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arXiv:0802.1910v1 [math.NT] 13 Feb 2008

On a theorem of V. Bernik in the metrical theory of

Diophantine approximation

by

V. Beresnevich (Minsk)1

1. Introduction.

denote the number of elements in a finite set S; the Lebesgue measure

of a measurable set S ⊂ R will be denoted by |S|; Pn will be the set of

integral polynomials of degree ? n.

denote the height of P, i.e. the maximum of the absolute values of its

coefficients; Pn(H) = {P ∈ Pn: H(P) = H}. The symbol of Vinogradov

≪ in the expression A ≪ B means A ? CB, where C is a constant. The

symbol ≍ means both ≪ and ≫. Given a point x ∈ R and a set S ⊂ R,

dist(x,S) = inf{|x−s| : x ∈ S}. Throughout, Ψ will be a positive function.

Mahler’s problem. In 1932 K. Mahler [10] introduced a classification

of real numbers x into the so-called classes of A,S,T and U numbers ac-

cording to the behavior of wn(x) defined as the supremum of w > 0 for

which

|P(x)| < H(P)−w

We begin by introducing some notation: #S will

Given a polynomial P, H(P) will

holds for infinitely many P ∈ Pn. By Minkovski’s theorem on linear forms,

one readily shows that wn(x) ? n for all x ∈ R. Mahler [9] proved that

for almost all x ∈ R (in the sense of Lebesgue measure) wn(x) ? 4n, thus

almost all x ∈ R are in the S-class. Mahler has also conjectured that for

almost all x ∈ R one has the equality wn(x) = n. For about 30 years the

progress in Mahler’s problem was limited to n = 2 and 3 and to partial

results for n > 3. V. Sprindzuk proved Mahler’s conjecture in full (see [12]).

A. Baker’s conjecture. Let Wn(Ψ) be the set of x ∈ R such that there

are infinitely many P ∈ Pnsatisfying

(1)

|P(x)| < Ψ(H(P)).

1The work has been supported by EPSRC grant GR/R90727/01

Key words and phrases: Diophantine approximation, Metric theory of Diophantine

approximation, the problem of Mahler

2000 Mathematics Subject Classification: 11J13, 11J83, 11K60

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A. Baker [1] has improved Sprindˇ zuk’s theorem by showing that

|Wn(Ψ)| = 0 if

∞

?

h=1

Ψ1/n(h) < ∞ and Ψ is monotonic.

He has also conjectured a stronger statement proved by V. Bernik [4] that

|Wn(Ψ)| = 0 if the sum

(2)

∞

?

h=1

hn−1Ψ(h)

converges and Ψ is monotonic. Later V. Beresnevich [5] has shown that

|R \ Wn(Ψ)| = 0 if (2) diverges and Ψ is monotonic. We prove

Theorem 1. Let Ψ : R → R+be arbitrary function (not necessarily

monotonic) such that the sum (2) converges. Then |Wn(Ψ)| = 0.

Theorem 1 is no longer improvable as, by [5], the convergence of (2) is

crucial. Notice that for n = 1 the theorem is simple and known (see, for

example, [8, p.121]). Therefore, from now on we assume that n ? 2.

2. Subcases of Theorem 1.

sets denoted by Wbig(Ψ), Wmed(Ψ) and Wsmall(Ψ) consisting of x ∈ R such

that there are infinitely many P ∈ Pnsimultaneously satisfying (1) and one

of the following inequalities

Let δ > 0. We define the following 3

(3)1 ? |P′(x)|,

(4)H(P)−δ? |P′(x)| < 1,

(5)

|P′(x)| < H(P)−δ

respectively. Obviously Wn(Ψ) = Wbig(Ψ)∪Wmed(Ψ)∪Wsmall(Ψ). Hence to

prove Theorem 1 it suffices to show that each of the sets has zero measure.

Since sum (2) converges, Hn−1Ψ(H) tends to 0 as H → ∞. Therefore,

(6) Ψ(H) = o(H−n+1) as H → ∞.

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3. The case of a big derivative. The aim of this section is to prove

that |Wbig(Ψ)| = 0. Let Bn(H) be the set of x ∈ R such that there exists a

polynomial P ∈ Pn(H) satisfying (3). Then

(7)Wbig(Ψ) =

∞ ?

N=1

∞ ?

H=N

Bn(H).

Now |Wbig(Ψ)| = 0 if |Wbig(Ψ)∩I| = 0 for any open interval I ⊂ R satisfying

(8)0 < c0(I) = inf{|x| : x ∈ I} < sup{|x| : x ∈ I} = c1(I) < ∞.

Therefore we can fix an interval I satisfying (8).

By (7) and the Borel-Cantelli Lemma, |Wbig(Ψ) ∩ I| = 0 whenever

(9)

∞

?

H=1

|Bn(H) ∩ I| < ∞.

By the convergence of (2), condition (9) will follow on showing that

(10)

|Bn(H) ∩ I| ≪ Hn−1Ψ(H)

with the implicit constant in (10) independent of H.

Given a P ∈ Pn(H), let σ(P) be the set of x ∈ I satisfying (3). Then

(11)Bn(H) ∩ I =

?

P∈Pn(H)σ(P).

Lemma 1. Let I be an interval with endpoints a and b. Define the following

sets I′′= [a,a + 4Ψ(H)] ∪ [b − 4Ψ(H),b]

sufficiently large H for any P ∈ Pn(H) such that σ(P) ∩ I′?= ∅, for any

x0∈ σ(P) ∩ I′there exists α ∈ I such that P(α) = 0, |P′(α)| > |P′(x0)|/2

and |x0− α| < 2Ψ(H)|P′(α)|−1.

and

I′= I \ I′′. Then for all

The proof of this Lemma nearly coincides with the one of Lemma 1 in

[5] and is left for the reader. There will be some changes to constants and

notation and one also will have to use (6).

Given a polynomial P ∈ Pn(H) and a real number α such that P′(α) ?=

0, define σ(P;α) = {x ∈ I : |x − α| < 2Ψ(H)|P′(α)|−1}. Let I′and I′′be

defined as in Lemma 1. For every polynomial P ∈ Pn(H), we define the set

ZI(P) = {α ∈ I : P(α) = 0 and |P′(α)| ? 1/2}.

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By Lemma 1, for any P ∈ Pn(H) we have the inclusion

(12)σ(P) ∩ I′⊂

?

α∈ZI(P)σ(P;α).

Given k ∈ Z with 0 ? k ? n, define

Pn(H,k) = {P = anxn+ ··· + a0∈ Pn(H) : ak= 0}

and for R ∈ Pn(H,k) let Pn(H,k,R) = {P ∈ Pn(H) : P − R = akxk}. It is

easily observed that

(13)Pn(H) =

n?

k=0

?

R∈Pn(H,k)Pn(H,k,R)

and

(14)#Pn(H,k) ≪ Hn−1

for every k.

Taking into account (11), (13), (14) and that |I′′| ≪ Ψ(H), it now becomes

clear that to prove (10) it is sufficient to show that for every fixed k and

fixed R ∈ Pn(H,k)

(15)

???

?

P∈Pn(H,k,R)

σ(P) ∩ I′??? ≪ Ψ(H).

Let k and R be fixed. Define the rational function˜R(x) = x−kR(x).

By (8), there exists a collection of intervals [wi−1,wi) ⊂ I (i = 1,...,s),

which do not intersect pairwise and cover I, such that˜R(x)′is monotonic

and does not change the sign on every interval [wi−1,wi). It is clear that s

depends on n only. Let ZI,R=?

and ZI,R∩ [wi−1,wi) = {α(1)

i

Pn(H,k,R), we obviously have the identity

P∈Pn(H,k,R)ZI(P), ki= #(ZI,R∩[wi−1,wi))

i,...,α(ki)

}, where α(j)

i

< α(j+1)

i

. Given a P ∈

xkP′(x) − kxk−1P(x)

x2k

=

?P(x)

xk

?′

=˜R(x)′.

Taking x to be α ∈ ZI(P) leads to

Now, by Lemma 1, |σ(P;α)| ≪ Ψ(H)|P′(α)|−1≪ Ψ(H)|˜R′(α)|−1.

P′(α)

αk

=˜R(α)′. By (8), |P′(α)| ≍ |˜R(α)′|.

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Using (12), we get

???

?

P∈Pn(H,k,R)

σ(P) ∩ I′??? ≪ Ψ(H)

s

?

i=1

ki

?

j=1

1

|˜R(α(j)

i)|

Now to show (15) it suffices to prove that for every i (1 ? i ? s)

(16)

ki

?

j=1

|˜R′(α(j)

i)|−1≪ 1.

Fix an index i (1 ? i ? s). If ki? 2 then we can consider two sequen-

tial roots α(j)

i

and α(j+1)

i

of two rational functions˜R + ai,j

respectively. For convenience let us assume that˜R′is increasing and posi-

tive on [wi−1,wi). Then˜R is strictly monotonic on [wi−1,wi), and we have

ai,j

k

?= ai,j+1

k

. It follows that |ai,j

Theorem and the monotonicity of˜R′, we get

k and˜R + ai,j+1

k

k− ai,j+1

k

| ? 1. Using The Mean Value

1 ? |ai,j

0− ai,j+1

0

| = |˜R′(α(j)

i) −˜R′(α(j+1)

i

)| =

= |˜R′(˜ α(j)

i)| · |α(j)

i

− α(j+1)

i

| ? |˜R′(α(j+1)

i

)| · |α(j)

i

− α(j+1)

i

|,

where ˜ α(j)

α(j+1)

i

i

is a point between α(j)

− α(j)

i

and α(j+1)

i

. This implies |˜R′(α(j+1)

i

)|−1?

i, whence we readily get

ki−1

?

j=1

|˜R′(α(j+1)

i

)|−1?

ki−1

?

j=1

?

α(j+1)

i

− α(j)

i

?

= α(ki)

i

− α(1)

i

? wi− wi−1.

The last inequality and |˜R′(α(1)

verified that (16) holds for every i with ki? 2 and is certainly true when

ki= 1 or ki= 0. This completes the proof of the case of a big derivative.

i)| ≍ |P′(α(1)

i)| ≫ 1 yield (16). It is easily

4. The case of a medium derivative.

val I satisfying (8). The statement |Wmed(Ψ)| = 0 will now follow from

|Wmed(Ψ) ∩ I| = 0. We will use the following

As above we fix an inter-

Lemma 2 (see Lemma 2 in [6]). Let α0,...,αk−1,β1,...,βk∈ R?{+∞}

be such that α0> 0, αj> βj? 0 for j = 1,...,k − 1 and 0 < βk< +∞.

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