Spectral boundary of complemented invariant subspaces in L^{p}(ℝ)

by

Zoltán Buczolich^{1} and Alexander Olevskiĭ^{2}

ABSTRACT. In this note we construct a compact set K of zero Hausdorff dimension which satisfies certain “arithmetic type” thickness property. The concept of this “arithmetic thickness” came from applications to Harmonic Analysis, introduced in a paper of V. Lebedev and A. Olevskiĭ. For example, there are no spectral sets whose “essential boundary” can contain the above set K.

1 Research supported by the Hungarian National Foundation for Scientific Research Grant No. T 016094 and FKFP B-07/1997

2 Research partly supported by BSF. AMS(MOS) subject classifications(1991). Primary 42A45, 28A78, 47A15, 47D30; Key words and phrases: spectral multipliers, complemented subspaces, thin sets.

Buczolich and Olevskiĭ Page 2

Introduction

We say that a Borel set Λ ⊆ ℝ (more precisely the corresponding equivalence class) is a spectral set for L^{p}(ℝ), 1 < p < ∞, if its indicator function belongs to the multiplier algebra:

(1)  1 1_{Λ} ∈ M_{p}(ℝ).

If Λ satisfies (1) then it yields a translation invariant complemented subspace E_{Λ} in L^{p}(ℝ) :

E_{Λ} =^{def}Clos {f ∈ L^{p}(ℝ) ∩ L^{2}(ℝ) : f̂|_{^{c}Λ} = 0},

where f̂ denotes the Fourier transform of f.

Conversely, every invariant complemented subspace can be obtained in this way; in a natural sense Λ is called the spectrum of E_{Λ}.

For p = 2 every Λ is a spectral set. For p ≠ 2 the situation is much more difficult. Some classical examples are known, like Λ = ℝ^{+}, which is spectral for any p ∈]1, ∞[, or more general ones, coming from the Littlewood-Paley decomposition.

All said above is well-known; see [LO1], or [O] for the precise definitions and references.

It was recently dicovered in [LO1] that for p ≠ 2 spectral sets cannot have complicated structure.

For example a Cantor set of positive Lebesgue measure is never spectral.

Theorem A. [LO1] If Λ is spectral for some p ≠ 2

(2)  the essential boundary ∂Λ has Lebesgue measure zero,

or, equivalently

(3)  Λ and ^{c}Λ both are (equivalent to) open sets.

In the above theorem we used the term essential boundary in the sense of [LO1], that is, ∂Λ = \overline{Λ^{d}} ∩ \overline{(^{c}Λ)^{d}}, where Λ^{d} denotes the Lebesgue density points of Λ and by \overline{A} we denote the closure of A.

In the sequel we will also use the concept of the (common) essential boundary of two disjoint measurable sets Λ and Λ′, which is denoted by ∂(Λ, Λ′) and x ∈ ℝ belongs to it iff any U(x) neighborhood of x contains portions of positive measure of both sets.

This result means that every spectral set Λ can be viewed as a union of some family of disjoint open intervals accumulated to a closed set F of measure zero - its essential boundary.

To obtain a full characterization of such families effectievely - this problem up to date seems to be hopeless.

In this note we are interested in what can be said about the essential boundary of a spectral set.

Definition. For a given p ≠ 2 a compact set K ⊆ ℝ is called a spectral-subboundary, (K ∈ SB) iff there exists a spectral set Λ such that ∂Λ ⊃ K.

Obviously, this notion means some kind of “thinness”. It is natural to compare it to metrical “thinness”. Theorem A tells that K ∈ SB implies that mK = 0, where m denotes the Lebesgue measure.

Is the converse implication true?

We prove that the answer is no, even if one replaces the Lebesgue measure with the much more sensitive Hausdorff one.

We follow the general approach of [LO1-2] which involves an analysis of possible distributions of equidistant nets through Λ.

Definition. We say that a given pair of disjoint sets Λ, Λ′ has the universal coloring property,

(4)  (Λ, Λ′) ∈ UC

if for any N ∈ ℕ and for any coloring of the set {1, ..., N} by two colors (say, by red and blue) there exists x_{0}, h ∈ ℝ such that

(5)  x_{k} =^{def}x_{0} + kh ∈ Λ or Λ′ if, k is blue, or red, respectively.

This notion plays a crucial role in [LO1-2]. It appeared there in a slightly stronger form: x_{k} was required to be a density point of the corresponding set; below we will use the UC property for open sets and hence this stronger assumption will also be satisfied.

Actually, Theorem A was obtained by the following implications:

(6)  m(∂(Λ, Λ′)) > 0 ⇒ (Λ, Λ′) ∈ UC ⇒ Λ is not a spectral set.

Next we introduce the following property of “arithmetical thickness” (A-thickness): Definition. We say that a compact set K is A-thick, that is K ∈ A, iff for any two

disjoint open sets Λ, Λ′

(7)  K ⊆ ∂(Λ, Λ′) ⇒ (Λ, Λ′) ∈ UC.

Our main result says that A-thick compact sets may be metrically small. We remark that in [K] several different concepts of thin sets in Harmonic Analysis were considered.

Main Results

Theorem 1. There exists a compact perfect set K ⊆ ℝ such that:

(i) dim_{H}K = 0, and (ii) K ∈ A.

We remark that dim_{H} denotes the usual Hausdorff dimension, and by a suitable refinement of our proof one can obtain compact sets with H_{h}(K) = 0, where H_{h} is the Hausdorff measure defined by using a function h (h(0) = 0, h(x) > 0, h is monotone increasing and continuous from the right for all t ≥ 0). In the special case when h(x) = x^{s}, that is, we use the s-dimensional Hausdorff measure we will use the notation H^{s}.

Theorem 2. There exists a compact perfect set K, satisfying property (i) and (iii) K ∉ SB.

Finally our last result gives some additional information about A-thickness:

Theorem 3. If K ∈ A then it contains arbitrarily long arithmetical progressions.

Reduction of Theorem 2 to Theorem 1

It is enough to show that property (ii) implies property (iii). Let K ∈ A. Suppose that for some p ≠ 2, K ∈ SB, that is, there exists a spectral set Λ such that ∂Λ ⊃ K. Due to Theorem A we can assume that Λ is open and ^{c}Λ is equivalent to an open set Λ′. It is clear that Λ′ is disjoint from Λ. Since Λ′ and ^{c}Λ are equivalent (Λ′)^{d} = (^{c}Λ)^{d} and hence K ⊆ ∂Λ = ∂(Λ, Λ′). Then K ∈ A implies that the pair (Λ, Λ′) satisfies the UC-property. According to (6) Λ is not a spectral set.

Proof of Theorem 1

We define K as the intersection of the nested closed sets E_{k}. Each E_{k} consists of finitely, say n_{k}, many closed intervals I_{j,k}, j = 1, ..., n_{k}.

We assume that for a fixed k the intervals I_{j,k} are indexed from left to right and they are disjoint. Denote the length of I_{j,k} by ℓ_{j,k}. We choose the intervals I_{j,k} in a way that ℓ_{j,k} is non-increasing in j, that is, ℓ_{j,k} ≥ ℓ_{j′,k} for j ≤ j′ when k is fixed. We denote by ℓ_{k} the length of the shortest interval I_{j,k}, clearly, ℓ_{k} = ℓ_{n_{k},k}. In our construction I_{j′,k−1} ∩ E_{k} will consist of ν_{j′,k} many equally spaced intervals of the form I_{j,k} and all these I_{j,k}’s will be of the same length denoted by η_{j′,k}.

If j′ < j′′ holds then the distance of the consecutive intervals I_{j,k} belonging to I_{j′′,k−1} ∩ E_{k} will be much less than the length of the I_{j,k}’s belonging to I_{j′,k−1} ∩ E_{k}.

Next we give a formal inductive definition of the sets E_{k}.

Put E_{1} = I_{1,1} = [0, 1].

Assume that k ≥ 2, and E_{k−1} = ∪_{j=1}^{n_{k−1}}I_{j,k−1} is defined and
ℓ_{1,k−1} ≥ ℓ_{2,k−1} ≥ ... ≥ ℓ_{n_{k−1},k−1} = ℓ_{k−1}.

We also assume that ℓ_{j,k−1} is an integer multiple of ℓ_{k−1} for j = 1, ..., n_{k−1}. Set

λ_{1,k} = \frac{ℓ_{k−1}}{2k^{k}} and ν_{1,k} = \frac{ℓ_{1,k−1}}{λ_{1,k}} .

Since ℓ_{1,k−1} is an integer multiple of ℓ_{k−1} it is easy to see that ν_{1,k} is an integer. Divide I_{1,k−1} into ν_{1,k} many intervals of length λ_{1,k} and denote these intervals by J′_{1}, ..., J′_{ν_{1,k}}. Finally in each interval J′_{j} choose a small closed interval J_{j} which has the same midpoint as J′_{j} and the length of J_{j} equals

η_{1,k} = \frac{λ_{1,k}}{2^{k}ν_{1,k}^{k−1}} .

We will define E_{k} ∩ I_{1,k−1} such that it will equal the union of the intervals J_{j}, that is, J_{j} = I_{j,k} for j = 1, ..., ν_{1,k}. Observe that ℓ_{k−1} is an integer multiple of η_{1,k}.

Now assume that j′ ≥ 2, the I_{j,k}’s are defined in the intervals I_{1,k−1}, ..., I_{j′−1,k−1} and the ones in I_{j′−1,k−1} are of length η_{j′−1,k}. We also assume that ℓ_{k−1} is an integer multiple of η_{j′−1,k}. Since ℓ_{j′,k−1} is an integer multiple of ℓ_{k−1} it will also be an integer multiple of η_{j′−1,k}. Set

λ_{j′,k} = \frac{η_{j′−1,k}}{k^{k}} and ν_{j′,k} = \frac{ℓ_{j′,k−1}}{λ_{j′,k}} ,

observe that ν_{j′,k} is an integer. Now we subdivide I_{j′,k−1} like we divided I_{1,k−1}, that is, divide I_{j′,k−1} into ν_{j′,k} many intervals of length λ_{j′,k} and denote them by J′_{1}, ..., J′_{ν_{j′,k}}. In each interval J′_{j} choose a small closed interval J_{j} which has the same midpoint as J′_{j} and the length of J_{j} equals

η_{j′,k} = \frac{λ_{j′,k}}{2^{k}ν_{j′,k}^{k−1}} .

Observe that λ_{j′,k} and η_{j′−1,k} are both integer multiples of η_{j′,k}. Finally, let E_{k} ∩ I_{j′,k−1} equal the union of the intervals J_{j}.

Repeating the above steps for a fixed k we can define E_{k} ∩ I_{j′,k−1} for all j′ = 1, ..., n_{k−1}. Continuing this process define E_{k} for all k.

Put K = ∩_{k=1}^{∞}E_{k}.

It is easy to see that K is a nowhere dense non-empty compact perfect set and for any k and j ≤ n_{k} we have K ∩ I_{j,k} ≠ ∅. It is also clear that

\frac{ℓ_{k−1}}{2k^{k}} = λ_{1,k} > η_{1,k} > ... > η_{n_{k−1},k} = ℓ_{k}.

This implies that for any k ≥ 1 we have

∑_{j=k+1}^{∞} ℓ_{j} < ℓ_{k}.

Since for k = 2, 3, ... the choice of λ_{j′,k} implies that each interval I_{j′,k−1} contains more than four intervals belonging to E_{k} it is always possible to find an x ∈ K which is in the first quarter of I_{j′,k−1}.

Next we verify that K is of zero Hausdorff dimension.

We need to verify that given any s > 0 and ϵ, δ > 0 one can find a cover of K by sets U_{i} such that d(U_{i}) < δ and ∑ d(U_{i})^{s} < ϵ.

It is enough to verify that for each s_{m} = 1/m a suitable cover can be found. Hence assume that m is fixed.

The intervals belonging to E_{k} cover K for any integer k. Let

S_{k} = ∑_{j=1}^{n_{k}} d(I_{j,k})^{\frac{1}{m}} .

Claim 1. For any k ≥ m + 1 we have S_{k} < S_{k−1}/2.

We prove Claim 1 later. First we show that this claim implies dim_{H}K = 0. Since S_{m} is finite, from Claim 1 it follows that S_{k} → 0 as k → ∞. Using d(I_{j,k}) = ℓ_{j,k} < ℓ_{k−1} and ℓ_{k} → 0 as k → ∞ we obtain that for any given ϵ, δ > 0 we can find k such that d(I_{j,k}) < δ for j = 1, ..., n_{k} and S_{k} = ∑_{j} d(I_{j,k})^{1/m} < ϵ. This implies H^{1/m}(K) = 0. This holds for all m; hence K is of zero Hausdorff dimension.

Proof of Claim 1. Assume that j′ ≤ n_{k−1} and I_{j′,k−1} ∩ E_{k} consists of the intervals J_{j}, j = 1, ..., ν_{j′,k}. Then d(J_{j}) = η_{j′,k} for all j and

∑ d(J_{j})^{\frac{1}{m}} = ν_{j′,k} · ηj′,k^{\frac{1}{m}} = ν_{j′,k} (\frac{λ_{j′,k}}{2^{k}ν_{j′,k}^{k−1}} )^{\frac{1}{m}} = A

using k ≥ m + 1 we can continue the estimation by

A < ν_{j′,k} \frac{λj′,k^{\frac{1}{m}}}{2νj′,k^{\frac{m−1}{m}}} = \frac{1}{2}νj′,k^{\frac{1}{m}}λj′,k^{\frac{1}{m}} = \frac{1}{2}ℓj′,k−1^{\frac{1}{m}} = \frac{1}{2}d(I_{j′,k−1})^{\frac{1}{m}} .

Recalling that the intervals J_{j} are those intervals I_{j,k} which are in I_{j′,k−1} the above result implies that

∑_{\{j:I_{j,k}⊆I_{j′,k−1}\}} d(I_{j,k})^{\frac{1}{m}} < \frac{1}{2}d(I_{j′,k−1})^{\frac{1}{m}} .

Summing this for all I_{j′,k−1} we obtain S_{k} ≤ S_{k−1}/2. This proves Claim 1.

We also need to verify that K ∈ A.

Denote by Ψ the union of the intervals contiguous to K and by s_{n} the set {1, ..., n} where n is a positive integer. Assume that Λ and Λ′ are disjoint open sets such that K ⊆ ∂(Λ, Λ′). Set G = Λ ∪ Λ′. Then G ⊆ Ψ is open and we can introduce the coloring function ϕ_{1} : G → {blue, red} by letting ϕ_{1} = red on Λ and ϕ_{2} = blue on Λ′. Since K ⊆ ∂(Λ, Λ′) the coloring ϕ_{1} of G is dense, by this we mean that both red and blue intervals are “dense” in K, that is, F ⊆ \overline{\{x : ϕ_{1}(x) = red\}} and K ⊆ \overline{\{x : ϕ_{2}(x) = blue\}}. To verify that K ∈ A we need to show that for any of the above given open sets Λ, and Λ′ the universal coloring property is satisfied, that is, we need to find x ∈ ℝ and h^{∗} ∈ ℝ such that x + ih^{∗} ∈ G for i = 1, ..., n and the two colorings ϕ_{1} and ϕ_{2} are compatible, that is, ϕ_{1}(x + ih^{∗}) = ϕ_{2}(i) for i = 1, ..., n.

Choose k_{0} > 2n^{2} and a j_{0} ≤ n_{k_{0}−1}. Then it is easy to see that ν_{j_{0},k} > k_{0}^{k_{0}} > n + 1 and hence I_{j_{0},k_{0}−1} ∩ E_{k_{0}} consists of more than n + 1 intervals of the form I_{j,k_{0}}. Choose a j_{1} such that

I_{j_{1},k_{0}}, I_{j_{1}+1,k_{0}}, ..., I_{j_{1}+n,k_{0}} ⊆ I_{j_{0},k_{0}−1}.

Let h_{k_{0}} = λ_{j_{0},k_{0}}. Recall that the intervals I_{j_{1}+i,k_{0}} are equally spaced in I_{j_{0},k_{0}−1} and I_{j_{1}+i,k_{0}} can be obtained from I_{j_{1},k_{0}} by a translation with ih_{k_{0}}. It is also clear that ℓ_{j_{1},k_{0}} = ℓ_{j_{1}+1,k_{0}} = ... = ℓ_{j_{1}+n,k_{0}} = η_{j_{0},k_{0}}.

Since the coloring of G is dense and the first quarter of I_{j_{1},k_{0}} contains points of K we can choose an x_{n} ∈ G such that ϕ_{1}(x_{n}) = ϕ_{2}(n) and [x_{n}, x_{n} + \frac{ℓ_{j_{1},k_{0}}}{2} ] ⊆ I_{j_{1},k_{0}}. Using ℓ_{k_{0}} ≤ ℓ_{j_{1},k_{0}} and the translation property of the intervals I_{j_{1}+i,k_{0}} we have

[x_{n} + ih_{k_{0}}, x_{n} + ih_{k_{0}} + \frac{ℓ_{k_{0}}}{2} ] ⊆ I_{j_{1}+i,k_{0}} ⊆ E_{k_{0}} for i = 0, ..., n.

Set ℓ′_{k} = ℓ_{k}/2 for k = 1, 2, ....

Now the following claim will imply that K ∈ A.

Claim 2. Assume that n, k_{0}, x_{n}, and h_{k_{0}} are given such that k_{0} > 2n^{2}, x_{n} ∈ G, the points x_{n} + ih_{k_{0}}, belong to different components of E_{k_{0}}, ϕ_{1}(x_{n}) = ϕ_{2}(n), and

[x_{n} + ih_{k_{0}}, x_{n} + ih_{k_{0}} + ℓ′_{k_{0}}] ⊆ E_{k_{0}} for i = 1, ..., n

then there exists x ∈ K and h ∈ ℝ such that

|h − h_{k_{0}}| < nℓ′_{k_{0}}
|x − (x_{n} + nh_{k_{0}})| < nℓ′_{k_{0}}

and

ϕ_{1}(x − ih) = ϕ_{2}(i) for i = 1, ..., n.

Proof of Claim 2 implies K ∈ A. The assumptions of the claim are satisfied; hence using x ∈ K, and h from the Claim let h^{∗} = −h and observe that ϕ_{1}(x + ih^{∗}) = ϕ_{2}(i), i = 1, ..., n shows that the coloring ϕ_{1} of G and ϕ_{2} are compatible.

We need the extra assumptions in the conclusion of Claim 2 to obtain a proof which is based on induction.

To prove Claim 2 we also need a subclaim:

Subclaim. Assume that k > 2n^{2}; x_{n}, and h_{k} are given such that x_{n} ∈ G, the points x_{n} + ih_{k} for i = 1, ..., n belong to different components of E_{k} and

[x_{n} + ih_{k}, x_{n} + ih_{k} + ℓ′_{k}] ⊆ E_{k} for i = 1, ..., n.

Then there exists h_{k+1} such that

h_{k} ≤ h_{k+1} ≤ h_{k} + n\frac{ℓ′_{k}}{k} ,

and the points

x_{n} + ih_{k+1} ∈ [x_{n} + ih_{k}, x_{n} + ih_{k} + ℓ′_{k}], i = 1, ..., n

belong to different components of E_{k+1}; furthermore if \overline{ℓ}_{n,k+1} denotes the length of the interval I_{j,k+1} ⊆ E_{k+1} which contains the point x_{n} + nh_{k+1} then

[x_{n} + ih_{k+1}, x_{n} + ih_{k+1} + \overline{ℓ}_{n,k+1}] ⊆ E_{k+1} for i = 1, ..., n.

Remark. Observe that \overline{ℓ}_{n,k+1} ≥ ℓ_{k+1} = 2ℓ′_{k+1} and hence the conclusion of the Subclaim implies that

[x_{n} + ih_{k+1}, x_{n} + ih_{k+1} + ℓ′_{k+1}] ⊆ E_{k+1}

also holds for i = 1, ..., n. This implies that we can repeat the application of the subclaim.

First we prove Claim 2 by using the Subclaim: We do induction on n.

Assume that k_{0} ≥ 2 is given. When n = 1 by our assumption [x_{n} + h_{k_{0}}, x_{n} + h_{k_{0}} + ℓ′_{k_{0}}] ⊆ E_{k_{0}}. Choose j_{1} such that x_{n} + h_{k_{0}} ∈ I_{j_{1},k_{0}}. Then there exists

λ < \frac{ℓ_{k_{0}}}{2k_{0}^{k_{0}}} = \frac{ℓ′_{k_{0}}}{k_{0}^{k_{0}}} < \frac{ℓ′_{k_{0}}}{2}

such that any subinterval of I_{j_{1},k_{0}} with length λ contains points of E_{k_{0}+1}, and hence points of K as well. Thus we can choose h ≥ h_{k_{0}} such that |h − h_{k_{0}}| < ℓ′_{k_{0}} and x_{n} + h = x ∈ K. Then |x − (x_{n} + h_{k_{0}})| = |h − h_{k_{0}}| < ℓ′_{k_{0}} also holds.

Assume that Claim 2 is valid for n − 1 ≥ 1 with all k_{0} > 2(n − 1)^{2}.

For the next step of the induction assume that k_{0} > 2n^{2} is given and the assumptions of Claim 2 are satisfied for n, k_{0}, x_{n} and h_{k_{0}}.

Since G is open and x_{n} ∈ G there exists ϵ > 0 such that ]x_{n} − ϵ, x_{n} + ϵ[⊆ G. Choose k_{1} > k_{0} such that ϵ/2n^{2} > ℓ′_{k_{1}}.

Keeping x_{n} fixed we can apply the Subclaim for k = k_{0}, k_{0} + 1, ..., k_{1} − 1 and we can choose h_{k_{1}} such that

h_{k_{0}} ≤ h_{k_{1}} ≤ h_{k_{0}} + n (\frac{ℓ′_{k_{0}}}{k_{0}} + \frac{ℓ′_{k_{0}+1}}{k_{0} + 1} + ... + \frac{ℓ′_{k_{1}−1}}{k_{1} − 1}) < h_{k_{0}} + 2n\frac{ℓ′_{k_{0}}}{k_{0}} ,

the points x_{n} + ih_{k_{1}} belong to different components of E_{k_{1}}, and

[x_{n} + ih_{k_{1}}, x_{n} + ih_{k_{1}} + \overline{ℓ}_{n,k_{1}}] ⊆ E_{k_{1}} for i = 1, ..., n.

Using that \overline{ℓ}_{n,k_{1}} ≥ 2ℓ′_{k_{1}} we also have

[x_{n} + ih_{k_{1}}, x_{n} + ih_{k_{1}} + 2ℓ′_{k_{1}}] ⊆ E_{k_{1}} for i = 1, ..., n.

Choose j_{1} such that x_{n} + h_{k_{1}} ∈ I_{j_{1},k_{1}}. Then there exists λ < ℓ′_{k_{1}}/k_{1}^{k_{1}} ≤ ℓ′_{k_{1}}/4 such that any subinterval of length λ in I_{j_{1},k_{1}} contains points of E_{k_{1}+1} and points of K as well. Since we have a dense coloring of G we can choose an x_{n−1} ∈ G such that ϕ_{1}(x_{n−1}) = ϕ_{2}(n − 1) and x_{n−1} ∈ [x_{n} + h_{k_{1}}, x_{n} + h_{k_{1}} + λ]. Then an easy computation shows that

[x_{n−1} + ih_{k_{1}}, x_{n−1} + ih_{k_{1}} + ℓ′_{k_{1}}] ⊆ [x_{n} + (i + 1)h_{k_{1}}, x_{n} + (i + 1)h_{k_{1}} + 2ℓ′_{k_{1}}] ⊆ E_{k_{1}}

holds for i = 1, ..., n − 1 and by Claim 2 applied for n − 1, k_{1}, x_{n−1}, and h_{k_{1}} there exists x ∈ K and h ∈ ℝ such that

|h − h_{k_{1}}| < (n − 1)ℓ′_{k_{1}}
|x − (x_{n−1} + (n − 1)h_{k_{1}})| < (n − 1)ℓ′_{k_{1}}

and

ϕ_{1}(x − ih) = ϕ_{2}(i) for i = 1, ..., n − 1.

Then

|h − h_{k_{0}}| ≤ |h − h_{k_{1}}| + |h_{k_{1}} − h_{k_{0}}| < (n − 1)ℓ′_{k_{1}} + 2n\frac{ℓ′_{k_{0}}}{k_{0}} < nℓ′_{k_{0}},

furthermore

|x − (x_{n} + nh_{k_{0}})| ≤ |x − (x_{n−1} + (n − 1)h_{k_{1}})| + |x_{n−1} + (n − 1)h_{k_{1}} − (x_{n} + nh_{k_{0}})| ≤

(n − 1)ℓ′_{k_{1}} + |x_{n−1} − (x_{n} + h_{k_{0}})| + (n − 1)|h_{k_{1}} − h_{k_{0}}| <

< (n − 1)ℓ′_{k_{1}} + 2n\frac{ℓ′_{k_{0}}}{k_{0}} + λ + (n − 1)2n\frac{ℓ′_{k_{0}}}{k_{0}} < nℓ′_{k_{0}},

where in the last estimation we used that k_{0} > 2n^{2}, ℓ′_{k_{1}} < ℓ′_{k_{0}}/k_{0}^{k_{0}} and λ < ℓ′_{k_{1}}/4. Finally, observe that

|x − nh − x_{n}| ≤ |x − (x_{n−1} + (n − 1)h_{k_{1}})| + |x_{n−1} − x_{n} − h_{k_{1}}| + n|h_{k_{1}} − h| <

(n − 1)ℓ′_{k_{1}} + λ + n(n − 1)ℓ′_{k_{1}} < ϵ

Thus x − nh ∈ (x_{n} − ϵ, x_{n} + ϵ) and hence ϕ_{1}(x − nh) = ϕ_{1}(x_{n}) = ϕ_{2}(n) this, together with the induction hypothesis implies

ϕ_{1}(x − ih) = ϕ_{2}(i) for i = 1, ..., n.

This completes the proof of Claim 2.

Finally we verify the Subclaim. We do again induction on n.

For n = 1 we have [x_{n} + h_{k}, x_{n} + h_{k} + ℓ′_{k}] ⊆ E_{k}. Choose j_{1} such that x_{n} + h_{k} ∈ I_{j_{1},k}. Then there exists λ_{1} ≤ ℓ′_{k}/(k + 1)^{(k+1)} such that during the construction of E_{k+1} the interval I_{j_{1},k} is being split into intervals of length λ_{1} and each such subinterval contains a component of E_{k+1}. Thus we can choose h_{k+1} such that

h_{k} ≤ h_{k+1} ≤ h_{k} + λ_{1} < h_{k} + \frac{ℓ′_{k}}{k}

and x_{n} + h_{k+1} is the left endpoint of a component of E_{k+1}. Then

[x_{n} + h_{k+1}, x_{n} + h_{k+1} + \overline{ℓ}_{n,k+1}] ⊆ E_{k+1},

where \overline{ℓ}_{n,k+1} is the length of the interval I_{j,k+1} which contains x_{n} + h_{k+1}. This completes the proof when n = 1.

Now assume that n ≥ 2, k > 2n^{2}, and the Subclaim is true for n − 1. Observe that using x_{n} ∈ G and the points x_{n} + ih, i = 1, ..., n − 1 the assumptions of the Subclaim (used for n − 1) are satisfied; hence by the induction hypothesis there exists h′_{k+1} such that

h_{k} ≤ h′_{k+1} ≤ h_{k} + (n − 1)\frac{ℓ′_{k}}{k} ,

and

x_{n} + ih′_{k+1} ∈ [x_{n} + ih_{k}, x_{n} + ih_{k} + ℓ′_{k}] for i = 1, ..., n − 1

and these points belong to different components of E_{k+1}; furthermore for i = 1, ..., n − 1 we have

[x_{n} + ih′_{k+1}, x_{n} + ih′_{k+1} + \overline{ℓ}_{n−1,k+1}] ⊆ E_{k+1},

where \overline{ℓ}_{n−1,k+1} is the length of the interval I_{j,k+1} which contains x_{n} + (n − 1)h′_{k+1}. Now, by our assumptions,

x_{n} + nh_{k} ≤ x_{n} + nh′_{k+1} ≤ x_{n} + nh_{k} + n_{n−1} \frac{ℓ′_{k}}{k} < x_{n} + nh_{k} + \frac{ℓ′_{k}}{2} .

Clearly, \overline{ℓ}_{n−1,k+1} < ℓ′_{k}/k^{k}. Hence

(8)  [x_{n} + nh′_{k+1}, x_{n} + nh′_{k+1} + \frac{ℓ′_{k}}{2} ] ⊆ [x_{n} + nh_{k}, x_{n} + nh_{k} + ℓ′_{k}] ⊆ E_{k}.

Since

x_{n} + (n − 1)h′_{k+1} < x_{n} + nh′_{k+1}

belong to different components of E_{k} there are j_{n−1} < j_{n} such that

x_{n} + (n − 1)h′_{k+1} ∈ I_{j_{n−1},k} and x_{n} + nh′_{k+1} ∈ I_{j_{n},k};

hence by (8)

(9)  [x_{n} + nh′_{k+1}, x_{n} + nh′_{k+1} + \frac{ℓ′_{k}}{2} ] ⊆ I_{j_{n},k} ⊆ E_{k}.

The definition of E_{k+1} implies that there exists λ_{n} and η_{j_{n},k+1} = \overline{ℓ}_{n,k+1} such that I_{j_{n},k} is divided into subintervals of length λ_{n}, each such subinterval contains a component of E_{k+1}, and this component is of length \overline{ℓ}_{n,k+1}. From j_{n−1} < j_{n} and from our construction it follows that λ_{n} < \overline{ℓ}_{n−1,k+1}/(k + 1)^{(k+1)}; hence we can choose h_{k+1} such that

h′_{k+1} ≤ h_{k+1} ≤ h′_{k+1} + λ_{n} < h′_{k+1} + \frac{\overline{ℓ}_{n−1,k+1}}{(k + 1)^{(k+1)}} < h′_{k+1} + \frac{ℓ′_{k}}{k} ≤ h_{k} + n\frac{ℓ′_{k}}{k}

and, using (9), we can also assume that x_{n} + nh_{k+1} is the left endpoint of a component of E_{k+1} ∩ I_{j_{n},k}. Then

[x_{n} + nh_{k+1}, x_{n} + nh_{k+1} + \overline{ℓ}_{n,k+1}] ⊆ E_{k+1}.

¿From

h′_{k+1} ≤ h_{k+1} ≤ h′_{k+1} + \frac{\overline{ℓ}_{n−1,k+1}}{(k + 1)^{(k+1)}} ,

and

\overline{ℓ}_{n,k+1} < λ_{n} < \frac{\overline{ℓ}_{n−1,k+1}}{(k + 1)^{(k+1)}}

it follows that

x_{n} + ih_{k+1} + \overline{ℓ}_{n,k+1} < x_{n} + ih′_{k+1} + \frac{(i + 1)\overline{ℓ}_{n−1,k+1}}{(k + 1)^{(k+1)}} < x_{n} + ih′_{k+1} + \overline{ℓ}_{n−1,k+1}

for i = 1, ..., n − 1. Thus x_{n} + ih′_{k+1} ≤ x_{n} + ih_{k+1} and

[x_{n} + ih′_{k+1}, x_{n} + ih′_{k+1} + \overline{ℓ}_{n−1,k+1}] ⊆ E_{k+1}

implies

[x_{n} + ih_{k+1}, x_{n} + ih_{k+1} + \overline{ℓ}_{n,k+1}] ⊆ E_{k+1}

for i = 1, ..., n − 1. Observe that above we have already established the last property for i = n as well. This completes the proof of the Subclaim and hence the proof of Theorem 1 is also complete.

Proof of Theorem 3

Given ℓ ∈ ℕ we denote by ϕ_{2,ℓ} the coloring of s_{4ℓ+2} = {1, ..., 4ℓ + 2} for which ϕ_{2,ℓ}(2i − 1) = red, ϕ_{2,ℓ}(2i) = blue for i ∈ {1, ..., 2ℓ + 1}.

We can assume that K is non-empty, compact and nowhere dense.

First we also assume that K is perfect and does not contain ℓ ≥ 3 long arithmetic progressions. We need to show that K ∉ A.

We denote by CK the set consisting of the intervals contiguous to K. If x ∈ Ψ = ℝ \ K then I_{K}(x) denotes the CK interval containing x.

Since the number ℓ is fixed, for ease of notation, we will just write ϕ_{2} instead of ϕ_{2,ℓ}. Our goal is to find open sets Λ and Λ′ such that K ⊆ ∂(Λ, Λ′), but (Λ, Λ′) ∉ UC. We will choose Λ and Λ′ by defining a coloring ϕ_{1} : Ψ → {blue, red}. This coloring will be constant on each CK interval. We will let Λ = {x ∈ Ψ : ϕ_{1}(x) = red}, and Λ′ = {x ∈ Ψ : ϕ_{1}(x) = blue}. It is clear that if we assume that ϕ_{1} is a dense coloring, that is, K ⊆ \overline{Λ} ∩ \overline{Λ′} then K ⊆ ∂(Λ, Λ′).

Since there are countably many intervals in CK we can choose an onto mapping ψ_{K} : Ψ → ℕ such that ψ_{K}(x) = ψ_{K}(y) iff I_{K}(x) = I_{K}(y).

We say that a pair (x; h) ∈ ℝ × (ℝ \ {0}) is admissible if the terms of the sequence {x+ih : i ∈ s_{4ℓ+2}} belong to Ψ and I_{K}(x+2ih) ≠ I_{K}(x+2jh) for i ≠ j, i, j ∈ {1, ..., 2ℓ+1}. The set of admissible pairs is denoted by P.

Observe that if ℓ and ϕ_{2} are fixed as above, ϕ_{1} is a coloring of Ψ, and ϕ_{1}(x + ih) = ϕ_{2}(i) for all i ∈ s_{4ℓ+2} then (x; h) ∈ P.

Now we turn to the definition of a suitable coloring ϕ_{1}.

Given η > 0 we denote by P_{η} the set of those (x; h) ∈ P for which |h| ≥ η. Clearly, P_{η_{0}} = ∅ for a sufficiently large η_{0}.

For (x; h) ∈ P we call the rank of (x; h) the number

R(x; h) = min{ψ_{K}(x + 2ih) : i = 1, ..., 2ℓ + 1}.

For any η > 0 we can consider the set R_{η} = {R(x; h) : (x; h) ∈ P_{η}}.

Claim. For any η > 0 the set R_{η} is bounded from above.

Proof of the Claim. If R_{η} is not bounded from above then using the compactness of K we can choose a sequence (x_{n}; h_{n}) ∈ P_{η} such that η_{0} > |h_{n}| ≥ η > 0, x_{n} → x ∈ ℝ, h_{n} → h ≠ 0 and ψ_{K}(x_{n} + 2ih_{n}) → ∞ for any i ∈ {1, ..., 2ℓ + 1}. Thus if i is fixed and ]a_{n}, b_{n}[= I_{K}(x_{n} + 2ih_{n}) then b_{n} − a_{n} → 0. Therefore lim_{n→∞} a_{n} = lim_{n→∞} x_{n} + 2ih_{n} = x + 2ih. Since a_{n} ∈ K we have x + 2ih ∈ K. This is true for all i ∈ {1, ..., 2ℓ + 1}, that is, K contains an arithmetic progression of length longer than ℓ, contradicting our assumption. This proves the Claim.

Now we define by induction the coloring ϕ_{1} on Ψ. In each step first we color by red finitely many CK intervals in order to “take care” of “blocking” sequences {x + ih} with (x; h) ∈ P_{η_{k}} with a suitable η_{k} > 0. Then to obtain a dense coloring we color some CK intervals by blue.

Step 1 of the coloring. Choose η_{1} > 0 such that

η_{1} < \frac{diameter (K)}{4(4ℓ + 2)}.

Take an M_{1} ∈ ℝ such that R_{η_{1}} < M_{1}. There are only finitely many intervals on which ψ_{K}(x) ≤ M_{1}. Denote the union of these intervals by U_{1}. Without limiting generality we can assume that U_{1} contains the two unbounded components of Ψ. Then ψ_{K}(x) > M_{1} for x ∈ ℝ \ (K ∪ U_{1}).

For x ∈ U_{1} let ϕ_{1}(x) = red.

Whenever (x; h) ∈ P_{η_{1}} we have R(x; h) ≤ M_{1}; and hence there exists an i ∈ {1, ..., 2ℓ + 1} such that x + 2ih ∈ U_{1}. Thus ϕ_{1}(x + 2ih) ≠ ϕ_{2}(2i) and hence the coloring ϕ_{2} of s_{4ℓ+2} is not compatible with the coloring of ϕ_{1}(x + 2ih).

Next, choose finitely many CK intervals which are disjoint from U_{1} such that, denoting by V_{1} their union, we have the following two properties satisfied:

α_{1}) If I_{1}, I_{2} ⊆ V_{1} are disjoint CK intervals then dist(I_{1}, I_{2}) > 2(4ℓ + 2)η_{1}; β_{1}) If x ∈ K then (x − 3(4ℓ + 2)η_{1}, x + 3(4ℓ + 2)η_{1}) ∩ V_{1} ≠ ∅.

It is easy to see that the choice of V_{1} can be made satisfying α_{1}) and β_{1}). For x ∈ V_{1} let ϕ_{1}(x) = blue.

Clearly α_{1}) says that blue intervals are not too close and β_{1}) says that they are not too far either.

The general steps of the coloring.

Assume now that we have already accomplished Step k of our definition. Thus we have the number η_{k} > 0 and the sets U_{k}, V_{k} such that U_{k} ∪ V_{k} ⊆ Ψ, ϕ_{1}(U_{k}) = {red}, ϕ_{1}(V_{k}) = {blue}, and the sets U_{k}, V_{k} consist of the union of finitely many CK intervals. If (x; h) ∈ P_{η_{k}} then there exists i ∈ {1, ..., 2ℓ + 1} such that x + 2ih ∈ U_{k}. Finally the following two properties are satisfied.

α_{k}) If I_{1}, I_{2} ⊆ V_{k} are disjoint CK intervals then dist(I_{1}, I_{2}) > 2(4ℓ + 2)η_{k}; β_{k}) If x ∈ F then (x − 3(4ℓ + 2)η_{k}, x + 3(4ℓ + 2)η_{k}) ∩ V_{k} ≠ ∅.

Step (k + 1) of the coloring.

Let η_{k+1} = η_{k}/2.

If h ∈ [η_{k+1}, η_{k}) and (x; h) ∈ P, that is, (x; h) ∈ P_{η_{k+1}} \ P_{η_{k}} then we define the k-rank of (x; h), denoted by R_{k}(x; h), by letting

R_{k}(x; h) = min{ψ_{K}(x + 2ih) : i = 1, ..., 2ℓ + 1, and x + 2ih ∉ V_{k}}.

Observe that assumption α_{k}) implies that for each (x; h) ∈ P_{η_{k+1}} \ P_{η_{k}} there is at most one i ∈ {1, ..., 2ℓ + 1} such that x + 2ih ∈ V_{k} and hence R_{k}(x; h) is well defined and finite for each (x; h) ∈ P_{η_{k+1}} \ P_{η_{k}}.

Assume that the set R_{η_{k+1}} = {R_{k}(x; h) : (x; h) ∈ P_{η_{k+1}} \ P_{η_{k}}} is not bounded from above. Again, like in an earlier similar argument, we can choose a sequence (x_{n}; h_{n}) ∈ P_{η_{k+1}} \ P_{η_{k}} such that x_{n} converges to an x ∈ ℝ, h_{n} → h with η_{k+1} ≥ |h| ≥ η_{k}, ψ_{K}(x_{n} + 2ih_{n}) → ∞ and x_{n} + 2ih_{n} ∉ V_{k} for any i ∈ {1, ..., 2ℓ + 1} \ {i^{∗}} where i^{∗} is a possible exception. Assume i ≠ i^{∗} and put ]a_{n}, b_{n}[ = I_{K}(x_{n} + 2ih_{n}). Again, lim_{n→∞} b_{n} − a_{n} = 0. Therefore, {x+2ih : i = 1, ..., 2ℓ+1, i ≠ i^{∗}} ⊆ K, but then K would contain an arithmetic progression of length at least ℓ, contradicting our assumption.

Thus R_{η_{k+1}} < M_{k+1} for a suitable constant M_{k+1}. We can also assume that M_{k+1} is so large that M_{k+1} > k + 1 and, whenever x ∈ K, then there exists an interval ]a, b[ ∈ CK such that ]a, b[ ∩]x − η_{k}, x + η_{k}[≠ ∅, ]a, b[ ∩V_{k} = ∅, and ψ_{K}( ]a, b[ ) is bounded by M_{k+1}.

Denote by U_{k+1}^{o} the set of points x for which x ∈ Ψ, ψ_{K}(x) ≤ M_{k+1}, x ∉ U_{k} ∪ V_{k} (that is we have not defined ϕ_{1} at x).

For x ∈ U_{k+1}^{o} put ϕ_{1}(x) = red, and let U_{k+1} = U_{k} ∪ U_{k+1}^{o}. Observe that U_{k+1} ∪ V_{k} contains all points x in Ψ for which ψ_{K}(x) ≤ M_{k+1}.

Now take (x; h) ∈ P_{η_{k+1}} \ P_{η_{k}}. Then R_{k}(x; h) ≤ M_{k+1} and hence there exists i ∈ {1, ..., 2ℓ+1} such that x+2ih ∉ V_{k} and ψ_{K}(x+2ih) ≤ M_{k+1}. This implies x+2ih ∈ U_{k+1} and hence ϕ_{1}(x + 2ih) = red ≠ ϕ_{2}(2i), that is, the coloring ϕ_{1} of x + 2ih is incompatible with the coloring ϕ_{2} of s_{4ℓ+2}.

Next, choose finitely many CK intervals which are disjoint from U_{k+1} such that denoting by V_{k+1} their union we have V_{k} ⊆ V_{k+1} and propositions α_{k+1}), and β_{k+1}) are satisfied (that is, we have propositions α_{k}) and β_{k}) satisfied with k replaced by k + 1). It is an easy exercise to show that we can make a suitable choice of V_{k+1}. For x ∈ V_{k+1} put ϕ_{1}(x) = blue.

By letting k → ∞ we can define a coloring ϕ_{1} of Ψ. If one takes x ∈ K, δ > 0 then it is easy to find k such that U_{k}∩]x − δ, x + δ[ ≠ ∅ and V_{k}∩]x − δ, x + δ[ ≠ ∅ and hence ϕ_{1} is dense.

Since P = P_{η_{0}} ⋃ ∪_{k=1}^{∞}(P_{η_{k+1}} \ P_{η_{k}}) our construction implies that for any (x; h) ∈ P there exists i ∈ {1, ..., 2ℓ + 1} such that

(10)  ϕ_{1}(x + 2ih) = red ≠ ϕ_{2}(2i).

However, our observation after the definition of P implies that for (x; h) ∉ P there always exists i ∈ {1, ..., 2ℓ + 1} such that (10) holds. This implies that the open sets Λ, and Λ′ defined by ϕ_{1} do not have the universal coloring property; hence K ∉ A. This concludes our proof for perfect K’s.

Denote by N_{ℓ} the set of those non-empty compact sets which does not contain ℓ long arithmetic progressions. If K is not perfect then we can use the following

Lemma. If the compact set K ∈ N_{ℓ} then there exists a perfect compact set K′ ∈ N_{ℓ} such that K ⊆ K′.

First we show that using this lemma we can complete the proof of Theorem 3. Indeed, if K ∈ N_{ℓ} then choose K′ according to the Lemma. By using our previous argument we can find open sets Λ and Λ′ such that K ⊆ K′ ⊆ ∂(Λ, Λ′) and (Λ, Λ′) ∉ UC. This implies that K ∉ A.

Proof of the Lemma. For each isolated point y_{n} of K we will choose a suitable neighborhood I_{n} =] y_{n} − r_{n}, y_{n} + r_{n}[ and denote by G the union of these neighborhoods. By choosing the r_{n}’s sufficiently small we can assume that I_{n} ∩ K = {y_{n}}. Since K ∈ N_{ℓ}, using induction on n, and at each step a compactness argument, we can assume that the radii, r_{n}, are so small that K ∪ G does not contain an ℓ long arithmetic progression such that each component of G contains at most one of its terms.

Choose a compact perfect set K_{0} for which 0 ∈ K_{0} ⊆ [0, 1] and K_{0} does not contain a 3 long arithmetic progression. Set

K′ = K ∪ ⋃_{n} (y_{n} + \frac{r_{n}}{3} K_{0}) .

It is not difficult to see that K′ satisfies the conditions of the Lemma, we leave the details to the reader.

Open Problems

1. Suppose that K is an independent compact set (see [K]). Is it true that K ∈ SB? Is it true that there exists a spectral set Λ such that ∂Λ = K?

2. Does the classical Cantor ternary set belong to SB?

3. Are the classes SB different for different p’s? (This question is motivated by 6.2.5 of [S].)

References

[K] J.-P. Kahane, Séries de Fourier absolument convergentes, Springer Verlag, NY, (1970).

[LO1] V. Lebedev and A. Olevskiĭ, Idempotents of Fourier Multiplier Algebra, Geometric and Functional Analysis, 4, No. 5 (1994)

[LO2] V. Lebedev and A. Olevskiĭ, Bounded groups of translation invariant operators, C. R. Acad. Sci. Paris, t. 322, Série I, p. 143-147, (1996).

[O] A. Olevskiĭ, Six Lectures on Translation-Invariant Operators and Subspaces, Workshop on Measure Theory and Real Analysis, Grado, Lecture Notes, (1997) (preprint).

[S] E.M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton University Press, Princeton NJ, (1970).

Department of Analysis, Eötvös Loránd University, Budapest, Rákóczi út 5, H-1088, Hungary email: buczo@ludens.elte.hu

Max-Planck-Institut für Mathematik, Bonn and School of Mathematical Sciences Tel Aviv University Ramat Aviv 69978 Israel email: olevskii@math.tau.ac.il