Characterization of equality in the correlation inequality for convex functions, the <span class="mathjax-formula">$U$</span>-conjecture

Characterization of equality in the correlation inequality for convex functions, the U-conjecture

Gilles Hargé

Equipe d’analyse et de probabilités, université d’Evry, rue du père Jarlan, bât. Maupertuis, 91025 Evry Cedex, France Received 16 June 2003; received in revised form 28 May 2004; accepted 14 June 2004

Available online 22 September 2004

Abstract

We give a characterization of the equality in Hu’s inequality (which is a correlation inequality between two convex functions inRⁿwith respect to the standard Gaussian measure). For this, we prove a new inequality which is slightly better than Hu’s inequality. Then, we obtain a result concerning the U-conjecture.

2004 Elsevier SAS. All rights reserved.

Résumé

On donne une caractérisation du cas d’égalité dans l’inégalité de Hu, qui est une inégalité de décorrélation entre deux fonc- tions convexes contre la mesure gaussienne dansRⁿ. Pour cela, on démontre une inégalité légèrement plus forte que l’inégalité de Hu. On obtient aussi un résultat concernant la conjecture U.

2004 Elsevier SAS. All rights reserved.

MSC: primary: 52A40, 60E15

Keywords: Gaussian measure; Ornstein–Uhlenbeck semigroup; Convexity; Correlation

1. Introduction

We denote byµnthe standard Gaussian measure onRⁿ. In 1973, Kagan, Linnik and Rao [11] considered the following question: ifP andQare two polynomials onRⁿ independent with respect toµn (such that ifX is a random vector onRⁿ of lawµn, thenP (X)andQ(X)are independent random variables), is it possible to find an orthogonal transformationU onRⁿ and an integerksuch thatP ◦U is a function of(x₁, . . . , x_k)andQ◦U is a function of(x_k+1, . . . , x_n)? If the answer is positive, we say that P andQare unlinked. This question is known as the U-conjecture. Kagan, Linnik and Rao gave a partial answer in [11]. Recently, Bhandari and Basu [5]

E-mail address: gharge@maths.univ-evry.fr (G. Hargé).

0246-0203/$ – see front matter 2004 Elsevier SAS. All rights reserved.

doi:10.1016/j.anihpb.2004.05.005

have shown that ifP andQare two convex, positive polynomials independent with respect toµ_nand ifP (0)=0 thenP andQare unlinked. Related to this problem, Bhandari and DasGupta [6] proved in 1994 that two convex and even functionsf andgare unlinked if they are uncorrelated (that is if

f g dµ_n=

f dµ_n

g dµ_n) under an additional hypothesis. This hypothesis is related to the Gaussian correlation conjecture which is still a conjecture in dimension greater than two (see [7] for further details and references on this conjecture), so, the result of Bhandari and DasGupta is not proved in the general case, that is, without this additional hypothesis. Recall now an inequality due to Hu [10] and which concerns two convex functionsf andginL²(µ_n)(see [9] or [8] for the second term in the inequality):

f g dµ_n

f dµ_n

g dµ_n+

xf dµ_n,

xg dµ_n

. (1.1)

If

xf dµ_n,

xg dµ_n0, this is a correlation inequality betweenf andg( is the usual scalar product on Rⁿ).

We will prove the following theorem, which shows in particular the result of Bhandari and DasGupta in the general case, and which gives a partial answer to the U-conjecture:

Theorem 1.1. Letf, g:Rⁿ→Rbe convex functions inL²(µ_n)for which equality holds in (1.1). Then, there exit an orthogonal transformationU onRⁿ, two vectorsα₁and α₂ inRⁿ, an integerk∈ {0, . . . , n}and two convex functionsf¯:R^k→RinL²(µ_k)andg¯:R^n−k→RinL²(µ_n−k)such that, for allxinRⁿ:

f (U x)= α₁, x + ¯f (x₁, . . . , x_k) and g(U x)= α₂, x + ¯g(x_k+1, . . . , x_n)

x=(x₁, . . . , x_n) .

Furthermore(α₁)_i=0 ifikand(α₂)_i=0 ifi > k(ifk=0 ork=n,we make obvious conventions).

Of course, the condition is sufficient. Actually, it is easy to see that ifϕ:Rⁿ→Ris a function inL²(µ_n)and if αbelongs toRⁿthenϕand the function:x→ α, xsatisfy (1.1) with equality.

Secondly, we will show a generalization of the result of Bhandari and Basu:

Theorem 1.2. Denote byX a random vector of lawµ_n. Let f, g:Rⁿ→Rbe convex functions inL²(µ_n). We assume thatf is an analytic function which verify, for allxinRⁿ,f (x)f (0).Iff (X)andg(X)are independent random variables thenf andgare unlinked.

The proof of Theorem 1.1 uses a new proof of Hu’s inequality and the Ornstein–Uhlenbeck semigroup. In fact, we will prove a new inequality which is slightly better than Hu’s inequality.

2. Reinforcement of Hu’s inequality

Two functionsf andg for which equality holds in (1.1) are not necessarily regular functions. Of course, it is possible to approximate such functions with regular functions but those last functions do not satisfy automat- ically (1.1) with equality. Nevertheless, that will be the case if we approximate f and g with the help of the Ornstein–Uhlenbeck semigroup. For this, we will show an inequality which is a reinforcement of Hu’s inequality (Theorem 2.1 below) and which concerns the Ornstein–Uhlenbeck semigroup.

For the last thirty years, semigroups have been used to prove various inequalities as correlation, concentration, Poincaré and log-Sobolev inequalities. Concerning correlation inequalities, it is possible to refer to the works of Pitt [17] (perhaps the first use of the Ornstein–Uhlenbeck semigroup to prove a correlation inequality), Bakry and Michel [3], Hu [10], or my former works [7,8]. There is also related work of Houdré, Pérez-Abreu and Surgailis [9]. Concerning concentration, Poincaré and log-Sobolev inequalities, see for example the surveys of Bakry and

Ledoux [1,2,12–14]. Furthermore, we will compare the following result with a work of Beckner [4] concerning Poincaré inequality.

We define the Ornstein–Uhlenbeck semigroup with the Mehler formula:

∀f ∈L²(µ_n), P_tf (x)=

f

e^−tx+

1−e^−2ty

dµ_n(y).

The Ornstein–Uhlenbeck operatorLis defined by:

D(L)=

f∈L²(µ_n),Ptf −f

t possesses a limit inL²(µ_n)whentgoes to 0 , Lf=lim

t→0

P_tf −f

t forf∈D(L).

It is well known thatD(L)= {f ∈L²(µ),f− x,∇f ∈L²(µ)}andLf=f− x,∇f(wheref− x,∇f is taken in distribution sense). Furthermore, we have:

∀f ∈D(L), ∀t >0, P_tf ∈D(L) and d

dtP_t(f )=LP_tf =P_tLf

inL²(µ_n) .

The properties ofP_t andLwe will use in the next theorems and remarks could be find in [15] or [16].

Theorem 2.1. Letf, g:Rⁿ→Rbe convex functions such thatf ∈L²(µ_n)andg∈L²(µ_n), then, for allt0:

f g dµn

f Ptg dµn+(1−e^−t)

xf dµn,

xg dµn

.

Remark 1. The inequality obtained in this theorem can be compared to a generalized Poincaré inequality due to Beckner [4]. He showed, forf inL²(µn):

f²dµn−

(Ptf )²dµn(1−e^−2t)

∇f ²dµn.

If we choosef =gin Theorem 2.1 and if we replacet by 2t, we obtain, for a convex functionf (using the fact thatP_t is a symmetric semigroup with respect toµ_n):

(1−e^−2t)

∇f dµ_n ²

f²dµ_n−

(P_tf )²dµ_n.

Remark 2. We deduce from Theorem 2.1, that iff andgare inL²(µ_n)and convex, then:

fg−P_tg

t dµ_n1−e^−t t

xf dµ_n,

xg dµ_n

.

So, we obtain, whent goes to 0 and ifg∈D(L):

−

f Lg dµ_n xf dµ_n,

xg dµ_n

.

We will prove this inequality before Theorem 2.1.

Remark 3. Recall that, forg∈L²(µ_n), lim_t→+∞P_tg=

g dµ_n (inL²(µ_n)). Consequently, if we lett goes to infinity in Theorem 2.1, we recover inequality (1.1).

Remark 4. In fact, the inequality of Theorem 2.1 is more accurate than inequality (1.1). Actually, Theorem 2.1 gives:

f g dµ_n

f P_tg dµ_n+(1−e^−t)

xf dµ_n,

xg dµ_n

.

Furthermore, if we apply inequality (1.1) tof andP_tg(which is convex with the help of Mehler formula), we obtain:

f Ptg dµn

f dµn

Ptg dµn+

xf dµn,

xPtg dµn

f dµn

g dµn+e^−t

xf dµn,

xg dµn

,

because

xP_tg dµ_n=

gP_tx dµ_n=e^−t

xg dµ_n. Consequently:

f g dµ_n

f P_tg dµ_n+(1−e^−t)

xf dµ_n,

xg dµ_n

f dµn

g dµn+

xf dµn,

xg dµn

.

Remark 5. Theorem 2.1 allows us to show that iff andgare convex functions and satisfy (1.1) with equality then P_tf andP_tgverify the same equality. Actually, we have:

f g dµ_n

f P_2tg dµ_n+(1−e^−2t)

xf dµ_n,

xg dµ_n

f dµn

g dµn+

xf dµn,

xg dµn

.

Iff andgsatisfy (1.1) with equality then:

f P2tg dµn=

f dµn

g dµn+e^−2t

xf dµn,

xg dµn

⇒

Ptf Ptg dµn=

Ptf dµn

Ptg dµn+

xPtf dµn,

xPtg dµn

.

Before given the proof of Theorem 2.1, we will show the following result:

Theorem 2.2. Iff, g:Rⁿ→Rare convex functions inL²(µ_n)and ifg∈D(L), then:

−

f Lg dµn

xf dµn,

xg dµn

.

Proof. We will prove this theorem forP_uf andP_ug(u >0) instead off andg. Then, we will deduce easily the result forf andgwith the following convergences inL²(µ_n):

lim

u→0P_uf=f, lim

u→0P_ug=g, lim

u→0LP_ug=lim

u→0P_uLg=Lg.

We will use the following properties ofP_t andL.

• ∀h∈L²(µ_n), ∀t >0, x →P_th(x) is a C^∞ function on Rⁿ, _∂x^∂

i(P_th)∈L²(µ_n) and ∀x ∈Rⁿ, ∀s 0,

∂

∂x_i(Ps+th)(x)=e^−tPs(_∂x^∂

i(Pth))(x).

• ∀(h₁, h₂)∈D(L)²,

h₁Lh₂dµ_n= −

∇h₁,∇h₂dµ_n. We notice thatPuf andPugare convex,C^∞and inD(L).

Defineθ (t )=

∇Puf, Pt∇Pugdµn. Becauseuis strictly positive,∇Puf and∇Pugare inL²(µn). Conse- quently:

tlim→∞θ (t )=

∇P_uf dµ_n,

∇P_ug dµ_n

.

Moreover:

θ(t )= n

i=1

∂Puf

∂xi

LP_t ∂Pug

∂xi

dµ_n= − n

i=1

∇ ∂Puf

∂xi

,∇

P_t

∂Pug

∂xi

dµ_n.

It is possible to justify this equality by saying thatP_t(^∂P_∂x^ug

i )belongs toD(L) and that ^∂P_∂x^uf

i belongs toD(L) (because ^∂P_∂x^uf

i =e^−u/2P_u/2(_∂x^∂

i(P_u/2f ))). We write:

P_t ∂Pug

∂xi

=e^t ∂

∂xi

P_tP_ug,

then, we obtain:

θ(t )= −e^t n

i=1

∇ ∂P_uf

∂x_i

,∇

∂P_t+ug

∂x_i

dµ_n= −e^t

Tr(HessP_ufHessP_t+ug) dµ_n. BecauseP_uf andP_t+ugare convex, we deduceθ(t )0. Consequently:

∇P_uf,∇P_ugdµ_n

∇P_uf dµ_n,

∇P_ug dµ_n

,

which gives:

−

P_uf LP_ug dµ_n

xP_uf dµ_n,

xP_ug dµ_n

. 2

Proof of Theorem 2.1. For two convex functionsf andginL²(µ_n), we define:

ξ(t )=

f P_tg dµ_n−

xf dµ_n,

xP_tg dµ_n

=

f P_tg dµ_n−e^−t

xf dµ_n,

xg dµ_n

.

It is sufficient to show thatξ is a decreasing function.

ξ(t )=

f LPtg dµn−

xf dµn,

xLPtg dµn

=

f LP_tg dµ_n−

xf dµ_n,

LxP_tg dµ_n

=

f LP_tg dµ_n+

xf dµ_n,

xP_tg dµ_n

.

We apply the previous theorem tof andP_tgto obtainξ(t )0. 2

3. Characterization of equality in Hu’s inequality

We begin to prove some elementary facts concerning convex functions. The following reasoning process is inspired by the one of Bhandari and DasGupta [6].

Lemma 3.1. Ifϕ:R→Ris a convex and non-constant function then limt→+∞ϕ(t )= +∞or limt→−∞ϕ(t )= +∞[6].

We deduce, by using the convexity ofϕ(x)− α, x: Ifϕ:Rⁿ→Ris convex and verify

∃α∈Rⁿ, ∃b∈R,∀x∈Rⁿ, ϕ(x)α, x +b, then:∀x∈Rⁿ,ϕ(x)= α, x +ϕ(0).

Lemma 3.2. Letf:Rⁿ→Rbe a convex function. We define:

E(f )=

h∈Rⁿ, ∃(a, b)∈R²,∀t∈R, f (t h)at+b . E(f )is a linear space contained inRⁿand

∃!α∈E(f ), ∀y₁∈E(f ),∀y₂∈Rⁿ, f (y₁+y₂)= α, y₁ +f (y₂).

Proof. It is obvious to see thatE(f )is a linear space. Let(e₁, . . . e_r)be an orthonormal basis ofE(f )andybe an element ofRⁿ. We denotea_i andb_i the numbers associated toe_i in the definition ofE(f ). We have:

f _r

i=1

x_ie_i+y

1

r+1 _r

i=1

f

(r+1)x_ie_i +f

(r+1)y

α, r

i=1

xiei

+ 1

r+1 _r

i=1

bi+f

(r+1)y ,

whereα=_r

i=1a_ie_i. For a fixedy, the map: (x₁, . . . x_r)→f (_r

i=1x_ie_i +y) is convex, so, we know from Lemma 3.1 that:

∀(x₁, . . . , x_r)∈R^r, f _r

i=1

x_ie_i+y

=

α, r

i=1

x_ie_i

+f (y).

Unicity ofαis obvious. 2

Remark 6. If we choosey₂=0 andh∈E(f )we see that∀t∈R, f (t h)=tα, h +f (0).

Lemma 3.3. Letϕ:Rⁿ→Rbe a convex function inL²(µ_n). Leth∈Rⁿ, we assume that the map:t→ϕ(t h)is a non-constant function. Let(e₁, . . . , e_k)be an orthonormal family inRⁿ(kn)and define:

˜ ϕ(x)=

ϕ

x+

k

i=1

yiei

dµk(y) wherey=(y₁, . . . , yk).

Then:

∃ε∈ {−1,1}, lim

εt→+∞ϕ(t h)˜ = +∞,

∃ε∈ {−1,1}, ∀s0, lim

εt→+∞(P_sϕ)(t h)= +∞.

Proof. We know from Lemma 3.1 that

∃ε∈ {−1,1}, lim

εt→+∞ϕ(t h)= +∞.

Lety∈Rⁿ:ϕ(₂^th)¹₂(ϕ(t h+y)+ϕ(−y))so lim_εt→+∞ϕ(t h+y)= +∞. Furthermore:

∃A >0, εt > A⇒ϕ t

2h

0.

With Fatou’s lemma, we write:

lim inf

εt→+∞

ϕ

t h+

k

i=1

yiei

+ϕ

− k

i=1

yiei

dµk(y)

lim inf

εt→+∞ ϕ

t h+

k

i=1

yiei

+ϕ

− k

i=1

yiei

dµk(y)

⇒ lim

εt→+∞ϕ(t h)˜ = +∞.

We can prove the result forP_sϕin the same way. 2

Lemma 3.4. Letf:Rⁿ→Rbe a convex function inL²(µ). Leth∈Rⁿands0. Like in the previous lemma, we associatef˜tof,then:

h /∈E(f )⇒

h /∈E(f )˜ andh /∈E(P_sf ) .

Proof. h /∈E(f )so, for everyainR, the mapt→f (t h)−atis non-constant. Definefa(x)=f (x)−ax, ^h

h ². f_ais convex and the mapt→f_a(t h)is non-constant, so, with the previous lemma:

∃ε∈ {−1,1}, lim

εt→+∞f˜_a(t h)= +∞. Moreover:

f_a(t h+_k

i=1y_ie_i) dµ_k(y)=

f (t h+_k

i=1y_ie_i) dµ_k(y)−at = ˜f (t h)−at. We deduce that the mapt→ ˜f (t h)−at is non-constant for alla. Soh /∈E(f ). We can prove the result for˜ P_sϕin the same way. 2 Remark 7. In fact, the following equalities are true:E(f )=E(f )˜ =E(P_sf ). Actually, we have for allβ∈ [0,√

2[ and becausef ∈L²(µ_n):

|f (βx)|dµ_n(x) <+∞. Leth∈E(f )then∃(a, b)∈R²,∀t∈R, f (t h)at+b. For y∈Rⁿandβ∈ ]1,√

2[, we write:t h+y=(1−_β¹)_β^β₋₁t h+_β¹βy, consequently:

f (t h+y)at+

1− 1 β

b+1

βf (βy).

We deduce:h∈E(f )˜ andh∈E(Psf ).

Now, we can prove Theorem 1.1.

Proof of Theorem 1.1. We choose(e₁, . . . , en)an orthonormal basis ofRⁿsuch that:

• (e1, . . . , er+k)is an orthonormal basis ofE(f )^⊥,

• (er+1, . . . , er+k)is an orthonormal basis ofE(f )^⊥∩E(g).

We use Lemma 3.2 to constructα₁associated tof andα₂associated tog. We obtain:

f _n

i=1

x_ie_i

=f _r+k

i=1

x_ie_i

+

α₁, n

i=r+k+1

x_ie_i

,

g _n

i=1

x_ie_i

=g _r

i=1

x_ie_i+ n

i=r+k+1

x_ie_i

+

α₂,

r+k

i=r+1

x_ie_i

.

We want to prover=0 so we assumer=0. We will use the fact thatr

i=1xiei∈/E(f )∪E(g)if(x1, . . . , xr)

=0.

Define (abuse of notations):

f (x˜ 1, . . . , xr)= ˜f _r

i=1

xiei

=

f _r+k

i=1

xiei

dµk(xr+1, . . . , xr+k),

˜

g(x1, . . . , xr)= ˜g _r

i=1

xiei

=

g _r

i=1

xiei+ n

i=r+k+1

xiei

dµn−r−k(xr+k+1, . . . , xn).

Recall that ifϕ:Rⁿ→Ris a function inL²(µ_n)and ifα∈Rⁿthenϕ and the mapx→ α, xsatisfy (1.1) with equality. So, becausef andgsatisfy (1.1) with equality, it is the case forf˜andg˜ againstµ_r inR^r. Furthermore (Lemma 3.4)_r

i=1x_ie_i∈/E(f )˜ ∪E(g)˜ if(x₁, . . . , x_r)=0. We deduceE(f )˜ =E(g)˜ = {0}. Using Remark 4 and becausef˜andg˜satisfy (1.1) with equality, we have:

∀u0,

f˜g dµ˜ r=

f P˜ ug dµ˜ r+(1−e^−u)

xf dµ˜ r,

xg dµ˜ r

,

(the Ornstein–Uhlenbeck semigroup we use here is the one ofR^r).

For the continuation of the proof, we only need the existence of one realu >0 such that this equality is verified.

The functionξ used in the proof of Theorem 2.1 is decreasing, so we deduce:

∀s∈ [0, u], ξ(s)=0.

Consequently, for alls∈ [0, u]:

−

f LP˜ _sg dµ˜ _r=

xf dµ˜ _r,

xP_sg dµ˜ _r

⇒ −

P_s/2f LP˜ _s/2g dµ˜ _r=

xf dµ˜ _r,

P_s/2xP_s/2g dµ˜ _r

=e^s/2

xf dµ˜ _r,

x P_s/2g dµ˜ _r

=

xP_s/2f dµ˜ _r,

x P_s/2g dµ˜ _r

⇒

∇P_s/2f ,˜ ∇P_s/2g˜dµ_r =

∇P_s/2f dµ˜ _r,

∇P_s/2g dµ˜ _r

.

We define, like in the proof of Theorem 2.2, the function:

θ (t )=

∇P_s/2f , P˜ _t∇P_s/2g˜dµ_r.

So we have, for everyt0,θ(t )=0, consequently:

∀t0,

Tr(HessPs/2f˜HessPt+s/2g) dµ˜ r=0.

Using the formula HessPt+s/2g˜=e^−2tPt(HessPs/2g), we obtain:˜

∀t0,

Tr

HessPs/2f P˜ t(HessPs/2g)˜

dµr =0.

Lettgoes to infinity, then:

Tr

HessPs/2f dµ˜ r

HessPs/2g dµ˜ r

=0. (3.1)

Matrices

HessP_s/2f dµ˜ _r and

HessP_s/2g dµ˜ _r are symmetric and positive. It is easy to see that if both are invertible then equality (3.1) is impossible. So, we can assume that

HessP_s/2f dµ˜ _r is not invertible. We deduce there exists an elementhofR^r,h=0, such that:

HessP_s/2f dµ˜ _r

h, h

=0

⇒ [HessP_s/2f˜]h, h

dµ_r =0

⇒ ∀x∈R^r,

[HessPs/2f˜](x)h, h

=0.

Defineς (λ)=P_s/2f (λh), we obtain, for all˜ λ,ς(λ)=0, consequently:

∃(a, b)∈R², ∀λ∈R, P_s/2f (λh)˜ =aλ+b.

Soh∈E(P_s/2f ). Then, we deduce from Lemma 3.4 that˜ h∈E(f )˜ but this is impossible becauseE(f )˜ = {0}. 2 Remark 8. Actually, we have proved that iff andgare two convex functions for which equality holds in Theo- rem 2.1 for a fixedt >0, then the conclusion of Theorem 1.1 remains valid.

Remark 9. We obtain in the proof of the theorem:α₁∈E(f )andα₂∈E(g).

We deduce immediately from Theorem 1.1:

Corollary 3.5. Letf, g:Rⁿ→Rbe two convex functions inL²(µ_n)such that

f g dµ_n=

f dµ_n

g dµ_n and

xf dµ_n,

xg dµ_n0. Then

xf dµ_n,

xg dµ_n =0 and there exist an orthogonal transformationU on Rⁿ, two vectorsα₁andα₂ inRⁿ, an integerk∈ {0, . . . , n}and two convex functionsf¯:R^k→RinL²(µ_k)and

¯

g:R^n−k→RinL²(µ_n−k)such that, for everyxinRⁿ:

f (U x)= α₁, x + ¯f (x₁, . . . , x_k) and g(U x)= α₂, x + ¯g(x_k+1, . . . , x_n)

x=(x₁, . . . , x_n) .

We have(α₁)_i=0 ifikand(α₂)_i=0 ifi > k.

Moreover, if

xf dµ_n=

xg dµ_n=0 then

xf dµ_n,

xg dµ_n =0 is verified and α₁=α₂=0 (conse- quently,f andgare unlinked).

Proof. We start with Hu’s inequality:

f g dµ_n

f dµ_n

g dµ_n+

xf dµ_n,

xg dµ_n

f dµ_n

g dµ_n.

We obtain:

xf dµ_n,

xg dµ_n =0. Then, we use Theorem 1.1 to constructU. Now, assume that

xf dµn=0. Denotex=(y₁, y₂)wherey₁=(x₁, . . . , xk)andy₂=(xk+1, . . . , xn),α₁= (0,α)¯ withα¯∈R^n−k. We obtain:

xf (U x) dµn(x)=(

y1f (y¯ 1) dµk(y1),α)¯ andα¯=0. 2

Remark 10. The second part of this corollary generalizes and proves the result of Bhandari and DasGupta [6] in any dimension (iff andgare even functions then

xf dµ_n=

xg dµ_n=0).

4. The U-conjecture

In the following, we denote byX=(X₁, . . . , Xn)a random vector of lawµn. We deduce from the previous corollary:

Theorem 4.1. Letf, g:Rⁿ→Rbe two convex functions inL²(µ_n). Assume that

xf dµ_n=0 and thatf (X)and g(X)are independent random variables thenf andgare unlinked.

Proof. The equality

xf dµ_n,

xg dµ_n =0 is verified. With notations and results of the first part of the previous corollary, we obtainα₁=0. So, it is possible to findUandf¯such that:

f (U x)= ¯f (x₁, . . . , x_k) and g(U x)=ax₁+ ¯g(x_k+1, . . . , x_n).

It is easy to see thatf (U X)andg(U X)are independent random variables. DenoteY =(X2, . . . , Xk)andZ= (X_k+1, . . . , X_n).f (X¯ ₁, Y )andaX₁+ ¯g(Z)are independent random variables, so it is the case forf (X¯ ₁, Y )and aX₁. Recall that in the proof of Theorem 1.1, we have obtained:

f (x¯ _r+1, . . . , x_r+k)=f _r+k

i=r+1

x_ie_i

,

where, forr+1ir+k,ei∈E(f )^⊥. Consequently,r+k

i=r+1xiei∈/E(f )if(xr+1, . . . , xr+k)=0. So, we have E(f )¯ = {0}. Letψ∈L²(µ₁)and let us assumea=0 andf¯depends onx₁.

Ef (X¯ ₁, Y )ψ (aX₁)

=Ef (X¯ ₁, Y ) E

ψ (aX₁)

=E

f (X¯ ₁, y) dµ_k−1(y) ψ (aX₁)

.

Define:f (x˜ ₁)=f (x¯ ₁, y) dµ_k−1(y). We haveE(f )˜ = {0}(Lemma 3.4). Furthermore:

Ef (X˜ ₁)ψ (aX₁)

=Ef (X˜ ₁) E

ψ (aX₁) .

We chooseψ (x₁)= ˜f (¹_ax₁). We obtain:f (X˜ ₁)=E(f (X˜ ₁))almost surely. So,f˜is a constant function, which is a contradiction withE(f )˜ = {0}. We deducea=0 orf¯does not depend onx₁. 2

The following result is to be compared to the one of Bhandari and Basu [5] who show that ifP andQare two convex, positive polynomials independent with respect toµand ifP (0)=0 thenP andQare unlinked.

Corollary 4.2. Letf, g:Rⁿ→Rbe two convex functions inL²(µ_n). We assume thatf andgare bounded below, that

xf dµ_n,

xg dµ_n =0 and that

f g dµ_n=

f dµ_n

g dµ_nthenf andgare unlinked.

Proof. We use notations and results of the first part of Corollary 3.5. Becausef andgare bounded below, it is easy to see thatα₁=α₂=0. 2

Now, we will prove Theorem 1.2. We begin with a lemma:

Lemma 4.3. Letf:Rⁿ→Rbe a convex function such that:

∀x∈Rⁿ, x=0⇒f (x) > f (0).

Forε >0, defineδ(ε)=sup{ x ,x∈Rⁿ,f (x)−f (0)ε}then limε→0δ(ε)=0.

Proof. Assume thatδ(ε)does not go to 0. We constructη >0 and a sequence(ε_p)_p1such that lim_p→+∞ε_p=0 andδ(ε_p) > ηfor allp. We associate toε_pan elementx_pofRⁿsuch that x_p > ηandf (x_p)−f (0)ε_p. We can assume that ^x_x^p

p goes toe(an element ofRⁿof norm equal to 1). Then:

f η

x_p x_p

1− η x_p

f (0)+ η

x_p f (x_p)f (0)+ε_p.

We obtain:f (ηe)f (0), sof (ηe)=f (0), but it is not possible becauseηe=0. 2

Proof of Theorem 1.2. We use here the same orthonormal basis as in the proof of Theorem 1.1. So, we construct α₁andα₂associated tof andg. Becausef is bounded below, we haveα₁=0. Consequently:

f _n

i=1

xiei

=f _r+k

i=1

xiei

,

g _n

i=1

x_ie_i

=g _r

i=1

x_ie_i+ n

i=r+k+1

x_ie_i

+

α₂,

r+k

i=r+1

x_ie_i

.

We assumer1. Define:

˜

g(x₁, . . . , x_r)=

g _r

i=1

x_ie_i+ n

i=r+k+1

x_ie_i

dµ_n−(r+k)(x_r+k+1, . . . , x_n),

f₁(x₁, . . . , x_r+k)=f _r+k

i=1

x_ie_i

.

We haveE(g)˜ = {0}(Lemma 3.4) andE(f₁)= {0}(because(e₁, . . . , e_r+k)is an orthonormal basis ofE(f )^⊥). Let ψ:R→Rbe measurable and bounded, we obtain:

˜

g(x₁, . . . , x_r)+

α₂,

r+k

i=r+1

x_ie_i

ψ (f₁)(x₁, . . . , x_r+k) dµ_r+k(x₁, . . . , x_r+k)

=

g(x)ψ◦f (x) dµn(x)

=

g(x) dµ_n(x)

ψ◦f (x) dµ_n(x)

= g(x˜ ₁, . . . , x_r)+

α₂,

r+k

i=r+1

x_ie_i

dµ_r+k(x₁, . . . , x_r+k)

×

ψ (f₁)(x₁, . . . , x_r+k) dµ_r+k(x₁, . . . , x_r+k).

Let considerY a random vector inR^r+k of lawµ_r+k and choose, forε >0, ψ (t )=1_[−ε,ε](t−f₁(0)). We write Y=(Y₁, Y₂)whereY₁∈R^r,Y₂∈R^k. So we have:

E

˜

g(Y₁)+ α₂, Y₂ 1_|f₁(Y )−f₁(0)|ε

P (|f₁(Y )−f₁(0)|ε)

=E

˜

g(Y₁)+ α₂, Y₂

=E

˜ g(Y₁)

.

If there existsx=0 such thatf₁(x)=f₁(0), and asf₁is convex, we obtain:

∀t∈ [0,1], f₁(t x)=f₁(0).

Butf₁is an analytic function, so:

∀t∈R, f₁(t x)=f₁(0).

Consequentlyx∈E(f1),which impliesx=0. We have obtained:∀x=0,f1(x) > f1(0). Now, we use Lemma 4.3 forf₁. We denote:δ(ε)=sup{ x ,x∈R^r+k,f₁(x)−f₁(0)ε}.

Eg(Y˜ ₁)+ α₂, Y₂ − ˜g(0) 1_{|f1(Y )−f1(0)|ε} P (|f1(Y )−f1(0)|ε)

Eg(Y˜ ₁)+ α₂, Y₂ − ˜g(0)1 _{Y δ(ε)} 1_{|f1(Y )−f1(0)|ε} P (|f₁(Y )−f₁(0)|ε)

.

We deduce:

lim

ε→0E

˜

g(Y₁)+ α₂, Y₂ 1_{|f1(Y )−f1(0)|ε} P (|f1(Y )−f1(0)|ε)

= ˜g(0).

Consequently:

E

˜ g(Y₁)

= ˜g(0).

The mapg˜is convex, that means that there is equality in Jensen’s inequality. We deduceg˜is an affine map, but this is impossible becauseE(g)˜ = {0}. Consequentlyr=0. Then, it is possible to find an orthogonal transformationU such that:

f (U x)= ¯f (x_r+1, . . . , x_r+k), g(U x)= ¯g(x_r+k+1, . . . , x_n)+ax_r+1.

Butf (U X)andg(U X)are independent random variables, so,a=0 orf¯does not depend onx_r+1(see the proof of Theorem 4.1). 2

Remark 11. The idea of using a function like 1_[−ε,ε](t−f₁(0))and the case of equality in Jensen’s inequality is due to Bhandari and Basu [5]. If we compare the proof given here to their proof, the novelty is the use ofE(g),the choice of the orthonormal basis, the extension to an entire functionf and the end of the proof (to provea=0 or f¯does not depend onx_r+1).

Remark 12. In the proof, we could use{x∈Rⁿ,∀t∈R,f (t x)=f (0)}instead ofE(f ). However, it is essential to work withE(g). We have to notice thatf (x)f (0)for allximplies{x∈Rⁿ,∀t∈R,f (t x)=f (0)} =E(f ) (by usingα1=0 and Remark 6).

References

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Lecture Notes in Math., vol. 1581, 1994, pp. 1–114.

[2] D. Bakry, Functional inequalities for Markov semigroups, preprint.

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188.

[4] W. Beckner, A generalized Poincaré inequality for Gaussian measures, Proc. Amer. Math. Soc. 105 (1989) 397–400.

[5] S.K. Bhandari, A. Basu, On the unlinking conjecture of independent polynomial functions, preprint.

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