Calc. Var. (2013) 46:461–503 DOI 10.1007/s00526-011-0490-z
Calculus of Variations
On Ricci coefficients of null hypersurfaces with time foliation in Einstein vacuum space-time: part I Qian Wang
Received: 22 July 2010 / Accepted: 15 December 2011 / Published online: 7 February 2012 © Springer-Verlag 2012
Abstract The main objective of this paper is to control the geometry of null cones with time foliation in Einstein vacuum spacetime under the assumptions of small curvature flux and a weak condition on the deformation tensor for the future directed unit normal T to each leaf. We establish a series of estimates on Ricci coefficients, which play a crucial role to prove the improved breakdown criterion in Wang (Commun. Pure Appl. Math. 65(1):0021–0076, 2012). Mathematics Subject Classification (2000)
35L
1 Introduction Consider a (3+1)-dimensional Einstein vacuum spacetime (M, g) foliated by t which are level hypersurfaces of a time function t monotonically increasing towards the future. Let D and ∇ denote the covariant differentiations with respect to g and the induced metric g on t respectively. We define on each t the lapse function n and the second fundamental form k by n := (−g(Dt, Dt))−1/2
and k(X, Y ) := −g(D X T, Y ),
where T denotes the future directed unit normal to t and X, Y ∈ T t . For any coordinate chart O ⊂ t0 with coordinates x = (x 1 , x 2 , x 3 ), let x 0 = t, x 1 , x 2 , x 3 be the transported coordinates obtained by following the integral curves of T. Under these coordinates the metric g takes the form g = −n 2 dt 2 + gi j d x i d x j , ∂t gi j = −2nki j .
(1.1)
Communicated by G. Huisken. Q. Wang (B) Department of Mathematics, Stony Brook University, Stony Brook, NY 11794, USA e-mail:
[email protected]
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1.1 Main result Consider an outgoing null cone H contained in (M, g) with vertex p verifying t ( p) = 0. Let St denote the intersections of H with t . Without loss of generality we may assume H = sup0
d ω (s). ds
Then D L L = 0. The affine parameter s of null geodesic is chosen such that s( p) = 0 and L(s) = 1. We will assume that the exponential map Gt : ω → ω (s(t)) is a global diffeomorphism from S2 to St for any t ∈ (0, 1]. Along H we introduce the null lapse function a −1 = −g(L , T). Then a > 0 on H with a( p) = 1. Moreover, along any null geodesic ω there holds dt (1.2) = n −1 a −1 , t ( p) = 0. ds We can define a conjugate null vector L on H with g(L , L) = −2 and such that L is orthogonal to the leafs St . Let N be the outward unit normal of St in t . Then in terms of L, L and a we have 1 1 a L + a −1 L , N = a L − a −1 L . T= (1.3) 2 2 We may choose an orthonormal frame (e A ) A=1,2 tangent to St . Then (e A ) A=1,2 , e3 = L, e4 = L form a null frame. Relative to this null frame we can define the null components of Riemannian curvature tensor R of the space-time (M, g) as follows α AB = R(L , e A , L , e B ),
βA =
1 R(e A , L , L, L), 2
1 1 (1.4) R(L, L , L, L), σ = R(L, L , L, L), 4 4 1 β A = R(e A , L, L, L), α AB = R(L, e A , L, e B ). 2 Moreover, we can also define the Ricci coefficients χ, χ , ζ , ζ and via the frame equations ρ=
D A L = χ AB e B − ζ A L , D L L = 2ζ A e A ,
D A L = χ AB e B + ζ A L,
D L L = 2ζ A e A − 2 L .
It is easy to see 1 1 / L e A + ζ L , DB e A = ∇ / B e A + χ AB e3 + χ e4 DL e A = ∇ A 2 2 AB / denotes the covariant derivative on St with respect to the induced metric γ . where ∇ We introduce the mass aspect functions μ and μ as follows
123
1 a2 μ = − D3 trχ + (trχ)2 − trχ, 2 4
(1.5)
1 μ = D4 trχ + trχ · trχ . 2
(1.6)
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where trχ denotes the trace of χ, i.e. trχ = γ AB χ AB . We will also use χˆ to denote the traceless part of χ. Similarly we can define trχ and χˆ . Let Trk = g i j ki j denote the trace of k. By setting λ = −Trk/3, then the traceless part of k can be written as kˆ := k + λg. Relative to the orthonormal frame {N , e A , A = 1, 2}, we may decompose kˆ along the null cone H by introducing the components η AB = kˆ AB A = kˆ AN Let ηˆ AB denote the traceless part of η. Since
δ AB η
δ = kˆ N N . AB
(1.7)
= −δ, it is easy to see
1 ηˆ AB = η AB + δγ AB . 2 / log n, −∇ N log n}. It We denote by π/ one of the following St tangent tensors {η, ˆ δ, , λ, −∇ is easy to check by definition that the Ricci coefficients ζ , ζ and ν := −L(a) verify / N log n + δ − λ, ν := −L(a) = −∇
(1.8)
/ A log a + A , ζ = ∇ / A log n − A . ζA = ∇ A
(1.9)
Let θ AB := g(∇ A N , e B ) be the second fundamental form of St in t . By definition of χ, χ and (1.3), it follows aχ AB = θ AB − k AB , a −1 χ AB = −θ AB − k AB ,
(1.10)
atrχ = trθ + δ + 2λ, a −1 trχ = −trθ + δ + 2λ.
(1.11)
Throughout this paper we will use γ (0) to denote the canonical metric on S2 . We will also use γt := γ (t, ω) to denote the induced metric of g on St parametrized by the normal coordinates ω = (ω1 , ω2 ) ∈ S2 . On each St we introduce the ratio of area elements √ |γt | vt (ω) := , ω ∈ S2 . (1.12) |γ (0) | Let |St | denote the area of St . We define the radius of St to be r (t) = (4π)−1 |St | and intro ◦ duce the normalized metric γ = r −2 γ . For scalar functions f on St we use f¯ := |S1t | St f dμγ p p to denote its average over St and define its L ω norm by f L p = St | f | p dμS2 . For any ω scalar functions f on H, we set 1
f := H
1 f nadμγ dt =
na f vt dμS2 dt. 0 |ω|=1
0 St
For any S-tangent tensor field F on H we define the norms 1 F 2L 2 (H)
=
|F|2 nadμγ dt 0 St
and 1 F 2L ∞ L 2 (H) := sup ω
t
ω∈S2
|F|2 nadt := sup
0
ω∈S2
|F|2 nadt.
(1.13)
ω
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For simplicity of presentation, we will suppress H in the definition of norms on H whenever there occurs no confusion. Thus, for any St tangent tensor F we can introduce the norm / L F L 2 + ∇ / F L 2 + r −1 F L 2 . N1 (F) = ∇ and define the curvature flux R(H) on H relative to t-foliation by 1 R(H)2 =
an(|α|2 + |β|2 + |ρ|2 + |σ |2 + |β|2 )dμγ dt. 0 St
The main result of this paper is the following Theorem 1.1 (Main Theorem) Let (M, g) be a smooth 3+1 Einstein vacuum spacetime foliated by t , the level hypersurfaces of a time function t with lapse function n. Consider an outgoing null hypersurface H = ∪0
(1.14)
with R0 sufficiently small. Then there hold the following estimates trχ − 2 R20 , s L ∞ (H) |a − 1| ≤ 21 , 1 2 |χ| ˆ nadt 0
1 |ζ |2 nadt 0
L∞ ω
L∞ ω
(1.16)
1 2 + |ζ | nadt 0
1 2 + |ν| nadt 0
R20 ,
(1.17)
R20 ,
(1.18)
L∞ ω
L∞ ω
/ / ∇trχ P 0 + μ P 0 + ∇trχ L 2 (H) + μ L 2 (H) R0 ,
(1.19)
N1 (χˆ ) + N1 (ζ ) + N1 trχ − r2 + N1 (trχ − (an)−1 antrχ ) R0 ,
(1.20)
trχ − r2 L 2 L ∞ + trχ − (an)−1 antrχ L 2 L ∞ R0 ,
(1.21)
t
ω
3/2 / sup |r ∇trχ| t≤1 + r
123
(1.15)
1/2
t
L 2ω
3 + sup r 2 |μ| t≤1
μ B0 R0 .
ω
0 / + r 1/2 ∇trχ B L 2ω
(1.22)
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The Besov norms P 0 and B0 appearing in the above statement will be defined by (4.1) and (4.2). Throughout this paper we will use the notation A B to mean A ≤ C · B for some appropriate universal constant C. It is worthy to remark that (1.15) is the most important estimate in the main theorem and none of the estimates among (1.15)–(1.21) can be proved independent of others. We will prove all of them simultaneously with a delicate bootstrap argument. 1.2 Application By a standard rescaling argument, see [7, p. 363], the estimates (1.15)–(1.22) in Theorem 1.1 can be rephrased as [12, Theorem 7 and Proposition 14], which are the crucial components in the proof of an improved breakdown criterion for Einstein vacuum spacetime in CMC gauge, stated as follows Theorem 1.2 [12, Theorem 1] Let (M, g) be a globally hyperbolic development of t0 foliated by the CMC level hypersurfaces of a time function t < 0. Then the space-time together with the foliation t can be extended beyond any value t∗ < 0 for which t∗
k L ∞ (t ) + ∇ log n L ∞ (t ) dt = K0 < ∞.
(1.23)
t0
Under the assumption (1.23) only, we have proved in [12, Sect. 3] that C −1 < n < C with C depending only on t ∗ , K0 and the Bel-Robinson energy on the initial slice t0 . The condition (1.14) has also been verified in [12, Theorems 5, 6] under the assumption (1.23) with the help of a bootstrap argument (see [12, BA1–BA3]), in particular involved with energy estimate for the geometric wave equation of the second fundamental form k. Therefore we can apply Theorem 1.1 to close the proof for Theorem 1.2. We recall that in [6,7] the estimates in Theorem 1.1 were obtained under the assumption sup k L ∞ (t ) + ∇ log n L ∞ (t ) = 0 < ∞ (1.24) [t0 ,t∗ )
combined with the same assumptions on R(H) and n, both of which can be obtained under (1.24) or the weaker assumption (1.23). The key improvement in Theorem 1.1 lies in that it relies on the weaker assumption N1 (π / ) ≤ R0 ,
(1.25)
which comes naturally as a consequence of (1.23). Due to the much weaker assumption (1.25), we no longer can adopt the approach contained in [5–7] to control Ricci coefficients and |a − 1|. For instance, let us consider the estimate on |a − 1|. As required in [7] and [12], we need to show that |a − 1| ≤
1 . 2
(1.26)
For q > 1, an assumption of ⎛ t ∗ ⎝ ( k q ∞
L ()
⎞ q1 + ∇
q log n L ∞ () )⎠
< 0 < ∞
(1.27)
t0
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by rescaling takes the following form on null cone H k L qt L ∞ (H) + ∇ log n L qt L ∞ (H) < R0 , x
x
which by (1.8) and (1.9) immediately gives ν L q L ∞ (H) + ζ L q L ∞ (H) R0 . t
t
x
x
(1.28)
/ L a and a( p) = 1, we can obtain (1.26) by integrating In view of the definition ν := −∇ along any null geodesic ω as long as R0 is sufficiently small. If q = 1, the above simple argument fails due to the fact that by recaling, there holds k L 1 L ∞ (H) + ∇ log n L 1 L ∞ (H ) < K0 t
t
x
x
which fails to be small. Thus, it is impossible to derive (1.28) immediately from such an assumption. Theorem 1.1 however provides the trace estimate for ν L ∞ L 2 (H) in (1.18) t x which is strong enough to guarantee |a − 1| ≤ 1/2. As remarked after the statement of Theorem 1.1, estimates such as ζ L ∞ L 2 (H) R0 · · · ω
t
/ are indispensable for establishing (1.15), (1.17) and estimates on μ and ∇trχ, all of which were employed to prove breakdown criterion in [7] and [12]. The above estimate for ζ can not directly follow from the assumption (1.27) with q < 2. It is well known that the embedding H 1 (St ) → L ∞ (St ) fails. Therefore the assumption (1.25), which is a consequence of (1.23) by energy estimate, can not control ζ , ν L 2 L ∞ by a t x simple Sobolev embedding. This forces us to estimate the weaker norm · L ∞ L 2 (H) , which, t x / can / and ∇ν according to our experience, would succeed only when special structures for ∇ζ be found. 1.3 Comparison to geodesic foliation Let us draw comparisons between geodesic foliation and time foliation on null hypersurfaces as follows. (1) In the case of geodesic foliation ([2,11]), the complete set of estimates in the main theorem can be obtained under the small curvature flux only. However in time foliation, (1.14) contains one more assumption (1.25). In order to understand the reason for assuming (1.25), let us sketch the approach to derive (1.18). We will prove and employ the sharp trace inequality (see Theorem 5.1) on null cones with time foliation, which lied in the heart of [2,4,3] for null hypersurfaces with geodesic foliation. To implement this idea, it is a must to control N1 (ν, ζ ). In view of (1.9) and (1.8), ν and ζ are combinations of elements of π, / therefore the assumption (1.25) guarantees the necessary control on ν, ζ . (2) The identity ζ + ζ = 0 only holds on null hypersurface H with geodesic foliation. Therefore, in the case of time foliation, the estimates for ζ are no longer identical to those for ζ . Note that ζ and ζ are on an equal footing in structure equations (2.4) and (2.15), we need N1 (·) and · L ∞ L 2 (H) estimates for both of them. ζ will be treated in the same fashion as t x in geodesic foliation, while the estimate for ζ L ∞ L 2 (H) requires further study on structure t x equations. Similar to (2.8) and (2.9), the quantity μ is connected to ζ by the Hodge system, (2.5) and (2.7), divζ = −ρˇ + 21 μ + · · · divζ = −ρˇ − μ + · · · curl ζ = σˇ , curl ζ = −σˇ .
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However μ fails to satisfy a similar transport equation as ( 2.15) for μ, consequently μ does not verify · P 0 estimate as μ, the treatment of ζ is therefore different from ζ . Note that if / , there holds the following decomposition for ∇ζ /LP + E / =∇ ∇ζ
(1.29)
with P and E satisfying appropriate estimates, we may rely on the sharp trace inequality to estimate ζ L ∞ L 2 . To obtain the important structure (1.29), we first derive a refined Hodge ω t system, (2.16) and (2.7), divζ = −ρˇ + L(aδ + 2aλ) + · · · (1.30) curl ζ = −σˇ . The pair of quantities (ρ, ˇ σˇ ) can be decomposed in the same way as in [2,11]. We set D1 : F → (divF, curl F) for any smooth St tangent 1-form F. (1.29) then will be obtained / L with ∇ / D1−1 , i.e. from (1.30) by commuting ∇ / D1−1 (aδ + 2aλ) + [∇ / L∇ / D1−1 , ∇ / L ](aδ + 2aλ) + ∇ / D1−1 (an(ρ, / )=∇ ˇ σˇ )) + · · · . ∇(anζ / L, ∇ / D1−1 ](aδ + 2aλ), We will decompose further in [13, Sect. 3] the new commutator [∇ −1 / D1 (an(ρ, ˇ σˇ )) and the commutators and curvature terms arising in control of the term ∇ χ, ˆ ζ L ∞ L 2 (H) . t x Recall that in order to control |a − 1| < 1/2 in (1.26), we need to estimate ν L ∞ L 2 , ω t which is a quantity that does not arise in the case of geodesic foliation. We again rely on sharp trace inequality to derive ν L ∞ L 2 , which requires another remarkable structure of the ω t form / =∇ / L P + E. ∇ν / i.e. (2.17), This will be done by deriving the transport equation for ∇a, 1 / − trχ ∇a / + ··· . / = −∇ L (∇a) ∇ν 2 To prove sharp trace inequality and to control P and E actually dominate this paper and its sequel [13]. Further comparison on technical details will be made in Sect. 2. 1.4 Organization of the paper This paper is organized as follows. In Sect. 2, we start with providing all the structure equations and making bootstrap assumptions. Then by using structure equations (2.1)–(2.18) and the Sobolev embedding, we establish a series of preliminary estimates including weakly ◦ 2 spherical property for the metric γ . In Sect. 3, we prove −α K L ∞ 2 0 + R0 with t Lx α ≥ 1/2 and establish a series of elliptic estimates. In Sect. 4, we briefly review the theory of geometric Littlewood Paley decomposition (GLP) and define Besov norms. We give the equivalence relation on Besov norms in Proposition 4.2 and the reduction argument in ◦ Lemma 4.1 based on the weakly spherical property for γ . With the help of these two arguments, in Sect. 5, we prove the sharp trace theorem, i.e. Theorem 5.1. In Sect. 6, by using Theorem 5.1 we establish the estimates of χˆ , ζ , ζ, ν L ∞ L 2 and close the proof of Theoω t rem 1.1 by assuming Propositions 6.2 and 6.3. The proofs of Propositions 6.3 and 6.2 are contained in [13].
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2 Preliminary estimates 2.1 Structure equations We collect the structure equations on the Ricci coefficients which are crucial for the proof of the main result. We first have d trχ + 1 (trχ)2 = −|χ| ˆ 2, (2.1) ds
d χˆ AB ds d ds ζ A d / ds ∇trχ
2
+ trχ χˆ AB = −α AB ,
(2.2)
= −χ AB ζ B + χ AB ζ B − β A ,
(2.3)
/ / / χˆ − (ζ + ζ )(|χ| = −χˆ · ∇trχ − 2χˆ · ∇ ˆ 2 + 21 (trχ)2 ), + 23 trχ ∇trχ d ds trχ
+ 21 trχtrχ = 2divζ − χˆ · χˆ + 2|ζ |2 + 2ρ,
(2.4) (2.5)
/ divχˆ = 21 ∇trχ + 21 trχ · ζ − χˆ · ζ − β,
(2.6)
curl ζ = 21 χˆ ∧ χˆ − σ,
(2.7)
divζ = −μ − ρˇ − |ζ |2 + 21 aδtrχ + aλtrχ
(2.8)
curl ζ = σˇ
(2.9)
In what follows, we record null Bianchi equations / L β A = divα − 2trχβ A + (2ζ A + ζ )α AB ∇ A
(2.10)
3
ζ ),
ζ − 21 trχ · χˆ + ζ ⊗ /⊗ L ρˇ + trχ · ρˇ = divβ + (ζ + 2ζ ) · β − 21 χˆ · (∇ 2
(2.11)
3
ζ ),
ζ + ζ ⊗ /⊗ L σˇ + trχ · σˇ = − curl β − (ζ + 2ζ ) ∧ β − 21 χˆ ∧ (∇ 2
(2.12)
/ L β + trχβ = −∇ρ / + (∇σ / ) + 2χˆ · β − 3(ζ ρ − ζ σ ) ∇
(2.13)
ˆ σˇ = σ − 21 χˆ ∧ χˆ . ρˇ = ρ − 21 χˆ · χ,
(2.14)
where
Moreover, there hold 1 / A trχ + trχζ A ) − trχ(χˆ · χˆ − 2ρ + 2ζ · ζ ) / + (ζ − ζ ) · (∇ L(μ) + trχμ = 2χˆ · ∇ζ 2 1 1 2 a trχ − a(δ + 2λ) |χ| ˆ 2 − aν(trχ)2 , (2.15) + 2ζ · χˆ · ζ + 4 2 L(aδ + 2aλ) + 23 aδtrχ = divζ + ρˇ + |ζ |2 + atrχ∇ N log n,
123
(2.16)
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/ L (∇a) / + 21 trχ ∇a / = −∇ν / − χˆ · ∇a / − (ζ + ζ ) · ν. ∇
(2.17)
The Gauss curvature K on each St verifies 1 1 K = − trχtrχ + χˆ · χˆ − ρ. 4 2 Furthermore, the following two commutation formulas hold:
(2.18)
(i) For any scalar functions U , d / A U + χ AB ∇ / A F + (ζ + ζ )F, / BU = ∇ ∇ ds d U. where F = ds (ii) For St tangent 1-form U A satisfying
dU A ds
d / AUB divU + χ AB ∇ ds = divF + (ζ + ζ ) · F +
(2.19)
= FA , there holds
1 trχζ A − χˆ AB ζ B + β A U A . 2
(2.20)
We will prove (2.16) and (2.17) and refer the reader to [1, Chapter 11] and [2, Sect. 2] for the derivation of all other formulae. Proof of (2.16) and (2.17) We first prove (2.16). In view of (2.5) and (2.1), L(a −1 trχ) = L(trχ )a −1 + trχ L(a −1 ) 1 −1 2 =a 2divζ + 2ρˇ + 2|ζ | − trχ · trχ − a −2 L(a)trχ, 2 1 L(atrχ) = L(trχ)a + trχ L(a) = −a |χ| ˆ 2 + (trχ)2 + L(a)trχ. 2 By using (1.11), we obtain 1 L(2δ + 4λ) = a −1 (2divζ + 2ρˇ + 2|ζ |2 ) − trχ(a −1 trχ + atrχ) + L log a(atrχ − a −1 trχ) 2 = a −1 2(divζ + ρˇ + |ζ |2 ) − trχ(δ + 2λ) + 2L(log a)trθ, Thus, with the help of (1.11) and (1.8), it follows that L(aδ) = a L(δ) + δL(a) 1 = (divζ + ρˇ + |ζ |2 ) − atrχ(δ + 2λ) + L(a)(trθ + δ) − 2a L(λ) 2 3 2 = divζ + ρˇ + |ζ | + atrχ − δ + ∇ N log n − L(2aλ). 2 Next we prove (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 St , 1 / L∇ / B a + trχ ∇ / B ν − χˆ BC · ∇ / C a − (ζ B + ζ )ν / B a = −∇ ∇ B 2 which is the desired formula.
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d d Let Dt denote dt along a null geodesics initiating from p. In view of (1.2), Dt = an ds . The following result is a normalized version of (2.19).
Proposition 2.1 For any smooth scalar function f , / f = −anχ · ∇ /f [Dt , ∇]
(2.21)
In view of (2.21), (2.20) and (2.6), there holds / 2 f + 2anβ · ∇ / f − 2anζ χˆ · ∇ /f / f = −antr χ / f − 2an χˆ · ∇ [Dt , ] / f − an ∇tr / χ∇ / f. − anζ trχ ∇ 2.2 Notations and Bootstrap assumptions In (1.13) we have introduced on H the norm F L ∞ L 2 for any S tangent tensor field F. For ω t q p the further analysis we need also the L t L x norm ⎛ ⎛ ⎞q/ p ⎞1/q 1 ⎜ ⎜ ⎟ ⎟ F L qt L xp := ⎝ ⎝ |F(t, ω)| p navt dμS2 ⎠ dt ⎠ 0
and the
p Lx L∞ t
|ω|=1
norm F L xp L ∞ t
⎛ ⎞1/ p ⎜ ⎟ := ⎝ sup vt |F| p dμS2 ⎠ , S2
t∈ω
where 1 ≤ p, q ≤ ∞. For ease of exposition, we will fix the following conventions / log n, λ, π/ denotes the collection of η, ˆ , δ, ∇ N log n, ∇ ι := trχ − r2 , V := trχ − 2s , κ := trχ − (an)−1 antrχ, A denotes the collection of χˆ , ζ, ζ , ν, / log a, π, A denotes the collection of A and χˆ , ∇ / / / χˆ ), or (μ, ∇ζ / ), The pair of quantities (M, D0 M) denotes either (∇trχ, ∇ R0 denotes the collection of α, β, ρ, σ, β, R¯ denotes the collection of R0 , trχ A, A · A, / A, R˜ denotes the collection of R¯ and ∇ Ht := ∪t ∈[0,t] St , with 0 < t ≤ 1, ◦ • S := St , γ := r −2 γ , γ (0) := γS2 , K := K − r12 .
• • • • • • • • •
It follows from [11, p. 288] and the comparison formulas in [2, Sect. 2] that Lemma 2.1 / r ∇trχ, / V, ∇a, r 2 μ → 0 as t → 0, limt→0 χ, ˆ ζ, ζ , ν L ∞ (St ) < ∞. In this paper we will prove the main result, Theorem 1.1, by a delicate bootstrap argument. We will make the following bootstrap assumption. Assumption 2.1 There exists a sufficiently small 0 < R0 < 0 < 1/2 such that ˆ ν, ζ, ζ L ∞ L 2 (H) ≤ 0 , |a − 1| ≤ V L ∞ (H) ≤ 0 , χ, ω
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t
1 . 2
(BA1)
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With the help of Lemma 2.1 we will show that the inequalities in (BA1) can be improved with 0 replaced by 20 + R0 , and |a −1| ≤ 1/4. When R0 is sufficiently small, 20 + R0 < 1 2 0 can be achieved. This enables us to obtain the desired estimates in Theorem 1.1 by the bootstrap principle. We emphasize that all the arguments in this paper rely on the inequalities in (BA1); for simplicity of presentation, we will not mention it explicitly. In this section we will start with deriving estimates for N1 (A) by establishing related estimates for / M = ∇trχ, μ which will be given in Propositions 2.4 and 2.5. Then we prove that κ and ι verify stronger estimates than A, which can be seen in (2.70) and Proposition 2.7. Finally ◦ we prove that (St , γ ) is weakly spherical. 2.3 Estimates for N1 (A), r 1/2 M L 2x L ∞ and M L 2 t We first recall a few results that have been proved in [2,3,10,12]. Proposition 2.2 [12] Under the assumption (BA1), there hold r C −1 ≤ vt /s 2 ≤ C, C −1 < < C, s where C is a positive constant.
(2.22)
It is easy to derive from |V | ≤ 0 in (BA1) and Proposition 2.2 that |strχ| + |r trχ| ≤ C and from |a − 1| ≤ 1/2 in (BA1) and
C −1
(2.23)
< n < C that
C −1 < an < C,
(2.24)
with C positive constants. With the help of Proposition 2.2 and the Sobolev embedding in 2-D slices S = St , there hold the following simple inequalities: 1 • Let Osc ( f ) := f − f¯ for any scalar function f on S where f¯ = |S| f dμγ , there S
holds the Poincare inequality / f L 2 (S) . r −1 Osc ( f ) L 2 (S) ∇
(Poin)
• For a scalar function f on S with vanishing mean, there holds the Sobolev inequality (see [3]) / 2 f L 1 (S) + ∇ / f L 2 (S) , f L ∞ (S) ∇
(GaNi)
which implies / 2 f L 2 (S) + r −1 ∇ / f L 2 (S) . r −1 f L ∞ (S) ∇ Consequently, for any scalar function on H whose restriction on each St has vanishing mean there holds / L 2 (H) . / 2 L 2 (H) + r −1 ∇ r −1 L 2 L ∞ (H) ∇ t
x
(2.25)
• Let F be a S tangent tensor field, then for 2 ≤ p < ∞ there holds (see [3]) 1− 2
2
/ F 2 p F p2 + r F L xp (S) ∇ L (S) L (S) x
x
−1+ 2p
F L 2x (S) .
(Sob)
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• For any S tangent tensor field F there hold (see [2,11]) + F L 4x L ∞ + F L 6 N1 (F), r −1/2 F L 2x L ∞ t t 1
r − 2 F L ∞ + r −1 F L 2 L ∞ N2 (F), t
(SobM1)
(SobM2)
x
where / Dt F L 2 + ∇ / 2 F L 2 . / L F L 2 + r −1 ∇ / F L 2 + ∇ N2 (F) := r −2 F L 2 + r −1 ∇ By interpolation, we have for b ≥ 4 and q ≥ 2 that 1
r − b F L b L 4 + r t
− q1 − 21
F L qt L 2 N1 (F). x
x
(SobIn)
Lemma 2.2 For V there holds the estimate V L ∞ 20 .
(2.26)
Proof This can be obtained by integrating along ω the equation (2.1), i.e. d 2 1 ˆ 2 V + V = − V 2 − |χ| ds s 2 with the help of Proposition 2.2, Lemma 2.1 and V L ∞ + χˆ L ∞ L 2 ≤ 0 in (BA1). ω
t
Lemma 2.3 For an S tangent tensor field F verifying /LF + ∇
p trχ F = G · F + H 2
with p ≥ 1 certain integer, if limt→0 r (t) p F = 0 and G L ∞ L 2 0 , then there holds ω
− 2p
t
|F| vt
t
p
vt 2 |H |nadt .
(2.27)
0
We will constantly use the Hardy-Littlewood inequality for scalar f on H, s 1 f L 2 | f | s s 0
(2.28)
L 2s
With the help of Lemmas 2.1 and 2.3, (2.2), (2.3), (BA1), (2.28) and (1.14) we obtain Lemma 2.4 There hold the estimates ˆ r −1 ζ L 2 + r −1/2 χ, ˆ r −1/2 ζ L 2x L ∞ R0 , r −1 χ, t / L χ, / L ζ L 2 R0 + 20 . ∇ ˆ ∇ In view of Lemma 2.4, (1.14) and (SobM1), by definition of elements of A we can obtain the following estimates for A. / L A L 2 20 + R0 . Proposition 2.3 There holds r −1 A L 2 + r −1/2 A L 2x L ∞ + ∇ t
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Null hypersurfaces with time foliation in Einstein vacuum space-time
473
For the detail proofs of Lemmas 2.3 and 2.4 and Proposition 2.3, we refer the readers to [12]. With the help of Proposition 2.3 and Lemma 2.3, we prove under the bootstrap assumption (BA1) the following result. Lemma 2.5 There hold the estimates / log s L ∞ L 2 0 , ∇
(2.29)
/ log s L 2 L 2 + s 1/2 ∇ / log s L 2 L ∞ 20 + R0 , ∇ ω t
(2.30)
ω
t
Osc 1 s
t
ω
L 2t L 2ω
1 1 2 + s Osc s
2 L∞ t Lω
20 + R0 ,
2 κ L 2 L 2 + r 1/2 κ L ∞ 2 0 + R0 . t Lω t
ω
(2.31)
(2.32)
Proof Apply (2.19) to U = s, we can derive the transport equation d 1 / B (s) + ζ A + ζ . / A (s) + trχ ∇ / A (s) = −χˆ AB ∇ ∇ A ds 2
(2.33)
Note that e A (s) → 0, as t → 0,1 in view of Lemma 2.3 with G = χˆ and (BA1), we can derive by integrating along a null geodesic ω initiating from the vertex, |s
1 −1 ∇(s)(t)| vt 2 s
−1 /
t
1
vt2 |ζ + ζ |nadt .
(2.34)
0 2 Taking L 2t norm first then L ∞ ω (S ), with the help of (BA1),
/ s −1 ∇(s) L ∞ L 2 ζ + ζ L ∞ L 2 0 ω
ω
t
t
which gives (2.29). By taking L 2t norm first then L 2ω (S2 ) and using (2.28), we can obtain from (2.34) that −1 2 / s −1 ∇(s) L 2 L 2 r (ζ + ζ ) L 2 0 + R0 , ω t
where for the last inequality we used r −1 A L 2 20 + R0 in Proposition 2.3. Similarly / L 2 L ∞ 20 + R0 . s −1/2 ∇s ω t Hence (2.30) is proved. Applying (Poin) to f = 1/s, (2.31) follows as a consequence of (2.30). By the definition of κ, we can derive 2 2 2 − (an)−1 an(trχ − ) + (1 − (an)−1 an) s s s 1 −1 −1 −1 + 2Osc (an) an − 2(an) s Osc (an). s
κ = trχ −
(2.35)
By (Poin), (2.24) and also in view of Propositions 2.2 and 2.3, we obtain / log(an) L 2 L 2 r −1 (ζ + ζ ) L 2 (H) 20 + R0 , (2.36) s −1 Osc (an) L 2 L 2 ∇ t
ω
t
ω
1 This initial condition can be easily checked by using the comparison formulas in [2, Sect. 2].
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Q. Wang
and similarly 1
1
2 s − 2 Osc (an) L ∞ 2 r 2 (ζ + ζ ) L ∞ L 2 0 + R0 . t Lω t ω
(2.37)
Using (2.24) and (2.36), the last term in (2.35) can be estimated as follows (an)−1 s −1 Osc (an) L 2 s −1 Osc (an) L 2 L 1 20 + R0 . t
ω
t
(2.38)
Combining (2.36), (2.38) and (2.26), we obtain 1 L 2 + 20 + R0 20 + R0 . r −1 κ L 2 r −1 Osc s where, for the last inequality, we employed (2.31). By (2.31), (2.37) and (2.26), we can get 1 1 1 −1 − 21 2 2 ∞ r κ L t L 2ω ≤ r Osc L ∞ r Osc (an) L ∞ 2 + (an) 1 + V L ∞ t Lω t Lω s 1
2 + s − 2 ((an)−1 an − 1) L ∞ 2 0 + R0 . t Lω
The proof is complete. Lemma 2.6
trχ − 2 2 + R0 , 0 r L 2t trχ −
(2.39)
2 20 + R0 . r L 2t L 2ω
(2.40)
Proof We can derive the transport equation d r 2 ˆ 2. r (tr χ − ) = (an)−1 antrχκ − r (an)−1 an|χ| ds r 2
(2.41)
by combining r d r = (an)−1 antrχ. ds 2 with d trχ = −(an)−1 antrχ · trχ + (an)−1 an ds
(2.42)
1 (trχ)2 − |χ| ˆ 2 2
which can be checked in view of the definition of tr χ and (2.1). Integrate (2.41) in t in view of r trχ − 2 → 0 as t → 0, 1 2 |trχ − | r r
s(t) r ˆ 2 ds(t ). (an)−1 antrχκ + r (an)−1 an|χ| 2 0
In view of (2.24), taking
L 2t
with the help of (2.28) yields
2 ˆ 2 L 2 L 1 . trχ − L 2 r κtrχ L 2 L 1 + r |χ| t ω t ω r t
123
(2.43)
Null hypersurfaces with time foliation in Einstein vacuum space-time
475
By (2.26) and (2.32), we can obtain r trχκ L 2 L 1 r V κ L 2 L 1 + 2 rs κ L 2 L 1 (20 + 1) κ L 2 L 1 20 + R0 . (2.44) t
ω
ω
t
t
ω
t
ω
By Proposition 2.3, we have 1/2 r |χ| ˆ 2 L 2 L 1 r 1/2 χ ˆ L∞ χˆ L 2 L 2 20 + R0 . 2 r t Lω ω
t
t
(2.45)
ω
(2.39) then follows by connecting (2.43), (2.44) and (2.45). Finally it is straightforward to see 2 1 2 + trχ − . trχ − = V − V + 2Osc r s r Hence
1 2 2 ∞ L 2 L 2 + tr χ − L 2 trχ − L 2 L 2 V L + Osc t ω r t ω s r t
which, with the help of (2.26), (2.31) and (2.39), implies (2.40). Lemma 2.7 Denote by R¯ one of the quantities, R0 , trχ A, A · A, there holds ¯ L 2 (H ) 20 + R0 . R t
(2.46)
Proof The estimate about R0 can be obtained directly from (1.14). With the help of Proposition 2.3 and (BA1) A · A L 2 (Ht ) A L 2x L ∞ A L ∞ L 2 20 + R0 . t ω
(2.47)
t
By (BA1) and Proposition 2.3, we have trχ A L 2 (Ht ) V · A L 2 (Ht ) + s −1 A L 2 (Ht ) r −1 A L 2 20 + R0 .
The estimate thus follows. Lemma 2.8 Let K = K −
1 , r2
then K L 2 (Ht ) 20 + R0 .
Proof In view of (2.18) and (1.11), K−
1 = r2
a 2 −1 r2
+
a2 ι 2r
+
ι(V + 2s )a 2 4
/ L a = −ν, (2.28) and (1.14) By ∇ 2 s −1 a − 1 r = 2aν r2 2 L (Ht ) 0
− atrχ(λ + 21 δ) − ρˇ
(2.48)
r −1 π / L 2 20 + R0 .
(2.49)
L 2ω L 2t
In view of (2.48), by (2.26) and r ≈ s in (2.22), also using (2.49), (2.40) and (2.46), 2 a −1 K − 1 ¯ L 2 2 + R0 r 2 2 + r −1 ι L 2 + R 0 L r 2 L 2 (Ht ) which is the desired estimate.
With Lemma 2.8, we can prove the following estimates.
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Lemma 2.9 Let D1 be the operator that takes any S tangent 1-form F to (divF, curl F). Let D2 be the operator that takes any S-tangent symmetric, traceless, 2-tenorfields F to divF. Denote by D one of the operators D1 , D2 . For any appropriate S tangent tensor fields F in 1 the domain of D, if r − 2 F L ∞ 2 0 , there holds t Lx / F L 2 (Ht ) + r −1 F L 2 (Ht ) D F L 2 (Ht ) + 20 + R0 . ∇
(2.50)
1
/ L ∞ L 2 0 , then For smooth scalar functions , if r − 2 ∇ t x / L 2 (Ht ) / 2 L 2 (Ht ) + r −1 ∇ / L 2 (Ht ) + 20 + R0 . ∇
(2.51)
Proof Let us prove (2.51) first. In view of Böchner identity on S, / 2 |2 + K |∇| / 2 = || / 2, |∇ S
we obtain
S
/ 2 |2 + r −2 |∇| / 2= |∇
St
/ 2− ||
St
/ 2. K |∇|
(2.52)
St 1
/ L ∞ L 2 0 , Noticing that on St , there holds by (Sob) and r − 2 ∇ t x / 2 L 2 (S ) ∇ / L 2 (S ) + r −1 ∇ / 2L 2 (S ) / 2L 4 (S ) ∇ ∇ x t x t x
x
t
/ 2 L 2 (S ) + 20 . 0 ∇ x t
t
(2.53)
Integrating (2.52) on 0 < t ≤ t, in view of (2.53) and Lemma 2.8, (2.51) follows by using Young’s inequality. For D1 : F → (divF, curl F) and D2 : F → divF we recall the identities (see [1, Proposition 2.2.1]) / F|2 + K |F|2 = |D1 F|2 , / F|2 + 2K |F|2 = 2 |D2 F|2 . |∇ |∇ S
S
S
Then (2.50) follows in the same way as (2.51).
S
By Lemma 2.9 and (2.46), we can derive the following / Lemma 2.10 For M = ∇trχ, μ, there holds D0 M L 2 (Ht ) 20 + R0 + M L 2 (Ht ) .
(2.54)
More precisely, / χ / ∇ ˆ L 2 (Ht ) 20 + R0 + ∇trχ L 2 (Ht ) , / L 2 (Ht ) 20 + R0 + μ L 2 (Ht ) . ∇ζ 1
20 + R0 in Proposition 2.3 and applying Lemma 2.9 to Proof By using r − 2 A L 2x L ∞ t F = χˆ , ζ , we can derive that
/ χ ∇ ˆ L 2 (Ht ) D2 χˆ L 2 (Ht ) + 20 + R0 , / L 2 (Ht ) D1 ζ L 2 (Ht ) + 20 + R0 . ∇ζ
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Null hypersurfaces with time foliation in Einstein vacuum space-time
477
In view of (2.6), (2.8) and (2.9), ¯ D1 ζ = (μ, 0) + R. ¯ / D2 χˆ = ∇trχ + R,
By using (2.46), (2.54) can be proved. / Recall that M = ∇trχ or μ. (2.4) and (2.15) can be symbolically 2 recast as /LM + ∇
p trχ M = χˆ · M + H1 + H2 + H3 2
(2.55)
where 3 ¯ ι · R, ¯ H2 = A · A · A, and H3 = r −1 R,
H1 = A · (D0 M + F), with
( p, F) =
/ (2, ∇trχ), if M = μ, / (3, 0), if M = ∇trχ.
We will establish the following estimates. / Proposition 2.4 Let M denote either ∇trχ or μ, there hold + M L 2 R0 + 20 , r 1/2 M L 2x L ∞ t
(2.56)
/ ∇ / χ ∇ζ, ˆ L 2 R0 + 20 .
(2.57)
Proof Noticing that (2.57) can be obtained immediately by combining the second estimate in (2.56) with Lemma 2.10, we first consider the second norm in (2.56). By (2.27), (2.55) and Lemma 2.1, integrating along the null geodesic ω initiating from the vertex, we obtain t p 3 − 2p 2 |M| ≤ vt |Hi |nadt = I1 (t) + I2 (t) + I3 (t). vt i=1 0
Consider H3 with the help of (2.28), ⎛ ⎞2 1 1 t p 2 − 2p + 21 1/2 vt2 |H3 |nadt ⎠ nadt vt I3 (t) nadt = ⎝vt 0
0
0
2 t − p+1 p−2 ¯ r |r R|nadt nadt r 0 0 1 ¯ 2 |r R| nadt , 1
0
where we used vt ≈
(r )2
≈
(t )2 .
2 This means signs and coefficients on the right side of the expression can be ignored. 3 In view of |a − 1| ≤ 1/2 in (BA1), the factors a m , m ∈ N can be ignored in H when we employ (2.55) to i
prove Proposition 2.4.
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By taking the L 1ω (S2 ) norm, with the help of (2.46) we then obtain ¯ L 2 R0 + 20 . I3 L 2 R
(2.58)
Similarly, we have ⎛ ⎞2 1 1 t p 2 p 1 − + 1/2 vt2 |H2 |nadt ⎠ nadt vt I2 (t) nadt = ⎝vt 2 2 0
0
1 ≤
0
⎛ p 1 ⎝vt− 2 + 2
t
⎞2 p 2 t
v |A| |A|nadt 2
⎠
nadt
0
0 4 1/2 A L 2 r A 2L ∞ . t t
By taking L 1ω , we obtain in view of (BA1) and Proposition 2.3 that I2 L 2 A 2L ∞ L 2 r 1/2 A L 2ω L ∞ 20 (20 + R0 ). t ω
(2.59)
t
Now consider H1 := A · (D0 M + F). We have ⎛ ⎞1/2 s(t) 1/2 − p+1 p vt |D0 M|2 + |F|2 ds(t )⎠ . vt I1 (t) A L 2 ⎝vt t
0
Taking
L 2ω
norm gives 1/2
vt
I1 L 2ω A L ∞ L 2 ( D0 M L 2 + F L 2 ). ω
t
Hence in view of (2.54) and (BA1) I1 L 2 0 ( M L 2 + F L 2 + 20 + R0 ).
(2.60)
/ Consequently, in view of (2.58), (2.59) and (2.60), if M = ∇trχ, we have 2 / / ∇trχ L 2 0 ∇trχ L 2 + 0 + R0 ,
(2.61)
2 / μ L 2 0 ( μ L 2 + ∇trχ L 2 ) + 0 + R0 .
(2.62)
and if M = μ
Noticing that 0 < 0 < 1/2, we conclude from (2.61) that 2 / ∇trχ L 2 0 + R0 .
(2.63)
From (2.62) and (2.63) it follows that μ L 2 20 + R0 . Now we prove the first estimate in (2.56). We have 3
3
vt4 I3 (t) ≤ vt4
− 2p
s(t) p 1 s(t) 3 − ¯ ¯ |ds(t ) vt2 2 | R|ds(t ) s 2 −p s p−2 | Rr 0
⎞ 21 ⎛ s(t) ¯ |2 ds(t )⎠ ⎝ | Rr 0
123
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Null hypersurfaces with time foliation in Einstein vacuum space-time
479
where we used s 2 ≈ vt , r ≈ s in Proposition 2.2 to derive the second inequality. Thus 3 4 ¯ L 2 (H) 20 + R0 . (2.64) sup |vt I3 (t)| R 2 t∈(0,1] Lω
We can proceed in a similar fashion for the other two terms, p 3 4− 2
3 4
vt I2 (t) ≤ vt
then
s(t) p s(t) 3 2 − p 2 vt |A| |A|ds(t ) s 2 s p |A|2 |A|ds(t )
0 2 r A L 2 r 1/2 A L ∞ , t t
3 4 sup |vt I2 (t)| 0
L 2ω
0
A 2L ∞ L 2 r 1/2 A L 2ω L ∞ 20 (20 + R0 ). t ω t
(2.65)
Similarly, 3
vt4 I1 (t) A L 2 r (|D0 M| + |F|) L 2 r 1/2 . t
Taking
t
using (2.57) for D0 M and (2.63) for F, also in view of (BA1) 3 4 2 2 (0 + R0 )0 . sup |vt I1 (t)| ( D0 M L 2 + F L 2 ) A L ∞ ω Lt 2 0
L 2ω ,
(2.66)
Lω
20 + R0 . Combining (2.64), (2.65) and (2.66), we conclude that r 1/2 M L 2x L ∞ t (H)
In the following we summarize the major estimates that have been obtained so far. / Proposition 2.5 Let M denote either μ or ∇trχ. There holds N1 (A) + r 1/2 M L 2x L ∞ + M L 2 20 + R0 t
(2.67)
Remark 2.1 By (SobM1), (Sob) and (1.14), 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 they can 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 to provide estimates for N1 (κ, ι) and κ, ι L 2 L ∞ . t ω With the help of (2.25) we first derive a simple consequence of (2.67). Lemma 2.11 r −1 Osc (an) L 2 L ∞ 20 + R0 . t
(2.68)
ω
Proof Applying (2.25) to = an − an yields / L2 . / 2 L 2 + r −1 ∇ r −1 L 2 L ∞ ∇ t
ω
/ log(an) and (2.24), it follows With the help of ζ + ζ = ∇ 1
1
1
/ L ∞ L 2 r − 2 ∇ / log(an) L ∞ L 2 = r − 2 (ζ + ζ ) L ∞ L 2 20 + R0 . r − 2 ∇ t t t x x x
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Q. Wang
In view of (2.51), we deduce / L 2 + 20 + R0 . r −1 L 2 L ∞ t
ω
/ log(an)) and (2.24), we obtain / Therefore, by using (an) = an(|ζ + ζ |2 + / log(an) L 2 + ζ + ζ 2L 4 + 20 + R0 r −1 L 2 L ∞ t
ω
N1 (ζ + ζ )(1 + N1 (ζ + ζ )) + 20 + R0 20 + R0 ,
where we employed (SobM1) and (2.67) for the last two inequalities. Proposition 2.6 There hold the estimates 1 /2 L 2 20 + R0 , r ∇ s Osc
(2.69)
1 , Osc (trχ) , κ, ι L 2 L ∞ 20 + R0 . t ω s
(2.70)
Proof Using (2.33) and the commutation formula in [1, Lemma 13.1.2], we obtain symbolically d 2 1 1 / 2 s = − ∇trχ / −∇ / χˆ · ∇s / / s + trχ ∇ / ∇s / + χˆ · ∇ / 2 s − trχ(ζ + ζ )∇s ∇ ds 2 2 / + (ζ + ζ ) · (ζ + ζ ) + ∇(ζ / + ζ ) + (χ · ζ + β)∇s, / − (ζ + ζ )χˆ · ∇s which can be rewritten as d 2 / 2 s = χˆ · ∇ / 2 s + ( R¯ + M + ∇ / χ) / + A · A + ∇(ζ / + ζ ). / s + trχ ∇ ˆ · ∇s ∇ ds Applying Lemma 2.3 to the above equation and using χ ˆ L ∞ L 2 (H) ≤ 0 in (BA1) and ω t 2 2 / s = 0, we obtain limt→0 r ∇ /2
|∇ s|
vt−1
s(t) / χˆ ) · ∇s / + A · A + ∇(ζ / + ζ )|ds(t ). vt |( R¯ + M + ∇ 0
Therefore, by Hölder inequality and (2.28), it follows ¯ L 2 + M L 2 + ∇ / L ∞ L 2 ( R / A L 2 ) + r (A · A + ∇ / A) L 2 L 2 . / 2 s L 2 L 2 s −1 ∇s ∇ ω t
ω
ω t
t
Thus, by using (2.46), (2.56), (2.29), (2.47) and (2.67), we obtain / 2 s L 2 L 2 20 + R0 . ∇
(2.71)
ω t
By a straightforward calculation, 1 /2 / −2 ∇s) / =∇ / 2 s − 2s −1 ∇s / · ∇s / = s 2 ∇(s −s 2 ∇ s we deduce / 2 (Osc (1/s)) L 2 ∇ / 2 s L 2 L 2 + ∇ / log s · ∇s / L2 L2 . r ∇ t
123
ω
t
ω
Null hypersurfaces with time foliation in Einstein vacuum space-time
481
By using (2.30) and (2.29), we obtain / log s L ∞ L 2 s ∇ / log s L 2 L ∞ 0 (20 + R0 ) / log s · ∇s / L 2 L 2 ∇ ∇ ω t t
ω
ω
t
which together with (2.71) gives / 2 (Osc (1/s)) L 2 20 + R0 . (2.72) r ∇ 1 1 / Osc s ) L 2 Now we apply (2.25) to = r Osc s . By using (2.30) we obtain ∇( 2 0 + R0 . In view of (2.72), we also conclude that 1 Osc (2.73) L 2 L ∞ 20 + R0 . t ω s By employing Osc (trχ) = Osc (V ) + 2Osc 1s and (2.26), it follows from (2.73) that Osc (trχ) L 2 L ∞ 20 + R0 . t
(2.74)
ω
In view of ι = Osc (trχ) + trχ − r2 , (2.74) together with (2.39) implies ι L 2 L ∞ 20 + R0 . t
ω
In view of (2.35), symbolically we have κ = V − (an)−1 anV + Osc
1 an + s −1 (an)−1 Osc (an) s an
+ (an)−1 s −1 Osc (an).
(2.75)
L 2t L ∞ ω
norm of κ in view of (2.75), by using (2.68) and (2.24), the last two terms can Taking be bounded by 20 + R0 . Due to (2.73) and (2.26), the L 2t L ∞ ω norms of the remaining three terms can be bounded by 20 + R0 . Thus we conclude κ L 2 L ∞ 20 + R0 . t
ω
Proposition 2.7 There holds N1 (ι) + N1 (κ) 20 + R0 . Proof In view of (2.1) and (2.42), we have 1 2 2 1 / L trχ − =− trχ − trχ − |χ| ˆ 2 − κ. ∇ r 2 r r / L (ι) L 2 20 + R0 . Together with By (BA1) and (2.40), (2.32) and (2.47), we have ∇ 2 (2.40) and (2.67), we conclude N1 (ι) 0 + R0 . In order to derive the estimate on N1 (κ), we note that 1 1 / L κ + trχ · κ = −|χ| ∇ ˆ 2 + (an)−2 |an χ| ˆ 2 + κ 2 − (an)−2 (an)2 κ 2 2 2 / L (an) − Osc ∇ / L (an) anκ . + (an)−2 antrχ Osc ∇
(2.76)
By using (2.32) and (2.23), we have trχ · κ L 2 20 + R0 . In order to estimate the terms on the right of (2.76), we first claim / L (an) L 2 20 + R0 . (2.77) r −1 Osc ∇ /∇ / L (an) = ∇(an / / L log(an)) = ∇ / Dt log(an), we have from (2.21) Indeed, observing that ∇ ∇ that / log(an) + anχ · ∇ / log(an) = an{∇ / L (ζ + ζ ) + χ · (ζ + ζ )}. /∇ / L (an) = Dt ∇ ∇
(2.78)
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Q. Wang
With the help of (Poin), (2.78) and (2.24), we derive / L (an) L 2 ∇ /∇ / L (an) L 2 trχ A L 2 + ∇ / L A L 2 + χˆ · A L 2 . r −1 Osc ∇ Thus (2.77) follows by virtue of (2.46) and Proposition 2.3. By employing (2.77), (2.24) and (2.23), we therefore obtain / L (an) L 2 r −1 Osc ∇ / L (an) L 2 20 + R0 . (an)−2 antrχ Osc ∇ Similarly, we have / L (an) anκ L 2 Osc ∇ / L (an) L 2 L 2 20 + R0 . (an)−2 Osc ∇ t
ω
Now consider other terms on the right of (2.76). By using (2.32) and (2.70) we obtain 2 κ 2 L 2 r κ L ∞ 2 κ L 2 L ∞ 0 + R0 , t Lω t
ω
and by using (SobM1) and (2.67) we have χˆ · χˆ L 2 χ ˆ 2L 4 20 + R0 . The other two terms can be estimated similarly in view of (2.24). Hence by taking the L 2 (H) / L κ and employing the equation (2.76) and the above estimates we obtain norm of ∇ / L κ L 2 20 + R0 . ∇
(2.79)
Finally, we obtain by definition / = ∇trχ / / log(an) · (an)−1 antrχ . ∇κ +∇ By using(2.23), (2.24) and (2.67) it yields −1 2 / L 2 (H) ∇trχ / ∇κ L 2 (H) + r (ζ + ζ ) L 2 (H) 0 + R0 .
Combining this estimate with (2.32) and (2.79), we can conclude N1 (κ) 20 + R0 .
Remark 2.2 By Proposition 2.7 and (2.70), we can regard κ and ι as elements of A. Without making the strong assumption (1.24) (see [7]), under the weaker condition (1.14) only, trχ − r2 no longer satisfies the L ∞ (H) estimate as (1.15) for trχ − 2s . Since St is / = 0 while ∇s / = 0. We will check the not a level set of affine parameter s, obviously, ∇r ◦ −2 −α γ weakly spherical property for = r γ . Integral operators , geometric Littlewood Paley decompositions Pk and Besov norms will then be defined by heat flow U (τ ) with respect ◦ to γ instead of s −2 γ . Due to the factor “an” in (2.42), the nontrivial evolution of r adds technical complexity. This issue can be settled by proving trχ − (an)−1 antrχ and trχ − r2 verify stronger estimates than χˆ . Examples of the application of (2.70) and Proposition 2.7 can be seen in the proof of (2.82), Proposition 3.3, Theorem 5.1, etc. 2.5 Weakly spherical surfaces ◦
◦
Let γ be the restriction metric on St , and define the rescaled metric γ on St by γ = r −2 γ . (0) Let γi j denote the canonical metric on S2 . Note that ◦
(0) lim γ i j = γi j ,
t→0
◦
(0) lim ∂k γ i j = ∂k γi j
t→0
(2.80)
where i, j, k = 1, 2. With its aid, using the bootstrap assumption (BA1), (2.67), we will ◦ prove that for each 0 < t ≤ 1 the leave (St , γ ) is a weakly spherical surface.
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Proposition 2.8 For the transport local coordinates (t, ω), the following properties hold ◦ true for all surfaces St of the time foliation on the null cone H: the metric γ i j (t) on each St verifies weakly spherical conditions i.e. ◦
(0)
γ i j (t) − γi j L ∞ 0 ω
(2.81)
◦
(0) ∂k γ i j (t) − ∂k γi j L 2ω L ∞ 0 t
Proof Since relative to the transport coordinate on H,
d ds γi j
(2.82)
= 2χi j ,
d ◦ (γ i j ) = an(κ · γi j + 2χˆ i j )r −2 . (2.83) dt Integrating (2.83) along null geodesic initiating from vertex, with the help of (2.80), (2.81) follows in view of (2.70) and (BA1). Integrating the following transport equation along a null geodesic initiating from vertex, ◦ ◦ ◦ d ∂k γ i j = an ∂k log(an)κ γ i j +∂k trχ γ i j dt ◦ + κ∂k γ i j +2∂k χˆ i j r −2 + 2∂k log(an)χˆ i j r −2 where i, j, k = 1, 2, with the initial condition given by (2.80), by κ L 2 L ∞ 20 + R0 in t ω (2.70) and a similar argument to Lemma 2.3, we can obtain ◦ s(t) ◦ γ (0) / k χˆ i j − · χˆ ) ∂k trχ · γ i j + r −2 (∇ ∂k i j (t) − ∂k γi j 0 ◦
(0)
+ ∂k log(an) γ i j κ + 2∂k log(an)χˆ i j r −2 + κ∂k γi j where represents Christoffel symbols, and · χˆ stands for the terms l = 1, 2. Then with the help of (2.81), ◦ γ (0) sup ∂k i j (t) − ∂k γi j 2 0
2
ds(t ),
l l=1 ki χˆl j
with
Lω
/ / L 2 L 2 χ ˆ L ∞ L 2 + ∇trχ L 2x L 1t + ∇ χˆ L 2x L 1t ω t ω t + r (ζ + ζ ) · κ L 2 L 1 + r |ζ + ζ ||χ| ˆ ω t
L 2ω L 1t
+ κ L 2 L 2 . t
ω
(2.84)
By (BA1), (2.32), (2.47), we obtain the terms in the line of (2.84) κ L 2 L 2 ( A L ∞ L 2 + 1) + A · A L 2 (H) 20 + R0 . t
ω
ω
t
Using Proposition 2.4, ◦ γ (0) sup ∂k i j (t) − ∂k γi j 0
20 + R0 + χ ˆ L ∞ L 2 L 2 L 2 . L 2ω
ω
t
ω t
Sum over all i, j, k = 1, 2, also using (2.81) ◦ (0) L 2ω ∂k (γ i j −γi j ) L 2ω + C, i, j,k=1,2
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where C is the constant such that the Christoffel symbol of γ (0) satisfies |∂γ (0) | ≤ C, (2.82) then follows by using χ ˆ L ∞ L 2 ≤ 0 < 1/2 in (BA1). ω
t
3 −α K L ∞ L 2x and elliptic estimates t
Define the operator a with a ≤ 0 such that for any S-tangent tensor fields F r −a F := (−a/2)
∞
a
a
τ − 2 −1 e−τ U (τ )Fdτ,
(3.1)
0 ◦
where denotes Gamma function and U (τ )F is defined on (St , γ ) by ∂ U (τ )F − γ◦ U (τ )F = 0, U (0)F = F. ∂τ The definition of a extends to the range a > 0 by defining for 0 < a ≤ 2m that
(3.2)
a F = a−2m · (r −2 I d − γ )m F. For any a ∈ R and any S-tangent tensor field F we can introduce the Sobolev norm F H a (S) := a F L 2 (S) . We record the basic properties of a in the following result (see [3]). Proposition 3.1 (1) 0 = I d and a · b = a+b for any a, b ∈ R. (2) For any S-tangent tensor field F and any a ≤ 0 r a a F L 2 (S) F L 2 (S) . (3) For any S-tangent tensor field F and any b ≥ a ≥ 0 a
1− a
b . r a a F L 2 (S) r b b F L 2 (S) and a F L 2 (S) b F Lb 2 (S) F L 2 (S)
(4) For any S-tangent tensor fields F and G and any 0 ≤ a < 1 a (F · G) L 2 (S) F L 2 (S) a G L 2 (S) + a F L 2 (S) G L 2 (S) . (5) For any S-tangent tensor field F there holds with 2 < p < ∞ and a > 1 −
2 p
F L p (S) a F L 2 (S) . Proposition 3.2 Under the assumption of (BA1) , if R0 > 0 is sufficiently small, then for all 1 2 ≤ α < 1 there hold 2 K α := −α (K − r −2 ) L ∞ 2 0 + R0 , t Lx
(3.3)
2 −α ρ ˇ L∞ 2 0 + R0 . t Lx
(3.4)
For smooth scalar functions f on H, with p > 2, α ≥ 1/2, define the good part of the commutator [−α , Dt ] f by −α
[
α
∞
, Dt ]g f := r Cα
α
τ 2 −1 e−τ [U (τ ), Dt ] f dτ, with Cα =
0
We first prove Proposition 3.2 by assuming the following result.
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1 . (α/2)
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Proposition 3.3 Let f be a smooth scalar function H, there holds 1− 2p
r −α [−α , Dt ]g f L 1 L 2 (1 + Iα t
x
)(20 + R0 ) f L 2 ,
1
1
where p > 2 and 1/2 ≤ α < 1, and Iα := 1 + K α1−α + K α2 . Proof of Proposition 3.2. If we can prove the estimates 2 ˇ L∞ −α (K + ρ) 2 0 + R0 , t Lx 1− 2p
2 ˇ L∞ −α ρ 2 (0 + R0 )(1 + Iα t Lx
then by the definition of Iα and noting We first prove (3.5). By (2.48),
1 1−α
),
(3.5) p > 2,
(3.6)
· (1 − 2p ) < 1, we can obtain (3.3) immediately.
a 2 − 1 a 2 ι ιtrχa 2 + + + trχ A, r2 2r 4 By Proposition 3.1 (2), (BA1), (2.22) Proposition 2.3 and (2.23), for α ≥ 1/2, K + ρˇ =
(3.7)
α+1 2 trχ A L ∞ −α (trχ A) L ∞ 2 r 2 0 + R0 . t Lx t Lω
By Proposition 3.1 (2) and (1.14), we have for α ≥ 21 −α a 2 − 1 −1 −α 2 α−1 2 (a − 1) L ∞ 2 r 2 ∞ 2 r (a − 1) L ∞ t Lω t Lω r2 Lt Lx t α−1 2 / / L a L 2 L 2 r a n ∇ L adt ∇ ω t ∞ 0
L t L 2ω
r −1 ν L 2 20 + R0 .
By Proposition 3.1 (2), (SobM1) and Proposition 2.7, we have for α ≥ 21 , −α a 2 ι 2 a 2 r α ι L ∞ 2 N1 (ι) 0 + R0 . t Lω r L ∞ 2 L t x Similarly, by (2.23), 2 1+α (a 2 trχι) ∞ α −α (a 2 trχι) L ∞ 2 r 2 0 + R0 . L t L 2ω r ι L ∞ t Lx t Lω
This finishes the proof of (3.5). Next we prove (3.6). Let W (t) = (−α ρ) ˇ 2 (t) − (−α ρ) ˇ 2 (0). Then t W (t) =
d ˇ 2 + antrχ(−α ρ) ˇ 2 dμγ dt (−α ρ) dt
0 St
t 2[Dt , −α ]g ρˇ · −α ρˇ + (antrχ + αantrχ)(−α ρ) = ˇ 2 0 St
+ 2−α Dt ρˇ · −α ρˇ dμγ dt .
(3.8)
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Let ϑ(t) be a smooth cut-off function with ϑ(0) = 1 and supported in [0, 21 ], |ϑ| ≤ 1, 2 2 2 (−α ρ(0)) ˇ = (ϑ−α ρ)(0) ˇ − (ϑ−α ρ)(1) ˇ
(3.9)
can be treated similar to W (t), then ˇ 2L ∞ L 2 −α ρ t
x
−α 2 [Dt , −α ]g ρ ˇ L 1 L 2 · −α ρ ˇ L∞ ρ) ˇ L 1 L 1 (3.10) 2 + (antrχ + αantrχ)( t Lx
t x t x 1 1 + −α Dt ρˇ · −α ρdμ ˇ γ dt + ϑ 2 −α Dt ρˇ · −α ρdμ ˇ γ dt (3.11) 0 S 0 S t
t
1 d 2 ϑ ϑ(−α ρ) ˇ + dt dμγ dt .
(3.12)
0 St
In view of (2.23) and Proposition 3.1 (2), 1
(|antrχ| + |antrχ |)(−α ρ) ˇ 2 dμγ dt r − 2 −α ρ ˇ 2L 2 ρ ˇ 2L 2 . 1
0 St d By | dt ϑ| 1 and Proposition 3.1 (2), for α ≥ 21 ,
(3.12) −α ρ ˇ 2L 2 ρ ˇ 2L 2 . We only need to estimate the first term in (3.11), and the second one will follow similarly. 1 −α −α ˇ γ dt −2α Dt ρ ˇ L 2 ρ ˇ L2 . (3.13) Dt ρˇ · ρdμ 0 S t
Assuming ˇ L 2 20 + R0 , −2α Dt ρ
(3.14)
then (3.13) (20 + R0 )2 , and it follows that 1− 2p
−α ρ ˇ 2L ∞ L 2 (20 + R0 )2 (1 + Iα t
x
2 2 ) −α ρ ˇ L∞ 2 + (0 + R0 ) . t Lx
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), Dt ρˇ +
3 ˜ / antrχ ρˇ = div(anβ) − ∇(an)β + an A · R˜ = div(anβ) + an A · R. 2
By Proposition 3.1, (2.23) and (1.14), with α ≥ 21 , −2α (antrχ ρ) ˇ L 2 r 2α trχ ρ ˇ L 2 ρ ˇ L 2 20 + R0 . By Proposition 3.1 and (1.14), for α ≥ 21 , −2α div(anβ) L 2 r 2α−1 anβ L 2 20 + R0 .
123
(3.15)
Null hypersurfaces with time foliation in Einstein vacuum space-time
487
By Proposition 3.1 (5) and Hölder inequality, we obtain 1
−2α
˜ 22 (an A · R) dt = L (S )
1
t
0
˜ · (an A · R)dμ ˜ −4α (an A · R) γ dt
0 St
˜ 2 4 an A · R ˜ 2 4/3 ≤ −4α (an A · R) L L L L t
−4α+ 21 +
x
t
x
˜ 2 2 A L ∞ L 4 R ˜ L2 . (an A · R) L L t x t
x
Consequently, by (SobM1), (2.67), (2.46) and Proposition 3.1 (2), t
1 ˜ 22 ˜ L 2 (H) (2 + R0 ), −2α (an A · R) dt r 2α− 2 − −2α (an A · R) 0 L (S ) t
0
which implies ˜ L 2 (H) 20 + R0 . −2α (an A · R)
(3.16)
Thus we complete the proof of (3.14). Proof of Proposition 3.3. By definition (3.1) and Proposition 2.1, r −α [−α , Dt ]g f ∞ τ α −1 −τ / Dt ]U (τ ) f − antrχ U / (τ ) f )dτ = Cα dτ τ 2 e r 2 U (τ − τ )([, 0
0
∞
τ
= Cα
α
dτ τ 2 −1 e−τ
0
/ 1 (τ ) + φ2 (τ )), r 2 U (τ − τ )(∇φ
(3.17)
0
where / (τ ) f, φ1 (τ ) = an(χˆ + κ) · ∇U / A + A · A + r −1 A) + ∇(anκ)} / / (τ ) f. φ2 (τ ) = {an(β + ∇ ∇U / / A, Noticing that κ can be regarded as an element of A, thus we have ∇(anκ) = an A· A+an ∇ and / (τ ) f, φ2 (τ ) = an(β + ∇ / A + A · A + r −1 A)∇U / (τ ) f. (3.18) φ1 (τ ) = an A · ∇U Let us set ∞ 1 = Cα
dτ τ
α 2 −1
e
−τ
τ
0
0
∞
τ
2 = C α 0
α
dτ τ 2 −1 e−τ
/ 1 (τ )dτ r 2 U (τ − τ )∇φ
r 2 U (τ − τ )φ2 (τ )dτ .
0
The difference between φ1 , φ2 in (3.18) and those in [10, p. 41] is the extra factor “an” in 2 / (3.18), which can be easily treated by using the estimates ∇(an) 4 0 + R0 and L∞ t Lx
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(2.24). Based on the estimate N1 (A) 20 + R0 which has been proved in (2.67) and Proposition 2.7, also using (2.24), we can proceed exactly as [2, Appendix] or [10, pages 41–43] to obtain for 21 ≤ α < 1 and p > 2, 1− 2p
1 L 1 L 2 + 2 L 1 L 2 f L 2 (1 + Iα t
t
x
x
)(20 + R0 ).
The proof is therefore complete. 3.1 L 2 estimates for Hodge operators Consider the following Hodge operators 4 on 2-surfaces S := St .
• The operator D1 takes any 1-forms F into the pairs of functions (divF, curl F). • The operator D2 takes any symmetric traceless 2-tensors F on S into the 1-forms divF. / +(∇σ / )
• The operator D1 takes the pairs of scalar functions (ρ, σ ) into the 1-forms −∇ρ on S. • The operator D2 takes 1-forms F on S into the 2-covariant, symmetric, traceless tensors − 21 L F γ , where / b Fa + ∇ / a Fb − (divF)γab . (L F γ )ab = ∇
Using Proposition 3.2 and Böchner identity, we can follow the same way as in [2,10] to obtain elliptic estimates for Hodge operators. Proposition 3.4 The following estimates hold on H, (i) Let D denote either D1 or D2 . The operator D is invertible on its range, for S tangent tensor F in the range of D, / D−1 F L 2 (S) + r −1 D−1 F L 2 (S) F L 2 (S) . ∇ / is invertible on its range and its inverse (−) / −1 verifies the estimate (ii) The operator (−) / ) / 2 (−) / −1 f L 2 (S) f L 2 (S) . / −1 f L 2 (S) + r −1 ∇(− ∇ (iii) The operator D1 is invertible as an operator defined for pairs of H 1 functions with mean zero (i.e. the quotient of H 1 by the kernel of D1 ) and its inverse D1−1 takes S-tangent L 2 1-forms F(i.e. the full range of D1 ) into pair of functions (ρ, σ ) with / + (∇σ / ) = F, verifies the estimate mean zero, such that −∇ρ / D1−1 F L 2 (S) F L 2 (S) , ∇ and by (i) and duality argument r −1 D1−1 F L 2 (S) F L 2 (S) . (iv) The operator D2 is invertible as an operator defined on the quotient of H 1 -vector fields by the kernel of D2 . Its inverse D2−1 takes S-tangent 2-forms Z which is in L 2 space into S tangent 1-forms F (orthogonal to the kernel of D2 ), such that D2 F = Z , verifies the estimate / D2−1 Z L 2 (S) Z L 2 (S) . ∇ 4 For various properties of these operators please refer to [1, p. 38] and [2, Sect. 4].
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As a consequence of (i)-(iv), let D−1 be one of the operators D1−1 , D2−1 , D1−1 or D2−1 . By duality argument, we have the following estimate for appropriate5 tensor fields F, D−1 divF L 2 (S) F L 2 (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 be one of the operators D1 , D2 and D1 . For F pairs of scalar functions in the first case, S-tangent one form for the second and third case, there hold / D−1 F) N1 (F). N2 (D−1 F) N1 (F) and N1 (∇
4 A brief review of theory of geometric Littlewood Paley ◦
With the help of the heat flow (3.2) on (St , γ ), a geometric Littlewood-Paley theory has been developed in [3]. Let m be smooth functions defined on [0, ∞) decaying sufficiently fast at infinity with a finite number of vanishing moments and set m k (τ ) := 22k m(22k τ ) for any integer k. Associated to m the geometric Littlewood-Paley (GLP) projections Pk introduced in [3] take the form ∞ Pk F :=
m k (τ )U (τ )Fdτ 0
for any St tangent tensor field F. For any interval I ⊂ Z we set PI = k∈I2 Pk . In particular, we set P≤0 = k≤0 Pk . For any S-tangent tensor field F we set Fn := Pn F. The following result collects the properties of GLP projections, for more properties please refer to [3,4,10]. Proposition 4.1 There exists an m as above such that the associated GLP projections Pk verify U (∞) + k∈Z Pk2 = I d. Moreover the GLP projections Pk verify the following properties. (1) (L p -boundedness) For any 1 ≤ p ≤ ∞, and any interval I ⊂ Z, PI F L p (S) F L p (S) (2) (Bessel inequality) For any tensorfield F on S, / F 2L 2 (S) Pk F 2L 2 (S) F 2L 2 (S) , 22k r −2 Pk F 2L 2 (S) ∇ k
k
(3) (Finite band property) For any 1 ≤ p ≤ ∞, / k F L p (S) 22k r −2 F L p (S) . P
(FBD)
/ P˜k , with P˜k the Moreover given m ∈ S we can find m˜ ∈ S such that 22k Pk = r 2 geometric Littlewood Paley projections associated to m, ˜ then / L p (S) . / Pk F = 2−2k r 2 P˜k F, Pk F L p (S) 2−2k r 2 F
(FD)
5 By “appropriate”, we mean the tensor F such that divF is in the space where D −1 is well-defined.
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In addition, there hold L 2 estimates ⎧ / Pk F L 2 (S) 2k r −1 F L 2 (S) ⎨ ∇ / F L 2 (S) 2k r −1 F L 2 (S) Pk ∇ ⎩ / ∇ P≤0 F L 2 (S) r −1 F L 2 (S) ,
(FBB)
and / F L 2 (S) Pk F L 2 (S) 2−k r ∇
(FB)
(4) (Bernstein Inequality) For appropriate tensor fields F on S, we have weak Bernstein inequalities ⎧ 2 −1 (1− 2p )k ⎪ p (S) r p P 2 F + 1 F L 2 (S) , k L ⎪ ⎪ ⎪ ⎪ ⎨ P F p r 2p −1 F ≤0 L (S) L 2 (S) (WB) 2 −1 (1− 2p )k ⎪ p 2 P F r + 1 F L p (S) 2 ⎪ k L (S) ⎪ ⎪ ⎪ 2 ⎩ −1 P≤0 F L 2 (S) r p F L p (S) . where 2 ≤ p < ∞ and 1p + p1 = 1, k ≥ 0. (5) For scalar functions f there hold the sharp Bernstein inequalities Pk f L ∞ (S) 2k r −1 f L 2 (S) , Pk f L 2 (S) 2k r −1 f L 1 (S) .
(SB)
Now we define for 0 ≤ θ ≤ 1 the Besov Bθ , P θ norms for S-tangent tensor fields F on
θ norm on S as follows: H and B2,1
F Bθ =
−θ (2k r −1 )θ Pk F L ∞ F L ∞ 2 + r 2, t Lx t Lx
(4.1)
(2k r −1 )θ Pk F L 2 L 2 + r −θ F L 2 L 2 ,
(4.2)
(2k r −1 )θ Pk F L 2x + r −θ F L 2x .
(4.3)
k>0
F P θ =
t
x
t
x
k>0
F B θ
2,1 (S)
=
k>0
Remark 4.1 For any a ∈ R and any S-tangent tensor field F 22ka r −2a Pk F 2L 2 (S) + r −2a P≤0 F 2L 2 . F 2H a (S) := a F 2L 2 (S) ≈ k>0
x
In certain situation, it is more convenient to work with the Besov norms defined by the classical Littlewood-Paley (LP) projections E k . Recall that (see [8,9]) for any scalar function f on R2 1 Ek f = ϕ(ξ/2k ) fˆ(ξ )ei xξ dξ, (2π)2 R2
where ϕ is a smooth function support in the dyadic shell { 21 ≤ |ξ | ≤ 2} verifying −k k∈Z ϕ(2 ξ ) = 1 when ξ = 0.
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Define for any 0 ≤ θ < 1 the B˜ θ and P˜ θ norms of any scalar function f on H by −θ (2k r −1 )θ E k f L ∞ f L ∞ f B˜ θ := 2 + r 2, t Lx t Lx k>0
f P˜ θ :=
(2k r −1 )θ E k f L 2 L 2 + r −θ f L 2 L 2 . t
x
t
x
k>0
Using Proposition 2.8 and (BA1) we can adapt [4, Proposition 3.28] to obtain the following lemma. Lemma 4.1 Under the bootstrap assumptions (BA1) , there exists a finite number of vector fields {X i }li=1 verifying the conditions / 0 X ) L 2 L ∞ 1, ∞ 1, r ∇(∇ X, r ∇0 X L ∞ t Lω x t / / L X = 0, ∞ (∇ − ∇0 )X L 2x L t 0 , ∇ where ∇0 represents the covariant derivative induced by the metric r 2 γ (0) . For appropriate 2 θ θ S-tangent tensor F ∈ L ∞ t L x , F ∈ B if and only if F · X i ∈ B , and C −1 F · X i Bθ ≤ F Bθ ≤ C F · X i Bθ , with 0 ≤ θ < 1, i
i
where C is a positive constant. The same results hold for the spaces P θ . Moreover N1 (F ⊗ X ) + F ⊗ X L ∞ L 2 N1 (F) + F L ∞ L 2 , ω t ω 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. j j ··· j
Definition 4.2 Let F be an (m, n) S-tangent tensor field on H and let Fi11i22···in m be the local components of F relative to {X i }li=1 . We define the B˜ θ and P˜ θ norms of F by j j ··· j j j ··· j Fi11i22···in m B˜ θ and F P˜ θ = Fi11i22···in m P˜ θ , F B˜ θ = where the summation is taken over all possible (i 1 · · · i n ; j1 · · · jm ). The equivalence between Bθ , P θ norms and B˜ θ , P˜ θ norms is given in the following result whose proof can be found in [10]. Proposition 4.2 Under the bootstrap assumptions (BA1) for arbitrary S-tangent tensor fields F on H there hold for 0 ≤ θ < 1, F B˜ θ ≈ F Bθ and F P˜ θ ≈ F P θ . Lemma 4.2 Let H be any St tangent tensor field and let f be any scalar function. Then / f L ∞ L 4 , f H P 0 H P 0 f L ∞ + H L 2 (H) r 1/2 ∇ t x
(4.4)
0 (S) norms. and the similar estimates hold for B0 and B2,1
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Proof By the GLP decomposition we can write H = H¯ + n∈N Pn2 H + P≤0 H , where H¯ = U (∞)H . Thus Pk ( f H
0
+
k>0
Pk ( f H¯ ) L 2 + f H L 2 .
k>0
Let Hn := Pn2 H . Consider the first term in (4.5). By (FB), (WB) and (FBB), / f Hn ) L 2 Pk ( f Hn ) L 2 2−k r Pk ∇( k>0,k>n>0
k>0,k>n>0
k>0,k>n>0
/ f · Hn ) L 2 + r Pk ( f · ∇ / Hn ) L 2 2−k r Pk (∇ k 1 / f · Hn 2 4/3 +2−k+n f L ∞ Hn L 2 2−k+ 2 r 2 ∇ L L t
k>0,k>n>0
k 1 / f L ∞ L 4 Hn L 2 +2−k+n f L ∞ Pn H L 2 2− 2 r 2 ∇ t x
x
k>0,k>n>0
/ f L ∞ L 4 + f L ∞ H P 0 , H L 2 r 1/2 ∇ t x and similarly,
/ f L ∞ L 4 + f L ∞ H P 0 . Pk ( f H≤0 ) L 2 H L 2 r 1/2 ∇ t x
k>0
Now consider the second term. By (FD), (FBB) and (WB), we obtain / n ) L 2 Pk ( f Hn ) L 2 2−2n r 2 Pk ( f H k>0,n≥k
k>0,n≥k
2
−2n
/ f∇ / Hn ) L 2 + r 2 Pk (∇ / f ·∇ / Hn ) L 2 r 2 Pk ∇(
k>0,n≥k
k 3 / f ·∇ / Hn 2 4/3 . 2−2n+k+n f L ∞ Pn H L 2 + 2−2n+ 2 r 2 ∇ L L
t
k>0,n≥k
x
By (FBB), we have 3
k
/ Hn L 2 / f ·∇ / Hn 2 4/3 2−2n+ 2 r 1/2 ∇ / f L ∞ L 4 r ∇ 2−2n+k/2 r 2 ∇ L L t x t
x
k
/ f L ∞ L 4 H L 2 , 2 2 −n r 1/2 ∇ t x thus
1
/ f L ∞ L 4 + f L ∞ H P 0 . Pk ( f Hn ) L 2 H L 2 r 2 ∇ t x
k>0,n≥k
/ H¯ = 0 and (SobM1), we can deduce At last, due to ∇ 1 1 / f L ∞ L 4 H¯ L 2 r 2 ∇ / f L ∞ L 4 H L 2 . Pk ( f H¯ ) L 2 2−k r 2 ∇ t t x x k>0
The proof is complete.
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4.1 Product estimates and Intertwine estimates Proposition 4.3 Let D be one of the operators D1 , D2 and D1 . Then for 1 < p ≤ 2 and any S-tangent tensor F on H there holds D−1 F L 2 (S) r
2− 2p
F L p (S) .
Proof For p = 2, the inequality follows immediately from Proposition 3.4. For p > 2, from (Sob) and Proposition 3.4 we infer for p > 2 satisfying 1p + p1 = 1 that 2
r p
−2
1− p2
2 p
/ D−1 F 2 r −1 D−1 F 2 + r −1 D−1 F L 2 (S) D−1 F L p (S) ∇ L (S) L (S) F L 2 (S)
Thus, by duality, we complete the proof.
Lemma 4.3 Let D denote one of the Hodge operators D1 , D2 , D1 and D2 , let D−1 denote the inverse of D. For Pk F with Pk the GLP projections associated to the heat equation (3.2), there hold for k > 0, 1 < p ≤ 2, D−1 Pk F L 2x 2−k r F L 2x and Pk D−1 F L 2x 2
−(2− 2p )k 2− 2p
r
F L xp .
Proof The first inequality can be proved by using (FB) in Proposition 4.1 and Proposition 3.4. The second can be proved by duality using the first inequality and (SobM1). The following result follows from the second estimate in Lemma 4.3 immediately. Proposition 4.4 Let D−1 denote either D1−1 , D1−1 , D2−1 , then for appropriate S-tangent tensor fields F on H and any 1 < p ≤ 2, D−1 F P θ r
2− 2p −θ
F L 2 L xp . t
Since the proof of the Hodge-elliptic estimate for geodesic foliation contained in [11, pages 295–301] only relied on 2 K L 2 + −α0 K L ∞ with α0 ≥ 1/2, 2 0 + R0 , t Lx
based on Lemma 2.8 and Proposition 3.2 the same proof also applies to the case of time foliation. Thus we can obtain the result on the Hodge-elliptic P σ estimates. Theorem 4.3 (Hodge-elliptic P σ -estimate) Let D denote either D1 , D2 or their adjoint operators D1 and D2 . Then for any S-tangent tensor fields ξ and F satisfying Dξ = F and any 21 > σ ≥ 0, / P σ F P σ + 0 D−1 F ∇ξ
q 1−q F L 2 L 2 , L bt L 2x t x
where 1/2 ≤ α0 < q < 1 − σ and b > 4. We now give a series of product estimates for Besov norms whose proofs can be found in [11, pages 302–304]. Lemma 4.4 For any S-tangent tensor fields F and G, 1
1
/ L 2 L 2 ) with b > 4, F · G P 0 N1 (F)( r − b G L b L 2 + r 2 ∇G t
x
t
x
(4.6)
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F · G P 0 N2 (r 1/2 F) G P 0 ,
(4.7)
1 / L 2 L 2 + G L ∞ L 2 . F · G P 0 N1 (r 2 F) ∇G t
ω
x
(4.8)
t
Corollary 1 Regard κ, ι also as elements of A, there hold 1
A · F P 0 (20 + R0 )N1 (r 2 F), (trχ, r −1 )F P 0 N1 (F),
(4.9)
1
an A · F P 0 (20 + R0 )N1 (r 2 F), an(trχ, r −1 )F P 0 N1 (F).
(4.10)
Proof We prove (4.9) first. Using (4.6) for b > 4, we have 1
1
1
/ F L 2 (H ) + r − b F L b L 2 ) N1 (A) · N1 (r 2 F) A · F P 0 N1 (A)( r 2 ∇ t
x
(4.11)
where the last inequality follows by using (SobIn). By the finite band property of GLP projections in Proposition 4.1, it is straightforward to obtain / F L 2 + r −1 F L 2 . r −1 F P 0 ∇
(4.12)
Noticing that trχ · F = ι · F + 2r −1 · F and N1 (ι) 20 + R0 in Proposition 2.7, the other inequality in (4.9) follows from (4.11) and (4.12) with A replaced by ι. Applying (4.4) to f = an and H = A · F, trχ F, r −1 F, and using (4.9) for H and the following inequalities 2 / ∇(an) an L ∞ 1, 4 ≈ ζ + ζ L ∞ L 4 0 + R0 , L∞ t Lx t x
we can derive (4.10).
5 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 admits a decom/ position of the form ∇(an F) = Dt P + E with tensors P and E of the same type as F, / F| < ∞, there holds the following sharp suppose limt→0 F L ∞ < ∞ and limt→0 r |∇ x trace inequality, F L ∞ L 2 N1 (F) + N1 (P) + E P 0 . ω
t
(5.1)
Remark 5.2 We will employ this theorem to estimate F L ∞ L 2 for F = ζ , ν, χˆ , ζ . By local ω t 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 ≥ 1 be any integer, for any S-tangent tensor fields F, H and G of the same type. There holds t −p p r r H · Gdt H P 0 (N1 (G) + G L ∞ L 2 ). (5.2) ω t 0
123
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Let us further assume lim r p F L ∞ = 0, lim G L ∞ < ∞, when p ≥ 2, x x
t→0
t→0
then there holds for p ≥ 1, t −p p r r Dt F · Gdt
N1 (F)N1 (G).
(5.3)
(5.4)
B0
0
Using a modified version of [4, Lemma 5.3], see [13, Appendix], and 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. t Proof of Theorem 5.1. We set ϕ(t) = 0 an|F|2 dt , then Dt ϕ = an|F|2 . Due to [2, Proposition 5.1] we have / B0 + r −1 ϕ L ∞ L 2 . ϕ L ∞ (H) ∇ϕ t x
(5.5)
−1 r −1 ϕ L ∞ F L 2 . 2 F L ∞ L 2 · r t Lx
(5.6)
It is easy to see ω
t
/ B0 . In view of (2.19), we obtain We now estimate ∇ϕ 1 / L ∇ϕ / + trχ ∇ϕ / / − (ζ + ζ )|F|2 . / = 2(an)−1 ∇(an ∇ · F) · F − χˆ · ∇ϕ 2 / with / Using the decomposition ∇(an · F) = Dt P + E and Lemma 4.1, we may pair ∇ϕ vector field X = X i to obtain 1 / · X ) = − r κ ∇ϕ / · X − r χˆ ∇ϕ / ⊗X / L (r ∇ϕ ∇ 2 / L P · F ⊗ X + 2(an)−1 E · F ⊗ X − |F|2 (ζ + ζ ) · X . (5.7) + r 2∇ We will not distinguish “⊗” with “·”, and also suppress X whenever there occurs no confusion. Note that under the transport coordinate (s, ω1 , ω2 ), we have ∂ϕ = ∂ωi
t
{2∂ωi F, F + |F|2 ∂ωi log(an)}nadt i = 1, 2,
0
∂ϕ by assumptions on initial condition, limt→0 ∂ω = 0. It then follows by integrating (5.7) i along a null geodesic ω from 0 to s(t) that, symbolically,
/ (∇ϕ)(t) = r −1
t
/ + r E · F}dt {r an(κ + χ) ˆ ∇ϕ
0
+ r −1
t
r Dt P · Fdt + r −1
0
Since by Proposition 4.2, / B0 ≈ ∇ϕ
t
anr |F|2 (ζ + ζ )dt .
(5.8)
0
/ / L∞ L2 , E k (∇ϕ) 2 + ∇ϕ L∞ t Lx t x
(5.9)
k>0
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/ L 2 L ∞ can be obtained in view of (5.8) and (SobM1). and the estimate of ∇ϕ x t / L2 / L 2 L ∞ ( κ L ∞ L 2 + χ ˆ L ∞ L 2 ) ∇ϕ ∇ϕ x t ω
ω
t
t
/ L P L 2 + N1 (A)N1 (F)) r −1 F L 2 + ( E L 2 + ∇
(5.10)
where the term on the right of (5.10) can be absorbed in view of κ, χˆ L ∞ L 2 0 , obtained ω t from (2.70) and (BA1). It remains to estimate the first term in (5.9). / ⊗ X via (5.8), we only need to Applying Littlewood Paley decomposition E k to ∇ϕ estimate the following terms t t E k r an(κ + χˆ )∇ϕ E k r E · Fdt / I1 = , I = , 2 ∞ ∞ k>0 k>0 2 2 0 L t L ω 0t Lt Lω t 2 E k r Dt P · Fdt E , I = r |F| (ζ + ζ )dt . I3 = 4 k ∞ ∞ k>0 k>0 0
0
L t L 2ω
L t L 2ω
Use (5.2) and Proposition 4.2, / P0 . ˆ L ∞ L 2 + N1 (κ) + κ L ∞ L 2 ) ∇ϕ I1 (N1 (χˆ ) + χ x
ω
t
t
Consequently, by (2.70), Proposition 2.7, (2.67) and χ ˆ L ∞ L 2 ≤ 0 in (BA1), ω
t
/ P0 . I1 0 ∇ϕ Apply (5.2) to I2 , I2 E P 0 (N1 (F) + F L ∞ L 2 ). ω
t
By (5.4) and Lemma 4.1 I3 N1 (P)(N1 (F) + F L ∞ L 2 ). ω
t
By (5.2), (4.8), (BA1) and (2.67), we have I4 (N1 (ζ + ζ ) + ζ + ζ L ∞ L 2 ) F · F P 0 ω
t
/ F L 2 + F L ∞ L 2 ). 0 N1 (F)( ∇ ω
t
Thus we conclude
/ B0 N1 (P) + E P 0 N1 (F) + F L ∞ L 2 + 0 ∇ϕ / P0 ∇ϕ ω
t
/ F L 2 + F L ∞ L 2 ). +0 N1 (F)( ∇ ω
t
/ B0 , yields / P 0 ∇ϕ This inequality, together with the fact that ∇ϕ / B0 (N1 (F) + F L ∞ L 2 ) N1 (P) + E P 0 + 0 N1 (F) . ∇ϕ ω
t
Combine the above inequality with (5.5) and (5.6), we get F 2L ∞ L 2 (N1 (F) + F L ∞ L 2 ) N1 (P) + E P 0 + N1 (F)0 ω
ω
t
ω
t
which implies (5.1) by Young’s inequality.
123
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+ F L ∞ L 2 · r −1 F L 2
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6 Main estimates In this section we will use Theorem 5.1 to complete the proof of Theorem 1.1 by improving the estimates in (BA1) We will frequently employ (2.67), (1.14), Proposition 2.7 and Remark 2.2, which schematically can be written as 1
1
/ / r 2 μ L 2x L ∞ N1 (A) + r 2 ∇trχ, + ∇trχ, μ L 2 + R0 L 2 20 + R0 , t
(6.1)
where ι, κ are regarded as elements of A. / Lemma 6.1 Let F = ∇trχ, μ, A · A, r −1 A, there holds / D−1 (an F) P 0 20 + R0 + F P 0 ∇ where D−1 is one of the operators
D1−1 ,
(6.2)
D2−1 , D1−1 .
Proof By Proposition 4.3 and (2.67), we have D−1 (an F) L b L 2 anr F L b L 2 r F L b L 2 20 + R0 . t
By (2.67), F L 2 (H)
20
t
x
t
x
x
+ R0 . We can infer from Theorem 4.3 and (4.4) that
/ D−1 (an F) P 0 an F P 0 + 20 + R0 F P 0 + 20 + R0 ∇
as desired. 6.1 Estimates for ν and |a − 1| Proposition 6.1 There hold the estimates ν L ∞ L 2 20 + R0 and |a − 1| ≤ 1/4. ω
t
Proof We rewrite (2.17) as / + err2 , and err2 = 21 trχ ∇a / + χˆ · ∇a / + A · ν. / =∇ / L (∇a) −∇ν Let us denote symbolically err2 = trχ · A + A · A. Then / + an / err2 −∇(anν) = Dt ∇a / log(an)ν + err2 = A · A + r −1 A. By (4.10) we can obtain with e rr2 = −∇ an err2 P 0 20 + R0 . / and E = an · e Applying Theorem 5.1 to P = ∇a rr2 , we have ν L ∞ L 2 N1 (ν) + N1 (P) + an err2 P 0 20 + R0 , ω
t
where we used N1 (ν) + N1 (P) 20 + R0 which follows from (2.67). d a and a( p) = 1, In view of ν := − ds t |a − 1| ≤
|ν|nadt ν L ∞ L 2 20 + R0 . ω
t
0
With 20 + R0 being sufficiently small, |a − 1| ≤ 1/4 can be achieved.
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6.2 Estimates for ζ , χˆ , ζ etc We will use the following two decomposition results whose proof will be given in [13]. Proposition 6.2 There holds the decomposition an(ρ, ˇ σˇ ) = Dt p + e with N1 ( p ) + e P 0 20 + R0 ,
(6.3)
where p = D1−1 β and e denotes a pair of functions on H. / Proposition 6.3 (1) Let F = ζ , there hold the decompositions ∇(na F) = Dt P + E, with P and E appropriate S tangent tensor fields verifying N1 (P) + E P 0 20 + R0 .
(2) Denote by F either χˆ or ζ and set M =
/ ∇trχ, if F = χˆ There holds the (μ, 0), if F = ζ.
decomposition / / D−1 (na M) ∇(na F) = Dt P + E + ∇
(6.4)
with P and E appropriate tensor fields verifying N1 (P) + E P 0 20 + R0 ,
lim r P L ∞ =0 x
t→0
(6.5)
In view of Theorem 5.1 and (6.1), it follows immediately from Proposition 6.3 (1) that ζ L ∞ L 2 20 + R0 . ω
t
/ For F = ζ, χˆ , (6.4) differs from the desired decomposition for ∇(an F) by the term / D−1 (an M). Now we prove ∇ / D−1 (an M) P 0 20 + R0 . As a consequence, this term ∇ can be incorporated into the error term E in the decomposition. Proposition 6.4 There holds the following estimate 1
2 2 / / χ ˆ L ∞ L 2 + ∇trχ P 0 + r ∇trχ B0 0 + R0 . ω
t
Proof By combining (6.4) with (2.4) we obtain d 3 / D−1 (an M)) − 21 (trχ)2 (ζ + ζ ), ˆ Dt P + E + ∇ M + trχ M = −χˆ · M − 2(an)−1 χ( ds 2 regarding ι as an element of A, symbolically, 3 / L M + trχ M = A · M + (an)−1 χˆ · (Dt P + E + ∇ / D−1 (an M)) ∇ 2 + (r −1 A + A · A) · A + r −2 A. Then 3 2 + r 3 an A(A · A + r −1 A) + ran A.
/ D−1 (an M))} Dt (r 3 M) = − r 3 anκ M + r 3 an{A · M + (an)−1 χ( ˆ Dt P + E + ∇
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/ Let us pair ∇trχ 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, and using lims→0 r ∇trχ = 0, we obtain M =r
−3
t 3 3 r an A · M + r χˆ · Dt P dt 0
+r
−3
t 3 / D−1 (an M) + an A · A + r −1 an A) + anr A dt . r A · (E + ∇ 0
ˆ L∞ < ∞, we have Using Lemma 4.1, (5.2), (5.4) and limt→0 χ x t −3 3 −1 −1 3 r / r A · an(M + A · A + r A) + E + ∇ D (an M) + r χ ˆ · D P t 0
/ D−1 (an M) P 0 (N1 (A) + A L ∞ L 2 )(N1 (P) + E P 0 + ∇ ω t
B0
+ an M P 0 + r −1 an A P 0 + an A · A P 0 ).
By Proposition 4.2, (2.28), (4.10) and (2.67), we obtain t −3 r anr A r −1 an A P 0 N1 (A) 20 + R0 . P0
0
Hence, in view of (4.4), (BA1), (4.10) and (6.2), we can obtain M P 0 N1 (P) + M P 0 + E P 0 + 20 + R0 N1 (A) + A L ∞ L 2 + 20 + R0 ω t 0 M P 0 + N1 (P) + E P 0 + 20 + R0 + 20 + R0 . (6.6) Since 0 < 0 < 1/2 is sufficiently small, the first term of (6.6) can be absorbed. Thus, in view of (6.5) we can obtain 2 / ∇trχ P 0 0 + R0 .
(6.7)
/ D−1 (an M) we obtain from (6.4), (6.2) and (6.7) the Therefore, by setting E˜ = E + ∇ decomposition ˜ P 0 20 + R0 . / ∇(an χˆ ) = Dt P + E˜ and N1 (P) + E
(6.8)
By Theorem 5.1 and (2.67), we conclude ˜ P 0 20 + R0 . χˆ L ∞ L 2 N1 (χ) ˆ + N1 (P) + E ω
t
as expected. 1 2 / Similar to the derivation of (6.7), we can get r 2 ∇trχ B0 0 + R0 .
Proposition 6.5 There holds the following estimate 1
ζ L ∞ L 2 + μ P 0 + r 2 μ B0 20 + R0 . ω
t
123
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Q. Wang
Proof Let M = (μ, 0), (2.15) can be written as d / D−1 (an M) − 2χˆ · ζ · ζ + (ζ − 2ζ )trχζ M + trχ M = 2(an)−1 χˆ Dt P + E + ∇ ds 1 2 1 / +∇trχ(ζ − ζ ) + trχ ρˇ + a trχ + A |χ| ˆ 2 − aν(trχ)2 . 4 2 Symbolically, d / D−1 (an M) + E + trχ ρˇ M + trχ M = (an)−1 χˆ · Dt P + ∇ ds / + r −1 atrχ A. +A · A · A + r −1 A + ∇trχ
(6.9)
In view of limt→0 r 2 μ = 0 in Lemma 2.1, we deduce M =r
−2
t
r
2
−1 anκ · M + χˆ · Dt P + A · F˜ + antrχ ρˇ + r a 2 ntrχ A dt (6.10)
0
/ / D−1 (an M). + r −1 A) + E + ∇ with F˜ = an(A · A + ∇trχ In view of Proposition 5.1, regarding κ as an element of A, t −2 2 anκ · M + χˆ · Dt P + A · F˜ dt r r 0
N1 (A) + A L ∞ L 2 ω
t
B0
˜ P0 . an M P 0 + N1 (P) + F
(6.11)
Note that by (4.10), (4.4), (6.2) and (6.1), we deduce ˜ P 0 an(A · A + r −1 A) P 0 + an ∇trχ / / D−1 (an M) P 0 F P 0 + E P 0 + ∇ 2 / ∇trχ P 0 + E P 0 + M P 0 + 0 + R0 .
˜ P 0 2 + R0 + M P 0 . Hence, by (6.1), (BA1), (4.4) and By (6.7) and (6.4), we have F 0 (6.4) (6.11) 0 ( M P 0 + 20 + R0 ). Assuming the following estimate for P 0 norm of the last two terms in (6.10) t −2 2 −1 r antrχ ρˇ + r a 2 ntrχ A dt r
P0
20 + R0 ,
(6.12)
0
since 0 < 0 < 1/2 can be chosen sufficiently small, we conclude that M P 0 20 + R0 .
(6.13)
/ D−1 (an M) P 0 20 + R0 . By (6.2) and (2.67), ∇ / / D−1 (an M). By In view of (6.4), we have ∇(anζ ) = Dt P + E , with E = E + ∇ Theorem 5.1, ζ L ∞ L 2 N1 (P) + E P 0 + N1 (ζ ) 20 + R0 . ω
123
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Null hypersurfaces with time foliation in Einstein vacuum space-time
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We need to show (6.12). By using (6.3) it suffices to establish the following two inequalities t −2 2 r r trχ D p dt t
20 + R0 ,
P t −2 2 −1 r r trχ e + r an A dt
0
(6.14)
0
20 + R0 .
(6.15)
P0
0
To prove (6.15), by Proposition 4.2, (2.28), also in view of (4.10) and (6.3), we can obtain t −2 −1 r r e + r an A dt
e P 0 + r −1 an A P 0 20 + R0 .
P0
0
By (5.2) and (6.3), we get t −2 2 r r ι · e dt 0
(N1 (ι) + ι L ∞ L 2 ) e P 0 20 + R0 , ω t B0
and similarly t −2 r r ι · an Adt 0
(N1 (ι) + ι L ∞ L 2 ) r −1 an A P 0 20 + R0 . ω t B0
The proof of (6.15) is complete. Now we prove (6.14). In view of Proposition 6.2, p = D1−1 β, we have lim r p = 0, trχ p P 0 N1 ( p ) R0 + 20 .
(6.16)
s→0
Using Proposition 4.2 and (6.16), also in view of (2.1), we derive t 1 2 r trχ D p dt t r2 0 P0 t 1 2 r trχ Dt p dt Ek r k>0
⎛
0
⎝ E k (r trχ p )
k>0
L 2t L 2ω
L 2t L 2ω
t 1 2 + r trχ Dt p dt r
+ E k r −1
0
t
L 2t L 2ω
⎞
/ L p L 2 L 2 Dt (r trχ ) p dt L 2t L 2 ⎠ + ∇ t 2
ω
x
0
/ L p L 2 L 2 trχ p P 0 + trχ p L 2 (H) + ∇ t x ⎛ t t ⎜ 2 2 −1 −1 2 2 p + r r antrχ trχ p dt + r r an (trχ ) +| χ| ˆ dt E E ⎝ k k 2 k>0 0
L t L 2ω
0
⎞ ⎟ ⎠. L 2t L 2ω
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Using (5.2), (4.10), (BA1) and (2.67), t 2 2 E k r −1 r an|χ| ˆ · p dt k>0 0
N1 (χˆ ) + χ ˆ L ∞ L 2 an χˆ · p P 0 ω
2 L∞ t Lω
(20 + R0 )N1 ( p ) 20 + R0 ,
where for the last inequality, we employed (6.3). By using (4.9) it is easy to see t 2 E k r −1 r antrχ trχ p dt 2 k>0 0
and k>0
t
trχ p P 0 N1 ( p ).
L t L 2ω
t E k r −1 r 2 an(trχ)2 p dt 2 0 L t L 2ω ⎛ t ⎜ −1 2 E ≤ r r anκtrχ p dt ⎝ k k>0
t −1 2 E + r r antrχ trχ p dt k
anκ
+ R0 + 1)N1 ( p )
L 2t L 2ω · r trχ p P 0 + trχ p P 0 (20 0
0
⎞ ⎟ ⎠ L 2t L 2ω
(6.17)
where we employed (4.9) and anκ · trχ p P 0 N1 ( p )(20 + R0 ).
(6.18)
To see (6.18), we first deduce with the help of (4.4) that anκ · r trχ p P 0 κ · r trχ p P 0 .
(6.19)
By (4.6) with G = r trχκ, we have for b > 4 that 1 1 / trχκ) L 2 (H) + r 1− b trχκ L b L 2 . κ · r trχ p P 0 N1 ( p ) r 2 ∇(r t
x
(6.20)
Since r κ L ∞ C, we have 1
3
3
2 / trχκ) L 2 (H) r 2 ∇trχκ / / L 2 (H) r 2 ∇(r L 2 (H) + r trχ ∇κ
/ / L 2 (H) ∇trχ L 2 (H) + ∇κ 20 + R0
where for the last inequality, we used (2.67) and Proposition 2.7. Moreover, by using (2.32) we have 1
1
r 1− b trχκ L b L 2 r − b κ L b L 2 20 + R0 . t
x
t
x
Thus, in view of (6.20) and (6.19), (6.18) is proved. This together with (6.16) gives (6.14). 1 Similar to the derivation of (6.13), we can obtain r 2 μ B0 20 + R0 . The proof is therefore complete.
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