Table Of ContentOPTIMAL PROBABILITY INEQUALITIES FOR RANDOM WALKS
RELATED TO PROBLEMS IN EXTREMAL COMBINATORICS
D.DZINDZALIETA †¶, T.JUSˇKEVICˇIUS ‡, AND M.SˇILEIKIS §¶
Dedicated to the memory of V.Bentkus
Abstract. LetSn=X1+···+Xn beasumofindependent symmetricrandomvariablessuch
2 that |Xi| ≤ 1. Denote by Wn = ε1+···+εn a sum of independent random variables such that
1 P{εi=±1}=1/2. Weprovethat
0
P{Sn∈A}≤P{cWk∈A},
2
where A is either an interval of the form [x,∞) or just a single point. The inequality is exact and
n
theoptimalvaluesofcandkaregivenexplicitly. ItimprovesKwapien´’sinequalityinthecaseofthe
a Rademacher series. We also provide a new and very short proof of the Littlewood-Offord problem
J
withoutusingSperner’sTheorem. Finally,anextensiontooddLipschitzfunctionsisgiven.
4
Key words. Concentrationinequalities,intersectingfamilies,randomwalks,tailprobabilities.
]
R
AMS subject classifications. 60E15,05D05.
P
.
h 1. Introduction. Let Sn = X1 + +Xn be a sum of independent random
···
t variables X such that
a i
m
X 1 and EX =0. (1.1)
i i
[ | |≤
2 Let Wn = ε1 + +εn be the sum of independent Rademacher random variables,
···
v i.e., such that P ε = 1 =1/2. We will refer to W as a simple random walk with
i n
2 n steps. { ± }
3
By a classical result of Hoeffding [6] we have the estimate
3
6
P S x exp x2/2n , x R. (1.2)
. n
1 { ≥ }≤ − ∈
1 (cid:8) (cid:9)
If we take S = W on the left-hand side of (1.2), then in view of the Central
n n
1
Limit Theorem we can infer that the exponential function on the right-hand side
1
: is the minimal one. Yet a certain factor of order x−1 is missing, since Φ(x)
v (√2πx)−1exp x2/2 for large x. ≈
i −
X Furthermore, it is possible to show that the random variable S is sub-gaussian
n
(cid:8) (cid:9)
r in the sense that
a
P S x cP √nZ x , x R,
n
{ ≥ }≤ ≥ ∈
(cid:8) (cid:9)
where Z is the standard normal random variable and c is some explicit positive con-
stant (see, for instance, [2]).
Perhaps the best upper bound for P S x was given by Bentkus [1], which
n
{ ≥ }
for integer x is optimal for martingales with differences X satisfying (1.1).
i
†VilniusUniversityInstituteofMathematicsandInformatics,Akademijos4,LT-08663,Vilnius,
Lithuania([email protected])
‡DepartmentofMathematicalSciences,UniversityofMemphis,Memphis,TN38152-3240,USA
([email protected])
§Department of Discrete Mathematics, Adam Mickiewicz University, Umultowska 87, 61-614
Poznan´, Poland([email protected])
¶Thisresearchwasfundedbyagrant(No. MIP-47/2010)fromtheResearchCouncilofLithuania
1
2 D.DZINDZALIETA,T.JUSˇKEVICˇIUSANDM.SˇILEIKIS
Although there are numerous improvements of the Hoeffding inequality, to our
knowledge there are no examples where the exact bound for the tail probability is
found. In this paper we give an optimal bound for the tail probability P S x
n
{ ≥ }
under the additional assumption of symmetry.
We henceforthreservethe notationS andW forrandomwalkswithsymmetric
n n
steps satisfying (1.1) and a simple random walk with n steps respectively.
Theorem 1.1. For x>0 we have
P W x if x +n 2Z,
n
P S x { ≥ } ⌈ ⌉ ∈ (1.3)
n
{ ≥ }≤(P Wn−1 x if x +n 2Z+1.
{ ≥ } ⌈ ⌉ ∈
The latter inequality can be interpreted by saying that among bounded random
walks the simple random walk is the most stochastic.
Kwapien´ proved (see [10]) that for arbitrary i.i.d. symmetric random variables
X and real numbers a with absolute value less than 1 we have
i i
P a X +...+a X x 2P X +...+X x , x>0.
1 1 n n 1 n
{ ≥ }≤ { ≥ }
The case n=2 with X =ε shows that the constant 2 cannot be improved.
i i
Theorem 1.1 improves Kwapien´’s inequality for Rademacher sequences. We be-
lieve that using the inequality in (1.3) with some conditioning arguments leads to
better estimates for arbitrarysymmetric randomvariables X under the assumptions
i
of Kwapien´’s inequality, but we will not go into these details in this paper.
We also consider the problem of finding the quantity
supP S =x ,
n
{ }
Sn
which can be viewed as a non-uniform bound for the concentration of the random
walk S at a point.
n
Theorem 1.2. For x>0 and k = x we have
⌈ ⌉
P S =x P W =k , (1.4)
n m
{ }≤ { }
where
min n,k2 , if n+k 2Z,
m= ∈
(min(cid:8)n 1(cid:9),k2 , if n+k 2Z+1.
− ∈
Equality in (1.4) is attained for(cid:8)S = xW(cid:9) .
n k m
We provide two different proofs for both inequalities. The first approach is based on
induction on the number of random variables ( 2). To prove Theorem 1.2 we also
§
need the solution of the Littlewood-Offord problem.
Theorem 1.3. Let a ,...,a be real numbers such that a 1. We have
1 n i
| |≥
maxP S (x k,x+k] P W ( k,k] .
n n
x∈R { ∈ − }≤ { ∈ − }
That is, the number of the choices of signs for which S lies in an interval of length
n
2k does not exceed the sum of k largest binomial coefficients in n.
Theorem 1.3 was first proved by Erdo˝s [5] using Sperner’s Theorem. We give a
very short solution which seems to be shorter than the original proof by Erdo˝s. We
only use induction on n and do not use Sperner’s Theorem.
EXTREMALRANDOMWALKS 3
Surprisingly, Theorems 1.1 and 1.2 can also be proved by applying results from
extremal combinatorics ( 3). Namely, we use the bounds for the size of intersecting
§
families of sets (hypergraphs) by Katona [7] and Milner [9].
UsingastrengtheningofKatona’sresultbyKleitman[8],weextendTheorem1.1
to odd 1-Lipschitz functions rather than just sums of the random variables X ( 4).
i
§
ItisimportanttonotethattheboundofTheorem1.1cannotbetrueforallLipschitz
functionssincetheextremalcaseisnotprovidedbyoddfunctions(forthedescription
of the extremal Lipschitz functions defined on general probability metric spaces see
Dzindzalieta [4]).
2. Proofs by induction on dimension. Wewillfirstshowthatitisenoughto
prove Theorems 1.1 and 1.2 in case when S is a linear combination of independent
n
Rademacher random variables ε with coefficients a 1.
i i
| |≤
Lemma 2.1. Let g :Rn R be a bounded measurable function. Then we have
→
sup Eg(X ,...,X )= sup Eg(a ε ,...a ε ),
1 n 1 1 n n
X1,...,Xn a1,...,an
where the supremum on the left-hand side is taken over symmetric independent ran-
dom variables X ,...,X such that X 1 and the supremum on the right-hand
1 n i
| | ≤
side is taken over numbers 1 a ,...,a 1.
1 n
− ≤ ≤
Proof. Define S =sup Eg(a ε ,...a ε ). Clearly
a1,...,an 1 1 n n
S sup Eg(X ,...,X ).
1 n
≤
X1,...,Xn
By symmetry of X ,...,X , we have
1 n
Eg(X ,...,X )= Eg(X ε ,...,X ε ).
1 n 1 1 n n
Therefore
Eg(X ,...,X )= E E[g(X ε ,...,X ε ) X ,...,X ] ES =S.
1 n 1 1 n n 1 n
| ≤
Thus, in view of Lemma 2.1 we will henceforth write S for the sum a ε + +
n 1 1
···
a ε instead of a sum of arbitrary symmetric random variables X .
n n i
Proof of Theorem 1.1. First note that the inequality is true for x (0,1] and all
∈
n. This is due to the fact that P S x 1/2 by symmetry of S and for all n
n n
{ ≥ } ≤
the right-handside of the inequality is givenby the tail of an odd number of random
signs, which is exactly 1/2. We can also assume that the largest coefficient a = 1
i
as otherwise if we scale the sum by a then the tail of the this new sum would be
i
at least as large as the former. We will thus assume, without loss of generality, that
0 a a ... a =1. Define a function I(x,n) to be 1 if x +n is even, and
1 2 n
≤ ≤ ≤ ≤ ⌈ ⌉
zero otherwise. Then we can rewrite the right-hand side of (1.3) as
P W +ε I(x,n) x ,
n−1 n
{ ≥ }
making an agreement ε 0.
0
≡
For x > 1 we argue by induction on n. Case n = 0 is trivial. Observing that
I(x 1,n)=I(x+1,n)=I(x,n+1) we have
−
P S x = 1P S x 1 + 1P S x+1
{ n+1 ≥ } 2 { n ≥ − } 2 { n ≥ }
1P W +ε I(x 1,n) x 1
≤ 2 { n−1 n − ≥ − }
+ 1P W +ε I(x+1,n) x+1
2 { n−1 n ≥ }
=P W +ε I(x,n+1) x .
n n+1
{ ≥ }
4 D.DZINDZALIETA,T.JUSˇKEVICˇIUSANDM.SˇILEIKIS
Proof of Theorem 1.3. We can assume that a a ... a 1. Without
1 2 n
≥ ≥ ≥ ≥
loss of generality we can also take a =1. This is because
n
P S (x k,x+k] P S /a (x k,x+k]/a
n n n n
{ ∈ − }≤ { ∈ − }
maxP S /a (x k,x+k] .
n n
≤ x∈R { ∈ − }
The claim is trivial for n=0. Let us assume that we have proved the statement
for 1,2,...,n 1. Then
−
P S (x k,x+k]
n
{ ∈ − }
=1P S (x k 1,x+k 1] + 1P S (x k+1,x+k+1]
2 { n−1 ∈ − − − } 2 { n−1 ∈ − }
=1P S (x k 1,x+k+1] + 1P S (x k+1,x+k 1]
2 { n−1 ∈ − − } 2 { n−1 ∈ − − }
1P W ( k 1,k+1] + 1P W ( k+1,k 1]
≤2 { n−1 ∈ − − } 2 { n−1 ∈ − − }
=1P W ( k 1,k 1] + 1P W ( k+1,k+1]
2 { n−1 ∈ − − − } 2 { n−1 ∈ − }
=P W ( k,k] .
n
{ ∈ − }
Note that we rearrangedthe intervals after the second equality so as to have two
intervals of different lengths and this makes the proof work.
BeforeprovingTheorem1.2,wewillobtainanupperboundforP S =x under
n
{ }
an additional condition that all a are nonzero.
i
Lemma 2.2. Let x>0, k = x . Suppose that 0<a a 1. Then
1 n
⌈ ⌉ ≤···≤ ≤
P W =k , if n+k 2Z,
n
P S =x { } ∈ (2.1)
n
{ }≤(P Wn−1 =k , if n+k 2Z+1.
{ } ∈
Proof. We first prove the lemma for x (0,1] and any n. By Theorem 1.3 we have
∈
n
P S =x 2−n . (2.2)
n
{ }≤ n/2
(cid:18)⌈ ⌉(cid:19)
On the other hand, if x (0,1], then k =1 and
∈
n
2−n =P W =1 , if n+1 2Z,
n
n (n+1)/2 { } ∈
2−n = (cid:18) (cid:19)
(cid:18)⌈n/2⌉(cid:19) 2−n n =P Wn−1 =1 , if n+1 2Z+1,
n/2 { } ∈
(cid:18) (cid:19)
where the second equality follows by Pascal’s identity:
n n 1 n 1 n 1
2−n =2−n − + − =21−n − =P W =1 .
n−1
n/2 n/2 n/2 1 n/2 { }
(cid:18) (cid:19) (cid:20)(cid:18) (cid:19) (cid:18) − (cid:19)(cid:21) (cid:18) (cid:19)
Let N = 1,2,... stand for the set of positive integers. Let us write B (x) for the
n
{ }
right-hand side of (2.1). Note that it has the following properties:
x B (x) is non-increasing; (2.3)
n
7→
x B (x) is constant on each of the intervals (k 1,k], k N; (2.4)
n
7→ − ∈
B (k)= 1B (k 1)+ 1B (k+1), if k =2,3,.... (2.5)
n 2 n−1 − 2 n−1
EXTREMALRANDOMWALKS 5
We proceed by induction on n. The case n=1 is trivial. To prove the induction
step for n 2, we consider two cases: (i) x=k N; (ii) k 1<x<k N.
≥ ∈ − ∈
Case(i). Fork =1 the lemmahas beenproved,soweassumethatk 2. By the
≥
inductional hypothesis we have
P S =k = 1P S =k a + 1P S =k+a
{ n } 2 { n−1 − n} 2 { n−1 n}
1B (k a )+ 1B (k+a ). (2.6)
≤ 2 n−1 − n 2 n−1 n
By (2.3) we have
B (k a ) B (k 1), (2.7)
n−1 n n−1
− ≤ −
and by (2.4) we have
B (k+a )=B (k+1). (2.8)
n−1 n n−1
Combining (2.6), (2.7), (2.8), and (2.5), we obtain
P S =k B (k). (2.9)
n n
{ }≤
Case (ii). For x (0,1] Lemma has been proved, so we assume k 2. Consider
∈ ≥
two cases: (iii) x/a k; (iv) x/a <k.
n n
≥
Case (iii). Define S′ = a′ε + +a′ ε , where a′ = ka /x, so that S′ = kS .
n 1 1 ··· n n i i n x n
Recall that a = max a , by the hypothesis of Lemma. Then a′ ka /x and the
n i i i ≤ n
assumptionx/a k imply that0<a′,...,a′ 1. Therefore,by (2.9)and(2.4)we
n ≥ 1 n ≤
have
P S =x =P S′ =k B (k)=B (x).
{ n } { n }≤ n n
Case (iv). Without loss of generality, we can assume that a =1, since
n
a a x
1 n
P S =x =P ε + + ε =
n 1 n
{ } a ··· a a
(cid:26) n n n(cid:27)
and k 1<x/a <k, by the assumption of the present case. Sequentially applying
n
−
the induction hypothesis, (2.4), (2.5), and again (2.4), we get
P S =x = 1P S =x 1 + 1P S =x+1
{ n } 2 { n−1 − } 2 { n−1 }
1B (x 1)+ 1B (x+1)
≤ 2 n−1 − 2 n−1
= 1B (k 1)+ 1B (k+1)
2 n−1 − 2 n−1
=B (k)=B (x).
n n
Proof of Theorem 1.2. Writing B (k) for the right-hand side of (2.1), we have,
n
by Lemma 2.2, that
n
P S =x maxB (k).
n j
{ }≤ j=k
If j+k 2Z, then B (k)=P W =k =B (k) and therefore
j j j+1
∈ { }
n
maxB (k)= max P W =k . (2.10)
j j
j=k k≤j≤n { }
k+j∈2Z
6 D.DZINDZALIETA,T.JUSˇKEVICˇIUSANDM.SˇILEIKIS
To finish the proof, note that the sequence P W =k = 2−j j , j = k,k+
{ j } (k+j)/2
2,k+4,... is unimodal with a peak at j =k2, i.e.,
(cid:0) (cid:1)
P W =k P W =k , if j k2,
j−2 j
{ }≤ { } ≤
and
P W =k >P W =k , if j >k2.
j−2 j
{ } { }
Indeed, elementary calculations yield that the inequality
j 2 j
2−j+2 − 2−j , j k+2,
(k+j)/2 1 ≤ (k+j)/2 ≥
(cid:18) − (cid:19) (cid:18) (cid:19)
is equivalent to the inequality j k2.
≤
3. Proofs based on results in extremal combinatorics. Let [n] stand for
the finite set 1,2,...,n . Consider a family of subsets of [n]. We denote by
{ } F |F|
the cardinality of . The family is called:
F F
(i) k-intersecting if for all A,B we have A B k.
∈F | ∩ |≥
(ii) an antichain if for all A,B we have A*B.
∈F
A well known result by Katona [7] (see also [3, p. 98, Theorem 4]) gives the exact
upper bound for a k-intersecting family.
Theorem 3.1 (Katona [7]). If k 1 and is a k-intersecting family of subsets
≥ F
of [n] then
n
n
, if k+n=2t,
j
j=t(cid:18) (cid:19)
|F|≤Xn n + n−1 , if k+n=2t 1. (3.1)
j t 1 −
j=t(cid:18) (cid:19) (cid:18) − (cid:19)
Notice that if k+n=2t,Xthen
n
n
=2nP W k . (3.2)
n
j { ≥ }
j=t(cid:18) (cid:19)
X
If k+n=2t 1, then using the Pascal’s identity n = n−1 + n−1 we get
− j j j−1
n n−1 (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
n n 1 n 1
+ − =2 − =2nP W k . (3.3)
n−1
j t 1 j { ≥ }
j=t(cid:18) (cid:19) (cid:18) − (cid:19) j=t−1(cid:18) (cid:19)
X X
The exact upper bound for the size of a k-intersecting antichain is given by the
following result of Milner [9].
Theorem 3.2 (Milner [9]). If a family of subsets of [n] is a k-intersecting
F
antichain, then
n n+k
, t= . (3.4)
|F|≤ t 2
(cid:18) (cid:19) (cid:24) (cid:25)
Note that we have
n
=2nP W =k , if n+k =2t, (3.5)
n
t { }
(cid:18) (cid:19)
EXTREMALRANDOMWALKS 7
and
n
=2nP W =k+1 , if n+k =2t 1. (3.6)
n
t { } −
(cid:18) (cid:19)
By Lemma 2.1 it is enough to prove Theorems 1.1 and 1.2 for the sums
S =a ε + +a ε ,
n 1 1 n n
···
where 0 a ,...,a 1. Denote as Ac the complement of the set A. For each
1 n
≤ ≤
A [n], write s = a a . We define two families of sets:
⊂ A i∈A i− i∈Ac i
= PA [n]:sP x , and = A [n]:s =x .
≥x A x A
F { ⊂ ≥ } F { ⊂ }
Proof of Theorem 1.1. We have
P S x =2−n .
n ≥x
{ ≥ } |F |
Let k = x . Since W takes only integer values, we have
n
⌈ ⌉
P W k =P W x and P W k =P W x .
n n n−1 n−1
{ ≥ } { ≥ } { ≥ } { ≥ }
Therefore, in the view of (3.1), (3.2), and (3.3), it is enough to prove that is
≥x
F
k-intersecting. Suppose thatthereareA,B suchthat A B k 1. Writing
≥x
∈F | ∩ |≤ −
σ = a , we have
A i∈A i
P sA =σA σAc =(σA∩B σAc∩Bc)+(σA∩Bc σAc∩B) (3.7)
− − −
and
sB =σB σBc =(σA∩B σAc∩Bc) (σA∩Bc σAc∩B). (3.8)
− − − −
Since
σA∩B σAc∩Bc σA∩B A B k 1<x,
− ≤ ≤| ∩ |≤ −
from (3.7) and (3.8) we get
min s ,s <x,
A B
{ }
which contradicts the fact s ,s x.
A B
≥
The following lemma implies Theorem 1.2. It also gives the optimal bound for
P S =x and thus improves Lemma 2.2.
n
{ }
Lemma 3.3. Let 0<a ,...,a 1 be strictly positive numbers, x>0, k = x .
1 n
≤ ⌈ ⌉
Then
P W =k , if n+k 2Z,
n
P S =x { } ∈
n
{ }≤(P Wn =k+1 , if n+k 2Z+1.
{ } ∈
Proof. We have
P S =x =2−n .
n x
{ } |F |
In the view of (3.4), (3.5), and (3.6), it is enough to prove that is a k-intersecting
x
F
antichain. To see that is k-intersecting it is enough to note that . To
x x ≥x
F F ⊂ F
8 D.DZINDZALIETA,T.JUSˇKEVICˇIUSANDM.SˇILEIKIS
show that is an antichain is even easier. If A,B and A ( B, then s s
x x B A
F ∈ F −
= 2 a >0, which contradicts the assumption that s =s =x.
i∈B\A i B A
Proof of Theorem 1.2. Lemma 3.3 gives
P
n
P S =x maxP W =k+1 I(k,j) ,
n j
{ }≤ j=k { − }
where again I(k,j)=I k+j 2Z . Note that if k+j 2Z we have
{ ∈ } ∈
P W =k 1/2P W =k +1/2P W =k+2
j j j
{ }≥ { } { }
=P W =k+1 , k >0.
j+1
{ }
Hence
n
maxP W =k+1 I(k,j) = max P W =k ,
j j
j=k { − } k≤j≤n { }
k+j∈2Z
the right-hand side being the same as the one of (2.10). Therefore, repeating the
argument following (2.10) we are done.
4. Extension to Lipschitz functions. One can extend Theorem 1.1 to odd
Lipschitz functions taken of n independent random variables. Consider the cube
C =[ 1,1]n withtheℓ1 metricd. Wesaythatafunctionf :C RisK-Lipschitz
n n
− →
with K >0 if
f(x) f(y) Kd(x,y), x,y C . (4.1)
n
| − |≤ ∈
We say that a function f : C R is odd if f( x) = f(x) for all x C . An
n n
→ − − ∈
example of an odd 1-Lipschitz function is the function mapping a vector to the sum
of its coordinates:
f(x ,...,x )=x + +x .
1 n 1 n
···
Note that the left-hand side of (1.3) can be written as P f(X ,...,X ) x .
1 n
{ ≥ }
As in Theorems 1.1 and 1.2, the crux of the proof is dealing with two-valued
random variables. The optimal bound for a k-intersecting family is not sufficient for
thiscase,thereforeweusethefollowinggeneralizationofTheorem3.1duetoKleitman
[8](see also[3, p. 102])whichwe state slightly reformulatedfor ourconvenience. Let
us define the diameter of a set family by diam =max A B .
A,B∈F
F F | △ |
Theorem 4.1 (Kleitman [8]). If k 1 and is a family of subsets of [n] with
≥ F
diam n k, then
F ≤ −
n
n
, if k+n=2t,
j
j=t(cid:18) (cid:19)
|F|≤Xn n + n−1 , if k+n=2t 1. (4.2)
j t 1 −
j=t(cid:18) (cid:19) (cid:18) − (cid:19)
X
To see that Theorem 4.1 implies Theorem 3.1, observe that A B k implies
| ∩ | ≥
A B n k.
| △ |≤ −
Theorem 4.2. Suppose that a function f : C R is 1-Lipschitz and odd. Let
n
→
X ,...,X be symmetric independent random variables such that X 1. Then, for
1 n i
| |≤
EXTREMALRANDOMWALKS 9
x>0, we have that
P W x , if n+ x 2Z,
n
P f(X ,...,X ) x { ≥ } ⌈ ⌉∈ (4.3)
1 n
{ ≥ }≤(P Wn−1 x , if n+ x 2Z+1.
{ ≥ } ⌈ ⌉∈
Proof. Applying Lemma 2.1 with the function
g(y ,...,y )=I f(y ,...,y ) x ,
1 n 1 n
{ ≥ }
we can see that it is enough to prove (4.3) with
X =a ε ,...,X =a ε
1 1 1 n n n
for any 1-Lipschitz odd function f. In fact, we can assume that a = = a = 1,
1 n
···
since the function
(x ,...,x ) f(a x ,...,a x )
1 n 1 1 n n
7→
is clearly 1-Lipschitz and odd.
GivenA [n],writef forf(2I (1) 1,...,2I (n) 1),whereI istheindicator
A A A A
⊆ − −
function of the set A. Note that
f f 2A B (4.4)
A B
| − |≤ | △ |
by the Lipschitz property. Consider the family of finite sets
= A [n]:f x ,
A
F { ⊆ ≥ }
so that
P f(ε ,...,ε ) x =2−n .
1 n
{ ≥ } |F|
Write k = x . Note that W and W take only integer values. Therefore by
n−1 n
⌈ ⌉
(3.2) and (3.3) we see that the right-hand side of (4.2) is equal, up to the power of
two, to the right-hand side of (4.3). Consequently, if diam n k, then Theorem
F ≤ −
4.1 implies (4.3). Therefore, it remains to check that for any A,B we have
∈ F
A B n k.
| △ |≤ −
Suppose that for some A,B we have f ,f x but A B n k+1. Then
A B
≥ | △ |≥ −
A Bc = (A B)c =n A B k 1,
| △ | | △ | −| △ |≤ −
and hence by (4.4) we have
fA fBc 2k 2. (4.5)
| − |≤ −
On the other hand we have that fBc x, as f is odd. Therefore
≤−
fA fBc 2x>2k 2,
− ≥ −
which contradicts (4.5).
10 D.DZINDZALIETA,T.JUSˇKEVICˇIUSANDM.SˇILEIKIS
Acknowledgement. We would like to thank Paul Balister for careful reading
and valuable remarks, which improved the exposition.
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