ON THE GAUSSIAN CONCENTRATION INEQUALITYAND ITS RELATION TO THE GAUSSIAN SURFACE

AREA (PRELIMINARY VERSION)

GALYNA LIVSHYTS

Abstract. Let γ2 be a standard Gaussian measure in Rn. Fora given measurable set Q in Rn, let HQ be a half space in Rn

such that γ2(Q) = γ2(HQ). The classical Gaussian concentrationinequality states that for all measurable sets Q ⊂ Rn and for allh > 0,

γ2(Q+ hBn2 ) ≥ γ2(HQ + hBn

2 ).

Under some minor constrains on the set Q we obtain an improve-ment of the latter inequality in a certain range of h, depending onthe Gaussian surface area of Q.

1. Introduction

We denote standard Gaussian measure on Rn by γ2. For a measur-able set Q ⊂ Rn,

γ2(Q) =

∫Q

e−|y|22 dy.

We recall that the Minkowski surface area of a convex set Q withrespect to the Standard Gaussian measure is defined to be

(1) γ2(∂Q) = lim infε→+0

γ2((Q+ εBn2 )\Q)

ε,

whereBn2 denotes Euclidian ball in Rn and “+” stands for the Minkowski

addition of sets.Sudakov, Tsirelson [8] and Borell [4] proved, that among all convex

sets of a fixed Gaussian measure, half spaces have the smallest Gaussiansurface area. On the other hand, it was shown by Ball [1], that theGaussian surface area of a convex set in Rn is asymptotically boundedfrom above by Cn

14 , where C is an absolute constant. Nazarov [7]

2010 Mathematics Subject Classification. Primary: 44A12, 52A15, 52A21.Key words and phrases. convex bodies, convex polytopes, Surface area, Gauss-

ian measures.1

2 GALYNA LIVSHYTS

proved the sharpness of Ball’s result:

(2) 0.28n14 ≤ max

Qγ2(∂Q) ≤ 0.64n

14 ,

where the maximum is taken over all convex sets Q in Rn.Let (X,µ) be a compact metric space with a Borel probability mea-

sure µ. Let B ⊂ X be a unit ball. The concentration function α(X, h)is defined to be

α(X, h) = supµ(A)≥ 1

2

(1− µ(A+ hB)) ,

where A is always a Borel subset of X (see [5], page 16).Analogously, for a measurable set Q ⊂ Rn we define a function

αQ(h) : R+ → R

by

αQ(h) := 1− γ2(Q+ hBn2 ).

It is well known (see, for example, [5], [2] or [3]) that for everymeasurable Q ⊂ Rn such that γ2(Q) ≥ 1

2,

(3) αQ(h) ≤ 1

2e−

h2

2 .

Moreover,

(4) γ2(Q+ hBn2 ) ≥ γ2(HQ + hBn

2 ),

where HQ is a half space such that γ2(Q) = γ2(HQ) (Theorem 1.2 in[5]).

In the present preprint we observe the relation of the estimates forαQ(h) with the Gaussian surface area γ2(∂Q). For some sets Q it allowsus to improve the inequality (4) for certain range of h. Namely, weprove the following

Theorem 1.1. For any convex set Q ⊂ Rn containing the origin and

for any 0 ≤ h ≤ 4√n√

πγ2(∂Q),

(5) γ2(Q+ hBn2 ) ≥ γ2(Q) +

√πγ2(∂Q)2

8√n

·(

1− e−√n√

πγ2(∂Q)h

).

Theorem 1.1 implies that for every convex set Q containing the ori-gin, and for every h > 0,

(6) αQ(h) ≤ 1− γ2(Q)−√πγ2(∂Q)2

8√n

·(

1− e−√n√

πγ2(∂Q)h

).

ON THE GAUSSIAN CONCENTRATION INEQUALITY 3

Let Q be a measurable set in Rn such that γ2(Q) = γ2(Hr), whereHr = {x ∈ Rn | 〈x, rθ〉 < 0} for a unit vector θ. The classical concen-tration (4) implies that for every set Q, and for every h > 0,

(7) αQ(h) ≤ 1− γ2(Q)− 1√2π

∫ r+h

r

e−t2

2 dt.

Claim 1. Let Q be a convex set in Rn such that γ2(Q) ≥ 12and

γ2(∂Q) ≥ 8√2π. Then the estimate (6) is stronger then (7) for all

h ∈ [0, cγ2(∂Q) log γ2(∂Q)√n

], where c is an absolute constant.

Proof. We observe that r ≥ 0 since γ2(Q) ≥ 12. Thus∫ r+h

r

e−t2

2 dt ≤∫ h

0

e−t2

2 dt.

Hence it suffices to show that

F (h) :=

√πγ2(∂Q)2

8√n

·(

1− e−√n√

πγ2(∂Q)h

)− 1√

2π

∫ h

0

e−t2

2 dt ≥ 0

on the interval [0, cγ2(∂Q) log γ2(∂Q)√n

]. We find

F ′(h) =γ2(∂Q)

8e−

√n√

πγ2(∂Q)h − 1√

2πe−

h2

2 .

By taking the logarithm we obtain that F ′(h) ≥ 0 if and only if

h2 − 2√n√

πγ2(∂Q)h+ 2 log

√2πγ2(∂Q)

8≥ 0,

which happens in particular if h ∈ [0, cγ2(∂Q) log γ2(∂Q)√n

] (here we used the

fact that γ2(∂Q) ≥ 8√2π

). We observe also that F (0) = 0. The function

F (h) is increasing on the interval [0, cγ2(∂Q) log γ2(∂Q)√n

], and thus positive.

Which implies the Claim. �

2. Proof of Theorem 1.1

Let Q be a convex set in Rn containing the origin. It was shown in[7] (page 3) that

(8) γ2(∂Q) ≤ maxy∈∂Q

√n√

π〈y, ny〉,

where ny stands for the normal vector at y.

4 GALYNA LIVSHYTS

The idea of the proof of the latter estimate is to consider a “polarcoordinate system” associated with the body Q and to write

1 = γ2(Rn) =1

(√

2π)n

∫Q

e−|y|22 dy =

(9)1

(√

2π)n

∫∂Q

∫ ∞0

D(y, t)e−|X(y,t)|

2 dtdσ(y),

where D(y, t) is the Jacobian of the change (y, t) → X(y, t). Theinequality (8) follows when X(y, t) = yt.

Another estimate useful for the proof was shown in [6] (equation(78)) following the idea of [7]:

(10) γ2(Q+ hBn2 ) ≥ γ2(Q) +

1

(√

2π)n

∫∂Q2

∫ h

0

e−|y+tny |2

2 dtdσ(y).

The idea of the proof of the latter fact is to consider X(y, t) = y + tnyand apply the argument similar to (9).

We prove the following

Lemma 2.1. Let Q be a convex set in Rn. Let ρ be any positive number.(i) For any h ∈ [0, 2ρ],

(11) γ2(Q+ hBn2 ) ≥ γ2(Q) +

(γ2(∂Q)−

√n√πρ

)· 1

2ρ· (1− e−2ρh).

(ii) For any h ≥ 2ρ,

γ2(Q+hBn2 ) ≥ γ2(Q)+

(γ2(∂Q)−

√n√πρ

)(1

2ρ· (1− e−4ρ2) + (h− 2ρ) · e−h2

).

Proof. Fix ρ > 0. We split the surface of the body into two parts:

S1 = {y ∈ ∂Q : 〈y, ny〉 ≥ ρ}and

S2 = {y ∈ ∂Q : 〈y, ny〉 < ρ}.By (8), γ2(S1) ≤

√n√πρ

. Thus,

(12) γ2(S2) ≥ γ2(∂Q)−√n√πρ.

The inequality (10) entails that

γ2(Q+ hBn2 ) ≥ γ2(Q) +

1

(√

2π)n

∫S2

∫ h

0

e−|y+tny |2

2 dtdσ(y),

since S2 ⊂ ∂Q.

ON THE GAUSSIAN CONCENTRATION INEQUALITY 5

We observe, that for any y ∈ S2,

|y + tny|2 = |y|2 + t2 + 2t〈y, ny〉 ≤ |y|2 + t2 + 2tρ.

Thus

γ2(Q+ hBn2 ) ≥ γ2(Q) +

1

(√

2π)n

∫S2

∫ h

0

e−|y|2+t2+2tρ

2 dtdσ(y) =

γ2(Q) + γ2(S2) ·∫ h

0

e−t2+2tρ

2 dt.

Using (12), we obtain that

(13) γ2(Q+ hBn2 ) ≥ γ2(Q) + (γ2(∂Q)−

√n√πρ

) ·∫ h

0

e−t2+2tρ

2 dt.

If h ≤ 2ρ, then t2 ≤ 2tρ for every t ∈ [0, h]. Thus in this case

(14)

∫ h

0

e−t2+2tρ

2 dt ≥∫ h

0

e−2tρdt =1

2ρ· (1− e−2ρh).

For h ≥ 2ρ we estimate∫ h

0

e−t2+2tρ

2 dt ≥∫ 2ρ

0

e−2tρdt+

∫ h

2ρ

e−t2

dt ≥

1

2ρ· (1− e−4ρ2) + (h− 2ρ)e−h

2

.(15)

Gluing (13) and (14) together we obtain the first part of the Lemma;the second part follows from (13) and (15). �

To finish the proof of Theorem 1.1 we plug ρ = 2√n√

πγ2(∂Q)into (11).

�Theorem 1.1 is one of the possible corollaries of Lemma 2.1 which

illustrates the use of the estimate; however, Lemma 2.1 may be ofseparate interest and imply other estimates which may be better indifferent ranges of h.

References

[1] K. Ball, The reverse isoperimetric problem for the Gaussian measure, DiscreteComput. Geometry, 10 (1993), 411-420.

[2] S. G. Bobkov, Spectral gap and concentration for some spherically symmetricprobability measures, Lect. Notes Math. 1807 (2003), 37-43.

[3] S. G. Bobkov, Gaussian concentration for a class of spherically invariant mea-sures, Journal of Mathematical Sciences, Vol. 167, No. 3 (2010), 326-339.

[4] C. Borell, The Brunn-Minkowski inequality in Gauss spaces, Invent. Math 30(1975), 207-216.

[5] M. Ledoux, M. Talagrand, The probability in Banach Space. Isoperymetry andprocesses, Springer-Verlag, Berlin 1991.

6 GALYNA LIVSHYTS

[6] G. V. Livshyts, Maximal surface area of a convex set in Rn with respect to logconcave rotation invariant measures, GAFA seminar notes, to appear.

[7] F. L. Nazarov, On the maximal perimeter of a convex set in Rn with respect toGaussian measure, Geometric Aspects of Func. Anal., 1807 (2003), 169-187.

[8] V. N. Sudakov and B. S. Tsirel’son, Extremal properties of half-spaces for spheri-cally invariant measures. Problems in the theory of probability distributions, II.Zap. Nauch. Leningrad Otdel. Mat. Inst. Steklov 41 (1974), 14-24 (in Russian).