STRONG CENTRAL LIMIT THEOREM FOR

STRONG CENTRAL LIMIT THEOREM FOR ISOTROPIC RANDOM WALKS IN Rd ˙ PIOTR GRACZYK, JEAN-JACQUES LOEB, AND TOMASZ ZAK Abstract. We prove an optimal Gaussian upper bound for the densities of isotropic random walks on Rd in spherical case (d ≥ 2) and ball case (d ≥ 1). In the one-dimensional case of symmetric Bernoulli distributions we prove an optimal Gaussian upper bound for the tails of the distribution functions. In all cases we deduce a strong version of the Central Limit Theorem for the isotropic random walks. We apply our results to get strong hipercontractivity inequalities and strong log-Sobolev inequalities.

1. Introduction Let σd be the normalized Lebesgue measure on the unit sphere S1d−1 . For d = 1 we put σ1 = 21 (δ−1 + δ1 ). Let X1 , X2 , ..., Xn be independent random vectors with the distribution σd . The sequence (Sn )n , where S0 = 0 and Sn = X1 + ... + Xn , describes an isotropic random walk in Rd . Evidently, the distribution of X1 + ... + Xn is given by σd∗n (written shortly σdn ). Similarly, we will consider µd , the uniform measure of the unit ball Bd in Rd , d ≥ 1 and the related isotropic random walk. Isotropic random walks were thoroughly examined; the history, useful facts and formulae concerning them can be found e.g. in the monograph of Hughes [12]. One of important applications of isotropic random walks is that in Rd , d ≥ 1, we can approximate the standard Gaussian measure by (normalized) convolution powers of σd or µd . Sn If Sñ = √ denotes the normalized random walk, by the Central Limit Theorem, the sequence n Sñ converges weakly to a Gaussian random vector Y . In Section 2 of this article we show that for d ≥ 2 the densities of Sñ are bounded by Cd gY where gY is the density of the limiting Gaussian distribution (Theorems 2.1 and 2.4) in both isotropic cases σd and µd , as well as in some other related cases, see Subsection 2.3. Gaussian bounds of convolutions of compactly supported symmetric measures have been proved by Hebisch and Saloff-Coste ([11]) but our optimal estimate by the limiting density gY cannot be deduced from [11]. As a corollary of this optimal estimate we prove that (1) Ef (Sñ ) → Ef (Y ) for all f ∈ L1 (gY dy), i.e. for any function f integrable with respect to the law of Y . It is an essential strengthening of the classical Central Limit Theorem for the sequence Xn . Sn In the case of σd for d = 1 the distributions of Sñ = √ are discrete measures hence in this n case we compare their tails with the Gaussian tail. In Section 3 we show an analogous Gaussian upper bound for tails. Theorem 3.1 generalizes the Hoeffding inequality and improves results of Talagrand [14] in the case of symmetric Bernoulli variables. Next we prove that (2) Ef (Sñ ) → Ef (Y ) for all monotonic f ∈ L1 (gY dy). We also show an example that without some additional assumption (such as monotonicity) the convergence in (2) may fail. 2000 Mathematics Subject Classification: 60G50, 60F05, 60B10, 47D06 Key words and phrases: Random walks, Central Limit Theorem, Gaussian estimates, Logarithmic Sobolev Inequality This research was partially supported by grants MNiSW N N201 373136 and ANR-09-BLAN-0084-01. 1

2

˙ P.GRACZYK, J.J.LOEB, AND T.ZAK

As an application, in Section 4 we show how to use our strong Central Limit Theorems to obtain direct proofs of strong hypercontractivity and strong logarithmic Sobolev inequalities for log-subharmonic functions and the Gaussian measure in Rd . This approach mirrors, to some extent, Gross’s proof of the Gaussian log-Sobolev inequality in [9]. 2. Isotropic random walks in Rd for d ≥ 2 2.1. Sphere case σd . If X = (X (1) , X (2) , ..., X (d) ) is a random vector with the distribution σd , then E(X) = (0, 0, ..., 0) and Cov(X (i) , X (j) ) = d1 δij , that is, the covariance matrix of X is Σ = d1 I; in order to justify the last assertion observe that if Y = (Y1 , ..., Yd ) is a standard Gaussian vector in Rd , then |YY | has the uniform distribution on S1d−1 . 2 d d/2 − d |x| e 2 on Rd . Define γd = N (0, d1 I)), the Gaussian measure with the density gd (y) = 2π Let X1 , X2 , ..., Xn be independent random vectors with the distribution σd . By the Central n √ tends weakly to γd . In the Limit Theorem the distribution of normalized sum Sñ = X1 +...+X n n language of convolutions this means that after normalization, σd tend weakly to γd . It turns out that for n > d + 2 the measures σdn are absolutely continuous with respect to the Lebesgue measure. Indeed, the characteristic function of σdn equals !n d/2−1 d 2 (ˆ σd (y))n = Γ( ) Jd/2−1 (|y|) 2 |y| where Jd/2−1 is a Bessel function of the first kind (see e.g. [12] (2.30)) When n > d + 2 this characteristic function is absolutely integrable, which implies that the density of σdn exists and is bounded and continuous. For fixed d let us denote by fn (x) the density of σdn . We notice that for n ≤ d + 2, the density fn can exist and be unbounded, see [12]. By the Local Central Limit Theorem ([4] Th.19.1), the densities fSñ of the normalized sums √ d √ Sn ˜ Sn = √ n) fn ( n x), tend uniformly to the Gaussian density gd (y) = , i.e. the functions ( n d |x|2 d/2 d e− 2 . In the main theorem of this section we show that the quotient of these densities 2π and of the limiting Gaussian density is bounded. Theorem 2.1. (Optimal Gaussian Bound for σdn ). Let d ≥ 2. There exists a constant Cd such that for all x ∈ Rd and n > d + 2 there holds d/2 √ d √ d |x|2 d (3) fSñ (x) = ( n) fn ( n x) ≤ Cd e− 2 . 2π Remark 2.2. For the random walk Sn = X1 + ... + Xn the inequality (3) reads as follows: for any x ∈ Rd and n > d + 2, d/2 d |x|2 d (4) fn (x) ≤ Cd e− 2n . 2π n The measures σd and γd are rotationally invariant and so are their densities. Denote fñ (r) = fn (|x|) for |x| = r. The estimates (3) and (4) are equivalent to the following inequality d r2 Cd (5) fñ (r) ≤ d/2 e− 2n . n To simplify the notation we write Cd for a modified constant. In order to prove Theorem 2.1, we will justify (5). Proof. From the Local Central Limit Theorem it follows that for any a > 0 there exists some ca > 0 such that (3) holds for all x with |x| ≤ a and n > d + 2, that is d/2 √ d √ d |x|2 d ( n) fn ( n x) ≤ ca e− 2 . 2π

STRONG CLT FOR ISOTROPIC RANDOM WALKS ON Rd

3

Equivalently, for all n > d + 2 d/2 √ d r2 d r ≤ a n. e− 2n , 2π n √ Hence it is enough to prove (5) for r ≥ a n. The maximal distance of Sn to the origin after n steps of the walk is less or equal to n so that fn (x) = 0 for |x| > n. Consequently, when n = d + 3, the inequality (5) holds for all r and a constant C. √ Let a = 3 d and Cd = max(ca , C). We will show by induction that (5) holds for this Cd . As we have noticed above, (5) is true for n = d + 3. We have Z (6) fn+1 (x) = fn (x − u)dσd (u). fñ (r) ≤ ca

S1d−1

The function fn depends only on |x − u| = (r2 + 1 − 2r cos(x, u))1/2 . The function fn+1 is also radial, so fn+1 (x) = fn+1 (0, . . . , |x|). For x0 = (0, . . . , |x|) we have cos(x0 , u) = ud . We obtain Z p fñ ( r2 + 1 − 2rud )dσd (u). fñ+1 (r) = S1d−1

Taking spherical coordinates in Rd with ud = R cos φ (cf. [12] p.61) we get Z |S1d−2 | π ˜ p 2 ˜ (7) fn+1 (r) = d−1 fn ( r + 1 − 2r cos φ) sind−2 φ dφ, |S1 | 0 where |S1d−1 | = 2π d/2 /Γ(d/2) is the measure of the unit sphere in Rd . Note that formula (7) is also true for d = 2 with |S10 | = 2. Suppose that for some n > d + 2 and all 0 ≤ r ≤ n the inequality (5) is true. Then, by this assumption, we have Z |S1d−2 | π ˜ p 2 ˜ fn ( r + 1 − 2r cos φ) sind−2 φ dφ ≤ fn+1 (r) = d−1 |S1 | 0 Z π d(r 2 +1−2r cos φ) Cd Γ(d/2) − 2n √ e sind−2 φ dφ = d/2 n πΓ((d − 1)/2) 0 Z π d(r 2 +1) dr cos φ Cd Γ(d/2) − 2n √ e e n sind−2 φ dφ = d/2 n πΓ((d − 1)/2) 0 n d/2−1 2 d(r +1) √ Cd Γ(d/2) dr − 2n d/2−1 = d/2 √ e π2 Γ((d − 1)/2) Id/2−1 = n dr n πΓ((d − 1)/2) d/2−1 Cd d − d(r2 +1) 2n dr Γ e 2n Id/2−1 , d/2 n 2 dr n because, by ([2] 9.6.18), we have Z π √ (8) ex cos θ sind−2 θ dθ = π2d/2−1 Γ((d − 1)/2)x1−d/2 Id/2−1 (x), 0

where Iα is a modified Bessel function of order α. √ We want to show that for r ≥ a n + 1 the last estimate for fñ+1 (r) is less than √ d r2 − 2(n+1) Cd e , i.e. to prove that for r ≥ a n+1 d/2 (n+1) d/2−1 d/2 d(r 2 +1) dr 2 d 2n dr n Γ Id/2−1 ≤ e 2n − 2(n+1) . 2 dr n n+1


4

But d(r2 + 1) dr2 dr2 d − = + , 2n 2(n + 1) 2n(n + 1) 2n hence it is enough to show that d/2 d/2−1 dr 2 2n dr n 1/n d e 2n(n+1) . Id/2−1 ≤ e Γ 2 dr n n+1 Now n 1/n n 1 e = 1 + + ... > 1, n+1 n+1 n √ hence it is enough to prove that for r ≥ a n + 1 d/2−1 dr 2 2n dr d Id/2−1 ≤ e 2n(n+1) . (9) Γ 2 dr n For this, we use Taylor expansions of both sides. By the well-known formula ([2] 9.6.10) ∞

X Iα (x) (x/2)2k = (x/2)α k!Γ(α + k + 1) k=0

(10)

and thus the left-hand side of (9) is equal to X X ∞ ∞ dr 2k ) ( 2n ( nr )2k ( d2 )2k d d LHS = Γ =Γ . 2 k=0 k!Γ(d/2 + k) 2 k=0 k!Γ(d/2 + k) The right-hand side of (9) is equal to RHS = e

dr 2 2n(n+1)

n r 2 d ) · 2 · n+1 (n

=e

=

∞ n k X ) ( nr )2k ( d2 )k ( n+1 k=0

k!

,

hence the difference " k k # ∞ X ( nr )2k ( d2 )k n Γ(d/2) d RHS − LHS = − . k! n+1 Γ(d/2 + k) 2 k=0 k Γ(d/2) n d k Denote the quantity in the square bracket: An,k = n+1 − Γ(d/2+k) . It is easy to check 2 1 that An,0 = 0 and An,1 = − n+1 . Now we show that for k = 2, 3, ..., we have An,k > 0. Indeed, for k = 2 2 2 2 2 n d n d Γ(d/2) 1 An,2 = − = − d d = n+1 Γ(d/2 + 2) 2 n+1 2 ( + 1) 2 2 2 2 n d/2 n d = − d − > 0, if n ≥ d + 2. n+1 n+1 d+2 +1 2 For k ≥ 3 we estimate: k k k Γ(d/2) d n (d/2)k−1 n An,k = − = − d > 0, n+1 Γ(d/2 + k) 2 n+1 ( 2 + 1)...( d2 + k − 1) 2 n d , and for k ≥ 3 this implies because for k = 2 we chose n such that n+1 > d+2 k−1 k (d/2)k−1 n 2 n ≤ ( ) < . n+1 n+1 ( d2 + 1)...( d2 + k − 1)


5

Finally, # 2 " 2 r 2 d 1 1 r 4 d n d RHS − LHS > − · + − = n+1 n 2 2! n 2 n+1 d+2 " ! # r 2 d 1 r 2 d n 2 d 1 − − . n 2 2 n 2 n+1 d+2 n+1 √ √ Now, if r > a n + 1 > a n then ! ! 2 2 1 r 2 d n d 1 da2 n d − > − . 2 n 2 n+1 d+2 n 4 n+1 d+2 −1 √ 2 n d When d ≥ 2 and n ≥ d + 2 we have n+1 − d+2 < 2d2 . Thus for a ≥ 3 d the following inequality holds ! 2 1 da2 n d 1 − > . n 4 n+1 d+2 n+1 We can now deduce the following strengthened Central Limit Theorem. Theorem 2.3. (Strong CLT for σd ) Let d ≥ 2, Y be a random Gaussian vector with law γd and f ∈ L1 (γd ). Then limn Ef (Sñ ) = Ef (Y ). √ √ n √ Proof. Let kn (x) = nfn ( n x) be the density of the normalized sum Sñ = X1 +...+X . By the n 1 d Local Central Limit Theorem, limn f knR= f gd . By Theorem R 2.1, |f kn | ≤ Cd |f gd | ∈ L (R ). By the Dominated Convergence Theorem f (x)kn (x) dx → g(x)dγd (x). 2.2. Ball case µd . Let (X1 , X2 , ..., Xd ) be a random vector with the uniform distribution µd on the unit ball Bd in Rd . When d ≥ 2, we have in polar coordinates (for 0 ≤ r ≤ 1 and s ∈ S1d−1 ) dµd (r, s) = drd−1 dr Now σd (ds) = interval.

ds |S1d−1 |

ds |S1d−1 |

is a probability measure on the unit sphere and so is drd−1 dr on the unit

R1 R1 1 d , hence E(Xi2 ) = d+2 and Because E(X12 + ... + Xd2 ) = 0 r2 · drd−1 dr = d 0 rd+1 dr = d+2 1 the covariance matrix of (X1 , X2 , ..., Xd ) equals d+2 Id. Observe that for d = 1 the formula 1 E(X12 ) = d+2 also holds. 1 n √ The weak limit of normalized sums Sñ = X1 +...+X is νd = N (0, d+2 I), the Gaussian measure n (d+2) |x|2 d/2 with the density hd (y) = d+2 e− 2 on Rd . The Gaussian measure approximating µn , 2π d/2 (d+2)R2 d+2 the law of Sn = X1 + ... + Xn , has the radial part g(R) = e− 2n . Denote by fn 2πn the density of µnd . Theorem 2.4. (Optimal Gaussian Bound for µnd ). Let d ≥ 1. There exists a constant Cd such that for all x ∈ Rd and n ≥ 1 there holds d/2 √ d √ (d+2) |x|2 d+2 (11) fSñ (x) = ( n) fn ( n x) ≤ Cd e− 2 . 2π

6


Proof. Similarly as in the proof of Theorem 2.1, by the Local CLT we see that it is sufficient to prove that there exist constants C, a > 0 such that √ (d+2)R2 C (12) fñ (R) ≤ d/2 e− 2n , R > a n, n ∈ N. n We set a = 2. The proof of the inequality (12) will proceed by induction. The starting point of induction n0 will be specified later. There is a constant C = C(n0 ) such that (12) holds for all R > 0 and n = 1, . . . , n0 . Analogously to (6) we have Z (13) fn+1 (x) = fn (x − u)dσd (u). Bd

Here we divide the proof into two cases d ≥ 2 and d = 1. Case d ≥ 2. Applying the polar coordinates, similarly as the formula (7), we obtain Z Z p |S1d−2 | 1 π ˜ fn ( r2 + R2 − 2rR cos φ) drd−1 sind−2 φ dφdr. fn+1 (R) = d−1 |S1 | 0 0 Using the assumption fñ (R) ≤

Cd nd/2

e−

(d+2)R2 2n

and the formula (8) we get (d+2)rR I Z 1 2 2 d/2−1 (d+2)(r +R ) n d Cd 2n drd−1 e− Γ fñ+1 (R) ≤ d/2 d/2−1 dr. n 2 (d+2)rR 0 2n

√

We want to prove that for n sufficiently big and all R > a n there holds the following inequality (d+2)rR Z 1 d/2 I d/2−1 (d+2)R2 (d+2)(r 2 +R2 ) n d n − 2(n+1) d−1 − 2n Γ dr e e . d/2−1 dr ≤ 2 n+1 (d+2)rR 0 2n

This is equivalent to the following (d+2)rR Z 1 d/2 I 2 d/2−1 (d+2)R2 (d+2)r n d n d−1 − 2n 2n(n+1) . Γ dr e dr ≤ e d/2−1 2 n+1 (d+2)rR 0 2n 2

When 0 < r < 1 and n > d + 2, we have 0 < (d+2)r < 12 so that for such r and n there holds 2n 2 (d+2)r 2 (d + 2)r2 1 (d + 2)r2 d + 2 (d + 2)r2 − 2n ≤1− e + ≤1− 1− . 2n 2 2n 4n 2n Thus, by positivity of the function Id/2+1 , it is enough to prove that ! (d+2)rR d/2 Z 1 d+2 2 I d/2−1 (d+2)R2 1 − 4n (d + 2)r n n d d−1 2n(n+1) . dr 1− dr ≤ e (14) Γ d/2−1 2 2n n+1 (d+2)rR 0 2n

have to evaluate two integrals involving Bessel functions. By ([8](5.52.1), page 624) R We p+1 x Ip (x) dx = xp+1 Ip+1 (x), hence (d+2)R Z 1 nI d n (d + 2)rR 2 rd/2 I d −1 dr = . 2 n (d + 2)R 0


7

Integrating by parts and using identity xIp−1 (x) − xIp+1 (x) = 2pIp (x) ([8](8.486.1), page 918), we get i h (d+2)R (d+2)R Z 1 + (d + 2)RI d +2 n dnI d +1 n n (d + 2)rR 2 2 rd/2+2 I d −1 dr = . 2 2 2 n (d + 2) R 0 Consequently, the left-hand side of (14) is 

 (d+2)R (d+2)R I I d d 2 +1 + 1)  n n  2 2 2 + (d + 2)(4n − d − 2) . 8n − 4n(d + 2) + (d + 2)  d/2 d/2+1  8n2 (d+2)R (d+2)R

Γ( d2

2n

Write g1 = ∞ X

Γ( d2 +1)

and g2 =

Γ( d2 +k+1)

(d+2)R 2n

Γ( d2 +k+2)

. By the formula (10) the last quantity is equal

2k

k!

k=0

Γ( d2 +1)

2n

8n2 − 4n(d + 2) + (d + 2)2 (d + 2)(4n − d − 2) g1 + g2 2 8n 8n2

.

The right-hand side of (14) has the following expansion: 2k (d+2)R d/2 d2 +k k ∞ 2 X (d+2)R 2n n n 2 2n(n+1) e . = n+1 k! n+1 d+2 k=0 Thus inequality (14) is equivalent to the following inequality for expansions: 2k (d+2)R ∞ X 2n Bn,k (15) ≥ 0, k! k=0 where for fixed d ≥ 2 we define the coefficients d2 +k k (d + 2)(4n − d − 2) n 2 8n2 − 4n(d + 2) + (d + 2)2 − g2 . Bn,k = − g1 2 n+1 d+2 8n 8n2 We have

d/2

4nd − d(d + 2) 8n2

d(d + 2)(d + 4) , as n → ∞. 48n3 √ We will show that Bn,k > 0 for all k ≥ 3 and that for R > a n there holds 2 4 (d + 2)R 1 (d + 2)R + Bn,2 > 0. (16) Bn,0 + Bn,1 2n 2 2n Bn,0 =

n n+1

−1+

∼ −

We have d/2+1 2 n 2 4n(d + 2) − (d + 2)2 4n − d − 2 Bn,1 = − 1− − 2 2 d+2 n+1 d+2 8n 2n (d + 4) √ and if we put R = a n in the second term of the inequality (16), then we get √ 2 (d + 2)a n a2 (d + 2)2 Bn,1 ∼ − , as n → ∞. 2n 2(d + 4)n2 √ Similarly, the third term of the inequality (16) for R = a n satisfies √ 4 1 (d + 2)a n a4 (d + 2)2 Bn,2 ∼ , as n → ∞. 2 2n 4(d + 4)n2


8

In order that inequality (16) holds for n > n0 (d), the parameter a must fulfill the condition d(d + 2)(d + 4) a2 (d + 2)2 a4 (d + 2)2 1 − − + > 0, n2 48n 2(d + 4) 4(d + 4) which is true for n > n0 (d), if only a4 (d + 2)2 a2 (d + 2)2 > , 4(d + 4) 2(d + 4) √ 4 2 and this is true for a > 2. In the bi–squared √ polynomial p(x) = Ax + Bx + C√from (16), we have A > 0 and p(0) = C < 0, hence p(a n) > 0 implies p(R) > 0 for all R > a n and the inequality (16) holds for such R. k d2 +k 2 n Finally let us show that Bn,k > 0 if k ≥ 3. Let ε > 0. We have Bn,k = d+2 [ n+1 − E] where (4n − d − 2) d + 2 d + 2k + 4 (d + 2)k−1 (d + 2)k−1 1+ · · < (1 + ε) , E= (d + 4)...(d + 2k) 4n n d + 2k + 2 (d + 4)...(d + 2k) because d+2 can be as small as we want, if n is big enough. This implies that it is sufficient to n prove that for some small ε and n big enough there holds d2 +k k−2 n d+2 d+2 (d + 2)k−1 > (1 + ε) > (1 + ε) . n+1 d+4 d + 2k (d + 4)...(d + 2k) d+2 Choose ε > 0 such that (1 + ε) d+4 < 1. Then for n big enough and k ≥ 3 there holds

n n+1

d2 +2 > (1 + ε)

d+2 n d+2 d+2 > (1 + ε) and > . d+4 d + 2k n+1 d+4

The inequality Bn,k > 0 for k ≥ 3 is proved. Case d = 1. In this case µ1 is the uniform distribution on the interval [−1, 1] with the density f (x) = 21 1[−1,1] (x). We will show that for a constant C1 C1 − 3x2 fn (x) = f ∗n (x) ≤ √ e 2n . 2πn 3x2

1 Suppose that for all x ∈ R and some n ∈ N there holds fn (x) ≤ √C2πn e− 2n . Then, by (6) Z 1 Z 1 3(x−u)2 du du C √ 1 e− 2n fn+1 (x) = fn (x − u) ≤ . 2 2 2πn −1 −1

We have to prove that Z

1

3x2 3(x−u)2 du C C1 √ 1 e− 2n ≤p e− 2(n+1) , 2 2πn 2π(n + 1)

−1

which is equivalent to 1

r 3x2 du n − 2(n+1) e ≤ e . 2 n+1 −1 But the left-hand side of the above is equal to Z

Z

1

e −1

−

3(x−u)2 − 2n

3(x−u)2 2n

3x2 du = e− 2n 2

Z

1

e −1

3xu n

3u2

e− 2n

du . 2


9 3u2

2

≤ 43 and we can estimate e− 2n from above For −1 ≤ u ≤ 1 and n ≥ 2 there holds 0 ≤ 3u 2n by three first terms of its Taylor expansion. This gives the following inequality Z 1 Z 1 3u2 du 3xu 3xu 3 3u2 du − e n 1 − (1 − ) e n e 2n ≤ = 2 4n 2n 2 −1 −1 2 sinh 3x 2n + 9x2 nx 3 3x 3x n = − 3 cosh . − 1− sinh 3x 3 4n 18x n 3x n n √ Now we have to show that for x > a n there holds 2 r 3x2 sinh 3x 2n + 9x2 nx 3x 3x n 3 n 2n(n+1) . − ≤ e sinh cosh − 1 − 3x 4n 18x3 n 3x3 n n+1 n We expand both sides in Taylor series and get RHS =

∞ X k=0

3x 2k n 6k k!

n n+1

k+ 21 , LHS =

∞ X

3x 2k n

h

1−

3 2n

+

k=0

3 n(2k+3)

+

9 8n2

−

9 4n2 (2k+3)

i

(2k + 1)!

2k

so that RHS − LHS =

P∞

k=0 Bn,k

( 3x n ) k!

, where

k+ 21

k! 3 9 9 1 3 Bn,k = − + + − 1− . 6k (2k + 1)! 2n n(2k + 3) 8n2 4n2 (2k + 3) 21 23 1 n n 9 27 − 1 + 2n − 16 1 − 10n In particular, Bn,0 = n+1 − 8n3 2 , Bn,1 = 16 n+1 + 40n and Bn,2 = 2 5 1 n 15 45 2 − 2! 1 − 14n + 56n . We observe that Bn,0 ∼ −5/(16n3 ), Bn,1 ∼ −1/(10n) and 2 36 n+1 5! Bn,2 ∼ 1/90. √ The rest of the proof is similar to the proof of the case d ≥ 2. Put x = a n and consider the fourth degree polynomial of a √ 2 √ 4 1 3a n 3a n wn (a) = Bn,0 + Bn,1 + Bn,2 . n 2 n n n+1

For large n we have 5 9a2 9a4 1 9a2 9a4 5 wn (a) ∼ − − + = 2 − − + . 16n3 10n2 10n2 n 16n 10 10 √ For all n, if a ≥ 2, then wn (a) > 0. It remains to prove that Bn,k > 0 for k ≥ 3. Observe first that 9 3 3 9 3 1 3 1 1 − + + − = −1 + + − 0. k 6 n+1 (2k + 1)! We must show that for n > n0 (d) and all k ≥ 3

n n+1

k+1/2 >

6k k! 6k = . (2k + 1)! (k + 1)(k + 2)...(2k + 1)


10

We check by simple calculations that this is true for k = 3, 4, 5. For instance for k = 3 and 3+1/2 63 n 18 > 4·5·6·7 n big enough n+1 = 70 . When k ≥ 6, we have k 6k 6 1 < . (k + 1)(k + 2)...(2k + 1) 7 2k + 1 21 n n 1 We notice that n+1 > 67 for n > 6 and n+1 > 2k+1 for n ≥ 1. This end the proof. In the same way as the Theorem 2.3 we prove Theorem 2.5. (Strong CLT for µd ) Let d ≥ 1. Let Y be a random Gaussian vector with law νd and f ∈ L1 (νd ). Then limn Ef (Sñ ) = Ef (Y ). 2.3. Convolutions. The Optimal Gaussian Bound is inherited√by convolutions of measures having this property. Below we write Sñ (X) = (X1 + . . . + Xn )/ n. Proposition 2.6. Let (Xi ), (Yj ) be independent random variables with laws µ and ν, with zero means and covariance matrices A and B respectively. Suppose that µ and ν satisfy the Optimal Gaussian Bound inequality: fSñ (X) ≤ C1 fN (0,A) ,

fSñ (Y ) ≤ C2 fN (0,B) .

Then fSñ (X+Y ) ≤ C1 C2 fN (0,A+B) , i.e. the measure µ ∗ ν satisfies the Optimal Gaussian Bound inequality. Proof. We have Sñ (X + Y ) = Sñ (X) + Sñ (Y ), so fSñ (X+Y ) = fSñ (X) ∗ fSñ (Y ) ≤ C1 C2 fN (0,A) ∗ fN (0,B) and the Proposition follows.

It is clear that Theorems 2.1 and 2.4 hold for uniform measures σd (r) and µd (r) on spheres and balls of radius r > 0, respectively. By Proposition 2.6 we obtain Corollary 2.7. Finite convolutions of measures σd (r) (when d ≥ 2) and µd (s) verify the Optimal Gaussian Bound inequality and the Strong Central Limit Theorem. 3. Isotropic random walk in R1 Let X1 , X2 , ... be independent random variables with the same symmetric Bernoulli distribution, P (Xn = 1) = P (Xn = −1) = 21 , and put Sn = X1 + ... + Xn . By µn we denote the Sn distribution of the normalized sum Sñ = √ . Then the distribution µn is given by the following n formula n j − n2 Sn − n2 j − n2 n 1 √ (17) µn =P √ =√ = , j = 0, 1, 2, 3, ..., n. j 2 n/2 n/2 n/2 Observe that one obtains the same law µn by normalizing sums of independent usual nBernoulli Y +...Y − random variables Yn , with P (Yn = 1) = P (Yn = 0) = 12 , i.e. µn is the law of 1 √n/2n 2 . An analog of Theorem 2.1 holds for tails of µn and N (0, 1). Let Y be the standard normal N (0, 1) random variable. If Φ is the distribution function of Y , then the tail Ψ(x) = 1 − Φ(x). Theorem 3.1. (Optimal Gaussian Bound for σ1n ) There exists C > 0 such that for all x > 0 and all n ∈ N Ψn (x) = P (Sñ > x) ≤ C P (Y > x) = CΨ(x).


11

Remark 3.2. The Theorem 3.1 strengthens the Hoeffding inequality (see e.g. [5], Prop. 1.3.5), 2 2 which states that P (Sñ > x) ≤ C e−x /2 . By the well-known estimate: Ψ(x) ∼ x1 e−x /2 for x → ∞ (see e.g. [6] VII, Lemma 2), the Theorem 3.1 is equivalent to the estimate 1 2 (18) P (Sñ > x) ≤ C e−x /2 , x > 0. x 1 The “missing factor” x in the Hoeffding inequality was observed by Talagrand ([14]). However, in the case of symmetric Bernoulli variables, our inequality (18) may be deduced from ([14] √ √ n (1.3)) only for x ≤ 2 and not for the full range x ≤ n. Theorem 3.1 suggests that Theorem 1.1 in [14] is true for K = 1 at least for symmetric binomial laws. Proof. Function Ψ(x) is strictly decreasing and Ψ(1) > 0, hence it is enough to prove the inequality for x > 1. Let us denote b(k, n, p) = nk pk (1 − p)n−k and put B(k, n, p) = Pk ν=0 b(ν, n, k). It is a standard exercise (see [6] Ex.VI.45 (10.9)) to show that Z p n−1 1 − B(k, n, p) = n tk (1 − t)n−k−1 dt. k 0 √

We will show that if p = 21 and k = b n2 + x 2n c then there exists a constant C such that for all n ∈ N and x > 1 there holds 1 − B k, n, 12 < CΨ(x). Moreover, because b(k, n, 12 ) = 0 for √ k > n, it is enough to prove the last inequality for all n ∈ N and 1 < x ≤ n. √ √ By (18) it is enough to show that for all n ∈ N, 1 < x ≤ n and k = b n2 + x 2 n c 1 C − x2 1 − B k, n, < e 2. 2 x In order to simplify the left–hand side of the last inequality we write R 1/2 k Z 1 2 t (1 − t)n−k−1 dt n n−1 1 − B(k, n, 21 ) n k 0 1 n = tk (1 − t)n−k−1 dt, = (n − k)2 n 1 b(k, n, 2 ) 0 k 2 so that it is enough to show that Z 1 2 C x2 1 n (n − k)2 tk (1 − t)n−k−1 dt ≤ e− 2 . (19) b k, n, 2 x 0 √

For k = b n2 + x 2 n c inequality (19) reads as √ Z 1 √ √ 2 n√ n x n C x2 bn +x 2nc n−b n + x 2 n c−1 2 2 (20) n − b + c t (1 − t) dt ≤ e− 2 . x n n b2 + 2 c 2 2 x 0 First we estimate the integral on the left-hand side of (20). Because for any real y there holds y − 1 < byc ≤ y, then Z 1 √ √ 2 x n x n n n tb 2 + 2 c (1 − t)n−b 2 + 2 c−1 dt ≤ 0

Z

1 2

√ n + x 2 n −1 2

√ n − x 2 n −1 2

Z

1 2

x√2 n

t dt. 1−t 0 0 n2 −1 n 1 2 −1 For t ∈ [0, 12 ] there holds 0 ≤ t(1 − t) ≤ 1/4, hence 0 ≤ t(1 − t) < . In order 4 R 1 t x√2 n to estimate the integral 02 1−t dt we use the Laplace method for estimating integrals of Rb type a exp (λS(x)) dx, when λ → ∞, see e.g. [13]. We have to estimate x√2 n Z 1 Z 1 √ 2 2 x n t t dt = e 2 ln 1−t dt, 1−t 0 0 t

(1 − t)

dt =

n

(t(1 − t)) 2 −1


12 √

t hence we take λ = x 2 n and S(t) = ln 1−t . If S(t) is C ∞ , maxt∈[a,b] S(t) is attained only at b and S 0 (b) 6= 0 — all these conditions are fulfilled in our case — then, by Laplace method, for Rb λ → ∞, there holds a exp (λS(t)) dt ∼ λS10 (b) , which in our case gives for some constant C1 and x > 1 x√2 n Z 1 2 t C1 dt ≤ √ . 1−t x n 0 This estimate gives the following n2 Z 1 √ √ 2 x n x n n 1 C2 bn + c n−b + c−1 t 2 2 (1 − t) 2 2 dt ≤ √ . x n 4 0

Substituting this estimate into (20), we see that it is enough to prove √ the following inequality: there exists a constant C3 such that for all n ∈ N and all x ∈ [1, n] there holds √ n2 2 n − b n2 + x 2 n c n√ 1 − x2 √ (21) ≤ C e . 3 b n2 + x 2 n c 4 n √

For further calculations it is convenient to write down the number k = b n2 + x 2in c as k = b n2 c+m h √ √ √ √ where m = b n2 + x 2 n c − b n2 c. Now if x ∈ [1, n], then m ∈ b 2n c, b n+1 n, c . When x = 2 which means that m = n − b n2 c = b n+1 c, the left-hand side of (21) is zero and the inequality 2 c − 1 (where holds trivially. Thus it is enough to show that for all n ∈ N and m = 1, 2, ..., b n+1 2 √ x n n n m = b 2 + 2 c − b 2 c), there holds n2 2 n − b n2 c − m 1 n − x2 √ . ≤ C e 3 4 b n2 c + m n √

√

Observe, that for even n we have m = b x 2 n c while for n odd m = b x 2n+1 c. For this reason the proof of (21) for even n is slightly different from that for odd n. If n is even then we have √

m = b x 2 n c, hence n is odd then m = (2m+1)2 − 2n

e

2 − x2

x)dx = P (g(Y ) > x)dx = P (Y x0 > u)g 0 (u)du g(x0 )

g(x0 )

x0

and (26) follows. The proof for symmetric binomial measures µn is analogous.

Theorem 3.4. (Strong CLT for σ1 ) If g ∈ L1 (γ) is in C 1 ([0, ∞)) and g is strictly monotonous on [x0 , ∞) for an x0 ≥ 0, then the DeMoivre–Laplace CLT holds for g: Z ∞ Z ∞ gdµn → gdγ, n→∞ 0

0

Proof. In Rthe proof weR suppose that g is strictly increasing on [x0 , ∞). By the Central Limit x x Theorem 0 0 gdµn → 0 0 gdγ and Ψn (x) → Ψ(x), n → ∞. In order to establish the convergence R∞ of integrals on [x0 , ∞), write the formula (27). The convergence of the term x0 g 0 Ψn dx to R∞ 0 g Ψdx follows by the Dominated Convergence Theorem if we use the Theorem 3.1 and the x0 integrability of g 0 Ψ with respect to the Lebesgue measure on R+ . An application of (26) ends the proof. In order to get a strong Central Limit Theorem from Theorem 3.1 in dimension one, some additional assumptions (such as monotonicity) must be put on f . This is showed by the following. Counterexample on the real line. For n = 4k 2 formula (17) reads as follows: µ4k2 ({2k}) = P

S4k2 − 2k 2 = 2k k

= P S4k2 = 4k

2

4k2 1 . = 2

Consider the following continuous function f : R → R: 2 2 for k = 1, 2, ... put f (k) = 34k and f (k ± 4−4k ) = 0. Now let f be linear and continuous 2 2 on intervals (k − 4−4k , k) and (k, k + 4−4k ). Finally put f (x) = 0 on the complement of 2 −4k2 ∪∞ , k + 4−4k ]. k=1 [k − 4 Let γ1 denote the standard Gaussian distribution on the real line. Because f is nonnegative and the standard Gaussian density on R is bounded by √12π , we have the following estimate: Z

∞

0