1 Introduction

The perturbation classes problem asks whether the perturbation classes for the upper semi-Fredholm operators \(P\Phi _+\) and the lower semi-Fredholm operators \(P\Phi _-\) coincide with the classes of strictly singular operators \({{\mathcal {S}}}{{\mathcal {S}}}\) and strictly cosingular operators \({{\mathcal {S}}}{{\mathcal {C}}},\) respectively. This problem was raised in [9] (see also [5, 19]), and it has a positive answer in some cases [11, 13,14,15, 21], but the general answer is negative in both cases [10, 8, Theorem 4.5]. However, it remains interesting to find positive answers in special cases because the definitions of \({{\mathcal {S}}}{{\mathcal {S}}}\) and \({{\mathcal {S}}}{{\mathcal {C}}}\) are intrinsic: to check that K is in one of them only involves the action of K,  while to check that K is in \(P\Phi _+\) or \(P\Phi _-\) we have to study the properties of \(T+K\) for T in a large set of operators.

In this paper, we consider two classes \(\Phi {\mathcal {S}}\) and \(\Phi {\mathcal {C}}\) introduced in [2] that satisfy

$$\begin{aligned}{} & {} {{\mathcal {S}}}{{\mathcal {S}}}(X,Y)\subset \Phi {\mathcal {S}}(X,Y)\subset P\Phi _+(X,Y) \quad \text {and}\\{} & {} {{\mathcal {S}}}{{\mathcal {C}}}(X,Y)\subset \Phi {\mathcal {C}}(X,Y)\subset P\Phi _-(X,Y). \end{aligned}$$

We study conditions on the Banach spaces X, Y so that some of these four inclusions become equalities, and we derive new positive answers to the perturbation classes problem for semi-Fredholm operators. When \(\Phi _+(X,Y)\ne \emptyset ,\) we show that \(\Phi {\mathcal {S}}(X,Y)= P\Phi _+(X,Y)\) when \(Y\times Y\) is isomorphic to Y,  \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)= \Phi {\mathcal {S}}(X,Y)\) when some quotients of X embed in Y (Theorem 3.1), and adding up we get conditions implying that \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)= P\Phi _+(X,Y)\) (Theorem 3.2). When \(\Phi _-(X,Y)\ne \emptyset \) we prove similar results (Theorems 4.1 and 4.2). We also state some questions concerning the classes \(\Phi {\mathcal {S}}\) and \(\Phi {\mathcal {C}}.\)

Notation. Along the paper, X,  Y and Z denote Banach spaces and \({\mathcal {L}}(X,Y)\) is the set of bounded operators from X into Y. We write \({\mathcal {L}}(X)\) when \(X=Y.\) Given a closed subspace M of X,  we denote \(J_M\) the inclusion of M into X,  and \(Q_M\) the quotient map from X onto X/M. An operator \(T\in {\mathcal {L}}(X,Y)\) is an isomorphism if there exists \(c>0\) such that \(\Vert Tx\Vert \ge c\Vert x\Vert \) for every \(x\in X.\)

The operator \(T\in {\mathcal {L}}(X,Y)\) is strictly singular, and we write \(T\in {{\mathcal {S}}}{{\mathcal {S}}},\) when there is no infinite dimensional closed subspace M of X such that the restriction \(TJ_M\) is an isomorphism; and T is strictly cosingular, and we write \(T\in {{\mathcal {S}}}{{\mathcal {C}}},\) when there is no infinite codimensional closed subspace N of Y such that \(Q_NT\) is surjective. Moreover, T is upper semi-Fredholm, \(T\in \Phi _+,\) when the range R(T) is closed and the kernel N(T) is finite dimensional; T is lower semi-Fredholm, \(T\in \Phi _-,\) when R(T) is finite codimensional (hence closed); T is Fredholm, \(T\in \Phi ,\) when it is upper and lower semi-Fredholm; and T is inessential, \(T\in {{\mathcal {I}}n},\) when \(I_X-ST\in \Phi \) for each \(S\in {\mathcal {L}}(Y,X).\)

2 Preliminaries

The perturbation class \(P{\mathcal {S}}\) of a class of operators \({\mathcal {S}}\) is defined in terms of its components:

Definition 2.1

Let \({\mathcal {S}}\) denote one of the classes \(\Phi _+,\) \(\Phi _-\) or \(\Phi .\) For spaces X, Y such that \({\mathcal {S}}(X,Y)\ne \emptyset ,\)

$$\begin{aligned} P{\mathcal {S}}(X,Y)= \{K\in {\mathcal {L}}(X,Y): \text { for each } T\in {\mathcal {S}}(X,Y), T+K\in {\mathcal {S}}\}. \end{aligned}$$

We could define \(P{\mathcal {S}}(X,Y)= {\mathcal {L}}(X,Y)\) when \({\mathcal {S}}(X,Y)\) is empty, but this is not useful. The components of \(P\Phi \) coincide with those of the operator ideal of inessential operators \({{\mathcal {I}}n}\) when they are defined [1], but given \(S\in {\mathcal {L}}(Y,Z)\) and \(T\in {\mathcal {L}}(X,Y),\) S or T in \(P\Phi _+\) does not imply \(ST\in P\Phi _+,\) and similarly for \(P\Phi _-\) [10]. However, the following result holds, and it will be useful for us.

Proposition 2.1

Suppose that \(K\in P\Phi _+(X,Y),\) \(L\in P\Phi _-(X,Y),\) \(S\in {\mathcal {L}}(Y)\) and \(T\in {\mathcal {L}}(X).\) Then, \(SK, KT\in P\Phi _+(X,Y)\) and \(SL, LT\in P\Phi _-(X,Y).\)

Proof

Suppose that \(K\in P\Phi _+(X,Y)\) and let \(S\in {\mathcal {L}}(Y)\) and \(U\in \Phi _+(X,Y).\) If S is bijective, then \(S^{-1}U\in \Phi _+(X,Y),\) hence \(U+SK= S(S^{-1}U +K)\in \Phi _+;\) thus \(SK \in P\Phi _+(X,Y).\) In the general case, \(S=S_1+S_2\) with \(S_1,S_2\) bijective; thus, \(SK= S_1K+S_2K \in P\Phi _+(X,Y).\)

The proof of the other three results is similar. \(\square \)

3 The perturbation class for \(\Phi _+\)

Given two operators \(S,T\in {\mathcal {L}}(X,Y),\) we denote by (S, T) the operator from X into \(Y\times Y\) defined by \((S,T)x=(Sx,Tx),\) where \(Y\times Y\) is endowed with the product norm \(\Vert (y_1,y_2)\Vert _1= \Vert y_1\Vert + \Vert y_2\Vert .\)

Inspired by the results of Friedman [6], the authors of [2] defined the following class of operators.

Definition 3.1

Suppose that \(\Phi _+(X,Y) \ne \emptyset \) and let \(K\in {\mathcal {L}}(X,Y).\) We say that K is \(\Phi \)-singular, and write \(K\in \Phi {\mathcal {S}},\) when for each \(S\in {\mathcal {L}}(X,Y),\) \((S,K)\in \Phi _+\) implies \(S\in \Phi _+.\)

The definition of \(\Phi {\mathcal {S}}(X,Y)\) is similar to that of \(P\Phi _+(X,Y),\) but the former one is easier to handle because the action of S and K is decoupled when we consider (S, K) instead of \(S+K.\)

With our notation, [6, Theorems 3 and 4] can be stated as follows:

Proposition 3.1

[2, Proposition 2.2] Suppose that \(\Phi _+(X,Y)\ne \emptyset .\) Then,

$$\begin{aligned} {{\mathcal {S}}}{{\mathcal {S}}}(X,Y)\subset \Phi {\mathcal {S}}(X,Y)\subset P\Phi _+(X,Y). \end{aligned}$$

Note that \({{\mathcal {S}}}{{\mathcal {S}}}\) is an operator ideal but \(P\Phi _+\) is not; \(P\Phi _+(X,Y)\) is a closed subspace of \({\mathcal {L}}(X,Y),\) and \(P\Phi _+(X)\) is an ideal of \({\mathcal {L}}(X).\)

Proposition 3.2

\(\Phi {\mathcal {S}}(X,Y)\) is closed in \({\mathcal {L}}(X,Y).\)

Proof

Let \(\{T_n\}\) be a sequence in \(\Phi {\mathcal {S}}(X,Y)\) converging to \(T\in {\mathcal {L}}(X,Y).\) Suppose that \(S\in {\mathcal {L}}(X,Y)\) and \((S,T) \in \Phi _+(X, Y \times Y).\) Note that the sequence \((S,T_n)\) converge to (S, T),  because \(\Vert (S,T_n)- (S,T)\Vert = \Vert T_n-T\Vert .\)

Since \( \Phi _+(X, Y \times Y)\) is an open set, there exists a positive integer N such that \((S,T_N)\in \Phi _+(X,Y\times Y).\) Then, \(T_N\in \Phi {\mathcal {S}}(X,Y)\) implies \(S \in \Phi _+(X,Y).\) Thus \(T\in \Phi {\mathcal {S}}(X,Y).\) \(\square \)

We state some basic questions on the class \(\Phi {\mathcal {S}}.\)

Question 3.1

Suppose that \(\Phi _+(X,Y)\ne \emptyset .\)

  1. (a)

    Is \(\Phi {\mathcal {S}}(X,Y)\) a subspace of \({\mathcal {L}}(X,Y)?\)

  2. (b)

    Is \(\Phi {\mathcal {S}}\) an operator ideal?

  3. (c)

    Is Proposition 2.1 valid for \(\Phi {\mathcal {S}}?\)

Answering a question in [6], an example of an operator \(K\in P\Phi _+\setminus \Phi {\mathcal {S}}\) was given in [2, Example 2.3], but we do not know if the other inclusion can be strict.

Question 3.2

Suppose that \(\Phi _+(X,Y)\ne \emptyset .\) Is \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)=\Phi {\mathcal {S}}(X,Y)?\)

A negative answer to Question 3.2 would provide a new counterexample to the perturbation classes problem for \(\Phi _+.\)

Let us see that the inclusions in Proposition 3.1 become equalities in some cases.

An infinite dimensional Banach space Y is isomorphic to its square, denoted \(Y\times Y\simeq Y,\) in many cases: \(L_p(\mu )\) and \(\ell _p\) \((1\le p\le \infty ),\) \(c_0,\) and C[0, 1]. On the other hand, James’ space J and some spaces of continuous functions on a compact like \(C[0,\omega _1]\) are not isomorphic to their square, where \(\omega _1\) is the first uncountable ordinal. See [4, 20].

Theorem 3.1

Suppose that the spaces X and Y satisfy \(\Phi _+(X,Y)\ne \emptyset .\)

  1. 1.

    If \(Y\times Y\simeq Y\), then \(\Phi {\mathcal {S}}(X,Y)= P\Phi _+(X,Y).\)

  2. 2.

    If every infinite dimensional subspace of X has an infinite dimensional subspace N such that X/N embeds in Y, then \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)=\Phi {\mathcal {S}}(X,Y).\)

Proof

(1) Let \(U:Y\times Y\rightarrow Y\) be a bijective isomorphism and let \(V,W\in {\mathcal {L}}(Y)\) such that \(U(y_1,y_2)=Vy_1+ Wy_2.\) If \(K\in P\Phi _+(X,Y),\) for each \(S\in {\mathcal {L}}(X,Y)\) such that \((S,K)\in \Phi _+\) we have \(U(S,K)=VS+ WK\in \Phi _+.\) By Proposition 2.1, \(WK\in P\Phi _+(X,Y).\) Then, \(VS\in \Phi _+,\) hence \(S\in \Phi _+.\) Thus we conclude that \(K\in \Phi {\mathcal {S}}(X,Y).\)

(2) Let \(K\in {\mathcal {L}}(X,Y),\) \(K\notin {{\mathcal {S}}}{{\mathcal {S}}}.\) By the hypothesis, there exists an infinite dimensional subspace N of X such that \(KJ_N\) is an isomorphism, and there is an isomorphism \(U:X/N\rightarrow Y.\) Then, \(S=UQ_N\in {\mathcal {L}}(X,Y)\) is not upper semi-Fredholm. We will prove that \(K\notin \Phi {\mathcal {S}}\) by showing that \((S,K)\in \Phi _+.\)

Recall that \(\Vert Q_Nx\Vert = \mathop \textrm{dist}(x,N).\) We can choose the isomorphism U so that \(\Vert Sx\Vert =\Vert UQ_Nx\Vert \ge \mathop \textrm{dist}(x,N)\) for each \(x\in X.\) Moreover, there is a constant \(c>0\) such that \(\Vert Kn\Vert \ge c\Vert n\Vert \) for each \(n\in N.\)

Let \(x\in X\) with \(\Vert x\Vert =1,\) and let \(0<\alpha <1\) such that \(c(1-\alpha )= 2\Vert K\Vert \alpha .\)

If \(\mathop \textrm{dist}(x,N)\ge \alpha \), then \(\Vert Sx\Vert \ge \alpha .\) Otherwise, there exists \(y\in N\) such that \(\Vert x-y\Vert <\alpha ;\) hence \(\Vert y\Vert > 1-\alpha .\) Therefore,

$$\begin{aligned} \Vert Kx\Vert \ge \Vert Ky\Vert -\Vert K(x-y)\Vert \ge c(1-\alpha )-\Vert K\Vert \alpha =\Vert K\Vert \alpha . \end{aligned}$$

Then, \(\Vert (S,K)x\Vert _1=\Vert Sx\Vert +\Vert Kx\Vert \ge \min \{\Vert K\Vert \alpha , \alpha \},\) hence (S, K) is an isomorphism; in particular \((S,K)\in \Phi _+,\) as we wanted to show. \(\square \)

In the known examples in which \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)\ne P\Phi _+(X,Y)\) in [8, 10], the space Y has a complemented subspace which is hereditarily indecomposable in the sense of [3, 16, 17]. Therefore, the question arises.

Question 3.3

Suppose that X and Y satisfy \(\Phi _+(X,Y)\ne \emptyset \) and \(Y\times Y\simeq Y.\)

Is \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)=P\Phi _+(X,Y)?\)

A Banach space X is subprojective if every closed infinite dimensional subspace of X contains an infinite dimensional subspace complemented in X. The spaces \(c_0,\) \(\ell _p\) (\(1\le p<\infty )\) and \(L_q(\mu )\) (\(2\le q<\infty )\) are subprojective [22]. See [7, 18] for further examples.

Corollary 3.1

Suppose that \(\Phi _+(X,Y)\ne \emptyset \) and the space X is subprojective. Then, \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)= \Phi {\mathcal {S}}(X,Y).\)

Proof

Every closed infinite dimensional subspace of X contains an infinite dimensional subspace N complemented subspace in X;  thus X/N is isomorphic to the complement of N. Since \(\Phi _+(X,Y)\ne \emptyset ,\) the quotient X/N is isomorphic to a subspace of Y and we can apply Theorem 3.1. \(\square \)

The next result is a refinement of Theorem 3.1 that is proved using the previous arguments.

Theorem 3.2

Suppose that \(\Phi _+(X,Y)\ne \emptyset ,\) \(Y\times Y\) embeds in Y and every infinite dimensional subspace of X has an infinite dimensional subspace N such that X/N embeds in Y. Then, \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)=P\Phi _+(X,Y).\)

Proof

Since \(Y\times Y\) embeds in Y,  there exist isomorphisms \(V,W\in {\mathcal {L}}(Y)\) such that \(R(V)\cap R(W)=\{0\}\) and \(R(V)+R(W)\) is closed. Hence, there exists \(r>0\) such that \(\Vert y_1+y_2\Vert \ge r(\Vert y_1\Vert +\Vert y_2\Vert )\) for \(y_1\in R(V)\) and \(y_2\in R(W),\) and clearly we can choose V, W so that \(r=1.\)

Let \(K\in {\mathcal {L}}(X,Y)\) with \(K\notin {{\mathcal {S}}}{{\mathcal {S}}}.\) Select an infinite dimensional subspace M of X such that \(KJ_M\) is an isomorphism, and let N be an infinite dimensional subspace of M such that there exists an isomorphism \(U:X/N\rightarrow Y.\) We can assume that \(\Vert Uz\Vert \ge z\) for each \(z\in X/N.\)

The operator \(S=VUQ_N \notin \Phi _+,\) and proceeding like in the proof of (2) in Theorem 3.1, we can show that \(S+WK\in \Phi _+.\) Then, \(WK \notin P\Phi _+,\) hence \(K \notin P\Phi _+,\) by Proposition 2.1. \(\square \)

Corollary 3.2

If X is separable,  \(Y\times Y\) embeds in Y,  and Y contains a copy of C[0, 1], then \({{\mathcal {S}}}{{\mathcal {S}}}(X,Y)=\Phi _+(X,Y).\)

Proof

It is well known that the space C[0, 1] contains a copy of each separable Banach space. \(\square \)

The class \(\Phi {\mathcal {S}}\) is injective in the following sense:

Proposition 3.3

Given an operator \(K \in {\mathcal {L}}(X, Y)\) and an (into) isomorphism \(L\in {\mathcal {L}}(Y,Y_0),\) if \(LK\in \Phi {\mathcal {S}}(X,Y_0)\), then \(K\in \Phi {\mathcal {S}}(X,Y).\)

Proof

Let \(K \in {\mathcal {L}}(X, Y)\) and let \(L \in {\mathcal {L}}(X,Y_0)\) be an isomorphism into \(Y_0\) such that \(LK \in \Phi {\mathcal {S}}(X,Y_0).\) Take \(S\in {\mathcal {L}}(X,Y)\) and suppose that \((S,K) \in \Phi _+(X, Y \times Y).\) Then, \((LS,LK)=(L\times L)(S,K)\in \Phi _+(X, Y_0 \times Y_0),\) where \((L\times L)\in {\mathcal {L}}(X\times X,Y_0 \times Y_0)\) is defined by \((L\times L)(x_1,x_2)=(Lx_1,Lx_2).\)

Since \(LK \in \Phi {\mathcal {S}}(X,Y_0)\) we obtain \(LS \in \Phi _+(X, Y_0).\) Therefore \(S \in \Phi _+(X, Y),\) hence \(K \in \Phi {\mathcal {S}}(X,Y).\) \(\square \)

4 The perturbation class for \(\Phi _-\)

Given two operators \(S,T\in {\mathcal {L}}(X,Y),\) we denote by [S, T] the operator from \(X\times X\) into Y defined by \([S,T](x_1,x_2) =Sx_1+Tx_2,\) where \(X\times X\) is endowed with the maximum norm \(\Vert (x_1,x_2)\Vert _\infty = \max \{\Vert y_1\Vert , \Vert y_2\Vert \}.\)

Definition 4.1

Suppose that \(\Phi _-(X,Y) \ne \emptyset \) and let \(K\in {\mathcal {L}}(X,Y).\) We say that K is \(\Phi \)-cosingular, and write \(K\in \Phi {\mathcal {C}},\) when for each \(S\in {\mathcal {L}}(X,Y),\) \([S,K]\in \Phi _-\) implies \(S\in \Phi _-.\)

Like in the case of \(\Phi {\mathcal {S}},\) the definition of \(\Phi {\mathcal {C}}(X,Y)\) is similar to that of \(P\Phi _-(X,Y),\) but the former one is easier to handle because the action of S and K is decoupled when we consider [S, K] instead of \(S+K.\)

Proposition 4.1

[2, Proposition 2.5] Suppose that \(\Phi _-(X,Y)\ne \emptyset .\) Then,

$$\begin{aligned} {{\mathcal {S}}}{{\mathcal {C}}}(X,Y)\subset \Phi {\mathcal {C}}(X,Y)\subset P\Phi _-(X,Y). \end{aligned}$$

Note that \({{\mathcal {S}}}{{\mathcal {C}}}\) is an operator ideal but \(P\Phi _-\) is not; \(P\Phi _-(X,Y)\) is a closed subspace of \({\mathcal {L}}(X,Y),\) and \(P\Phi _-(X)\) is an ideal of \({\mathcal {L}}(X).\)

Proposition 4.2

\(\Phi {\mathcal {C}}(X,Y)\) is closed in \({\mathcal {L}}(X,Y).\)

Proof

Let \(\{T_n\}\) be a sequence in \(\Phi {\mathcal {C}}(X,Y)\) converging to \(T\in {\mathcal {L}}(X,Y).\) Suppose that \(S\in {\mathcal {L}}(X,Y)\) and \([S,T] \in \Phi _-(X\times X,Y).\) Note that the sequence \([S,T_n]\) converge to [S, T].

Since \(\Phi _-(X\times X,Y)\) is an open set there exists a positive integer N such that \([S,T_N] \in \Phi _-(X \times X, Y).\) Hence, \(T_N \in \Phi {\mathcal {C}}(X,Y)\) implies \(S \in \Phi _-(X,Y).\) \(\square \)

Question 4.1

Suppose that \(\Phi _-(X,Y)\ne \emptyset .\)

  1. (a)

    Is \(\Phi {\mathcal {C}}(X,Y)\) a subspace of \({\mathcal {L}}(X,Y)?\)

  2. (b)

    Is \(\Phi {\mathcal {C}}\) an operator ideal?

  3. (c)

    Is Proposition 2.1 valid for \(\Phi {\mathcal {C}}?\)

Answering a question in [6], an example of an operator \(K\in P\Phi _-\setminus \Phi {\mathcal {C}}\) was given in [2], but we do not know if the other inclusion can be strict.

Question 4.2

Suppose that \(\Phi _-(X,Y)\ne \emptyset .\) Is \({{\mathcal {S}}}{{\mathcal {C}}}(X,Y)=\Phi {\mathcal {C}}(X,Y)?\)

A negative answer to Question 4.2 would provide a new counterexample to the perturbation classes problem for \(\Phi _-.\)

Next we will show that the inclusions in Proposition 4.1 become equalities in some cases.

Theorem 4.1

Suppose that the spaces X and Y satisfy \(\Phi _-(X,Y) \ne \emptyset .\)

  1. 1.

    If \(X\times X \simeq X \), then \(\Phi {\mathcal {C}}(X,Y)=P \Phi _-(X, Y).\)

  2. 2.

    If every infinite codimensional closed subspace of Y is contained in an infinite codimensional closed subspace N which is isomorphic to a quotient of X, then \(\Phi {\mathcal {C}}(X,Y)= {{\mathcal {S}}}{{\mathcal {C}}}(X, Y).\)

Proof

(1) Let \(U:X \rightarrow X \times X\) be a bijective isomorphism and let \(U_1,U_2 \in {\mathcal {L}}(X)\) such that \(U(x)=(U_1x,U_2x).\) If \(K \in P\Phi _-(X, Y),\) for each \(S\in {\mathcal {L}}(X,Y)\) such that \([S,K] \in \Phi _-(X \times X, Y)\) we have \([S,K]U=SU_1+KU_2 \in \Phi _-(X, Y \times Y).\) By Proposition 2.1, \(KU_2 \in P\Phi _-(X, Y).\) Then, \(SU_1 \in \Phi _-(X, Y),\) hence \(S \in \Phi _-(X, Y).\) Thus we conclude that \(K \in \Phi {\mathcal {C}}(X,Y).\)

(2) Let \(K\in {\mathcal {L}}(X,Y)\) \(K\notin {{\mathcal {S}}}{{\mathcal {C}}}.\) Then, there exists an infinite codimensional closed subspace M of Y such that \(Q_NK\) is surjective. By the hypothesis, there exist an infinite codimensional closed subspace N such that \(M\subset N\) and a surjective operator \(V\in {\mathcal {L}}(X,M).\)

Observe that \(S=J_NV\in {\mathcal {L}}(X,Y)\) is not in \(\Phi _-.\) We prove that \(K\notin \Phi {\mathcal {C}}\) by showing that \([S,K] \in \Phi _-(X\times X, Y).\)

Indeed, note that \(R(S)=N.\) Moreover \(Q_NK\) surjective implies \(R(K)+N=Y.\) Since \(R([S,K])= R(S)+ R(K),\) [S, K] is surjective. \(\square \)

In the known examples in which \({{\mathcal {S}}}{{\mathcal {C}}}(X,Y)\ne P\Phi _-(X,Y)\) in [8, 10], the space X has a complemented subspace which is hereditarily indecomposable. Therefore, the question arises.

Question 4.3

Suppose that \(\Phi _-(X,Y)\ne \emptyset \) and \(X\times X\simeq X.\)

Is \({{\mathcal {S}}}{{\mathcal {C}}}(X,Y)=P\Phi _-(X,Y)?\)

A Banach space X is superprojective if each of its infinite codimensional closed subspaces is contained in some complemented infinite codimensional subspace. The spaces \(c_0,\) \(\ell _p\) \((1< p<\infty )\) and \(L_q(\mu )\) (\(1< q \le 2)\) are superprojective. See [7, 12] for further examples.

Corollary 4.1

Suppose that \(\Phi _-(X,Y) \ne \emptyset \) and the space Y is superprojective. Then, \(\Phi {\mathcal {C}}(X,Y)= {{\mathcal {S}}}{{\mathcal {C}}}(X, Y).\)

Proof

Every closed infinite codimensional subspace of Y is contained in an infinite codimensional complemented subspace N. Since \(\Phi _-(X,Y) \ne \emptyset ,\) there exists \(T\in {\mathcal {L}}(X,Y)\) with \(R(T)\supset N,\) and composing with the projection P on Y onto N we get \(R(PT)=N,\) and we can apply Theorem 4.1. \(\square \)

The following result is a refinement of the previous results in this section.

Theorem 4.2

Suppose that \(\Phi _-(X,Y) \ne \emptyset ,\) there exists a surjection from \(X \times X\) onto X,  and every closed infinite codimensional subspace of Y is contained in a closed infinite codimensional subspace N which is isomorphic to a quotient of X. Then, \(P\Phi _-(X,Y)= {{\mathcal {S}}}{{\mathcal {C}}}(X,Y).\)

Proof

\(K\in {\mathcal {L}}(X,Y),\) \(K\notin {{\mathcal {S}}}{{\mathcal {C}}}.\) Then, there exists an infinite codimensional subspace N of Y such that \(Q_NK\) is surjective. By hypothesis, we can assume that there exists a surjective operator \(U:X\rightarrow N.\) Then, \(S=J_NU\in {\mathcal {L}}(X,Y)\) is not in \(\Phi _-.\) Moreover, [S, K] is surjective: \(R([S,K])= R(S)+R(K)= N+R(K)=Y,\) hence \([S,K]\in \Phi _-(X\times X,Y).\)

Let \(V:X \rightarrow X \times X\) be a surjection and let \(V_1,V_2 \in {\mathcal {L}}(X)\) such that \(V(x)=(V_1x,V_2x)\) for each \(x\in X.\) Then, \([S,K]V= SV_1+KV_2\in \Phi _-(X,Y).\) Since \(SV_1\notin \Phi _-\), we get \(KV_2\notin P\Phi _-;\) hence \(K\notin P\Phi _-\) by Proposition 2.1. \(\square \)

Corollary 4.2

If Y is separable,  there exists a surjection from \(X\times X\) onto X,  and X has a quotient isomorphic to \(\ell _1\), then \({{\mathcal {S}}}{{\mathcal {C}}}(X,Y)=\Phi _-(X,Y).\)

Proof

It is well known that every separable Banach space is isomorphic to a quotient of \(\ell _1.\) \(\square \)

The class \(\Phi {\mathcal {C}}(X,Y)\) is surjective in the following sense:

Proposition 4.3

Given \(K\in {\mathcal {L}}(X,Y)\) and a surjective operator \(Q\in {\mathcal {L}}(Z,X),\) if \(KQ \in \Phi {\mathcal {C}}(Z,Y)\), then \(K \in \Phi {\mathcal {C}}(X,Y).\)

Proof

Let \(S\in {\mathcal {L}}(X,Y)\) such that \([S,K] \in \Phi _-(X \times X, Y),\) and let \(Q:Z \rightarrow X\) be a surjective operator. Then, the operator \(Q\times Q\in {\mathcal {L}}(Z \times Z,X\times X)\) defined by \((Q\times Q)(a,b)=(Qa,Qb)\) is surjective. Thus, \([S,K](Q\times Q)= [SQ,KQ]\) is in \(\Phi _-(Z \times Z,Y).\) Since \(KQ \in \Phi {\mathcal {C}}(Z,Y)\), we obtain \(SQ \in \Phi _-(Z, Y),\) hence \(S \in \Phi _-(X,Y),\) and we conclude \(K \in \Phi {\mathcal {C}}(X,Y).\) \(\square \)

The dual space \((X\times X, \Vert \cdot \Vert _\infty )^*\) can be identified with \((X^*\times X^*, \Vert \cdot \Vert _1)\) in the obvious way. Hence, the conjugate operator \([S,T]^*\) can be identified with \((S^*,T^*).\) Indeed, for \(x^*\in X^*\) and \(x\in X\), we have

$$\begin{aligned} \langle [S,T]^* x^*, x\rangle= & {} \langle x^*, [S,T]x\rangle = \langle x^*, Sx+Tx\rangle \\= & {} \langle S^*x^*+T^*x^*, x\rangle = \langle (S^*,T^*)x^*, x\rangle . \end{aligned}$$

As a consequence, \([S,T]\in \Phi _-\) if and only if \((S^*,T^*)\in \Phi _+.\) Similarly, \((S,T)^*\) can be identified with \([S^*,T^*].\)

The following result describes the behavior of the classes of \(\Phi \)-singular and \(\Phi \)-cosingular operators under duality.

Proposition 4.4

Let \(K\in {\mathcal {L}}(X,Y).\)

  1. 1.

    If \(K^* \in \Phi {\mathcal {S}}(Y^*, X^*)\), then \(K \in \Phi {\mathcal {C}}(X,Y).\)

  2. 2.

    If \(K^* \in \Phi {\mathcal {C}}(Y^*, X^*)\), then \(K \in \Phi {\mathcal {S}}(X,Y).\)

Proof

(1) Let \(S\in {\mathcal {L}}(X,Y)\) such that \([S,K] \in \Phi _-(X \times X, Y).\) Then, \([S,K]^* \in \Phi _+.\) Since \([S,K]^*\equiv (S^*,K^*),\) we have \((S^*,K^*) \in \Phi _+(Y^*, X^* \times X^*),\) and from \(K^* \in \Phi {\mathcal {S}}(Y^*, X^*)\) we obtain \(S^* \in \Phi _+(Y^*, X^*);\) therefore, \(S \in \Phi _-(X, Y),\) and hence \(K \in \Phi {\mathcal {C}}(X,Y).\)

The proof of (2) is similar. \(\square \)