Definition 1. Let p(·) be a variable exponent and θ ≥ 0. A measurable function a is called a (1, p(·), θ) -atom (resp. (2, p(·), θ) -atom, (3, p(·), θ) -atom) if there exists a stopping time τ such that

(a1)$$, ∀ n ≤ τ

(a2)$$

Theorem 4. Let $$ and θ ≥ 0. If the martingale $$, then there exists a sequence of triplet $$ so that for each n ≥ 0,

Conversely, if the martingale f has a decomposition of (14), then

Proof. Let $$. Now consider the stopping time for each k ∈ ℤ:

It is easy to see that the sequence of these stopping times is nondecreasing. For each stopping time τ, denote $$. It is easy to write that

For each k ∈ ℤ, let $$. If μ_k ≠ 0, we set

If μ_k = 0, we set $$ for each n ∈ ℕ. For each fixed k ∈ ℤ, $$ is a martingale. Since $$, we get

We can easily check that $$ is a bounded L₂-martingale. Hence, there exists an element a^k ∈ L₂ such that $$. If n ≤ τ_k, then $$, and $$. Consequently, it implies that a^k is really a (1, p(·), θ)-atom.

Denote Λ_k≔{τ_k<∞}. Knowing that {τ_k<∞} = {s(f) > 2^k} and τ_k is nondecreasing for each k ∈ ℤ, we obtain Λ_k+1⊆Λ_k. Now, we point out that

Indeed, for a fixed x₀ ∈ Ω, there is k₀ ∈ ℤ so that $$ and $$, then we have

This means

For the converse part, according to the definition of (1, p(·), θ)-atom, we easily conclude

This implies

Taking over all the admissible representations of (14) for f, we obtain the desired result.

Next, we will characterize Q_p(·),θ and D_p(·),θ by atoms, respectively. The proof is similar to the proof of Theorem 4. For the completeness of this paper, we provide some details.

Theorem 5. Suppose $$ and θ ≥ 0. If the martingale $$ (resp. D_p(·),θ), then there exists a sequence of triplet $$ (resp. $$) so that for each n ∈ ℕ,

Conversely, if the martingale $$ has admissible representation (27), then f ∈ Q_p(·),θ (resp. D_p(·),θ) and

Proof. The proof follows the ideas in Theorem 4, so we omit some details. Suppose $$ (resp. D_p(·),θ). We define stopping times as follows:

Let $$ and $$ be defined as in the proof of Theorem 4. And replace Λ_k = {τ_k<∞} = {s(f) > 2^k} by the Λ_k = {τ_k<∞} = {λ_∞ > 2^k}. Then, we obtain that $$ and (28) still hold.

For the converse part, write

Clearly, $$ is a nonnegative, nondecreasing, and adapted sequence with S_n+1(f) ≤ λ_n (resp.∣f_n | ≤λ_n). Thus, we get

Remark 6. Suppose $$ and θ ≥ 0. We conclude that the sum $$ in Theorem 4 converges to f in $$ as M⟶−∞, N⟶∞. Indeed, it follows by the subadditive of the operator s, we get, for any M, N ∈ ℤ with M < N,

Definition 7. For $$ and θ ≥ 0, the generalized BMO martingale space is defined by

Remark 8. If θ = 0, BMO_p(·),θ degenerates to the variable exponent BMO martingale space BMO_p(·) introduced and studied in [12]. If θ = 1 and p(·) ≡ p, BMO_p(·),θ becomes the grand BMO martingale space BMO_p) studied in [26].

Lemma 9 (Hölder’s inequality, see [34]). Let $$ satisfy

Then, there exists a constant C such that for all f ∈ L_p(·) and g ∈ L_q(·), we have fg ∈ L_r(·) and

Lemma 10 (see [27].)Suppose $$ satisfying (43).

• (1)

  For each $$, we get

• (2)

  Let $$ satisfy (43). If r(·) satisfies

Theorem 11. Suppose that $$ satisfies (43) and θ ≥ 0. Then, for every f ∈ BMO₁, there has

Proof. If $$ satisfies (43), then we clearly get that p(·) − η also satisfies (43) for 0 < η < p₋ − 1. It follows from Lemmas 9 and 10 that

This is equivalent to the following inequality:

Hence, we have

Taking the supremum over all stopping times, we deduce

Conversely, from the definition of L_p(·),θ, we get

It follows from Lemma 9 that

Hence, by (38), we deduce that

From what has been discussed above, we draw the conclusion that

Theorem 11 improves the recent results [12, 26], respectively. More precisely, if we consider the case θ = 0, then the following result holds:

Corollary 12. If p(·) satisfies (43) with 1 < p₋ ≤ p₊ < ∞, then for f ∈ BMO₁,

Corollary 13 (see [26].)Suppose 1 < p < ∞, then for f ∈ BMO₁,