On Ricci coefficients of null hypersurfaces with time foliation in Einstein vacuum space-time
aa r X i v : . [ m a t h . A P ] J un ON RICCI COEFFICIENTS OF NULL HYPERSURFACES WITHTIME FOLIATION IN EINSTEIN VACUUM SPACE-TIME
QIAN WANG
Abstract.
The main objective of this paper is to control the geometry of nullcones with time foliation in Einstein vacuum spacetime under the assumptionsof small curvature flux and a weaker condition on the deformation tensor for T . We establish a series of estimates on Ricci coefficients, which plays a crucialrole to prove the improved breakdown criterion in [12]. Introduction
Consider a (3+1)-dimensional Einstein vacuum spacetime ( M , g ) foliated by Σ t which are level hypersurfaces of a time function t monotonically increasing towardsthe future. Let D and ∇ denote the covariant differentiations with respect to g andthe induced metric g on Σ t respectively. We define on each Σ t the lapse function n and the second fundamental form k by n := ( − g ( D t, D t )) / and k ( X, Y ) := − g ( D X T , Y ) , where T denotes the future directed unit normal to Σ t and X, Y ∈ T Σ t . For anycoordinate chart O ⊂ Σ t with coordinates x = ( x , x , x ), let x = t, x , x , x bethe transported coordinates obtained by following the integral curves of T . Underthese coordinates the metric g takes the form(1.1) g = − n dt + g ij dx i dx j , ∂ t g ij = − nk ij . Main result.
Consider an outgoing null cone contained in ( M , g ), whosevertex is denoted by p and intersections with Σ t are denoted by S t . The null vector l ω in T p M parametrized with ω ∈ S , is normalized by g ( l ω , T p ) = −
1. We denoteby Γ ω ( s ) the outgoing null geodesic from p withΓ ω (0) = p, dds Γ ω (0) = l ω and define the null vector field L by L (Γ ω ( s )) = dds Γ ω ( s ) . Then D L L = 0. The affine parameter s of null geodesic is chosen such that s ( p ) = 0and L ( s ) = 1 . Let H = ∪ Primary 54C40, 14E20; Secondary 46E25, 20C20. we can always suppose t ( p ) = 0 and 0 ≤ t ≤ t, x ) of ( M , g ). The normalized time function, for simplicity, is still denoted by t . L is orthogonal to the leafs S t . In addition we can choose an orthonormal frame( e A ) A =1 , tangent to S t such that ( e A ) A =1 , , e = L , e = L form a null frame, i.e. g ( L, L ) = − , g ( L, L ) = g ( L, L ) = g ( L, e A ) = g ( L, e A ) = 0 , g ( e A , e B ) = δ AB . Let a − = −h L, T i with a ( p ) = 1. It follows that along any null geodesic Γ ω , thereholds(1.2) dtds = n − a − , t ( p ) = 0 . Let N be the outward unit normal of S t on Σ t . Then(1.3) T = 12 (cid:0) aL + a − L (cid:1) , N = 12 (cid:0) aL − a − L (cid:1) . We define the Ricci coefficients χ, χ, ζ, ζ, ̟ via the frame equations D A L = χ AB e B − ζ A L, D A L = χ AB e B + ζ A L, D L L = 2 ζ A e A , D L L = 2 ζ A e A − ̟L. Thus we also have D L e A = / ∇ L e A + ζ A L, D B e A = / ∇ B e A + 12 χ AB e + 12 χ AB e where / ∇ denotes the covariant derivative restricted on S t .Let λ = − Tr k , where Tr k = g ij k ij . We decompose ˆ k ij := k ij + λg ij , thetraceless part of k , relative to the orthonormal frame { N, e A , A = 1 , , } along nullcone H by introducing the following components(1.4) η AB = ˆ k AB ǫ A = ˆ k AN δ = ˆ k NN . Denote by ˆ η AB the traceless part of η . Since δ AB η AB = − δ , it is easy to seeˆ η AB = η AB + 12 δ AB δ. We denote by /π one of the following S t tangent tensors { ˆ η, δ, ǫ, λ, − / ∇ log n, −∇ N log n } .It is easy to check by definition that the Ricci coefficients ζ, ζ, ν verify ν := − L ( a ) = − / ∇ N log n + δ − λ, (1.5) ζ A = / ∇ A log a + ǫ A , ζ A = / ∇ A log n − ǫ A . (1.6)Let us define θ AB := h D A N, e B i . By definition of χ, χ and (1.3), it follows that aχ AB = θ AB − k AB , a − χ AB = − θ AB − k AB , (1.7) a tr χ = tr θ + δ + 2 λ, a − tr χ = − tr θ + δ + 2 λ. (1.8)We define the null components of Riemannian curvature tensor relative to t -foliation, α AB = R ( L, e A , L, e B ) , β A = 12 R ( e A , L, L, L ) ,ρ = 14 R ( L, L, L, L ) , σ = 14 ⋆ R ( L, L, L, L ) , (1.9) β A = 12 R ( e A , L, L, L ) , α AB = R ( L, e A , L, e B ) . We define also the mass aspect functions µ and µ as follows µ = − D tr χ + a χ ) − ̟ tr χ, (1.10) µ = D tr χ + 12 tr χ · tr χ. (1.11)Denote γ t := γ ( t, ω ) the induced metric of g on S t , relative to normal coordinates ω = ( ω , ω ) in the tangent space at p . Define the radius function of S t to be r ( t ) = p (4 π ) − | S t | and define the metric ◦ γ by ◦ γ = r − γ . We denote by γ (0) thecanonical metric on S . On each S t we introduce the ratio of area elements(1.12) v t ( ω ) := p | γ t | p | γ (0) | , ω ∈ S . For smooth scalar functions f , the average of f on S t is defined by ¯ f := | S t | R S t f dµ γ . For any scalar functions f , Z H f := Z Z S t f nadµ γ dt = Z Z | ω | =1 naf v t dµ S dt. We define L pω norm for smooth functions f on S t by k f k pL pω = R S t | f | p dµ S , anddefine L norm on null cone H for any smooth function f by k f k L ( H ) = Z Z S t | f | nadµ γ dt. For simplicity, we will suppress the H in the definition of norms on H wheneverthere occurs no confusion. Define for any S t tangent tensor F the norm on HN ( F ) = k / ∇ L F k L + k / ∇ F k L + k r − F k L . Define R ( H ), curvature flux on H relative to t -foliation, by R ( H ) = Z Z S t an ( | α | + | β | + | ρ | + | σ | + | β | ) dµ γ dt. For any S t -tangent tensor field F we define the norm k F k L ∞ ω L t ( H ) by k F k L ∞ ω L t := sup ω ∈ S Z | F | nadt := sup ω ∈ S Z Γ ω | F | nadt. The main result of this paper is the following Theorem 1.1 ( Main Theorem). Let ( M , g ) be a smooth 3+1 Einstein vacuumspacetime foliated by Σ t , the level hypersurfaces of a time function t with lapse func-tion n . Consider an outgoing null hypersurface H = ∪ By a standard rescaling argument, see [7, page 363], the esti-mates (1.14)-(1.21) in Theorem 1.1 can be rephrased as [12, Theorem 7, Proposition14], which are the crucial components in the proof of an improved breakdown cri-terion for Einstein vacuum spacetime in CMC gauge, stated as follows Theorem 1.2. [12, Theorem 1] Let ( M , g ) be a globally hyperbolic developmentof Σ t foliated by the CMC level hypersurfaces of a time function t < . Then thespace-time together with the foliation Σ t can be extended beyond any value t ∗ < for which, (1.22) Z t ∗ t (cid:0) k k k L ∞ (Σ t ) + k∇ log n k L ∞ (Σ t ) (cid:1) dt = K < ∞ . Under the assumption (1.22) only, we have proved in [12, Section 3] that C − 1, an assumption of(1.26) (cid:18)Z t ∗ t ( k k k qL ∞ (Σ) + k∇ log n k qL ∞ (Σ) ) (cid:19) q < Λ < ∞ by rescaling takes the following form on null cone Hk k k L qt L ∞ x ( H ) + k∇ log n k L qt L ∞ x ( H ) < R , which by (1.5) and (1.6) immediately gives(1.27) k ν k L qt L ∞ x ( H ) + k ζ k L qt L ∞ x ( H ) . R . In view of the definition ν := − / ∇ L a and a ( p ) = 1, we can obtain (1.25) by inte-grating along any null geodesic Γ ω as long as R is sufficiently small.If q = 1, the above simple argument fails due to the fact that by recaling, thereholds k k k L t L ∞ x ( H ) + k∇ log n k L t L ∞ x ( H ) < K which fails to be small. Thus, it is impossible to derive (1.27) immediately fromassumption. Theorem 1.1 however provides the trace estimate for k ν k L ∞ x L t ( H ) in(1.17) which is strong enough to guarantee | a − | ≤ .As remarked after the statement of Theorem 1.1, estimates such as k ζ k L ∞ ω L t ( H ) . R · · · are indispensable for establishing (1.14), (1.16) and estimates on µ and / ∇ tr χ , allof which were employed to prove breakdown criterion in [7] and [12]. The aboveestimate for ζ can not directly follow from the assumption (1.26) with q < H ( S t ) ֒ → L ∞ ( S t ) fails. Therefore theassumption (1.24), which is a consequence of (1.22) by energy estimate, can nei-ther control k ζ, ν k L t L ∞ x immediately nor by Sobolev embedding. This forces us toestimate the weaker norm k · k L ∞ x L t ( H ) , which, according to our experience, wouldsucceed only when special structures for / ∇ ζ and / ∇ ν can be found.1.3. Comparison to geodesic foliation. Let us draw comparisons between geo-desic foliation and time foliation on null hypersurfaces as follows.(1) In the case of geodesic foliation ([2, 11]), the complete set of estimates inthe main theorem can be obtained under the small curvature flux only. However intime foliation, (1.13) contains one more assumption (1.24). In order to understandthe reason for assuming (1.24), let us sketch the approach to derive (1.17). Wewill prove and employ the sharp trace inequality (see Theorem 5.1) on null coneswith time foliation, which lied in the heart of [2, 4, 3] for null hypersurfaces withgeodesic foliation. To implement this idea, it is a must to control N ( ν, ζ ). Inview of (1.5) and (1.6), ν and ζ are combinations of elements of /π , therefore theassumption (1.24) guarantees the necessary control on ν, ζ . QIAN WANG (2) The identity ζ + ζ = 0 only holds on null hypersurface H with geodesicfoliation. Therefore, in the case of time foliation, the estimates for ζ are no longeridentical to those for ζ . Note that ζ and ζ are on an equal footing in structureequations (2.4) and (2.15), we need N ( · ) and k · k L ∞ x L t ( H ) estimates for both ofthem. ζ will be treated in the same fashion as in geodesic foliation, while theestimate for k ζ k L ∞ x L t ( H ) requires further study on structure equations. Similar to(2.8) and (2.9), the quantity µ is connected to ζ by the Hodge system, (2.5) and(2.7), (cid:26) div ζ = − ˇ ρ − µ + · · · curl ζ = ˇ σ, (cid:26) div ζ = − ˇ ρ + µ + · · · curl ζ = − ˇ σ. However µ fails to satisfy a similar transport equation as ( 2.15) for µ , µ conse-quently does not verify k ·k P estimate as µ , the treatment of ζ is therefore differentfrom ζ . Note that if there holds the following decomposition for / ∇ ζ ,(1.28) / ∇ ζ = / ∇ L P + E with P and E satisfying appropriate estimates, we may rely on the sharp traceinequality to estimate k ζ k L ∞ ω L t . To obtain the important structure (1.28), we firstderive a refined Hodge system, (2.16) and (2.7),(1.29) (cid:26) div ζ = − ˇ ρ + L ( aδ + 2 aλ ) + · · · curl ζ = − ˇ σ. The pair of quantities (ˇ ρ, ˇ σ ) can be decomposed in the same way as contained in[2, 11]. We set D : F → (div F, curl F ) for any smooth S t tangent 1-form F .(1.28) then will be obtained from (1.29) by commuting / ∇ L with / ∇D − . The newcommutator [ / ∇ L , / ∇D − ]( aδ + 2 aλ ) will be decomposed in Proposition 6.1 togetherwith other commutators arising in control of k ˆ χ, ζ k L ∞ x L t ( H ) .Recall that in order to control | a − | < / k ν k L ∞ ω L t , which is a quantity that does not arise in the case of geodesic foliation.We again rely on sharp trace inequality to derive k ν k L ∞ ω L t , which requires anotherremarkable structure of the form, / ∇ ν = / ∇ L P + E. This will be done by deriving the transport equation for / ∇ a , i.e. (2.17), / ∇ ν = −∇ L ( / ∇ a ) − 12 tr χ / ∇ a + · · · . To prove sharp trace inequality and to control P and E actually dominate thearticle. Further comparison on technical details will be made in Section 2.1.4. Organization of the paper. This paper is organized as follows. In Section 2,we start with providing all the structure equations and making bootstrap assump-tions. Then by using structure equations (2.1)-(2.18) and Sobolev embedding, weestablish a series of preliminary estimates including weakly spherical property forthe metric ◦ γ . In Section 3, we prove k Λ − α K k L ∞ t L x . ∆ + R with α ≥ / ◦ γ . With thehelp of these two arguments, in Section 5, we prove the sharp trace theorem, i.e. Theorem 5.1. In order to obtain the structure required in Theorem 5.1, the com-mutators involved are decomposed in Proposition 6.1, which is the main purposeof Section 6. In Section 7, we estimate k ˆ χ, ζ, ζ, ν k L ∞ ω L t by using Theorem 5.1 andProposition 6.1. In Section 8, we prove dyadic Sobolev inequalities and (6.76) inTheorem 6.1. 2. Preliminary estimates Structure equations. The proof of Main Theorem relies crucially on thefollowing set of structure equations. We will prove (2.16) and (2.17) and refer thereader to [1, Chapter 11] and [2, Section 2] for the derivation of all other formulae. d tr χds + 12 (tr χ ) = −| ˆ χ | , (2.1) d ˆ χ AB ds + tr χ ˆ χ AB = − α AB , (2.2) dds ζ A = − χ AB ζ B + χ AB ζ B − β A , (2.3) dds / ∇ tr χ + 32 tr χ / ∇ tr χ = − ˆ χ · / ∇ tr χ − χ · / ∇ ˆ χ − ( ζ + ζ )( | ˆ χ | + 12 (tr χ ) ) , (2.4) dds tr χ + 12 tr χ tr χ = 2div ζ − ˆ χ · ˆ χ + 2 | ζ | + 2 ρ, (2.5) div ˆ χ = 12 / ∇ tr χ + 12 tr χ · ζ − ˆ χ · ζ − β, (2.6) curl ζ = 12 ˆ χ ∧ ˆ χ − σ, (2.7) div ζ = − µ − ˇ ρ − | ζ | + 12 aδ tr χ + aλ tr χ (2.8) curl ζ = ˇ σ (2.9)In what follows, we record null Bianchi equations / ∇ L β A = div α − χβ A + (2 ζ A + ζ A ) α AB (2.10) L ˇ ρ + 32 tr χ · ˇ ρ = div β + ( ζ + 2 ζ ) · β − 12 ˆ χ · ( / ∇ b ⊗ ζ − 12 tr χ · ˆ χ + ζ b ⊗ ζ ) , (2.11) L ˇ σ + 32 tr χ · ˇ σ = − curl β − ( ζ + 2 ζ ) ∧ β − 12 ˆ χ ∧ ( / ∇ b ⊗ ζ + ζ b ⊗ ζ ) , (2.12) / ∇ L β + tr χβ = − / ∇ ρ + ( / ∇ σ ) ⋆ + 2ˆ χ · β − ζρ − ζ ⋆ σ )(2.13)where ˇ ρ = ρ − 12 ˆ χ · ˆ χ, ˇ σ = σ − 12 ˆ χ ∧ ˆ χ. (2.14) QIAN WANG Moreover, there hold L ( µ ) + tr χµ = 2 ˆ χ · / ∇ ζ + ( ζ − ζ ) · ( / ∇ A tr χ + tr χζ A ) − 12 tr χ ( ˆ χ · ˆ χ − ρ + 2 ζ · ζ )+ 2 ζ · ˆ χ · ζ + (cid:18) a tr χ − a ( δ + 2 λ ) (cid:19) | ˆ χ | − aν (tr χ ) , (2.15) L ( aδ + 2 aλ ) + 32 aδ tr χ = div ζ + ˇ ρ + | ζ | + a tr χ ∇ N log n, (2.16) / ∇ L ( / ∇ a ) + 12 tr χ / ∇ a = − / ∇ ν − ˆ χ · / ∇ a − ( ζ + ζ ) · ν. (2.17)The Gauss curvature K on each S t verifies(2.18) K = − 14 tr χ tr χ + 12 ˆ χ · ˆ χ − ρ. The following two commutation formulas hold:(i) For any scalar functions U ,(2.19) dds / ∇ A U + χ AB / ∇ B U = / ∇ A F + ( ζ + ζ ) F, where F = dds U .(ii) For S t tangent 1-form U A satisfying dU A ds = F A , there holds(2.20) dds div U + χ AB / ∇ A U B = div F + ( ζ + ζ ) · F + ( 12 tr χζ A − ˆ χ AB ζ B + β A ) U A . Proof of (2.16). In view of (2.5) and (2.1), L ( a − tr χ ) = L (tr χ ) a − + tr χL ( a − )= a − (2div ζ + 2 ˇ ρ + 2 | ζ | − 12 tr χ · tr χ ) − a − L ( a )tr χ,L ( a tr χ ) = L (tr χ ) a + tr χL ( a ) = − a ( | ˆ χ | + 12 (tr χ ) ) + L ( a )tr χ By using (1.8), L (2 δ + 4 λ ) = a − (2div ζ + 2 ˇ ρ + 2 | ζ | ) − 12 tr χ ( a − tr χ + a tr χ ) + L log a ( a tr χ − a − tr χ )= a − ζ + ˇ ρ + | ζ | ) − tr χ ( δ + 2 λ ) + 2 L (log a )tr θ, we can obtain with the help of (1.8) and (1.5) L ( aδ ) = aL ( δ ) + δL ( a )= (div ζ + ˇ ρ + | ζ | ) − a tr χ ( δ + 2 λ ) + L ( a )(tr θ + δ ) − aL ( λ )= div ζ + ˇ ρ + | ζ | + a tr χ ( − δ + ∇ N log n ) − L (2 aλ )which gives the desired formula. (cid:3) Proof of (2.17). Recall that by L ( a ) = − ν , combined with (2.19)[ / ∇ L , / ∇ ] a = − χ · / ∇ a − ( ζ + ζ ) ν we may obtain relative to orthornomal frame on S t , / ∇ L / ∇ B a + 12 tr χ / ∇ B a = − / ∇ B ν − ˆ χ BC · / ∇ C a − ( ζ B + ζ B ) ν This completes the proof. (cid:3) Let D t denote ddt along a null geodesics initiating from p . In view of (1.2), D t = an dds . In comparison with geodesic foliation, the term ( ζ + ζ ) / ∇ L U in (2.19)is no longer trivial due to ζ + ζ = 0. This term however can be avoided if weconsider the commutator [ D t , / ∇ ] U instead. Similarly, in view of [1, Lemma 13.1.2],it is simpler to consider [ D t , / ∇ ] F than [ / ∇ L , / ∇ ] F for S -tangent tensor fields F . Proposition 2.1. For any smooth scalar function f , (2.21) [ D t , / ∇ ] f = − anχ · / ∇ f In view of (2.21), (2.20) and (2.6), there holds [ D t , / ∆] f = − antrχ / ∆ f − an ˆ χ · / ∇ f + 2 anβ · / ∇ f − anζ ˆ χ · / ∇ f − anζ tr χ / ∇ f − an / ∇ trχ / ∇ f. Combining [11, P.288] with the comparison formulas in [2, Section 2], we have Lemma 2.1. V, / ∇ a, r / ∇ tr χ, r µ → as t → , lim t → k ˆ χ, ζ, ζ, ν k L ∞ ( S t ) < ∞ For S tangent tensor fields F on H , we introduce the following norms. For1 ≤ p, q ≤ ∞ we define the L qt L px norm on Hk F k L qt L px := Z Z | ω | =1 | F ( t, ω ) | p nav t dµ S ! qp dt q and the L px L ∞ t norm k F k pL px L ∞ t := Z S sup t ∈ Γ ω ( v t | F | p ) dµ S . Notations and Bootstrap assumptions. We fix the following conventions • /π denotes the collection of ˆ η, ǫ, δ, ∇ N log n, / ∇ log n, λ, • ι := tr χ − r , V := tr χ − s , κ := tr χ − ( an ) − an tr χ, • A denotes the collection of ˆ χ, ζ, ζ, ν, • A denotes the collection of A and ˆ χ, / ∇ log a, /π, • The pair of quantities ( M, D M ) denotes either ( / ∇ tr χ, / ∇ ˆ χ ) , or ( µ, / ∇ ζ ) , • R denotes the collection of α, β, ρ, σ, β , • ¯ R denotes the collection of R , tr χA, A · A, • ˜ R denotes the collection of ¯ R and / ∇ A , • H t := ∪ t ′ ∈ [0 ,t ] S t ′ , with 0 < t ≤ , • S := S t , ◦ γ := r − γ , γ (0) := γ S , K := K − r . Assumption 2.1. We make the following bootstrap assumption: (BA1) k V k L ∞ ( H ) ≤ ∆ , k ˆ χ, ν, ζ, ζ k L ∞ ω L t ( H ) ≤ ∆ , | a − | ≤ where we can assume that < R < ∆ < / . The goal is to improve the inequalities in BA1 with the ∆ replaced by ∆ + R ,and | a − | ≤ . When R is sufficiently small, ∆ + R < ∆ can be achieved.We will start with deriving estimates for N ( A ) by establishing related estimatesfor M = / ∇ tr χ, µ , which will be contained in Propositions 2.4 and 2.5. Then weprove that κ and ι verify stronger estimates than A , which can be seen in (2.84)and Proposition 2.7. At last we prove ( S t , ◦ γ ) is weakly spherical.2.3. Estimates for N ( A ) , k r / M k L x L ∞ t and k M k L . Recall a few results thathave been proved in [12] and [2, 3, 10]. Proposition 2.2. [12] Under the assumption BA1, there hold (2.22) C − ≤ v t /s ≤ C, C − < rs < C, where C is a positive constant. It is easy to derive from | V | ≤ ∆ in BA1 and Proposition 2.2 that(2.23) | s tr χ | + | r tr χ | ≤ C and from | a − | ≤ in BA1 and C − < n < C that(2.24) C − < an < C, with C positive constants.With the help of Proposition 2.2, there hold the following simple inequalities bySobolev embedding in 2-D slices S = S t . • Let O sc ( f ) := f − ¯ f for any smooth function f on S where ¯ f = | S | R S f dµ γ , there holds the Poincare inequality( Poin) k r − O sc ( f ) k L ( S ) . k / ∇ f k L ( S ) . • For a smooth function Ω on S with vanishing mean, there holds the follow-ing Sobolev inequality (see [3])( GaNi) k Ω k L ∞ ( S ) . k / ∇ Ω k L ( S ) + k / ∇ Ω k L ( S ) , which implies r − k Ω k L ∞ ( S ) . k / ∇ Ω k L ( S ) + r − k / ∇ Ω k L ( S ) , (2.25) k r − Ω k L t L ∞ x . k / ∇ Ω k L + k r − / ∇ Ω k L . (2.26) • Let F be a S tangent tensor field, (see [3])( Sob) k F k L px ( S ) . k / ∇ F k − p L x ( S ) k F k p L x ( S ) + k r − p F k L x ( S ) , with 2 < p < ∞ . • Let F be a S tangent tensor field, there hold (see [2, 11]) k r − / F k L x L ∞ t + k F k L x L ∞ t + k F k L . N ( F ) , ( SobM1) ( SobM2) k r − F k L ∞ + k r − F k L t L ∞ x . N ( F ) , where N ( F ) := k r − F k L + k r − / ∇ L F k L + k r − / ∇ F k L + k / ∇D t F k L + k / ∇ F k L . (2.27) By interpolation,(2.28) k r − b F k L bt L x + k r − q − F k L qt L x . N ( F ) , with b ≥ , q ≥ . Lemma 2.2. (2.29) k V k L ∞ . ∆ Proof. This can be obtained by integrating along Γ ω the equation (2.1), i.e. dds V + 2 s V = − V − | ˆ χ | with the help of Proposition 2.2, Lemma 2.1 and k V k L ∞ + k ˆ χ k L ∞ ω L t ≤ ∆ inBA1. (cid:3) Lemma 2.3. For a S tangent tensor field F verifying (2.30) / ∇ L F + p tr χF = G · F + H with p ≥ certain integer, if lim t → r ( t ) p F = 0 and k G k L ∞ ω L t . ∆ , then thefollowing estimate holds (2.31) | F | . v − p t Z t v t ′ p | H | nadt ′ . We will constantly use the Hardy-Littlewood inequality for scalar f on H ,(2.32) (cid:13)(cid:13)(cid:13)(cid:13) s Z s | f | (cid:13)(cid:13)(cid:13)(cid:13) L s . k f k L s With the help of Lemmas 2.1 and 2.3, (2.2), (2.3), BA1, (2.32) and (1.13) we obtain Lemma 2.4. (2.33) k r − ˆ χ, r − ζ k L + k r − / ˆ χ, r − / ζ k L x L ∞ t . R , (2.34) k / ∇ L ˆ χ, / ∇ L ζ k L . R + ∆ . In view of Lemma 2.4, (1.13) and ( SobM1 ), by definition of elements of A , wecan summarize the estimates for A , Proposition 2.3. k r − A k L + k r − / A k L x L ∞ t + k / ∇ L A k L . ∆ + R . For the proofs of Lemmas 2.3 and 2.4 and Proposition 2.3, see [12].With the help of Proposition 2.3 and Lemma 2.3, we prove the following resultunder the assumption of BA1. Lemma 2.5. (2.35) k / ∇ log s k L ∞ ω L t . ∆ , (2.36) k / ∇ log s k L t L ω + k s / / ∇ log s k L ω L ∞ t . ∆ + R , (2.37) kO sc ( 1 s ) k L t L ω + k s O sc ( 1 s ) k L ∞ t L ω . ∆ + R , (2.38) k κ k L t L ω + k r / κ k L ∞ t L ω . ∆ + R . Proof. Apply (2.19) to U = s , we can derive the transport equation(2.39) dds / ∇ A ( s ) + 12 tr χ / ∇ A ( s ) = − ˆ χ AB / ∇ B ( s ) + ζ A + ζ A . Note that e A ( s ) → , as t → in view of Lemma 2.3 with G = ˆ χ and BA1, wecan derive by integrating along a null geodesic Γ ω initiating from vertex,(2.40) | s − / ∇ ( s )( t ) | . s v − t Z t v t ′ | ζ + ζ | nadt ′ . Taking L t norm first then L ∞ ω ( S ), with the help of BA1, k s − / ∇ ( s ) k L ∞ ω L t . k ζ + ζ k L ∞ ω L t . ∆ which gives (2.35).By taking L t norm first then L ω ( S ), we can obtain from (2.40) by using (2.32)that k s − / ∇ ( s ) k L ω L t . k r − ( ζ + ζ ) k L . ∆ + R , where for the last inequality we employed k r − A k L . ∆ + R in Proposition 2.3.Similarly k s − / / ∇ s k L ω L ∞ t . ∆ + R . Hence (2.36) is proved.Applying ( Poin ) to f = s , (2.37) follows as a consequence of (2.36).According to definition, we can derive κ = tr χ − s − ( an ) − an (tr χ − s ) + 2 s (1 − ( an ) − an )+ 2 O sc ( 1 s )( an ) − an − an ) − s − O sc ( an ) . (2.41)By ( Poin ), (2.24) and also in view of Propositions 2.2 and 2.3, we obtain(2.42) k s − O sc ( an ) k L t L ω . k / ∇ log( an ) k L t L ω . k r − ( ζ + ζ ) k L ( H ) . ∆ + R , and similarly(2.43) k s − O sc ( an ) k L ∞ t L ω . k r ( ζ + ζ ) k L ∞ t L ω . ∆ + R . Using (2.24) and (2.42), the last term in (2.41) can be estimated as follows k ( an ) − s − O sc ( an ) k L t . k s − O sc ( an ) k L t L ω . ∆ + R . (2.44)Combining (2.42), (2.44) and (2.29), we obtain k r − κ k L . k r − O sc ( 1 s ) k L + ∆ + R . ∆ + R . where, for the last inequality, we employed (2.37).By (2.37), (2.43) and (2.29), we can get k r κ k L ∞ t L ω ≤ k r O sc ( 1 s ) k L ∞ t L ω + k ( an ) − r − O sc ( an ) k L ∞ t L ω + | V | + k s − (( an ) − an − k L ∞ t L ω . ∆ + R . The proof is complete. (cid:3) This initial condition can be easily checked by using the comparison formulas in [2, Section2]. Lemma 2.6. (cid:13)(cid:13)(cid:13)(cid:13) tr χ − r (cid:13)(cid:13)(cid:13)(cid:13) L t . ∆ + R , (2.45) (cid:13)(cid:13)(cid:13)(cid:13) tr χ − r (cid:13)(cid:13)(cid:13)(cid:13) L t L ω . ∆ + R . (2.46) Proof. We can derive the transport equation(2.47) dds ( r ( trχ − r )) = ( an ) − r an tr χκ − r ( an ) − an | ˆ χ | . by combining(2.48) dds r = ( an ) − r an tr χ. with dds tr χ = − ( an ) − an tr χ · tr χ + ( an ) − an ( 12 (tr χ ) − | ˆ χ | )which can be checked in view of the definition of trχ and (2.1).Integrate (2.47) in t in view of r tr χ − → t → | tr χ − r | . r Z s ( t )0 { (cid:12)(cid:12)(cid:12) ( an ) − r an tr χκ (cid:12)(cid:12)(cid:12) + r ( an ) − an | ˆ χ | } ds ( t ′ ) . In view of (2.24), taking L t with the help of (2.32) yields(2.49) k tr χ − r k L t . k rκ tr χ k L t L ω + k r | ˆ χ | k L t L ω . By (2.29) and (2.38), we can obtain k r tr χκ k L t L ω . k rV κ k L t L ω + k rs κ k L t L ω . (∆ + 1) k κ k L t L ω . ∆ + R . (2.50)By Proposition 2.3, we have(2.51) k r | ˆ χ | k L t L ω . k r / ˆ χ k L ∞ t L ω k r / ˆ χ k L t L ω . ∆ + R . (2.45) then follows by connecting (2.49), (2.50) and (2.51).Note that it is straightforward to have(2.52) tr χ − r = V − V + 2 O sc ( 1 s ) + tr χ − r , hence k tr χ − r k L t L ω . k V k L ∞ + kO sc ( 1 s ) k L t L ω + k trχ − r k L t which implies (2.46) with the help of (2.29), (2.37) and (2.45). (cid:3) Lemma 2.7. Denote by ¯ R one of the quantities, R , tr χA, A · A , there holds (2.53) k ¯ R k L ( H t ) . ∆ + R . Proof. The estimate about R can be obtained directly from (1.13). With the helpof Proposition 2.3 and BA1(2.54) k A · A k L ( H t ) . k A k L x L ∞ t k A k L ∞ ω L t . ∆ + R . By BA1 and Proposition 2.3, we have k tr χA k L ( H t ) . k V · A k L ( H t ) + k s − A k L ( H t ) . k r − A k L . ∆ + R . The estimate thus follows. (cid:3) Lemma 2.8. Let K = K − r , then k K k L ( H t ) . ∆ + R . Proof. In view of (2.18) and (1.8), K − r = a − r + a ι r + ι ( V + s ) a − a tr χ ( λ + 12 δ ) − ˇ ρ (2.55)By / ∇ L a = − ν , (2.32) and (1.13)(2.56) (cid:13)(cid:13)(cid:13)(cid:13) a − r (cid:13)(cid:13)(cid:13)(cid:13) L ( H t ) = (cid:13)(cid:13)(cid:13)(cid:13) r − Z s aν (cid:13)(cid:13)(cid:13)(cid:13) L ω L t . k r − /π k L . ∆ + R . In view of (2.55), by (2.29) and r ≈ s in (2.22), also using (2.56), (2.46), (2.53), (cid:13)(cid:13)(cid:13)(cid:13) K − r (cid:13)(cid:13)(cid:13)(cid:13) L ( H t ) . (cid:13)(cid:13)(cid:13)(cid:13) a − r (cid:13)(cid:13)(cid:13)(cid:13) L + k r − ι k L + k ¯ R k L . ∆ + R which is the desired estimate. (cid:3) With Lemma 2.8, we can prove the following estimates. Lemma 2.9. Let D be the operator that takes any S tangent 1-form F to ( div F, curl F ) .Let D be the operator that takes any S -tangent symmetric, traceless, 2-tenorfields F to div F . Denote by D one of the operators D , D . For any appropriate S tangenttensor fields F in the domain of D , if k r − F k L ∞ t L x . ∆ , there holds (2.57) k / ∇ F k L ( H t ) + k r − F k L ( H t ) . kD F k L ( H t ) + ∆ + R . For smooth scalar functions Ω , if k r − / ∇ Ω k L ∞ t L x . ∆ , then (2.58) k / ∇ Ω k L ( H t ) + k r − / ∇ Ω k L ( H t ) . k / ∆Ω k L ( H t ) + ∆ + R . Proof. Let us prove (2.58) first. In view of B¨ochner identity on S ,(2.59) Z S | / ∇ Ω | + K | / ∇ Ω | = Z S | / ∆Ω | , we obtain(2.60) Z S t ′ | / ∇ Ω | + r − | / ∇ Ω | = Z S t ′ | / ∆Ω | − Z S t ′ K | / ∇ Ω | . Noticing that on S t ′ , there holds by ( Sob ) and k r − / ∇ Ω k L ∞ t L x . ∆ , k / ∇ Ω k L x ( S t ′ ) . k / ∇ Ω k L x ( S t ′ ) k / ∇ Ω k L x ( S t ′ ) + r − k / ∇ Ω k L x ( S t ′ ) . ∆ k / ∇ Ω k L x ( S t ′ ) + ∆ . (2.61)Integrating (2.60) on 0 < t ′ ≤ t , in view of (2.61) and Lemma 2.8, (2.58) follows byusing Young’s inequality.For D : F → (div F, curl F ) and D : F → div F we recall the identities (see [1,Proposition 2.2.1]) Z S | / ∇ F | + K | F | = Z S |D F | , Z S | / ∇ F | + 2 K | F | = 2 Z S |D F | . Then (2.57) follows in the same way as (2.58). (cid:3) By Lemma 2.9 and (2.53), we can derive the following Lemma 2.10. For M = / ∇ tr χ, µ , there holds (2.62) kD M k L ( H t ) . ∆ + R + k M k L ( H t ) . More precisely, k / ∇ ˆ χ k L ( H t ) . ∆ + R + k / ∇ tr χ k L ( H t ) , k / ∇ ζ k L ( H t ) . ∆ + R + k µ k L ( H t ) . Proof. By using k r − A k L x L ∞ t . ∆ + R in Proposition 2.3 and applying Lemma2.9 to F = ˆ χ, ζ , we can derive that k / ∇ ˆ χ k L ( H t ) . kD ˆ χ k L ( H t ) + ∆ + R , k / ∇ ζ k L ( H t ) . kD ζ k L ( H t ) + ∆ + R . In view of (2.6), (2.8) and (2.9), D ˆ χ = / ∇ tr χ + ¯ R, D ζ = ( µ, 0) + ¯ R. By using (2.53), (2.62) can be proved. (cid:3) Recall that M = / ∇ tr χ or µ . (2.4) and (2.15) can be symbolically recast as(2.63) / ∇ L M + p χM = ˆ χ · M + H + H + H where H = A · ( D M + F ) , H = A · A · A, and H = r − ¯ R, ι · ¯ R, with(2.64) ( p, F ) = (2 , / ∇ tr χ ) if M = µ (3 , 0) if M = / ∇ tr χ. We will establish the following estimates Proposition 2.4. Let M denote either / ∇ tr χ or µ , there hold (2.65) k r / M k L x L ∞ t + k M k L . R + ∆ , (2.66) k / ∇ ζ, / ∇ ˆ χ k L . R + ∆ . Proof. Noticing that (2.66) can be obtained immediately by combining the secondestimate in (2.65) with Lemma 2.10, we first consider the second norm in (2.65).By (2.31), (2.63) and Lemma 2.1, integrating along the null geodesic Γ ω initiatingfrom vertex, we obtain(2.67) | M | ≤ X i =1 (cid:12)(cid:12)(cid:12)(cid:12) v − p t Z t v p t ′ | H i | nadt ′ (cid:12)(cid:12)(cid:12)(cid:12) = I ( t ) + I ( t ) + I ( t ) . This means signs and coefficients on the right side of the expression can be ignored. In view of | a − | ≤ / a m , m ∈ N can be ignored in H i when we employ(2.63) to prove Proposition 2.4. Consider H with the help of (2.32), Z (cid:12)(cid:12)(cid:12) v / t I ( t ) (cid:12)(cid:12)(cid:12) nadt = Z (cid:18) v − p + t Z t v p t ′ | H | nadt ′ (cid:19) nadt (2.68) . Z (cid:12)(cid:12)(cid:12)(cid:12) r − p +1 Z t r ′ p − | r ′ ¯ R | nadt ′ (cid:12)(cid:12)(cid:12)(cid:12) nadt . (cid:12)(cid:12)(cid:12)(cid:12)Z | r ′ ¯ R | nadt ′ (cid:12)(cid:12)(cid:12)(cid:12) where we employed v t ′ ≈ ( r ′ ) ≈ ( t ′ ) . Then by taking L ω ( S ), with the help of (2.53), we obtain(2.69) k I k L . k ¯ R k L . R + ∆ . Similarly, we have Z (cid:12)(cid:12)(cid:12) v / t I ( t ) (cid:12)(cid:12)(cid:12) nadt = Z (cid:18) v − p + t Z t v p t ′ | H | nadt ′ (cid:19) nadt ≤ Z (cid:18) v − p + t Z t v p t ′ | A | | A | nadt ′ (cid:19) nadt . k A k L t k r / A k L ∞ t . By taking L ω , we obtain in view of BA1 and Proposition 2.3 that(2.70) k I k L . k A k L ∞ ω L t k r / A k L ω L ∞ t . ∆ (∆ + R ) . Now consider H := A · ( D M + F ). We have (cid:12)(cid:12)(cid:12) v / t I ( t ) (cid:12)(cid:12)(cid:12) . k A k L t v − p +1 t Z s ( t )0 v pt ′ (cid:0) |D M | + | F | (cid:1) ds ( t ′ ) ! / . (2.71)Take L ω , k v / t I k L ω . k A k L ∞ ω L t ( kD M k L + k F k L ) . Hence in view of (2.62) and BA1(2.72) k I k L . ∆ ( k M k L + k F k L + ∆ + R ) . Consequently, in view of (2.69), (2.70) and (2.72), if M = / ∇ tr χ , we have k / ∇ tr χ k L . ∆ k / ∇ tr χ k L + ∆ + R , (2.73)and if M = µ k µ k L . ∆ ( k µ k L + k / ∇ tr χ k L ) + ∆ + R . (2.74)From (2.73), noticing that 0 < ∆ < / 2, we conclude(2.75) k / ∇ tr χ k L . ∆ + R . k µ k L . ∆ + R then follows from (2.74) by using (2.75). Now we consider the first estimate in (2.65). v t I ( t ) ≤ v − p t Z s ( t )0 v p − t ′ | ¯ R | ds ( t ′ ) . s − p Z s ( t )0 s p − | ¯ Rr | ds ( t ′ ) . Z s ( t )0 | ¯ Rr | ds ( t ′ ) ! where we used s ≈ v t , r ≈ s in Proposition 2.2 to derive the second inequality.Thus(2.76) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) sup t ∈ (0 , | v t I ( t ) | (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ω . k ¯ R k L ( H ) . ∆ + R . We can proceed in a similar fashion for the other two terms, v t I ( t ) ≤ v − p t Z s ( t )0 v p t ′ | A | | A | ds ( t ′ ) . s − p Z s ( t )0 s p | A | | A | ds ( t ′ ) . r k A k L t k r / A k L ∞ t , then(2.77) (cid:13)(cid:13)(cid:13)(cid:13) sup Proposition 2.5. Let M denote either µ or / ∇ tr χ . There holds (2.79) N ( A ) + k r / M k L x L ∞ t + k M k L . ∆ + R Remark . By ( SobM1 ),( Sob ) and (1.13), it is easy to check ( a m A ) and ( a m A )with m ∈ N verify the same estimates as A and A respectively. Consequently theycan also be regarded as elements of A and A respectively.2.4. More Estimates for κ and ι . The main purpose of this subsection is toprovide estimates for N ( κ, ι ) and k κ, ι k L t L ∞ ω . We first derive a simple consequencefrom (2.79) with the help of (2.26). Lemma 2.11. (2.80) k r − O sc ( an ) k L t L ∞ ω . ∆ + R 08 QIAN WANG Proof. Apply (2.26) to Ω = an − an ,(2.81) k r − Ω k L t L ∞ ω . k / ∇ Ω k L + k r − / ∇ Ω k L . Note that with the help of ζ + ζ = / ∇ log( an ) and (2.24), there holds k r − / ∇ Ω k L ∞ t L x . k r − / ∇ log( an ) k L ∞ t L x = k r − ( ζ + ζ ) k L ∞ t L x . ∆ + R . in view of (2.58), we deduce k r − Ω k L t L ∞ ω . k / ∆Ω k L + ∆ + R . We obtain in view of / ∆( an ) = an ( | ζ + ζ | + / ∆ log( an )), (2.24) k r − Ω k L t L ∞ ω . k / ∆ log( an ) k L + k ζ + ζ k L + ∆ + R . N ( ζ + ζ )(1 + N ( ζ + ζ )) + ∆ + R (2.82) . ∆ + R where we employed ( SobM1 ) and (2.79) for the last two inequalities. (cid:3) Proposition 2.6. k r / ∇ ( 1 s ) k L . ∆ + R , (2.83) kO sc ( 1 s ) , O sc ( tr χ ) , κ, ι k L t L ∞ ω . ∆ + R . (2.84) Proof. Using (2.39), in view of the commutation formula in [1, Lemma 13.1.2],symbolically, we obtain dds / ∇ s + tr χ / ∇ s = − / ∇ tr χ / ∇ s + ˆ χ · / ∇ s − 12 tr χ ( ζ + ζ ) / ∇ s − / ∇ ˆ χ · / ∇ s − ( ζ + ζ ) ˆ χ · / ∇ s + ( ζ + ζ ) · ( ζ + ζ ) + / ∇ ( ζ + ζ ) + ( χ · ζ + β ) / ∇ s. We then rewrite it as dds / ∇ s + tr χ / ∇ s = ˆ χ · / ∇ s + ( ¯ R + M + / ∇ ˆ χ ) · / ∇ s + A · A + / ∇ ( ζ + ζ ) . Apply Lemma 2.3 to the above equation with the help of k ˆ χ k L ∞ ω L t ( H ) ≤ ∆ in BA1and lim t → r / ∇ s = 0,(2.85) | / ∇ s | . v − t Z s ( t )0 v t ′ | ( ¯ R + M + / ∇ ˆ χ ) · / ∇ s + A · A + / ∇ ( ζ + ζ ) | ds ( t ′ ) . Hence, by H¨older inequality and (2.32), we have k / ∇ s k L ω L t . k s − / ∇ s k L ∞ ω L t ( k ¯ R k L + k M k L + k / ∇ A k L )+ k r ( A · A + / ∇ A ) k L ω L t . (2.86)By (2.53), (2.65), (2.35), (2.54) and (2.79), we obtain(2.87) k / ∇ s k L ω L t . ∆ + R . By a straightforward calculation, − s / ∇ ( 1 s ) = s / ∇ ( s − / ∇ s ) = / ∇ s − s − / ∇ s · / ∇ s we deduce k r / ∇ ( O sc (1 /s )) k L . k / ∇ s k L t L ω + k / ∇ log s · / ∇ s k L t L ω . By (2.36) and (2.35)(2.88) k / ∇ log s · / ∇ s k L t L ω . k / ∇ log s k L ∞ ω L t k s / ∇ log s k L ω L ∞ t . ∆ (∆ + R ) . Combined with (2.87),(2.89) k r / ∇ ( O sc (1 /s )) k L . ∆ + R . Now we apply (2.26) to Ω = r O sc ( s ). By (2.36), k / ∇ ( O sc ( s )) k L . ∆ + R .Also in view of (2.89), we conclude that(2.90) kO sc ( 1 s ) k L t L ∞ ω . ∆ + R . By O sc (tr χ ) = O sc ( V ) + 2 O sc ( s ) and (2.29), it follows from (2.90) that(2.91) kO sc (tr χ ) k L t L ∞ ω . ∆ + R . In view of ι = O sc (tr χ ) + tr χ − r , (2.91) together with (2.45) implies k ι k L t L ∞ ω . ∆ + R . In view of (2.41), symbolically,(2.92) κ = V − ( an ) − anV + O sc ( 1 s ) anan + s − ( an ) − O sc ( an )+( an ) − s − O sc ( an ) . Taking L t L ∞ ω of κ in view of (2.92), by (2.80) and (2.24), the last two terms arebounded by ∆ + R . Due to (2.90) and (2.29), L t L ∞ ω of the remaining three termsare bounded by ∆ + R . Thus we conclude k κ k L t L ∞ ω . ∆ + R . (cid:3) Proposition 2.7. N ( ι ) + N ( κ ) . ∆ + R . Proof. In view of (2.1) and (2.48), we have / ∇ L (tr χ − r ) = − 12 (tr χ − r )tr χ − | ˆ χ | − r κ. By BA1 and (2.46), (2.38) and (2.54), we have k / ∇ L ( ι ) k L . ∆ + R . Togetherwith (2.46) and (2.79), we conclude N ( ι ) . ∆ + R .Now we take L ( H ) norm of dds κ with the help of dds κ + tr χ · κ = −| ˆ χ | + ( an ) − | an ˆ χ | + 12 κ − 12 ( an ) − ( an ) κ (2.93) + ( an ) − (cid:16) an tr χ O sc ( / ∇ L ( an )) − O sc ( / ∇ L ( an )) anκ (cid:17) . By (2.38) and (2.23), k tr χκ k L . ∆ + R . For the terms on the right of (2.93),we first claim(2.94) k r − O sc ( / ∇ L ( an )) k L . ∆ + R Indeed, / ∇ / ∇ L ( an ) = / ∇ ( an / ∇ L log( an )) = / ∇D t log( an )by (2.21),(2.95) / ∇ / ∇ L ( an ) = D t / ∇ log( an ) + anχ · / ∇ log( an ) = an { / ∇ L ( ζ + ζ ) + χ · ( ζ + ζ ) } . By ( Poin ), (2.95) and (2.24) k r − O sc ( / ∇ L ( an )) k L . k / ∇ / ∇ L ( an ) k L . k tr χA k L + k / ∇ L A k L + k ˆ χ · A k L (2.96) (2.94) follows by using (2.53) and Proposition 2.3.Using (2.94), (2.24) and (2.23) k ( an ) − an tr χ O sc ( / ∇ L ( an )) k L . k r − O sc ( / ∇ L ( an )) k L . ∆ + R . Similarly k ( an ) − O sc ( / ∇ L ( an )) anκ k L . kO sc ( / ∇ L ( an )) k L t L ω . ∆ + R . Now consider other terms on the right of (2.93), by (2.38) and (2.84) k κ k L . k rκ k L ∞ t L ω k κ k L t L ∞ ω . ∆ + R , by ( SobM1 ) and (2.79) k ˆ χ · ˆ χ k L . k ˆ χ k L . ∆ + R and the other two terms can be estimated similarly in view of (2.24). Hence(2.97) k / ∇ L κ k L . ∆ + R . At last, we obtain by definition / ∇ κ = / ∇ tr χ + / ∇ log( an ) · ( an ) − an tr χ. By using(2.23), (2.24) and (2.79), k / ∇ κ k L ( H ) . k / ∇ tr χ k L ( H ) + k r − ( ζ + ζ ) k L ( H ) . ∆ + R . Combined with (2.38) and (2.97), we can conclude N ( κ ) . ∆ + R . (cid:3) Remark . By Proposition 2.7 and (2.84), we can regard κ and ι as elements of A . Without making the strong assumption (1.23) (see [7]), under the weaker condi-tion (1.13) only, tr χ − r no longer satisfies the L ∞ ( H ) estimate as (1.14) for tr χ − s .Since S t is not a level set of affine parameter s , obviously, / ∇ r = 0 while / ∇ s = 0.We will check the weakly spherical property for ◦ γ = r − γ . Integral operators Λ − α ,geometric Littlewood Paley decompositions P k and Besov norms will then be de-fined by heat flow U ( τ ) with respect to ◦ γ instead of s − γ . Due to the factor “ an ”in (2.48), the nontrivial evolution of r adds technical complexity. This issue can besettled by proving tr χ − ( an ) − an tr χ and tr χ − r verify stronger estimates than ˆ χ .Examples of the application of (2.84) and Proposition 2.7 can be seen in the proofof (2.100), Proposition 3.3, Theorem 5.1, etc.2.5. Weakly spherical surfaces. Let γ be the restriction metric on S t , and definethe rescaled metric ◦ γ on S t by ◦ γ = r − γ . Let γ (0) ij denote the canonical metric on S . Note that(2.98) lim t → ◦ γ ij = γ (0) ij , lim t → ∂ k ◦ γ ij = ∂ k γ (0) ij where i, j, k = 1 , 2. With its aid, using the bootstrap assumption BA1, (2.79), wewill prove that for each 0 < t ≤ S t , ◦ γ ) is a weakly spherical surface. Proposition 2.8. For the transport local coordinates ( t, ω ) , the following propertieshold true for all surfaces S t of the time foliation on the null cone H : the metric ◦ γ ij ( t ) on each S t verifies weakly spherical conditions i.e. k ◦ γ ij ( t ) − γ (0) ij k L ∞ ω . ∆ (2.99) k ∂ k ◦ γ ij ( t ) − ∂ k γ (0) ij k L ω L ∞ t . ∆ (2.100) Proof. Since relative to the transport coordinate on H , dds γ ij = 2 χ ij ,(2.101) ddt ( ◦ γ ij ) = an ( κ · γ ij + 2 ˆ χ ij ) r − . Integrating (2.101) along null geodesic initiating from vertex, with the help of (2.98),(2.99) follows in view of (2.84) and BA1.Integrating the following transport equation along a null geodesic initiating fromvertex, ddt ∂ k ◦ γ ij = an (cid:16) ∂ k log( an ) κ ◦ γ ij + ∂ k tr χ ◦ γ ij + κ∂ k ◦ γ ij +2 ∂ k ˆ χ ij r − + 2 ∂ k log( an ) ˆ χ ij r − (cid:17) where i, j, k = 1 , 2, with the initial condition given by (2.98), by k κ k L t L ∞ ω . ∆ + R in (2.84) and a similar argument to Lemma 2.3, we can obtain (cid:12)(cid:12)(cid:12) ∂ k ◦ γ ij ( t ) − ∂ k γ (0) ij (cid:12)(cid:12)(cid:12) . (cid:12)(cid:12)(cid:12) Z s ( t )0 (cid:16) ∂ k tr χ · ◦ γ ij + r − ( / ∇ k ˆ χ ij − Γ · ˆ χ )+ ∂ k log( an ) ◦ γ ij κ + 2 ∂ k log( an ) ˆ χ ij r − + κ∂ k γ (0) ij (cid:17) ds ( t ′ ) (cid:12)(cid:12)(cid:12) , where Γ represents Christoffel symbols, and Γ · ˆ χ stands for the terms P l =1 Γ lki ˆ χ lj with l = 1 , 2. Then with the help of (2.99), (cid:13)(cid:13)(cid:13)(cid:13) sup 2, also using (2.99) k Γ k L ω . X i,j,k =1 , k ∂ k ( ◦ γ ij − γ (0) ij ) k L ω + C, where C is the constant such that the Christoffel symbol of γ (0) satisfies | ∂γ (0) | ≤ C ,(2.100) then follows by using k ˆ χ k L ∞ ω L t ≤ ∆ < / (cid:3) k Λ − α K k L ∞ t L x and elliptic estimates Define the operator Λ a with a ≤ S -tangent tensor fields F (3.1) Λ a F := r − a Γ( − a/ Z ∞ τ − a − e − τ U ( τ ) F dτ, where Γ denotes Gamma function and U ( τ ) F is defined on ( S t , ◦ γ ) by(3.2) ∂∂τ U ( τ ) F − ∆ ◦ γ U ( τ ) F = 0 , U (0) F = F. The definition of Λ a extends to the range a > < a ≤ m thatΛ a F = Λ a − m · ( r − Id − ∆ γ ) m F. We record the basic properties of Λ a in the following result (see [3]). Proposition 3.1. (i) Λ = Id and Λ a · Λ b = Λ a + b for any a, b ∈ R . (ii) For any S -tangent tensor field F and any a ≤ r a k Λ a F k L ( S ) . k F k L ( S ) . (iii) For any S -tangent tensor field F and any b ≥ a ≥ r a k Λ a F k L ( S ) . r b k Λ b F k L ( S ) and k Λ a F k L ( S ) . k Λ b F k ab L ( S ) k F k − ab L ( S ) . (iv) For any S -tangent tensor fields F and G and any ≤ a < k Λ a ( F · G ) k L ( S ) . k Λ F k L ( S ) k Λ a G k L ( S ) + k Λ a F k L ( S ) k Λ G k L ( S ) . (v) For any S -tangent tensor field F there holds with < p < ∞ and a > − p k F k L p ( S ) . k Λ a F k L ( S ) . (vi) For any a ∈ R and any S -tangent tensor field F k F k H a ( S ) := k Λ a F k L ( S ) Proposition 3.2. Under the assumption of BA1 , if R > is sufficiently small,then for all ≤ α < there hold K α := k Λ − α ( K − r − ) k L ∞ t L x . ∆ + R , (3.3) k Λ − α ˇ ρ k L ∞ t L x . ∆ + R . (3.4)For smooth scalar functions f on H , with p > α ≥ / 2, define the good partof the commutator [Λ − α , D t ] f by[Λ − α , D t ] g f := r α C α Z ∞ τ α − e − τ [ U ( τ ) , D t ] f dτ, with C α = 1Γ( α ) . We first prove Proposition 3.2 by assuming the following result. Proposition 3.3. Let f be a smooth scalar function H , there holds k r − α [Λ − α , D t ] g f k L t L x . (1 + I − p α )(∆ + R ) k f k L , where p > and / ≤ α < , and I α := 1 + K − α α + K α . Proof of Proposition 3.2. It suffices to prove the following estimates k Λ − α ( K + ˇ ρ ) k L ∞ t L x . ∆ + R , (3.5) k Λ − α ˇ ρ k L ∞ t L x . (∆ + R )(1 + I − p α ) , p > . (3.6)By (3.5) and (3.6), according to the definition of I α , with − α · (1 − p ) < 1, we canobtain (3.3).Let us first prove (3.5). By (2.55),(3.7) K + ˇ ρ = a − r + a ι r + ι tr χa χA, By Proposition 3.1 (ii), BA1, (2.22) Proposition 2.3 and (2.23), for α ≥ / k Λ − α (tr χA ) k L ∞ t L x . k r α +1 tr χA k L ∞ t L ω . ∆ + R . By Proposition 3.1 (ii) and (1.13), we have for α ≥ (cid:13)(cid:13)(cid:13)(cid:13) Λ − α ( a − r ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L x . k r − Λ − α ( a − k L ∞ t L ω . k r α − ( a − k L ∞ t L ω . (cid:13)(cid:13)(cid:13)(cid:13) r α − Z t a n / ∇ L adt ′ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L ω . k / ∇ L a k L ω L t . k r − ν k L . ∆ + R . By Proposition 3.1 (ii), ( SobM1 ) and Proposition 2.7, we have for α ≥ , (cid:13)(cid:13)(cid:13)(cid:13) Λ − α ( a ιr ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L x . k a r α ι k L ∞ t L ω . N ( ι ) . ∆ + R . Similarly, by (2.23), k Λ − α ( a tr χι ) k L ∞ t L x . k r α ( a tr χι ) k L ∞ t L ω . k r α ι k L ∞ t L ω . ∆ + R . This finishes the proof of (3.5).Next we prove (3.6). Let W ( t ) = (Λ − α ˇ ρ ) ( t ) − (Λ − α ˇ ρ ) (0). Then W ( t ) = Z t Z S t ′ (cid:18) ddt (Λ − α ˇ ρ ) + an tr χ (Λ − α ˇ ρ ) (cid:19) dµ γ dt ′ = Z t Z S t ′ (cid:16) D t , Λ − α ] g ˇ ρ · Λ − α ˇ ρ + ( an tr χ + αan tr χ )(Λ − α ˇ ρ ) + 2Λ − α D t ˇ ρ · Λ − α ˇ ρ (cid:17) dµ γ dt ′ . (3.8)Let ϑ ( t ) be a smooth cut-off function with ϑ (0) = 1 and supported in [0 , ], | ϑ | ≤ − α ˇ ρ (0)) = ( ϑ Λ − α ˇ ρ )(0) − ( ϑ Λ − α ˇ ρ )(1) 24 QIAN WANG can be treated similar to W ( t ), then k Λ − α ˇ ρ k L ∞ t L x . k [ D t , Λ − α ] g ˇ ρ k L t L x · k Λ − α ˇ ρ k L ∞ t L x + k ( an tr χ + αan tr χ )(Λ − α ˇ ρ ) k L t L x (3.10) + Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z S t ′ Λ − α D t ˇ ρ · Λ − α ˇ ρdµ γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt ′ + Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z S t ′ ϑ Λ − α D t ˇ ρ · Λ − α ˇ ρdµ γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt ′ (3.11) + Z Z S t ′ (cid:12)(cid:12)(cid:12)(cid:12) ϑ ddt ϑ (Λ − α ˇ ρ ) (cid:12)(cid:12)(cid:12)(cid:12) dµ γ dt ′ . (3.12)In view of (2.23) and Proposition 3.1 (ii), Z Z S t ′ ( | an tr χ | + | an tr χ | )(Λ − α ˇ ρ ) dµ γ ′ dt ′ . k r − Λ − α ˇ ρ k L . k ˇ ρ k L . By | ddt ϑ | . α ≥ ,(3.12) . k Λ − α ˇ ρ k L . k ˇ ρ k L . We only need to estimate the first term in (3.11), and the second one will followsimilarly. Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z S t ′ Λ − α D t ˇ ρ · Λ − α ˇ ρdµ γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt ′ . k Λ − α D t ˇ ρ k L k ˇ ρ k L . (3.13)Assuming(3.14) k Λ − α D t ˇ ρ k L . ∆ + R , then (3.13) . (∆ + R ) , and it follows that k Λ − α ˇ ρ k L ∞ t L x . (∆ + R ) (1 + I − p α ) k Λ − α ˇ ρ k L ∞ t L x + (∆ + R ) . Thus (3.6) is proved. (3.4) follows as an immediate consequence.To prove (3.14), we rely on the transport equation derived by (2.11), D t ˇ ρ + 32 an tr χ ˇ ρ = div( anβ ) − / ∇ ( an ) β + anA · ˜ R = div( anβ ) + anA · ˜ R. By Proposition 3.1, (2.23) and (1.13), with α ≥ , k Λ − α ( an tr χ ˇ ρ ) k L . k r α tr χ ˇ ρ k L . k ˇ ρ k L . ∆ + R . By Proposition 3.1 and (1.13), for α ≥ ,(3.15) k Λ − α div( anβ ) k L . k r α − anβ k L . ∆ + R . By Proposition 3.1 (v) and H¨older inequality, we obtain Z k Λ − α ( anA · ˜ R ) k L ( S t ′ ) dt ′ = Z Z S t ′ Λ − α ( anA · ˜ R ) · ( anA · ˜ R ) dµ γ dt ′ ≤ k Λ − α ( anA · ˜ R ) k L t L x k anA · ˜ R k L t L / x . k Λ − α + + ( anA · ˜ R ) k L t L x k A k L ∞ t L x k ˜ R k L . Consequently, by ( SobM1 ), (2.79), (2.53) and Proposition 3.1 (ii), Z t k Λ − α ( anA · ˜ R ) k L ( S t ′ ) dt ′ . k r α − − Λ − α ( anA · ˜ R ) k L ( H ) (∆ + R ) , which implies(3.16) k Λ − α ( anA · ˜ R ) k L ( H ) . ∆ + R . Thus we complete the proof of (3.14). (cid:3) Proof of Proposition 3.3. By definition (3.1) and Proposition 2.1, r − α [Λ − α , D t ] g f = C α Z ∞ dτ τ α − e − τ Z τ r U ( τ − τ ′ )([ / ∆ , D t ] U ( τ ′ ) f − an tr χ / ∆ U ( τ ′ ) f ) dτ ′ = C α Z ∞ dτ τ α − e − τ Z τ r U ( τ − τ ′ )( / ∇ φ ( τ ′ ) + φ ( τ ′ )) , (3.17)where φ ( τ ′ ) = an ( ˆ χ + κ ) · / ∇ U ( τ ′ ) f,φ ( τ ′ ) = { an ( β + / ∇ A + A · A + r − A ) + / ∇ ( anκ ) } / ∇ U ( τ ′ ) f. Noticing that κ can be regarded as an element of A , thus we have / ∇ ( anκ ) = anA · A + an / ∇ A , and φ ( τ ′ ) = anA · / ∇ U ( τ ′ ) f, φ ( τ ′ ) = an ( β + / ∇ A + A · A + r − A ) / ∇ U ( τ ′ ) f. (3.18)Let us set Φ = C α Z ∞ dτ τ α − e − τ Z τ r U ( τ − τ ′ ) / ∇ φ ( τ ′ ) dτ ′ Φ = C α Z ∞ dτ τ α − e − τ Z τ r U ( τ − τ ′ ) φ ( τ ′ ) dτ ′ . The difference between φ , φ in (3.18) and those in [10, Page 41] is the extra factor“ an ” in (3.18), which can be easily treated by using the estimates k / ∇ ( an ) k L ∞ t L x . ∆ + R and (2.24). Based on the estimate N ( A ) . ∆ + R which has beenproved in (2.79) and Proposition 2.7, also using (2.24), we can proceed exactly as[2, Appendix] or [10, pages 41–43] to obtain for ≤ α < p > k Φ k L t L x + k Φ k L t L x . k f k L (1 + I − p α )(∆ + R ) . The proof is therefore complete. (cid:3) L estimates for Hodge operators. Consider the following Hodge opera-tors on 2-surfaces S := S t • The operator D takes any 1-forms F into the pairs of functions (div F, curl F ). • The operator D takes any symmetric traceless 2-tensors F on S into the1-forms div F . • The operator ⋆ D takes the pairs of scalar functions ( ρ, σ ) into the 1-forms − / ∇ ρ + ( / ∇ σ ) ⋆ on S . • The operator ⋆ D takes 1-forms F on S into the 2-covariant, symmetric,traceless tensors − d L F γ , where( d L F γ ) ab = / ∇ b F a + / ∇ a F b − (div F ) γ ab . Using Proposition 3.2 and B¨ochner identity, we can follow the the same way as[2, 10] to obtain elliptic estimates for Hodge operators. For various properties of these operators please refer to [1, Page 38] and [2, Section 4]. Proposition 3.4. The following estimates hold on H , (i) Let D denote either D or D . The operator D is invertible on its range,for S tangent tensor F in the range of D , k / ∇D − F k L ( S ) + k r − D − F k L ( S ) . k F k L ( S ) . (ii) The operator ( − / ∆) is invertible on its range and its inverse ( − / ∆) − verifiesthe estimate k / ∇ ( − / ∆) − f k L ( S ) + k r − / ∇ ( − / ∆) − f k L ( S ) . k f k L ( S ) . (iii) The operator ⋆ D is invertible as an operator defined for pairs of H func-tions with mean zero (i.e. the quotient of H by the kernel of ⋆ D ) andits inverse ⋆ D − takes S -tangent L F (i.e. the full range of ⋆ D )into pair of functions ( ρ, σ ) with mean zero, such that − / ∇ ρ + ( / ∇ σ ) ⋆ = F ,verifies the estimate k / ∇ ⋆ D − F k L ( S ) . k F k L ( S ) , and by (i) and duality argument k r − ⋆ D − F k L ( S ) . k F k L ( S ) . (iv) The operator ⋆ D is invertible as an operator defined on the quotient of H -vector fields by the kernel of ⋆ D . Its inverse ⋆ D − takes S -tangent2-forms Z which is in L space into S tangent 1-forms F (orthogonal tothe kernel of D ), such that ⋆ D F = Z , verifies the estimate k / ∇ ⋆ D − Z k L ( S ) . k Z k L ( S ) . As a consequence of (i)-(iv), let D − be one of the operators D − , D − , ⋆ D − or ⋆ D − . By duality argument, we have the following estimate for appropriate tensorfields F , kD − div F k L ( S ) . k F k L ( S ) . Using Proposition 3.2 and Lemma 2.8 and following the similar argument in [2,Proposition 4.24 and Lemma 6.14] we can obtain Lemma 3.1. Let D one of the operators D , D and ⋆ D . For F pairs of scalarfunctions in the first case, S -tangent one form for the second and third case, therehold N ( D − F ) . N ( F ) and N ( / ∇D − F ) . N ( F ) . A brief review of theory of geometric Littlewood Paley Consider S the collection of smooth functions on [0 , ∞ ) vanishing sufficientlyfast at ∞ and verifying the vanishing moment property Z ∞ τ k ∂ k m ( τ ) dτ = 0 , k + k ≤ N. We set m k ( τ ) := 2 k m (2 k τ ) for some smooth function m ∈ S . Recall from [3] thegeometric Littlewood-Paley (GLP) projections P k associated to m which take theform P k F := Z ∞ m k ( τ ) U ( τ ) F dτ By “ appropriate” , we mean the tensor F such that div F is in the space where D − iswell-defined. for any S t tangent tensor field F , where U ( τ ) F is defined by the heat flow (3.2) on( S t , ◦ γ ) . Proposition 4.1. There exists m ∈ S such that the GLP projections P k asso-ciated to m verify U ( ∞ ) + P k ∈ Z P k = Id . By f we denote a scalar function and F a S -tangent tensor field on H , the GLP projections P k associated to arbitraryinduced function m verify the following properties: (i) ( L p -boundedness) For any ≤ p ≤ ∞ , and any interval I ⊂ Z , k P I F k L p ( S ) . k F k L p ( S ) (ii) (Bessel inequality) For any tensorfield F on S , X k k P k F k L ( S ) . k F k L ( S ) , X k k r − k P k F k L ( S ) . k / ∇ F k L ( S ) (iii) (Finite band property) For any ≤ p ≤ ∞ , k ≥ , (4.1) k / ∆ P k F k L p ( S ) . k r − k F k L p ( S ) . Moreover given m ∈ S we can find ˜ m ∈ S such that k P k = / ∆ ˜ P k , with ˜ P k the geometric Littlewood Paley projections associated to ˜ m , then (4.2) P k F = 2 − k ˜ P k / ∆ F, k P k F k L p ( S ) . − k r k / ∆ F k L p ( S ) . In addition, there hold L estimates (4.3) k / ∇ P k F k L ( S ) . k r − k F k L ( S ) k P k / ∇ F k L ( S ) . k r − k F k L ( S ) k / ∇ P ≤ F k L ( S ) . r − k F k L ( S ) , and (4.4) k P k F k L ( S ) . − k r k / ∇ F k L ( S ) (iv) (Bernstein Inequality) For appropriate tensor fields F on S , we have weakBernstein inequalities (4.5) k P k F k L p ( S ) . r p − (cid:16) (1 − p ) k + 1 (cid:17) k F k L ( S ) , k P ≤ F k L p ( S ) . r p − k F k L ( S ) k P k F k L ( S ) . r p − (cid:16) (1 − p ) k + 1 (cid:17) k F k L p ′ ( S ) k P ≤ F k L ( S ) . r p − k F k L p ′ ( S ) . where ≤ p < ∞ and p + p ′ = 1 , k ≥ . (v) With the help of Proposition 3.2, we have the sharp Bernstein inequalities (4.6) k P k f k L ∞ ( S ) . k r − k f k L ( S ) , k P k f k L ( S ) . k r − k f k L ( S ) . We will also use the notations for any S -tangent tensor field F ,(4.7) F n := P n F, F ≤ := X k ≤ P k F. For more properties, one can refer to [3, 4] and [10]. Now we define for 0 ≤ θ ≤ B θ , P θ norms for S -tangent tensor fields F on H and B θ , norm on S as follows: k F k B θ = X k> k (2 k r − ) θ P k F k L ∞ t L x + k r − θ F k L ∞ t L x , (4.8) k F k P θ = X k> k (2 k r − ) θ P k F k L t L x + k r − θ F k L t L x , (4.9) k F k B θ , ( S ) = X k> k (2 k r − ) θ P k F k L x + k r − θ F k L x . (4.10) Remark . For any a ∈ R and any S -tangent tensor field F k F k H a ( S ) := k Λ a F k L ( S ) ≈ X k> ka r − a k P k F k L ( S ) + r − a k P ≤ F k L x . In certain situation, it is more convenient to work with the Besov norms definedby the classical Littlewood-Paley (LP) projections E k . Recall that (see [8, 9]) forany scalar function f on R we can define E k f = 1(2 π ) Z R ϕ ( ξ/ k ) ˆ f ( ξ ) e ixξ dξ, where ϕ is a smooth function support in the dyadic shell { ≤ | ξ | ≤ } andsatisfying P k ∈ Z ϕ (2 − k ξ ) = 1 when ξ = 0.Define for any 0 ≤ θ < B θ and ˜ P θ norms of any scalar function f on H by k f k ˜ B θ := X k> k (2 k r − ) θ E k f k L ∞ t L x + k r − θ f k L ∞ t L x , (4.11) k f k ˜ P θ := X k> k (2 k r − ) θ E k f k L t L x + k r − θ f k L t L x . (4.12)Using Proposition 2.8 and BA1 we can adapt [7, Proposition 3.28] to obtain thefollowing lemma. Lemma 4.1. Under the bootstrap assumptions (BA1) , there exists a finite numberof vector fields { X i } li =1 verifying the conditions (cid:26) k X, r ∇ X k L ∞ t L ∞ ω . , k r / ∇ ( ∇ X ) k L x L ∞ t . , k ( / ∇ − ∇ ) X k L x L ∞ t . ∆ , / ∇ L X = 0 , where ∇ represents the covariant derivative induced by the metric r γ (0) . Forappropriate S -tangent tensor F ∈ L ∞ t L x , F ∈ B θ if and only if F · X i ∈ B θ , and C − X i k F · X i k B θ ≤ k F k B θ ≤ C X i k F · X i k B θ , with ≤ θ < , where C is a positive constant. The same results hold for the spaces P θ . Moreover N ( F ⊗ X ) + k F ⊗ X k L ∞ ω L t . N ( F ) + k F k L ∞ ω L t , where ⊗ stands for either a tensor product or a contraction. Lemma 4.1 allows us to define Besov norms for arbitrary S -tangent tensor fields F on H by the classical LP projections. See [3, Page 2] for the finite band and sharp Berstein inequalities of E k . Definition 4.2. Let F be an ( m, n ) S -tangent tensor field on H and let F j j ··· j m i i ··· i n be the local components of F relative to { X i } li =1 . We define the ˜ B θ and ˜ P θ normsof F by k F k ˜ B θ = X k F j j ··· j m i i ··· i n k ˜ B θ and k F k ˜ P θ = X k F j j ··· j m i i ··· i n k ˜ P θ , where the summation is taken over all possible ( i · · · i n ; j · · · j m ).The equivalence between B θ , P θ norms and ˜ B θ , ˜ P θ norms is given in the followingresult whose proof can be found in [10]. Proposition 4.2. Under the bootstrap assumptions BA1 for arbitrary S -tangenttensor fields F on H there hold for ≤ θ < , k F k ˜ B θ ≈ k F k B θ and k F k ˜ P θ ≈ k F k P θ . Lemma 4.2. Let H be any S t tangent tensor and let f be a smooth function, k f H k P . k H k P k f k L ∞ + k H k L ( H ) k r / / ∇ f k L ∞ t L x . (4.13) and the similar estimates hold for B and B , ( S ) .Proof. By GLP decomposition, H = ¯ H + P n ∈ N P n H + P ≤ H , where ¯ H = U ( ∞ ) H . k f H k P ≤ X k> k P k ( f H Product estimates and Intertwine estimates.Proposition 4.3. Let D be one of the operators D , D and ⋆ D . Then for
For p = 2, the inequality follows immediately from Proposition 3.4. For p > 2, from ( Sob ) and Proposition 3.4 we infer for p ′ > p + p ′ = 1 that k r p − ⋆ D − F k L p ′ ( S ) . k / ∇ ⋆ D − F k − p ′ L ( S ) k r − ⋆ D − F k p ′ L ( S ) + k r − ⋆ D − F k L ( S ) . k F k L ( S ) Thus, by duality, we complete the proof. (cid:3) Lemma 4.3. Let D denote one of the Hodge operators D , D , ⋆ D and ⋆ D , let D − denote the inverse of D . For P k F with P k the GLP projections associated tothe heat equation (3.2), there hold for k > , < p ≤ , kD − P k F k L x . − k r k F k L x and k P k D − F k L x . − (2 − p ) k r − p k F k L px . Proof. The first inequality can be proved by using (4.4) in Proposition 4.1 andProposition 3.4. The second can be proved by duality with the help of the firstinequality and ( SobM1 ). (cid:3) The following result follows from the second estimate in Lemma 4.3 immediately. Proposition 4.4. Let D − denote either D − , ⋆ D − , D − , then for appropriate S -tangent tensor fields F on H and any < p ≤ , (4.15) kD − F k P θ . k r − p − θ F k L t L px . Since the proof of the Hodge-elliptic estimate for geodesic foliation contained in[11, pages 295–301] only relied on k K k L + k Λ − α K k L ∞ t L x . ∆ + R , with α ≥ / . Hence, based on Lemma 2.8 and Proposition 3.2, the same proof also applies to thecase of time foliation. We can obtain the result on the Hodge-elliptic P σ estimates. Theorem 4.3 (Hodge-elliptic P σ -estimate) . Let D denote either D , D or theiradjoint operators ⋆ D and ⋆ D . Then for any S -tangent tensor fields ξ and F satisfying D ξ = F and any > σ ≥ , (4.16) k / ∇ ξ k P σ . k F k P σ + ∆ kD − F k qL bt L x k F k − qL t L x , where / ≤ α < q < − σ and b > . We now give a series of product estimates for Besov norms. Proofs can be seenin [11, pages 302–304]. Lemma 4.4. For any S -tangent tensor fields F and G , k F · G k P . N ( F )( k r − b G k L bt L x + k r / ∇ G k L t L x ) with b > , (4.17) k F · G k P . N ( r / F ) k G k P , (4.18) k F · G k P . N ( r F ) (cid:0) k / ∇ G k L t L x + k G k L ∞ ω L t (cid:1) . (4.19) Corollary 1. Regard κ, ι also as elements of A , there hold k A · F k P . (∆ + R ) N ( r F ) , k ( tr χ, r − ) F k P . N ( F ) , (4.20) k anA · F k P . (∆ + R ) N ( r F ) , k an ( tr χ, r − ) F k P . N ( F ) . (4.21) Proof. Let us prove (4.20) first. Using (4.17) for b > 4, we have k A · F k P . N ( A )( k r / ∇ F k L ( H ) + k r − b F k L bt L x ) . N ( A ) · N ( r F )(4.22)where the last inequality follows by using (2.28).By finite band property of GLP in Proposition 4.1, it is straightforward to obtain(4.23) k r − F k P . k / ∇ F k L + k r − F k L . Noticing that tr χ · F = ι · F + 2 r − · F and N ( ι ) . ∆ + R in Proposition 2.7,the other inequality in (4.20) follows by (4.22) and (4.23) with A replaced by ι .Applying (4.13) to f = an and H = A · F, tr χF, r − F , (4.21) can be derived byusing (4.20) for H and the following inequalities for f (4.24) k / ∇ ( an ) k L ∞ t L x ≈ k ζ + ζ k L ∞ t L x . ∆ + R , k an k L ∞ . . (cid:3) Sharp trace theorem The purpose of the section is to prove Theorem 5.1 (Sharp trace theorem) . Let F be an S -tangent tensor which admitsa decomposition of the form / ∇ ( anF ) = D t P + E with tensors P and E of the sametype as F , suppose lim t → k F k L ∞ x < ∞ and lim t → r | / ∇ F | < ∞ , there holds thefollowing sharp trace inequality, (5.1) k F k L ∞ ω L t . N ( F ) + N ( P ) + k E k P . Remark . We will employ this theorem to estimate k F k L ∞ ω L t for F = ζ, ν, ˆ χ, ζ .By local analysis, the two initial assumptions can be checked for the four quantities.The following result gives the important inequalities to prove Theorem 5.1. Proposition 5.1. Let p ≥ be any integer, for any S -tangent tensor fields F , H and G of the same type. There holds (cid:13)(cid:13)(cid:13)(cid:13) r − p Z t r ′ p H · Gdt ′ (cid:13)(cid:13)(cid:13)(cid:13) B . k H k P ( N ( G ) + k G k L ∞ ω L t ) . (5.2) Let us further assume (5.3) lim t → r p k F k L ∞ x = 0 , lim t → k G k L ∞ x < ∞ , when p ≥ , then there holds for p ≥ , (cid:13)(cid:13)(cid:13)(cid:13) r − p Z t r ′ p D t F · Gdt ′ (cid:13)(cid:13)(cid:13)(cid:13) B . N ( F ) N ( G ) . (5.4)Using a modified version of [4, Lemma 5.3], (see in Lemma 8.1 in Appendix),also using Proposition 4.2, Proposition 5.1 follows by repeating the procedure in [4]and [10, Appendix]. We omit the detail of the proof of Proposition 5.1. Proof of Theorem 5.1. We set ϕ ( t ) = R t an | F | dt ′ , then D t ϕ = an | F | . Due to [2,Proposition 5.1] we have(5.5) k ϕ k L ∞ ( H ) . k / ∇ ϕ k B + k r − ϕ k L ∞ t L x . It is easy to see(5.6) k r − ϕ k L ∞ t L x . k F k L ∞ ω L t · k r − F k L . We now estimate k / ∇ ϕ k B . In view of (2.19), we obtain(5.7) / ∇ L / ∇ ϕ + 12 tr χ / ∇ ϕ = 2( an ) − / ∇ ( an · F ) · F − ˆ χ · / ∇ ϕ − ( ζ + ζ ) | F | . Using the decomposition / ∇ ( an · F ) = D t P + E and Lemma 4.1, we pair / ∇ ϕ withvector field X = X i to obtain / ∇ L ( r / ∇ ϕ · X ) = − rκ / ∇ ϕ · X − r ˆ χ / ∇ ϕ ⊗ X + r (cid:0) / ∇ L P · F ⊗ X + 2( an ) − E · F ⊗ X − | F | ( ζ + ζ ) · X (cid:1) . (5.8)We will not distinguish “ ⊗ ” with “ · ”, and also suppress X whenever there occursno confusion. Note that under the transport coordinate ( s, ω , ω ), we have ∂ϕ∂ω i = Z t { h ∂ ω i F, F i + | F | ∂ ω i log( an ) } nadt ′ i = 1 , , by assumptions on initial condition, lim t → ∂ϕ∂ω i = 0. It then follows by integrating(5.8) along a null geodesic Γ ω from 0 to s ( t ) that, symbolically,( / ∇ ϕ )( t ) = r − Z t { r ′ an ( κ + ˆ χ ) / ∇ ϕ + r ′ E · F } dt ′ + r − Z t r ′ D t P · F dt ′ + r − Z t anr ′ | F | ( ζ + ζ ) dt ′ . (5.9)Since by Proposition 4.2,(5.10) k / ∇ ϕ k B ≈ X k> k E k ( / ∇ ϕ ) k L ∞ t L x + k / ∇ ϕ k L ∞ t L x , and the estimate of k / ∇ ϕ k L x L ∞ t can be obtained in view of (5.9) and ( SobM1 ). k / ∇ ϕ k L x L ∞ t . ( k κ k L ∞ ω L t + k ˆ χ k L ∞ ω L t ) k / ∇ ϕ k L (5.11) + ( k E k L + k / ∇ L P k L + N ( A ) N ( F )) k r − F k L where the term on the right of (5.11) can be absorbed in view of k κ, ˆ χ k L ∞ ω L t . ∆ ,obtained from (2.84) and BA1. It remains to estimate the first term in (5.10).Applying Littlewood Paley decomposition E k to / ∇ ϕ ⊗ X via (5.9), we only needto estimate the following terms I = X k> (cid:13)(cid:13)(cid:13)(cid:13) E k Z t r ′ an ( κ + ˆ χ ) / ∇ ϕ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L ω , I = X k> (cid:13)(cid:13)(cid:13)(cid:13) E k Z t r ′ E · F dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L ω ,I = X k> (cid:13)(cid:13)(cid:13)(cid:13) E k Z t r ′ D t P · F dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L ω , I = X k> (cid:13)(cid:13)(cid:13)(cid:13) E k Z t r ′ | F | ( ζ + ζ ) dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L ω . Use (5.2) and Proposition 4.2, I . ( N ( ˆ χ ) + k ˆ χ k L ∞ x L t + N ( κ ) + k κ k L ∞ ω L t ) k / ∇ ϕ k P . Consequently, by (2.84), Proposition 2.7, (2.79) and k ˆ χ k L ∞ ω L t ≤ ∆ in BA1, I . ∆ k / ∇ ϕ k P . Apply (5.2) to I , I . k E k P ( N ( F ) + k F k L ∞ ω L t ) . By (5.4) and Lemma 4.1 I . N ( P )( N ( F ) + k F k L ∞ ω L t ) . By (5.2), (4.19), BA1 and (2.79), we have I . ( N ( ζ + ζ ) + k ζ + ζ k L ∞ ω L t ) k F · F k P . ∆ N ( F )( k / ∇ F k L + k F k L ∞ ω L t ) . Thus we conclude k / ∇ ϕ k B . ( N ( P ) + k E k P ) (cid:16) N ( F ) + k F k L ∞ ω L t (cid:17) + ∆ k / ∇ ϕ k P + ∆ N ( F )( k / ∇ F k L + k F k L ∞ ω L t ) . This inequality, together with the fact that k / ∇ ϕ k P . k / ∇ ϕ k B , yields k / ∇ ϕ k B . ( N ( F ) + k F k L ∞ ω L t ) ( N ( P ) + k E k P + ∆ N ( F )) . Combine the above inequality with (5.5) and (5.6), we get k F k L ∞ ω L t . ( N ( F ) + k F k L ∞ ω L t ) ( N ( P ) + k E k P + N ( F )∆ )+ k F k L ∞ ω L t · k r − F k L which implies (5.1) by Young’s inequality. (cid:3) Error estimates We will employ the following conventions: • ˇ R denotes either the pair (ˇ ρ, − ˇ σ ) or β • D − ˇ R denotes either D − (ˇ ρ, − ˇ σ ) or ⋆ D − β • D − ˇ R denotes either D − D − (ˇ ρ, − ˇ σ ) or D − ⋆ D − β • D − D t ˇ R denotes either ⋆ D − D t β or D − D t (ˇ ρ, − ˇ σ ) • C ( ˇ R ) denotes [ D t , D − ](ˇ ρ, − ˇ σ ) or [ D t , ⋆ D − ] β • D − D t ˇ R denotes D − D − D t (ˇ ρ, − ˇ σ ) or D − ⋆ D − D t β • D − C ( ˇ R ) denotes D − [ D t , D − ](ˇ ρ, − ˇ σ ) or D − [ D t , ⋆ D − ] β • F denotes D − ˇ R or ( aδ + 2 aλ ). • D − F denotes either D − ˇ R or D − ( aδ + 2 aλ ).6.1. Commutation formula. We will study error terms which arise from com-muting D t with Hodge operators. Regard ι also an element of A , symbolically, thecommutation formula and its good part can be written as follows[ D t , / ∇ ] F = an (cid:18) ( A + 1 r ) / ∇ F + ( A + 1 r ) · A · F + β · F (cid:19) , (6.1) [ D t , / ∇ ] g F := an (cid:18) ( A + 1 r ) / ∇ F + ( A + 1 r ) · A · F (cid:19) . (6.2)Due to the nontrivial factor “ an ” in (6.1), the treatment in [11, Section 6] hasto be modified. We rewrite equations (2.11), (2.12) and (2.13) as L (ˇ ρ, − ˇ σ ) = D β + r − ˇ R + A · ˜ R, (6.3) / ∇ L β = ⋆ D ( ρ, σ ) + r − ˇ R + A · ˜ R, (6.4)where ˜ R := R + / ∇ A + A · A + r − A. We will consider the commutators(6.5) C ( ˇ R ) = ( C ( ˇ R ) , C ( ˇ R ) , C ( ˇ R ))given in [2, Definition 6.3] which, by using the above conventions, can be writtensymbolically as C ( ˇ R ) = / ∇D − [ D t , D − ] ˇ R,C ( ˇ R ) = / ∇ [ D t , D − ] D − ˇ R,C ( ˇ R ) = [ D t , / ∇ ] D − ˇ R. Corresponding to (6.3) and (6.4), we introduce the error terms Err := D − D t (ˇ ρ, − ˇ σ ) − anβ and g Err := ⋆ D − D t β − an ( ρ, σ ) . (6.6)Denote by F either Err or g Err . Symbolically, F has the form F = D − { an ( r − ˇ R + A · ˜ R ) } . We then infer from (6.3) and (6.4) the symbolic expression(6.7) D − D t ˇ R = anR + F . For simplicity, we use ( aδ + 2 aλ ) to denote the pair of quantities ( aδ + 2 aλ, By using (4.15) with θ = 0 and p = , Proposition 3.4 and the H¨older inequalitywe infer that(6.8) k F k P . k r A · ˜ R k L t L x + k ˇ R k L . ∆ + R . The purpose of this section is to prove Proposition 6.1. There hold the following decomposition for commutators, C ( ˇ R ) = D t P + E, [ D t , / ∇D − ]( aδ + 2 aλ ) = D t P ′ + E ′ , where P, P ′ and E, E ′ are S tangent tensors verifying N ( P ) + N ( P ′ ) + k E k P + k E ′ k P . ∆ + R . lim t → (cid:0) r k P k L ∞ ( S ) + r k P ′ k L ∞ ( S ) (cid:1) = 0 . Proof of Proposition 6.2: Part I. In order to prove Proposition 6.1, let usconsider the structure of commutators. We first use (6.1) to write (cid:0) C ( ˇ R ) , / ∇ [ D t , D − ]( aδ + 2 aλ ) (cid:1) = / ∇ [ D t , D − ] g F + / ∇D − ( anβ · D − F ) , (6.9) (cid:0) C ( ˇ R ) , [ D t , / ∇ ] D − ( aδ + 2 aλ ) (cid:1) = [ D t , / ∇ ] g D − F + anβ · D − F , (6.10)where[ D t , D − ] g F := D − (cid:0) an ( A + r − ) · / ∇D − F + an ( A + r − ) · A · D − F (cid:1) . The terms / ∇ [ D t , D − ] g F and [ D t , / ∇ ] g D − F are the “good” parts in the corre-sponding commutators and will be proved to be P bounded (see (6.13)-(6.16)).The terms / ∇D − ( anβ · D − F ) and anβ · D − F in (6.9) and (6.10) can not bebounded in P norm and will be further decomposed in Section 6.7.Let us first establish (6.11), (6.13)-(6.16) in two steps in the following result Proposition 6.2. For the error terms C ( ˇ R ) , C ( ˇ R ) , C ( ˇ R ) , C ( ˇ R ) , etc, therehold k C ( ˇ R ) k P . ∆ + R , (6.11) k C ( ˇ R ) k P . ∆ + R , (6.12) C ( ˇ R ) = / ∇D − ( anβ · D − ˇ R ) + err, (6.13) C ( ˇ R ) = anβ · D − ˇ R + err (6.14) / ∇ [ D t , D − ]( aδ + 2 aλ ) = / ∇D − ( anβ · D − ( aδ + 2 aλ )) + err (6.15) [ D t , / ∇ ] D − ( aδ + 2 aλ ) = anβ · D − ( aδ + 2 aλ ) + err (6.16) with k err k P . ∆ + R . We will rely on (2.79), ( SobM1 ), (1.13), Proposition 2.7 and Remark 2.2, i.e.(6.17) k A k L x L ∞ t + k A k L + N ( A ) + k r / ∇ tr χ k L x L ∞ t + k / ∇ tr χ k L + k R k L . ∆ + R , where ι, κ are regarded as elements of A . Step 1. We first prove (6.11). In view of (6.1),(6.18) C ( ˇ R ) = D − ( an { ( A + r − )( / ∇D − ˇ R ) + ( A + r − ) · A · D − ˇ R + β · D − ˇ R } ) . Using (4.15) with θ = 0 and p = , Proposition 3.4, also with the help of (6.17)and H¨older inequality, we can estimate the various terms in (6.18) to get(6.19) k C ( ˇ R ) k P . ∆ + R + ∆ · N ( D − ˇ R ) . By the definition of N ( D − ˇ R ) and Proposition 3.4 it follows that N ( D − ˇ R ) . R + ∆ + kD − D t ˇ R k L t L x + k C ( ˇ R ) k L t L x . While it follows from (6.7), (6.8) and (1.13) that kD − D t ˇ R k L . k F k P + k anR k L . ∆ + R . Combining the above three inequalities and using the smallness of ∆ we obtain(6.11).In the above proof, together with Lemma 3.1, (2.79) and Remark 2.1 we havealso verified the following Proposition 6.3. kD − D t ˇ R k L . R + ∆ , (6.20) k [ D t , D − ] ˇ R k L . R + ∆ , (6.21) N ( F ) . R + ∆ , (6.22) N ( / ∇D − F ) . R + ∆ , N ( D − F ) . R + ∆ , (6.23) where F denotes either D − ˇ R or ( aδ + 2 aλ ) . Step 2. We will prove (6.13)-(6.16). Let us first establish the following Lemma 6.1. Denote by D − one of the operators among D − , D − and ⋆ D − .For any S t tangent tensor H , b > k r − − b D − ( an / ∇D − H ) k L bt L x . N ( D − H ) k ζ + ζ k L ∞ t L x + N ( r − D − H ) . Proof. Using Propositions 3.4 and 4.3, k r − − b D − ( an / ∇D − H ) k L bt L x . k r − − b (cid:0) |D − ( / ∇ ( an ) D − H ) | + |D − / ∇ ( an D − H ) | (cid:1) k L bt L x . k r − − b / ∇ ( an ) D − H k L bt L / x + k r − − b D − H k L bt L x . Since by using (2.28), we obtain k r − − b / ∇ ( an ) D − H k L bt L / x . k r − − b D − H k L bt L x k / ∇ ( an ) k L ∞ t L x . N ( D − H ) k ζ + ζ k L ∞ t L x , and k r − − b D − H k L bt L x . N ( r − / D − H ). Then (6.24) follows. (cid:3) The proof of (6.13)-(6.16) can be completed by using (6.22) combined with thefollowing result. Lemma 6.2. Denote by D − either D − or D − . For appropriate S -tangent tensorfield F , there hold k [ D t , / ∇ ] g D − F k P + k / ∇ [ D − , D t ] g F k P . N ( F ) , (6.25) k [ D t , / ∇D − ] g F k P . N ( F ) . (6.26) Proof. Observe that k [ D t , / ∇D − ] g F k P . k [ D t , / ∇ ] g D − F k P + k / ∇ [ D − , D t ] g F k P , it suffices to prove (6.25) only.We first derive in view of (6.1) by using Lemma 4.4 with 4 < b < ∞ , k [ D t , / ∇ ] g D − F k P . N ( / ∇D − F ) (cid:16) k r / ∇ ( anA ) k L t L x + k r − b anA k L bt L x (cid:17) + N ( D − F ) (cid:0) k anA · A k P + k r − anA k P (cid:1) (6.27) + k r − an / ∇D − F k P . (6.28)By (4.21) and (6.17), we obtain k r − anA k P . ∆ + R , k anA · A k P . ∆ N ( A ) . ∆ + R . Then the term in (6.27) can be bounded by (∆ + R ) N ( D − F ).We then consider the term in (6.28). In view of (4.13) and Theorem 4.3, k r − an / ∇D − F k P . k r − F k P + ∆ k r − D − F k qL bt L x k r − F k − qL ( H ) . Since by Proposition 4.3 and ( SobM1 ), we obtain, k r − D − F k L bt L x . k F k L bt L x . N ( F ) , also by using (4.23), we deduce k r − an / ∇D − F k P . N ( F ) + ∆ N ( F ) . By (6.17), it is easy to check k r / ∇ ( anA ) k L t L x + k r − b anA k L bt L x . ∆ + R . Consequently, we conclude that(6.29) k [ D t , / ∇ ] g D − F k P . (∆ + R )( N ( / ∇D − F ) + N ( D − F )) + N ( F ) . We then infer from (6.29) and Lemma 3.1 that(6.30) k [ D t , / ∇ ] g D − F k P . N ( F ) . Next, we prove for S -tangent tensor fields F on H the following inequality holds (6.31) k [ D t , D − ] g F k L bt L x . N ( F ) with 4 < b < ∞ . Indeed, by using Proposition 4.3 with p = 4 / 3, Proposition 3.4, ( SobM1 ), (6.17),(6.24) and Lemma 3.1 we then derive that k [ D t , D − ] g F k L bt L x . k r / A · / ∇D − F k L bt L / x + k r / A · A · D − F k L bt L / x + k r − D − ( an / ∇D − F ) k L bt L x + k r − / A · D − F k L bt L / x . k A k L bt L x k / ∇D − F k L ∞ t L x + kD − F k L ∞ t L x × ( k A · A k L bt L x + kk r − A k L bt L x )+ N ( r − / D − F ) + ∆ N ( D − F ) . N ( / ∇D − F ) + N ( D − F ) . N ( F ) . We will improve the right hand side of (6.31) to be N ( D − F ) in the next section. The combination of (6.31), (6.30) and (4.16) gives k / ∇ [ D − , D t ] g F k P . k [ D t , / ∇ ] g D − F k P + ∆ k [ D t , D − ] g F k qL bt L x · k [ D t , / ∇ ] g D − F k − qL . N ( F ) . where in the above inequalities, b > α < q < (cid:3) Proof of Proposition 6.2: Part II. We record the following result, which,for completeness, will be proved in Appendix by using Lemma 8.1. Lemma 6.3. For any smooth S -tangent tensor field F and for exponent ≤ q ≤ ∞ ,we have the following inequality for k > k r − − q P k F k L qt L x . − k − q k N ( F ) , (6.32) k r − q F k k L qt L x . − kq N ( F )(6.33)We will complete the proof of Proposition 6.2 by studying error type terms inProposition 6.5Let us first establish the following result with the help of Lemma 6.3. Proposition 6.4. Let D − denote either D − or ⋆ D − . For any S -tangent tensorfields F and G on H there holds k r − b D − ( anF · / ∇ G ) k L bt L x . N ( F ) N ( G ) , with < b < ∞ . Proof. Note that based on Lemma 8.1 in Appendix, the inequalities [11, Lemma5.1,(6.25),(6.26)] still hold true. We only need to modify the case l < n < m of theproof in [11, P.310-311].When l < n < m , anF n · / ∇ G m = / ∇ ( anF n · G m ) − / ∇ ( an ) F n · G m − an / ∇ F n · G m , thus we need to consider the three terms I lnm : = k r − b P l D − ( an / ∇ F n · G m ) k L bt L x I lnm : = k r − b P l D − / ∇ ( anF n · G m ) k L bt L x I lnm : = k r − b P l D − ( / ∇ ( an ) F n · G m ) k L bt L x . Using C − < an < C , following the same procedure in [11], we can get(6.34) X Let D be one of the operators D , ⋆ D or D . In the following result, we useProposition 6.4 to estimate the error type terms. Proposition 6.5. For S -tangent tensors G on H verifying N ( G ) < ∞ , set E ( G ) := r − D − ( anA · G ) or D − ( anA · A · G ) , E ( G ) := D − ( an / ∇ A · G ) or D − ( an / ∇ G · A ) , The following estimates hold k r − b E ( G ) k L bt L x + k r − b E ( G ) k L bt L x . (∆ + R ) N ( G ) where < b < ∞ .Proof. Using Proposition 4.3 with p = , ( SobM1 ) and (6.17), we get k r − b D − ( anA · A · G ) k L bt L x . k r − b A · A · G k L bt L / x . k A k L ∞ t L x k G k L bt L x . (∆ + R ) N ( G ) , by using Proposition 3.4, (2.28), (6.17) and ( SobM1 ), we have k r − − b D − ( anA · G ) k L bt L x . k r − b A · G k L bt L x . k r − b A k L bt L x k G k L ∞ t L x . (∆ + R ) N ( G ) . For E ( G ), we infer from Proposition 6.4 and (6.17) that k r − b E ( G ) k L bt L x . N ( G ) N ( A ) . (∆ + R ) N ( G ) . We thus obtain the desired estimates. (cid:3) By analyzing the expression of β and C ( F ) := [ D t , D − ] F , we have Corollary 2. The following inequalities hold for any S -tangent tensor F , k r − b D − ( anβ · F ) k L bt L x . N ( F )(∆ + R )(6.35) k r − b C ( F ) k L bt L x . N ( r − D − F )(6.36) where < b < ∞ . Proof. Using Codazzi equation (2.6), i.e. anβ = an / ∇ A + an ( A · A + r − A ) , we infer D − ( anβ · F ) = E ( F ) + E ( F ) . Whence (6.35) follows from Proposition 6.5.Similarly, using (6.1) we can write C ( F ) = E ( D − F ) + E ( D − F ) + r − D − ( an / ∇D − F ) . For the last term, using (6.24) with H = F , we infer k r − − b D − ( an / ∇D − F ) k L bt L x . N ( r − D − F )(∆ + R + 1) . The desired estimate then follows from Proposition 6.5. (cid:3) Proof of (6.12) in Proposition 6.2. Combining Proposition 4.3, (6.36) and (6.22)we derive for D − either D − or D − , 4 < b < ∞k r − b D − C ( ˇ R ) k L bt L x . k r − b C ( ˇ R ) k L bt L x . N ( r D − ˇ R ) . ∆ + R . (6.37)Observe that C ( ˇ R ) can be written symbolically in the form C ( ˇ R ) = / ∇D − C ( ˇ R ) . By Hodge-elliptic estimate (4.16) with < q < < b < ∞ , (6.11), (6.21)and (6.37) k C ( ˇ R ) k P . k C ( ˇ R ) k P + ∆ kD − C ( ˇ R ) k qL bt L x k C ( ˇ R ) k − qL ( H ) . ∆ + R + ∆ kD − C ( ˇ R ) k qL bt L x (∆ + R ) − q . ∆ + R . This is the desired estimate. (cid:3) A preliminary estimate for ( ρ, σ ) . Define¯ ρ = 1 | S t | Z S t ρdµ γ and ¯ σ = 1 | S t | Z S t σdµ γ . We have Lemma 6.4. (6.38) | r ¯ ρ | + | r ¯ σ | . ∆ + R , (6.39) | r ¯ˇ ρ | + | r ¯ˇ σ | . ∆ + R . Proof. By [2, Eq. (41)], i.e.(6.40) dds ρ + 32 tr χρ = F where F = div β − ˆ χ · α + ( ζ + 2 ζ ) · β , the transport equation for ¯ ρ can be obtainedas follows dds (¯ ρ ) = / ∇ L ( r − Z S ρ )= − an tr χ ( an ) − ¯ ρ + r − ( an ) − Z S (cid:0) / ∇ L ρ + tr χρ (cid:1) andµ γ = − ( an ) − an tr χ ¯ ρ + ( an ) − (cid:18) − an tr χρ + anF (cid:19) , and dds ( r ¯ ρ ) = 3 r an tr χ ( an ) − ¯ ρ + r ( − ( an ) − an tr χ ¯ ρ + ( an ) − (cid:18) − an tr χρ + anF (cid:19)) = − r ( an ) − an tr χ ( ρ − ¯ ρ ) + r ( an ) − anF . Integrating the above identity in t , due to lim t → r θ ρ = 0 for any θ > 0, we canobtain r ¯ ρ ( t ) = r − Z t r (cid:18) − an tr χ · O sc ( ρ ) + anF (cid:19) dt ′ . In view of (2.23), (1.13) and Proposition 2.2, we obtain r − Z t (cid:12)(cid:12)(cid:12) r an tr χ · O sc ( ρ ) (cid:12)(cid:12)(cid:12) ≤ k r O sc ( ρ ) k L t L ω k r ′ k L (0 ,t ] r − . k ρ k L ( H ) . R . By integration by part on S = S t , we can obtain r anF = Z S (cid:18) an (div β − 12 ˆ χ · α ) + an ( ζ + 2 ζ ) · β (cid:19) dµ γ = Z S an ( ζβ − 12 ˆ χ · α ) dµ γ = Z S anA · R dµ γ . Hence by (1.13) and Proposition 2.3, (cid:12)(cid:12)(cid:12)(cid:12) r − Z t r ′ anF dt ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ r − k r ′ k L (0 ,t ] k ( r ′ ) A · R k L t L ω ≤ k R k L k rA k L ∞ t L ω . ∆ + R . Following the same procedure as above, we can obtain the same estimate for ¯ σ inview of [2, Eq. (42)].Note that | r ( ˆ χ · ˆ χ, ˆ χ ∧ ˆ χ ) | . k r A k L ∞ t L ω . (∆ + R ) (6.41)(6.39) follows by connecting (6.38) with (6.41). (cid:3) Proposition 6.6. For < b < ∞ , there hold k r − b − D − ( an (ˇ ρ, − ˇ σ )) k L bt L x . ∆ + R , (6.42) k r − b − D − ( an ( ρ, − σ )) k L bt L x . ∆ + R . (6.43) Proof. Let H = (ˇ ρ − ¯ˇ ρ, − ˇ σ + ¯ˇ σ ). In view of D D − H = H , D − ( an (ˇ ρ, − ˇ σ )) = D − ( an D D − H ) + D − ( an (¯ˇ ρ, − ¯ˇ σ )) . By Proposition 4.3, k r − b − D − ( an (¯ˇ ρ, − ¯ˇ σ )) k L bt L x . k r − b an (¯ˇ ρ, − ¯ˇ σ ) k L bt L ω . Hence, in view of (6.39) and (1.13) k r − b ( an (¯ˇ ρ, − ¯ˇ σ )) k L bt L ω . k r an (¯ˇ ρ, − ¯ˇ σ ) k − b L ∞ t L ∞ ω k ran (¯ˇ ρ, − ¯ˇ σ ) k b L t L ω . ∆ + R . (6.44)By Leibnitz rule, Proposition 4.3 and (2.28), we have k r − b − D − ( an D D − H ) k L bt L x . k r − b − D − D ( an D − H ) k L bt L x + k r − b − D − ( / ∇ ( an ) D − H ) k L bt L x . k r − b − D − H k L bt L x + k r − b ( ζ + ζ ) D − H k L bt L / x . N ( D − H ) + k r − b D − H k L bt L x k ζ + ζ k L ∞ t L x . N ( D − H )(1 + ∆ + R ) . ∆ + R For the last two inequalities, we employed (6.17) and (6.22). (6.43) can be obtained by combining (6.42) with the estimate for r − D − ( anA · A )in view of the estimate for E ( A ) in Proposition 6.5. (cid:3) L bt L x estimates for D − E G . For arbitrary S -tangent tensor field F , we de-note by E G either [ D t , D − ](ˇ ρ, − ˇ σ ) · F or Err · F . In what follows, we establishestimates for kD − E G k L bt L x with 4 < b < ∞ , which will be employed for theHodge-elliptic P estimates of error terms arising in the decomposition procedurein Section 6.7. Proposition 6.7. Denote by D either D or D , for appropriate S -tangent tensorfields F , the following estimates hold (6.45) k r − b D − E G k L bt L x . (∆ + R ) N ( F ) . More precisely, k r − b D − ( Err · F ) k L bt L x . (∆ + R ) N ( F )(6.46) k r − b D − ([ D t , D − ](ˇ ρ, − ˇ σ ) · F ) k L bt L x . (∆ + R ) N ( F ) . (6.47) where Err is defined in (6.6) and < b < ∞ . In order to prove Proposition 6.7, we may use the error type terms introducedin Proposition 6.5 to rewrite (6.6) in view of (2.11) and (2.12) as(6.48) Err = D − ( an tr χ (ˇ ρ, − ˇ σ )) + E ( A ) + E ( A ) . Proof of Proposition 6.7. (6.47) can be obtained by using Proposition 4.3, (6.36),(6.22) and ( SobM1 ) as follows, k r − b D − ( C ( ˇ R ) · F ) k L bt L x . k r − b C ( ˇ R ) k L bt L x k F k L ∞ t L x . (∆ + R ) N ( F ) , and similarly, by using Proposition 6.5, for i = 1 , k r − b D − ( E i ( A ) · F ) k L bt L x . k r − b E i ( A ) k L bt L x k F k L ∞ t L x . (∆ + R ) N ( F ) . Thus, in order to prove (6.46), in view of (6.48), it remains to show k r − b D − (cid:0) D − ( an tr χ (ˇ ρ, − ˇ σ )) · F (cid:1) k L bt L x . N ( F )∆ . For this estimate, we proceed as follows. Let H = (ˇ ρ − ¯ˇ ρ, − ˇ σ + ¯ˇ σ ), then H = D D − H . k r − b D − (cid:0) D − ( an tr χ (ˇ ρ, − ˇ σ )) · F (cid:1) k L bt L x . k r − b D − ( D − ( an tr χ D D − H ) F ) k L bt L x + k r − b D − (cid:0) D − ( an tr χ (¯ˇ ρ, − ¯ˇ σ )) · F ) k L bt L x By I and I , we denote the two terms on the right of the inequality. UsingProposition 4.3, ( SobM1 ), (2.28) and (6.22), I . k r − b D − ( an tr χ D − H · F ) k L bt L x + k r − b D − ( D − ( / ∇ ( an tr χ ) D − H ) · F ) k L bt L x . k r − b D − H · F k L bt L x + k r − b + D − ( / ∇ ( an tr χ ) D − H ) k L bt L x k F k L ∞ t L x . k F k L ∞ t L x k r − b D − H k L bt L x + k r − b / ∇ ( an tr χ ) k L bt L x kD − H k L ∞ t L x k F k L ∞ t L x . N ( F ) N ( D − H ) . (∆ + R ) N ( F ) where we employed k r − b / ∇ ( an tr χ ) k L bt L x . k r / ∇ ( an tr χ ) k − b L ∞ t L x k / ∇ ( an tr χ ) k b L . ∆ + R . By Proposition 4.3, (6.44) and ( SobM1 ), I . k r − b + D − ( an tr χ (¯ˇ ρ, − ¯ˇ σ )) k L bt L x k F k L ∞ t L x . k r − b an tr χ (¯ˇ ρ, − ¯ˇ σ ) k L bt L x k F k L ∞ t L x . k r − b (¯ˇ ρ, − ¯ˇ σ ) k L bt L ω N ( F ) . (∆ + R ) N ( F ) . The proof is complete. (cid:3) L bt L x estimates for / ∇ L D − F . We will establish the following Proposition 6.8. Denote by D − F either D − ˇ R or D − ( aδ + 2 aλ ) . There holds k r − b D t D − Fk L bt L x . R + ∆ , < b < ∞ . Case 1: F = D − ˇ R . We denote by D − F either D − Err or ⋆ D − g Err . To prove Proposition 6.8, wewill rely on (6.49) in the following result. Proposition 6.9. For F = ( Err, g Err ) with Err and g Err given by (6.6), there hold (6.49) k r − b D − F k L bt L x . ∆ + R , < b < ∞ , (6.50) k / ∇D − F k P . ∆ + R . Assuming (6.49), now we prove Proposition 6.8. Proof of Proposition 6.8 for Case 1. In view of the formula D t D − ˇ R = [ D t , D − ] D − ˇ R + D − [ D t , D − ] ˇ R + D − D t ˇ R, we only need to show for 4 < b < ∞ , there hold k r − b [ D t , D − ] D − ˇ R k L bt L x . ∆ + R (6.51) k r − b D − [ D t , D − ] ˇ R k L bt L x . ∆ + R (6.52) k r − b D − D t ˇ R k L bt L x . ∆ + R . (6.53)(6.51) follows from (6.36) with F = D − ˇ R , by using the fact that N ( r − D − ˇ R ) . N ( D − ˇ R ) . ∆ + R . (6.52) was proved in (6.37).It only remains to prove (6.53). Consider first the case D − D t ˇ R = D − D − D t (ˇ ρ, − ˇ σ ).By anβ = / ∇ ( anA ) + an ( A · A + r − A ) , ( SobM1 ) and (6.17), k r − b D − ( anβ ) k L bt L x . k r − b D − ( / ∇ ( anA ) + an ( A · A + r − A )) k L bt L x . k r − b A k L bt L x + k r − b +1 A · A k L bt L x . N ( A ) + N ( A ) . ∆ + R . Then by (6.49), we obtain k r − b D − ( Err + anβ ) k L bt L x . ∆ + R . In view of the definition of Err in (6.6), we have(6.54) k r − b D − D t (ˇ ρ, − ˇ σ ) k L bt L x . ∆ + R . Using ⋆ D − D t β = an ( ρ, σ ) + g Err, Proposition 6.6 and (6.49), k r − b D − D t β k L bt L x . k r − b D − ( an ( ρ, σ )) k L bt L x + k r − b D − F k L bt L x . ∆ + R (6.55)In view of (6.54) and (6.55), (6.53) is proved. (cid:3) To prove Proposition 6.9, we will rely on the following result. Lemma 6.5. Let D − denote one of the operators D − , D − or ⋆ D − . For anyappropriate S -tangent tensor field G , there holds (6.56) kD − ( an ˇ ρ · G ) k L bt L x . k Λ − α ˇ ρ k L ∞ t L x N ( G ) where α ≥ and < b < ∞ .Proof. We can adapt the proof for [11, Lemma 4.4] in view of k / ∇ ( an ) k L x L ∞ t . ∆ + R and an < C , to derive the following estimates for S tangent tensor fields F , k Λ − α ( an ˇ ρ · F m ) k L ( S ) . k Λ − α ˇ ρ k L ∞ t L x m r − k P m F k L ( S ) , (6.57) k P m ( an ˇ ρ · D − P l F ) k L ( S ) . k Λ − α ˇ ρ k L ∞ t L x αm r − α k P l F k L ( S ) , (6.58)where α > α ≥ / D − denotes either D − or ⋆ D − .Set Ω nl := D − P l ( an ˇ ρ · P n G ) , with l, n ∈ N . We now prove(6.59) X l,n> k Ω nl k L bt L x . k Λ − α ˇ ρ k L ∞ t L x N ( G ) , and lower frequency terms can be treated similarly. By duality argument, Propo-sition 3.1 (iii) and Lemma 4.3,(6.60) kD − Λ α P l F k L ( S ) . ( − α ) l r − α k F k L ( S ) . We first prove (6.59) for the case 0 < n < l . With the help of (6.60) and (6.57), k Ω nl k L x . − (1 − α ) l n r − α k Λ − α ˇ ρ k L ∞ t L x k P n G k L x . Take L bt norm for 4 < b < ∞ , and (6.32) in Lemma 6.3 k Ω nl k L bt L x . − (1 − α ) l n ( − b ) r + b − α k Λ − α ˇ ρ k L ∞ t L x N ( G ) . Since we can choose α < α < + b , we deduce X 1. By (6.58), h Ω nl , F i = h P l ( an ˇ ρP n G ) , P l⋆ D − F i . αn r − α k Λ − α ˇ ρ k L ∞ t L x k P n G k L x . Hence, by (6.32) in Lemma 6.3, k Ω nl k L bt L x . αn − ( + b ) n r − α + + b k Λ − α ˇ ρ k L ∞ t L x N ( G ) . Consequently, X Proof of Proposition 6.9. (6.50) can be derived by using (6.49), Theorem 4.3 and(6.8).Now we consider (6.49). By letting F = 1 in (6.46), we obtain for 4 < b < ∞ that k r − b D − Err k L bt L x . ∆ + R . Thus we only need to consider D − g Err .By definition of g Err in (6.6), in view of (2.6), we rewrite g Err symbolically asfollows(6.61) g Err = ⋆ D − ( an tr χβ + anA · ( / ∇ A + A · A + r − A )) + ⋆ D − ( an ( ζ · ρ − ζ ⋆ σ )) . By Propositions 4.3 and 6.5, k r − b D − ( anA · ( / ∇ A + r − A + A · A )) k L bt L x . ∆ + R . According to (6.61), we consider W = k r − b D − ( an tr χβ ) k L bt L x , and U = k r − b D − ( anζ · ˇ ρ ) k L bt L x , V = k r − b D − ( anζ ⋆ ˇ σ ) k L bt L x . By β = ⋆ D ⋆ D − β , also using Propositions 3.4 and 4.3, (2.28), (6.17) and (6.22) W . k r − b D − (cid:0) ( ⋆ D ( an tr χ ⋆ D − β )) − / ∇ ( an tr χ ) ⋆ D − β (cid:1) k L bt L x . k r − b ( an tr χ ⋆ D − β ) k L bt L x + k r − b / ∇ ( an tr χ ) ⋆ D − β k L bt L / x . k r − b ⋆ D − β k L bt L x + k r − b ⋆ D − β k L bt L x k r / ∇ ( an tr χ ) k L ∞ t L x . N ( r / ⋆ D − β )( k r / ∇ ( an tr χ ) k L ∞ t L x + 1) . ∆ + R . By (2.7), clearly ˇ σ = curlζ . Thus by Propositions 4.3 and 6.5, we obtain V = k r − b D − E ( ζ ) k L bt L x . N ( ζ ) N ( ζ ) . (∆ + R ) . By Proposition 3.4 and (6.56), we have U . k r − b D − ( an ˇ ρ · ζ ) k L bt L x . k r − b +1 D − ( an ˇ ρ · ζ ) k L bt L x . N ( ζ ) k Λ − α ˇ ρ k L ∞ t L x . (∆ + R ) where we employed (2.79) and (3.4) to obtain the last inequality. (cid:3) Case 2: F = ( aδ +2 aλ ) . We first give a slightly stronger result than Proposition6.8 for Case 2. Proposition 6.10. There holds for < b < ∞ , (6.62) k r − − b D t D − ( aδ + 2 aλ ) k L bt L x . ∆ + R . Proof. It is easy to see k r − − b D t D − ( aδ + 2 aλ ) k L bt L x . k r − − b D − D t ( aδ + 2 aλ ) k L bt L x + k r − − b [ D t , D − ]( aδ + 2 aλ ) k L bt L x . (6.63)By (2.16) and (2.7),(6.64) anζ = D − D t ( aδ + 2 aλ ) − D − ( an (ˇ ρ, ˇ σ )) + D − err where, symbolically, err := an ( a/π tr χ + A · A ). k r − − b D − D t ( aδ + 2 aλ ) k L bt L x . k r − − b ( anζ ) k L bt L x + k r − − b D − ( an (ˇ ρ, ˇ σ )) k L bt L x (6.65) + k r − − b D − err k L bt L x . (6.66)By (2.28) and Proposition 6.6, the two terms on the right of (6.65) can be boundedby N ( ζ ) + ∆ + R . ∆ + R . For (6.66), by Proposition 3.4 and (6.17), also in view of (2.23) and (2.28), wededuce k r − − b D − err k L bt L x . k r − b err k L bt L x . k r − b a/π tr χ k L bt L x + k r − b A · A k L bt L x . k r − − b /π k L bt L x + k A · A k L ∞ t L x . N ( /π ) + k A k L ∞ t L x . ∆ + R . (6.67)Thus, we proved(6.68) k r − − b D − D t ( aδ + 2 aλ ) k L bt L x . ∆ + R . Repeat the derivation for (6.36), also using Lemma 3.1, k r − − b [ D − , D t ]( aδ + 2 aλ ) k L bt L x . (∆ + R ) N ( r − D − ( aδ + 2 aλ )) . (∆ + R ) N ( a/π ) . (∆ + R ) . Hence (6.62) is proved. (cid:3) Lemma 6.6. (6.69) k err k P + k / ∇D − err k P . ∆ + R , Proof. By (4.21), we have(6.70) k an · A · A k P . ∆ N ( A ) . ∆ + R Using (4.21) and (4.13), we obtain(6.71) k a n tr χ/π k P . N ( a/π ) . ∆ + R . Thus the first inequality of (6.69) is proved. The second one can be proved bycombining (6.70), (6.71), (6.67) and Theorem 4.3. (cid:3) Decomposition for commutators. In order to prove Proposition 6.1, itremains to decompose the “bad” terms which have not been treated in Proposition6.2, i.e anβ · D − F , / ∇D − ( anβ · D − F ) , with F either D − ˇ R or ( aδ + 2 aλ ) . In view of (6.23) and Proposition 6.8, theassumptions in the following theorem are satisfied with F = D − F . Then theproof of Proposition 6.1 is complete by using the following Theorem 6.1. Assume that F is an S -tangent tensor field of appropriate order on H verifying N ( F ) < ∞ and k r − b / ∇ L F k L bt L x < ∞ with < b < ∞ . Then we have (i) There exists a 1-form E such that (6.72) anβ = D t D − ˇ R + E with k E k P . ∆ + R (ii) There exists a decomposition anβ · F = D t P + E , where P and E are tensorfields of the same type as anβ · F with (6.73) lim t → k P k L ∞ x = 0 and the estimates N ( P ) . ∆ N ( F ) , k E k P . ∆ · (cid:0) N ( F ) + k r − b / ∇ L F k L bt L x (cid:1) . (6.74)(iii) There exist tensors ¯ P and ¯ E verifying (6.74) so that (6.75) / ∇D − ( anβ · F ) = D t ¯ P + ¯ E, where D denote either D or D , and (6.76) lim t → k ¯ P k L ∞ x < ∞ Proof. In view of (6.6), we have(6.77) anβ = D t D − ˇ R + C ( ˇ R ) + Err. This proves (i) by noting that E := Err + C ( ˇ R ) satisfies k E k P . ∆ + R inview of (6.8) and (6.11).Now we prove (ii). We have from (6.77) that anβ · F = ( D t D − ˇ R + Err + C ( ˇ R )) · F = D t ( D − ˇ R · F ) + E B + E G , where E B := −D − ˇ R · D t F and E G := ( Err + C ( ˇ R )) · F. By (4.18), (6.11) and (6.8) we obtain(6.78) k E G k P . N ( F ) (cid:0) k Err k P + k C ( ˇ R ) k P (cid:1) . (∆ + R ) N ( F ) . By (4.17) and (6.22) we have k E B k P . N ( D − ˇ R )( k r − b / ∇ L F k L bt L x + k r / ∇D t F k L t L x ) . ( R + ∆ )( N ( F ) + k r − b / ∇ L F k L bt L x ) . Now we set P := D − ˇ R · F and E := E B + E G , (6.79) In Theorem 6.1 and the following proofs, ˇ R = (ˇ ρ, − ˇ σ ) and C ( ˇ R ) = [ D t , D − ](ˇ ρ, − ˇ σ ), sincethe other case that ˇ R = β will not come up here. from the above estimates we have k E k P . (∆ + R )( N ( F ) + k r − b / ∇ L F k L bt L x ) . In order to estimate N ( P ), let us estimate k E k L first. By using H¨older’sinequality and ( Sob ), we can obtain k E B k L = kD − ˇ R · / ∇ L F k L . kD − ˇ R k L ∞ t L x k / ∇ L F k L t L x . N ( D − ˇ R )( k / ∇D t F k L + k r − / ∇ L F k L ) , and by using k E G k L ( H ) . k E G k P and (6.78) we can obtain k E G k L . (∆ + R ) N ( F ) . Therefore(6.80) k E k L . (∆ + R ) N ( F ) . Now we show(6.81) N ( P ) . N ( F )(∆ + R ) . With the help of D t P = anβ · F − E and (6.80) we can estimate k / ∇ L P k L asfollows k / ∇ L P k L . k β · F k L t L x + k E k L . (∆ + R ) N ( F ) . Similar to [2, Section 6.12], we get k∇ P k L t L x . (∆ + R ) N ( F ) . We completethe proof of (6.74).By (6.79) k P k L ∞ x ≤ k F k L ∞ x kD − ˇ R k L ∞ x . r kD − ˇ R k L ∞ x N ( F )(6.82)Since N ( F ) < ∞ and lim t → kD − ˇ R k L ∞ x < ∞ , (6.73) follows by letting t → P := D F , then we can apply (ii) to constructrecurrently two sequences of S -tangent tensor fields { P i } and { E i } such that(6.83) anβ · D − P i − = D t P i + E i where D − denote either D − or D − and N ( P i ) ≤ C (∆ + R ) N ( D − P i − ) , (6.84) k E i k P ≤ C (∆ + R ) (cid:16) N ( D − P i − ) + k r − b / ∇ L D − P i − k L bt L x (cid:17) . (6.85)Such P i and E i = E Bi + E Gi can be constructed as in the proof of (ii). Then for i = 1 , · · · , P i = D − ˇ R · D − P i − , P = D F (6.86) E Bi := −D − ˇ R · D t D − P i − , E Gi := ( Err + C ( ˇ R )) · D − P i − . (6.87)In particular, P and E have been given by (6.79).With the above definition of P k and E k , using (6.83) / ∇D − ( anβ · F ) = D t ¯ P k + / ∇D − ( D t P k ) + ¯ E k , where ¯ P k = / ∇D − ( P + ... + P k − ) + P + ... + P k (6.88) ¯ E k = [ / ∇D − , D t ] g ( P + ... + P k − ) + / ∇D − ( E + ... + E k )+ E + ... + E k . By using Lemma 3.1, it is easy to see from (6.84) that(6.89) N ( P k ) ≤ ( C (∆ + R )) k N ( F ) . Moreover we have Proposition 6.11. For { P k } ∞ k =1 and { E k } ∞ k =1 there hold (6.90) k r − b / ∇ L D − P k k L bt L x . (∆ + R )( N ( D − P k − ) + k / ∇ L D − P k − k L bt L x ) , (6.91) k / ∇D − E k k P . k E k k P +(∆ + R )( N ( D − P k − )+ k / ∇ L D − P k − k L bt L x ) . We will prove this result at the end of this section. We observe that Lemma 3.1,(6.90), (6.84) and (6.85) clearly imply(6.92) k E k k P ≤ ( C (∆ + R )) k (cid:16) N ( F ) + k r − b / ∇ L F k L bt L x (cid:17) . It follows from (6.89), (6.91), (6.92) and (6.26) that N ( ¯ P k − ¯ P j ) ≤ N ( F ) X j ≤ m ≤ k − ( C (∆ + R )) m . ( C (∆ + R )) j N ( F ) , and k ¯ E k − ¯ E j k P ≤ (cid:0) N ( F ) + k r − b / ∇ L F k L bt L x (cid:1) X j ≤ m ≤ k − ( C (∆ + R )) m . ( C (∆ + R )) j (cid:0) N ( F ) + k r − b / ∇ L F k L bt L x (cid:1) . Therefore { ¯ P k } forms a Cauchy sequence relative to the norm N ( · ), while { ¯ E k } forms a Cauchy sequence relative to the P norm. Denote by ¯ P and ¯ E theircorresponding limits, we have N ( ¯ P ) . (∆ + R ) N ( F ) and k ¯ E k P . (∆ + R ) (cid:0) N ( F ) + k r − b / ∇ L F k L bt L x (cid:1) . We also observe that for sufficiently small ∆ , k / ∇D − ( anβ · F ) − D t ¯ P k − ¯ E k k L = k / ∇D − ( D t P k ) k L . N ( P k ) . Letting k → + ∞ , we get k / ∇D − ( anβ · F ) − D t ¯ P − ¯ E k L t L x = 0 . Hence / ∇D − ( anβ · F ) = D t ¯ P + ¯ E . This completes the proof of (6.75) in (iii). (6.76)will be proved in Appendix.Now we conclude this section by proving Proposition 6.11. We first prove (6.91).By using (4.16) we have k / ∇D − E k k P . k E k k P + (∆ + R ) kD − E k k qL bt L x k E k k − qL , where 4 < b < ∞ and 1 / < q < < b < ∞ that(6.93) k r − b D − E k k L bt L x . (∆ + R ) (cid:16) N ( D − P k − ) + k / ∇ L D − P k − k L bt L x (cid:17) . By the construction of P k and E k , it suffices to show it for k = 1. To this end, inview of E = E G + E B , we can complete the proof by using Proposition 6.7 for k r − b D − E G k L bt L x and the estimate k r − b D − E B k L bt L x . k r − b E B k L bt L / x . kD − ˇ R k L ∞ t L x kD t F k L bt L x . N ( D − ˇ R ) kD t F k L bt L x . (∆ + R ) k / ∇ L F k L bt L x . which follows from Proposition 4.3, H¨older inequality, ( SobM1 ) and (6.22).In order to prove (6.90), we first note that(6.94) k r − b / ∇ L D − P k k L bt L x . k r − b [ D t , D − ] P k k L bt L x + k r − b D − D t P k k L bt L x . By using (6.36), the first term on the right hand side of (6.94) can be estimated as k r − b C ( P k ) k L bt L x . N ( r − D − P k ) . N ( r D − P k ) . N ( P k ) , while by using (6.83), (6.93) and (6.35), the second term can be estimated as k r − b D − D t P k k L bt L x . k r − b D − ( anβ · D − P k − − E k ) k L bt L x . k r − b D − ( anβ · D − P k − ) k L bt L x + k r − b D − E k k L bt L x . (∆ + R )( N ( D − P k − ) + k / ∇ L D − P k − k L bt L x ) . Therefore (6.90) is proved. (cid:3) Main estimatesLemma 7.1. Let F = / ∇ tr χ, µ, A · A, r − A , there holds (7.1) k / ∇D − ( anF ) k P . ∆ + R + k F k P where D − is one of the operators D − , D − , ⋆ D − .Proof. By Proposition 4.3 and (2.79), we have kD − ( anF ) k L bt L x . k anrF k L bt L x . k rF k L bt L x . ∆ + R . By (2.79), k F k L ( H ) . ∆ + R . We can infer from Theorem 4.3 and (4.13) that k / ∇D − ( anF ) k P . k anF k P + ∆ + R . k F k P + ∆ + R . as desired. (cid:3) Now we improve BA1 with the help of Theorem 5.1.7.1. Estimates for ν and | a − | .Proposition 7.1. k ν k L ∞ ω L t . ∆ + R , | a − | ≤ . Proof. We rewrite (2.17) as follows − / ∇ ν = / ∇ L ( / ∇ a ) + err , and err = 12 tr χ / ∇ a + ˆ χ · / ∇ a + A · ν. Let us denote symbolically err = tr χ · A + A · A, hence − / ∇ ( anν ) = D t / ∇ a + an f err with f err = − / ∇ log( an ) ν + err = A · A + r − A. By (4.21) we can obtain k an f err k P . ∆ + R . Applying Theorem 5.1 to P = / ∇ a and E = an · f err , we have k ν k L ∞ ω L t . N ( ν ) + N ( P ) + k an f err k P . ∆ + R , where in view of (2.79), N ( ν ) + N ( P ) . ∆ + R .In view of ν := − dds a and a ( p ) = 1, | a − | ≤ Z t | ν | nadt ′ . k ν k L ∞ ω L t . ∆ + R . With ∆ + R being sufficiently small, | a − | ≤ can be achieved. Then Propo-sition 7.1 follows. (cid:3) Estimate for ζ .Proposition 7.2. k ζ k L ∞ ω L t . ∆ + R . Proof. By (6.6) / ∇D − ( an ( ρ, σ )) = / ∇D − ⋆ D − D t β − / ∇D − e Err = D t / ∇D − ˇ R + C ( ˇ R ) + / ∇D − F . By Proposition 6.1, there exists P and E such that C ( ˇ R ) = D t P + E .Let ˜ P = P + / ∇D − ˇ R and ˜ E = / ∇D − F + E . Then by (6.23) and (6.50) / ∇D − ( an ( ρ, σ )) = D t ˜ P + ˜ E, N ( ˜ P ) + k ˜ E k P . ∆ + R . In view of (6.64), / ∇ ( anζ ) = / ∇D − ( D t ( aδ + 2 aλ )) + / ∇D − ( an ( ρ, σ )) + / ∇D − err , = D t / ∇D − ( aδ + 2 aλ ) + [ / ∇D − , D t ]( aδ + 2 aλ ) + / ∇D − err + D t ˜ P + ˜ E. (7.2)In view of Proposition 6.1, there exists P ′ an E ′ such that[ / ∇D − , D t ]( aδ + 2 aλ ) = D t P ′ + E ′ , with N ( P ′ ) + k E ′ k P . ∆ + R . Let P ′′ = P ′ + / ∇D − ( aδ + 2 aλ ), we conclude that D t / ∇D − ( aδ + 2 aλ ) + [ / ∇D − , D t ]( aδ + 2 aλ ) = D t P ′′ + E ′ . By (6.23), N ( P ′′ ) . ∆ + R . Hence, we obtain / ∇ ( anζ ) = D t P + E , where E = E ′ + / ∇D − err + ˜ E and P = P ′′ + ˜ P . Also using Lemma 6.6 for k / ∇D − err k P , we can conclude that k E k P . ∆ + R , N ( P ) . ∆ + R . Proposition 7.2 then follows by using Theorem 5.1 and N ( ζ ) . ∆ + R . (cid:3) Estimate for ˆ χ .Proposition 7.3. k ˆ χ k L ∞ ω L t . ∆ + R Proof. First, by (2.6),div( an ˆ χ ) = 12 an / ∇ tr χ + 12 an tr χζ − anβ + anζ ˆ χ = anM + anA · A + r − anζ − anβ from (6.6), anβ = D − D t (ˇ ρ, − ˇ σ ) − F with F = Err. Hence div( an ˆ χ ) = anM + anA · A + r − anζ − D − D t (ˇ ρ, − ˇ σ ) + F This gives an ˆ χ = −D − D − D t (ˇ ρ, − ˇ σ ) + D − ( F + anM + anA · A + r − anζ ) . Set D − = D − D − and D − = D − , we obtain after taking covariant derivatives(7.3) / ∇ ( an ˆ χ ) = − / ∇D − D t ˇ R + F + / ∇D − ( anM ) , where F = / ∇D − ( F + anA · A + r − anζ ) and M = / ∇ tr χ .By (6.50) and (7.1),(7.4) k F k P . ∆ + R . Consider the first term on the right of (7.3). By using the notations in (6.5), wecan write / ∇D − D t ( ˇ R ) = D t ( / ∇D − ˇ R ) + C ( ˇ R ) . where, by Proposition 6.1, there exist tensors P ′ and E ′ so that C ( ˇ R ) = D t P ′ + E ′ and(7.5) N ( P ′ ) + k E ′ k P . ∆ + R , lim t → r k P ′ k L ∞ x = 0 . Thus (7.3) becomes(7.6) / ∇ ( an ˆ χ ) = D t P + / ∇D − ( anM ) + E where P = / ∇D − ˇ R + P ′ and E = F + E ′ . By using (6.23),(7.4) and (7.5)(7.7) N ( P ) + k E k P . ∆ + R , lim t → r k P k L ∞ x = 0By combining (7.6) with (2.4) we obtain dds M + 32 tr χM = − ˆ χ · M − an ) − ˆ χ ( D t P + E + / ∇D − ( anM )) − 12 (tr χ ) ( ζ + ζ ) , regarding ι as an element of A , symbolically, / ∇ L M + 32 tr χM = A · M + ( an ) − ˆ χ · ( D t P + E + / ∇D − ( anM ))+ ( r − A + A · A ) · A + r − A. Then D t ( r M ) = − r anκM + r an { A · M + ( an ) − ˆ χ ( D t P + E + / ∇D − ( anM )) } + r anA ( A · A + r − A ) + ranA. Let us pair / ∇ tr χ with vector fields X i in Lemma 4.1, which is still denoted by M .Regarding κ also as an element of A , integrating in t , in view of lim s → r / ∇ tr χ = 0, M = r − Z t r ′ anA · M + r ′ ˆ χ · D t P + r − Z t r ′ A · ( E + / ∇D − ( anM ) + anA · A + r ′− anA ) + anr ′ Adt ′ . Using Lemma 4.1, (5.2) and (5.4) in view of lim t → k ˆ χ k L ∞ x < ∞ , we can obtain k r − Z t r ′ A · (cid:0) an ( M + A · A + r − A ) + E + / ∇D − ( anM ) (cid:1) + r ′ ˆ χ · D t P k B . ( N ( A ) + k A k L ∞ ω L t )( N ( P ) + k E k P + k / ∇D − ( anM ) k P + k anM k P + k r − anA k P + k anA · A k P ) . By Proposition 4.2, (2.32), (4.21) and (2.79), we obtain k r − Z t anr ′ A k P . k r − anA k P . N ( A ) . ∆ + R . Hence, in view of (4.13), BA1, (4.21) and (7.1), we can obtain k M k P . (cid:0) N ( P ) + k M k P + k E k P + ∆ + R (cid:1) (cid:16) N ( A ) + k A k L ∞ ω L t (cid:17) + ∆ + R . ∆ (cid:0) k M k P + N ( P ) + k E k P + ∆ + R (cid:1) + ∆ + R . (7.8)Since 0 < ∆ < / k / ∇ tr χ k P . ∆ + R . Thus, by setting ˜ E = E + / ∇D − ( anM ) we obtain from (7.6), (7.1) and (7.9) thedecomposition(7.10) / ∇ ( an ˆ χ ) = D t P + ˜ E and N ( P ) + k ˜ E k P . ∆ + R . By Theorem 5.1 and (2.79), we conclude k ˆ χ k L ∞ ω L t . N ( ˆ χ ) + N ( P ) + k ˜ E k P . ∆ + R . as expected. (cid:3) Similar to the derivation of (7.9), we can get(7.11) k r / ∇ tr χ k B . ∆ + R . Estimate for ζ .Proposition 7.4. k ζ k L ∞ ω L t . ∆ + R . Proof. By using (2.8) and (2.9),div( anζ ) = anζ · ζ − anµ − an ˇ ρ + 12 a nδ tr χ, curl ( anζ ) = an ˇ σ + an ( ζ + ζ ) ∧ ζ. Symbolically, D ( anζ ) = anA · A − an ( µ, − an (ˇ ρ, − ˇ σ ) + an ( aδ tr χ, . Hence anζ = −D − ( an (ˇ ρ, − ˇ σ )) + D − ( anA · A ) − D − ( an ( µ, D − ( an ( r − A, . Let J be the involution ( ρ, σ ) → ( − ρ, σ ) , F = g Err is given by (6.6), / ∇ ( anζ ) = / ∇D − · J · ⋆ D − D t β + / ∇D − · J · F − / ∇D − ( anµ, / ∇D − ( an · A · A ) + / ∇D − ( r − anA, . Set D − = D − · J · ⋆ D − and D − = D − . By using (6.5), we get / ∇ ( anζ ) = D t / ∇D − β + C ( ˇ R ) + / ∇D − ( anM ) + F where M = ( µ, 0) and F = / ∇D − (cid:0) F + an ( A · A + r − A ) (cid:1) .By (6.50) and (7.1), we derive k F k P . ∆ + R .In view of Proposition 6.1, for some tensors ˜ P and ˜ E such that C ( ˇ R ) = D t ˜ P + ˜ E .With E = ˜ E + F , P = ˜ P + / ∇D − β , we can write / ∇ ( anζ ) = D t P + / ∇D − ( anM ) + E, and(7.12) N ( P ) + k E k P ≤ ∆ + R , lim t → r k P k L ∞ x = 0Let M = ( µ, dds M + tr χM = 2( an ) − ˆ χ (cid:0) D t P + E + / ∇D − ( anM ) (cid:1) − χ · ζ · ζ + ( ζ − ζ )tr χζ + / ∇ tr χ ( ζ − ζ ) + tr χ ˇ ρ + (cid:18) a tr χ + A (cid:19) | ˆ χ | − aν (tr χ ) . Symbolically, dds M + tr χM =( an ) − ˆ χ · (cid:0) D t P + / ∇D − ( anM ) + E (cid:1) + tr χ ˇ ρ + A · (cid:0) A · A + r − A + / ∇ tr χ (cid:1) + r − a tr χA. (7.13)In view of lim t → r µ = 0 in Lemma 2.1, we deduce(7.14) M = r − Z t r ′ (cid:16) anκ · M + ˆ χ · D t P + A · ˜ F + an tr χ ˇ ρ + r ′− a n tr χA (cid:17) dt ′ with ˜ F = an ( A · A + / ∇ tr χ + r − A ) + E + / ∇D − ( anM ) . In view of Proposition 5.1, regarding κ as an element of A , (cid:13)(cid:13)(cid:13) r − Z t r ′ (cid:16) anκ · M + ˆ χ · D t P + A · ˜ F (cid:17) dt ′ (cid:13)(cid:13)(cid:13) B (7.15) . (cid:16) N ( A ) + k A k L ∞ ω L t (cid:17) (cid:16) k anM k P + N ( P ) + k ˜ F k P (cid:17) . Note that by (4.21), (4.13), (7.1) and (6.17), we deduce k ˜ F k P . k an ( A · A + r − A ) k P + k an / ∇ tr χ k P + k E k P + k / ∇D − ( anM ) k P . k / ∇ tr χ k P + k E k P + k M k P + ∆ + R . By (7.9) and (7.12) k ˜ F k P . ∆ + R + k M k P . Hence, by (6.17), BA1, (4.13) and (7.12)(7.15) . ∆ ( k M k P + ∆ + R ) . Assuming the following estimate for P norm of the last two terms in (7.14)(7.16) (cid:13)(cid:13)(cid:13) r − Z t r ′ (cid:16) an tr χ ˇ ρ + r ′− a n tr χA (cid:17) dt ′ (cid:13)(cid:13)(cid:13) P . ∆ + R , since 0 < ∆ < / k M k P . ∆ + R . By (7.1) and (2.79), k / ∇D − ( anM ) k P . ∆ + R .In view of (7.12), we have / ∇ ( anζ ) = D t P + E ′′′ , with E ′′′ = E + / ∇D − ( anM ).By Theorem 5.1, k ζ k L ∞ ω L t . N ( P ) + k E ′′′ k P + N ( ζ ) . ∆ + R . We now prove (7.16). With the help of (6.4), by letting p ′ := ⋆ D − β, e ′ = [ ⋆ D − , D t ] β − g Err + anA · A, also noting that (6.11) gives k [ D t , ⋆ D − ] ˇ R k P . ∆ + R , combined with (6.8),(4.21) and (6.17), we can get the following decomposition(7.18) an (ˇ ρ, ˇ σ ) = D t p ′ + e ′ with N ( p ′ ) + k e ′ k P . ∆ + R . (7.16) can be derived by establishing the following inequalities (cid:13)(cid:13)(cid:13)(cid:13) r − Z t r ′ tr χ D t p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) P . ∆ + R , (7.19) (cid:13)(cid:13)(cid:13)(cid:13) r − Z t r ′ tr χ (cid:16) e ′ + r ′− anA (cid:17) dt ′ (cid:13)(cid:13)(cid:13)(cid:13) P . ∆ + R . (7.20)To prove (7.20), by Proposition 4.2, (2.32), also in view of (4.21) and (7.18), wecan obtain (cid:13)(cid:13)(cid:13)(cid:13) r − Z t r ′ (cid:16) e ′ + r ′− anA (cid:17) dt ′ (cid:13)(cid:13)(cid:13)(cid:13) P . k e ′ k P + k r − anA k P . ∆ + R . By (5.2) and (7.18), we get (cid:13)(cid:13)(cid:13)(cid:13) r − Z t r ′ ι · e ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) B . ( N ( ι ) + k ι k L ∞ ω L t ) k e ′ k P . ∆ + R , and similarly (cid:13)(cid:13)(cid:13)(cid:13) r − Z t r ′ ι · anAdt ′ (cid:13)(cid:13)(cid:13)(cid:13) B . ( N ( ι ) + k ι k L ∞ ω L t ) k r − anA k P . ∆ + R . The proof of (7.20) is complete.Now we prove (7.19). Recall that p ′ = ⋆ D − β , then(7.21) lim s → rp ′ = 0 , k tr χp ′ k P . N ( p ′ ) . R + ∆ . Using Proposition 4.2 and (7.21), also in view of (2.1), we derive (cid:13)(cid:13)(cid:13)(cid:13) r Z t r ′ tr χ D t p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) P . X k> (cid:13)(cid:13)(cid:13)(cid:13) E k r Z t r ′ tr χ D t p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω + (cid:13)(cid:13)(cid:13)(cid:13) r Z t r ′ tr χ D t p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω . X k> (cid:18) k E k ( r ′ tr χp ′ ) k L t L ω + k E k r − Z t D t ( r ′ tr χ ) p ′ dt ′ k L t L ω (cid:19) + k / ∇ L p ′ k L t L x . k tr χp ′ k P + k tr χp ′ k L ( H ) + k / ∇ L p ′ k L t L x + X k> (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ an tr χ tr χp ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω + (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ an (cid:0) (tr χ ) + | ˆ χ | (cid:1) p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω ! . Using (5.2), (4.21), BA1 and (2.79), X k> (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ an | ˆ χ | · p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L ∞ t L ω . (cid:16) N ( ˆ χ ) + k ˆ χ k L ∞ ω L t (cid:17) k an ˆ χ · p ′ k P . (∆ + R ) N ( p ′ ) . ∆ + R , where for the last inequality, we employed (7.18).It is easy to see by (4.20) X k> (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ an tr χ tr χp ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω . k tr χp ′ k P . N ( p ′ ) . and X k> (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ an (tr χ ) p ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω ≤ X k> (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ anκ tr χp ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω + (cid:13)(cid:13)(cid:13)(cid:13) E k r − Z t r ′ an tr χ tr χp ′ dt ′ (cid:13)(cid:13)(cid:13)(cid:13) L t L ω ! . k anκ · r tr χp ′ k P + k tr χp ′ k P . (∆ + R + 1) N ( p ′ )(7.22)where we employed (4.20) and(7.23) k anκ · tr χp ′ k P . N ( p ′ )(∆ + R ) . To see (7.23), we first deduce with the help of (4.13) that(7.24) k anκ · r tr χp ′ k P . k κ · r tr χp ′ k P . By (4.17) with G = r tr χκ (7.25) k κ · r tr χp ′ k P . N ( p ′ ) (cid:16) k r / ∇ ( r tr χκ ) k L ( H ) + k r − b tr χκ k L bt L x (cid:17) where b > 4. Since k rκ k L ∞ . C and k r / ∇ ( r tr χκ ) k L ( H ) . k r / ∇ tr χκ k L ( H ) + k r tr χ / ∇ κ k L ( H ) . k / ∇ tr χ k L ( H ) + k / ∇ κ k L ( H ) . ∆ + R where for the last inequality, we employed (2.79) and Proposition 2.7. And by(2.38), we have k r − b tr χκ k L bt L x . k r − b κ k L bt L x . ∆ + R . Thus in view of (7.25) and (7.24), (7.23) is proved. In view of (7.21), (7.19) isproved. (cid:3) Similar to (7.17), we can obtain k r µ k B . ∆ + R . Appendix Dyadic sobolev inequalities. We start with proving Lemma 6.3 and a fewuseful consequences, which will be used to prove (6.76) in the second subsection.Let us still regard κ and ι as elements of A . Using Propositions 2.5, 2.7, Lemma2.8 and (1.13), more precisely k K k L t L x + k β k L t L x + N ( A ) . ∆ + R , we can adapt the approach in [4, Lemma 5.3] and [10, Chapter 9] to derive Lemma 8.1. For any smooth S t tangent tensor fields F and all q < sufficientlyclose to q = 2 , k r − q [ P k , D t ] F k L qt L x + 2 − k k r − q / ∇ [ P k , D t ] F k L qt L x . − k + N ( F ) , (8.1) k r − [ P k , D t ] F k L t L x + 2 − k k r / ∇ [ P k , D t ] F k L t L x . − k + N ( F ) . (8.2)Now we are ready to prove Lemma 6.3. Proof of Lemma 6.3. The result is trivial when q = 2. So we only need to considerthe case q > 2. It is easy to get the following estimate k r − − q P k F k L qt L x . k r − P k F k q L t L x k r − P k F k q − q L ∞ t L x . Moreover we have by integrating along an arbitrary null geodesic, k r − P k F k L x L ∞ t . k P k ( D t F ) k L t L x k r − P k F k L t L x + k r − [ P k , D t ] F · P k F k L t L x + k r − P k F k L t L x . (8.3)We can obtain the following estimate with q ′ + q = 1 and 2 < q < ∞k r − [ P k , D t F ] · P k F k L t L x . k r − q [ P k , D t ] F k L q ′ t L x k r − q − P k F k L qt L x . − k + N ( F ) k r − q − P k F k L qt L x (8.4)where we employed Lemma 8.1 to derive the last inequality. Combining the above estimates we obtain k r − − q P k F k qL qt L x . k r − P k F k L t L x (cid:16) k P k ( D t F ) k L t L x k r − P k F k L t L x + 2 − k + N ( F ) k r − q − P k F k L qt L x + k r − P k F k L t L x (cid:17) q − (8.5)Using (4.4), we get for any 0 ≤ α < − qq +1 , k r − − q P k F k L qt L x . − k ( + q ) (1 + 2 − αk ) N ( F ) . Combine (8.3), (8.4) and (6.32) with 2 < q < ∞ , by using (4.4) we obtain k r − P k F k L ∞ t L x . − k N ( F ) . To see (6.33), we derive by ( Sob ) and (4.3) k r − q F k k L qt L x . k r − q / ∇ F k k L qt L x k r − − q F k k L qt L x + k r − − q F k k L qt L x . (2 k + 1) k r − − q P k F k L qt L x . Combined with (6.32), (6.33) follows. (cid:3) Lemma 8.2. Let D − denote one of the operators D − , D − and ⋆ D − . Therehold the following estimates for appropriate S tangent tensor G , kD − P k G k L ∞ t L x . r − k N ( G ) , k > , (8.6) k / ∇D − P k G k L ∞ t L x . N ( G ) (cid:16) − k r k K k L x (cid:17) , k > . (8.7) Proof. (8.6) can be obtained by using Lemma 4.3 and (6.32). Now consider (8.7).For any S tangent tensor fields F , define k F k H x = k / ∇ F k L ( S ) + k r − F k L ( S ) . Recall by B¨ochner identity contained in [3], there holds(8.8) k / ∇ F k L x . k / ∆ F k L x + k K · F k L x + k K k L x k / ∇ F k L x + r − k F k H x , and by Sobolev embedding(8.9) k F k L ∞ x . r p k / ∇ F k p L x k F k p − p H x + k F k H x , < p < ∞ . It is easy to observe from [1, P.38] that symbolically(8.10) / ∆ = ⋆ DD ± ( K + r − Id ) . Hence with 2 < p < ∞ ,(8.11) k / ∇ F k L x . k ⋆ DD F k L x + k K k pp − L x r p − k F k H x + k K k L x k / ∇ F k L x + r − k F k H x . For F = D − H , using Proposition 3.4 k / ∇ D − H k L x . k ⋆ D H k L x + k K k pp − L x r p − kD − H k H x + k K k L x k / ∇D − H k L x + r − kD − H k H x . k ⋆ D H k L x + k K k pp − L x r p − k H k L x + (cid:0) k K k L x + r − (cid:1) k H k L x Let H = P k G , by (4.3), Proposition 3.4, Lemma 4.3, we obtain, k / ∇ D − P k G k L x . k ⋆ D P k G k L x + ( k K k L x + r − ) k P k G k L x . (cid:0) (2 k + 1) r − + k K k L x (cid:1) k P k G k L x . By ( Sob ), Proposition 3.4, (6.32) k / ∇D − P k G k L x . k / ∇ D − P k G k L x k / ∇D − P k G k L x + r − k / ∇D − P k G k L x . (1 + 2 − k r k K k L x ) N ( G ) . (cid:3) Proof of (6.76). We first prove (6.76) by assuming the following results. Lemma 8.3. Denote by D − G one of the terms D − G , D − G and ⋆ D − G forappropriate S tangent tensor G . Let F = D − ˇ R · D − G , we have k / ∇ F k B , . N ( G ) (cid:16) ∆ + R + c r ( kD − ˇ R k L ∞ x + k ˇ R k L x + r kD ˇ R k L ∞ x ) (cid:17) where c depends on k rK k L ∞ x + K α . Lemma 8.4. Let D − denotes one of the operators D − , D − and ⋆ D − . For S tangent tensor H , there hold (8.12) k / ∇D − H k B , . k ⋆ D H k B , + k H k L ω + c r N ( H ) , (8.13) k / ∇D − H k L ∞ x ≤ k ⋆ D H k B , + ( c r θ + 1)( k / ∇ H k L x + k H k L ω + c r N ( H )) . where c is depending on the quantity r k K k L ∞ x + K α + r k / ∇ K k L x , and θ > isvery close to . Let us set P = P ∞ i =1 P i and ˜ P = / ∇D − P. Since ¯ P = ˜ P + P − P and in view of(6.73) that lim t → k P k L ∞ ω = 0, (6.76) can be proved by establishing(8.14) lim t → k P k L ∞ x = 0 , and lim t → k ˜ P k L ∞ x < ∞ . In view of (6.86), we have by ( SobM2 ), Lemma 3.1 and (6.89), k P i k L ∞ ω ≤ kD − ˇ R k L ∞ ω kD − P i − k L ∞ ω . r kD − ˇ R k L ∞ ω N ( D − P i − ) . r N ( P i − ) kD − ˇ R k L ∞ ω . r ( C (∆ + R )) i − N ( F ) kD − ˇ R k L ∞ ω . Summing over i ≥ 1, with (∆ + R ) sufficiently small, k P k L ∞ ω . r N ( F ) kD − ˇ R k L ∞ ω . Noting that lim t → kD − ˇ R k L ∞ ω < ∞ and N ( F ) < ∞ ,lim t → k P k L ∞ x = 0 . It remains to prove the second part of (8.14). By (8.13), there holds(8.15) k ˜ P k L ∞ x ≤ k / ∇ P k B , + ( c r θ + 1) (cid:16) k / ∇ P k L x + k P k L ω + c r N ( P ) (cid:17) . Recall the definition of P i in (6.86). For each P i = D − ˇ R · D − P i − , by Lemma8.3, there holds k / ∇ P i k B , . c N ( P i − ) r (cid:0) kD − ˇ R k L ∞ x + k ˇ R k L x + r kD ˇ R k L ∞ x (cid:1) + N ( P i − )(∆ + R ) . In view of (6.89), summing over i ≥ k / ∇ P k B , . X i ≥ k / ∇ P i k B , . c X i ≥ (cid:0) C (∆ + R ) (cid:1) i N ( F ) r (cid:0) kD − ˇ R k L ∞ x + k ˇ R k L x + r kD ˇ R k L ∞ x (cid:1) + X i ≥ ( C (∆ + R )) i N ( F )(∆ + R ) . Hence k / ∇ P k B , . c N ( F ) r (cid:0) kD − ˇ R k L ∞ x + k ˇ R k L x + r kD ˇ R k L ∞ x (cid:1) + N ( F )(∆ + R ) . (8.16)By (6.86), ( SobM1 ), Lemma 3.1, (6.89) and (6.22), k P i k L ω . r − kD − ˇ R k L x kD − P i − k L x . N ( D − ˇ R ) r − N ( D − P i − ) . N ( P i − )(∆ + R ) . ( C (∆ + R )) i − (∆ + R ) N ( F ) . Therefore(8.17) k P k L ω . (∆ + R ) N ( F ) . It is easy to derive from P = P i ≥ P i and (6.89) that(8.18) N ( P ) . (∆ + R ) N ( F ) . Thus in view of (8.15), the combination of (8.16), (8.17) and (8.18) implieslim t → k ˜ P k L ∞ x . N ( F )(∆ + R ) < ∞ , as desired. Proof of Lemma 8.3. Let us compute k / ∇ F k L x first. k / ∇ F k L x = k / ∇ ( D − ˇ R · D − G ) k L x . k / ∇D − ˇ R k L x kD − G k L ∞ x + kD − ˇ R k L x k / ∇ D − G k L x . By ( SobM2 ), ( SobM1 ), Proposition 3.4, Lemma 3.1 and (6.22), k / ∇ F k L x . r k ˇ R k L x N ( D − G ) + N ( D − ˇ R ) N ( / ∇D − G ) . (cid:16) r k ˇ R k L x + N ( D − ˇ R ) (cid:17) N ( G ) . (cid:16) r k ˇ R k L x + ∆ + R (cid:17) N ( G ) . Now we prove(8.19) X k> k P k / ∇ ( D − ˇ R · D − G ) k L x . c N ( G ) r (cid:0) kD − ˇ R k L ∞ x + k ˇ R k L x + r k / ∇ ˇ R k L ∞ x (cid:1) . Indeed, we will employ GLP decompositions to write G = X m> P m G + P ≤ G + U ( ∞ ) G. For simplicity, we consider the high frequency terms P m> P m G . The other twoterms can be treated similar to Case 2 . Case 1: k < m . By (4.3) and (8.6), k P k / ∇ ( D − ˇ R · D − G m ) k L x . k − m r N ( G ) kD − ˇ R k L ∞ x . Thus we obtain X k> X m>k k P k / ∇ ( D − ˇ R · D − G m ) k L x . r N ( G ) kD − ˇ R k L ∞ x . Case 2: k > m . We decompose further such that(8.20) P k / ∇ ( D − ˇ R · D − G m ) = P k / ∇ ( X n> P n + P ≤ )( D − ˇ R · D − G m ) . For simplicity we consider the high frequency terms, and the low frequency termscan be treated similarly. We can adapt the proof for [11, Proposition 4.5] to obtainthe following inequality for S tangent tensor field F and 1 > α > α ≥ k P k / ∇ P n F k L x . (cid:16) min( k,n ) − | n − k | r − + 2 min( k,n ) − (1 − α ) max( k,n ) K α r − α + 2 −| k − n | k K k αL x K α (cid:17) k P n F k L x . (8.21)Let I nm = k P n ( D − ˇ R · D − G m ) k L x , we have k P k / ∇ P n ( D − ˇ R · D − G m ) k L x . (cid:16) min( k,n ) − | n − k | r − + 2 min( k,n ) − (1 − α ) max( k,n ) K α r − α + 2 −| k − n | k K k αL x K α (cid:17) I nm . (8.22)Now we estimate I nm . Let us first consider the case that n > m > 0. ByProposition 4.1 (iii), we have I nm . r − n k ˜ P n / ∆( D − ˇ R · D − G m ) k L x . r − n (cid:16) k / ∆ D − ˇ R · D − G m k L x + k ˜ P n ( / ∇D − ˇ R · / ∇D − G m ) k L x + k / ∆ D − G m · D − ˇ R k L x (cid:17) . By (8.6) and (8.10), we have k / ∆ D − ˇ R · D − G m k L x . (cid:0) k / ∇ ˇ R k L ∞ x + k K k L ∞ x kD − ˇ R k L ∞ x + r − kD − ˇ R k L ∞ x (cid:1) − m r N ( G ) . By (4.5), Propositions 3.4 and (8.7), k ˜ P n ( / ∇D − ˇ R · / ∇D − G m ) k L x . n r − k / ∇D − ˇ R k L x k / ∇D − G m k L x . n r − k ˇ R k L x N ( G ) (cid:16) − m r k K k L x (cid:17) . By (8.10), (4.3) and (6.32), (8.6), we obtain k / ∆ D − G m · D − ˇ R k L x . kD − ˇ R k L ∞ x (cid:0) k ⋆ D G m k L x + k K k L ∞ x kD − G m k L x + r − kD − G m k L x (cid:1) . kD − ˇ R k L ∞ x (cid:16) m r − + k K k L ∞ x − m r + r − − m (cid:17) N ( G ) . Hence I nm . r − n N ( G ) (cid:16) m r − kD − ˇ R k L ∞ x + 2 n r − k ˇ R k L x (cid:0) − m r k K k L x (cid:1) + k / ∇ ˇ R k L ∞ x − m r + k K k L ∞ x kD − ˇ R k L ∞ x − m r (cid:17) . (8.23)Combined with (8.22), summing over k, n, m > k > n > m and n > k > m , we summarize the results as follows X k,n,m> ,k>m,n>m k P k / ∇ P n ( D − ˇ R · D − G m ) k L x . c N ( G ) r (cid:0) kD − ˇ R k L ∞ x + k ˇ R k L x + r k / ∇ ˇ R k L ∞ x (cid:1) and c depends on k r K k L ∞ x + K α + k K k L x .It remains to estimate I nm when k > m > n . By (8.6),(8.24) I nm . kD − ˇ R · D − G m k L x . − m r N ( G ) kD − ˇ R k L ∞ x . Combined with (8.21)(8.25) X k,n,m> ,k>m>n k P k / ∇ P n ( D − ˇ R · D − G m ) k L x ≤ c N ( G ) r kD − ˇ R k L ∞ x (cid:3) Recall the following expression holds symbolically for any S tangent tensor F ,(see [3],[11, page 300]),(8.26) [ / ∇ , / ∆] F = / ∇ ( K · F ) + K · / ∇ F. Proof of Lemma 8.4. Assuming (8.12), we first prove (8.13). For simplicity let usset ˜ H = / ∇D − H . We have by [4, Proposition 3.20 (x)],(8.27) k ˜ H k L ∞ x . X k> k r − k P k ˜ H k L x + r θ c ( k / ∇ ˜ H k L x + k r − ˜ H k L x ) , where c depends on k K k L x , and θ > k ˜ H k B , . k ⋆ D H k B , + k H k L ω + r c N ( H ) . We then estimate k / ∇ ˜ H k L x , by applying (8.11) to F = D − H . By using Proposition3.4 and ( SobM1 ) we can obtain k / ∇ ˜ H k L x . k ⋆ D H k L x + r − k H k L x + r c N ( H ) , (8.29)with c depending on k K k L x .By combining (8.29), (8.28) and (8.27) and Proposition 3.4, (8.13) follows. Now consider (8.12). Using GLP projections, we need to prove X k,m> k r − k P k / ∇ P m D − H k L x . k ⋆ D H k B , + c r N ( H )(8.30) X k,m> k r − k P k / ∇ P ≤ D − H k L x . c r N ( H ) , (8.31)where c depends on k K k L x + r k / ∇ K k L x + r k K k L ∞ x .The proof of (8.31) is similar to the following Case 1 of the treatment for I km := 2 k r − k P k / ∇ P m D − H k L x , thus we will give the proof of (8.30) only. Case 1: k > m . By (4.2),(8.32) I km ≤ − k r (cid:0) k P k / ∇ / ∆ P m D − H k L x + k P k [ / ∆ , / ∇ ] P m D − H k L x (cid:1) . Let us denote the two terms on the right by I km and I km respectively. In view of(8.10), (4.3) and Lemma 4.3 we have I km . − k r k P k / ∇ P m ( ⋆ D H + K D − H + r − D − H ) k L x . m − k k P m⋆ D H k L x + 2 − k r k H k L x + 2 − k r k P k / ∇ P m ( K · D − H ) k L x (8.33)We only need to employ (8.21) to estimate the last term of (8.33).2 − k r k P k / ∇ P m ( K · D − H ) k L x . (cid:16) − | m − k | + 2 −| m − k | − (1 − α ) k K α r − α + 2 − k −| k − m | r k K k αL x K α (cid:17) k P m ( K D − H ) k L x . (8.34)For the last two terms, in view of k > m , ( SobM2 ) and Lemma 3.1 X k>m (cid:16) −| m − k | − (1 − α ) k K α r − α + 2 − k −| k − m | r k K k αL x K α (cid:17) k P m ( K D − H ) k L x . K α (cid:16) r − α + r k K k αL x (cid:17) k K k L x r N ( H ) . (8.35)Let us decompose K = P n P n K + ¯ K and consider the high frequency term for thepurpose of simplicity. With the help of Proposition 3.4, the proof contained in [11,pages 299–300] implies for m, n > k P m ( K n D − H ) k L x . − | m − n | k P n K k L x (cid:0) kD − H k L ∞ x + k H k L x (cid:1) . Therefore the first term on the right of (8.34) can be estimated as follows, X k>m X n> − | m − k | k P m ( K n D − H ) k L x . X k>m X n> − | m − k |− | m − n | k P n K k L x (cid:0) kD − H k L ∞ x + k H k L x (cid:1) . k K k B , r N ( H )where we employed ( SobM2 ), Lemma 3.1 and ( SobM1 ) to obtain the last inequal-ity. It is easy to check by (4.4) that k K k B , . k K k L x + r k / ∇ K k L x . Consequently, X k>m I km . k ⋆ D H k B , + c r N ( H ) . Now we consider I km with the help of (8.26) and ( SobM2 ), also using (4.3) andLemma 4.3 I km . − k r (cid:0) k P k / ∇ ( K · P m D − H ) k L x + k P k ( K · / ∇ P m D − H ) k L x (cid:1) . − k r (cid:16) k / ∇ K k L x k P m D − H k L ∞ x + k K k L ∞ x k / ∇ P m D − H k L x + r − k P k ( / ∇ P m D − H ) k L x (cid:17) . − k (cid:16) r k / ∇ K k L x r N ( D − H ) + r k K k L ∞ x k H k L x + r − k H k L x (cid:17) . Also using Lemma 3.1 and ( SobM1 ) X k>m> I km . (cid:0) r k / ∇ K k L x + r k K k L ∞ x (cid:1) r N ( H ) + r − k H k L x . Case 2: k < m . Consider I km in this case by (4.2) and (4.1). I km ≤ k − m r k P k / ∇ / ∆ P m D − H k L x ≤ k − m r (cid:0) k P k / ∆ / ∇ P m D − H k L x + k P k [ / ∇ , / ∆] P m D − H k L x (cid:1) ≤ k − m r − k P k / ∇ P m D − H k L x + 2 k − m r k P k [ / ∇ , / ∆] P m D − H k L x . (8.37)Let us denote by I km the first term in the line of (8.37) and by I km the secondterm. Consider I km first. By (4.3), (4.2) and (8.10) I km . k − m r − k P m D − H k L x . k − m k P m / ∆ D − H k L x . − | k − m | (cid:0) k P m⋆ D H k L x + k P m ( K D − H ) k L x + r − k P m D − H k L x (cid:1) By (4.4), Proposition 3.4, ( SobM2 ) and Lemma 3.1, k P m ( K D − H ) k L x . − m r (cid:0) k / ∇ K k L x kD − H k L ∞ x + k K k L ∞ x k H k L x (cid:1) . − m r N ( H ) (cid:0) k / ∇ K k L x + k K k L ∞ x (cid:1) . (8.38)Also using Lemma 4.3, I km . − | k − m | (cid:16) k P m⋆ D H k L x + r − − m k H k L x + 2 − m r N ( H )( k / ∇ K k L x + k K k L ∞ x ) (cid:17) . Hence, we obtain X k,m> ,k SobM1 ), X k,m> ,m>k I km . r − k H k L x + r N ( H ) · r k K k L ∞ x . 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