FURTHER NUMBER THEORY

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1 FURTHER NUMBER THEORY SANJU VELANI. Dirichlet s Theorem and Continued Fractions Recall a fundamental theorem in the theory of Diophantine approximation. Theorem. (Dirichlet 842). For any real number x and N N, there exists p, q Z with q N such that - important consequence. x p q ( q ) qn 2 Theorem.2 (Dirichlet). For any real number x there exists infinitely many rationals p q (q > 0) such that x p q < q 2 Remark. () Any real number can be approximated by rationals with rate (denominator) 2. (2) If we impose condition that p is reduced (i.e. (p, q) = ) then statement requires x q ( ) to be irrational; indeed suppose x = a p. Then, b q x p q = a b p q = aq pb bq aq pb bq Thus q < b for (.) x p q < q 2 But there are only a finite number of rationals with q < b of interest. (3) In view of (2), imposing condition that the rationals are reduced gives a criterion for determining whether a given real number x is rational or irrational.

2 2 SANJU VELANI Key part of the course is to investigate numbes for which the rate of approximation in Dirichlet (i.e. (denominator) 2 ) can be improved by some power. More precisely, for τ 2, let W(τ) denote the set of real numbers x for which there exists infinitely many rationals p q (q > 0) such that x p q < q τ τ 2 gives us that W(2) = R, which is just the theorem above. τ > 2 - studying W(τ) brings into play serious measure theory - in particular the world of fractals. Lets rephrase Thereom 2: given any irrational x we can find an infinite sequence of good rationals pn which approximate x with rate at least. qn 2 Question: Is there a natural way to find these good approximates? Yes - the convergents pn associated with a continued fraction expansion of x satsifies the crucial inequality x p n < qn 2 Let x be irrational. Then x has a unique representation as an infinite continued fraction, i.e. given x there exists a unique sequence of integers a 0, a, a 2,... where a 0 Z and a i for all i (called the partial quotients of x), such that a 0 + = [a 0 ; a, a 2,...]. a + a To find the partial quotients use the continued fraction algorithm. Example.. The Golden Ratio has the simplest continued fraction expansion 5 + = [;,,,...] := []. 2 The above is an example of a quadratic irrational; i.e. a root of an algebraic polynomial of degree 2, such as f(x) = ax 2 + bx + c (a, b, c Z, a 0). So we have that 5+ 2 is a root of f(x) = x 2 x and it has a periodic continued fraction expansion. Definition: If an infinite continued fraction contains a block b, b 2,..., b n which repeats indefinitely then the continued fraction expansion is called periodic. So, we write

3 FURTHER NUMBER THEORY 3 for [a 0, a,...,a m, b,...,b n ] An elegant statement is [a 0, a,...,a m, b,...,b n, b,...,b n,...]. Theorem.3. The continued fraction expansion of irrational x is periodic if an only if x is a quadratic irrational. Remark. If x is periodic, i.e. x = [a 0,..., a m, b,...,b n ] then the partial quotients in its continued fraction expansion are bounded by a constant K(x) - namely the max{ a 0, a,...,a m, b,...,b n }. Theorem.3 quadratic irrationals have bounded partial quotients - the set Bad. Given x = [a 0, a,...] consider its nth convergent p n and satisfying p n = [a 0, a, a 2,...,a n ] n 0. p 0 = a 0 q 0 = p = a a 0 + q = a. p n = a n p n + p n 2 = a n + 2. Lemma.. for all n wtih strict inequality when n > (Exercise). Lemma.2. (p n, ) = for all n The key properties of convergents are Theorem.4. If pn Corollary.. The convergents pn (n 0) is the nth convergent to irrational x then x p n < + (n 0) satisfy x p n < q 2 compare from Theorem.2 (just use Lemma.)..

4 4 SANJU VELANI Theorem.5. If pn (n 0) is the nth convergent to irrational x then x p n > (+ + ). Upshot: Theorem.4 and.5 the approximation by convergents is trapped by the denominators of the convergents. Theorem.6 (Best Approximation). Let pn (n 0) be the nth convergent of irrational x. If b, the rational a satisfies b x p n x a. b Every other rational with the same or smaller denominator differs from x by a greater amount. 2. Hurwitz and Bad The theory of continued fractions not only enables us to effectively prove Dirichlet s Theorem (Theorem.2) - it enables us to improve it. Theorem 2. (Hurwitz 89). For any real number x there exists infinitely many rationals p (q > 0) such that q (2.) x p q <. 5q 2 So the rate of q 2 is replaced by the rate of 5q 2. The continued fraction theory for x irrational, at least one of the three consecutive convergents satisfy (2.). p n, p n+ +, p n+2 +2 Question: Can 5 be replaced by something smaller? Theorem 2.2. The constant 5 in Hurwitz is best possible. Proof. Assume that 5 can be replaced by something smaller, in particular by 5 + ǫ (ǫ > 0, arbitrary). Consider x = 5+ 2, root of the polynomial f(t) = t 2 t = (t x )(t x 2 )

5 FURTHER NUMBER THEORY 5 where x 2 = 5. Assume there exists a sequence of rationals p i 2 q i satisfying x p i q i < (. 5 + ǫ)qi 2 Then for i sufficiently large x 2 p i q i x 2 x = 5 which is just the triangle inequality. It follows that + x p i q i < ( 5+ǫ)q i 2 <ǫ ( ) f pi < q i ( 5 + ǫ 5 + ǫ)qi 2 ( ) pi q2 i f <. q i However the L.H.S. is a positive integer - a contradiction - there are no integers in (0, ). The proof shows that if we replace 5 by anything smaller (say 5+ǫ ) then Hurwitz statement is false for the Golden Ratio, i.e. with x = 5+ and ǫ > 0 arbitrary there are at 2 most a finite number of rationals p such that q x p q < ( 5 + ǫ)q 2. Alternatively, there exists a constant say c(x) such that x p q > c(x) for all p q 2 q Q. In fact we can take c(x) = 5+ǫ for the Golden Ratio. So, 5+ 2 has the simplest continued fraction and is the worst approximable irrational. 2.. Badly Approximable Numbers Definition: A real number x is said to be badly approximable if there exists a constant c(x) > 0 such that x p q > c(x) for all p q 2 q Q. In view of Theorem 2.2, 0 < c(x) < 5.

6 6 SANJU VELANI Let Bad denote the set of badly approximable numbers. Clearly Bad since 5 + γ = Bad 2 and indeed tγ Bad for t Z \ {0} (Exercise). A connection between Bad and the theory of continued fractions is given by Theorem 2.3. An irrational x Bad if and only if the partial quotients a i in its continued fraction expansion [a 0 ; a, a 2,...] are bounded; i.e. there exists a constant K(x) > 0 such that a i K(x) for all i. Corollary 2.. Bad is an uncountable set. Corollary 2.2. Let x be a quadratic irrational. Then x Bad. Proof. This is just a consequence of Theorem.3. Remark. A famous conjecture states that the quadratic irrationals are the only algebraic irrationals in Bad. For example does 2 3 have bounded partial quotients? Recall from chapter Theorems.4 and.5. Then we can look at the proof of Theorem 3. Proof. x Bad x has bounded partial quotients. This is an exercise. Now examine and see whether the continued fraction expansion is also bounded. Suppose x = [a 0 ; a, a 2,...] with a i K. Let p Q with q. Then there exists n such that q q <. The convergents are the best approximates (Theorem.6), so However x p q x p n > (+ + ) = ( ). qn 2 qn = a n+ + n + = a n+ + + (Lemma, Chapter ) q }{{} n a n+ + 2 K + 2. Thus

7 FURTHER NUMBER THEORY 7 If n = then If n 2, x p q > qn 2 (K + 2) = q 2 (K + 2) > q 2 (K + 2) q 2 qn 2 ( qn q 0 = q a K > K +. ) 2 (q ). Upshot: We have that i.e. x Bad. = a n + 2 = a n + q n 2 K + x p q > q 2 (K + 2)(K + ) 2 Particular goal is to investigate the size of Bad. m(bad) = 0, i.e. zero Lebesgue measure. Although it is small in terms of Lebesgue measure it is maximal in terms of dimension as dim(bad) = 3. Algebraic and Transcendental Numbers An algebraic number is a number x which satisfies an algebraic equation. An equation of the form a n x n + a n x n + + a x + a 0 = 0 with each a i Z and at least one of these a i is non-zero. A number which is not algebraic is called transcendental. Example 3.. We see the following numbers are algebraic: Every rational a is algebraic because it satisfies bx a = 0. b 2 is algebraic because it satisfies x 2 2 = 0. i = is algebraic because it satisfies x 2 + = 0.

8 8 SANJU VELANI An algebraic number satisfies an infinite number of algebraic equations of different degrees. For example x = 2 satisfies x 2 2 = 0 x 4 4 = 0 x 6 8 = 0 and so on. If x satisfies an algebraic equation of degree n but none of lower degree then we say that x is of degree n. Example is of degree 2 and any rational is of degree. The set of algebraic numbers is countable while the set of transcendental numbers is uncountable. It s much harder to prove that a given number is transcendental than to prove that it is irrational. Transcendental irrational. (If ζ is transcendental and rational then ζ is algebriac, which is a contradiction). Aim: is to produce as many examples of transcendental numbers as we like. First a useful notion: Definition: A real number x is said to be approximable by rationals to order n if there exists a positive constant c depending only on x such that x p q < c has infinitely many rational solutions p (q > 0) with (p, q) =. q Dirichlet s Theoreom implies that an irrational is approximable to order 2. Exercise: Any rational is approximable to order but not to any higher order. Liouville used the following to construct transcendental numbers. Theorem 3. (Liouville 844). Suppose x is a real algebraic number of degree n. Then there exists a constant c(x) > 0 such that x p q > c(x) for every rational p (q > 0) distinct from x. q Corollary 3.. A real algebraic number of degree n is not approximable to order n + or higher. Proof. Exercise - indeed n + can be replaced by n + ǫ (ǫ > 0 is arbitrary).

9 FURTHER NUMBER THEORY 9 Example 3.3. Let β = 0! + 0 2! + 0 3! + = 0.! 2! 000 3! ! 0... Now let N be arbitrary. For n > N let Now examine β n = n i= 0 i! = p n 0 n! = p n. 0 < β p n = β β n because n > N. = 0 (n+)! + 0 (n+2)! + = 0 (n+)! ( + 0 (n+)! 0 (n+2)! + 0 (n+)! 0 (n+3)!) + ) = 0 (n+)! ( + < 0 (n+)! i=0 } 0 (n+2) {{} < 0 0 i }{{} 0 = } 0 (n+3)(n+2) {{} < 00 + < 2 0 (n+)! = 2q (n+) n Upshot: For any n > N there exists pn Q such that β p n < 2. q N n 0 (n+4)(n+3)(n+2 } {{ } < 000 < 2q N n ) + Now Liouville implies β is not algebraic of degree less than N. But N is arbitrary, so β is not algebraic of any degree, which implies β is transcendental. Remark. Clearly we could have replaced 0 by other integers to obtain other transcendental numbers. The general principle of the construction is simply that a number defined by a sufficiently rapid sequence of rational approximations is necessarily transcendental. Definition: A real number ζ is said to be Liouville number if for every m N there exists p m q m Q (q m > ) such that ζ p m < q m m. In view of Liouville s Theorem q m

10 0 SANJU VELANI Corollary 3.2. Any Liouville number is transcendental. Indeed the number is a Liouville number. β = i= 0 i! Remark. Let L denote the set of Liouville numbers. It follows from definition that if ζ L then there exists infinitely many rationals p such that q ζ p q < q τ where τ > 0 is arbitrary. i.e. L W(τ) - see Chapter. Recall: W(2) = R and it now follows that for τ > 2 we have W(τ).. If ζ L, then the rate of approximation in Dirichlet s Theorem (i.e. (denominator) 2 ) can be improved by an arbitrary power. In some sense Liouville numbers are at the opposite end of the spectrum to badly approximable numbers. A real number x is said to be very well approximable if w W(τ) for some τ > 2. Clearly, L W where W is the set of very well approximable numbers. Theorem 3.2 (Mahler s Conjecture - 984). Let K denote the middle third Cantor set. Clearly K := { x [0, ] : x } a i 3 i, with a i [0, ] i= set of points in [0, ] whose base 3 expansions are free of the digit. ζ = 2 3 n! K n= and is Liouville, i.e. ζ K 2. Mahler s Conjecture states that there exists very well approximable numbers other than Liouville numbers in the middle third Cantor set. i.e. (W \ L) K. This was solved in 2007 (Maths Annalen, vol 338) and we will return to the proof later. Proof. (Liouville) Let ζ be a real, algebraic number of degree n, i.e. ζ is a root of f(x) = a n x n + a n x n + + a x + a 0 with a i Z and a n 0.

11 Note. f is a real continuous function. FURTHER NUMBER THEORY Fix an interval around ζ say I λ [ζ λ, ζ + λ] (with λ (0, )) such that I λ contains no other root of f except ζ (λ = λ(ζ)). Now there exists c > 0 (c (λ)) such that f (x) c for all x I λ. Let c(ζ) = min{ c, λ} and consider any rational p (q > 0) then q () If p I q λ then ζ p c(ζ) q λ c(ζ) q, n. (2) If p I q λ and p ζ then q ( ) 0 p f = a np n + a n p n q + + a p + a 0 q for q > 0. By the Mean Value Theorem there exists x (ζ, p ) such that q ( ) f p f(ζ) f q (x) = p ζ q i.e. ( ) ( ) p p f = q q ζ f (x). Hence, ( ζ p p f q) q = f c(ζ)qn (x) c qn 3.. Roth s Theorem Let ζ be real algebraic number of degree n 2. Then, Liouville implies that (3.) ζ p q < q τ has only finitely many rational solutions if τ > n. Theorem 3.3 (Thue - 909). He showed that (3.) holds if τ > 2 n + Theorem 3.4 (Siegel - 92). He showed that (3.) holds if τ > 2 n Theorem 3.5 (Dyson - 947, Gelfond - 952). They showed independently that (3.) holds if τ > 2n

12 2 SANJU VELANI Theorem 3.6 (K Roth - 955). Roth finally showed that (3.) holds if τ > 2, for which he was awarded the fields medal in 958. Roth s result is the best possible. Note. Liouville s Theorem implies that if x L then x is transcendental. Roth s Theorem implies that if x W(τ) for τ > 2 then x is transcendental. 4. Naturally Occurring Irrationals and Transcendental Numbers Example (irrational and algebraic), e, π, e + π, sin 2. In Year 3, we proved that e is irrational - this was quite straight forward and made use of the fact that We need to be slightly more sophisticated to prove that π is irrational. Lemma 4.. For n N, let e = r=0 r!. f(x) = xn ( x) n. n! Then f(x) and all its derivatives f (m) (x) take integer values at x = 0. Remark. Some remarks about f(x): () Since, f( x) = f(x), the conclusion remains true at x =. (2) For 0 < x < we have 0 < f(x) < n! Proof. To be done as an exercise. Theorem 4.. π 2 is irrational, which trivially implies that π is irrational. Proof. Suppose that π 2 = a b for a, b N. With f as in Lemma 4., write G(x) = b n( π 2n f(x) π 2n 2 f (2) (x) + π 2n 4 f (4) (x) + ( ) n f (2n) (x) ). In view of the Lemma, we have G(0) and G() are integers. Also note that d dx (G (x) sin(πx) πg(x) cos(πx)) = G (x) sin(πx) + πg (x) cos(πx) πg (x) cos(πx) + π 2 G(x) sin(πx) = ( G (x) + π 2 G(x) ) sin(πx) = b n( ) π 2n+2 f(x) + ( ) n f (2n+2) (x) sin(πx) }{{} =0

13 because f is a polynomial with leading term x 2n. Hence, integrating throughout we obtain π b FURTHER NUMBER THEORY 3 = b n π (2n+2) f(x) sin(πx) = π 2 a n f(x) sin(x). [ G a n (x) sin(πx) f(x) sin(πx) dx = π = G() + G(0). But, by our second remark of Lemma 4. we have that 0 < π 0 a n f(x) sin(πx) < < n! dx < πan n! 0 ] G(x) cos(πx) 0 the last inequality holds for n large. A contradiction as there are no integers in (0, ). The proof can be adapted to show that e y is irrational for every rational y 0. Hermite in 873 proved the e is transcendental and π was proved to be transcendental by Lindemann in 882. Theorem 4.2 (Lindemann-Weierstrauss, 885). α,..., α n are algebraic and β,...,β n are algebraic distinct such that not all β i are zero then α e β + + α n e βn 0. Equivalently if α,...,α n are linearly independent over Q then for all polynomials p Z[X,...,X n ] such that p(α,..., α n ) 0. Corollary 4.. π is transcendental. We have that e iπ = e 0 = e 0 + e iπ = 0. Assume that π is algebraic, which implies iπ is algebraic by the closure of the algebraic numbers. By the Lindermann-Weierstrauss theorem we have that e 0 + e iπ 0 if π is algebraic contradiction. From this we can get that e α, sin α, cosα, tan α, sinh α, cosh α, tanh α are all transcendental for α algebraic. This is because they can all be written as a linear combination of e α Also, log α is transcendental for α an algebraic number. Hilbert s 7th Problem: If α is an algebraic number such that α 0, and β is algebraic irrational then is α β transcendental? For example, <

14 4 SANJU VELANI 2 2, ( 2) 3 3,...,e π ( ) i. this was solved around by Gelfond and Schneider. 5. Metric Diophantine Approximation - the Lebesgue Theory ψ : R + is a real positive function such that ψ(r) 0 as r. We say that R approximable functions. W(ψ) = { x I x p } q ψ(q) for inf. many (p, q) Z N x is ψ-well approximable {x} W(ψ). We aim to determine the size of W(ψ). We say that almost all numbers in an interval (could be R in which case almost all real numbers) have a certain property P if the set of numbers lacking the property P has measure zero. Example 5.. Almost all real numbers are non-integers. W(ψ) is a lim sup set. Let E i be a sequence of subsets in R n. The lim sup set denoted by E associated with this sequence is defined as follows E = ( ) E i j= i=j = lim sup i It simply corresponds to the set of points in R n that lie in infinitely many of the set E i (i N). lim sup sets are fundamental objects in metric number theory. We now show that W(ψ) is a lim sup set. For fixed q N, let E i. Note. We have that A q (ψ) = q B p=0 ( ) p I q, ψ(q) and we have equality if ψ(q) < 2q. m ( A q (ψ) ) 2qψ(q)

15 FURTHER NUMBER THEORY 5 Then W(ψ) is simply the set of numbers in I which lie in infinitely many sets A q (ψ) with q =, 2,... i.e. a lim sup set. Now notice that W(ψ) = A q (ψ) m= q=m W(ψ) A q (ψ) q=m i.e. the balls (or rather intervals) form a cover for W(ψ). Thus, m ( W(ψ) ) ( ) m A q (ψ) q=m m ( A q (ψ) ) q=m ( ) 2 qψ(q) q=m Now suppose qψ(q) <. Then given ǫ > 0, there exists m 0 such that for all m m 0 and qψ(q) < ǫ 2. q= q=m It follows from ( ), that m m 0 and hence But ǫ > 0 is arbitrary whence m ( W(ψ) ) ǫ. m ( W(ψ) ) = 0 Theorem 5.. Let ψ : R + be a real positive function such that qψ(q) <. Then m ( W(ψ) ) = 0.

16 6 SANJU VELANI This theorem is in fact a simple consequence of a general result in probability theory. 5.. The convergence Borel-Cantelli Lemma Let (Ω, µ) be a measure space with µ(ω) < and let E i (i N) be a family of measurable sets in Ω such that Then µ(e i ) <. i= µ(e ) = 0 where E = lim sup i E i, i.e. the set of points in Ω which lie in infinitely many E i (i N). Note. Theorem is a trivial consequence. Let Ω = I = [0, ], µ = m and E q = A q (ψ). Then, if m ( A q (ψ) ) < then m ( W(ψ) ) = 0. Proof. Let ǫ > 0, since µ(e i ) < then there exists N 0 such that for all N N 0 we have But, for each N N we have Thus But ǫ > 0 is arbitrary whence ( ) µ E i µ(e i ) < ǫ. i N i=n E E i. i N ( ) µ(e ) µ E i < ǫ. i N µ(e ) = 0 A simple consequence of Theorem is the following corollary. Corollary 5.. For τ > 2 we have m ( W(τ) ) = 0. Recall that W(τ) is W(ψ) with ψ(q) = q τ. Then for τ > 2. qψ(q) = q τ <

17 Also, note that m ( W(ψ) ) = 0 when FURTHER NUMBER THEORY 7 or (ǫ > 0). ψ(r) = ψ(r) = q 2 log q +ǫ q 2 log q(log log q) +ǫ. In view of Theorem, what can we say about m ( W(ψ) ) when qψ(q) =? A fundamental thereom in metric number theory is Theorem 5.2 (Khintchine s Theorem (924)). Let ψ : R + be a monotonic function such that ψ(r) 0 as r. Then m ( W(ψ) ) { 0 if q= = qψ(q) < if q= qψ(q) = Remark. We note that () Khintchine s Thereom is an elegant zero-full law. (2) = m(i) (3) m ( W(ψ) ) = when ψ(q) = q 2 log q or ψ(q) = q 2 log log q To prove the full measure result the divergence of the measure sum isn t in general enough. Convergence of the Borel-Cantelli Lemma says that if µ(e ) < the µ(e )=0. However it s not the case that if µ(e i ) = then µ(e ) = µ(ω) or indeed µ(e ) > 0. Example 5.2. For i N let E i = [0, ], then we have i m(e i ) = i m(e i ) = i = but clearly E = {0}, i.e. m(e ) = 0 The problem here is that the sets E i overlap too much - in fact they are nested. We require that the sets E i (i N) are in some sense independent. If we had independence, i.e. for i j

18 8 SANJU VELANI (5.) µ(e i E j ) = µ(e i )µ(e j ) and µ(e i ) =, then we could conclude that µ(e ) = µ(ω). In applications (5.) rarely holds. More realistic is quasi-independence, i.e. for i j there exists a constant C such that (5.2) µ(e i E j ) Cµ(E i )µ(e j ). Then if µ(e i ) = we have µ(e ) C > 0. So, by weakening independence to (5.) we only get positive measure for E - this is no problem if we already know (by some other means) that µ(e ) that µ(e ) = 0 or µ(ω). The following result shows that we can weaken (5.2) and its this result that is often used to establish full measure statements The Divergence Borel-Cantelli Lemma Let (Ω, µ) be a probability measure space (i.e. µ(ω) = ) and let E i (i N) be a family of measurable sets in Ω such that Then for all Q sufficiently large µ(e i ) =. i= i,j Q ( Q µ(e i E j ) C µ(e i ) quasi-independence on average, where C is a constant. The key to Khintchine s Theorem is establishing that the sets A q (ψ) are quasi-independent on average; it s here where the monotonicity assumption is required. This is not required for the convergence part. Removing monotonicity from Khintchine s Theorem is a serious problem and is called the Duffin-Schaeffer conjecture. Recall, W(ψ) is invariant under integer/rational translations. Khintchine s Theorem implies i= ) 2

19 FURTHER NUMBER THEORY 9 Corollary 5.2. Let ψ : R + be a monotonic function such that ψ(r) 0 as r and qψ(q) =. Then almost all real numbers are ψ-well approximable, i.e. m ( {x R x is not ψ approximable} ) = 0. Corollary 5.3. m(bad) = 0 Proof. Take the function ψ(q) = q 2 log q { and B(ψ) = x R c(x) > 0 s.t. x p q > c(x)ψ(q) for all p } q (q > 0). By Corollary 2 we have m ( B(ψ) ) = 0 but Bad B(ψ) so m(bad) = 0. In view of Corollary 5. we have that m ( W(τ) ) = 0 for τ > 2. We d expect to the size of W(τ) to decrease as τ increases (i.e. as rate of approximation increases). For example we d expect W(300) to be smaller than W(3) - clearly W(300) W(3) but Lebesgue measure can t distinguish between these two sets - both are zero. Upshot: require a finer measurement of size that can distinguish between sets of Lebesgue measure zero. 6. Metric Diophantine Approximation: the Hausdorff Theory Start by generalising Lebesgue measure and the notion of integer dimension. 6.. Hausdorff Measure and Dimension Let F be a non-empty subset in R n (no harm in assuming F is uncountable)!! DIAGRAM!! We cover our set F by balls B i of diameter d i ρ (with ρ > 0). If {B i } is a countable collection of balls B i with diameter d i ρ that cover F, i.e. F i B i then {B i } is said to be a ρ-cover of F. For s 0 and ρ > 0 let

20 20 SANJU VELANI (6.) H s ρ (F) = inf { i d s i } {B i} is a ρ-cover of F As ρ decreases the class of allowed covers of F in (6.) is reduced. Therefore H s ρ(f) increases as ρ decreases and so approaches a limit as ρ 0. Then (6.2) H s = lim ρ 0 H s ρ (F), which exists for any subset F of R n ; can be 0, positive, finite or. The quantity H s (F) is the s-dimensional Hausdorff measure of the set F. When s = n an integer then H n is comparable to n-dimensional Lebesgue measure m n, i.e. for F R n (6.3) c m n (F) F n c 2 m n (F). where c, c 2 are constants (greater than 0) depending only on n. Some Analysis of (6.3) Let ρ <. If t > s and {B i } is a ρ-cover of F then d t i = d s i dt s i < ρ t s d s i i i because d i ρ. So, only taking infimum we have H t ρ(f) ρ t s H s ρ(f). On letting ρ 0, we see that if H s (F) < then H t (F) = 0 for t > s. Furthermore for s = 0 and F uncountable we have that H 0 (F) is infinite. Hence, there exists a critical value of s at which H s (F) jumps from to 0 as s increases. This critical value of s is called the Hausdorff dimension of F written simply as dim F. Remarks By definition of dim F we have that { H s 0 if s > dim F (F) = if s < dim F. If s = dim F, then H s (F) may be zero or infinite or may satisfy F is said to be an s-set. 0 < H s (F) <

21 Let I = [0, ]. Then dim I = and 0 if s > H s (I) = if s = if s < FURTHER NUMBER THEORY 2 I is an s-set. To calculate dim F (say dim F = α), its usually the case that we establish upper and lower bounds separately. dim F α dim F α (upper bound) (lower bound). Upper bounds for dim F are usually easy we can exploit a natural cover of F. Example 6.. Let K be the standard middle Cantor set. Then dim K log 2 log 3. Recall K = n=0k n where K n is a union of 2 n balls (intervals) of diameter 3 n. Thus, for each n N we have E n K and so the intervals in E n form a ρ = 3 n cover of K. It follows that as n (i.e. ρ 0) if In other words ( ) n 2 Hρ s (K) 2n (3 n ) s = 0 3 s 2 3 < s > log 2 s log 3. H s (K) = 0 if s > log 2 log 3. It follows from the definition of Hausdorff dimension that dim K log 2 log 2. In fact, dim K = log 3 log 3 dim K = inf{s H s (K) = 0} and moreover To prove H log 2 log 3 (K) =. dim F log 2 log 3

22 22 SANJU VELANI we need to work with arbitrary covers of K. Let {B i } be an arbitrary ρ-cover with ρ <. K - is bounded and closed (intersection of closed intervals), i.e. K is compact. Hence without loss of generality we can assume that {B i } is finite. For each B i, let k be the integer such that (6.4) 3 (k+) d i < 3 k. Then the B i intersects at most one interval of E k separated by at least 3 k. If j k then B i intersects at most (6.5) 2 j k = 2 j 3 sk 2 j 3 s d s i intervals of E j, where s = log 2 log 3 enough such that and the final inequality holds from (6.4). Now j large 3 (j+) d i for all B i {B i } can be done since the number of {B i } <. Then j k for each B i and (6.5) invalid. Furthermore {B i } must intersect all 2 j intervals of E j {B i } is a cover for k, i.e. 2 j = #{I E j B i I } i #{I E j B i I } i 2 j 3 s d s i by (6.5), i.e. for any arbitrary cover {B i } we have that for s = log 2 log 3 d s i 3 s = 2 which implies H s (K) 2 dim K log 2 log 3 Even in the simple Cantor set example, the lower bound for dimk is much more involved than the upper bound which is easy. This is usually the case and the number theoretic sets W(ψ) and W(τ) are no exception The Jarnik-Besicovitch Theorem Recall, that W(ψ) = lim sup q A q (ψ) where Also, for each m N A q (ψ) = q B p=0 ( ) p q, ψ(q) I.

23 FURTHER NUMBER THEORY 23 W(ψ) A q (ψ) q=m and the balls in here form a cover for W(ψ). ψ : R + is a real, positive function such that ψ(r) 0 as r. Suppose for the moment that ψ is monotonic. Now fix m N. Now fix m N and let ρ = ρ(m) = 2ψ(m). Then the balls in {A q (ψ)} q m form a ρ cover of W(ψ). Thus, H s ρ ( ) W(ψ) q ( 2ψ(q) ) s q=m The right hand side 0 as m (i.e. ρ = 2ψ(m) 0) if qψ s (q) < q= i.e. H s( W(ψ) ) = 0 if the s-volume sum converges. Actually, monotonicity on ψ can be removed (exercise) and we have proved the following theorem. Theorem 6.. Let ψ : R + be a function such that ψ(r) 0 as r. Then H s( W(ψ) ) = 0 if qψ s (q) <. When s = then H m one dimensional Lebesgue measure, and then the theorem is just the convergent part of Khintchines theorem (Theorem 5.2). Corollary 6.. q= { } dim W(ψ) inf s qψ s (q) <. which just follows from Theorem 6. and the definition of Hausdorff dimension. Corollary 6.2. For τ 2 we have dim W 2 τ. Proof. We have that qψ s (q) = q= q= q q τs < q= because ψ(q) = q τ and s > 2 τ The divergence result (Theorem 6.) and upper bound dimension result (Corollary 6.2) just exploit the natural cover of the lim sup set! The substance of the next result is the lower bound. Theorem 6.2 (The Janik-Besicovitch Theorem). For τ 2 we have dim ( W(τ) ) = 2 τ.

24 24 SANJU VELANI From this we see that W(300) is smaller than W(3). Excellent result but gives no information about H s as s = dim ( W(τ) ) the critical exponent.!! DIAGRAM!! Does there exist a Hausdorff measure version of Khintchines Theorem (a Lebesgue measure statement)? Yes! Theorem 6.3 (Jarnik s Theorem). Let s (0, ) and ψ : R + be a monotonic function such that ψ(r) 0 as r. Then H s( W(ψ) ) { 0 if q= = qψs (q) < if q= qψs (q) = An Important Remark In Jarnik s original statement various technical conditions on ψ and indirectly s were imposed, which prevented s = note even as stated the theorem is false when s =. This is because H s( W(ψ) ) m ( W(ψ) ) =, which is not equal to infinity in the divergence case. However, these technical conditions have recently been removed (2006) and gives the clear cut statement. Moreover it allows us to combine Khintchine and Jarnik to obtain a unifying statement Theorem 6.4 (Khintchine-Jarnik 2006). Let s (0, ] and ψ : R + be a monotonic function such that ψ(r) 0 as r. Then H s( W(ψ) ) { 0 if q= = qψs (q) < H s( I) if q= qψs (q) =. Corollary 6.3. For τ > 2 H 2 τ ( W(τ) ) =. 7. A Mass Transference Principle The goal is to show that the Hausdorff theory for W(ψ) is in fact a consequence of the Lebesgue theory. In particular Dirichlet s theorem implies the Jarnik-Besicovitch theorem. Khintchine s Theorem (Lebesgue Statement) implies Jarnik s Theorem (Hausdorff statement). Start by giving the heuristics. () Dirichlet Jarnik-Besicovitch. We know that m ( W(2) ) = H ( W(2) ) = H (I). We want to renormalize H H 2 τ preserves H and ball B dim W(τ) 2 τ (τ 2) - the hard part of the Jarnik-Besicovitch theorem (note that H 2 τ (I) > 0 and H 2 τ (I) = if τ > 2).

25 FURTHER NUMBER THEORY 25 (2) Khintchine Jarnik. This is the divergent part of each theorem in the main subtance. So in Jarnik we re given qψ s (q) = s (0, ). Want to show H ( W(ψ) ) =. We have that qψ s (q) = H ( W(ψ) ) = H (I). We again renormalize H H s, which preserves H and ball B. Then we obtain a statement such as H s( W(ψ) ) = H s (I) then for s < we have H s (I) = 7.. The Mass Transference Principle (a simplified version) Let Ω be a compact subset of R n such that (7.) B δ (B) r(b) δ (δ > 0) for any ball B = B(x, r) in Ω with x Ω and r r 0. A consequence of (7.) in that 0 < H δ (Ω) < and dim Ω = δ we return to this later. Given a ball B and s 0, let and so B δ = B. B s = B ( ) x, r s δ Theorem 7. (Mass Transfer Principle (2006)). Let {B i } i N be a sequence of balls in Ω with radii r(b i ) 0 as i. If H δ (lim sup B s i ) = Hδ (Ω) then Corollary 7.. For τ > 2, dim W(τ) 2 τ. H s (lim sup B δ i ) = H s (Ω) Proof. Dirichlet and the Mass Transfer Principle (with Ω = [0, ], δ =, s = 2 and τ B p = B(p, q q q τ ) in [0, ]) Corollary 7.2. Let s (0, ) and ψ : R + be a monotonic function such that ψ(r) 0 as r. Then H s( W(ψ) ) = 0 if qψ s (q) =. Proof. Khintchine and the Mass Transfer Principle (with Ω = [0, ], δ =, B p q = B(p q, ψ(q))

26 26 SANJU VELANI Recall 8. A problem of Kurt Mahler Remember, from chapter 3, that a real number is very well approximable if x W(τ) for τ > 2. If W is the set of very well approximable numbers then m(w) = 0. Also L W, the set of Liouville numbers, which have dim L = 0. We regard K to represent the middle third Cantor set. Clearly Mahler s conjecture Mahler conjectured that ζ = 2 3 n! K L. n= (W \ L) K, i.e. the Liouville numbers are not the only well approximable numbers in K. 8.. Premliminaries - The Cantor set and measure Rationals of the form p are special!! 2 2 n of all possible rationals p in I lie in K. Let 3 n 3 n µ be the natural measure on K each of the 2 n intervals I n of length 3 n in E n has mass 2 n, i.e. µ(i n ) = 2 n = (3 n ) γ where γ = log 2. Assume x K then µ(b) µ(i log 3 k+) 3 γ r γ. Clearly, ( ) µ(e n ) = µ = 2 n 2 n =. I n ξ n I n µ is supported on K and µ(k) =. Furthermore, for any subset A of [0, ] we have A i I i and I i e where ξ is the collection of all basic intervals in Cantor construction We want to know { } µ(a) = inf µ(i i ) : A U i I i and I i ξ. i Let B = B(x, r) be a ball (interval) of radius r < and let k be the integer such that 3 (k+) r < 3 k. Then B intersects at most one interval in ξ k!! DIAGRAM!!

27 FURTHER NUMBER THEORY 27 Then i.e. µ ( B(x, r) ) µ(i k ) = 2 k = (3 k ) γ = 3 γ (3 (k+) ) γ 3 γ r γ (8.) µ ( B(x, r) ) 3 γ r γ for arbitrary balls. Now suppose that x K, then B contains at least one interval in ξ k+ and so i.e. µ ( B(x, r) ) µ(i k+ ) = 2 (k+) = (3 (k+) ) γ 3 γ r γ (8.2) 3 γ r γ µ ( B(x, r) ) 3 γ r γ. Indeed (8.) is all that is required to show that Mass Distribution Principle dim K γ = log 2 log 3. Let µ be a measure supported on a subset F of R n and suppose for some γ there exists constants c > 0 and δ > 0 such that for all balls B with diameter d δ. Then µ(b) cd δ In particular, dimf γ. H γ (F) µ(f). c Proof. If {B i } is a ρ-cover of F with ρ < δ then ( ) 0 < µ(f) = µ B i i µ(b i ) i c i d γ i. Taking infimum, Hρ(F) γ µ(f). Thus H γ (F) µ(f) because H γ c c ρ increases as ρ 0.

28 28 SANJU VELANI The Strategy towards Mahler Rationals with denominator powers of 3 are special for K; let and consider the set W A (ψ) = { x I A = {3 n n 0,, 2,... } x p } q < ψ(q)for inf. many (p, q) Z A. This is the set of ψ-well approximable numbers with denominators of rational approximation restrcited to A. Here ψ : R + is a monotonic function such that ψ(r) 0 as r and without loss of generality assume ψ(r) < 2r otherwise W A (ψ) = I. Now consider the set W A (ψ) K and for n N let with p 3 n K. By definition, A n (ψ) = 0 p 3 n B = 0 p 3 n B ( p )) 3 n, ψ(3n K ( p )) 3 n, ψ(3n K W A (ψ) K = lim sup A n (ψ). n Now, with µ = to the Cantor measure, we have µ ( A n (ψ) ) 2 2 n 3 γ ψ(3 n ) γ by. Indeed since the balls are centred at points of K we have that µ(a n (ψ)) 2 n ψ(3 n ) γ = ( 3 n ψ(3 n ) ) γ by with γ log 2. In view of Borel-Cantelli convergence Lemma (Chapter 5) we have log 3 that µ ( W A (ψ) K ) = 0 if n= ( 3 n ψ(3 n ) )γ. The following is a Khintchine type theorem for W A (ψ) K with respect to µ.

29 FURTHER NUMBER THEORY 29 Theorem 8.. Let ψ : R + be a monotonic function with ψ(r) as r 0. Then µ ( W A (ψ) K ) { 0 if ( 3 n ψ(3 n ) ) γ < = = µ(k) if ( 3 n ψ(3 n ) ) =. Essentially this is exercise 9 on the problem sheet. Now µ H γ, which is γ-dimensional Hausdorff measure. Indeed H γ (K) = = µ(k). So in view of we have H γ( B(x, r) ) r γ for x K. Thus, we are in the position to apply the Mass Transference Principle (Chapter 7) with Ω = K, δ = γ and B p 3 n K = B ( p, ψ(3 n ) ) - 3 n this yields the divergent part of the following (the convergent part is easy). Theorem 8.2. Let ψ : R + be a monotonic function with ψ(r) 0 on r and s (0, γ) H s( W A (ψ) K ) { 0 if 3 nγ ψ(3 n ) s < = = µ(k) if 3 nγ ψ(3 n ) s =. (note H s (K) = if s < γ). Theorem 2 strongly implies Mahler. With ψ(r) = r τ (τ > 0) let s write W A (τ) for W A (ψ). Then theorem 2 implies dim ( W A K ) { = inf s } 3 nτ 3 nτs < = ( ) 3 τs+γ n Corollary 8.. For τ we have in particular, for τ > dim ( W A (τ) K ) = γ τ H γ τ ( WA (τ) K ) =. For τ > 2, every point in W A (τ) is by definition very well approximable, i.e. Thus, for any τ > 2 W A (τ) W K. We can make τ arbitrarily close to 2, so dim(w K) dim ( W A (τ) K ) = γ τ. Now, dim L = 0; then dim(w K) γ 2. dim ( (W \ L K) γ 2