Range-kernel weak orthogonality of some elementary operators

Definition 1

If X is a complex Banach space, then for any elements x , y ∈ X , we say that x is orthogonal to y , noted by x ⊥ y , iff for all α , β ∈ C there holds

for all α , β ∈ C or R .

Theorem 2

Let A , B be operators in B ( ℋ ) . If δ A , B is Fuglede, then the range of δ A , B (resp. the range of δ A , B ∣ C p ) is orthogonal to the kernel of δ A , B (resp. the kernel of δ A , B ∣ C p ) for all 1 ≤ p ≤ ∞ .

Definition 3

[4] If E : X → Y and T : Y → Z are bounded linear operators between Banach spaces and 0 < k ≤ 1 ,

We say that T is weakly orthogonal to E , written T ∠ E , or equivalently

For 0 < k ≤ 1 , we say that ( E , T ) has a 1 k -gap if T ∠ k E .

Theorem 4

[14,15,16] Let A , B be normal operators, C , D ∗ be hyponormal operators in B ( ℋ ) such that [ A , C ] = [ B , D ] = 0 and J = B ( ℋ ) or J = C p : 1 ≤ p ≤ ∞ , then

1. If ker A ∩ ker C = { 0 } = ker B ∗ ∩ ker D ∗ , then for all X ∈ B ( ℋ ) such that E ˜ ( X ) , E ( X ) ∈ J

   S ∈ ker E ∩ J ⇒ min { ∥ E ( X ) + S ∥ J , ∥ E ˜ ( X ) + S ∥ J } ≥ ∥ S ∥ J .

2. If ker A ∩ ker C ≠ { 0 } or ker B ∗ ∩ ker D ∗ ≠ { 0 } , then there exists k verifying 0 < k < 1 such that

   ∀ X ∈ B ( ℋ ) , ∀ S ∈ ker E ∩ J : ∥ E ( X ) + S ∥ J ≥ k ∥ S ∥ J , where E ( X ) ∈ J .

3. If J = C 2 ( ℋ ) with its inner product ⟨ X , Y ⟩ = tr ( Y ∗ X ) and E is defined on C 2 ( ℋ ) , then for all S ∈ ker E , we get

   ∥ E ( X ) + S ∥ 2 2 = ∥ E ( X ) ∥ 2 2 + ∥ S ∥ 2 2 ; ∥ E ˜ ( X ) + S ∥ 2 2 = ∥ E ˜ ( X ) ∥ 2 2 + ∥ S ∥ 2 2 .

Lemma 5

[13] Let A , B be commuting operators in B ( ℋ ) with no trivial kernel.

1. Let ξ be the elementary Fuglede operator defined by ξ ( X ) = A X A ∗ − B X B ∗ , then

   1. if ker A ∩ ker B = { 0 } , then ker A reduces A and ker B reduces B .

   2. if [ A , B ∗ ] = 0 and ker A ∩ ker B ≠ { 0 } , then ker A ∩ ker B reduces A and B .

2. Let A ∈ B ( ℋ ) , B ∈ B ( K ) , and E ( X ) = A X B ; X ∈ B ( K , ℋ ) , then E is the Fuglede operator if and only if ker A reduces A and ker B ∗ reduces B ∗ .

Lemma 6

[9] Let T be an operator represented by block matrix as T = ( T i , j ) i , j = 1 n .

1. If T ∈ B ( ℋ ) , then 1 n 2 ∑ i , j ∥ T i , j ∥ 2 ≤ ∥ T ∥ 2 ≤ ∑ i , j ∥ T i , j ∥ 2

2. If T ∈ C p ( T ) ; 1 ≤ p < ∞ , then

   1 n p − 2 ∥ T ∥ p p ≤ ∑ i , j ∥ T i , j ∥ p p ≤ ∥ T ∥ p p for all 2 ≤ p < ∞ , ∥ T ∥ p p ≤ ∑ i , j ∥ T i , j ∥ p p ≤ 1 n p − 2 ∥ T ∥ p p for all 1 ≤ p ≤ 2 .

Proposition 7

Let ℋ , K be Hilbert spaces and A ∈ B ( ℋ ) , B ∈ B ( K ) , and E ∈ B ( B ( K , ℋ ) ) such that E ( X ) = A X B . If E is the Fuglede operator, then E is w -orthogonal, E ∠ E ˜ , and the inequality

holds if

1. J = B ( K , ℋ ) with k = 1 / 2 or

2. J = C p ( K , ℋ ) with k = 1 2 2 − p p , if 1 ≤ p ≤ 2 and k = 1 2 p − 2 p , if  2 ≤ p < ∞ .

Proof

If A and B ∗ are injective operators, then E is injective. So there is nothing to prove.

Suppose that A or B ∗ is the non-injective operator. From Lemma 5(2) and with respect to the decompositions:

we obtain A = A 1 ⊕ A ; B = B 1 ⊕ 0 . Let X = ( X i , j ) i , j = 1 , 2 : K → ℋ . Then

From the injectivity of A 1 and B 1 ∗ it yields that any S = ( S i j ) i , j = 1 , 2 in ker E can be written as

where the operators S 12 , S 21 , and S 22 are arbitrary.

Hence, for all S ∈ ker E and all X ∈ B ( K , ℋ ) , by Lemma 5(2), we have

Also, for all S ∈ ker E ∩ C p ( K , ℋ ) and all X ∈ B ( K , ℋ ) such that E ( X ) ∈ C p ( K , ℋ ) , we get

1. if 1 ≤ p ≤ 2 , then

   min { ∥ E ( X ) + S ∥ p , ∥ E ˜ ( X ) + S ∥ p } = A 1 X 11 B 1 S 12 S 21 S 22 p ≥ 1 2 2 − p p ∥ S 12 ∥ p p + ∥ S 21 ∥ p p + ∥ S 22 ∥ p p 1 / p ≥ 1 2 2 − p p ∥ S ∥ p .

2. if 2 ≤ p < ∞ , then

   min { ∥ E ( X ) + S ∥ p , ∥ E ˜ ( X ) + S ∥ p } = A 1 X 11 B 1 S 12 S 21 S 22 p ≥ 1 2 p − 2 p ∥ S 12 ∥ p p + ∥ S 21 ∥ p p + ∥ S 22 ∥ p p 1 / p ≥ 1 2 p − 2 p ∥ S ∥ p . □

Corollary 8

If ker A reduces A and ker B ∗ reduces B ∗ , then E is w -orthogonal, E E ˜ and satisfies the relation (6).

Lemma 9

Let Δ be the elementary operator defined on B ( ℋ ) by Δ ( X ) = A X B − X , where A , B ∈ B ( ℋ ) . If Δ is a Fuglede operator, then Δ is orthogonal and Δ ⊥ Δ ˜ .

Proof

The proof is the same as the one in Theorem 4.□

Proposition 10

Let A , B be doubly commuting operators in B ( ℋ ) and

If ker A reduces A , ker B reduces B and ξ orthogonal, then ker A ∩ ker B = { 0 } .

Proof

We consider the following three cases:

1. Let us suppose that N = ker A ∩ ker B ≠ { 0 } .

2. If ker A ≠ ker B and ker B ⊈ ker A , then with respect to the decomposition

   ℋ = ( ker B ) ⊥ ⊕ ( ker B ⊖ N ) ⊕ N

   and from the hypothesis it yields

   A = A 1 ⊕ A 2 ⊕ 0 , B = B 1 ⊕ 0 , and S = ( S i j ) i , j = 1 , … , 3 ,

   where A 2 is an injective operator. Hence,

   S ∈ ker ξ ⇒ A 1 S 11 A 1 ∗ = B 1 S 11 B 1 ∗ ; A 1 S 12 A 2 ∗ = A 2 S 21 A 1 ∗ = 0 ; S 22 = 0

   and the other entries are arbitrary. Choosing X and S as follows:

   X = 0 ⊕ ( e ⊗ e ) ⊕ 0 , S = 0 ⊕ 0 R R ∗ C ,

   where e is a non-zero vector in ℋ , R is an operator of rank one, and C is a self-adjoint operator of rank one. Then

   ξ ( X ) + S = 0 ⊕ A 2 e ⊕ A 2 e R R ∗ C .

   Applying Lemma (2.4) [15], we get

   ∥ ξ ( X ) + S ∥ p < 0 R R ∗ C p = ∥ S ∥ p ( p ≠ 2 ) .

3. If ker A ≠ ker B and ker A ⊈ ker B , we proceed similarly as in the first case, it suffices to replace A by B and B by A in the preceding argument.

4. If ker A = ker B , then with respect to the decomposition

   ℋ = ( ker B ) ⊥ ⊕ ker B ,

   we get

   A = A 1 ⊕ 0 , B = B 1 ⊕ 0 .

   Letting S = ( S i j ) i , j = 1 , … , 3 . Then

   S ∈ ker ξ ⇒ A 1 S 11 A 1 ∗ = B 1 S 11 B 1 ∗ ,

   and the other entries are arbitrary. Choosing X and S as

   X = ( e ⊗ e ) ⊕ 0 , S = 0 R R ∗ C ,

   where e and R are as in (i). Then

   ξ ( X ) + S = A 1 e ⊗ A 1 e − B 1 e ⊗ B 1 e R R ∗ C .

   If A 1 e ⊗ A 1 e = B 1 e ⊗ B 1 e for all e ∈ ℋ , then it follows from this fact and the injectivity of A 1 and B 1 that A 1 = α B 1 with ∣ α ∣ = 1 and hence ξ = 0 , which is a contradiction with the assumption ξ ≠ 0 . We use Lemma (2.4) [15] to complete the proof for p ∉ { 1 , 2 } .

5. In the case p = 1 , let us assume that A 1 e ⊗ A 1 e − B 1 e ⊗ B 1 e is an operator of rank 2 and has eigenvalues λ 1 , λ 2 with ∣ λ 1 ∣ = ∣ λ 2 ∣ for all e ∈ ℋ , then we can check that

   ∣ λ 1 ∣ = ∣ λ 2 ∣ ⇔ ∥ A 1 e ∥ = ∥ B 1 e ∥ and ⟨ A 1 e , B 1 e ⟩ = 0 ∀ e ∈ ℋ .

   If ker B ∗ ⊆ ker B , then by the injectivity of B 1 ∗ , it follows that A 1 = 0 , which is a contradiction since A ≠ 0 .

6. If ker B ∗ ⊈ ker B , with respect to the decomposition

   ℋ = ( ker B 1 ∗ ) ⊥ ⊕ ker B 1 ∗ ⊕ ker B

   we get

   A = A 1 ⊕ A 2 ⊕ 0 , B = B 1 B 2 0 0 ⊕ 0 , and S = ( S i j ) i , j = 1 , … , 3 .

   Since A 1 is injective and S ∈ ker ξ , we obtain

   A 1 S 11 A 1 ∗ = B 1 S 11 B 1 ∗ ; S 12 = S 21 = S 22 = 0

   and the other entries are arbitrary.

7. We rewrite S on the following decomposition

   ℋ = ( ker B 1 ∗ ) ⊥ ⊕ ker B ⊕ ker B 1 ∗

   and choose S 11 = S 23 = S 32 = 0 , S 13 = R , S 31 = R ∗ , S 34 = C , and X = e ⊗ e ⊕ 0 ( R , C , e are as in (i)). Then

   ξ ( X ) + S = A 1 e ⊗ A 1 e − B 1 e ⊗ B 1 e R R ∗ C ⊕ 0 .

   If A 1 e ⊗ A 1 e − B 1 e ⊗ B 1 e is an operator of rank two and has eigenvalues λ 1 , λ 2 with ∣ λ 1 ∣ = ∣ λ 2 ∣ for all e ∈ ℋ , then from the previous argument we conclude that A 1 = 0 . On the other hand, if B 2 ≠ 0 , then from [ A , B ] = 0 it follows that A 2 = 0 , also a contradiction with the fact that A ≠ 0 .

8. If B 2 = 0 ( ker B ∗ reduces B ∗ ), then B = B 1 ⊕ 0 , A = 0 ⊕ A 2 ⊕ 0 ( A 2 is injective), and S = ( S i j ) i , j = 1 , … , 3 . Hence, it follows from A S A ∗ = B S B ∗ that

   S = 0 S 12 S 13 S 21 0 S 23 S 31 S 32 S 33

   and the other entries are arbitrary.

9. To conclude the proof we can argue similarly as in the first case (i).□

Corollary 11

Let ( A 1 , A 2 ) , ( B 1 , B 2 ) be 2-tuples of doubly commuting operators in B ( ℋ ) and E : B ( ℋ ) → C p ( ℋ ) ; ( 1 ≤ p ≤ ∞ , p ≠ 2 ) be the elementary operator defined by E ( X ) = A 1 X B 1 − A 2 X B 2 such that ker A 1 , ker A 2 , ker B 1 ∗ , and ker B 2 ∗ reduce A 1 , A 2 , B 1 ∗ , and B 2 ∗ , respectively. If E is orthogonal, then

Proof

Consider the space ℋ ⊕ ℋ and the following decompositions

Then, for all X ∈ B ( ℋ ) , we have A Y A ∗ − B Y B ∗ = A 1 X B 1 − A 2 X B 2 , ker A = ker A 1 ∩ ker B 1 ∗ , and ker B = ker A 2 ∩ ker B 2 ∗ . Hence,

So to achieve the proof, it suffices to apply the previous proposition.□

Theorem 12

Let A 1 , A 2 , N 1 , and N 2 be operators in B ( ℋ ) such that ( N 1 , N 2 ) is 2-tuple normal and [ A 1 , N 1 ] = [ A 2 , N 2 ] = 0 . Let E ( X ) = A 1 X A 2 − N 1 X N 2 such that E ( X ) , E ˜ ( X ) ∈ J . If E is the Fuglede operator, then E is w -orthogonal and E ∠ E ˜ . Furthermore,

1. If J = C p ( ℋ ) : ( 1 ≤ p < ∞ , p ≠ 2 ) , then E is orthogonal if and only if ker A 1 ∩ ker N 1 = { 0 } = ker A 2 ∗ ∩ ker N 2 ∗ ;

2. If ker A 1 ∩ ker N 1 = { 0 } = ker A 2 ∗ ∩ ker N 2 ∗ , then E is orthogonal and E ⊥ E ˜ ;

3. If ker A 1 ∩ ker N 1 ≠ { 0 } or ker A 2 ∗ ∩ ker N 2 ∗ ≠ 0 , then for all X ∈ B ( ℋ ) and all S ∈ ker E ∩ J ,

   min { ∥ E ( X ) + S ∥ J , ∥ E ˜ ( X ) + S ∥ J } ≥ k ∥ S ∥ J ,

   where

   1. J = B ( ℋ ) : k = 1 2 n ;

   2. J = C p ( ℋ ) : k = 1 2 4 − 2 p p , if 1 ≤ p ≤ 2 and k = 1 2 2 p − 4 p , if 2 ≤ p < ∞ .

Proof

1. The implication ( ⇒ ) follows from Corollary 8.

2. Let ξ : B ( ℋ ) → J be the elementary operator defined by ξ ( X ) = A X A ∗ − N X N ∗ , where A , N ∈ B ( ℋ ) , N is normal with [ A , N ] = 0 . Assume that ξ is the Fuglede operator.

3. Suppose that N is invertible and set D = N − 1 A . ξ is Fuglede implies that ξ D is Fuglede, where ξ D is the elementary operator induced by D . By Lemma 2.4 [16], we have

   ∥ ξ ( X ) + S ∥ J = ∥ A X A ∗ − N X N ∗ + S ∥ J = ∥ D ( N X N ∗ ) D ∗ − N X N ∗ + S ∥ J ≥ ∥ S ∥ J .

   Similarly, it can be shown that ∥ ξ ˜ ( X ) + S ∥ J ≥ ∥ S ∥ J for any operator S ∈ ( ker ξ ) ∩ J .

4. Suppose that N is injective, set Δ n = { λ ∈ C : ∣ λ ∣ ≤ 1 n ; n ∈ N } and μ ( Δ n ) denotes the corresponding spectral projection, where P n = I − μ ( Δ n ) converges strongly to I .

5. From [ A , N ] = 0 and by Fuglede-Putnam’s theorem it follows that [ A , N ∗ ] = 0 and therefore P n ℋ reduces both A and N . Let

   ℋ = P n ℋ ⊕ ( I − P n ) ℋ ,

   then A = A 1 n ⊕ A 2 n ; N = N 1 n ⊕ N 2 n and P n = I ⊕ 0 , where N 1 n is invertible. Hence,

   P n ( ξ ( X ) + S ) P n = [ A 1 n X 11 A 1 n ∗ − N 1 n X 11 N 1 n ∗ + S 11 ] ⊕ 0

   for all X = ( X i j ) i , j = 1 , 2 ∈ B ( ℋ ) and S = ( S i j ) i , j = 1 , 2 ∈ ker ξ implying

   ∥ ξ ( X ) + S ∥ J ≥ ∥ P n ( ξ ( X ) + S ) P n ∥ J = ∣ A 1 n X 11 A 1 n ∗ − N 1 n X 11 N 1 n ∗ + S 11 ∥ J ≥ ∥ S 11 ∣ J .

   Since P n ( ξ ∗ ( X ) + S ) P n = [ A 1 n ∗ X 11 A 1 n − N 1 n ∗ X 11 N 1 n + S 11 ] ⊕ 0 , then

   ∥ ξ ∗ ( X ) + S ∥ J ≥ ∥ P n ( ξ ∗ ( X ) + S ) P n ∥ J = ∥ A 1 n ∗ X 11 A 1 n − N 1 n ∗ X 11 N 1 n + S 11 ∥ J ≥ ∥ S 11 ∥ J .

   On the other hand, for all positive integer n , we have

   ∥ S 11 ∥ J = ∥ P n S P n ∥ ≤ min { ∥ ξ ( X ) + S ∥ J , ∥ ξ ∗ ( X ) + S ∥ J } < ∞

   and thus sup n ∥ P n S P n ∥ < ∞ for all S ∈ ker ξ ∩ J and X ∈ B ( ℋ ) . It follows by Lemma 3 [14] that

   min { ∥ ξ ( X ) + S ∥ J , ∥ ξ ˜ ( X ) + S ∥ J } ≥ ∥ S ∥ J .

6. Suppose that N is an arbitrary normal operator.

7. If ker A ∩ ker N = { 0 } , then ℋ may be decomposed as

   ℋ = [ ( ker N ) ⊥ ⊖ ker A ] ⊕ ker A ⊕ ker N .

   Since ξ is Fuglede and by Lemma 5(1.(i)), we have ker A reduces A , A = A 11 ⊕ 0 ⊕ A 22 and N = N 11 ⊕ N 22 ⊕ 0 .

8. For X = ( X i j ) i , j = 1 , 2 , 3 ∈ B ( ℋ ) , set ξ 1 ( X 11 ) = A 11 X 11 A 11 ∗ − N 11 X 11 N 11 ∗ and S = ( S i j ) i , j = 1 , 2 , 3 ∈ ker ξ , then

   ξ 1 ( S 11 ) = 0 and A 11 S 13 A 22 ∗ = A 22 S 31 A 11 ∗ = A 22 S 33 A 22 ∗ = 0 ,

   N 11 S 12 N 22 ∗ = N 22 S 21 N 11 ∗ = N 22 S 22 N 22 ∗ = 0 .

   Since ξ is Fuglede, then S ∈ ker ξ ˜ and

   (7) ξ 1 ∗ ( S 11 ) = 0 and A 11 ∗ S 13 A 22 = A 22 ∗ S 31 A 11 = A 22 ∗ S 33 A 22 = 0 ,

   N 11 ∗ S 12 N 22 = N 22 ∗ S 21 N 11 = N 22 ∗ S 22 N 22 = 0 .

   Since N 11 , N 22 , A 11 , and A 22 are injective, we get from (ii) that ξ 1 is orthogonal, ξ 1 ⊥ ξ 1 ˜ and S 13 = S 31 = S 33 = S 12 = S 22 = S 21 = 0 and any operator S ∈ ker ξ has the form

   S = S 11 0 0 0 0 S 23 0 S 32 0 ,

   where S 23 and S 32 are arbitrary.

9. Let S 23 = U 23 ∣ S 23 ∣ , S 32 = U 32 ∣ S 32 ∣ be the polar decompositions of S 23 and S 32 , respectively, let V be the operator defined by

   V = I ⊕ 0 U 32 ∗ U 23 ∗ 0 .

   Then

   ∥ ξ ( X ) + S ∥ J ≥ ∥ V ( ξ ( X ) + S ) ∥ J = ξ 1 ( X 11 ) + S 11 ∗ ∗ ∗ ∣ S 32 ∣ ∗ ∗ ∗ ∣ S 23 ∣ .

   Applying Lemma 6, we get

   1. if J = B ( ℋ ) :

      ∥ ξ ( X ) + S ∥ ≥ max { ∥ ξ 1 ( X 11 ) + S 11 ∥ , ∥ S 32 ∥ , ∥ S 23 ∥ } ≥ max { ∥ S 11 ∥ , ∥ S 32 ∥ , ∥ S 23 ∥ } = ∥ S ∥ .

   2. if J = C p ( ℋ ) ; ( 1 ≤ p < ∞ ) :

      ∥ ξ ( X ) + S ∥ p ≥ ( ∥ ξ 1 ( X 11 ) + S 11 ∥ p p + ∥ S 32 ∥ p p + ∥ S 23 ∥ p p ) 1 / p ≥ ( ∥ S 11 ∥ p p + ∥ S 32 ∥ p p + ∥ S 23 ∥ p p ) 1 / p = ∥ S ∣ p .

   By the same method, we have that ξ ⊥ ξ ˜ .

10. If M = ker A ∩ ker N ≠ { 0 } and ℋ is decomposed as

    ℋ = ( ker N ) ⊥ ⊕ [ ker N ⊖ M ] ⊕ M ,

    then by Lemma 5(ii) and the fact that ξ is Fuglede, it follows that M reduces A , and hence

    A = A 11 ⊕ A 22 ⊕ 0 , N = N 11 ⊕ 0 .

    For X = ( X i j ) i , j = 1 , 2 , 3 , we set ξ 1 ( X ) = A 11 X 11 A 11 ∗ − N 11 X 11 N 11 ∗ and let S = ( S i j ) i , j = 1 , 2 , 3 ∈ ker ξ . From the injectivity of A 11 and A 22 , we obtain

    ξ 1 ( S 11 ) = 0 and S 12 = S 21 = S 22 = 0 .

    By simple computation, we get

    ∥ ξ ( X ) + S ∥ J = ξ 1 ( X 11 ) + S 11 ∗ S 13 ∗ ∗ S 23 S 31 S 32 S 33 J .