A geometric approach to correlation inequalities in the plane

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www.imstat.org/aihp 2014, Vol. 50, No. 1, 1–14

DOI:10.1214/12-AIHP494

A geometric approach to correlation inequalities in the plane

A. Figalli

^a

, F. Maggi

^b

and A. Pratelli

^c

aDepartment of Mathematics, The University of Texas at Austin, 1 University Station C1200, Austin TX 78712, USA.

E-mail:[email protected]

bDipartimento di Matematica “U. Dini,” Università di Firenze, Viale Morgagni 67/A, 50134 Firenze, Italy. E-mail:[email protected] cDipartimento di Matematica “F. Casorati,” Università di Pavia, via Ferrata 1, 27100 Pavia, Italy. E-mail:[email protected]

Received 22 July 2011; revised 16 February 2012; accepted 27 April 2012

Abstract. By elementary geometric arguments, correlation inequalities for radially symmetric probability measures are proved in the plane. Precisely, it is shown that the correlation ratio for pairs ofwidth-decreasing setsis minimized within the class of infinite strips. Since open convex sets which are symmetric with respect to the origin turn out to be width-decreasing sets, Pitt’s Gaussian correlation inequality (the two-dimensional case of the long-standing Gaussian correlation conjecture) is derived as a corollary, and it is in fact extended to a wide class of radially symmetric measures.

Résumé. En utilisant des arguments géométriques élémentaires, on démontre des inégalités de corrélation pour des mesures de probabilité à symétrie radiale. Plus précisément on montre que, parmi la famille des ensembles width-decreasing, le ratio de corrélation est minimisé par des bandes. Comme les ouverts convexes symétriques appartiennent à cette famille, on retrouve comme corollaire le résultat de Pitt sur la validité de la conjecture de corrélation gaussiennne en dimension 2, qui est étendue dans ce papier à une large classe de mesures à symétrie radiale.

MSC:Primary 60E15; secondary 52A40; 62H05

Keywords:Correlation inequalities; Gaussian correlation conjecture; Radially symmetric measures

1. Introduction

We address the minimization of the correlation ratio of a radially symmetric probability measureμonR², providing in particular a new and elementary proof of Pitt’s correlation inequality for the planar Gaussian measure. Let us say that a setS⊂R²is a strip (symmetric with respect to the origin) if there existν∈S¹andh >0 such that

S=

x∈R²: |x·ν|< h .

Two stripsSandSare orthogonal if they are associated to orthogonal vectorsνandνinS¹. Next we introduce the family ofwidth-decreasing setsas the class of those planar, open sets, which are symmetric with respect to the origin, which contain the origin, and whose angular-length (roughly speaking) decreases at least as the angular-length of a strip. Precisely, if we setBr= {x∈R²:|x|< r}, then an open setEis a width-decreasing set if 0∈E,x∈Eimplies

−x∈E, and, for everyr >0 such that 0<H¹(E∩∂B_r) <2πrwe have H¹(E∩∂B_s)≤H¹(S∩∂B_s) ∀s > r,

providedSis a strip withH¹(S∩∂Br)=H¹(E∩∂Br). We are thus in the position to state our main result.

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Theorem 1. Let μbe a radially symmetric probability measure onR².IfE andF are two width-decreasing sets inR²,then there exist two orthogonal stripsS_EandS_F such that

μ(E∩F )

μ(E)μ(F )≥ μ(S_E∩S_F)

μ(S_E)μ(S_F). (1.1)

In other words,the correlation ratio ofμis minimized,among pairs of width-decreasing sets,over pairs of orthogonal strips.

Remark 1. LetKndenote the family of open,convex sets inRⁿ,n≥2,which are symmetric with respect to the origin.

Theorem2,in theAppendix,shows that every set in K2 is a width-decreasing set. Conversely, non-convex width decreasing sets are easily constructed:for example,ifR_π_/2denotes the counter-clockwise rotation by ninety degrees around the origin,E∈K2is an ellipse of axesb > a >0,and ifc=(a+b)/2,thenE=(E∩B_c)∪R_π_/2(E\B_c) is a width-decreasing set(sinceE is a width-decreasing set andH¹(E∩∂B_r)=H¹(E∩∂B_r)for everyr=c) which,clearly,is not convex.

Let nowγ_n=(2π)⁻^n/2e^−|^x^|²^/2dxdenote the standard Gaussian measure onRⁿ. In the caseμ=γ₂, Fubini’s theorem implies that the right-hand side of (1.1) is equal to 1. By combining these two facts with Theorem1, we provide a new justification of the planar Gaussian correlation inequality (see (1.4) below), that is completely alternative to Pitt’s semi-group approach [6], and it is based only on elementary geometric considerations. In fact, the tensorization property of the Gaussian measure is not necessary to obtain non-trivial correlation inequalities in the plane. For example, we can deduce from Theorem1the following class of correlation inequalities, which extend Pitt’s inequality to a wide class of radially symmetric probability measures.

Corollary 2. LetV:[0,∞)→Rbe a Lipschitz function,such that V(r)

r is decreasing on(0,∞), (1.2)

and thatμ=e⁻^{V (}^|^z^|⁾dzis a probability measure onR².Then

μ(E∩F )≥μ(E)μ(F ) (1.3)

for every pair of width-decreasing setsEandF inR².

Observe that, by choosingE=F=R², we immediately check the sharpness of (1.3).

The paper is divided in three sections. In Section2we define the class of width-decreasing sets, prove Theorem1, and discuss the equality cases in (1.1) under some additional assumption onμ(cf. Remark3). In Section3we prove Corollary2, and also address the case whenV(r)/risincreasingon(0,∞), see Corollary7(in particular, we obtain non-trivial correlation inequalities for all probability measuresμonR²of the formμ=c_pe^−|^z^|^pdz,p >0). Finally, in theAppendixwe show that every set inK2is a width-decreasing set (Theorem2).

We finally recall that the Gaussian correlation conjecture postulates the validity of the inequality

γ_n(E∩F )≥γ_n(E)γ_n(F ) (1.4)

for every pair of setsE, F ∈Kn,n≥2. As pointed out to us by Michael Loss, in [2], Theorem 3.2, Christer Borell proves then-dimensional Gaussian correlation inequality (1.4) for every pair of setsE, F∈Bn, where, by definition, E∈BnifEis open, 0∈E,x∈Eimplies−x∈E, and, for everyr >0 such that 0<Hⁿ⁻¹(E∩∂Br) < nωnrⁿ⁻¹, we have

Hⁿ⁻¹(E∩∂B_s)≤Hⁿ⁻¹(C∩∂B_s) ∀s > r,

providedC= {x∈Rⁿ: |x−(x·ν)ν|< h}(ν∈Sⁿ⁻¹,h >0) is such thatHⁿ⁻¹(E∩∂Br)=Hⁿ⁻¹(C∩∂Br). If n≥3, the classesBn andKndo not coincide (nor are contained one into the other), although, in some sense, they

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have a considerably large intersection. Concerning the planar case, it is easily seen that the classB2coincides with the class of width-decreasing sets inR², so that, in particular, Theorem2in theAppendiximpliesK2⊂B2. In fact, the inclusionK2⊂B2is also proved by Borell in Section 4 of his paper. However, for the sake of clarity, we have opted to include an elementary geometric proof of this result in our appendix. We also remark that the argument used by Borell in order to prove (1.4) in the classBn makes essential use of the tensorization property ofγ_n, and thus, in the planar case, it does not seem suitable to recover the more general Theorem1.

It is to be noticed that there are many other interesting results about correlation inequalities in various settings, see for instance Khatri [4], Sydak [8], Schechtman, Schlumprecht, Zinn [7], Harge [3], Kolesnikov [5], and the lecture notes [1].

2. Width decreasing sets and planar correlation inequalities

In this section we prove Theorem1. We begin with some definitions and terminology. A probability measureμonR² isradially symmetricif for every Borel setE⊂R²andθ∈(0,2π)we have

μ(E)=μ R_θ(E)

,

where R_θ:R²→R² denotes the counter-clockwise rotation around the origin by the angle θ. By a standard dis- integration argument, we see that if μis a radially symmetric probability measure and E,F are Borel sets inR², then

H¹(E∩∂B_r)≤H¹(F∩∂B_r) ∀r >0 ⇒ μ(E)≤μ(F ). (2.1)

(Here and in the sequel,H¹denotes the 1-dimensional Hausdorff measure of a set.)

Given an open setE⊂R²which contains the origin, the (normalized)angular-length function ofE,θE:(0,∞)→ [0,π/2], is defined as

θE(r)=1 4

H¹(E∩∂B_r)

r , r >0.

SinceEis open and contains the origin, the functionθ_E is always lower semicontinuous on(0,∞), and constantly equal toπ/2 in a neighborhood of 0.

We now reformulate the notion of width-decreasing set defined in the introduction in terms of angular-length functions: we say thatEis awidth-decreasing set, ifEis open, symmetric with respect to the origin, contains the origin, and, for everyr >0 such thatθE(r) <π/2, the angular-length function ofEis bounded from above on(r,∞) by the angular-length function of a stripSsuch thatθ_E(r)=θ_S(r), see Fig.1. More concisely, we ask that, for any r∈(0,∞),

θ_E(r) <π

2 ⇒ θ_E≤θ_S on(r,∞) (2.2)

for every strip S withθ_E(r)=θ_S(r). It is easily seen that if E is a width-decreasing set, then its angular-length functionsθE is decreasing.

Proof of Theorem1. Without loss of generality we can assume that bothEandFare different fromR²(otherwise the result is trivial). Recalling that by assumptionEandF are open, we can immediately check thatθ_E+θ_F:(0,∞)→ [0,π]is a decreasing, lower semicontinuous function satisfyingθ_E(0⁺)+θ_F(0⁺)=πandθ_E(+∞)+θ_F(+∞)=0.

Hence, if we set r₀=inf

r >0: θ_E(r)+θ_F(r)≤π 2

,

thenr₀∈(0,∞)andθ_E(r₀)+θ_F(r₀)≤π/2. Ifθ_E(r₀)=π/2 thenθ_F(r₀)=0, and soF⊂E. In this case, we setS_E to be any strip such thatE⊂S_E, and we setS_F to be any strip orthogonal toS_E, so to find

μ(E∩F ) μ(E)μ(F )= 1

μ(E)≥ 1

μ(S_E)≥ μ(S_E∩S_F)

μ(S_E)μ(S_F). (2.3)

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Fig. 1. Width decreasing sets. IfE∩∂Bris a proper subset of∂BrandSis any strip withH¹(E∩∂Br)=H¹(S∩∂Br), then for everys≥r we haveH¹(E∩∂Bs)≤H¹(S∩∂Bs).

Fig. 2. A squareEand a (qualitative picture) of its vertical double-cap symmetrizationE^∗.E^∗is obtained by rearranging the connected compo- nents ofE∩∂Brinto pairs of opposite circular arcs, with center on the vertical axis. The horizontal double-cap symmetrizationE_∗ofEis obtained by aπ/2-rotation ofE^∗.

The caseθ_E(r₀)=0,θ_F(r₀)=π/2 is settled by symmetry. Hence we are left to consider the case that θ_E(r₀)+θ_F(r₀)≤π

2, 0< θ_E(r₀) < π

2, 0< θ_F(r₀) <π

2. (2.4)

In this case we are going to replaceEby itsvertical double-cap symmetrizationE^∗, defined as (see Fig.2) E^∗=

r>0

re^iθ:

θ−π 2

< θ_E(r)

∪

re^iθ: θ−3π

2

< θ_E(r)

,

and, simultaneously, to replaceF by itshorizontal double-cap symmetrizationF_∗, defined as F_∗=

r>0

re^iθ: |θ|< θ_F(r)

∪

re^iθ: |θ−π|< θ_F(r) ,

where we writere^iθ=(rcosθ, rsinθ ). By construction, it is clear thatθ_E=θ_E∗andθ_F=θ_F_∗. Moreover, by (2.1), μ(E)=μ

E^∗

, μ(F )=μ(F_∗). (2.5)

Since H¹

E^∗∩F_∗∩∂B_r

=4 max

0, θE(r)+θ_F(r)−π 2

≤H¹(E∩F ∩∂B_r), (2.6)

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again by (2.1) we have μ

E^∗∩F_∗

≤μ(E∩F ), (2.7)

so that

μ(E∩F )

μ(E)μ(F )≥ μ(E^∗∩F_∗)

μ(E^∗)μ(F_∗). (2.8)

We now observe the crucial fact that the notion of width-decreasing set is invariant under both our symmetrizations, being a property only of the angular-length function. Hence, bothE^∗andF_∗are width-decreasing.

Since the circular slices ofE^∗are pairs of opposite vertical caps, the fact thatE^∗is a width-decreasing set and the property 0< θ_E(r₀) <π/2 force the inclusions

E^∗\B_r₀⊂S_E, (2.9)

SE∩Br₀⊂E^∗∩Br₀, (2.10)

whereS_Eis the vertical strip such thatθ_S_E(r₀)=θ_E(r₀)(see Fig.3). More precisely,S_E=S(ν, h)for ν=e₁, h=r₀sin

θ_E(r₀) . On the other hand, forF we find

F_∗\B_r₀⊂S_F, (2.11)

SF ∩Br₀⊂F_∗∩Br₀, (2.12)

whereS_F is the horizontal strip such thatθ_S_F(r₀)=π/2−θ_E(r₀)(observe that for general width-decreasing sets it can happen thatπ/2−θ_E(r₀) > θ_F(r₀), see Fig.3). If we now set

E=

E^∗∩B_r₀

∪(S_E\B_r₀), F=(F_∗∩B_r₀)∪(S_F\B_r₀),

then by constructionE^∗∩F_∗=E∩F(recall thatθE(r0)+θF(r0)≤π/2, soE^∗∩∂Br₀andF_∗∩∂Br₀ are disjoint).

Moreover, by (2.9) and (2.11),E^∗⊂EandF_∗⊂F. Thus, μ(E^∗∩F_∗)

μ(E^∗)μ(F_∗)≥ μ(E∩F )

μ(E)μ( F ) . (2.13)

Fig. 3. A pair of setsEandFsuch thatE=E^∗,F=F_∗, andθ_E(r₀)+θ_F(r₀) <π/2. On the right, the setE=(E^∗∩Br₀)∪(S_E\Br₀).

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Let us now notice the trivial inequality b

a ≥b−c

a−c ∀0< c < b≤a. (2.14)

Sinceθ_E(r₀)+θ_F(r₀)≤π/2 we haveB_r₀\S_E⊂S_F∩B_r₀. Hence, recalling thatE\B_r₀=S_E\B_r₀andS_F∩B_r₀⊂F, E\S_E=(E∩B_r₀)\S_E⊂E∩S_F ∩B_r₀⊂E∩F .

Therefore we may apply (2.14) to obtain μ(E∩F )

μ(E)μ( F ) ≥ μ(E∩F ) −μ(E\S_E)

(μ(E)−μ(E\S_E))μ(F )= μ(F∩S_E)

μ(S_E)μ(F ). (2.15)

Similarly, from the inclusionF\SF ⊂F∩SEand by applying again (2.14), we conclude μ(F∩S_E)

μ(SE)μ(F )≥ μ(F∩S_E)−μ(F\S_F)

μ(SE)(μ(F ) −μ(F\SF))= μ(S_E∩S_F)

μ(SE)μ(SF). (2.16)

Combining together (2.8), (2.13), (2.15) and (2.16), we finally get (1.1), so the proof is completed.

Remark 3 (A necessary condition for equality in (1.1)). Let us now discuss the sharpness of(1.1)under the assump- tion that

μ(A) >0 for every open setA⊂R².

We claim that,in this case,the inequality sign in(1.1)is strict unless E^∗andF_∗are orthogonal strips.

To verify this,let us recall that in proving(1.1)we have considered three separate cases:F ⊂E,E⊂F,or else.

In the first case,F ⊂E,we denoted byS_E any strip containingE,byS_F any strip orthogonal toS_E,and then we deduced(1.1)from the chain of inequalities(2.3),

μ(E∩F ) μ(E)μ(F )= 1

μ(E)≥ 1

μ(S_E)≥ μ(S_E∩S_F) μ(S_E)μ(S_F).

The first inequality sign is strict unlessμ(S_E\E)=0;the second inequality is then necessarily strict(unless we are in the trivial caseE=SE=R²).Therefore,in the caseF ⊂E, (1.1)is always a strict inequality,and the same holds true in the symmetric caseE⊂F.Let us now assume thatEF =∅.In this case,if(1.1)holds as an equality,we deduce from(2.13)thatμ(E^∗)=μ(E)andμ(F_∗)=μ(F ). By(2.9), (2.11)and by our assumptions onμ,this implies that

E^∗\B_r₀ =S_E\B_r₀, F_∗\B_r₀=S_F\B_r₀.

At the same time,thanks to(2.15)and (2.16),the equality sign in(1.1)implies μ(E\S_E)=μ(F\S_F)=0,that finally gives

E^∗=S_E, F_∗=S_F, using(2.10)and(2.12).

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3. Extensions of Pitt’s correlation inequality

In this section we present some classes of radially symmetric measures such that the right-hand side of (1.1) admits an explicit, non-trivial lower bound, and, in particular, we prove Corollary2. Precisely, we consider a Borel function V:[0,∞)→R, we set

f (z)=e⁻^{V (}^|^z^|⁾, z∈R², (3.1)

so thatf >0 onR², and we work with the measure

μ=f (z)dz. (3.2)

We shall assume as usual thatμis a probability measure onR², that is, we shall assume that 2π

_∞

0

e⁻^{V (r)}rdr=

R²f (z)dz=μ R²

=1. (3.3)

We begin with the following lemma, that, in combination with Theorem 1, will readily imply Corollary2. In the following, we will denote the generic point ofR²asz=(x, y).

Lemma 4. Letf, V , μbe as in(3.1), (3.2),and(3.3).Assume that

f (x, y)f (a, b)≥f (a, y)f (x, b) ∀0≤x≤a,0≤y≤b. (3.4)

Then inf

μ((−x, x)×(−y, y))

μ((−x, x)×R)μ(R×(−y, y)): x, y >0

=1. (3.5)

Proof. The fact that the infimum is less than or equal to 1 is easily seen by lettingx, y→ ∞. Let us now prove the converse inequality.

We define the function F (a, b)=μ

(0, a)×(0, b)

=1 4μ

(−a, a)×(−b, b)

, a, b >0.

SinceF >0, if we setH (a, b)=log(F (a, b)),a, b >0, thanks to (3.4) we get

∂H

∂a(a, b)= _b

0 f (a, y)dy F (a, b) ,

∂²H

∂a ∂b(a, b)=f (a, b)F (a, b)−b

0 f (a, y)dya

0 f (x, b)dx F (a, b)²

= _a

0 dx_b

0[f (a, b)f (x, y)−f (a, y)f (x, b)]dy

F (a, b)² ≥0.

In particular, for everyy >0,(∂H /∂b)(·, y)is increasing on(0,∞), so that H (a, b)−H (a, y)≥H (x, b)−H (x, y) ∀0≤x≤a,0≤y≤b, that is,

F (a, b)F (x, y)

F (a, y)F (x, b)≥1 ∀0≤x≤a,0≤y≤b. (3.6)

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Notice now thatF is separately increasing in both its variables and F (x, y)=1

4μ

(−x, x)×(−y, y)

∀0≤x≤ ∞,0≤y≤ ∞.

Therefore, by lettinga, b→ ∞in (3.6), we get (3.5).

Proof of Corollary2. By Lemma4we have only to check thatf (defined fromV by (3.1)) satisfies (3.4), which amounts in proving that

V

a²+b²

−V

x²+b²

≤V

a²+y²

−V

x²+y²

(3.7) for every 0≤x≤a≤ ∞and 0≤y≤b≤ ∞. In fact, by (1.2) we have that

y→ ∂

∂xV

x²+y²

=V(

x²+y²)

x²+y² x is decreasing on(0,∞),

from which we easily deduce (3.7).

Lemma 5. Letf, V , μbe as in(3.1), (3.2),and(3.3),and assume thatV is continuous at0and that

f (x, y)f (a, b)≤f (a, y)f (x, b) ∀0≤x≤a,0≤y≤b. (3.8)

Then inf

μ((−x, x)×(−y, y))

μ((−x, x)×R)μ(R×(−y, y)): x, y >0

= e⁻^{V (0)} (

Re⁻^{V (t )}dt )². (3.9)

Remark 6. Notice that,thanks to(3.3),the continuity ofV at0ensures that

Re⁻^{V (t)}dt <∞. Proof. The lower bound for the infimum in (3.9) is easily seen by lettingx, y→0⁺.

To prove the converse inequality, let us defineF andHas in the proof of Lemma4. Having assumed (3.8) in place of (3.4), instead of (3.6) we now get that

F (a, b)F (x, y)

F (a, y)F (x, b)≤1 ∀0< x≤a,0< y≤b. (3.10)

In particular we find that, whenever 0≤x≤a, 0≤y≤b, F (x, b)

F (x,∞)F (∞, b) ≥ F (a, b)F (x, y) F (x,∞)F (∞, b)F (a, y)

= F (a, b)(y_x

0 f (t,0)dt+o(y)) F (x,∞)F (∞, b)(y_a

0 f (t,0)dt+o(y)), that is, lettingy→0⁺,

F (x, b)

F (x,∞)F (∞, b)≥ F (a, b)_x

0 f (t,0)dt F (x,∞)F (∞, b)a

0 f (t,0)dt ∀0≤x≤a, b >0.

We now leta→ ∞to find that F (x, b)

F (x,∞)F (∞, b)≥

_x

0 f (t,0)dt F (x,∞)_∞

0 f (t,0)dt ∀x, b >0. (3.11)

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We finally notice that, again by (3.8), d

dx x

0 f (t,0)dt

F (x,∞) =f (x,0)x 0 dt_∞

0 f (t, s)ds−x

0 f (t,0)dt_∞

0 f (x, s)ds F (x,∞)²

= _x

0 dt_∞

0 [f (x,0)f (t, s)−f (t,0)f (x, s)]ds

F (x,∞)² ≥0,

so that, in particular,

x>0inf x

0 f (t,0)dt F (x,∞) = lim

x→0⁺

x

0 f (t,0)dt

F (x,∞) = f (0,0) _∞

0 f (0, s)ds. Combining (3.11) with

f (0,0)=e⁻^{V (0)},

_∞

0

f (t,0)dt= _∞

0

f (0, s)ds=

Re⁻^{V (t )}dt,

we conclude the proof of (3.9).

Corollary 7. LetV:[0,∞)→Rbe a Lipschitz function such that V(r)

r is increasing on(0,∞), (3.12)

andμ=e⁻^{V (}^|^z^|⁾dzis a probability measure onR².Then μ(E∩F )≥ e⁻^{V (0)}

(

Re⁻^{V (t )}dt )²μ(E)μ(F ) (3.13)

for every pair of width-decreasing setsEandF inR².

Proof. Arguing as in the proof of Corollary2, we now see that (3.12) implies the validity of (3.8). In particular,

combining Lemma5with Theorem1, we immediately deduce (3.13).

Appendix: Planar symmetric convex sets are width-decreasing sets

In this section we prove thatK2is a subset of the family of width-decreasing sets. This result was first proved by Borell [2]. For the sake of clarity, we include here a new elementary geometric proof of this fact.

Theorem 2. A planar convex set,symmetric with respect to the origin,is a width-decreasing set.

We first present two simple lemmas, that shall be used in the proof of the theorem. Let us first note that any stripS of widthh(i.e.,S= {x∈R²: |x·ν|< h}for someν∈S¹), satisfies

θ_S(s)=

π/2 ifs∈(0, h),

arcsin(h/s) ifs≥h, (A.1)

where arcsin :[0,1] → [0,π/2] is the inverse function of the sinus on[0,π/2]. In particular (2.2) is equivalent to check that, ifθE(r) <π/2, then

θ_E(s)≤arcsin

rsin(θE(r)) s

∀s≥r, (A.2)

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since any stripSwithθ_E(r)=θ_S(r)has widthh=rsin(θE(r)). From these remarks we easily obtain the following criterion for a set to be width-decreasing.

Lemma 8. LetEbe an open set inR²,symmetric with respect to the origin,containing the origin,and such that θE(r) <π

2 ⇒ there existsδ >0such thatθE≤θSon[r, r+δ) (A.3) for every stripSwithθ_E(r)=θ_S(r).ThenEis a width-decreasing set.

Proof. It is clear from (A.3) that θ_E is a decreasing function. In fact, with the help of (A.2), we easily deduce from (A.3) that, if we set,

D⁺θ_E(r)=lim sup

ε→0⁺

θ_E(r+ε)−θ_E(r)

ε , (A.4)

then, for everyr >0 such thatθ_E(r) <π/2, we have D⁺θE(r)≤ −tan(θE(r))

r ≤0. (A.5)

Let nowr >0 be such thatθ_E(r) <π/2, letSbe a strip such thatθ_E(r)=θ_S(r), and set r0=sup

s≥r: θE≤θSon(r, s) .

We have to prove thatr₀= ∞. Assume on the contrary thatr₀<∞. Sinceθ_E is decreasing,θ_E(r₀)≤θ_E(r) <π/2.

By (A.3) (applied withr₀in place ofr), there existsδ >0 such that

θE≤θ_S, on(r0, r0+δ), (A.6)

whereSis any strip such thatθ_S(r₀)=θ_E(r₀). The stripShas widthh=rsin(θE(r)), while the stripShas width h=r₀sin(θE(r₀)). By (A.5), the maps→ssin(θE(s))is decreasing, therefore

θ_S≤θS, on(0,∞). (A.7)

By (A.6) and (A.7) we conclude thatθ_E≤θ_Son(r, r₀+δ), against the maximality ofr₀. Thusr₀= ∞and the lemma

is proved.

Given three pointsP , Q, Rin the plane, we denote by(P QR)the angle atQdefined by the pointsP andR.

Lemma 9. Givenr >0,θ∈(0,π/2)andα∈(0,π/2−θ ),letP =re^iθ,Q=reⁱ⁽^π−^θ⁻^α).The angleβ between the linesL_{P Q}andL_OP depicted in Fig.4,satisfies

β=θ+α

2. (A.8)

Fig. 4. The situation in Lemma9.

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Proof. Since|P| = |Q|, the angles(OP Q)and(OQP )are both equal toβ. This implies that 2β+(QOP )=π,

from which the formula forβfollows immediately observing that (QOP )=π−θ−(α+θ ),

see Fig.4.

We are now in position to show Theorem2.

Proof of Theorem2. LetE∈K2, letr >0 be such that θ_E(r) <π/2, and let S=S(h, ν) be any strip of width h=rsin(θE(r)), so thatθE(r)=θS(r). We consider two cases.

CaseI.IfE∩∂B_r consists of a pair of disjoint open circular arcs.

In this case, by convexity, we easily find that, for a suitableν∈S¹, the stripS=S(h, ν)satisfies S∩B_r⊂E∩B_r.

Then, again by convexity,E\Br⊂S\Br, and thus H¹(E∩∂B_s)≤H¹(S∩∂B_s) ∀s≥r.

In particular,θ_E(s)≤θ_S(s)for everys≥r, as required.

CaseII.IfE∩∂B_r consists ofN≥2pairs of disjoint open circular arcs.

Thanks to Lemma8(see (A.5)), we are left to prove that D⁺θ_E(r) <−tan(θE(r))

r , (A.9)

whereD⁺θ_E(r)is defined as in (A.4) (note that we need the strict sign in (A.9) to obtain the validity of (A.3) for someδ >0). By assumption, we know that

E∩∂B_r= N

h=1

I_h∪J_h, N≥2,

whereI_h andJ_h are open circular arcs in∂B_r, opposite to each other (i.e.J_h= {−x: x∈I_h}). Since E is open, convex, and symmetric with respect to the origin, we find that, for everyε >0 sufficiently small

E∩∂B_r₊_ε= N

h=1

I_h^ε∪J_h^ε, (A.10)

whereI_h^εandJ_h^εare opposite open circular arcs in∂B_r₊_ε, with I_h^ε⊆r+ε

r I_h, J_h^ε⊆r+ε r J_h.

We are going to prove (A.9) from the following upper bound forH¹(I_h^ε)in terms ofH¹(I_h): whenever 1≤h≤N, H¹

I_h^ε

≤H¹(I_h)+ε r

H¹(I_h)−2rtan θ_E(r)

+o(ε), (A.11)

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asε→0. Before coming to the proof of (A.11), let us see why it does imply (A.9). From (A.10) we find that θ_E(r+ε)−θ_E(r)=1

4

H¹(E∩∂B_r₊_ε)

r+ε −H¹(E∩∂B_r) r

= 1 2r

N

h=1

H¹

I_h^ε 1−ε

r

−H¹(I_h)

+o(ε)

≤ 1 2r

N

h=1

−2εtan θ_E(r)

+o(ε)= −ε rNtan

θ_E(r) +o(ε).

Dividing byεand lettingε→0, we immediately find (A.9) (note that we can achieve the strict sign thanks to the fact thatN≥2).

We are thus left with proving (A.11). Without loss of generality, we can argue onI₁andI₁^ε. Up to a rotation we can assume that

I₁=

re^it: |t|< θ

, J₁=

−re^it: |t|< θ , whereθsatisfies

0< θ < θ_E(r) (A.12)

(in particular,θ <π/2, and the pointP =re^iθ belongs to the relative boundary ofI₁in∂B_r). Let nowϕ∈(0,π)be such that the pointQ=re^iϕ satisfies

Q∈

E∩∂B_r⁺

\(I₁∪J₁),

where∂B_r⁺=∂B_r∩ {x₂>0}. It is clear thatϕcannot be too close to 0 or toπ. More precisely, if we defineαso that ϕ=π−θ−α,

the fact that H¹

E∩∂B_r⁺

\(I₁∪J₁)

=r

2θE(r)−2θ , gives us the estimate

α >2θE(r)−2θ, (A.13)

see Fig.5. Now, letLP Qdenote the line passing throughP andQ, and let

P_ε=(r+ε)e^{iθ (ε,ϕ)}, (A.14)

be the point satisfying

P_ε∈L_{P Q}∩∂B_r₊_ε, lim

ε→0⁺

P_ε=P ,

(see Fig.5). SinceEis convex and open, we have the inclusion I₁^ε⊂

(r+ε)e^it: |t|< θ (ε, ϕ) ,

from which we derive the following upper bound forH¹(I₁^ε):

H¹ I₁^ε

≤2(r+ε)θ (ε, ϕ).

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Fig. 5. To prove (A.11) we use the boundH¹(I₁^ε)≤2(r+ε)θ (ε, ϕ), and we show thatθ (ε, ϕ)=θ−εγ+O(ε²), withγ≥tan(θ_E(r))/r.

Let us now estimateθ (ε, ϕ).

Set v =e^iθ,w=e^i(θ^−π^/2), and define the points P_ε=P +εv, P_ε=P_ε +εγ rw, where γ >0 is such that P_ε∈L_{P Q}. Let us observe that|P_ε−P_ε| ≤Cε², or equivalently

(r+ε)e^{iθ (ε,ϕ)}=P_ε=P +εv+εγ rw+O ε²

, from which we get

(r+ε)tan

θ−θ (ε, ϕ)

=εγ r+O ε²

,

see Fig.5. Hence, using that tan(δ)=δ+O(δ²)forδsmall, we easily obtain θ (ε, ϕ)=θ−εγ+O

ε² .

Now, if we defineβ=(P_εP P_ε), then we haveγ r=tan(β), and by Lemma9 β=θ+α

2.

By (A.13), this implies the crucial estimate γ≥tan(θE(r))

r . (A.15)

Collecting all together and recalling thatH¹(I₁)=2rθ, we finally obtain H¹

I₁^ε

≤2(r+ε)θ (ε, ϕ)=2(r+ε)(θ−εγ )+O ε²

=2rθ+ε(2θ−2rγ )+O ε²

=H¹(I1)+ε r

H¹(I1)−2r²γ +O

ε² ,

which combined with (A.15) leads to (A.11), and hence to the proof of the theorem.

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Acknowledgments

We thank Franck Barthe, Almut Burchard, Eric Carlen, Michael Loss, and Frank Morgan for stimulating at various stages the writing of this paper. The authors acknowledge the hospitality at the Mathematisches Forschungsinstitut Oberwolfach during the 2010 workshop “Calculus of Variations,” where the present work was conceived. This work has been partially supported by the NSF Grant DMS-0969962, the ERC Starting Grant n. 258685 ANOPTSETCON, and the ERC Advanced Grant n. 226234 ANTEGEFI.

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