Rademacher functions

Transcription

1 Rademacher functions Jordan Bell Department of Mathematics, University of Toronto July 6, 4 Binary expansions Define S : {, } N [, ] by Sσ σ k k, σ {, }N. For example, for σ and σ, σ 3,..., for σ and σ, σ 3,..., Sσ ; Sσ Let σ {, } N. If there is some n N such that σ n and σ k for all k n +, then defining σ k k n τ k k n k n, we have Sσ n σ k k + kn+ n k σ k k + n Sτ. One proves that if either i there is some n N such that σ n and σ k for all k n + or ii there is some n N such that σ n and σ k for all k n +, then S Sσ contains exactly two elements, and that otherwise S Sσ contains exactly one element. In words, except for the sequence whose terms are only or the sequence whose terms are only, S Sσ contains exactly two elements when σ is

2 eventually or eventually, and S Sσ contains exactly one element otherwise. We define ɛ : [, ] {, } N by taking ɛt to be the unique element of S t if S t contains exactly one element, and to be the element of S t that is eventually if S t contains exactly two elements. For k N we define ɛ k : [, ] {, } by ɛ k t ɛt k, t [, ]. Then, for all t [, ], t Sɛt which we call the binary expansion of t. Rademacher functions ɛ k t k, For k N, the kth Rademacher function r k : [, ] {, +} is defined by r k t ɛ k t, t [, ]. We can rewrite the binary expansion of t [, ] in as r k t k Define r : R {, +} by k ɛkt k rx [x], ɛ k t k t. where [x] denotes the greatest integer x. Thus, for x < we have rx, for x < we have rx, and r has period. Lemma. For any n N, r n t [n t] r n t, t [, ] In the following theorem we use the Rademacher functions to prove an identity for trigonometric functions. Theorem. For any nonzero real x, 3. cos x k sin x x. Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, p. 4,

3 Proof. Let n N and let c,..., c n R. The function c k r k is constant on each of the intervals [ s n, s + n, s n. 3 There is a bijection between n {, +} n and the collection of intervals 3. Without explicitly describing this bijection, we have giving exp We have i c k r k t dt exp e ix t dt e i n s s+ n s n exp n exp i δ n δ n e iδ kc k e ic k + e ic k, c k r k t dt ix ix e ixt i c k r k t dt δ k c k cos c k. 4 r k t k converges uni- Using we check that the sequence of functions n formly on [, ] to t, and hence using 5 we get r k t exp ix k dt e e ix ix ix + sin x ix x. 5 e ix t dt sin x x as n. Combining this with 4, which we apply with c k x k, we get as n, proving the claim. cos x k sin x x 3

4 We now give an explicit formula for the measure of those t for which exactly l of r t,..., r n t are equal to. We denote by µ Lebesgue measure on R. We can interpret the following formula as stating the probability that out of n tosses of a coin, exactly l of the outcomes are heads. Theorem 3. For n N and l n, µ{t [, ] : r t + + r n t l n} n n. l Proof. Define φ : [, ] R by But for m Z, hence Therefore { µ t [, ] : φt π π π φt π e ix l n+ n r kt dx e imx dx δ m, { } r k t l n { m m, n r kt l n n r kt l n. π π φtdt π π π π 6 e ix l n+ n r kt dxdt e ixl n e ixl n cos n xdx; e ix n r kt dtdx the last equality uses 4 with c x,..., c n x. Furthermore, writing cos n x n e ix + e ix n n k n e ixk n, k Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, pp

5 we calculate using 6 that π π proving the claim. n n e ixl n cos n xdx n e ixl n e ixk n dx k π k n n n e ixk l dx k π k n n n δ k l, k k n n n δ k,l k k n n, l We now prove that the expected value of a product of distinct Rademacher functions is equal to the product of their expected values. 3 Theorem 4. If k,..., k n are positive integers and k < < k n, then r k t r kn tdt. Proof. Write J r k t r kn tdt and define which satisfies φx + φx r ks k x, x R, s r ks k x + ks k s r ks k x φx. 3 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 85, Solution 3.. s 5

6 Hence, as φ has period and r has period, But, as φ has period, rxφxdx J k k k r k tφ k tdt r k tφ k tdt φxdx hence J, proving the claim. k k j k j rxφxdx j+ j rxφxdx. φxdx rxφxdx rxφxdx φxdx For each n N, if f is a function defined on the integers we define n I n f f r k t dt. Lemma 5. For any n N, Proof. Using Theorem 4 we get I n x n, I n x 4 3n n. I n x n. n r k t dt r k t + j k r k t dt r j tr k tdt φxdx, 6

7 Using Theorem 4 we get, since r j t 4 r j t and r j t 3 r j t, n 4 I n x 4 r k t dt r k t n + j j, k, l all distinct j, k, l, m all distinct 4 n + nn. n r j t 3 r k t + j k j r j t r k tr l t r j tr k tr l tr m tdt 4 n r j t r k t j k j Our proof of the next identity follows Hata. 4 Lemma 6. For any n N, I n x π cos n x x dx. Proof. For n N and c,..., c n R, n exp i c k r k t dt cos c k r k t dt+i n sin c k r k t dt, and since 4 tells us that the left-hand side of the above is real, it follows that we can write 4 as n cos c k r k t dt cos c k. 7 Suppose that ξ is a positive real number. Using t xξ and doing integration by parts, cos xξ cos t x dx ξ t dt ξ cos t t ξ ξ π. sin t dt t + ξ sin t dt t 4 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 88, Solution

8 It is thus apparent that for any real ξ, cos xξ x dx ξ π. For any n N, applying the above with ξ n r kt we get I n x π r k t π dt π π π x cos x n r kt x dxdt cos x r k t dtdx x I ncos x dx. Applying 7 with c k x for each k, this is equal to π x cos x dx cos n x π x dx, completing the proof. We use the above formula for I n x to obtain an asymptotic formula for I n x. 5 Theorem 7. Proof. By Lemma 6, I n x n. π I n x π For ɛ <, define φ ɛ : [, π R by φ ɛ x cos n x x dx. x + log cos x. ɛ We also define α ɛ arccos ɛ, β ɛ α ɛ cos n x x dx, 5 Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, p., Masayoshi Hata, Problems and Solutions in Real Analysis, p. 88, Solution

9 and for σ >, K ɛ,σ αɛ exp x nx σ Let < ɛ <. Until the end of the proof, at which point we take ɛ, we shall keep ɛ fixed. For < x < α ɛ we have, using arccos ɛ ɛ, hence φ x x + log cos x < ɛ + log ɛ ɛ + log ɛ <, On the other hand, φ ɛx cos x < exp x. dx. x tan x, φ ɛ x ɛ ɛ sec x, so φ ɛ φ ɛ and φ ɛ t > for all t < α ɛ, giving and hence φ ɛ x >, exp x < cos x. ɛ Collecting what we have established so far, for < x < α ɛ we have exp x < cos x < exp x. ɛ This shows that and therefore K ɛ, ɛ On the other hand, so K ɛ, αɛ αɛ exp nx αɛ ɛ x dx K ɛ, ɛ + β ɛ π I n x. exp x nx dx αɛ K ɛ, + β ɛ π I n x. Now summarizing what we have obtained, we have cos n x x dx, cos n x x dx, K ɛ, + β ɛ π I n x K ɛ, ɛ + β ɛ. 8 9

10 For σ >, doing the change of variable t n σ x, αɛ exp K ɛ,σ nx σ x dx n n σ αɛ σ As n, the right-hand side of this is asymptotic to n e t n σ t dt π. σ Dividing 8 by n and taking the limsup then gives π I n x π lim sup n n ɛ, or lim sup n indeed β ɛ depends on n, but β ɛ < ɛ yields lim sup n I n x n π ɛ ; e t dt. α ɛ, which does not depend on n. Taking I n x n π. On the other hand, taking the liminf of 8 divided by n gives π I n x π lim inf n n, or lim inf n I n x n π. Combining the limsup and the liminf inequalities proves the claim. Lemma 8. For any ξ R and n N, I n e ξ x < I n e ξx + I n e ξx cosh ξ n. t We will use the following theorem to establish an estimate similar to but weaker than the law of the iterated logarithm. 6 Theorem 9. For any ɛ >, for almost all t [, ], log n exp n+ɛ r k t dt <. n n 6 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 89, Solution 3.7.

11 Proof. Define f n : [, ], by f n t log n exp n+ɛ r k t. n Applying Lemma 8 with ξ log n n, f n tdt n +ɛ cosh n log n. It is not obvious, but we take as given the asymptotic expansion n log n cosh n 8 n 3 log n 45 + log n3 + 8 log n4 + On 3/, n and using this, n log n n +ɛ cosh log n n n +ɛ + O n +ɛ Thus n f n tdt n n n +ɛ + On <. n +ɛ + On. Because each f n is nonnegative, using this with the monotone convergence theorem gives the claim. Theorem. For almost all t [, ], lim sup n n r kt n log n. Proof. Let ɛ >. By Theorem 9, for almost all t [, ] there is some n t such that n n t implies that f n t <, where we are talking about the functions f n defined in the proof of that theorem; certainly the terms of a convergent series are eventually less than. That is, for almost all t [, ] there is some n t such that n n t implies that taking logarithms, log n ɛ log n + r k t n <, and rearranging, n r kt n log n < + ɛ + ɛ. For each s N, let E s be those t [, ] such that lim sup n n r kt n log n > + s.

12 For each s, taking < ɛ < s we get that almost all t [, ] do not belong to E s. That is, for each s, the set E s has measure. Therefore E s has measure. That is, for almost all t [, ], for all s N we have t E s, i.e. n lim sup r kt + n n log n s, and this holding for all s N yields completing the proof. 3 Hypercubes lim sup n E s n r kt n log n, Let m n be Lebesgue measure on R n, and let Q n [, ] n. 7 Theorem. If f C[, ], then x + + x n lim f n Q n n Proof. Define X n : Q n R by dm n x f. We have X n x + + x n, x Q n. n Q n X n dm n x n x k dx k n, 7 Masayoshi Hata, Problems and Solutions in Real Analysis, p. 6, Solution..

13 and we define V n Q n X n dm n x Q n n n n n x k n + x j x k n X n + 4 dm nx j k 3 + n 3n + n 4n 4 n. x kdx k + n j k j n j k j 4 4 x j dx j x k dx k + 4 Suppose that c n is a sequence of positive real numbers tending to, and define J n J n c to be those x Q n such that X nx c n. Then V n Q n X n dm n x X n dm n x J n c ndm n x J n c nm n J n, so Take c n n /3, giving m n J n V n c n n c. n m n J n n /3. Let ɛ >. Because f is continuous, there is some δ > such that t < δ implies that ft f < ɛ; furthermore, we take δ such that f δ 6 < ɛ. 3

14 Set N > δ 3. For n N and x Q n \ J n, X nx < c n n /3 N /3 < δ, and so This gives us fx n xdm n x f Q n which proves the claim. fx n x f < ɛ. fx n x f dm n x Q n fx nx f dm n x J n + fx nx f dm n x Q n\j n f dm n x + ɛdm n x J n Q n\j n f m n J n + ɛ n /3 f + ɛ < f δ + ɛ 6 < ɛ, 4