Chinese Journal of MathematicsVolume 2013 (2013), Article ID 834637, 4 pageshttp://dx.doi.org/10.1155/2013/834637

Research Article

Characterizations of Strong Strictly Singular Operators

C. Ganesa Moorthy¹ and C. T. Ramasamy²

¹Department of Mathematics, Alagappa University, Karaikudi 630 005, India²Department of Mathematics, H. H. The Rajah’s College, Pudukkottai, Tamil Nadu 622001, India

Received 26 August 2013; Accepted 12 October 2013

Academic Editors: W. Klingenberg, X. Tang, and C. Yin

Copyright © 2013 C. Ganesa Moorthy and C. T. Ramasamy. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

A new class of operators called strong strictly singular operators on normed spaces is introduced. This class includes the class of precompact operators, and is contained in the class of strictly singular operators. Some properties and characterizations for these operators are derived.

1. 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 , 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 (cf: [1]). A linear transformation has a bounded inverse if and only if for all , for some . 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 (see [2]). This gives a motivation to define a new class of operators called strong strictly singular operators.

Write , where is the null space of , as the quotient space endowed with the quotient norm. Let denote the quotient map from onto . The 1-1 operator induced by is defined by . Note that is 1-1 and linear with range same as the range of . If is 1-1 then . Also, and . So, we have the following conclusions.(a) is precompact on if and only if is precompact on .(b) is compact on if and only if is compact on .

2. 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.

Proof. Let be strong strictly singular. Suppose is an infinite dimensional subspace of such that the restriction of on has a bounded inverse. Then, there is a positive constant such that for every . Since is 1-1 on , we have . Let us consider that . Then, . Also, for given , there are and such that and . This shows that is not strong strictly singular. This contradiction shows that is strictly singular, whenever is strong strictly singular.

The converse of Lemma 2 is not true. For example, there is an onto continuous linear operator 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.

Lemma 3. is strong strictly singular if and only if is strictly singular.

Proof. Suppose that is any subspace of such that, for all , we have , for some fixed . Then, for any fixed , for given , there is an element such that and This proves that if is strong strictly singular, then is strictly singular.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 . Then This proves that if 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 so is. Hence is strictly singular, and hence is strong strictly singular.

Note that the inclusion map , , is strong strictly singular but not precompact [3, page 170].

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.

Proof. Let us consider a subspace of such that and such that for each given , there is a satisfying and , for some fixed . Let . Then is 1-1 on . Moreover, for given , there are and such that , and . For this , there is a such that and . For this , let us write , with . Then, we have . Thus, we have for every . Since is strictly singular, . So, . This proves that is a strong strictly singular operator.

Corollary 7. Let be a strictly singular operator on a Banach space such that , for some closed subspace of . Then is strong strictly singular.

Corollary 8. Let be a strictly singular operator such that . Then is strong strictly singular.

3. 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.

Theorem 9 (cf: [1]). Suppose . The following four statements are equivalent. (i) is strictly singular.(ii)For every infinite dimensional subspace , there exists an infinite dimensional subspace such that is precompact on .(iii)Given and given , an infinite dimensional subspace of , there exists an infinite dimensional subspace such that restricted to has norm not exceeding .(iv)Given 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.

Theorem 10. Suppose . The following four statements are equivalent. (i) is strong strictly singular.(ii)For every infinite dimensional subspace satisfying and , there exists an infinite dimensional subspace of satisfying and such that is precompact on .(iii)Given and given , an infinite dimensional subspace of , satisfying and , there exists an infinite dimensional subspace of satisfying and such that restricted to has norm not exceeding .(iv)Given and given , an infinite dimensional subspace of satisfying and , there exists an infinite dimensional subspace of satisfying and 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 with the property that every infinity dimensional subspace of contains a vector such that and . Then there exists a subspace with 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.(b)For a given subspace with , there is no positive constant such that for any subspace of with , there is an element with .

Proof. : Suppose that fails to be true. Then there is a subspace of satisfying and there is a positive constant such that for every subspace of satisfying there is an element such that . Then, there is a subspace of such that and restricted to is a homeomorphism; which follows from Theorem 11. So, is not strictly singular. Hence, is not strong strictly singular. This proves .: Trivial.

4. Properties

Proposition 13. Let be a sequence of strong strictly singular operators such that for some operator . Suppose that , for all . Then is also strong strictly singular.

Proof. Note that and each is strictly singular, by Lemma 3. So 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. (i)There is an infinite dimensional subspace of such that .(ii)There is an infinite dimensional subspace of such that .(iii)There are infinite dimensional subspaces and of and , respectively, and there is an onto isomorphism such that .

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 , then we select as so that (i) is true. If , then find a subspace of such that and is 1-1 on . In this case (iii) is true with . Similarly, we can show that either (ii) or (iii) is true if .

Remark 15. It is possible to select and such that cardinalities of bases for and are equal to the cardinality of a basis for .

Theorem 16. Let , be bounded linear operators between normed spaces. Define a bounded linear operator by . Then one has the following. (i)If and are precompact (compact), then is precompact (compact).(ii)If and are strictly singular operators, then is a strictly singular operator.(iii)If and are strong strictly singular operators, then is a strong strictly singular operator.

Proof. (i) Is obvious. To prove (ii), suppose that and are strictly singular operators. Let be infinite dimensional subspace of . If there is an infinite dimensional subspace of such that , then we can find an infinite dimensional subspace of such that is precompact on . In this case, there is an infinite dimensional subspace of such that is precompact on . Similarly, if there is an infinite dimensional subspace of such that , then is precompact on an infinite dimensional subspace of . Suppose that there are infinite dimensional subspaces and of and , respectively, and an onto linear isomorphism such that . Find an infinite dimensional subspace of such that is precompact on . Find an infinite dimensional subspace of such that is precompact on . Then, is precompact on an infinite dimensional subspace of . This proves (ii). To prove (iii), suppose that and are strong strictly singular operators. Note that and . So the natural mapping from to coincides with the natural mapping from to . Since, by (ii), the later is strictly singular and the former is also strictly singular. So, is strong strictly singular. This proves (iii).

The following corollary is well known.

Corollary 17. If , are two strictly singular (precompact) operators. Then is strictly singular (precompact).

Proof. It is follows from the continuous mappings from into , from into and strictly singular (precompact) mapping from into .

Lemma 18. Let be a strong strictly singular operator. Let be a 1-1 continuous operator. Then is strong strictly singular.

Proof. Note that . Since the natural mapping is strictly singular, the mapping is strictly singular. Then, by Lemma 3, is strong strictly singular.

Corollary 17, Lemma 18, and Proposition 13 lead to natural questions about limits of sequences of strong strictly singular operators and sum of two strong strictly singular operators.

References

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D. van Dulst, “On strictly singular operators,” Compositio Mathematica, vol. 23, no. 2, pp. 169–183, 1971.