CHARACTERIZATIONS OF PARABOLIC MUCKENHOUPT CLASSES

. This paper extends and complements the existing theory for the parabolic Muck-enhoupt weights motivated by one-sided maximal functions and a doubly nonlinear parabolic partial diﬀerential equation of p -Laplace type. The main results include characterizations for the limiting parabolic A ∞ and A 1 classes by applying an uncentered parabolic maximal function with a time lag. Several parabolic Calder´on–Zygmund decompositions, covering and chaining arguments appear in the proofs.


Introduction
This paper discusses parabolic Muckenhoupt weights sup where R ± (γ) are space-time rectangles with a time lag γ ≥ 0, see Definition 2.1.This class of weights was introduced by Kinnunen and Saari in [13,14].The main results in [13,14] are characterizations of weighted norm inequalities for the centered forward in time parabolic maximal functions, self-improving phenomena related to parabolic reverse Hölder inequalities, factorization results and a Coifman-Rochberg type characterization of parabolic BM O.This paper complements and extends these results.Instead of the centered parabolic maximal function in [13,14], the corresponding uncentered maximal function gives a more streamlined theory, see Section 6. Observe that the centered and uncentered maximal functions are not comparable in the parabolic case.
There are many characterizations for the standard Muckenhoupt A ∞ condition, but little is known in the parabolic case.This paper is an attempt to create the missing theory and several interesting open questions remain.Theorem 4.1 gives new characterizations for the parabolic Muckenhoupt A ∞ class in terms of quantitative absolute continuity with a time lag.Section 5 gives a new proof for the parabolic reverse Hölder inequality [14,Theorem 5.2].A complete theory for the parabolic Muckenhoupt A 1 class, including factorization and characterization results for the full range of the time lag, is obtained in Section 7. Theorem 3.1 complements [14, Proposition 3.4 (iv) and (vii)] and shows that the results are independent of the time lag and the distance between the upper and lower parts of the rectangles.Several parabolic Calderón-Zygmund decompositions, covering and chaining arguments appear in the proofs.
On the other hand, parabolic Muckenhoupt weights are related to the doubly nonlinear parabolic equation for some positive constants C 0 and C 1 with 1 < p < ∞.In particular, this class of partial differential equations includes the doubly nonlinear p-Laplace equation with A(x, t, u, Du) = |Du| p−2 Du.
In the natural geometry of (1.1), we consider space-time rectangles where the time variable scales to the power p. Observe that solutions can be scaled, but constants cannot be added.If u(x, t) is a solution, so does u(λx, λ p t) with λ > 0. The main challenge of (1.1) is the double nonlinearity both in time and space variables.Trudinger [27] showed that a scale and location invariant parabolic Harnack's inequality holds true for nonnegative weak solutions to (1.1) in parabolic rectangles, see also Gianazza and Vespri [9], Kinnunen and Kuusi [10] and Vespri [28].This implies that nonnegative solutions to (1.1) are parabolic Muckenhoupt weights with γ > 0. We note that Harnack's inequality is not true with γ = 0 which can be seen from the heat kernel already when p = 2.

Definition and properties of parabolic Muckenhoupt weights
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 sides parallel to the coordinate axes and equally long, that is, Q = Q(x, L) = {y ∈ R n : |y i − x i | ≤ L, i = 1, . . ., n} with x ∈ R n and L > 0. The point x is the center of the cube and L is the side length of the cube.Instead of Euclidean cubes, we work with the following collection of parabolic rectangles in R n+1 .Definition 2.1.Let 1 < p < ∞, x ∈ R n , L > 0 and t ∈ R. A parabolic rectangle centered at (x, t) with side length L is and its upper and lower parts are where 0 ≤ γ < 1 is the time lag.
Note that R − (γ) is the reflection of R + (γ) with respect to the time slice R n × {t}.The spatial side 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 R ± for R ± (0).The top of a rectangle R = R(x, t, L) is Q(x, L) × {t + L p } and the bottom is Q(x, L) × {t − L p }.The λ-dilate of R with λ > 0 is denoted by λR = R(x, t, λL).
The integral average of f ∈ L 1 (A) in measurable set A ⊂ R n+1 , with 0 < |A| < ∞, is denoted by This section discusses basic properties of parabolic Muckenhoupt weights.We begin with the definition of the uncentered parabolic maximal functions.The differentials dx dt in integrals are omitted in the sequel.Definition 2.2.Let 0 ≤ γ < 1 and f be a locally integrable function.The uncentered forward in time and backward in time parabolic maximal functions are defined by A locally integrable nonnegative function w is called a weight.We give definitions for parabolic Muckenhoupt classes A + q and A + 1 .Definition 2.3.Let 1 < q < ∞ and 0 ≤ γ < 1.A weight w belongs to the parabolic Muckenhoupt class A + q (γ) if [w] A + q (γ) = sup where the supremum is taken over all parabolic rectangles R ⊂ R n+1 .If the condition above holds with the time axis reversed, then w ∈ A − q (γ).
Definition 2.4.Let 0 ≤ γ < 1.A weight w belongs to the parabolic Muckenhoupt class for every parabolic rectangle R ⊂ R n+1 .If the condition above holds with the time axis reversed, then w ∈ A − 1 (γ).The class A + 1 (γ) can be characterized in terms of the parabolic maximal functions.Proposition 2.5.Let 0 ≤ γ < 1.A weight w is in A + 1 (γ) if and only if there exists a constant C such that for almost every (x, t) ∈ R n+1 .Moreover, we can choose C = [w] A + 1 (γ) .The statement also holds for A − 1 (γ) with M γ+ .Proof.Assume that (2.1) holds.Then for almost every (x, t) ∈ R + (γ), and thus by taking the essential infimum over every (x, t) ∈ R + (γ) we have w > Cw(x, t).
For every ε > 0 there is a rectangle R whose spatial corners and the bottom time coordinate have rational coordinates such that (x, is open, and thus there is a positive distance between t and the bottom of w > Cw(x, t).
Hence, we may assume that the spatial corner points and the bottom time coordinate of the parabolic rectangles R satisfying (2.2) are rational.The A + 1 (γ) condition and (2.2) imply that w(y, s).
Let {R i } i∈N be an enumeration of parabolic rectangles in R n+1 with rational spatial corners and rational bottom time coordinates.Define w(y, s) for i ∈ N. Then |E i | = 0 for every i ∈ N and the argument above shows that E ⊂ i∈N E i .Thus, we have |E| = 0 and the claim (2.1) follows with C = [w] A + 1 (γ) .The next lemma tells that the class of parabolic Muckenhoupt weights is closed under taking maximum and minimum.
Proof.Let u = max{w, v}.For 1 < q < ∞, we have By taking supremum over all parabolic rectangles R ⊂ R n+1 , we obtain the first claim.The corresponding claim for the minimum follows similarly.
Next we discuss a duality property of the parabolic Muckenhoupt weights.Here q ′ = q q−1 denotes the conjugate exponent of q.

Time lag and distance between upper and lower parts
The following theorem asserts that we can change the time lag in the upper and lower parts of parabolic rectangles and also the distance between them.In particular, the definition of A + q (γ), 1 ≤ q < ∞, does not depend on 0 < γ < 1.
Note that for τ = 1 we have S − (α) = R − (α).We could also consider translated upper parts of parabolic rectangles but this can be included in the translation of lower parts.

Proof. Assume that
where . This proves the case τ = 1.By Hölder's inequality, it follows that Then iterating the previous inequality, we get w.
This shows the claim whenever β = 0.If β > 0, we partition S − (α) into subrectangles U − i (α) with spatial side length β 1/p L and time length β(1 − α)L p such that the overlap of {U − i (α)} i is bounded by 2 n+1 .This can be done by dividing each spatial edge of S − (α) into ⌈β −1/p ⌉ equally long subintervals with an overlap bounded by 2, and the time interval of S − (α) into ⌈β −1 ⌉ equally long subintervals with an overlap bounded by 2. We observe that every w.
Therefore, we have w.
We conclude that By letting q → 1, we obtain the same conclusion for A + 1 (γ).We prove the other direction.Let R 0 ⊂ R n+1 be an arbitrary parabolic rectangle.Without loss of generality, we may assume that the center of R 0 is the origin.Let m be the smallest integer with Then there exists 0 ≤ ε < 1 such that We partition R + 0 (γ) = Q(0, L) × (γL p , L p ) by dividing each of its spatial edges into 2 m equally long intervals and the time interval into ⌈(1 − γ)2 mp /(1 − α)⌉ equally long intervals.Denote the obtained rectangles by U + i,j with i ∈ {1, . . ., 2 mn } and j ∈ {1, . . ., ⌈(1 − γ)2 mp /(1 − α)⌉}.The spatial side 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 (α) 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 (α) 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 where 1 ≤ θ ≤ √ n depends on the angle between x i and the spatial axes and is chosen such that the center of Q k is on the boundary of Q k−1 .We have , where b ∈ {1, . . ., 2 m } depends on the distance of Q i to the center of Q 0 = Q(0, L).The number of cubes in the spatial chain {Q ′ k } Ni k=0 is Next, we also take the time variable into consideration in the construction of the chain.Let , for k ∈ {0, . . ., N i }, be the upper and the translated 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 We show that the translated 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 the time length of R 0 \ R + 0 (γ).This follows from the computation This implies that P − N ⊂ R + 0 .Denote these rectangles P + N and P − N by R + and R − , respectively.These are the central rectangles 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 ) and P − k+1 = P + k+1 − (0, τ (1 + α)l p ) for k ∈ {N i , . . ., N − 1}.For every j ∈ {2, . . ., ⌈(1 − γ)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 − γ)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), that is, 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 and 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 and 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, modify the definitions of P + k for k ∈ {1, . . ., 2 m−1 } by and then add M j rectangles defined by 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 By choosing 0 ≤ β j < 1 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 1 for k ∈ {1, . . ., 2 m−1 }, and thus for every k ∈ {1, . . ., N + M j }.Fix U + i,j .Denote δ = 1/(q − 1).Hölder's inequality and the assumption (3.1) imply that . By iterating the inequality above, we obtain Thus, we have We can apply a similar chaining argument in the reverse time direction for R − 0 (γ) with the exception that we also extend (and modify if needed) every chain such that the corresponding central rectangles coincide with R + and R − .A rough upper bound for the number of rectangles needed for the additional extension is given by Thus, the constant s above is two times larger in this case.Then again by iterating Hölder's inequality and the assumption (3.1), we obtain where c 2 = c 2s−1 0 . By combining (3.2) and (3.3), we conclude that Since R 0 was an arbitrary parabolic rectangle in R n+1 , it holds that w ∈ A + q (γ).This completes the proof for q > 1. Letting q → 1 in the argument above, we obtain the claim for q = 1.

Characterizations of parabolic A ∞
This section discusses several characterizations of the parabolic Muckenhoupt A ∞ condition in terms of quantitative and qualitative measure conditions.A connection to a parabolic Gurov-Reshetnyak condition is also included.See [15,24] for the Gurov-Reshetnyak class.
Theorem 4.1.Let 0 < γ < 1 and w be a weight.The following conditions are equivalent.
(i) w ∈ A + q (γ) for some 1 < q < ∞. (ii) There exist constants K, δ > 0 such that (iii) For every 0 < α < 1 there exists 0 < β < 1 such that for every parabolic rectangle R and for every measurable set (iv) There exist 0 < α, β < 1 such that for every parabolic rectangle R and for every measurable set (v) There exist 0 < α, β < 1 such that for every parabolic rectangle R we have (vi) There exists 0 < ε < 1 such that for every parabolic rectangle R we have The proof is presented in subsections below.
4.1.Quantitative measure condition.We show that (i) ⇔ (ii) in Theorem 4.1.The following theorem also holds in the case p = 1.
Theorem 4.2.Let 0 ≤ γ < 1 and w be a weight.Then w ∈ A + q (γ) for some 1 < q < ∞ if and only if there exist constants K, δ > 0 such that Proof.Assume first that w ∈ A + q (γ).Let E be a measurable subset of R + (γ).By Hölder's inequality, we have Then we prove the other direction.The assumption is equivalent with where K > 0, q = δ −1 > 0 and E is a measurable subset of R + (γ).Since the ratio of the Lebesgue measure of R + (γ) to the Lebesgue measure of E is always greater than or equal to 1, we may assume without loss of generality that the exponent q is strictly greater than 1.Denote and hence we get where q ′ = q q−1 is the conjugate exponent of q.Letting 0 < ε < q ′ − 1 and applying Cavalieri's principle allows us to evaluate Thus, we obtain where c ε = 1 + εK q ′ /(q ′ − 1 − ε).By taking the supremum over all parabolic rectangles, we conclude that w ∈ A + 1+1/ε and thus the proof is complete.
To prove the reverse implication from (iv) to (i), we need the following auxiliary result.
Lemma 4.3.Let 0 < γ, α, β < 1. Assume that w satisfies the qualitative measure condition, that is, for every parabolic rectangle R and for every measurable set Then we have the following properties.
(i) For every parabolic rectangle R and every measurable set where C ≥ 1 depends on p, γ, β and θ.
Proof.(i) This is simply the contraposition of the qualitative measure condition.
(ii) Let θ > 0 and R ⊂ R n+1 be a fixed parabolic rectangle of side length L. Choose m ∈ N such that 1) .
We partition R − (γ) into subrectangles R − 0,i (γ) with spatial side length L/2 m and time length (1 − γ)L p /2 pm such that the overlap of {R − 0,i (γ)} i is bounded by 2. This can be done by dividing each spatial edge of R − (γ) into 2 m equally long pairwise disjoint intervals, and the time interval of R − (γ) into ⌈2 pm ⌉ equally long subintervals such that their overlap is bounded by 2.
Our plan is to shift every rectangle R − 0,i (γ) forward in time by multiple times of (1 + γ)L p /2 pm until the shifted rectangles are contained in R − (γ) + (0, θL p ).To this end, choose N ∈ N such that We first move every rectangle R − 0,i (γ) forward in time by (N −1)(1+γ)L p /2 pm .Then we shift once more by the distance (1 + γ)L p /2 pm those rectangles that are not yet subsets of R − (γ) + (0, θL p ). Denote so obtained shifted rectangles by R − N,i (γ).Observe that the choice of N and (4.1) ensures that all shifted rectangles R − N,i (γ) are contained in R − (γ) + (0, θL p ).By the construction and the bounded overlap of R − 0,i (γ), the overlap of R − N,i (γ) is bounded by 4. Then we apply (i) for E = R + 0,i (γ) and continue applying (i) for shifted rectangles total of N times to obtain Therefore, we conclude that ) and the bounded overlap of R − N,i (γ).Since C is an increasing function with respect to θ, the claim follows.
Lemma 4.4.Let 0 < γ < 1 and w > 0 be a weight.Assume that there exist 0 < α, β < 1 such that for every parabolic rectangle R and every measurable set E ⊂ R + (γ) for which w(E) < βw(R − (γ)) it holds that |E| < α|R + (γ)|.Then there exist τ ≥ 1 and c = c(n, p, γ, α, β) such that for every parabolic rectangle R = R(x, t, L) ⊂ R n+1 and λ ≥ (w U − ) −1 we have where We partition S + 0 by dividing each spatial edge into 2 equally long intervals.If we divide the time interval of S + 0 into ⌈2 p ⌉ equally long intervals.Otherwise, we divide the time interval of S + 0 into ⌊2 p ⌋ equally long intervals.We obtain subrectangles S + 1 of S + 0 with spatial side length L 1 = l x (S + 1 ) = l x (S + 0 )/2 = L/2 and time length either For every S + 1 , there exists a unique rectangle R 1 with spatial side length L 1 = L/2 and time length 2L p 1 = 2L p /2 p such that R 1 has the same top as S + 1 .We select those rectangles S + 1 for which and denote the obtained collection by we subdivide S + 1 in the same manner as above and select all those subrectangles S + 2 for which 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 side into 2 equally long intervals.If we divide the time interval of we divide the time interval of S + i−1 into ⌊2 p ⌋ equally long intervals.We obtain subrectangles S + i .For every S + i , there exists a unique rectangle R i with spatial side length L i = L/2 i and time length 2L p i = 2L p /2 pi such that R i has the same top as S + i .Select those S + i for which and denote the obtained collection by 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 (4.2) holds, then we have On the other hand, if (4.3) holds, then This gives a lower bound 2 pi for every S + i .Suppose that (4.3) is satisfied at the ith step.Then we have an upper bound for the time length of S + i , since On the other hand, if (4.2) is satisfied, then In this case, (4.3) has been satisfied at an earlier step i ′ with i ′ < i.We obtain by using the upper bound for S + i ′ .Thus, we have for every Note that ε 0 = 0. We have a collection {S + i,j } i,j of pairwise disjoint rectangles.However, the rectangles in the corresponding collection {U − i,j } i,j may overlap.Thus, we replace it by a subfamily { U − i,j } i,j of pairwise disjoint rectangles, which is constructed in the following way.At the first step, choose {U − 1,j } j and denote it by { U − 1,j } j .Then consider the collection {U − 2,j } j where each U − 2,j either intersects some U − 1,j or does not intersect any U − 1,j .Select the rectangles U − 2,j that do not intersect any U − 1,j , and denote the obtained collection by { U − 2,j } j .At the ith step, choose those U − i,j that do not intersect any previously selected U − i ′ ,j , i ′ < i.Hence, we obtain a collection { U − i,j } i,j of pairwise disjoint rectangles.Observe that for every U − i,j there exists U − i ′ ,j with i ′ < i such that (4.4) pr x (U − i,j ) ⊂ pr x ( U − i ′ ,j ) and pr t (U − i,j ) ⊂ 3pr t ( U − i ′ ,j ).Here pr x denotes the projection to R n and pr t denotes the projection to the time axis.Note that S + i,j is spatially contained in U − i,j , that is, pr x S + i,j ⊂ pr x U − i,j .In the time direction, we have Therefore, by (4.4) and (4.5), it holds that (4.6) i,j For the rest of the proof and to simplify the notation, let U − i = U − i,j and U − i−1 = U − i−1,j ′ be fixed, where U − i was obtained by subdividing the previous

Then we have
We construct a chain of rectangles from U − i to U − i−1 .Define the first element of the chain by where a 0 denotes the time coordinate of the bottom of V 0 .For the rest of the chain, we consider separately the spatial variable and the time variable.We start with the spatial variable.Let for k ∈ {0, . . ., N }.We have for every m ∈ {1, . . ., n}, since pr x (R i ) ⊂ pr x (R i−1 ).Here x Ri,m denotes the mth coordinate of for every m ∈ {1, . . ., n}.Observe that ).We move on to the time variable where the chain is constructed as follows.Let We define the elements of the chain by With this choice, we must add one more rectangle into the chain for the chain to end at where b i is chosen such that the bottom of V N +1 coincides with the bottom of Here K i measures the time length between the bottom of S + i and the bottom of S + i−1 , and K i ∈ {0, . . ., ⌈2 p ⌉ − 1} or K i ∈ {0, . . ., ⌊2 p ⌋ − 1} depending on the partition level i.If the bottom of S + i intersects the bottom of S + i−1 , then K i = 0.Moreover, d i is time distance from the bottom of U − i to the bottom of U − i−1 or equivalently V N +1 , that is, Equation (4.8) is equivalent with By the choice of τ , we deduce that from which we obtain With this choice of b i , we have and thus Hence, we may apply Lemma 4.3 (i) to get By the definition of V k , we can apply Lemma 4.3 (i) for each pair of By combining the previous estimates, where If (x, t) ∈ S + 0 \ i,j S + i,j , then there exists a sequence of subrectangles up to a set of measure zero.Using this together with (4.6) and (4.9), we obtain , where c = c 1 c 2 /2 n+p .This completes the proof.
The following theorem shows that the qualitative measure condition in Theorem 4.1 (iv) implies the parabolic Muckenhoupt condition in Theorem 4.1 (i).
Theorem 4.5.Let 0 < γ < 1 and w be a weight in R n+1 .Assume that there exist 0 < α, β < 1 such that for every parabolic rectangle R and every measurable set E ⊂ R + (γ) for which w(E) < βw(R − (γ)) it holds that |E| < α|R + (γ)|.Then w ∈ A + q (γ) for some q > 1. Proof.We prove the claim first for weights satisfying that w −1 is bounded.Let R ⊂ R n+1 be a parabolic rectangle.Let ε > 0 to be chosen later.Denote B = (w U − ) −1 and ρ = 1 − α.Applying Cavalieri's principle with Lemma 4.4, we obtain where c is the constant from Lemma 4.4.By choosing ε > 0 to be small enough, we can absorb the integral over R + (γ) of the second term to the left-hand side to get . Hence, we have (4.10) where γ) with spatial side length L 1 = L/2 and time length (1 − γ)L p 1 = (1 − γ)L p /2 p .This can be done by dividing each spatial edge of R τ,− 0 (γ) into two equally long pairwise disjoint intervals, and the time interval of R τ,− 0 (γ) into ⌈2 p τ (1 + γ)/(1 − γ)⌉ equally long intervals such that their overlap is bounded by 2. Thus, the overlap of R + 1,j (γ) is bounded by 2. Then consider R τ,− 1,j (γ) and cover it in the same way as before by M rectangles R + 2,j (γ) with spatial side length such that their overlap is bounded by 2 for fixed R τ,− i−1,j (γ).We observe that the time distance between the bottom of U − i,j and the bottom of R + 0 (γ) is at most As in the proof of Lemma 4.4, we construct a chain from each Then we have

Define its elements by
Observe that The time distance between the bottom of V 0 and the bottom of Hence, the maximum possible distance between the bottom of V 0 and the bottom of V N is (4.12) By combining (4.11) and (4.12), we obtain an upper bound for the time length from the bottom of R + 0 (γ) to the bottom of V N .Based on this, we fix U σ,− 0 by choosing σ such that We add one more rectangle V N +1 into the chain so that the chain would end at U σ,− 0 .Let where b i is chosen such that the bottom of V N +1 intersects with the bottom of U σ,− 0 .Then U σ,− 0 is contained in V N +1 .Next we find an upper bound for b i .By recalling the definition of τ (4.7) from the proof of Lemma 4.4, we observe that the shortest possible time length from the bottom of R + 0 (γ) to the bottom Therefore, the time distance between the bottom of V N and the bottom of U σ,− 0 is less than By this, we obtain an upper bound for b By the definition of V k , we can apply Lemma 4.3 (i) for each pair of where C is the constant from Lemma 4.3 (ii).Note that β −N ≤ β −1+(n+p)i log α 2 .We conclude that We iterate (4.10) to obtain We observe that II tends to zero if ε < 1 c1M as N → ∞ since w −ε is bounded by the initial assumption.For the inner sum of the first term I, we apply (4.13) to get Thus, it follows that We estimate the sum by whenever ε is small enough, for example, ε < 2 n+p /(c 1 ξM ).Then it holds that for small enough ε.We conclude that (4.14) where We have shown that (4.14) holds for weights satisfying that w −1 is bounded.For general w, we consider truncations max{w, 1/k}, k ∈ N, and apply (4.14) with Fatou's lemma as k → ∞.Hence, (4.14) holds for general weights as well.Let q = 1 + 1/ε.Applying Theorem 3.1, we conclude that w ∈ A + q (γ).This completes the proof.
Theorem 4.6.Let 0 ≤ γ < 1 and w be a weight.The following conditions are equivalent.
(i) Assume that there exists 0 < ε < 1 such that Then for ε < λ < 1 we have (ii) Assume that there exist 0 < α, β < 1 such that Then we have Proof.We first show that (i) holds.
For the other direction, we set } and This completes the proof.

Parabolic reverse Hölder inequality
In this section, we show that parabolic Muckenhoupt weights satisfy the parabolic reverse Hölder inequality.The lemma below is the main ingredient in a new proof of [14, Theorem 5.2].
Proof.Let R 0 = R(x 0 , t 0 , L) = Q(x 0 , L) × (t 0 − L p , t 0 + L p ).By considering the function w(x + x 0 , t + t 0 ), we may assume that the center of R 0 is the origin, that is, We partition S − 0 by dividing each spatial edge [−L, L] into 2 equally long intervals and the time interval of S − 0 into ⌈2 p /(1 − γ)⌉ equally long intervals.We obtain subrectangles S − 1 of S − 0 with spatial side length l x (S − 1 ) = l x (S − 0 )/2 = L/2 and time length For every S − 1 , there exists a unique rectangle R 1 with spatial side length l x = L/2 and time length l t = 2L p /2 p such that R 1 has the same bottom as S − 1 .We select those rectangles S − 1 for which Select all those subrectangles S − 2 for which We continue this selection process recursively.At the ith step, we partition unselected rectangles S − i−1 by dividing each spatial side into 2 equal parts, and if we divide the time interval of we divide the time interval of S − i−1 into ⌈2 p ⌉ equal parts.We obtain subrectangles S − i .For every S − i , there exists a unique rectangle R i with spatial side length l x = L/2 i and time length l t = 2L p /2 pi such that R i has the same top as S − i .Select those S − i for which we continue the selection process in S − i .This manner we obtain a collection {S − i,j } i,j of pairwise disjoint rectangles.
Observe that if (5.1) holds, then we have On the other hand, if (5.2) holds, then Hence, we get an upper bound for the time length i .Suppose that (5.2) is satisfied at the ith step, that is, Then we have a lower bound for the time length of S − i , since On the other hand, if (5.1) is satisfied, then There are two alternatives at the ith step of the selection process: Either (5.2) has not yet been satisfied at the earlier steps or (5.2) has been satisfied at a step i ′ with i ′ ≤ i.In the first case, we have In the second case, we obtain the same estimate 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 , we observe that This implies R i ⊂ R 0 for every i ∈ N. By the construction of the subrectangles S − i , we have 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 maximal 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 { 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 { 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 i, j there exists i ′ , j ′ with i ′ < i such that (5.5) 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 In the time direction, we have Therefore, by (5.5) and (5.6), it holds that (5.7) up to a set of measure zero.By (5.3), (5.4) and (5.7), we obtain We have Then by the assumption (qualitative measure condition) it holds that from which we obtain since w S − j > λ.Thus, we can conclude that With the previous lemma, we show that the parabolic Muckenhoupt condition implies the parabolic reverse Hölder inequality, compare with [14, Theorem 5.2].Theorem 5.2.Let 1 ≤ q < ∞, 0 < γ < 1 and w ∈ A + q (γ).Then there exists ε > 0 and c > 0 such that Proof.By Theorem 4.1, we see that the assumptions of Lemma 5.1 are satisfied and thus it can be applied.We prove the claim first for bounded functions.Thus, assume that w is bounded.Let R ⊂ R n+1 be a parabolic rectangle.Let ε > 0 to be chosen later.Applying Cavalieri's principle with Lemma 5.1, we obtain By choosing ε > 0 to be small enough, we can absorb the integral over R − of the second term to the left-hand side to get Hence, we have ( where with spatial side length l x = L/2 1/p and time length l t = L p /2.This can be done by dividing each spatial edge of R − 0 into two equally long intervals that may overlap each other, and the time interval of R − 0 into two equally long pairwise disjoint intervals.Observe that the overlap of R − 1,j is bounded by M/2 = 2 n .Then consider R + 1,j and cover it in the same way as before by M rectangles R − 2,j with spatial side length l x = L/2 2/p and time length l t = L p /2 2 .At the ith step, cover R + i−1,j by M rectangles R − i,j with spatial side length l x = L/2 i/p and time length l t = L p /2 i such that their overlap is bounded by M/2.Note that every R i,j is contained in R 0 .Then iterating (5.8) we obtain We observe that II tends to zero if for the inner sum of the first term I we have Thus, it follows that We estimate the sum by , whenever ε is small enough, for example .
Then it holds that ), see the proof of Theorem 3.1, we conclude that where .
Hence, the claim holds for bounded functions.For unbounded w, we consider truncations min{w, k}, k ∈ N, and apply the claim with the monotone convergence theorem as k → ∞.This completes the proof.
Corollary 5.3.Let 1 ≤ q < ∞, 0 < γ < 1 and w ∈ A + q (γ).Let 0 ≤ α < 1 and τ > 0. Then there exist ε > 0 and c > 0 such that i with spatial side length ρL and time length ρ p L p such that the overlap of {R − i } i is bounded by 2 n+1 .This can be done by dividing each spatial edge of R − (α) into ⌈ρ −1 ⌉ equally long subintervals with an overlap bounded by 2, and the time interval of R − (α) into ⌈ρ −p ⌉ equally long subintervals with an overlap bounded by 2. We observe that every R + i is contained in S + 0 (α).Then Theorem 5.2 implies that there exists a constant C 1 such that where . By iterating the previous argument and Hölder's inequality, we obtain w, where w.
The parabolic reverse Hölder inequality implies the following self-improving property for parabolic Muckenhoupt weights.

Weighted norm inequalities for parabolic maximal functions
The next theorem is a weak type estimate for the uncentered parabolic maximal function.For the corresponding result for the centered parabolic maximal function, see [12,14,17].The proof is based on [8].Theorem 6.1.Let 1 ≤ q < ∞, 0 ≤ γ < 1 and w be a weight.Then w ∈ A + q (γ) if and only if there exists a constant C = C(n, p, q, γ, [w] A + q (γ) ) such that for every λ > 0 and f ∈ L 1 loc (R n+1 ).Proof.Assume that the weak type estimate holds.We observe that it is enough to prove the claim for w that is bounded from below.Let f = w 1−q ′ χ R + (γ) and 0 < λ < (w for every z ∈ R − (γ), we have By letting λ → (w 1−q ′ ) R + (γ) , we obtain By taking supremum over all parabolic rectangles R ⊂ R n+1 , we conclude that w ∈ A + q (γ).To prove the other direction, we make the following assumptions.First we may assume f to be bounded and compactly supported function.We note that it suffices to prove the claim for the following restricted maximal function and then let ξ → 0 to obtain the conclusion.Next we can reduce the statement to showing that since by summing up we get Moreover, it is sufficient to prove that for L > 0 we have since we may let L → ∞ to obtain the claim.Denote E = B(0, L) ∩ {λ < M γ+ ξ f ≤ 2λ}.By Lemma 2.6, it is enough to show the claim for truncations max{w, a}, a > 0. Hence, assume that w is bounded from below by a.
By the definition of E, for every z ∈ E there exists a parabolic rectangle R z such that z is the center point of R − z (γ), l(R z ) ≥ ξ and and thus the side length of R z is bounded from above.Denote P z = 5R z and consider P ± z (α) with α = γ/5 p .We observe that the top time coordinates of P − z (α) and R − z (γ) coincide and similarly the bottom time coordinates of P + z (α) and R + z (γ).In particular, we have R ± z (γ) ⊂ P ± z (α).Since z∈E P z is a bounded set, the absolute continuity of the integral implies that there is 0 < ε < 1 such that We choose R z k which top has a largest time coordinate and denote it by R z k 1 .Then we consider R z k which top has a second largest time coordinate.
Otherwise, we select R z k and denote it by R z k 2 .Suppose that R z k l , l = 1, . . ., m − 1, have been chosen.Consider R z k with a next largest top time coordinate and select it if z k is not contained in any previously chosen P − z k l (α) and denote it by R z km .In this manner we obtain a finite collection {R z k l } l .We proceed to the second selection.Denote R z k l with a largest side length by R 1 .Suppose that R j , j = 1, . . ., i − 1, have been chosen.Then consider R z k l with a next largest side length and select it if P − z k l (α) P − j (α) for every j = 1, . . ., i − 1, and denote it by R i .Hence, we obtain a finite collection {R i } i .Note that R − i (γ) R − j (γ) for i = j.It follows that every z k is contained in some P − i (α) and we have Therefore, we obtain (6.1) By the construction, we observe that for given k ∈ Z the corresponding Fix i and denote Since R − j (γ) of approximately same side length are pairwise disjoint, we obtain where C 1 = 2 3n+p+4 .We proceed to estimate In particular, we select those rectangles to F k that are approximately of side length 2 −k and do not intersect the previously chosen rectangles.Define F = ∞ k=k0 F k .For every m ∈ J 2 i \ F , there exists j ∈ F such that R − j (γ) ∩ R − m (γ) = ∅ and l(R m ) < l(R j ).Moreover, recall that R − m (γ) R − j (γ) by the selection process.Thus, we have Then a similar argument as for J 1 i can be applied for the second sum above.More precisely, let We have By the previous estimate, the definition of F and F k and the pairwise disjointness of R − j (γ) in F k , we obtain Combining the estimates for J 1 i and J 2 i , we get We extract from {R + i (γ)} i a collection of subsets F i that have bounded overlap.Fix i and denote the number of indices in J i by N .If N ≤ 2c, we choose Then it follows that 2c Thus, for every F i it holds that (6.2) 2 Moreover, for every z ∈ F i we have Next we observe that R + i (γ) of approximately same side length have bounded overlap.Fix z ∈ R n+1 and denote where S z is a rectangle centered at z with side lengths l x (S z ) = 2 −k+1 and l t (S z ) = 2 −kp+1 .Since R − i (γ) are pairwise disjoint, we obtain We show that F i have a bounded overlap.Fix z ∈ i F i .Consider a rectangle R i0 ∈ {R i } i with a largest side length such that z ∈ F i0 .Let 2 −k0−1 < l(R i0 ) ≤ 2 −k0 .By the previous estimate and (6.3), we have where C 2 = 2 3n+p+9 /(1 − γ).
Observe that in the case γ = 0, we do not need to use Theorem 3.1.This completes the proof.

2 i
2), Theorem 3.1 with the constant C and the bounded overlap of F i to get w(E) ≤
and denote the obtained collection by {S − 1,j } j .If we subdivide S − 1 by dividing each spatial edge [−L, L] into 2 equally long intervals.Ifl t (S − 1 ) ⌊2 p ⌋ ≤ (1 − γ)L p 2 2p ,we divide the time interval of S − 1 into ⌊2 p ⌋ equally long intervals.Otherwise, we divide the time interval of S − 1 into ⌈2 p ⌉ equally long intervals.We obtain subrectangles S − 2 of S − 1 with spatial side length l x (S − 2 ) = l x (S − 1 )/2 and time length either