Spectral Gaps for Reversible Markov Processes with Chaotic Invariant

arXiv:1812.03874v1 [math-ph] 10 Dec 2018

Spectral Gaps for Reversible Markov Processes with Chaotic Invariant Measures: The Kac Process with Hard Sphere Collisions in Three Dimensions Eric A. Carlen1 , Maria C. Carvalho2 , and Michael Loss3 1. Department of Mathematics, Hill Center, Rutgers University 110 Frelinghuysen Road Piscataway NJ 08854 USA and CMAF, University of Lisbon, 1649-003 Lisbon, Portugal 2. Department of Mathematics and CMAF, University of Lisbon, 1649-003 Lisbon, Portugal 3. School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332 USA

December 6, 2018

Abstract We develop a method for producing estimates on the spectral gaps of reversible Markov jump processes with chaotic invariant measures, and we apply it to prove the Kac conjecture for hard sphere collision in three dimensions. Mathematics subject classification numbers: 60G07, 39B62, 35Q20 Key Words: spectral gap, kinetic theory

Contents 1 Introduction 1.1 The Kac collision process . . . . . . . . . . . . . . . 1.2 The generator of the Kac process . . . . . . . . . . . 1.3 The conjugate Kac process and its generator . . . . 1.4 The link between the Kac process and its conjugate 1.5 Proof of Theorem 1.6 . . . . . . . . . . . . . . . . . . 1.6 Proof of the main theorem . . . . . . . . . . . . . . . 1

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Work of Eric Carlen is partially supported by U.S. National Science Foundation grant DMS 0901632 Work of Maria Carvalho is partially supported by Funda¸c˜ ao para a Ciˆencia e Tecnologia grants UID/MAT/04561/2013 and SFRH/BSAB/113685/2015 3 Work of Michael Loss is partially supported by U.S. National Science Foundation grant DMS-1600560 c 2011 by the authors. This paper may be reproduced, in its entirety, for non-commercial purposes.

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2 Estimates for the conjugate process 2.1 Quantitative chaos . . . . . . . . . . 2.2 The operators W (α) and P (α) . . . . . bN,0 . . . . . . . . 2.3 The spectrum of L bN,α , α ∈ (0, 2] . . 2.4 The spectrum of L

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14 16 19 22 23

b N,1 for large N . 3 A sharper lower bound on ∆ 3.1 The trial function decomposition . . . . . . . 3.2 Lower bound on DN,1 (f, f ) . . . . . . . . . . eN,1 (g, g) and D eN,1 (s, s) . . 3.3 Lower bounds on D 3.4 Proof of Theorem 3.1 . . . . . . . . . . . . . .

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25 25 27 32 33

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A Some computational proofs

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b N,2 B Quantitiative estimates on ∆ 35 B.1 An explicit bound for N ≥ 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 B.2 An explicit bound for N = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1

Introduction

In a seminal paper of 1956, Mark Kac [11] introduced a family of continuous time reversible Markov √ √ jump processes on the sphere S N −1 ( N ) of radius N in RN . This family of processes, and its generalizations, have drawn the attention of many researchers. Kac was motivated by a connection, in the large N limit, to the non-linear Boltzmann equation. The connection arrises through a particular “asymptotic independence” property of sequences {dµN }, where dµN is a probability measure on S N −1 . This property is possessed, in particular, by the sequence {dσN } of uniform √ √ probability measures on S N −1 ( N ). Let ~v = (v1 , . . . , vN ) denote a generic point on S N −1 ( N ) of √ 2 radius N . Let φ be any bounded continuous function on Rk and dγ = (2π)−1/2 e−v /2 dv be the unit Gaussian probability measure on R. As is well known, going back at least to Mehler [13], Z Z lim φ(v1 , . . . , vk )dγ ⊗k . φ(v1 , . . . , vk )dσN = √ N →∞ S N−1 ( N )

Rk

As long as one only looks at coordinates belonging to a fixed, finite set, in the large N limit, the coordinates in this set are asymptotically independent. The main result of [11] concerned sequences √ of probability measures {dµN } on S N −1 ( N ) with the property that, for some probability density f on R with zero mean and unit variance, lim

Z

√ N →∞ S N−1 ( N )

φ(v1 , . . . , vk )dµN =

Z

Rk

φ(v1 , . . . , vk )

k Y

f (vj )dvj ,

j=1

in which case the sequence {dµN } was said by Kac to be f (v)dv chaotic. He proved that chaoticity was propagated in time by solutions of the forward Kolmogrov equations associated to the Kac processes. Moreover, if {dµN (t)} is the sequence of laws at time t starting from an f (v)dv chaotic sequence, {dµN (t)} is f (t, v)dv chaotic where f (t, v) is the solution of the Kac- Boltzmann equation with initial data f (v). (The Kac Boltzmann equation is a simple model of the Boltzmann equation

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for a gas in one dimension.) He also made a conjecture, that went unsolved for a long time, concerning the spectral gap of the generator of this family of processes. Since the processes are reversible, their generators are self adjoint, and it is not hard to see that the null space is spanned by the constants. Kac conjectured a gap ∆N separating 0 from the rest of the spectrum that is bounded below uniformly in N . That is, limN →∞ ∆N > 0. This was finally proved by Janvresse in 2000 [10], and shortly afterwards the exact value of ∆N for all N was determined in [3]. A few years after his original work, Kac returned to these problems [12], but this time for a physically realistic model of a gas in three dimensions undergoing “hard sphere” collision that conserve energy an momentum. As he showed, this physical model would have, through propagation of chaos, a direct connection to the actual Boltzmann equation for hard sphere collisions, and not only a toy model of it. However, in the physical model, the rates at which different pairs of molecules collide depend on their velocities: The rates are not bounded away from 0, and there is no bound from above that is uniform in N . It is much harder to estimate spectral gaps for the generators of jump processes with rates that are not bounded from below, and the lack of un upper bound that is uniform in N makes it much harder to prove propagation of chaos. In this paper, we prove the Kac conjecture for the Kac model with hard sphere collisions in R3 . We do so using a method that has three essential components. These are: (1) The introduction of a conjugate process, in which at each step all but one of the velocities are updated. The rates in this process are still not bounded below, but they depend only on the one velocity that is left fixed during the jump. There is also a simple connection between the spectral gaps of the original process and the conjugate process, and the central problem becomes the determination of the spectral gap for the conjugate process. (2) Quantitative estimates on the chaoticity of the sequence of invariant measures: We prove and apply estimates quantitatively expressing the near independence of any finite set of coordinates for large N (3) A trial function decomposition: We decompose any trial function f for the spectral gap problem into 3 pieces, f = s + g + h that are mutually orthogonal in the L2 space for the invariant measure, and, due to quantitative chaos estimates, are nearly orthogonal with respect to the inner product given by the Dirichlet form of the conjugate process. Each of these pieces has a particular special structure that facilitates the proof of estimates of the type we seek. The first two components have been present in our work on Kac type models since our early papers [3, 4] on the models (as in [11]) with uniform jump rates, though in the early papers, the conjugate process is not considered explicitly as a process. However, the connection between its spectral gap and the spectral gap for the Kac process has been central to the approach from the beginning. Work by two of us and Jeff Geronimo [6] dealt with the quantitative chaos estimates need for the three dimensional energy and momentum conserving collision considered here, but applied them to “Maxwellian molecules” models which, unlike to hard sphere model, has rates that are bounded below. There too, the approach yielded the exact value of the the spectral gap for a wide class of Maxwellian molecules models. Finally in [5] we proved the Kac conjecture for a “hard sphere” model with one dimensional velocities, and introduced a somewhat simpler version of component (3), the trial function decomposition. In application to kinetic theory, as explained in [5], the spectral gap in the symmetric sector; i.e., for functions that are invariant under permutation of coordinates, is especially impor-

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tant. It is this quantity that may be related to the spectral gap for the linearized Boltzmann equation, and one would like to have explicit estimates on this gap. Therefore, in [5] we worked hard to render all estimates as sharp and explicit as possible, and to treat only the symmetric sector for which fewer estimates were required. It was clear to us at the time we wrote [5] that we had a general method that would prove the existence of a spectral gap, uniformly in N , for the physical three dimensional hard sphere Kac model, and we announced this in several lectures. The result is quoted in reference 9 of [14], as a personal communication, and used in the development of the quantitative treatment of propagation of chaos that is provided there. After our paper [5] appeared with the details provided only for the symmetric sector and the one dimensional model, St´ephane Mischler and Cl´ement Mouhot asked us several times to provide the details. This paper answers their request, and moreover, in the course of preparing this answer, it has provided a clearer picture of the how the method explained [5] in can be extended and applied to more complicated models, such as the main example treated here. The method to be explained here may be applied to a wide class of sequences of reversible Markov jump processes whose sequence of invariant measures satisfies certain “quantitative chaos” estimates that are specified here. The method is not at all restricted to the treatment of the symmetric sector, and perhaps had we explained the method in [5] without obscuring it behind the detail of so many explicit computations, necessary for the precise quantitative estimates obtained there, this would have been clear some years ago. Therefore, in the present paper, we prove the Kac conjecture for hard sphere collisions in three dimensions without any symmetry condition in as simple a manner as possible to provide a clear view of the method. To do this, we make use of constants C that change from line to line but are independent of N that are not explicitly evaluated here, but easily could be – at the expense of more pages and less clarity. In addition to the applications to quantitative propagation of chaos developed in [14], uniform bounds on the spectral gap are important in certain problems concerning the hydrodynamic limits of certain kinetic models, as explained in [9]. These authors considered a model one dimensional model essentially equivalent to the one considered in [5], and asked for the spectral gap. Sasada [16] provided the answer to the question they raised, noting that she could not simply apply the result of [5] as it applied to the symmetric sector only. This is true, but as shown here, the method used in [5] may readily extended to answer a much broader range of questions. Much beautiful work has been done on the question of estimating spectral gaps for Kac type processes, and we refer to the papers of Sasada [16] and Caputo [1, 2], in addition to our own papers cited here, for significant contributions. However, it is not clear to us that any of the other methods that have been developed for this class of models applies to the main example at hand which is considerably more complex that the models considered most other work.

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1.1

The Kac collision process

For N ∈ N, p ∈ R3 and E > |p|2 , let SN,E,p be the set consisting of N –tuples ~v = (v1 , . . . , vN ) of vectors vj in R3 with N N 1 X 1 X |vj |2 = E and vj = p . N N j=1

j=1

In what follows, a point ~v ∈ SN,E,p specifies the velocities of a collection of N particles with mass 2, so that E is the kinetic energy per particle, and p is one-half the momentum per particle. We consider the Markov jump process on SN,E,p that was introduced by Mark Kac [12] describing a random binary collision process for the N particles. The collision mechanism conserves energy and momentum, and thus if the process starts on SN,E,p , it will remain on SN,E,p for all time. Recall that a random variable T with values in (0, ∞) is exponential with parameter λ in case Pr(T ≥ t) = e−λt . When the collision process begins, associated to each pair (vi , vj ), i < j, is an exponential random variable Ti,j with parameter !−1 N λi,j = N |vi − vj |α , (1.1) 2 where 0 ≤ α ≤ 2, and α = 1 is the case of main interest: As explained in [12], (1.1) is motivated by a connection between the Kac process and the Boltzmann equation, and α = 1 corresponds to “hard-sphere collisions”. Ti,j represents the waiting time for particles i and j to collide, and the set of these random times is taken to be independent. The first collision occurs at time T = min{Ti,j } . i 21 N , and for 1 + O( N12 ). Thus, for all α ∈ [0, 2], for large N , the expected waiting time |vj |2 ≤ 21 N , λj (~v ) = 2N for a jump is very close to 1/N , and this one jump will bring N − 1 of the particles very close to equilibrium. If the expected waiting time were exactly 1/N and the jump took all N particles to equilibrium, the spectral gap would be exactly 1 − 1/N . This is not misleading; we shall show that for the conjugate Kac process, the spectral gap is indeed 1 − 1/N plus lower order corrections. To write down the generator, introduce the conditional expectation operators Pk , k = 1, . . . , N , defined as follows: For any function φ in L2 (SN ), and any k with 1 ≤ k ≤ N , define Pk (φ) to be the orthogonal projection of φ onto the subspace of L2 (SN ) consisting of square integrable functions that depend on ~v through vk alone. That is, Pk (φ) is the unique element of L2 (SN ) of the form f (vk ) such that Z Z f (vk )g(vk )dσN (1.18) φ(~v )g(vk )dσN = SN

SN

for all continuous functions g on R3 . In probabilistic language, Pk φ is the conditional expectation of φ given vk : Pk φ = E{φ : vk } . (1.19) The generator of the conjugate Kac process is then given by α/2 N  1 X N 2 − (1 + |vk |2 )N b [f − Pk f ] , LN,α f = N (N − 1)2

(1.20)

k=1

which is the analog of (1.10). Define the quadratic form DN,α by

b N,α f iL2 (σ ) . DN,α (f, f ) = −hf, L N

A simple computation using (1.12) shows that

α/2  2 N Z  2  N − (1 + |vk |2 )N 1 X f − f Pk f dσN . DN,α (f, f ) = 2 N (N − 1) SN k=1

(1.21)

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The spectral gap for the conjugate Kac process is the quantity defined by n o b N,α = inf DN,α (f, f ) : f ∈ L2 (SN ) , hf, 1iL2 (S ) = 0 and kf k2 2 ∆ = 1 . L (S ) N N

(1.22)

The following theorem bears out the heuristic discussion in Remark 1.3

b N,α > 0. Moreover, there is a constant C 1.4 THEOREM. For all N ≥ 3, and all α ∈ [0, 2], ∆ independent of N such that b N,α ≥ 1 − 1 − C . (1.23) ∆ N N 3/2

1.5 Remark. The constant C is large enough that the first statement does not follow from (1.23) which is only a meaningful bound when N is large enough that the right side is positive.

1.4

The link between the Kac process and its conjugate

The following theorem provides the link between the Kac process and its conjugate: 1.6 THEOREM. ∆N,α ≥

N b N,α . ∆N −1,α ∆ N −1

(1.24)

Before proving Theorem 1.6, we recall some explicut formulas that will be useful here and elsewhere. The proof of Theorem 1.6 uses the methods intorduced in [3, 4, 5]. The estimation of ∆N,α in terms of ∆N −1,α is based on a parameterization of SN in terms of SN −1 × B where B is the unit ball. For each k = 1, . . . , N , define πk : SN → B by 1 vk . (1.25) N −1 P P 2 (Note that because of the constraints N vj = 0 and N j=1 j=1 |vj | = N , the largest value of |vk | on √ SN is N − 1.) Define a map T1 : SN −1 × B → SN as follows:   √ 1 1 T1 (~y , v) = v, . . . , β(v)yN −1 − √ v , (1.26) N − 1v , β(v)y1 − √ N −1 N −1 πk (~v ) = √

where

N (1 − |v|2 ) . (1.27) N −1 The subscript 1 in T1 indicates that the vector v from B went into the first place. We likewise define T2 , . . . , TN by placing this coordinate in the corresponding position. In the coordinates (~y , v) on SN induced by any of the maps Tk , one has the integral factorization formula # Z "Z Z β 2 (v) =

φ(~v )dσN =

B

SN

where dνN (v) =

SN−1

φ(Tk (~y , v))dσN −1 dνN (v) .

|S 3N −7 | (1 − |v|2 )(3N −8)/2 dv . |S 3N −4 |

(1.28)

(1.29)

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Also, note that for i 6= k, j 6= k, Ri,j,σ (Tk (~y , v)) = Tk (Ri,j,σ (~y ), v) .

(1.30)

We now have the means to relate EN,α to EN −1,α . For each k = 1, . . . , N , define the conditional Dirichlet form EN,α (f, f |vk ) on L2 (SN , σN ) by

!−1 N −1 EN,α (f, f |vk ) = (N − 1) × 2   Z X Z yi − yj α |yi −yj | b σ · [f (Tk (πk (~v ), y) − f (Ri,j,σ Tk (πk (~v ), y))]2 dσdσN −1 (y) . |y − y | 2 i j SN−1 S

i 3,   C (1.34) ∆N,α ≥ 1 − 3/2 ∆N −1,α N The main result will follow easily from this, and a bound on ∆2,α , and our next task is to prove such a bound. Because |vi − vj | can be arbitrarily small on SN for any N > 2, for any given C > 0, there will be functions f ∈ L2 (σN ) that satisfy hf, 1iL2 (σN ) = 0 and kf k2L2 (σN ) = 1 such that f (~v )LN,α f (~v ) < Cf (~v )LN,0 f (~v ) for some ~v ∈ SN . This precludes a simple and direct comparison of the Dirichlet forms EN,α and EN,0 . For N = 2, things are much better: Then by definition of S2 := S2,1,0 , for all (v1 , v2 ) ∈ S2 , v2 = −v1 , and |v1 | = |v2 | = 1, so that |v1 − v2 | = 2 everywhere on S2 . That is, for N = 2, there is no significant difference between α = 0 and α > 0. For α = 0 and a number of choices of b, ∆2 has been computed in [6]. It is easy to apply the methods there to compute ∆2 for b(x) = 2|x|, the hard-sphere choice. 1.7 THEOREM (Spectral gap for N = 2 and hard sphere collisions ). With b(x) = 2|x|, ∆2,1 =

10 . 3

(1.35)

More generally, as long as b is even and continuous on [−1, 1] and satisfies (1.6), ∆2,α > 0 for all α ∈ [0, 2]. Proof. As shown in [6], the eigenvalues of L2,0 are given by   Z 1 1 Pn (x)b(x)dx − 1 λn = 2 2 −1

(1.36)

where Pn denote the nth degree Legendre polynomial with the normalization Pn (1) = 1. (This standard notation is used in this proof only; elsewhere in the paper Pn denotes one of the projection operators defined in (1.19).) Since Pn has the parity of n, and since b is even, λn = −2 for all odd values of n.

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Making direct computations, one finds   1 5 λ2 = 2 −1 =− 6 3



 1 25 λ4 = 2 − − 1 = − . 24 12

and

Next, using the classical bound (see Theorem 7.3.3 in [17]) |Pn (x)|2
0 for all such b. Since |Pn (x)| < |Pn (1)| = 1 for all n > 0 and all −1 < x < 1, as long as b is continuous, (1.36) shows that for each n > 0, λn < 0. (All that is really required is that the probability measure 12 b(t)dt is not concentratedat ±1.) Then application of (1.37) as in [6, Section 2] shows that for some finite consntat C, |λn | ≥ (1 − Cn−1/2 ). Thus, |λn | ≥ 12 for all but fintely many n and now supn>0 λn < 0 follws from λn < 0 for all n. We are now ready to prove the main theorem: Proof of Theorem 1.2. Since ∆2,α > 0 by Theorem 1.7, Theorem 1.6 and the first part of Theorem 1.4 yield 3 b 3,α > 0 , ∆3,α ≥ ∆2,α ∆ 2 and then the obvious iteration yields ∆N,α > 0 for all N ≥ 2. To go further and prove that inf N ≥2 ∆N,α > 0, we use the second part of Theorem 1.4:  ∞  Y C −3/2 1 − 3/2 > 0 and for all N ≥ N0 Let N0 be such that 1 − CN0 > 0. Then K0 := j j=N 0

∆N,α ≥ K0 ∆N0 ,α .

(1.38)

1.8 Remark. As we shall see, it is possible to explicitly compute the constant C in Theorem 1.4. To keep the presentation free of clutter, we have not carried this through here, but it would be a simple if tedious exercise to track the constants step by step. As for the first part of Theorem 1.4, it b N,α for all N ≥ 4, and we do so below. The case N = 3 is easy to give an explicit lower bound on ∆ b 3,α > 0. However we do is more difficult, and we use a simple compactness argument to prove ∆ b 3,α . Thus, the method we employ to prove Theorem 1.2 sketch a method for explicitly estimating ∆ can be used to prove explicit bounds.

It remains to prove Theorem 1.4, and we prepare the way for this in the next section. Throughout the rest of the paper, we are concerned soley with the conjugate Kac process. All of the analysis that directly involves the Kac process itself is complete at this point.

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2

Estimates for the conjugate process

It is in the proof of Theorem 1.4 that new ideas are required to deal with the non-uniform jump rates of the conjugate process, and we beging with a heuristic discussion of these ideas. As in the case α = 0, we rely in part on the fact that the invariant measure σN (of both processes) is chaotic in the sense of Kac. More specifically, it is γ chaotic where dγ = (2π/3)−3/2 e−3|v|

2 /2

dv

is the isotropic Gaussian distribution on R3 with unit variance. This means that for any k ∈ N and any bounded continuous function ψ(v1 , . . . , vk ) function of R3k , Z Z ψ(v1 , . . . , vk )dγ ⊗k . ψ(v1 , . . . , vk )dσN = lim N →∞ SN

R3k

That is, as long as k is much less than N , the random variables v1 , . . . , vk are nearly independent, and by symmetry this is true of any set of k distinct coordinate functions on SN . The notion of chaos was also introduced by Kac in [11], and the main result of that paper was the for the model with one dimensional velocities and α = 0, chaos is propagated by the dynamics. Propagation of chaos for α > 0 is much harder, and this was only proved later by Sznitman [18], also in 3 dimensions. b N,0 has a special structure that facilitates the study of the spectral In case α = 0, the range of L b N,0 . The range is the space AN defined as follows: gap for L

2.1 DEFINITION. Let AN denote the subspace of L2 (SN ) that is the closure of the span on functions of the form N X ϕj (vj ) (2.1) f (~v ) = j=1

for bounded continuous functions ϕ1 , . . . , ϕN in R3 such that

R

SN

f dσN = 0.

b N,α . Nonetheless, as we explain, the gap When α 6= 0, AN is not an invariant subspace of L may be bounded using a trial function decomposition based on AN , and for this the approximate independence that comes along with the chaoticity of σN is essential. To see how this works, suppose that one replaces the state space SN with R3N , and replaces the conjugate Kac process with the “conditional jump to uniform” with respect to dγ ⊗N . In this case, with the invariant measure being a product measure, the corresponding conditional expectation operators Pk will all commute. One might therefore expect that the operators Pk figuring in the b N,α almost commute for large N . Suppose that they exactly commute, or, definition (1.20) of L what is the same thing, that the coordinate functions v1 , . . . , vk are exactly independent. R R R P ϕj (vj ) by ϕj (vj ) − SN ϕj (vj )dσN , Since 0 = SN f dσN = N j=1 SN ϕj (vj )dσN = 0, replacing R we may assume without loss of generality in (2.1) that SN ϕj (vj )dσN = 0 for each j. Granted the exact independence, we would then have that for k 6= j, Pk ϕ(vj ) = 0, while Pk ϕk (vk ) = ϕk (vk ).

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Thus, for f ∈ AN , f − Pk f = DN,α (f, f ) = =

P

j6=k

ϕj (vj ), and then, again using the independence,

α/2 X  2 N Z 1 X N − (1 + |vk |2 )N ϕ2j (vj )dσN 2 N (N − 1) j6=k k=1 SN  ! Z α/2 Z  2 N 2 X X N − (1 + |vk | )N 1 dσN  ϕ2j (vj )dσN  N (N − 1)2 SN SN

(2.2)

j6=k

k=1

As we shall show below, the integral over the rate, which is evidently independent of k, is bounded below by 1 − C/N 2 for some constant C that is independent of N . Thus, we would have DN,α (f, f ) ≥



1−

C N2



N

N −1X kϕj (vj )k22 = N j=1



1−

C N2



N −1 kf k22 N

which is even better than (1.23). Notice that the independence of the coordinate functions allows us to effectively replace the non-uniform rates in the conjugate process by their averages – and then we are back in a constant rate setting. We shall show below that the approximate independence is enough to obtain such a bound. Of course one must consider trial functions that are not in AN , and for trial functions f that are in A⊥ N , things are better still. Such functions are in the null space of Pk for each k. Therefore, for f ∈ A⊥ N , we would have from (1.21) α/2 ! Z N  1 X N 2 − (1 + |vk |2 )N f 2 dσN , (2.3) DN,α (f, f ) = N (N − 1)2 SN k=1

which is a significant simplification of (1.21). It is shown below (see Lemma 2.10 and Remark 2.11) that for some constant C independent of N , α/2  N  1 X N 2 − (1 + |vk |2 )N α 1 C ≥ 1 − − 2 . 2 N (N − 1) 2 N N k=1

Combining this with (2.3) would then yield   α 1 C 1− − 2 kf k22 . DN,α (f, f ) ≥ 2 N N For α > 0, this is much stronger than than (1.23), and for this bound we do not even use the approximate independence. b N,α , one has to show that for g ∈ AN and h ∈ A⊥ , Since AN is not an invariant subspace for L N D(g, h) is small. We shall show, again using the approximate independence, that |D(g, h)| ≤

C kgk2 khk2 . N 3/2

It is the estimate in this step that is responsible for the N 3/2 term in (1.34). A more refined argument, like the one provided for this step in [5] for the model with one dimensional velocities, would presumably improve N 3/2 to N 2 , but since we have elected not to keep track of constants, there is no point in pursuing this here.

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Our proof will closely follow these heuristics, but of course we will have to carefully control the departures from exact independence wherever it was used above. To do this, we shall have to further split the components in AN into two pieces that are well-behaved for different reasons. The next subsection presents the main “quantitative chaos” estimates that we need to control the effects of the weak dependence of the coordinate function for large N . It is important that some of the results turn out to be meaningful even for small N , such as N = 3.

2.1

Quantitative chaos

A number of the quantitative chaos bounds that we need may be expressed in terms of the Koperator that we now define: Let B denote the unit ball in R3 , and let νN be given by (1.29) so that for any function ψ on the unit ball, and any k, Z Z ψ(πk (~v ))dσ . f (v)dνN = B

SN

We define the correlation operator K on L2 (B, νN ) by Z ψ1∗ (π1 (~v ))ψ2 (π2 (~v ))dσ . hψ1 , Kψ2 iL2 (B,νN ) = SN

K is evidently self adjoint. The spectrum of K is determined in [6], where the following facts are proven: 2.2 LEMMA. The operator K is compact on L2 (B, νN ). The constant function 1 is an eigenvector with eigenvalue 1, the three components of v and |v|2 − 1/(N − 1) are eigenfunctions with eigenvalue −1/(N − 1), and for N ≥ 3. The given functions span these eigenspaces, and all other eigenvalues 5N −3 . Therefore, for all ψ1 , ψ2 ∈ L2 (B, µN ) that are orthogonal to are not larger in value than 3(N −1)3 the constants, the three components of v and v 2 − 1, Z 5N − 3 ∗ ψ1 (π1 (~v ))ψ2 (π2 (~v ))dσ ≤ kψ2 ◦ π1 k2 kψ2 ◦ π2 k2 . (2.4) 3(N − 1)3 SN

Equivalently, for all functions ψ ∈ L2 (B, νN ), that are orthogonal to 1, the three components of v and v 2 , 5N − 3 kKψk2 ≤ kψk2 . (2.5) 3(N − 1)3

It will be convenient in what follows to think of K as an operator on the subspace H of ˜ k (~v )) where N ) spanned by functions of the form ϕ(vk ) for some k. Note that ϕ(vk ) = ϕ(π √ ˜ k (~v )). The spectrum of K thought of this way is, ϕ(v) ˜ = ϕ( N − 1v). Define Kϕ(vk ) = (K ϕ)(π of course, the same, but the eigenfunctions change. For example, now |v|2 − 1 is an eigenfunction with eigenvalue −1/(N − 1). In this notation, we have that for any function ξ on R3 so that ξ(v1 ) is in L2 (σN ), E{ξ(v1 ) | vN = v} = Kψ(vN ) . L2 (σ

This slight abuse of notation will simplify many formulas that follow without introducing any ambiguity.

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CCL December 6, 2018

Since K is compact on H, the subspace of L2 (SN ) specified in the paragraph above, there is a orthonormal basis of H consisting of ejgenvectors of K. This orthonormal eigenbasis is also determined explicitly in [6], but all we need to know is that is can be written as {ηι }ι≥0 where √ η0 (v) = 1 , ηj (v) = 3ej · v for j = 1, 2, 3, and η4 (v) = |v|2 − 1 . Let κι denote the eigenvalue corresponding to ηι , so that Kηι = κι ηι . Our first application of Lemma 2.2 concerns the norm of functions in AN : P 2.3 THEOREM. Let N ≥ 3, and let f = N j=1 ϕj (vj ) ∈ AN be orthogonal to 1 Then, 

5N − 3 1− 3(N − 1)3 R

X N k=1

kϕk k22

≤

kf k22

X  N 5N − 3 ≤ 1+ kϕk k22 , 3(N − 1)2

(2.6)

k=1

where kϕk k22 denotes SN |ϕk (vk )|2 dσN . In particular, if HN,k denotes the subspace of L2 (SN ) consisting of functions of the form ϕ(vk ), the map T :

N M k=1

HN,k → AN

by T (ϕ1 (v1 ), . . . , ϕN (vN ) =

N X

ϕk (vk )

k=1

is bounded with a bounded inverse.     5N −3 5N −3 2.4 Remark. Define cN := 1 − 3(N and C = 1 + N 3 2 −1) 3(N −1) . Note the different exponents in the denominator, and that cN > 0 for all N ≥ 3. Also note that cN = 1 − O(1/N ), and CN = 1 + O(1/N ), and then we can rewrite (2.6) as cN

N X k=1

kϕk k22 ≤ kf k22 ≤ CN

N X k=1

kϕk k22 ,

and of course we would have equality here with cN = CN = 1 if the coordonate functions were exactly indoenendent. Lemma 2.3 gives a quantitative expression of the fact that for large N , the coordinate functions are approximately pairwise independent. Proof of Theorem 2.3. As noted above, we may assume that each ϕj is orthogonal to the constants. We expand each ϕj in the eigenbasis of K as follows: ϕj (vj ) =

∞ X

aj,ι ηι (vj ) .

(2.7)

ι=1

Then evidently kϕj k22 = Z

SN

∞ X ι=1

|aj,ι |2 , and for j 6= k,

ϕ∗j (vj )ϕk (vk )dσN

=

∞ X

ι,ι′ =1

a∗j,ι ak,ι′ hηj,ι , Kηk,ι′ iH

=

∞ X

ι,=1

κι a∗j,ι ak,ι .

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CCL December 6, 2018

Therefore, when f is given by (2.1) and (2.7),     N ∞ ∞ N N N X X X X X X 2 2 ∗ 2 (1 − κι )  aj,ι  |aj,ι | + κι |aj,ι | + κι ℜaj,ι ak,ι  = kf k2 = j=1 ι=1 ι=1 j=1 j=1 j6=k,j,k=1

Observe that for ι = 1, same way, we see that

N X

PN j

aj,1

j

N Z √ X = 3 j

SN

ϕj (vj )e1 · v)dσN =

Z

SN

(2.8)

f (~v )e1 · v)dσN = 0. In the

aj,1 = 0 for ι = 2, 3, 4. Hence we have the identity

2 N X 2 2 aj,ι . κι |aj,ι | + (1 − κι ) kf k2 = j=1 ι=5 ι=1 j=1 ∞ X

N X

∞ X

(2.9)

5N −3 By Lemma 2.2, min{1 − κι } = 3(N , and thus we have the lower bound. For the upper bound, −1)3 we go back to (2.8) for ι > 4, and write (using κι = −1/(N − 1) for ι = 1, 2, 3, 4)   N ∞ N 4 X N X X X X N  (2.10) |aj,ι |2 + κι ℜa∗j,ι ak,ι  |aj,ι |2 + kf k22 = N −1 ι=5

ι=1 j=1

Then since

N X

j6=k,j,k=1 ∞ X ι=5

ℜa∗j,ι ak,ι ≤ (N − 1)

 N X  |aj,ι |2 + κι j=1

Since for all N ≥ 3,

N N −1

N X j=1

N X

j6=k,j,k=1

 ≤ 1+

5N −3 3(N −1)2

j=1

|aj,ι |2 , using Lemma 2.2 once more yields 

ℜa∗j,ι ak,ι  

j6=k,j,k=1

≤



5N − 3 1+ 3(N − 1)2

X ∞ X N ι=5 j=1

|aj,ι |2 .

, this gives us the upper bound.

There is another type of quantitative chaos estimate that we need. For any functions ξ on R3 such that ξ(vk ) ∈ L2 (σN ) for some (and hence all) k, consider the conditional expectation E{ξ(vk ) | vj = v} = Kξ(v)

(2.11)

for j 6= k. If the coordinate functions were exactly independent, this would simply be the expectation of ξ(vk ), which is a finite constant. It turns out that when ξ(vk ) is a polynomial in |vk |2 , the conditional expectation is at least bounded – not only on SN , which is trivial, but the bound is independent of N . Here is one such estimate: 2.5 LEMMA. For ψ(v) = |v|8 , there is a constant C < ∞ such that kKψk∞ ≤ C for all N . Proof of Lemma 2.5. The formula (1.26) gives us r 8 Z N − v2 1 ~y − p Kψ(v) = E{ψ(v1 ) | vN = v} = v dσN −1 N − 1 N (N − 1) SN−1 ! 4 Z  |v|8 N − v2 8 7 |~y | dσN −1 − 4 ≤ 2 N −1 N (N − 1)4 SN−1

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CCL December 6, 2018

It is evident that

R

SN−1

|~y |8 dσN −1 is bounded uniformly in N , and in fact, lim

Z

N →∞ SN−1

|~y |8 dσN −1 = (2π/3)−3/2

Z

R3

|y|8 e−3|y|

2 /2

.

In the remainder of this section we collect the other estimates of this type that we need. Their proofs, whihch are more intricate but still largely computational, are presented in Appendix A. 2.6 LEMMA. There is a finite constant C such that for all N > 3 and all v such that v = vN for some ~v ∈ SN , C (2.12) |E{|v1 |4 | vN = v} − S(v)| ≤ N where N 2 + |v|4 − 2N |v|2 . (2.13) S(v) = (N − 1)2 2.7 LEMMA. There is a finite constant C such that for all N > 3 and all (v, w) such that (v, w) = (vN −1 , vN ) for some ~v ∈ SN , |E{|v1 |4 | (vN −1 , vN ) = (v, w)} − S(v, w)| ≤ where S(v, w) =

2.2

C N

N 2 + |v|4 + |v|4 + 2N |v|2 + 2N |w|2 + 2|v|2 |w|2 . (N − 2)2

(2.14)

(2.15)

The operators W (α) and P (α) .

It is useful to introduce some notation before beginning the work of producing lower bounds on b N,α . Define the self adjoint operator P (α) by DN,α (f, f ), and hence ∆ P

(α)

α/2 N  1 X N 2 − (1 + |vk |2 )N Pk . = N (N − 1)2

(2.16)

k=1



α/2 N 2 − (1 + |vk |2 )N Notice that for each k, both Pk and the multiplication operator are com(N − 1)2 muting and self adjoint, so that P (α) itself is indeed self adjoint, and even non-negative. Also, since each Pk is a projection, we have α/2  2 N Z 1 X N − (1 + |vk |2 )N |Pk f |2 dσN = hf, P (α) f iL2 (SN ,σN ) . N (N − 1)2 SN

(2.17)

k=1

Next define the function W (α) by W

(α)

α/2 N  1 X N 2 − (1 + |vk |2 )N . = N (N − 1)2 k=1

(2.18)

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CCL December 6, 2018

Then

α/2  2 Z N Z 1 X N − (1 + |vk |2 )N 2 W (α) f 2 dσN , f dσ = 2 N (N − 1) SN SN

(2.19)

k=1

and we can write:

DN,α (f, f ) :=

Z

SN

W (α) f 2 dσN − hf, P (α) f iL2 (SN ,σN ) .

(2.20)

Equivalently, by the computations just below (1.33),  2 α/2 N Z 1 X N − (1 + |vk |2 )N DN,α (f, f ) = [f − Pk f ]2 dσN . N (N − 1)2 SN

(2.21)

k=1

from which is follows that DN,α (f, f ) ≥ 0 for all f since for each k, (1 + |vk |2 ) ≤ N . It follows that DN,α (f, f ) = 0 if and only if f − Pk f = 0 almost everywhere for each k, and then in this case kf k22

− hf, P

(0)

f iL2 (SN )

N Z 1 X |f − Pk f |2 dσN = 0 . = DN,0 (f, f ) = N SN k=1

Evidently, P (0) is a contraction, and 1 is an eigenvalue of multiplicity one, and the eigenspace is spanned by the constant function 1 [6]. This proves: 2.8 LEMMA. For all N ≥ 2 and all α ∈ [0, 2], and all non-zero f ∈ L2 (SN ) that are orthogonal to the constants, DN,α (f, f ) > 0. We use the following lemma proved in [5, Lemma 3.5]: 2.9 LEMMA. For all 0 < α < 2 and all x > −1, (1 + x)α/2 ≥ 1 + α2 x − (1 − α2 )x2 .

(2.22)

2.10 LEMMA. For all N , all 0 < α ≤ 2, and for all ~v ∈ SN , 1 − (1 − α2 )

N ((N − 1)2 + 1) − (N − 1)4

α 2

1 N +1 + (1 − α2 ) ≤ W (α) (~v ) ≤ 2 (N − 1) (N − 1)3  α/2 1 1− . (2.23) (N − 1)2

Furthermore, for all ~v ∈ SN , W (0) (~v ) = 1

and

W (2) (~v ) = 1 −

1 . (N − 1)2

(2.24)

 N N  α 1 X 1 1 X N 2 − (1 + |vk |2 )N 2 Proof. Because . Then since x 7→ x 2 =1− |vk | = 1, 2 2 N N (N − 1) (N − 1) k=1 k=1 is concave on R+ for 0 ≤ α ≤ 2, Jensen’s inequality yields the upper bound. To prove the lower bound, we use the inequality (2.22). Then writing N 2 − (1 + |vk |2 )N N (1 − |vk |2 ) − 1 = 1 + , (N − 1)2 (N − 1)2

(2.25)

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CCL December 6, 2018

and applying (2.22) and the identity W

(α)

(~v ) ≥ 1 −

α 2

PN

2 k=1 vk

= N , we find that

2 N  X 1 N (1 − |vk |2 ) − 1 α 1 . − (1 − 2 ) (N − 1)2 N (N − 1)2 k=1

Expanding the square on the right and applying W (α) (~v ) ≥ 1 − The maximum of PN

α 2 (N

N X k=1

PN

2 k=1 vk

= N twice more, we find

N 2 X 1 N α 1−N α ) ) − (1 − − (1 − |vk |4 . 2 (N − 1)4 2 (N − 1)4 − 1)2

(2.26)

k=1

|vk |4 on SN is no greater than the maximum of the convex function

2 k=1 xk on the convex set of (x1 , . . . , xN ) satisfying

0 ≤ xj ≤ N − 1

for all

j = 1, . . . , N

and

N X

xj = N .

(2.27)

j=1

The extreme points are obtained by permuting the coordinates of (N − 1, 1, 0, . . . , 0). Evaluating the sum at such a point yields the stated bound, The final statement is obvious. 2.11 Remark. Lemma 2.10 shows that for large N , W

(α)

   α 1 1 (~v ) ≥ 1 − 1 − +O . 2 N N2

(2.28)

The fact that the coefficient of 1/N is no less than −1 is essential for the result that we shall prove. We are particularly concerned with the case α = 1, and shall provide all the details in this case only. For α = 1, the lower bound simplifies further to W (1) (~v ) ≥ CN := 1 −

1 2N

1 1 1 1 − 12 + 21 − 21 2 3 −1 (N − 1) (N − 1) (N − 1)4

(2.29)

It is easily seen that for all N ≥ 2, CN increases as N increases. For small N , we have the explicit values 64 21 and C4 = . C3 = 32 81 We now turn to P (α) . By (2.25), for each k for all vk , 

N 2 − (1 + |vk |2 )N wN,α (vk ) := (N − 1)2 where xN (vk ) = Note that −1 ≤ xN (vk ) ≤

1 . Then N −1

α/2

= (1 + xN (vk ))α/2 ,

N 1 − |vk |2 . N − 1 (N − 1)2

wN,α (vk ) ≤



N N −1

α/2

(2.30)

(2.31)

(2.32)

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CCL December 6, 2018

and by (2.30) and the bounds from Lemma 2.9, 1 + α2 x + ( α2 − 1)x2 ≤ (1 + x)α/2 ≤ 1 + α2 x, |wN,α (vk ) − 1| ≤

1 3N N2 2 + |v | + |vk |4 k N − 1 (N − 1)2 (N − 1)4

(2.33)

for all α ∈ [0, 2], where we have made estimates to simplify the right had side. Thus, while W (α) is only constant for α = 0, 2, it is nearly constant for all α ∈ (0, 2) when N is large. However, its range, and hence the spectrum of the multiplication operator specified by W (α) , is a closed interval of positive length. 2.12 LEMMA. For all α ∈ [0, 2], the null space of P (α) is independent of α. If h belongs to the null space of P (0) , then Pk h = 0 for each k = 1, . . . N . For all α ∈ [0, 2], the closure of the range of P (α) is the subspace AN of L2 (SN ) defined in Definition 2.1. Proof. Since P (α) ≥ 0, h belongs to the null space of P (α) is and only if hh, P (α) hi = 0. But 0 = hh, P

(α)

N Z 1 X wN,α (vk )|Pk h|2 dσ . hi = N SN k=1

Since wN,α (vk ) ≥ 0 almost everywhere, it must be the case that |Pk h|2 vanishes identically. Thus is h is in the null space of P (α) , Pk h = 0 for each k, and h is in the null space of P (0) . Conversely if h is in the null space of P (0) , then Pk h = 0 for each k, and then clearly P (α) h = 0. Since each P (α) ≥ 0, the closure of its range is the orthogonal complement of its null space. Since the null space does not depend on α, neither does the range. Evidently, AN is the closure of the range of P (0) .

2.3

bN,0 The spectrum of L

b N,α for Already in our paper [4] we have proved results that specify the exact spectral gap of L α = 0. This case is especially ammenable for several reasons. First, since W (0) = 1, b N,0 f = f − P (0) f , L

and hence the problem is to determine the spectrum of P (0) . Second, there is an orthonormal basis of L2 (SN ) consisting of eigenfunctions of P (0 . This is the case because each Pk is an average of rotations, so the finite dimensional spaces spanned by spherical harmonics of given maximal degree are invariant under P (0) , and therefore one can study the spectrum of P (0) by studying the eigenvalue equation P (0) f = λf . This is the approach we took in our previous work. However, this approach cannot work even for α = 2, the next simplest case: In this case, P (2) has an interval of continuous spectrum, as we shall see below. Therefore, we now give another argument that b N,0 that does extend to α = 2 at least. determines the spectral gap of L 2.13 LEMMA. For all N ≥ 3,

b N,0 = 1 − 3N − 1 = 1 − 1 + O ∆ 3(N − 1)2 N



1 N2



.

(2.34)

and the second largest eigenvalue of P (0) , µ(0) , is given by µ(0) =

3N − 1 . 3(N − 1)2

(2.35)

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CCL December 6, 2018

Proof. The range of P (0) is AN , and it suffices to determine the spectrum of P (0) as an operator N X ϕj (vj ) ∈ AN , we compute on AN . For f (~v ) = j=1

    N N N X X X 1 ϕk (vk ) + P (0)  ϕj (vj ) = Kϕj (vk ) . N j=1

By this computation, with T :

j6=k,j=1

k=1

LN

j=1 HN,j

→ AN defined as in Theorem 2.3,

T −1 P (0) T = M(0)

(0)

where M(0) = [Mi,j ] is the N × N block matrix operator on (0)

Mi,j = I

if i = j

(0)

LN

and Mi,j = K

j=1 HN,j

given by

if i 6= j .

Note that M(0) is unitarily equivalent to the block matrix operator in        

I + (N − 1)K 0 0 ··· 0 I −K 0 ··· 0 0 I − K ··· .. .. .. .. . . . . 0 0 0 0

0 0 0 .. . I −K



LN

j=1 HN,j

given by

      

It follows that either λ is an eigenvalue of I + (N − 1)K or else λ is an eigenvalue of I − K. Thus, the second largest eigenvalue of M(0) , and hence P (0) , is either 1 + (N − 1)κ, where κ is the second largest eigenvalue of K, or else 1 − κ where κ is the least eigenvalue of K. From the information on the spectrum of K provided in Lemma 2.2, one immediately deduces (2.35), and then (2.34) follws directly.

2.4

bN,α , α ∈ (0, 2] The spectrum of L

After α = 0, the next simplest case is α = 2 since then at least W (2) is constant; as we have seen W (2) = 1 − (N − 1)−2 . It follows that 1 is an eigenfunction for P (2) with eigenvalue 1 − (N − 1)−2 , b N,2 . and it spans the eigenspace. That is, 1 spans the null space of L

b N,2 > 0. 2.14 LEMMA. For all N ≥ 3, ∆

Proof. The range of P (2) is AN , and as with α = 0, M(2) := T −1 P (2) T has a simple block matrix structure:     N N N X X X 1 ϕj (vj ) ϕj (vj ) = P (2)  wN,2 (vk )Pk  N j=1 j=1 k=1   N N X X 1 wN,2 (vk ) ϕk (vk ) + Kϕj (vk ) = N k=1

j6=k,j=1

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CCL December 6, 2018

By Theorem 2.3, M(2) = T −1 P (2) T = W(2) (I + C),  wN,2 (v1 ) 0  0 w  N,2 (v2 ) 1  W(2) = . .. N ···  0

bf I is the identity on

LN

where ··· ··· .. .

0 0 .. .

0

wN,N (vN )

···

j=1 HN,j ,

and C is  0  K  1   ... C= N   K K

given by K 0

K K

··· ··· ···

··· K K

··· ··· .. . 0 K

K K .. .



   ,  



    ,   K  0

Since M(2) and P (2) are similar, they have the same spectrum, and in particular,, the spectrum of M(2) is real. (This is also evident from the identity M(2) = W(2) (I + C), and the fact that for bounded operators A and B on any Hilbert space, AB and BA have the same spectrum.) Since the range of wN,2 is [0, (N − 1)−1 ], this interval is the spectrum of W(2) . Note that C, and hence W(2) C is compact. By Weyl’s lemma, the essential spectrum of T P (2) T −1 , and hence of P (2) , is the essential spectrum of W(2) , which is the interval [0, (N − 1)−1 ]. Hence any spectrum of P (2) in (N − 1)−1 , 1 − (N − 1)−2 ) consists of isolated eigenvalues, and the isolated eigenvlaues can only accumulate at a point in [0, (N − 1)−1 ]. In particular, 1− (N − 1)−2 cannot be an accumulation point, and hence P (2) has a spetral gap below its top eigenvalue 1 − (N − 1)−2 . This proves that b N,2 > 0 for all N ≥ 3. ∆ For α ∈ (0, 2), W (α) is not constant – although for large N it is nearly constant. This means bN,α simply by determining the spectrum that for such α, one cannot determine the spectrum of L (α) b of P , and moreover, AN is not invariant under LN,α . However, there is a simple comparisson b N,α that we seek. that one can make between DN,α and DN,2 that provides the bound on ∆

2.15 LEMMA. For all N ≥ 3, and all α ∈ [0, 2], 1−α/2  b N,2 > 0 . b N,α ≥ N − 1 ∆ ∆ N

Proof. By (2.32), for all f and k, and all α ∈ (0, 2),  α/2     2 N − 1 1−α/2 N 2 − (1 + |vk |2 )N N − (1 + |vk |2 )N 2 [f − Pk f ] ≥ [f − Pk f ]2 (N − 1)2 N (N − 1)2 It follows immediately that

DN,α (f, f ) ≥ and then that b N,α ≥ ∆





N −1 N

N −1 N

1−α/2

1−α/2

DN,2 (f, f ) ,

b N,2 > 0 . ∆

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CCL December 6, 2018

At this point, we have proved the first part of Theorem 1.4, and all that remains is to prove the second part.

3

b N,1 for large N . A sharper lower bound on ∆

In this section we obtain lower bounds on DN,1 (f, f ) for f orthogonal to the constants that become sharper and sharper as N increases. To keep the computations simple, we do this explicitly for α = 1, though the method applies to all α ∈ (0, 2). We shall prove the following, which is simply a specialization of Theorem 1.4: 3.1 THEOREM. There is a constant C independent of N such that whenever f is orthogonal to the constants,   C 1 − 3/2 kf k22 . DN,1 (f, f ) ≥ 1 − (3.1) N N The bound (3.1) is meaningless for N such that the right side is negative. However, no matter what C is, there is an N0 ∈ N such that for all N ≥ N0 , the right side is positive. From that point on, we have what we need for our induction. The rest of this section is devoted to the proof of Theorem 3.1

3.1

The trial function decomposition

We begin by specifying a trial function decomposition that we shall use. Let AN be the subspace of L2 (σN ) defined in Definition 2.1. For any f ∈ L2 (SN ) orthogonal to the constants, define p and h to be the orthogonal projections of f onto AN and A⊥ N respectively. Then since 1 ∈ AN , h is orthogonal to the constant, and then p = f − h is orthogonal to the constants. By Lemma 2.12, h is the component of f in the null space of P (α) for each α ∈ [0, 2], and hence hf, P (α) f i = hp, P (α) pi . Z

SN

W

(α) 2

f dσ − hf, P

(α)

f iL2 (SN ) =

Z

SN

(3.2)

W (α) f 2 dσ − hp, P (α) piL2 (SN ) .

(3.3)

Since p ∈ AN , there is are N function ϕ1 , . . . , ϕN of a single variable such that ϕj (vj ) ∈ L2 (SN ) for each j, and N X φj (vj ) . (3.4) p(~v ) = j=1

Since

N Z X j=1

SN

φj (vj )dσN =

Z

SN

p(~v )dσN = 0, replacing each φj (vj ) by φj (vj ) −

R

SN

φj (vj )dσN does

not change the sum in (3.4), and we may assume without loss of generality that for each j = 1, . . . , N , φj (vj ) is orthogonal to the constants. Now notice that for any j, the three components of vj and |vj |2 − 1 are a set of 4 mutually orthogonal functions, and each is orthogonal to the constants. We make a further decomposition of φj (vj ) as follows:

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CCL December 6, 2018

3.2 DEFINITION. Let p be a function given by a sum of the form (3.4) where for each j, φj (vj ) is orthogonal to the constants, define ψj (vj ) to be the orthogonal projection of φj (vj ) onto the span of the three components of vj and |vj |2 − 1, and define ϕj (vj ) = φj (vj ) − ψj (vj ) . Then each ψj has the form ψj (vj ) = aj · vj + bj (|vj |2 − 1) (3.5) for some aj ∈ R3 and some bj ∈ R. Define g(~v ) =

N X

ϕj (vj ) and

s(~v ) =

N X

ψj (vj )

(3.6)

j=1

j=1

Finally the trial function decomposition of any f ∈ L2 (σN ) that is orthogonal to the constants is given by f =g+s+h (3.7) where h is the component of f in the null space of P (α) , p is the component of f in the closure of the range of P (α) , and p = g + s is the decomposition of p defined in (3.7). 3.3 Remark. It is easy to see that when p is symmetric under coordinate permutations, one can take the functions ϕj in (3.4) to be all the same. In particular, each ψj has the form ψj (vj ) = a · vj + b(|vj |2 − 1) for some fixed a ∈ R3 and b ∈ R. Then     N N X X vj  + b  (|vj |2 − 1) = 0 s(~v ) = a ·  j=1

j=1

on account of the constraints on the momentum and energy. Hence when p is symmetric s = 0. Hence in this case, the trial function decomposition simplifies to f = g + h, as in [5].

Each of the components g, s and h have their own special properties that we shall repeatedly use. P (1) A very useful feature of g(~v ) = N j=1 ϕj (vj ) is that, on account of Lemma 2.2, for j 6= k and 2 any function ξ such that ξ(vk ) ∈ L (σN ) |hϕj (vj ), ξ(vk )i| ≤

C kϕj k2 kξk2 . N2

In particular, the different ϕj (vj ) are nearly orthogonal. P 4 (2) A very useful feature of s(~v ) = N ), and for a j=1 ψj (vj ) is that each ψj (vj ) belongs to L (σN R constant C independent of N , kψj k4 ≤ kψj k2 . This is essentially because the integrals ST |v|2m dσN are bounded uniformly in N for each m. In particular, if we wish to estimate the L2 (σN ) norm of |vk |2 ψk (vk ), we can apply Schwarz’s inequality to bound this by Ckψk k4 , and then, changing C, to Ckψk k2 . For the components of g we cannot do this, and would have to use the point-wise bound |vk |2 ≤ N − 1 instead, By Lemma 2.2, for j 6= k, Z Z 1 ψj (vj )ψk (vk )dσN = − ψj (vj )ψk (vj )dσN , N − 1 ST ST which is a weaker “almost orthogonality” property than we have for the components of g. NonetheP 2 less, we shall still have that for large N , ksk22 ≈ N j=1 kψj k2 , as follows from Theorem 2.3.

(3) A very useful feature of h(~v ) is that Pk h = 0 for each k, and in particular, P (1) h = 0.

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CCL December 6, 2018

3.2

Lower bound on DN,1 (f, f )

For α = 1, the lower bound (2.26) simplifies to

Define

f (1) (~v ) := 1 + W (1) (~v ) ≥ W eN,1 (f, f ) = D

Z

SN

N X 1 N 1 − |vk |4 . (N − 1)3 2 (N − 1)4

(3.8)

k=1

f (1) f 2 dσN − hf, P (1) f i . W

(3.9)

eN,1 (f, f ). By (3.8), DN,1 (f, f ) ≥ D Now let f be orthogonal to the constants, and let f = g+s+h be the trial function decomposition of f as specified above. This notation will be used throughout this subsection. Note that eN,1 (f, f ) = D eN,1 (g, g) + D eN,1 (s, s) + D eN,1 (h, h) D eN,1 (g, h) + 2D eN,1 (s, h) + 2D eN,1 (g, s) + 2D

The next lemma says that g, s and h are almost mutually orthogonal with respect to the inner eN,1 , and hence the last three terms above make a negligible contribution. This product given by D decouples the contributions of g, s and h, which may then be analyzed separately, taking advantage of their different helpful properties. 3.4 LEMMA. There is a constant C independent of N such that for any f ∈ L2 (σN ) that is orthogonal to the constants, if f = g + s + h is the trial function decomposition as specified above, then C 2|DN,1 (g, h)| + 2|DN,1 (s, h)| + 2|DN,1 (g, s)| ≤ 3/2 kf k22 . N (1) Proof. Since P h = 0, and since g and h are orthogonal, Z Z 1 N (1) f e |vk |4 ghdσN W ghdσN = − DN,1 (g, h) = 2 (N − 1)4 SN SN N Z X 1 N = − |vk |4 ϕk (vk )hdσN (3.10) 2 (N − 1)4 S N k=1 N Z X N 1 |vk |4 ϕj (vj )hdσN dσN (3.11) − 2 (N − 1)4 SN j6=k

The integral in (3.10) vanishes since Pk h = 0. To estimate the integral in (3.11), Z (wN,1 (vk ) − 1)ϕj (~v )h(~v )dσN ≤ k(wN,1 (vk ) − 1)ϕj (vj )k2 khk2 . SN

Next consider the integral in (3.11). It will be convenient to introduce the notation ξ(x) = x8 for the eighth power. Then, with this definition, the Schwarz inequality, and then application of the K operator, Z 1/2 Z 8 2 4 |vk | ϕ (vj )dσ |vk | ϕj (vj )hdσ ≤ khk2 SN

= khk2

Z

SN

SN

Kξ(vj )ϕ2j (vj )dσ

1/2

.

(3.12)

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CCL December 6, 2018

By Lemma 2.5 below, there is a constant C so that, independent of N , kKξk∞ ≤ C, Therefore, Z 4 ≤ Ckhk2 kϕj k2 . h|v | ϕ (v )dσ (3.13) j j k SN

Using this in(3.11) gives us

Z N (N − 1)4 SN

N X k=1

|vk |4

!

and then since Lemma 2.3 gives us N X j=1

kϕk2 ≤



  N X N ghdσ ≤ Ckhk2  kϕj k2  , (N − 1)3

(3.14)

j=1

5N − 3 1− 3(N − 1)2

−1/2

√

N kgk2 ,

we have that the left hand side of (3.14) is bounded by NC3/2 kgk2 khk2 for a constant C independent of N . So far we have shown that DN,α (g, h) ≥ −CN −3/2 . The same reasoning then shows that DN,α (g, −h) ≥ −CN −3/2 . Hence |DN,α (g, h)| ≤ CN −3/2 . In the same way we show that DN,α (s, h) ≥ −CN −3/2 . Then as before, we also have DN,α (s, −h) ≥ −CN −3/2 , and we conclude that |DN,α (s, h)| ≤ CN −3/2 . Finally, we consider DN,α (s, g). This time we must also estimate hs, P (1) gi. Because the span of {ηj,ℓ (vj )} is invariant under P (0) , hs, P (1) gi = hs, (P (1) − P (0) )gi

(3.15)

1 N

(3.16)

=

+

+

1 N

N X

h(wN,1 (vk ) − 1)ψk (vk ), ϕk (vk )i

k=1 N X N X k=1 j6=k

h(wN,1 (vk ) − 1)ψj (vj ), ϕk (vk )i

(3.17)

N N 1 XX h(wN,1 (vk ) − 1)ψj (vj ), Pk ϕj (vℓ )i N

(3.18)

k=1 j6=k

+

N N 1 XX h(wN,1 (vk ) − 1)ψk (vk ), Pk ϕj (vℓ )i N

(3.19)

1 N

(3.20)

k=1 j6=k

+

N X

j6=k,k6=ℓ,ℓ6=j

We estimate (3.16) as follows: N 1 X h(wN,1 (vk ) − 1)ψk (vk ), ϕk (vk )i ≤ N k=1

h(wN,1 (vk ) − 1)ψj (vj ), Pk ϕℓ (vℓ )i

N N 1 X C X kψk k2 kϕk k2 kwN,1 − 1k4 kψk k4 kϕk k2 ≤ 2 N N k=1

C ≤ ≤ 2 (ksk22 + kgk22 ) . N

k=1

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CCL December 6, 2018

Since Pk ψj (vj ) = − N 1−1 ψj (vk ), the argument used to estimate (3.16) shows that the absolute value of the sum in (3.17) is bounded above by N N C XX C kψj k2 kϕk k2 ≤ 2 (ksk22 + kgk22 ) , N3 N k=1 j6=k

as we found for (3.16). Since for k 6= j, kPk ϕj k2 ≤ CN −2 kϕj k2 , the argument used to estimate (3.16) shows that the absolute value of the sum in (3.18) is bounded above by N N C XX C kψj k2 kϕj k2 ≤ 3 (ksk22 + kgk22 ) , N4 N k=1 j6=k

even better than the previous bounds. For the terms in (3.19) we have the bound N N C XX C kψk k2 kϕj k2 ≤ 3 (ksk22 + kgk22 ) . 4 N N k=1 j6=k

Finally, for the terms in (3.20), we have the bound C N5

N X

j6=k,k6=ℓ,ℓ6=j

Altogether, we have |hs, P (1) gi| ≥

kψj k2 kϕℓ k2 ≤

C 2 N 2 (ksk2

C (ksk22 + kgk22 ) . N3

+ kgk22 ).

eN,1 (g, g) and D eN,1 (s, s). We now turn to the estimation of D

3.5 LEMMA. There is a constant C independent of N ≥ 3 such that for all g and s as above, hg, P

(1)

1/2  2 N Z 1 X C N − (1 + |vk |2 )N gi ≤ ϕ2k (vk )dσ + 2 kgk22 . 2 N (N − 1) N SN

(3.21)

k=1

and

1/2  2 N Z 1 X N − (1 + |vk |2 )N hs, P si ≤ ψk2 (vk )dσ . 2 N (N − 1) k=1 SN P Proof. Note first of all that Pk g = ϕk (vk ) + j6=k Kϕj (vk ), and thus (1)

hg, P (1) gi = =

+

N Z 1 X N k=1 SN N Z 1 X N SN



N 2 − (1 + |vk |2 )N (N − 1)2

1/2

|Pk g|2 dσN

1/2 N 2 − (1 + |vk |2 )N ϕ2k (vk )dσN (N − 1)2 k=1 1/2 Z  2 N 2 XX N − (1 + |vk |2 )N ϕk (vk )Kϕj (vk )dσN N (N − 1)2 SN 

(3.22)

(3.23)

k=1 j6=k

+

1/2 Z  2 N N − (1 + |vk |2 )N 1 XXX Kϕj (vk )Kϕℓ (vk )dσN N (N − 1)2 SN k=1 j6=k ℓ6=k

(3.24)

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CCL December 6, 2018

By the Schwarz inequality and (2.5), the sum of integrals in (3.23) is bounded above by N C XX C kϕj k2 kϕk k2 ≤ 2 kgk22 . 3 N N k=1 j6=k

Similarly, by (2.5) and Lemma 2.3, the sum of integrals in (3.24) is bounded above by N C X C kϕj k2 kϕk k2 ≤ 3 kgk22 . 4 N N j,k=1

Using the two bounds we have just derived on (3.23) and (3.24) respectively, yields (3.21). 1 X ψj (vk ). By the constriant equations To prove (3.22), note that Pk s(~v ) = ψk (vk ) − N −1 j6=k

specifying SN , proved.

N X

ψj (vk ) = 0.

Therefore, Pk s(~v ) =

j=1

N −2 ψk (vk ). N −1

Since

N −2 < 1, (3.22) is N −1

3.6 LEMMA. There is a constant C such that for all N and all g and s as above, N X k=1

and

N X k=1

4 2

|vk | g dσ ≤

4 2

|vk | s dσ ≤

N Z X k=1

N Z X k=1

SN

SN

ϕk (vk )2 |vk |4 dσ + CN kgk22 ,

(3.25)

ψk (vk )2 |vk |4 dσ + CN ksk22 ,

(3.26)

Proof. Z

N X

SN k=1

|vk |4 g2 dσ = =

N X

i,j,k=1 N Z X

ϕi (vi )ϕj (vj )|vk |4 dσ

SN

ϕk (vk )2 |vk |4 dσ

k=1 N XZ X

+ 2

k=1 j6=k

+

N XZ X k=1 j6=k

+

X

SN

SN

Z

ϕj (vj )ϕk (vk )|vk |4 dσ

ϕj (vj )2 |vk |4 dσ

i6=j,j6=k,k6=i SN

ϕi (vi )ϕj (vj )|vk |4 dσ

(3.27)

(3.28)

(3.29) (3.30)

By Lemma 3.7, Lemma 3.8 and Lemma 3.9 the terms in (3.27), (3.28) and (3.29) add up to no more than CN kgk22 , which proves (3.25). The same argument using the same lemmas proves (3.26).

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CCL December 6, 2018

3.7 LEMMA. There is a constant C such that for all N and all g and s as above, 2

N XZ X k=1 j6=k

and 2

SN

ϕj (vj )ϕk (vk )|vk |4 dσ ≤ CN kgk22 .

(3.31)

ψj (vj )ψk (vk )|vk |4 dσ ≤ Cksk22 .

(3.32)

N XZ X k=1 j6=k

SN

Proof. For j 6= k, using the pointwise bound |vk |4 ≤ (N − 1)2 and then (2.5), Z (5N − 3)(N − 1)2 ϕj (vj )ϕk (vk )|vk |4 dσ ≤ (N − 1)2 kKϕj k2 kϕk k2 ≤ kϕj k2 kϕk k2 . 3(N − 1)3 SN Then by Theorem 2.3, (3.31) follows. Next, Z ψj (vj )ψk (vk )|vk |4 dσ ≤ kKψj k2 k|vk |4 ψk k2 ≤ SN

(3.33)

1 C kψj k2 Ckψk k4 ≤ kψj k2 Ckψk k2 . (3.34) N −1 N

Then by Theorem 2.3 again, (3.32) follows. 3.8 LEMMA. There is a constant C such that for all N and all g and s as above, N XZ X

ϕj (vj )2 |vk |4 dσN ≤ CN kgk22 .

(3.35)

N XZ X

ψj (vj )2 |vk |4 dσN ≤ CN ksk22 .

(3.36)

k=1 j6=k

and

k=1 j6=k

SN

SN

Proof. By Lemma 2.6, there is a finite constant C independent of N such that N XZ X k=1 j6=k

SN

2

4

ϕj (vj ) |vk | dσN ≤ N

N Z X j=1

SN

N

X N 2 + |vj |4 − 2N |vj |2 2 kϕj k22 . ϕj (vj )dσN + CN 2 (N − 1) j=1

Then by Lemma 2.3, (3.35) follows. The same analysis yields (3.36) . 3.9 LEMMA. There is a constant C such that for all N and all g and s as above, Z X ϕi (vi )ϕj (vj )|vk |4 dσ ≤ Ckgk22 .

(3.37)

i6=j,j6=k,k6=i SN

and

X

Z

i6=j,j6=k,k6=i SN

ψi (vi )ψj (vj )|vk |4 dσ ≤ Cksk22 .

(3.38)

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CCL December 6, 2018

Proof. By Lemma 2.7, there is a finite constant C independent of N (but changing from line to line) such that Z X ϕi (vi )ϕj (vj )|vk |4 dσ ≤ CN 2 kϕi k2 kϕj k2 + i6=j,j6=k,k6=i SN

(N − 2)

XZ i6=j

SN

N 2 + |vi |4 + |vj |4 + 2N |vi |2 + 2N |vj |2 + 2|vi |2 |vj |2 ϕi (vi )ϕj (vj )dσN . (N − 2)2

Note that for i 6= j, Z N 2 + |vi |4 + |vj |4 + 2N |vi |2 + 2N |vj |2 C ϕi (vi )ϕj (vj )dσN ≤ kϕi k2 kϕj k2 N −2 N SN since in each term we may either replace ϕi by Kϕi or ϕj by Kϕj , and this gives a factor of CN −2 . P Then by Lemma 2.3, i6=j kϕi k2 kϕj k2 ≤ CN kgk22 . The remaining terms must be handled differently. For j = 1, . . . , N , let ξj denote the function ξj (v) = |vj |2 ϕj (vj ), and note that ξj is orthogonal to the constants. Therefore, Z 1 |vi |2 |vj |2 ϕi (vi )ϕj (vj )dσN = hξi , Kξj i N −2 SN N − 2 1 1 ≤ kξi k2 kξj k2 ≤ Ckϕi k2 kϕj k2 N −2N −1 P Then by Lemma 2.3, i6=j kϕi k2 kϕj k2 ≤ CN kgk22 , and (3.37) follows. Next, Z X ψi (vi )ψj (vj )|vk |4 dσ ≤ CN 2 kψi k2 kψj k2 + i6=j,j6=k,k6=i SN

(N − 2)

XZ i6=j

SN

N 2 + |vi |4 + |vj |4 + 2N |vi |2 + 2N |vj |2 + 2|vi |2 |vj |2 ψi (vi )ψj (vj )dσN . (N − 2)2

The main term is Z X N2 X N2 N2 X ψi (vi )ψj (vj )dσN = hψ1 , Kψj i = kψi k2 kψj k2 , N −2 N −2 (N − 2)(N − 1) SN i6=j

i6=j

i6=j

and simple estimates show that all remaining terms are smaller.

3.3

eN,1 (g, g) and D eN,1 (s, s) Lower bounds on D

eN,1 (g, g) and D eN,1 (s, s). We first define a quadratic form F on We are now ready to estimate D L2 (σN ) as follows: For all functions r in L2 (σN ), define F(r, r) :=

N 1 2 (N − 1)4

Z

N X

|vk |4 r 2 dσ + hr, P (1) ri

(3.39)

eN,1 (s, s) ≥ ksk22 − F(s, s) . D

(3.40)

SN k=1

3.10 LEMMA. For all g and s as above, eN,1 (g, g) ≥ kgk22 − F(g, g) D

and

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CCL December 6, 2018

Proof. This is immediate from (2.26), (3.9) and the definition of F. 3.11 LEMMA. There is a finite constant C independent of N such that for all g and s as above, with F defined by (3.39)   C 1 + 2 kgk22 . (3.41) F(g, g) ≤ N N

and

F(s, s) ≤



C 1 + 2 N N



ksk22 .

(3.42)

Proof. By Lemma 3.5 and Lemma 3.6, # "Z  1/2 N Z N X 1 X C 1 N2 N 2 − (1 + |vk |2 )N 4 F(g, g) ≤ + |vk | ϕ(vk )2 dσ + 2 kgk22 2 4 N (N − 1) 2 (N − 1) N SN SN k=1

k=1

Define y := Then 0 ≤ y ≤ N/(N − 1), and F(g, g) ≤ where w(y) =



N |v1 |2 . (N − 1)2 Z

w(y)ϕ2 (v1 )dσN

(3.43)

SN

N −y N −1

1/2

1 + y2 2

p Simple calculations show that w(y) ≤ N/(N − 1) for all 0 ≤ y ≤ N/(N − 1), and in fact, for N ≥ 7, w(y) is monotone decreasing on this interval. Then (3.43) becomes   p 1 1 2 kϕk22 . F(g, g) ≤ N/(N − 1)kϕk2 ≤ 1 + 2N −1 Now (3.41) follows directly from Lemma 2.3. The proof of (3.41) is the same.

3.4

Proof of Theorem 3.1

Proof of Theorem 3.1. By Lemma 3.4, eN,1 (f, f ) ≥ D eN,1 (g, g) + D eN,1 (s, s) + D eN,1 (h, h) − CN −3/2 kf k2 s . DN,1 (f, f ) ≥ D

By Lemma 3.10 and Lemma 3.11,

eN,1 (g, g) + D eN,1 (s, s) ≥ D



C 1 − 2 1− N N



(kgk22 + khk22 ) .

Since P (1) h = 0, the lower bound (2.28) yields   C 1 e − 2 khk22 . DN,1 (h, h) ≥ 1 − 2N N

adding the estimates completes the proof since kf k22 = kgk22 + ksk22 + khk22 .

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CCL December 6, 2018

Appendix A

Some computational proofs

Proof of Lemma 2.6. By (1.26), E{|v1 |4 | vN = v} =  Z η 4 (v)|~y |4 + SN−1

 |v|4 η 2 (v) η 2 (v) 2 2 |~y · v| + +2 |~y | dσN −1 , (A.1) (N − 1)2 (N − 1)4 (N − 1)2

where η 2 (v, w) = Define MN :=

R

SN−2

N − |v|2 − |v|2 /(N − 1) , N −1

|~y |4 dσN −2 which is bounded uniformly in N : lim

Z

4

N →∞ SN−1

−3/2

|~y | dσN −1 = (2π/3)

Z

R3

|y|4 e−3|y|

2 /2

.

Then the right hand side of (A.1) becomes MN η 4 (v) +

|v|4 η 2 (v) 1 2 2 η (v)|v| + + 2 3(N − 1)2 (N − 1)4 (N − 1)2

(A.2)

Note that for some constant C independent of N , |v|4 η 2 (v) C 1 2 2 η (v)|v| + + 2 ≤ . 2 4 2 3(N − 1) (N − 1) (N − 1) N

(A.3)

Next, η 4 (v) = +

N 2 + |v|4 − 2N |v|2 (N − 1)2 |v|4 (N − |v|2 )|v|2 +2 . 4 (N − 1) (N − 1)3

(A.4)

Again, for some constant C independent of N , |v|4 (N − |v|2 )|v|2 C . + 2 ≤ (N − 1)4 (N − 1)3 N Proof of Lemma 2.7. By a simple adaptation of (1.26), E{|v1 |4 | (vN −1 , vN ) = (v, w)} =   Z β 2 (v, w) |v + w|4 β 2 (v, w) 2 4 4 2 β (v, w)|~y | + |~y · (v + w)| + +2 |~y | dσN −2 , (A.5) (N − 2)2 (N − 2)4 (N − 2)2 SN−2 where β 2 (v, w) =

N − |v|2 − |w|2 − |v + w|2 /(N − 2) , N −2

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CCL December 6, 2018

which is non-negative on the allowed values for (v, w). Note that β 2 (v, w) ≤ N/(N − 2). Define R MN := SN−2 |~y |4 dσN −2 which is bounded uniformly in N : Z Z 2 4 −3/2 |y|4 e−3|y| /2 . |~y | dσN −2 = (2π/3) lim N →∞ SN−2

R3

Then the right hand side of (A.5) becomes MN β 4 (v, w) +

1 |v + w|4 β 2 (v, w) 2 2 β (v, w)|v + w| + + 2 3(N − 2)2 (N − 2)4 (N − 2)2

(A.6)

Note that for some constant C independent of N , |v + w|4 β 2 (v, w) C 1 2 2 β (v, w)|v + w| + + 2 ≤ . 2 4 2 3(N − 2) (N − 2) (N − 2) N

(A.7)

Next, β 4 (v, w) = +

N 2 + |v|4 + |w|4 + 2N |v|2 + 2N |w|2 + 2|v|2 |w|2 (N − 2)2 |v + w|4 (N − |v|2 − |w|2 )|v + w|2 + 2 . (N − 2)4 (N − 2)3

(A.8)

Again, for some constant C independent of N , (N − |v|2 − |w|2 )|v + w|2 C |v + w|4 + 2 ≤ . (N − 2)4 (N − 2)3 N

Appendix B B.1

b N,2 Quantitiative estimates on ∆

An explicit bound for N ≥ 4

By (2.32), P (α) defined in (2.16) satisfies 0≤P

(α)

≤



N N −1

α/2

P (0)

(B.1) (0)

for all α ∈ [0, 2]. As we have seen, the second largest eigenvalue of P (0) , denoted µN , is given by (0)

µN =

3N − 1 . 3(N − 1)2

It follows from (B.1) and (B.3) that for all f orthogonal to the constants, α/2  3N − 1 N (α) kf k22 , hf, P f i ≤ N −1 3(N − 1)2 for all α ∈ [0, 2]. For α = 2, we have hf, P (2) f i ≤

N (3N − 1) kf k22 . Note that 3(N − 1)3

1 1 1 5 2 N (3N − 1) = + + , 3 2 3(N − 1) N − 1 3 (N − 1) 3 (N − 1)3

(B.2)

(B.3)

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CCL December 6, 2018

1 which evidently decreases monotonically as N increases. Next, since W (2) (~v ) = 1 − (N −1) 2 , we have that for all f orthogonal to the constants   1 1 2 b N,2 f i = hf, (W (2) − P (2) )f i ≥ 1 − 1 − 8 − kf k22 . (B.4) − hf, L N − 1 3 (N − 1)2 3 (N − 1)3

b 3,2 f i ≥ − 1 kf k2 . But already for N = 4, it For N = 3, this yields only the useless bound −hf, L 2 4 yields b 4,2 f i ≥ 28 kf k2 . −hf, L 2 81 Since the right hand side of (B.4) increases as N increases, this, together with the comparison from Lemma 2.15, proves: B.1 THEOREM. For all N ≥ 4,

and for all α ∈ (0, 2), b N,α ≥ ∆

b N,2 ≥ 1 − ∆ 

N −1 N

1 1 1 8 2 − − >0, 2 N − 1 3 (N − 1) 3 (N − 1)3

1−α/2 

1 1 8 2 1 − − 1− 2 N − 1 3 (N − 1) 3 (N − 1)3



>0.

At this point, the only estimate we lack for a fully quantitative result is a quantitative estimate b 3,2 . on ∆

B.2

An explicit bound for N = 3

b 3,2 = 3 − ν3 where By what has been explained earlier, ∆ 4 n ν3 = sup hf, P (2) f iL2 (S3 ) : kf k2 = 1 ,

hf, 1iL2 (SN ) = 0

o

,

(B.5)

b 3,2 > 0, or, what is the same ν3 < 3 . and by Lemma 2.14, ∆ 4 b 3,2 ≥ 1 . Therefore, we need only consider the possibility that ν3 > 1 , If ν3 ≤ 12 , then evidently ∆ 4 2 and as we have seen, in in this case ν3 is an eigenvalue of P (2) , and necessarily ν3 < 34 . In seeking the second largest eigenvalue of P (2) , we need only consider functions f of the form f (~v ) =

N X

ϕ(vj )

(B.6)

j=1

or f (~v ) = ϕ(v1 ) − ϕ(v2 ) ,

(B.7)

where in the second case we have taken advantage of the the symmetry of P (2) to assume without loss of generality that f is antisymmetric under interchange of v1 and v2 . B.2 LEMMA. For N = 3, the largest eigenvalue of P (2) on the orthogonal complement of the b 3,2 ≥ 0.015, or else the gap eigenfucntion symmetric sector is no greater than 0.735. Thus, either ∆ is symmetric.

37

CCL December 6, 2018

Proof of Lemma B.2. For later use, we begin the proof for N ≥ 3, and specialize to N = 3 later. Let f be given by (B.7), where we may assume that ϕ is orthogonal to the constants. Then 1 1 wN,2 (v1 )(1 − K)ϕ(v1 ) − wN,2 (v2 )(1 − K)ϕ(v2 ) = λ(ϕ(v1 ) − ϕ(v2 ) . N N Multiplying by ϕ(v1 ) and integrating, Z 1 wN,2 (v1 )|(1 − K)ϕ(v1 )|2 = λhϕ, (1 − K)ϕi . N SN

(B.8)

By (2.30), wN,2 (v) = Using (B.9) in (B.8) yields

N N − |vk |2 . N − 1 (N − 1)2

(B.9)

1 1 hϕ, (1 − K)2 ϕi − h(1 − K)ϕ, |v|2 (1 − K)ϕi = λhϕ, (1 − K)ϕi . (B.10) N −1 (N − 1)2 √ Now write 1 − Kϕ = ψ + ζ where ψ is orthogonal to the constants, the three components of v and |v|2 . Then ζ is an eigenvector of K with eigenvalue −1/(N − 1), and hence 1 1 hϕ, (1 − K)2 ϕi = hψ, (1 − K)ψi , N −1 N −1

and

(B.11)

√ √ N −2 k|v|ζk22 h(1 − K)ϕ, |v|2 (1 − K)ϕi = h 1 − Kψ, |v|2 1 − Kψi + (N − 1)2 √ √ N −1 k|v|ζk22 (B.12) − 2k|v| 1 − Kψk2 N − 2   √ N −2 1 √ ≥ 1− k|v|ζk22 , h 1 − Kψ, |v|2 1 − Kψi + (1 − t) t (N − 1)2 for all t > 0, where we have used the arithmetic-geometric mean inequality. At this point we specialize to N = 3, and carry out some explicit computations that could be done for all N ≥ 3, but are then more cumbersome. P Write ζ = 4j=1 aj ηj (v) where, as before, η(j)v) = e·v for j = 1, 2, 3, and where η4 (v) = |v|2 −1. One readily computes that 4 X |aj |2 k|v|ηj k22 (B.13) k|v|ζk22 = and that

Z

S3

|v1 |4 dσ3 =

5 and 4

Z

j=1

7 |v1 |6 dσ3 = . From here ti follows that 4 S3

k|v|ηj k22 =

5 4

for

j = 1, 2, 3

and k|v|η4 k22 = 1 .

√ √ Using this in (B.13) finally yields k|v|ζk22 ≥ kζk22 , and evidently h 1 − Kψ, |v|2 1 − Kψi ≤ √ 3k 1 − Kψk22 . Therefore, for 0 < t < 1, we have from (B.12) that   √ 1 2 h(1 − K)ϕ, |v| (1 − K)ϕi ≥ 3 1 − k 1 − Kψk22 + (1 − λ)kζk22 . λ

38

CCL December 6, 2018

Using this estimate in (B.10) yields   √ 3 1 1 √ 2 2 k 1 − Kψk2 + kζk2 − 3 1 − k 1 − Kψk22 − (1 − t)kζk22 ≥ λ(kψk22 + kζk22 ) . 2 4 t The second most negative eigenvalue of K for N = 3 is − 83 ; see [6, Section 8], where this √ 2 eigenvalue is denoted κ1,2 . It follows that k 1 − Kψk22 ≤ 11 8 kψk2 . Therefore,     1 3 3 −3+ − 1 + t kζk22 ≥ λ(kψk22 + kζk22 ) . kψk22 + 16 t 4 Choosing t = 0.985, we have that 0.735(kψk22 + kζk22 ) ≥ λ(kψk22 + kζk22 ). The remaining task is to bound the second largest eigenvalue of P (2) in the symmetric sector. We begin considering general N ≥ 3 and shall specialize to N = 3 later. Let f be given as in (B.6). Then P (2) f = λf becomes N N X 1 X ϕ(vj ) . wN,2 (vk )(ϕ(vk ) + (N − 1)Kϕ(vk )) = λ N

(B.14)

j=1

k=1

By Theorem 2.3,

1 wN,2 (v)(ϕ(v) + (N − 1)Kϕ(v)) = λϕ(v) . (B.15) N we have that Therefore, multiplying both sides of (B.15) by ϕ(v) and integrating, we obtain Z 1 ϕ(v1 )wN,2 (v1 )(ϕ(v1 ) + (N − 1)Kϕ(v1 ))dσN = λkϕk22 . (B.16) N SN By (2.30), (B.16) becomes Z 1 hϕ, Kϕi − ϕ(v1 )|v1 |2 Kϕ(v1 ))dσN = N − 1 SN   Z 1 1 |v1 |2 ϕ2 (v1 ))dσN . (B.17) kϕk22 + λ− N −1 (N − 1)2 SN

Define an operator M by M φ(v) = |v|2 (1 + (N − 1)K)φ(v) . Then (B.17) becomes 

1 λ− N −1



=

hϕ, Kϕi − (N − 1)−2 hϕ, M ϕi , kϕk22

Thus λ − (N − 1)−1 can be computed by computing the supremum of the right hand side as ϕ ranges over functions that are orthogonal to 1, the three components of v and |v|2 . Note that M commutes with rotations so the different angular momentum sectors are mutually orthogonal, and can be considered separately. In each sector, by the usual recursion relations for orthogonal polynomials, the matrix representing M in the eigenbasis of K is tri-diagonal and explicitly computable, and the bounds proved in [6, Section 8] can be used to limit the number of angular momentum sectors that need to be considered. Hence one could obtain explicit bounds this way.

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39

References [1] P. Caputo, Spectral gap inequalities in product spaces with conservation laws, In Stochastic Analysis on Large Interacting Systems T. Funaki and H. Osada, eds., Math. Soc. Japan, Tokyo. (2004), 5388. [2] P. Caputo, On the spectral gap of the Kac walk and other binary collision processes, ALEA Lat. Am. J. Probab. Math. Stat., 4, (2008), 205-222. [3] E. A. Carlen, M. C. Carvalho and M. Loss: Many body aspects of approach to equilibrium, in Journes Equations aux derivees partielles, Nantes, 5-9 Juin 2000. [4] E. A. Carlen, M. C. Carvalho and M. Loss: Determination of the spectral gap for Kac’s master equation and related stochastic evolution, Acta Mathematica 191, (2003) 1-54 [5] E. A. Carlen, M. C. Carvalho and M. Loss: Spectral gap for the Kac model with hard sphere collisions. J. Funct. Anal. 266 (2014), no. 3, 1787-1832 [6] E. A. Carlen, J. Geronimo and M. Loss: Determination of the spectral gap in the Kac model for physical momentum and energy conserving collisions, SIAM Jour. Mathematical Analysis 40, no. 1, (2008) 327-364 [7] W. Feller, An introduction to Probability Theory and its Applications, Volume I, Wiley and sons, New York, 1950 [8] G. Giroux and R. Ferland, Global spectral gap for DirichletKac random motions. J. Stat. Phys. 132, (2008) 561-567. [9] A. Grigo, A. Khanin and D. Sz´ asz, Mixing rates of particle systems with energy exchange, Nonlinearity 25 (2012), 2349-2376. [10] E. Janvresse, Spectral Gap for Kac’s model of Boltzmann Equation, Annals. of Prob., 29 (2001) 288-304. [11] M. Kac, Foundations of kinetic theory, Proc. 3rd Berkeley symp. Math. Stat. Prob., J. Neyman, ed. Univ. of California, vol 3, (1956) 171–197. [12] M. Kac Probability and Related Topics in Physical Sciences, Interscience Publ. LTD., London, New York (1959) ¨ die Entwicklung einer Function von beliebig vielen Variablen nach [13] F.G. Mehler Uber Laplaschen Functionen h¨ oherer Ordnungn. Crelle’s Journal 66 (1866), 161–176. [14] S. Mischler and C. Mouhot, Kac’s Program in Kinetic Theory, Invent. Math., 193 (2013) 1-147. [15] M. Sasada, On the spectral gap of the Kac walk and other binary collision processes on ddimensional lattice, In Symmetries, Integrable Systems and Representations, K. Iohara et al., eds. Springer Proc. Math. Stat. 40, 543560. Springer, Heidelberg, 2013.

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