Range-Kernel orthogonality and elementary operators on certain Banach spaces

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

Definition 2

If E : X → Y and T : Y → Z are bounded linear operators between Banach spaces. T is called orthogonal to E provided

Theorem 3

Let K be a closed linear subset of X and x ∉ K , then

Proof

Let φ ∈ D ( x ) such that ⟨ φ , y ⟩ = 0 , for all y ∈ K . Then,

and

that is, ∥ x ∥ ≤ ∥ x + y ∥ , for all y ∈ K . Hence, x ⊥ K .

For the converse, let x ∉ K such that x ⊥ K . Then, for all y ∈ K , x and y are linearly independent vectors. Let L be the closed subspace spanned by K and { x } , L = [ K , { x } ] , and define the function φ on L by φ ( α x + β y ) = α ∥ x ∥ 2 for all y ∈ K and all α , β ∈ C . Clearly φ is linear (by the assumption that K is a linear subset of X ). To prove the continuity of φ , let z ∈ L , then z = α x + β y and φ ( z ) = α ∥ x ∥ 2 . By the definition of ⊥ and the assumption that x ⊥ K , we derive that ∥ z ∥ ≥ ∥ α x ∥ . If α ≠ 0 , we have

If α = 0 , then ∣ φ ( z ) ∣ = ∣ φ ( β y ) ∣ = 0 . Hence,

Therefore, φ is continuous on L and ∥ φ ∥ = ∥ x ∥ . By Hahn-Banach’s theorem there is a continuous linear functional φ ˜ on X such that φ ˜ ∣ L = φ and ∥ φ ˜ ∥ = ∥ φ ∥ , where φ ˜ ∣ L is the restriction of φ ˜ on L . It follows, by the definition of φ and φ ˜ ∣ L = φ , that K ⊆ ker φ ˜ and φ ˜ ∈ D ( x ) .□

Corollary 4

1. ∀ x , y ∈ X   x ⊥ y ⇔ ∃ φ ∈ D ( x ) : φ ( y ) = 0 .

2. ∀ x ∈ X , ∀ φ ∈ D ( x ) : x ⊥ ker φ .

Remark 5

It is clear that if { x n } n is a sequence in a subset K converging to y and x ⊥ x n , for all n , then x ⊥ y . Hence, x ⊥ K ⇒ x ⊥ K ¯ .

Lemma 6

Let X be a Banach space.

1. If K and L are closed subspaces of X and K ⊆ L ⊥ r , then K ⊕ L is closed.

2. Let T ∈ B ( X ) and s ∈ X . Then

   1. s ⊥ ran T ⇔ ∃ φ ∈ D ( s ) : φ ∈ ker T † ;

   2. If T is orthogonal, then ker T ⊕ ran ( T ) ¯ is a closed subspace of X .

Proof

1. Let z ∈ X : z = lim n ( x n + y n ) and z n = ( x n + y n ) ; for all n ≥ 1 , where x n ∈ K and y n ∈ L . From K ⊆ L ⊥ r , we get

   ∥ z n − z n + p ∥ = ∥ y n − y n + p + x n − x n + p ∥ ≥ ∥ x n − x n + p ∥ , for all n , p .

   Then, { x n } n is a Cauchy sequence, hence lim n x n exists in K . Setting x = lim n x n , we get lim n y n = z − x ∈ L , and therefore, z ∈ K ⊕ L .

2. 1. By Theorem 3 and Remark 5, s ⊥ ran T ⇔ ∃ φ ∈ D ( s ) , ran ( T ) ⊆ ker φ , that is, φ ( T x ) = ( T † φ ) x = 0 , for all x ∈ X . Hence, s ⊥ ran T ⇔ ∃ φ ∈ D ( s ) : φ ∈ ker T † .

   2. It is a direct consequence of assertions (1) and (2)(i).□

Theorem 7

Let T and f : X → X be bounded maps not necessarily linear or additive, and a ∈ X . Then the following assertions hold:

1. Suppose that the relation f ( x ) + T ( y ) = f ( y ) + T ( x ) holds for all x , y ∈ X ; if f ( a ) ⊥ T ( x ) , for all x ∈ X , then the map F f defined by (6) has a global minimum at a ;

2. If we suppose that T is linear and f ( x ) = T ( x ) + f ( 0 ) , for all x ∈ X , then F f defined by (6) has a global minimum at a if and only if f ( a ) ⊥ ran T if and only if there is φ ∈ D ( f ( a ) ) such that φ ∈ ker T † , where T † is the duality map of T . Moreover, if f ( a ) is a smooth point (or the space X is smooth), then the existence of φ is unique.

Proof

1. It follows from Corollary 1 that for all x ∈ X , f ( a ) ⊥ T ( x ) ⇔ ∃ φ ∈ X † : φ ( f ( a ) ) = ∥ f ( a ) ∥ 2 = ∥ φ ∥ 2 and φ ( T ( x ) ) = 0 . Then, by the relation defined in (i), we get φ ( f ( a ) ) = φ ( f ( a ) + T ( x ) ) = φ ( f ( x ) + T ( a ) ) = φ ( f ( x ) ) . So that ∥ φ ( f ( a ) ) ∥ = ∥ f ( a ) ∥ ∥ φ ∥ ≤ ∥ φ ∥ ∥ f ( x ) ∥ . Hence, ∥ f ( a ) ∥ ≤ ∥ f ( x ) ∥ , for all x ∈ X .

2. Since the maps f , T satisfy the relation cited in (i), the sufficient condition follows from (i). By linearity of T , we get f ( a ) + λ T ( x ) = f ( a + λ x ) for all x ∈ X , λ ∈ C . Hence, F f has a global minimum at a implies ∥ f ( a ) + λ g ( x ) ∥ = ∥ f ( a + λ x ) ∥ ≥ ∥ f ( a ) ∥ . The other equivalence follows from Lemma 1.

   If f ( a ) is a smooth point (or the space X is smooth), then f ( a ) has only one functional support (for the functional support of smooth points, see [10]) and therefore D ( f ( a ) ) has one element (the map D is a single-valued function if X is smooth).□

Lemma 8

[8] If ℐ is a separable ideal of compact operators in B ( ℋ ) equipped with unitary invariant norm, then its dual ℐ † is isometrically isomorphic to an ideal of compact operators J not necessarily separable as follows:

Corollary 9

Let A ∈ C p ( ℋ ) ; ( 1 ≤ p < ∞ ), T ∈ B ( C p ( ℋ ) ) and f , F f are defined as in Theorem 7(ii) and where f ( A ) is given by its polar decomposition f ( A ) = u ∣ f ( A ) ∣ . Then, the following assertions hold

1. If A ∈ C p ( ℋ ) ; ( 1 < p < ∞ ) , then F f has a global minimizer at A if and only if f ( A ) ⊥ ran T if and only if ∣ f ( A ) ∣ p − 1 u ∗ ∈ ker T † ;

2. If A ∈ C 1 ( ℋ ) and f ( A ) is a smooth point, then F f has a global minimizer at A if and only if f ( A ) ⊥ ran T if and only if u ∗ ∈ ker T † when f ( A ) is injective (or u ∈ ker T † when f ( A ) ∗ is injective).

Proof

From Theorem 7, we have that F f has a global minimizer at A if and only if there exists φ ∈ D ( f ( A ) ) such that φ ∈ ker T † ,

1. If A ∈ C p ( ℋ ) ; ( 1 < p < ∞ ) , then by the isomorphism in (8), it follows that

   φ ∈ C p ( ℋ ) † ⇔ ∃ R ∈ C q : 1 < q < ∞ ( ℋ ) ; 1 p + 1 q = 1 : ϕ R = φ ; ∥ φ ∥ = ∥ R ∥ and φ ( X ) = tr R X for all X ∈ C p ( ℋ ) .

   Hence, by the smoothness of C p : 1 < p < ∞ ( ℋ ) , F f has a global minimizer at A if there is a unique operator R such that φ ( f ( A ) ) = tr ( f ( A ) R ) = ∥ f ( A ) ∥ p 2 = ∥ R ∥ q 2 and tr ( T † ( R ) X ) = 0 , for all X ∈ C p : 1 < p < ∞ ( ℋ ) . To finish the proof, it suffices to take R = ∥ f ( A ) ∥ p 2 − p ∣ f ( A ) ∣ p − 1 u ∗ and the rank one operator X = e ⊕ e , e ∈ ℋ . So that with simple calculation, we get ⟨ T ∗ ( R ) e , e ⟩ = 0 , for all e ∈ ℋ and thus T † ( R ) = T † ( ∣ f ( A ) ∣ p − 1 u ∗ ) = 0 . Conversely, T † ( ∣ f ( A ) ∣ p − 1 u ∗ ) = 0 ⇒ T † ( R ) = 0 and then t r ( T † ( R ) X ) = 0 , for all X ∈ C p : 1 < p < ∞ ( ℋ ) .

2. It is well known that C 1 ( ℋ ) is not reflexive, even not smooth space and its dual C 1 ( ℋ ) † is isometrically isomorphic to B ( ℋ ) . This isomorphism is given by

   (9) ϕ : B ( ℋ ) ⇔ C 1 ( ℋ ) † , R ↦ ϕ R : ϕ R ( X ) = tr ( X R ) .

   Then,

   φ ∈ C 1 ( ℋ ) † ⇔ ∃ R ∈ B ( ℋ ) : ϕ R = φ ; ∥ φ ∥ = ∥ R ∥ and φ ( X ) = tr R X for all X ∈ C 1 ( ℋ ) .

   So that if f ( A ) is a smooth point, then F f has a global minimizer at A iff there is a unique operator R such that

   φ ( f ( A ) ) = tr ( f ( A ) R ) = ∥ f ( A ) ∥ 1 2 = ∥ R ∥ 2

   and

   tr ( T † ( R ) X ) = 0 , ∀ ∈ C 1 ( ℋ ) .

   Since f ( A ) is smooth, then by Holub [11], either f ( A ) or f ( A ) ∗ is injective, thus either u or u ∗ is an isometry, i.e., u u ∗ = I or u ∗ u = I . So it suffices to take R = ∥ f ( A ) ∥ 1 u ∗ or R = ∥ f ( A ) ∥ 1 u , which is the unique operator required in both cases f ( A ) or f ( A ) ∗ is injective.□

Lemma 10

Let K be a closed subspace of X . If X is reflexive and K ⊥ r = { 0 } , then K = X .

Proof

If K ≠ X , then there exists φ ∈ X † : K ⊆ ker φ . Since D ( φ ) is not empty, then there is f ∈ X † † with f ( φ ) = ∥ φ ∥ 2 = ∥ f ∥ 2 . Let J be the natural injection between X and X † † , i.e.,

So by the reflexivity of X , J is a bijection and then there is 0 ≠ x ∈ X such that J ( x ) = f . Hence, φ ( x ) = ∥ x ∥ 2 = ∥ φ ∥ 2 . Thus, φ ∈ D ( x ) and by application of Theorem 1, we get 0 ≠ x ∈ K ⊥ r = { 0 } , a contradiction.□

Proposition 11

Let X be a reflexive, smooth and strictly convex Banach space and T ∈ B ( X ) , if T † is orthogonal. Then

Proof

Let s ∈ X such that s ⊥ ran T . Then, by Lemma 6(2.i) and the smoothness of X , there is a unique φ s ∈ D ( s ) such that φ s ∈ ker T † . Again, by assumptions and Lemma 6(2.i), there is ψ φ s ∈ D ( φ s ) such that ψ φ s ∈ ker T † † . Let J be the natural injection between X and X † † as defined in (10). We see that J ( s ) φ s = ∥ φ s ∥ 2 and ∥ J ( s ) ∥ = ∥ φ s ∥ , which means that J ( s ) ∈ D ( φ s ) . By the reflexivity of X , J is a bijection. Hence, there is c ∈ X : ψ φ S = J ( c ) and ∥ J ( c ) ∥ = ∥ c ∥ = ∥ φ s ∥ , J ( c ) φ s = φ s ( c ) = ψ φ s ( φ s ) = ∥ φ s ∥ 2 . Then φ s ∈ D ( s ) ∩ D ( c ) and since X is strictly convex, we get c = s . Thus, J ( c ) = J ( S ) ∈ ker T † † . Therefore, ( T † † J ( s ) ) φ = J ( s ) ( T † φ ) = ( T † φ ) s = φ ( T s ) = 0 , for all φ ∈ X † , that is, s ∈ ker T .□

Remark 12

If X is a reflexive separable Banach space and T † is orthogonal, then the implication (11) holds with respect to suitable norm in X . Indeed, if X is separable, then there is an equivalent norm, which is smooth and strictly convex in X .

Corollary 13

Let X be a reflexive, smooth and strictly convex Banach space and T ∈ B ( X ) . If T and T † are orthogonal, then

Proof

If T is orthogonal, then, by Definition 2, it follows that

and the reverse implication of (12) follows by Proposition 1. Let us prove the decomposition (13): let y ∈ X such that y ∈ ( ker T ⊕ ran ( T ) ¯ ) ⊥ r , then there is φ y ∈ D ( y ) such that φ y ( s ⊕ T X ) = 0 , for all s ∈ ker T and all x ∈ X . For s = 0 , it follows, by Lemma 6(2.i), that Y ⊥ ran T , and by Proposition 1, ran T ⊆ ker T . So that, we can choose x = 0 and s = y , this yields φ y ( y ) = 0 . This means y ⊥ y , and hence y = 0 . Finally, the decomposition (13) follows from Lemma 10.□

Lemma 14

Let ℋ , K be Hilbert spaces, A ∈ B ( ℋ ) , B ∈ B ( K ) and E ∈ B ( B ( K , ℋ ) ) such that E ( X ) = A X B . If A and B ∗ are injective operators, then E is injective.

Proof

If A X B = 0 with A injective, then X B = 0 = B ∗ X ∗ = 0 implies X ∗ = 0 = X since B ∗ is injective. Thus, E is injective.□

Lemma 15

Let A = ( A 1 , A 2 , … , A n ) and B = ( B 1 , B 2 , … , B n ) with A i , B i be operators in B ( ℋ ) such that

If E is the elementary operator defined on C p : 1 ≤ p < ∞ ( ℋ ) by

then

1. ker E = ker E ˜ ⇒ ker E † = ker E † ˜ ;

2. If E ( S ) = 0 = E ˜ ( S ) for some compact operator S , then [ ∣ S ∣ , B i ] = 0 for all 1 ≤ i ≤ n .

Proof

1. E † ( S ) = 0 ⇔ E ˜ ( S ∗ ) = 0 . From the equality, ker E = ker E ˜ , it follows that E † ( S ) = 0 ⇔ E ( S ∗ ) = 0 ⇔ E † ˜ ( S ) = 0 .

2. See [12].□

Proposition 16

Let E be an elementary operator defined on C p : 1 < p < ∞ ( ℋ ) , then

in one of the following cases:

1. E ( X ) = A X B with A ∗ and B injective operators;

2. E ( X ) = A X B − C X D , where A , B normal operators, D , C ∗ hyponormal operators with [ A , C ] = [ B , D ] = 0 and ker A ∗ ∩ ker C ∗ = { 0 } = ker B ∩ ker D .

Proof

1. The duality adjoint E † is defined by E † ( X ) = B X A and using Lemma 6, we get E † is injective and then is orthogonal. So the result follows by Proposition 1.

2. We have E † ( X ) = B X A − D X C and applying Duggal’ result [6], we get E † is orthogonal, and then by Proposition 1, the proof is complete.□

Theorem 17

Let E be an elementary operator defined on C p : 1 < p < ∞ ( ℋ ) and the map F E : C p : 1 < p < ∞ ( ℋ ) → R + defined by F E ( X ) = ∥ E ( X ) + S ∥ p . Then the following assertions:

1. F E has a global minimizer at S ∈ C p : 1 < p < ∞ ( ℋ ) ;

2. S ⊥ ran E ;

3. S ∈ ker E

1. E ( X ) = A X B such that A , A ∗ , B and B ∗ are injective operators;

2. E ( X ) = A X B − C X D ; such that ( A , C ) and ( B , D ) are 2-tuples of commuting normal operators and ker A ∩ ker C = { 0 } = ker B ∩ ker D ;

3. E ( X ) = ∑ i = 1 , … , n A i X B i − X ; such that ∑ i = 1 n A i A i ∗ ≤ 1 , ∑ i = 1 n A i ∗ A i ≤ 1 , ∑ i = 1 n B i B i ∗ ≤ 1 , ∑ i = 1 n B i ∗ B i ≤ 1 and ker E = ker E ˜ .

Proof

From Theorem 2, we know that ( i ) ⇔ ( i i ) for any operator E . When the elementary operator E takes the form (a) or (b) we have, by Proposition 16, that ( i i ) ⇒ ( i i i ) and with respect to the hypothesis cited in our theorem, we argue similarly as in Proposition 16 in the reverse sense, to get ( i i i ) ⇒ ( i i ) .

The case when E takes the form (3) (as an application of the previous section, we give a simpler and shorter proof of ( i i i ) ⇔ ( i i ) than given by Duggal [13]). We start by proving that E is orthogonal.

Let 0 ≠ S = u ∣ S ∣ ∈ ker E , then ∑ i = 1 , … , n A i u ∣ S ∣ B i = u ∣ S ∣ = ∑ i = 1 , … , n A i ∗ u ∣ S ∣ B i ∗ and by Lemma 6(ii), we get ∑ i = 1 , … , n A i ∗ u B i ∗ ∣ S ∣ = u ∣ S ∣ p − 1 , multiply both sides of the equation by ∣ S ∣ p − 2 , hence by the commutativity derived from Lemma 6(ii), we obtain ∑ i = 1 , … , n A i ∗ u ∣ S ∣ p − 1 B i ∗ = u ∣ S ∣ p − 1 . Then, it follows by taking the adjoint that ∑ i = 1 , … , n B i ∣ S ∣ p − 1 u ∗ A i = ∣ S ∣ p − 1 u ∗ , which means that ∣ S ∣ p − 1 u ∗ ∈ ker E † and by Corollary 4(i), we get that E is orthogonal.