Sidon’s problem on thin bases

Một phần của tài liệu Số học tổ hợp của GS Vũ Hà Văn and GS Tao (Trang 32 - 35)

1.3 The exponential moment method

1.3.1 Sidon’s problem on thin bases

We now apply Chernoff’s inequality to the study of thin bases in additive combi- natorics.

Definition 1.11 (Bases) LetBNbe an (infinite) set of natural numbers, and letkZ+. We define thecounting function rk,B(n) for anynNas

rk,B(n) := |{(b1, . . . ,bk)∈ Bk:b1+ ã ã ã +bk=n}|.

We say thatBis abasis of order kif every sufficiently large positive integer can be represented as sum ofk(not necessarily distinct) elements ofB, or equivalently if rk,B(n)≥1 for all sufficiently largen. Alternatively, Bis a basis of orderkif and only ifN\k Bis finite.

Examples 1.12 The squares N∧2= {0,1,4,9, . . .}are known to be a basis of order 4 (Legendre’s theorem), while the primes P= {2,3,5,7, . . .} are con- jectured to be a basis of order 3 (Goldbach’s conjecture) and are known to be a basis of order 4 (Vinogradov’s theorem). Furthermore, for anyk≥1, the kth powersNk= {0k,1k,2k, . . .} are known to be a basis of order C(k) for some finite C(k) (Waring’s conjecture, first proven by Hilbert). Indeed in this case, the powerful Hardy–Littlewood circle method yields the stronger result that

rm,Nk(n)=m,k(nmk−1) for all largen, ifm is sufficiently large depending on k (see for instance [379] for a discussion). On the other hand, the powers ofk kN= {k0,k1,k2, . . .}and the infinite progressionkãN= {0,k,2k, . . .}are not bases of any order whenk>1.

The functionrk,Bis closely related to the density of the setB. Indeed, we have the easy inequalities

nN

rk,B(n)≤ |B∩[0,N]|k

nk N

rk,B(n) (1.21) for anyN ≥1; this reflects the obvious fact that ifn=b1+ ã ã ã +bkis a decompo- sition of a natural numbernintoknatural numbersb1, . . . ,bk, thennNimplies thatb1, . . . ,bk∈[0,N], and converselyb1, . . . ,bk∈[0,N] impliesnk N. In particular ifBis a basis of orderkthen

|B∩[0,N]| = (N1/k). (1.22) Let us say that a basis B of orderkisthinifrk,B(n)=O(logn) for all large n. This would mean that|B∩[0,N]| =N1/k+ok(1), thus the basis B would be nearly as “thin” as possible given (1.22). In the 1930s, Sidon asked the question of whether thin bases actually exist (or more generally, any basis which is “high quality” in the sense thatrk,B(n)=no(1)for alln). As Erd˝os recalled in one of his memoirs, he thought he could provide an answer within a few days. It took a little bit longer. In 1956, Erd˝os [92] positively answered Sidon’s question.

Theorem 1.13 There exists a basis BZ+of order2so that r2,B(n)=(logn) for every sufficiently large n. In particular, there exists a thin basis of order2.

Remark 1.14 A very old, but still unsolved conjecture of Erd˝os and Tur´an [98]

states that if BNis a basis of order 2, then lim supn→∞r2,B(n)= ∞.In fact, Erd˝os later conjectured that lim supn→∞r2,B(n)/logn>0 (so that the thin basis constructed above is essentially as thin as possible). Nothing is known concerning these conjectures (though see Exercise 1.3.10 for a much weaker result).

Proof Define1a setBZ+randomly by requiring the eventsnB(fornZ+) to be jointly independent with probability

P(nB)=min

C logn

n ,1

1Strictly speaking, to make this argument rigorous one needs an infinite probability space such as Wiener space, which in turn requires a certain amount of measure theory to construct. One can avoid this by proving a “finitary” version of Theorem 1.13 to provide a thin basis for an interval [1,N] for all sufficiently largeN, and then gluing those bases together; we leave the details to the interested reader. A similar remark applies to other random subsets ofZ+which we shall construct later in this chapter.

whereC>0 is a large constant to be chosen later. We now show thatr2,B(n)= (logn) for all sufficiently largenwith positive probability (indeed, it is true with probability 1). Writing

r2,B(n)=

i+j=n

I(iB)I(jB)=

1≤i<n/2

I(iB)I(niB)

+O(1) we see that it suffices to show that the probability

P

1≤i<n/2

I(iB)I(niB)=(logn) for all but finitely manyn

is positive (if the constants in the () notation are chosen appropriately). By the Borel–Cantelli lemma (Lemma 1.2) and the convergence of∞

n=1 1

n2, it thus suffices to show that

P

1≤i<n/2

I(iB)I(niB)=(logn)

=1−O 1

n2 for all largen.

By linearity of expectation (1.3), we have forn>1 E

1≤i<n/2

I(iB)I(niB)

=

1≤i<n/2

C2 logi

i

log(ni)

ni +OC(1)

=

C2ln1/2n n1/2

1≤i<n/2

ln1/2i i1/2

+OC(1)

=(C2logn)+OC(1). In particular, by choosingClarge enough, we may take

32 lognE

1≤i<n/2

I(iB)I(niB)

κlogn for alln>1 and someκ >32.

Observe that the restrictioni <n/2 ensures that the boolean random variables I(iB)I(niB) are jointly independent. If we now apply Corollary 1.9 with :=1/2, we conclude that

P

1≤i<n/2

I(iB)I(niB)≤ κ 2logn

≤2/n2,

and

P

1≤i<n/2

I(iB)I(niB)≤ 3κ 2 logn

≤2/n2.

The claim follows.

It is quite natural to ask whether Theorem 1.13 can be generalized to arbitraryk.

Using the above approach, in order to obtain a basisBsuch thatrk,B(n)=(logn), we should set P(nB)=cn1/k−1ln1/kn for all sufficiently largen. As before, we have

rk,B(n)=

x1+ããã+xk=n

I(x1 ∈B)ã ã ãI(xkB). (1.23) Although rk,B(n) does have the right expectation (logn), we face a major problem: the variablesI(x1∈ B), . . . ,I(xkB) withk>2 are no longer inde- pendent. In fact, a typical numberxappears in quite many ( (nk−2)) solutions of x1+ ã ã ã +xk =n. This dashes the hope that one can use Theorem 1.8 to conclude the argument.

It took a long time to overcome this problem of dependency. In 1990, Erd˝os and Tetali [97] successfully generalized Theorem 1.13 for arbitraryk:

Theorem 1.15 For any fixed k, there is a subset BN such that rk,B(n)= (logn)for all sufficiently large n. In particular, there exists a thin basis of order k for any k.

We shall discuss this theorem later in a later section. Let us now turn instead to another application.

Một phần của tài liệu Số học tổ hợp của GS Vũ Hà Văn and GS Tao (Trang 32 - 35)

Tải bản đầy đủ (PDF)

(532 trang)