# Characterizations of Strong Strictly Singular Operators

Authors: C. Ganesa Moorthy; C. T. Ramasamy
Journal: Chinese Journal of Mathematics
 (2013)

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## 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.


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## Body

## 1. Introduction

The spacesX and Y will denote normed spaces, and T:X→Y will denote a bounded linear mapping from a normed space X into a normed space Y 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 in X. If T(BX) is totally bounded in Y, where BX is the open unit ball in X, then T is called a precompact operator. If T(BX)¯, closure of T(BX), is compact in Y, then T is called a compact operator. Every precompact operator is strictly singular (cf: [1]). The collection of all strictly singular (precompact) operators from X into Y forms a closed subspace of the normed space ℬ(X,Y), the collection of all bounded linear operators from X into Y (cf: [1]). The collection of strictly singular (precompact) operators on X (from X into X) forms a closed ideal of the normed algebra ℬ(X)(=ℬ(X,X)) (cf: [1]). A linear transformation T:X→Y has a bounded inverse if and only if ∥x∥≤c∥Tx∥ for all x∈X, for some c>0. It is easy to see as a consequence of the open mapping theorem that a continuous linear transformation T from a Banach space X into a Banach space Y has closed range, if and only if for given x∈X, there is an element y∈X such that Tx=Ty and ∥y∥≤c∥Ty∥, for some fixed c>0 (see [2]). This gives a motivation to define a new class of operators called strong strictly singular operators.WriteX^=X/N(T), where N(T) is the null space of T, as the quotient space endowed with the quotient norm. Let πT denote the quotient map from X onto X^. The 1-1 operator T^:X^→Y induced by T is defined by  T^(x+N(T)=πT(x))=Tx. Note that T^ is 1-1 and linear with range same as the range of T. If T is 1-1 then T=T^. Also, BX^=πT(BX) and T(BX)=T^(BX^). So, we have the following conclusions.(a)
T is precompact on X if and only if T^ is precompact on X^.(b)
T is compact on X if and only if T^ is compact on X^.


## 2. Definition

Definition 1.
An operatorT:X→Y is said to be strong strictly singular  if for any subspace M of X such that dim T(M)=∞ and the null space N(T)⊂M there is no positive number c with the property: for a given x∈M, there is an element y∈M such that Tx=Ty and ∥y∥≤c∥Ty∥.Lemma 2.
Every strong strictly singular operator is strictly singular.Proof.
LetT be strong strictly singular. Suppose  M0 is an infinite dimensional subspace of X such that the restriction of T on M0 has a bounded inverse. Then, there is a positive constant c such that ∥x∥≤c∥Tx∥ for every x∈M0. Since T is 1-1 on M0, we have N(T)∩M0={0}. Let us consider that M=M0+N(T). Then, dimT(M)=dimT(M0)=∞. Also, for given x∈M, there are y∈M0 and z∈N(T) such that x=y+z and ∥y∥≤c∥Ty∥=c∥Tx∥. This shows that T is not strong strictly singular. This contradiction shows that T is strictly singular, whenever T is strong strictly singular.The converse of Lemma2 is not true. For example, there is an onto continuous linear operator T:ℓ1→ℓ2 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.
T is strong strictly singular if and only if T^ is strictly singular.Proof.
Suppose thatM^0 is any subspace of X^ such that, for all x+N(T)∈M^0, we have ∥x+N(T)∥≤c∥T^(x+N(T))∥, for some fixed c>0. Then, for any fixed c′>c, for given x∈πT-1(M^o), there is an element y∈πT-1(M^0) such that x+N(T)=y+N(T) and
(1)∥y∥≤(c′c)∥x+N∥≤c′∥T^(x+N(T))∥=c′∥T^(y+N(T))∥=c′∥T(y)∥,Tx=Ty.
This proves that if T is strong strictly singular, then T^ is strictly singular.
On the other hand, suppose thatM is any subspace of X such that N(T)⊂M and such that for a given x∈M there is an element y∈M such that Tx=Ty and ∥y∥≤c∥Ty∥, for fixed c>0. Then
(2)∥x+N(T)∥≤∥y∥≤c∥Ty∥=c∥T^(y+N(T))∥=c∥T^(x+N(T))∥,foranyx+N(T)∈πT(M).
This proves that if T^ is strictly singular, then T is strong strictly singular. This completes the proof.Corollary 4.
IfT is 1-1 and strictly singular, then T is strong strictly singular.Corollary 5.
Every precompact operator is strong strictly singular.Proof.
IfT:X→Y is precompact, then T^ so is. Hence T^ is strictly singular, and hence T is strong strictly singular.Note that the inclusion mapI:ℓp→ℓq, 1≤p<q<∞, is strong strictly singular but not precompact [3, page 170].Although the converse of Lemma2 is not true in general, it is true partially which is seen from the next theorem.Theorem 6.
LetT:X→Y be a strictly singular operator. Suppose that X=N(T)⊕Z, for some subspace Z of X, and the corresponding projection P:X→Z defined by p(y+z)=z with y∈N(T) and z∈Z is continuous. Then T is strong strictly singular.Proof.
Let us consider a subspaceM of X such that N(T)⊂M and such that for each given x∈X, there is a y∈M satisfying Tx=Ty and ∥y∥≤c∥Tx∥, for some fixed c>0. Let M0=P(M). Then T is 1-1 on M0. Moreover, for given x∈M0, there are y0∈M and z0∈N(T) such that y0=x+z0, and Tx=Ty0. For this y0∈M, there is a y∈M such that ∥y∥≤c∥Ty0∥ and Ty=Ty0. For this y∈M, let us write y=x+z1, with z1∈N(T). Then, we have ∥x∥=∥Py∥≤∥P∥∥y∥≤c∥P∥∥Ty0∥=c∥P∥∥Tx∥. Thus, we have ∥x∥≤c∥P∥∥Tx∥ for every x∈M0. Since T is strictly singular, dimM0<∞. So, dimT(M)=dimM0<∞. This proves that T is a strong strictly singular operator.Corollary 7.
LetT:X→Y be a strictly singular operator on a Banach space X such that X=N(T)⊕Z, for some closed subspace Z of X. Then T is strong strictly singular.Corollary 8.
LetT:X→Y be a strictly singular operator such that dim(N(T))<∞. Then T is strong strictly singular.


## 3. Characterization

A known classical characterization of strictly singular operators is given in Theorem9. 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 X into X itself.Theorem 9 (cf: [1]).
SupposeT∈ℬ(X,Y). The following four statements are equivalent.  (i)
T is strictly singular.(ii)
For every infinite dimensional subspaceM⊂X, there exists an infinite dimensional subspace N⊂M such that T is precompact on N.(iii)
Givenϵ>0 and given M, an infinite dimensional subspace of X, there exists an infinite dimensional subspace N⊂M such that T restricted to N has norm not exceeding ϵ.(iv)
Givenϵ>0 and given M, an infinite dimensional subspace of X, there exists an infinite dimensional subspace N⊂M such that T restricted to N is precompact and it has norm not exceeding ϵ.Using Lemma3 and Theorem 9, one can easily verify the following.Theorem 10.
SupposeT∈ℬ(X,Y). The following four statements are equivalent.  (i)
T is strong strictly singular.(ii)
For every infinite dimensional subspaceM⊂X satisfying N(T)⊂M and dimT(M)=∞, there exists an infinite dimensional subspace N of X satisfying N(T)⊂N⊂M and dimT(N)=∞ such that T is precompact on N.(iii)
Givenϵ>0 and given M, an infinite dimensional subspace of X, satisfying N(T)⊂M and dimT(M)=∞, there exists an infinite dimensional subspace N of X satisfying N(T)⊂N⊂M and dimT(N)=∞ such that T restricted to N has norm not exceeding ϵ.(iv)
Givenϵ>0 and given M, an infinite dimensional subspace of X satisfying N(T)⊂M and dimT(M)=∞, there exists an infinite dimensional subspace N of X satisfying N(T)⊂N⊂M and dimT(N)=∞ such that T restricted to N is precompact and its has norm not exceeding ϵ.The following interesting theorem on automatic continuity is due to van Dulst [5].Theorem 11 (see [5]).
LetT:X→Y be a linear transformation (need not be continuous). Suppose that there exists a constant c>0 with the property that every infinity dimensional subspace of X contains a vector x such that ∥x∥=1 and ∥Tx∥<c. Then there exists a subspace L⊂X with dimX/L<∞ such that TL, the restriction of T to L, is continuous.Corollary 12.
LetT:X→Y be a bounded linear transformation. Then the following statements are equivalent.(a)
T is strong strictly singular.(b)
For a given subspaceM⊂X with dimT(M)=∞, there is no positive constant c such that for any subspace M1 of M with dimT(M1)=∞, there is an element x∈M1+N(T) with ∥x∥≤c∥Tx∥.Proof.
(
a
)
⇒
(
b
): Suppose that (b) fails to be true. Then there is a subspace M of X satisfying dimT(M)=∞ and there is a positive constant c such that for every subspace M1 of M satisfying dimT(M1)=∞ there is an element x∈M1+N(T) such that ∥x∥≤c∥Tx∥. Then, there is a subspace M^0 of πT(M) such that dimπT(M)/M^0<∞ and T^ restricted to M^0 is a homeomorphism; which follows from Theorem 11. So, T^ is not strictly singular. Hence, T is not strong strictly singular. This proves (a)⇒(b).
(
b
)
⇒
(
a
): Trivial.


## 4. Properties

Proposition 13.
LetTn:X→Y be a sequence of strong strictly singular operators such that ∥Tn-T∥→0 for some operator T:X→Y. Suppose that N(T)=N(Ti)=N(Tj), for all i,j. Then T is also strong strictly singular.Proof.
Note that∥T^n-T^∥→0 and each T^n is strictly singular, by Lemma 3. So T^ is strictly singular. Hence T is strong strictly singular, by Lemma 3.The following technical lemma is on vector spaces.Lemma 14.
LetM be an infinite dimensional subspace of X×Y, where X and Y are vector spaces over the same field. Then at least one of the following is true.(i)
There is an infinite dimensional subspaceX1 of X such that X1×{0}⊂M.(ii)
There is an infinite dimensional subspaceY1 of Y such that {0}×Y1⊂M.(iii)
There are infinite dimensional subspacesX1 and Y1 of X and Y, respectively, and there is an onto isomorphism L:X1→Y1 such that {(x,Lx):x∈X1}⊂M.Proof.
LetMX, MY be the projections of M on X and Y, respectively. Then MX is of infinite dimension, or MY is of infinite dimension. Suppose that MX is of infinite dimension. Let SX be a basis of MX. To each x∈SX, find an element Lx∈MY such that (x,Lx)∈M. Extend the map x↦Lx to L:MX→MY as a linear mapping. If dimL(MX)<∞, then we select X1 as kerL so that (i) is true. If dimL(MX)=∞, then find a subspace X1 of MX such that L(X1)=L(MX) and L is 1-1 on X1. In this case (iii) is true with Y1=L(MX). Similarly, we can show that either (ii) or (iii) is true if dimMY=∞.Remark 15.
It is possible to selectX1 and Y1 such that cardinalities of bases for X1 and Y1 are equal to the cardinality of a basis for M.Theorem 16.
LetT1:X1→Y1, T2:X2→Y2 be bounded linear operators between normed spaces. Define a bounded linear operator T=T1×T2:X1×X2→Y1×Y2 by T(x1,x2)=(T1x1,T2x2). Then one has the following.(i)
IfT1 and T2 are precompact (compact), then T is precompact (compact).(ii)
IfT1 and T2 are strictly singular operators, then T is a strictly singular operator.(iii)
IfT1 and T2 are strong strictly singular operators, then T is a strong strictly singular operator.Proof.
(i) Is obvious. To prove (ii), suppose thatT1 and T2 are strictly singular operators. Let M be infinite dimensional subspace of X1×X2. If there is an infinite dimensional subspace X1′ of X1 such that X1′×{0}⊂M, then we can find an infinite dimensional subspace X1′′ of X1′ such that T1 is precompact on X1′′. In this case, there is an infinite dimensional subspace X1′′×{0} of M such that T is precompact on X1′′×{0}. Similarly, if there is an infinite dimensional subspace X2′ of X2 such that {0}×X2′⊂M, then T is precompact on an infinite dimensional subspace of M. Suppose that there are infinite dimensional subspaces X1′ and X2′ of X1 and X2, respectively, and an onto linear isomorphism L:X1′→X2′ such that {(x1,Lx1):x1∈X1′}⊂M. Find an infinite dimensional subspace X1′′ of X1′ such that T1 is precompact on X1′′. Find an infinite dimensional subspace X2′′ of L(X1′′) such that T2 is precompact on X2′′. Then, T is precompact on an infinite dimensional subspace L-1(X2′′)×X2′′ of M. This proves (ii). To prove (iii), suppose that T1 and T2 are strong strictly singular operators. Note that N(T)=N(T1)×N(T2) and (X1×X2)/N(T)=X1/N(T1)×X2/N(T2). So the natural mapping from (X1×X2)/N(T) to Y1×Y2 coincides with the natural mapping from X1/N(T1)×X2/N(T2) to Y1×Y2. Since, by (ii), the later is strictly singular and the former is also strictly singular. So, T is strong strictly singular. This proves (iii).The following corollary is well known.Corollary 17.
IfT1:X→Y, T2:X→Y are two strictly singular (precompact) operators. Then T1+T2 is strictly singular (precompact).Proof.
It is follows from the continuous mappingsx↦(x,x) from X into X×X, (y1,y2)↦y1+y2 from Y×Y into Y and strictly singular (precompact) mapping (x1,x2)↦(T1x1,T2x2) from X×X into Y×Y.Lemma 18.
LetT:X→Y be a strong strictly singular operator. Let S:Y→Z be a 1-1 continuous operator. Then S∘T is strong strictly singular.Proof.
Note thatN(T)=N(S∘T). Since the natural mapping S1:X/N(T)→Z is strictly singular, the mapping S1:X/N(S∘T)→Z is strictly singular. Then, by Lemma 3, S∘T is strong strictly singular.Corollary17, 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.



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