Lebesgue Constants

DOI 10.1007/s00041-009-9115-8

Triangular Dirichlet Kernels and Growth of L

Lebesgue Constants

Marshall Ash

Received: 30 September 2009 / Published online: 24 December 2009

© Springer Science+Business Media, LLC 2009

Abstract LetP be a polygon inZ²and consider the mapping of anL¹(T²)function into the partial sum of its Fourier series determined by the dilate ofP by the integer N. If the image space is endowed with theL^p norm, 1< p <∞, then the operator norm will be given by theL^pnorm of

(m,n)∈N Pe^{2π i(mx+ny)}. The size of this oper- ator norm is shown to beO(N^2(1−1/p))when the polygon is a triangle. The estimate is independent of the shape of the triangle. For aksided polygon the corresponding estimate isO(kN^2(1−1/p)).

Keywords Lebesgue constant·Dirichlet kernels in two dimensions·Dirichlet kernels for polygons

Mathematics Subject Classification (2000) Primary 42B15·42A05·Secondary 42A45

1 Introduction

In one dimension there is a very well known estimate [9]

₁

0

N n=1

e^{2π nix} dx 4

π²lnN. (1)

This integral is called the Lebesgue constant. In higher dimensions there are as many Lebesgue constants as there are generalizations of an interval of integers,

Communicated by Tom Körner.

M. Ash (

DePaul University, Chicago, IL, USA e-mail:[email protected]

{1,2, . . . , N}. One such generalization is the set of two dimensional integer lattice points lying in the right triangle

Kθ,N=

x∈E²:x1+θ⁻¹x2≤N, x1≥0, x2≥0 ,

whereEis the real numbers andθ∈(0,1]. In [8], A.A. Yudin and V.A. Yudin produce an estimate similar to relation (1) for this case. Reference [6] is also relevant here.

Whenp∈(1,∞), theL¹relation (1) has a well knownL^panalogue, namely ₁

0

N n=1

e^{2π nix}

p

dx 2

π _∞

0

sinu u

^pdu N^p−1. (2) A sharp form of this appears as Lemma 1 of [1]. In Theorem1we extend what Yudin and Yudin did for the triangleKθ,N and theL¹norm to the case of the same triangle K_θ,N and theL^pnorm, for every finitep >1. The extension is routine except at one point where the easily verified inequality

d dx

_N

n=1

e^{2π nix}

≤CN

|x|, 0<|x| ≤1 2 proves to be useful.

An immediate corollary for an estimate of anL¹or anL^pLebesgue constant for a triangleKθ,N is a similar estimate for any two dimensional polygon. In theL¹case, results for all polyhedrons in all dimensions are already known. See Chap. 6 of [5] for this. In Corollary2we state theL^p result for polygons which follows immediately from Theorem1. Also results for all dimensions which include those found here when restricted to dimension two have been published recently [3]. However in contrast to the proof given here, the methods employed in [3] are not at all number theoretic.

Also the bounds established in [3], when specialized to two dimensions, are less precise than the bounds found here.

My original interest in this question came from studying a fundamental question from functional analysis: Is a commuting with translation operator of weak restricted type(2,2)necessarily bounded onL²(Z)? The affirmative answer to that question had a strong connection to this question: Given a subset of[0,1]of very small mea- sure, can one find a sum of exponentials with coefficients 0 or 1 whoseL^p norm is largely concentrated on that subset? The result of this paper played an important role in the positive resolution of the second question. See [2] for explanations of these questions and their connections.

2 Results

Forn=(n₁, n₂)∈Z², lete(nx)denote exp(2π i(n1x₁+n₂x₂)).

Theorem 1 LetEbe the real numbers,θ∈(0,1]and K_θ,N =

x∈E²:x₁+θ⁻¹x₂≤N, x₁≥0, x2≥0 .

Then for each finitep >1 and arbitraryN≥1,

T²

n∈K_θ,N∩Z²

e (nx)

p

dx≤CpN^2p−2

uniformly with respect toθandN.

Corollary 2 LetP be aksided polygon inE²of diameterN. Then

T²

n∈P∩Z²

e (nx)

p

dx≤kCpN^2p−2.

The corollary is almost immediate since the polygon can be decomposed intok−2 triangles. See Theorem 2 in [8] for the details. So we pass immediately to the proof of the theorem.

Proof Upon settingM= N, we have

_θ,N(x)=

n∈Kθ,N∩Z²

e (nx)= M n1=0

e (n₁x₁)

θ (N−n1) n2=0

e (n₂x₂)

since the upper edge of the triangle defining K is given by x₁+θ⁻¹x₂=N, or x₂=θ (N−x₁). The inner sum is a geometric series sincee(n₂x₂)=e(x₂)ⁿ² so that we have

θ (N−n₁) n₂=0

e (n₂x₂)=e ((θ (N−n1) +1) x2)−1 e (x₂)−1

and

θ,N(x)= M n1=0

e (n1x1)e ((θ (N−n₁) +1) x2)−1 e (x₂)−1 . Since

(θ (N−n1) +1)=θ (N−n1)+1− {θ (N−n1)}, where here, and hereafter,{· · · }will be reserved to mean fractional part;

θ,N(x)= M n1=0

e (n1x1)e ((θ (N−n₁)+1) x2− {θ (N−n₁)}x₂)−1

e (x₂)−1 .

But

e ((θ (N−n1)+1) x2− {θ (N−n1)}x2)−1

=e ((θ (N−n1)+1) x2)−1

+e ((θ (N−n1)+1) x2− {θ (N−n1)}x2)

−e ((θ (N−n₁)+1) x2)

=e ((θ (N−n1)+1) x2)−1

=e ((θ (N−n1)+1) x2) (e (− {θ (N−n1)}x2)−1) , so

e (n₁x₁) (e ((θ (N−n₁)+1) x2− {θ (N−n₁)}x₂)−1)

=e (n₁x₁) (e ((θ (N−n₁)+1) x2)−1)

+e (n1(x1−θ x2)+(θ N+1)x2) (e (− {θ (N−n1)}x2)−1) , and

_θ,N(x)= M n₁=0

e (n₁x₁)e ((θ (N−n1)+1) x2)−1 e (x2)−1

+ M n1=0

e (n₁(x₁−θ x₂)+(θ N+1)x2)e (− {θ (N−n1)}x2)−1 e (x2)−1

=J1(x)+J2(x) . We first estimate

I1=

T²|J1(x)|^pdx.

For the Dirichlet kernelD_n(x)=_n

ν=0e(νx)=e(^nx₂)^sin(π(n_{sinπ x}^+1)x), we use the esti- mates

D1 |D_n(x)| ≤n+1 n,

D2 |D_n(x)| ≤_|2x¹_| |x|⁻¹if 0<|x| ≤1/2, D3 |D_n(x)| ≤π n(n+1) n²,

D4 |D_n(x)| n|x|⁻¹if 0<|x| ≤1/2.

Estimate D2 uses sinπ x ≥2x on [0,1/2]. Estimate D3 follows from |D_n| =

|_n

ν=12π iνe(νx)| ≤2π_n

ν=1ν. For D4, differentiate the closed form ofD_n fac- tor oute(^nx₂)and then take|e(^nx₂)| =1 into account to get for 0<|x| ≤1/2,

D_n(x)=

π insinπ (n+1) x

sinπ x −πcosπ x

sin²π xsinπ (n+1) x + cosπ (n+1) x

sinπ x π (n+1)

≤ π n 1

2x +

π 1

(2x)²π (n+1) x +

1

2xπ (n+1) n

|x|.

Returning to the estimate ofI₁, from

e (n1x1) (e ((θ (N−n1)+1) x2)−1)

=e ((θ N+1) x2) e (n₁(x₁−θ x₂))−e (n₁x₁) and 1/|e (x₂)−1| |x₂|⁻¹follows

I1

T²|x2|^−p|e ((θ N+1) x2) D_M(x1−θ x2)−D_M(x1)|^pdx. (3) Since

|e ((θ N+1) x2) D_M(x₁−θ x₂)−D_M(x₁)|

= |e ((θ N+1) x2) DM(x1−θ x2)−DM(x1−θ x2) +D_M(x₁−θ x₂)−D_M(x₁)|

≤ |e ((θ N+1) x2)−1| |DM(x1−θ x2)| + |DM(x1−θ x2)−DM(x1)|

= |2 sinπ (θ N+1) x2| · |D_M(x₁−θ x₂)| + |D_M(x₁−θ x₂)−D_M(x₁)|. (4) So to boundI1, it suffices to boundIAandIB, where

IA=

T²|x2|^−p 2 sin1

2(θ N+1) x2

^p|DM(x1−θ x2)|^pdx, and IB=

T²|x2|^−p|DM(x1−θ x2)−DM(x1)|^pdx.

First, the change of variablex₁=x1−θ x2,x₂=x2has Jacobian 1, so I_A=

T|D_M(x₁)|^pdx₁

T|x₂|^−p|2 sinπ (θ N+1) x2|^pdx₂.

The first integral is O(N^p−1) by Lemma 2 of [1], and the substitution u = π(θ N+1)x2shows that the second integral is

_∞

0

− _∞

π (θ N+1)

u π (θ N+1)

^−p|2 sinu|^p du π (θ N+1)

=c_p(θ N+1)^p−1−E_p,θ(N ) where c_p =_∞

0 (^2πsin_u ^u)^{p du}_π and |E_p,θ(N )| (θ N +1)^p−1_∞

π(θ N+1)u^−pdu=

π^−p+1

p−1 . So uniformly inθ,

I_A=O_p

N^2p−2

.

We now turn to the problem of estimatingI_B. For estimatingI_Bwe change nota- tion from(x1, x2)to(x, y)and fromdxtodxdy. We must estimate

IB= _1/2

−1/2

_1/2

−1/2

|y|^−p|DM(x−θy)−DM(x)|^pdxdy.

We make two reductions. First it is enough to integrate over[0,1/2]². This is because symmetry considerations show that _1/2

0

_1/2

0 =₀

−1/2

₀

−1/2 and that the integrals over the other two quadrants are both equal to

1/2 0

1/2 0

|y|^−p|D_M(x+θy)−D_M(x)|^pdxdy.

The estimation of this last integral is very similar to, but slightly simpler than the other one, since the set where x andy are big but x−θy is small needs special consideration in the first case, while the set wherexandyare big butx+θyis small is empty in the second case. Thus we only need to estimate

1/2 0

1/2 0

|y|^−p|D_M(x−θy)−D_M(x)|^pdxdy.

We may also assume thatθ=1, for making the substitutionx=x, y=θygives θ^p−1

θ/2 0

θ/2 0

|y|^−p|D_M(x−y)−D_M(x)|^pdxdy,

which is certainly dominated by _1/2

0

_1/2

0

|y|^−p|DM(x−y)−DM(x)|^pdxdy,

uniformly forθ∈(0,1].

Partition[0,1/2]²into five parts,R, S, T , U, V. A picture is useful:U= [0,_N²] × [0,_N¹] is the lower left corner,V is the remainder of the bottom strip of thickness N⁻¹,S is the remainder of the left strip of thicknessN⁻¹. The remaining square with lower left corner(N⁻¹, N⁻¹)and upper right corner(1/2,1/2)is made up of T, a thin diagonal strip whose upper boundary lies on the line y−x =N⁻¹ and whose lower boundary lies on the liney−x= −N⁻¹and ofR, a union of an upper left triangle and a lower right triangle.

First, sinceSis whereN⁻¹< yand 0< x < N⁻¹, then

S

_1/2

N⁻¹

1 y^p

N⁻¹ 0

(N+N )^pdx

dy

y^−p+1|^N1/2⁻¹

N^pN⁻¹=O

N^2p−2

.

Second,Ris where all ofx, y,and|x−y|are> N⁻¹so that

R

_1/2

N⁻¹

1 y^p

1/2 N⁻¹

1

|x−y|+1 x

p

dx

dy 1/2

N⁻¹

1 y^p

1/2 N⁻¹

1 t^pdt

dy+

1/2 N⁻¹

1 y^p

1/2 N⁻¹

1 x^pdx

dy

=2

y^−p+1|^N_1/2⁻¹ x^−p+1|^N_1/2⁻¹

=O

N^2p−2

.

Notice that here and elsewhere we usef +g^pp≤2^p(f^pp+ g^pp), which we sometimes write in the formf+g^pp f^pp+ g^pp.

Third,T is where|x−y|< N⁻¹, butx > N⁻¹andy > N⁻¹so that

T

T

1

y^p N+1

x

p

dx dy _1/2

N⁻¹

1 y^p

Ly

N^pdt

dy+ _1/2

N⁻¹

1 y^p

Ly

1 x^pdx

dy

=2

y^−p+1|^N_1/2⁻¹ N^p−1+N^p−1 =O

N^2p−2

,

where for the first integral we use|L_y| = |{(x, y)∈T}| =O(N⁻¹)and for the second one we use inf{x:(x, y)∈Ly}> N⁻¹.

Next,Uis wherex <2N⁻¹, y∈(0, N⁻¹), so the estimate

|DM(x−y)−DM(x)| ≤supD_M |y| yields

U

_N−1

0

1 y^p

2N⁻¹ 0

N²y

p

dx

dy

=O

N^2pN⁻¹N⁻¹ =O

N^2p−2

.

What remains isV, wherex >2N⁻¹, y < N⁻¹and we must estimate

V

= _N−1

0

1/2 2N⁻¹

1

y^p|D_M(x−y)−D_M(x)|^pdx

dy.

Apply the mean value inequality|D_M(x−y)−D_M(x)| ≤sup|D_M||y|and estimate (D4) to get

V = _N−1

0

1/2 2N⁻¹

1

y^p|DM(x−y)−DM(x)|^pdx

dy _N−1

0

1/2 2N⁻¹

1 y^p

M^py^p inf0<t <y|x−t|^pdx

dy

M^p _N−1

0

dy _∞

2N⁻¹

1

x^pdx N^pN⁻¹N^p−1=N^2p−2, sincex >2N⁻¹andy < N⁻¹implies that

0<t <yinf |x−t| =x−y≥x 2x.

This completes the proof thatI_BisO(N^2p−2). So we conclude that I1=Op

N^2p−2

. (5)

It remains to show that I₂=

T²|J₂(x)|^pdx=O

N^2p−2

.

Since

|J2(x)| = |e ((θ N+1)x2)|

M n1=0

e (n1(x1−θ x2))e (− {θ (N−n1)}x2)−1 e (x2)−1

=

M n₁=0

e (n₁(x₁−θ x₂))e (− {θ (N−n1)}x2)−1 e (x2)−1

, I2

T²

1

|x₂|^p

M n1=0

e (n (x1−θ x2)) (e (− {θ (N−n)}x2)−1)

p

dx

≤

T²

1

|x2|^p

M n=0

e (n (x₁−θ x₂)) ∞ s=1

(−2π ix2)^s({θ (N−n)})^s s!

p

dx

≤

T²

_∞

s=1

(2π )^s s!

M n=0

e (n (x1−θ x2)) ({θ (N−n)})^s

p

dx

=

T²

_∞

s=1

(2π )^s s!

M n=0

e (nx1) ({θ (N−n)})^s

p

dx.

Denoting(

T²|f|^p)^1/pbyfpthe last inequality is I₂^1/p

∞ s=1

(2π )^s s!

M n=0

e (nx1) ({θ (N−n)})^s p

so by Minkowski’s inequality, I₂^1/p

∞ s=1

(2π )^s s!

M n=0

e (nx1) ({θ (N−n)})^s

p

which can be rewritten as I₂^1/p

∞ s=1

(2π )^s s!

⎛

⎝

T²

M n=0

e (nx₁) ({θ (N−n)})^s

p

dx

⎞

⎠

1/p

=^∞

s=1

(2π )^s s!

⎛

⎝

T

M n=0

e (nx) ({θ (N−n)})^s

p

dx

⎞

⎠

1/p

= ∞ s=1

(2π )^s s!

Hp,θ,N(s)1/p

.

In the penultimate step, we used

T²|f (x1)|dx1dx2=1 0dx2

1

0|f (x1)|dx1=1· 1

0|f (x)|dx. We now have to find a good estimate forH_p,θ,N(s).

If we knew that

H_θ,N(s) N^p−1ln^ps+N^p−1ln^pN+ s^p N^3p, then we would have

I₂^1/p ∞ s=1

(2π )^s s!

N^1−1/plns+N^1−1/plnN+ s N³

= _∞

s=1

(2π )^slns s!

N^1−1/p+ _∞

s=1

(2π )^s s!

N^1−1/plnN

+ _∞

s=1

(2π )^s (s−1)!

N⁻³

=O_p

N^1−1/plnN

, (6)

I2 Op

N^p−1ln^pN

(7) since the infinite sums are constants for eachp. Finally estimates (5) and (7) prove

the theorem.

Lemma 3

H_p,θ,N(s) N^p−1ln^ps+N^p−1ln^pN+ s^p N^3p. Proof Let us set=M⁻⁴and

ϕ(u)=

u^s 0≤u≤1−,

1

(1−)^s(1−u) 1−≤u≤1, and extend it to be 1-periodic. Expandϕinto its Fourier series:

ϕ(u)= ˆ

ϕ(ν) e (νx) .

The Fourier coefficients satisfy

(1) | ˆϕ(ν)| ≤1, ν∈Z,

(2) | ˆϕ(ν)| ≤ |ν|⁻¹Varϕ |ν|⁻¹, ν=0, (3) | ˆϕ(ν)| ≤ |ν|⁻²Varϕ s⁻¹|ν|⁻², ν=0.

To see (1), notice thatϕ is nonnegative and continuous and has its maximum value on[0,1]atx=1− and that value is(1−)^s which is bounded by 1. To see (2), notice thatϕ is monotone increasing on[0,1−]and monotone (linearly) decreasing on[1−,1]so its total variation is bounded by (2). To see (3), notice that

ϕ(u)=

su^s−1 0< u <1−,

−¹(1−)^s 1− < u <1,

so the total variation ofϕ is the rise ofs(1−)^s−1from 0⁺to(1−)⁻ plus the jump ofs(1−)^s−1−(−¹(1−)^s)atx=1−plus the jump of¹(1−)^satx=1.

All three of these quantities are s⁻¹.

Further setA= {n∈Z∩ [0, M] : {θ (N−n)} ≤1−}and letA¯=Z∩ [0, M] \A be the complement ofAin{0,1,2, . . . , M}. Then

H_p,θ,N(s)=

T

M n=0

e (nx) ({θ (N−n)})^s

p

dx

T

n∈A

e (nx) ({θ (N−n)})^s

p

dx

+

T

n∈ ¯A

e (nx) ({θ (N−n)})^s

p

dx

=G1+G2.

Sinceϕ(θ (N−n))= {(θ (N−n))}^s for{(θ (N−n))} ≤1−,we have

G₁=

T

n∈A

ϕ(θ (N−n)) e (nx)

p

dx,

G^1/p₁ =

n∈A

ϕ(θ (N−n)) e (nx) p

=

n∈A

ν∈Z

ˆ

ϕ(ν) e (νθ (N−n)) e (nx) p

≤

ν∈Z

ϕˆ(ν)

e (νθ N )

n∈A

e (n (x−νθ )) p

.

But observing|e(νθ N )| =1, making the variable changex=x−νθ, notingA¯= {0,1, . . . , M} \A, and applying Minkowski’s inequality gives

e (νθ N )

n∈A

e (n (x−νθ )) p

=

n∈A

e (n (x−νθ )) p

=

n∈A

e (nx) p

=

M n=0

e (nx)−

n∈ ¯A

e (nx) p

≤

M n=0

e (nx) p

+

n∈ ¯A

e (nx) p

,

so

G₁≤2^p

ν∈Z

ϕˆ(ν)p

T

M n=0

e (nx)

p

dx+

T

n∈ ¯A

e (nx) ^pdx

.

We use the estimates

|DM|^p=O(N^p−1t )and ∞

ν=−∞

ϕˆ(ν)=ϕˆ(0)+

1≤|ν|≤^s

+

|ν|>^s

1+

1≤|ν|≤^s

1

|ν|+

|ν|>^s

1

|ν|²

1+ln s

+1/

s

=1+ln(sM⁴)+ 1 sM⁴ ln(sN ) ,

and our estimate forG1becomes G1 ln(sN )

⎛

⎝N^p−1+

T

n∈ ¯A

e (nx)

p

dx

⎞

⎠. (8) Turning toG₂, forn∈ ¯A,{(θ (N−n))}>1−, so that

{(θ (N−n))}^s> (1−)^s and

1− {(θ (N−n))}^s=1− {(θ (N−n))}^s

<1−(1−)^s

≤s.

Therefore

G2=

T

n∈ ¯A

e (nx)

1+ {θ (N−n)}^s−1

p

dx

=

T

n∈ ¯A

e (nx)+

n∈ ¯A

e (nx)

{θ (N−n)}^s−1

p

dx

≤2^p

⎛

⎝

T

n∈ ¯A

e (nx)

p

dx+((M+1) s)^p

⎞

⎠.

But=M⁻⁴so

G2

T

n∈ ¯A

e (nx)

dx+ s^p

N^3p. (9)

Our final assertion is that

T

n∈ ¯A

e (nx)

p

dx N^p−1. (10)

If (10) holds, then from (8)

H_p,θ,N(s)≤ G₁+G₂ ln^p(sN )

N^p−1+N^p−1 + s^p

N^3p+N^p−1 N^p−1ln^ps+N^p−1ln^pN+ s^p

N^3p,

since ln^psN=(lns+lnN )^p ln^ps+ln^pN, which finishes the lemma and hence the theorem. So all that remains is the verification of relation (10).

The idea of the rest of the proof is to show that the elements ofA¯lie in an arith- metic progression (or possibly in 2 of them), from which (10) is obvious. Explicitly, ifA¯= {a+b,2a+b, . . . , ka+b} ∪ {c+d,2c+d, . . . , kc+d}, then

T

n∈ ¯A

e (nx)

p

dx≤2^p

T

k j=1

e ((j a+b) x)

p

dx

+2^p

T

j=1

e ((j c+d) x)

p

dx

=2^p ₁

0

k j=1

e ((j a) x)

p

dx+2^p ₁

0

j=1

e ((j c) x)

p

dx

=2^p1 a

_a

0

k j=1

e (jy)

p

dy+2^p1 c

_c

0

j=1

e (jy)

p

dy

=2^p ₁

0

k j=1

e (jy)

p

dy+2^p ₁

0

j=1

e (jy)

p

dy

=O_p(N^p−1).

Thus it only remains to prove the following lemma.

Lemma 4 Let N be a positive real number and letθ∈(0,1]. Let M= Nand =M⁻⁴. ThenA¯= {n∈ {0,1, . . . , M} :1− <{θ (N−n)}<1}is an arithmetic progression or the union of two arithmetic progressions.

Proof We distinguish two cases,θirrational andθrational.

Case 1.θirrational.

Let{Q_k:k=1,2, . . .}be the sequence of denominators of the convergents of the continued fraction expansion ofθ. Define the positive integerk0by

Q_k0≤M < Q_k0+1.

Letn1, n2∈ ¯A. Then from{θ (N−n1)}>1−and{θ (N−n2)}>1−we get θ (n₁−n₂) =min

∈Z|θ (n₁−n₂)−|

≤ |(θ (N−n₂)−θ (N−n₁))−(k₂−k₁)|

= |{θ (N−n2)} − {θ (N−n1)}|

where{θ (N−n_i)} =θ (N−n_i)−k_i for integersk_i. Thus θ (n1−n2)< = 1

M⁴, (11)

where · denotes the distance to the nearest integer.

Since|n1−n2| ≤M, there is anm,1≤m≤Msuch that

θ m< . (12)

From elementary number theory we will need the following three facts.

LetP_k/Q_kandP_k+1/Q_k+1be successive convergents ofθ. Then

Ifm < Qk+1, then for all∈Z, |θ m−| ≥ |θ Qk−Pk|, (13)

|θ Q_k−P_k|> 1 Qk+Qk+1

, (14)

θ− P_k Qk

< 1 QkQk+1

. (15)

See Theorem 7.13 of [7] for the first fact, Theorem 13 in Khinchin’s book [4] for the second, and Theorem 7.11 of [7] for the third.

Next we estimateθ mfrom below. Sincem≤M < Q_k+1, from facts (13) and (14) we get

θ m =min

∈Z|θ m−|> 1

Q_k0+Q_k0+1. (16) Putting (16) together with (12), gives

1

M⁴ = > 1

Qk₀+Qk₀+1≥ 1 M+Qk₀+1

.

This is only possible ifQ_k0+1isO(M⁴); explicitly, from this we get M+Qk₀+1> M⁴,

Q_k0+1> M⁴−M, so that as soon asM≥2,

Q_k0+1≥1

2M⁴. (17)

We show that ifn₁, n₂∈ ¯A, then n₁−n₂≡0 (mod Q_k0). Indeed let n₂−n₁= qQk0+r, where 0< r < Qk0. Then

θ (n₁−n₂) =θ

qQ_k0+r

≥ θ r −θ qQ_k0

≥ θ r −qθ Q_k0. (18) For the last step, note that for any norm and any integerq,

qa = a+ · · · +a

≤ a + · · · + a =qa. We also have

θ r ≥ 1

2Qk₀; (19)

since applying facts (13) and (14) withk=k₀−1 gives θ r =min

∈Z|θ r−|

> 1 Q_k0−1+Q_k0

> 1 Q_k0+Q_k0.

Also

θ Qk₀= Qk₀

P_k0 Q_k0 +

θ Qk₀−Qk₀

P_k0 Q_k0

= Pk₀+

θ− P_k0

Q_k0 Qk₀

=

θ− P_k0 Q_k0 Qk₀

≤Qk₀

θ− P_k0

Q_k0

≤Qk₀

θ− P_k0

Q_k0

< Qk₀

1

Q_k0Q_k0+1 = 1 Q_k0+1. Taking this and inequality (19) into account,

θ r −qθ Q_k0≥ 1

2Qk₀ −q 1 Qk₀+1

.

Combining this with (18) we have

θ (n₁−n₂) ≥ 1 2Qk0

−q 1 Q_k0+1

.

TakingQ_k0≤M,q≤Mand (17) into account, from this we get θ (n1−n2) ≥ 1

2M−M 1

1/2M⁴

= 1/2M³−2M

M⁴ .

As soon asM≥3, this is contrary to

θ (n₁−n₂)< = 1 M⁴.

Consequently all then_i are congruent modQ_k0 and there is a singler, 0≤r < Q_k0 so that to everyn∈ ¯Athere corresponds aq so that

n=qQk₀+r.

Let us consider the sequencexq,0≤q≤t= ^M_Q⁻_k^r

0 , where for each integerq,xq= {θ (N−(qQk₀+r))}is on the circle of circumference 1. Ifq1=q2, thenθirrational implies thatθ (N−(q₂Q_k0+r))−θ (N−(q₁Q_k0+r))=θ (q₁−q₂)Q_k0 is not an integer, so thatx_q1=x_q2.