A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
C. V IOLA
Diophantine approximation in short intervals
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 6, n
o4 (1979), p. 703-717
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C. VIOLA
(*)
1. - Introduction.
A
simple
consequence of some basicinequalities
in theelementary theory
of continued fractions is that
or that
(see [2]),
if andonly if $
is ao badly approximable
» irrationalnumber,
i.e. one
having
boundedpartial quotients
in its continued fractionexpansion.
Similar bounds
frequently
occur e.g. in Fourieranalysis, and
one may need conditionsensuring
the existence ofsuch $ satisfying
suitable additionalassumptions. Clearly
the bestpossible
constantimplied by
theC-symbol
in
(1)
will beobtained,
for anysufficiently large Q, when $
isequivalent
tounder a unimodular
transformation,
i.e.(*) Istituto di Matematica, Universita di Pisa.
Pervenuto alla Redazione il 29 Settembre 1978 ed in forma definitiva il 22
Mag-
gio 1979.
For any fixed « E R and
any 6
>0,
the interval(« - 6,
«+ 6)
containsinfinitely
manysuch
and it is natural toexpect
that there will be at least onesatisfying
with the
appropriate
constant Cindependent of $,
as soon asprovided
the functionf(3)
of thelength
of the interval tendsrapidly enough
to
infinity
as 6 - 0.The
problem
offinding
suitable a E R and functionsf( 3 )
of suitablegrowth,
such that for
every 6
in some range 0 6 L1 thereexists $
=$(3)
asabove,
seems not to have been stated sofar,
and is theobject
of thepresent
paper. This arises in a natural way in the formulation of the Dirichlet pro- blem with
approximate
data for thevibrating string equation ([5]).
2. - Definitions and statement of results.
Let 0
4
oo and letf(3)
> 0 be a function in the interval 06
L1 such thatDEFINITION.
V(I, d )
=!7(f( ð), d )
is the setof
real numbers asatisfying
the
following
condition :For any
ð,
06 4,
thereexists $
such thatfor
allintegers
Here
11 x 11
denotes the distance from the real number x to the nearestinteger.
. Thusmax
(sin (nq$)) -2
=(sin (a
min11 q$ 11 ))-2 .
1-a-Q 1aQ ’
From the
theory
of continued fractions(inequality (4) below)
one seesthat the constant
(3 - Ý5)/2
is bestpossible,
andcorresponds
to a suitablechoice
of $
in theequivalence
class of(Ý5 + 1)/2. Replacing (3 - A/6)/2
by
some smallerpositive
constant in the above definition would allow oneto
take $
in otherequivalence classes,
but would not affect the set/7(/, 4 ).
If
Lll Ll2,
thenobviously
Similarly,
iff(6) g(6)
thenThe set
Y(f, L1)
isclearly
defined modulo 1. In otherwords,
for every
integer
m.Our theorem 1 is:
We must therefore
analyse
the structure ofY(f, Zt)
when/(!)
=o(1/6) (6
-0).
Theorem 1 is in factessentially
bestpossible
if onerequires
!7(f, A)
=R,
for we prove(corollary 1)
thatFor any d > 0 and any
function /(!)
=o(llb) (6 --> 0),
every numbercx,e is
irrational.This result is obtained as a
corollary
of afairly general
statement(theorem 2),
from which we also derive(corollary 2):
Let
X f# Q
and let co > 1.If
there existinfinitely
manyp/q
EQ
such thatthen, for
any L1 > 0 and anyfunction
we haveIn the
opposite
direction we prove(theorem 3) :
If
for
some constantsH > 0,
(o > 1 andfor
allp /q
EQ,
thenFor convenience we introduce the
following
DEFINITION. Let
oc 0 Q.
Thetype of
DC is the upper bound SZof
the setof
rosatisfying
f or infinitely many plq E Q.
If a has
type Q,
then1 S2 c)o. By
Roth’stheorem,
everyalgebraic
irrational number has
type
1.Also, by
well known results in the metrictheory
of continued fractions([4]),
or as asimple
consequence ofcorollary
3and theorem 4
below,
we have that almost all numbers(i.e.
allexcept
a setof measure
zero)
havetype
1.From the above results we
immediately
obtain(corollaries
3 and6):
I f
a hastype
Q >1,
thenf or
any haveand any
K,
d > 0 weIf
oc hastype
thenfor
any there exists such thatThe latter statement shows that the sets
g(Kb-O, c)o), where -1
01,
are
« large ». However,
since thedependence
of the constantK(o, ot)
on xis
highly erratic,
it is desirable to evaluate the measure ofQ(K6-6, 00)
forfixed
0,
K. We have(theorem 4) :
For any 0 such
that 2
0 1 and any .K >0,
theLebesgue
measureof
the set
exceeds
In
particular,
for any fixed6, 2
81,
and any 8 >0,
the abovemeasure exceeds 1- e
provided K > C(O) (1 Is) 1-0.
It is
plain
from the above results that 0= 2
is a critical value for the setsS°(K6-°, L1).
In this case we prove(corollaries 5
and7) :
If
ot has unboundedpartial quotients
in its continuedfraction expansion then, for
anyK, L1
>0,
This
implies
inparticular
thatY(Kb-il J)
is a set of measure zero forevery
K,
4 > 0(see [4]).
If 0153 f# Q
has boundedpartial quotients a;A,
thenwith
On
combining
these results we have(corollary 8):
U Y(Kb-17 oo) is
the setof
irrational numbers with boundedpartial quotients.
.g > o
The method of theorems 2 and 3
applies
also to numbers a of infinitetype.
In this case, the definition oftype
can benaturally
modifiedby introducing
anypositive,
continuous anddecreasing
function17(x)
of thereal variable x > 1
satisfying 11 qa 11 > 77 (q)
for allintegers q > 1,
andIIqncxll 11 27 (q,,,)
forinfinitely
manyintegers qn>l.
We let1p(x)
=77 (x) Ix
and denoteby 1p-l
the inverse function.The
following
isagain
an immediate consequence of theorem 2:Let
a 0 Q, let V(x) > 0
be adecreasing
continuousf unction of x> 1,
andlet
pn/qn
EQ
be a sequencesatisfying
Then, for
any L1 > 0 and anyf unction f(ð)
=o(l/ð1p-l(ð)) (b
--->0),
wehave
On the other
hand,
anargument
similar to theproof
of theorem 3yields (theorem 5):
Let
q (x)
> 0 be adecreasing
continuousfunction of
x>lsatisfying
for
everyinteger
q:>1,
and letV (x) = q (x) Ix.
ThenThis
gives (corollary 9):
For every
a 0 Q
there exist and such that3. - Notation.
Vinogradov’s symbol
« indicatesinequality containing
anunspecified positive
constant factor.As
previously stated, llxli
is the distance from x to the nearestinteger.
We denote
by [a, b]
the closed intervala x b
orb c x c a, by (a, b)
the
corresponding
openinterval,
andsimilarly
for[a, b).
Asusual,
thesimple
continued fraction
where the
partial quotients
ao, al , a2 , ... areintegers satisfying ai > I
fori > 1,
is denoted
by
We follow Cassels’ tract
[1]
for the notation ofconvergents:
ifwe let
and
so that the
convergents to ;
areAlso,
we denoteby $;
=[ai,
ai+1,...]
thecomplete quotient
of the con-tinued fraction
(3) corresponding
to thepartial quotient
ai.We shall use in the
sequel
severalelementary
results about continued fractions andFarey
sequences, withoutquoting
themexplicitly. They
canall be found e.g. in
[3], [4].
4. - Proofs.
LEMMA 1. Let
plq, rls
E[0, 1]
befractions satisfying
ps - qr= ±
1.Then there
exists $
E(p/q, r/s)
such thatfor
allintegers Q > max fq, s}.
PROOF.
If q
= s =1,
so that we takeThe sequence
(qi)
of the denominators of theconvergents to $
is the Fibonacci sequenceql==q2:=l,q,=2,....
For anyinteger Q>max{q,8}=l
there exists
i > 2
such thatqi Q qi+l -
Thenmin and
Hence
If assume and let
where
(If q
s, we should define andproceed similarly).
Then rls and p/q
are two consecutiveconvergents
to$, whence $ c- (plq, rls), and, for a
suitablem>2,
where
The sequence
(qi) for $
is now such that 81 q. = q. Hence for anyinteger Q >
maxfq, sl
= q = qm thereexists i > m
such that qiQ
qi+1.Again
and
(4)
follows.LEMMA 2. Let a E
[0, 1)..F’or
0 6 d let n =n(b)
be the leastinteger
> f(6), :Fn
theFarey
sequenceof
order n, and letpjq
andrls
be two succes-sive terms
of :F n
such thatIf
for
every thenPROOF. We have and
If $
is as in lemma1,
and for any
integer Q > f(6)
we haveQ > n > max fq, s), whence, by
lemma1,
THEOREM 1:
PROOF.
By (2)
it suffices to prove that everyxe[0yl) belongs
toY(1-16, oo ) .
Weapply
lemma 2 with L1 = oo,f(b)
=1/6.
If n,p /q
andr /s
are as in lemma
2,
we havefor if
q = 1
then s = n, if s=l then q = n, and otherwiseq>2, s>2,
qs>q+
s > n. The theorem follows from lemma 2.THEOREM 2, Let a E R and
f (b)
be such that there exist sequences6,,
-7 0and
p n /qn
EQ satisfying
Then, for
anyPROOF. We will show that for any constant C > 0 and any
sufficiently large n,
ifthen
Q min Ilk11
C for someQ > f(6n). By (5)
andi)
we obtainIf we let
then, by i)
andii),
Hence,
for anyinteger Q satisfying
we have
by (6)
COROLLARY 1. For any L1 > 0 and any
function f satisfying
we have
PROOF. We take in theorem
2,
and choosefor all n. The
assumptions
of theorem 2 are then verified for any sequence3n
- 0. Hencep Iq 0 :7(f, d ).
COROLLARY 2. Let
oc 0 Q
and let co > 1.If
there existinfinitely
manyp /q
EQ’v uch
thatthen, for
any d > 0 and anyfunction
we havePROOF. Let be a sequence
satisfying
We may
apply
theorem 2 with the choice since andCOROLLARY 3.
If
a hastype
Q >1,
thentor
any any constantsK,
L1 > 0 we haveand
PROOF.
By corollary
2 with 1 coD
such thatCOROLLARY 4. Let .
have unbounded
partial quotients
ai, let(ain)
be asubsequence of (ai)
such thatand let
(qi)
denote the sequenceof
the denominatorsof
theconvergents
to a.I f f( ð) satisfies
then, f or
anyPROOF. We have whence
Since
for all i,
we mayapply
theorem 2 to the sequences andCOROLLARY 5.
If Lx 0 Q
has unboundedpartial quotients then, for
anyconstants
K, L1
>0,
PROOF. We
apply corollary
4 to the function The abovecondition
becomesand follows
immediately
from(7).
THEOREM 3. Let
If
for
some constants andf or
all thenPROOF. As in the
proof
of theorem 1 we may assumeex E (0, 1).
Wenow
apply
lemma 2 with If n,p /q
andr/s
are as in lemma
2,
we have andOn
multiplying by
we obtainBy
Holder’sinequality,
Therefore
whence
and the theorem follows from lemma 2.
COROLLARY 6.
If
a hastype
Q c>o, thentor
any exists K =g ( o, a )
> 0 such thatthere
PROOF.
By
theorem 3 with 0 =a)/(l + w),
so thatcv > SZ,
and .K = 2°H.COROLLARY 7.
If
has bounded
partial quotients, aiA,
thenwith
PROOF. For we have
We may therefore
apply
theorem 3 with cv =1 and Hencewith COROLLARY 8.
partial quotients.
I is the set
of
irrational numbers with boundedPROOF.
By
corollaries1, 5
and 7.THEOREM 4. For any 0 such
that -1
0 1 andf or
any K >0,
where I" denotes the
Lebesgue
measure.REMARK. It is clear from the
following proof
that the constantOf(20 - 1 ) occurring
above could bereplaced by
where C
is the Riemann zeta-function.PROOF. Let D be the set of those 6 > 0 that are either rational or such that
f(3)
= K3-° is aninteger.
D is countable and dense in(0, o).
Forany 6
>0,
letEð
be the set ofirrational $ satisfying
for all
integers Clearly
Therefore g(f,c5o)
n[0, 1)
is a Borel set and hence isLebesgue-measurable.
On
writing
we have to show
that,
for any w > 1 andany H
>0,
By
theorem3,
if a E[0, 1)
and there exists such thatHence
Trivially
and
(8)
follows.THEOREM 5. Let be a
decreasing
continuousfunction of
x> 1satisfying
f or
everyinteger
and let ThenPROOF
(*).
We may assume0153 E (0, 1).
Weapply
lemma 2 with d ==
V(1), f(6)
=2/by,-1(6).
If n,plq
andr/s
are as in lemma2,
with q,>s,then q + s > n and q
>n/2,
whenceWe have
(*)
I am indebted to D.Zagier
for some remarks which led to thepresent proof
of theorem 5.
and therefore
Since q
isdecreasing,
thisyields
whence
and
again
the theorem follows from lemma 2.COROLLARY 9. I’or
every (x 0 Q
there exist and such thatPROOF.
By
theorem 5.REFERENCES
[1] J. W. S. CASSELS, An Introduction to
Diophantine Approximation, Cambridge
Tracts in Mathematics and Mathematical
Physics,
vol. 45,(Cambridge
1957).[2] G. H. HARDY - J. E. LITTLEWOOD, Some
problems of Diophantine approxima-
tion : a series
of
cosecants, Collected papers of G. H.Hardy,
vol. I, OxfordUniversity
Press (1966), pp. 237-252.[3] G. H. HARDY - E. M. WRIGHT, An Introduction to the
Theory of
Numbers,fourth edition, Oxford
University
Press (1960).[4] A. YA. KHINCHIN, Continued
fractions (English
translation. TheUniversity
ofChicago
Press, 1964).[5] G. PAPI FROSALI, On the