PARABOLIC JOHN–NIRENBERG SPACES WITH TIME LAG

. We introduce a parabolic version of the so-called John–Nirenberg space that is a generalization of functions of parabolic bounded mean oscillation. Parabolic John–Nirenberg inequalities, which give weak type estimates for the oscillation of a function, are shown in the setting of the parabolic geometry with a time lag. Our arguments are based on a parabolic Calder´on–Zygmund decomposition and a good lambda estimate. Chaining arguments are applied to change the time lag in the parabolic John–Nirenberg inequality.


Introduction
Functions of bounded mean oscillation BMO and their generalization, the so-called John-Nirenberg space denoted by JN q with 1 < q < ∞, were introduced by John and Nirenberg in [6].Let Ω be a domain in R n .A function f ∈ L 1 loc (Ω) belongs to the John-Nirenberg space JN q (Ω), 1 where the supremum is taken over countable collections {Q i } i∈N of pairwise disjoint subcubes of Ω.A particularly useful result for BMO is the John-Nirenberg lemma which gives an exponential decay estimate for the mean oscillation of a function in BMO.Moreover, the John-Nirenberg lemma for JN q gives a weak type estimate for the oscillation of a function in JN q implying that the John-Nirenberg space is a subset of weak L q .The corresponding time-dependent theory was initiated by Moser in [16,17] to study the regularity of parabolic partial differential equations, and parabolic BMO was explicitly defined by Fabes and Garofalo in [5].The proof of the parabolic John-Nirenberg lemma requires genuinely new ideas compared to the time independent case.The theory of parabolic BMO has been further studied in [1,7,[9][10][11]19].See also [14] for the onedimensional case.
The purpose of this paper is to introduce and study a parabolic version of the John-Nirenberg space in the parabolic geometry with a time lag 0 < γ < 1.This geometry is motivated by certain parabolic nonlinear partial differential equations, see [1,7,[9][10][11]19].Let Ω T = Ω × (0, T ) be a space-time cylinder in R n+1 .A function f ∈ L 1 loc (Ω T ) belongs to the parabolic John-Nirenberg space P JN + q (Ω T ), 1 where the supremum is taken over countable collections {R i } i∈N of pairwise disjoint space-time subrectangles of Ω T .Here R i is a parabolic rectangle in the parabolic p-geometry with 1 < p < ∞ and 0 < γ < 1 is the time lag, see Section 2 for notation.In a similar way to the standard setting, this space is a generalization of parabolic BMO in a sense that parabolic JN q contains parabolic BMO, and parabolic BMO is obtained as the limit of parabolic John-Nirenberg spaces as q → ∞.A one-sided John-Nirenberg space has been considered by Berkovits [2] in the case p = 1, but the geometry is not related to nonlinear parabolic partial differential equations.For recent developments in the theory of John-Nirenberg spaces in the standard setting, we refer to [3,4,8,12,13,15,18,[21][22][23][24].
Our main results are parabolic John-Nirenberg inequalities for parabolic JN q .The main challenges are the time lag between the two mean oscillations in the definition and the parabolic geometry.The proof of the parabolic John-Nirenberg inequality in Theorem 3.1 is based on a parabolic Calderón-Zygmund decomposition and a good lambda inequality.Theorem 4.1 shows that by applying suitable parabolic chaining arguments we may improve the obtained John-Nirenberg lemma by changing the time lag.This allows us to conclude that the parabolic John-Nirenberg space is independent of the size of the time lag, see Corollary 4.2.For more about chaining techniques in the parabolic geometry, see [7,19,20].

Definition and properties of parabolic John-Nirenberg spaces
The underlying space throughout is R n+1 = {(x, t) : x = (x 1 , . . ., x n ) ∈ R n , t ∈ R}.Unless otherwise stated, constants are positive and the dependencies on parameters are indicated in the brackets.The Lebesgue measure of a measurable subset A of R n+1 is denoted by |A|.A cube Q is a bounded interval in R n , with edges parallel to the coordinate axes and equally long, that is, The point x is the center of the cube and L is the edge length of the cube.Instead of Euclidean cubes, we work with the following collection of parabolic rectangles in R n+1 .
and its upper and its lower parts are where −1 < γ < 1 is called the time lag.
Note that R − (γ) is the reflection of R + (γ) with respect to the time slice R n × {t}.The spatial edge length of a parabolic rectangle R is denoted by l x (R) = L and the time length by l t (R) = 2L p .For short, we write The positive and the negative parts of a function f are denoted by Let Ω ⊂ R n be an open set and T > 0. A space-time cylinder is denoted by Ω T = Ω × (0, T ).It is possible to consider space-time cylinders Ω × (t 1 , t 2 ) with t 1 < t 2 , but we focus on Ω T .We recall the definition of the parabolic BMO.The differentials dx dt in integrals are omitted in the sequel.
If the condition above holds with the time axis reversed, then f ∈ PBMO − γ,r (Ω T ).This section discusses basic properties of the parabolic John-Nirenberg space.We begin with the definition.Definition 2.3.Let Ω ⊂ R n be a domain, T > 0, 0 ≤ γ < 1, 1 < q < ∞ and 0 < r < q.A function f ∈ L r loc (Ω T ) belongs to P JN + q,γ,r (Ω T ) if where the supremum is taken over countable collections {R i } i∈N of pairwise disjoint parabolic subrectangles of Ω T .If the condition above holds with the time axis reversed, then f ∈ P JN − q,γ,r (Ω T ).We shall write P JN + q and f whenever parameters are clear from the context or are not of importance.Observe that f ∈ L r loc (Ω T ) belongs to P JN + q,γ,r (Ω T ) if and only if for every collection {R i } i∈N of parabolic rectangles there exist constants c i ∈ R, that may depend on R i , with where The next lemma shows that for every parabolic rectangle R, there exists a constant c R , depending on R, for which the infimum in the definition above is attained.In the sequel, this minimal constant is denoted by c R .The proof is analogous to that of the parabolic BMO in [9], and thus is omitted here.Lemma 2.4.Let Ω T ⊂ R n+1 be a space-time cylinder, 0 ≤ γ < 1, 1 < q < ∞ and 0 < r < q.Assume that f ∈ P JN + q,γ,r (Ω T ).Then for every parabolic rectangle R ⊂ Ω T , there exists a constant c R ∈ R, that may depend on R, such that In particular, it holds that We list some properties of the parabolic John-Nirenberg space below.In particular, P JN + q is closed under addition and scaling by a positive constant.On the other hand, multiplication by negative constants reverses the time direction.The proof is similar to that of the parabolic BMO in [9], and thus is omitted here.Lemma 2.5.Let 0 ≤ γ < 1, 1 < q < ∞ and 0 < r < q.Assume that f and g belong to P JN + q,γ,r (Ω T ) and let P JN ± q = P JN ± q,γ,r (Ω T ).Then the following properties hold.
Remark 2.6.For 1 ≤ r < q, the constants in (ii) and (iv) can be avoided by considering the norm sup , which is an equivalent norm with our definition.However, the current definition will be convenient in the proof of the John-Nirenberg lemma below.
Properties (i) and (iv) of Lemma 2.5 imply that every P JN + q,γ,r function can be approximated pointwise by bounded P JN + q,γ,r functions.
Remark 2.7.If f ∈ PBMO + γ,r (Ω T ) with |Ω T | < ∞, then f ∈ P JN + q,γ,r (Ω T ) for every 1 < q < ∞.More precisely, we have The parabolic John-Nirenberg space is a generalization of the parabolic BMO in the sense that a function is in PBMO + γ,r if and only if the P JN + q,γ,r norm of the function is uniformly bounded as q tends to infinity.
Proof.We may assume that as q → ∞.Hence, we have We can interchange the order of taking the supremum and the limit since is an increasing function of q which can be seen by Hölder's inequality.Thus, we conclude that lim

Parabolic John-Nirenberg inequality
This section discusses a John-Nirenberg inequality for parabolic John-Nirenberg spaces.The argument is based on a parabolic Calderón-Zygmund decomposition and applies a parabolic sharp maximal function to obtain a good lambda estimate.For short, we suppress the variables (x, t) in the notation and, for example, write λ q for every λ > 0.
Proof.The beginning of the proof is similar to the John-Nirenberg lemma for parabolic BMO in [9].However, the rest of the proof uses several new ideas when dealing with weak type estimates for parabolic JN q compared to the exponential decay estimate for parabolic BMO.Let R 0 = R = R(x 0 , t 0 , L) = Q(x 0 , L) × (t 0 − L p , t 0 + L p ).By considering the function f (x + x 0 , t + t 0 ), we may assume that the center of R 0 is the origin, that is, R 0 = Q(0, L) × (−L p , L p ).By (i) of Lemma 2.5, we may assume that c R0 = 0. We note that it is sufficient to prove the first inequality of the theorem since the second inequality follows from the first one by applying it to the function −f (x, −t).
Let f = f P JN + q,γ,r (R0) .We claim that for every λ > 0. Let m be the smallest integer with 3 + α ≤ 2 pm+1 (α − γ), that is, we divide the time interval of S + 0 into ⌊2 pm ⌋ equally long intervals.Otherwise, we divide the time interval of S + 0 into ⌈2 pm ⌉ equally long intervals.Here ⌊•⌋ and ⌈•⌉ are the floor and ceiling functions, respectively.We obtain subrectangles S + 1 of S + 0 with spatial edge length l x (S + 1 ) = l x (S + 0 )/2 m = L/2 m and time length either For every S + 1 , there exists a unique rectangle R 1 with spatial edge length l x = L/2 m and time length l t = 2L p /2 mp such that R 1 has the same top as S + 1 .We select those rectangles S + 1 for which λ < c R1 and denote the obtained collection by {S + 1,j } j .If λ ≥ c R1 , we subdivide S + 1 in the same manner as above and select all those subrectangles S + 2 for which λ < c R2 to obtain family {S + 2,j } j .We continue this selection process recursively.At the ith step, we partition unselected rectangles S + i−1 by dividing each spatial edge into 2 m equally long intervals.If we divide the time interval of we divide the time interval of S + i−1 into ⌈2 pm ⌉ equally long intervals.We obtain subrectangles S + i .For every S + i , there exists a unique rectangle R i with spatial edge length l x = L/2 mi and time length l t = 2L p /2 pmi such that R i has the same top as S + i .Select those S + i for which λ < c Ri and denote the obtained collection by {S + i,j } j .If λ ≥ c Ri , we continue the selection process in S + i .In this manner we obtain a collection {S + i,j } i,j of pairwise disjoint rectangles.Observe that if (3.1) holds, then we have On the other hand, if (3.2) holds, then This gives an upper bound 2) is satisfied at the ith step.Then we have a lower bound for the time length of S + i , because On the other hand, if (3.1) is satisfied, then In this case, (3.2) has been satisfied at an earlier step i ′ with i ′ < i.We obtain by using the lower bound for S + i ′ .Thus, we have for every S + i .By using the lower bound for the time length of S + i and the choice of m, we observe that ).This implies for a fixed rectangle S + i−1 and for every subrectangle By the construction of the subrectangles S + i , we have (3.4) We have a collection {S + i,j } i,j of pairwise disjoint rectangles.However, the rectangles in the corresponding collection {S − i,j } i,j may overlap.Thus, we replace it by a subfamily { S − i,j } i,j of pairwise disjoint rectangles, which is constructed in the following way.At the first step, choose {S − 1,j } j and denote it by { S − 1,j } j .Then consider the collection {S − 2,j } j where each S − 2,j either intersects some S − 1,j or does not intersect any S − 1,j .Select the rectangles S − 2,j that do not intersect any S − 1,j , and denote the obtained collection by { S − 2,j } j .At the ith step, choose those S − i,j that do not intersect any previously selected S − i ′ ,j , i ′ < i.Hence, we obtain a collection { S − i,j } i,j of pairwise disjoint rectangles.Observe that for every S − i,j there exists S − i ′ ,j with i ′ ≤ i such that (3.6) pr x (S − i,j ) ⊂ pr x ( S − i ′ ,j ) and pr t (S − i,j ) ⊂ 3pr t ( S − i ′ ,j ).Here pr x denotes the projection to R n and pr t denotes the projection to the time axis.
Rename {S + i,j } i,j and { S − i,j } i,j as {S + i } i and { S − j } j , respectively.Let S(λ) = i S + i .Note that S + i is spatially contained in S − i , that is, pr x S + i ⊂ pr x S − i .In the time direction, we have Therefore, by (3.6) and (3.7), it holds that Let λ > δ > 0 and consider the collection i is contained in some S + k , we get the partition Fix k ∈ N. We have S + j ⊂ S + k for every j ∈ J k , where S + k was obtained by subdividing a previous where A > 1 will be specified later.Then for j ∈ M k we have By summing over j ∈ M k , we obtain for every (x, t) ∈ R k − , where M ♯ f is the parabolic sharp maximal function defined by It holds that for every (x, t) ∈ k∈K R k − .Thus, we obtain By a similar argument to the Vitali covering theorem, we obtain a countable collection {R i } i of pairwise disjoint parabolic rectangles such that Then we have , where c 2 = 5 n+p 2 q r +2 c 1 /(1 − γ).On the other hand, if k / ∈ K, we have Summing over all indices k / ∈ K and applying (3.4) together with (3.5), we conclude that where By combining the cases k ∈ K and k / ∈ K, we obtain It is left to consider the case j / ∈ M k .Using (3.5), we have for every (x, t) ∈ R j and thus for every (x, t) ∈ k j / ∈M k R j .Then arguing as before, we obtain . By using (3.8) and combining the cases j ∈ M k and j / ∈ M k , we conclude that By choosing A = 2 q r +1 c 3 and replacing λ and δ by 2 1 r λ and λ, respectively, we have Let .
For 0 < λ ≤ λ 0 , we have Assume then that λ > λ 0 .There exists an integer N ∈ N such that 2 r λ 0 ) q for every K = 0, 1, . . ., N , where c 5 = 2 q r +1 A q r c 4 .We prove the claim by induction.First, note that the claim holds for K = 0, because Then assume that the claim holds for K ∈ {0, 1, . . ., N − 1}, that is, We show that this implies the claim for K + 1.By using (3.9) for 2 K r λ 0 we observe that Therefore, the claim holds for K + 1.If (x, t) ∈ S + 0 \ S(λ), then there exists a sequence {S + l } l∈N of subrectangles containing (x, t) such that c R l ≤ λ and |S + l | → 0 as l → ∞.Then by (3.5) and Lemma 2.4 it holds that The Lebesgue differentiation theorem [9, Lemma 2.3] implies that f (x, t) r + ≤ λ r for almost every (x, t) ∈ S + 0 \ S(λ).It follows that {(x, t) ∈ S + 0 : f (x, t) + > λ} ⊂ S(λ) up to a set of measure zero.We conclude that where C = 2 q r c 5 .This completes the proof.

Chaining arguments and the time lag
Weak type estimates in Theorem 3.1 together with chaining arguments enable us to change the time lag in the parabolic John-Nirenberg inequality.
Proof.The beginning of the proof is similar to [9,Theorem 4.1].Let R 0 = R and f = f P JN + q,γ,r (R0) .Without loss of generality, we may assume that the center of R 0 is the origin.Since f ∈ P JN + q,γ,r (R 0 ), Theorem 3.1 holds for any parabolic subrectangle of R 0 and for any γ < α < 1.Let m the smallest with Then there exists 0 ≤ ε < 1 such that We partition R + 0 (ρ) = Q(0, L) × (ρL p , L p ) by dividing each of its spatial edges into 2 m equally long intervals and the time interval into ⌈(1 − ρ)2 mp /(1 − α)⌉ equally long intervals.Denote the obtained rectangles by U + i,j with i ∈ {1, . . ., 2 mn } and j ∈ {1, . . ., ⌈(1 − ρ)2 mp /(1 − α)⌉}.The spatial edge length of U + i,j is l = l x (U + i,j ) = L/2 m and the time length is For every U + i,j , there exists a unique rectangle R i,j that has the same top as U + i,j .Our aim is to construct a chain from each U + i,j to a central rectangle which is of the same form as R i,j and is contained in R 0 .This central rectangle will be specified later.First, we construct a chain with respect to the spatial variable.Fix U + i,j .Let P 0 = R i,j and where 1 ≤ θ ≤ √ n depends on the angle between x i and the spatial axes and is chosen such that the center of for every k ∈ {0, . . ., N i }, Next, we also take the time variable into consideration in the construction of the chain.Let and P − k = P + k − (0, (1 + α)l p ), for every k ∈ {0, . . ., N i }, be the upper and the lower parts of a parabolic rectangle respectively.These will form a chain of parabolic rectangles from U + i,j to the eventual central rectangle.Observe that every rectangle P Ni coincides spatially for all pairs (i, j).Consider j = 1 and such i that the boundary of Q i intersects the boundary of Q 0 .For such a cube Q i , we have b = 1, and thus N = N i = L l − 1.In the time variable, we travel from t 1 the distance (N + 1)(1 + α)l p + (1 − α)l p = (1 + α)Ll p−1 + (1 − α)l p .We show that the lower part of the final rectangle P − N is contained in R 0 .To this end, we subtract the time length of U + i,1 from the distance above and observe that it is less than half of the time length of R 0 \ (R + 0 (ρ) ∪ R − 0 (σ)).This follows from the computation This implies that P − N ⊂ R + 0 (ρ − (ρ + σ)/2).Denote this rectangle P N by R = R ρ .This is the central rectangle where all chains will eventually end.
Let j = 1 and assume that i is arbitrary.We extend the chain {P k } Ni k=0 by N − N i rectangles into the negative time direction such that the final rectangle coincides with the central rectangle R.More precisely, we consider Q ′ k+1 = Q ′ Ni , P + k+1 = P + k − (0, (1 + α)l p ) and P − k+1 = P + k+1 − (0, (1 + α)l p ) for k ∈ {N i , . . ., N − 1}.For every j ∈ {2, . . ., ⌈(1 − ρ)2 mp /(1 − α)⌉}, we consider a similar extension of the chain.The final rectangles of the chains coincide for fixed j and for every i.Moreover, every chain is of the same length N + 1, and it holds that Then we consider an index j ∈ {2, . . ., ⌈(1 − ρ)2 mp /(1 − α)⌉} related to the time variable.The time distance between the current ends of the chains for pairs (i, j) and (i, 1) is Our objective is to have the final rectangle of the continued chain for (i, j) to coincide with the end of the chain for (i, 1), namely, with the central rectangle R. To achieve this, we modify 2 m−1 intersections of P + k and P − k+1 by shifting P k and also add a chain of M j rectangles traveling to the negative time direction into the chain {P k } N k=0 .We shift every P k , k ∈ {1, . . ., 2 m−1 }, by a β j -portion of their temporal length more than the previous rectangle was shifted, that is, we move each P k into the negative time direction a distance of kβ j (1 − α)l p .The values of M j ∈ N and 0 ≤ β j < 1 will be chosen later.In other words, for every k ∈ {1, . . ., 2 m−1 }, we modify the definitions of P + k by ), and then add M j rectangles defined by for every k ∈ {N, . . ., N + M j − 1}.Note that there exists 1 ≤ τ < 2 such that We would like to find such 0 ≤ β j < 1 and M j ∈ N that which is equivalent with With this choice all final rectangles coincide.Choose M j ∈ N such that we have Observe that 0 ≤ β j ≤ 1 2 for every j.For measures of the intersections of the modified rectangles, it holds that for every k ∈ {1, . . ., 2 m−1 }, and thus , where in the last inequality we used that P k are pairwise disjoint separately for even k and for odd k.We have We observe that The first sum of the right-hand side can be estimated by Theorem 3.1 as follows i,j λ q .
To estimate the second sum above, assume that λ ≥ 2B f /|P + 0 | 1 q .This implies that for every i, j.Thus, We can apply a similar chaining argument in the reverse time direction for R − 0 (σ) with the exception that we also extend (and modify if needed) every chain such that the central rectangle R σ coincides with R = R ρ .A rough upper bound for the number of rectangles needed for the additional extension is given by Thus, the constant B above is 2 1 r − 1 q times larger in this case.This proves the second inequality of the theorem.
To show the second inequality, we make the restriction 0 < r ≤ 1 so that we can apply Theorem 4.1.This is not an issue because after establishing the second inequality for 0 < r ≤ 1 we can use the first inequality to obtain the whole range 0 < r ≤ s.Cavalieri's principle and Theorem 4.1 imply that Cλ s−q−1 f q P JN + q,γ,r (Ri) dλ + s q,γ,r (Ri) + |R + i (ρ)| 1− s q f s P JN + q,γ,r (Ri) ≤ Cq q − s |R + i (ρ)| 1− s q f s P JN + q,γ,r (Ri) .