _k ^⊥ E ^⊥ + P _k ^⊥ E

= P _k ^⊥ E ^⊥ ⊕ P _k ^⊥ E

= P _k ^⊥ (E ^⊥ ⊕ E) = P _k ^⊥ . Therefore A _E E ^⊥ = E ^⊥ A

on H.

For each E in L,

E ^⊥ A

E ^⊥ = A _E E ^⊥ E ^⊥ = A _E E ^⊥ = E ^⊥ A

So A

is an operator in AlgL. ¤

3. The equation Ax = y in AlgL

Assume that x and y are vectors in H and A is an operator in AlgL such that Ax = y. Then kE ^⊥ yk = kE ^⊥ Axk = kE ^⊥ AE ^⊥ xk ≤ kAkkE ^⊥ xk for all E ∈ L. If, for convenience, we adopt the convention that a fraction whose numerator and denominator are both zero is equal to zero, then the above inequality may be stated in the form

sup

E∈L

kE ^⊥ yk

kE ^⊥ xk ≤ kAk _.

We consider the above fact when L is a subspace lattice without the commutative condition.

Lemma 3.1. Let X and Y be operators in B(H) and let x and y be vectors in H. Then the following are equivalent:

(1) There exists an operator A in B(H) such that AXx = Y y.

(2) kY yk ≤ λ

kXxk, where λ

= inf{λ : kY yk ≤ λkXxk}.

Moreover, if condition (2) holds, we may choose an operator A such that kAk = λ

.

Proof. Assume that

kY yk ≤ λ

kXxk,

where λ

= inf {λ : kY yk ≤ λkXxk}. Let M = {αXx : α ∈ C}. Then M is a linear manifold. Define A : M → H by A(αXx) = αY y. Then A is well-defined by the assumption. Define Ag = 0 for all g in M ^⊥ . Then A is an operator acting on H and AXx = Y y.

If Xx 6= 0, then

kAk = kY yk kXxk ≤ λ

Since kY yk = kAXxk ≤ kAkkXxk, λ

≤ kAk. So kAk = λ

.

The proof of the converse is obvious. ¤

Lemma 3.2. Let x

, x

, . . . , x n and y

, y

, . . . , y n be vectors in H and let X and Y be operators in B(H). Then the following are equivalent:

(1) There exists an operator A in B(H) such that AXx i = Y y i for i = 1, 2, . . . , n.

(2) sup{k X n i=1

α _i Y y _i k/k X n i=1

α _i Xx _i k : α _i ∈ C} = λ < ∞.

Moreover, if condition (2) holds, we may choose an operator A such that

kAk = λ.

Proof. Assume that sup{k

X n i=1

α i Y y i k/k X n

i=1

α i Xx i k : α i ∈ C} = λ < ∞.

Then

(∗∗) k

X n i=1

α _i Y y _i k ≤ λk X n i=1

α _i Xx _i k for α _i ∈ C.

Let N _X = { X n i=1

α _i Xx _i : α _i ∈ C }. Then N _X is a linear manifold. Define A : N _X → H by

A(

X n i=1

α i Xx i ) = X n i=1

α i Y y i .

Then A is well-defined by (∗∗). Define Ag = 0 for all g in N _X ^⊥ . Then A is an operator in B(H) and AXx _i = Y y _i (i = 1, 2, . . . , n). If P _n

i=1 α i Xx i 6= 0, then kAk = sup{k

X n i=1

α _i Y x _i k/k X n i=1

α _i Xx _i k : α _i ∈ C} ≤ λ.

Since k

X n i=1

α _i Y y _i k = kA(

X n i=1

α _i Xx _i )k ≤ kAkk X n i=1

α _i Xx _i k, λ ≤ kAk.

So kAk = λ. The proof of the converse is obvious. ¤ We can extend this fact to the countable case.

Lemma 3.3. Let x

, x

, . . . and y

, y

, . . . be vectors in H and X and Y be operators in B(H). Then the following are equivalent:

(1) There exists an operator A in B(H) such that AXx _n = Y y _n for n ∈ N.

(2) sup{k X n i=1

α _i Y y _i k/k X n i=1

α _i Xx _i k : α _i ∈ C and n ∈ N} = λ < ∞.

Moreover, if condition (2) holds, we may choose an operator A such that kAk = λ.

From Lemmas 3.1, 3.2 and 3.3, we can obtain generalized results on

the interpolation problems in AlgL.

Theorem 3.4. Let L be a subspace lattice on H and let x and y be vectors in H. Let P _x be the projection onto sp(x). If P _x E = EP _x for each E ∈ L, then the following are equivalent:

(1) There exists an operator A in AlgL such that Ax = y.

(2) sup{kE ^⊥ yk/kE ^⊥ xk : E ∈ L} = K

< ∞.

Moreover, if condition (2) holds, we may choose an operator A such that kAk = K

.

Proof. Assume that sup ©

kE ^⊥ yk/kE ^⊥ xk : E ∈ L ª

= K

< ∞. By Lemma 3.1, for each E in L, there exists an operator A _E in B(H) such that A E (E ^⊥ x) = E ^⊥ y and kA E k ≤ K

. In particular, if E = 0, then we have an operator A

in B(H) such that A

x = y and kA

k ≤ K

. Then for each E ∈ L, A _E E ^⊥ x = E ^⊥ y = E ^⊥ A

x. Hence A _E E ^⊥ = E ^⊥ A

on sp(x). And for each E in L, A _E E ^⊥ h = 0 for h in sp(x) ^⊥ .

For, let E be in L and h be in sp(x) ^⊥ . Then

­ E ^⊥ h, E ^⊥ x ®

= ­

h, E ^⊥ x ®

= ­

h, E ^⊥ P _x x ®

= ­

h, P _x E ^⊥ x ®

= ­

P _x h, E ^⊥ x ®

= 0.

Hence E ^⊥ h ∈ sp(E ^⊥ x) ^⊥ for h ∈ sp(x) ^⊥ . Since A E g = 0 for all g in sp(E ^⊥ x) ^⊥ , A _E E ^⊥ h = 0.

By the definition of the operator A

, E ^⊥ A

h = 0 for h ∈ sp(x) ^⊥ . Hence A _E E ^⊥ h = 0 = E ^⊥ A

h for h ∈ sp(x) ^⊥ . So A _E E ^⊥ = E ^⊥ A

. Let E be in L. Then

E ^⊥ A

E ^⊥ = A _E E ^⊥ E ^⊥ = A _E E ^⊥ = E ^⊥ A

So A

is an operator in AlgL. Since A

x = y and E ^⊥ A

E ^⊥ = E ^⊥ A

, kE ^⊥ yk = kE ^⊥ A

xk = kE ^⊥ A

E ^⊥ xk ≤ kA

kkE ^⊥ xk _.

So K

≤ kA

k. Therefore kA

k = K

.

The proof of the converse is obvious. ¤

Theorem 3.5. Let L be a subspace lattice on H and let x

, x

, . . . , x _n and y

, y

, . . . , y _n be vectors in H. Let M = {

X n i=1

α _i x _i : α _i ∈ C}

and P _M be the projection onto M. If P _M E = EP _M for each E ∈ L,

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{k X n i=1

α _i E ^⊥ y _i k/k X n i=1

α _i E ^⊥ x _i k : E ∈ L, α _i ∈ C} = K

< ∞.

Moreover, if condition (2) holds, we may choose an operator A such that kAk = K

.

Proof. Suppose that

sup{k X n i=1

α _i E ^⊥ y _i k/k X n i=1

α _i E ^⊥ x _i k : E ∈ L, α _i ∈ C} = K

< ∞.

Let E be in L. Then by Lemma 3.2, there exists an operator A E in B(H) such that A _E E ^⊥ x _i = E ^⊥ y _i for i = 1, 2, . . . , n. If E = 0, then A

x i = y i for i = 1, 2, . . . , n. So A E E ^⊥ x i = E ^⊥ y i = E ^⊥ A

x i for each i = 1, 2, . . . , n. Hence

A E E ^⊥ ( X n

i=1

α i x i ) = X n i=1

α i A E E ^⊥ x i = X n i=1

α i E ^⊥ y i

= X n i=1

α _i E ^⊥ A

x _i

= E ^⊥ A

( X n i=1

α i x i ).

Therefore A _E E ^⊥ = E ^⊥ A

on M.

Let h ∈ M ^⊥ . Then

¿ E ^⊥ h,

X n i=1

E ^⊥ x _i À

= X n i=1

­ h, E ^⊥ x _i ®

= X n i=1

­ h, E ^⊥ P _M x _i ®

= X n i=1

­ h, P M E ^⊥ x i

® = X n i=1

­ P M h, E ^⊥ x i

® = 0.

So E ^⊥ h ∈ N ^⊥ , where N = sp{E ^⊥ x

, E ^⊥ x

, . . . , E ^⊥ x _n }. Since A _E g = 0

for all g in N ^⊥ , A _E E ^⊥ h = 0.

Hence E ^⊥ A

= A E E ^⊥ for all E in L.

For each E ∈ L,

E ^⊥ A

E ^⊥ = A E E ^⊥ E ^⊥ = A E E ^⊥ = E ^⊥ A

.

Hence A

∈AlgL. Moreover, since A

x _i = y _i for i = 1, 2, . . . , n and E ^⊥ A

E ^⊥ = E ^⊥ A

,

k X n i=1

α _i E ^⊥ y _i k = k X n i=1

α _i E ^⊥ A

x _i k

= k X n i=1

α i E ^⊥ A

E ^⊥ x i k

= kE ^⊥ A

( X n i=1

α _i E ^⊥ x _i )k ≤ kA

kk X n i=1

α _i E ^⊥ x _i k.

So K

≤ kA

k. Hence kA

k = K

. ¤

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Young Soo Jo

Department of Mathematics Keimyung University Daegu 704-701, Korea E-mail: [email protected] Joo Ho Kang

Department of Mathematics Daegu University

Daegu 712-714, Korea E-mail: [email protected] Dong Wan Park

Department of Mathematics

Keimyung University

Daegu 704-701, Korea

E-mail: [email protected]