Q SPACES OF SEVERAL REAL VARIABLES

MATTS ESSÉN, SVANTE JANSON, LIZHONG PENG, AND JIE XIAO

Abstract. For α ∈ (−∞, ∞), let Q_{α}(ℝ^{n}) be the space of all measurable functions with

sup[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy < ∞,

where the supremum is taken over all cubes I with the edge length ℓ(I) and the edges parellel to the coordinate axes in ℝ^{n}. If α ∈ (−∞, 0), then Q_{α}(ℝ^{n}) = BMO(ℝ^{n}), and if α ∈ [1, ∞), then Q_{α}(ℝ^{n}) = {constants}. In the present paper, we discuss the case α ∈ [0, 1). These spaces are new subspaces of BMO(ℝ^{n}) containing some special Besov spaces. We characterize functions in Q_{α}(ℝ^{n}) by means of the Poisson extension, p-Carleson measures, mean oscillation and wavelet coefficients, and give a dyadic counterpart. Finally, we pose some open problems.

Contents

1. Introduction 1 2. Basic properties 3 3. Poisson extension 10 4. Generalized Carleson measures 16 5. Mean oscillation 19 6. Wavelets 25 7. The dyadic counterpart 30 8. Some problems 34 References 35

1. Introduction

For each p ∈ (0, 1), the space Q_{p} was introduced in [3] as the Banach space of all analytic functions f in the unit disk △ satisfying

sup_{w∈△} ∫∫ _{△} |f′(z)|^{2}[g(z, w)]^{p}dm(z) < ∞,  (1.1)

Date: November 3, 1999. 1991 Mathematics Subject Classification. 46E15. Key words and phrases. Q_{α}(ℝ^{n}), Poisson extension, Carleson measure, Local analysis,

Wavelets, Dyadic counterpart.

This research was carried out when J. Xiao visited the Department of Mathematics, Uppsala University during 1998. He was supported by a grant from the department and from the Swedish Institute, as well as from the NNSF and SEC of China.

where g(z, w) = log |(1 − \overline{w}z)/(w − z)| is the Green function of △ and m is the Lebesgue measure. As in [3], Q_{p} is a proper subspace of BMOA (obtained by taking p = 1 in (1.1)) and Q_{p_{1}} ⊊ Q_{p_{2}} if 0 < p_{1} < p_{2} < 1.

Essén and Xiao [5] showed that an analytic function f in the Hardy space H^{2} on the unit disk belongs to Q_{p} if and only if its boundary values on the unit circle ∂△ satisfy

sup_{I} |I|^{−p} ∫_{I} ∫_{I} \frac{|f(e^{iθ}) − f(e^{iφ})|^{2}}{|e^{iθ} − e^{iφ}|^{2−p}} dθ dφ < ∞,  (1.2)

where the supremum is taken over all subarcs I ⊂ ∂△. Janson [9] used (1.2) to define a dyadic analogue Q_{p}^{d} of Q_{p} and to prove that Q_{p} is the intersection of Q_{p}^{d} and BMOA. Observe that (1.2) makes sense even if f is not analytic and it becomes possible to consider an extension to Harmonic Analysis over Euclidean spaces. This is our starting point.

Throughout this paper, we always let ℝ^{n} be n-dimensional Euclidean space, and let ℝ_{+}^{n+1} be the upper half space based on ℝ^{n}. A cube means always a cube in ℝ^{n} with edges parallel to the coordinate axes. We denote the edge length of a cube I by ℓ(I), and the Lebesgue measure of I by |I|; thus ℓ(I) = |I|^{1/n}. Also, for t > 0, tI means the cube which has the same center as I and the edge length tℓ(I). We let |x| denote the usual Euclidean norm for x ∈ ℝ^{n}.

For α ∈ (−∞, ∞), in analogy with (1.2), we define Q_{α}(ℝ^{n}) to be the space of all measurable functions on ℝ^{n} that satisfy

∥f∥_{Q_{α}(ℝ^{n})}^{2} = sup_{I}[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy < ∞,  (1.3)

where I ranges over all cubes in ℝ^{n}. Note that we have changed the parameter from p in (1.2) to α in (1.3); the relation between them is p = 1−2α (for n = 1 as in (1.2)). We will henceforth use Q_{α}(ℝ^{n}) as defined by (1.3) exclusively, hopefully avoiding any possible confusion.

Note that ∥f∥_{Q_{α}(ℝ^{n})} = 0 if and only if f is constant a.e.; we thus regard Q_{α}(ℝ^{n}) as a Banach space of functions modulo constants. (It is immediate that ∥f∥_{Q_{α}(ℝ^{n})} is a norm; completeness also is easily verified, see Section 2.)

Remark 1.1. Since every cube I is contained in a cube J with dyadic edge length (i.e. ℓ(J) ∈ {2^{k} : k ∈ ℤ}) such that ℓ(J) < 2ℓ(I), it is obvious that we obtain an equivalent definition, with an equivalent norm, if we consider only cubes of dyadic edge lengths in (1.3).

Similarly, one can consider balls instead of cubes (with ℓ(I) replaced by the radius).

We first observe that if α = −\frac{n}{2}, then Q_{α}(ℝ^{n}) = BMO(ℝ^{n}), which can be defined for example as the space all functions in L_{loc}^{1}(ℝ^{n}) satisfying

∥f∥_{BMO(ℝ^{n})}^{2} = sup_{I} |I|^{−1} ∫_{I} |f(x) − f(I)|^{2} dx < ∞,  (1.4)

where the supremum is taken over all cubes I in ℝ^{n} and

f(I) = |I|^{−1} ∫_{I} f(x) dx

stands for the mean value of f over the cube I, cf. [10]. In fact, we will prove below that Q_{α}(ℝ^{n}) = BMO(ℝ^{n}) for all α ∈ (−∞, 0).

It is well known that the important space BMO(ℝ^{n}) can be described by Poisson integrals, Carleson measures, wavelet coefficients and dyadic cubes. The purpose of this paper is to give analogues for Q_{α}(ℝ^{n}), α ∈ (0, 1), which are given in Sections 2–4 and 6–7. In Section 7 we also consider a dyadic version of Q_{α}(ℝ^{n}). In Section 5, we will provide a local analysis of Q_{α}(ℝ^{n}) which sheds further light on the relation between Q_{α}(ℝ^{n}) and BMO(ℝ^{n}). Finally, in Section 8 we will pose some open problems.

Remark 1.2. One can similarly define Q_{α} spaces of functions defined on a cube in ℝ^{n} or a torus 𝕋^{n}. Many of the results below extend to these situations, but we leave the details to the reader.

Remark 1.3. It is also possible to study Q_{p} spaces in several complex variables, see [2] and [15].

Some notations. Throughout the paper, α is a fixed number in (−∞, ∞); usually we assume α ∈ (0, 1). C and c will denote unspecified positive constants, possibly different at each occurence; the constants may depend on α and the dimension n, but not on the functions or cubes involved. (They may sometimes depend on other fixed parameters, for example, in Section 6, on the choice of wavelets.) We write X ≍ Y, meaning cX ≤ Y ≤ CX.

We sometimes consider dyadic cubes: Let D_{0} = D_{0}(ℝ^{n}) be the set of unit cubes whose vertices have integer coordinates, and let, for any integer k ∈ ℤ, D_{k} = D_{k}(ℝ^{n}) = {2^{−k}I : I ∈ D_{0}}; then the cubes in D = ⋃_{−∞}^{∞} D_{k} are called dyadic. Furthermore, if I is any cube, we let D_{k}(I), k ≥ 0, denote the set of the 2^{kn} subcubes of edge length 2^{−k}ℓ(I) obtained by k successive bipartitions of each edge of I. Moreover, put D(I) = ⋃_{0}^{∞} D_{k}(I).

χ_{E} is the characteristic function of the set E.

We let, for x ∈ ℝ^{n}, |x|_{∞} be the l^{∞}-norm on ℝ^{n}: |(x_{1}, …, x_{n})|_{∞} = max_{k} |x_{k}|.

2. Basic properties

This section is devoted to some simple properties of Q_{α}(ℝ^{n}) and to relations between Q_{α}(ℝ^{n}) and the Besov spaces.

We first observe that, by simple changes of variables in (1.3), ∥f∥_{Q_{α}(ℝ^{n})} is not affected by translations or dilations of ℝ^{n}, i.e. by replacing f(x) by f(x − x_{0}), x_{0} ∈ ℝ^{n} or f(tx), t > 0; if we use the norm defined using balls, cf. Remark 1.1, the same holds for rotations. Thus,

Theorem 2.1. Q_{α}(ℝ^{n}) is invariant under translations, rotations and dilations, and thus under all similarities of ℝ^{n}; moreover, there exists an equivalent norm on the space such that all similarities preserve the norm.

We note the following alternative characterization of Q_{α}(ℝ^{n}).

Lemma 2.2. Let −∞ < α < ∞. Then f ∈ Q_{α}(ℝ^{n}) if and only if

sup_{I}[ℓ(I)]^{2α−n} ∫_{|y|<ℓ(I)} ∫_{I} |f(x + y) − f(x)|^{2} dx\frac{dy}{|y|^{n+2α}} < ∞.  (2.1)

Proof. If the double integrals in (1.3) and (2.1) are denoted by A(I) and B(I), respectively, then by the change of variable y → x + y and simple geometry one obtains A(I) ≤ B(\sqrt{n}I) and B(I) ≤ A(3I).

The following properties indicate that we only need to pay attention to the case α ∈ [0, 1) for n > 1, and to the case α ∈ [0, 1/2] for n = 1.

Theorem 2.3.

(i) Q_{α}(ℝ^{n}) is decreasing in α, i.e. Q_{α}(ℝ^{n}) ⊇ Q_{β}(ℝ^{n}) if α ≤ β. (ii) If n ≥ 2 and α ≥ 1, or if n = 1 and α > 1/2, then Q_{α}(ℝ^{n}) contains only

functions that are a.e. constant, and thus Q_{α}(ℝ^{n}) = {0} (as a Banach space).

(iii) If α < 0, then Q_{α}(ℝ^{n}) = BMO(ℝ^{n}).

Remark 2.4. The inclusion in (i) is strict if α < β except in the cases 1 ≤ α < β (n ≥ 2), 1/2 < α < β (n = 1), and α < β < 0, where equality holds by (ii) or (iii); see Example 2.10 and Remarks 2.8 and 2.11 below. In particular, if n ≥ 2 and 0 ≤ α < 1 or n = 1 and 0 ≤ α ≤ 1/2, then {0} ⊊ Q_{α}(ℝ^{n}) ⊊ BMO(ℝ^{n}).

Proof. (i). Suppose α < β. If f ∈ Q_{β}(ℝ^{n}), then for any cube I in ℝ^{n}, we have

∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy

= ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2β}} |x − y|^{2(β−α)} dx dy

≤ C[ℓ(I)]^{2(β−α)} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2β}} dx dy

≤ C[ℓ(I)]^{n−2α}∥f∥_{Q_{β}(ℝ^{n})}^{2},

that is to say, f ∈ Q_{α}(ℝ^{n}). So, Q_{β}(ℝ^{n}) ⊆ Q_{α}(ℝ^{n}).

(ii). First assume α > n/2. If f ∈ Q_{α}(ℝ^{n}), then by (1.3), for any cube I in ℝ^{n},

∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy ≤ [ℓ(I)]^{n−2α}∥f∥_{Q_{α}(ℝ^{n})}^{2}.

Since n − 2α < 0, letting I grow to ℝ^{n} in the last inequality produces

∫_{ℝ^{n}} ∫_{ℝ^{n}} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy = 0.

So, f is constant a.e. on ℝ^{n}.

Secondly, assume α ≥ 1 and assume f ∈ Q_{α}(ℝ^{n}). Assume first that f ∈ C^{1}(ℝ^{n}), and that f is non-constant. By considering either the real or imaginary part, we may further assume that f is real. Then there exists a point x_{0} with ∇f(x_{0}) ≠ 0, and by the rotation invariance (Theorem 2.1) we may assume that ∇f(x_{0}) is directed along the positive x_{1}-axis. Then there exist δ > 0 and a small cube I about x_{0} on which ∂f/∂x_{1} > 2δ and |∂f/∂x_{k}| < δ, k ≥ 2. Let D be the cone {x : |x_{2}| + ⋯ + |x_{n}| < x_{1} < ℓ(I)/2}; then if x, y ∈ I and x−y ∈ D, by the mean value theorem, f(x)−f(y) > δ(x_{1}−y_{1}). Consequently,

∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy

≥ ∫_{I/2} ∫_{z∈D} \frac{δ^{2}z_{1}^{2}}{|z|^{n+2α}} dz dx

= cδ^{2}|I| ∫_{0}^{ℓ(I)/2} \frac{z_{1}^{2}}{z_{1}^{1+2α}} dz_{1} = ∞,

a contradiction. Thus, if f ∈ Q_{α}(ℝ^{n}) ∩ C^{1}(ℝ^{n}), then f is constant.

Now, Q_{α}(ℝ^{n}) is translation invariant by Theorem 2.1, and it follows by Minkowski’s inequality that if f ∈ Q_{α}(ℝ^{n}) and g ∈ L^{1}(ℝ^{n}), then f ∗ g ∈ Q_{α}(ℝ^{n}). In particular, if g ∈ C_{0}^{∞}(ℝ^{n}), then f ∗ g ∈ Q_{α}(ℝ^{n}) ∩ C^{1}(ℝ^{n}), and thus (assuming α ≥ 1) f ∗ g is constant. Finally, choosing a sequence g_{n} ≥ 0 with ∫ g_{n} = 1 and supp g_{n} shrinking to 0, f ∗ g_{n} → f a.e., and it follows that f is a.e. constant, which completes the proof of (ii). (Note that the first case uses large cubes and the second case small cubes to show that f has to be constant.)

(iii). Case 1: −n/2 ≤ α < 0. On the one hand, by (i), Q_{α}(ℝ^{n}) ⊆ Q_{−n/2}(ℝ^{n}) = BMO(ℝ^{n}). On the other hand, if f ∈ BMO(ℝ^{n}) and I is a cube, then for every y ∈ ℝ^{n} with |y| < ℓ(I)

∫_{I} |f(x + y) − f(x)|^{2} dx
≤ ∫_{I} 2(|f(x + y) − f(2I)|^{2} + |f(x) − f(2I)|^{2}) dx
≤ 4 ∫_{2I} |f(x) − f(2I)|^{2} dx ≤ C|I|∥f∥_{BMO(ℝ^{n})}^{2}

and thus, since α < 0,

∫_{|y|<ℓ(I)} ∫_{I} |f(x + y) − f(x)|^{2} dx\frac{dy}{|y|^{n+2α}}

≤ C|I|∥f∥_{BMO(ℝ^{n})}^{2} ∫_{|y|<ℓ(I)} \frac{dy}{|y|^{n+2α}}

= C[ℓ(I)]^{n−2α}∥f∥_{BMO(ℝ^{n})}^{2}.

By Lemma 2.2, f ∈ Q_{α}(ℝ^{n}) and hence Q_{α}(ℝ^{n}) = BMO(ℝ^{n}) for α ∈ [−n/2, 0).

Case 2: α ∈ (−∞, −n/2]. In this case, BMO(ℝ^{n}) ⊆ Q_{α}(ℝ^{n}) is known by (i). Now, let f ∈ Q_{α}(ℝ^{n}) and let I be a cube. If x, y ∈ I, then the set \{z ∈ I : min(|x − z|, |y − z|) > \frac{1}{8}ℓ(I)\} has measure at least \frac{1}{2}|I|, and thus, using −2α − n ≥ 0,

∫_{I} min(|x − z|^{−2α−n}, |y − z|^{−2α−n}) dz > cℓ(I)^{−2α−n}|I| = cℓ(I)^{−2α}.

Consequently,

\begin{matrix}|I|^{−2} ∫_{I} ∫_{I} |f(x) − f(y)|^{2} dx dy \\ ≤ Cℓ(I)^{2α−2n} ∫_{I} ∫_{I} ∫_{I} min(|x − z|^{−2α−n}, |y − z|^{−2α−n})|f(x) − f(y)|^{2} dx dy dz \\ ≤ Cℓ(I)^{2α−2n} ∫_{I} ∫_{I} ∫_{I} min(|x − z|^{−2α−n}, |y − z|^{−2α−n}) \\ × (|f(x) − f(z)|^{2} + |f(y) − f(z)|^{2}) dx dy dz \\ ≤ Cℓ(I)^{2α−n}(∫_{I} ∫_{I} \frac{|f(x) − f(z)|^{2}}{|x − z|^{2α+n}} dx dz + ∫_{I} ∫_{I} \frac{|f(y) − f(z)|^{2}}{|y − z|^{2α+n}} dy dz) \\ = Cℓ(I)^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{2α+n}} dx dy \end{matrix} (2.2)

Hence the left hand side of (2.2) is bounded as I ranges over all cubes, which means that f ∈ BMO(ℝ^{n}), cf. (5.2). Thus Q_{α}(ℝ^{n}) ⊆ BMO(ℝ^{n}), and we have shown Q_{α}(ℝ^{n}) = BMO(ℝ^{n}).

We may now easily verify that Q_{α}(ℝ^{n}) is a Banach space.

Theorem 2.5. Q_{α}(ℝ^{n}) is complete, and thus a Banach space.

Proof. Let {f_{m}} be a Cauchy sequence in Q_{α}(ℝ^{n}). By Theorem 2.3 and its proof, Q_{α}(ℝ^{n}) ⊆ BMO(ℝ^{n}) with the inclusion map bounded. Hence, {f_{m}} is a Cauchy sequence in BMO(ℝ^{n}) too, and f_{m} → f in BMO(ℝ^{n}) for some f. It follows easily, using Fatou’s lemma, that for every k ≥ 1, ∥f − f_{k}∥_{Q_{α}(ℝ^{n})} ≤ lim sup_{m→∞} ∥f_{m} − f_{k}∥_{Q_{α}(ℝ^{n})}, which implies that f_{k} → f in Q_{α}(ℝ^{n}) too.

The following result relates Q_{α} spaces defined in different dimensions.

Theorem 2.6. Let −∞ < α < ∞. Let f be a function on ℝ^{n}, n ≥ 1, and define F on ℝ^{n+1} by F (x, t) = f(x), x ∈ ℝ^{n}, t ∈ ℝ. Then F ∈ Q_{α}(ℝ^{n+1}) ⟺ f ∈ Q_{α}(ℝ^{n}).

Proof. A cube in ℝ^{n+1} can be written I × [a, a + ℓ(I)] for a cube I in ℝ^{n} and a real number a. Thus,

\begin{matrix}∥F ∥_{Q_{α}(ℝ^{n+1})}^{2} \\ = sup_{I,a}[ℓ(I)]^{2α−(n+1)} ∫_{I} ∫_{I} ∫_{a}^{ℓ(I)+a} ∫_{a}^{ℓ(I)+a} \frac{|f(x) − f(y)|^{2}}{|(x, t) − (y, u)|^{n+1+2α}} dt du dx dy. \end{matrix}

The multiple integral is independent of a, so we may take a = 0. Moreover, letting s = t − u, and assuming α > −n/2, as we may by Theorem 2.3(iii),

∫_{0}^{ℓ(I)} ∫_{0}^{ℓ(I)} \frac{dt du}{|(x, t) − (y, u)|^{n+1+2α}}

≍ ∫_{0}^{ℓ(I)} ∫_{0}^{ℓ(I)} \frac{dt du}{|x − y|^{n+1+2α} + |t − u|^{n+1+2α}}

≤ ℓ(I) ∫_{−∞}^{∞} \frac{ds}{|x − y|^{n+1+2α} + |s|^{n+1+2α}}

= Cℓ(I)|x − y|^{−n−2α},

while a similar opposite inequality follows by considering only t and u with |t − u| < |x − y|. Consequently,

∥F ∥_{Q_{α}(ℝ^{n+1})}^{2} ≍ sup_{I}[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy = ∥f∥_{Q_{α}(ℝ^{n})}^{2}.

Connection with Besov spaces. Denote the homogeneous Besov spaces on ℝ^{n} by Λ_{α}^{p,q}(ℝ^{n}). We refer to e.g. [16] for a general definition; if 0 < α < 1 and 1 ≤ p, q < ∞, then Λ_{α}^{p,q}(ℝ^{n}) consists of all measurable functions f such that

∫_{ℝ^{n}} [∫_{ℝ^{n}} |f(x + y) − f(x)|^{p} dx]^{q/p} \frac{dy}{|y|^{n+qα}} < ∞,  (2.3)

and if 0 < α < 1 and 1 ≤ p < q = ∞, then Λ_{α}^{p,q}(ℝ^{n}) consists of the functions f such that

∥f∥_{Λ_{α}^{p,∞}(ℝ^{n})} = sup_{y∈ℝ^{n}} |y|^{−α} [∫_{ℝ^{n}} |f(x + y) − f(x)|^{p} dx]^{1/p} < ∞.  (2.4)

Theorem 2.7. Let n ≥ 2 and 0 < α < 1, or n = 1 and 0 < α < 1/2.

(i) If q ≤ 2, then Λ_{α}^{n/α,q}(ℝ^{n}) ⊆ Q_{α}(ℝ^{n}). (ii) If β > α and q ≤ ∞, then Λ_{β}^{n/β,q}(ℝ^{n}) ⊆ Q_{α}(ℝ^{n}).

Remark 2.8. In the case n = 1 and α = 1/2, it is seen by (1.3) and (2.3) that

Q_{1/2}(ℝ) = \{f : ∫_{ℝ} ∫_{ℝ} \frac{|f(x) − f(y)|^{2}}{|x − y|^{2}} dx dy < ∞\} = Λ_{1/2}^{2,2}(ℝ),

which coincides with the Sobolev space L_{1/2}^{2}(ℝ). Thus (i) holds in this case too, while (ii) holds only for q ≤ 2.

In particular, Q_{1/2}(ℝ) contains non-constant functions, and it is thus clear that the inclusion Q_{1/2}(ℝ) ⊃ Q_{β}(ℝ), β > 1/2, is strict (cf. Theorem 2.3 and Remark 2.4).

Proof. (i). Since Λ_{α}^{n/α,q}(ℝ^{n}) ⊂ Λ_{α}^{n/α,2}(ℝ^{n}) [16, Chapter 3, Theorem 4], we may assume that q = 2. Thus, assume that f ∈ Λ_{α}^{n/α,2}(ℝ^{n}). By Hölder’s inequality with exponents n/2α and n/(n − 2α) we get, for any cube I in ℝ^{n},

\begin{matrix}∫_{|y|<ℓ(I)} ∫_{I} |f(x + y) − f(x)|^{2} dx \frac{dy}{|y|^{n+2α}} \\ ≤ |I|^{\frac{n−2α}{n}} ∫_{ℝ^{n}} [∫_{I} |f(x + y) − f(x)|^{n/α} dx]^{2α/n} \frac{dy}{|y|^{n+2α}} , \end{matrix}

which gives f ∈ Q_{α}(ℝ^{n}) by (2.3) and Lemma 2.2.

(ii). Since Λ_{β}^{n/β,q}(ℝ^{n}) ⊂ Λ_{γ}^{n/γ,q}(ℝ^{n}) for β > γ [16, Chapter 3, Theorem 5], we may assume that α < β < 1, and in the case n = 1 further β ≤ 1/2. If f ∈ Λ_{β}^{n/β,q}(ℝ^{n}) ⊆ Λ_{β}^{n/β,∞}(ℝ^{n}), then for any cube I in ℝ^{n} we apply Hölder’s inequality with exponents n/2β and n/(n − 2β) together with (2.4) to get

∫_{|y|<ℓ(I)} ∫_{I} |f(x + y) − f(y)|^{2} dx\frac{dy}{|y|^{n+2α}}

≤ |I|^{\frac{n−2β}{n}} ∫_{|y|<ℓ(I)} [∫_{I} |f(x + y) − f(x)|^{n/β} dx]^{2β/n} \frac{dy}{|y|^{n+2α}}

≤ ℓ(I)^{n−2β} ∫_{|y|<ℓ(I)} |y|^{2β−n−2α} dy ∥f∥_{Λ_{β}^{n/β,∞}(ℝ^{n})}^{2}

≤ Cℓ(I)^{n−2α}∥f∥_{Λ_{β}^{n/β,∞}(ℝ^{n})}^{2}.

So, f ∈ Q_{α}(ℝ^{n}) and the result follows.

Remark 2.9. The inclusions in Theorem 2.7 are the only possible for these α. First, if Λ_{β}^{p,q}(ℝ^{n}) ⊆ Q_{α}(ℝ^{n}), then the inclusion mapping is bounded by the closed graph theorem, and since for any t > 0, the norm of the dilation f_{t}(x) = f(tx) satisfies ∥f_{t}∥_{Q_{α}(ℝ^{n})} = ∥f∥_{Q_{α}(ℝ^{n})} by Theorem 2.1, while ∥f_{t}∥_{Λ_{β}^{p,q}(ℝ^{n})} = t^{β−n/p}∥f∥_{Λ_{β}^{p,q}(ℝ^{n})} (for a suitable choice of norm), we see that necessarily β − n/p = 0 and thus p = n/β.

Secondly, the following example shows that Λ_{β}^{n/β,q}(ℝ^{n}) ⊈ Q_{α}(ℝ^{n}) for β < α or β = α and q > 2.

Example 2.10. Let φ ∈ S(ℝ^{n}) be a fixed function such that φ̂ has support in the unit ball and φ ≠ 0 on the cube [−3π, 3π]^{n}. (Such functions are easily constructed, for example as a dilation φ_{1}(δx) with φ̂_{1} ∈ C_{0}^{∞}, φ̂_{1} ≥ 0 and δ small.)

Let (a_{k})_{1}^{∞} be a sequence real numbers with ∑_{k=1}^{∞} a_{k}^{2} < ∞, and define

g(x) = ∑_{k=1}^{∞} a_{k} exp(2^{k}x_{1}i),

where x_{1} is the first coordinate of x. Let f = φg. It is then easily seen, by the definition [16, p. 51], that for every p, q and α

∥f∥_{Λ_{α}^{p,q}(ℝ^{n})} ≍ ∥(2^{kα}a_{k})_{1}^{∞}∥_{ℓ^{q}} = (∑_{k=1}^{∞} 2^{qkα}|a_{k}|^{q})^{1/q}

(with the usual modification if q = ∞).

In particular, if ∑_{k} 2^{2kα}|a_{k}|^{2} < ∞, then f ∈ Λ_{α}^{n/α,2}(ℝ^{n}), and thus f ∈ Q_{α}(ℝ^{n}) by Theorem 2.7, provided n ≥ 2 and 0 < α < 1 or n = 1 and 0 < α ≤ 1/2.

Conversely, if f ∈ Q_{α}(ℝ^{n}), then, choosing I = [−π, π]^{n} in Lemma 2.2,

∫_{|y|<2π} ∫_{I} |f(x + y) − f(x)|^{2} dx \frac{dy}{|y|^{n+2α}} < ∞.

Since

f(x + y) − f(x) = φ(x + y)(g(x + y) − g(x)) + g(x)(φ(x + y) − φ(x)),

and |φ(x + y)| ≥ c > 0 for x ∈ I and |y| < 2π,

|g(x + y) − g(x)|^{2}
≤ C|φ(x + y)(g(x + y) − g(x))|^{2}
≤ C|f(x + y) − f(x)|^{2} + C|g(x)|^{2}|φ(x + y) − φ(x)|^{2}.

Moreover,

|φ(x + y) − φ(x)| ≤ C|y|,

and thus

\begin{matrix}∫_{|y|<2π} ∫_{I} |g(x)|^{2}|φ(x + y) − φ(x)|^{2} dx \frac{dy}{|y|^{n+2α}} \\ ≤ C ∫_{ℝ^{n}} |g(x)|^{2} ∫_{|y|<2π} |y|^{2−2α−n} dy < ∞. \end{matrix}

Consequently, using Parseval’s formula and writing y = (y_{1}, y′), y′ ∈ ℝ^{n−1},

∞ > ∫_{|y|<1} ∫_{x∈I} |g(x + y) − g(x)|^{2} dx \frac{dy}{|y|^{n+2α}}

= (2π)^{n} ∫_{|y|<1} ∑_{k=1}^{∞}|a_{k}(e^{2^{k}y_{1}i} − 1)|^{2} \frac{dy}{|y|^{n+2α}}

≥ c ∑_{k=1}^{∞} |a_{k}|^{2} ∫_{|y′|<y_{1}<2^{−k}} |2^{k}y_{1}|^{2} \frac{dy}{y_{1}^{n+2α}}

≥ c ∑_{k=1}^{∞} |a_{k}|^{2} ∫_{0}^{2^{−k}} 2^{2k}y_{1}^{1−2α} dy_{1}

= c ∑_{k=1}^{∞} 2^{2kα}|a_{k}|^{2}.

We have thus shown that, for α as in Theorem 2.7,

f ∈ Q_{α}(ℝ^{n}) ⟺ ∑_{k=1}^{∞} 2^{2kα}|a_{k}|^{2} < ∞.

Choosing e.g. a_{k} = 2^{−kα}k^{−1}, we obtain a function f ∈ Q_{α}(ℝ^{n}) \ Q_{β}(ℝ^{n}) for every β > α, justifying the claim in Remark 2.4. Similarly, if β < α or β = α and q ≥ 2, suitable choices of a_{k} yields f ∈ Λ_{β}^{n/β,q}(ℝ^{n}) \ Q_{α}(ℝ^{n}) as asserted in Remark 2.9.

Remark 2.11. For α ≤ 0, it may be shown that if f is as in Example 2.10, then

f ∈ Q_{α}(ℝ^{n}) ⟺ \begin{cases}∑_{1}^{∞} |a_{k}|^{2} < ∞, & α < 0, \\ ∑_{1}^{∞} k|a_{k}|^{2} < ∞, & α = 0. \end{cases}

In particular, this shows that Q_{0}(ℝ^{n}) ⊊ Q_{α}(ℝ^{n}) = BMO(ℝ^{n}), α < 0.

3. Poisson extension

In this section, we discuss differences and similarities between Q_{α}(ℝ^{n}), α ∈ (0, 1), and BMO(ℝ^{n}) with respect to Poisson extensions to ℝ_{+}^{n+1}.

Let f be any measurable function on ℝ^{n} that satisfies

∫_{ℝ^{n}} \frac{|f(x)|}{1 + |x|^{n+1}} dx < ∞.  (3.1)

Its Poisson integral is defined by

f(x, t) = ∫_{ℝ^{n}} P_{t}(x − y)f(y) dy,  (3.2)

where

P_{t}(x) = \frac{c_{n}t}{(t^{2} + |x|^{2})^{\frac{n+1}{2}}} , c_{n} = \frac{Γ(\frac{n+1}{2})}{π^{\frac{n+1}{2}}}.

Define the gradient of f(x, t) by

|∇f(x, t)|^{2} = |\frac{∂f(x, t)}{∂t}|^{2} + ∑_{j=1}^{n} |\frac{∂f(x, t)}{∂x_{j}} |^{2} ,

and the Carleson box based on a cube I by

S(I) = I × (0, ℓ(I)] = {(x, t) ∈ ℝ_{+}^{n+1} : x ∈ I, t ∈ (0, ℓ(I)]}.

We extend a lemma of Stegenga [18] from one dimension to higher dimensions.

Lemma 3.1. Let I and J be cubes in ℝ^{n} centered at x_{0} with ℓ(J) ≥ 3ℓ(I) and let f ∈ L_{loc}^{1}(ℝ^{n}). For α ∈ (0, 1), there is a constant C independent of f, I and J such that

\begin{matrix}∫_{S(I)} |∇f(x, t)|^{2}t^{1−2α} dx dt \\ ≤ C ∫_{J} ∫_{J} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy \\ + C[ℓ(I)]^{n+2(1−α)} (∫_{ℝ^{n}\ \frac{2}{3} J} |f(x) − f_{J}|\frac{dx}{|x − x_{0}|^{n+1}} )^{2} . \end{matrix}  (3.3)

Proof. Without loss of generality, we assume that x_{0} is the origin. Also, let φ be a function with 0 ≤ φ ≤ 1 such that φ = 1 on \frac{2}{3}J, supp φ ⊂ \frac{3}{4}J, and

|φ(x) − φ(y)| ≤ C[ℓ(J)]^{−1}|x − y|,  x, y ∈ ℝ^{n}.  (3.4)

Write φ̃ = 1 − φ. Then we have

f = f_{J} + (f − f_{J})φ + (f − f_{J})φ̃ = f_{1} + f_{2} + f_{3}.

We also have

f(x, t) = f_{1}(x, t) + f_{2}(x, t) + f_{3}(x, t)

for the corresponding Poisson integrals. In the integral with the gradient square, f_{1} contributes nothing since it is constant.

We have

\frac{∂f(x, s)}{∂s} = ∫_{ℝ^{n}} \frac{∂P_{s}(y)}{∂s}[f(x + y) − f(x)] dy

and therefore by the elementary estimates

|\frac{∂P_{s}(y)}{∂s}| ≤ cs^{−n−1}, |\frac{∂P_{s}(y)}{∂s}| ≤ c|y|^{−n−1},

we see that

∥\frac{∂f(·, s)}{∂s}∥_{L^{2}(ℝ^{n})}

≤ Cs^{−n−1} ∫_{|y|≤s} ∥f(· + y) − f∥_{L^{2}(ℝ^{n})} dy

+ C ∫_{|y|>s} ∥f(· + y) − f∥_{L^{2}(ℝ^{n})} \frac{dy}{|y|^{n+1}} .

12 MATTS ESSÉN, SVANTE JANSON, LIZHONG PENG, AND JIE XIAO

Next set y = rξ ∈ ℝ^{n}, with r = |y|, and |ξ| = 1. Then with

Ω(r) = ∫_{|ξ|=1} ∥f(· + rξ) − f∥_{L^{2}(ℝ^{n})}dξ

we write

\begin{matrix}A = ∥\frac{∂f(x, s)}{∂s}∥_{L^{2}(ℝ^{n})} \\ ≤ Cs^{−n−1} ∫_{0}^{s} Ω(r)r^{n−1}dr + C ∫_{s}^{∞} Ω(r)r^{−2}dr. \end{matrix}

Therefore, by Hardy’s inequalities [19, p. 272],

∫_{0}^{∞} s^{1−2α}A^{2}ds ≤ C ∫_{0}^{∞}[Ω(r)]^{2}r^{−1−2α}dr.

Note that by Hölder’s inequality,

[Ω(r)]^{2} ≤ C ∫_{|ξ|=1} ∥f(· + rξ) − f∥_{L^{2}(ℝ^{n})}^{2}dξ.

Substituting this in the above leads to the bound

\begin{matrix}C ∫_{|ξ|=1} ∫_{0}^{∞} ∥f(· + rξ) − f∥_{L^{2}(ℝ^{n})}^{2}r^{−1−2α}drdξ \\ = C ∫_{ℝ^{n}} [∫_{ℝ^{n}} |f(x + y) − f(x)|^{2} dx] \frac{dy}{|y|^{n+2α}} . \end{matrix}

In the same way we can prove

\begin{matrix}∫_{0}^{∞} t^{1−2α}∥\frac{∂f(·, t)}{∂x_{j}}∥_{L^{2}(ℝ^{n})}^{2}dt \\ ≤ C ∫_{ℝ^{n}} [∫_{ℝ^{n}} |f(x + y) − f(x)|^{2} dx] \frac{dy}{|y|^{n+2α}} . \end{matrix}

Now, we obtain ∫_{S(I)} |∇f_{2}(x, t)|^{2}t^{1−2α} dx dt

≤ ∫_{0}^{∞} [∫_{ℝ^{n}} |∇f_{2}(x, t)|^{2}t^{1−2α} dx] dt

≤ C ∫_{ℝ^{n}} ∫_{ℝ^{n}} \frac{|f_{2}(x) − f_{2}(y)|^{2}}{|x − y|^{n+2α}} dx dy

= ∫∫ _{x,y∈J} ⋯ + ∫∫ _{x∉J,y∈ \frac{3}{4} J} ⋯ + ∫∫ _{y∉J,x∈ \frac{3}{4} J} ⋯ = B_{1} + B_{2} + B_{3}.

As to B_{1}, we have using (3.4)

|f_{2}(x) − f_{2}(y)| ≤ |f(x) − f(y)| + C[ℓ(J)]^{−1}|x − y||f(y) − f_{J}|.

Thus, we need only estimate

[ℓ(J)]^{−2} ∫_{J} ∫_{J} \frac{|f(y) − f_{J}|^{2}}{|x − y|^{n+2α−2}} dx dy

= [ℓ(J)]^{−2} ∫_{J} |f(y) − f_{J}|^{2} [∫_{J} |x − y|^{2−n−2α} dx] dy

≤ C[ℓ(J)]^{−2α} ∫_{J} |f(y) − f_{J}|^{2} dy

= C[ℓ(J)]^{−2α−n} ∫_{J} ∫_{J} |f(x) − f(y)|^{2} dx dy

≤ C ∫_{J} ∫_{J} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy,

cf. (5.2), which gives the estimate for B_{1}. The B_{2} and B_{3} terms are handled similarly as the last estimate, using f_{2}(x) = 0 for x ∉ J.

Moreover, |∇f_{3}(x, t)|

≤ ∫_{ℝ^{n}} |∇P_{t}(x − y)|f_{3}(y) dy

≤ C ∫_{ℝ^{n}\ \frac{2}{3} J} \frac{|f(y) − f_{J}|}{(t + |x − y|)^{n+1}} dy.

If (x, t) ∈ S(I) then for y ∈ ℝ^{n} \ \frac{2}{3}J,

\frac{1}{(t + |x − y|)^{n+1}} ≤ \frac{C}{|y|^{n+1}} ,

and hence

∫_{S(I)} |∇f_{3}(x, t)|^{2}t^{1−2α} dx dt

≤ (∫_{S(I)} t^{1−2α} dx dt) (∫_{ℝ^{n}\ \frac{2}{3} J} \frac{|f(x) − f_{J}|}{|x|^{n+1}} dx)^{2}

≤ C[ℓ(I)]^{n+2(1−α)} (∫_{ℝ^{n}\ \frac{2}{3} J} \frac{|f(x) − f_{J}|}{|x|^{n+1}} dx)^{2} .

Combining the above inequalities, we obtain (3.3).

With Lemma 3.1, we can characterize functions of Q_{α}(ℝ^{n}) in terms of the Poisson integral. Note that setting α = 0 in (3.5) yields a characterization of BMO(ℝ^{n}) [6].

Theorem 3.2. Let α ∈ (0, 1) and let f ∈ L_{loc}^{2}(ℝ^{n}) satisfy (3.1). Then f ∈ Q_{α}(ℝ^{n}) if and only if

∫_{S(I)} |∇f(x, t)|^{2}t^{1−2α} dx dt ≤ M[ℓ(I)]^{n−2α}  (3.5)

for some M < ∞ and all cubes I ⊂ ℝ^{n}.

Proof. First, suppose that f ∈ Q_{α}(ℝ^{n}). Then, by Theorem 2.3, f ∈ BMO(ℝ^{n}) with ∥f∥_{BMO(ℝ^{n})} ≤ C∥f∥_{Q_{α}(ℝ^{n})}. For convenience, we may assume that the cube I has the origin as its center. Now, let J = 3I. Then we have

∫_{ℝ^{n}\ \frac{2}{3} J} |f(x) − f_{J}|\frac{dx}{|x|^{n+1}}

≤ ∑_{k=0}^{∞} ∫_{3^{k}J\3^{k−1}J} |f(x) − f_{J}|\frac{dx}{|x|^{n+1}}

≤ C ∑_{k=0}^{∞}[3^{k}ℓ(J)]^{−(n+1)} ∫_{3^{k}J} |f(x) − f_{3^{k}J}| dx + C ∑_{k=0}^{∞}(3^{k}ℓ(J))^{−1}|f_{3^{k}J} − f_{J}|

≤ \frac{C}{ℓ(I)} [∥f∥_{BMO(ℝ^{n})} + ∑_{k=1}^{∞} \frac{k}{3^{k}} ∥f∥_{BMO(ℝ^{n})}]

≤ C[ℓ(I)]^{−1}∥f∥_{BMO(ℝ^{n})}.

So,

\begin{matrix}[ℓ(I)]^{n+2(1−α)} (∫_{ℝ^{n}\ \frac{2}{3} J} |f(x) − f_{J}| dx/|x − x_{0}|^{n+1})^{2} \\ ≤ C[ℓ(I)]^{n−2α}∥f∥_{Q_{α}(ℝ^{n})}^{2}. \end{matrix}

This inequality, the definition of Q_{α}(ℝ^{n}) and Lemma 3.1 imply (3.5), which proves one implication.

Conversely, suppose that (3.5) holds for f. We claim that then

∫_{|y|<ℓ(I)} [∫_{I} |f(x + y) − f(x)|^{2} dx] \frac{dy}{|y|^{n+2α}} ≤ CM[ℓ(I)]^{n−2α}.  (3.6)

First, by the triangle inequality,

|f(x + y) − f(x)|
≤ |f(x + y) − f(x + y, |y|)|
+ |f(x + y, |y|)) − f(x, |y|)|
+ |f(x, |y|) − f(x)|
= A_{1} + A_{2} + A_{3}.

Q SPACES OF SEVERAL REAL VARIABLES 15

For A_{3}, we employ Minkowski’s inequality to get

(∫_{I} |A_{3}|^{2} dx)^{\frac{1}{2}}

≤ ∫_{0}^{|y|} [∫_{I} |\frac{∂f(x, t)}{∂t}|^{2} dx]^{\frac{1}{2}} dt

≤ ∫_{0}^{|y|} [∫_{I} |∇f(x, t)|^{2} dx]^{\frac{1}{2}} dt.

By Hardy’s inequality [19, p. 272], we further have

∫_{|y|<ℓ(I)} \frac{1}{|y|^{n+2α}} (∫_{I} |A_{3}|^{2} dx) dy

≤ C ∫_{0}^{ℓ(I)} \{∫_{0}^{s} [∫_{I} |∇f(x, t)|^{2} dx]^{\frac{1}{2}} dt\}^{2} s^{−1−2α}ds

≤ C ∫_{0}^{ℓ(I)} [∫_{I} |∇f(x, t)|^{2} dx] t^{1−2α}dt

≤ CM[ℓ(I)]^{n−2α}.

Since, for |y| < ℓ(I), ∫_{I} |A_{1}|^{2} dx = ∫_{I+y} |A_{3}|^{2} dx ≤ ∫_{3I} |A_{3}|^{2} dx, we similarly obtain

∫_{|y|<ℓ(I)} \frac{1}{|y|^{n+2α}} (∫_{I} |A_{1}|^{2} dx) dy ≤ CM[ℓ(I)]^{n−2α}.

It remains to handle A_{2}. It is clear that

|A_{2}| ≤ ∫_{0}^{|y|} |∇f(x + te_{y}, |y|)|dt, e_{y} = y/|y|.

If |y| ≤ ℓ(I), then an application of Minkowski’s inequality gives

(∫_{I} |A_{2}|^{2} dx)^{\frac{1}{2}}
≤ ∫_{0}^{|y|} [∫_{I} |∇f(x + te_{y}, |y|)|^{2} dx]^{\frac{1}{2}} dt
≤ C ∫_{0}^{|y|} [∫_{3I} |∇f(x, |y|)|^{2} dx]^{\frac{1}{2}} dt
= C|y| [∫_{3I} |∇f(x, |y|)|^{2} dx]^{\frac{1}{2}} .

Hence,

∫_{|y|≤ℓ(I)} \frac{1}{|y|^{n+2α}} (∫_{I} |A_{2}|^{2} dx) dy

≤ ∫_{S(3I)} |∇f(x, t)|^{2}t^{1−2α} dx dt

≤ CM[ℓ(I)]^{n−2α}.

Putting these estimates on A_{1}, A_{2} and A_{3} together, we see that (3.6) holds, and thus, by Lemma 2.2, f belongs to Q_{α}(ℝ^{n}).

4. Generalized Carleson measures

Theorem 3.2 shows that it is natural to introduce a generalized Carleson measure. Let S(I) = I × (0, ℓ(I)] be the Carleson box based on the cube I ⊂ ℝ^{n}. Given p > 0 and a positive Borel measure μ on ℝ_{+}^{n+1}, we say that μ is a p-Carleson measure if

μ(S(I)) ≤ M[ℓ(I)]^{pn},

for some M < ∞ and all cubes I ⊂ ℝ^{n}. Of course, the case p = 1 gives the classical Carleson measures. It follows from Theorem 3.2 that whenever α ∈ (0, 1), f ∈ Q_{α}(ℝ^{n}) if and only if |∇f(x, t)|^{2}t^{1−2α} dxdt is a (1 − 2α/n)- Carleson measure. In fact, the second half of the proof of Theorem 3.2 shows more generally that if there exists a differentiable extension F (x, t) of f to ℝ_{+}^{n+1} which satisfies the condition that |∇F (x, t)|^{2}t^{1−2α} dxdt is a (1−2α/n)-Carleson measure, then f is in Q_{α}(ℝ^{n}).

We next want to give a characterization of Q_{α}(ℝ^{n}) by means of an integral on ℝ_{+}^{n+1}. From now on, denote by δ(x) the distance of the point x ∈ ℝ_{+}^{n+1} to the boundary ∂ℝ_{+}^{n+1}. Also, ỹ stands for the symmetric point of y ∈ ℝ_{+}^{n+1} with respect to ℝ^{n}, that is to say, if y = (y_{1}, …, y_{n+1}), then ỹ = (y_{1}, …, −y_{n+1}).

Lemma 4.1. Let μ be a Borel measure on ℝ_{+}^{n+1} and p ∈ (0, ∞). Then μ is a p-Carleson measure if and only if

sup_{y∈ℝ_{+}^{n+1}} ∫_{ℝ_{+}^{n+1}} (\frac{δ(y)}{|x − ỹ|^{n+1}} )^{p} dμ(x) < ∞.  (4.1)

Proof. Sufficiency. Take y to be the center of the Carleson box S(I). If x ∈ S(I), then |x − ỹ| < Cℓ(I) and hence

∫_{ℝ_{+}^{n+1}} (\frac{δ(y)}{|x − ỹ|^{n+1}} )^{p} dμ

≥ ∫_{S(I)} (\frac{δ(y)}{|x − ỹ|^{n+1}} )^{p} dμ

≥ \frac{cμ(S(I))}{[ℓ(I)]^{pn}}.

Thus, if (4.1) holds, then μ is a p-Carleson measure.

Necessity. Fix y = (y′, y_{n+1}) ∈ ℝ_{+}^{n+1}. Let I ⊂ ℝ^{n} be the cube with center y′ and edge length δ(y), and for each positive integer m = 1, 2, …, define E_{m} to be the Carleson box S(2^{m}I). When x ∈ E_{1}, we have |x − ỹ| ≥ δ(y). Also, if x ∈ E_{m+1}\E_{m}, then c2^{m}δ(y) ≤ |x−ỹ| ≤ C2^{m+1}δ(y). Thus, if μ is a p-Carleson measure,

∫_{ℝ_{+}^{n+1}} (\frac{δ(y)}{|x − ỹ|^{n+1}} )^{p} dμ(x)

= (∫_{E_{1}} + ∑_{m=1}^{∞} ∫_{E_{m+1}\E_{m}}) (\frac{δ(y)}{|x − ỹ|^{n+1}} )^{p} dμ(x)

≤ C \frac{μ(E_{1})}{δ(y)^{np}} + C ∑_{m=1}^{∞} \frac{μ(E_{m+1})}{2^{mp(n+1)}[δ(y)]^{np}}

≤ C.

The last constant C is independent of y. We are done.

For convenience, we will from now on use the same notation to denote f on ℝ^{n} and its Poisson extension to the upper half space ℝ_{+}^{n+1}. So, we have

Theorem 4.2. Let α ∈ (0, 1) and let f ∈ L_{loc}^{2}(ℝ^{n}) satisfy (3.1). Then f ∈ Q_{α}(ℝ^{n}) if and only if its Poisson integral f(x) = f(z, t) on ℝ_{+}^{n+1} satisfies

sup_{y∈ℝ_{+}^{n+1}} ∫_{ℝ_{+}^{n+1}}(\frac{δ(y)}{|x − ỹ|^{n+1}} )^{1−2α/n}|∇f(x)|^{2}[δ(x)]^{1−2α} dx < ∞.  (4.2)

Proof. If 1 − 2α/n > 0, then the proof is immediate by Theorem 3.2 and Lemma 4.1.

It remains only to treat the (less interesting) case n = 1 and α ≥ 1/2. First, if n = 1 and α = 1/2, then the integral in (4.2) is independent of y and equals ∫_{ℝ_{+}^{n+1}} |∇f(x)|^{2} dx, which is finite if and only if f ∈ Q_{1/2}(ℝ), cf. Remark 2.8.

Finally, the case n = 1 and α > 1/2 is trivial. Then f ∈ Q_{α}(ℝ) only if f is a.e. constant by Theorem 2.3(ii). Similarly, if (4.2) holds, it follows by using Fatou’s lemma as y → 0 that ∇f(x) = 0 in ℝ_{+}^{n+1}, and thus f is a.e. constant.

Green potential. Observe that Q_{p} was first defined by the Green potential. Can we characterize Q_{α}(ℝ^{n}) in terms of the Green function for a half space? When n = 1, the Green function of the upper half plane is

G(x, y) = log \frac{|x − ỹ|}{|x − y|}.  (4.3)

When n > 1, the Green function of the upper half space ℝ_{+}^{n+1} is given by

G(x, y) = \frac{1}{|x − y|^{n−1}} − \frac{1}{|x − ỹ|^{n−1}} .  (4.4)

(cf. [1, p. 65]). Indeed, this function is related to the quantity introduced in Lemma 4.1.

Lemma 4.3. Let x, y ∈ ℝ_{+}^{n+1}. Then

(i) G(x, y) ≥ \frac{2δ(x)δ(y)}{|x−ỹ|^{n+1}}, n ≥ 1.

(ii) G(x, y) ≤ C \frac{δ(x)δ(y)}{|x−ỹ|^{2}|x−y|^{n−1}} , n ≥ 2.

(iii) G(x, y) ≤ \frac{−4 log a}{1−a^{2}} \frac{δ(x)δ(y)}{|x−ỹ|^{2}}, a ≤ \frac{|x−y|}{|x−ỹ|}, n = 1.

Proof. See [1, p. 68] and [7, p. 289].

Remark 4.4. It was proved in [12] that f ∈ BMO(ℝ^{n}) if and only if

sup_{y∈ℝ_{+}^{n+1}} ∫_{ℝ_{+}^{n+1}} G(x, y)|∇f(x)|^{2} dx < ∞.  (4.5)

The following characterization of Q_{α}(ℝ^{n}) by means of the Green potential will employ Lemma 4.3 and Remark 4.4.

Theorem 4.5. Let α ∈ (0, 1) and let f ∈ L_{loc}^{2}(ℝ^{n}) satisfy (3.1). Then f ∈ Q_{α}(ℝ^{n}) if and only if its Poisson integral f(x) = f(z, t) satisfies

sup_{y∈ℝ_{+}^{n+1}} ∫_{ℝ_{+}^{n+1}}[δ(x)]^{2( \frac{1}{n} −1)α}[G(x, y)]^{1−2α/n}|∇f(x)|^{2} dx < ∞.  (4.6)

Proof. If (4.6) holds and 1 − 2α/n ≥ 0, then Lemma 4.3(i) and Theorem 4.2 imply f ∈ Q_{α}(ℝ^{n}). The exceptional case n = 1 and α > 1/2 follows as in the proof of Theorem 4.2.

Conversely, suppose f ∈ Q_{α}(ℝ^{n}). We have

∫_{ℝ_{+}^{n+1}}[δ(x)]^{2( \frac{1}{n} −1)α}[G(x, y)]^{1−2α/n}|∇f(x)|^{2} dx

= \{∫_{|x−y|/|x−ỹ|≤ \frac{1}{2}} + ∫_{|x−y|/|x−ỹ|> \frac{1}{2}} \} [δ(x)]^{2( \frac{1}{n} −1)α}[G(x, y)]^{1−2α/n}|∇f(x)|^{2} dx

= I_{1}(y) + I_{2}(y).

Since |x − y| ≤ \frac{1}{2}|x − ỹ| implies |x − y| ≤ 2δ(x) and, by (4.3) or (4.4), G(x, y) ≥ c/|x − y|^{n−1},

I_{1}(y)

≤ C ∫_{|x−y|/|x−ỹ|≤ \frac{1}{2}} \frac{|x − y|^{\frac{2(n−1)α}{n}}}{[δ(x)]^{\frac{2α(n−1)}{n}}} G(x, y)|∇f(x)|^{2} dx

≤ C ∫_{|x−y|≤2δ(x)} [\frac{|x − y|}{δ(x)}]^{\frac{2α(n−1)}{n}} G(x, y)|∇f(x)|^{2} dx

≤ C ∫_{ℝ_{+}^{n+1}} G(x, y)|∇f(x)|^{2} dx.

We know that Q_{α}(ℝ^{n}) ⊂ BMO(ℝ^{n}), and it follows from the last estimate and Remark 4.4 that sup_{y∈ℝ_{+}^{n+1}} I_{1}(y) < ∞.

As to I_{2}(y), we apply Lemma 4.3(ii) or (iii) to derive

I_{2}(y)

≤ C ∫_{|x−y|/|x−ỹ|≥ \frac{1}{2}} |∇f(x)|^{2} [\frac{δ(x)δ(y)}{|x − ỹ|^{2}|x − y|^{n−1}} ]^{1−2α/n} [δ(x)]^{2α( \frac{1}{n} −1)}dx

≤ C ∫_{|x−y|/|x−ỹ|≥ \frac{1}{2}} |∇f(x)|^{2}[δ(x)]^{2α( \frac{1}{n} −1)} [\frac{δ(x)δ(y)}{|x − ỹ|^{n+1}} ]^{1−2α/n} dx

≤ C ∫_{ℝ_{+}^{n+1}} |∇f(x)|^{2}[δ(x)]^{1−2α} [\frac{δ(y)}{|x − ỹ|^{n+1}} ]^{1−2α/n} dx.

With the help of Theorem 4.2, we deduce that sup_{y∈ℝ_{+}^{n+1}} I_{2}(y) < ∞.

5. Mean oscillation

In this section we give an alternative characterization of Q_{α}(ℝ^{n}) in terms of the (square) mean oscillation over cubes. We follow the one-dimensional case given in [9].

We define, for any cube I and an integrable function f on I,

f(I) = \frac{1}{|I|} ∫_{I} f(x)dx,

the mean of f on I, and

Φ_{f}(I) = \frac{1}{|I|} ∫_{I} |f(x) − f(I)|^{2}dx,

the square mean oscillation of f on I. Obviously, Φ_{f}(I) < ∞ ⇔ f ∈ L^{2}(I); we may extend the definition to all measurable functions f on I by letting Φ_{f}(I) = ∞ when f ∉ L^{2}(I). Recall that f ∈ BMO(ℝ^{n}) if and only if sup_{I} Φ_{f}(I) < ∞ [10]. Note the well-known identities

\frac{1}{|I|} ∫_{I} |f(x) − a|^{2}dx = Φ_{f}(I) + |f(I) − a|^{2}  (5.1)

for any complex number a, and

\frac{1}{|I|^{2}} ∫_{I} ∫_{I} |f(x) − f(y)|^{2} dx dy = 2Φ_{f}(I).  (5.2)

Furthermore, if I ⊂ J, then by (5.1),

Φ_{f}(I) ≤ \frac{1}{|I|} ∫_{I} |f(x) − f(J)|^{2}dx ≤ \frac{|J|}{|I|}Φ_{f}(J),  (5.3)

and, similarly,

|f(I) − f(J)|^{2} ≤ \frac{|J|}{|I|}Φ_{f}(J).  (5.4)

We define, for any cube I and a measurable function f on I, recalling the notation D_{k}(I) for the successive dyadic partitions of I,

Ψ_{f,α}(I) = ∑_{k=0}^{∞} ∑_{J∈D_{k}(I)} 2^{(2α−n)k}Φ_{f}(J).  (5.5)

We will prove below that Q_{α}(ℝ^{n}) can be characterized by sup_{I} Ψ_{f,α}(I) < ∞. We begin with some simple preliminary lemmas.

Lemma 5.1. Let −∞ < α < ∞. For any cube I and f ∈ L^{2}(I), with J ranging over the 2^{n} subcubes in D_{1}(I),

Φ_{f}(I) = 2^{−n} ∑_{J∈D_{1}(I)} Φ_{f}(J) + 2^{−n} ∑_{J∈D_{1}(I)} |f(J) − f(I)|^{2}  (5.6)

and

Ψ_{f,α}(I) ≍ ∑_{J∈D_{1}(I)} Ψ_{f,α}(J) + ∑_{J∈D_{1}(I)} |f(J) − f(I)|^{2}.  (5.7)

Proof. By (5.1), Φ_{f}(I)

= |I|^{−1} ∑_{J∈D_{1}(I)} ∫_{J} |f − f(I)|^{2}

= 2^{−n} ∑_{J∈D_{1}(I)}(Φ_{f}(J) + |f(J) − f(I)|^{2}),

which is (5.6).

Next, this and the definition (5.5) yield, since D_{k}(I) = ⋃_{J∈D_{1}(I)} D_{k−1}(J) for k ≥ 1,

Ψ_{f,α}(I)

= Φ_{f}(I) + ∑_{k=1}^{∞} ∑_{J∈D_{1}(I)} ∑_{K∈D_{k−1}(J)} 2^{(2α−n)k}Φ_{f}(K)

= Φ_{f}(I) + ∑_{J∈D_{1}(I)} 2^{2α−n}Ψ_{f,α}(J)

≍ ∑_{J∈D_{1}(I)}(Ψ_{f,α}(J) + Φ_{f}(J) + |f(J) − f(I)|^{2}),

which yields (5.7), since Ψ_{f,α}(J) + Φ_{f}(J) ≍ Ψ_{f,α}(J).

Lemma 5.2. If α < 0, then Ψ_{f,α}(I) ≍ Φ_{f}(I).

Proof. By Lemma 5.1 and induction, ∑_{J∈D_{k}(I)} 2^{−nk}Φ_{f}(J) ≤ Φ_{f}(I), and hence

Φ_{f}(I) ≤ Ψ_{f,α}(I) ≤ ∑_{k=0}^{∞} 2^{2αk}Φ_{f}(I).

Lemma 5.3. Let −∞ < α < ∞. Then, for any cube I and f ∈ L^{2}(I),

Ψ_{f,α}(I) ≤ C[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy.

Proof. By (5.5) and (5.2),

\begin{matrix}Ψ_{f,α}(I) \\ = ∑_{k=0}^{∞} ∑_{J∈D_{k}(I)} 2^{(2α−n)k} \frac{1}{2}(2^{−nk}|I|)^{−2} ∫_{J} ∫_{J} |f(x) − f(y)|^{2} dx dy \\ = ∫_{ℝ^{n}} ∫_{ℝ^{n}} κ_{I}(x, y)|f(x) − f(y)|^{2} dx dy, \end{matrix}  (5.8)

where

κ_{I}(x, y) = \frac{1}{2} ∑_{k=0}^{∞} ∑_{J∈D_{k}(I)} 2^{(2α+n)k}|I|^{−2}χ_{J}(x)χ_{J}(y).  (5.9)

First assume that α > −n/2. Since x, y ∈ J ∈ D_{k}(I) implies |x − y|_{∞} ≤ ℓ(J) = 2^{−k}ℓ(I), and thus 2^{k} ≤ ℓ(I)/|x − y|_{∞} ≤ Cℓ(I)/|x − y|, we then have

κ_{I}(x, y)

≤ ∑_{2^{k}≤ℓ(I)/|x−y|_{∞}} 2^{(2α+n)k}|I|^{−2}

≤ C(\frac{ℓ(I)}{|x − y|})^{2α+n}|I|^{−2}

= C[ℓ(I)]^{2α−n}|x − y|^{−2α−n};

furthermore κ_{I}(x, y) = 0 unless x, y ∈ I. Consequently, by (5.8), the inequality holds.

The case α ≤ −n/2 follows by Lemma 5.2 and (2.2).

We will prove the converse inequality for α < 1/2 in Lemma 5.8 below, but first we give a slightly weaker, but also more general, converse.

Lemma 5.4. Let −∞ < α < ∞. For any cube I and f ∈ L_{loc}^{2}(ℝ^{n}),

[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy

≤ \frac{C}{|I|} ∫_{|t|_{∞}<ℓ(I)} Ψ_{f,α}(I + t) dt + CΨ_{f,α}(I)

≤ C sup_{|t|_{∞}<ℓ(I)} Ψ_{f,α}(I + t).

Proof. By (5.8) and Fubini’s theorem,

\begin{matrix}\frac{1}{|I|} ∫_{|t|_{∞}<ℓ(I)} Ψ_{f,α}(I + t) dt \\ = ∫_{ℝ^{n}} ∫_{ℝ^{n}} \frac{1}{|I|} ∫_{|t|_{∞}<ℓ(I)} κ_{I+t}(x, y) dt |f(x) − f(y)|^{2} dx dy. \end{matrix}

This and (5.8) show that it suffices to verify

\begin{matrix}\frac{1}{|I|} ∫_{|t|_{∞}<ℓ(I)} κ_{I+t}(x, y) dt + κ_{I}(x, y) \\ ≥ cℓ(I)^{2α−n}|x − y|^{−2α−n},  x, y ∈ I. \end{matrix}  (5.10)

First, suppose that x, y ∈ I with |x−y|_{∞} ≤ \frac{1}{2}ℓ(I) and let l ≥ 0 be such that

2^{−l−2}ℓ(I) < |x − y|_{∞} ≤ 2^{−l−1}ℓ(I).

Then, by (5.9), and noting that x ∉ I + t and thus κ_{I+t}(x, y) = 0 when |t|_{∞} > ℓ(I),

\frac{1}{|I|} ∫_{|t|_{∞}<ℓ(I)} κ_{I+t}(x, y) dt

≥ \frac{1}{|I|} ∫_{ℝ^{n}} \frac{1}{2} ∑_{J∈D_{l}(I+t)} 2^{(2α+n)l}|I|^{−2}χ_{J}(x)χ_{J}(y) dt

= \frac{2^{(2α+n)l}}{2|I|^{3}} ∑_{J∈D_{l}(I)} ∫_{ℝ^{n}} χ_{J+t}(x)χ_{J+t}(y) dt

≥ c|x − y|^{−2α−n}ℓ(I)^{2α−2n} ∑_{J∈D_{l}(I)} ∫_{ℝ^{n}} χ_{J+t}(x)χ_{J+t}(y) dt.

Now, χ_{J+t}(x)χ_{J+t}(y) = χ_{J−x}(−t)χ_{J−y}(−t). Thus the final integral equals the volume of (J −x)∩(J −y), which for each J is a rectangular box with edges at least ℓ(J) − |x − y|_{∞} ≥ \frac{1}{2}ℓ(J), and thus volume at least 2^{−n}|J|. Consequently, the sum over J is at least 2^{−n}|I|, and (5.10) holds for |x − y| ≤ \frac{1}{2}ℓ(I).

Finally, if x, y ∈ I with |x − y| > \frac{1}{2}ℓ(I), then, taking k = 0 in (5.9),

κ_{I}(x, y) ≥ \frac{1}{2}|I|^{−2} ≥ cℓ(I)^{2α−n}|x − y|^{−2α−n}

and (5.10) holds in this case too.

As an immediate consequence of Lemmas 5.3 and 5.4, we obtain our alternative characterization of Q_{α}(ℝ^{n}).

Theorem 5.5. Let −∞ < α < ∞. Then Q_{α}(ℝ^{n}) equals the space of all measurable functions f on ℝ^{n} such that sup_{I} Ψ_{f,α}(I) is finite, where I ranges over all cubes in ℝ^{n}. Moreover, the square root of this supremum is a norm on Q_{α}(ℝ^{n}), equivalent to ∥f∥_{Q_{α}(ℝ^{n})} as defined above.

In order to prove a full converse to the inequality in Lemma 5.3, we begin with two further preliminary lemmas, which also may have independent interest.

Lemma 5.6. Let α < 1/2. Let I, I′ and I′′ be three cubes of equal size, |I| = |I′| = |I′′|, such that I′ and I′′ are adjacent and I ⊂ I′ ∪ I′′. Then, for any f ∈ L^{1}(I′ ∪ I′′),

Φ_{f}(I) ≤ Φ_{f}(I′) + Φ_{f}(I′′) + |f(I′) − f(I′′)|^{2},  (5.11)
Ψ_{f,α}(I) ≤ C(Ψ_{f,α}(I′) + Ψ_{f,α}(I′′) + |f(I′) − f(I′′)|^{2}).  (5.12)

Proof. It follows from (5.1) that

Φ_{f}(I)
≤ |I|^{−1} ∫_{I} |f(x) − (f(I′) + f(I′′))/2|^{2} dx
≤ |I|^{−1} ∫_{I′∪I′′} |f(x) − (f(I′) + f(I′′))/2|^{2} dx
= Φ_{f}(I′) + Φ_{f}(I′′) + \frac{1}{2}|f(I′) − f(I′′)|^{2},

proving (5.11).

For (5.12), we assume for simplicity that I′ = [0, 1)^{n} and I′′ = I′ + e_{1}, where e_{1} is the unit vector (1, 0, …, 0); this is no loss of generality by homogeneity. Note that by assumption then I = I′ + te_{1} for some t ∈ [0, 1]. For each j ≥ 0, let D_{j}^{∗} = D_{j}(I′) ∪ D_{j}(I′′) be the set of the 2^{nj+1} dyadic cubes with side 2^{−j} contained in I′ ∪ I′′. If I_{j} ∈ D_{j}(I), then I_{j} ⊂ J ∪ (J + 2^{−j}e_{1}) for some J ∈ D_{j}^{∗}, and thus by (5.11) applied to J,

Φ_{f}(I_{j}) ≤ Φ_{f}(J) + Φ_{f}(J + 2^{−j}e_{1}) + |f(J) − f(J + 2^{−j}e_{1})|^{2}.

The 2^{nj} different choices of I_{j} ∈ D_{j}(I) yield different J ∈ D_{j}^{∗}, and summing over all j and I_{j} we thus obtain,

\begin{matrix}Ψ_{f,α}(I) \\ = ∑_{j=0}^{∞} ∑_{I_{j}∈D_{j}(I)} 2^{(2α−n)j}Φ_{f}(I_{j}) \\ ≤ 2 ∑_{j=0}^{∞} ∑_{J∈D_{j}^{∗}} 2^{(2α−n)j}Φ_{f}(J) + ∑_{j=0}^{∞} ∑_{J∈D_{j}^{0}} 2^{(2α−n)j}|f(J) − f(J + 2^{−j}e_{1})|^{2}, \end{matrix} (5.13)

where D_{j}^{0} = {J ∈ D_{j}^{∗} : J + 2^{−j}e_{1} ∈ D_{j}^{∗}}.

The first double sum on the right hand side of (5.13) is just Ψ_{f,α}(I′) + Ψ_{f,α}(I′′). In order to estimate the final sum, consider a cube J ∈ D_{j}^{0} for some j ≥ 0. Let I^{∗} be the smallest dyadic cube that contains J ∪(J +2^{−j}e_{1}), and let the edge length of I^{∗} be 2^{−j+m}, where m ≥ 1. Moreover, for 0 ≤ l ≤ m, let J_{l} and K_{l} be the dyadic cubes of edge length 2^{−j+l} that contain J and J + 2^{−j}e_{1}, respectively; thus J = J_{0} ⊂ J_{1} ⊂ ⋯ ⊂ J_{m} = I^{∗} and J + 2^{−j}e_{1} = K_{0} ⊂ ⋯ ⊂ K_{m} = I^{∗}. Using the Cauchy–Schwarz inequality and (5.6), we obtain

\begin{matrix}|f(J) − f(J + 2^{−j}e_{1})|^{2} \\ ≤ (∑_{l=1}^{m} |f(J_{l−1}) − f(J_{l})| + ∑_{l=1}^{m} |f(K_{l}) − f(K_{l−1})|)^{2} \\ ≤ (2 ∑_{l=1}^{∞} l^{−2})(∑_{l=1}^{m} l^{2}|f(J_{l}) − f(J_{l−1})|^{2} + ∑_{l=1}^{m} l^{2}|f(K_{l}) − f(K_{l−1})|^{2}) \\ ≤ C ∑_{l=1}^{m} l^{2}(Φ_{f}(J_{l}) + Φ_{f}(K_{l})). \end{matrix} (5.14)

If J ∪ (J + 2^{−j}e_{1}) ⊆ I′ or I′′, then |I^{∗}| ≤ 1 and m ≤ j. If J ⊆ I′ and J + 2^{−j}e_{1} ⊆ I′′, however, then I^{∗} = [0, 2)^{n} and m = j + 1; in this case we modify (5.14) by observing that J_{j} = I′ and K_{j} = I′′ which by the same argument yields

\begin{matrix}|f(J) − f(J + 2^{−j}e_{1})|^{2} \\ ≤ C ∑_{l=1}^{j} l^{2}(Φ_{f}(J_{l}) + Φ_{f}(K_{l})) + C|f(I′) − f(I′′)|^{2}. \end{matrix}  (5.15)

We now keep j ≥ 0 fixed and sum (5.14) or (5.15) for J ∈ D_{j}^{0}. We observe that the cubes J_{l} and K_{l} that appear belong to D_{j−l}^{∗}, with 1 ≤ l ≤ j. Moreover, each dyadic cube J′ = ∏_{1}^{n}[a_{i}, b_{i}) in D_{j−l}^{∗} appears as a J_{l} or a K_{l} only for the J ∈ D_{j}^{∗} that are adjacent to either the face x_{1} = a_{1}, the face x_{1} = b_{1} or the mid plane x_{1} = (a_{1} + b_{1})/2, and there are thus at most 3 · 2^{(n−1)l} such J. Consequently, since (5.15) is used for 2^{(n−1)j} cubes J,

\begin{matrix}∑_{D_{j}^{0}} |f(J) − f(J + 2^{−j}e_{1})|^{2} \\ ≤ C ∑_{l=1}^{j} ∑_{J∈D_{j−l}^{∗}} 2^{(n−1)l}l^{2}Φ_{f}(J) + C2^{(n−1)j}|f(I′) − f(I′′)|^{2}. \end{matrix}

Summing over j we finally obtain, substituting j = k + l and observing that ∑_{l=1}^{∞} l^{2}2^{(2α−1)l} < ∞,

∑_{j=0}^{∞} ∑_{J∈D_{j}^{0}} 2^{(2α−n)j}|f(J) − f(J + 2^{−j}e_{1})|^{2}

≤ C ∑_{l=1}^{∞} ∑_{k=0}^{∞} ∑_{J∈D_{k}(I′)∪D_{k}(I′′)} 2^{(2α−n)k+(2α−n)l}2^{(n−1)l}l^{2}Φ_{f}(J)

+ C ∑_{j=0}^{∞} 2^{(2α−n)j}2^{(n−1)j}|f(I′) − f(I′′)|^{2}

= CΨ_{f,α}(I′) + CΨ_{f,α}(I′′) + C|f(I′) − f(I′′)|^{2}, which by (5.13) completes the proof of (5.12).

Lemma 5.7. Let −∞ < α < 1/2. If I and J are any cubes with I ⊂ J and ℓ(I) = \frac{1}{2}ℓ(J), then Ψ_{f,α}(I) ≤ CΨ_{f,α}(J).

Proof. Suppose, for notational convenience, that J = [0, 2)^{n} and that I = ∏_{1}^{n}[a_{i}, a_{i} + 1), with 0 ≤ a_{i} ≤ 1. Let w(I) = Card{a_{i} : 0 < a_{i} < 1}; we prove the result by induction on w(I).

If w(I) = 0, then I is one of the subcubes in D_{1}(J), and the result follows by Lemma 5.1. Otherwise, choose i such that 0 < a_{i} < 1 and let I′ and I′′ be the cubes obtained from I by replacing [a_{i}, a_{i} + 1) by [0, 1) and [1, 2), respectively, keeping the other coordinates unchanged. Then I′ and I′′ are adjacent, I ⊂ I′ ∪ I′′ ⊂ J and w(I′) = w(I′′) = w(I) − 1. By Lemma 5.6, Ψ_{f,α}(I) ≤ C(Ψ_{f,α}(I′) + Ψ_{f,α}(I′′) + |f(I′) − f(I′′)|^{2}). Since Ψ_{f,α}(I′), Ψ_{f,α}(I′′) ≤ CΨ_{f,α}(J) by the induction hypothesis and |f(I′) − f(I′′)|^{2} ≤ CΦ_{f}(J) ≤ CΨ_{f,α}(J) by (5.4), the result follows.

Lemma 5.8. If −∞ < α < 1/2, then, for any cube I and f ∈ L^{2}(I),

Ψ_{f,α}(I) ≍ [ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy.

Proof. One inequality is shown in Lemma 5.3.

To show the other, we may assume that f is extended to ℝ^{n} with f constant = f(I) outside I. If t ∈ ℝ^{n} with |t|_{∞} < ℓ(I), then I + t is contained in a cube J with ℓ(J) = 2ℓ(I) such that I ∈ D_{1}(J). Since the extended f is constant on the 2^{n} − 1 other cubes in D_{1}(J), Lemma 5.1 yields Ψ_{f,α}(J) ≤ CΨ_{f,α}(I), and then Lemma 5.7 yields

Ψ_{f,α}(I + t) ≤ CΨ_{f,α}(J) ≤ CΨ_{f,α}(I).

Finally, Lemma 5.4 yields

[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy ≤ CΨ_{f,α}(I).

Remark 5.9. We do not know whether any of Lemmas 5.6, 5.7 and 5.8 holds for α ≥ 1/2.

6. Wavelets

The purpose of this section is to observe that the well known characterization of BMO(ℝ^{n}) in terms of an orthonormal wavelet basis, see Meyer [13, p. 154], extends to Q_{α}(ℝ^{n}). In the sequel, we let

ψ_{j,k̂,l}(x) = 2^{jn/2}ψ_{l}(2^{j}x − k̂), j ∈ ℤ, k̂ ∈ ℤ^{n}, l = 1, …, 2^{n} − 1.

be a 1-regular orthonormal wavelet basis as in [13, Chapter 3]. We adopt the shorter notation ψ_{j,k̂,l} = ψ_{λ}, with λ ∈ Λ = ℤ × ℤ^{n} × {1, …, 2^{n} − 1}. For simplicity we consider only the case of wavelets of compact support, and assume thus that the wavelets satisfy the conditions in [13, p. 108]; in particular supp ψ_{λ} ⊆ mI(λ), where m is a constant (fixed throughout this section) and for every λ = (j, k̂, l) ∈ Λ we use I(λ) to denote the dyadic cube

I(λ) = {x : 2^{j}x − k̂ ∈ [0, 1)^{n}}.  (6.1)

We write ℓ(λ) = ℓ(I(λ)).

Recall that D is the set of all dyadic cubes in ℝ^{n}. For I ∈ D and a sequence a = (a(λ))_{λ∈Λ}, let

T_{a,α}(I) = |I|^{−1} ∑_{I(λ)⊆I}(\frac{ℓ(I)}{ℓ(λ)})^{2α}|a(λ)|^{2}.

Lemma 6.1. If α > 0, then for every dyadic cube I and sequence a,

T_{a,α}(I) ≍ ∑_{k=0}^{∞} 2^{(2α−n)k} ∑_{J∈D_{k}(I)} T_{a,0}(J).

Proof. The right hand side equals

∑_{J∈D(I)} ∑_{I(λ)⊆J}(\frac{ℓ(J)}{ℓ(I)})^{n−2α}|J|^{−1}|a(λ)|^{2}

= ℓ(I)^{2α−n} ∑_{I(λ)⊆I} |a(λ)|^{2} ∑_{\begin{matrix}J∈D(I) \\ J⊇I(λ) \end{matrix}} ℓ(J)^{−2α}

≍ ℓ(I)^{2α−n} ∑_{I(λ)⊆I} |a(λ)|^{2}ℓ(λ)^{−2α}.

Observe that the wavelet coefficients of functions in BMO(ℝ^{n}) are characterized by

sup_{I∈D(ℝ^{n})}T_{a,0}(I) < ∞

[13]. This extends to Q_{α}(ℝ^{n}) as follows.

Theorem 6.2. Let 0 < α < 1. If f ∈ Q_{α}(ℝ^{n}), then the sequence of its wavelet coefficients

a(λ) = (f, ψ_{λ}) = ∫_{ℝ^{n}} f(x)\overline{ψ_{λ}(x)} dx,

satisfies

sup_{I∈D} T_{a,α}(I) < ∞.  (6.2)

Conversely, every sequence a(λ) satisfying (6.2) is the sequence of wavelet coefficients of a unique f ∈ Q_{α}(ℝ^{n}); moreover, ∥f∥_{Q_{α}(ℝ^{n})} ≍ sup_{I∈D} T_{a,α}(I)^{1/2}.

Proof. First, let f ∈ Q_{α}(ℝ^{n}) and I ∈ D. For J ∈ D_{k}(I), we write

f = f_{mJ} + (f − f_{mJ})χ_{mJ} + (f − f_{mJ})χ_{ℝ^{n}\mJ} = f_{1} + f_{2} + f_{3}.

Since supp ψ_{λ} ⊆ mI(λ), (f_{3}, ψ_{λ}) = 0 if I(λ) ⊆ J. On the other hand, the integral of each wavelet ψ_{λ} is zero. So (f, ψ_{λ}) = (f_{2}, ψ_{λ}), and, using (5.2),

∑_{I(λ)⊆J} |(f, ψ_{λ})|^{2}

≤ ∑_{λ} |(f_{2}, ψ_{λ})|^{2}

= ∥f_{2}∥_{L^{2}(ℝ^{n})}^{2}

= |mJ|Φ_{f}(mJ)

= \frac{1}{2|mJ|} ∫_{mJ} ∫_{mJ} |f(x) − f(y)|^{2} dx dy.

This gives that for J ∈ D_{k}(I),

T_{a,0}(J)

= \frac{1}{|J|} ∑_{I(λ)⊆J} |a(λ)|^{2}

≤ \frac{1}{|J||mJ|} ∫_{mJ} ∫_{mJ} |f(x) − f(y)|^{2} dx dy.

Using Lemma 6.1, we obtain in the same manner as for Lemma 5.3

T_{a,α}(I)

≤ C ∑_{k=0}^{∞} 2^{(2α−n)k} ∑_{J∈D_{k}(I)} T_{a,0}(J)

≤ C ∫_{mI} ∫_{mI} |f(x) − f(y)|^{2} ∑_{k=0}^{∞} ∑_{J∈D_{k}(I)} 2^{(2α−n)k}|J|^{−2}χ_{mJ}(x)χ_{mJ}(y) dx dy

≤ Cℓ(I)^{2α−n} ∫_{mI} ∫_{mI} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy

≤ C∥f∥_{Q_{α}(ℝ^{n})}^{2}.

Thus (6.2) follows.

Conversely, suppose that (6.2) holds; multiplying f by a constant, we may assume that T_{a,α}(I) ≤ 1 for every dyadic cube I. In particular, T_{a,0}(I) ≤ T_{a,α}(I) ≤ 1 for every dyadic cube I, and thus, by [13, Section 5.6],

f = ∑_{λ} a(λ)ψ_{λ} ∈ BMO(ℝ^{n}),

with the sum converging e.g. in the weak^{∗} topology on BMO(ℝ^{n}). We will show that f ∈ Q_{α}(ℝ^{n}), with ∥f∥_{Q_{α}(ℝ^{n})} ≤ C.

Fix a (not necessarily dyadic) cube I of dyadic edge length and consider a subcube J ∈ D(I). Let Λ_{0}(J) = {λ ∈ Λ : mI(λ) ∩ J ≠ ∅} and partition this set into

Λ_{1} = Λ_{1}(J) = {λ ∈ Λ_{0}(J) : |I(λ)| ≤ |J|},
Λ_{2} = Λ_{2}(J) = {λ ∈ Λ_{0}(J) : |J| < |I(λ)| ≤ |I|},
Λ_{3} = Λ_{3}(J) = {λ ∈ Λ_{0}(J) : |I| < |I(λ)|}.

Since ψ_{λ} = 0 on J unless λ ∈ Λ_{0} we have, on J, f = f_{1} + f_{2} + f_{3}, where

f_{i} = ∑_{λ∈Λ_{i}} a(λ)ψ_{λ}.

Hence, using the Cauchy–Schwarz inequality,

Φ_{f}(J) ≤ 3(Φ_{f_{1}}(J) + Φ_{f_{2}}(J) + Φ_{f_{3}}(J)).  (6.3)

We treat the three terms separately. First,

Φ_{f_{1}}(J) ≤ \frac{1}{|J|}∥f_{1}∥_{L^{2}}^{2} = \frac{1}{|J|} ∑_{λ∈Λ_{1}} |a(λ)|^{2}.  (6.4)

Secondly, |∇ψ_{λ}| ≤ Cℓ(λ)^{−1}|I(λ)|^{−1/2}, and thus

|f_{2}(x) − f_{2}(y)| ≤ C ∑_{λ∈Λ_{2}} |a(λ)|ℓ(λ)^{−1}|I(λ)|^{−1/2}|x − y|.

Consequently, letting ε = 1 − α (for example) and using the Cauchy–Schwarz inequality,

Φ_{f_{2}}(J)

≤ Cℓ(J)^{2}( ∑_{λ∈Λ_{2}} |a(λ)|ℓ(λ)^{−1}|I(λ)|^{−1/2})^{2}

≤ Cℓ(J)^{2} ∑_{λ∈Λ_{2}} |a(λ)|^{2}ℓ(λ)^{−2}|I(λ)|^{−1}(\frac{ℓ(λ)}{ℓ(J)})^{ε} ∑_{λ∈Λ_{2}}(\frac{ℓ(J)}{ℓ(λ)})^{ε}.

If λ ∈ Λ_{2}, then I(λ) is a dyadic cube contained in a cube with the same center as J and edge length (m + 1)ℓ(λ) + ℓ(J) ≤ (m + 2)ℓ(J). Hence, for each k ≥ 1, there are at most (m + 2)^{n} such cubes I(λ) with ℓ(λ) = 2^{k}ℓ(J). Moreover, there is constant (viz. 2^{n} − 1) number of different λ for each such cube, and thus Card{λ ∈ Λ_{2} : ℓ(λ) = 2^{k}ℓ(J)} ≤ C for each k ≥ 1. Consequently,

∑_{λ∈Λ_{2}}(\frac{ℓ(J)}{ℓ(λ)})^{ε} ≤ ∑_{k=1}^{∞} C2^{−kε} ≤ C,

and thus

Φ_{f_{2}}(J) ≤ C ∑_{λ∈Λ_{2}} |a(λ)|^{2}(\frac{ℓ(J)}{ℓ(λ)})^{2−ε}|I(λ)|^{−1}.  (6.5)

Thirdly, we similarly have |f_{3}(x) − f_{3}(y)|

≤ C ∑_{λ∈Λ_{3}} |a(λ)|ℓ(λ)^{−1}|I(λ)|^{−1/2}|x − y|

≤ C|x − y| ∑_{λ∈Λ_{3}} ℓ(λ)^{−1}

using

|a(λ)||I(λ)|^{−1/2} ≤ T_{a,α}^{1/2}(I(λ)) ≤ 1.

Again, there is a bounded number of terms for each ℓ(λ), and now ℓ(λ) = 2^{k}ℓ(I), k ≥ 1; hence

|f_{3}(x) − f_{3}(y)| ≤ C|x − y|ℓ(I)^{−1}

and thus

Φ_{f_{3}}(J) ≤ Cℓ(J)^{2}ℓ(I)^{−2}.  (6.6)

Consequently, by (6.3), (6.4), (6.5) and (6.6),

Φ_{f}(J)

≤ C \frac{1}{|J|} ∑_{λ∈Λ_{1}} |a(λ)|^{2}

+ C ∑_{λ∈Λ_{2}} |a(λ)|^{2}(\frac{ℓ(J)}{ℓ(λ)})^{2−ε}|I(λ)|^{−1}

+ Cℓ(J)^{2}ℓ(I)^{−2}.

Summing over J ∈ D(I) we obtain, cf. (5.5),

\begin{matrix}Ψ_{f,α}(I) \\ = ∑_{J∈D(I)}(\frac{ℓ(J)}{ℓ(I)})^{n−2α}Φ_{f}(J) \\ ≤ C ∑_{J∈D(I)} ∑_{λ∈Λ_{1}(J)} |a(λ)|^{2}ℓ(J)^{−2α}ℓ(I)^{−n+2α} \\ + C ∑_{J∈D(I)} ∑_{λ∈Λ_{2}(J)} |a(λ)|^{2}ℓ(J)^{n−2α+2−ε}ℓ(λ)^{ε−2−n}ℓ(I)^{2α−n} \\ + C ∑_{J∈D(I)}(\frac{ℓ(J)}{ℓ(I)})^{n−2α+2}. \end{matrix}  (6.7)

The final sum equals

∑_{j=0}^{∞} 2^{nj}(2^{−j})^{n−2α+2} = ∑_{j=0}^{∞} 2^{−(2−2α)j} = C.

In the two double sums, we interchange the order of summation. If λ occurs there, then ℓ(λ) ≤ ℓ(I) and mI(λ) ∩ I ≠ ∅; thus, if we let E(I) be the set of dyadic dubes I′ of the same size as I with mI′ ∩ I ≠ ∅, it follows that I(λ) ∈ D(I′) for some I′ ∈ E(I).

Fix one such λ, with ℓ(λ) = 2^{−k}ℓ(I). For each j ≤ k, there are at most C cubes J ∈ D_{j}(I) with λ ∈ Λ_{1}(J), each contributing 2^{2αj}|I|^{−1}|a(λ)|^{2} to the first double sum in (6.7). Similarly, for each j > k, there are at most C2^{(j−k)n} cubes J ∈ D_{j}(I) with λ ∈ Λ_{2}(J), each contributing 2^{−j(n−2α+2−ε)−k(ε−2−n)}|I|^{−1}|a(λ)|^{2} to the second double sum. Together these yield at most

\begin{matrix}C|I|^{−1}|a(λ)|^{2}(∑_{j=0}^{k} 2^{2αj} + ∑_{j=k+1}^{∞} 2^{−(2−2α−ε)(j−k)+2αk}) \\ ≤ C2^{2αk}|I|^{−1}|a(λ)|^{2}. \end{matrix}

As a consequence, (6.7) yields

Ψ_{f,α}(I)

≤ C ∑_{I′∈E(I)} ∑_{k=0}^{∞} ∑_{I(λ)∈D_{k}(I′)} 2^{2αk}|I|^{−1}|a(λ)|^{2} + C

= C ∑_{I′∈E(I)} T_{a,α}(I′) + C ≤ C.

We have proved that Ψ_{f,α}(I) ≤ C for every cube I of dyadic edge length. Since the same estimate applies to every translate I + t, Lemma 5.4 shows that for every cube I of dyadic edge length,

[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{|x − y|^{n+2α}} dx dy ≤ C,

finally, Remark 1.1 yields f ∈ Q_{α}(ℝ^{n}) and ∥f∥_{Q_{α}(ℝ^{n})} ≤ C.

Uniqueness of f follows from the uniqueness in BMO(ℝ^{n}); if f, g ∈ Q_{α}(ℝ^{n}) ⊂ BMO(ℝ^{n}) have the same wavelet coefficients, then they define the same linear functional on H^{1}(ℝ^{n}) and thus f = g as elements of BMO(ℝ^{n}) (i.e. modulo constants), see again [13, Section 5.6].

7. The dyadic counterpart

We define a dyadic analogue of Q_{α}(ℝ^{n}) and give some results relating the two spaces.

Once again, recall that D is the set of all dyadic cubes in ℝ^{n}, and define the dyadic distance δ(x, y) between two points in ℝ^{n} by

δ(x, y) = inf{ℓ(I) : x, y ∈ I ∈ D}.

(The dyadic distance is infinite between points in different octants.) Note that

|x − y| ≤ \sqrt{d} δ(x, y).  (7.1)

The space Q_{α}^{d}(ℝ^{n}) is defined as the space of all (measurable) functions f on ℝ^{n} such that

∥f∥_{Q_{α}^{d}(ℝ^{n})}^{2} = sup_{I∈D}[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{δ(x, y)^{n+2α}} dx dy < ∞.  (7.2)

Note that ∥f∥_{Q_{α}^{d}(ℝ^{n})} = 0 if and only if f is a.e. constant in each octant. Q_{α}^{d}(ℝ^{n}) is a Banach space if we regard it as a space of functions modulo such functions.

We have analogues of Lemma 5.8 and Theorem 5.5.

Lemma 7.1. Let −∞ < α < ∞. Then, for any cube I ∈ D and f ∈ L^{2}(I),

Ψ_{f,α}(I) ≍ [ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{δ(x, y)^{n+2α}} dx dy.

Proof. Suppose first α > −n/2. If I is a dyadic cube and x, y ∈ I, then x, y ∈ J for some J ∈ D_{k}(I) if and only if δ(x, y) ≤ 2^{−k}ℓ(I), and thus, by (5.9),

\begin{matrix}κ_{I}(x, y) \\ = \frac{1}{2} ∑_{2^{k}≤ℓ(I)/δ(x,y)} 2^{(2α+n)k}|I|^{−2} \\ ≍ (\frac{ℓ(I)}{δ(x, y)})^{2α+n}|I|^{−2} \\ = ℓ(I)^{2α−n}δ(x, y)^{−2α−n}. \end{matrix}  (7.3)

The result follows by (5.8).

If α ≤ −n/2, this argument yields the inequality

Ψ_{f,α}(I) ≥ C[ℓ(I)]^{2α−n} ∫_{I} ∫_{I} \frac{|f(x) − f(y)|^{2}}{δ(x, y)^{n+2α}} dx dy.

The opposite inequality follows by Lemma 5.3 and (7.1).

Theorem 7.2. Let −∞ < α < ∞. Then f ∈ Q_{α}^{d}(ℝ^{n}) if and only if

sup_{I∈D} Ψ_{f,α}(I) < ∞.

Moreover, sup_{I∈D}[Ψ_{f,α}(I)]^{1/2} is a norm on Q_{α}^{d}(ℝ^{n}), equivalent to ∥f∥_{Q_{α}^{d}(ℝ^{n})} as defined above.

Proof. Immediate by Lemma 7.1.

We have also an analogue of Theorem 2.3. Let BMO^{d}(ℝ^{n}) be the dyadic BMO space defined by {f ∈ L_{loc}^{2}(ℝ^{n}) : sup_{I∈D} Φ_{f}(I) < ∞}.

Theorem 7.3.

(i) Q_{α}^{d}(ℝ^{n}) is decreasing in α, i.e. Q_{α}^{d}(ℝ^{n}) ⊇ Q_{β}^{d}(ℝ^{n}) if α < β. (ii) If α > n/2, then Q_{α}^{d}(ℝ^{n}) contains only functions that are a.e. constant

in each octant, i.e. Q_{α}^{d}(ℝ^{n}) = {0} (as a Banach space). (iii) If α < 0, then Q_{α}^{d}(ℝ^{n}) = BMO^{d}(ℝ^{n}).

Proof. (i) follows directly by δ(x, y) ≤ ℓ(I) for x, y ∈ I ∈ D.

For (ii), suppose that f ∈ Q_{α}^{d}(ℝ^{n}), with α > n/2. If J is a dyadic cube, and J_{k} the dyadic cube containing J with ℓ(J_{k}) = 2^{k}ℓ(J), then Ψ_{f,α}(J_{k}) ≥ 2^{(2α−n)k}Φ_{f}(J), and by letting k → ∞ we obtain Φ_{f}(J) = 0, J ∈ D.

Finally, (iii) follows by Lemmas 7.1 and 5.2.

Remark 7.4. The inclusion in (i) is strict unless n/2 < α < β or α < β < 0, as can be seen by the following example.

Example 7.5. For a cube I, let I_{−} and I_{+} denote its left and right halves, separated by a hyperplane x_{1} = a through the center of I. Let {a_{k}}_{1}^{∞} be a sequence with ∑_{1}^{∞} |a_{k}|^{2} < ∞ and define f = ∑_{k=1}^{∞} ∑_{J∈D_{k}(I_{0})} a_{k}(χ_{J_{+}} − χ_{J_{−}}), where I_{0} is any fixed cube. It is then easily seen (cf. Theorem 7.11 below) that

f ∈ Q_{α}^{d}(ℝ^{n}) ⟺ \begin{cases}∑_{1}^{∞} |a_{k}|^{2} < ∞, & α < 0, \\ ∑_{1}^{∞} k|a_{k}|^{2} < ∞, & α = 0, \\ ∑_{1}^{∞} 2^{2αk}|a_{k}|^{2} < ∞, & 0 < α ≤ n/2, \\ every a_{k} = 0, & n/2 < α. \end{cases}

In particular, Q_{n/2}^{d}(ℝ^{n}) ≠ {0}, so there is no cut-off at α = 1 as for the non-dyadic spaces (Theorem 2.3).

Remark 7.6. If one instead considers Q_{α}^{d} defined on a cube, cf. Remark 1.2, then the cut-off at α = n/2 in Theorem 7.3(ii) disappears, and the space is non-trivial for arbitrarily large α.

Relations between Q_{α}(ℝ^{n}) and Q_{α}^{d}(ℝ^{n}). We first observe that if we consider all cubes I with dyadic edge lengths in Theorem 7.2, instead of just dyadic cubes, we obtain instead Q_{α}(ℝ^{n}).

Lemma 7.7. Let −∞ < α < ∞. Then Q_{α}(ℝ^{n}) equals the space of all functions f on ℝ^{n} such that sup_{I} Ψ_{f,α}(I) < ∞, where I ranges over the set of all cubes in ℝ^{n} with dyadic edge lengths. Moreover, ∥f∥_{Q_{α}(ℝ^{n})} ≍ sup_{I} Ψ_{f,α}(I)^{1/2}.

Proof. Immediate by Remark 1.1 and Lemmas 5.3 and 5.4.

Let τ_{t} denote the translation operator τ_{t}f(x) = f(x − t).

Theorem 7.8. Let −∞ < α < ∞. Then f ∈ Q_{α}(ℝ^{n}) if and only if τ_{t}f ∈ Q_{α}^{d}(ℝ^{n}) for all t ∈ ℝ^{n} and sup_{t} ∥τ_{t}f∥_{Q_{α}^{d}(ℝ^{n})} < ∞. Moreover, ∥f∥_{Q_{α}(ℝ^{n})} ≍ sup_{t} ∥τ_{t}f∥_{Q_{α}^{d}(ℝ^{n})}.

Proof. Since every cube of dyadic edge length is the translate of a dyadic cube, Lemma 7.7 and Theorem 7.2 show that

∥f∥_{Q_{α}(ℝ^{n})}^{2}

≍ sup_{t∈ℝ^{n}} sup_{I∈D} Ψ_{f,α}(I − t)

= sup_{t∈ℝ^{n}} sup_{I∈D} Ψ_{τ_{t}f,α}(I)

≍ sup_{t∈ℝ^{n}} ∥τ_{t}f∥_{Q_{α}^{d}(ℝ^{n})}^{2},

and the result follows.

Theorem 7.9. Let −∞ < α < 1/2. Then Q_{α}(ℝ^{n}) = Q_{α}^{d}(ℝ^{n}) ∩ BMO(ℝ^{n}).

Proof. The inclusion Q_{α}(ℝ^{n}) ⊆ Q_{α}^{d}(ℝ^{n}) ∩ BMO(ℝ^{n}) follows directly from the definitions, (7.1) and Theorem 2.3(iii).

Conversely, suppose that f ∈ Q_{α}^{d}(ℝ^{n})∩BMO(ℝ^{n}). Let I be a cube of dyadic edge length. Then there exist a family of 2^{n} dyadic cubes I_{i} of the same size as I, such that the union ⋃_{i} I_{i} is a cube J with ℓ(J) = 2ℓ(I) and I ⊂ J. (J is not necessarily a dyadic cube.) By Lemma 5.7 and (5.5),

Ψ_{f,α}(I) ≤ CΨ_{f,α}(J)

= C ∑_{i=1}^{2^{n}} Ψ_{f,α}(I_{i}) + CΦ_{f}(J)

≤ C∥f∥_{Q_{α}^{d}(ℝ^{n})}^{2} + C∥f∥_{BMO(ℝ^{n})}^{2}.

Hence Ψ_{f,α}(I) is bounded uniformly for all cubes I of dyadic edge length, and the result follows by Lemma 7.7.

Remark 7.10. We do not know whether the condition α < 1/2 in Theorem 7.9 is necessary. The theorem fails at least for 1 ≤ α ≤ n/2, since then, cf. Example 7.5, χ_{I_{+}} − χ_{I_{−}} ∈ Q_{α}^{d}(ℝ^{n}) ∩ BMO(ℝ^{n}) for any cube I, while Q_{α}(ℝ^{n}) = {0} by Theorem 2.3.

The Haar system. Let h_{λ}, λ ∈ Λ, be the n-dimensional Haar system. Functions in BMO^{d}(ℝ^{n}) can be described by the Haar system, and the characterization coincides with one given above of BMO(ℝ^{n}) using a wavelet basis [13, p. 157]. Similarly, for Q_{α}^{d}(ℝ^{n}) we have

Theorem 7.11. Let α > 0. If f ∈ Q_{α}^{d}(ℝ^{n}), then the sequence of its Haar coefficients a(λ) = (f, h_{λ}), λ ∈ Λ, satisfies

sup_{I∈D} T_{a,α}(I) < ∞.  (7.4)

Conversely, every sequence a(λ) satisfying (7.4) is the sequence of Haar coefficients of a unique f ∈ Q_{α}^{d}(ℝ^{n}).

Proof. If a(λ) = (f, h_{λ}) and I is a dyadic cube, then

(f − f(I))χ_{I} = ∑_{I(λ)⊆I} a(λ)h_{λ}

and thus

Φ_{f}(I) = |I|^{−1} ∑_{I(λ)⊆I} |a(λ)|^{2} = T_{a,0}(I).

It follows by the definition (5.5) and Lemma 6.1 that

Ψ_{f,α}(I) ≍ T_{a,α}(I),

and the result follows by Theorem 7.2.

Let U be the isometry of L^{2}(ℝ^{n}) given by U(ψ_{λ}) = h_{λ}. Then Theorems 6.2 and 7.11 show that the operator U can be extended to an isomorphism between Q_{α}(ℝ^{n}) and Q_{α}^{d}(ℝ^{n}).

Corollary 7.12. Let 0 < α < 1. Then the Banach spaces Q_{α}(ℝ^{n}) and Q_{α}^{d}(ℝ^{n}) are isomorphic; more precisely, the map U(∑ a(λ)ψ_{λ}) = ∑ a(λ)h_{λ} is an isomorphism (with the sums interpreted formally or as converging in suitable weak topologies).

8. Some problems

In this section, we would like to mention some open problems.

John–Nirenberg type inequality. The John–Nirenberg distribution inequality [10] says that if f ∈ BMO(ℝ^{n}) then for any cube I and any t > 0,

m_{I}(t) = |\{x ∈ I : |f(x) − f(I)| > t\}| ≤ C|I| exp(\frac{−ct}{∥f∥_{BMO(ℝ^{n})}}).  (8.1)

We hope to explore the following

Problem 8.1. Let α ∈ (0, 1). Give a John–Nirenberg type inequality for Q_{α}(ℝ^{n}).

With (8.1) and the local behaviour of Q_{α}(ℝ^{n}), we may conjecture that if f ∈ Q_{α}(ℝ^{n}) then

∑_{k=0}^{∞} 2^{(2α−n)k} ∑_{J∈D_{k}(I)} \frac{m_{J}(t)}{|J|} ≤ Ct^{−1} exp(\frac{−ct}{∥f∥_{Q_{α}(ℝ^{n})}}).  (8.2)

Duality. Corresponding to the important result BMO(ℝ^{n})=[H^{1}(ℝ^{n})]^{∗}, a problem of Q_{α}(ℝ^{n})-type is

Problem 8.2. Let α ∈ (0, 1). Find a predual of Q_{α}(ℝ^{n}).

Fefferman–Stein type decomposition. Fefferman and Stein [6] proved that f ∈ BMO(ℝ^{n}) if and only if there are φ_{j} ∈ L^{∞}(ℝ^{n}) such that

f = φ_{0} + ∑_{j=1}^{n} R_{j}(φ_{j}),

where R_{j}φ = φ ∗ (x_{j}/|x|^{n+1}) are the Riesz transforms. So, we naturally pose

Problem 8.3. Let α ∈ (0, 1). Give a Fefferman–Stein type decomposition of Q_{α}(ℝ^{n}).

Note that the space Q_{p}(∂△) has a decomposition of Fefferman–Stein type [14, Theorem 1.2].

Quasiconformal homeomorphism. Reimann [17] proved that if φ is an ACL-homeomorphism from ℝ^{n}, n ≥ 2, to itself, then the composition operator C_{φ}, defined by C_{φ}f = f ∘ φ, is bounded on BMO(ℝ^{n}) if and only if φ is quasiconformal. On the line, it was noted by P. Jones that the quasi-conformal ACL-homeomorphisms φ are those which satisfy the A_{∞}-condition for log φ′ (cf. p.15 in [4]). With Theorem 2.1, we therefore have the following

Problem 8.4. Let α ∈ (0, 1). Prove or disprove C_{φ} is bounded on Q_{α}(ℝ^{n}).

Quasiconformal extension. We can introduce the concept of Q_{α}(Ω)—a Q space on a general domain Ω ⊆ ℝ^{n}, by considering only cubes I ⊆ Ω in (1.3). Jones’s theorem [11] tells us that for a Jordan domain Ω ⊂ R^{2}, BMO(Ω) = BMO(R^{2})|_{Ω} (the restriction of BMO(ℝ^{n}) to Ω) if and only if ∂Ω (the boundary of Ω) is a quasi-circle. So, our question is

Problem 8.5. Let α ∈ (0, 1). Find a geometric property of ∂Ω such that Q_{α}(Ω) = Q_{α}(R^{2})|_{Ω}.

Further problems can be posed via [4].

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Department of Mathematics, Uppsala University, PO Box 480, S-751 06 Uppsala, Sweden

E-mail address: matts.essen@math.uu.se

Department of Mathematics, Uppsala University, PO Box 480, S-751 06 Uppsala, Sweden

E-mail address: svante.janson@math.uu.se

Department of Mathematics, Peking University, Beijing 100871, P. R. China

E-mail address: lzpeng@sxx0.math.pku.edu.cn

Department of Mathematics, Peking University, Beijing 100871, P. R. China

E-mail address: jxiao@sxx0.math.pku.edu.cn