Fourier Analysis of the Parity-Vector Parameterization of the Generalized Collatz px+1 maps
aa r X i v : . [ m a t h . G M ] F e b Fourier Analysis of the Parity-VectorParameterization of the Generalized Collatz px + 1 maps Maxwell C. Siegel ∗ Sunday, February 23, 2020
Abstract
Let p be an odd prime, and consider the map H p which sends an in-teger x to either x or px +12 depending on whether x is even or odd. Thevalues at x = 0 of arbitrary composition sequences of the maps x and px +12 can be parameterized over the -adic integers ( Z ) leading to a con-tinuous function from Z to Z p which the author calls the “characteristicfunction” (or “numen”) of H p . Lipschitz-type estimates are given for thecharacteristic function when p − is a power of and is a primitive rootmod p , and it is shown that the set of periodic points of H p is equal to theset of (rational) integer values attained by the characteristic function over Z . Additionally, although the pre-image of R under the characteristicfunction has zero Haar measure in the Z , by pre-composing the charac-teristic function with an appropriately selected self-embedding of Z , onecan perform Fourier analysis of the aforementioned composite. Using thisapproach, explicit upper bounds are computed for the absolute value of aperiodic point of H p whose parity vector contains at least ⌈ log p ⌉ − sbetween any two consecutive s. Keywords: Collatz Conjecture, p -adic analysis, -adic integers, Harmonic anal-ysis, parity vector, Pontryagin duality, px + 1 map, x + 1 map, x + 1 map. Introduction p be an odd prime, and consider the map H p : N → N (where N k def = { n ∈ Z : x ≥ k } ) defined by: H p ( x ) def = ( x if n ≡ px +12 if n ≡ (1) ∗ University of Southern California, Dornsife School of Letters & Sciences, Department ofMathematics. Department of Mathematics 3620 S. Vermont Ave., KAP 104 Los Angeles, CA90089-2532. E-mail: [email protected] Declarations of interest: none This research did not receive any specific grant from funding agencies in the public,commercial, or not-for-profit sectors. a r ≡ b denotes the congruence of a and b mod r . As is well known,of central importance to these maps—among them, the infamous Collatz or x + 1 map—is the parity vector of a given input x . This is the sequence [ x ] , [ H p ( x )] , (cid:2) H ◦ p ( x ) (cid:3) , . . . of the values mod of the iterates of x under H p .This paper’s approach is to view inputs as being subordinate to their parityvectors, and to that end, one can, in essence, parameterize the essential featuresof H p ’s dynamics as a function on the space of all possible parity vectors.To do this, we define the maps: h ( x ) def = x h ( x ) def = px + 12 Next, for each n ∈ N , let { , } n denote the set of all n -tuples whose entriesare s and s. We denote an arbitrary such tuple as j = ( j , . . . , j n ) . Givenany such tuple, we write | j | to denote the length of j , so that any tuple can bewritten as j = (cid:0) j , . . . , j | j | (cid:1) . Writing: J def = ∞ [ n =1 { , } n (2)to denote the set of all tuples of s and s of finite length, we then consider the composition sequence h j defined by: h j = h j ◦ · · · ◦ h j | j | , ∀ j ∈ J (3)Note the near-equivalence of j with parity vectors. Indeed, given any x, n ∈ N ,there exists a unique j ∈ J of length n so that H ◦ np ( x ) = h j ( x ) . Said j is theparity vector for the first n iterates of x under H p , albeit written in reverseorder.Since the h j s are affine linear maps (of the form ax + b ), so is the compositionsequence h j . As such, for any j ∈ J , we have: h j ( x ) = h ′ j (0) x + h j (0) = p ( j ) | j | x + h j (0) , ∀ j ∈ J (4)where: ( j ) def = number of s in j (5)As the number of s in j tends to infinity, we can identify j with the -adicinteger z whose -adic digits are precisely the entries of j . In doing so, the ratio p ( j ) / | j | then tends to in p -adic absolute value regardless of the value of x ,allowing us to realize the map j h j (0) as a continuous function χ p : Z → Z p .This, the author calls the characteristic function or (when treating χ p as arandom variable) the numen of H p .In Section 1 of this paper, χ p , its formulae, and formulae for periodic points ofthe H p maps are constructed. While many of these results have been implicitly2ecognized in previous work many times over—for instance, in the propertiesof the Syracuse random variable defined by Tao (Tao. 2019)— χ p provides aunifying perspective on the matter. 12 in Lemma 1 gives the (relatively)closed-form formula for χ p . Using this and the function: B ( t ) = t − ⌈ log ( t +1) ⌉ (6)the main result of Section 1 ( Theorem 2 ) is that x ∈ Z is a periodic pointof H p if and only if x = χ p ( B ( t )) for some t ∈ N . Additionally, in the thirdsubsection, we provide an explicit Lipschitz-type estimate on χ p : Z → Z p which (as discussed in the paper’s Conclusion) was the initial impetus for thepresent research.Next, in Section 2 we work with χ p as a real-valued function on Z , using theHaar probability measure on Z to do Fourier Analysis. While the set S p ⊆ Z of all -adic integers at which χ p takes values in R has—as would be expected—measure zero ( Theorem 4 ), pre-composing χ p with homeomorphisms η : Z → U (where U is some subset of S p which is ours to choose) provides a work-around.In this perspective, the choice of U corresponds to a choice of conditions on theparity vectors of periodic points of H p (in Z ). By cleverly selecting U and η , the Fourier coefficients of χ p ◦ η can then be analyzed or even explicitlycomputed, thereby providing a method for establishing control on the size ofperiodic points of H p whose parity vectors satisfy the conditions encoded by U . As an application of this “ L method”, in the second subsection of Section2, —we show that H p can have only finitely many periodic points in N whoseparity vectors always have at least ⌈ log p ⌉ − consecutive s between any two s ( Theorem 6 ). As Lagarias and others have done, we begin by reinterpreting j in terms of binaryexpansions of integers (Lagarias. 1985). In writing a tuple j = (cid:0) j , . . . , j | j | (cid:1) ∈ J ,we say a tuple i ∈ J was obtained by “adding m terminal zeroes to j ” whenever: i = j , . . . , j | j | , , . . . , | {z } m times As such, we institute the following definitions:
Definition 1 :I. We define an equivalence relation ∼ on J by the property that, for all i , j ∈ J , i ∼ j if and only if one of i or j can be obtained by adding finitely manyterminal zeroes to the other. 3 emark 1 : Observe that the value of h j (0) is independent of the number ofterminal zeroes in j : h i (0) = h j (0) , ∀ i , j ∈ J such that i ∼ j II. Let J / ∼ be the set of equivalence classes of J under the equivalencerelation ∼ defined above. Then, we define the map β : J / ∼→ N by: β ( j ) def = | j | X ℓ =1 j ℓ ℓ − (7) Remark 2 :i. β ( j ) is well-defined with respect to ∼ , since adding terminal zeroes to j does not change the value of β ( j ) .ii. Observe that β ( i ) = β ( j ) occurs if and only if i ∼ j .iii. The map β is well-known in Collatz literature, where—following Lagarias’convention—it is usually denoted by Q (Lagarias. 1985).III. The characteristic function (or numen ) of H p , denoted χ p : N → Q ,is defined by: χ p ( t ) def = h β − ( t ) (0) , ∀ t ∈ N (8)where β − ( t ) is any element j of J so that β ( j ) = t . Remark 3 : By
Remark 2 , χ p is well-defined, since the choice of a pre-imageof t under β does not affect the value of χ p ( t ) . Definition 2 :I. For all t ∈ N , let ( t ) denote the number of s in the binary/ -adicexpansion of t . We extend to Z by defining ( t ) = ∞ for all t ∈ Z \ N .II. For all t ∈ N , let λ ( t ) denote the number of digits ( s and s) in thebinary/ -adic expansion of t . Using the convention that λ (0) = 0 , it followsthat: λ ( t ) = ⌈ log ( t + 1) ⌉ , ∀ t ∈ N (9)We extend λ to Z by defining λ ( t ) = ∞ for all t ∈ Z \ N . Note also that λ ( t ) = ⌊ log t ⌋ + 1 for all t ∈ N . Remark 4 : Note that and λ satisfy the functional equations: (2 m t + k ) = ( t ) + ( k ) , ∀ t ∈ N , ∀ m ∈ N , ∀ k ∈ { , . . . , m − } (10) λ (2 m t + k ) = λ ( t ) + m, ∀ t ∈ N , ∀ m ∈ N , ∀ k ∈ { , . . . , m − } (11)In particular: (2 t ) = ( t ) (2 t + 1) = ( t ) + 1 t ∈ N , and: λ (2 t ) = λ (2 t + 1) = λ ( t ) + 1 for all t ∈ N .Next, as is well known, every positive integer has a unique binary expansionof the form P Kk =1 n k for some K ∈ N and some strictly increasing sequenceof non-negative integers ≤ n < n < · · · < n K < ∞ . We can extendthis notation to include by defining the sum to be when K = 0 . Usingthis notation to denote an arbitrary non-negative integer, χ p then satisfies theformula: Proposition 1 : χ p K X k =1 n k ! = K X k =1 p k − n k +1 (12)where the sum on the right is defined to be whenever K = 0 . Remark 5 : This formula is well-known in Collatz literature; see (Tao. 2019)for but one example. Indeed, as remarked by Tao in his blog post about his pa-per, it follows by induction that, as defined, χ p is injective (Ibid). However, thisinjectivity does not necessarily hold when we consider the value of χ p modulo p N for some N ∈ N . This line of thinking leads to the Lipschitz-type estimatesto be discussed in short shrift.Proof: We proceed by induction on K . When K = 0 , χ p (0) = h β − (0) (0) .Since β − (0) is the set of all tuples of finite length whose entries are zero, wechoose the one-element tuple as the representative, which gives: χ p X k =1 n k ! = χ p (0) = h (0) = 02 = 0 = X k =1 p k − n k +1 and so, the formula holds for K = 0 .Next, letting K = 1 , χ (2 n ) = h β − (2 n ) (0) . Note that the n + 1 tuple j = , . . . , | {z } n times , then satisfies: β ( j ) = n +1 X ℓ =1 j ℓ ℓ − = 0 · − + 0 · − + · · · + 0 · n − | {z } n terms +1 · ( n +1) − = 2 n and so: χ p (2 n ) = h β − (2 n ) (0)= h , . . . , | {z } n times , (0) = h ◦ n ( h (0)) = 12 n p · n +1 = X k =1 p k − n k +1 K = 1 .Now, proceeding to the inductive step, suppose the formula holds for anarbitrary K ≥ . So, let t be any non-negative integer with at most K non-zero binary digits. Without loss of generality, suppose t has K non-zero binarydigits, with t = P Kk =1 n k for some n k . Note that λ ( t ) = n K + 1 . Then, bythe inductive hypothesis, χ p ( t ) = P Kk =1 p k − nk +1 . Next, let n K +1 be any integergreater than n K . Then: t + 2 n K +1 = t + 0 · n K +1 + · · · + 0 · n K +( n K +1 − n K − | {z } n K +1 − n K − terms +2 n K +1 Hence: χ p ( t + 2 n K +1 ) = (cid:16) h β − ( t ) ◦ h ◦ n K +1 − n K − ◦ h (cid:17) (0)= h β − ( t ) (cid:18) · n K +1 − n K − (cid:19) = h β − ( t ) (cid:18) n K +1 − n K (cid:19)(cid:16) h β − ( t ) ( x ) = h ′ β − ( t ) (0) x + h β − ( t ) (0) (cid:17) ; = h ′ β − ( t ) (0)2 n K +1 − n K + h β − ( t ) (0) | {z } χ ( t ) χ p ( t ) = K X k =1 p k − n k +1 ! ; = h ′ β − ( t ) (0)2 n K +1 − n K + K X k =1 p k − n k +1 Since h β − ( t ) ( x ) is an affine linear map which is given by a composition sequenceof the affine linear maps h and h , its derivative at will be the product ofthe derivatives at of each of the maps in the composition sequence. Since h ′ (0) = 1 / and h ′ (0) = p/ , it follows that h ′ β − ( t ) (0) = p t ) λ ( t ) = p K nK +1 .Hence: χ p K +1 X k =1 n k ! = χ p ( t + 2 n K +1 )= h ′ β − ( t ) (0)2 n K +1 − n K + K X k =1 p k − n k +1 = p K / n K +1 n K +1 − n K + K X k =1 p k − n k +1 = K +1 X k =1 p k − n k +1 Thus, the K th case implies the ( K + 1) th case. As such, by induction, theformula holds for all K ∈ N . 6.E.D.Using some extra bits of terminology, we can give χ p a more closed-formexpression. Definition 3 : Let t ∈ Z . Then, we write { β k ( t ) } k ∈{ ,..., ( t ) } to denote theunique, strictly increasing sequence of non-negative integers so that the -adicexpansion of t is given by: t = ( t ) X k =1 β k ( t ) (13)The above sum is defined to be whenever t = 0 . Lemma 1 : χ p can be extended to a continuous function from Z to Z p ,which is given by the formula: χ p ( t ) = ( t ) X k =1 p k − β k ( t )+1 (14)Proof: When t ∈ N , this is just Proposition 1 restated using the notation ( t ) and β k ( t ) . On the other hand, given any sequence { t n } n ∈ N ⊆ N con-verging -adically to t ∈ Z \ N as n → ∞ , observe that ( t n ) → ∞ . Sincethe k th term of 14 has a p -adic magnitude of p − k , and so, χ p ( t n ) is then asequence of rational numbers converging to the limit χ p ( t ) ∈ Z p . Thus, χ p canindeed by -adically interpolated to a continuous map from Z to Z p .Q.E.D. Theorem 1 : χ p is the unique continuous function from Z to Z p satisfyingthe functional equations: χ p (2 t ) = 12 χ p ( t ) (15) χ p (2 t + 1) = pχ p ( t ) + 12 for all t ∈ Z .Proof: Let t = P Kk =1 n k ∈ N . Then 12implies: χ p K X k =1 n k ! = χ p K X k =1 n k +1 ! = K X k =1 p k − n k +2 = 12 K X k =1 p k − n k +1 = χ p K X k =1 n k ! χ p K X k =1 n k ! + 1 ! = χ p (cid:0) n +1 + 2 n +1 + · · · + 2 n K +1 (cid:1) = 12 + p n +2 + p n +2 + · · · + p K n K +2 = 12 (cid:18) p (cid:18) n +1 + p n +1 + · · · + p K − n K +1 (cid:19)(cid:19) = 12 pχ p K X k =1 n k !! Hence 15 hold true for all t ∈ N . By the -adic continuity of χ p , the functionalequations then hold true for all t ∈ Z .Finally, let f : Z → Z p be any continuous function satisfying: f (2 t ) = 12 f ( t ) f (2 t + 1) = pf ( t ) + 12 for all t ∈ Z . Setting t = 0 gives f (0) = f (0) , which forces f (0) = 0 .Consequently, f (1) = f (2 · pf (0)+12 = , and so f (2 n ) = n f (cid:0) (cid:1) = n .Then, for any n > n : f (2 n + 2 n ) = f (cid:0) n (cid:0) n − n + 1 (cid:1)(cid:1) = 12 n f (cid:0) n − n + 1 (cid:1) = 12 n pf (cid:0) n − n − (cid:1) + 12= 12 n +1 + p n +1 f (cid:0) n − n − (cid:1) = 12 n +1 + p n +1 n − n − f (1)= 12 n +1 + p n +1 Continuing in this manner, by induction, it then follows that: f K X k =1 n K ! = K X k =1 p k − n k +1 = χ p K X k =1 n K ! and thus, that f ( t ) = χ p ( t ) holds for all t ∈ N . The -adic continuity of f andthe density of N in Z then forces f ( t ) = χ p ( t ) for all t ∈ Z .Q.E.D. 8 .2 Rational integer values of χ p are periodic points of H p As the title indicates, in this subsection we prove that x ∈ Z is a periodic pointof H p if and only if x = χ p ( z ) for some z ∈ Z ( Theorem 2) . To begin, using β − t ( t ) in lieu of j , we can write 4 as: h β − ( t ) ( x ) = h ′ β − ( t ) (0) x + h β − ( t ) (0) = p ( t ) x λ ( t ) + χ p ( t ) , ∀ x ∈ Z , ∀ t ∈ N (16)Our next proposition is nearly tautological—but its consequences are anythingbut. Proposition 2 : Let x ∈ Z be a periodic point of H p (i.e., H ◦ np ( x ) = x forsome n ∈ N ). There exists a t ∈ N so that: χ p ( t ) = (cid:18) − p ( t ) λ ( t ) (cid:19) x (17)Additionally, for any j ∈ J for which h j ( x ) = x , it must be that: χ p ( β ( j )) = (cid:18) − p ( β ( j )) λ ( β ( j )) (cid:19) x (18)Proof: If x ∈ N is a periodic point of H p (say, of period n ), there is then aunique j ∈ J of length n so that h j ( x ) = H ◦ np ( x ) = x . As such, for t = β ( j ) : x = h j ( x ) h β − ( t ) ( x ) = p ( t ) x λ ( t ) + χ p ( t ) m (cid:18) − p ( t ) λ ( t ) (cid:19) x = χ p ( t ) By the same reasoning, for any j ∈ J for which h j ( x ) = x , it must be that: χ p ( β ( j )) = (cid:18) − p ( β ( j )) λ ( β ( j )) (cid:19) x holds true, as desired.Q.E.D. Lemma 2 : Every periodic point of H p in N is of the form: χ p (cid:18) t − λ ( t ) (cid:19) (19)for some t ∈ N . In particular, if x is the periodic point of H p , then: x = χ p (cid:18) β ( j )1 − λ ( β ( j )) (cid:19) (20)9here j is any element of J such that h j ( x ) = x .Proof: For a periodic point x , let j ∈ J be the shortest tuple for which h j ( x ) = x , and let j n denote the concatenation of n copies of j . No matter howmany copies of j we concatenate, the associated composition sequence leaves x fixed: h j n ( x ) = h ◦ n j ( x ) = h ◦ n − j ( h j ( x )) = h ◦ n − j ( x ) = · · · = x As such, 18 yields: χ p ( β ( j n )) = (cid:18) − p ( β ( j n )) λ ( β ( j n )) (cid:19) x Now, observe that: β ( j n ) = n − X m =0 | j | m (cid:16) j + j · · · + j | j | | j |− (cid:17)| {z } β ( j ) = 1 − | j | n − | j | β ( j ) Letting t denote β ( j ) , define: t n def = β ( j n ) , ∀ n ∈ N (21)Since | j | = λ ( t ) , we can then write: t n = 1 − nλ ( t ) − λ ( t ) t (22)Additionally, since j n is the concatenation of n copies of j , we have that: ( β ( j n )) = ( j n ) = ( j ) n = ( t ) nλ ( β ( j n )) = | j n | − n | j | − nλ ( t ) − As such: χ p ( β ( j n )) = (cid:18) − p ( β ( j n )) λ ( β ( j n )) (cid:19) x m χ p (cid:18) − nλ ( t ) − λ ( t ) t (cid:19) = − (cid:18) p ( t ) λ ( t ) (cid:19) n ! x (23)Finally, since: lim n →∞ − nλ ( t ) − λ ( t ) t Z = t − λ ( t ) , ∀ t ∈ N (where Z = denotes equality in Z ) and since χ p : Z → Z p is continuous, thisgives: χ p (cid:18) t − λ ( t ) (cid:19) = lim n →∞ χ p (cid:18) − nλ ( t ) − λ ( t ) t (cid:19) = lim n →∞ − (cid:18) p ( t ) λ ( t ) (cid:19) n ! x Z p = x Z p = denotes equality in Z p ). This proves every periodic point of H p in Z is of the form 19 for some t ∈ N ; in particular, t = β ( j ) , where j is any elementof J satisfying h j ( x ) = x , which gives us 20.Q.E.D. Definition 4 : Let B : N → Z be defined by: B ( t ) def = t − λ ( t ) = t − ⌈ log ( t +1) ⌉ , ∀ t ∈ N (24) Proposition 3 : B extends to a continuous function from Z to Z . Inparticular:I. The restriction of B to Z \ N is the identity map: B ( t ) = t for all t ∈ Z \ N . (Hence, with respect to the Haar probability measure of Z , B isequal to the identity map almost everywhere on Z ).II. B (2 n −
1) = − for all n ∈ N .Proof: As a sequence { t n } n ∈ N of positive integers converges -adically toa limit t ∈ Z , one of two things can happen. If λ ( t n ) remains bounded as n → ∞ , then t must be an element of N , and so, lim n →∞ B ( t n ) = B ( t ) . Theother possibility is that λ ( t n ) → ∞ as n → ∞ , in which case, λ ( t n ) tends to in Z , and hence: lim n →∞ B ( t n ) = lim n →∞ t n − λ ( t n ) Z = lim n →∞ t n − t which shows that B ( t ) = t for all t ∈ Z \ N . Thus, we can continuously (and,therefore, uniquely) extend B to a function on Z , and this extension equals theidentity map for all t ∈ Z \ N , a set of full measure in Z .Finally, for (II), note that n − n − − = P n − k =0 k , and so, λ (2 n −
1) = n ,and so: B (2 n −
1) = 2 n − − λ (2 n − = 2 n − − n = 2 n − − n = − as claimed.Q.E.D.Using B ( t ) , we can re-interpret 20 as a functional equation relating χ p ( B ( t )) and χ p ( t ) . Lemma 3 : χ p ( B ( t )) = χ p ( t )1 − p t ) λ ( t ) , ∀ t ∈ N (25)Proof: Since B (0) = χ p (0) = χ p ( B (0)) = 0 , observe that the identity holds11or t = 0 . So, letting t ∈ N , we have that: χ p ( B ( t )) Z p = χ p ∞ X m =0 ( t ) X k =1 mλ ( t )+ β k ( t )+1 Z p = ( t ) X k =1 p k − β k ( t )+1 + ( t ) X k =1 p k + ( t ) − λ ( t )+ β k ( t )+1 + ( t ) X k =1 p k +2 ( t ) − λ ( t )+ β k ( t )+1 + · · · Z p = ∞ X m =0 ( t ) X k =1 p k + m ( t ) − mλ ( t )+ β k ( t )+1 Z p = ∞ X m =0 (cid:18) p ( t ) λ ( t ) (cid:19) m | {z } converges if ( t ) ≥ × ( t ) X k =1 p k − β k ( t )+1 | {z } χ p ( t ) ( if t = 0) ; Z p = χ p ( t )1 − p t ) λ ( t ) Thus, the identity holds for all t ∈ N .Q.E.D. Remark 6 : Lemma 3 shows that B ( N ) is a subset of Q ∩ Z mapped into Q by χ p . It would be of interest to know if the “converse” holds true; that is, is B ( N ) the pre-image of Q ∩ Z p under χ p ? Theorem 2 : An integer x is a periodic point of H p if and only if: x = χ p ( B ( t )) = χ p ( t )1 − p t ) λ ( t ) = ( t ) X k =1 λ ( t ) − β k ( t ) − p k − λ ( t ) − p ( t ) (26)for some t ∈ N .Proof: Note that H p (0) = 0 for all odd p ≥ , and that χ p ( B (0)) = χ p (0) =0 . So, it suffices to prove the theorem for x ∈ Z \ { } .I. Let x be a non-zero periodic point of H p . Then, by Lemma 2 , there is a t ∈ N so that χ p ( t ) = (cid:16) − p t ) λ ( t ) (cid:17) x . Hence: x = χ p ( t )1 − p t ) λ ( t ) = χ p ( B ( t )) as desired.II. Let t ∈ N , and suppose that χ p ( B ( t )) ∈ Z . Letting x denote χ p ( B ( t )) —and noting that we then have x = χ p ( t )1 − p t )2 λ ( t ) —by 16, we can write: h β − ( t ) ( x ) = p ( t ) x λ ( t ) + χ p ( t ) = p ( t ) x λ ( t ) + x (cid:18) − p ( t ) λ ( t ) (cid:19) = x j ∈ β − ( t ) , it then follows that h j ( x ) = x . Claim : Let t ∈ N , and suppose there is a j ∈ β − ( t ) so that h j ( x ) = x .Then, x is a periodic point of H p .Proof of Claim : Let t and j be as given, By way of contradiction, supposethat x was not a periodic point of H p . Since the composition sequence h j fixes x , our counterfactual assumption on x forces there to be an n ∈ { , . . . , | j |} sothat: H ◦ n ( x ) = h j ,...,j n ( x ) but for which: H ◦ n +1 ( x ) = h j ,...,j n ,j n +1 ( x ) (Note that t ∈ N then guarantees that | j | ≥ .) So, at some point in applyingto x the maps in the composition sequence defining h j , we must have appliedthe “wrong” map; that is, h j ,...,j n ( x ) was odd (resp. even), but j n +1 = 0 (resp. j n +1 = 1 ), causing the composition sequence h j to diverge from the “natural”path corresponding to the motions of x under iterations of H .Noting that for any m ∈ N : h (2 m + 1) = 2 m + 12 ∈ Z (cid:20) (cid:21) \ Z h (2 m ) = 2 pm + 12 ∈ Z (cid:20) (cid:21) \ Z we see that applying the “wrong” map at any step in the composition sequencesends us from Z to an element of Z (cid:2) (cid:3) (the dyadic rational numbers) of theform a , where a is an odd integer. So, letting m ∈ N and a ∈ Z + 1 bearbitrary, we have that: (cid:12)(cid:12)(cid:12) h (cid:16) a m (cid:17)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) a m +1 (cid:12)(cid:12)(cid:12) = 2 (cid:12)(cid:12)(cid:12) a m (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) h (cid:16) a m (cid:17)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) p a m + 12 (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) pa + 2 m +1 m +1 (cid:12)(cid:12)(cid:12)(cid:12) St ∆ = | pa | | m +1 | = 2 (cid:12)(cid:12)(cid:12) a m (cid:12)(cid:12)(cid:12) where St ∆ denotes an application of the strong triangle inequality, using thefacts that a and p are odd and m ≥ .Since a dyadic rational number q is a non-integer if and only if q = a m forsome a ∈ Z + 1 and m ∈ N , this shows that, for any such q , | h ( q ) | = | h ( q ) | = 2 | q | . So, | h i ( q ) | = 2 | i | | q | for all such q and any i ∈ J . Hence: h i ( q ) / ∈ Z , ∀ i ∈ J , ∀ q ∈ Z (cid:20) (cid:21) \ Z (27)In other words, once a composition sequence of h and h outputs a non-integerdyadic rational, there is no way to apply composition sequences of h and h toarrive back at an integer output . 13n summary, our assumption that x ∈ Z was not a periodic point of H p forces h j ,...,j n ,j n +1 ( x ) = H ◦ n +1 ( x ) , and hence, forces h j ,...,j n ,j n +1 ( x ) to be anon-integer dyadic rational number. But then, Z contains x = h j ( x ) = h j n +2 ,...,j | j | (cid:0) h j ,...,j n ,j n +1 ( x ) (cid:1) / ∈ Z which is not an integer. This is impossible. Consequently, for our j , h j ( x ) = x forces x to be a periodic point of H p . This proves the Claim . X Since t was an arbitrary positive-integer for which χ p ( B ( t )) ∈ Z , this showsthat χ p ( B ( t )) is a periodic point of H p in Z for any t ∈ N for which χ p ( B ( t )) ∈ Z , which then proves Theorem 2 .Q.E.D.The following table displays the first few values of the various functionsconsidered thus far. As per
Theorem 2 , note that the integer values attainedby χ ( B ( t )) and χ ( B ( t )) are all periodic points of the shortened collatz map H —this includes fixed points at negative integers, as well. t ( t ) λ ( t ) χ p ( t ) χ p ( B ( t )) χ ( B ( t )) χ ( B ( t ))0 0 0 0 0 0 01 1 1
12 12 − p − −
14 14 − p −
13 2 2 p p − p − −
18 18 − p
15 13 p p − p − − p p − p − − p + p p + p − p − −
116 116 − p
113 111 p
16 8+ p − p −
10 2 4 p
16 4+ p − p −
111 3 4 p + p
16 8+4 p + p − p − −
12 2 4 p
16 2+ p − p −
13 3 4 p + p
16 8+2 p + p − p − −
14 3 4 p + p
16 4+2 p + p − p − By looking at χ p ( B ( t )) , we can see definite patterns emerge, such as thefact that: χ p ( B (2 n − − p , ∀ n ∈ N and, more significant, that χ ( B ( t )) is more often positive than negative, butthat the opposite holds for p = 5 (and, heuristically, for all odd integers p ≥ ).Most significant, in the author’s opinion, however, is the example of: χ p ( B (10)) = 4 + p − p = 14 − p
14n showing that every integer periodic point x of H p is of the form: x = χ p ( B ( t )) = χ p ( t )1 − p t ) λ ( t ) = ( t ) X k =1 λ ( t ) − β k ( t ) − p k − λ ( t ) − p ( t ) for some t ∈ N , Theorem 2 demonstrates that (cid:12)(cid:12) λ ( t ) − p ( t ) (cid:12)(cid:12) = 1 is a suf-ficient condition for χ p ( B ( t )) to be a periodic point of H p . However, as the t = 10 case shows, this is not a necessary condition, because there could bevalues of t where the numerator and denominator of χ p ( B ( t )) share a commondivisor that, when cancelled out, reduce χ p ( B ( t )) to an integer, despite the po-tentially large absolute value of λ ( t ) − p ( t ) for that value of t . In fact, thanksto P. Mihăilescu’s resolution of Catalan’s Conjecture , it would seem thatestimates on the archimedean size of λ ( t ) − p ( t ) do not help us understand χ p ( B ( t )) a whit better! Mihăilescu’s Theorem (Cited in (Cohen. 2008)): The only choice of x, y ∈ N and m, n ∈ N for which: x m − y n = 1 are x = 3 , m = 2 , y = 2 , n = 3 (that is, − = 1 ).With Mihăilescu’s Theorem , it is easy to see that, for any odd integer p ≥ , (cid:12)(cid:12) λ ( t ) − p ( t ) (cid:12)(cid:12) will never be equal to for any t ≥ . Consequently, forall prime p ≥ , any rational integer value of χ p ( B ( t )) (and hence, any peri-odic point of H p ) of the form χ p ( B ( t )) , where t ≥ , must occur as a result of P ( t ) k =1 λ ( t ) − β k ( t ) − p k − being a multiple of of λ ( t ) − p ( t ) ∈ Z \ {− , } . Thiswould suggest that classical techniques of transcendence theory (ex: Baker’sMethod) or additive number theory (ex: the Circle Method) need not botherto show up if and when the call is ever sent out to gather the miraculous math-ematical army needed to conquer the Collatz Conjecture. Instead, it seemsthat the key—at least for proving the (non-)existence of periodic points—is tounderstand the roots of the polynomials: ( t ) X k =1 λ ( t ) − β k ( t ) − x k − , ( t ) X k =1 p k − x λ ( t ) − β k ( t ) − , λ ( t ) − x ( t ) , x λ ( t ) − y ( t ) , etc.over the finite fields ( Z /q Z , for prime q ). To that end, we should likely move tostudying the subgroup of bijections of Z /q Z (i.e., the subgroup of the symmetricgroup S q ) generated by the maps h ( x ) = x and h ( x ) = px +12 . χ p Finally, to address the original impetus for writing this paper, we prove anexplicit Lipschitz estimate for χ p . A good rule of thumb for the H p maps isthat their “ -ness” is generally regular and well-behaved; it is in their “ p -ness”15hat their peculiar irregularities show themselves. Case in point, as mentionedpreviously, χ p is not, in general, injective modulo p N for arbitrary N . Not onlythat, but, the values of χ p are not equally distributed across the residue classesmod any given p N . Indeed, as 14 shows, χ p ( t ) is a unit of is Z p (and hence, itsvalue mod p N is always co-prime to p ) for all non-zero t ∈ Z .As such, it would be desirable to establish control over the distribution of χ p ( t ) mod p m as t varies over { , . . . , n − } , for any given integers n ≥ and m ∈ { , . . . , n } . Our next result consists of exactly such a relation, stated hereas a system of congruences between the locations of the binary digits of s and t . Theorem 3 : Let p be an odd prime so that:I. is a primitive root mod p .II. There is an integer j ∈ N so that p = 2 j + 1 .Then, given any integers m, s, t ∈ N , the congruence χ p ( s ) p m ≡ χ p ( t ) issatisfied whenever the congruences : β k ( s ) p − ≡ β k ( t ) (28) β k ( s ) p m − k ≡ β k ( t ) hold for all k ∈ { , . . . , m } . More generally: | χ p ( s ) − χ p ( t ) | p ≤ max k ≥ p − (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1+ ν p ( β k ( s ) − β k ( t ))) − k +1 , ∀ s , t ∈ Z (29)Here ν p ( m ) is the p -adic valuation of m , and (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) is an Iversonbracket , which evaluates to whenever the enclosed statement is true andevaluates to whenever the enclosed statement is false.To give the proof of Theorem 3 , we need a formula for the p -adic magnitudeof m − . For that, we need to know the multiplicative order of modulo p k ,for k ∈ N . Proposition 4 : Let p be an odd prime number such that:I. is a primitive root mod p .II. p = 2 j + 1 for some j ∈ N .Then: ord p k (2) = ( p − p k − = 2 j p k − , ∀ k ∈ N (30)Proof: Let k ∈ N , and let ϕ denote Euler’s Totient Function . Then, theorder of the multiplicative group (cid:0) Z /p k Z (cid:1) × is: Note that β k ( s ) p m − k ≡ β k ( t ) always holds true for k = m , because the integers β m ( s ) and β m ( t ) , being integers, are necessarily congruent to one another mod p m − m = 1 . (cid:0) p k (cid:1) = p k − p k − = ( p − p k − = 2 j p k − Since is a primitive root mod p :ord p (2) = (cid:12)(cid:12)(cid:12) ( Z /p Z ) × (cid:12)(cid:12)(cid:12) = ϕ ( p ) = p − j Now, since the projection mod p (the map written by [ · ] p ) defines a grouphomomorphism from (cid:0) Z /p k Z (cid:1) × to ( Z /p Z ) , ord p (cid:16) [ x ] p (cid:17) (the order of the imageof x ∈ (cid:0) Z /p k Z (cid:1) × under the mod p homomorphism) must divide ord p k ( x ) . Assuch, ord p (2) | ord p k (2) .So, arguing by induction, suppose that ord p k (2) = 2 j p k − . Then, using themod p k projection from (cid:0) Z /p k +1 Z (cid:1) × onto (cid:0) Z /p k Z (cid:1) × , it must be that ord p k (2) | ord p k +1 (2) . Since:ord p k +1 (2) | (cid:12)(cid:12)(cid:12)(cid:0) Z /p k +1 Z (cid:1) × (cid:12)(cid:12)(cid:12) = ϕ (cid:0) p k +1 (cid:1) = 2 j p k In particular, note that the
Chinese Remainder Theorem then implies that (cid:0) Z /p k +1 Z (cid:1) × is then isomorphic to the direct product of a group of order p − j and a group of order p k . Since ord p k (2) is assumed to be j p k − , and since is a primitive root mod p , this then forces ord p k +1 (2) = 2 j p k . Hence, 30 followsby induction.Q.E.D. Proposition 5 : Let p be as in Proposition 4 . Then: | m − | p = p − (cid:20) m p − ≡ (cid:21) ( ν p ( m )+1) , ∀ m ∈ N (31)Proof: We need to determine the largest power of p = 2 j + 1 that divides m − . To do this, let k ∈ N so that p k | (2 m − . By elementary grouptheory, this forces m to be a multiple of ord p k (2) . By Proposition 4 , we knowthat: ord p k (2) = ( p − p k − = 2 j p k − So, in order that p k | (2 m − occur, we need for m j p k − ≡ . As such, | m − | p = p − κ , where κ is the largest positive integer for which m j p k − ≡ . Ifno such κ exists, then | m − | p = 1 .To that end, note that, since j ∈ N , the congruence m j p κ − ≡ implies m j ≡ . As such, m being a multiple of j is a necessary condition for theexistence of a κ so that m j p κ − ≡ . Additionally, if m is a multiple of j , then k = 1 satisfies m j p k − ≡ . As such, if m is a multiple of j , then κ necessarily17xists and is ≥ . Consequently, for m ≥ : | m − | p = ( if j ∤ mp − max { k ∈ N :2 j p k − | m } if j | m = ( if j ∤ mp − max { k ∈ N : p k | pm j } if j | m = ( if j ∤ mp − ν p ( pm j ) if j | m = p − [ j | m ] ν p ( pm j ) (cid:0) j = p − ν p (cid:0) j (cid:1) = 0 (cid:1) ; = p − (cid:20) m p − ≡ (cid:21) ( ν p ( m )+1) Q.E.D.
Proof of Theorem 3 : Using 14, it follows that, for all s , t ∈ Z : | χ p ( s ) − χ p ( t ) | p = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 p k − (cid:18) β k ( t )+1 − β k ( s )+1 (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p ( Strong ∆ -Ineq. ) ; ≤ max k ≥ (cid:12)(cid:12)(cid:12)(cid:12) p k − (cid:18) β k ( t )+1 − β k ( s )+1 (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) p = max k ≥ (cid:12)(cid:12)(cid:12)(cid:12) p k − (cid:18) β k ( s ) − β k ( t ) β k ( t )+ β k ( s )+1 (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) p ( p is odd ) ; = max k ≥ (cid:12)(cid:12) β k ( s ) − β k ( t ) (cid:12)(cid:12) p p − k = max k ≥ z }| {(cid:12)(cid:12)(cid:12) β k ( t ) (cid:12)(cid:12)(cid:12) p (cid:12)(cid:12) β k ( s ) − β k ( t ) − (cid:12)(cid:12) p p − k ( β k ( s ) − β k ( t ) ∈ N , ∀ k ) ; = max k ≥ p − (cid:20) β k ( s ) − β k ( t ) p − ≡ (cid:21) ( ν p ( β k ( s ) − β k ( t ))+1) p k − = max k ≥ p − (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1+ ν p ( β k ( s ) − β k ( t ))) − k +1 This gives the sought-after Lipschitz-type estimate 29.To prove the congruences condition, fix m, s, t ∈ N . Then, in order for χ ( s ) and χ ( t ) to be congruent mod p m as elements of Z p , it suffices that: | χ p ( s ) − χ p ( t ) | p ≤ p − m , and thus, that: p − (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1+ ν p ( β k ( s ) − β k ( t ))) − k +1 ≤ p − m , ∀ k ∈ N (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1 + ν p ( β k ( s ) − β k ( t ))) + k − ≥ m, ∀ k ∈ N Now, if k ≥ m + 1 , then: (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1 + ν p ( β k ( s ) − β k ( t ))) + k − ≥ m + 1) − m and so, the inequality automatically holds for k ≥ m +1 . Thus, to have χ p ( s ) p m ≡ χ p ( t ) , it suffices that: (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1 + ν p ( β k ( s ) − β k ( t ))) + k − ≥ m holds for k ∈ { , . . . , m } . Adding − k + 1 to both sides then gives: (cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1 + ν p ( β k ( s ) − β k ( t ))) ≥ m − k + 1 Finally, suppose there is a k ∈ { , . . . , m } for which β k ( s ) is not congruent to β k ( t ) mod p − . Then: z }| {(cid:20) β k ( s ) p − ≡ β k ( t ) (cid:21) (1 + ν p ( β k ( s ) − β k ( t ))) ≥ m − k + 1 m k ≥ m + 1 which is impossible. As such, it must be that β k ( s ) p − ≡ β k ( t ) holds forall k ∈ { , . . . , m } whenever χ p ( t ) p m ≡ χ p ( s ) . Thus, we have that if ν p ( β k ( s ) − β k ( t )) ≥ m − k + 1 and β k ( s ) p − ≡ β k ( t ) hold true for all k ∈{ , . . . , m } , it must be that χ ( t ) p m ≡ χ ( s ) . Since: ν p ( β k ( s ) − β k ( t )) ≥ m − k + 1 m β k ( s ) − β k ( t ) p m − k ∈ Z m β k ( s ) p m − k ≡ β k ( t ) we then have that the validity of the congruences 28 for all k ∈ { , . . . , m } isthen sufficient to guarantee χ p ( t ) p m ≡ χ p ( s ) .Q.E.D. 19 Fourier Analysis of the Characteristic Function L Method
While the outermost pair of equalities in 26 are already known, (see (Tao.2019)), the parameterization of the quantity: χ p ( B ( t )) = χ p ( t )1 − p t ) λ ( t ) = 12 p ( t ) X k =1 λ ( t ) − β k ( t ) p k λ ( t ) − p ( t ) as a function of t ∈ N —and, more generally, the treatment of χ p as a functionon Z —appears to be novel.It seems natural enough to want to study χ p over Z , seeing how—as wehave seen—the values of t for which χ p ( t ) is a non-zero periodic point of H p (namely, those values in some subset of B ( N )) are all -adic integers which are not elements of N . Moreover, over Z , the distinction between χ p and χ p ◦ B disappears: χ p ( B ( z )) = χ p ( z ) , ˜ ∀ z ∈ Z where ˜ ∀ means “for almost every”—since, as we saw in Proposition 3 , thefunction B is almost-everywhere equivalent on Z to the identity map of Z .In analyzing χ p as a function over Z , we speak of elements z of Z as being routes . Viewing -adic integers as infinite sequences of s and s (correspondingto their -adic digits), the s and s represent a composition sequence of the evenand odd branches of H p ( h and h ). The value of χ p over a particular route z tells us the image of the corresponding composition sequence at , which—aswe have seen—is, in general, a p -adic integer. With this in mind, we institutethe following piece of terminology: Definition : Let U ⊆ Z . We say an x ∈ Z is routed through U whenever χ − p ( { x } ) ⊆ U (that is, whenever U contains the pre-image of x under χ p ). Remark 7 : Thus, “every positive integer periodic point of H p routed through U ” means “every periodic point x ∈ N of H p so that χ − p ( { x } ) ⊆ U ”.Of course, the foremost obstacle in attempting to analyze χ p as a functionover Z is that, as a function taking values in Z p , it is not clear exactly how wecan go about utilizing the tools of Fourier analysis or even p -adic analysis in thissituation. The latter (Mahler’s Theorem, p -adic measures, the Amice transform,etc.)—to the author’s knowledge—only works when the functions being studeidgo from one p -adic space to another—though χ p is a function from Z to Z p ,where p = 2 . On the other hand, the former requires functions taking values in C or subsets thereof, and—as a function taking values in Z p —most of χ p ’s outputs From this point onward, the only measure we shall speak of is the Haar measure on Z .Mostly keeping with the notation of the author’s paper [cite myself] we write {·} to denotethe -adic fractional part, d z to denote the Haar probability measure on Z , ˆ Z to denote thePontryagin dual of Z —the -Prüfer group, which we view additively, identifying it with thegroup Z (cid:2) (cid:3) / Z of dyadic rational numbers under addition modulo . C . That being said, it remains to beseen if something useful can be done with χ p as a function from Z to Z p . Inthe present paper, we shall focus on the application of Fourier-analytic tools tothe case where χ p is restricted to be real-valued. Definition 4 : Let S p denote the set of all z ∈ Z such that χ p ( z ) is a realnumber, in addition to being a p -adic integer. We also write: S + p def = { z ∈ S p : χ p ( z ) ≥ } (32) S − p def = { z ∈ S p : χ p ( z ) < } (33)Finally, in an admitted abuse of notation, we write S p ( z ) (resp. S + p ( z ) , S − p ( z ) )to denote the indicator function of S p (resp. S + p , S − p ). Proposition 6 : The following identities hold true for all z ∈ Z :I. S p (2 z ) = S p ( z ) (34) S p (2 z + 1) = S p ( z ) II. S + p (2 z ) = S + p ( z ) S + p ( z ) S + p (2 z + 1) = S + p ( z ) (35)Proof:I. By the functional equations 15, observe that if z ∈ S p , then χ p ( z ) =2 χ p (2 z ) ∈ R , and so, z ∈ S p . Conversely, if z ∈ S p , then χ p (2 z ) = χ p ( z ) ∈ R ,and so, z ∈ S p . Thus, S p (2 z ) = S p ( z ) (i.e., z ∈ S p if and only if z ∈ S p ).Similarly, if z + 1 ∈ S p , then χ p ( z ) = χ p (2 z +1) − p ∈ R , and so z ∈ S p .Conversely, if z ∈ S p then χ p (2 z + 1) = pχ p ( z )+12 ∈ R . Thus, S p (2 z + 1) = S p ( z ) (i.e., z ∈ S p if and only if z + 1 ∈ S p ).II. Let z ∈ S + p . Then, χ p (2 z ) = χ p ( z ) ≥ , and so, z ∈ S + p . On the otherhand, suppose z ∈ S + p . Then: ≤ χ p (2 z ) = 12 χ p ( z ) ≤ χ p ( z ) and so, z ∈ S + p . Thus, S + p (2 z ) = S + p ( z ) .Similarly, if z ∈ S + p . Then: χ p (2 z + 1) = pχ p ( z ) + 12 ≥ > and so, z + 1 ∈ S + p . Thus, S + p + 1 ⊆ S + p . In fact: S + p + 1 ⊆ S + p ∩ (cid:26) z : χ p ( z ) ≥ (cid:27) S + p + 1 ⊆ Z + 1 , this can be written as: S + p + 1 ⊆ (cid:26) z ∈ S + p : χ p ( z ) ≥ (cid:27) ∩ (2 Z + 1) On the other hand, suppose z + 1 ∈ S + p . Then, ≤ χ p (2 z + 1) = pχ p ( z )+12 ,and so, χ p ( z ) = χ p (2 z +1) − p . Consequently, z ∈ S + p (and hence, z +1 ∈ S + p +1 )if and only if χ p (2 z +1) − p ≥ , which occurs if and only if χ p (2 z + 1) ≥ . So: (cid:26) z ∈ S + p : χ p ( z ) ≥ (cid:27) ∩ (2 Z + 1) ⊆ S + p +1 ⊆ (cid:26) z ∈ S + p : χ p ( z ) ≥ (cid:27) ∩ (2 Z + 1) and so: (cid:18)(cid:26) z ∈ S + p : χ p ( z ) ≥ (cid:27)(cid:19) ∩ (2 Z + 1) = 2 S + p + 1 In terms of indicator functions, we have proven: h z ≡ i (cid:20) χ p ( z ) ≥ (cid:21) S + p ( z ) = S + p (cid:18) z − (cid:19) , ∀ z ∈ Z Replacing z with z + 1 then gives: S + p ( z ) = h z + 1 ≡ i (cid:20) χ p (2 z + 1) ≥ (cid:21) S + p (2 z + 1)= (cid:20) χ p (2 z + 1) ≥ (cid:21) S + p (2 z + 1)= (cid:20) pχ p ( z ) + 12 ≥ (cid:21) S + p (2 z + 1)= [ pχ p ( z ) ≥ S + p (2 z + 1)= [ χ p ( z ) ≥ S + p (2 z + 1)= S + p ( z ) S + p (2 z + 1) as desired.Q.E.D.As would be expected, S p has measure zero in Z , as we now prove. Theorem 4 : S p has zero measure in Z .Proof: Note that the map θ : Z → Z defined by: θ ( z ) = h z ≡ i z h z ≡ i z − is the left-shift operator. Viewing -adic integers as infinite strings of s and s, θ takes the inputted string z and deletes its left-most entry (the s digit of z ).22oreover, it is well-known that the left-shift θ is an ergodic map on Z . Notingthat the functional equations 34: S p (2 z ) = S p ( z ) S p (2 z + 1) = S p ( z ) imply: S p ( z ) = h z ≡ i S p (cid:16) z (cid:17) + h z ≡ i S p (cid:18) z − (cid:19) = S p ( θ ( z )) , ∀ z ∈ Z we have that S p is invariant under pre-composition by θ . Since θ is erogdicon Z , this forces S p ( z ) to be equal to a constant c almost everywhere on Z .Since S p ( z ) is an indicator function, this forces S p ( z ) to be either equal to almost everywhere or almost everywhere. Observing that χ p ( z ) / ∈ R for any z ∈ Z \ Q (because such “irrational” z clearly force χ p ( z ) to be an “irrational” p -adic integer, which is never a rational integer), and since Z \ Q has full measurein Z , this then precludes S p ( z ) from equalling almost everywhere on Z .Consequently, S p ( z ) is zero almost everywhere on Z , and hence, the set S p then has measure zero in Z .Q.E.D.One of the thorniest properties of the dynamics of the H p maps over Z isthat the sets of relevance ( Q , N , etc.) to conjectures such as Collatz have zeromeasure in Z . Fortunately, when it comes to χ p and S p , there is a work-around.The general idea is as follows:(1) Choose a subset U of S p which is parameterized by way of some suffi-ciently nice function η : Z → U .(2) Use the properties of U & η and/or the properties of χ p | U (the restrictionof χ p to U )—or, equivalently, the properties of χ p ◦ η —to estimate or otherwisecontrol the values of χ p on U . In doing so, one then establishes control over theintegers routed through U .Using Fourier analysis over Z , one sure-fire way of accomplishing this agendais by showing that the Fourier transform of χ p ◦ η (that is, the “Fourier coeffi-cients” of χ p ◦ η ) are in L (cid:16) ˆ Z (cid:17) . This is the “ L -method”, detailed below. Theorem 5 (“The L Method”) : Let U ⊆ S p , and let η : Z → U be ahomeomorphism onto U . Then, write: χ p ; η ( z ) def = χ p ( η ( z )) , ∀ z ∈ Z (36)and write: ˆ χ p ; η ( t ) def = Z Z χ p ; η ( z ) e − πi { t z } d z (37)for all t ∈ ˆ Z (if any) for which the integral exists. Remark 8 : Here, {·} is the -adic fractional part, and the integral is withrespect to the -adic Haar probability measure.23ith these hypotheses, if ˆ χ p ; η ∈ L (cid:16) ˆ Z (cid:17) , then H p has at most finitely manyperiodic points which are routed through U . In particular, every such periodicpoint must be less than or equal to k ˆ χ p ; η k L ( ˆ Z ) .Proof: Let η be as given—suppose η is a homeomorphism from Z onto somesubset U of S p , and suppose that ˆ χ p ; η ∈ L (cid:16) ˆ Z (cid:17) . Then, the Fourier series: X t ∈ ˆ Z ˆ χ p ; η ( t ) e πi { t z } converges absolutely to χ p ; η ( z ) on Z , which shows that χ p ; η is a bounded,continuous of z on Z . In particular: sup z ∈ Z χ p ; η ( z ) = sup z ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X t ∈ ˆ Z ˆ χ p ; η ( t ) e πi { t z } (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X t ∈ ˆ Z | ˆ χ p ; η ( t ) | = k ˆ χ p ; η k L ( ˆ Z ) Since, sup z ∈ Z χ p ; η ( z ) = sup z ∈ U χ p ( z ) , we then have that sup z ∈ U χ p ( z ) ≤ k ˆ χ p ; η k L ( ˆ Z ) .Hence, any positive integer value attained by χ p ( y ) on U must be less than orequal to k ˆ χ p ; η k L ( ˆ Z ) . Since Theorem 2 guarantees that every integer valueattained by χ p on Z is a periodic point of H p , this shows that any periodicpoint of H p routed through U must be ≤ k ˆ χ p ; η k L ( ˆ Z ) . Consequently, H p canhave only finitely many such periodic points.Q.E.D. L Method
While it might be overly optimistic to hope that an approach as simple asthe one detailed in
Theorem 5 could be of any use, the author has found atleast one tractable application of this method, to be detailed in this subsection.With respect to future work, what remains to be seen is the extent to whichthe method and its conclusions can be refined (particularly with regards to thechoice of η and U ).The case we shall deal with relies on a simple (though relatively weak, andcertainly non-novel) criterion for guaranteeing that χ p ( z ) is a positive real num-ber. To begin with, as we saw in the proof of Lemma 2 , we had the equality23: χ p (cid:18) − nλ ( t ) − λ ( t ) t (cid:19) = − (cid:18) p ( t ) λ ( t ) (cid:19) n ! x where n ∈ N , where x was some periodic point of H p , and where t was somenon-negative integer. Setting n = 1 gives us: χ p ( t ) = (cid:18) − p ( t ) λ ( t ) (cid:19) x Theorem 2 , it must be that the above x is equal to χ p ( B ( t )) .Thus, 23 becomes: χ p (cid:18) − nλ ( t ) − λ ( t ) t (cid:19) = − (cid:18) p ( t ) λ ( t ) (cid:19) n ! χ p ( t ) (38)As we saw in Lemma 2 , when t > , (cid:12)(cid:12) p ( t ) / λ ( t ) (cid:12)(cid:12) p ≤ /p , and hence, the right-hand side of 38 converges to in p -adic absolute value as n → ∞ . However, notethat if t > is such that p ( t ) / λ ( t ) < , then the right-hand side of 38 willconverge to in Archimedean absolute value as well. Moreover, the p -adic andArchimedean limits are the same . Thus, by controlling the value of p ( t ) / λ ( t ) as t converges -adically to some limit t , we can explicitly construct subsets of Z on which χ p will be non-negative real valued. The next few definitions andresults (at least one of which is already known in a different context) take careof the details of doing this. Definition 5 : Let q be a prime number, and let { t n } n ∈ N be a sequence in Q ∩ Z q (that is, a sequence of rational numbers which are also q -adic integers),and let t ∈ Q ∩ Z q . Then, we say that the t n s are doubly convergent to t / converge doubly to t in Z q and R whenever the t n s converge to t in both q -adic and Archimedean absolute value: lim n →∞ | t − t n | = lim n →∞ | t − t n | q = 0 Definition 6 : Let r p : N → Q be defined by: r p ( n ) def = p ( n ) λ ( n ) = p ( n ) ⌈ log ( n +1) ⌉ Remark 9 :I. Recall that for every t ∈ Z , we write β k : Z → N to denote the uniqueexponent-giving functions so that:i. For all t , β k ( t ) < β k +1 ( t ) holds for all k ∈ { , . . . , ( t ) } .ii. For all t ∈ Z , t = P ( t ) k =1 β k ( t ) in Z .iii. For t ∈ N , r p ( t ) = p t ) β t )( t )+1 .II. For all t ∈ N and all k ∈ { , . . . , ( t ) } : t = ( t ) X j =1 β j ( t ) 2 βk ( t )+1 ≡ k X j =1 β j ( t ) | {z } [ t ] βk ( t )+1 where [ t ] βk ( t )+1 is the unique integer in (cid:8) , . . . , β k ( t )+1 − (cid:9) so that t βk ( t )+1 ≡ [ t ] βk ( t )+1 . Hence: r p ([ t ] βk ( t )+1 ) = r p k X j =1 β j ( t ) = p k − β k ( t )+1 , ∀ t ∈ Z (39)25II. For all t ∈ N , we can write: χ p ( t ) = ( t ) X k =1 p k − β k ( t )+1 = 1 p ( t ) X k =1 r p ([ t ] βk ( t )+1 ) (40)where both sums are defined to be when t = 0 . Lemma 4 : Let t = P ∞ n =1 β n ( t ) be an arbitrary -adic integer, and let: t n def = [ t ] βn ( t )+1 (41)Then, the sequence { χ p ([ t ] n ) } n ∈ N converges doubly in Z p and R to the limit χ p ( t ) if and only if the sequence of ratios: r p ( t n ) = p n β n ( t )+1 , ∀ n ∈ N is summable (in R ); that is, if and only if: ( t ) X n =1 r p ( t n ) = ( t ) X n =1 r p ([ t ] βn ( t )+1 ) < ∞ Moreover, when this series converges, it follows that χ p ( t ) = p P ( t ) n =1 r p ( t n ) ,and that χ p ( t ) is a positive real number.Proof: Let t and the t n s be as given. . Since the t n s converge -adically to t , the continuity of χ p : Z → Z p then guarantees that the χ p ( t n ) s converge to χ p ( t ) in Z p , irrespective of the summability of the r p ( t n ) s.Next, 40 gives: χ p ( t n ) = 1 p ( t n ) X k =1 r p ([ t n ] βk ( tn )+1 ) = 1 p n X k =1 r p ( t k ) since t n has n s in its -adic expansion, and since t n m ≡ t k for all m ∈ { , . . . , n } and all k ∈ { , . . . , n } . Thus, for all integers m, n ∈ N with n > m : | χ p ( t n ) − χ p ( t m ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p n X k =1 r p ( t k ) − p m X k =1 r p ( t k ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 p n X k = m +1 r p ( t k ) As such, the χ p ( t n ) s are Cauchy in R if and only if the sequence of partial sums P nk =1 r p ( t k ) are Cauchy in R . Thus, he χ p ( t n ) s converge to a limit in R —and that limit is equal to p P ∞ k =1 r p ( t k ) —if and only if the series P ∞ k =1 r p ( t k ) converges in R .Finally, note that since: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p n +1 X k =1 r p ( t k ) − p n X k =1 r p ( t k ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p = (cid:12)(cid:12)(cid:12)(cid:12) p r p ( t n +1 ) (cid:12)(cid:12)(cid:12)(cid:12) p = (cid:12)(cid:12)(cid:12)(cid:12) p n β n +1 ( t )+1 (cid:12)(cid:12)(cid:12)(cid:12) p = 1 p n
26t follows by the felicitous properties of p -adic convergence that the partialsums of the r k s converge to p P ∞ k =1 r p ( t k ) in Z p . Thus, the summability of P nk =1 r p ( t k ) over R is both necessary and sufficient to guarantee that χ p ( t n ) converges to χ p ( t ) = p P ∞ k =1 r p ( t k ) in both R and Z p as n → ∞ . Moreover,this limit is a positive real number, since p P ∞ k =1 r p ( t k ) is then a convergentseries of positive real numbers.Q.E.D. Remark 10 :I. Observe also that χ p ( t n ) converges doubly to for any sequence { t n } n ∈ N ⊆ N converging to in Z .II. In light of Lemma 4 , we say χ p converges doubly at z ∈ Z wheneverthe sequence { χ p ([ z ] n ) } n ∈ N converges doubly to χ p ( z ) in Z p and R .III. Lemma 4 also proves that S + p is then the set of all t ∈ Z such that χ p ([ t ] n ) converges doubly to χ p ( t ) in Z p and R . Proposition 7 : Let t be an arbitrary -adic integer which is not a non-negative real integer. Then:I. lim sup n →∞ ( β n +1 ( t ) − β n ( t )) > ln p ln 2 is a necessary condition for the limit lim n →∞ χ p ([ t ] n ) = χ p ( t ) to converge doubly in Z p and R .II. lim inf n →∞ ( β n +1 ( t ) − β n ( t )) > ln p ln 2 is a sufficient condition for the limit lim n →∞ χ p ([ t ] n ) = χ p ( t ) to converge doubly in Z p and R . Remark 11 : That is, a necessary condition that a -adic integer t be anelement of S + p is that there are infinitely many s in the -adic digits of t whichare separated from the next by at least l ln p ln 2 m − consecutive s. Likewise,a sufficient condition for t to be an element of S + p is that all but finitely many s in the -adic digits of t are separated from the next by at least l ln p ln 2 m − consecutive s.Proof: As a short-hand, write β n in lieu of β n ( t ) , and, once again, let t n = [ t ] βn +1 . Then: r p ( t n +1 ) = p n +1 β n +1 +1 = p β n +1 − β n p n β n +1 = p β n +1 − β n r p ( t n ) Thus, by the ratio test, P ∞ n =0 r p ( t n ) will diverge if: < lim inf n →∞ r p ( t n +1 ) r p ( t n +1 ) = p lim inf n →∞ β n +1 − β n m lim sup n →∞ ( β n +1 − β n ) < ln p ln 2 On the other hand, the series will converge whenever:27 > lim sup n →∞ r p ( t n +1 ) r p ( t n ) = p lim sup n →∞ β n +1 − β n m lim inf n →∞ ( β n +1 − β n ) > ln p ln 2 Consequently, by
Lemma 4 , as a sufficient condition to guarantee that P ∞ n =0 r p ( t n ) converges in R , the condition “ lim inf n →∞ ( β n +1 − β n ) > ln p ln 2 ” is then sufficient to ensure that χ p ([ t ] m ) converges to χ p ( t ) doubly in R and Z p . Likewise,the condition lim sup n →∞ ( β n +1 − β n ) ≥ ln p ln 2 is then necessary for the series toconverge doubly.Q.E.D.As mentioned previously, these convergence criteria are rather weak, andso, improving them (tightening the necessary condition and/or loosening thesufficient condition) is likely the first step toward refining the results to be givenbelow. That being said, the sufficient condition lim inf n →∞ ( β n +1 ( t ) − β n ( t )) > ln p ln 2 is of particular interest to us, because the set of all t ∈ Z for which it holds can , in fact, be continuously parameterized over Z .The trick to using the L method appears to be cleverly constructing subsetsof S p which can be continuously parameterized over Z by some function η ,where η satisfies functional equations that are compatible with the functionalequations 15 satisfied by χ p . As it turns out, the set of all t ∈ Z in whichall but finitely many s are separated from the next by at least l ln p ln 2 m − consecutive s is homeomorphic to Z , and the homeomorphism satisfies exactlythe kind of functional equations needed to make the Fourier coefficients of χ p ’sprecomposition with that homeomorphism explicitly computable. Definition 7 :I. For every integer κ ≥ , define the function τ κ : Z → Z by: τ κ ∞ X n =0 a n κn ! def = ∞ X n =0 a n κn (42)That is, τ inserts κ − zeros between every two s in the -adic digits of itsinput. Remark 12 : As pointed out by Alain Robert (Robert. p. 172), the functions τ κ are injective and continuous, satisfying the κ -Hölder condition: | τ κ ( x ) − τ κ ( y ) | = | x − y | κ , ∀ x , y ∈ Z (43)II. Let D κ def = τ κ ( Z ) denote the image of Z under τ κ . That is, D κ is the setof all -adic integers whose -adic digits contain at least κ − zeroes betweenany two s. With these definitions, using the L method, we can then prove:28 heorem 6 : Let p be an odd integer ≥ , and let κ be any positive integerfor which the following statements all hold true: κ ≥ ⌈ log p ⌉ (44) κ − > p Then, H p can have only finitely many periodic points in N that are routedthrough D κ . In particular, any such periodic point x satisfies: x ≤ κ − (cid:0) κ +1 − − p (cid:1) (2 κ − p −
1) (2 κ +1 − p − (45) Proposition 8 : For all D κ has measure zero in Z for all κ ≥ .Proof: Let κ ≥ . In an abuse of notation, let: D κ ( z ) def = [ z ∈ D κ ] = ( if z ∈ D κ if z / ∈ D κ (46)denote the indicator function of D κ . Now, fix y ∈ D κ . Then, there is a unique z ∈ Z so that y = τ κ ( z ) . Then the functional equations 53, give: κ y = τ κ (2 z )2 κ y + 1 = τ κ (2 z + 1) and hence, show that y ∈ D κ implies both κ y and κ y + 1 are in D κ . That is: [2 κ y ∈ D κ ] = [ y ∈ D κ ] = [2 κ y + 1 ∈ D κ ] m D κ (2 κ y ) = D κ ( y ) = D κ (2 κ y + 1) (47)identities which hold for all y ∈ Z .Next, observe that the -adic digits of any j ∈ { , . . . , κ − } contains a which is next to strictly less than κ − consecutive s. Consequently, no such j is an element of D κ . Since y ∈ D κ implies κ y is an element of D κ whosefirst κ -adic digits are s, we then have that for all j ∈ { , . . . , κ − } , κ y + j contains a string of strictly less than κ − consecutive s among its first κ digits.Thus, y ∈ D κ implies κ y + j / ∈ D κ for all j ∈ { , . . . , κ − } . More generally,it then follows that for all j ∈ { , . . . , κ − } , the number κ z + j is not in D κ for any z ∈ Z . As such: D κ (2 κ z + j ) = 0 , ∀ z ∈ Z , ∀ j ∈ { , . . . , κ − } (48)Using the above facts about the indicator function for D κ , let: ˆ D κ ( t ) = Z Z D κ ( z ) e − πi { t z } d z (49)29e its Fourier transform. Splitting Z up mod κ yields: ˆ D κ ( t ) = κ − X j =0 Z κ Z + j D κ ( z ) e − πi { t z } d z By 48, the integrand is identically for j ∈ { , . . . , κ − } . This leaves us with: ˆ D κ ( t ) = Z κ Z D κ ( z ) e − πi { t z } d z + Z κ Z +1 D κ ( z ) e − πi { t z } d z Making changes of variables in the two integrals ( x = z / κ , y = ( z − / κ )gives: ˆ D κ ( t ) = 12 κ Z Z D κ (2 κ x ) e − πi { κ t x } d x + 12 κ Z Z D κ (2 κ y + 1) e − πi { t (2 κ y +1) } d y (47) ; = 12 κ Z Z D κ ( x ) e − πi { κ t x } d x + 12 κ Z Z D κ ( y ) e − πi { t (2 κ y +1) } d y = 12 κ Z Z D κ ( x ) e − πi { κ t x } d x | {z } ˆ D κ (2 κ t ) + e − πit κ Z Z D κ ( y ) e − πi { κ t y } d y | {z } ˆ D κ (2 κ t ) = 1 + e − πit κ ˆ D κ (2 κ t ) Thus, we have the functional equation: ˆ D κ ( t ) = 1 + e − πit κ ˆ D κ (2 κ t ) , ∀ t ∈ ˆ Z (50)Setting t = 0 and using the fact that ˆ D κ (0) is | D κ | —the measure of D κ —we have that ˆ D κ (0) = κ − ˆ D κ (0) , and thus, that (cid:0) − − κ (cid:1) ˆ D κ (0) = 0 whichshows that D κ has measure zero for all κ ≥ .Q.E.D. Proposition 8 : If κ ≥ ⌈ log p ⌉ , then S + p contains D κ .Proof: Fix κ ≥ ⌈ log p ⌉ , and let z ∈ D κ be arbitrary. Note that z can bewritten as z = P ∞ n =1 β n ( z ) . Since τ κ ( z ) = P ∞ n =1 κβ n ( z ) and since κ ≥ ⌈ log p ⌉ ,we have that: κβ n +1 ( z ) − κβ n ( z ) ≥ ⌈ log p ⌉ β n +1 ( z ) − ⌈ log p ⌉ β n ( z ) ≥ ⌈ log p ⌉ (1 + β n ( z )) − ⌈ log p ⌉ β n ( z )= ⌈ log p ⌉ > ln p ln 2 Lemma 4 , it must be that: χ p ( τ κ ( z )) = lim n →∞ ( [ τ κ ( z )] n ) X k =1 p k − β k ( [ τ κ ( z )] n ) +1 = lim n →∞ β n X k =1 p k − β k (cid:16) [ τ κ ( z )] βn +1 (cid:17) +1 = lim n →∞ β n X k =1 p k − κβ k ( z )+1 ( κ ≥ ⌈ log p ⌉ ) ; ≤ lim n →∞ β n X k =1 p k − ⌈ log p ⌉ β k ( z )+1 = ∞ X k =1 p k − ⌈ log p ⌉ β k ( z )+1 < ∞ Thus, χ p ( τ κ ( z )) is doubly convergent in Z p and R whenever κ ≥ ⌈ log p ⌉ , andso, the convergent series: χ p ( τ κ ( z )) = ∞ X k =1 p k − κβ k ( z )+1 is a positive real number for κ ≥ ⌈ log p ⌉ , as desired. Thus, D κ ⊆ S p for all κ ≥ ⌈ log p ⌉ .Q.E.D. Definition 8 : Let: ω p,κ def = χ p ◦ τ κ (51) ˆ ω p,κ ( t ) def = Z Z χ p ( τ κ ( z )) e − πi { t z } (52)By the L method, to prove Theorem 6 , it suffices to show that ˆ ω p,κ ∈ L (cid:16) ˆ Z (cid:17) whenever κ satisfies 44. To do this, we first establish functional equations54 for ω p,κ . Then, we show that 54 has a unique solution in L ( Z ) , andthat the Fourier coefficients of this solution must be given by 57 and 58. Thisthen completely determines ω p,κ and ˆ ω p,κ . We can then directly show that ˆ ω p,κ ∈ L (cid:16) ˆ Z (cid:17) . Proposition 9 : For all κ ∈ N , the following functional equations hold truefor all z ∈ Z : τ κ (2 z ) = 2 κ τ κ ( z ) (53) τ κ (2 z + 1) = 2 κ τ κ ( z ) + 1 κ ∈ N be arbitrary, and write z = P ∞ n =0 z n n ∈ Z . Then:I. τ κ (2 z ) = τ κ ( P ∞ n =1 z n − n ) = P ∞ n =1 z n − κn = P ∞ n =0 z n κn + κ = 2 κ τ κ ( z ) II. τ κ (2 z + 1) = τ κ ∞ X n =1 z n − n ! = 1 + ∞ X n =1 z n − κn = 1 + ∞ X n =0 z n κn + κ = 1 + 2 κ τ κ ( z ) Q.E.D.
Proposition 10 : Let κ be arbitrary. Then, for all z ∈ Z : ω p,κ (2 z ) = ω p,κ ( z )2 κ (54) ω p,κ (2 z + 1) = pω p,κ ( z ) + 2 κ − κ Proof:I. ω p,κ (2 z ) = χ p ( τ κ (2 z )) = χ p (2 κ τ κ ( z )) = 2 − κ χ p ( τ κ ( z )) = 2 − κ ω p,κ ( z ) II. ω p,κ (2 z + 1) = χ p ( τ κ (2 z + 1))= χ p (2 κ τ κ ( z ) + 1)= pχ p (cid:0) κ − τ κ ( z ) (cid:1) + 12= p − κ χ p ( τ κ ( z )) + 12= pχ p ( τ κ ( z )) + 2 κ − κ = pω p,κ ( z ) + 2 κ − κ Q.E.D.
Lemma 5 : Let p be an odd integer ≥ , and let κ ≥ be arbitrary. Let f ∈ L ( Z ) satisfy the functional equations: f (2 z ) = f ( z )2 κ (55) f (2 z + 1) = pf ( z ) + 2 κ − κ z ∈ Z , and let: ˆ f ( t ) = Z Z f ( z ) e − πi { t z } d z denote the Fourier transform of f . Then:I. ˆ f ( t ) = 1 + pe − πit κ +1 ˆ f (2 t ) + e − πit (2 t ) , ∀ t ∈ ˆ Z (56)II. The values of ˆ f ( t ) are given by: ˆ f (0) = 2 κ − κ +1 − − p ˆ f (cid:18) (cid:19) = 12 1 − κ κ +1 − − p ˆ f ( t ) = ˆ f (cid:18) (cid:19) log | t | Y n =0 p (cid:0) e − πit (cid:1) n κ +1 , ∀ t ∈ ˆ Z \ { } where the product over n is defined to be when t = (i.e., when the upperlimit of the product is log | / | = log = − ). Remark 13 : This shows that if there exists a solution f ∈ L ( Z ) to thesystem of functional equations 56, then it must be unique , and is given by thefunction with the Fourier coefficients described in (II).Proof: 33. Let f be as given. Then, for all t ∈ ˆ Z : ˆ f ( t ) = Z Z f ( z ) e − πi { t z } d z = Z Z f ( z ) e − πi { t z } d z + Z Z +1 f ( z ) e − πi { t z } d z (cid:18) x = z y = z − (cid:19) ; = 12 Z Z f (2 x ) e − πi { t (2 x ) } d x + 12 Z Z f (2 y + 1) e − πi { t (2 y +1) } d y = 12 κ +1 Z Z f ( x ) e − πi { t (2 x ) } d x + 12 κ +1 Z Z (cid:0) pf ( y ) + 2 κ − (cid:1) e − πi { t y + t } d y = 12 κ +1 Z Z f ( x ) e − πi { t x } d x | {z } ˆ f (2 t ) + pe − πit κ +1 Z Z f ( y ) e − πi { t y } d y | {z } ˆ f (2 t ) + e − πit Z Z e − πi { t y } d y | {z } (2 t ) = 1 + pe − πit κ +1 ˆ f (2 t ) + e − πit (2 t ) and so: ˆ f ( t ) = 1 + pe − πit κ +1 ˆ f (2 t ) + e − πit (2 t ) which is 56.II. Letting t = 0 in 56 gives: ˆ f (0) = 1 + p κ +1 ˆ f (0) + 14 m ˆ f (0) = 2 κ − κ +1 − − p Next, letting t = in 56 and using the fact that ˆ f (0) = κ − κ +1 − − p and that ˆ f is periodic with period gives: ˆ f (cid:18) (cid:19) = 1 − p κ +1 ˆ f (0) −
14 = 1 − p κ +1 κ − κ +1 − − p −
14 = 12 1 − κ κ +1 − − p Now, note that in 56: ˆ f ( t ) = 1 + pe − πit κ +1 ˆ f (2 t ) + e − πit (2 t ) (2 t ) vanishes for all t ∈ ˆ Z with | t | ≥ (i.e., for t = 0 , ). Consequently, we have: ˆ f ( t ) = 1 + pe − πit κ +1 ˆ f (2 t ) , ∀ | t | ≥ Using this functional equation, it follows that for any t with | t | = 2 m ≥ = 4 ,there will be m − terms in the product defining ˆ f ( t ) , ending with a constantterm of ˆ f (cid:0) (cid:1) . Since m − | t | − | t | this can be written as: ˆ f ( t ) = ˆ f (cid:18) (cid:19) log | t | Y n =0 p (cid:0) e − πit (cid:1) n κ +1 , ∀ | t | ≥ When t = , log | t | = − . As such, if we adopt the convention that theproduct over n is defined to be when t = , we then have that: ˆ f ( t ) = ˆ f (cid:18) (cid:19) log | t | Y n =0 p (cid:0) e − πit (cid:1) n κ +1 , ∀ t ∈ ˆ Z \ { } as desired.Q.E.D. Lemma 6 : Let p be an odd integer ≥ , let κ be a positive integer, and let g : ˆ Z → C be the function defined by: g (0) = 2 κ − κ +1 − − pg (cid:18) (cid:19) = 12 1 − κ κ +1 − − pg ( t ) = g (cid:18) (cid:19) log | t | Y n =0 p (cid:0) e − πit (cid:1) n κ +1 , ∀ t ∈ ˆ Z \ { } where the product is when t = .With these hypotheses, if p < κ − , then g ∈ L (cid:16) ˆ Z (cid:17) . In particular: k g k L ( ˆ Z ) ≤ κ − (cid:0) κ +1 − − p (cid:1) (2 κ − p −
1) (2 κ +1 − p − Proof: We have that: k g k L ( ˆ Z ) = | g (0) | + (cid:12)(cid:12)(cid:12)(cid:12) g (cid:18) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) g (cid:18) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) X t ∈ ˆ Z | t | ≥ log | t | Y n =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p (cid:0) e − πit (cid:1) n κ +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X t ∈ ˆ Z | t | ≥ log | t | Y n =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p (cid:0) e − πit (cid:1) n κ +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X t ∈ ˆ Z | t | ≥ (cid:18) p + 12 κ +1 (cid:19) − | t | We evaluate the sum over t by writing n t ∈ ˆ Z : | t | ≥ o as the union of n t ∈ ˆ Z : | t | = 2 m o over m ∈ N : X t ∈ ˆ Z | t | ≥ (cid:18) p + 12 κ +1 (cid:19) − | t | = ∞ X m =2 X t ∈ ˆ Z | t | = 2 m (cid:18) p + 12 κ +1 (cid:19) m − = ∞ X m =2 (cid:18) p + 12 κ +1 (cid:19) m − X t ∈ ˆ Z | t | = 2 m The t -sum on the far right is simply the number of elements of n t ∈ ˆ Z : | t | = 2 m o ,which is m − (the number of irreducible fractions in (0 , with a denominatorof m ). As such: X t ∈ ˆ Z | t | ≥ log | t | Y n =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p (cid:0) e − πit (cid:1) n κ +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ∞ X m =2 (cid:18) p + 12 κ +1 (cid:19) m − m − = ∞ X m =1 (cid:18) p + 12 κ (cid:19) m Thus, the condition κ − > p is sufficient to guarantee: k g k L ( ˆ Z ) ≤ | g (0) | + (cid:12)(cid:12)(cid:12)(cid:12) g (cid:18) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) g (cid:18) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X m =1 (cid:18) p + 12 κ (cid:19) m < ∞ Using g (0) = κ − κ +1 − − p , g (cid:0) (cid:1) =
12 1 − κ κ +1 − − p , and summing the geometric seriesthen yields: k g k L ( ˆ Z ) ≤ κ − (cid:0) κ +1 − − p (cid:1) (2 κ − p −
1) (2 κ +1 − p − as desired.Q.E.D.In summary all that we have shown: Theorem 7 : Let p be an odd integer ≥ , let κ be an integer ≥ ⌈ log p ⌉ sothat κ − > p . Then:I. The Fourier coefficients of ω p,κ are given by: ˆ ω p,κ (0) = 2 κ − κ +1 − − p (57)36 ω p,κ ( t ) = ˆ ω p,κ (cid:18) (cid:19) log | t | Y n =0 p (cid:0) e − πit (cid:1) n κ +1 , ∀ t ∈ ˆ Z \ { } (58)where the product is defined to be when t = .II. ˆ ω p,κ ∈ L (cid:16) ˆ Z (cid:17) , with: k ˆ ω p,κ k L ( ˆ Z ) ≤ κ − (cid:0) κ +1 − − p (cid:1) (2 κ − p −
1) (2 κ +1 − p − Proof: Use
Lemmata 5 & 6 along with the
Fourier Inversion Theorem for the Fourier Transform over Z and ˆ Z to compute the formulae for theFourier coefficients of ω p,κ . The p < κ − condition then guarantees the L summability of ˆ ω p,κ and the specified upper bound on its L norm.Q.E.D. Proof of Theorem 6 : Letting κ satisfy 44, since ˆ ω p,κ is in L (cid:16) ˆ Z (cid:17) , by the L method (here, η is τ κ ) and it must be that any periodic point of H p in routedthrough D κ is ≤ k ˆ ω p,κ k L ( ˆ Z ) ≤ κ − ( κ +1 − − p ) (2 κ − p − κ +1 − p − , which then forces there tobe only finitely many such periodic points. Since D κ ⊆ S + p , this shows thatthere can be only finitely many periodic points of H p in N which are routedthrough D κ , because any such periodic point must be ≤ κ − ( κ +1 − − p ) (2 κ − p − κ +1 − p − .Q.E.D. The original impetus for this paper was to share the Lipschitz estimate (