Concentration of mass on the Schatten classes
O. Guédon
a,∗, G. Paouris
b,1aUniversité Pierre et Marie Curie, Équipe d’Analyse Fonctionnelle, Institut de Mathématiques de Jussieu, boîte 186, 4, Place Jussieu, 75252 Paris cedex 05, France
bLaboratoire d’Analyse et de Mathématiques Appliquées, Université de Marne-la-Vallée, 5, Bd Descartes, Champs-sur-Marne, 77454 Marne-la-Vallée cedex 2, France
Received 20 May 2005; received in revised form 7 November 2005; accepted 6 January 2006 Available online 7 July 2006
Abstract
Let 1p∞andB(Spn)be the unit ball of the Schatten trace class of matrices onCnor onRn, normalized to have Lebesgue measure equal to one. We prove that
λ
T ∈B Snp
: THS n c1t
exp
−c2tnkp
for everyt1, wherekp=min{2,1+p/2},c1, c2>1 are universal constants andλis the Lebesgue measure. This concentration of mass inside a ball of radius proportional tonfollows from an almost constant behaviour of theLqnorms (with respect to the Lebesgue measure onB(Spn)) of the Hilbert–Schmidt operator norm ofT. The same concentration result holds for every classical ensembles of matrices like real symmetric matrices, Hermitian matrices, symplectic matrices or antisymmetric Hermitian matrices.
The result is sharp whenp=1 andp2.
©2006 Elsevier Masson SAS. All rights reserved.
Résumé
Pour tout 1p∞, soitB(Spn)la boule unité des classes de Schatten à trace, normalisée pour avoir un volume égal à 1. Nous prouvons que pour toutt1,
λ
T ∈B Snp
: THS n c1t
exp
−c2tnkp
où kp=min{2,1+p/2}, c1, c2>1 sont des constantes universelles etλ désigne la mesure de Lebesgue. Ce phénomène de concentration du volume à l’intérieur d’une boule euclidienne de rayon proportionnel àndécoule d’un comportement presque constant des normesLq (par rapport à la mesure de Lebesgue sur B(Spn)) de la norme Hilbert–Schmidt d’un opérateurT. Le même phénomène de concentration est valable pour tous les ensembles classiques de matrices : les matrices symétriques réelles, les matrices hermitiennes, les matrices symplectiques ou encore les matrices hermitiennes antisymétriques. De plus, le résultat est optimal lorsquep=1 etp2.
©2006 Elsevier Masson SAS. All rights reserved.
* Corresponding author.
E-mail addresses:guedon@math.jussieu.fr (O. Guédon), grigoris_paouris@yahoo.co.uk (G. Paouris).
1 G. Paouris acknowledges the financial support provided through the European Community’s Human Potential Programme under contract MRTN-CT-2004-511953 (Phenomena in High Dimensions).
0246-0203/$ – see front matter ©2006 Elsevier Masson SAS. All rights reserved.
doi:10.1016/j.anihpb.2006.01.002
MSC:46B07; 47B10
Keywords:Schatten–Von Neumann classes; Isotropic measure; Kahane Khinchine inequalities; Concentration inequalities
0. Introduction
LetK be a symmetric convex body inRd, and let E∈Ell= {E an ellipsoid, |E| =ωd}where|E| denotes the volume ofEandωdis the volume of the Euclidean unit ball inRd. For everyq1 consider theLqnormalized norm of the Euclidean norm associated toE:
Iq(K,E)= 1
|K|1+q/d
K
xqEdλ(x) 1/q
,
whereλis the Lebesgue measure onRd.
Conjecture (C).There exist a universal constantC >0and a functionφ(d)tending to infinity withd, which satisfy the following: for every centrally symmetric convex body K in Rd there is an ellipsoid E∈Ell such that, for all qφ(d),
1
|K|
K
xqEdλ(x) 1/q
C
1
|K|
K
x2Edλ(x) 1/2
.
Equivalently, this means thatIq(K,E)CI2(K,E).
The goal of this paper is to verify conjecture (C) for the case of Schatten trace classes of matrices and their subspaces which are of particular interest. Observe that for everyα >0 and everyq1,Iq(αK,E)=Iq(K,E)and for every linear transformationT ∈SLn(R),
Iq
T (K), B2d
=Iq K,
T∗T−1
B2d
. (1)
By (1), it is clear that in order to check the conjecture (C), our goal is to find a position of the symmetric convex body Kfor which the desired estimates can be obtained withE=B2d.
The isotropic constantLKof a symmetric convex bodyKis defined by LK=min I2(K,E)/√
d; Ean ellipsoid, |E| =ωd
. (2)
It is well known that the hyperplane conjecture is equivalent to the fact thatLKis uniformly bounded for all symmetric convex bodies (see [13]). We say that a symmetric convex bodyKis in isotropic position when the minimum at Eq. (2) is attained by the Euclidean unit ballB2d. It is natural to expect that the conjecture (C) will be satisfied for a convex body in isotropic position.
By Borell’s inequality [6], we know that for any symmetric convex bodyK∈Rd, Iq
K, B2d
CqI2 K, B2d
.
Moreover, ifKis a symmetric convex body ofRd in isotropic position, Alesker [1] proved that for everyq2, Iq
K, B2d
C√
qI2 K, B2d
.
For general symmetric convex bodies, this is the best known estimate but it is very far from the conjecture. If we translate this result in terms of concentration of measure, it says that for everyt1,
λ x∈K: x2Ct I2
K, B2d
exp
−t2 whereKdenotes the homothetic image ofKof volume 1.
It turns out that the claimed conjecture is equivalent to some type of concentration of mass inequality for con- vex bodies. This is part of Theorem 1.1 in [15], where other equivalent formulations of the question are studied.
In the particular case of isotropic 1-unconditional convex bodies K (1-unconditional means that for every point
x =(x1, . . . , xd)∈K and(ε1, . . . , εd)∈ {−1,1}d, the point(ε1x1, . . . , εdxd)belongs to K), Bobkov and Nazarov [5] proved a sharp concentration inequality for the Euclidean norm with respect to the normalized Lebesgue measure onK:
Theorem.[5]There exists a universal constantC >0such that for every isotropic1-unconditional convex bodyK inRd, ifKdenotes the homothetic image ofKof volume1, then for everyt1,
λ x∈K: x2Ct I2
K, B2d
exp
−t√ d
.
But it is well known (see [13]) that sup{LK, K1-unconditional convex bodies}is finite. So that in this case, there exists a constantc >0 such thatI2(K, B2d)c√
dfor alld1. The previous result states that the mass of an isotropic 1-unconditional convex body is highly concentrated inside a Euclidean ball of radiusc√
d. This shows that, for this class of bodies, the conjecture (C) is true withφ(d)proportional to√
d. (See Lemma 8.)
Our purpose is to show that for allp1, the unit balls of the Schatten trace classes of matricesB(Spn)of every clas- sical ensemble of matrices (real matrices, complex matrices, real symmetric matrices, complex Hermitian matrices, symplectic matrices or antisymmetric Hermitian matrices) share the same property, where the appropriate Euclidean norm is the Hilbert–Shmidt norm of an operator (which is the norm associated to the unit ballB(S2n)). As observed in [11], in the case of real and complex matrices, these unit balls are in isotropic position. However, we believe that this is not the case for every classical ensemble of matrices. For example, it is known that the situation may be different for real symmetric matrices as indicated in the paper [4]. We prove here:
Theorem 1.There exist universal constantsc1, c2>0such that for everyn1, for every1p∞and2q c1min{n2, n1+p/2},
1
|B(Spn)|
B(Snp)
TqHSdλ(T ) 1/q
c2 1
|B(Spn)|
B(Spn)
T2HSdλ(T ) 1/2
.
The same result holds if we replaceB(Spn)withB(Spn)∩E whereE is the subspace defining one of the classical ensembles of matrices.
More generally, there are universal constantsC, C>0such that for everyn1, for every1p∞, for every 2qc1min{n2, n1+p/2}and every spaceEof classical ensemble of matrices(real matrices, complex matrices, real symmetric matrices, complex Hermitian matrices, symplectic matrices or antisymmetric Hermitian matrices),
CnIq
B Spn
∩E, B S2n
∩E Cn.
Remark.Theorem 1 states that the conjecture (C) is valid forK=B(Spn)∩Ewithφ(dn) min{dn, dn1/2+p/4}where dn n2is the corresponding dimension of the convex bodyK(see Section 1).
The next theorem is a translation of the previous result in terms of concentration of measure.
Theorem 2.There exist universal constants c1, c2>0 such that if 1p∞ and B(Spn)is the unit ball of the Schatten trace class of matrices, normalized to have Lebesgue measure one, then
λ T ∈B Spn
: THSc1nt
exp
−c2t nkp for everyt1, wherekp=min{2,1+p/2}.
The same result holds if we replaceB(Spn)withB(Spn)∩E, the unit ball ofSpn∩Enormalized to have Lebesgue measure one, whereEis the subspace defining one of the classical ensembles of matrices.
The proof relies on the same method for general matrices in the real or complex case and for different classical ensembles of matrices. The computation of the integrals Iq will use a method similar to the one used in [11] to prove the uniform boundedness of the isotropic constants of the unit balls ofSnp(for real or complex matrices). For
all particular ensemble of matrices, classical change of variables reduces the computation of the integralsIq to the computation of integrals overBpnwith different type of densities of the form
1i<jn
xia−xjab n
i=1
|xi|c,
wherea, bare positive integers andcis a nonnegative integer. A usual trick in convexity makes the computation of the integralsIq tractable. Surprisingly, some evaluations are possible and give a very sharp comparison betweenIq
andI2wheneverq is not too large (depending onnandp).
The organization of the paper is as follows. In Section 1, we indicate that the desired estimate of Theorem 1 can be reduced to an estimate of integrals overRn. In Section 2 we make the computations that complete the proof of Theorem 1. Finally in Section 3 we explain why Theorem 1 and Theorem 2 are equivalent.
0.1. Notations
Forn1, we work onRnorCnwhich are equipped with their canonical Euclidean structure. The spacesMn(R) andMn(C)ofn×nmatrices with real or complex entries are equipped with the associated Euclidean structure defined byT2HS=tr(TT )for anyT ∈Mn, embedded inRn2 or R2n2 endowed with the Lebesgue measure denoted by dT. For everyx∈Rn, let
xp= n
i=1
|xi|p 1/p
for 1p <∞andx∞=max
in|xi|.
For any matrixT ∈Mn(R)or T ∈Mn(C), let s(T )=(s1(T ), . . . , sn(T ))be the non-increasing rearrangement of the singular values ofT i.e. the eigenvalues of(T∗T )1/2. For every 1p∞, we defineσp(T )= s(T )p. LetSpn be the Schatten trace class of matrices on then-dimensional Euclidean space equipped with the normσp, and denote
B Spn
= T ∈Mn; σp(T )1 .
Of course,σ2(T )= THS= TB(S2n). We denote by|A|the Lebesgue measure of any Borel setA⊆Spn, when it is finite and when 0<|A|<+∞, we denoteA˜=λAwhereλis chosen such that| ˜A| =1.
Whenever we writea bwe mean that there exist universal constantsc1, c2>0 such thatc1abc2a. 1. Reduction to integrals overRn
A functionF:Rn→Ris said to be a symmetric function if for every permutationπon{1, . . . , n}, F (x1, . . . , xn)=F (xπ(1), . . . , xπ(n)).
The functionF:Rn→Ris said to be positively homogeneous of degreek∈Rif for everyα∈R∗+, F (αx1, . . . , αxn)=αkF (x1, . . . , xn).
Following the book of Mehta [12], it is natural to consider the spaces of real self-adjoint matrices, complex Hermitian matrices, antisymmetric Hermitian matrices, symplectic matrices and we refer to this book for the definitions of these ensembles. We will denote byEthe subspace ofMnthat defines these ensemble (whatever is the field of the considered manifolds, real or complex). In each of these cases, it is well known [12] that there existsa, bpositive integer, a nonnegative integercand a constantcndepending only onn, a, b, csuch that for any symmetric continuous functionF:Rn→R,
T∈B(Spn)∩E
F
s1(T ), . . . , sn(T )
dT =cn
Bnp
F (x)fa,b,c(x)dx (3)
where
fa,b,c(x)=
1j <in
xia−xjab× n
i=1
|xi|c and Bpn= x∈Rn, xp1 .
A similar type of change of variables is discussed and explained in the work of Saint-Raymond [17]. He proved that the same result holds when we consider the space E of complex (or real) matrices [17] with a=2, b=2, c=1 (ora=2,b=1,c=0). Sincefa,b,c is positively homogeneous of degreeabn(n−1)/2+cn, definedn by dn=abn(n−1)/2+(c+1)n, then in each of these examples,dn=dimE. We will use the following lemma which is exactly in the same spirit as Lemma 1 in [11].
Lemma 1.For any symmetric continuous functionF:Rn→R, positively homogeneous of degreek, for everyp1, one has
T∈B(Snp)∩E
F
s1(T ), . . . , sn(T )
dT = cn
(1+(dn+k)/p)
Rn
F (x)e−xppfa,b,c(x)dx.
Proof. We recall here a proof of this result for reason of completeness. It follows from Fubini’s theorem:
Rn
F (x)e−xppfa,b,c(x)dx=
Rn
F (x)fa,b,c(x) +∞
xpp
e−tdt
dx
= +∞
0
e−t
{x,xpt1/p}
F (x)fa,b,c(x)dx
dt
=
1+dn+k p
Bpn
F (x)fa,b,c(x)dx
sinceF is positively homogeneous of degreekandfa,b,cof degreedn−n. We conclude using (3). 2
Next, we need some estimates about the volume of the setsB(Spn)∩E. Exact computations of these volumes are done in the paper of Saint-Raymond [17] for the unit balls of matrices ofSpn with real entries or complex entries. In the next proposition, we give a proof of an estimate of(|B(Snp)∩E|)1/dnvalid for every ensembles of matrices that we consider in this paper. Recall that in each case, the dimension ofE,dn, is equivalent ton2(up to constants depending only onaandb). We will need the following classical estimate about the operator norm of a random Gaussian operator inE.
Lemma 2.LetEbe one of the classical ensemble considered in this paper. LetGE be a Gaussian vector inE. There exists a constantc2such that for every integern,
EGEB(S∞n)c√ n.
Proof. By Chevet’s inequality [9], it is well known thatEGMn(R)B(Sn∞)2√
2nwhenGMn(R)is an×nGaussian matrix with independent real GaussianN(0,1)entries. In the case of complex entries, we decompose the matrix as sum of its real and imaginary part and use triangle inequality to get thatEGMn(C)B(Sn∞)4√
2n. In the case of quaternionic entries, we separate it in a sum of four terms and get thatEGMn(H)B(Sn∞)8√
2n. We can also refer to the survey [10] where better constants are obtained. Now, letΣ be one of the field R,C, orHthen for every ensemble E, it is clear that if we denote byGE a Gaussian vector in EthenGE=PE(GMn(Σ ))wherePE is the orthogonal projection onE. Since an orthogonal projection has norm 1, the estimate of the lemma follows. 2
We are now able to state the proposition.
Proposition 3.For everyp1, up to constants depending only onaandb, B
Spn
∩E1/dn dn−1/4−1/2p n−1/2−1/p. (4)
Proof. LetKbe a symmetric convex body inRdandB2dthe Euclidean ball inRdthen by Urysohn’s inequality [16], we get that
|K|
|B2d| 1/d
c
√dEGKo
whereGis a Gaussian vector inRd with independentN(0,1)entries,Ko= {y∈Rd,∀x∈K,x, y1}denotes the polar ofK andcis a universal constant. Since|B2d|1/d 1/√
d, we get that for every symmetric convex body K⊂Rd,
|K|1/dc2EGKo
d .
Moreover, by reverse Santaló inequality [7] there exist universal constantc1such that |B2d|
|K| 1/d
c1 |Ko|
|B2d| 1/d
.
Using again Urysohn’s inequality forKoand the fact that|B2d|1/d 1/√
d, there is a universal constantc2such that c2
EGK |K|1/d.
Therefore, we get that for every symmetric convex bodyK⊂Rd, c2
EGK |K|1/dc2EGKo
d .
We will apply this inequality in the situation we are interested in. LetK=B(Spn)∩EandB2d=B2dn=B(S2n)∩E.
Remark that the norm inS∞n corresponds to the operator norm. By Hölder’s inequality, for any matrixT ∈E, for everyp1,
TB(Spn)n1/pTB(S∞n) and T(B(Snp)∩E)oTB(Sp∗n )
where 1/p+1/p∗=1. LetGE be a Gaussian vector inEthen, by Lemma 2, there exists a constantc >0 such that for everyn,EGEB(S∞n)c√
n. Therefore,
EGEKo=EGE(B(Spn)∩E)oEGEB(Sp∗n )n1/p∗EGEB(S∞n)cn3/2−1/p where 1/p+1/p∗=1, and
EGEK=EGEB(Spn)n1/pEGEB(S∞n)cn1/2+1/p. Sincedn n2, this proves the estimate(4). 2
We are now able to explain how the estimates of theLq norms in Theorem 1 is related to the estimate of integrals with respect to the measureMa,b,c,pdefined by
Ma,b,c,p(f )=
Rn
f (x)fa,b,c(x)e−xppdx.
Lemma 4.LetEbe any of the classical space of matrices on the real or complex field(the space of real, complex matrices, real self-adjoint matrices, complex Hermitian matrices, antisymmetric Hermitian matrices or symplectic matrices), and leta,bandcthe corresponding integers associate to the ensemble defined by(3)and letdn=dimE= abn(n−1)/2+(c+1)n. Then for every1p <∞, if we defineMa,b,c,pas the measure of densityfa,b,c(x)e−xpp, for every1qdn,
Iq
B Spn
∩E, B S2n
∩E
n1/2−1/p
Ma,b,c,p(xq2) Ma,b,c,p(1)
1/q
where the constants in this equivalence depend only ona,b,c.
Proof. For notational convenience, we will writeIq(B(Spn)∩E)instead of Iq
B Spn
∩E, B Sn2
∩E .
Using Lemma 1 for the functionsF =1 andF (x)= xq2, we get B
Spn
∩E=
T∈B(Spn)∩E
dT = cn
(1+dn/p)
Rn
fa,b,c(x)e−xppdx,
and Iq
B Spn
∩E
=B Spn
∩E−1/dn B
Spn
∩E|−1
T∈B(Snp)∩E
n
i=1
σi(T )2q/2
dT 1/q
=B Spn
∩E−1/dnB Spn
∩E−1 cn
(1+(dn+q)/p)
Rn
xq2fa,b,c(x)e−xppdx 1/q
=B Spn
∩E−1/dn
cn(1+dn/p) cn(1+(dn+q)/p)
1/q
Rnxq2fa,b,c(x)e−xppdx
Rnfa,b,c(x)e−xppdx 1/q
.
We have proved in(4)that|B(Spn)∩E|1/dn n−1/2−1/p and since (1+dn/p)
(1+(dn+q)/p) 1/q
dn−1/p
we get the claimed estimate (becausedn n2). 2
For the measuresMa,b,c,p, we can provide exact computations of some integrals which are presented in the next section. The result which is of interest for our problem is as follows and will be proved in the next section.
Lemma 5.For every integers a,b,c, there are positive constants c1 and c2 such that for everyn∈N, for every 1p2, define the measuresMa,b,c,pas above, then, for every2qc1n1+p/2,
Ma,b,c,p(xq2) Ma,b,c,p(1)
1/q
c2n1/2+1/p.
Proof of Theorem 1. It is well known (see [13]) that for every symmetric convex bodyK⊂Rd, 1
|K|1+2/d
K
x22dλ(x) 1/2
1
|B2d|1+2/d
B2d
x22dλ(x) 1/2
C√
d
whereB2dis the Euclidean ball associated to · 2andCis a universal constant.
Whenp2, the assertion of the theorem follows from the following observation:
For every symmetric convex bodyK⊂Rd, Id K, B2d
max
x∈Kx2. (5)
Indeed, Id
K, B2d
=
K
xd2dx 1/d
Sd−1
K
x, θddxdσ (θ ) 1/d
.
Sinceθ→(
K|x, θ|ddx)1/d defines a norm onRd, it is well known using concentration of measure on the sphere [14] that
Id(K,E)
max
θ∈Sd−1
K
x, θddx 1/d
and a classical use of Brunn–Minkowski inequality (see [6]) proves that for every symmetric convex bodyK,
K
x, θddx 1/d
cmax
x∈K
x, θ which finishes the proof of(5).
LetEbe one of the classical ensemble of matrices,dn=dimEwheredn=abn(n−1)/2+(c+1)nanda,b,c are defined in(3). LetK=B(Spn)∩EandE=B(S2n)∩E=B2dn.
Ifp2 then by(5),Idn(K,E) maxT∈KTHS and by(4), we conclude that there are constantsC, C>0 (depending only ona,b,c) such thatCnI2(K,E)Idn(K,E)Cnwhich proves the assertion of the theorem.
If 1p2, then by Lemma 4 1
|K|1+q/dn
K
TqHSdλ(T ) 1/q
n1/2−1/p
Ma,b,c,p(xq2) Ma,b,c,p(1)
1/q
.
By Lemma 5, we get that there existc1andc2depending only ona,b,csuch that for everyn, for every 2q c1n1+p/2,
C dn
1
|K|1+q/dn
K
TqHSdλ(T ) 1/q
c2n.
This ends the proof of the theorem. 2
2. Computation of integrals with respect to the measuresMa,b,c,p
First we need some precise results concerning the computation of the integrals of positively homogeneous functions with respect to the measuresMa,b,c,p. We shall prove a generalization of Lemma 3 in [11], which is based on a method developed by Aomoto [3] for the study of Jacobi polynomials associated to Selberg integrals. We start by recalling the notations. Leta, b∈N∗,c∈Nandfa,b,c:Rn→Rdefined by
fa,b,c(x)=fa,b,c(x1, . . . , xn)=
1j <in
xia−xjab× n
i=1
|xi|c.
Denote byMp,a,b,c=Mpthe measure inRnwith density fa,b,c,p(x1, . . . , xn)=fa,b,c(x)exp
− n
i=1
|x|pi
i.e. for everyf:Rn→R, Mp(f )=
Rn
f (x)fa,b,c(x)e−ni=1|x|pi dx.
Denote by dn =abn(n−1)/2+(c+1)n. Observe that fa,b,c is a positively homogeneous function of order abn(n−1)/2+cn. Note also that for every function g:Rn→Rpositively homogeneous of order s >−n such that
Sn−1|g(u)|dσ (u) <+∞,for everyp1,
Rn|g(x)|exp(−xpp)dxis finite.
Lemma 6.Letβ0,s−dn−βandf:Rn→R+a positively homogeneous function of ordersandC1onRn\{0}.
Then
(β+c+1)Mp
n
i=1
|xi|β
f (x)
=pMp
xββ++ppf (x)
−Mp n
i=1
|xi|βxi∂f
∂xi
(x)
−abMp
n
i=1
k=i
|xi|βxia xia−xka
f (x)
.
Proof. For fixedx2, . . . , xnwe defineg1:R→R+by
g1(x1)= |x1|βx1f (x1, x2, . . . , xn)fa,b,c(x1, x2, . . . , xn)exp
−|x1|p exp
− n
j=2
|xj|p
.
Sincef is positively homogeneous, limx1→+∞g1(x1)=0 and limx1→−∞g1(x1)=0. Without loss of generality, we can assume thatx2< x3<· · ·< xnthereforeg1 is continuous onR\ {0, x2, . . . , xn}and has finite left and right limits at each of these points of discontinuity. Observe also thatg1(x)=0 for everyx∈ {0, x2, . . . , xn}. Integration by parts proves the following equality, for almost all(x2, . . . , xn)with respect to the Lebesgue measure inRn−1:
∞
−∞
|x1|βf (x)exp
− n
j=1
|xj|p
fa,b,c(x)dx1
= − 1 β+1
∞
−∞
|x1|βx1 ∂
∂x1
f (x)exp
− n
j=1
|xj|p
fa,b,c(x)
dx1.
For every x1∈ {/ x2, . . . , xn}, we can assume that there exists 2mn−1 such thatx1∈(xm, xm+1)(the cases x1< x2andx1> xnare handled similarly). Then,
1j <in
xia−xjab=
2jm
xja−x1ab
m+1jn
x1a−xjab
2j <in
xia−xjab
and it is easy to check that
∂fa,b,c(x)
∂x1 =
ab
k1
x1a−1 x1a−xak
+ c
x1
fa,b,c(x).
Combining with the integral identity above, we get that for almost all(x2, . . . , xn)
(β+1) ∞
−∞
|x1|βf (x)exp
−xpp
fa,b,c(x)dx1
=p ∞
−∞
|x1|β+pf (x)exp
−xpp
fa,b,c(x)dx1− ∞
−∞
|x1|βx1
∂f
∂x1(x)
exp
−xpp
fa,b,c(x)dx1
− ∞
−∞
ab
k1
|x1|βx1a x1a−xka
+c|x1|β
f (x)exp
−xpp
fa,b,c(x)dx1.
Integration with respect tox2, . . . , xnand the definition ofMpgive (β+c+1)Mp
|x1|βf (x)
=pMp
|x1|β+pf (x)
−Mp
|x1|βx1∂f
∂xi
(x)
−abMp
k1
|x1|βx1a x1a−xka
f (x)
.
This is obviously valid for everyi=1, . . . , nand summing these equalities we get (β+c+1)Mp
n
i=1
|x1|β
f (x)
=pMp
xββ++ppf (x)
−Mp n
i=1
|xi|βxi∂f
∂xi(x)
−abMp n
i=1
k=i
|xi|βxia xia−xka
f (x)
. 2
Corollary 7.With the notations defined at the beginning of this section, we have:
(a) iff is a positively homogeneous function of degrees, integrable onSn−1ands >−dnthen (dn+s)Mp
f (x)
=pMp
xppf (x)
; (b) for anyq >−dn−2,
pMp
xpp++22xq2
−qMp
x44xq2−2 ζ(a,2)
dn n +2
Mp
xq2+2 whereζ(1,2)=32 andζ(a,2)=1for everya2;
(c) for anyq >−dn−p, pMp
x2p2pxq2
−qMp
xpp++22xq2−2
ζ(a,p) dn
n +p
Mp
xppxq2
whereζ(a,p)=max{a2a+p,1}.
Remark.Simple observations in the proof below show that the two inequalities established in (b) and (c) can be reversed with other constants (independent of the dimensionn).
Proof. (a) Letg:R→Rdefined by g(t )=
Rn
f (x)fa,b,c(x)exp
−txpp
dx
thengis clearly aC1function onRand g(t )= −
Rn
xppf (x)fa,b,c(x)exp
−txpp
dx.
However, sincef is positively homogeneous of degrees andfa,b,c is positively homogeneous of degreeabn(n− 1)/2+cntherefore, by the change of variabley=t1/pxwe get that
g(t )=t−(s+dn)/pg(1).
Taking the derivative at 1 proves (a).
For (b) and (c) we will use Lemma 6, forf (x)= xq2. So observe that ∂x∂f
i =qxixq2−2. One can check that for any positive integeraandβ >0, ifζ =ζ(a,β)=max{a2a+β,1}then,
∀x1=x2, |x1|βx1a− |x2|βx2a x1a−x2a ζ
|x1|β+ |x2|β .
So, we get n
i=1
k=i
|xi|βxia xia−xka =1
2 n
i=1
k=i
|x1|βx1a− |x2|βx2a x1a−x2a
n−1
2 ζxββ. To prove assertion (b) we apply Lemma 6 withβ=2. So we get
(c+3)Mp
xq2+2
pMp
xpp++22xq2
−qMp
x44xq2−2
−abζ(a,2)n−1 2 Mp
xq2+2 . To prove assertion (c) we apply Lemma 6 withβ=p. So we get
(c+p+1)Mp
xppxq2
pMp
x2p2pxq2
−qMp
xpp++22xq2−2
−abζ(a,p)n−1 2 Mp
xppxq2
. 2
We conclude this section by proving the key estimate of Lemma 5.