Mixed martingale Hardy spaces

In this paper we consider the martingale Hardy spaces defined with the help of the mixed $L_{\pv}$-norm. Five mixed martingale Hardy spaces will be investigated: $H_{\pv}^{s}$, $H_{\pv}^S$, $H_{\pv}^M$, $\cP_{\pv}$ and $\cQ_{\pv}$. Several results are proved for these spaces, like atomic decompositions, Doob's inequality, boundedness, martingale inequalities and the generalization of the well-known Burkholder-Davis-Gundy inequality.


Introduction
The mixed Lebesgue spaces were introduced in 1961 by Benedek and Panzone [2].They considered the Descartes product (Ω, F , P) of the probability spaces (Ω i , F i , P i ), where Ω = d i=1 Ω i , F is generated by d i=1 F i and P is generated by d i=1 P i .The mixed L − → p -norm of the measurable function f is defined as a number obtained after taking successively the L p 1 -norm of f in the variable x 1 , the L p 2 -norm of f in the variable x 2 , . .., the L p d -norm of f in the variable x d .Some basic properties of the spaces L − → p were proved in [2], such as the well known Hölder's inequality or the duality theorem for L − → p -norm (see Lemma 2).The boundedness of operators on mixed-norm spaces have been studied for instance by Fernandez [28] and Stefanov and Torres [35].Using the mixed Lebesgue spaces, Lizorkin [31] considered Fourier integrals and estimations for convolutions.Torres and Ward [37] gave the wavelet characterization of the space L − → p (R n ).For more about mixed-norm spaces the papers [1,7,8,17,19,20,23,24,34,35] are referred.
The classical martingale Hardy spaces (H s p , H S p , H * p , P p , Q p ) have a long history and the results of this topic can be well applied in the Fourier analysis.In the celebrated work of Burkholder and Gundy [6] it was proved that the L p norms of the maximal function and the quadratic variation, that is the spaces H M p and H S p , are equivalent for 1 < p < ∞.In the same year, Davis [9] extended this result for p = 1.In the classical case, Herz [18] and Weisz [39] gave one of the most powerful techniques in the theory of martingale Hardy spaces, the so-called atomic decomposition.Some boundedness results, duality theorems, martingale inequalities and interpolation results can be proved with the help of atomic decomposition.Details for the martingale Hardy spaces can be found in Burkholder and Gundy [4], [5], [6], Garsia [11], Long [32] or Weisz [38,39].For the application of martingale Hardy spaces in Fourier analysis see Gát [12,13], Goginava [14,15] or Weisz [39,40].
In this paper we will introduce five mixed martingale Hardy spaces: In Section 3, Doob's inequality will be proved, that is, we will show that In Section 4, we give the atomic decomposition for the five mixed martingale Hardy spaces.Using the atomic decompositions and Doob's inequality, the boundedness of general σ-subadditive operators from H s − → p to L − → p , from P − → p to L − → p and from Q − → p to L − → p can be proved (see Theorems 7 and 8).With the help of these general boundedness theorems, several martingale inequalities will be proved in Section 5 (see Corollary 19).We will show, that if the stochastic basis (F n ) is regular, then the five martingale Hardy spaces are equivalent.As a consequence of Doob's inequality, the well-known Burkholder-Davis-Gundy inequality can be shown.Moreover, if the stochastic basis is regular, then the so-called martingale transform is bounded on L − → p .
We denote by C a positive constant, which can vary from line to line, and denote by C p a constant depending only on p.The symbol A ∼ B means that there exist constants α, β > 0 such that αA ≤ B ≤ βA.

Martingale Hardy spaces
Suppose that the σ-algebra The expectation and conditional expectation operators relative to F n are denoted by E and E n , respectively.An integrable sequence f = (f n ) n∈N is said to be a martingale if For n ∈ N, the martingale difference is defined by then f is called an L − → p -bounded martingale and it will be denoted by f ∈ L − → p .The map ν : Ω → N ∪ {∞} is called a stopping time relative to For a martingale f = (f n ) and a stopping time ν, the stopped martingale is defined by Let us define the maximal function, the quadratic variation and the conditional quadratic variation of the martingale f relative to (Ω, F , P, (F n ) n∈N ) by The set of the sequences (λ n ) n∈N of non-decreasing, non-negative and adapted functions with λ ∞ := lim n→∞ λ n is denoted by Λ.With the help of the previous operators, the mixed martingale Hardy spaces can be defined as follows:

Doob's inequality
In this section, we will prove that the maximal operator M is bounded on the space then there is a function g ∈ L − → p such that for all n ∈ N, f n = E n g.For k ∈ {1, . . ., d} and m ∈ N, let us denote where m stands in the k-th position.The conditional expectation operator relative to F ∞,...,∞,m,∞,...,∞ is denoted by E ∞,...,∞,m,∞,...,∞ .We need the following maximal operators: for an integrable function f , let For 0 < p < ∞ and w > 0, the weighted space L p (w) consists of all functions f , for which We need the following lemma.
Lemma 3 Let ϕ be a positive function.Then for all r > 1, M is bounded from Proof.It is easy to see, that the operator M is bounded from L ∞ (Mϕ) to L ∞ (ϕ).We will prove that M is bounded from L 1 (Mϕ) to L 1,∞ (ϕ) as well, where L 1,∞ (ϕ) denotes the weak-L 1 (ϕ) space.From this, it follows by interpolation (see e.g.[3]) that for all r > 1, the operator M is bounded from L r (Mϕ) to L r (ϕ), in oder words, (1) holds.Let ̺ > 0 arbitrary and let ν ̺ := inf {n ∈ N : which finishes the proof.

Now we prove that
Proof.We will prove the theorem by induction in d.If d = 1, then the function f has only 1 variable and the theorem holds (see, e.g., Weisz [39]).Suppose that the theorem is true for some fixed d ∈ N and for all 1 < − → p = (p 1 , . . ., p d ) < ∞ and f ∈ L − → p .For a function f with d variables and for the vector (p 1 , . . ., p k ), let us denote Using this notation, the condition of the induction can be written in the form We will show that if 1 < p d+1 < ∞, then for all f ∈ L (p 1 ,...,p d+1 ) , where f has d + 1 variable and the maximal operator M d+1 is taken in the variable ), and therefore So we get that Here the function M d+1 (T ∞ f ) has d variables: x 2 , . . ., x d+1 and the maximal operator is taken over the d-th variable, that is over x d+1 .Therefore by induction we have that (2) can be estimated by which means that so the theorem holds for p 1 = ∞.Now let choose a number r for which 1 < r < min{p 2 , . . ., p d , p d+1 }.It will be shown that It is easy to see that .
Let − → q := p 2 r , . . ., p d+1 r .Then the vector − → q has d coordinates and 1 We can suppose that ϕ > 0. Then = Since 1 < r < ∞, applying Lemma 3 for the variable x d+1 , we have that for all fixed x 1 , . . ., x d , Hence (4) can be estimated by Here M d+1 ϕ is a function with d variables, the vector ( − → q ) ′ has d coordinates such that 1 < ( − → q ) ′ < ∞ and the maximal operator is taken in the d-th coordinate, that is over the variable x d+1 .So, by induction we get that Therefore we can estimate (5) by Consequently, Combining the results ( 3) and ( 6) we get by interpolation that for all 1 < p 1 < ∞ Using induction, the proof is complete.
Remark 1 Using the proof of the previous theorem, this result can be generalized for − → p -s, for which its first k coordinates are ∞, but the others are strongly between 1 and ∞, that is, for − → p -s, with for some k ∈ {1, . . ., d}.Now, we can generalize the well-known Doob's inequality.Using the previous theorem, we get that the maximal operator M is bounded on and the proof is complete.
The following corollary is well-known for classical Hardy spaces with − → p = (p, . . ., p).
Proof.We prove the theorem for two dimensions and for the exponent − → p := (p, ∞), where 1 < p < ∞.The proof is similar for higher dimensions.Let us define the following sequence of functions Then for an arbitrary fixed y and for all fixed y / ∈ [2 −n , 1), the previous integral is 0. From this follows that for all n ∈ N, Hence we get that for all y ∈ [0, 2 −n ), [0,1) and therefore Remark 2 This counterexample proves also that M 2 is not bounded on L (p,∞) .Moreover, the counterexample shows also that the classical Hardy-Littlewood maximal operator considered in Huang et al. [21] is not bounded on L (p,∞) (cf.Lemma 3.5 in [21] and Lemma 4.8 in [34]).

Atomic decomposition
First of all we need the definition of the atoms.For − → p , a measurable function a is called an (s, − → p )-atom (or (S, − → p )-atom or (M, − → p )-atom) if there exists a stopping time τ such that Now we can give the atomic decomposition of the space and only if there exist a sequence (a k ) k∈Z of (s, − → p )-atoms and a sequence (µ k ) k∈Z of real numbers such and where 0 < t ≤ 1 and the infimum is taken over all decompositions of the form (8).
Proof.Let f ∈ H s − → p and let us define the following stopping time: Obviously f n can be written in the form Let If µ k = 0, then let a k n = 0.If n ≤ τ k , then a k n = 0 and naturally Moreover, (a k n ) is L 2 -bounded (see [39]), therefore there exists almost everywhere, by the dominated convergence theorem (see e.g.[2]) we get that Moreover, for all x ∈ Ω and for all 0 < t ≤ 1, Since the sets O k \ O k+1 are disjoint, we have Conversely, if f has a decomposition of the form (8), then and so for all 0 < t ≤ 1, , which proves the theorem.
For the classical martingale Hardy space H s p , this result is due to the second author (see [39]).For the spaces Q − → p and P − → p we can give similar decompositions.
and only if there exist a sequence (a k ) k∈Z of (M, − → p )-atoms (or (S, − → p )-atoms) and a sequence (µ k ) k∈Z of real numbers such that (8) holds and , where 0 < t ≤ 1 and the infimum is taken over all decompositions of the form (8).
Proof.Let f ∈ P − → p and (λ n ) n∈N be a sequence such that |f n | ≤ λ n−1 and λ ∞ = sup n λ n ∈ L − → p .Let the stopping time τ k be defined by and µ k and a k n be given by (10).Then again f n = k∈Z µ k a k n and we can prove as before that Conversely, assume that for some µ k and a k n , the martingale (f n ) can be written in the form (8). For n ∈ N, let us define It is clear, that (λ n ) is a nonnegative adapted sequence and for all n ∈ N, Therefore, for all 0 < t ≤ 1, The case of Q − → p is similar.
The stochastic basis (F n ) is said to be regular, if there exists R > 0 such that for all A ∈ F n there exists B ∈ F n−1 for which A ⊂ B and P(B) ≤ R P(A).If the stochastic basis is regular, then atomic decomposition can also be proved for the remainder two martingale Hardy spaces, H M − → p and H S − → p .
Theorem 6 Let 0 < − → p < ∞ and suppose that the stochastic basis ) if and only if there exist a sequence (a k ) k∈Z of (M, − → p )-atoms (or (S, − → p )-atoms) and a sequence (µ k ) k∈Z of real numbers such that (8) holds and , where 0 < t < min{p 1 , . . ., p d , 1} and the infimum is taken over all decompositions of the form (8).
Proof.We will prove the theorem only for H M − → p .The case of H S − → p is similar.Suppose that f ∈ H M − → p and for k ∈ Z, let us define the stopping time regular and the set I k,j := {̺ k = j} ∈ F j , there exist I k,j ∈ F j−1 such that I k,j ⊂ I k,j and P(I k,j ) ≤ R P(I k,j ).Let us define an other stopping time Then τ k is non-decreasing and using Lemma 4, which means that lim k→∞ P({τ k < ∞}) = 0, that is lim k→∞ τ k = ∞ almost everywhere and therefore lim k→∞ Let µ k and a k n be defined again as in (10).Then a k = (a k n ) is again an (M, − → p )-atom.For all 0 < t < min{p 1 , . . ., p d , 1}, , and therefore, using Lemma 2 and Hölder's inequality, , where we have used that {̺ k < ∞} = {Mf > 2 k }.So we have that where the right hand side can be estimated by f , similarly as in the proof of Theorem 4.
Conversely, if f has a decomposition of the form (8), then can be proved similarly as in Theorem 4.
Lemma 4 Let 0 < − → p < ∞ and suppose that the stochastic basis (F n ) is regular.If ̺ k and τ k are the stopping times defined in the proof of Theorem 6, then Proof.It is enough to prove that for some 0 < ε < min{1, p 1 , . . ., p d }, the inequality holds.Notice that . By Lemma 2, there exists a function g with g ( − → p /ε) ′ ≤ 1, such that Using Hölder's inequality with − → p /ε < r < ∞ and the regularity of (F n ), we obtain By Lemma 1, we get that where we have used that ( − → p /ε) ′ > 1 and therefore M is bounded on L ( − → p /ε) ′ .The proof is complete.
Corollary 2 If the stochastic basis (F n ) is regular, then

Martingale inequalities
We will prove the analogous version of the classical martingale inequalities (see e.g., Weisz [39]) for the five mixed martingale Hardy spaces.To this end, we need the following boundedness results.
Let X be a martingale space, Y be a measurable function space.Then the operator U : Theorem 7 Let 0 < − → p < ∞ and suppose that the σ-sublinear operator T : H s r → L r is bounded, where − → p = (p 1 , . . ., p d ) and r > p i (i = 1, . . ., d).If for all (s, − → p )-atom a where τ is the stopping time associated with the (s, − → p )-atom a, then for all f ∈ H s − → p , Proof.By the σ-sublinearity of T and the atomic decomposition of H s − → p given in Theorem 4, we have By Lemma 2, there exists a function g ∈ L ( − → p /t) ′ with g ( − → p /t) ′ ≤ 1, such that Since {τ k < ∞} ∈ F τ k , using the fact that a k = a k χ {τ k <∞} and equation ( 11), we have T a k χ {τ k <∞} = T a k χ {τ k <∞} .Since t < r, the previous expression can be estimated by Hölder's inequality Here, by the boundedness of T and by the fact that a k is an (s, − → p )-atom, we get where A ∈ F τ k .This implies that Hence, ( 12) can be estimated by Since − → p < r, therefore ( − → p /t) ′ /(r/t) ′ > 1, so by the boundedness of M (see Theorem 2), we obtain that By (9), we have and we get that T is bounded from The following theorem can be proved similarly.If all (S, − → p )-atoms (resp.(M, − → p )-atoms) a satisfy (11), then for all f ∈ Q − → p (resp. It is easy to see that for all (s, − → p )-atoms a, (S, − → p )-atoms a or (M, − → p )-atoms a and A ∈ F τ , s(aχ A ) = s(a)χ A , S(aχ A ) = S(a)χ A and M(aχ A ) = M(a)χ A .This means that the operators s, S and M satisfy condition (11).
Let f ∈ H s − → p .The σ-sublinear operator M is bounded from H s 2 to L 2 (see e.g.Weisz [39]), that is Mf 2 ≤ c f H s 2 .So we can apply Theorem 7 with the choice r = 2 and − → p := (p 1 , . . ., p d ), where p i < 2 and we get that The operator S is also bounded from H s 2 to L 2 (see [39]), hence using Theorem 7 we obtain From the definition of the Hardy spaces it follows immediately that By the Burkholder-Gundy and Doob's inequality, for all 1 < r < ∞, S(f ) r ≈ M(f ) r ≈ f r (see Weisz [39]).Using this, inequality (15) and Theorem 8, we have For and the second inequality in (16), we get that Similarly using the first inequality in ( 16), we have that for all 0 < − → p < ∞.

Corollary 3
We have the following martingale inequalities: Theorem 9 If the stochastic basis (F n ) is regular, then the five Hardy spaces are equivalent, that is with equivalent quasi-norms.
Proof.We know (see e.g.Weisz [39]) that S n (f ) ≤ R  (20), we obtain that Q − → p = H s − → p .Since the stochastic basis (F n ) is regular, the theorem follows from Corollary 2 and from (18).
for some k ∈ {1, . . ., d}.Then for all (f n ) n∈N non-negative, measurable function sequence Proof.From Lemma 2, we know that there exists a function g ∈ L Using Hölder's inequality and Theorem 2, we have and the proof is complete.
As an application of the previous theorem, we get the following martingale inequality.
Proof.Indeed, using Theorem 10 with the choice Using Theorem 10, we have that At the same time, which proves the theorem.

1 ,Lemma 6 .≤.
which implies that |d k g| ≤ 4λ k−1 .ThenS n (g) ≤ S n−1 (g) + |d n g| ≤ S n−1 (f ) + S n−1 (h) + 4λ n−1 ≤ λ n−1 + 2λ n−1 + 2 n−1 k=1 E k−1 (λ k − λ k−1 ) + 4λ n−1 and therefore g Q− → p ≤ c λ ∞ − → p .Choosing λ n := S n (f ), we get thath G− → p ≤ c f H S − → p and g Q− → p ≤ c f H S − → p, which proves the theorem.A similar lemma can be proved for H M − → p in the same way.Suppose that 1 < − → p < ∞ or − → p satisfies(21).If f ∈ H M − → p , then there exists h ∈ G − → p and g ∈ P − → p such that f = h + g andh G− → p ≤ c f H M − → p and g P− → p ≤ c f H M − → pNow we are ready to generalize the well known Burkholder-Davis-Gundy inequality.Theorem 11 If 1 < − → p < ∞ or − → p satisfies (21), then the spaces H S − → p and H M − → p are equivalent, that is H S − → p = H M − → p with equivalent norms.Proof.Let f ∈ H S − → p .By Lemma 5, there exists h ∈ G − → p and g ∈ Q − → p such that f = h + g and h G− → p ≤ c f H S − → p and g Q− → p ≤ c f H S h G− → p + c g Q− → p ≤ c f H S − → p The reverse inequality f H S − → p ≤ c f H M − → pcan be proved similarly.For a martingale f , the martingale transform is defined by(T f ) n := ∞ k=1 b k−1 d k f,where b k are F k -measurable and |b k | ≤ 1.The martingale transform is bounded onL − → p , if 1 < − → p < ∞.Theorem 12 If 1 < − → p < ∞, then for all f ∈ L − → p , T f − → p ≤ c f − → p .Proof.Because of |b k | ≤ 1, it is clear that S(T f ) ≤ S(f ).By Theorem 11, the spaces H M − → p and H S − → p are equivalent.Therefore using Theorem 2,