Irregularities of uniform distribution

Acta Mathematica Academiae Scientiarum Hungaricae Tomus 37 (1 - -3 ) , (1981), 185--221.

IRREGULARITIES OF UNIFORM DISTRIBUTION

By M. STEWART (Boston)

w 1. Introduction

Let X be the unit interval with endpoints identified (a group via addition mod 1). Quantities greater than 1 will be considered as lying in X upon reducing them rood 1. Now fix 0 irrational in X and I ~ X any subinterval, and define S,(O, I) =

n - - I

= ~ xz(jO). By the Kronecker--Weyl theorem, Jim n-l(S,(O, I--x))= III, where- j = O n - - ~

]. ] denotes the length of an interval, and the limit is uniform in x. We say / - (JOlj~_l is uniformly distributed rood 1 (u.d. rood 1). The Weyl Criterion (effectively art approximation argument) then shows that for any f in C(X) and x in X,

n - - 1

{n-X Z f(x+jO)} converges to fxf(Y)#(dY), where # denotes Saa r measure on X. j = 0

We point out a consequence of the u.d. rood 1 of {jO}. Define the rotation Ro: X ~ X by Ro(x)=x+O, and let ~ be the Borel field on X. Another approximation argument reveals that the only R0-invariant functions in LI(X, ~, #) are the constants. Therefore if AC~, then RoA=A forces #(A)--O or # (At )=0 . We then say that R o is ergodic with respect to #. R o is then irreducible in the sense that X does not decompose into measurable R0-invariant subsets in a nontrivial way. The reason that the ergodicity of Ro is weaker than the u.d. rood 1 of {jO} is that the u.d. mod 1 forces # to be the only R~-invariant Borel probability measure on X. For let v also be R0-invariant, and pick fEC(X). Then ,(f)=ff(z)d,=

n - - 1

= f f ( x +jO)dv for all jEN. Therefore v( f )= f li~m (n-~s~_ ~ f (x +jO))dv=f(p(f))dv=

= # ( f ) . As a result, Ro is ergodic with respect to #, and therefore R o is termed uniquely ergodie. The results of this paper have to do with the u.d. mod 1 of sequences: arising from {S,(O, I - x ) } and the ergodicity with respect to normalized Haar measure (but not the unique ergodicity) of corresponding transformations called skew products, of which we shall give examples later.

To start, take c~ and fl in R and consider {S.~--nil}, where, form now on, S,, denotes S.(O, I -x ) . Let exp (2niy) be denoted e(y). If we define f : X-+C by

n - - 1

f(x) = e(ctZx(x)-fi) and f('~ = [[ f(x+iO), then f(")(x) = e(S.a-nfl). i = 0

Choosing, say f i=0, we have a functionfo(x)=e(az1(x)) with fo(n)(x)=e(S.O). Finally, let ~.=g(f("~), ~.,0=#(f0(")), and define

( l . I ) r~(O) = { t - - I / 1 : lira [a.,0l = 1}. IIn011 ~ 0

Acta Mathematica Academiae Sc~entiarum Hungaricae, 37, I98I

186 M. STEWART

Clearly any fixed fl would have given rise to the same F,(O). Also, the location of [ in X is immaterial in (1.1); only t=[I l matters.

The motivation for this definition is the following: Suppose for some x in X that

N

(1.2) lim N -~ ~ e(S,c~-nfl) = 0 N ~ o o n = l

fails. Then two things are true: i) {S,,e-n[l} is not u.d. mod i - - this follows from the Weyl Criterion; ii) there is a nontrivial measurable function h on X with h(x)=f(x)h(x+ 0). (Part ii) actually is equivalent to the failure of (1.2) for some x; see, e.g., [12, p. 779].) Since [h(x)l=[h(x+O)I, the ergodicity of Ro implies that lhl is constant; let that constant be 1. If * denotes complex conjugation, we have h(x)h*(x+nO)=f(")(x) for n>0 . Then by the continuity of L 2 translation

(1.3)

with the consequence that (1.4)

which is equivalent to (1.1). be shown in Section 4.

lira c~n = 1, IInOII ~ 0

l i m I~.l = l , I[ nO I[ ~ 0

That (1.3) is equivalent to h(x)=f(x)h(x+O) will

It has long been known that the analysis of 0 as a continued fraction could unlock the rood 1 distribution properties of {jO} and sequences related to {S,}; e.g., see [3]. Such analysis was furthered in [10], helping discover, for instance, values of 0 and t and the existence of intervals I of length t for which

N - - 1

lim I N -1 Z (S, rood 2)} fails even to exist. (Note that this is equivalent with the

nonexistence of the limit in (1.2) for (c~, f l )=( �89 0).) By way of the continued fraction expansion of 0, Theorem 1 describes the arithmetic nature of numbers t satisfying (1.1), while Theorem 2 describes the arithmetic nature of those t causing (1.2) to fail. (To date there seems to be no good conjecture corresponding to Theorem 2 for ~ irrational.)

Therefore, let O=[al, a2, ...] as a continued fraction, with convergents ~al, az, . . . , ak]=Pk/qk in lowest terms. The set of partial quotients of 0 is the set {a,}, i>=1. Also let [Ix[I--rain In-x] ; It "l[ gives rise to the standard Euclidean

?1

metric on X. Then define

i = I i = i ~ - ~

(t.6) F~ = (tEro(o): ]im= (bio 0 mod 1 = 0},

,(1.7) F}(O) = t]tEF~ Z Ibi[qiHqiO[] < ~ and ~ ilbi~l, < co}. i=l i=l

Convergence of the sum of (1.5) is understood with respect to H" []. This brings as to the first two results, proved for ~ = �89 in [10] and sketched for general a in [t 1].

THEOREM 1. I f 0 is irrational, then F~(O)C=F~(O)C=F~ for all a~Z.

~ c t a Mathernat ica Acadern iae Sc~en t iarum Hungar icae , 37, 1981

I R R E G U L A R I T I E S OF U N I F O R M D I S T R I B U T I O N 187

THEOREM 2. Suppose o~ is rational, ~=p/q in lowest terms (q# l ) ; and let A be the group of complex qth roots of unity. Also, define f(x)=e(~Zto, t)(x)). I f tEF~(qO), then for some f e d there is a measurable solution g (A-valued a.e.) to g(x)=f f (x )g(x +O).

In particular, if we put ~=l/q and ~-=e(j/q) for a fixed j between 1 and q, then we have from (1.2) that { (S , -n j ) rood q} does not assume all values between 1 and q with the relative frequency 1/q as n tends to ~. (Indeed, let 0 have unbounded partial quotients, and suppose tEF~(qO)-ZO, I=[0, t). Then there are an uncountable number of x such that the average value

N

lira U -1 ~ ' [(S,(0, I - x ) - n j ) r o o d q] N~r n = l

need not even exist; the proof is that of [10, Theorem 1]. It follows that, under N

these assumptions, lira N -1 Z ( S , rood q) does not exist either.) N~co n = l

Theorem 2 also has an ergodic-theoretic consequence. Form the space Y= X • A and let T: Y-~ Y be the skew product T(x, ],)-- (x + O, ~3f(x)), with 6 and f , as above; we have T"(x, y)=(x+nO, y(3f)r Then X has q ergodic components; that is, X decomposes into the q disjoint, measurable sets {(x, ~ig*(x))}xeX , where ~/ runs through A, and on each of these sets T is ergodic with respect to Haar measure. But again, if 0 has unbounded partial quotients and t~ZO, then the orbits under T are dense in X [10, p. 6]. Thus we see that such minimal (every orbit dense) skew products can fail to be ergodic. See also [7] for this type of result.

We now state some corollaries of the right-hand inclusion of Theorem 1; these were noted previously in [12].

THEOREM. For every irrational 0 and t-~[l] with t~ZO it is true for almost all ~ that (1.2) holds for all ft.

PROOF. Let A = { ~ : f o r s o m e tiER, (1.2) fails}. Then if aEA, we have {(b:0 rood 1} approaches 0, where the b, come from (1.5) for t. But the set of

causing {(b~) rood 1} to tend to 0 has measure 0 by the Riemann--Lebesgue lemma.

Note that for almost all ~ the previous result is true for all n~, nEZ-{0}. Using the Weyl Criterion, one obtains

COROLLARY. I f t(~ZO, then for almost all ~ the sequence {Sn~-n[3} is u.d. rood 1 for all [3.

Setting /3=0 it follows that {S,,~} is u.d. rood 1 a.e. (~), a fact which is contained in [14, Theorem 21]. Now setting f i= t~ we obtain the following:

THEOREM. I f 0 is irrational and t~ZO, then for almost all ~ the sequence {(S,,-nt)~} is u.d. rood 1.

This result implies that for 0 irrational and t~ZO, the sequence {S, , -nt} is unbounded. For if not, then there could not exist numbers ~ arbitrarily close to 0 for which {(S,-nt)o~} would be dense, much less u.d. rood 1.

Therefore we obtain a new proof of a result of KeSTEN [3].

A c t a M a ~ h e m a t ~ c a AcaderP.$ae S c ~ e n t i a r u m HuTzgar~cae, 37, 1981

188 M. STEWART

THEOREM. Let 0 be irrational, Ic=X any interval. I f there exists x in X such that { S , ( O , I - x ) - n ] I I } is bounded, then IIIEZO.

Kesten's theorem raises some ergodic-theoretic questions concerning certain skew products called cylinder flows, one of which questions is answered here. Again fix 0 irrational and I=C X a subinterval of length t. Let E be the closed subgroup of R (not X) generated by - t and 1 - t . Put Y = X • and define the skew product T: Y ~ Y by T(x, y)=(x+O, Y+(Z1(X)--t)). Then T"(x, 3:)= =(x+nO, y + ( S , - n t ) ) , so that if t~ZO, Kesten's theorem says T has no bounded orbits. Since T clearly preserves Haar measure on Y, it is therefore natural to ask whether T is ergodic. The following result shows that this is almost always the case, giving specific conditions on t ensuring ergodicity.

THEOREM 4. Let 0 be irrational, i) I f t~ F~ and i f 1, t, and 0 are rationally independent, then T is ergodie with respect to Haar measure on Y; ii) i f t is rational, then T is ergodic with respect to Haar measure on Y.

The proof follows that of [1, Theorem 4] in the main, but it also draws upon S CrIMIDT'S notion [8, Sects. 2, 3] of an essential value of a cocyle associated to an ergodic transformation. (In our context the cocycle is the "skewing function" Zx(x) - t , and the ergodic transformation is Ro.) Also of importance in the proof is the following number-theoretic result. Recall that {qk} is the sequence of denominators of the convergents of 0.

THEOrtI~M 3. Let be 0 irrational. Then t~ F~ if and only i f lira sup ([[qitH-

- � 8 9 l[q,O][) >0 . This result is itself of interest, for it comes as a bit of a surprise when compared

with the following fact:

PROPOSITION ([13]). Given a sequence {ri} of positive integers with ri+l/ri >- ~2/2:>2 for all i, there are an uncountable number o f t in X with [Ir~tl[<2/2

for all i.

Now from Theorem 3, suppose that 0 has bounded partial quotients, so that F~ and let denominators {qis} be chosen with lim (llqistll- �89

J = c > 0 . The qis can be chosen to satisfy qis+l/q~s:>2/2>2 by taking every other one of them if necessary, and the quantities �89 are bounded below by a positive number - - see Proposition 2.1 i) and vi). As a result of Theorem 3 we may sharpen the fact that Hq~stl]<2/2 for uncountably many t to O<e<[]q,stl]<2/2 for all but a countable number of these t, where e depends on 0 alone.

Let us give the right-hand side of the equivalence in Theorem 3 the name condition A. It is motivated by conditions A and A" of [1]; there it is simply noted that for each 0 the conditions are satisfied for almost all t.

Theorem 4 was previously known only in some special cases. For example, SC~MIDT dealt with t = �89 and 0 with bounded partial quotients in [9], while CONZE and KEANE in [1,15] treated t = � 8 9 0 irrational (and other cases via his conditions A, A').

The author wishes to express thanks to Professors William Veech and Frank Jones of Rice University for their friendship and guidance in the preparation of this work.

Ac~a M a t h e m a ~ i c a A c a d e m i a e S c i e n t t a r u m H u n g a r i e a e , 37, I981

IRREGULARITIES OF U N I F O R M DISTRIBUTION 189

w 2. Continued fractions I

X and 0 are as in the introduction, and we have 0<1 . Since 0 -1>1 , we may extract the greatest integer a~ in 0 -1. Then O-l-a~ is irrational and less than 1, so we may repea t the process of inverting and drawing out the greatest integer. Since 0 is irrational, this process is infinite, and we write 0 = [a~, a~ . . . . ] or O=l/al+l/a~+ . . . . Call a n the n th partial quotient of 0' and let the rational number [a~, az . . . . . an] be called the n tra convergent of 0, equal to p,/q, in lowest terms. For uniformity we adopt the following conventions: p 0 : O and q0=l , p _ l = l and q_~=0; llqoOIl=O and Ilq_x0[l=l.

We then have the following relations for all i = > 1 ; for proofs, see, e.g., [4].

PROPOSITION 2.1. i) p i = a i p i _ l + p i _ 2 ; q i = a i q i _ l + q i _ 2 . ii) O<pdql, i odd; O>pJqt, i even.

iii) q~OE(O, �89 i even; q~OE(�89 1), i odd. (We will express this re!ationship by saying that qiO and qi+xO have opposite senses.)

iv) Ilq~Ol[=lq~O-pd= min [Iq01l. l ~--q<ql + 1

v) ]O-p/q~l<(q~q~+~)-L

vi) q~lIq~O[lE(1/(a~+~+2), 1/a~+l); q,+ll[q~O[IE(�89 1).

vii) I f [qO-pi<l/2q and (p, q ) = l , then P=Pi and q=qi, some j.

viii) I f rE{l, 2 . . . . , ai} , then [Ivqi-lOIl =vllq~-lOl].

ix) I f vE{1, 2, ..., at}, then II(vqz-l+q~-~)Ol[ =llq,-~0ll-vllq,-lO[[.

Since the following find use often in the sequel, we present their proofs.

PROPOSITION 2.2. i) ~ a t 4 . 2 j + l q i + 2 j O = - - q i _ l O. j=o

ii) ~ at+~j+itlqi+2jOl] =llq~-lOll. j = 0

PROOF. From Proposition 2.1i) we have ak+lq~=qk+l--qk_l, SO that ai+2j+lqi+2jO=qi+~i+lO--qi+.,.j_lO. Therefore the sum in i) telescopes to -qi-lO, as desired. For ii) Proposition 2.1ix) implies that []q~+~OH=ll(a~+lqk+qk-OOll= =llqk-~OII--a~+lllqkOII. (Compare this with Proposition 2 .1 i ) . ) Therefore a~+~j+lllq:+zjOIl=llqe+v_iOll-Hqe+~j+lOll, so that the sum in ii) telescopes to llqt-101].

m

PROPOStTION 2.3. i) I f N= Z ciqi is a positive integer with ci nonnegative t = 0

integers for all i, then NO= ,~ C~llq, O[I- Z e, llq, OII. /=o i = 0

/even / o d d

ii) I f t= ~ b~qiO with biE{O, 1 . . . . . ai+1} for all i, then [Itll<=lIq~OIl. i = r + l

Acta Mathematica Aeaclemiae Seient iarum Hungaricae, 37, 1981

190 M. STEWART

PROOF. From Proposition 2.1 iii) and the definitions it follows that for all in m

i>=O, q,O=(-1)*l[q, Oll. Therefore :NO= ~ ciqiO= ~ (-1)*e, llq, O[I, which yields i). / = 0 / = 0 �9

As for ii), we note that Proposition 2.1 iii) also says that the contribution of successive

terms in z~ b, qiO, if nonzero, have opposite senses. Therefore i = r + l

= max 1

max [ ~ a~+ 2j 1[ q, + 2j-101l, ~,j=l

{j~--1 b,+ej-l[lq,+2j-lOl[, ~ b,+2sllq,+ejOH) ~- j = l

a,+2j+xllq~+mjO[I) = max(llq,0ll, llq~+~0[I) = ]]q,01l. j = l

PROPOSITION 2.4. Each interval [r/qk, (r + l)/qk), O<=r<qk, contains an element of {iO}, O<=i<qk.

PROOF. From Proposition 2.1iv) we have that ]iO--(iPk)/qkl=(i/qk)ilq~Otl, O<-i<qk. Thus if k is even (resp. odd), the first qk multiples of 0 all sit just to the right (resp. left) of {i/qk}. (This uses Proposition 2.1 iii) and v); each of the multiples of 0 are within 1/qk+ ~ (on the right if k is even, on the left if k is odd) of the nearest i/qk.)

COROLLARY 2.5. i) I f k is even (resp. odd), then the element of {iO}, O<=i<qk, closest to J/qk, O<=J<qk, on the right (resp. lef t ) is ijO, where j=--ijPk mod qk.

qk--1

ii) ~ ' liO--(ipUqk)l=�89 i = 9

PROI'OSITION 2.6. For nEZ, IIn~ll =lln" I1~11 I1-

PRooF. lln" 11~11 II =lln (min ]m-~])ll =lln(m0-~)ll =11 +(nmo-n~)ll =lln~ll,

w 3. F~(O), F~ and the canonical representation

In this section we produce two infinite series representations for certain elements t of X. The first series appears in (1.5) and is valid under the assumption that t satisfies (1.4). The second series is valid for t~ZO, and it is of a more elementary nature. Its primary use is in Theorem 3, but we shall note some similarities it bears with the former expansion.

Therefore let us assume first that t satisfies (1.4). The following was developed in [10], but the reader is warned that the notation there is quite different from ours.

Let {m j} be the sequence of (not necessarily positive) multiples of 0 that best approximate t mod 1; i . e . , ]ImjO-t[[= min I]mO-tll. It follows from

lml<lms+ll [10, Lemma 11] that for large j, mj+l--mj=ejqkj, e I = + 1 . Then if k is also large and qk occurs in this fashion bk times, the occurrences must be successive, say qk=qkj=qkj+l . . . . . qkj+b~--l' and 8j=Sj+l=:. .=ej+bk_V Therefore we

Acta Mathema~ica Acaclemiae Scient iarum Hungarieae, 37, 1981

I R R E G U L A R I T I E S O F U N I F O R M D I S T R I B U T I O N 191

er k

may write t :-mO+ z~ ~jbjqjO. For k>=jo let nk denote the quantity m + z~ ejbjqj . J=Jo J=Jo

In particular, each nk is some mj. It also follows that if • is preassigned and positive, then there is an index Jl such that if j>J l and bj+l~O, then ]nji<6qj+l. Let k > j be the next index for which b ~ O . Then also qk<[nkl<6qk+l, and Ink ]/bk q k : 11 + nj/bk qk[, where [nj/bkq k[ < ~(qj + 1/qk)'< 3. Therefore k~ l~,m ~o t nk/bkqk I ~---

=1. We also have {Ink[qkO}-*O; as a result,

(3.1) lira b k qk [I qk Oil = O. k - - ~

Therefore, to prove the right-hand inclusion of Theorem 1, it is sufficient to prove that {(bk~) mod 1} tends to 0.

We have that F~(O)___ F~ In [10] it was also noted that i) F~(O) is a subgroup of X (Lemma 13 ; the proof is for ct = �89 but works in general), and ii) any two expansions for t, say,

t = mlO+ ~ biqiO and t = m20+ ~ ciqiO with lim biqil[qiO[[ -- lim ciqel[qiO][ = 0

satisfy bi=ci for all i~io, some i0 (Lemma 14). This series representation for tEF~ enables us to prove

LEMMA 3.2. I f tEF~ then lira Ilqktll=0. k ~

co

PROOF. Write t=mO+ z~b~qiO as in (1.5), and pick ~5>0. Let k 0 be so large i = 1

that k~_ko forces ]b~IqkllqkOl[-<~ and Inkl<~qk+~. Then

f:? El) +qk,i=k Ib~l "llq~O <= 6+~( l+qk/qk+l+. . . ) .

Now since q~+2>2q~ by Proposition 2.1 i), the last series in parenthesis splits into

the series Y~qk/qk+2i-~ and ~ qk/qk+2i, each of which is dominated by 2. Therefore

IIq~tll<5/~, and the lemma is proved. Note that this lemma proves the " i f" statement of Theorem 3. We are left

with proving that tr F~ satisfies condition A (which in turn will make an equivalence out of Lemma 3.2). To this end we introduce another infinite series representation. Now, instead of assuming tEF~(O), we assume t~ZO.

LEMMA 3.3. One may expand any positive integer N as

(3.4) N = Z, ciq i, i = 0

where m depends on N, and for all j we have i) 0 = c j = a j + ~ (we also require O<=co<a~) and ii) /f c j+l=aj+~, then c j=0 . I f we also require cm>O, the expansion (3.4) is unique.

A c t a M a t h e m a t i c a AcacIemiae S c i e n t i a r u m Hungar icae , 37, 198I

~92 M. STEWART

PROOF. N = I may be written as lqo, and any N<q~ may be written as Nqo. N=q~ may be written as lqa, and these are all obviously unique.

j - - 1 Now we claim that for all j no number N~qj can be expanded as ~ , ciqi

i = 0 subject to the given conditions. By the preceding paragraph, we are done for j : 1. Assume now the claim is true for all j~k . Then

ciq i <= max ai+~qk+max ~ , ciql, (ak+l--1)qk+max ciq~ = i = 0 i = 0

= max (ak+lqg+(qk-1-- 1), ag+lq k - 1) = qk+l-- 1.

With the claim established, we show that the expansion (3.4) exists and is unique for qj<N<=qj+~, all j=>0. We are done for j = 0 . Suppose we know the result for j<=k, and let qg N=qk+~. If N=qk+l, then in view of the above claim, the unique expansion is lqk+~, so we may assume that qk<N<q~+l. Write N-:NlqI,+Ne with N~ and N2 nonnegative, and N1 maximal. Then N2<qk,

k--1

:so let its unique expansion be N2= Z c~qi. Then a valid expansion for N is i = 0

k--1 k- -1

Naqk+Y~ clq i. It is also unique, since if N:Noqk+ Z. diqi with No<Nx, then i = 0 i ~ O

~, diq~=M is a valid expansion for M~qk , which is a contradiction. The lemma ,i= 0

is proved. In (3.4) let ci=c~(N), and extend to all nonnegative integers by defining

c~(N)=0, i>m. Then for every positive integer N we have a well-defined infinite sequence of nonnegative integers, of which only a finite number are positive. I f Nr the sequences {c~(N)}, {ci(M)} are d is t inct . Now define functions bi on Z+0 by setting bi(NO)=ci(N).

LEMMA3.5. I f t~ZO, then lim bi(NO)=bi(t) exists. I f bi+l(t)=ai+2, N O ~ t

:then bi(t)=O.

PROOF. We induct on i. First define M(c )= {NO: co(N)=c }, O<:c<=a~-l. Then

M(0) ~ I(0) = (~_, a2iq2~-lO, Z a21+lq~iO] = (-qo O, ~ ql O~ f = l ]

M(c) c= 1(c) = [[cqo+(a~-l)qdO+ Z. a~q~.O, eqoO+ Za~,+lq~iO] = k i=2 i ~ l ]

= ((-q~+(c-1)qo)O, (-ql+cqo)O), 1 <- c <= a~--l, where I ( 0 ) = X if a l = l . The estimates for the endpoints of I(c) are greatest possible in view of Proposition 2.3 i), and the infinite sums are evaluated by Pro- position 2.2 i). 1(0) contains 0, and moving from left to right across [0, 1) we encounter I(1), 1(2) . . . . . l ( a z -1 ) in order. That the right endpoint of I(a~-1) is the left endpoint of 1(0) is clear from Proposition 2.1 i). Thus the I(c) are disjoint

a l - - l _ _

for O<=c<=a~-l, and [.J I(c)=X. Finally the common endpoints of the 1(c) are i = 0

.Ac ta M a t h e m a t , ica A c a d e m i a e S c f e n t S a r u m H u n g a r i c a e , .77, i981


all in Z0. Therefore lim bo(NO) must exist, for if {N,O} and {MiO} converge NO~t

to t and bo(Mi) =-m, bo(N~) ~n with m ~n, then t must be an endpoint of an I(c), contradicting t ~ Z0.

To see that lim bl(NO) must exist, define M(c(~ ~i), NO~t

i=O, 1}. Immediately we have

a 2 a2--1

M(c ~~ = 0 M(c ~~ c ~ for c ~~ = 0, and M(c ~ = U M(c ~~ c(1)) for c (~ >0. c(1)=0 c(1)=0

Using estimates analogous to those above, one may find disjoint intervals I(c ~~ c o)) a

containing M(c ~~ c ~1)) such that U 1(cr176 c(1))=1(c(~ where a=a~ or a 2 - 1 c(D=O

according to c(~ or c(~ The endpoints of each I(c (~ c (~)) are again in ZO, so that, as before, lira (b~(NO)) must exist for tf[ZO, but b~(t)=a2 forces

NO~t

bo(t) =0. The inductive step is now obvious, and we omit it. The lemma is done.

LEMMA 3.6. I f b,(NO)---b,(NO), O<=i<-R, then llNO-~OlI<=2/q~,

PROOF. NO = ~ b,(NO)q,O and -N0-- ~ , bi(NO)qiO, so that i = 0 i=O

IINO-N0[[= ] =RZ,+ t ' b,(O)q,O= , =~R+~ b,(NO)q,O[[<2 max [], =R~+ ~ c ,q ,O<2/qR,

by Proposition 2.3 ii).

LEMMA 3.7. i) For all tr we have

(3.8) t = ~ b,(t)q,O, i = 0

with b,+l(t)=ai+~ forcing b,( t )=0, and bi(t)C{O, 1, ..., a/+l} for all i. ii) This expansion is unique with respect to these properties.

PROOF. Given e>0, choose an integer R > 0 with 2/qR<e/2. Then choose N > 0 with [[NO-tl[<e/2 and b~(t)=bi(NO), O<=i<=R. We have

[ i t -< NOlJ+[NO--,=Zoo ( )q0[[ e / 2 + / 2 = l i t - - b i t i < e = 8 .

Letting e tend to 0, i) obtains. As for ii), suppose t = ~ b,q,O with the same i = 0

R

properties. Let NR= ~ b,q,. Uniqueness in Lemma 3.3 implies that ~i=ci(NR), i = 0

O<-_i<=R. Thus b~(NrO)=~. As R - ~ and N~O~t, continuity of bi at t forces b,~b,(t), all i. That is, bi=b,(t), all i.

DErINmON. The expansion (3.8) is called the canonical representation for t r ZO. (See also [2, 6] for other applications of this representation.)

13 A c t a M a t h e m a t i c a Acac lemiae S c i e n t i a r u m Hungar icae , 37, 1981

194 M. STEWART

Suppose 0 has unbounded partial quotients and tEF~ so that t has

both the representation mO+ ~ biq, O from (1.5) and the canonical representation i ~ l

7_~eiqiO. Note that {biq,]]qiO[[}-*O is equivalent to {bi/ai+l}-*O, by Proposition i = 0

2.1 vi). Therefore, by the uniqueness of each of these representations, they are eventually the same if b~_->0 for all large i. One might ask more generally if there

is a simple criterion for t = ~ elq~O (canonically) to belong to F~ Lemma 3.2 i ~ 0

is a beginning to the answer. Further, the cases into which Theorem 3 falls for 0 with unbounded partial quotients all restrict the behavior of {e;} slightly. How- ever, it is hard to imagine a concise criterion for the solution. For example, by Case A2 of Section 10 it follows that {e~/a~+~} can cluster only at 0 and 1, but one easily constructs counterexamples to the converse.

w 4. Compact abelian group extensions of R o

In this section we prove Theorem 2, assuming Theorem 1. (The proof of Theorem 1 is contained in the two following sections.) Theorem 2 gives conditions under which a certain functional equation has a measurable solution; the con- sequential failing of ergodicity of a corresponding skew product was noted in the introduction. (Henceforth ergodicity with respect to Haar measure is implied.) Before the proof we present two results of a similar nature which show that Theorem 2 fits into a rather broad scheme. In the following, G is a compact abelian group and f : X ~ G is a measurable function. On the product space Y=X)<G define the skew product T: Y ~ Y by T(x, ~) = (x + O, (f(x)). T preserves Haar measure, so one may ask if T is ergodic. A classic result in this direction is the following (see, e.g., [2, Lemma 2.1]):

PROPOSITION 4.1. With notations as above, let G=X, and let f be any measurable fimction. Then T is ergodic if and only if there is no nontrivial measurable solution g to (4.2) g (x) -- f k (X) g (X + O) for any k ~ Z - {0}.

PROOF. The characters { ~ } , kEZ, span L2(G). Therefore, suppose T is not ergodic. Then there is a function hCL2(X• with Th=h everywhere, h not constant a.e. (x, ~). (Here Th (x) = h (Tx).) Then we write h (x, ~) = ~ , ek (x)~ k,

Z

with CkEL2(X) for all k, s o that~ek(x)(k=~Ck(X+O)(kfk(x). Necessarily ek(x)= =ek(x+O)fk(x) for some nonzero k and nontrivial ek(x), so that (4.2) is satisfied. Conversely, if (4.2) is satisfied, then h(x, ~)=g(x)~ k is nonconstant, measurable, and T-invariant.

Virtually the same result is true, and the same proof works, if G=A = the complex qt~ roots of unity. Here r the character group of G equals {r l<=k<=q, and the requirement that (4.2) fail for k ~ Z - { 0 } is replaced by the failure of (4.2) for k~ {1, ..., q - l } . In turn, if q is prime, it is equivalent to require

A c t a M a t h e m a t i c a A c a d e m i a e S c i e n t i a r u m Hungar icae , 37, I981

IRREGULARITIES O F UNIFORM DISTRIBUTION ] 95

(4.2) to fail for k = l alone. Specializing f(x)=e((p/q)Zto, o(x)) for (p, q ) = l , we obtain

PROI'OSITION [10]. Let a be rational, a=p/q in lowest terms, and let G=A = =the group of complex qth roots of unity. Set f(x)=e(c~ZEo, t)(x)). Then i) T is ergodic if and only if (4.2) has no nontrivial solution for kC{1, 2 . . . . , q - l } ; ii) i f q is prime, then T is ergodie if there is no nontrivial solution to (4.2) for k= 1 ; iii) for any q, a nontrivial solution to (4.2) exists for k= 1 if and only if t satisfies (1.3).

In iii) above, (1.3) obviously holds if there is a solution to (4.2) for k = 1. That the converse is true will follow from the line proof of Theorem 2.

For Theorem 2 we continue with all the notations of the above proposition, i t - - 1

as well as tCF~(qO). Set ~=qO, and let f(")(x)= ] I f ( x + i O , a,,=#(f(")). With

~ so defined, (1.4) is true for t. (This uses Theorem 1, which we are assuming.) Recall that Theorem 2 states that if t~F~(O, then there exist 5CA and g: X ~ A measurable with (4.2') 6f(x) g(x+ 0) = g(x).

LEMMA 4.3. I f t satisfies (1.4), then there is a continuous function h(x, .) from X to L2(X) (with a measurable representation h ( . , .) on X • such that for all x and w it is trite for almost all y that h(x, y) is in A and

(4.4) h(x+w, y) = h(x, y+ qw)h(w, y).

PROOf. Define h0: {n~},~_l~L~(X) by h0(n~, . )=fcq,)( . ) . Then

(4.5) ho((m+n)~, y) = h0(m~, y+ nqOho(n~, y).

Now given e>0, let W be a neighbourhood of 0 in X such that [ a , l > l - e whenever m~E W. Then a compactness argument shows that there exist yEA and q>0, r/ depending on q and e, and r/ tending to 0 as e tends to 0, with l~{x:f~")(x)~y}<~?. Then, by (4.5), l t {x: f (q ' ) (x)=l}>l-ql l . This implies IIl-h0(m~, o)11~<21/~, so that h0 is uniformly continuous from {n~} to L~(X). Therefore there is a continuous extension h from X to L~(X). However, as defined, /~ may not be measurable in X• To remedy this, let {P"}, n ~ l , be a sequence of partitions of X satisfying p ~ . p 2 ~ . . . , and lim (mesh P " ) - 0 . Let {P/'},

n ~ r

l~i<=d,, be the distinct elements of P", and pick m~CP~ for l<-i<=d,. Next, a.

set h,(x, y) = Z ZeT(x)ho (mT~, Y). Finally, arrange that II h(x)-/~(x')l13< 1/n whenever i= 1

x and x' lie in the same Pp. Then {h,(-, -)} is a Cauchy sequence in L2(X• with limit h, say. By altering it on a set of measure 0 if necessary, the function h is now seen to verify all the requirements of the lemma.

RE~ARK. The converse of Lemma 4.3 is true, as the proof of [10, Proposition 6] shows.

t3" A e t a Mathema~ica Acaclem~ae Sc~en t farum Hungar tcae , 37, 1981

196 M. STEWART

Theorem 2 now follows quickly. We have a.e. (x) q--1 q--1

h(~, x) = l-[ f(x+i~) = t~ h(O, x+i~) i = 0 i=O

q q

h(~, x+~) = I[f(x+i~) = l I h(O, h+i~), f = l i = l

so that f(x)/f(x+q~)=h(O, x)/h(O, x+q~). This says f(x)/h(O, x) is invariant under Rq;, which is ergodic on X since q~ is irrational. Therefore, for some 3CA, 6f(x)=h(O, x) a.e. (x). Now pick 2 in X with q2--O. Putting

q--1

(4.6) g(x) = 17 h-(i+l)(L x+iO), i = 0

one immediately has g measurable and g(x)/g(x+O)=h(O, x)=6f(x) a.e. (x), which is (4.2"). Theorem 2 is proved.

ReMaRK. That (1.3) implies the existence of a solution to (4.2) for k = 1 in the proposition from [10] previously mentioned now follows by proving the existence of h: X~L2(X) satisfying

(4.7) h(x+w, y) = h(x, y+w)h(w, y)

a.e. (y) for fixed x and w. To do so, define h0: {nO},,~I~L~(X) by ho(x,y)= n--1

=f(") (y)= I[f(x+iO), and argue as in Lemma 4.3. Then set x=O in (4.7). This i = 0

formula must hold a.e. (w) for almost all y by the Fubini theorem. Fixing such a y, h(w+O, y)=f(y+w)h(w, y). Then (4.2) holds for g(w)=h*(w-y, y), and g is nontrivial.

Now we offer an explicit construction of both g and 6 in Theorem 2. Assume rt--J.

tEF~(~). Since (4.2') holds, we have tEF~(O). (Here f(")(x)= I[f(x+iO).) By i = 0

Theorem 1, which we are assuming, tCF~ so that t=kO+ ZejbjqjO, as in j = l

(1.5). Let {Qk} be the denominators of the convergents of ~, and recall that :llqjOll <l/(2qqj) for large j by (3.1). For all j one of the following applies.

Case I. qj=rhj with r[q and r maximally so. Then lhj~-q/r(pj)l=q/rHqjOl[< <q/(2hj). Then (hj., q/r)=1 forces h i to be a denominator of a convergent of ~, by 2.1 vii).

Case II. (qj, q ) = l . Then lqj~--qpi[<l/(2qi). In this case, q1 is a denominator for ~.

REMARK. Clearly Case II fits into Case I with r = 1, but the distinction between r = l and r > l will prove notationally convenient.

Since tEF~(~), t=h~+ Z6jcjQj~. But from the above we also have j = l

t = kO+ Z ~j(rbflq)(h~) + Z ~j(bi/q)(qj~). I n

A c t a Mathe~7~atica A c a d e m i a e S c i e n t i a r u m H u n g a r i c a e , 37, 1981

IRREGULARITIES OF U N I F O R M DISTRIBUTION 197

By the remark preceding Lemma 3.2, these last two series for t eventually agree. In particular, if Case lI occurs, then q2]bj, large j. If this happens, we write b~=qdj, dj=O(q). Now

(4.8) t = mqO+ ~ e~biq~O, i = 1

where qZlbj, large j, whenever qj is a denominator for ~. As in Section 3, we set J

nj=qm+ ,~ eib~qi. i = 1

NOTATION. ~p/x) will denote e(c~Zo.~o)(x)). Ao=qmO, A i=njO for j=>l. We have the following relationship, which is straightforward to check:

(4.9) (p,+b(X) = (q~,(X)~Pb(X-- a))/q~b(-- a). qn--1 qn--1

Suppose e = +_+_ 1. If we let w (qne) denote ]-]" ~o~* ( - ieO) = f f 9* ( -ieO), thell i=O i=1

by (4.9) we have qn--1

%.~(x) :- w(qn~) 1[ q,~(x-iW). i = 0

Now put

(4.10) qn-- 1

H~.~(x) = 1[ ~o~(x-ieO), i = 0

(4.10 g..~(x) = H~.~(x+ �89

Then by (4.9) for an infinite sum (namely the sum (4.8)), we have

f(x) = (p,,,(x) F[ [9,jbs,j(x--a~_a)/q~,sbjqs(--Aj_l)]. y=1

(This product converges for all x # 0 , t, because 9a~ multiplied by the k th partial product yields ~o,~ by (4.9), and clearly {q3,,,.(x)}~f(x) for x # 0 , t.) Next, let m=ne, with n>0 , e= ___1, One may also immediately verify

LEMMA4.12. 9~,,(x)=w(qne)gq,,(x+O)/gq,.(x). Now write

n--1

gJ = g~jb~qj, Hj = H, jbjoi , wj(n)= 1[ (p*~(--Aj_l-iejO). i = 0

Repeatedly using Lemma 4.12, we obtain

(4. t 3) f (x) = w (qne)gq,, (x + O)/g~,,, (x) ~ [gj (x - A j -1 + O)/gj (x - A j_ a)] (wj (bj q j)). j = l

Therefore, formally defining

(4.14) g* (x) = g~,~ (x) / ] gj ( x - Aj _ 1), j = l

(4.15) 6 = w* (qn~) / I w~. (bj @, j = l

A c t a 5 , I a t h e m a t i c a A c a d e ~ i a e S c i e n t i a r u m H u n g a r i c a e , 37, 198~

198 M. S T E W A R T

we may formally write (4.2') from (4.13). To validate this construction, we will show that (4.14) converges a.e. (x). Then, since (4.13) and (4.14) converge for a common value of x such that (4.14) also converges for x+O, it will follow immediately that (4.15) converges. That (4.14) converges in L ~ and a.e. (x) is a consequence of

LEM~A4.16. I f t6F~(~), then for alt large j with bj#O

[ I n j - l l h ~ 2b~qjllqjOl].

PgooF. For large j, if bj #0 , then either Case I or Case II holds. Assume Case II holds; Case I is treated similarly upon replacing qj with hj and dj with b i.

Group the factors of (4.10) q at a time as follows: if O~=i~=qj-1, the factors in a group have subscripts {i, i+qi .... , i+(q-1)qi}, ..., { i+(dj-q)qj .. . . , i+ + ( d i - 1 ) q j}. The corresponding products will be equal to 1 except on a set of measure at most qllqjOl]. There being diq ~ such groups, the Iemma follows.

oo

Since t6F~(~), we have ~b~q~llqjO[l<~. But then, using Lemma 4.16 j = l

--I[[1-JIH~t I ~<-2~bjq~ilqjOl[<~.~ Therefore (4.14) converges in L ~ and a.e.(x).

As noted previously, the expansion for 6 must be valid. (There is a straightforward proof of this fact as well.) The construction of g and 6 is complete.

w s. r,(0) ~ r o(0)

Choose tEF~(O); that is, t satisfies (1.4). We assume ~E(0,1), since the inclusion is true simultaneously for all co+n, nCZ. Also, by (3.1) we have F~(O)C=

~F~ so that already any t satisfying (1.4) has an expansion t=mO+ ~ ~jbjqjO j = l

lira (bjc 0 rood 1~0. with lira bjqj[]qjO[]=O. It is enough to prove ~-~ j ~

To this end define ~p~ and ~ (a in X) functions from X to C by 9 , ( x ) = = e(c~Xto, o~ (x)), ~, (x) = 9* (x) = e(c~X[~, 1) (x)). By convention, q0(x) ~- 1 ~ ~ko(X ). We

n--: t

have f(x)=~ot(x). Also define 9~"~(x)= l[q~_~o(x)~/t_io(X). Since q~") and f(") i ~ 0

have the same discontinuities and the same jumps at these discontinuities, we have that q~(")=e(-cO(fc")(O))*f ~'~ by comparing values at x--0. It follows that if fl,, = p(cp(")), then (5.1) lira ]fl,,l = 1.

n 0 ~ 0

For n > 0 let 6, denote the length of the shortest interval of constancy for f("), and define z, and L ( I L [ < n ) by [[•O-t[[=z,,= rain [[iO-t[[. It is proved

li[<n IJ . l

in [10] that as nO~O, n6,,~O. Letting 2=sgn (j,), define ~(")=/-/~ola0~b(u_,)0 i = 1

and A = {n: 6,=z,}. It is also proved in [10] that

(5.2) llq~ (") - Ot_s.,,o ~(")11t <- 2 ( n - 1)z,,

A c t a MaShemat i ea Acaclerniae Se ien t~arum Hungar f eae , 37, 198I

IRREGULARITIES OF UNIFORM DISTRIBUTION 199

which tends to 0 as n ~ , n~A. Defining 7,=/~(~k(")), we obtain from (5.2)

(5.3) lim I~.1 = 1. nO~O, n ~ A

The following lemma is stated without proof.

LEMMA 5.4 [10, Lemma 2]. Given 6 > 0 and 0 < a < z < l , there exists an index Jl such that if J>=Jl and n~(aqj+l, zq~+l), then n~A and ILlollqflll<&

Now let bj # 0. Note that if j is large, only one integer r can satisfy lrqi--njl < k

<6qj, where nl,=m+ ~ e i b j was defined in Section 3. That integer r must j = l

equal ejbj, large j, since {ejbjqj-njI=lnj_ll<6q~. Therefore the fact that {(bj~) mod 1} tends to 0 will be a corollary of the following lemma. Recall that {ink} is the sequence of multiples of 0 that best approximate t.

LEMMA 5.5. I f 0 < e < : l and 3>0, there exists an index j~ such that if J>=J2, qf<]mkl<eqj+l, and [lmkO--t[l= min IlhO-tl[, then the sequence of integers

I~l<~qj+l dj satisfying (5.6) Imk--d r qjl < 206qj

also satisfy !im {(die 0 mod 1}=0.

PROOF. Suppose 6 < m i n (�89 c 0. In Lemma 5.4 let "c=~, a=6, and use 6/2 for 6. Then, if [-] denotes the geates t integer function, Jt,qj+lj = Iraqi, so that either lrnkl<6qj+~ or Imd.llqjON<6/2. The latter is impossible, however, if lmkl>=6qj+~, so we may assume qj<[rnkl<6qj+l. Let ~>0 be such that IInO[l<e forces n6n<6, and nEA forces 1~.1>1-6. For large enough j the interval (6qj+~,~qj+~) contains an integer n with lIn0ll<e. By Lemma 5.4, nCA and ]lmkO--tl] =6,=z,,. Since n+]mkl<qj+~ , we see that llqj011 is the shortest interval of constancy for 0 ("). Define a=>l by Imkl=aqj+r, where O<=r<qj. Clearly a = bj or b j - 1 for large j. The proof now breaks into two cases.

Case I. ~ rational, e =p/q in lowest terms.

SubcaseA. a~O(q). Let ~/k=sgnrnk; then t/k=sgnjn, n having been chosen r

above. Define ~(0")= l~Ph~h,k--,, where p~ and ~p~ denote P~o and ~0 , h = l

respectively. Also let a=bq. The Imkl p-discontinuities of ~(") gather themselves into r bunches with

(bq + 1) in a bunch and (qk--r) bunches with bq in a bunch. The first r bunches of discontinuities are located at ~lk(i+hqj)O, h---0, 1, ..., bq-1, i=1 , 2 . . . . . r. The other bunches are located at qk(i+hqj)O, h=0 , 1, . . . ,bq-1, i = r + l , ..., q3" If r = 0 , then there simply are no bunches of the former kind. Now assume j. is odd; j even is treated similarly by switching the words left and right in what follows. Immediately to the right of a f-bunch, is the corresponding 0-bunch; between this bunch and the next f -bunch on the right the value of ~b(") is 1. Therefore, any x with ~p(")(x)#~o(n)(x) lies between the shorter arc joining the endpoints of one of the 2r bunches of ~0 (n). As a result {x: O(n)#O0(") } has measure less than

A e t a M a t h e m a t i c s AcacIemiae ScSen t iarum H~ngar icae , 37, i981

200 M. STEWART

21mkl-IlqjOll <2a. If we let a~-")=/Z(~o(~)), then for the n previously chosen we have

( 5 . 7 ) I @ ) l > 1 -

Now the number of factors of ~(o ") is 2r--=21mk--~lkaqj I. Were ]lq2-1011 the shortest distance between discontinuities of ~(0 "), we would have }mk-~aqi } �9 �9 lIqj_lOll<5,V2, or ]mk--~lkaqjOl<5~qj , which implies (5.6). While []q2-10[I is

not the shortest length, �88 bounds the shortest length from below; hence (5.6) follows.

To compute the shortest length of constancy for r (~) first note that anything shorter than Ilqj-lOll must come from a term such as ~oc~a_,. The range of c-d+r~ is M = { n - m + l .. . . , n + m - 1 } . The cardinality of M is less than 2qj, so that in M there are at most two numbers of the form uq) and two of the form vqj+qj_l.

We may further require (raising J2 if necessary) that nE[�88 1/3 qi+1] and similarly for all m CM (also lowering ~ if necessary). Note that lind" Ilqj 011 < ~ forces qjllq2Oll <~, or qj<2~qj+l. From uqj> �88 qj+l we obtain IluqjOll >�88 (qj+dqj)llqjO[I > > l/(8qj)= �88 l]qj-~Oll. From vq]<~qj+~ we obtain I] (vqj+qj_OOll=l lq~_~Oll - - vllqj 011 > 1/(2@ - 1/(3@ = 1/(6q j) > 1/311 q j-~ 0II. All of this implies (5.6).

Subcase B. a - - l ( q ) . The q~-discontinuities of ~(") bunch together at 11k(i+hq~)O, h=0 , 1 . . . . , a , i=1, . . . , r , and at rlk(i+hqj)O, h=0 , 1 . . . . , a - l , i=r+ 1, ..., qj. (The same convention is adopted as before in case r=0. ) If we now

q2 define ~1(")= / / Oh~q)h~k_, and 6 (~)- ,,z,t,(,)'~ 1 - -~wl ~, we similarly arrive at [~")1>1-56.

h = r + l The rest of the argument proceeds as at the end of Subcase A.

Subcase C. a=-c(q), cE {0, -1}. We show this subcase to be impossible. The discontinuities of ~(~) are just as in Subcase B. Therefore, defining

h = l / X h = r + l ]

we again have ~ ) and ~b (") agreeing outside the bunches. Letting J~")':F(~b(~n)). we obtain (5.8) I~")l > 1-56.

Now from Subcase A we have that the 2q~ intervals between bunches o f discontinuities of ~2 (~) have measure greater than �88 each. Let the union of these intervals be divided into two sets U and V, U containing the qJ intervals between each of the q~ bunches and the corresponding ~ _ , bunches, and V containing the other intervals. Now p(U)>�88 and /z(V)>l/8, as well. But ~b2(")lv=~k(")lv=l and @)lo=~<")lo assumes two values (e(act) and e((a+l)a)), neither of which are equal to 1, by assumption. This contradicts (5.8), however, for small enough 6.

Case II. c~ irrational. Suppose we fix e > 0 and an integer j3>0 so that j>J3 forces lejb#q~-n~[<eqj. Then, writing lm, l=aqj+r with O<=r<qj, it follows that r/qj<e if a=b# and r /q~>l - e if a=b#- l . Define s = q ~ - r (resp. r) if a=bj (resp. b# - l ) . We have s / q j > l - e . Recall from Subcase C above: the function ~b~") and the sets U, V. Then ~(") assumes the value e(b~a) o~

A c t a Mathema~Aca Acadern iae Sc~entiaruvn Hungar icae , 37, 1981

I R R E G U L A R I T I E S OF UNIFORM[ D I S T R I B U T I O N 201

s of the qj intervals of U (each of length greater than 1/(Sqj) from Subcase A), the value e(Be), B=bj+_l, on the other q j - s intervals of U (each of length less. than 7/(8q~)), and the value 1 on V (a set of measure at least 1/8). Also, from Subcase A, it is clear that I z (UUVy<20 . On U U V therefore, ~(") is a convex combination 2~e (b f 0 + 2~e(Bc 0 + (1 - 2~- 2~) 1, where the middle term is negligible, and the other two terms are not negligible. As a result, (5.8) forces lira (bje)=O

j ~ o o

(rood 1). This completes the lemma, which in turn proves that F,(O)~=F~

w 6. II--1

Since f(")(x) = I~ f ( x + iO), we have that i = 0

(6.1) .f(m +.) (x) = f (m) (x + nO)f (") (x).

Clearly if two of {lam+.[, It%l, In, I} are near i, then so must be the third. We wiI1 establish (1.4) for tCF~(O) by proving first that

(6.2) lim I~,l = 1. II n01I ~ 0

In~i<n<q~+x-inul

A general value of n will not lie in an interval as in (5.2). However for large h depending on n, n+(bh+l)q a does, because ]nhf<(bh+l)q h for large h. More-- over, II(bh+l)qhOl] =llrnOl[ is small for large h, so that if llnOIt is small, (6.1) and (6.2) will prove (1.4).

Therefore, suppose !nk[<n<qk+~--Inkl, in which case nEA and 6,=llnkO--tll. If k is large, the difference ill (5.2) is small in L~-norm. Therefore it is sufficient to prove (6.2) for t~,] rather than for [a,].

We introduce notation, the significance of which will become clear. Define ~j, rlj,j>=2 by ~j= �89 and ~b=a/2(ej-ej_~). Then if k>=h>2 define. ekh=l~(ek+eh_l) and ~]kh=l//2(,~.g--~h_l). Also set

ktg -- ~k, O'k = ak bk qk

ilk-1 ~- ak-{-~k, tYk-1 : ak +Sk-l. bk-l qk-l"-}-llk

,U h = : O ' h + l + ~ h + l ~ a h = f f h + l + S h b h q h + t l h + l

f l h - 1 ~- tYh + ~kh, O'h- 1 = qk + rlkh and

~'lk ~- (DItk "'" ~ak

Th = q~uh "'" (P~h

To = (P,~-I "'" ~a,,,_~,

where all subscripts in the definition of the T i are understood to be multiplied by 0. Note that if e j=sgn(nj) for j > h - 1 , then q~k.'. (Pq~=Tk... ThT0, and Tj contains l+]%-p~]=bjq~ terms. Also, cr ,=nk--n,_a+~/,+l+. . .+r/k. I f r / j#0 and ~/i=0 for jo<i-<j, ~/j0#0, then ~/jo=-~/j. Therefore To has l~h-a-~-~l + 1<= t~-~1 + 1 terms.

Acta Mathematica Aeademiae Scientiarurn Hungaricae, 37, 1981

2 0 2 M. S T E W A R T

If Inkl<n<qk+l--lnkl, let Vj;n (for j = 0 and h<=j<=k) have a similar definition with $~_, replacing 9~ for each a. The relation [[qhO[[<[[nO[J<j[qh_xO[] defines h; h_->2 if [In0][ is small. Define (,=/~(~0kg0;,,). Pairing terms ~o~$~_, with integrals of norm ~l-2[[n01[, we conclude that ~_-->1--2([nh_~]+1). [[nO[[ => >=l--2(]nh-1]+l).f[qh-lO[], so that as [[n0][-*0,~,-~l. This reduces (6.2)further

to proving

(6.3) lim [[ T ~176 ToTo.,,([~ = 0, II nO [[ ~ 0

[nkl'<n<qk + 1-- Ink[

where the __+ sign varies with n, and h is as above. Assume tEF~(O), t :mO+

+ ~/~ ejbjqjO. We consider two cases. j = l

Case I. ~ rational, ~:p/q in lowest terms. Recall that ][T(")-ToT0;,,[]~ k

Z []I-TyTj, , [[ , that Ty has b~q~ factors, and that b~-O(q) for large j satisfy- j = h

:ing h<-j<=k. Grouping terms as in Lemma 5.5, we obtain [#(~])[~l-bjqy[]q~O[], k

.and similarly for Ty;,,. Therefore, the error in (6.3) is less than 2~b]q~[[qyO][, j = h

which tends to 0 as h-~ o0 (and this happens as ][nO[[-~0).

Case II. ~ irrational. This time do not break up Ty and Tj;,,. Simply notice -that ~y~y; , takes the values 1 and e(b~oO except on a set of measure bjqj][qyO[[.

k k k

The error in (6.3)is therefore ~[ll-~t"j~'j;n[l<=~2bjqflqjO[l+~[1-e(b:~)[, j = h j = h j = h

which tends to 0 as I[n0]l-~0. The inclusion is proved, finishing the proof of Theorem 1.

w 7. Noncompact abelian group extensions of Ro

Again let I ~ X be an interval of length t, and let E be the closed subgroup ,of R (not X) generated by - t and 1 - t . Set Y = X • and (o(x)=zx(x)-t. Then define the skew product T: Y-+ Y by T(x, y)-=(x+O, y+q~ (x)). If for

n--1 n > 0 we define ~o(n)(x)= ~ , ~o(x+iO), then Tn(x, y):(x+nO, y+9(n)(x)). In

i = 0

fact, ~o (") can be defined for n_<- 0 by setting q~(0) (x) --- 0 and (o (~) (x)--- ~ ~o(x- iO) i = 1

for n<0 . Then the preceding formula for T~(x, y) is valid for all nEz. In the language of [8], ~o is a cocycle for R 9 on X. This is, 9 is a Borel

function from X to E such that for all m, nEZ and xEX, ~o(m+'O(X)=~O(m)(x+nO)+ +~0(n)(x). We say ~EE is an essential value of q~ if given any Borel set A in X o f positive measure and 5>0, there exists an integer n with #(A(1R~"A(~ (~ {x: [~o(")(x)--~I<e})>0. Trivially 0 is an essential value for ~o. SCI-IMIDT proves

the following in [8], and we refer to them in the next section in the proof of Theorem 4.

EV 1. The set of essential values of ~o is a closed subgroup of E, denoted E(cp).

EV 2. I f ~ is the set of measurable T-invariant sets in Y, then E(~o) may be ,characterized as {~EE: O~B=B for all BE~r where O~(x, y)--(x, y+7) .

Ac ta M a t h e m a t i e a Acac temiae Sc ien t~arum Hungar~cae, 37, 1981


EV 3. T is ergodic if and only if E(q0=E. (This is immediate from EV 2.)

EV4. I f (o denotes the eoeycle (o(")(x)=9(")(x)+E(ep) from X to E/E(q O, then 0 is the only essential value of (o.

Relation EV 4 sa3s that if we can find 7EE(q~) with ~ nonzero, then we can effectively reduce T to a "compact skew product" T: XXE/~Z~XXE/czZ, and T will be ergodic if and only if T is ergodic. (See the method of proof of [9, Theorem 2.8].)

As in the introduction, we say t satisfies condition A if lim sup (llqd][-

- �89 q~flq, Oll) > 0 . We now give three lemmas to be used in the proof of Theorem 4.

LEMMA 7.1. For every integer q with qt not an integer, there exist sets B and q--1

B" in X such that i) l~(B)>=llqtl[, lz(B')>=llqtl[ ; ii) the function ~(x)= ~ q~(x+(j/q)) j = 0

is constant on B and on B'; i i i ) f o r every xEB and' x'EB', [q/(x)-~(x')l=l. PROOf. ~(x) is invariant under translation by 1/q and is constant on the

elements of the partition ~ defined by the points {-j /q, t - j /q}, j=O, 1 . . . . , q - 1 . Define t - h /q as the left endpoint of the element of @ whose right endpoint is 1. Then put

q--1 B = ~ [t-(h-bj)/q, - j /q) , B" = BL

j = 0

Since each element of ~ has length either llqtll/q or (1-1lqtll)/q, i)checks. Since O(x)=O(x+l/q), i i )checks and reveals that for every x in B and x' in B', [O(x)--O(x')l is constant. To compute the constant, set x = t and x ' = a n y element s in the element of @ just to the left of the one containing t. O(t) samples the function q~ at the points {t+j/q}, O~j<q. For some r, then, l<-r<=q-1, r of these points lie in [0, t) and q - r of them lie in It, 1). Thus [O(t ) -O(s)[= = Ir(1 - t) + ( q - r ) ( - t) - [(r + 1)(1 - t) + ( q - r - 1 ) ( - t)]]= 1, and the lemma is done.

LEMMA 7.2. Under condition A there exists e0>0 such that for infinitely many k there exist sets Ck, C~ in X with i) P(Ck)>eo,#(C~)>eo; ii) o(qk)(x) is constant on Ck and on C~; iii) for every XECk and x'EC~, ko(qk)(x)--(o(qk)(x')l=l.

qk-- 1

PROOF. Define 0k(X): ~ ' ~p(X+j/qk), and let Ek={x: ~p(ak)(x)=~bk(x)}. Let j=O

k be even; the proof for k odd is similar. Let ii0 be defined as in Corollary 2.5, qk--1

and set Dk= U [--ijO,--J/qk), D ~ = t + D k : { t + x : xCD~}. Also assume B and j = 0

B" have been taken from Lemma 7.1 with q:qk. Then set Ck=B•E k and C~=B'NEk. To prove the result it is enough to show that the measures of Ck and C;, are greater than []qkt [l --1/2 qkl[qkOI]. We treat two cases.

Case I. [[qktH >(qk--1)llqkOI], In this instance one has that DkC:B and D~C:B'; in fact, BNEk=B~D~ and B'NEk=B' \D~. The desired result follows from Corollary 2.5 ii).

A c t a M a t h e m a t i c a Acadern~ae S c i e n t i a r u m Hungar icae , 37, 198~

204 M. STEWART

Case lI. We may write ~qkl[qkOl[<llqetll~(qk--1)][qkOl[. Let [t--ck/qk[: = rain )--i/qk[, and let the integer mk be determined by []qktl[/HqkO[lC(me--1, me].

i~i~--q k

Assume t--ck/qk is negative; the proof for t--Ck/qk positive is similar. Again a straightforward computation gives

It(B~(BOEk)) = P (71i- [ - iO, -- ipk/qk) ) < 1./2 qkl]qkO,]

and

It(B'~(B'AEk))= # [-iO, --ipk/qk) --I t [--iO, tq-(i--Ck)/q k < �89 qkl[qkOll. t 0 "=

The result follows, finishing the proof. The following result was proved by Koksma. We present a proof as well (see

[5, p. 143]).

LEMMA 7.3. Let xl<=x2<=...NxN be N points in (0,1) and define x0=0, XN+I=I. Let f be of bounded variation on [0, 1]. Then

x . - t t ~ ar , n = l

where D}, the discrepancy of {xi}, l<=i<=N, is equal to max(max lx i - i /NI , i

lxi+l-i/Nl).

PROOt~. The first step is to prove 1 N N x?+l

(7.4) -~ ~___,f(x,) - f f ( t ) dt = ~ t ( t - n/N) df(t). = X n : O ~n

Now the right side of (7.4) equals

f tdf(t)- (f(xn+,)--f(xo))n = tf(t)101- f ( t ) d t + f (x ,+, ) - f (1) , X n = 0 X =

which is the left side of (7.4). Now for n fixed between 0 and N we have [t-nlNl<=max(]x~-n/N[,

]x,+~-n/N1)<=D~v for x,<=t<=x,+,, so that the desired result follows immediately. For our purposes we fix any xEX and let {x~}, l<=i<=N, be {x+jO}, l~j<=qk �9

Also put f = o. We obtain

COROLLARY 7.5. For all x in X and k > 0 we have ig(q~)(x)]<=2.

w 8. P r o o f o f T h e o r e m 4

Notations are as in the previous section. Also, we are assuming Theorem 3, so that, in particular, the results of Lemma 7.2 are true. Let A ~ Y be strictly invariant (TA =A) and of positive measure. We will show that A = Y up to a null set. In what follows, equations involving sets may only be true modulo a null set; the context will indicate this.

A c t a M a t h e m a t i c a A e a c l e m i a e S c ~ e n t i a r u m H u n g a r $ c a e , 37, 1981

I R R E G U L A R I T I E S OF UI 'q lFORM D I S T R I B U T I O N 205

We begin with the proof of part i). By Lemma 7.2 we pick a subsequence Q ' = {q~<q~<...} of the sequence of denominators for 0 such that

,) (8 .1) (x) = => s0

/ > 1 �9 t where I=hI= �89 and ~k-*~, 1r and s0>0 is as in the statement of the lemma. Now define the following sets for xEZ and ~ [ - 2 , 2].

D a, = {x: lim'xnfl~0<q~)(x)--~l' = 0}, Be,(x ) = {limit points of {9{*a)(x)}, k => 1}.

Clearly ~BQ,(X) if and only if x~D~,. Also, since for x#O, t,

lim inf lm<qa (x)- 4 a (x + 0)1 = lim inf lep ( x ) - 9 (x + q~, 0) l = 0 k ~ k+oo

it is true that Ba,(x)=Bo,(x+O ) for x # 0 , t. By (8.1), p{y: q~(~)(Y)=~k for infinitely many k}->s0 . As a result, #(D~,)=>s0 . Therefore # (Da , ) - I , since x-*B o, (x) is easily seen to be a measurable function. We have proved

LEMMA 8.2. There exists 7 '~ [ -2 , 2], I~ ' [~ �89 with #(D~,)=I.

Let the c~' named in this lemma be fixed now for the rest of the proof of part i). For fi<E define Ba= {x: (x, fi)6A}. We will show that C = {(x, fi)< Y: (x, fi)~4

and (x, f i+~ ' )~A} has zero measure. Indeed, suppose C has positive measure. C is T-invariant, so C E s e in the language ofEV 2 (in Section 7). But o~,C(~C= ~ , so that o~,C#C. Therefore c~' is not an essential value of 9, by EV 2. But this contradicts

L E M M A 8 . 3 . I f #(D~,)=I, then ~'CE(~o).

PROOF. Let Z be a set of positive measure in X, and fix e >0. Choose a subsequence Q " ~ Q" such that " " Zz(X+qkO)-~Zz(X) a.e. (x). Note that the qk and c~,', along with ~', satisfy (8.1) just as did the full sequence.

Now for some k l > 0 there is a set Z I ~ Z of positive measure with Zl+qs qtt

~ Z ( m o d 0 ) for all k>=k~. That is, for all k~k~, we have Z l c Z N R o kZ.

Define, for s fixed as above, Z1, j= {xCZI: Iq~(~)(x)-c(l<s }. Z1 ~ U~ ZI, j; j = l

indeed, almost every x in Z1 belongs to infinitely many ZI, j, since /~(D~,,)=I. Therefore infinitely many Zx, j have positive measure; pick one of them, say Z~,h, with h>=kx. Immediately, Zl, hc=(ZnRoq~Zn{x:l~r so that a'EE(~o), as claimed.

Therefore C has measure 0 in Y, or BaC=Bp+,,, almost all ft. Using the fact that - e ' is also an essential value, we get Bp ~ Ba_ ~, for almost all fl, and hence Ba=BB+ ~, for almost all ft. Thus, defining ~(x)={fl: (x, fl)~A}, we may regard ~b(x) as a subset of E/e'Z, defined only up to a null set.

Now if U and V are measurable sets in E/a'Z, define d(U, V)=2(UAV), where )~ denotes Lebesgue measure on E/~'Z. Note that liminfd(q~(x+qkO), r a.e.

By Lemma 7.2 there is a fixed constant eo>0 and for infinitely many k sets C k, C~ of measure greater than eo in X with q~(qO(x)~2 k on C k and ~p(qk~ (x) _=_

A c t s Mathemat ics Academiae Scient iarum Hungaricae, 37, 1981

206 M. STEWART

=2k__+l on C~, where the ___ need not be determined. Since we have fb(x+qkO)= = ~(x)+~(qk)(x) for all k and x, we have the following relations for any preassigned e>0 : first, for each large enough k given by Lemma 7.2, there is a subset Ck of Ck of positive measure for which d(~(x)+2k, 4~(x))<e. For these k, d (~(x)+2k , ~ (x ) )<e a.e. (x) by the ergodicity of R 0. Similarly, using sets C; and C:,, we have d.(r , ~(x))-<~ a.e. (x) for large k. Therefore d(~b(x)+2k, r a.e. (x), so that ~ ( x ) = ~ ( x ) + l a.e. (x). This proves that 1 also is an essential value of ~0, by EV 2. Therefore by EV 1 Z G E((o).

Let F=E/Z. Define T~: XXF-~XXF by T~(x, y)=(x+O, (y+ep(x)) rood 1)= =(x§ and define the cocycle ~01 from X to F by (o~(x)=cp(x)+Z. As remarked before, T is ergodic on Y if and only if T1 is ergodic on XXF. But 2"1 is ergodic trivially, since !, 0, and t are rationally independent. Theorem 4 i) is proved.

The following example shows the necessity of the assumption that 1, 0 and c o

t are rationally independent. Let 0=[3, 2]=[3, 2, 3, 2 . . . . ], and set t = ~ q~i-~O. i = l

Then t~ZO=F~ this is because 0 has bounded partial quotients. But

2t = ~ 2q~_10 = 1 - O. i = l

Now we prove part ii); the proof is essentially that of [1, Theorem 4]. Let t=p/q in lowest terms, and again suppose A ~ Y is T-invariant and of positive measure. We show A = Y up to a null set. To this end define Bj={x: (x,j/q)EA}, Cj, k=BjVIBj+ k. Let {qki} be a subsequence of the denominators of 0 with q:(qki" This is possible since q, and q-+~ are relatively prime for all i. Also assume the qk, have been picked so that Zc(x+qk, O)~zc(x ) a.e.(x), where C=Cj, k~. Now define

Da = {x: (p(%)(x) = h/q for infinitely many i}.

It is easily shown that each D h is R0-invariant; being measurable, it is of measure 0 o r 1.

Now ~o(qk* ) (x)=r--qk, (p/q) for some r between 0 and qk,, so that q:(qep(qk)(x) for every i and x. As a result, D h is empty for q[h. Therefore by Corollary 7.5

2q we have X = ~ D j, so that some Dj has measure 1. Henceforth let j be fixed

j = --2q q~J

with this property. Then a proof along the lines of Lemma 8.2 shows that Bk=Bk+ j for all k; that is, Ck, j has measure 0 for all k. Now if j = ___ 1, we are done, for then A=Y. So suppose otherwise; we nowhave I j ]~2 and q,[j.

Define ~P(x)={k/q: (x, k/q)CA}. As in the proof of part i) we shall consider 4~(x)C= {i/q}~=~ for all x. But now the fact that ~(x+O)=~(x)+q~(x) (modj/q) shows that card (4~(x))=card (~(x+O)). Since g(x)=card (~(x)) is measurable, it is a.e. constant, say g(x)==-rn a.e.(x), m ~ 0 . More is true. Consider the values of

(x) as lying in the set of jth complex roots of 1. Almost every r is m of these j points. But again, by ergodicity, there can be only one configuration for these m points (up to rotation).

Now there are only j possible configurations, all rotations of one another, for ~(x). Cyclically number these configurations as 0, 1,2, . . . , j - l , and let

Acta M a t h e m a t i c a A e a d e m i a e S e i e n t i a r u m Hungar icae , 37, 1981

I R R E G U L A R I T I E S O F U N I F O R M D I S T R I B U T I O N 207"

G(x) be the number assigned to the configuration for ~(x). Also define O(x)= =qrp(x) (modj) . Then G(x) is a measurable function from X to {0, 1 . . . . , j - l } and satisfies G(x+O)=G(x)+~k(x) (rood j). But this contradicts the following result.

THEOREM [1, Theorem 1]. Let F be the cyclic group o f order j, and let ~b(x) be defined as above. Then there is no measurable F-valued solution to f(x+O)-= =f (x )+O(x) .

Because of this contradiction we are forced to assume A-- Y up to a null set, finishing ii) and Theorem 4.

It remains to demonstrate Theorem 3.

w 9. Continued fractions II

We record here some calculations to be used in the following two sections, As in [3, Section 1] we introduce the notation

at = [at; ai+l, a i+2 , . . .] = a i + l / a i + t ( i ~ l), q~ = ' r i = " " a lq i - l+q i -2 , q~-l/qi" We then have

�9 t t t - - 1 PRO~OSlTION 9.1. For all j>-O, i) aj+lqj=qj+l, ii) Ilqj01] =(qj+l) �9

PROOF. Both statements follow by induction on j. First we have q~=a;qo+ +q_~=a[=l/O, which proves the initial step for both i) and ii). Now assume i) holds for all j < k ; we seek to verify qk+l=ak+lqk: " " But qk+a=a~+l' " (qk + (qk-1/ak+l)), SO that we are reduced to showing q~--qk=qk+~/a~+l . By the respective definitions of qk and q~, we have q~--qk=(a~--ak) qk_l=qk_l/a~+l which was required. Finally, assume ii) holds for j < k . Then

liqkOl[ = llqk-z01l--akllqk--~Oll 1" " a " " = " " = = lqk-1-- k/qk II--(ak/ak)l/qk-1 P �9 �9

= (1/ak+l)(1/qk) = 1/qk+l, and the proposition is proved.

�9 - -1 COROLLARY 9.2. For all j>-'O, i) @lqjOl]=(aj+~+r~) , ii) q~ltqj+xOll= = (a~+~)-l(a~+l +r~)-L

LEMMA 9.3. For all j >=-1, qj+~llqjOII +qjl[qj+~Oll=l.

PROOF. If j = - - l , we have qo=llq_~Ol[=l and q_ l=0 , which give the result. For j_->0 we may write

qj+~ I1 qj 011 + qj II qj+l 0[I = qj+i/qi+l+ q/(aj+~ q j+l) =

= (J+~qJ+l-t-qj)/(aj+2qj+J) = qj+Jqj+2 =- I.

LEMMA 9.4. Let the canonical representation for tE X \ Z O be t= ~ blqiO. Then / = 0

for large j, J

i) 0 < ~ biqi<qj+l, 1 = 0

Acta Ma~hematica Acaclemiae Sc ient iarum Hungaricae, 37, I981

2 0 8 M. S T E W A R T

ii) ~ biqiO lies in the shorter arc of X joining --qj-lO and -qjO. Also, i = j

b~qiO lies in the same element of the partition {[0, �89 [�89 1)} of X as does

q~O, where bk is the first nonzero term in {bj, bj+1 .... }.

PROOF. Part i) is immediate from the construction of the canonical representation, especially Lemma 3.3. As for ii), the shorter arc in X joining qjO and qj+lO

.contains 0 for large j, by 2.1 iii). Thus we have that ~b~q~O lies in the shorter i = j

arc joining ~(maxbj+si)qj+2:0 and 2(maxbj+si+Oqj+si+~O. But these last i = 0 i = 0

two quantities are -qj_~O and -qjO, respectively, by 2.2 ii). Further, suppose b~>0, k>=j, and k minimally so. Then bk+l<ak+2, by canonicity. Therefore

b~q~O lies in the shorter arc joining bkqkO+ Z (max bk+2i)qk+siO and bkqkO+ ~ = j i = 1

+(ak+~-l)qe+lO+~(max bk+zi+Oqk+2~+lO. The endpoints of this arc are i = l

beqkO--qk+~O and bkqkO--qk+~O--qkO, and both of these quantities lie in the same element of {[0, �89 [�89 1)} as does qkO. The lemma is verified.

LEMMA 9.5. Let 0 have bounded partial quotients. Then there exists constants J,~l depending only on 0 such that for all i~1, i) a[a/+~>2+6, ii) ri>6, iii) q~/q[ < l-~1.

PROOF. Choose 6 = ( l + M ) -1, q - - (2+M) -~, where M=sup{aj}. Then �9 �9 �9 > �9 aiai+x= 1 +aiai+l----- 1+a ,+1>2+~. Next rf=qi_z/q~---q,_l/(aiqi_~+qt_~)=l/(af+

+ri_l) >~. The proof ofiii) is just as straightforward,and it is omitted.

COROLLARY 9.6. I f 0 has boundedpartial quotients, then for some c >0 depending only on 0 it is true for all i>=O that liq~+#tl/Hq~Oll<�89

PROOF. By Proposition 9.1, []q~+2Ol[/]lqiOll =(a;+2a;+~) -x. Now apply Lemma 9.5 i).

w 10. Proof of Theorem 3

Pick t~F~ and write t canonically as t= ~ b,q,O. Define the coefficients i = 0

of t to be the sequence {bi}. We seek an infinite subsequence S = {qis} of the denominators of 0 with

(10.1) lira (11 q~jtl] - 1/2q~ s I[ q~s0ll) = c > 0.

The evaluation of ][qet[] is made possible by the canonical representation; the proof examines a collectively exhaustive set of cases for the coefficients of t.

i - - 1 co

NOTATION. I ~ = ~ bjqjO, IIi=bfqillqiOl], IIIi= ~ bjqjO. j = O j = i + l

Ac~a M a t h e m a t i c a A c a d e m i a e S c i e n t i a r u m Hungavicae , 87, 1981


Particular cases being considered for the coefficients of t will be represented i n a table as follows:

ij

... A B C ...

... X Y Z ...

This notation indicates that {q,} has been found with b,_l=_A and a,.=-X, bl.-~B and ai.+~-Y, bi.+l-C ~nd aij+z-Z, etc. Each t~ble will be of 'finite

. 1 J J " V " " "

length. Any other restrictions on A, B, X, , , etc. will be stated in parenthesis. Additional notation to deal with {a0}~ ~ will be introduced later.

Case A1. ij

L M (M=>4, L~{O, 1, M-1 , M}).

Here S = {q,j} satisfies (10.1). To see this, note that

: lI q~j t II = il ( I , , ) q,j 0 + It 0 + q,j (IIIij)[I.

Now [llI,~[I =l[Lq~jlIq,jO[111 by Proposition 2.6, and the sense of I15 is that of %0. (See Proposition 2.1 iii) for the definition of sense.) By Lemm~t 9.4 ii) qij(IIIij) has at most the magnitude qijllqij+~Oll in the sense of q~O, and it has at most the magnitude qi.llqi Oil in the sense opposed to qi 0. Finally, by Proposition 2 6 and Lemma 9.4 i~, II'I~,q,,01l =I.,jllq,,0ll < (0, q,,llq,,Oll)/and the sense is that of qb0. From these estimates we obtain

qo tE ((L- 1) %11 q,jOll, (Z + 1) q,all q,, Oll + qfj]l q,a+1011). Thus in order to verify (10.1) it is enough to find a constant c > 0 such that for all large j we have the following relations :

(10.2a) (Z + 1)q,, II qi.iOt[ + qi a [1 qi,+l 0ll < 1 - �89 I[ q0 0ll - e

(10.2b) ( t - 1)%11%011 > �89 11%011+c.

Using Corollary 9.2, (10.2a) reduces to

1 ' " ' -1/(2(M'+ri,))-c ( L + ) / ( M +%)+ll(a,,+,)(M +%)< 1

M " a" 1 ' where M ' is an abuse of notation meaning = i cq-l=aiiq-loy lai~+=. Therefore we seek L+3/2+l/a~.+2<M'+ri.--e. The arbitrary constant is new, but it is

�9 $ J # �9

stmply greater than the old constant by the factor M +r~., which is bounded above ' . . . . 2

and below. We simplify the above inequality to L§ Now since L<=M--2 by assumption and % > 0 for all j , it is clear that c = ~ will suffice. As for (10.2b), we require

(L - 3/2)/(M' + ris ) > c, L - - 3/2 > c.

This is also true for c = �88 since L=>2. Case A1 is verified.

Case A2. {a0+~}-+~o and {biJaia+x}-+=, 0 < = < 1 . In this instance S={q,;} is a valid choice. To prove this we may estimate IlI~a[l<qJqisOll<l/a~+, as m

i 4 a e t a M a t h e m a t i e a A c a d e m i a e S c i e n t i a r u m Hungar~eae, 37, 1981

210 M. STEWART

Case A1. But 1/ais+l tends to 0. Likewise i[IIIisll represents an addition of no more than q~jllq;j+~OII or a subtraction of no more than qJqejOl], both of which tend to 0. Therefore

lim [Iqist[I = lim I l l I j = lim l[b,j%llq~flllll = lim Ilb~s/%+~l] = II~ll > 0. j . .~ j~,~ g-~o �9 J - } c o

Case A2 is verified. Hence if {aij+~} tends to % then {bi./ai+l} can now cluster �9 3 J .

only at 0 and 1. Therefore we add to the table notation the possibilities

& ix 0 ~o oo c o

The first of these indicates {ai:+l}-" oo and {biJais+~}~O, while the other indicates {a /s+l}~o and {bij/a,~+l}~l.

Before proceeding we introduce an important principle. Recall that F~ is a subgroup of X. By Lemma 3.1, if toCY~ then {[[qd0[l}~0. Thus, if toCF~ then {[]qd[[} and {[[qi(t• have the same cluster points. In what follows we often shall have occasion to add or subtract elements of F~ to t to simplify calculations of 1]qitI[; by the above remark, this is permissible. In fact we shall abuse notation and perform these F~ without changing the notation for the coefficients of t. Put another way, we shall switch from t to t + to for the remainder of the proof but continue the notation ~ ' b~qiO for the new value t+to.

Now we make the first F~ Let I = {i: bi(~ {0, 1, ai+~-1, a~}}. Then {ai+~}iEi tends to infinity, for otherwise Case A1 would apply. Now let

-11 = {iEI: bJai+~ ~- 1/2}, /2 = / " , , / 1 .

We may assume {bJai+,}~Ex,.-,-1 and {bJa~+l},e,,~O, else Case A2 would apply. Now note that to= ~ ' (ai+l-bi-1)qiO-Y~biqiO is in F~ because each of the

11 12

sums belong to F~ which is a group. Adding to to t, we may now assume bi~ {0, 1, ai+, - l , a,+~} for all i. Also the representation is still canonical.

Case B1.

B4.

ij B2. b B3. /s

0 1 0 0 L - - 1 0 r 1 oo L ( L ~ 2 ) L (L_~3) ~o L ~ o ( L ~ 2 )

iy B5. i x B6. i s

oo L - 1 oo co 1 L - 1 oo

oo L ~ ( L ~ 2 ) r L (L_~3) L o o ( L g 3 )

For these six cases S = {q~j} satisfies condition A (10.1), except for a lone subcase of B1, namely

Case B7. ij 0 1 0 0 o o

L 1 1 oo (L_-_>2).

For B7 S = {q0+l} is valid. Cases B1--B6 are treated exactly similarly; therefore, we present the calculations only for B1 and B7.

A c t a M a t h e m a t t c a A c a d e m ~ a e Sc ien t iarurn Hungar i cae , 37, 1981

I R R E G U L A R I T I E S O F U N I F O R M D I S T R I B U T I O N 2 1 1

For B1 we again write

II qi~ t ll --- ll(Ii) qij 0 + qij qls 0 + qis (IIIij) l[.

By Lemma 9.4i) and the assumption b, 1--0 we have 0<I i < a s , Also, the . . . . . j - - j -i j - - "

contribution of qi- (IIIzj) is an addltxon of at most q~ ]]qi +1011 or a subtraction a . J J

of at most qi:l[ q~j + 2 0I[. Therefore we require both of the following:

(10.3a) (q,~+ q~-0[1% 0[I + q~j II q~+x01l < 1 - �89 q~ II q~jOll - c

(10.3b) q,j 11 qij 011 - q~j II qij + 2 0[I > �89 q~: [1 q,j 011 + c.

First, (10.3a) reduces to

(3/2 + %)I(L'+ %) + ( l / M 3 ( I / (L '+ %)) < 1 - c

3 / 2 + % + 1/M" < L ' + r i j - c = L + 1]M" + r i j - c .

(We continue with the abuse of notation introduced in the verification of (10.2a).) The last inequality becomes 3 / 2 < L - e , which is true since L=>2. Next (10.3b) reduces to

(1//2 -- l l (M' ai'+z))/(L" + r,j) > c, �89 -- l l (M' a's+,) > c

which is true except when M = I , a~:+3-~1, and { a i j + 4 } ~ . But this is the given for Case B7.

As for Case B7, we must verify each of the following:

(10.4a) (q~a+q~-Ollqi:+lOll+q~j+xllq~j+~OH < 1-�89

(10.41)) qfj 11 qo+~011 + q~+l II q~+2 011 > �89 %+1 [I qij+a 011 + c.

Greatest (resp. smallest) estimates for I~j+l and III~j+~ have been made in (10.4a) (resp. (10.4b)) above. For details, see Lemma 9.4 and the previously established cases. Upon dividing both sides of (10.4a) by %+~I1%+~011 we obtain

qis-1/ qi~+ l + ri~+ l + 1/ai~+3 -< ai~+2 + r i s+l - - 1//2 --C

or, since aij+2-1, q~j_llqij+l<l/2-c. But this is true since L=ai~+a~_2. For (10.4b) to be true we seek

ri.~+l+ 1/ai'+3 > 1/2+c.

This is guaranteed by the fact that {a~'j+a}---1.

Case C1. ij C2. ij

1 1 1 M - 1 L M ( L , M > = 2 ) L M ( L , M _ ~ 2 )

C3. ij C4. ij L - 1 1 L - 1 M - 1

L M (L,M-->2) L M ( L , M _ ~ 2 )

C5. ij C6. ij

L 1 L M - 1 L M(M_-->2) L M (M_~2)

14" Acta Mathemattca Aeademtae Seientiarura Hungartcae, 37, 198I

212 M. STEWART

For each of these cases it can be shown that either S = {q~} or {q~j+l} :satisfies condition A, without determining which. The demonstration is given only for Case C1, the other cases using the same method of proof but only slightly different estimates of Iij, III~j, etc. . .

For C1, in order that S = {q~j} it is sufficient that both of the following are true: (10.5a) (I~j+q~s)llqtflll-q~jllq,j+lO[I--- qijl[III~j+lll < 1 - 1/2qjq~,Ol[-e

(10.5b) (I~s+q~)llq~jO[l-q~,l[qt,+lOll+_q~,,l[III~,+ll[ > �89 l[q~sOl[+c. :, i

The determination of the above ambiguous signs depends upon the parity of the index of the first positive term in {b~}i>~j+l, and may vary with j, but a t least the two signs must agree for all j .

On the other hand, to have S = {q~j+l} it is enough to establish these:

(10.5c) (I~j+qt~+q~+l)llq~j+iOll+-q,j+ll[III~j+l][ < 1-�89

(10.5d) (Iij+ qij+ q0+l ) I1 qej+1011-+ q~j+llllII~j+ll[ > �89 q~j+lll q~j+10l] --c.

For each j the ambiguous sign in (10.5c) and (10.5d) must be the negative of the one in (10.5a, b).

Rename these last four inequalities a , b, c, and d for short. We show that one of each of the following pairs of inequalities must be true: a and c, a and d, b and e, b and d. When this is done, then it will follow that either S = {q~j} or {qi~+l} is valid.

As for the pair a and c, multiply a on both sides by q~,+l, and multiply through c by q~j. Adding these inequalities and invoking Lemma 9.3, we obtain

(10.6) (I!~+q~) < qt,+l+q~,-�89 ~. If (10.6) is true, then either a or c must be true. Estimating Ii,+q~<q~,+~ ~, we rewrite (10.6) as : (10.7) q~s+l(ll q~+~0[I + [I q~;0[I) < 2 - c .

N o w q~s+l[Iq~.,O[l=q~.~+l/q~,+l by Proposition 9.1ii), and the fact that a~+~= _--M<oo implies that q~,+l/q;j+~<l-c, some positive c, for all j. Also, llq~s+1011 < [lqi~0[I. Together these facts verify (10.7).

N o w look at the pair b and d. Multiplying b (resp. d) on both sides by qis+l (resp. qis), adding, and collecting terms, we obtain

(10.8) (I~+ qf) > �89 qf~+101[ + [I qi~OII) + cq~.,.

Note that the constant in'(10.8) differs from that in (10.6). Now estimating Is. >0, we arrive at 00.7), which has been verified alreadyl Therefore either b or d i~olds also.

For the pair a and d form the linear combination a([lq~+lO[l)+d(l[q~jO[]) (it should be clear what this means) to obtain, after clearing terms

(10.9) llIII~+lll+�89 .l[qe~+~0[I < (2-c)[Iqo+lOl[. Estimate �89 (q~s + qis + 1) [I q~s 011 < q~+ ~11 qls 011 < 1 -- c; what results is

(10.10) ,.i :: ' [llII0+,ll < IIq,j+~0[I.

A c ta ~ Ma.~kemc~cct , A e a d e m i a t z se~en t ia~ 'um. t t u n g a T t c a e , 37, 1981

I R R E G U L A R I T I E S O F U N I F O R M D I S T R I B U T I O N 213

Since bi.+l=-l, canonicity forces max lllIIi.+lll=llqi~+lOll-llqij+201]. Thus (10.10) is verifie~i, and either a or d must hold. ~

For the pair b and c form the linear combination bl[q,s+lOII +cl[q~j0[I and clear terms. We seek to verify

(10.11) 1[ q,~+101[ + I[III,~+IH + 1/2(q~+ q,s+l)[I q,j 011. I] q~+~ 0[I < [1 qfjO]l - eli q,:+lOl[. Estimating lllIIi +111--IIq~ +~011 (since it is the qi.+zO sense of IIIi.+1 that makes

J a a . a

b and c both most difficult to achieve) and simplifying, we arrive at (10.7) again. Hence (10.11) is verified. Thus either b or c must hold, establishing C1. Cases C2--C6 are each upheld by an exactly similar proof.

Now suppose bif~l,a~s+l--M>=2. Then

0 1 Moo

is forced, else one of the B-series or C-series cases applies. Similarly, if ab+l--M>=2 and bij=-M-1, then

ij

~ M - 1 0 M

is forced, for similar reasons. (These two above situations may both occur if b~s-1 and ai.+1~2.)

J .

In hght of these severe limitations on the coefficients of t, define Q = {bi=0 or 1}, R={bi=a~+i"l or a~+~}. We wish to have QfqR=O, but this is impossible if a i+~=l o r 2, or if tables

i_A_/ i_.j_/ i_L

0 1 1 1 1 2

exist. As for the pair (bls, a~ +0=(1, 2), the sequence {ij} splits into two disjoint subsequences {lj} and {m~} with {bt~+l} tending to o~ and {b,,~+~} tending to 0. The above splitting is not unique; select any such splitting. Then put {bt:}= ~ Q and {b,,,j}==R. Finally, if a i s + ~ l , then put bij=O in Q and bi~=l in R. Now Q and R are disjoint.

A study begins of the strings of consecutive coefficients of t in Q and in R.

w 11. Proof of Theorem 3 (continued)

First we examine the strings of elements of O Suooose ~b, ~c t, trot b. . ~ t~ and bi~+lEQ for only finitely many j each. We look at the nonzero elements bi. so described. But if bi -=1, as we must have, then [ai +1/ must tend to co,

J J ~. j J

else an appropriate case for the B-series or C-series must apply. Then by a suitable F~ of t (namely a subtraction of Z..qijO), we may assume that if {bij}c= Q and if bi~-lEQ and bis+lEQ for only finitely many j each, then bb=-0.

A c t a M a t h e m a ~ c a Acac lern iae S c i e n t ~ a r u m H u n g a r ~ c a e , 37, 1981

214 M. STEWART

Next suppose strings {bi ,bi .+l , . . . ,bk.} occur in Q with ks>is for all j . l J d J �9 �9

Let Q ={bi . . . . . . bk. lJ. Then bt.-=l, at + ~ M - > 2 with {bt ] C Q ' is impossible, J a--�9 d J - - J = 0 for reasons exactly simdar to those above. Making therefore another F (0)-subtrac-

tion, we may arrange Q ' = {0}. Now, Q, is composed of the following strings: Q1-- {isolated b~s--0}, Q2 = {strings of zeroes bij, his+l, ..., bnj}, Q3 = {strings b~s, .... bkj with bls . . . . . bkj_~=0 and bk.=l }.

Now we examine the strings of elements of R. Again, first suppose there are isolated elements of R; that is, {bi.}C_R, but b~j_IER and b~.+ICR for only finitely many j each. Then it cannot- occur that b ~ . = - a i . + ~ - f - M - l > = l , for

�9 �9 �9 J 3 .

then a B-series or C-series case apphes. Thus for {bi~} we have bij=a~j+~, or ff bt .=ai j+l -1 on a subsequence, then { a v + l } ~ on that subsequence.

The other elements of R run in stri~ags, say {b~., ..., bg.}. We subdivide these �9 �9 / �9 - J I ! . �9 strings into R ={strings with b i s : a i + ~ = l } and R =('other strings}. In R

�9 . . 3 . �9 I f

we have no integer repeated mfimtely often m .(j {aij+~ . . . . . a~s}, and m R no J

integer appears infinitely often in U {%+~, ..., %}. Again the reasoning is that J

a previous case would apply otherwise�9 Thus by appropriate F~ we can arrange the following:

i) in R', if k s > i s + l , then bij . . . . . bk j_ l=0 and bkj=a~j+l;

ii) in R', if k s = i s + l , then b ~ = l and bt, j=ak~+~--I (and {akj+~}~o);

iii) in R", bo=als+l- l ({ai j+l}-- , -~) , b~j+a . . . . . bk.i_l.=O , and bks=akj+l

(resp. ak~+~' l ) if k j > i s + l (resp. ks=is+l) ) .

Now the original definitions of Q and R have become meaningless, but we can partition the coefficients of t, using table notation in the following sets:

Q 1. ij R 1. ij ij 0 M co

ao+ 1 m Q2. ij k s R2. ij kj i s kj

0 . . . 0 a i j + l . . . a k j + l

Q3. ij l(,j R3.

0 0 1

a i l + l � 9 1 4 9 ak j + l

R4.

R5.

0 . . . 0 oo l oo

1 . . . . . l oo

i s kS O . . . O M 1 . . . . . M

i s ks is ks ~ 0 ... O M ~ M - 1 . . . . . M ~ M

ij kj ij kj ~oO. . .O ~ oo oo o o o o c ~ o o o

' . ! .

~c ta Mathematica Academiae Scient iarum Hungaricae, 37s 1981


The notation suggests infinite numbers of strings of perhaps varying lengths. In Q3 {ak~+l} must cluster at at least one M, 2<=M< co. Finally (this is obvious) every Q grouping is followed by an R-grouping and vice versa. There follows a series of arguments showing which groupings may appear consecutively.

First, every Q3 group not followed by an R1 group may be eliminated. For if either an R2 or R3 group follows Q3 infinitely often, then there is a choice for S by Case B1. Further, if R4 or R5 follows Q3 infinitely often, then there is a F~ - alteration that permits a regrouping so as to destroy the Q3 groups. We show the details for Q3--R4; those for Q 3 1 R 5 go similarly. We have either

ij

0 . . . 0 1 ~ o 0 . . . 0 M

a i j + 1 . . . a k j akj+ 1 oo ... ~ M o r

i j

0 . . . 0 1 ~ M - 1 .

a i j + l . . . a k j aki+l ~ M

In the former instance add ~ q k j + l O - - Z q k ~ + . a O ; in the latter add ~qk~+~O. Then simplifying via qn+~=a,+~qn+qn_~, we obtain either

b 0 . 0 M

a i j + 1 . . . a k j + 1 co . . . co M o r

i j

0 O M ,

aij+l �9 �9 akj+~ co M

both of which may be regrouped as Q2--R1. The claim is established. Now that Q3 must be followed by R1, each of R3 and R5 and the first of each

R2 and R4 can be broken into strings of Q1, Q2, and/or R1, because in these instances the previous Q strings must have ended with a zero. This greatly reduces the list of R strings to include, after renaming, only the following:

R1. /~ i i R2. b /~ R3. ij M ,o~ ~ o M - 1 ~o oo 1 oo M co oo M ( M ~ 2 ) co co 1 oo

Next, R3 may be eliminated. To do so, simply allow those b i = a i + l = l which belong to an R3 grouping to belong to Q instead, repudiating the previous definition. Now in reference to the new list of groupings, in Q3, {a~j+l} may cluster at 1. An R1 must still follow a Q3, however.

Finally, Q3 can be excluded from happening infinitely often. For every Q3 string is followed by an R1; to the right of that R1 may occur some Q1--R1 pairs, but the number of such pairs is always finite. (If the number of pairs were not finite, a suitable F~ would put tEZOC=F~ a contradiction.) Therefore

A c t a M a t h e m a t i c a A c a d e m i a e S c i e n t i a r u m H u n g a r i c a e , 37, 1981

216 M. STEWART

eventually either an R1 is followed by a Q2 (or the next Q3 - - the proof is the same and is only given for R1--Q2), or a Q1 is followed by an R2. In table notation we have one of the following:

b . . .0 1 ao+2 0 ao+ 4 ... 0 ao+2l ~ 0 0 ...

ao+l ao+2 ai.i+3 aij+s ... ab+2t ~ a i j + 2 l ~ + l . . .

b ... 0 1 a0+z 0 a 0 + 4 �9 0 ai~+21j 0 oo M - 1 or 0o

a 0 a i j + l ai.i+2 . . . aii+21 i a i ~ + 2 / ~ + 1 o o M c o

In the R1 occurrences in the above two tables an appropriate r~ was made in order to insure the bi=a~+~. Of course canonicity here has been lost, but using the recurrence relation q,,=a,,q,,-~+q,,-2 we may reduce these tables to

ij

... 0 . . .0 1 0 . . .

a i j + l . . . a i j + 2 l j + 1 a i j+21j+2

. . . 0 . . . . . . . . . . . 0 M o r oo

a i i + l . . . a i j+21j+ 1 oo M co

In the latter case we split the table into a Q2--R1 pair. I f the former table occurs, we are done by Case B1 if {als+2t.+l} clusters at any M, I < M < ~. Further, we

�9 J . . . .

may ignore any subsequence of {a0+2r.+j } tending to mfimty, since the correspond- . 3 . �9

mg .~qo+ztjO represents a F~ contribution to t, and so may be subtracted. Therefore we may assume a~j+21.+l=l, which allows either a Q2- -R1- -Q1 or

�9 $ . . . . .

a Q2- -R1- -Q2 split o f th e table. Q3 has been eliminated from happening infimtely often�9

Henceforth b~EQ forces bi=0. Also we may now arrange that biCR1 forces b~=ai+~ without destroying canonicity. To do this add Z q o O to t, where {b~j} represents those elements of R1 equal to a0+1-1 . Since the previous Q-string now must end with a zero, canonicity is indeed not destroyed. ,Relettering for the last time, we have the following list:

Q. zeroes R1. isolated bi=ai+l

R2. ij R3. ij

o o M - 1 co

M (M = > 2) o~

The study of Q and R is complete, and we now concentrate on the lengths of strings of zeroes in Q.

A c t a Wlathemat ica A c a d e m i a e S c i e n t i a r u m Hungar icae , 37, 1981


w 12. Proof of Theorem 3 (conclusion)

We begin by assuming that the lengths of the zero-strings in Q are even infinitely often. Let these strings be denoted by {bl~., bzs+l . . . . . bmj}. Also let the left ends of the union of the R2 and R3 strings be denoted by {b,~}. It could happen also that {ah+l, a,1+1, %+1 . . . . } tends to infinity, which we now assume.

Now if the number of strings in Q of odd length is finite, it is easy to infer that ~q~O+t=rnO+~q, , jO-~ql jO, or tEF~ Thus we must have also an infinite number of odd lengths; in fact, an infinite number of odd lengths _->3. For if almost all the odd lengths are equal to 1, the above calculation still holds, and tEF~ Let these odd zero-strings of length =>3 be denoted by {b b, bij+ 1, ..., bkj}; we have k j - i j even and positive for all j . Given all the previous assumptions, it cannot be that {aa+l, a k l + l , ai~+i, ...} also tends to infinity. For then let {uj}= {ij}U {/i} in increasing order, and let {vj}={kj}U {m j} also in increasing order. Then t = ~qoj O- ~q , j O- Y~q,~ 0 + toO, a contradiction.

Therefore either {%+1} or {akj+~} is bounded. Suppose {a~ j+ l}~ , and restrict to a constant subsequence of {akj+~}. Then an argument as in Case C1 show that either S = {qkj} or {qkj-1} is valid, or that tEF~ The details are omitted.

Thus {a~+t} has a constant subsequence; restrict to it now, say, a~j+!-M. We treat the following cases.

Case D1. ij Case D2. ij Case D3. ij ~ o 0 0 0 o o L - 1 0 0 0 0 L 0 0 0 ~ M o~ L M L M

For Case D1 we have S = {qi~} if M=>2, unless we further have

ij

o o 0 0 0 ~ o ~ M 1 1

whereupon S={qij+2 }. If M=I , then tCF~ unless {ai;+~} has a constant subsequence. Then S=(q i+ l} on this subsequence satisfies condition A. The

e ~ J v rification of each of these choices for S is routine and is omitted. For Case D2 the choices are all the same as those for D1 except the last one.

Here, if M = I and {aij+~} has a constant subsequence, S={qlj-x} on that subsequence is permitted. Again the verification is omitted, and D3 remains.

Case D3 offers the same choices as D2 in case {as. 1} tends to infinity There- fore, on a subsequence if necessary, {ai._l}_~K. Next suppose {a..+2} has a sub- . . . . J , ~ J ,

sequence tending to mfimty. Restrict to this subsequence. Then if M = I , set S={qij-1}. If M ~ 2 , set S={qij} unless

ij

0 0 1 0 0 0 1 1 n oo ( M ~ 2 )

A c t a M a t h e m a t i c a Academ~ae S c i e n t i a r u m Hungar~cae, 37, 198I

218 M. STEWART

occurs, whereupon S = {qjj-1} is valid. Again the verifications are omitted. In fact, the above argument ~s enough to verify that (J {a,~+l . . . . , a~j+l} is bounded.

1

We are left with the following subcases of D3,

Case D4. ij k~ Case D5. i i k i

O L 0 ... O O R O L 0 ... O 0 ~ K L M . . . P Q R K L M . . . P Q ~o

The notation is ambiguous here if {k s - i s} is unbounded; in this case we require O O " < " ~ " < " < nly {a,~+a+l} t be constant for 0 = h = 3 and {a~,+h+x} constant for - 3 = h = 0 .

Case D5 allows S = {q,~} or {q,~-l}; the verification is again omitted, being exactly similar to that for Case C1. Only D4 is left; we rewrite it as

(12.1) ij kj

O L O ... O N K L M 1 M ~ ... NeN1 N

In this setting it is true that either S={q~ 1}, {q~j}, or {qij+l} satisfies condmon A. The proof revolves writing the three pmrs of mequahties

(12.2a) (Ii~_,-t-Lqej_a)l[qij_lOll + q~s_l(III~j_0 < 1 - �89 (12.2b) (Ii~_l+Lqi~_O[lqij_lOl[ + q~j_l(III~j_0 > �89

(12.2c) (I,j_~+Zq~j_l)llq,jOII- q,~(III~j_a) < 1 - � 8 9

(12.2d) (Itj_z+Z%-Ollq~jOf}- %(IIIe~_l) > 1 - � 8 9

(12.2e) (Ie~-l+Lq~-Ollq~+lOl[ + q~s+l(I[I~j-0 < 1 - �89

(12.2f) (Ie~-a+Lq~-~)llq~+~O[[ + q~j+l(III~j-0 > �89 +c

and determining when either (a and b) or (c and d) or (e and f) are true. Omitting the calculations, the strategy is as follows. First, both b and e are always true, as is at least one of a and c. Thus either a or d or f is required, and sufficient conditions for that are any of the following (Case D4):

(12.3a) L = 1 and MI~<2M~

(12.3b) L ~ 2 and M 2 ~ 2

(12.3c) L => 2, M = 1, and M1 <= 2(M~+l) / (2 -M~) .

Thus in the presence of any one of the above conditions either S = {qz.-1}, �9 . $ .

{q0} or {q~+l} satisfies (10.1). To treat the outstanding cases, the following device will be useful.

Note that l j + l - m s is even for all j by the definitions, and that certain bm +2n+~ are equal to a,, +~,+, and all other b~ are zero in the range {bins+l,

. / J - . . . .

bmj+~ . . . . . btj+l-1}. Define s to be the real number obtained by setting the remaining b,,,+2,+1 equal to am~+~,+2 and setting all other bi=O (i>=O). Then t + s = m O + +'.~qtjO--~q,,~O, so that t+s~rO(o). Therefore, by Lemma 3.2, {llqi~t[[} and

Ac~a M a t h e m a t i c a A c a d e m i a e S c i e n t i a r u m Hungar icae , 37, 1981


{liqiJII} have the same cluster points. This says that to verify condition A we may use s and t interchangeably.

Under the assumptions made in the first paragraph of this section there are five exceptional cases left; we continue to refer to (12.1) for each of them.

Case D5. L>=2, M2-----1, MI>2(M~+I)/(2--M~), and the odd zero-strings

are infinitely often of length 3.

Case D6. L = > 2, M~ = 1, M1 > 2(M~" + 1)/(2 - M~), and the odd zero-strings are infinitely often longer than 3.

Cases D7 D9. L = I and MI > 2M~.

Case D5. We show that either S-- {q~j-1} or {qij} is valid by verifying either (12.2a) or (12.2d). (By the previous argument, this is the only pair left to treat.) We replace MI>2(M~+I)/(2--M~) by the evidently weaker condition M1~5. Referring to s instead of t (see the second previous paragraph for the definition of s), we are done in light of (12.3) for s unless M3~3 (which is weaker than either M3>2(N'+l) / (2-N" ) for N-----1 or M2>2N" for N~2) . We seek to demonstrate

(%-1 + %)1[ %-10It. II % 0tl < (2 - e) (ll qii 01[ -II % +~, 011),

(qb-1 + qb)[] q,,-10[I < ( 2 - c)/M~

((L + l)qb-a + qb_~)/(L' q~j_l + qb_2) < (2-c)/M~,

(L + l)M~qb_~ + M~q,j_2 < (2-c)L ' q , f ~ +(2-c)q,s_ z.

Since M o=1 this last inequality is true if (L+I)M~<(2--c)L'. Now estimate M~<4/3, because /142=1 and M3=>3. We obtain (4/3)(L+l)<(2-c)L', and this is true for all L_->2 no matter how large M~ is.

Case D6. We proceed as in D5; it remains to show only

(q~j-l + q,)ll qfj-lO[l " II q,jOll < ( 2 - c)([[ qzjO[I -It qe~+a 0ll)

in order to prove either~ S = {%-1} or {qij} satisfies (10.1). Now I1%+a011/11%011 =

=I/(M~M~M~M~)<5--c. (This is seen by estimating Mz-~...-=Ms=I and

proceeding as in Lemma 9.5 i).) Therefore the desired inequality becomes

((L + 1) q~j_a + qi~- 2)/(L'q,,_l + q,~- 2) < (8 - c)/5, 5 (L + 1) < (8 - c) L',

which is true for all L--_>2 and all M~. In Cases DT----D9 we may assume L-~I and M1=>3. For each of these cases

either S = {q~j-1} or {q~} is valid, for (12.2d) can itself be verified. The cases are described as follows:

Case D7. kj--ii_~2.

Case D8. k j - i j ~ 4 infinitely often, and blj_z~ab_o " (or bb=_a~j_z-l=L-1 and {bi~_,}-~ oo).

A c t a MaShernat~ca A c a d e m i a e S c i e n t i a r u m Hungar icae , 37, 198!

220 M. STEWART

Case D9. kj-ij>=4 infinitely often, and bi~_3=_O. We present the proof only for D7, the others being of a similar nature.

For D7, first suppose {a~j+3}~. If so, then either S={qk:} or {qk~-~}, the proof is as in Case C1 and is omitted. Therefore suppose {akj+~} is constant, restricting to a subsequence if necessary. Now the zero-strings beginning at bkj+2 must all be odd eventually, by virtue of the initial assumptions of this section. Upon switching to s instead of t, (12.3) now holds for s unless M~=I and M3->3. Now to show (12.2d) it is enough to verify

q,j-1ilq~jOll-%11%+2011 > �89 �89 > c.

This inequality is true because M~M~>4+c, and qij_2/qij<%-c~ for some ca, e2>0.

With the D-series established it is true that in the presence of the outstanding assumptions the odd zero-strings of length _->3 are only finite in number. Thus the even lengths can no longer satisfy {ate+t, am~+~, a12+l . . . . } ~ o , else tEF~

If {a l ,+ l }~ , restrict to a constant subsequence of {a~+~}, say amj+~-M. If M~=2, ~elect S={qm. }. If M : I , then either S={qm.} or {qmj-~}, or ~EF~ the proof again is as in'C1. Therefore we suppose, on ~ subsequence if necessary, that {atj+l}=.M. We treat three major cases.

Case El. lj rnj Case E2. l i rnj co 0 . . . 0 r L - 1 0 ... 0

M ~ L M (L->2)

Case E3. l i my

O L O . . .O L M

First, in each of El--E3 we assume that m j - l j ~ 3 infinitely often. In Case El, if M_->2, put S={qt~}. If M = I , then t~F~ unless {a~ ~_~}

has a constant subsequence. E i t h e r S:{qli} o r {q/j+1} on this subsequence satisfies condition A. The verifications are omitted. In fact, from here to the end of the proof only the proper choices for S will be shown, the proofs being standard.

In Case E2, M=>2 again permits S= {qls}. If M = 1, the only case in which S cannot be {qt~-~} is

lj ~ o L - 1 0 0 0

L 1 1 ~ (L~2) . Here select S= {qt~}.

In Case E3, if {alj_a} ~ ~ is forced, the same choices hold as in E2. Therefore suppose, on a subsequence if necessary, that {a~_a}=-K. Then S= {q~} if M~2, except for the lone subcase

lj

0 0 1 0 0 . . . 1 1 M (M=>2),

A c t a M a t h e m a t i c a Aeacle~ziae S c i e n t i a r u m Hungar icae , 37, 1981


where S={qtj_l} is valid. Now if M = I , choose S={qt;_l} unless

li 0 L 0 0 0 K L 1 1 o ~

occurs. Here one of {q~} or {qzs-1} suffices for S. Now return to El--E3 under the assumption that mj =l j § 1, all j. For Case El,

S--{qtj} if M~2, but if M=I is forced, it can be proved that t=mO§ or tCF~ For CaseE2 again select S~--{ql~ } if M~2, and put S={qb_l} if M = 1. For Case E3 either S= {qlj-1} or {qtj}, as in Case C1. The assumption that {ats+a} has a bounded sequence has led to a determination of S in every instance.

At last we may contradict the opening assumption of this section; namely, that there exists an infinite number of zero-strings in Q of even length. Then the prior entire argument for odd strings (the D-series) goes through again. (The claims therein are the same, but the proofs that depended originally on there being an infinite number of even strings become even more transparent in the presence of only finitely many such strings.) And with this last observation the proof of the theorem is complete.

References

[1] J. P. CONZE, l~quirdpartition et ergodieitd de transformations cylindriques, preprint (Universit6 de Rennes, 1976).

[2] H. FtmSTENBERG, Strict ergodicity and transformations of the torus, Amer. J. Math., 88 (1961), 573--601.

[3] H. KEST~N, On a conjecture of Erd6s and Sziisz related to uniform distribution mod 1, Acta Arith., 12 (1966), 193--212.

[4] A. YA. KHINCnlN, Continued Fractions, University of Chicago Press (Chicago, 1964). [5] L. Kun'ZRS and H. NIEDERREITER, Uniform Distribution of Sequences, John Wiley and Sons

(New York, 1974). [6] J. LESCA, Sur la rrpartition modulo 1 des suites (n~), Seminaire Delange--Pisot--Poitou

(the6rie des nombres) 8e annee (1966/67), no. 2. [7] E. SATAEV, On the number of invariant measures for flows on orientable surfaces, Izv. Akad.

Nauk. USSR, 9 (1975), 813--830. [8] K. SCrlMIDT, Cohomology and skew products of ergodic transformations, preprint (University

of Warwick, 1974). [9] K. SC~IMIDr, A cylinder flow arising from irregularity of distribution, preprint (University of

Warwick, 1975). [10] W. A. VEECI~, Strict ergodicity in zero-dimensional dynamical systems and the Kronecker--

Weyl theorem mod 2, Trans. Amer. Math. Soc., 140 (1969), 1--33. [11] W. A. V~EcrI, Well distributed sequences of integers, Trans. Amer Math. Soc., 161 (1971), 63--70. [12] W. A. VEECH, Topological dynamics, Bull. Amer. Math. Soc., 83 (1977), 775--830. [13] T. VUAVARAOHAVAN, On the fractional parts of powers of a number IV, J. Indian Math. Soc.,

12 (1948), 33--39. [14] H. WEX'L, (3bet die Gleichverteilung yon Zahlen rood Eins, Math. Ann., 77 (1916), 313--352. [15] J. P. CONZ~ and M. KEANE, Ergodicite d'un flot cylindrique, preprint (Universit6 de Rennes).

(Received October 1, 1979)

Added in proof (May 14, 1981). The author has been informed that part i) of Theorem 4 has been strengthened by Ishai Oren; namely, T is ergodic whenever 1, t, and 0 are rationally independent. The proof follows quite similar lines. DEPARTMENT OF MATHEMATICS NORTHEASTERN UNIVERSITY BOSTON, MASSACHUSETTS 02115 USA

Acta rdathematiea Academiae Sc$entiarum Hungaricae, 37, 1981

Irregularities of uniform distribution

Documents

Transcript of Irregularities of uniform distribution