Characterizations of Strong Strictly Singular Operators

The spacesX andYwill denote normed spaces, andT : X → Y will denote a bounded linear mapping from a normed spaceX into a normed spaceY in this paper. Completeness is assumed only when it is specifically stated. An operator T is called strictly singular if it does not have a bounded inverse on any infinite dimensional subspace contained inX. If T(B X ) is totally bounded in Y, where B X is the open unit ball in X, then T is called a precompact operator. If T(B X ), closure of


Introduction
The spaces  and  will denote normed spaces, and  :  →  will denote a bounded linear mapping from a normed space  into a normed space  in this paper.Completeness is assumed only when it is specifically stated.An operator  is called strictly singular if it does not have a bounded inverse on any infinite dimensional subspace contained in .If (  ) is totally bounded in , where   is the open unit ball in , then  is called a precompact operator.If (  ), closure of (  ), is compact in , then  is called a compact operator.Every precompact operator is strictly singular (cf: [1]).The collection of all strictly singular (precompact) operators from  into  forms a closed subspace of the normed space B(, ), the collection of all bounded linear operators from  into  (cf: [1]).The collection of strictly singular (precompact) operators on  (from  into ) forms a closed ideal of the normed algebra B()(= B(, )) (cf: [1]).A linear transformation  :  →  has a bounded inverse if and only if ‖‖ ≤ ‖‖ for all  ∈ , for some  > 0. It is easy to see as a consequence of the open mapping theorem that a continuous linear transformation  from a Banach space  into a Banach space  has closed range, if and only if for given  ∈ , there is an element  ∈  such that  =  and ‖‖ ≤ ‖‖, for some fixed  > 0 (see [2]).This gives a motivation to define a new class of operators called strong strictly singular operators.Write X = /(), where () is the null space of , as the quotient space endowed with the quotient norm.Let   denote the quotient map from  onto X.The 1-1 operator T : X →  induced by  is defined by T( + () =   ()) = .Note that T is 1-1 and linear with range same as the range of .If  is 1-1 then  = T. Also,  X =   (  ) and (  ) = T( X).So, we have the following conclusions.
(a)  is precompact on  if and only if T is precompact on X.(b)  is compact on  if and only if T is compact on X.

Definition
Definition 1.An operator  :  →  is said to be strong strictly singular if for any subspace  of  such that dim () = ∞ and the null space () ⊂  there is no positive number  with the property: for a given  ∈ , there is an element  ∈  such that  =  and ‖‖ ≤ ‖‖.Lemma 2. Every strong strictly singular operator is strictly singular.
The converse of Lemma 2 is not true.For example, there is an onto continuous linear operator  : ℓ 1 → ℓ 2 which is strictly singular [1, page 89, III.3.7] but not strong strictly singular, because it has a closed range.
A simple characterization for strong strictly singularity is the following lemma.
On the other hand, suppose that  is any subspace of  such that () ⊂  and such that for a given  ∈  there is an element  ∈  such that  =  and ‖‖ ≤ ‖‖, for fixed  > 0. Then ( This proves that if T is strictly singular, then  is strong strictly singular.This completes the proof.

Corollary 4.
If  is 1-1 and strictly singular, then  is strong strictly singular.
Corollary 5. Every precompact operator is strong strictly singular.
Proof.If  :  →  is precompact, then T so is.Hence T is strictly singular, and hence  is strong strictly singular.
Although the converse of Lemma 2 is not true in general, it is true partially which is seen from the next theorem.Theorem 6.Let  :  →  be a strictly singular operator.Suppose that  = () ⊕ , for some subspace  of , and the corresponding projection  :  →  defined by (+) =  with  ∈ () and  ∈  is continuous.Then  is strong strictly singular.
Corollary 7. Let  :  →  be a strictly singular operator on a Banach space  such that  = () ⊕ , for some closed subspace  of .Then  is strong strictly singular.

Characterization
A known classical characterization of strictly singular operators is given in Theorem 9. A new similar characterization for strictly singular operators is found in the paper [4], when the operators are 1-1 and they are from a Banach space  into  itself.
(i)  is strictly singular.
(ii) For every infinite dimensional subspace  ⊂ , there exists an infinite dimensional subspace  ⊂  such that  is precompact on .
(iii) Given  > 0 and given , an infinite dimensional subspace of , there exists an infinite dimensional subspace  ⊂  such that  restricted to  has norm not exceeding .
(iv) Given  > 0 and given , an infinite dimensional subspace of , there exists an infinite dimensional subspace  ⊂  such that  restricted to  is precompact and it has norm not exceeding .
Using Lemma 3 and Theorem 9, one can easily verify the following.
(i)  is strong strictly singular.
(iii) Given  > 0 and given , an infinite dimensional subspace of , satisfying () ⊂  and dim () = ∞, there exists an infinite dimensional subspace  of  satisfying () ⊂  ⊂  and dim () = ∞ such that  restricted to  has norm not exceeding .
(iv) Given  > 0 and given , an infinite dimensional subspace of  satisfying () ⊂  and dim () = ∞, there exists an infinite dimensional subspace  of  satisfying () ⊂  ⊂  and dim () = ∞ such that  restricted to  is precompact and its has norm not exceeding .
The following interesting theorem on automatic continuity is due to van Dulst [5].
Theorem 11 (see [5]).Let  :  →  be a linear transformation (need not be continuous).Suppose that there exists a constant  > 0 with the property that every infinity dimensional subspace of  contains a vector  such that ‖‖ = 1 and ‖‖ < .Then there exists a subspace  ⊂  with dim / < ∞ such that   , the restriction of  to , is continuous.
Corollary 12. Let  :  →  be a bounded linear transformation.Then the following statements are equivalent.
(a)  is strong strictly singular.
Proof.(a) ⇒ (b): Suppose that (b) fails to be true.Then there is a subspace  of  satisfying dim () = ∞ and there is a positive constant  such that for every subspace  1 of  satisfying dim ( 1 ) = ∞ there is an element  ∈  1 +() such that ‖‖ ≤ ‖‖.Then, there is a subspace M0 of   () such that dim   ()/ M0 < ∞ and T restricted to M0 is a homeomorphism; which follows from Theorem 11.So, T is not strictly singular.Hence,  is not strong strictly singular.This proves (a) ⇒ (b).(b) ⇒ (a): Trivial.
Proof.Note that ‖ T − T‖ → 0 and each T is strictly singular, by Lemma 3.So T is strictly singular.Hence  is strong strictly singular, by Lemma 3.
The following technical lemma is on vector spaces.
Lemma 14.Let  be an infinite dimensional subspace of  × , where  and  are vector spaces over the same field.Then at least one of the following is true.
1→  1 such that {(, ) :  ∈  1 } ⊂ .Proof.Let   ,   be the projections of  on  and , respectively.Then   is of infinite dimension, or   is of infinite dimension.Suppose that   is of infinite dimension.Let   be a basis of   .To each  ∈   , find an element  ∈   such that (, ) ∈ .Extend the map   →  to  :   →   as a linear mapping.If dim (  ) < ∞, then we select  1 as ker  so that (i) is true.If dim (  ) = ∞, then find a subspace  1 of   such that ( 1 ) = (  ) and  is 1-1 on  1 .In this case (iii) is true with  1 = (  ).Similarly, we can show that either (ii) or (iii) is true if dim   = ∞.Remark 15.It is possible to select  1 and  1 such that cardinalities of bases for  1 and  1 are equal to the cardinality of a basis for .Let  1 :  1 →  1 ,  2 :  2 →  2 be bounded linear operators between normed spaces.Define a bounded linear operator  =  1 ×  2 :  1 ×  2 →  1 ×  2 by ( 1 ,  2 ) = ( 1  1 ,  2  2 ).Then one has the following.To prove (ii), suppose that  1 and  2 are strictly singular operators.Let  be infinite dimensional subspace of  1 ×  2 .If there is an infinite dimensional subspace   1 of  1 such that   1 × {0} ⊂ , then we can find an infinite dimensional subspace   1 of   1 such that  1 is precompact on   1 .In this case, there is an infinite dimensional subspace   1 × {0} of  such that  is precompact on   1 × {0}.Similarly, if there is an infinite dimensional subspace   2 of  2 such that {0} ×   2 ⊂ , then  is precompact on an infinite dimensional subspace of .Suppose that there are infinite dimensional subspaces   1 and  2 of  1 and  2 , respectively, and an onto linear isomorphism  :   1 →   2 such that {( 1 ,  1 ) :  1 ∈   1 } ⊂ .Find an infinite dimensional subspace   1 of   1 such that  1 is precompact on   1 .Find an infinite dimensional subspace   2 of (  1 ) such that  2 is precompact on   2 .Then,  is precompact on an infinite dimensional subspace  −1 (  2 ) ×   2 of .This proves (ii).To prove (iii), suppose that  1 and  2 are strong strictly singular operators.Note that (i) If  1 and  2 are precompact (compact), then  is precompact (compact).(ii)If  1 and  2 are strictly singular operators, then  is a strictly singular operator.(iii)If  1 and  2 are strong strictly singular operators, then  is a strong strictly singular operator.Proof.(i) Is obvious.