ON THE KLAINERMAN-MACHEDON CONJECTURE OF THE QUANTUM BBGKY HIERARCHY WITH SELF-INTERACTION XUWEN CHEN AND JUSTIN HOLMER Abstract. We consider the 3D quantum BBGKY hierarchy which corresponds to the N -particle Schr¨ odinger equation. We assume the pair interaction is N 3β−1 V (N β •). For interaction parameter β ∈ (0, 32 ), we prove that, as N → ∞, the limit points of the solutions to the BBGKY hierarchy satisfy the space-time bound conjectured by Klainerman-Machedon [37] in 2008. This allows for the application of the Klainerman-Machedon uniqueness theorem, and hence implies that the limit is uniquely determined as a tensor product of solutions to the Gross-Pitaevski equation when the N -body initial data is factorized. The first result in this direction in 3D was obtained by T. Chen and N. Pavlovi´c [11] for β ∈ (0, 14 ) and subsequently by X. Chen [15] for β ∈ (0, 27 ]. We build upon the approach of X. Chen but apply frequency localized Klainerman-Machedon collapsing estimates and the endpoint Strichartz estimate to extend the range to β ∈ (0, 23 ). Overall, this provides an alternative approach to the mean-field program by Erd¨ os-Schlein-Yau [23], whose uniqueness proof is based upon Feynman diagram combinatorics.
Contents 1. Introduction 1.1. Organization of the paper 1.2. Acknowledgements 2. Proof of the Main Theorem 3. Estimate of the Potential Part 3.1. The simpler β ∈ (0, 25 ) case 3.2. Proof of Theorem 3.1 4. The Xb Norms and a Few Strichartz Estimates 4.1. Various Forms of Collapsing Estimates 4.2. A Strichartz Estimate for P P (k) 5. Conclusion Appendix A. The Topology on the Density Matrices Appendix B. Proof of Estimates (2.7) and (2.9) References
2 6 7 7 10 11 13 18 19 25 28 28 29 31
2010 Mathematics Subject Classification. Primary 35Q55, 35A02, 81V70; Secondary 35A23, 35B45. Key words and phrases. BBGKY Hierarchy, N -particle Schr¨odinger Equation, Klainerman-Machedon Space-time Bound, Quantum Kac’s Program. 1
2
XUWEN CHEN AND JUSTIN HOLMER
1. Introduction The quantum BBGKY hierarchy refers to a sequence of trace class operator kernels n oN (k) , where t ∈ R, xk = (x1 , x2 , · · · , xk ) ∈ R3k , x0k = (x01 , x02 , · · · , x0k ) ∈ R3k , γ N (t, xk ; x0k ) k=1 which are symmetric, in the sense that (k)
(k)
γ N (t, xk , x0k ) = γ N (t, x0k , xk ), and (k)
(k)
γ N (t, xσ(1) , · · · xσ(k) , x0σ(1) , · · · x0σ(k) ) = γ N (t, x1 , · · · xk , x01 , · · · x0k ),
(1.1)
for any permutation σ, and satisfy the quantum BBGKY linear hierarchy of equations which written in operator form is i i h 1 X h (k) (k) (k) (1.2) i∂t γ N + 4xk , γ N = VN (xi − xj ) , γ N N 16i<j6k k i h N −k X (k+1) + Trk+1 VN (xj − xk+1 ) , γ N N j=1
with prescribed initial conditions (k)
(k)
γ N (0, xk , x0k ) = γ N,0 (xk , x0k ). Here 4xk denotes the standard Laplacian with respect to the variables xk ∈ R3k , the operator VN (x) represents multiplication by the function VN (x), where VN (x) = N 3β V (N β x)
(1.3)
is an approximation to the Dirac δ function, and Trk+1 means taking the k + 1 trace, for example, Z (k+1) (k+1) Trk+1 VN (xj − xk+1 ) γ N = VN (xj − xk+1 ) γ N (t, xk , xk+1 ; x0k , xk+1 )dxk+1 . We devote this paper to proving the following theorem. Theorem 1.1 (Main theorem). Assume the interaction parameter β ∈ (0, 2/3) and the pair n oN 6 (k) is a solution to the interaction V ∈ L1 ∩W 2, 5 + . Suppose that the sequence γ N (t, xk ; x0k ) k=1
quantum BBGKY hierarchy (1.2) subject to the energy condition: there is a C (independent of N and k) such that for any k > 0, there is a N0 (k) such that
(k) (k) (1.4) ∀ N > N0 (k) , sup S γ N 2 6 C k Lx,x0
t∈R
where S
(k)
k E Y
D = ∇xj ∇x0j . Then, for every finite time T , every limit point Γ = j=1
γ
(k) ∞ k=1
of
{ΓN }∞ N =1
n oN ∞ (k) = γN k=1
N =1
in
L
k>1
C ([0, T ] , L1k ) with respect to the product
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
3
topology τ prod (defined in Appendix A) satisfies the Klainerman-Machedon space-time bound: there is a C independent of j, k such that Z T
(k)
R Bj,k+1 γ (k+1) (t) 2 dt 6 C k , (1.5) L x,x0
0
where
L1k
2
3k
is the space of trace class operators on L (R ), R
(k)
k Y ∇ x =
∇x0 , and j j
j=1
Bj,k+1 = Trk+1 δ (xj − xk+1 ) , γ
(k+1)
.
In particular, this theorem establishes a positive answer to Conjecture 1 by Klainerman and Machedon in 2008 for β ∈ (0, 2/3). Conjecture 1 (Klainerman-Machedon [37]). Under condition (1.4), for β ∈ (0, 1], every ∞ (k) ∞ limit point Γ = γ of {ΓN }N =1 satisfies space-time bound (1.5). k=1 The quantum BBGKY hierarchy (1.2) is generated from the N -body Hamiltonian evolution ψ N (t) = eitHN ψ N (0) with the N -body Hamiltonian 1 X (1.6) HN = −4xN + N 3β V (N β (xi − xj )) N 16i<j6N where the factor 1/N is to make sure that the interactions are proportional to the number of particles, and the pair interaction N 3β V (N β (xi − xj )) is an approximation to the Dirac δ function which matches the Gross-Pitaevskii description of Bose-Einstein condensation that the many-body effect should be modeled by a strong on-site self-interaction. Since ψ N ψ N is n oN (k) 0 a probability density, we define the marginal densities γ N (t, xk ; xk ) by k=1 Z (k) 0 γ N (t, xk ; xk ) = ψ N (t, xk , xN −k )ψ N (t, x0k , xN −k )dxN −k , xk , x0k ∈ R3k . n oN (k) Then we have that γ N (t, xk ; x0k )
satisfies the the quantum BBGKY hierarchy (1.2) if
k=1
(k)
we do not distinguish γ N as a kernel and the operator it defines.1 Establishing the N → ∞ limit of hierarchy (1.2) justifies the mean-field limit in the GrossPitaevskii theory. Such an approach was first proposed by Spohn [43] and can be regarded as a quantum version of Kac’s program. We see that, as N → ∞, hierarchy (1.2) formally converges to the infinite Gross-Pitaevskii hierarchy Z X k (k) (k) (1.7) i∂t γ + 4xk , γ = V (x)dx Trk+1 δ (xj − xk+1 ) , γ (k+1) . j=1
When the initial data is factorized γ
(k)
(0, xk ; x0k )
=
k Y
¯ 0 (xj ), φ0 (xj )φ
j=1 1From
here on out, we consider only the β > 0 case. For β = 0, see [21, 38, 40, 42, 30, 31, 13, 6] .
4
XUWEN CHEN AND JUSTIN HOLMER
hierarchy (1.7) has a special solution (1.8)
γ
(k)
(t, xk ; x0k )
=
k Y
¯ xj ), φ(t, xj )φ(t,
j=1
if φ solves the cubic NLS Z i∂t φ = −4x φ +
(1.9)
V (x)dx |φ|2 φ.
Thus such a limit process shows that, in an appropriate sense, (k)
lim γ N =
N →∞
k Y
¯ xj ), φ(t, xj )φ(t,
j=1
hence justifies the mean-field limit. Such a limit in 3D was first proved in a series of important papers [20, 22, 23, 24, 25] by Elgart, Erd¨os, Schlein, and Yau.2 Briefly, the Elgart-Erd¨os-Schlein-Yau approach3 can be described as the following: Step A. Prove that, with respect to the topology τ prod defined in Appendix A, the sequence L ∞ 3k 1 ). {ΓN }N =1 is compact in the space k>1 C [0, T ] , L R ∞ Step B. Prove that every limit point Γ = γ (k) k=1 of {ΓN }∞ N =1 must verify hierarchy (1.7). Step C. Prove that, in the space in which the limit points from Step B lie, there is a unique solution to hierarchy (1.7). Thus {ΓN }∞ N =1 is a compact sequence with only one limit point. Hence ΓN → Γ as N → ∞. In 2007, Erd¨os, Schlein, and Yau obtained the first uniqueness theorem of solutions [23, Theorem 9.1] to the hierarchy (1.7). The proof is surprisingly delicate – it spans 63 pages and uses complicated Feynman diagram techniques. The main difficulty is that hierarchy (1.7) is a system of infinitely coupled equations. Briefly, [23, Theorem 9.1] is the following: Theorem 1.2 (Erd¨os-Schlein-Yau uniqueness [23, Theorem 9.1]). There is at most one (k) ∞ nonnegative symmetric operator sequence γ that solves hierarchy (1.7) subject to the k=1 energy condition ! k Y 1 − 4 xj γ (k) 6 C k . (1.10) sup Tr t∈[0,T ]
j=1
In [37], based on their null form paper [36], Klainerman and Machedon gave a different proof of the uniqueness of hierarchy (1.7) in a space different from that used in [23, Theorem 9.1]. The proof is shorter (13 pages) than the proof of [23, Theorem 9.1]. Briefly, [37, Theorem 1.1] is the following: 2Around 3See
the same time, there was the 1D work [1]. [5, 29, 41] for different approaches.
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
5
Theorem 1.3 (Klainerman-Machedon (k) ∞ uniqueness [37, Theorem 1.1]). There is at most one symmetric operator sequence γ that solves hierarchy (1.7) subject to the space-time k=1 bound (1.5). For special cases like (1.8), condition (1.10) is actually sup kh∇x i φkL2 6 C,
(1.11)
t∈[0,T ]
while condition (1.5) means Z (1.12) 0
T
|∇x | |φ|2 φ 2 dt 6 C. L
When φ satisfies NLS (1.9), both are known. In fact, due to the Strichartz estimate [33], (1.11) implies (1.12), that is, condition (1.5) seems to be a bit weaker than condition (1.10). The proof of [37, Theorem 1.1] (13 pages) is also considerably shorter than the proof of [23, Theorem 9.1] (63 pages). It is then natural to wonder whether [37, Theorem 1.1] simplifies Step C. To answer such a question it is necessary to know whether the limit points in Step B satisfy condition (1.10), that is, whether Conjecture 1 holds. Away from curiosity, there are realistic reasons to study Conjecture 1. While [23, Theorem 9.1] is a powerful theorem, it is very difficult to adapt such an argument to various other interesting and colorful settings: a different spatial dimension, a three-body interaction instead of a pair interaction, or the Hermite operator instead of the Laplacian. The last situation mentioned is physically important. On the one hand, all the known experiments of BEC use harmonic trapping to stabilize the condensate [2, 19, 7, 34, 44]. On the other hand, different trapping strength produces quantum behaviors which do not exist in the Boltzmann limit of classical particles nor in the quantum case when the trapping is missing and have been experimentally observed [26, 45, 18, 32, 17]. The Klainerman-Machedon approach applies easily in these meaningful situations ([35, 9, 14, 15, 16, 27]). Thus proving Conjecture 1 actually helps to advance the study of quantum many-body dynamic and the mean-field approximation in the sense that it provides a flexible and powerful tool in 3D. The well-posedness theory of the Gross-Pitaevskii hierarchy (1.7) subject to general initial datum also requires that the limits of the BBGKY hierarchy (1.2) lie in the space in which the space-time bound (1.5) holds. See [8, 10, 11]. As pointed out in [20], the study of the Hamiltonian (1.6) is of particular interest when β ∈ (1/3, 1]. The reason is the following. In physics, the initial datum ψ N (0) of the Hamiltonian evolution eitHN ψ N (0) is usually assumed to be close to the ground state of the Hamiltonian 1 X HN,0 = −4xN + ω 20 |xN |2 + N 3β V (N β (xi − xj )). N 16i<j6N The preparation of the available experiments and the mathematical work [39] by Lieb, Seiringer, Solovej and Yngvason confirm this assumption. Such an initial datum ψ N (0) is localized in space. We can assume all N particles are in a box of length 1. Let the effective radius of the pair interaction V be a, then the effective radius of VN is about
6
XUWEN CHEN AND JUSTIN HOLMER
3 a/N β . Thus every particle in the box interacts with a/N β × N other particles. Thus, for β > 1/3 and large N , every particle interacts with only itself. This exactly matches the Gross-Pitaevskii theory that the many-body effect should be modeled by a strong on-site self-interaction. Therefore, for the mathematical justification of the Gross-Pitaevskii theory, it is of particular interest to prove Conjecture 1 for self-interaction (β > 1/3) as well. To the best of our knowledge, the main theorem (Theorem 1.1) in the current paper is the first result in proving Conjecture 1 for self-interaction (β > 1/3). For β 6 1/3, the first progress of Conjecture 1 is the β ∈ (0, 1/4) work [11] by T. Chen and N. Pavlovi´c and then the β ∈ (0, 2/7] work [15] by X.C. As a matter of fact, the main theorem (Theorem 1.1) in the current paper has already fulfilled the original intent of [37], namely, simplifying the uniqueness argument of [23], because [23] deals with β ∈ (0, 3/5). Conjecture 1 for β ∈ [2/3, 1] is still open. 1.1. Organization of the paper. In §2, we outline the proof of Theorem 1.1. The overall pattern follows that introduced by X.C.[15], who obtained Theorem 1.1 for β ∈ (0, 72 ]. Let (k) P≤M be the Littlewood-Paley projection defined in (2.1). Theorem 1.1 will follow once it is established that for all M ≥ 1, there exists N0 depending on M such that for all N ≥ N0 , there holds (1.13)
(k)
(k+1)
kP≤M R(k) BN,j,k+1 γ N
(t)kL1T L2
x,x0
6 Ck
where BN,j,k+1 is defined by (2.3). Substituting the Duhamel-Born expansion, carried out to coupling level K, of the BBGKY hierarchy, this is reduced to proving analogous bounds on the free part, potential part, and interaction part, defined in §2. Each part is reduced via the Klainerman-Machedon board game. Estimates for the free part and interaction part were previously obtained by X.C. [15]. For the estimate of the interaction part, one takes K = ln N , the utility of which was first observed by T. Chen and N. Pavlovi´c [11]. The main new achievement of our paper is the improved estimates on the potential part, which are discussed in §3. We make use of the endpoint Strichartz estimate in place of the Sobolev inequality employed by X.C [15]. The Strichartz estimate is phrased in terms of Xb norms. We also introduce frequency localized versions of the Klainerman-Machedon collapsing estimates, allowing us to exploit the frequency localization in (1.13). Specifically, the op(k) (k) (k+1) erator P≤M does not commute with BN,j,k+1 , however, the composition P≤Mk BN,j,k+1 P∼Mk+1 enjoys better bounds if Mk+1 Mk . We prove the Strichartz estimate and the frequency localized Klainerman-Machedon collapsing estimates in §4. Frequency localized space-time techniques of this type were introduced by Bourgain [4, Chapter IV, §3] into the study of the well-posedness for nonlinear Schr¨odinger equations and other nonlinear dispersive PDE. (k) In X.C. [15], (1.13) is obtained without the frequency localization P≤M for β ∈ (0, 72 ]. In Theorem 3.2, we prove that this estimate still holds without frequency localization for β ∈ (0, 25 ) by using the Strichartz estimate alone. This already surpasses the self-interaction threshold β = 13 . For the purpose of proving Conjecture 1, the frequency localized estimate (1.13) is equally good, but allows us to achieve higher β.
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
7
1.2. Acknowledgements. J.H. was supported in part by NSF grant DMS-0901582 and a Sloan Research Fellowship (BR-4919). 2. Proof of the Main Theorem We establish Theorem 1.1 in this section. For simplicity of notation, we denote k·kLp [0,T ]L2
x,x0
by k·kLp L2 T
x,x0
and denote k·kLpt (R)L2
x,x0
by k·kLpt L2 0 . Let us begin by introducing some notax,x
i i tion for Littlewood-Paley theory. Let P≤M be the projection onto frequencies ≤ M and PM the analogous projections onto frequencies ∼ M , acting on functions of xi ∈ R3 (the ith i0 coordinate). We take M to be a dyadic frequency range 2` ≥ 1. Similarly, we define P≤M i0 and PM , which act on the variable x0i . Let (k) P≤M
(2.1)
=
k Y
0
i i P≤M P≤M .
i=1
To establish Theorem 1.1, it suffices to prove the following theorem. Theorem 2.1. Under the assumptions of Theorem 1.1, there exists a C (independent of k, M, N ) such that for each M ≥ 1 there exists N0 (depending on M ) such that for N > N0 , there holds (2.2)
(k)
(k+1)
kP≤M R(k) BN,j,k+1 γ N
(t)kL1T L2
x,x0
6 Ck
where (2.3)
(k+1)
BN,j,k+1 γ N
i h (k+1) . = Trk+1 VN (xj − xk+1 ) , γ N
We first explain how, assuming Theorem 2.1, we can prove Theorem 1.1. Passing to the (k) weak* limit γ N → γ (k) as N → ∞, we obtain (k)
kP≤M R(k) Bj,k+1 γ (k+1) kL1T L2
x,x0
6 Ck
Since this holds uniformly in M , we can send M → ∞ and, by the monotone convergence theorem, we obtain kR(k) Bj,k+1 γ (k+1) kL1T L2
x,x0
6 Ck
which is exactly the Klainerman-Machedon space-time bound (1.5). This completes the proof Theorem 1.1, assuming Theorem 2.1. The rest of this paper is devoted to proving Theorem 2.1. Without loss of generality, we prove estimate (2.2) for k = 1, that is (2.4)
(1)
(2)
kP≤M R(1) BN,1,2 γ N kL1T L2
x,x0
6C
for N > N0 (M ). We are going to establish estimate (2.4) for a sufficiently small T which depends on the controlling constant in condition (1.4) and is independent of N and M, then a bootstrap argument together with condition (1.4) give estimate (2.4) for every finite time at the price of a larger constant C.
8
XUWEN CHEN AND JUSTIN HOLMER
We start by rewriting hierarchy (1.2) as (2.5)
(k) γ N (tk )
= U
(k)
(k) (tk )γ N,0
Z
tk
+
(k) (k)
U (k) (tk − tk+1 )VN γ N (tk+1 )dtk+1
0
+
N −k N
tk
Z
(k+1) (k+1) γ N (tk+1 )dtk+1
U (k) (tk − tk+1 )BN
0
with the short-hand notation: −it4
0
x k, U (k) = eit4xk e h i X 1 (k) (k) (k) VN γ N = VN (xi − xj ), γ N N 16i<j6k
(k+1) (k+1) γN
BN
=
k X
(k+1)
BN,j,k+1 γ N
.
j=1
We omit the i in front of the potential term and the interaction term so that we do not need to keep track of its exact power. (2) Writing out the kth Duhamel-Born series of γ N by iterating hierarchy (2.5) k times, we have (2) γ N (t2 )
= U
(2)
(2) (t2 )γ N,0
Z +
t2
(2) (2)
U (2) (t2 − t3 )VN γ N (t3 )dt3
0
N −2 N
Z
t2
(3) (3)
U (2) (t2 − t3 )BN γ N (t3 )dt3 0 Z N − 2 t2 (2) (2) (3) (3) (2) = U (t2 )γ N,0 + U (t2 − t3 )BN U (3) (t3 )γ N,0 dt3 N 0 Z t2 (2) (2) + U (2) (t2 − t3 )VN γ N (t3 )dt3 0 Z Z t3 N − 2 t2 (2) (3) (3) (3) + U (t2 − t3 )BN U (3) (t3 − t4 )VN γ N (t4 )dt4 dt3 N 0 0 Z Z t3 N − 2 N − 3 t2 (2) (3) (4) (4) + U (t2 − t3 )BN U (3) (t3 − t4 )BN γ N (t4 )dt4 dt3 N N 0 0 = ... +
After k iterations4 (2.6)
4Henceforth,
(2)
γ N (t2 ) = F P (k) (t2 ) + P P (k) (t2 ) + IP (k) (t2 )
the k’s appearing in our formulas are the coupling level which is distinct from the k in the statement of Theorem 2.1 (which has been fixed at k = 1).
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
9
where the free part at coupling level k is given by (2)
F P (k) = U (2) (t2 )γ N,0 + !Z Z tj−1 j k t2 Y X N +1−l (3) (j) ··· U (2) (t2 − t3 )BN · · · U (j−1) (tj−1 − tj )BN N 0 0 j=3 l=3 (j) × U (j) (tj )γ N,0 dt3 · · · dtj , the potential part is given by ! Z t2 j k X Y N + 1 − l (2) (2) U (2) (t2 − t3 )VN γ N (t3 )dt3 + P P (k) = N 0 j=3 l=3 Z tj−1 Z t2 (3) (j) ··· U (2) (t2 − t3 )BN · · · U (j−1) (tj−1 − tj )BN × 0 0 Z tj (j) (j) (j) U (tj − tj+1 )VN γ N (tj+1 )dtj+1 dt3 · · · dtj , × 0
and the interaction part is given by IP
(k)
=
k Y N +1−l N l=3
···U
(k)
!Z
(tk −
t2
Z
tk
(3)
U (2) (t2 − t3 )BN · · ·
··· 0
0
(k+1) tk+1 )BN
(k+1) γ N (tk+1 )
By (2.6), to establish (2.4), it suffices to prove
(1) (1) (k) (2.7)
P≤M R BN,1,2 F P
6C
(2.8)
(1) (1)
P≤M R BN,1,2 P P (k)
6C
(2.9)
(1) (1) (k) P R B IP
≤M
N,1,2
6C
L1T L2x,x0
L1T L2x,x0
L1T L2x,x0
dt3 · · · dtk+1 .
for some C and a sufficiently small T determined by the controlling constant in condition (1.4) (j) and independent of N and M. We observe that BN has 2j terms inside so that each summand (2) of γ N (t2 ) contains factorially many terms (∼ k!). We use the Klainerman-Machedon board game to combine them and hence reduce the number of terms that need to be treated. Define (3)
(j+1) (j+1)
JN (tj+1 )(f (j+1) ) = U (2) (t2 − t3 )BN · · · U (j) (tj − tj+1 )BN
f
,
where tj+1 means (t3 , . . . , tj+1 ) , then the Klainerman-Machedon board game implies the lemma. Lemma 2.1 (Klainerman-Machedon board game). One can express Z t2 Z tj ··· JN (tj+1 )(f (j+1) )dtj+1 0
0
10
XUWEN CHEN AND JUSTIN HOLMER
as a sum of at most 4j−1 terms of the form Z JN (tj+1 , µm )(f (j+1) )dtj+1 , D
or in other words, Z Z t2 ··· 0
tj
JN (tj+1 )(f
0
(j+1)
)dtj+1 =
XZ m
JN (tj+1 , µm )(f (j+1) )dtj+1 .
D
Here D ⊂ [0, t2 ]j−1 , µm are a set of maps from {3, . . . , j+1} to {2, . . . , j} satisfying µm (3) = 2 and µm (l) < l for all l, and JN (tj+1 , µm )(f (j+1) ) = U (2) (t2 − t3 )BN,2,3 U (3) (t3 − t4 )BN,µm (4),4 · · · · · · U (j) (tj − tj+1 )BN,µm (j+1),j+1 (f (j+1) ). Proof. Lemma 2.1 follows the exact same proof as [37, Theorem 3.4], the KlainermanMachedon board game, if one replaces Bj,k+1 by BN,j,k+1 and notices that BN,j,k+1 still commutes with eit4xi e−it4xi whenever i 6= j. This argument reduces the number of terms by combining them. In the rest of this paper, we establish estimate (2.8) only. The reason is the following. On the one hand, the proof of estimate (2.8) is exactly the place that relies on the restriction β ∈ (0, 2/3) in this paper. On the other hand, X.C. has already proven estimates (2.7) and (2.9) as estimates (6.3) and (6.5) in [15] without using any frequency localization. For completeness, we include a proof of estimates (2.7) and (2.9) in Appendix B. Before we delve into the proof of estimate (2.8), we remark that the proof of estimates (2.7) and (2.9) is independent of the coupling level k and we will take the coupling level k to be ln N for estimate (2.9).5 3. Estimate of the Potential Part In this section, we prove estimate (2.8). To be specific, we establish the following theorem. Theorem 3.1. Under the assumptions of Theorem 1.1, there exists a C (independent of k, M1 , N ) such that for each M1 > 1 there exists N0 (depending on M1 ) such that for N > N0 , there holds
(1) (1) (k) P R B P P
1 2 6C
≤M1 N,1,2 LT Lx,x0
where P P (k) is given by (2.7). In this section, we will employ the estimates stated and proved in Section 4. Due to the technicality of the proof of Theorem 3.1 involving Littlewood-Paley theory, we prove a simpler β ∈ (0, 52 ) version first to illustrate the basic steps in establishing Theorem 3.1. We then prove Theorem 3.1 in Section 3.2. 5The
technique of taking k = ln N for estimate (2.9) was first observed by T.Chen and N.Pavlovi´c [11].
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
11
3.1. The simpler β ∈ (0, 25 ) case. Theorem 3.2. For β ∈ (0, 52 ), we have the estimate
(1) (k)
R BN,1,2 P P 1 2
LT Lx,x0
6C
for some C and a sufficiently small T determined by the controlling constant in condition (1.4) and independent of N. Proof. The proof is divided into four steps. We will reproduce every step for Theorem 3.1 in Section 3.2. Step I. By Lemma 2.1, we know that Z t2 (2) (2) (k) U (2) (t2 − t3 )VN γ N (t3 )dt3 PP = 0 ! j k X Y N +1−l + (3.1) N j=3 l=3 Z XZ × JN (tj , µm ) m
D
tj
U (j) (tj −
(j) (j) tj+1 )VN γ N (tj+1 )dtj+1
0
P where m has at most 4j−3 terms inside. For the second term, we iterate Lemma 4.2 to prove the following estimate6:
Z
(1)
(j+1)
R BN,1,2
J (t (3.2) , µ )(f )dt N m j+1 j+1
L1T L2x,x0
D
T
Z = 0
Z
(1) (2)
R BN,1,2 U (t2 − t3 )BN,2,3 · · · dt3 . . . dtj+1
Z 6
[0,T ]j
6 T
1 2
dt2
L2x,x0
D
(1)
R BN,1,2 U (2) (t2 − t3 )BN,2,3 · · · 2 dt2 dt3 ...dtj+1 L Z
Z [0,T ]j−1
(1)
R BN,1,2 U (2) (t2 − t3 )BN,2,3 ... 2 2 dt2 L
21 dt3 ...dtj+1
(Cauchy-Schwarz) Z
(2)
1
R BN,2,3 U (3) (t3 − t4 )... dt3 ...dtj+1 (Lemma 4.2) 6 CT 2 [0,T ]j−1
(Iterate j − 2 times) ...
1 6 (CT 2 )j−1 R(j) BN,µm (j+1),j+1 f (j+1) L1 L2 T
6This
x,x0
also helps in proving estimates (2.7) and (2.9)–see Appendix B
!
dtj
12
XUWEN CHEN AND JUSTIN HOLMER
Applying relation (3.2), we have
(1)
R BN,1,2 P P (k)
L1T L2x,x0
t2
Z
(1)
6 R BN,1,2
U
(2)
(t2 −
0
+
k X
j−3
4
1 2
(CT ) t2
U
(2)
(t2 −
0
j=3
U
(j)
(j) (j) tj+1 )VN γ N (tj+1 )dtj+1
(tj −
L1T L2x,x0
0
Z
(1)
6 R BN,1,2 +
tj
Z
(j−1)
R BN,µm (j),j
j−2
j=3
k X
(2) (2) t3 )VN γ N (t3 )dt3
1 2 LT Lx,x0
(2) (2) t3 )VN γ N (t3 )dt3
1 2 LT Lx,x0
Z
j−2 (j−1) (CT ) R BN,µm (j),j 1 2
tj
U
(j)
(tj −
(j) (j) tj+1 )VN γ N (tj+1 )dtj+1
.
L1T L2x,x0
0
Inserting a smooth cut-off θ(t) with θ(t) = 1 for t ∈ [−T, T ] and θ(t) = 0 for t ∈ [−2T, 2T ]c into the above estimate, we get
(1)
R BN,1,2 P P (k)
L1T L2x,x0 t2
Z
(1)
6 R BN,1,2 θ(t2 )
(2)
U
(t2 −
0
(2) (2) t3 )θ(t3 )VN γ N (t3 )dt3
1 2 LT L x,x0
Z k X
(j−1) 1 j−2
BN,µm (j),j θ(tj ) + (CT 2 ) R
tj
U
(j)
(tj −
(j) (j) tj+1 )θ(tj+1 )VN γ N (tj+1 )dtj+1
L1T L2x,x0
0
j=3
Step II. The Xb space version of Lemma 4.2, Lemma 4.3, then turns the last step into
(1) (k) R B P P
N,1,2
L1T L2x,x0
Z 6 Ckθ(t2 ) 0
+C
k X
t2
(2) (2) U (2) (t2 − t3 )R(2) θ(t3 )VN γ N (t3 ) dt3 kX (2)
1 2+
1 2
j−2
(CT )
Z kθ(tj ) 0
j=3
tj
(j) (j) U (j) (tj − tj+1 )R(j) θ(tj+1 )VN γ N (tj+1 ) dtj+1 kX (j)
1 2+
Step III. We then proceed with Lemma 4.1 to get
(1) (k) R B P P
N,1,2
L1T L2x,x0
k X 1 (2) (2) (j) (j) 6 CkR(2) θ(t3 )VN γ N (t3 ) kX (2) + C (CT 2 )j−2 kR(j) θ(tj+1 )VN γ N (tj+1 ) kX (j) . −1 2+
j=3
1+ −2
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
13
Step IV. Now we would like to utilize Lemma 4.6. We first analyse a typical term to demonstrated the effect of Lemma 4.6. To be specific, we have (2) (2) kR(2) θ(t3 )VN γ N (t3 ) kX (2) 1+ −2
C (2) kVN (x1 − x2 )θ(t3 )R(2) γ N (t3 )kX (2) 1+ N −2 (2) C + k (VN )0 (x1 − x2 )θ(t3 ) |∇x2 | ∇x01 ∇x02 γ N (t3 )kX (2) 1+ N −2 (2) C + k (VN )00 (x1 − x2 )θ(t3 ) ∇x01 ∇x02 γ N (t3 )kX (2) N −1 2+ C (2) 6 kVN kL3+ kθ(t3 )R(2) γ N kL2t L2 0 3 x,x N (2) 1 C + kVN0 kL2+ kθ(t3 ) h∇x1 i 2 h∇x2 i ∇x01 ∇x02 γ N kL2t L2 0 3 x,x N C (2) + kVN00 kL 65 + kθ(t3 )S (2) γ N kL2t L2 0 3 x,x N (2) 6 CkS (2) γ N kL22T L2 0
6
x,x
since kVN /N kL3+ kVN0 /N kL2+ , and kVN00 /N kL 56 + are uniformly bounded in N for β ∈ (0, 25 ). In fact, kVN /N kL3+ 6 N 2β−1 kV kL3+ kVN0 /N kL2+ 6 N
5β −1 2
kV 0 kL2+
kVN00 /N kL 56 + 6 N
5β −1 2
kV 00 kL 65 +
6
6
where by Sobolev, V ∈ W 2, 5 + implies V ∈ L 5 + ∩ L6+ and V 0 ∈ L2+ . Using the same idea for all the terms, we end up with
(1) (k)
R BN,1,2 P P 1 2 LT Lx,x0
6 CT
1 2
(2) 2 kS (2) γ N kL∞ 2T Lx,x0
k X 1 (j) 2 + CT (CT 2 )j−2 j 2 kS (j) γ N kL∞ 2T L 1 2
x,x0
j=3 (j)
(j 2 terms inside VN ) ∞ X 1 1 1 2 6 CT 2 C + CT 2 (CT 2 )j−2 C j (Condition (1.4)) j=3
6 C < ∞. This concludes the proof of Theorem 3.2. 3.2. Proof of Theorem 3.1. To make formulas shorter, let us write (k)
(k)
R6Mk = P6Mk R(k) , (k)
since P6Mk and R(k) are usually bundled together.
14
XUWEN CHEN AND JUSTIN HOLMER
3.2.1. Step I. By (3.1),
(1)
R6M1 BN,1,2 P P (k)
L1T L2x,x0
Z
(1)
6 R6M1 BN,1,2
t2
U
(2)
(t2 −
0
(2) (2) t3 )VN γ N (t3 )dt3
1 2 LT Lx,x0
Z k X
X
(1)
(j) + JN (tj , µm )(f )dtj
R6M1 BN,1,2 j=3
where f
(j)
tj
Z
L1T L2x,x0
D
m
(j) (j)
U (j) (tj − tj+1 )VN γ N (tj+1 )dtj+1
= 0
where
P
m
has at most 4j−3 terms inside. By Minkowski’s integral inequality,
Z
(1)
(j)
R JN (tj , µm )(f )dtj
6M1 BN,1,2
L1T L2x,x0
D
T
Z
Z
R(1) BN,1,2 U (2) (t2 − t3 )BN,2,3 · · · dt3 . . . dtj 6M1
= 0
Z 6
[0,T ]j−1
dt2
L2x,x0
D
(1) (2)
R6M1 BN,1,2 U (t2 − t3 )BN,2,3 · · ·
dt2 dt3 . . . dtj
L2
By Cauchy-Schwarz in the t2 integration, Z 21 Z
2 1
(1) (2) 6T2
R6M1 BN,1,2 U (t2 − t3 )BN,2,3 · · · dt2 dt3 . . . dtj L2
[0,T ]j−1
By Lemma 4.4,
X M1 1−ε Z
(2)
(3) 6 Cε T
R6M2 BN,2,3 U (t3 − t4 ) · · · dt3 . . . dtj M2 [0,T ]j−1 M >M 1 2
2
1
Iterating the previous step (j − 3) times, 1−ε
X 1 M1 M2 Mj−2
(j−1) j−2 6 (Cε T 2 ) ···
R6Mj−1 BN,µm (j),j f (j) 1 2 M M M LT Lx,x0 2 3 j−1 Mj−1 >···>M2 >M1 1−ε
X 1 M1
(j−1) j−2 (j) 2 = (Cε T )
R6Mj−1 BN,µm (j),j f 1 2 Mj−1 LT Lx,x0 M >···>M >M 2
j−1
1
where the sum is over all M2 , . . . , Mj−1 dyadic such that Mj−1 > · · · > M2 > M1 . Hence
(1)
R6M1 BN,1,2 P P (k) 1 2 LT Lx,x0
Z
(1) 6
R6M1 BN,1,2
t2
U
(2)
(t2 −
+
j=3
1
(Cε T 2 )j−2
L1T L2x,x0
0
k X
(2) (2) t3 )VN γ N (t3 )dt3
X Mj−1 >···>M1
M11−ε
(j−1) (j) R B f
1 2 . 6Mj−1 N,µm (j),j 1−ε LT Lx,x0 Mj−1
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
15
We then insert a smooth cut-off θ(t) with θ(t) = 1 for t ∈ [−T, T ] and θ(t) = 0 for t ∈ [−2T, 2T ]c into the above estimate to get
(1) (k)
R6M1 BN,1,2 P P 1 2 LT Lx,x0
Z
(1) 6
R6M1 BN,1,2 θ(t2 )
t2
U
(2)
(2) (2) t3 )θ(t3 )VN γ N (t3 )dt3
(t2 −
L1T L2x,x0
0
+
k X
1
X
(Cε T 2 )j−2
j=3
Mj−1 >···>M1
M11−ε 1−ε Mj−1
(j−1) (j) ˜ R B θ(t ) f
6Mj−1 N,µm (j),j j
,
L1T L2x,x0
where the sum is over all M2 , . . . , Mj−1 dyadic such that Mj−1 > · · · > M2 > M1 , and Z tj (j) (j) (j) (j) ˜ f = U (tj − tj+1 ) θ(tj+1 )VN γ N (tj+1 ) dtj+1 0
3.2.2. Step II. Using Lemma 4.5, the Xb space version of Lemma 4.4, we turn Step I into
(1)
R6M1 BN,1,2 P P (k) 1 2 LT Lx,x0
X M 1−ε 1 6 1−ε M 2 M >M 2
1
Z
θ(t2 )
t2
U
(2)
(t2 − t3 )
0
dt3
(2)
X1
2+
k X 1 + (Cε T 2 )j−2 j=3
X Mj >Mj−1 >···>M1
M11−ε
(j) ˜(j)
θ(tj )R6Mj f (j) X1 Mj1−ε + 2
3.2.3. Step III. Lemma 4.1 gives us
(1) (k)
R6M1 BN,1,2 P P
L1T L2x,x0
where
(2) (2) (2) R6M2 θ(t3 )VN γ N (t3 )
≤A+B
X M 1−ε
(2)
(2) (2) 1 A= 1−ε R6M2 θ(t3 )VN γ N (t3 ) (2) X 1 M2 − + M >M 2
1
2
and k X 1 B= (Cε T 2 )j−2 j=3
X Mj >Mj−1 >···>M1
M11−ε
(j)
(j) (j) R θ(t )V γ (t )
(j) j+1 j+1 6Mj N N 1−ε X 1 Mj − + 2
3.2.4. Step IV. We focus for a moment on B. Applying Lemmas 3.1 and 3.2, we can carry M out the sum in M2 ≤ · · · ≤ Mj−1 at the expense of a factor M1j . Hence B.
k X
1 2
(Cε T )
j−2
j=3
X M1 1−2
(j)
(j) (j)
R6Mj θ(tj+1 )VN γ N (tj+1 ) (j) Mj X 1 − + M >M j
1
2
where the sum is over dyadic Mj such that Mj ≥ M1 . Applying (4.25), k X X M1 1−2 1 1 (j) j−2 2 B. (Cε T 2 ) j min(Mj2 , N 2β )N 2 β−1 kθ(tj+1 )S (j) γ N (tj+1 ) kL2t L2 0 j+1 x,x M j j=3 M >M j
1
16
XUWEN CHEN AND JUSTIN HOLMER
Rearranging terms B.
k X
1
(j)
(Cε T 2 )j−2 j 2 kθ(tj+1 )S (j) γ N (tj+1 ) kL2t
j+1
1 1−2 N 2 β−1 L2xx0 M1
j=3
X
(· · · )
Mj >M1
where X
X
(· · · ) =
Mj >M1
min(Mj1+2 , Mj−1+2 N 2β ) .
Mj >M1
We carry out the sum in Mj by dividing into Mj 6 N β (for which min(Mj1+2 , Mj−1+2 N 2β ) = Mj1+2 ) and Mj > N β (for which min(Mj1+2 , Mj−1+2 N 2β ) = Mj−1+2 N 2β ). This yields X min(Mj1+2 , Mj−1+2 N 2β ) Mj >M1
X
.
+
N β >Mj >M1
.
X
X
(...)
Mj >M1 ,Mj >N β
X
Mj1+2 +
N β >Mj >1
Mj−1+2 N 2β
Mj >N β
. N β+2 . Hence B .
k X
1
(j)
(Cε T 2 )j−2 j 2 kθ(tj+1 )S (j) γ N (tj+1 ) kL2t
j+1
3 1−2 N 2 β−1+2 L2x,x0 M1
j=3
.
3 M11−2 N 2 β−1+2
k X
1
1
(j)
2 (Cε T 2 )j−2 T 2 j 2 kS (j) γ N kL∞ t L
x,x0
j=3
.
3 M11−2 N 2 β−1+2
k X
1
1
(Cε T 2 )j−2 j 2 T 2 C j
by Condition (1.4)
j=3 3
. CM11−2 N 2 β−1+2 , for T small enough. Therefore, for β < 2/3, there is a C independent of M1 and N s.t. given a M1 , there is N0 (M1 ) which makes B 6 C, for all N > N0 . This completes the treatment of term B for β < 2/3. Term A is treated similarly (without the need to appeal to Lemmas 3.1, 3.2 below). Whence we have completed the proof of Theorem 3.1 and thence Theorem 2.1. Lemma 3.1.
X
M1 ≤M2 ≤···≤Mj−2 ≤Mj−1
1 ≤
(log2
Mj−1 M1
+ j − 3)j−3
(j − 3)!
,
where the sum is in M2 , . . . , Mj−2 over dyads, such that M1 ≤ M2 ≤ M3 ≤ · · · ≤ Mj−2 ≤ Mj−1 .
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
Proof. This is equivalent to
17
X i1 ≤i2 ≤···≤ij−2 ≤ij−1
(ij−1 − i1 + j − 3)j−3 , 1 ≤ (j − 3)!
where the sum is taken over integers i2 , . . . , ij−2 such that i1 ≤ i2 ≤ · · · ≤ ij−2 ≤ ij−1 . We use the estimate (for p ≥ 0, ` ≥ 0) q X
(i + `)p ≤
i=0
(q + ` + 1)p+1 , p+1
which just follows by estimating the sum by an integral. First, carry out the sum in i2 from i1 to i3 to obtain ! i3 X X X = 1 ≤ (i3 − i1 + 1). i1 ≤i3 ≤···≤ij−1
i1 ≤i3 ≤···≤ij−1
i2 =i1
Next, carry out the sum in i3 from i1 to i4 , X
i4 X
i1 ≤i4 ≤···≤ij−1
i3 =i1
X
iX 4 −i1
i1 ≤i4 ≤···≤ij−1
i3 =0
≤
≤
X
≤
i1 ≤i4 ≤···≤ij−1
! (i3 − i1 + 1) ! (i3 + 1)
(i4 − i1 + 2)2 . 2
Continue in this manner to obtain the claimed bound. Lemma 3.2. For each α > 0 (possibly large) and each > 0 (arbitrarily small), there exists t > 0 (independent of M ) sufficiently small such that ∀ j ≥ 1, ∀ M ,
we have
tj (α log M + j)j ≤ M j!
Proof. We use the following fact: for each σ > 0 (arbitrarily small) there exists t > 0 sufficiently small such that x 1 x + 1 ≤ eσ (3.3) ∀ x > 0, t x To apply this fact to prove the lemma, use Stirling’s formula to obtain j tj (α log M + j)j α log M + j j ≤ (et) j! j Define x in terms of j by the formula j = α(log M )x. Then x α log M 1 x = (et) +1 x
18
XUWEN CHEN AND JUSTIN HOLMER
Applying (3.3), ≤ eσα log M = M σα
4. The Xb Norms and a Few Strichartz Estimates Define the norm Z
(k)
2
kα kX (k) =
hτ + |ξ k | −
b
2 |ξ 0k | i2b
1/2 2 (k) 0 0 ˆ (τ , ξ k , ξ k ) dτ dξ k dξ k α
We will use the case b = 21 + of the following lemma. Lemma 4.1. Let
1 2
(4.1)
< b < 1 and θ(t) be a smooth cutoff. Then
Z t
(k) (k)
θ(t) U (t − s)β (s) ds . kβ (k) kX (k)
b−1
(k)
0
Xb
Proof. The estimate reduces to the space-independent estimate
Z t
0 0
for 12 < b ≤ 1 (4.2) θ(t) h(t ) dt
b . khkHtb−1 , 0
Ht
def
Indeed, taking h(t) = hxk x0k (t) = U (k) (−t)β (k) (t, xk , x0k ), applying the estimate (4.2) for fixed xk , x0k , and then applying the L2xk x0 norm to both sides, yields (4.1). Now we prove estimate k (4.2). Let P≤1 and P≥1 denote Littlewood-Paley projections ontoR frequencies |τ |R . 1 and t |τ | & 1 respectively. Decompose h = P≤1 h + P≥1 h and use that 0 P≥1 h(t0 ) = 21 (sgn(t − t0 ) + sgn(t0 ))P≥1 h(t0 ) dt0 to obtain the decomposition Z t h(t0 ) dt0 = H1 (t) + H2 (t) + H3 (t), θ(t) 0
where Z H1 (t) = θ(t) H2 (t) = H3 (t) =
t
P≤1 h(t0 ) dt0
0 1 θ(t)[sgn ∗P≥1 h](t) dt0 2 Z +∞ 1 θ(t) sgn(t0 )P≥1 h(t0 ) dt0 . 2 −∞
We begin by addressing term H1 . By Sobolev embedding (recall boundedness of the Hilbert transform for 1 < p < ∞,
1 2
< b ≤ 1) and the Lp → Lp
kH1 kHtb . kH1 kL2t + k∂t H1 kL2/(3−2b) . t
Using that kP≤1 hk
L∞ t
. khkHtb−1 , we thus conclude
kH1 kHtb . (kθkL2t + kθkL2/(3−2b) + kθ0 kL2/3−2b )khkHtb−1 . t
t
Next we address the term H2 . By the fractional Leibniz rule, kH2 kHtb . khDt ib θkL2t k sgn ∗P≥1 hkL∞ + kθkL∞ khDt ib (sgn ∗P≥1 h)kL2t . t t
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
19
However, ˆ )kL1 . khk b−1 . k sgn ∗P≥1 hkL∞ . khτ i−1 h(τ Ht τ t On the other hand, ˆ )kL2 . khk b−1 . khDt ib sgn ∗P≥1 hkL2t . khτ ib hτ i−1 h(τ Ht τ Consequently, kH2 kHtb . (khDt ib θkL2t + kθkL∞ )khkHtb−1 . t For term H3 , we have kH3 kHtb
Z
. kθkHtb
+∞
sgn(t )P≥1 h(t ) dt 0
−∞
0
0
.
L∞ t
However, the second term is handled via Parseval’s identity Z Z 0 0 0 ˆ ) dτ , sgn(t )P≥1 h(t ) dt = τ −1 h(τ t0
|τ |≥1
from which the appropriate bounds follow again by Cauchy-Schwarz. Collecting our estimates for H1 , H2 , and H3 , we have
Z t
0 0
θ(t) h(t ) dt
b . Cθ khkHtb−1 ,
0
Ht
where Cθ = kθkL2t + kθ0 kL2/(3−2b) + khDt ib θkL2t + kθkL2/(3−2b) + kθkL∞ t t
t
4.1. Various Forms of Collapsing Estimates. Lemma 4.2. There is a C independent of j, k, and N such that, (for f (k+1) (xk+1 , xk+1 ) independent of t)
(k)
R BN,j,k+1 U (k+1) (t)f (k+1) 2 2 6 C kV k 1 R(k+1) f (k+1) 2 . L L L L x,x0
t
x,x0
Proof. One can find this estimate as estimate (A.18) in [11] or a special case of Theorem 7 of [15]. For more estimates of this type, see [35, 28, 12, 14, 3, 27]. We have the following consequence of Lemma 4.2. Lemma 4.3. There is a C independent of j, k, and N such that (for α(k+1) (t, xk+1 , xk+1 ) dependent on t) kR(k) BN,j,k+1 α(k+1) kL2t L2
x,x0
6 CkR(k+1) α(k+1) kX (k+1) 1+ 2
Proof. Let fτ(k+1) (xk+1 , x0k+1 ) = Ft7→τ (U (k+1) (−t)α(k+1) (t, xk+1 , x0k+1 )) where Ft7→τ denotes the Fourier transform in t 7→ τ . Then Z (k+1) 0 α (t, xk+1 , xk+1 ) = eitτ U (k+1) (t)f (k+1) (xk+1 , x0k+1 ) dτ τ
20
XUWEN CHEN AND JUSTIN HOLMER
By Minkowski’s inequality (k)
kR BN,j,k+1 α
(k+1)
Z kL2t L2
x,x0
≤ τ
kR(k) BN,j,k+1 U (k+1) (t)f (k+1) kL2t L2
x,x0
dτ
By Lemma 4.2, Z ≤
kR(k+1) f (k+1) kL2
x,x0
τ b
dτ
For any b > 21 , we write 1 = hτ i−b hτ i and apply Cauchy-Schwarz in τ to obtain ≤ khτ ib R(k+1) f (k+1) kL2
τ ,x,x0
= kR(k+1) α(k+1) kX (k+1) b
Lemma 4.4. For each ε > 0, there is a Cε independent of Mk , j, k, and N such that (k)
kR(k) P6Mk BN,j,k+1 U (k+1) (t)f (k+1) kL2t L2 0 x,x 1−ε
X Mk
(k+1) (k+1) (k+1) 6 Cε kV kL1 P6Mk+1 f
R
2 Mk+1 Lx,x0 M >M k+1
k
where the sum on the right is in Mk+1 , over dyads such that Mk+1 > Mk . In particular, we have Lemma 4.2 back if we carry out the sum and let Mk → ∞. Proof. It suffices to take k = 1 and prove (2) −1
(1)
kR P≤M1 BN,1,2 (R ) U
(2)
(t)f
(2)
kL2t L2
x1 x01
X M1 1−ε (2) kP≤M2 f (2) kL2t L2 0 ≤ Cε kV kL1 x2 x2 M 2 M >M 2
1
where the sum is over dyadic M2 such that M2 > M1 . For convenience, we take only “half” of the operator BN,1,2 : For α(2) (t, x1 , x2 , x01 , x02 ), define Z (2) 0 def ˜ (BN,1,2 α )(t, x1 , x1 ) = VN (x1 − x2 )α(2) (t, x1 , x2 , x01 , x2 ) dx2 x2
Note that
(2) −1 (2) (2) ˜ R BN,1,2 (R ) U (t)f b(τ , ξ 1 , ξ 01 ) ZZ 0 Vc N (ξ 2 + ξ 2 )|ξ 1 | d (2) (ξ − ξ − ξ 0 , ξ , ξ 0 , ξ 0 ) dξ dξ 0 = δ(· · · ) 1 2 2 2 2 1 2 2 0 0 f 0 |ξ − ξ − ξ ||ξ ||ξ | ξ 2 ,ξ 2 1 2 2 2 2 (1)
where 2
2
2
δ(· · · ) = δ(τ + |ξ 1 − ξ 2 − ξ 02 | + |ξ 2 |2 − |ξ 01 | − |ξ 02 | ) Divide this integration into two pieces: ZZ ZZ 0 = (· · · ) dξ 2 dξ 2 + |ξ 2 |≤|ξ 02 |
|ξ 02 |≤|ξ 2 |
(· · · ) dξ 2 dξ 02
In the first term, decompose the ξ 02 integration into dyadic intervals, and in the second term, decompose the ξ 2 integration into dyadic intervals: X ZZ X ZZ 0 20 2 = PM2 (· · · ) dξ 2 dξ 2 + PM (· · · ) dξ 2 dξ 02 2 M2 ≥M1
|ξ 2 |≤|ξ 02 |
M2 ≥M1
|ξ 02 |≤|ξ 2 |
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
21 0
1 Observe that, in the first integration, we can insert for free the projection P≤3M P1 P2 2 ≤M1 ≤M2 0 0 1 and in the second integration, we can insert P≤3M P1 P2 . 2 ≤M1 ≤M2 X ZZ 0 0 1 P≤3M P 1 P 2 P 2 (· · · ) dξ 2 dξ 02 = 2 ≤M1 ≤M2 M2 |ξ 2 |≤|ξ 02 |
M2 ≥M1
+
X ZZ M2 ≥M1
0
|ξ 02 |≤|ξ 2 |
0
1 P≤3M P 1 P 2 P 2 (· · · ) dξ 2 dξ 02 2 ≤M1 ≤M2 M2
Then for each piece, we proceed as in Klainerman-Machedon [37], performing CauchySchwarz with respect to measures supported on hypersurfaces and applying the L2τ ξ ξ0 norm 1 1 to both sides of the resulting inequality.7 In this manner, it suffices to prove the following 2 estimates, uniform in τ 0 = τ − |ξ 01 | : 2(1−ε) ZZ |ξ 1 |2 M1 0 (4.3) δ(· · · ) , 2 2 dξ 2 dξ 2 ≤ Cε |ξ 02 |∼M2 , M2 |ξ 1 − ξ 2 − ξ 02 | |ξ 2 |2 |ξ 02 | |ξ 2 |≤M2
(recall that |ξ 1 | . M1 M2 ) and also 2(1−ε) ZZ M1 |ξ 1 |2 0 dξ 2 dξ 2 ≤ Cε . (4.4) δ(· · · ) 2 0 2 0 2 |ξ 2 |∼M2 , M 2 |ξ − ξ − ξ | |ξ | |ξ | 1 2 2 2 2 |ξ 0 |≤M 2
2
In both (4.3) and (4.4), 2
2
δ(· · · ) = δ(τ 0 + |ξ 1 − ξ 2 − ξ 02 | + |ξ 2 |2 − |ξ 02 | ). By rescaling ξ 2 7→ M2 ξ 2 and ξ 02 7→ M2 ξ 02 , (4.3) and (4.4) reduce to, respectively, the following. (4.5) ZZ |ξ 1 |2 def 0 dξ 2 dξ 02 ≤ Cε |ξ 1 |2(1−ε) , δ(· · · ) for |ξ 1 | 1, I(τ , ξ 1 ) = 2 0 2 0 2 |ξ 02 |∼1, |ξ 1 − ξ 2 − ξ 2 | |ξ 2 | |ξ 2 | |ξ |≤2 2
(4.6) for |ξ 1 | 1,
0
0
def
I (τ , ξ 1 ) =
|ξ 1 |2
ZZ δ(· · · ) |∼1,
|ξ 2 |ξ 02 |≤2
|ξ 1 − ξ 2 −
2 ξ 02 |
2
|ξ 2 |
2 |ξ 02 |
dξ 2 dξ 02 ≤ Cε |ξ 1 |2(1−ε) .
To be precise, the ξ 1 in estimates (4.5) and (4.6) is ξ 1 /M2 in estimates (4.3) and (4.4). We shall obtain the upper bound |ξ 1 |2 log |ξ 1 |−1 for both (4.5), (4.6). First, we prove (4.6). Begin by carrying out the ξ 02 integral to obtain Z 1 H 0 (τ 0 , ξ 1 , ξ 2 ) 0 0 2 I (τ , ξ 1 ) = |ξ 1 | dξ 2 2 1 2 ≤|ξ 2 |≤2 |ξ 1 − ξ 2 ||ξ 2 | 2 where H 0 (τ 0 , ξ 1 , ξ 2 ) is defined as follows. Let P 0 be the truncated plane defined by P 0 (τ 0 , ξ 1 , ξ 2 ) = ξ 02 ∈ R3 | (ξ 02 − λω) · ω = 0 , |ξ 02 | ≤ 2 where ω= 7Notice
that Vc N
L∞
ξ1 − ξ2 , |ξ 1 − ξ 2 |
λ=
τ 0 + |ξ 1 − ξ 2 |2 + |ξ 2 |2 2|ξ 1 − ξ 2 |
6 kVN kL1 = kV kL1 i.e. Vc N is a dummy factor.
22
XUWEN CHEN AND JUSTIN HOLMER
Now let (4.7)
0
0
H (τ , ξ 1 , ξ 2 ) =
dσ(ξ 02 )
Z ξ 02 ∈P 0 (τ 0 ,ξ 1 ,ξ 2 )
2
2
|ξ 1 − ξ 2 − ξ 02 | |ξ 02 |
where the integral is computed with respect to the surface measure on P 0 . Since |ξ 1 − ξ 2 | ∼ 1, |ξ 2 | ∼ 1, we have the following reduction Z 0 0 2 I (τ , ξ 1 ) . |ξ 1 | H 0 (τ 0 , ξ 1 , ξ 2 ) dξ 2 1 ≤|ξ 2 |≤2 2
We now evaluate H 0 (τ 0 , ξ 1 , ξ 2 ). Introduce polar coordinates (ρ, θ) on the plane P 0 with respect to the “center” λω, and note that |ξ 1 − ξ 2 − ξ 02 |2 = ||ξ 1 − ξ 2 |ω − ξ 02 |2 = |(|ξ 1 − ξ 2 | − λ)ω − (ξ 02 − λω)|2 = (|ξ 1 − ξ 2 | − λ)2 + |ξ 02 − λω|2
(4.8)
= (|ξ 1 − ξ 2 | − λ)2 + ρ2 = α 2 + ρ2 where α = |ξ 1 − ξ 2 | − λ =
|ξ 1 |2 − 2ξ 1 · ξ 2 − τ 0 2|ξ 1 − ξ 2 |
Also, (4.9)
2
2
2
|ξ 02 | = |(ξ 02 − λω) + λω| = |ξ 02 − λω| + λ2 = ρ2 + λ2
Using (4.8) and (4.9) in (4.7), H 0 (τ 0 , ξ 1 , ξ 2 ) =
Z √4−λ2 (ρ2
0
2πρ dρ + α2 )(ρ2 + λ2 )
p The restriction to 0 ≤ ρ ≤ 4 − λ2 arises from the fact that the plane P 0 must sit within the ball |ξ 02 | ≤ 2. In particular, H 0 (τ , ξ 1 , ξ 2 ) = 0 if |λ| ≥ 2 since then the plane P 0 is located entirely outside the ball |ξ 02 | ≤ 2. Since |λ| ≤ 2, we have |α| ≤ 3 and |τ 0 | ≤ 10. We consider the three cases: (A) |λ| ≤ 41 (which implies |α| ≥ 14 ), (B) |α| ≤ 14 (which implies |λ| ≥ 14 ), and (C) |λ| ≥ 41 and |α| ≥ 41 . Case (C) is the easiest since clearly |H 0 (τ 0 , ξ 1 , ξ 2 )| ≤ C. Let us consider case (B). Then √ ! Z 2 Z √2 ρ dρ dν 2 H 0 (τ , ξ 1 , ξ 2 ) . = = log 1 + 2 2 2 α2 ρ=0 ρ + α ν=0 ν + α Substituting back into I 0 , I 0 (τ 0 , ξ 1 ) . |ξ 1 |2
√ ! 2 log 1 + 2 dξ 2 α |ξ 2 |≤2
Z
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
Since |α| ≤
√
3, it follows that8 √ log(1 +
2
α2
23
) ≤ c + | log |α|| ≤ c + | log |(|ξ 1 |2 − 2ξ 1 · ξ 2 − τ 0 )|| τ 0ξ1 1 )| = c + | log 2|ξ 1 · (ξ 2 − ξ 1 + 2 2|ξ 1 |2 τ 0ξ1 1 = c + | log |ξ 1 · (ξ 2 − ξ 1 + )| 2 2|ξ 1 |2
Hence
1 τ 0ξ1 | log |ξ 1 · (ξ 2 − ξ 1 + )| dξ 2 2 2|ξ 1 |2 |ξ 2 |≤2
Z I (τ , ξ 1 ) . |ξ 1 | 1 + 0
0
2
0
τ ξ1 Denoting by B(µ, r) the ball of center µ and radius r, the substitution ξ 2 7→ ξ 2 + 21 ξ 1 − 2|ξ |2
yields, with µ = 12 ξ 1 − 0
1
τ 0 ξ1 , 2|ξ 1 |2
0
2
I (τ , ξ 1 ) . |ξ 1 |
Z | log |ξ 1 · ξ 2 || dξ 2
1+ B(µ,2)
2
. |ξ 1 | By rotating coordinates so that tion of µ,
ξ ξ1
−1
log |ξ 1 |
Z
ξ | log | 1 · ξ 2 || dξ 2 + |ξ 1 | B(µ,2)
= (1, 0, 0), and letting µ0 denote the corresponding rota-
Z −1 I (τ , ξ 1 ) . |ξ 1 | log |ξ 1 | + 0
0
2
B(µ0 ,2)
| log |(ξ 2 )1 | dξ 2
where (ξ 2 )1 denotes the first coordinate of the vector ξ 2 . Since |τ 0 | ≤ 10, it follows that |µ0 | . |ξ 1 |−1 and we finally obtain I 0 (τ 0 , ξ 1 ) . |ξ 1 |2 log |ξ 1 |−1 as claimed, completing Case (B). Case (A) is similar except that we begin with the bound Z 2 2πρ dρ 0 0 H (τ , ξ 1 , ξ 2 ) . 2 2 ρ=0 ρ + λ This completes the proof of (4.6). Next, we prove (4.5). In the integral defining I(τ 0 , ξ 1 ), we have the restriction 21 ≤ |ξ 02 | ≤ 2 and |ξ 2 | ≤ 2. Note that if 14 ≤ |ξ 2 | ≤ 2, then the argument above that provided the bound for I 0 (τ 0 , ξ 1 ) applies. Hence it suffices to restrict to |ξ 2 | ≤ 14 , from which it follows that |ξ 1 − ξ 2 − ξ 02 | ∼ 1. Begin by carrying out the ξ 02 integral to obtain Z 1 H(τ 0 , ξ 1 , ξ 2 ) 2 0 (4.10) I(τ , ξ 1 ) = |ξ 1 | dξ 2 2 2 |ξ 2 |≤2 |ξ 1 − ξ 2 ||ξ 2 | 8The
first step is simply: if x ≥ δ > 0, then log(1 + x) ≤ log x + log(1 + 1δ ). The second step uses that |ξ 1 − ξ 2 | ∼ 1, which follows since |ξ 1 | 1 and |ξ 2 | ∼ 1.
24
XUWEN CHEN AND JUSTIN HOLMER
where H(τ 0 , ξ 1 , ξ 2 ) is defined as follows. Let P be the truncated plane defined by 1 P (τ 0 , ξ 1 , ξ 2 ) = { ξ 02 ∈ R3 | (ξ 02 − λω) · ω = 0 , ≤ |ξ 02 | ≤ 2 } 2 where ξ − ξ2 τ 0 + |ξ 1 − ξ 2 |2 + |ξ 2 |2 ω= 1 , λ= |ξ 1 − ξ 2 | 2|ξ 1 − ξ 2 | Now let Z dσ(ξ 02 ) 0 H(τ , ξ 1 , ξ 2 ) = 0 2 0 2 ξ 02 ∈P (τ 0 ,ξ 1 ,ξ 2 ) |ξ 1 − ξ 2 − ξ 2 | |ξ 2 | where the integral is computed with respect to the surface measure on P . Since |ξ 1 −ξ 2 −ξ 02 | ∼ 1 and |ξ 02 | ∼ 1, we obtain H(τ 0 , ξ 1 , ξ 2 ) ≤ C. Substituting into (4.10), we obtain Z dξ 2 0 2 I(τ , ξ 1 ) . |ξ 1 | 2 |ξ 2 |≤ 41 |ξ 1 − ξ 2 ||ξ 2 | ! Z Z dξ dξ 2 2 + . |ξ 1 |2 2 2 2|ξ 1 |≤|ξ 2 |≤ 41 |ξ 1 − ξ 2 ||ξ 2 | |ξ 2 |≤2|ξ 1 | |ξ 1 − ξ 2 ||ξ 2 | In the first integral, we change variables ξ 2 = |ξ 1 |η, and in the second integral, we use the bound |ξ 1 − ξ 2 |−1 ≤ 2|ξ 2 |−1 to obtain ! Z Z dξ dη 2 . |ξ 1 |2 log |ξ 1 |−1 . |ξ 1 |2 + ξ1 3 2 1 |ξ | |η|≤2 | |ξ | − η||η| 2|ξ 1 |≤|ξ 2 |≤ 4 2 1
This completes the proof of (4.5). Lemma 4.5. For each ε > 0, there is a Cε independent of Mk , j, k, and N such that
X Mk 1−ε
(k+1) (k+1) (k+1) (k) (k) (k+1) kR P6Mk BN,j,k+1 α kL2t L2 0 6 Cε R P6Mk+1 α
(k) . x,x M X1 k+1 + M >M k+1
k
2
where the sum on the right is in Mk+1 , over dyads such that Mk+1 > Mk . Proof. The proof is exactly the same as deducing Lemma 4.3 from Lemma 4.2. We include the proof for completeness. Let fτ(k+1) (xk+1 , x0k+1 ) = Ft7→τ (U (k+1) (−t)α(k+1) (t, xk+1 , x0k+1 )) where Ft7→τ denotes the Fourier transform in t 7→ τ . Then Z (k+1) 0 α (t, xk+1 , xk+1 ) = eitτ U (k+1) (t)f (k+1) (xk+1 , x0k+1 ) dτ τ
By Minkowski’s inequality kR
(k)
(k) P6Mk BN,j,k+1 α(k+1) kL2t L2 0 x,x
Z ≤ τ
(k)
kR(k) P6Mk BN,j,k+1 U (k+1) (t)f (k+1) kL2t L2
x,x0
By Lemma 4.4, ≤ Cε
X Mk+1 >Mk
Mk Mk+1
1−ε Z τ
(k+1)
kR(k+1) P6Mk+1 f (k+1) kL2
x,x0
dτ
dτ
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
25
For any b > 12 , we write 1 = hτ i−b hτ ib and apply Cauchy-Schwarz in τ to obtain X Mk 1−ε (k+1) ≤ Cε khτ ib R(k+1) P6Mk+1 f (k+1) kL2 0 τ ,x,x Mk+1 Mk+1 >Mk
X Mk 1−ε
(k+1) (k+1) (k+1) = Cε P6Mk+1 α
(k) .
R Mk+1 X1 + M >M k+1
k
2
4.2. A Strichartz Estimate for P P (k) . Lemma 4.6. Assume γ (k) (t, xk ; x0k ) satisfies the symmetric condition (1.1). Let (4.11)
β (k) (t, xk ; x0k ) = V (xi − xj )γ (k) (t, xk ; x0k )
Then we have the estimates: (4.12)
kβ (k) kX (k)
. kV k
kβ (k) kX (k)
. kV kL3+ kγ (k) kL2t L2 0 , x
1+ −2
(4.13)
1+ −2
(4.14)
kβ (k) kX (k)
1+ −2
6+
Lx5
kh∇xi ih∇xj iγ (k) kL2t L2 0 , x,x
x,x
1 2
. kV kL2+ kh∇xi i γ (k) kL2t L2 0 . x x,x
Proof. It suffices to prove Lemma 4.6 for k = 2. Since we will need to deal with Fourier transforms in only selected coordinates, we introduce the following notation: F0 denotes Fourier transform in t, Fj denotes Fourier transform in xj , and Fj 0 denotes Fourier transform in x0j . Fourier transforms in multiple coordinates will denoted as combined subscripts – for example, F010 = F0 F10 denotes the Fourier transform in t and x01 .9 Let T denote the translation operator (T f )(x1 , x2 ) = f (x1 + x2 , x2 ) Suppressing the x01 , x02 dependence, we have (4.15)
(F12 T β (2) )(t, ξ 1 , ξ 2 ) = (F12 β (2) )(t, ξ 1 , ξ 2 − ξ 1 )
Also (4.16)
e−2itξ1 ·ξ2 (F12 T β (2) )(t, ξ 1 , ξ 2 ) = F1 (F2 T β (2) )(t, x1 − 2tξ 2 , ξ 2 ) (ξ 1 )
Now (F012 β (2) )(τ − |ξ 2 |2 + 2ξ 1 · ξ 2 , ξ 1 , ξ 2 − ξ 1 )
(4.17)
9We
= (F012 T β (2) )(τ − |ξ 2 |2 + 2ξ 1 · ξ 2 , ξ 1 , ξ 2 ) by (4.15) it|ξ |2 −2itξ ·ξ (2) 1 2 (F )(t, ξ 1 , ξ 2 ) (τ ) = F0 e 2 e 12 T β it|ξ |2 = F0 e 2 F1 (F2 T β (2) )(t, x1 − 2tξ 2 , ξ 2 ) (ξ 1 ) (τ ) by (4.16) 2 = F01 eit|ξ2 | (F2 T β (2) )(t, x1 − 2tξ 2 , ξ 2 ) (τ , ξ 1 )
are going to apply the endpoint Strichartz estimate on the non-transformed coordinates. We do not know currently the origin of such an technique. The only other place we know about it is [13, Lemma 6].
26
XUWEN CHEN AND JUSTIN HOLMER
By changing variables ξ 2 7→ ξ 2 − ξ 1 and then changing τ 7→ τ − |ξ 2 |2 + 2ξ 1 · ξ 2 , we obtain kβ (2) kX (2)
1+ −2
(2) 1 2 2 = kβˆ (τ , ξ 1 , ξ 2 , ξ 01 , ξ 02 )hτ + |ξ 1 |2 + |ξ 2 |2 − |ξ 01 | − |ξ 02 | i− 2 + kL2
τ ξ1 ξ2 ξ01 ξ02
(2) 1 2 2 = kβˆ (τ − |ξ 2 |2 + 2ξ 1 · ξ 2 , ξ 1 , ξ 2 − ξ 1 , ξ 01 , ξ 02 )hτ + 2 |ξ 1 |2 − |ξ 01 | − |ξ 02 | i− 2 + kL2
τ ξ1 ξ2 ξ01 ξ02
Applying the the dual Strichartz (see (4.19) below), the above is bounded by h i −1 . kF01 (F012 β (2) )(τ − |ξ 2 |2 + 2ξ 1 · ξ 2 , ξ 1 , ξ 2 − ξ 1 ) (t, x1 )k 2 2 56 + 2 Lξ Lt Lx1 L 2
x01 x02
Utilizing (4.17), the above is equal to = k(F2 T β (2) )(t, x1 − 2tξ 2 , ξ 2 )k
6+
L2t L2ξ Lx51 L2 0
x1 x02
2
Change variable in x1 7→ x1 + 2tξ 2 to obtain = k(F2 T β (2) )(t, x1 , ξ 2 )k
6+
L2t L2ξ Lx51 L2 0
x1 x02 2
2
Now note that from (4.11), we have (F2 T β (2) )(t, x1 , ξ 2 ) = V (x1 )(F2 T γ (2) )(t, x1 , ξ 2 ) It follows that k(F2 T β (2) )(t, x1 , ξ 2 )k
6+
L2t L2ξ Lx51 L2 0
x x0
2
1 2
(2)
= V (x1 ) k(F2 T γ )(t, x1 , ξ 2 )kL2 0 0 x1 x2
6+
L2t L2ξ Lx51 2
≤ kV kL 56 + k(F2 T γ (2) )(t, x1 , ξ 2 )kL2t L2ξ
(4.18)
2
2 L∞ x1 L 0
x1 x02
≤ kV kL 56 + k(F2 T γ (2) )(t, x1 , ξ 2 )kL2t L2
L∞ ξ2 x01 x02 x1
and continue with Sobolev, Plancherel, etc. We also need to remark that we split γ (2) into the piece where |ξ 1 | ≥ |ξ 2 | and the piece where |ξ 2 | ≥ |ξ 1 |. The above represents the treatment of the case |ξ 1 | ≤ |ξ 2 |. This proves estimate (4.12). Using H¨older exponents (3+, 2, 56 +) and (2+, 3, 65 +) in (4.18) yields estimates (4.13) and (4.14). It remains to prove the following dual Strichartz estimate (here σ (2) (t, x1 , x01 , x02 ), note that the x2 coordinate is missing): (4.19)
2
2
1
khτ + 2 |ξ 1 |2 − |ξ 01 | − |ξ 02 | i− 2 + σ ˆ (2) (τ , ξ 1 , ξ 01 , ξ 02 )kL2τ L2
ξ1 ξ01 ξ02
. kσ (2) k
6+
L2t Lx51 L2 0
x1 x02
The estimate (4.19) is dual to the equivalent estimate (4.20)
kσ (2) kL2t L6− 2 x L 0 1
x1 x02
2
2
1
. khτ + 2 |ξ 1 |2 − |ξ 01 | − |ξ 02 | i 2 − σ ˆ (2) (τ , ξ 1 , ξ 01 , ξ 02 )kL2τ L2
ξ1 ξ01 ξ02
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
27
To prove (4.20), we prove (4.21)
kσ (2) kL2t L6x
x1 x02
1
2
2
L2 0 1
. khτ + 2 |ξ 1 |2 − |ξ 01 | − |ξ 02 | i 2 + σ ˆ (2) (τ , ξ 1 , ξ 01 , ξ 02 )kL2τ L2
ξ1 ξ01 ξ02
The estimate (4.20) follows from the interpolation of (4.21) and the trivial equality kσ (2) kL2t L2x
2
2
1
ˆ (2) (τ , ξ 1 , ξ 01 , ξ 02 )kL2τ L2 = khτ + 2 |ξ 1 |2 − |ξ 01 | − |ξ 02 | i0 σ
L2 0
ξ1 ξ01 ξ02
x1 x02
Thus proving (4.19) is reduced to proving (4.21), which we do now. Let (4.22)
def
0
0
φτ (x1 , x01 , x2 ) = F0 [U 1 (−2t)U 1 (−t)U 2 (−t)σ (2) (t, x1 , x01 , x02 )](τ )
Then note φτ is independent of t and Z 0 0 0 (2) 0 σ (t, x1 , x1 , x2 ) = eitτ U 1 (2t)U 1 (t)U 2 (t)φτ (x1 , x01 , x02 )dτ Thus Z
(2)
kσ kL2t L6x
1
L2 0
x1 x02
. τ
Z . τ
Z . τ
0
0
kU 1 (t)U 2 (t)U 1 (2t)φτ (x1 , x01 , x02 )kL2t L6x kU 1 (2t)φτ (x1 , x01 , x02 )kL2t L6x kU 1 (2t)φτ (x1 , x01 , x02 )kL2 0
L2 0
dτ
L2t L6x1
dτ
1
x1 x02
x1 x02
1
L2 0
x1 x02
Now apply Keel-Tao [33] endpoint Strichartz estimate to obtain Z . kφτ (x1 , x01 , x02 )kL2 0 0 L2x dτ x1 x2
τ
. khτ i
1 + 2
1
φτ (x1 , x01 , x02 )kL2τ L2
x1 x01 x02
It follows from (4.22) that 2
2
1
ˆ (2) (τ , ξ 1 , ξ 01 , ξ 02 )kL2 = khτ + 2 |ξ 1 |2 − |ξ 01 | − |ξ 02 | i 2 + σ
τ ξ1 ξ01 ξ02
which completes the proof of (4.21). Corollary 4.1. Let β (k) (t, xk , x0k ) = N 3β−1 V (N β (xi − xj ))γ (k) (t, xk , x0k ) Then for N ≥ 1, we have (4.23)
5 k |∇xi | ∇xj β (k) kX (k) . N 2 β−1 kh∇xi ih∇xj iγ (k) kL2t L2
xx0
−1 2+
and 1
kβ (k) kX (k) . N 2 β−1 kh∇xi ih∇xj iγ (k) kL2t L2
(4.24)
xx0
1+ −2
(k)
(k)
Consequently, (R≤M = P≤M R(k) ) (4.25)
1
(k)
kR≤M β (k) kX (k) . N 2 β−1 min(M 2 , N 2β )kS (k) γ (k) kL2t L2 1+ −2
xx0
dτ
28
XUWEN CHEN AND JUSTIN HOLMER
Proof. Estimate (4.23) follows by applying either (4.12), (4.13), or (4.14) according to whether two derivatives, no derivatives, or one derivative, respectively, lands on N 3β−1 V (N β (xi − xj )). Estimate (4.24) follows by applying (4.12). Finally, (4.25) follows from (4.23) and (4.24), as follows. Let Y Q= |∇x` | 1≤`≤k `6=i,j
Then (k)
kR≤M β (k) kX (k) ≤ M 2 kQβ (k) kX (k) −1 2+
−1 2+
The Q operator passes directly onto γ (k) , and one applies (4.24) to obtain 1
(k)
kR≤M β (k) kX (k) . N 2 β−1 M 2 kS (k) γ (k) kL2t L2
(4.26)
xx0
−1 2+
On the other hand, (k) kR≤M β (k) kX (k) ≤ kQ |∇xi | ∇xj β (k) kX (k) 1+ −2
−1 2+
The Q operator passes directly on γ (k) , and one applies (4.23) to obtain (4.27)
5
(k)
kR≤M β (k) kX (k) . N 2 β−1 kS (k) γ (k) kL2t L2
xx0
−1 2+
Combining (4.26) and (4.27), we obtain (4.25). 5. Conclusion In this paper, we have established a positive answer to Conjecture 1 by Klainerman and Machedon [37] in 2008 for β ∈ (0, 2/3). This is the first progress in proving Conjecture 1 for self-interaction (β > 1/3). Moreover, our main theorem (Theorem 1.1) has already fulfilled the original intent of [37], namely, simplifying the uniqueness argument of [23] which deals with β ∈ (0, 3/5). Conjecture 1 for β ∈ [2/3, 1] is still open. Appendix A. The Topology on the Density Matrices In this appendix, we define a topology τ prod on the density matrices as was previously done in [20, 21, 22, 23, 24, 25, 35, 9, 14, 15] Denote the spaces of compact operators and trace class operators on L2 R3k as Kk and L1k , respectively. Then (Kk )0 = L1k . By the fact that Kk is separable, we select a dense (k) (k) countable subset {Ji }i>1 ⊂ Kk in the unit ball of Kk (so kJi kop 6 1 where k·kop is the operator norm). For γ (k) , γ˜ (k) ∈ L1k , we then define a metric dk on L1k by ∞ X (k) dk (γ (k) , γ˜ (k) ) = 2−i Tr Ji γ (k) − γ˜ (k) . i=1
A uniformly bounded sequence topology if and only if
(k) γN
∈ L1k converges to γ (k) ∈ L1k with respect to the weak* (k)
lim dk (γ N , γ (k) ) = 0. N
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
29
For fixed T > 0, let C ([0, T ] , L1k ) be the space of functions of t ∈ [0, T ] with values in L1k which are continuous with respect to the metric dk . On C ([0, T ] , L1k ) , we define the metric dˆk (γ (k) (·) , γ˜ (k) (·)) = sup dk (γ (k) (t) , γ˜ (k) (t)). t∈[0,T ]
We can then define a topology τ prod on the space ⊕k>1 C ([0, T ] , L1k ) by the product of topologies generated by the metrics dˆk on C ([0, T ] , L1k ) . Appendix B. Proof of Estimates (2.7) and (2.9) (2)
Proof of Estimate (2.7). Utilizing Lemma 2.1 and estimate (3.2) to the free part of γ N , we obtain
(1) (1)
P≤M R BN,1,2 F P (k) (t2 ) 1 2 LT Lx,x0
6 R(1) BN,1,2 F P (k) (t2 )
L1T L2x,x0
(2) 6 CT R(2) γ N,0 1 2
L2x,x0
(2) (2) 6 CT R γ N,0 1 2
L2x,x0
(2) 6 CT R(2) γ N,0 1 2
L2x,x0
Z k X X
(1)
(j) (j)
R B J (t )dt + , µ )(U (t )γ N,1,2 N j m j j N,0
j=3
+
k X X j=3
+C
L1T L2x,x0
D
m
1 2
C(CT )
j−2
(j−1) (j) (j) BN,µm (j+1),j+1 U (tj )γ N,0
R
L1T L2x,x0
m
∞ X
1
(j) 4j−2 (CT 2 )j−1 R(j) γ N,0
L2x,x0
j=3
.
Via condition (1.4), we can choose a T small enough such that the series in the above estimate converge. Whence, we have shown estimate (2.7). Proof of Estimate (2.9). We proceed like the proof of estimate (2.7) and end up with
(1) (1) (k)
P≤M R BN,1,2 IP 1 2 LT Lx,x0
(1) (k) 6 R BN,1,2 IP
L1T L2x,x0
Z X
(k+1) (1)
R BN,1,2
6 J (t , µ )(γ (t ))dt N k+1 m k+1 k+1 N
m
L1T L2x,x0
D
X 1
(k+1) 6 C(CT 2 )k−1 R(k) BN,µm (k+1),k+1 γ N (tk+1 )
L1T L2 x,x0
m
We then investigate
(k) (k+1)
R BN,µm (k+1),k+1 γ N
L1T L2x,x0
.
.
30
XUWEN CHEN AND JUSTIN HOLMER
1 , we have Set µm (k + 1) = 1 for simplicity and look at BN,1,k+1 Z 2 (k) 1 (k+1) R BN,1,k+1 γ N (t) dxk dx0k 2 Z Z (k) (k+1) 0 R dxk dx0k = V (x − x )γ (t, x , x ; x , x )dx N 1 k+1 k k+1 k+1 k+1 k N
2 ! ! k Z Z k Y Y 0 (k+1) 0 0 6 C VN (x1 − xk+1 ) ∇ xj ∇xj γ N (t, xk , xk+1 ; xk , xk+1 )dxk+1 dxk dx0k j=2 j=1 2 Z Z (k+1) (k) 0 +C VN (x1 − xk+1 )R γ N (t, xk , xk+1 ; xk , xk+1 )dxk+1 dxk dx0k = C(I + II). We estimate I and II as following:
= 6
6 6
I 2 ! k ! Z Z k Y Y 0 (k+1) 0 0 ∇xj VN (x1 − xk+1 ) ∇xj γ N (t, xk , xk+1 ; xk , xk+1 )dxk+1 dxk dx0k j=1 j=2 Z Z 2 dxk dx0k |VN0 (x1 − xk+1 )| dxk+1 2 ! k ! Z Y k Y 0 (k+1) ∇ x × ∇xj γ N (t, xk , xk+1 ; x0k , xk+1 ) dxk+1 (Cauchy-Schwarz) j j=1 j=2 2 ! k ! Z Z Y k Y 0 2 (k+1) ∇ x N 5β kV 0 kL2 ∇xj γ N (t, xk , xk+1 ; x0k , xk+1 ) dxk+1 dxk dx0k j j=2 j=1 Z 5β 0 2 CN kV kL2 dxk dx0k 2 ! k ! Z k D E Y Y
0 (k+1) ∇x × ∇xk+1 ∇x0k+1 ∇xj γ N (t, xk , xk+1 ; x0k , x0k+1 ) dxk+1 dx0k+1 j j=2
j=1
(Trace Theorem)
2 2 6 CN 5β kV 0 kL2 S (k+1) γ (k+1) L∞ L2 T
x,x0
and 2 Z Z (k+1) (k) 0 VN (x1 − xk+1 )R γ II = (t, xk , xk+1 ; xk , xk+1 )dxk+1 dxk dx0k N
2 6 CN 3β kV k2L2 S (k+1) γ (k+1) L∞ L2 (Same method as I), T
x,x0
6
where V ∈ W 2, 5 + implies that V ∈ H 1 by Sobolev. Accordingly, Z 2
2 (k) 1 (k+1) 2 R BN,1,k+1 γ N (t) dxk dx0k 6 CN 5β kV kH 1 S (k+1) γ (k+1) L∞ L2 T
x,x0
.
PROOF OF THE KLAINERMAN-MACHEDON CONJECTURE WITH HIGH β
31
Thence
(1) (1)
P≤M R BN,1,2 IP (k)
L1T L2x,x0
6
X
1
(k+1) C(CT 2 )k−1 R(k) BN,µm (k+1),k+1 γ N (tk+1 )
L1T L2 x,x0
m k−1
6 4
1 2
C(CT )
k−1
T
CN
5β 2
2 kV kH 1 S (k+1) γ (k+1) L∞ L2 T
1
6 C kV kH 1 (T 2 )k+2 N
5β 2
x,x0
C k . (Condition (1.4))
As in [11, 15], take the coupling level k = ln N , we have
5β 1
(1) (1) (k)
P≤M R BN,1,2 IP 1 2 6 C kV kH 1 (T 2 )2+ln N N 2 N c . LT Lx,x0
Selecting T such that T 6 e−(5β+2C) , ensures that 1
(T 2 )ln N N and thence
5β 2
N c 6 1,
(1) (1)
P≤M R BN,1,2 IP (k)
L1T L2x,x0
6 C,
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[email protected] Department of Mathematics, Brown University, 151 Thayer Street, Providence, RI 02912 E-mail address:
[email protected]