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On the Convexified Sauer–Shelah Theorem
Journal of Combinatorial Theory, Series B TB1736 journal of combinatorial theory, Series B 69, 183192 (1997) article no. TB961736
On the Convexified SauerShelah Theorem* Stanislaw J. Szarek Department of Mathematics, Case Western Reserve University, Cleveland, Ohio 44106-7058
and Michel Talagrand Equipe d 'Analyse-Tour 46, U.A. au C.N.R.S. no. 754, Universite Paris VI, 4 Place Jussieu, 75230 Paris Cedex 05, France; and Department of Mathematics, Ohio State University, 231 West 18th Avenue, Columbus, Ohio 43210 Received August 19, 1994
Let A be a subset of [0, 1] n. Given =>0, we can find a subset I of [1, ..., n] such that the convex hull in R I of the projection of A onto [0, 1] I contains the cube [12&=, 12+=] I, and that card I n&K(n=+- n log(2 ncard A)), where K>0 is a universal constant. 1997 Academic Press
1. INTRODUCTION For a subset I of [1, ..., n], let us denote by P I the natural projection from [0, 1] n to [0, 1] I. The SauerShelah lemma [Sa], [Sh] asserts that given a subset A of [0, 1] n and an integer k with card A> i k ( ni ), there exists I/[1, ..., n] with card I>k and P I (A)=[0, 1] I. This result has proved to be of considerable use in analysis and probability. One drawback of this result is however the fact that, even when card A 2 n&1, we cannot guarantee that card I>n2. Moreover, even if, say, card A2 n > .999, it is not, for large n, possible to guarantee that card In .501. Yet it is at times valuable to find subsets I of [1, ..., n] such that n&card I is small, yet P I (A) is ``big.'' One result of this nature and an application was discovered in [S-T], and we introduce the necessary notation to explain this result. For a subset B of [0, 1] I, we denote by conv B the convex hull of B, when B is seen as a subset of R I. We say that B/[0, 1] I is symmetric if (x i ) i # I # B O (1&x i ) i # I # B. * Work partially supported by NSF grants. E-mail: sjs13po.cwru.edu. E-mail: talagranmath.ohio-state.edu.
183 0095-895697 25.00 Copyright 1997 by Academic Press All rights of reproduction in any form reserved.
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Theorem 1.1 [S-T]. Consider a symmetric subset A of [0, 1] n with card A 2 n&1, and consider =>0. Then there exists I/[1, ..., n] with card I n(1&4=) and conv P I (A)#[12&=, 12+=] I. It is a natural question to investigate what happens when card AR2 n&1 (or if A is not symmetric). The method of [S-T], that relies on the SauerShelah lemma, apparently does not yield estimates of card I of the correct order. Obtaining such estimates is the purpose of the present paper. An improvement of Theorem 1.1 in the symmetric case can be found in [A]; for another related sharp result see [G]. A first observation is that a straightforward push down argument reduces the problem to the case when A is hereditary, i.e., x # A, y # [0, 1] n;
\i n, y i x i O y # A.
Lemma 1.2. For each subset A of [0, 1] n, there is an hereditary subset B of [0, 1] n, with card B=card A, and such that, for each = 0, each I/[1, ..., n] we have [12&=, 12+=] I /conv P I (B) O [12&=, 12+=] I /conv P I (A). Proof. Let us fix j n. For x # [0, 1] n, we denote by x the sequence obtained by replacing the j th coordinate of x by zero. We define, for x # A T j (x)=x
if x A
T j (x)=x
if x # A
We define T j (A)=[T j (x); x # A]. We leave to the reader the easy proof that T j is one to one on A, so that card T j (A)=card A. We also leave to the reader to check that after iterating this operation for j=1, ..., n, the resulting set is hereditary. So it is enough to prove that [12&=, 12+=] I /3 conv P I (A) O [12&=, 12+=] I /3 conv P I (T j (A)). It should be obvious that we can assume j # I. If [12&=, 12+=] I /3 conv P I (A), we can find (' i ) i # I # [&1, 1] I with (12+' i =) i # I conv P I (A),
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CONVEXIFIED SAUERSHELAH THEOREM
185
and the HahnBanach theorem asserts that there are numbers (: i ) i # I such that \x # A, : : i x i < : : i ( 12 +' i =). i#I
i#I
Consider :~ i given by :~ i =: i if i{j and :~ j = |: j |. Define '~ i similarly. We show that \y # T j (A), : :~ i y i < : :~ i ( 12 +'~ i =) i#I
i#I
(and this finishes the proof ). We note that : i ' i :~ i '~ i for all i. Thus it suffices to find, for any y # T j (A), and x # A such that x i =y i if i{j and |: j | y j &: j x j 1 & 2 (|: j | &: j )=: j . If y j =0, we take (any) x such that T j (x)=y. If y j =1, we observe, using the definition of T j , that both y and y belong to A, and we choose x=y or x=y depending on whether : j 0 or : j <0. K Given a number s, we denote by (s) I the element of R I that has all its coordinates equal to s. Lemma 1.3.
If A is hereditary, then
[s&=, s+=] I /conv P I (A) (s+=) I /conv P I (A). Proof. In fact, if x # conv P I (A), and 0 y i x i for all i # I, we have y # conv P I (A). K The problem now becomes, given r, to find I with card I as large as possible and (r) I # conv P I (A). A key ingredient of the proof will be the ability to measure the size of a subset of [0, 1] I not only by its cardinality or, equivalently, by its normalised counting measure, but also by its size with respect to the measures + p, I =((1&p) $ 0 +p$ 1 ) I, where $ t is the unit mass concentrated at t. (For I=[1, ..., n], we write + p, I =+ p, n .) Thus it will be natural to state a result in the line of Theorem 1.1 not only when one controls the cardinality of A, but when one controls + p, n(A). Before we state our result, it is instructive to motivate it by considering the standard example where, for some q # R,
{
=
A= (x i ) i n ; : x i q . i n
Here
{
=
P I (A)= (x i ) i # I ; : x i q . i#I
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SZAREK AND TALAGRAND
If card I=n&k and if (r) I # conv P I (A), we have r q(n&k), i.e., k (nr&q)r, so that k
n(r&p) np&q + . r r
(1.1)
We now observe, using standards estimates on the tail of the binomial distribution, that (at least when np2 q np) log 1+ p(A) is of order (np&q) 2np(1&p). Thus for p r 2p (which is the interesting range) we must have k
1 n(r&p) n(1&p) log 1+ p(A) + . K p p
\
+
(1.2)
(Of course the statement says nothing when the right hand side is greater than or equal to n which happens, in particular, if rp is large.) The meaning of our main result is that the example presented above is essentially the worst case. Theorem 1.4. Assume p 12, and consider an heriditary subset A of [0, 1] n. Consider p r 1. Then we can find I/[1, ..., n] with (r) I # conv P I (A) and n&card I K
\
n(r&p) n + log 1+ p(A) . p p
+
In this statement, as well as in the rest of the paper, K (with a subcript or not) denotes an effectively computable universal (i.e. independent of p, n, r, A or any other parameters) constant, the value of which may vary between occurences. Corollary 1.5. Consider a subset A of [0, 1] n. Then we can find I/[1, ..., n] with [12&=, 12+=] I /conv P I (A) and
\
card I n&K n=+ n log
2n . card A
+
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CONVEXIFIED SAUERSHELAH THEOREM
Comments. 1. Theorem 1.4 provides a value of n&card I of the correct order. Finding the exact best possible value is of course a more challenging question. 2. The reason that we assume p 12 is simply that the case p=12 is by far the most interesting, and that the case p 12 has an identical proof. Minor modifications are required to handle the case p>12. Let us now comment on the proof of Theorem 1.4. To construct I we will remove points of [1, ..., n] one at a time. When choosing which point j to remove first, the natural idea is to insure that P J (A) is ``large,'' where J is the complement of j. The first thought would be to require that + p, J (P J (A)) is substantially larger than + p, n(A). However, examples in [K-K-L] show that this is impossible. Rather, we will require a control of +p$, J (P J (A)) where p$ is somewhat larger than p. (The reader should observe that, when A is hereditary, increasing p decreases + p(A).) This is the object of the main Lemma proved in Section 2. The proof of Theorem 1.4 is then completed in Section 3.
2. THE MAIN LEMMA Lemma 2.1. bi =
|
Consider an hereditary set A/[0, 1] n, and for i n, set
\
n
( p&x i ) d+ p, n(x) = : A
+
` p xj(1&p) 1&xj ( p&x i ) .
x # A j=1
Assume p 23. Consider j n such that b j =max i n b i . Set J= [1, ..., n]&[ j]. Then, for p$=p(1+(1Kn)) we have log + p$, J (P J (A)) log + p, n(A)+
bj . 2+ p, n(A)
Comment. Thus we achieve that P J (A) is larger than A in the sense that + p$, J (P J (A)) is somewhat larger than + p, n(A) for p$=p(1+(1Kn)). Before we start the proof, we need an auxiliary result. Lemma 2.2. +p$, n K+ p, n . Proof.
Assume p p$ p(1+(13n)), p 23 . Then (1K) + p, n
Consider x # [0, 1] n. Set l= i n x i , so that + p, n([x])=p l(1&p) n&l; + p$, n([x])=p$ l(1&p$) n&l
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SZAREK AND TALAGRAND
and + p, n([x]) p = + p$, n([x]) p$
l
\ +\
1&p 1&p$
+
n&l
.
Now 1 p 1 (1+13n) p$ 1
1&(23) 2 1&p 1+ . 1&p$ 1&(23)(1+13n) n
Thus 1 + ([x]) 2 1+ p, n n (1+13n) + p$, n([x]) n
n
\ +.
K
Proof of Lemma 2.1. Step 1. To simplify the notation, we assume without loss of generality that j=n. We set +=+ p, n , +~ =+ p, J . We set A 0 =[x=(x i ) i n&1 # [0, 1] n&1; x 0 # A] A 1 =[x=(x i ) i n&1 # [0, 1] n&1; x 1 # A]
,
where x 0 denotes concatenation. We note that since A is hereditary, A 1 /A 0 and A 0 =P J (A). We set a i =+~(A i ) for i=0, 1; note that a 0 a 1 and +(A)=(1&p) a 0 + pa 1 a 0. We have b n =(1&p) a 0p+pa 1( p&1)=p(1&p)(a 0 &a 1 ). Also +~(A 0 )&+(A)=a 0 &+(A)=p(a 0 &a 1 ) and thus +~(A 0 )=+(A)+
bn bn =+(A) 1+ . 1&p (1&p) +(A)
\
Now, p(1&p)(a 0 &a 1 ) bn = p. +(A) (1&p) a 0 +pa 1
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+
(2.1)
CONVEXIFIED SAUERSHELAH THEOREM
189
Since log(1+x(1&p)) x for 0 x p<1, we have proved that log +~(A 0 ) log +(A)+
bn . +(A)
(2.2)
Step 2. The difference between (2.2) and what we want is that the left-hand side involves +~ =+ p, J rather than + p$, J . Increasing p to p$ will decrease log + p, J (A 0 ), and we have to show that the resulting ``loss'' will not wipe out the ``gain'' witnessed by (2.2). For s 1, let us set & s =+ s, J . The key point is Margulis' formula [M] that gives an expression for (d& s ds)(A 0 ). Consider, for each i
}
}
(2.3)
We have, by the same computation as that leading to (2.1), + p, n&2(C i )=
1 p(1&p)
|
A0
( p&x i ) d+ p, n&1(x).
Now bi =
|
( p&x i ) d+ p, n(x)=(1&p) A
+p
|
A1
(1&p)
|
A0
( p&x i ) d+ p, n&1(x)
( p&x i ) d+ p, n&1(x)
|
A0
( p&x i ) d+ p, n&1(x)
Thus, by (2.3), (2.4), we get (since 1&p 13)
}
K Knb n d& s 0 (A ) : b i ds p i
}
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(2.4)
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SZAREK AND TALAGRAND
since b i b n for i
K 1 nb n
K 1 nb n
} ds log & (A ) } p+~(A ) p+(A) 0
s
0
since +(A) +~(A 0 ). Thus log + s, J (A 0 ) log & s(A 0 ) log & p(A 0 )& log & p(A 0 )&
K 1 nb n (s&p) p+(A)
bn bn =log +~(A 0 )& 2+(A) 2+(A)
if s p(1+1(2K 1 n)). Combining this with (2.2) concludes the proof.
K
3. PROOF OF THEOREM 1.4 We denote by K 2 the constant of Lemma 2.1. By induction over q n2, we construct a decreasing sequence J q of subsets of [1, ..., n], with card J q =n&q, such that, setting
\
A q =P Jq(A), p(0)=p, p(q)=p(q&1) 1+
1 , + q =+ p(q), Jq , K 2(n&q+1)
+
we have b i (A q ) i # Jq 2+ q(A q )
log + q+1(A q+1 ) log + q(A q )+sup where b i (A q )=
|
( p q &x i ) d+ q(x). Aq
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(3.1)
CONVEXIFIED SAUERSHELAH THEOREM
191
The construction is done by successive applications of Lemma 2.1. This is possible, since we can assume K 2 large enough that
\
1+
1 K 2 n2
q
+ \
1+
1 K 2 n2
+
n2
4 3
(and hence p(q) 23 since p 12). Consider now m n4. Then, since log + q(A q ) 0 for each q, by summation of the inequalities (3.1) for q 2m we see that we can find m q 2m such that b i (A q ) 2 |log + p, n(A)|. + m i # Jq q(A q )
sup Thus, if i # J q , we have
|
(x i &p(q)) d+ q(x)=&b i (A q ) &
Aq
2 |log + p, n(A)| + q(A q ) m
and thus y i =:
1 + q(A q )
|
x i d+ q(x) p(q)& Aq
2 |log + p, n(A)|. m
(3.2)
We observe that the point ( y i ) i # Jq belongs to conv A q , since the definition of y i exhibits this point as the barycenter of a measure on A q . Also, we note that conv A q is hereditary. Thus (3.2) implies that (s) Jq # conv A q , where s=p(q)&
2 |log + p, n(A)|. m
Now
\
p(q) p 1+
1 K2 n
q
+ \
p 1+
q pm , p+ K2 n K2 n
+
so that s p+
pm 2 & |log + p, n(A)|. K2 n m
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Thus, given r p, we get that s r provided m
K(r&p) n +K p
n
p log +
1 . p, n(A)
Since n&card I=q 2m, the proof is complete.
REFERENCES S. Alesker, Remark on SzarekTalagrand theorem, Combin. Probab. Comput., to appear. [G] A. Giannopoulos, A proportional DvoretzkyRogers factorization result, Proc. Amer. Math. Soc. 124, No. 1 (1996), 233241. [K-K-L] J. Kahn, G. Kalai, and N. Linial, The influence of variables on Boolean functions, in ``Proceedings 29th IEEE FOCS, 1988,'' pp. 5880. [M] G. A. Margulis, Probabilistic characteristics of graphs with large connectivity, Problems Inform. Transmission (1977). [Sa] N. Sauer, On the density of families of sets, J. Combin. Theory Ser. A 13 (1972), 145147. [Sh] S. Shelah, A combinatorial problem: Stability and order for models and theories in infinitary languages, Pacific J. Math. 41 (1972), 247261. [S-T] S. Szarek and M. Talagrand, An ``isomorphic'' version of the SauerShelah lemma and the BanachMazur distance to the cube, in ``Geometric Aspects of Functional Analysis,'' Lecture Notes in Mathematics, Vol. 1376, pp. 105112, Springer-Verlag, BerlinNew York, 1989. [A]
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On the Convexified SauerShelah Theorem* Stanislaw J. Szarek Department of Mathematics, Case Western Reserve University, Cleveland, Ohio 44106-7058
and Michel Talagrand Equipe d 'Analyse-Tour 46, U.A. au C.N.R.S. no. 754, Universite Paris VI, 4 Place Jussieu, 75230 Paris Cedex 05, France; and Department of Mathematics, Ohio State University, 231 West 18th Avenue, Columbus, Ohio 43210 Received August 19, 1994
Let A be a subset of [0, 1] n. Given =>0, we can find a subset I of [1, ..., n] such that the convex hull in R I of the projection of A onto [0, 1] I contains the cube [12&=, 12+=] I, and that card I n&K(n=+- n log(2 ncard A)), where K>0 is a universal constant. 1997 Academic Press
1. INTRODUCTION For a subset I of [1, ..., n], let us denote by P I the natural projection from [0, 1] n to [0, 1] I. The SauerShelah lemma [Sa], [Sh] asserts that given a subset A of [0, 1] n and an integer k with card A> i k ( ni ), there exists I/[1, ..., n] with card I>k and P I (A)=[0, 1] I. This result has proved to be of considerable use in analysis and probability. One drawback of this result is however the fact that, even when card A 2 n&1, we cannot guarantee that card I>n2. Moreover, even if, say, card A2 n > .999, it is not, for large n, possible to guarantee that card In .501. Yet it is at times valuable to find subsets I of [1, ..., n] such that n&card I is small, yet P I (A) is ``big.'' One result of this nature and an application was discovered in [S-T], and we introduce the necessary notation to explain this result. For a subset B of [0, 1] I, we denote by conv B the convex hull of B, when B is seen as a subset of R I. We say that B/[0, 1] I is symmetric if (x i ) i # I # B O (1&x i ) i # I # B. * Work partially supported by NSF grants. E-mail: sjs13po.cwru.edu. E-mail: talagranmath.ohio-state.edu.
183 0095-895697 25.00 Copyright 1997 by Academic Press All rights of reproduction in any form reserved.
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Theorem 1.1 [S-T]. Consider a symmetric subset A of [0, 1] n with card A 2 n&1, and consider =>0. Then there exists I/[1, ..., n] with card I n(1&4=) and conv P I (A)#[12&=, 12+=] I. It is a natural question to investigate what happens when card AR2 n&1 (or if A is not symmetric). The method of [S-T], that relies on the SauerShelah lemma, apparently does not yield estimates of card I of the correct order. Obtaining such estimates is the purpose of the present paper. An improvement of Theorem 1.1 in the symmetric case can be found in [A]; for another related sharp result see [G]. A first observation is that a straightforward push down argument reduces the problem to the case when A is hereditary, i.e., x # A, y # [0, 1] n;
\i n, y i x i O y # A.
Lemma 1.2. For each subset A of [0, 1] n, there is an hereditary subset B of [0, 1] n, with card B=card A, and such that, for each = 0, each I/[1, ..., n] we have [12&=, 12+=] I /conv P I (B) O [12&=, 12+=] I /conv P I (A). Proof. Let us fix j n. For x # [0, 1] n, we denote by x the sequence obtained by replacing the j th coordinate of x by zero. We define, for x # A T j (x)=x
if x A
T j (x)=x
if x # A
We define T j (A)=[T j (x); x # A]. We leave to the reader the easy proof that T j is one to one on A, so that card T j (A)=card A. We also leave to the reader to check that after iterating this operation for j=1, ..., n, the resulting set is hereditary. So it is enough to prove that [12&=, 12+=] I /3 conv P I (A) O [12&=, 12+=] I /3 conv P I (T j (A)). It should be obvious that we can assume j # I. If [12&=, 12+=] I /3 conv P I (A), we can find (' i ) i # I # [&1, 1] I with (12+' i =) i # I conv P I (A),
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CONVEXIFIED SAUERSHELAH THEOREM
185
and the HahnBanach theorem asserts that there are numbers (: i ) i # I such that \x # A, : : i x i < : : i ( 12 +' i =). i#I
i#I
Consider :~ i given by :~ i =: i if i{j and :~ j = |: j |. Define '~ i similarly. We show that \y # T j (A), : :~ i y i < : :~ i ( 12 +'~ i =) i#I
i#I
(and this finishes the proof ). We note that : i ' i :~ i '~ i for all i. Thus it suffices to find, for any y # T j (A), and x # A such that x i =y i if i{j and |: j | y j &: j x j 1 & 2 (|: j | &: j )=: j . If y j =0, we take (any) x such that T j (x)=y. If y j =1, we observe, using the definition of T j , that both y and y belong to A, and we choose x=y or x=y depending on whether : j 0 or : j <0. K Given a number s, we denote by (s) I the element of R I that has all its coordinates equal to s. Lemma 1.3.
If A is hereditary, then
[s&=, s+=] I /conv P I (A) (s+=) I /conv P I (A). Proof. In fact, if x # conv P I (A), and 0 y i x i for all i # I, we have y # conv P I (A). K The problem now becomes, given r, to find I with card I as large as possible and (r) I # conv P I (A). A key ingredient of the proof will be the ability to measure the size of a subset of [0, 1] I not only by its cardinality or, equivalently, by its normalised counting measure, but also by its size with respect to the measures + p, I =((1&p) $ 0 +p$ 1 ) I, where $ t is the unit mass concentrated at t. (For I=[1, ..., n], we write + p, I =+ p, n .) Thus it will be natural to state a result in the line of Theorem 1.1 not only when one controls the cardinality of A, but when one controls + p, n(A). Before we state our result, it is instructive to motivate it by considering the standard example where, for some q # R,
{
=
A= (x i ) i n ; : x i q . i n
Here
{
=
P I (A)= (x i ) i # I ; : x i q . i#I
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SZAREK AND TALAGRAND
If card I=n&k and if (r) I # conv P I (A), we have r q(n&k), i.e., k (nr&q)r, so that k
n(r&p) np&q + . r r
(1.1)
We now observe, using standards estimates on the tail of the binomial distribution, that (at least when np2 q np) log 1+ p(A) is of order (np&q) 2np(1&p). Thus for p r 2p (which is the interesting range) we must have k
1 n(r&p) n(1&p) log 1+ p(A) + . K p p
\
+
(1.2)
(Of course the statement says nothing when the right hand side is greater than or equal to n which happens, in particular, if rp is large.) The meaning of our main result is that the example presented above is essentially the worst case. Theorem 1.4. Assume p 12, and consider an heriditary subset A of [0, 1] n. Consider p r 1. Then we can find I/[1, ..., n] with (r) I # conv P I (A) and n&card I K
\
n(r&p) n + log 1+ p(A) . p p
+
In this statement, as well as in the rest of the paper, K (with a subcript or not) denotes an effectively computable universal (i.e. independent of p, n, r, A or any other parameters) constant, the value of which may vary between occurences. Corollary 1.5. Consider a subset A of [0, 1] n. Then we can find I/[1, ..., n] with [12&=, 12+=] I /conv P I (A) and
\
card I n&K n=+ n log
2n . card A
+
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CONVEXIFIED SAUERSHELAH THEOREM
Comments. 1. Theorem 1.4 provides a value of n&card I of the correct order. Finding the exact best possible value is of course a more challenging question. 2. The reason that we assume p 12 is simply that the case p=12 is by far the most interesting, and that the case p 12 has an identical proof. Minor modifications are required to handle the case p>12. Let us now comment on the proof of Theorem 1.4. To construct I we will remove points of [1, ..., n] one at a time. When choosing which point j to remove first, the natural idea is to insure that P J (A) is ``large,'' where J is the complement of j. The first thought would be to require that + p, J (P J (A)) is substantially larger than + p, n(A). However, examples in [K-K-L] show that this is impossible. Rather, we will require a control of +p$, J (P J (A)) where p$ is somewhat larger than p. (The reader should observe that, when A is hereditary, increasing p decreases + p(A).) This is the object of the main Lemma proved in Section 2. The proof of Theorem 1.4 is then completed in Section 3.
2. THE MAIN LEMMA Lemma 2.1. bi =
|
Consider an hereditary set A/[0, 1] n, and for i n, set
\
n
( p&x i ) d+ p, n(x) = : A
+
` p xj(1&p) 1&xj ( p&x i ) .
x # A j=1
Assume p 23. Consider j n such that b j =max i n b i . Set J= [1, ..., n]&[ j]. Then, for p$=p(1+(1Kn)) we have log + p$, J (P J (A)) log + p, n(A)+
bj . 2+ p, n(A)
Comment. Thus we achieve that P J (A) is larger than A in the sense that + p$, J (P J (A)) is somewhat larger than + p, n(A) for p$=p(1+(1Kn)). Before we start the proof, we need an auxiliary result. Lemma 2.2. +p$, n K+ p, n . Proof.
Assume p p$ p(1+(13n)), p 23 . Then (1K) + p, n
Consider x # [0, 1] n. Set l= i n x i , so that + p, n([x])=p l(1&p) n&l; + p$, n([x])=p$ l(1&p$) n&l
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SZAREK AND TALAGRAND
and + p, n([x]) p = + p$, n([x]) p$
l
\ +\
1&p 1&p$
+
n&l
.
Now 1 p 1 (1+13n) p$ 1
1&(23) 2 1&p 1+ . 1&p$ 1&(23)(1+13n) n
Thus 1 + ([x]) 2 1+ p, n n (1+13n) + p$, n([x]) n
n
\ +.
K
Proof of Lemma 2.1. Step 1. To simplify the notation, we assume without loss of generality that j=n. We set +=+ p, n , +~ =+ p, J . We set A 0 =[x=(x i ) i n&1 # [0, 1] n&1; x 0 # A] A 1 =[x=(x i ) i n&1 # [0, 1] n&1; x 1 # A]
,
where x 0 denotes concatenation. We note that since A is hereditary, A 1 /A 0 and A 0 =P J (A). We set a i =+~(A i ) for i=0, 1; note that a 0 a 1 and +(A)=(1&p) a 0 + pa 1 a 0. We have b n =(1&p) a 0p+pa 1( p&1)=p(1&p)(a 0 &a 1 ). Also +~(A 0 )&+(A)=a 0 &+(A)=p(a 0 &a 1 ) and thus +~(A 0 )=+(A)+
bn bn =+(A) 1+ . 1&p (1&p) +(A)
\
Now, p(1&p)(a 0 &a 1 ) bn = p. +(A) (1&p) a 0 +pa 1
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+
(2.1)
CONVEXIFIED SAUERSHELAH THEOREM
189
Since log(1+x(1&p)) x for 0 x p<1, we have proved that log +~(A 0 ) log +(A)+
bn . +(A)
(2.2)
Step 2. The difference between (2.2) and what we want is that the left-hand side involves +~ =+ p, J rather than + p$, J . Increasing p to p$ will decrease log + p, J (A 0 ), and we have to show that the resulting ``loss'' will not wipe out the ``gain'' witnessed by (2.2). For s 1, let us set & s =+ s, J . The key point is Margulis' formula [M] that gives an expression for (d& s ds)(A 0 ). Consider, for each i
}
}
(2.3)
We have, by the same computation as that leading to (2.1), + p, n&2(C i )=
1 p(1&p)
|
A0
( p&x i ) d+ p, n&1(x).
Now bi =
|
( p&x i ) d+ p, n(x)=(1&p) A
+p
|
A1
(1&p)
|
A0
( p&x i ) d+ p, n&1(x)
( p&x i ) d+ p, n&1(x)
|
A0
( p&x i ) d+ p, n&1(x)
Thus, by (2.3), (2.4), we get (since 1&p 13)
}
K Knb n d& s 0 (A ) : b i ds p i
}
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(2.4)
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SZAREK AND TALAGRAND
since b i b n for i
K 1 nb n
K 1 nb n
} ds log & (A ) } p+~(A ) p+(A) 0
s
0
since +(A) +~(A 0 ). Thus log + s, J (A 0 ) log & s(A 0 ) log & p(A 0 )& log & p(A 0 )&
K 1 nb n (s&p) p+(A)
bn bn =log +~(A 0 )& 2+(A) 2+(A)
if s p(1+1(2K 1 n)). Combining this with (2.2) concludes the proof.
K
3. PROOF OF THEOREM 1.4 We denote by K 2 the constant of Lemma 2.1. By induction over q n2, we construct a decreasing sequence J q of subsets of [1, ..., n], with card J q =n&q, such that, setting
\
A q =P Jq(A), p(0)=p, p(q)=p(q&1) 1+
1 , + q =+ p(q), Jq , K 2(n&q+1)
+
we have b i (A q ) i # Jq 2+ q(A q )
log + q+1(A q+1 ) log + q(A q )+sup where b i (A q )=
|
( p q &x i ) d+ q(x). Aq
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(3.1)
CONVEXIFIED SAUERSHELAH THEOREM
191
The construction is done by successive applications of Lemma 2.1. This is possible, since we can assume K 2 large enough that
\
1+
1 K 2 n2
q
+ \
1+
1 K 2 n2
+
n2
4 3
(and hence p(q) 23 since p 12). Consider now m n4. Then, since log + q(A q ) 0 for each q, by summation of the inequalities (3.1) for q 2m we see that we can find m q 2m such that b i (A q ) 2 |log + p, n(A)|. + m i # Jq q(A q )
sup Thus, if i # J q , we have
|
(x i &p(q)) d+ q(x)=&b i (A q ) &
Aq
2 |log + p, n(A)| + q(A q ) m
and thus y i =:
1 + q(A q )
|
x i d+ q(x) p(q)& Aq
2 |log + p, n(A)|. m
(3.2)
We observe that the point ( y i ) i # Jq belongs to conv A q , since the definition of y i exhibits this point as the barycenter of a measure on A q . Also, we note that conv A q is hereditary. Thus (3.2) implies that (s) Jq # conv A q , where s=p(q)&
2 |log + p, n(A)|. m
Now
\
p(q) p 1+
1 K2 n
q
+ \
p 1+
q pm , p+ K2 n K2 n
+
so that s p+
pm 2 & |log + p, n(A)|. K2 n m
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SZAREK AND TALAGRAND
Thus, given r p, we get that s r provided m
K(r&p) n +K p
n
p log +
1 . p, n(A)
Since n&card I=q 2m, the proof is complete.
REFERENCES S. Alesker, Remark on SzarekTalagrand theorem, Combin. Probab. Comput., to appear. [G] A. Giannopoulos, A proportional DvoretzkyRogers factorization result, Proc. Amer. Math. Soc. 124, No. 1 (1996), 233241. [K-K-L] J. Kahn, G. Kalai, and N. Linial, The influence of variables on Boolean functions, in ``Proceedings 29th IEEE FOCS, 1988,'' pp. 5880. [M] G. A. Margulis, Probabilistic characteristics of graphs with large connectivity, Problems Inform. Transmission (1977). [Sa] N. Sauer, On the density of families of sets, J. Combin. Theory Ser. A 13 (1972), 145147. [Sh] S. Shelah, A combinatorial problem: Stability and order for models and theories in infinitary languages, Pacific J. Math. 41 (1972), 247261. [S-T] S. Szarek and M. Talagrand, An ``isomorphic'' version of the SauerShelah lemma and the BanachMazur distance to the cube, in ``Geometric Aspects of Functional Analysis,'' Lecture Notes in Mathematics, Vol. 1376, pp. 105112, Springer-Verlag, BerlinNew York, 1989. [A]
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