Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 1, 27–35

On Restriction Properties of Multiplication Operators

A. Plichko and V. Shevchik

Abstract. A multiplication operator A acting in a rearrangement-invariant function space E is considered. Infinite dimensional subspaces X of E for which the restriction A|X is an isomorphism are described. Applications to multiplied trigonometric sequences in Banach function spaces are given.

Keywords: Multiplication operators, rearrangement-invariant functions spaces AMS subject classification: 46 B, 47 B

1. Introduction

Let (T, Σ, μ) be a finite measure space and let E be a rearrangement-invariant function space defined on T (see [4: p. 118]). We consider the multiplication operator by a bounded measurable function a = a(t) (t ∈ T) given by

Ax = ax (x ∈ E), (Ax)(t) = a(t) x(t) (t ∈ T).

Obviously, A is a bounded linear operator acting in E. In general A is not invertible and not compact. In order to investigate properties of the operator A we restrict it to some infinite dimensional subspaces of E, where A has a more simple nature. We consider the following two kinds of such subspaces:

1. Infinite dimensional subspaces X ⊂ E such that the restriction A|X is an isomorphism, i.e. inf{∥Ax∥ : x ∈ X with ∥x∥ = 1} > 0.

2. Infinite dimensional subspaces X ⊂ E such that the restriction A|X is compact.

We give a description of subspaces of both kinds. Clearly, properties of such restrictions are helpful to understand mapping properties of A at all.

In addition, let us make the following observation:

Suppose X ⊂ E is an infinite dimensional subspace of E such that A|X is an isomorphism. If a sequence {x_{n}} is a basis or unconditional basis of X, then the sequence {ax_{n}} is a basic sequence or unconditional basic sequence in E, respectively. We will show that this observation is useful in order to describe properties of the sequences {a(t) cos nt} and {a(t) sin nt} in some Banach function spaces.

A. Plichko: Ped. Univ., Dept. Math., Str. Shevchenko, Korovograd 316050, Ukraine V. Shevchik: Friedrich-Schiller-Univ., Math. Inst., Ernst-Abbe.-Pl. 1-4, D-07743 Jena e-mail: shevchik@minet.uni-jena.de ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

The paper is organized as follows.

In Section 2 we explain some notations and formulate a simple statement on general properties of a multiplication operator A by a bounded measurable function. In Section 3 we study subspaces X of E such that A|X is an isomorphism (in the sense explained above). Subspaces of even and of odd functions on a compact symmetric domain T ⊂ ℝ^{n} are of special interest. We find conditions on the function a under which A|X are isomorphisms (Proposition 3.4 is the main result here). In Section 4 we consider multiplied trigonometric sequences, i.e. sequences of the form {a(t) cos nt} and {a(t) sin nt} where a is a continuous function. Using results of Section 3 we answer questions on basic properties of such sequences in spaces L_{p}(−π, +π) (1 < p < +∞). A similar problem is investigated in the multidimensional case. Namely, we find conditions under which a multiplied n-dimensional trigonometric sequence on the cube K = [−π, +π]^{n} is an unconditional basic sequence in L_{2}(K). We also study multiplied lacunary trigonometric sequences. Under some assumptions on a we show that such sequences are unconditonal basic sequences in L_{p}(−π, +π) (1 ≤ p < +∞). The investigation is based on studying normed sequences {x_{n}} ⊂ E such that ∥Ax_{n}∥ → 0 as n → +∞. Finally, Section 5 is devoted to the study of subspaces X of E such that A|X is compact. The case E = L_{2}(0, 1) and a(t) = t is considered separately. The question whether A|X is strictly singular is discussed. We give an example of a subspace X ⊂ L_{p}(0, 1) (1 < p < 2) such that A|X, where Af(t) = tf(t), is strictly singular but non-compact. We also discuss spectral properties of a compact selfadjoint operator that corresponds to a compact restriction of a multiplication operator in L_{2}(0, 1).

In the case when E = L_{2}(0, 1) and a is a continuous function, some of the results of this paper were announced in [5].

Acknowlegement. The second named author has been supported by DFG, grant Nr. Tr 374/2.

2. Notation. Simplest properties of multiplication operators

We use the notation supp a = {t ∈ T : a(t) ≠ 0} and γ(a) = {t ∈ T : a(t) = 0} = T \ supp a. By χ_{σ} we denote the characteristic function of a set σ ∈ Σ. For δ > 0 let σ_{δ} = {t : |a(t)| ≥ δ} and χ_{δ} := χ_{σ_{δ}}. If X is a Banach space, then S(X) denotes the unit sphere of X. By ”subspace” we mean a closed infinite dimensional subspace.

Let X, Y be Banach spaces and T : X → Y a linear bounded operator. Recall that T is called strictly singular if, for any subspace Z ⊂ X, T |Z is not an isomorphism, i.e. inf{∥T x∥ : x ∈ S(X)} = 0.

Proposition 2.1. The following statements are obvious:

1. A is injective on E if and only if μ(γ(a)) = 0. If μ(γ(a)) ≠ 0, then dim (Ker A) = ∞.

2. A maps E isomorhically onto E if and only if there exists δ > 0 such that μ(T \ σ_{δ}) = 0.

3. A is compact if and only if A is strictly singular, and A is strictly singular if and only if A is the zero operator, i.e. μ(supp a) = 0.

4. A has closed range if and only if there exists δ > 0 such that μ(supp a \ σ_{δ}) = 0. If in this case μ(T \ supp a) ≠ 0, then dim (ImA) = ∞.

3. Restrictions of A being isomorphisms

In this section we consider subspaces X of E for which A|X is an isomorphic map. We will use the following lemma.

Lemma 3.1. Let {x_{k}} ⊂ S(E) and ∥Ax_{k}∥ → 0 as k → +∞. Then, for every δ > 0,

∥χ_{δ}x_{k}∥ → 0  (k → +∞).  (3.1)

Proof. Indeed, ∥χ_{δ}x_{k}∥ ≤ ∥\frac{1}{δ}aχ_{δ}x_{k}∥ ≤ ∥\frac{1}{δ}ax_{k}∥ = \frac{1}{δ}∥Ax_{k}∥ → 0 as k → +∞, and the proof is complete

Proposition 3.2. Let X ⊂ E be a subspace. The following conditions are equivalent:

1. A|X is an isomorphism.

2. There exist δ, ε > 0 such that ∥χ_{δ}x∥ ≥ ε for every x ∈ S(X).

3. There exists δ > 0 such that P_{δ}|X is an isomorphism where P_{δ}x = χ_{δ}x.

Proof. 2 ⇔ 3 follows from the definitions and 2 ⇒ 1 follows from Lemma 3.1. To show 1 ⇒ 2 let A|X be an isomorphism. Then there exists δ > 0 such that ∥Ax∥ = ∥ax∥ ≥ 2δ for every x ∈ S(X). Put c = sup{|a(t)| : t ∈ T }. Then

2δ ≤ ∥ax∥ ≤ ∥axχ_{δ}∥ + ∥axχ_{T \σ_{δ}}∥ ≤ c∥xχ_{δ}∥ + δ∥xχ_{T \σ_{δ}}∥ ≤ c∥xχ_{δ}∥ + δ.

Hence ∥xχ_{δ}∥ ≥ \frac{δ}{c} = ε, and the proof is complete

Proposition 3.3. Let T be a closed domain in ℝ^{n} with Lebesgue measure μ and X a subspace of a rearrangement-invariant space E defined on T. Suppose a : T → ℝ is a continuous function. Then the following conditions are equivalent:

1. A|X is an isomorphism.

2. There exist a closed set σ ⊂ supp a and ε > 0 such that ∥χ_{σ}x∥ ≥ ε for every x ∈ S(X).

3. There exists a closed set σ ⊂ supp a such that P_{σ}x = χ_{σ}x is an isomorphism in X.

Proof. It is sufficient to change slightly the proof of Proposition 3.2. Namely, in the proof of 1 ⇒ 2 we note that the set σ_{δ} is closed and in the proof of 2 ⇒ 1 in view of the compactness of T we have for the closed subset σ ⊂ supp a that δ = inf{|a(t)| : t ∈ σ} > 0

Proposition 3.4. Let T be a compact symmetric domain in ℝ^{n} with Lebesgue measure μ and a : T → ℝ a continuous function. Suppose that X ⊂ E is a subspace consisting of even or odd functions. The restriction A|X is an isomorphism if and only if

a(t) = 0 ⟹ a(−t) ≠ 0.  (3.2)

Remark that condition (3.2) implies a(θ) ≠ 0 where θ = (0, ..., 0) ∈ ℝ^{n}.

Proof of Proposition 3.4. Suppose that condition (3.2) holds. First we show that there exists δ > 0 such that, for the set σ_{1} = {t ∈ T : |a(t)| ≤ δ}, σ_{δ} ⊃ σ = −σ_{1}. Indeed, suppose the contrary. Then there exists a sequence {t_{k}} ∈ T such that a(t_{k}) → 0 and a(−t_{k}) → 0 as k → ∞. Using the compactness of T and the continuity of a we find a point t_{0} ∈ T such that a(t_{0}) = a(−t_{0}) = 0. This contradicts (3.2). For every function f from the rearrangement-invariant space E we have ∥f∥ = ∥f^{−}∥ where f^{−} is defined by f^{−}(t) = f(−t). Thus for even functions x we have

∥xχ_{σ}∥ = ∥x^{−}χ_{σ}∥ = ∥xχ_{σ}^{−}∥ = ∥xχ_{σ_{1}}∥.

Similarly, for odd functions x we have

∥xχ_{σ}∥ = ∥ − x^{−}χ_{σ}∥ = ∥x^{−}χ_{σ}∥ = ∥xχ_{σ_{1}}∥.

In view ∥x∥ ≤ ∥xχ_{δ}∥ + ∥xχ_{σ_{1}}∥ ≤ 2∥xχ_{δ}∥ we have in both cases

∥Ax∥ = ∥ax∥ ≥ ∥axχ_{δ}∥ ≥ δ∥xχ_{δ}∥ ≥ \frac{δ}{2}∥x∥.

Therefore A|X is an isomorphism.

To prove the ”only if” part suppose that (3.2) fails, that is there exists t_{0} ∈ T such that a(t_{0}) = a(−t_{0}). We consider two cases:

a) t_{0} ≠ 0. We denote by Q_{k} ∈ ℝ^{n} the cube with center t_{0} and side length \frac{1}{k}. Put σ_{k} = Q_{k} ∩ T. It is obvious that μ(σ_{k}) ≠ 0. Denote χ_{k} := χ_{σ_{k}} and Put x_{k} = χ_{k} + χ_{k}^{−} and y_{k} = χ_{k} − χ_{k}^{−}. The function x_{k} is even and the function y_{k} is odd. In view of the coninuity of the function a,

∥A (\frac{x_{k}}{∥x_{k}∥})∥ ≤ sup_{t∈σ_{k}} |a(t)| \frac{∥χ_{k}∥}{∥x_{k}∥} + sup_{t∈−σ_{k}} |a(t)| \frac{∥χ_{k}^{−}∥}{∥x_{k}∥} → 0  (k → ∞).  (3.3)

Since σ_{k} ∩ −σ_{k} = ∅ for every k > k_{0}, we have ∥y_{k}∥ ≥ ∥χ_{k}∥ > 0 for k > k_{0}. Therefore the proof of

A (\frac{y_{k}}{∥y_{k}∥}) → 0  (k → +∞)  (3.4)

is similar to that of (3.3).

b) t_{0} = 0. Let σ_{k} be the cube centered at (\frac{1}{k} , ..., \frac{1}{k}) with side length \frac{1}{k}. In this case the proof of (3.3) - (3.4) is the same as that in the case a). It follows from (3.3) - (3.4) that A|X is not an isomorphism. This completes the proof

4. Multiplied trigonometric sequences

Recall that a sequence {x_{n}} ⊂ X in a Banach space X is said to be a basic sequence if it is a basis of its closed linear span (see [3: p. 1]).

Proposition 4.1. Let a be a continuous function on [−π, +π] such that condition (3.2) is fulfilled. Then {a(t) cos nt} and {a(t) sin nt} are basic sequences in the space L_{p}(−π, +π) (1 < p < +∞) and unconditional basic sequences in the space L_{2}(−π, +π).

Proof. It follows immediately from Proposition 3.4 and the well known property that the trigonometric sequence is a basis in the space L_{p}(−π, +π) (1 < p < +∞, p ≠ 2) and an unconditonal basis in the space L_{2}(−π, +π). We also use the simple statement that each isomorphism maps a basic sequence into a basic sequence and an unconditional basic sequence into an unconditional basic sequence

Remark 4.2. It follows from known results of the theory of rearrangement-invari-ant spaces that Proposition 4.1 is also valid for rearrangement-invariant spaces on [−π, +π] with non-trivial Boyd indices (see [4: p. 130]).

Remark 4.3. Proposition 3.4 implies that {t cos nt}_{n∈ℕ} and {t sin nt}_{n∈ℕ} are basic sequences in L_{p}(0, 2π) (1 < p < +∞, p ≠ 2) and unconditonal basic sequences in L_{2}(0, 2π). This contrasts to properties of the sequence {te_{n}(t)}_{n≥1}, where e_{1}(t) = 1, e_{2}(t) = cos t, e_{3}(t) = sin t, ... - the trigonometric sequence. It is easy to show that {te_{n}(t)}_{n≥1} is not deficiently minimal in L_{2}(0, 2π), i.e. it is not minimal after a deletion of a finite number of elements (see [6: p. 121]).

Let T = K = [−π, +π]^{n}. Then the set of all possible different n-products

sin(k_{1}t_{i_{1}}) ⋯ sin(k_{s}t_{i_{s}}) cos(k_{s+1}t_{i_{s+1}}) ⋯ cos(k_{n}t_{i_{n}})  (4.1)

where 0 ≤ k_{1}... < +∞, 1 ≤ i_{1}, ..., i_{n} ≤ n, 0 ≤ s ≤ n and t_{i_{j}} ∈ [−π, +π] forms an orthogonal basis of the space L_{2}(K). Elements (4.1) with even s generate a subspace of even functions, elements (4.1) with odd s generate a subspace of odd functions.

The proof of the following assertion is similar to that of Proposition 4.1.

Proposition 4.4. Let a be a continuous function on the cube K such that condition (3.2) is fulfilled. Then the products of the function a by functions (4.1) in the case of even or odd s form unconditional basic sequences in the space L_{2}(K).

Let us recall some definitions. A sequence {x_{k}} of elements of a rearrangementinvariant space E is said to be

disjoint if μ{t : x_{k}(t)x_{l}(t) ≠ 0} = 0 (k ≠ l)

almost disjoint if, for a disjoint (corresponding) sequence \{y_{k}\} ⊂ E, \frac{∥x_{k}−y_{k}∥}{∥x_{k}∥} → 0 as k → +∞.

A rearrangement-invariant space E is said to have an absolutely continuous norm if, for every decreasing sequence of measurable sets {σ_{k}} such that ∩σ_{k} = ∅ and for every x ∈ E, ∥χ_{σ_{k}}x∥ → 0 as k → +∞.

Lemma 4.5. Let E be a rearrangement-invariant space with absolutely continuous norm and μ(γ(a)) = 0. Assume that a sequence {x_{k}} ⊂ S(E) satisfies (3.1) for some δ > 0. Then {x_{k}} contains an almost disjoint subsequence {x_{k_{i}}} such that for a corresponding disjoint sequence {y_{i}}

sup{|a(t)| : t ∈ supp y_{i}} → 0  (i → +∞).  (4.2)

Proof. First we show that for every ε, δ > 0 and every k there exist 0 < δ′ < \frac{δ}{2} and k′ > k such that for the characteristic function χ_{δ′,δ} of the set σ_{δ} ∪ (T \ σ_{δ′})

∥χ_{δ′,δ} x_{k′}∥ < ε.  (4.3)

Indeed, using (3.1) we choose k′ > k such that ∥χ_{δ} x_{k′}∥ < \frac{ε}{2}. Since the norm is absolutely continuous we can choose 0 < δ′ < \frac{δ}{2} such that ∥χ_{T \δ′} x_{k′}∥ < \frac{ε}{2}. Now we have

∥χ_{δ′,δ} x_{k′}∥ ≤ ∥χ_{δ} x_{k′}∥ + ∥χ_{T \δ′} x_{k′}∥ < ε.

We construct a required almost disjoint sequence using an inductive process. In the first step we put k = 1 and, using the absolute continuity of the norm, choose δ_{1} such that ∥χ_{T \σ_{δ_{1}}} x_{1}∥ < 1. Suppose that the (i − 1)st step is done. Using (4.3) we choose k_{i} > k_{i−1} and 0 < δ_{i} < \frac{δ_{i−1}}{2} such that ∥χ_{δ_{i},δ_{i−1}} χ_{k_{i}}∥ < \frac{1}{i}. As a correspondent disjoint sequence {y_{i}} we take y_{i} = χ_{{t:δ_{i}≤|a(t)|≤δ_{i−1}}} x_{k_{i}}. By construction, {y_{i}} is disjoint and, since δ_{i} → 0 if i → ∞, condition (4.2) holds. Futhermore,

\frac{∥x_{k_{i}} − y_{i}∥}{∥x_{k_{i}}∥} = ∥x_{k_{i}} − χ_{\{t: δ_{i}≤|a(t)|<δ_{i−1}\}} x_{k_{i}}∥ = ∥χ_{δ_{i},δ_{i−1}} x_{k_{i}}∥ < \frac{1}{i} → 0

as i → +∞, and the proof is complete

The following proposition is a simple corollary of Lemmas 3.1 and 4.5.

Proposition 4.6. Let E be a rearrangement-invariant space with absolutely continuous norm and μ(γ(a)) = 0. Suppose also that ∥Ax_{k}∥ → 0 (k → +∞) for some {x_{k}} ⊂ S(E). Then {x_{k}} contains an almost disjoint subsequence such that a corresponding disjoint sequence satisfies condition (4.2).

Corollary 4.7. Let E be a rearrangement-invariant space with absolutely continuous norm and μ(γ(a)) = 0. Suppose also that a subspace X ⊂ E does not contain any almost disjoint sequence.Then A|X is an isomorphism.

Corollary 4.8. Let X be a subspace included into all spaces L_{p}(μ) (1 ≤ p < +∞). Suppose that the L_{p}-norms are equivalent on X and μ(γ(a)) = 0. Then the restriction A|X is an isomorphism in every space L_{p}(μ) (1 ≤ p < +∞).

Proof. Assume the contrary. Proposition 4.6 yields that X contains an almost disjoint sequence {x_{k}}. By virtue of well known stability properties of basic sequences, {x_{k}} contains a subsequense {x_{k_{i}}} such that it is equivalent to the standard basis of l_{p} in the space L_{p} and to the standard basis of l_{q} in the space L_{q}. But it is a classical result that the standard bases of l_{p} and l_{q} are not equivalent

Let us remind that a sequence {n_{k}} of positive integers is said to be lacunary if inf_{k} \frac{n_{k+1}}{n_{k}} = λ > 1.

Corollary 4.9. Suppose T = [−π, +π], 1 ≤ p < +∞, μ(γ(a)) = 0 and that the sequence {n_{k}} is lacunary. Then {ae_{n_{k}}} with {e_{n}} being the trigonometric sequence is an unconditional basic sequence in L_{p}(−π, +π).

Proof. It follows immediately from a known property of lacunary sequences in L_{p} (1 ≤ p < +∞): the L_{p}-norms on the linear span of a lacunary sequence are equivalent

5. Restrictions of A with the compact mapping property

In this section we study subspaces X of a rearrangement-invariant space E such that A|X are compact mappings. The symbol x_{n}→^{w}x means weak convergence.

Proposition 5.1. Let X be a subspace of a reflexive rearrangement-invariant space E and μ(γ(a)) = 0. The following conditions are equivalent:

1. A|X is compact.

2. For every {x_{k}} ∈ S(X) such that x_{k}→^{w}0 and for every δ > 0 we have ∥χ_{δ}x_{k}∥ → 0 as k → +∞.

3. Every weakly zero sequence {x_{k}} ⊂ S(X) contains an almost disjoint subsequence such that for a corresponding disjoint sequence {y_{i}} condition (4.2) holds.

Proof. Since in a reflexive Banach space a compact operator maps weakly zero convergent sequences to sequences converging strongly to zero, 1 ⇒ 2 follows from Lemma 3.1. The proof of 2 ⇒ 3 follows from Lemma 4.5. 3 ⇒ 1: Let {x_{k}} ⊂ S(X) and x_{k}→^{w}0 as k → +∞. Choose an almost disjoint subsequence {x_{k_{i}}} of the sequence {x_{k}} such that for a correspondent disjoint sequence {y_{i}} (4.2) holds. Then ∥Ay_{i}∥ → 0. Therefore ∥Ax_{k_{i}}∥ → 0 as i → +∞. This means that A|X is compact

Let us formulate an alternative version of Proposition 5.1 adapted to the Hilbert space L_{2}(0, 1) and the multiplication by the independent variable t.

Proposition 5.2. Let E = L_{2}(0, 1) and a(t) = t. Suppose X is a subspace of L_{2}(0, 1). The following conditions are equivalent:

1. A|X is compact.

2. For every othonormal sequence {x_{n}} ⊂ X and for every δ ∈ (0, 1), ∫_{δ}^{1} |x_{k}(t)|^{2} dt → 0 as k → +∞.

3. Every orthonormal sequence {x_{k}} ⊂ X contains an almost disjoint subsequence {x_{k_{i}}} such that, for a corresponding disjoint sequence {y_{i}}, supp y_{i} ⊂ [δ_{i}, δ_{i−1}] where {δ_{i}} is a decreasing sequence of real numbers with δ_{i} → 0 as i → +∞.

Remark 5.3. Proposition 5.2 fails if its conditions are only fulfilled for some orthonormal sequences. Namely, it follows from the results of [2] that there exists an orthonormal basis {x_{k}} of L_{2}(0, 1) such that ∥tx_{k}(t)∥ → 0 as k → +∞. But the multiplication operator by the independent variable t is non-compact in L_{2}(0, 1).

It is well known that every strictly singular operator in a Hilbert space is compact (see [1]). We have noted (Proposition 2.1) that a strictly singular multiplication operator in a rearrangement-invariant space E is compact. What about restrictions of a multiplication operator A acting in a rearrangement-invariant space E? Is every strictly singular restriction compact? In general the answer is ”no”.

Example 5.4. Put E = L_{p}(−1, +1) (1 < p < 2). By {r_{n}} we denote the sequence of Rademacher functions defined on [0, 1] and extended by zero on [−1, 0). Let χ_{n} be the characteristic function of the interval (−\frac{1}{2^{n}} , −\frac{1}{2^{n+1}} ) and χ̃_{n} the corresponding normalized function. We consider the sequence {x_{n}} given by x_{n} = r_{n} + χ̃_{n}. It is easy to show that {x_{n}} in the space L_{p}(−1, +1) (1 < p < 2) is equivalent to the standard basis of l_{p}. Let X be the subspace of L_{p}(−1, +1) spanned by {x_{n}}. Now we consider the multiplication operator by the function a(t) = t acting in L_{p}(−1, +1) (1 < p < 2). It is also easy to see that the sequence {Ax_{n}} is equivalent to the standard basis of the space l_{2}. But it is well known that the natural imbedding of l_{p} into l_{2} is strictly singular and non-compact.

To close this section, we consider spectral properties of some compact operator connected with a multiplication operator in L_{2}(0, 1).

Let a be a continuous function such that a(t_{0}) = 0 for some t_{0} ∈ [0, 1]. Given a subspace X ⊂ L_{2}(0, 1), we denote by P_{X} the orthogonal projection onto X. Suppose that A|X is compact. We denote by B_{X} = P_{X}AP_{X} the compact selfadjoint operator acting in L_{2}(0, 1).

Proposition 5.5. Let {λ_{n}} be a sequence of real numbers such that

max_{t∈[0,1]} a(t) > λ_{1} > λ_{2} > ...  and  lim_{k→∞} λ_{n} = 0.

Then there exists a subspace X ⊂ L_{2}[0, 1] such that λ_{n} (n ∈ ℕ) are eigenvalues of the compact selfajoint operator B_{X}.

Proof. It is obvious that for the characteristic function χ_{σ} of σ ∈ Σ and for the orthogonal projection P_{σ} corresponding to the one-dimensional subspace generated by χ_{σ}

P_{σ}AP_{σ}χ_{σ} = \frac{⟨a(t), χ_{σ}(t)⟩}{∥χ_{σ}∥} = ∫_{σ} a(t) dt /\sqrt{∫_{σ} dt}

Using this observation, the continuity of the function a and an induction process it is easy to construct a sequence of disjoint sets σ_{n} such that, for the orthogonal projection P_{X} onto the subspace X = cl span {χ_{σ_{n}}}, P_{X}AP_{X}(χ_{σ_{n}}) = λ_{n}χ_{σ_{n}}. This means that B_{X}(χ_{σ_{n}}) = λ_{n}χ_{σ_{n}}

Restriction Properties of Multiplication Operators 35

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Received 29.04.1998