aa r X i v : . [ m a t h . D S ] M a y SUBSTITUTIONS AND -DISCREPANCY OF { nθ + x } DAVID RALSTON
Abstract.
The sequence of 1 / { x + iθ mod 1 } is realizedthrough a sequence of substitutions on an alphabet of three symbols; particularattention is paid to x = 0. The first application is to show that any asymptoticgrowth rate of the discrepancy sums not trivially forbidden may be achieved.A second application is to show that for badly approximable θ and any x therange of values taken over i = 0 , , . . . n − n ),a stronger conclusion than given by the Denjoy-Koksma inequality. Introduction
Given an irrational θ and some x ∈ [0 ,
1) = S (all addition in S is takenmodulo one), let(1) f ( x ) = χ [0 , / ( x ) − χ [1 / , ( x ) . With θ fixed, the 1 / -discrepancy sums of the sequence { x + iθ } are given by S n ( x ) = n − X i =0 f ( x + iθ ) . Two results are classical in this setting, for any irrational θ and for all x :(2) S n ( x ) ∈ o ( n ) , S n ( x ) / ∈ O (1) . The first restriction is due to unique ergodicity of the underlying rotation, and thesecond is a theorem of Kesten [5].We will use standard continued fraction notation; partial quotients are denoted a i ( θ ), and convergents are denoted p i ( θ ) /q i ( θ ). When θ is clear from context wewill simply write a i , p i and q i . The distance from x to the nearest integer is denoted k x k . As θ ∈ (0 ,
1) without loss of generality, we will assume that a ( θ ) = 0 andomit this term, writing simply θ = [ a , a , a , . . . ] = 1 a + 1 a + 1 a + . . . . All necessary background in continued fractions may be found in [6]. The
Gaussmap will be denoted by γ , and acts as the non-invertible shift on the sequence of Date : November 12, 2018.2010
Mathematics Subject Classification.
Primary: 11K38, Secondary: 37E20, 37B10.
Key words and phrases. discrepancy, irrational rotation, renormalization, substitution. partial quotients:(3) γ ( θ ) = 1 θ mod 1 , γ ([ a , a , . . . ]) = [ a , a , . . . ] . Our goal is to investigate what behavior is possible for the sequence S n ( x ) withinthe constraints of (2). Because the sequence S n is not monotone, however, it willbe more convenient to consider the following sequences, which track the maximal and minimal discrepancies, as well as the range of values taken: M n ( x ) = max { S i ( x ) : i = 1 , . . . , n − } , (4) m n ( x ) = min { S i ( x ) : i = 1 , . . . , n − } , (5) ρ n ( x ) = M n ( x ) − m n ( x ) + 1 . (6)It is worth clarifying that m n is taken as a minimum over integers , and as such cangenerally be expected to be negative. It is a matter of later convenience that i = 0is not considered: for example, M (0) = m (0) = S (0) = 1.We will develop a renormalization procedure through which the sequence ofvalues f ( x + iθ ) can be determined from a sequence of substitutions. Let θ < / A = [0 , / B = [1 / , − θ ), C = [1 − θ, A to be closed and B to be open), we will say that we make a change of endpoints of the intervals A , B , and C . Our central result is the following: Theorem 1.1.
Given any irrational θ and any x ∈ [0 , , there is a sequence ofwords ω i (some of which may be empty) and substitutions σ i (infinitely many arenot identity) both defined on the alphabet { A, B, C } , given by a dynamic processdepending on x and θ , such that the infinite word given by (7) ω σ ( ω σ ( ω σ ( . . . ))) encodes the orbit of x up to at most two errors. Alternately, the coding is exact upto a change of endpoints of the intervals A , B and C . The dependence of σ i on θ and ω i on ( x, θ ) is explicit.There is one special point x ( θ ) for which all ω i may be taken to be the emptyword, in which case the infinite word (8) lim n →∞ ( σ ◦ σ ◦ · · · ◦ σ n − ) ( ω ) will encode the orbit of x ( θ ) regardless of the choice of nonempty word ω . The orbitof zero can alternately be determined by (9) lim n →∞ (cid:0) σ ′ ◦ σ ′ ◦ · · · ◦ σ ′ n − (cid:1) ( ω ′ n − ) , where σ ′ n are either substitutions or a different map. This distinction and the word ω ′ n are explicitly presented. We will include some remarks regarding the point x ( θ ) (including a completecharacterization of those θ for which x ( θ ) = 0 in Proposition 4.3), as well as provingthat the sequence of substitutions σ i is eventually periodic if and only if θ is aquadratic surd (Proposition 4.4).As (0 , / ⊂ A and (1 / , ⊂ ( B ∪ C ), any change of endpoints is completelyirrelevant to the asymptotic growth rates of M n ( x ), m n ( x ), and ρ n ( x ). WhileTheorem 1.1 provides a way to produce the orbit of an arbitrary point, computationof the words ω i is a nontrivial task. However, for the special point x ( θ ) and for 0, UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } the process is much simpler. We will show that given any growth condition thatdoes not violate (2), such behavior is seen to be possible: Theorem 1.2.
Suppose that { c n } and { d n } are two increasing sequences of positivereal numbers, both in o ( n ) , the differences ∆ c n = c n +1 − c n are in O (1) (similarly for { ∆ d n } ), and at least one of { c n } , { d n } is divergent. Thenthere is a dense set of θ such that if { c n } is divergent, then lim sup n →∞ M n (0) c n = 1 , while if { c n } is bounded then so is M n (0) . Similarly, if { d n } is divergent, then lim sup n →∞ | m n (0) | d n = 1 , while if { d n } is bounded then so is m n (0) . A closely related result concerns the sequence of values M n ( x ) / | m n ( x ) | : Theorem 1.3.
Let ≤ r ≤ r ≤ ∞ . Then there is a dense set of θ such that theset of accumulation points of the sequence (cid:26) M n (0) | m n (0) | : n = 0 , , , . . . (cid:27) is the interval [ r , r ] . We will also include a partial rederivation of [2, Theorem 1] in Corollary 5.3: acharacterization of those θ for which S n ( θ ) ≥ n ≥ a i ( θ ) aredrawn from a finite set (such θ are said to be badly approximable or of finite type ),then S n ( x ) ∈ O (log n ). Theorem 1.4. If θ is of finite type, then for all x we have ρ n ( x ) ∼ log n , meaningthat the ratio is bounded away from both zero and infinity. Corollary 1.5. If θ is of finite type, then | S n ( x ) | / ∈ o (log n ) for every x , and m n ( x ) ∈ o (log n ) = ⇒ M n ( x ) ∼ log n, and vice-versa. If A ∪ B represents a single interval, then as S has been partitioned into twointervals of length θ and 1 − θ , the analogous problem would be to encode the Sturmian sequences , and generating Sturmian sequences using a sequence of sub-stitutions is intimately related to continued fraction expansions for numbers: seefor example [3, Chapter 6]. The study of substitutions as they relate to discrep-ancy sequences of different intervals has been initiated before [1], in this paper ourapproach is different: • the interval [0 , /
2] is not dynamically defined , i.e. not dependent on θ (although it is fixed), • we develop an approach for all θ (not just quadratic surds, though theprocess is nicest in this setting), • we generate the orbit of any starting point x (though x = 0 is one particu-larly nice case that we investigate). DAVID RALSTON Symbol Spaces, Encodings, and Substitutions
All background material pertaining to common definitions in symbolic dynamicsand substitution systems may be found in [3, Chapter 1]; we present here only ashort summary of specific notation used herein. Let A = { A, B, C } , and denote by A ∗ the free monoid on A . Given ω ∈ A ∗ , we denote ω = ( ω ) ( ω ) . . . ( ω ) n − , and say that ω is a word of length n with letters ( ω ) i drawn from the alphabet A . Note that ω i will refer to a sequence of words indexed by i , while ( ω ) i willdenote the individual letters of a fixed word ω . This similarity is a potential sourceof confusion, but the latter notation is much more common in this work: we willrarely refer to specific letters in a given word.Denote by | ω | the length of ω . Elements in A ∗ multiply by concatenation, andwe adopt power notation for this operation: ( AB ) = ABABAB , for example. Theempty word (the identity under concatenation) we denote ∅ . A factor of ω (of finiteor infinite length) is some finite word ψ of length n such that there is some i forwhich ( ψ ) j = ( ω ) i + j , j = 0 , , . . . , n − . If i = 0 then we say ψ is an left factor of ω , and we say ψ is a right factor of ω if( ψ ) n − = ( ω ) | ω |− . The factor ψ will be called proper if ψ / ∈ { ω, ∅} .Any map σ : A → A ∗ may be extended to a map on A ∗ be requiring it to bea homomorphism. The following is nonstandard but natural. Endow A N with thecylinder topology, and let a finite word ω ∈ A ∗ represent a clopen set: the set ofall elements of A N with left factor ω . We may then further extended σ to a mapon A N by defining σ ( ω ) = ∞ \ i =0 σ (( ω ) ( ω ) . . . ( ω ) i − ) . In all of these situations we refer to σ as a substitution .Given a sequence of words ω , ω , . . . such that ω i is a left factor of ω i +1 , if ∞ \ i =0 ω i = { x } , then we say that x ∈ A N is the limit of the words ω i .Now consider the space S = [0 ,
1) with the map R θ ( x ) = x + θ mod 1 for someirrational θ . Suppose that X is partitioned into three intervals A , B , and C . Thengiven a word ω , we say that ω encodes the orbit of x if for all i ≤ | ω | − ω ) i = A ⇐⇒ x + iθ ∈ A, and similarly for B and C . Given a partition, then, to each x ∈ S we may identifyan infinite word ω ∈ Ω: the infinite word which encodes the (forward) orbit of x .Let D be the discontinuities of ( f ◦ R iθ )( x ) for i = 0 , , , . . . : D = {− iθ, − iθ + 1 / } , i = 0 , , , . . . . For each x ∈ D , then, we replace x ∈ S with two points, a right and left limit,denoted x + and x − . We set R θ (0 + ) = R θ (1 − ) = θ, UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } and similarly for (1 / ± ; while this makes the rotation two-to-one at these points,note that with respect to the alphabet A , the symbolic coding for the forward orbitof θ + and θ − are identical, so we do not distinguish them. We still denote our spaceby S . We may now make each of A , B and C closed, although we have made S totally disconnected.Given an irrational θ , partition S = [0 + , − ] according to Table 1 and in a slightabuse of notation let S be the set of all words which encode orbits with respect tothese conventions. θ < / θ > / A = h + , − i C = [0 + , (1 − θ ) − ] B = h
12 + , (1 − θ ) − i B = h (1 − θ ) + , − i C = [(1 − θ ) + , A = h
12 + , − i Table 1.
The partition S = A ∪ B ∪ C depending on θ .The following lemma is immediate, and immediately explains the apparent am-biguity in the statement of Theorem 1.1: Lemma 2.1. If ω is an infinite word encoding the orbit of a point x ∈ S underrotation by θ , then ω encodes the orbit of some x ∈ S without the introduction of D with at most two errors. Alternately the coding is exact up a change of endpointsof the intervals A , B and C .Proof. The orbit of any point can hit the endpoints of A , B and C at most twice. (cid:3) The Renormalization Procedure
Recall γ , the Gauss map (3); we define a similar map.(10) g ([ a , a , a , . . . ]) = [ a , a , . . . ] = γ ( θ ) ( a = 0 mod 2)[1 , a , a , . . . ] = γ ( θ ) ( a = 1 mod 2 , a = 1)[ a + 1 , a , . . . ] = 1 − θ ( a = 1) . Note that if θ > /
2, then necessarily g ( θ ) < /
2. It will be convenient to define(11) E ( x ) = max { n ≤ x : n ∈ Z , n = 0 mod 2 } . The triplet { X, µ, T } refers to a compact probability space { X, µ } and a contin-uous transformation T on X which preserves µ . Given irrational θ , we denote(12) θ n = g n ( θ ) , δ n = 1 − E ( a ( θ n )) θ n , I n = { S , µ, R θ n } . Note that δ n = 1 if and only if θ > /
2; otherwise δ n < / I n into intervals A , B and C according to Table 1, and recall thatby convention we have disconnected each I n such that all iterates of the character-istic functions of A , B and C under R iθ n are continuous. Given { X, µ, T } and a set S ⊂ X , the return time to S is given by n ( x ) = min { n > T n ( x ) ∈ S } . DAVID RALSTON
As irrational rotations are minimal, n ( x ) will be defined for all x ∈ S if S is aninterval of positive length. The induced system on S is defined by { S, µ | S , T | s } , where T | S ( x ) = T n ( x ) ( x ) for all x ∈ S . Define I ′ n +1 ⊂ I n by I ′ n +1 = [0 + , δ − n ] . Finally, define the substitutions σ n = σ ( θ n ) according to Table 2, and define thefunctions ϕ n = ϕ ( θ n ) according to:(13) ϕ ( x ) = ( − x ( a ( θ ) = 1) δ − n x ( a ( θ ) = 1)Case Substitution a = 2 k, a = 1 A → ( A k +1 B k − C )( A k B k − C ) a − B → ( A k B k C )( A k B k − C ) a − C → ( A k B k C )( A k B k − C ) a a = 2 k, a = 1 A → ( A k B k C )( A k B k − C ) a B → ( A k +1 B k − C )( A k B k − C ) a C → ( A k +1 B k − C )( A k B k − C ) a − a = 2 k + 1 A → A k B k CB → A k +1 B k − CC → Aa = 1 A → AB → BC → C Table 2.
The substitution σ as a function of θ . Lemma 3.1.
Suppose that θ < / , E ( a ( θ )) = 2 k , and (1 − kθ ) + ≤ x ≤ (cid:18) − ( k − θ (cid:19) − . Then the orbit of x begins A k B k − C .Proof. The assumption θ < / S according to Table 1 aswell as guaranteeing that k ≥
1. Note that the lower inequality certainly guaranteesthat 12 − kθ < x ≤ (cid:18) − ( k − θ (cid:19) − , which tells us that x + iθ ≤ (1 / − for i = 0 , , . . . ( k − x + kθ > /
2. Sothe coding of the orbit of x begins with exactly A k before seeing either B or C . Aswe know (1 − kθ ) + ≤ x < − (2 k − θ, we know that we have x + (2 k − θ <
1, while x + 2 kθ ≥ + . Therefore, oncewe have accounted for the points x + iθ for i = 0 , , . . . , k −
1, the terms i = k, k + 1 , . . . , (2 k −
1) must all belong to either B or C . That C is an interval oflength exactly θ guarantees that exactly the final term is C . The rest of the terms(if there are any) are therefore B . (cid:3) UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } Proposition 3.2.
We have the measurable and continuous isomorphism n I ′ n +1 , µ | I ′ n +1 , ( R θ n ) | I n +1 o ϕ n −−→ (cid:8) I n +1 , µ, R θ n +1 (cid:9) . Furthermore, for all x ∈ A ⊂ I n +1 , the word σ n ( A ) encodes the orbit of ϕ − ( x ) through its return to I ′ n +1 (the encoding is with respect to the partition A , B , C in I n ), and similarly for B and C .Proof. In the case that θ n > /
2, then θ n +1 = 1 − θ n and I ′ n +1 = [0 + , − ]. However,by referring to Table 1, we see that the intervals A , B and C exactly reflect thereversal of orientation given by ϕ n ( x ) = 1 − x , and the substitution σ n is identity.So we proceed on the assumption that θ n < /
2: in I n we have A = [0 + , / − ] , B = [1 / + , (1 − θ ) − ] , C = [(1 − θ ) + , − ] . Then ϕ n is scalar multiplication by δ − n , so there are only two things to show: • The first-return map ( R θ n ) | I ′ n +1 is rotation by θ n +1 , after rescaling by ϕ n ,and • the substitution σ n encodes the correct information.There are three cases to consider: a ( θ n ) = 1 mod 2, or a ( θ n ) = 0 mod 2 withthe sub-cases a ( θ n ) = 1 or = 1. Assume for now that a ( θ n ) = 0 mod 2 and a ( θ n ) = 1.As a ( θ n ) = 0 mod 2 and a ( θ n ) = 1, we have g ( θ n ) = γ ( θ n ) > /
2, so in I n +1 we have C = [0 + , (1 − θ n +1 ) − ] , B = [(1 − θ n +1 ) + , / − ] , A = [1 / + , − ] , with corresponding preimages in I ′ n +1 scaled by δ n . We will first verify that theintervals have the desired return times (which may be read from the length of thewords σ n ( A ), σ n ( B ) and σ n ( C )) and that the induced map is indeed rotation by θ n +1 (up to scale δ n ). As E ( a ( θ n )) = a ( θ n ) we have δ n = k q ( θ n ) · θ n k , from which it follows that the return time of 0 is n (0) = q = a a + 1 , and one may now verify that the entire interval ϕ − n ( C ) has this return time; thepreimage of the right endpoint of C under ϕ n is exactly 1 − ( q + q ) θ n . Theremaining points in I ′ n +1 have return time q + q and the induced map is a rotationby q θ n on [0 + , δ − n ]; see Figure 1. • q , , ϕ − n ( C ) • − ( q + q ) θ n ϕ − n ( A ∪ B ) q + q iiiiii t t iiiiiiiii q , , • k q ( θ n ) · θ n k q + q hhhhhhh s s hhhhhhh • • k q ( θ n ) θ n k • k q ( θ n ) · θ n k Figure 1.
Return times for the case a ( θ n ) = 0 mod 2 , a ( θ n ) = 1 . DAVID RALSTON
At this point we may verify that the rotation is by g ( θ n ), up to scale: k q ( θ n ) · θ n k δ n = q ( θ n ) · θ n − p ( θ n )1 − q ( θ n ) · θ n = ( a a + 1) θ n − a − a θ n = a (cid:16) a − θ n (cid:17) + 1 θ n − a = 1 − a γ ( θ n ) γ ( θ n )= γ ( θ n ) . Now suppose that x ∈ ϕ − ( B ), and for convenience denote E ( a ) = a = 2 k .Clearly, the orbit of x begins with a point in A (in I n , as A = [0 + , / − ] contains[0 + , δ − n ]). As x < / − kθ n , however, we have(1 − kθ n ) + ≤ x + θ n ≤ ((1 / − ( k − θ n ) − , so by Lemma 3.1, we may concatenate the word A k B k − C to this initial A . Since2 k = a , we now have x + θ n + (2 kθ n ) < x + θ n ≤ ((1 / − ( k − θ n ) − . Either we have returned to I ′ n +1 , in which case we are done, or we have not, inwhich case we apply Lemma 3.1 again, repeating until we return to I ′ n +1 , whichmust take a total of q + q = a ( a + 1) + 1 steps.For those points in the interval ϕ − n ( a ), note that the only discontinuity of R iθ n for i = 0 , , . . . , q to distinguish the orbits compared to points in ϕ − n ( A ) is thepoint 1 / − kθ , which will change the single term x + kθ from an ‘ A ’ to a ‘ B ’. Pointsin ϕ − n ( C ) are considered identically to those in ϕ − n ( B ), noting that the shorterreturn time requires one fewer concatenation of A a B a − C .The other cases are similarly considered; the case a ( θ n ) = 0 mod 2, a ( θ n ) = 1is nearly identical, while for the case a ( θ n ) = 1 mod 1 , = 1 we have δ n > θ n , sothe return time of 0 + is one, explaining the much shorter substitution σ n ( C ) = A in this case. (cid:3) Denote the iterated pull-back of I n into I by(14) ˜ I n = (cid:0) ϕ − ◦ · · · ◦ ϕ − n − (cid:1) ( I n ) . Corollary 3.3.
We have the measurable and continuous isomorphism n ˜ I n , µ | ˜ I n , ( R θ ) | ˜ I n o ( ϕ n − ◦···◦ ϕ ) −−−−−−−−−→ { I n , µ, R θ n } . Furthermore, for any x ∈ A ⊂ I n , the word ( σ ◦ · · · ◦ σ n − ) ( A ) encodes the orbitof (cid:0) ϕ − ◦ · · · ◦ ϕ − n − (cid:1) ( x ) in I through its return to ˜ I n , and similarly for B , C . Proof of Theorem 1.1
The proof of (8) is immediate in light of Corollary 3.3; the point x ( θ ) is givenby x ( θ ) = ∞ \ i =0 ˜ I i , UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } where the ˜ I i were defined in (14). This intersection is nonempty as the sets arenested closed intervals in the compact space S . The length of ˜ I n is given by δ · δ · · · δ n − , and we have already remarked that for θ n < /
2, we have δ n < /
2. As no twosuccessive terms in the sequence θ , θ , . . . may be larger than one half, the lengthtends to zero, and the intersection is either a singleton or a pair { x − , x + } . Inthe latter scenario, however, both x − and x + would have identical coding of theirforward orbits. As we did not ‘split’ the points iθ or iθ + 1 / i > S , this is not possible.As all non-identity substitutions map each letter to a word beginning in A , andall non-identity substitutions map A to a word of length at least three, and no twoconsecutive substitutions may be identity, it follows that the sequence of words( σ ◦ σ ◦ · · · ◦ σ n − ) ( ω )has a limit regardless of the choice of nonempty ω , and Corollary 3.3 shows thatthis word must encode the orbit of x ( θ ) in the disconnected version of S . Lemma2.1 finishes the proof of this portion of Theorem 1.1.Let us now turn our attention to constructing the orbit of an arbitrary x ∈ S .Define x = x + iθ, i ∈ { j ≥ x + jθ ∈ I ′ } , and let ω be the word which encodes the orbit of x through its arrival to x ; if x ∈ I ′ , we may set ω to be the empty word (though we are not required to doso). We now pass to the system I , letting ( x ∈ I ) = ϕ ( x ∈ I ′ ). We set x to be a point in I ′ which is in the orbit of x , and let ω be the word encodingthis finite portion of the orbit, then pass to I , etc. Equation (7) now follows fromProposition 3.2 so long as infinitely many ω n = ∅ . We only have the option ofletting all but finitely many ω n be empty if x is a preimage of x ( θ ); we have alreadyremarked in this case that the limiting word may be found handily.A potential source of confusion at this point is the desire to claim that x ( θ ) = 0,as we always construct I ′ n +1 = [0 + , δ − n ]. However, ϕ n ( x ) = 1 − x for those n suchthat θ n > /
2. So ϕ − n ◦ ϕ − n +1 pulls back I n +2 to the interval [(1 − δ n +1 ) + , − ] ⊂ I n .Those θ for which x ( θ ) = 0 will be addressed in Proposition 4.3. Proposition 4.1.
Without loss of generality, ω n may be required to either be empty,or a proper right factor of either σ n ( A ) , σ n ( B ) , or σ n ( C ) .Proof. The images of R iθ n (cid:0) I ′ n +1 (cid:1) cover all of I n through the return times, so any x may be viewed as returning to I ′ n +1 via a right factor of one of these words. If thereturn is through the entire word σ n ( A ), we would have begun with x n ∈ I ′ n +1 andcould have set ω n = ∅ . (cid:3) Remark.
One could alternately require that ω n be nonempty by allowing all nonemptyright factors of σ n ( A ), σ n ( B ), and σ n ( C ); instead of ω n = ∅ for x ∈ I ′ n +1 , let ω n be σ applied to the letter encoding whichever interval in I n +1 contains ϕ n ( x ).In order to construct the orbit of zero we will side-step this computation alto-gether: Lemma 4.2.
Suppose that θ n > / . Let Ω encode the orbit of + in the system I n ,and Υ encode the orbit of + in the system I n +1 . Then for all i ≥ , (Ω) i = (Υ) i .For i = 0 , (Ω) = C while (Υ) = A . Proof.
The isomorphism ϕ n ( x ) = 1 − x and the identity substitution σ n ensuresthat Ω is identical to the coding of the orbit of 1 − in I n +1 . As the forward orbitof 0 under rotation by the irrational θ n does not hit any other endpoints of theintervals A , B , and C , we have that the orbit of 1 − and 0 + in the system I n +1 areidentical after this initial term. (cid:3) With this lemma in mind, then, define the map Ψ( ω ) on both A ∗ and A N :(15) (Ψ ω ) i = ( C ( i = 0) ω i ( i = 0) . Define the maps σ ′ n = σ ′ ( θ n ):(16) σ ′ ( θ ) = ( σ ( θ ) ( θ < / θ > / . Then (9) follows if we appropriately choose the words ω ′ n to accurately encode somestring of the initial orbit of 0 + in I n . Then the resulting word (cid:0) σ ′ ◦ σ ′ ◦ · · · ◦ σ ′ n − (cid:1) ( ω ′ n )will accurately represent the initial orbit of 0 + , but it is no longer guaranteed thatthe length of this word increases! For example, if θ = [3 , , , , , . . . ], then wewill alternate between σ ′ n being Ψ and a substitution which maps C → A . Setting ω ′ n = A for all those n for which θ n < / A Ψ −→ C σ −→ A Ψ −→ C σ −→ · · · Define(17) ω ′ n = A k +1 B k − C ( a ( θ n ) = 2 k ) A k +1 B k ( a ( θ ) = 2 k + 1)Ψ( ω ′ n +1 ) ( a ( θ ) = 1) . The reader may verify that the word ω ′ n does accurately encode some initialportion of the orbit of 0 + depending on the parity of a ( θ n ). Note that wheneverΨ is applied, it affects only the first letter of its input. From this it follows that if ω = ( ω ) ν , then(18) (cid:0) σ ′ ◦ · · · ◦ σ ′ n − (cid:1) ( ω ) = (cid:0) σ ′ ◦ · · · ◦ σ ′ n − (cid:1) (( ω ) ) ( σ ◦ · · · ◦ σ n − ) ( ν ) . As ω ′ n always has length larger than one, our previous reasoning now guaranteesthat the length of Ω ′ n diverges, establishing (9) and completing the proof.Before moving on to the study of the growth rates of discrepancy sums, wepresent a few observations about this process. Proposition 4.3.
Those θ for which x ( θ ) = 0(= 0 + ) are exactly the set (19) H = { θ : a i − ( θ ) = 0 mod 2 , i = 1 , , . . . } . Proof.
We leave the reader to verify that H is exactly the set of θ for which g n ( θ ) < / n . For those θ ∈ H , then, we always have I ′ n +1 = [0 + , δ − n ], where δ n <
1, and we never need apply the isomorphism ϕ n ( x ) = 1 − x . That is,0 ∈ (cid:0) ϕ − ◦ · · · ◦ ϕ − n − (cid:1) ( I n )for all n : 0 = x ( θ ). UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } On the other hand, if n is the first index such that θ n > /
2, we must have ϕ n ( x ) = 1 − x . As θ n +1 < /
2, however, it follows that within I n , we have ϕ − n ◦ ϕ − n +1 ( I n +2 ) = [(1 − δ n +1 ) + , − ] , from which it follows that 0 / ∈ (cid:0) ϕ − ◦ · · · ◦ ϕ − n +1 (cid:1) ( I n +2 ) . (cid:3) Proposition 4.4.
The sequence of substitutions σ n is eventually periodic if andonly if θ is a quadratic surd.Proof. Clearly the sequence σ n is eventually periodic if and only if the orbit of θ under g is eventually periodic. From the definition (10) of g we have for all i ≥ a i ( θ n +1 ) = a i + k ( θ n ) : k = a ( θ n ) = 1 mod 2 , = 1)1 ( a ( θ n ) = 1)2 ( a ( θ n ) = 0 mod 2)So, if a i ( θ ) are eventually periodic (Gauss’ criteria for quadratic surds), we musthave infinitely many n such that for all i ≥ j, ka i ( θ n k ) = a i ( θ n j ) . Suppose that a period of a i ( θ ) is given by the terms α , . . . , α N , and assume withoutloss of generality that for i ≥ a i ( θ n k ) = α i mod N . Then a ( θ n k ) is either 1, α , or α + 1. Since the collection n k was infinite, onevalue must be taken twice, giving a period in the orbit g ( θ ).On the other hand, assume that θ j = θ j + nk for n = 0 , , . . . and k = 0. From(20) it follows that a i ( θ ) is eventually periodic. (cid:3) Remark.
The periods under g and γ need not be the same, nor is one necessarilylonger than the other. For example, the golden mean has period one under γ butperiod two under g , while θ = [2 , , , , . . . ] has period two under γ and period oneunder g . Furthermore, the sequence σ n is purely periodic if and only if θ n = θ for some n = 0, which is not the same as the partial quotients of θ being purelyperiodic. Consider for example θ = [3 , , , , . . . ], whose partial quotients are clearlynot purely periodic, but satisfies θ = θ .5. The Arithmetic of Our Substitutions
Let θ < /
2, so that f ( x ) = ( +1 ( x ∈ A ) − x ∈ B ∪ C ) . For θ > / ω ∈ A n , define (consistent with existing notation) S ( ω ) = n − X i =0 ( χ A − χ B ∪ C ) ω i ,M ( ω ) = max { S ( ω . . . ω j − ) : j = 1 , , . . . , n } ,m ( ω ) = min { S ( ω . . . ω j − ) : j = 1 , , . . . , n } . Note that we do not include the empty word in determining M ( ω ), m ( ω ). Proposition 5.1.
Suppose | ω | = n = 0 , ω = C , M ( ω ) ≥ , ω does not have CC , CB or BA as factors, and σ is a substitution given by Table 2, depending on θ . If a ( θ ) = 0 mod 2 and a ( θ ) = 1 , or if a ( θ ) = 1 , then: S ( σ ( ω )) = S ( ω ) , M ( σ ( ω )) = M ( ω ) + E ( a ) , m ( σ ( ω )) = m ( ω ) . On the other hand, if a ( θ ) = 0 mod 2 and a ( θ ) = 1 , then S ( σ ( ω )) = − S ( ω ) , M ( σ ( ω )) = − m ( ω ) + E ( a ) , m ( σ ( ω )) = − M ( ω ) . Finally, if a ( θ ) = 1 mod 2 , = 1 , and either • ( ω ) n − = C , or • ( ω ) n − = C , but there is some j = n such that S (( ω ) ( ω ) . . . ( ω ) j − ) = m ( ω ) ,then also S ( σ ( ω )) = − S ( ω ) , M ( σ ( ω )) = − m ( ω ) + E ( a ) , m ( σ ( ω )) = − M ( ω ) . If a ( θ ) = 1 mod 2 , ( ω ) n − = C and S (( ω ) . . . ( ω ) j − ) > m ( ω ) for all j = n , then S ( σ ( ω )) = − S ( ω ) , M ( σ ( ω )) = − m ( ω ) − E ( a ) , m ( σ ( ω )) = − M ( ω ) . Proof.
The prohibition on CB , CC and BA being factors of ω are necessary for ω to encode the orbit of any point under rotation by any θ , so this condition is notprohibitive in our setting.In all cases, the statements regarding the value S ( σ ( ω )) follow from exam-ining S ( σ ( x )) for each x ∈ A ; the reader may consult Table 2 to verify that S ( σ ( x )) = ± S ( x ) as described, and the statement then follows from the fact that σ is a homomorphism. We will turn our attention, then, to the statements regarding m ( σ ( ω )) and M ( σ ( ω )). All cases but the last are considered similarly with thepossible sign-change outlined above in mind.For example, suppose that a = 0 mod 2 and a = 1. Let ω = υψ , where υ isthe largest left factor of ω such that S ( υ ) = M ( ω ) −
1: note that as M ( ω ) ≥ M ( ω ), we have ( ψ ) = A .As S ( σ ( υ )) = S ( υ ) = M ( ω ) − M ( σ ( A )) = E ( a ) + 1, we know that M ( σ ( ω )) ≥ M ( σ ( υ ) ψ ) = M ( ω ) + E ( a ) . Assume on the other hand that σ ( ω ) = σ ( υ ) νψ, S ( σ ( υ ) ν ) > M ( ω ) + E ( a ) , and υ is of maximal length to allow such a decomposition. Note that ν = ∅ as S ( σ ( υ )) = S ( υ ) ≤ M ( ω ). As υ is a proper factor, it is followed by a letter, andby maximality on the length of υ , ν is a proper left factor of either σ ( A ), σ ( B ), or σ ( C ), and E ( a ) = 0. If υ is followed by A in ω , S ( σ ( υ )) = S ( υ ) ≤ M − . On the other hand, S ( ν ) ≤ E ( a )+1 = M ( σ ( A )), contradicting the value S ( σ ( υ ) ν ).The possibility of υ followed by B or C are similarly considered; the larger possible S ( σ ( υ )) = M ( ω ) is countered by S ( ν ) ≤ E ( a ) in these cases.The ambiguity in the situation when a ( θ ) = 1 mod 2, = 1 is due to the substi-tution σ ( A ) = C , which does not achieve an intermediate sum of E ( a ) (as does σ ( B )). On the assumption that there is some proper left factor ψ of ω such that UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } S ( ψ ) = m ( ω ), however, we know that the letter which follows ψ must be A ; similarcomputations to the above then apply. If the only left factor of ω which achieves asum of m ( ω ) is in fact ω itself, then if the final letter of ω is B we again have noproblem.Assume, then, that S ( ω ) = m ( ω ), there is no proper left factor with this sum,and ω ends with the letter C . As M ( ω ) ≥ C (that is, ω = C ). If this letter is A , then the left factor ψ such that ω = ψAC has the minimal sum as its sum (even if it is empty), andthe preceding reasoning applies. Therefore ω must be of the form ψBC (recall that CC is not a factor): considering σ ( B ) following S ( σ ( ψ )) = − m ( ω ) − (cid:3) For convenience, denote σ ( n ) = σ ◦ σ ◦ · · · ◦ σ n − , (21) σ ′ ( n ) = σ ′ ◦ σ ′ ◦ · · · ◦ σ ′ n − . (22)Recall (17) and define for n ≥ n = σ ( n ) ( A ) , Ω ′ n = σ ′ ( n ) ( ω ′ ( n )) . Define p n to track the parity of how many θ i > / p n = n − X i =1 χ (1 / , ( θ i ) ! mod 2 . We now have all the tools necessary to precisely study the sequences M n ( y ) and m n ( y ) for y ∈ { x ( θ ) , } : Proposition 5.2.
Assume that θ < / . Then S (Ω n ) = ( − p n , S (Ω ′ n ) = 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) M (Ω n ) − X i ≤ n − p i =0 E ( a ( θ i )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ , M (Ω ′ n ) = 1 + X i ≤ np i =0 E ( a ( θ i )) , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m (Ω n ) − − X i ≤ n − p i =1 E ( a ( θ i )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ , m (Ω ′ n ) = 1 − X i ≤ np i =1 E ( a ( θ i )) . Proof.
The word Ω n in (23) is formed by successive substitutions acting on theword A ; as such, it will always begin with A , so M (Ω n ) ≥
1. We immediately seethat all S (Ω n ) = ± p n by applying Proposition 5.1 insuccession. The ambiguous case in Proposition 5.1 arose when ω was a word whichhad a nonnegative maximal sum (as do all Ω n ) and whose minimum sum is onlyachieved as its total sum, with C as a terminal factor. Furthermore, we wouldneed θ n to have first partial quotient odd and larger than one. For this to happenwith the restriction that all S (Ω n ) = ± S (Ω n ) = − A ), and therefore S (Ω n − ) = 1.This scenario also require that M (Ω n − ) = 1 (otherwise m (Ω n ) < − ≤ S (Ω n )); sothis situation can only occur in our scenario when Ω n − = A : this possible error of one may only appear once in the sequence of arithmetic computations from repeatedapplication of Proposition 5.1 .We leave to the reader the verification that the parity of p n exactly dictateswhether substitutions will add to the maximal values or subtract from the minimalvalues; refer to Proposition 5.1 again.Let us now consider Ω ′ n . Note that σ ′ j = Ψ exactly when θ j > /
2, exactly when σ j − has the property that S ( σ j − ( ω )) = − S ( ω ). Clearly we have S (Ψ( ω )) = S ( ω ) − ω begins with A . Also note that if S ( ω ) = 1, then if m ( ω ) = 1we must have ω = A : it is never possible in our construction for ω to terminatewith C , S ( ω ) = 1 , and m (Ψ( ω )) = S (Ψ( ω )) is the only time this value is reached. Our choice of ω ′ ( n ) always begins with A and has S ( ω ′ ( n )) = 1, and for those σ n such that S ( σ n ( A )) = −
1, the reader may verify that S ( σ n (Ψ( ω ))) = 2 − S ( ω )by applying Proposition 5.1. While this change will change the sum of +1 to − S (Ω ′ n ) = 1 . Furthermore, as m ( ω ′ n ) = 1 for all ω ′ n , if we do apply Ψ (so m (Ψ ω ) = − σ , we see M ( σ (Ψ ω )) ≥ − m (Ψ ω ) + E ( a ) − ≥ E ( a ) − ≥ , so we may always apply Proposition 5.1 without worrying about the possible errorof one. (cid:3) Corollary 5.3 ([2], Theorem 1, case k = 2) . We have S n ( θ ) ≥ for all n ≥ ifand only if x ( θ ) = 0 .Proof. By viewing the ergodic sums as an additive cocycle, for all n > S n ( θ ) = S n +1 (0) −
1, so we have by Proposition 5.2: S | Ω ′ n |− ( θ ) = 0 , M | Ω ′ n |− ( θ ) = X i ≤ np i =0 E ( a ( θ i )) , m | Ω ′ n |− ( θ ) = − X i ≤ np i =1 E ( a ( θ i )) . So S n ( θ ) ≥ n if and only if p i = 0 mod 2 for all i such that θ i < / p i = 0 mod 2 for all i . A direct inductive argument showsthat p i = 0 for all i if and only if a i − ( θ ) = 0 mod 2 by considering the action of g (10), which corresponds by Proposition 4.3 to x ( θ ) = 0. (cid:3) Remark.
Using that σ are all homomorphisms, a more constructive version of (7)is ω σ (1) ( ω ) σ (2) ( ω ) · · · σ ( n ) ( ω n ) · · · , which allows a more direct way of computing the word through successive compu-tation of the words ω n (given the starting point x ). Lemma 5.4.
We always have (cid:12)(cid:12)(cid:12) σ ( n ) ( A ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) σ ( n ) ( B ) (cid:12)(cid:12)(cid:12) , and if we define the matrices M i = M ( θ i ) according to Table 3, then M n − M n − · · · M M (cid:20) (cid:21) = (cid:20) | σ ( n ) ( A ) || σ ( n ) ( C ) | (cid:21) . UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } Proof.
The first claim follows directly from the following observation: for all sub-stitutions σ , the words σ ( A ) and σ ( B ) are always of the same length and alwayscontain the same number of letters drawn from { A, B } . That is, within (cid:0) ϕ − n − ◦ · · · ◦ ϕ − (cid:1) ( A ∪ B ) ⊂ ˜ I n the return time under R θ to ˜ I n is constant, and similarly on the pullback of C .One need only count the number of C and { A, B } within σ n ( C ) and σ n ( { A, B } ) toconstruct the relevant matrices. (cid:3) Case M ( θ ) a ( θ ) = 0 mod 2, a ( θ ) = 1 (cid:20) ( a − a + 1 a ( a − a + a a + 1 (cid:21) a ( θ ) = 0 mod 2, a ( θ ) = 1 (cid:20) ( a − a + a a + 1( a − a + 1 a (cid:21) a ( θ ) = 1 mod 2, = 1 (cid:20) a − (cid:21) a ( θ ) = 1 (cid:20) (cid:21) Table 3.
The matrices M ( θ ) used to determine return times inthe induced systems. Lemma 5.5. | Ω n | ≤ | Ω ′ n | ≤ | Ω n +1 | . Proof.
The lower inequality is direct in light of (18), recalling that ( ω ′ n ) = A . Theupper bound follows from Lemma 5.4, noting that while ω ′ n may or may not be a leftfactor of σ n ( A ), it does contain the same number of { A, B } versus C as a proper leftfactor of σ n ( A ). Furthermore, the only substitutions for which | σ ( C ) | > | σ ( A ) | arethose corresponding to a = 0 mod 2, a = 0; such substitutions are not followedby Ψ. That is, (cid:12)(cid:12)(cid:12) σ ′ ( n ) ( A ) (cid:12)(cid:12)(cid:12) ≤ | Ω n | , completing the proof of the upper bound. (cid:3) Example 5.6.
Let θ = √ , , , . . . ]. Then as θ is a quadratic irra-tional, the sequence of substitutions σ i is eventually periodic by Proposition 4.4.As g ( θ ) = θ , the sequence of substitutions is periodic with period one, given by σ : A → AACACB → ABCACC → ABCACAC
The point x ( θ ) = 0 by Proposition 4.3, so applying Theorem 1.1, the orbit of zerois given by the sequencelim n →∞ σ n ( A ) = AACACAACACABCACACAACACABCACAC . . .
The self-similar structure of the sequence of ergodic sums S n (0) is not exact (as σ ( B ) = σ ( C )), but nonetheless highly regular. This regularity was noticed by D.Hensley in [4, Figure 3.4]. We give several plots of S n (0) for different values of n (a) N = 5, σ ( A ) = AACAC (b) N = 29, σ ( A ) (c) N = 169, σ ( A ) (d) N = 33461, σ ( A ) Figure 2.
Plots of S i (0) for different ranges of 0 ≤ i ≤ N , where θ = √ − x ( θ ) will be seenfor any quadratic irrational θ in light of Proposition 4.4.For quadratic irrational θ / ∈ H , computation of the point x ( θ ) is not too difficult: Example 5.7.
Let θ = [1 , , . . . ] be the golden mean. Recall that S will be parti-tioned such that A = [(1 / + , − ] as θ > /
2. As g ( θ ) = θ , and a = 1 correspondsto the identity substitution, the only non-identity substitution generated is σ : A → ABCACB → AACACC → AAC
So, the orbit of x ( θ ) is given bylim n →∞ σ n ( A ) = ABCACAACACAACABCACAAC . . . , while the orbit of 0 is given byΨ( σ ( . . . Ψ( AAC ))) =
CACABCACAACABCACAACAC . . . .
To compute the point x ( θ ), we need to determine the intervals ˜ I n . For those θ n = [2 , , , . . . ] we have δ n = 1 − θ n = 1 − − θ ) = 2 θ − . Denote this quantity by δ for convenience. For this particular θ we do not ever havetwo consecutive θ n < /
2, so the intervals I ′ n +1 ⊂ I n strictly alternate between[0 + , δ − ] and [(1 − δ ) + , − ] (for those n = 0 mod 2; for odd n we have θ n > / I ′ n +1 = I n ). So the sequence of preimages ˜ I n (recall again (14)) is given by (cid:2) + , − (cid:3) , (cid:2) (1 − δ ) + , − (cid:3) , (cid:2) (1 − δ ) + , (1 − δ + δ ) − (cid:3) , . . . whose intersection is given by the geometric series x ( θ ) = ∞ X i =0 ( − i δ i = 11 + (2 θ −
1) = 12 θ .
UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } (a) x = 0, with orbit CACABCACAAC . . . (b) x = x ( θ ) = 1 / (2 θ ), with orbit ABCACAACAC . . .
Figure 3.
Plots of S i ( x ) for 0 ≤ i ≤ θ is the goldenmean for the two given values of x . Note that as θ > /
2, we have A → − B, C → +1.See Figure 3 for both of these orbits.One particularly striking corollary of Proposition 5.2 is the following, which doesnot seem to be apparent from any other technique: Corollary 5.8. If θ is a quadratic irrational, then lim n →∞ M n (0) | m n (0) | ∈ Q ∗ , where Q ∗ = Q ∪ {∞} , and p/ ∞ for any positive integer p . If θ n = θ n + k is aminimal period under the orbit of g and p n + k = p n + 1 , then the ratio tends to one. Furthermore, for any nonnegative p/q ∈ Q ∗ , there is a quadratic irrational θ suchthat the above ratio has limit p/q .Proof. We have already shown that g n ( θ ) is eventually periodic for such θ in Propo-sition 4.4. It follows from Proposition 5.2 that M n (0) and m n (0) see a periodicsequence of adjustments by bounded integer amounts, which must therefore haverational limit. If one period reflects a change in the parity of p , it will always befollowed by the mirrored changes in M n , m n , producing a limit of one.To produce quadratic irrationals with the desired limit, if q = 0 then θ ∈ H willsuffice ( m n (0) ≡
1, and M n (0) must therefore diverge), and for p = 0 any θ suchthat a ( θ ) = 1 and g ( θ ) ∈ H will suffice (here M n (0) ≡ p/q with neitherzero, just set θ = [2 p, , , q − , , , p − , , , q − , , , . . . ] , and verify that we will first add p to M n (0), then subtract q from m n (0), etc. (cid:3) Proof of Theorem 1.2
Let c n and d n be divergent monotone sequences in o ( n ) with bounded differences∆ c n , ∆ d n ; we will construct a dense set of θ such thatlim sup n →∞ M n (0) c n = lim sup n →∞ | m n (0) | d n = 1 . Any irrational θ is completely determined by its sequence of partial quotients,which is equivalent to its orbit under g , and its orbit under g is completely deter-mined by the sequence of values a ( θ i ) ( a = 1 mod 2) , a ( θ i ) , a ( θ i ) ( a = 0 mod 2) . Suppose, then, that the first finitely many partial quotients of θ are prescribed, suchthat the first n values of θ i are fixed. Without loss of generality, insert an additionalsingle term if necessary so that p n = 0 (recall (24)). We are now completely free tochoose k to construct ω ′ n (refer to (17)). If we denote M (Ω ′ n ) = M, m (Ω ′ n ) = m, | Ω ′ n | = L n , it follows from Proposition 5.2 that once we choose k , we will have M (Ω ′ n +1 ) = M + k, m (Ω ′ n +1 ) = m. Denote by L n +1 ( k ) = | Ω ′ n +1 | as a function of k .Assume first that M < c L n , so we wish to increase the maximal sum comparedto the sequence c n . Then let a ( θ n ) be odd, so ω ′ ( n + 1) = A k +1 B k . From (18) and the previous observation that | σ ( n ) ( A ) | = | σ ( n ) ( B ) | , it follows that L n +1 ( k ) = | ˜ ω | + 2 k | σ ( n +1) ( A ) | , where ˜ ω = σ ′ ( n +1) ( A ) . Consider, then, the proper left factors A i of ω ′ ( n + 1) for i = 1 , , . . . , k + 1.Applying Proposition 5.1, the new maximal sum M + k is achieved at a time N ,where | ˜ ω | + ( k − | σ ( n ) ( A ) | ≤ N ≤ | ˜ ω | + k | σ ( n ) ( A ) | . UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } As c n ∈ o ( n ), we may choose k ≥ M + kc ( | ˜ ω | + k | σ ( n ) ( A ) | ) ≥ . If, however, we had M ≥ c L n , then we would wish to not greatly increase M compared to c n . In this case, let θ n = [2 , k, , . . . ], and pass directly to consideringthe word σ ′ ( n +1) ( C ) = σ ′ ( n ) ( A k +1 B k − C ) , as C is always a left factor of ω ′ n +1 = Ψ( ω ′ n +2 ) in this case. Then the maximal sumreached for this word is M + 1, but its length is (similarly to before) L n +1 ( k ) = | ˜ ω | + 2 k | σ ( n ) ( A ) | . We are now in the position of being able to increase the length of the word without increasing the maximal sum of M + 1, so as c n is divergent, choose k ≥ M + 1 c ( | ˜ ω | + k | σ ( n ) ( A ) | ) ≤ . After applying g twice (to skip past the next θ k > / m ( n ). Continuing in thisfashion, then, we construct a dense set of θ (as the initial string of partial quotientswas arbitrary). That the lim sups are actually one follows from the minimal choiceof k and that ∆ c n , ∆ d n are bounded.To prove the analogous statements where one of M n , m n is desired to remainbounded, one need only repeat the same arguments using θ n ∈ H (recall (19)) sothat the value p n is eventually constant.The statement of Theorem 1.2 applies as well to M n ( x ( θ )) and m n ( x ( θ )); theproof is simpler, in fact, as the map Ψ is not a concern, and the possible errorof one from Proposition 5.2 is not an asymptotic concern. This process is highlyamenable to diagonalization techniques. For example: Corollary 6.1.
Given a countable collection of sequences c ( i ) n and d ( i ) n , all of whichare divergent and in o ( n ) , such that c (1) n ≤ c (2) n ≤ . . . , d (1) n ≥ d (2) n ≥ . . . , there is a dense set of θ for which c ( i ) n ∈ o ( M n (0)) , | m n (0) | ∈ o ( d ( i ) n ) for all i .Proof. Apply Theorem 1.2 after using a diagonalization process to construct c n , d n , both monotone, divergent, and in o ( n ) such c ( i ) n ∈ o ( c n ) , d n ∈ o ( d ( i ) n ) . (cid:3) Many permutations of the above corollary are possible. For example, we mayconstruct a dense set of θ such that the discrepancy sums grow in both directionsfaster than any n − ǫ (but necessarily in o ( n ), of course!), or such that the discrep-ancy sums are bounded below, but M n (0) grows slower than all iterated logarithms(but necessarily divergent, of course!), etc. See Figure 4 for an example where (a) θ exhibiting very slow growth of M n (0); this portion of the graph will repeat 2 times with noadditional growth. (b) γ ( θ ) exhibiting very fast growth of M n (0); this sawtooth pattern will continue to climb byrepeating itself E (2 ) / Figure 4.
Two different extreme growth rates for θ and γ ( θ ).for both θ and γ ( θ ) we have m n ≥
1, but M n ( θ ) / ∈ o ( n − ǫ ) for any ǫ > M n ( γ ( θ )) ∈ o (log ( i ) n ) for all i . In Figure 4 we set θ = [2 , , , , , , , . . . ] . Using diagonalization techniques one may similarly find a dense set of θ suchthat lim sup i →∞ M n i ( j ) (0) c ( j ) n i ( j ) = 1 UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } for an arbitrary collection of divergent sequences c ( j ) n in o ( n ) for different subse-quences n i ( j ) → ∞ depending on j , and similarly for the | m n (0) | and a collectionof sequences d ( j ) n .Truly, beyond the constraints of (2), any asymptotic behavior desired is possible.7. Proof of Theorem 1.3
Suppose that(25) lim inf n →∞ M n (0) | m n | = r , lim sup n →∞ M n (0) | m n (0) | = r . That the set of accumulation points of the sequence is the entire closed interval[ r , r ] is direct and is left to the reader. Let an arbitrary finite string of partialquotients a , . . . , a N be given which determine θ i for i = 0 , , . . . , n −
1, and forconvenience again assume without loss of generality that p n = 0.Now let c n and d n be arbitrary integer-valued strictly increasing sequences suchthat ∆ c n and ∆ d n are in O (1) andlim inf n →∞ c n d n = ρ , lim sup n →∞ c n d n = ρ . Furthermore, assume that c > M (Ω ′ n ) = M and d > | m (Ω ′ n ) | = m .Continue the continued fraction expansion of θ in the following way: θ n = [2( c − M ) + 1 , d − m ) , c − c ) , d − d ) , . . . ] . Then Ω ′ n will see the sequence of M (Ω ′ n +2 k ) = c k and m (Ω ′ n +2 k ) = − d k ; thebounded differences ∆ c n and ∆ d n ensure that the limiting behavior is the same asthe limiting behavior along the subsequence of times | Ω ′ n | . Example 7.1.
Suppose that θ = [1 , , , , . . . ]. Then we begin computing thesequence of values M n (0) and | m n (0) | according to Proposition 5.2:(26) θ = [1 , , , , . . . ] p = 0 E ( a ) = 0 ( M, | m | ) = (1 , θ = [3 , , , , . . . ] p = 1 E ( a ) = 1 ( M, | m | ) = (1 , θ = [1 , , , , . . . ] p = 1 E ( a ) = 0 ( M, | m | ) = (1 , θ = [4 , , , , . . . ] p = 0 E ( a ) = 2 ( M, | m | ) = (3 , θ = [5 , , , , . . . ] p = 0 E ( a ) = 2 ( M, | m | ) = (5 , θ = [1 , , , . . . ] p = 0 E ( a ) = 0 ( M, | m | ) = (5 , θ = [7 , , , , . . . ] p = 1 E ( a ) = 3 ( M, | m | ) = (5 , θ k through θ k +4 , we will subtract 2 k + 1 from m while adding 2(2 k + 2) to M . We thereforehave ρ = ρ = 2, or lim n →∞ M n (0) | m n (0) | = 2 . See Figure 5 for this θ . Figure 5.
A specific θ for which M n (0) / | m n (0) | has limit two;refer to (26) and note the changes to M , m .8. Proof of Theorem 1.4
Lemma 8.1.
Suppose that f ( x ) is a step function on S with k < ∞ disconti-nuities, and denote V ( f ) the variation of f . Define S n ( x ) , M n ( x ) and m n ( x ) asbefore. As we have not restricted f to be integer-valued, define ρ N ( x ) = ( M N − m N ) ( x ) . Let n be such that q n ≤ N < q n +1 . Then for any x, y ∈ S : ρ N ( y ) ≤ ρ q n +2 ( x ) + a n +1 V ( f ) . Proof.
Consider the set { x + iθ } for i = 0 , , . . . , q n −
1. Choose 0 ≤ j < q n such that x + jθ is closest to y . Then the distance between x + jθ and y is no larger than q − n .For each discontinuity d i there are therefore at most a n +1 preimages of d i within thisinterval for time L = 0 , , . . . , q n +1 −
1. It follows that f ( x + ( j + i ) θ ) = f ( y + iθ )for all but at most k · a n +1 of i = 0 , , . . . , N < q n +1 . As j + i is less than q n + q n +1 ≤ q n +2 , the lemma follows. (cid:3) Assume that a i ( θ ) ≤ M for all i . Then (continuing with existing notation) wesee that for some C > θ (27) C n − ≤ | Ω ′ n | ≤ ( M + 1) n +2 . The lower bound is due to the exponential decay in the length of the interval ˜ I n (any C < I n +1 is less than half as large as ˜ I n at least halfthe time, with the n − I ′ = I , or θ > / a i ( θ ) ≤ M .while at the same time,(28) n − ≤ ρ | Ω ′ n | (0) ≤ nM ′ n = Ω ′ n +1 (corresponding to those θ n > /
2) and for the rest, ρ (Ω n +1 ) ≥ ρ (Ω n ) + 1, as UBSTITUTIONS AND -DISCREPANCY OF { nθ + x } E ( a ) ≥ θ n < /
2. The upper bound follows as E ( a i ( θ )) ≤ M/ i . Now, for any N let k be chosen such that | Ω k | ≤ N ≤ | Ω k +1 | . From (27): kC ≤ log | Ω ′ k | ≤ log( N ) ≤ log | Ω ′ k +1 | ≤ kC , for two constants C and C which do not depend on k . From (28):( k + 1) M ≥ ρ | Ω ′ k +1 | (0) ≥ ρ N (0) ≥ ρ | Ω ′ k | (0) ≥ k − , so ρ n (0) ∼ log( n ). The full theorem now follows from Lemma 8.1. Acknowledgements
The author is greatly indebted to many people for support and helpful conver-sations over the development of this paper. The original impetus for studying thisproblem came from a problem posed by M. Boshernitzan while the author was aPh.D. student at Rice University, while a rudimentary form of Theorem 1.2 arosefrom discussions at PRIMA 2008 during a visit supported by the University of NewSouth Wales. The author is currently supported by the Center for Advanced Studiesat Ben Gurion University of the Negev, where Barak Weiss has provided invaluablesuggestions on improving the clarity of an early draft. Of course, any mistakes orunclear passages in the current form are entirely the author’s responsibility.
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