Some extremal problems for linear operators in Banach spaces

M. I. Ostrovskii Department of Mathematics The Catholic University of America Washington, D.C. 20064

                        e-mail: ostrovskii@cua.edu

September 30, 2003

1 Introduction

Let X be a Banach space, L(X) the space of all bounded linear operators on X. For A ∈ L(X) we denote by ∥A∥ its norm, by σ(A) the spectrum of A, and by ρ(A) the spectral radius, that is ρ(A) = sup{|t| : t ∈ σ(A)}. We say that A ∈ L(X) is a V - operator if there is B ∈ L(X) with ∥B∥ = 1 and ρ(AB) = ∥A∥. If ∥A∥ = 1 this means that 1 ∈ σ(AB), for some B of norm one. If a Banach space X is such that all operators in L(X) are V -operators, we say that X has the V -property, or is a V -space.

The notion of a V -operator arises in the theory of operator fractional linear transformations and related topics. Note that a fractional linear transformation

T ↦ (T − C)(I_{X} − AT)^{−1}

is defined on the closed unit ball of L(X) iff either ∥A∥ < 1 or ∥A∥ = 1 and A is not a V -operator.

The purpose of my talk is to describe what is known about V - operators and V -spaces and to discuss some open problems.

For technical reasons it is convenient to extend the definitions to the operators acting between Banach spaces. We say that A ∈ L(X, Y) is a V -operator if there is B ∈ L(Y, X) such that ∥B∥ = 1 and ρ(AB) = ∥A∥. If all A ∈ L(X, Y) are V -operators, we say that the pair (X, Y) is a V -pair.

2 V -spaces

We say that an operator A attains its norm if there is a unit vector x such that ∥Ax∥ = ∥A∥.

Lemma 1 If an operator attains its norm, then it is a V - operator.

Proof. Let A ∈ L(X, Y), x ∈ X, ∥x∥ = 1, ∥Ax∥ = ∥A∥. Choose a norm-one functional f ∈ Y^{∗} such that f(Ax) = ∥A∥ and define an operator B ∈ L(Y, X) by By = f(y)x. Then AB(Ax) = ∥A∥Ax. Hence ∥A∥ ∈ σ(AB).

Corollary 1 Any finite-dimensional space is a V -space.

Remark. Lemma 1 implies a more general fact: If a Banach space X is such that each operator from L(X) attains its norm, then X is a V -space. However, it seems to be unknown whether there are infinite dimensional spaces with this property. (I shall make some more comments on this problem at the end of my talk.)

Some V -operators do not attain their norms. This statement is an immediate consequence of the following result: If p ∈ (1, ∞), then l_{p}-direct sums of finite-dimensional spaces have the V -proper-ty.

Recall, that if {X_{n}}_{n=1}^{∞} is a sequence of Banach spaces, then their l_{p}-direct sum is the space of all sequences x = {x_{n}}_{n=1}^{∞}, for which x_{n} ∈ X_{n} for all n and the norm ||x|| = (∑_{n=1}^{∞} ||x_{n}||^{p})^{1/p} is finite. The l_{p}-direct sum is denoted by (∑_{n=1}^{∞} ⊕X_{n})_{p}.

Proposition 1 (V.Khatskevich–V.Shulman–M.O., [5]) Let {X_{n}}_{n=1}^{∞} be any sequence of finite dimensional normed spaces and let p ∈ (1, ∞). Then X = (∑_{n=1}^{∞} ⊕X_{n})_{p} is a V - space.

Proof. Let A ∈ L(X). We need to show that A is a V -operator. It is easy to see that it is enough to do this for ||A|| = 1.

Let {x_{n}}_{n=1}^{∞} be a sequence satisfying ||x_{n}|| = 1 and ||A(x_{n})|| → 1. The sequence {x_{n}}_{n=1}^{∞} contains a weakly convergent subsequence, we may assume that {x_{n}}_{n=1}^{∞} is itself weakly convergent. Let x be its weak limit.

We consider two cases (x ≠ 0 and x = 0). Case 1. x ≠ 0.

In this case we shall show that ||Ax|| = ||x||, so A attains its norm, and we are done.

Using the fact that the sequences {x_{n} − x}_{n=1}^{∞} and {A(x_{n} − x)}_{n=1}^{∞} are weakly null and the standard argument about blockbases in l_{p}-direct sums we get ∥Ax∥^{p} = ∥Ax_{n}∥^{p} − ∥A(x_{n} − x)∥^{p} + o(1) ≥ ∥Ax_{n}∥^{p} − ∥x_{n} − x∥^{p} + o(1) = 1 − ∥x_{n} − x∥^{p} + o(1) = ∥x_{n}∥^{p} − ∥x_{n} − x∥^{p} + o(1) = ∥x∥^{p} + o(1) that is ∥Ax∥ ≥ ∥x∥.

Case 2. x = 0.

In this case the sequence {x_{n}} is weakly null and we may assume without loss of generality that vectors x_{n} are of the form

x_{n} = ∑_{k=m(n−1)}^{m(n)−1} u_{k},

where u_{k} ∈ X_{k} and 1 ≤ m(0) < m(1) < ⋯ < m(n) < . . ..

Passing to a subsequence if necessary, we may assume that there exists a sequence {y_{n}} in X such that

lim_{n→∞} ||y_{n} − A(x_{n})|| = 0,

||y_{n}|| = 1, and the vectors y_{n} are of the form

y_{n} = ∑_{k=s(n−1)}^{s(n)−1} w_{k},

where w_{k} ∈ X_{k} and 1 ≤ s(0) < s(1) < ⋯ < s(n) < . . ..

Let W be the closed linear span of the sequence {y_{n}}_{n=1}^{∞}. Let P_{n} be a projection of norm 1 from (∑_{k=s(n−1)}^{s(n)−1} ⊕X_{k})_{p} onto the linear span of the vector y_{n}. Combining such projections we get a projection P : X → W of norm 1.

It is easy to see that the operator D : W → X given by D(y_{n}) = x_{n} satisfies ||D|| = 1. Let B = DP. Then ||B|| = 1 and

||(AB)y_{n} − y_{n}|| = ||A(x_{n}) − y_{n}|| → 0.

Hence 1 ∈ σ(AB).

Proposition 1 can be strengthened in the following way.

Theorem 1 (M. O. [6]) Let 1 < p < ∞ and let {X_{n}}_{n=1}^{∞} be a sequence of finite dimensional normed spaces. Then each subspace of X = (∑_{n=1}^{∞} ⊕X_{n})_{p} is a V -space.

The proof is somewhat more complicated than the proof of Proposition 1 because the space does not have to contain vectors x_{n} and y_{n} of the form described above. Of course, we can approximate everything as well as we would like. However, because the V -property is an isometric property, it requires some effort to straighten everything out. I am not going to discuss the details.

Theorem 1 gives the hope that the V -property is subspace-here-ditary. At the moment it is unclear whether it is the case in general. It is easy to check that the V -property is heredited by 1-comple-mented subspaces.

In my opinion the most interesting question related to this discussion is: Problem. Does there exist a space with few operators having the V -property?

3 A method for finding non-V -operators

Recall that an operator A ∈ L(X, Y) is called strictly singular if its restriction to any infinite-dimensional subspace of X is not an isomorphism. It is well known that the set S(X, Y) of all strictly singular operators form a subspace in L(X, Y).

Definition 1 We say that an operator A ∈ L(X, Y) has a strictly singular hump if there is K ∈ S(X, Y) with ∥A + K∥ < ∥A∥.

The following result is the main tool for finding non-V -operators.

Proposition 2 Strictly singular hump criterion (V.Khatskevich–V.Shulman–M.O., [5]). An operator having a strictly singular hump is a V -operator iff it attains its norm.

Proof. Suppose that A has a strictly singular hump and is a V -operator. We may assume that ∥A∥ = 1 and 1 ∈ σ(AB) for some B ∈ L(Y, X) with ∥B∥ = 1. Let K ∈ S(X, Y) be such that ||A + K|| < 1. Then ∥AB + KB∥ < 1 and

I_{Y} − AB = (I_{Y} − (AB + KB)) + KB.

It is clear that the operator I_{Y} − (AB + KB) ∈ L(Y) is invertible and the operator KB is strictly singular. By a result of T. Kato (1958) strictly singular perturbation of an invertible operator is a Fredholm operator of zero index. Hence the operator I_{Y} − AB is a Fredholm operator of zero index. By our assumption it is not invertible, therefore ker(I_{Y} − AB) ≠ {0}. Let y ∈ ker(I_{Y} − AB), ||y|| = 1. It is easy to see that ||ABy|| = ||By|| = 1, so A attains its norm. The converse implication is obvious.

First we use the following special case of the Strictly Singular Hump Criterion.

Corollary 2 If a strictly singular operator does not attain its norm, then it is not a V -operator.

Corollary 3 Each V -space is reflexive.

Proof. If X is not reflexive then, by the well-known theorem of R. C. James, there is a norm-one functional f ∈ X^{∗} with f(x) < ∥x∥, for each non-zero x ∈ X. Choose y ∈ X, ∥y∥ = 1, and set Ax = f(x)y. Then A is a strictly singular operator which does not attain its norm. By Corollary 2 A is not a V -operator. Remark. One can prove Corollary 3 in a more straightforward way.

Corollary 4 If p < q, then (l_{p}, l_{q}) is not a V -pair.

Proof. It is well known that for p < q the formal identity map I ∈ L(l_{p}, l_{q}) is strictly singular. Let D be the operator on l_{q} defined by De_{k} = (1 − 1/k)e_{k}, where {e_{k}}_{k=1}^{∞} is the unit vector basis in l_{q}. Then A = DI is strictly singular, ∥A∥ = 1, and ∥Ax∥ < ∥x∥ for x ≠ 0, because ∥Dx∥ < ∥x∥. It remains to apply Corollary 2.

Corollary 5 The space X = R ⊕_{r} l_{p} is not a V -space if p ≠ r.

Proof. We consider the case r < p only (later, in Corollary 11, we shall see that the V -property is self-dual). Define an operator K on X by

K(b, a_{1}, a_{2}, …) = (b − a_{1}, a_{1}, a_{2}/2, a_{3}/3, …).

Clearly K is compact. Straightforward verification shows that

(1) ∥I − K∥ = 2^{\frac{1}{r}−\frac{1}{p}} > 1.

(2) I − K does not attain its norm.

By Proposition 2 the statements (1) and (2) imply that X is not a V -space (we apply the Strictly Singular Hump Criterion with A = I − K).

Part of the verification:

(I − K)(b, a_{1}, a_{2}, …) = (a_{1}, 0, \frac{1}{2}a_{2}, \frac{2}{3}a_{3}, \frac{3}{4}a_{4}, …).

Hence

||(I − K)(0, 1, 0, …, 0, 1, 0, …)|| = ||(1, 0, …, \frac{k − 1}{k}, …)|| =

(1 + (\frac{k − 1}{k})^{r})^{1/r} → 2^{1/r}.

On the other hand

||(0, 1, 0, …, 0, 1, 0, …)|| = 2^{1/p}.

Using the Strictly Singular Hump Criterion and some well-known result on the structure of Banach spaces we can find many other examples of non-V -spaces. I list them without proofs.

Corollary 6 If p, q, r ∈ [1, ∞], then l_{p} ⊕_{r} l_{q} is a V -space if and only if p = q = r ∈ (1, ∞).

Corollary 7 If p ≠ 2, then L_{p}(0, 1) is not a V -space.

By C_{p} (1 ≤ p < ∞) we denote the Schatten-von Neumann ideals of L(H), where H is a separable Hilbert space.

Corollary 8 If 1 ≤ p < ∞ and p ≠ 2, then C_{p} is not a V - space.

At the moment Proposition 2 and its straightforward generalizations (which I shall discuss later) are the only existing tools to establish that an operator is not a V -operator. In particular, Proposition 2 can be used to establish the following result.

Theorem 2 (M. O. [6]) Let X be a separable infinite dimensional Banach space with a symmetric basis. If X has the V -property, then X is isometric to l_{p}, 1 < p < ∞.

We need a result of Bohnenblust (1940) [1]. We are going to use the version of his result dealing with unconditional bases. For a space with a fixed basis {e_{i}} by support of a vector x we mean the subset of N corresponding to non-zero coefficients of x.

Theorem 3 Let X be a Banach space with an unconditional basis {e_{i}}_{i=1}^{∞}. If for any two pairs (x_{1}, y_{1}) and (x_{2}, y_{2}) of disjointly supported vectors the equalities

||x_{1}|| = ||x_{2}|| and ||y_{1}|| = ||y_{2}||

imply

||x_{1} + y_{1}|| = ||x_{2} + y_{2}||,

then X is c_{0} or l_{p} (1 ≤ p < ∞).

Proof of Theorem 2. Suppose that X is a reflexive Banach space with a symmetric basis {e_{i}}_{i=1}^{∞}, and X is not isometric to l_{p}, 1 < p < ∞. It is enough to show that these assumptions imply that X does not have the V -property.

By the Bohnenblust’s Theorem 3 there exist two pairs (x_{1}, y_{1}) and (x_{2}, y_{2}) of disjointly supported vectors in X such that

||x_{1}|| = ||x_{2}|| and ||y_{1}|| = ||y_{2}||  (1)

but

||x_{1} + y_{1}|| < ||x_{2} + y_{2}||  (2)

Since the inequality is preserved under small perturbations of the vectors we may and shall assume that all vectors are finitely supported.

Also we may assume that y_{1} = y_{2} = y. In fact, since X has a symmetric basis we may assume that all vectors x_{1}, x_{2}, y_{1}, and y_{2} are disjoint. We consider three pairs (x_{1}, y_{1}), (x_{1}, y_{2}), and (x_{2}, y_{2}). All three pairs satisfy the condition (1) (after the corresponding relabeling). The inequality (2) implies that at least one of the following is true:

||x_{1} + y_{1}|| < ||x_{1} + y_{2}||

or

||x_{1} + y_{2}|| < ||x_{2} + y_{2}||.

Relabeling the vectors in the first case we get the claim.

Since the basis {e_{i}} is symmetric, we may assume that the supports of x_{1} and x_{2} are contained in {1, …, k}, and the support of y is contained in {k + 1, … }.

Let Z_{1} = lin{e_{1}, …, e_{k}} and Z_{2} = cl(lin{e_{k+1}, … }). Then

X = Z_{1} ⊕ Z_{2}.

Let R_{1} : Z_{1} → Z_{1} be a one-dimensional operator given by

R_{1}(x) = \frac{f(x)x_{2}}{||x_{2}||},

where f ∈ X^{∗} is such that ||f|| = 1 and f(x_{1}) = ||x_{1}|| = ||x_{2}||. Let R_{2} : Z_{2} → Z_{2} be an operator of the form

R_{2} ( ∑_{i=k+1}^{∞} a_{i}e_{i}) = ( ∑_{i=k+1}^{∞}(1 − ε_{i})a_{i}e_{i}) ,

where {ε_{i}}_{i=1}^{∞} are selected in such a way that 1 > ε_{i} > ε_{i+1} > 0, lim_{i→∞} ε_{i} = 0, and

||x_{1} + y|| < ||x_{2} + Ry||.  (3)

Let R : Z_{1} ⊕ Z_{2} → Z_{1} ⊕ Z_{2} be given as

R = R_{1} ⊕ R_{2}.

Then

R(x_{1} + y) = x_{2} + Ry,

by (3) we get ||R|| > 1 It is also clear, that R = I + K, where K is a compact operator. Therefore R is an operator with a strictly singular (even compact) hump.

Therefore to finish the proof of the theorem it remains to show that R does not attain its norm.

It is done using in the following way. We prove Claim. For each vector z ∈ X satisfying ||Rz|| > ||z|| there exists a vector z̃ such that ||z̃|| = ||z|| and ||Rz̃|| > ||Rz||.

Proof. We can find z̃ in the following way. For z = ∑_{i=1}^{∞} a_{i}e_{i} we let

z̃ = ∑_{i=1}^{k} a_{i}e_{i} + ∑_{i=k+1}^{∞} a_{i+1}e_{i+1}.

The equality ||z|| = ||z̃|| follows because the basis {e_{i}}_{i=1}^{∞} is symmetric.

To show that ||Rz̃|| > ||Rz|| we observe that R cannot attain its norm on a vector from Z_{1}, because ||R|| > 1 and ||R_{1}|| = 1.

Therefore ∑_{i=k+1}^{∞} a_{i+1}e_{i+1} ≠ 0. Now the inequality ||Rz̃|| > ||Rz|| follows using routine calculations (the main idea is that the shifted coordinates are multiplied by larger numbers, because (1 − ε_{i+1})/(1 − ε_{i}) > 1, see the definition of R_{2} and R). I omit the details.

4 On duality

Using the James’s theorem on norm-attaining functionals in the same way as before we get.

Corollary 9 If (X, Y) is a V -pair, then X is reflexive.

Proposition 3 If A ∈ L(X, Y) is a V -operator, then the same is true for A^{∗}.

Proof. Let B ∈ L(Y, X) be such that ||B|| = 1 and ρ(AB) = ∥A∥. Then ||B^{∗}|| = 1 and ρ(A^{∗}B^{∗}) = ρ(B^{∗}A^{∗}) = ρ((AB)^{∗}) = ||A|| = ||A^{∗}||.

Corollary 10 If (X, Y) is a V -pair and Y is reflexive, then (Y^{∗}, X^{∗}) is also a V -pair.

Proof. By Proposition 9 the space X is reflexive. Hence any operator in L(Y^{∗}, X^{∗}) is adjoint to an operator in L(X, Y), and we may apply Proposition 3.

Corollary 11 A Banach space X is a V -space iff X^{∗} is a V - space.

5 Remarks and open problems

Remark. Analysis of the proof of the Strictly Singular Hump Criterion shows that the only property of strictly singular operators that we need can be described in the following way:

We say that K ∈ L(X, Y) is an N-operator if for every D ∈ L(Y, X) the operator KD has the following property: if (I_{Y} − KD) ∈ L(Y) is not invertible, then ker(I_{Y} − KD) ≠ {0}.

There exist N-operators which are not strictly singular. For example, every strictly cosingular operator between reflexive spaces is an N-operator.

We can define operators with N-humps and prove the following generalization of Proposition 2:

• An operator having an N-hump is a V -operator iff it attains its norm.

I do not know any interesting applications of this generalization of Proposition 2.

The most natural related problem is: Problem. To discover other approaches to finding non-V - operators.

In connection with the fact that norm-attaining operators are V -operators it is natural to recall the following, to the best of my knowledge still open, problem. Problem. Does there exist an infinite dimensional Banach space X such that each A ∈ L(X) attains its norm? Related results and observations:

• R. C. James’s on characterization of reflexivity implies that if X is such that each A ∈ L(X) attains its norm, then X and L(X) are reflexive spaces.

Note. To show that L(X) is reflexive, we use the identification L(X) = (X ˆ⊗_{π}X^{∗})^{∗}. Recall that this identification is valid for any reflexive X without any approximation property assumptions.

• J. R. Holub [3] proved that if X has the approximation property, then the reflexivity of L(X) implies that X is finite dimensional.

• N. J. Kalton [4, Theorem 2] proved that L(X) cannot be reflexive for nonseparable X.

• Hence the only possible candidates for X in the problem are separable reflexive spaces without 1-complemented subspaces having the approximation property.

• If the answer to the problem is “Yes”, then it will be natural to try to show that the V -property is not subspace-hereditary examining subspaces of the spaces on which all of the operators attain their norms (the mentioned result of Holub implies that those of infinite dimensional subspaces which have the approximation property have operators which are not norm attaining).

The last result that I would like to mention is:

• (M. O. [6]) For each infinite dimensional Banach space X there exists an equivalent norm ||| · ||| on X such that the space (X, ||| · |||) is not a V -space.

Sketch of the proof. By Proposition 2 it is enough to find an operator with a strictly singular hump which does not attain its norm.

We do this in the following way. We let P be a projection onto a subspace of codimension 1. It is an exercise on renorming theory to construct a new norm ||| · ||| in X such that |||P ||| = 2 and this norm is not attained. Since there exists a one-dimensional operator Q in X such that P + Q = I_{X} (and |||I_{X}||| = 1), the operator P is a not-norm-attaining operator with a strictly singular hump in X. Hence (X, ||| · |||) is not a V -space.

I have mentioned this result because it can be used to get a proof of the fact that for infinite dimensional X with the approximation property (A. P.) L(X) cannot be reflexive.

We need the well-known fact, that for X with the A. P. the space X ˆ⊗_{π}X^{∗} is isometric to the space of nuclear operators on X. Using this fact and the description of the duality we get, that for X with the A. P. L(X) is reflexive if and \underline{only if} all operators in L(X) attain their norms.

Now, if L(X) is reflexive, then L(X, ||| · |||) is reflexive, hence, all operators in (X, ||| · |||) attain their norms (because (X, ||| · |||) also has A. P.). We get a contradiction.

G. Godefroy and P. Saphar [2, p. 693]: “If there exists a reflexive Banach space X such that L(X) = K(X) ⊕ span(I_{X}), and X is without the C. A. P., then K(X) is reflexive.”

References

[1] F. Bohnenblust, An axiomatic characterization of L_{p}-spaces, Duke Math. J., 6 (1940), 627–640.

[2] G. Godefroy and P. Saphar, Duality in spaces of operators and smooth norms on Banach spaces, Illinois J. Math., 32 (1988), no. 4, 672–695.

[3] J. R. Holub, Reflexivity of L(E, F), Proc. Amer. Math. Soc., 39 (1973), 175–177.

[4] N. J. Kalton, Spaces of compact operators, Math. Ann., 208 (1974), 267–278.

[5] V. A. Khatskevich, M. I. Ostrovskii, and V. S. Shulman, Extremal problems for operators in Banach spaces arising in the study of linear operator pencils, Integral Equations and Operator Theory, to appear.

[6] M. I. Ostrovskii, Extremal problems for operators in Banach spaces arising in the study of linear operator pencils, II, Integral Equations and Operator Theory, to appear.