Duffin-Schaeffer Conjecture. Aug. 9, 2010

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1 Aug. 9, 2010

2 Itroductio Diophatie approximatio. For α Q c [0, 1), there are ifiitely may itegers m, with (m, ) = 1 such that Let ψ : N [0, ) α m < 1 2 A = {α [0, 1) : α m < ψ(), (m, ) = 1 i.o.} Questio : What is the Lebesque measure of A?

3 Lebesque Measure µ o [0, 1) µ(i ) = b a, for I = [a, b], (a, b), [a, b), or (a, b], so µ[0, 1) = 1 For A = k I k, If it is disjoit, µ(a) = k µ(i k) I geeral, µ(a) k µ(i k) A is of measure zero, µ(a) = 0 if for ay ɛ, there is a fiite or coutable family of itervals I k such that A I k ad µ(i k ) ɛ. µ(a) = 1 if µ(a c ) = 0. A property holds for a.e. x if the set of x it fails to hold the property is a set of measure zero.

4 Lebesque Measure µ o [0, 1) Example Coutable set A = {r k } is of measure zero. Cosider I k = (r k ɛ, r 2 k+1 k + ɛ ) ad µ(i 2 k+1 k ) = ɛ 2 k from Diophatie approximatio, for almost every α, there are ifiitely may pairs of m, with (m, ) = 1 α m < 1 2

5 Khitchie s theorem(1924) Let ψ : N [0, ) ad suppose that ψ() o-icreasig. If ψ() =, the µ(a ) = 1,i.e, for a.e. α, there are ifiitely may pairs of m, with (m, ) = 1 α m < ψ() However, if ψ() <, the µ(a ) = 0,i.e, for a.e. α, there are oly fiitely may pairs of m, with (m, ) = 1 α m < ψ()

6 Examples of Khitchie s theorem ψ() = 1 log, The µ(a ) = 1 i.e, for a.e. α, there are ifiitely may pairs of m, with (m, ) = 1 α m < 1 2 log ψ() = 1 (log ) 1+ɛ, The µ(a ) = 0 i.e, for a.e. α, there are oly fiitely may pairs of m, with (m, ) = 1 α m < 1 2 (log ) 1+ɛ

7 A Note that A = {α : α A i.o} A = {α : m s.t. α m < ψ() = 1 m I (m, ) (m,)=1 where I (m, ) = So ( m ψ() µ(a ) m ad equailty holds for ψ() 1 2, ), m + ψ() µ(i m, ) = 2 ψ() φ() with(m, ) = 1}

8 Probability µ(a) : the probability of x A, pickig x [0, 1) Borel-Catelli lemma (a) If µ(a ) <, the µ(a ) = 0 (b) If µ(a ) = ad µ(a i A j ) = µ(a i )µ(a j ). the µ(a ) = 1. Erdős-Réyi theorem µ(a ) diverges. The, µ(a ) lim sup N ( N 1 µ(a )) 2 1 i,j N µ(a i A j ) Corollary of BC Lemma If ψ() φ() coverges, for almost every α, there are oly fiitely may m, with (m, ) = 1 α m < ψ().

9 Duffi-Schaeffer cojecture(1941) If ψ() φ() diverges, the µ(a ) = 1, that is, for almost every α, there are ifiitely may m, with (m, ) = 1 α m < ψ() Remark Duffi-Schaeffer : ψ() 1 2 so, µ(a ) = ψ() φ() Polligto-Vaugha : overcome this difficulty.

10 Cassel s Zero Oe Law(1950) Let ψ : N [0, ). B = {α [0, 1) : α m < ψ() The, µ(b) is either 0 or 1. i.o } Remark The proof is based o that Tx = 2x (mod1) is ergodic. (T is ergodic if T 1 A = A, the µ(a) is 0 or 1.)

11 Gallagher s Ergodic theorem(1961) Let ψ : N [0, ). A = {α [0, 1) : α m < ψ(), (m, ) = 1 i.o.} The, µ(a ) is either 0 or 1. Remark The proof uses the fact that for all prime p, T (x) = px + s p (mod 1) is ergodic,

12 Duffi-Schaeffer Theorem Theorem Suppose that ψ() 1 2 ad ψ() φ() diverges. I additio, lim sup N ( N The µ(a ) = 1 1 ) ( N 1 ψ() φ() ψ()) c > 0 1

13 Lemma Let M, N be positive itegers ad A be a positive umber. Let k be the umber of solutios for 0 < mn M A with 1 m M ad 1 N. The k 2A Proof) Let M = M (M,N) ad N = N (M,N). 0 < mn M A (M, N) This solutio satisfies mn = a (mod M ) for 1 a A (M,N), which has oly oe solutio i m (mod M ) for each a. So, there are (M, N) umber of solutios i m for each value of a. Hece A k 2 (M, N) = 2A (M, N)

14 Outlie of Proof 1 Show that µ(a A t ) 8ψ()ψ(t), for t. k : umber of itersectios of I (m, ) ad I (s, t) L : maximum legth of itersectio of I (m, ) ad I (s, t) k 4 max(tψ(), ψ(t)) I (m, ) I (s, t) 0 < m s t < ψ() + ψ(t) t 0 < mt s tψ() + ψ(t) 2 max(tψ(), ψ(t)) L mi of two itervals = 2 mi( ψ(), ψ(t) t ) µ(a A t ) k L 4 max(tψ(), ψ(t)) 2 mi( ψ(), ψ(t) ) t 8ψ()ψ(t)

15 Outlie of Proof 2 apply Erdős-Réyi theorem From ψ() 1 2, µ(a ) = 2 ψ() φ() ad µ(a ) = ( 1 i,j N µ(a N ) 2 i A j ) 8 1 ψ() + N 1 ψ() ( N µ(a ) lim sup 1 µ(a )) 2 N 1 i,j N µ(a i A j ) ( lim sup 1 N ) 2 ( N 2 ψ() 8 φ() ψ()) > 0 the Zero Oe Law µ(a ) =

16 Corollary of Duffi-Schaeffer Theorem Recall that Hece, φ(p) p p:prime 1 p φ(p) p 1 p = = 1 1 p p For almost every α, there are ifiitely may prime p ad iteger m such that α m p < 1 p 2

17 Further Result Erdős(1970) ψ() = c or 0 Valler(1978) ψ() c Polligto ad Vaugha (1990) k dimesioal versio (k 2) ( ) k ψ() ψ() diverges. The, for a.e. x = (x 1,, x k ) [0, 1) k max( x 1 a 1,, x k a k ) < ψ() with (a 1 a k, ) = 1 have ifitely may solutios.

18 Refereces Metric Number Theory, Gly Harma (1998) Oxford Uiversity Press Duffi, R.J. ad Schaeffer, A.C. (1941). Khitchie s problem i metric Diophatie approximatio Duke J.,8,