A Scaling Property of Farey Fractions

European Journal of Mathematics (2016) 2:383–417 DOI 10.1007/s40879-016-0098-0

RESEARCH ARTICLE

A scaling property of Farey fractions

Matthias Kunik1

Received: 18 September 2015 / Revised: 26 January 2016 / Accepted: 28 January 2016 / Published online: 25 February 2016 © Springer International Publishing AG 2016

Abstract The Farey sequence of order n consists of all reduced fractions a/b between 0 and 1 with positive denominator b less than or equal to n. The sums of the inverse denominators 1/b of the Farey fractions in prescribed intervals with rational bounds have a simple main term, but the deviations are determined by an interesting sequence of polygonal functions fn.Forn →∞we also obtain a certain limit function, which describes an asymptotic scaling property of functions fn in the vicinity of any ﬁxed fraction a/b and which is independent of a/b. The result can be obtained by using only elementary methods. We also study this limit function and especially its decay behaviour by using the Mellin transform and analytical properties of the Riemann zeta function.

Keywords Farey sequences · Riemann zeta function · Mellin transform · Hardy spaces

Mathematics Subject Classiﬁcation 11B57 · 11M06 · 44A20 · 42B30

1 Introduction

The Farey sequence Fn of order n consists of all reduced fractions a/b with natural denominator b n in the unit interval [0, 1], arranged in order of increasing size. Dropping just the restriction 0 a/b 1 gives the inﬁnite extended Farey sequence Fext n of order n. We use these notations from Deﬁnitions 2.1 and 2.3 and present an overview of new results in each section.

B Matthias Kunik [email protected]

1 IAN, Universität Magdeburg, Gebäude 02, Universitätsplatz 2, 39106 Magdeburg, Germany 123 384 M. Kunik

If the denominator b of a reduced fraction a/b is small compared to n, then the Fext Farey sequence n has a regular behaviour in a sufﬁciently small vicinity of a/b.To illustrate this we consider the Farey sequence F100 in the vicinity of 0/1. In this case we have 50 subsequent fractions

1 1 1 1 1 < < < < ···< . (1) 100 99 98 97 51

These are followed by 16 blocks consisting of two fractions 1/(q +2), 2/(2q +3), ranging from q = 48 down to q = 33, namely

1 2 1 2 1 2 1 2 < < < < ···< < < < . (2) 50 99 49 97 36 71 35 69

After a small “transient section”, consisting of the three fractions

1 2 3 < < , (3) 34 67 100 the reader will find further regular patterns with bigger blocks in F100. In Sect. 2 we determine the general law behind these examples, which describes Fext the local behaviour of n for large values of n in the vicinity of a fixed reduced fraction a/b. For this purpose we derive a simple explicit representation formula for the corresponding neighbouring fractions in terms of the function ξ+(q) in Lemma 2.5. Fext This lemma enables a simple representation and calculation of sections in n like (1) Fext and (2), whereas general transient sections in n like (3) are defined and determined in Lemma 2.7. These results are summarized in Theorem 2.8, the main result of Sect. 2, which is needed for Sect. 4. In Sect. 3 we study three families of 1-periodic functions

∗, , : R → R n n fn which are related to the Farey sequence Fn. Now we deﬁne these functions and explain how they describe the structure of the Farey sequence. In addition, we obtain interesting relations between the functions n and fn, the prime number theorem and the Riemann hypothesis. ∗ The 1-periodic functions n are determined in the unit interval. With the Farey fractions in Deﬁnition 2.1 and for aα−1/bα−1 t < aα/bα, α = 1,...,N, we put + + ∗( ) =−bα bα−1 − aα aα−1 . n t t 2 bα + bα−1

∗( ) = ∗( ) = / ∗ In addition we require n 1 n 0 1 2. Then the function n is linear between two subsequent Farey fractions in Fn, and has a jump of height 1/bα at each Farey fraction aα/bα ∈ Fn. Using only elementary properties of Farey sequences, we obtain in Theorem 3.2 the following L2(0, 1)-estimate: 123 A scaling property of Farey fractions 385

Ψ* 20 0.5

0.4

0.3

0.2

0.1 (t) 0 * 20 Ψ −0.1

−0.2

−0.3

−0.4

−0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t axis ∗ Fig. 1 The jumps of 20 are centered around zero

1 ∗ ∗ log n 2 = (t)2 dt = O , n →∞. (4) n 2 n n 0

∗ The plot of 20 in Fig. 1 shows that the jump discontinuities are centered symmetrically 2 ∗ around zero. This holds in general and enables the estimation of the L -norm of n . Using β(t) = t −t−1/2, the 1-periodic functions n : R → R introduced in Deﬁnition 3.4 may be rewritten as

β(jt) μ(k) (t) =− , t ∈ R. n j k jn kn/j

∗ F Like n , the function n is linear between two subsequent Farey fractions in n, and also has a jump of height 1/bα at each Farey fraction aα/bα ∈ Fn. A plot of 20 is given in Fig. 2. We derive the general Theorem 3.8 to estimate the L2-norm of arbitrary linear combinations of β(j ·). Its application to n gives absolute constants c1, c2 > 0 such that there holds for all n 3, n μ( ) 2 c1 1 k 2 n log n log log n j2 k 2 j=1 kn/j n (5) 1 μ(k) 2 c log n log log n . 2 j2 k j=1 kn/j 123 386 M. Kunik

Ψ 20 0.5

0.4

0.3

0.2

0.1 (t) 0 20 Ψ −0.1

−0.2

−0.3

−0.4

−0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t axis

Fig. 2 The jumps of 20 are not centered around zero

2 Here h2 is the L (0, 1)-norm of any square integrable function h. ∗ Recall that n and n both have its jumps of height 1/b at each Farey fraction a/b. Thus it is natural to deﬁne the following continuous function fn : R → R for each n ∈ N by ( ) = ∗( ) − ( ). fn t n t n t

The elementary properties of these polygonal functions fn are collected in Theo- rem 3.6. Using the notations from Deﬁnition 2.1 and

n aα ϕ(k) ξα = ∈ Fn, sn = , bα k k=1 we especially obtain that fn gives control of the partial sums of the inverse denominators 1/b of the Farey fractions a/b ∈ Fn, namely

α 1 1 1 fn(ξα) = snξα + 1 + − for all α = 0,...,N. 2 bα bβ β=0

Aplotof f20 is given in Fig. 3. It follows from (5), (4) and the prime number theorem in Prachar [15, Kapitel III, 2 Satz 5.1] that the L -norms of n and fn tend to zero for n →∞, see Corollary 3.9. It also results Corollary 3.10, a variant of the Franel–Landau theorem for the charac- terization of the Riemann hypothesis, see [3, Chapter 12.2], [4,8,10,13]. Section 4 contains the main result, Theorem 4.6, which was obtained by using only elementary properties of Farey fractions, namely from Theorem 2.8. Theorem 4.6 123 A scaling property of Farey fractions 387

f = Ψ* – Ψ 20 20 20 0.1

0.08

0.06

0.04

0.02

(t) 0 20 f −0.02

−0.04

−0.06

−0.08

−0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t axis

Fig. 3 The polygonal function f20 describes the behaviour of the Farey sequence and the function fn with large order n in the vicinity of any ﬁxed fraction in terms of a limit function g as follows: Assume / ∈ Fext that a b n and put a x a g , (n, x) = b f + − f , x ∈ R. a b n b bn n b

Then for n →∞the sequence of functions ga,b(n, ·) converges uniformly on each interval [−x∗, x∗], x∗ > 0 fixed, towards a continuous, piecewise smooth and odd limit function g : R → R, which is independent of the choice of the fraction a/b. The finer shades in the distribution of the Farey fractions can be described in terms of the polygonal functions fn, and especially in the vicinity of any fixed fraction a/b they are determined by the asymptotic scaling property of the functions fn in Theorem 4.6. In Sect. 5, Theorem 5.4, we have derived an integral representation for the limit function g by using analytical properties of the Riemann zeta function, the Mellin transform and the theory of Hardy spaces. This integral representation and Fourier analysis is used in order to justify in Corollary 5.5 that the limit function g decays to zero.

2 Farey sequences of order n ∈ N = n ϕ( ) F Deﬁnition 2.1 For n and N k=1 k the Farey sequence n of order n consists of all reduced and ordered fractions

0 = a0 < a1 < a2 < ···< aN = 1 1 b0 b1 b2 bN 1 123 388 M. Kunik with 1 bα n for α = 0, 1,...,N. = = 5 ϕ( ) = + = Example 2.2 For n 5 and N k=1 k 10 we have the following N 1 11 fractions: 0 < 1 < 1 < 1 < 2 < 1 < 3 < 2 < 3 < 4 < 1 . 1 5 4 3 5 2 5 3 4 5 1 Fext F Deﬁnition 2.3 The Farey sequence n is the extension of n consisting of all reduced fractions a/b with a ∈ Z and b ∈ N, b n.

The following theorem provides two important and well-known properties of the Farey Fext sequences n , which can be obtained by employing results from the textbook [6, Chapter III]. Theorem 2.4 (basic properties of Farey fractions) (a) Assume that a, a ∈ Z and b, b , n ∈ N with (a, b) = (a , b ) = 1. Then the / < / Fext reduced fractions a b a b are consecutive fractions in n if and only if

a b − ab = 1, b n, b n and b + b > n.

(b) Assume that a/b < a∗/b∗ are consecutive reduced fractions in the extended Fext Farey sequence b of order b. If b n, then the two numbers ∗ + ( − ∗)/ a < a a n b b b b∗ + b(n −b∗)/b

Fext are consecutive reduced fractions in the extended Farey sequence n of order n.

Lemma 2.5 Assume that a/b < a∗/b∗ are consecutive reduced fractions in the Fext ∈ N extended Farey sequence b of order b n, where n . We deﬁne

a∗ + aq ξ+(q) = , q 0, (6) b∗ + bq and for α ∈ N the quantities

⎧ ∗ ⎪ n − b ⎨⎪ ,α= 1, ∗ b Qα = ∗ (7) ⎪ n − αb + b ⎩ ,α 2, αb as well as ∗ ∗ n + 1 − αb − b qα = . (8) αb Then the inequalities ∗ ∗ − 0 qα+1 q Qα 1 123 A scaling property of Farey fractions 389

determine exactly those integer numbers q 0 for which ξ+(q +q ) gives a section of Fext F = / = / n if q takes all values of α, starting with q 1 1 and running down to q 0 1.

Proof We ﬁrst note that ∗ qα 0 for all α ∈ N (9) simply follows from 1 b∗ b n. Assume that 0 q = a /b < q = a /b 1 are consecutive reduced fractions in Fα. Now we form the two fractions

∗ ∗ aa + (a +aq)b aa + (a +aq)b ξ+(q + q ) = <ξ+(q + q ) = (10) ba + (b∗+bq)b ba + (b∗+bq)b for an integer number q 0, and note that all matrices in ∗+ + ( ∗+ ) + ( ∗+ ) bb bq · a a = ba b bq b ba b bq b aa∗+aq b b aa + (a∗+aq)b aa + (a∗+aq)b have determinant one. We conclude that the fractions in (10) are already reduced.

Using Theorem 2.4(a), we conclude that ξ+(q +q ) and ξ+(q +q ) are consecutive Fext fractions of n if and only if the following three inequalities are satisﬁed:

∗ ba + (b +bq)b n, (11) ∗ ba + (b +bq)b n, (12) ∗ b(a +a ) + (b +bq)(b +b ) n + 1. (13)

The left-hand side in (13) takes its smallest value for a = 1, b = α, a = 0, b = 1, provided that q < q are any two consecutive fractions in Fα. Hence (13) is satisﬁed if and only if the integer q satisﬁes the inequality

∗ . q qα+1 (14)

The largest value on the left-hand side in (11) and (12) results for α = 1 with a = b = 1 and a = 0, b = 1, respectively, and for α 2 with a = α − 1, b = α.In any case inequalities (11) and (12) are satisﬁed for the integer number q if and only if

∗ q Qα − 1, (15)

provided that q < q are any two consecutive fractions in Fα. Inequalities (9), (14) and (15) conclude the proof of the lemma. √ Lemma 2.6 Assume that n, b, A ∈ N are such that n 3b and A n/(2b) − 1/4. Using the quantities in Lemma 2.5, we obtain

∗ < ∗ α = ,..., . 0 qα+1 Qα for all 1 A 123 390 M. Kunik

∗ ∗ Proof For α = 1wehaveQα =(n −b )/b 2 due to n 3b, and a simple ∗ < ∗ α estimation shows that 1 q2 Q1 in this case. Now assume that 2. In order to ∗ < ∗ α show qα+1 Qα for all A we only have to guarantee that

n + 1 − (α +1)b∗ − b n − αb∗ + b + 1 < − 1. (α +1)b αb

This inequality is satisﬁed if and only if n + b 1 1 α< + − , (16) 2b 16b2 4b √ and the right-hand side in (16) is larger than n/(2b) − 1/4.

Fext , , ∈ N Lemma 2.7√ (transient sections of n ) Assume that n b A are such that n 3b /( ) − / ∗ = ∗ ∗ = ∗ − and A n 2b 1 4. Then we have either Qα+1 qα+1 or Qα+1 qα+1 1 for all α = 1,...,A − 1. Fext In the latter case we call the section of n in the interval ξ ( ∗ ), ξ ( ∗ − ) + qα+1 + qα+1 1

Fext = ∗ − a transient section (of n ) with index q qα+ 1. The transient section with 1 index q contains all fractions ξ+(q +q ) with q ∈ Fα ∩ (0, 1], and besides these only fractions of the form ξ+(q +a/(α +1)), where a ∈[1,α] is relatively prime to α + 1.

Proof For 1 α

n − (α +1)b∗ + b n + 1 − (α +1)b∗ − b η+ = ,η− = , (α +1)b (α +1)b and obtain

∗ ∗ 2b − 1 Qα+ =η+, qα+ =η−,δ= η+ − η− = ∈ (0, 1), 1 1 (α +1)b because b 1 and α + 1 2. Case 1: There is an index k ∈ N such that

k − 1 <η− k η+ < k + 1.

=η =η ∗ = ∗ <δ< In this case we have k − + and hence Qα+1 qα+1. Regarding 0 1 we only have to study the following case. Case 2: There is an integer number k 0 with k <η− <η+ < k + 1. Here we η = + η = ∗ = ∗ − obtain that − k 1 and + k, i.e. Qα+1 qα+1 1. Only in this case we = ∗ − have to justify the stated properties of the transient section with index q qα+1 1. 123 A scaling property of Farey fractions 391

∗ ∗ From Lemma 2.6 we ﬁrst obtain that qα+ < Qα since 1 α

a a 0 q = < q = 1 b b

in Fα, and we will also admit that q = 0/1. Then the mediant ξ formed with ξ+(q +q ) and ξ+(q +q ) is given by the reduced fraction

( + ) + ( ∗+ )( + ) ξ = a a a a aq b b b(a +a ) + (b∗+bq)(b +b ) with the denominator

B = b(a +a ) + (b∗+bq)(b +b ).

Fext = ∗ − The transient section of n with index q qα+1 1 violates (14) and thus contains at least one mediant ξ of this form with

B n, (17)

ξ ∈ Fext such that n and

ξ ( ∗ ) ξ<ξ ( ∗ − ). + qα+1 + qα+1 1

For any such mediant ξ we will prove next that

b + b = α + 1. (18)

We assume that b + b α + 2. Lemma 2.6 and α + 1 A imply that

∗ ∗ ∗ ∗ n + 1 − (α +2)b − b q = qα+ − 1 = Qα+ > qα+ , 1 1 2 (α +2)b which contradicts (17):

∗ B b(a +a ) + (b +bq)(α+2) ∗ ∗ > b(a +a ) + b (α +2) + n + 1 − (α +2)b − b

= n + 1 + b(a +a −1) n + 1. 123 392 M. Kunik

We have shown (18) and hence obtain

∗ B = b(a +a ) + (b +bq)(α+1) ∗ ∗ b + b (α +1) + n + 1 − (α +1)b − b − b(α +1) = n + 1 − b(α +1).

We summarize these inequalities in

n + 1 − b(α +1) B n. (19)

Fext It remains to show that the transient section of n with index q contains besides the fractions ξ+(q +q ) with q ∈ Fα ∩(0, 1] only some mediants ξ = ξ+(q +a/(α +1)), where a = a + a ∈[1,α] is relatively prime to b + b = α + 1. For this purpose we will use the inequality x 1 3 1 − < x − − for all x 3. (20) 2 4 4 2 = ∗ − We regard that q qα+1 1 implies

b∗(α +1) + n + 1 − (α +1)b∗ − b n + 1 − b(α + 2) b∗+bq − b = , α + 1 α +1 and note the equivalence n + 1 − b(α +2) 1 n 3 1 b(α +1) ⇐⇒ α + 1 − + − + . α + 1 2 b 4 b Using (20) we obtain with x = n/b 3 for all α = 1,...,A − 1, n 1 n 3 1 1 n 3 1 α + 1 − < − − < − + − + , 2b 4 b 4 2 2 b 4 b and hence ∗ b +bq b(α +1).

The denominator B of the reduced fraction representing ξ+(q +q ) satisﬁes

∗ B = ba + (b +bq)b b(α +1),

and the denominator B of the reduced fraction representing ξ+(q +q ) the inequality

∗ B = ba + (b +bq)b b(α +1).

The last two inequalities and (19) imply that

B + B > n, B + B > n, 123 A scaling property of Farey fractions 393 which concludes the proof of the lemma. The following theorem is a slightly reformulated summary of the preceding lemmas. Fext , , ∈ N Theorem 2.8 (local structure√ of n right to a/b) Assume that n b A are such that n 3b and A n/(2b) − 1/4. We use the notations of the function ξ+ in (6), ∗ ∗ ( ∗ , ∗ ]⊇( ∗ , ∗ ] of the quantities Qα in (7) and qα in (8). Then we have Qα+1 Qα qα+1 Qα α = ,..., − ( ∗ , ∗ ] for all 1 A 1, and each interval qα+1 Qα has at least length one. We Fext = Fext ∩[ξ ( ∗ ), ξ ( ∗ )) deﬁne the section n,α n + Qα + Qα+1 . ∗ = ∗ Fext ξ (ξ) (a) In the case Qα+1 qα+1, the section n,α consists exactly of the fractions + ξ ∈ Fext ∩ ( ∗ , ∗ ] with α Qα+1 Qα . ∗ = ∗ ∗ = ∗ − Fext (b) In the case Qα+1 qα+1, we have Qα+1 qα+1 1, and n,α contains all ξ (ξ) ξ ∈ Fext ∩ ( ∗ , ∗ ] = ∗ − fractions + with α Qα+1 Qα . For the index q qα+1 1 we Fext obtain a transient section in n,α which also contains some fractions of the form ξ+(q +a/(α+1)), where a ∈[1,α] is relatively prime to α + 1. Finally, besides these fractions of the form ξ+(ξ) and ξ+(q +a/(α+1)) there is no other fraction Fext in n,α.

∗, F 3 A family of functions n n and fn related to n

Deﬁnition 3.1 We use Deﬁnition 2.1 and put for aα−1/bα−1 t < aα/bα, α = 1,...,N: + + ∗( ) =−bα bα−1 − aα aα−1 , n t t 2 bα + bα−1 ∗( ) = ∗( ) = / and in addition n 1 n 0 1 2. Then we have the 1-periodic extension ∗ : R → R n . ∗ F The function n is linear between two subsequent Farey fractions in n, and has a jump of height 1/bα at each Farey fraction aα/bα ∈ Fn. 2 ∗ Theorem 3.2 (L -estimate of n )

1 log n ∗2 = ∗(t)2dt = O , n →∞. n 2 n n 0

Proof For a t b and s, q ∈ R we consider f (t) = s ·t + q. Then

b b − a f (t)2 dt = f (a)2 + f (a) f (b) + f (b)2 dt. (21) 3 a

We put ηα = (1/bα−1 + 1/bα)/2forα = 1,...,N,use(21) and obtain

N η2 N ∗2 = 1 α − 1 1 . n 2 (22) 3 bα bα− 4 b2 b2 α=1 1 α=1 α α−1 123 394 M. Kunik

From the symmetry relation bN−α = bα, which is valid for α = 1,...,N, and from Theorem 2.4(a) it follows for n > 1 that

N n 1 1 1 = 2 . (23) b2 b2 j2 (k − j)2 α=1 α α−1 j=(n+1)/2 n+1k<2 j (k, j)=1

The following inequality is valid for all natural numbers a b:

b 1 dx = 1 − 1 . m2 x2 a b a

We use it for n > 1toestimate

N n j 1 1 1 2 2 2 2 2 α= bα bα−1 =( + )/ j = − + m 1 j n 1 2 m n j 1 1 1 1 4 2 − + (24) j2 n − j j n2 n/2j

We approximate the right-hand side in (24) by the integral

n−1 2x − n dx 2 1 1 log n 2 · = log (n − 1) − + = O n − x x3 n2 n2 n2 (n − 1)2 n2 n/2 for n →∞. It follows from (23) that

N 1 log n = O for n →∞. (25) b2 b2 n2 α=1 α α−1

Finally we make use of the inequalities n < 2ηα 2 bα bα−1 and η2 2 α n 2n α α− 3 3 2 2 b b 1 bα bα−1 bα bα−1 in order to conclude with (25) and (22) the desired norm inequality. 123 A scaling property of Farey fractions 395

∗2 ∗ Remark 3.3 The estimate of n 2 is possible, because the jumps of n are symmetrically centered around zero. This is illustrated for n = 20 in Fig. 1. Note that (25) also follows from [5, Theorem 1]. Next we deﬁne a sequence of functions n, which ∗ have the same height of the jumps at each Farey fraction like n .

Deﬁnition 3.4 For k ∈ N and t ∈ R we make use of the 1-periodic function kt kt 1 qk(t) =− μ(d) − − d d 2 d|k in order to deﬁne for n ∈ N the 1-periodic function

n 1 = q . n k k k=1

Aplotof20 is presented in Fig. 2. We note that in general these jumps are not symmetrically centered around zero. Note that qk(t) has a jump of height +1 at each reduced fraction t = j/k but is ∗ continuous elsewhere. Comparing n with n , we obtain that the function n is linear between two subsequent Farey fractions in Fn and also has a jump of height 1/bα at each Farey fraction aα/bα ∈ Fn. ∗ Therefore, the difference between n and n is free from jumps, which justiﬁes the following notion.

Deﬁnition 3.5 For n ∈ N we deﬁne the polygonal function fn : R → R as

( ) = ∗( ) − ( ). fn t n t n t

Aplotof f20 is depicted in Fig. 3.

Theorem 3.6 (elementary properties of fn)

(a) fn : R → R is continuous and odd. (b) fn(0) = fn(1/2) = fn(1) = 0. (c) The function fn is linear between two subsequent Farey fractions ξα−1 = / <ξ = / F α = ,..., aα−1 bα−1 α aα bα in n for 1 N. = n ϕ( )/ η = η = (d) If we deﬁne the sum sn k=1 k k, 0 0 and the quantities α (1/bα−1 + 1/bα)/2 for α = 1,...,N, then we obtain

α fn(ξα) = sn ξα − ηβ for all α = 0,...,N. β=0

∗, Proof Statement (c) simply follows from the deﬁnitions of n n and fn, and (b) from ∗ (a). To show (a) we ﬁrst conclude that fn is continuous, since n and n both have the 123 396 M. Kunik same height 1/bα of the jump at aα/bα ∈ Fn.Fort ∈ (0, 1)\Fn we have n(1−t) = −n(t). The Farey fractions ξα = aα/bα ∈ Fn satisfy symmetry conditions

bN−α = bα, aN−α = bα − aα,ξN−α = 1 − ξα for all α = 0,...,N.

∗( − ) =−∗( ) ∈ ( , )\F These conditions also imply that n 1 t n t for t 0 1 n. We obtain that fn(−t) = fn(1−t) =−fn(t) for t ∈[0, 1] and hence for all t ∈ R, since fn is continuous on R. Finally note that statement (d) is equivalent to sn bα + bα−1 fn(ξα) − fn(ξα−1) = − ηα = sn − (ξα − ξα−1) (26) bα bα−1 2 for all α = 1,...,N, because fn is continuous with fn(0) = 0. Now (26) follows ∗ ( ) =−( + )/ ( ) =− ξ < <ξ from n t bα bα−1 2 and n t sn for α−1 t α. The following result can be checked by using 1fork = 1, μ(d) = 0fork > 1. d|k

Lemma 3.7 Let λ: N → C be a completely multiplicative function, i.e. λ(1) = 1 and λ(mk) = λ(m)λ(k) for all m, k ∈ N. We deﬁne the two divisor matrices Aλ,n = n×n n×n (a jk) j,k=1,...,n ∈ C and Bλ,n = (bkm)k,m=1,...,n ∈ C with entries ⎧ ⎨ k λ for j|k, = a jk ⎩ j 0 for j |k, ⎧ ⎨ m m μ λ for k|m, = bkm ⎩ k k 0 for k |m.

Then the matrix Bλ,n is the inverse of Aλ,n, i.e. n 1 for j = m, a jkbkm = δ jm = 0 for j = m. k=1

This lemma is needed for the proof of the next theorem. Theorem 3.8 (L2-estimates for spaces of discrete functions) There are constants T c1, c2 > 0 such that the following two estimates are valid for all x = (x1,...,xn) ∈ Rn and all n 3:

(a) | |2 n 2 c1 x 2 xkβ(k ·) c2|x | log n log log n log n log log n k=1 2 123 A scaling property of Farey fractions 397

with the 1-periodic sawtooth function β : R → R given by

1 β(t) = t −t− . 2

(b) | |2 n 2 c1 B1,n x 2 xk qk c2|B1,n x | log n log log n. log n log log n k=1 2 Here |y| denotes the Euclid-norm of any vector y ∈ Rn.

Proof We make use of Lemma 3.7. First we show that (b) follows from (a), then we T n prove (a). For a given vector x = (x1,...,xn) ∈ R , we put

T y = (y1,...,yn) =−B1,n x with y j =− μ(α)xαj for j = 1,...,n, αn/j and obtain n n n y j β(j ·) =− μ(α)xαj β(j ·) = xk qk. j=1 j=1 αn/j k=1 Hence (b) follows from (a) if we replace x by y in (a). To prove (a) we deﬁne the multiplicative function h : N → N by h =|μ|∗I , where ∗ denotes the Dirichlet convolution of arithmetical functions and I : N → N the identity, I (k) = k for all k ∈ N. For every arithmetical function λ: N → C we deﬁne the diagonal matrix Dλ,n = n×n (λ( j)δjk) j,k=1,...,n ∈ C . An elementary, but lengthy calculation yields n 2 1 √ 2 T n xkβ(k ·) = D ϕ / , A1/I,n x , x = (x1,...,xn) ∈ R . (27) 12 h I n k=1 2

Regarding that n ϕ( ) ( ) √ 2 k h k 2 D ϕ / , x = x h I n k2 k k=1 T n for all x = (x1,...,xn) ∈ R with 6 ϕ(k)h(k) 1 = 1 − 1 π 2 k2 p2 p|k we obtain 6 |x |2 D√ x 2 |x |2, x = (x ,...,x )T ∈ Rn. (28) π 2 ϕh/I,n 1 n 123 398 M. Kunik

For the harmonic series we have

n 1 H(n) = = O(log n) for n →∞. (29) k k=1

Next we note the following inequality between multiplicative functions: 1 1 k ∗1 (k) = for all k ∈ N, (30) I d ϕ(k) d|k which follows from the following evaluation at prime powers pα > 1: ∞ α 1 α 1 1 p p ∗1 (p ) = = = . I d p j p − 1 ϕ(pα) d|pα j=0

From Prachar [15, Kapitel I, Satz 5.1] we obtain

k C log log k for all k 3 (31) ϕ(k) with an absolute constant C > 0. Now we apply the Cauchy–Schwarz inequality to ⎛ ⎞ 2 2 ⎝ 1 xαβ ⎠ |A / , x | = √ · √ , 1 I n β β αn βn/α and obtain

x2 2 1 αβ |A / , x | · 1 I n β β αn βn/α βn/α 2 xαβ 1 H(n) = H(n) x2. β d k αn βn/α kn d|k

Using (29), (30), (31), we obtain a constant C > 0, such that we have for n 3 and x ∈ Rn, 2 2 |A1/I,n x | C|x | log n log log n. (32) β( ·) +···+ β( ·)2 We conclude the upper bound for x1 1 xn n 2 in (a) from (27), (28) and (32). Since |μ| 1, we have componentwise

|B1/I,n( j, k)| |A1/I,n( j, k)| for all j, k = 1,...,n, 123 A scaling property of Farey fractions 399 and hence 2 2 |B1/I,n x | C|x | log n log log n (33) ∈ Rn β( ·)+···+ β( ·)2 for n 3 and x . The lower bound for x1 1 xn n 2 in (a) follows −1 = > from Lemma 3.7 with B1/I,n A1/I,n and (33), (28), (27) for appropriate c1 0 and all n 3: 2 2 2 |x | = B1/I,n A1/I,n x C|A1/I,n x | log n log log n 1 √ 2 D ϕ / , A /I,n x log n log log n 12c h I n 1 1 n 2 log n log log n = xkβ(k ·) . c 1 k=1 2 For the next two corollaries we make use of log n f + O for n →∞, (34) n 2 n 2 n which follows from the deﬁnition of fn and Theorem 3.2.

2 Corollary 3.9 (decay of L -norms of functions fn)

lim fn2 = 0. n→∞

Proof Put xk = 1/k for the components of x in Theorem 3.8(b) and form the vector

T y n = (y1,...,yn) = B1,n x (35) with components 1 μ(k) y = for j = 1,...,n. (36) j j k kn/j From Prachar [15, Kapitel III, Satz 5.1] we obtain uniformly in x 1

∞ √ √ μ(k) − − ξ dξ = O e c log x + O e c log (37) k ξ kx x > μ( )/ with an absolute constant c 0. The sums kx k k are bounded for x 1, which gives for n →∞ μ( ) 2 2 1 k 1 |y n| = + O √ . √ j2 k n j n kn/j 123 400 M. Kunik

From (37) we obtain for any ﬁxed α>0 and x 2 that μ(k) c (α) , k (log x)α kx and the right-hand side in Theorem 3.8(b) converges to zero,

2 lim n = 0. n→∞ 2

Now (34) concludes the proof of the corollary.

Corollary 3.10 (variant of the Franel–Landau Theorem) The Riemann hypothesis is equivalent to the following statement: For all ε>0 there is a constant c(ε) > 0 such that c(ε) f for all n ∈ N. n 2 n1/2−ε ( ) = μ( ) Proof We put M x kx k and start with

− μ(k) M(n) n 1 M(k) = + for all n ∈ N. (38) k n k(k +1) kn k=1

The prime number theorem,

∞ M(n) μ(k) lim = = 0 for all n ∈ N, n→∞ n k k=1 allows us to form the limit n →∞in (38),

∞ M(k) = 0, k(k +1) k=1 which gives the following representation with an absolutely convergent series on the right-hand side:

∞ μ(k) M(n) M(k) = − for all n ∈ N. (39) k n k(k +1) kn k=n

From [14] it is known that Riemann’s hypothesis is equivalent to the statement |M(x)| c(ε)x1/2+ε for x 1 and all ﬁxed ε>0, and the proof follows with (39), Theorem 3.8(b) and (34)using(35) and (36) in the proof of Corollary 3.9. 123 A scaling property of Farey fractions 401

4 Behaviour of functions fn near Farey fractions

Our aim is to describe the behaviour of fn in the vicinity of a ﬁxed reduced fraction / ∈ Fext ∈ N a b n for large values of the Farey index n .

Lemma 4.1 6 1 ϕ(k) log t − = O for t →∞. π 2 t k t kt

Proof Using that ϕ( ) − μ( ) k = p 1 = d , k p d p|k d|k we have for all n ∈ N

n ϕ(k) μ(d) n μ(d) 6n = = = − R k d d d π 2 n kn kn d|k d=1 with the error term μ(d) n n μ(d) R = n + − n d2 d d d d>n dn and the estimate

∞ dx 1 |R | n + 2 + log n for all n ∈ N, (40) n x2 d n dn which concludes the proof of the lemma.

Lemma 4.2 Assume that the continuous function f :[1, ∞) → (0, ∞) is convex and monotonically decreasing. Deﬁne the primitive function F :[1, ∞) → (0, ∞) by ( ) = x ( ) ∈ N F x 1 f t dt. Then we have for all q± with q− q+,

1 F(q+) − F(q−) + ( f (q−) − f (q+)) f (q) 2 q−q

Proof We prove the equivalent inequalities

1 F(q+) − F(q−) − ( f (q−) − f (q+)) f (q) F(q+) − F(q−). 2 q−

Summation with respect to q shows that it is sufﬁcient to consider the special case q+ = q− + 1: We obtain from the monotonicity of f that F(q− + 1) − F(q−) f (q− + 1) = f (q). q−

From the convexity of f we have

1 F(q− + 1) − F(q−) ( f (q−) + f (q− + 1)), 2 and hence

1 F(q− + 1) − F(q−) − ( f (q−) − f (q− + 1)) f (q− + 1) = f (q). 2 q−

Lemma 4.3 With the conditions and notations from Theorem 2.8 we deﬁne for α = 1,...,A − 1 the section Fext = Fext ∩ ξ ( ∗ ), ξ ( ∗ ) n,α n + Qα + Qα+1 and the sum a 1 Sα, = . n b b / ∈Fext a b n,α ( ) = /( ∗+ ) = ( ∗ ) = ( ∗ ) Put x+ q n b bq for q 0,xα x+ Qα ,xα+1 x+ Qα+1 . Then we obtain (α + ) 2 α ϕ( ) − 1 b 2 − 2 − b ( − ) a − k xα+1 xα+1 xα xα+1 xα bSα,n log 2 n2 2n b k xα k=1 bα bα (xα+ − xα) + . n 1 n + 1 − b(α +1)

Proof We make use of Theorem 2.8. For the partial sum

a = 1 Sα,n b b a /b =ξ+(ξ) ξ∈Fext∩( ∗ , ∗ ] α Qα+1 Qα

+ of Sα,n(a/b) and with Fα ={ξ ∈ Fα : 0 <ξ 1} we obtain the representation 123 A scaling property of Farey fractions 403

∗ Qα−1 a 1 Sα, = (41) n b ba + (b∗+bq)b + = ∗ a /b ∈Fα q Qα+1 and the estimates a a a α Sα,n Sα,n Sα,n + . (42) b b b n + 1 − b(α +1)

The right-hand side of (42) results from the fractions ξ+(q +a/(α +1)) in the transient sections, using (19) in the proof of Lemma 2.7 and the fact that ϕ(α+1) α is the number of integers a ∈[1,α], which are relatively prime to α + 1. We obtain from (41) and (42) that

Q∗ −1 a 1 α 1 α Sα,n + b b b∗+bq n + 1 − b(α +1) + = ∗ a /b ∈Fα q Qα+1 ∗ − α ϕ( ) Qα 1 α = k 1 + . ∗+ + − (α + ) = k = ∗ b bq n 1 b 1 k 1 q Qα+1

Now we apply Lemma 4.2 with f (q) = 1/(b∗+bq) and get ! α ϕ( ) ∗+ ∗ a k 1 b bQα + 1 − 1 Sα,n log ∗ ∗ ∗ ∗ ∗ ∗ b k b b +bQ b +bQ b +bQα k=1 α+1 α+1 (43) α + . n + 1 − b(α +1)

∗ Regarding x+(q) = n/(b +bq) we obtain from (43) the right-hand side of the desired inequality. For the proof of the left-hand side we start from

1 = 1 · 1 ba + (b∗+bq)b b (b∗+bq) 1 + a /b · b/(b∗+bq)

1 − a · b . b (b∗+bq) b 2 (b∗+bq)2

Then (41) and (42) yield a a Sα, Sα, n b n b ∗ ∗ Qα−1 Qα−1 (44) 1 1 − a b . b b∗+bq b 2 (b∗+bq)2 + = ∗ + = ∗ a /b ∈Fα q Qα+1 a /b ∈Fα q Qα+1 123 404 M. Kunik

We apply again Lemma 4.2 and infer that

∗ − ! Qα 1 α ϕ( ) − 1 1 k 1 xα+1 + xα+1 xα ∗ log (45) b b +bq k b xα 2n + = ∗ = a /b ∈Fα q Qα+1 k 1 as well as

∗ − Qα 1 α ϕ( ) a b 1 + 1 k · − xα + xα+1 b 2 (b∗+bq)2 2 2 k n n / ∈F+ q=Q∗ k=1 a b α α+1 ! (46) + b − b . ( ∗+ ∗ )2 ( ∗+ ∗ )2 b bQα+1 b bQα

For (46) we note that

− α ϕ( ) a = b a = 1 1 = 1 + 1 k . 2 2 + b b 2 b 2 2 = k a /b ∈Fα a /b ∈Fα a /b ∈Fα k 1

We use ! α ϕ( ) 1 + 1 k · b − b ∗ ∗ 2 ∗ ∗ 2 2 2 k (b +bQ ) (b +bQα) k=1 α+1 α + 1 b 2 2 · xα+ − xα , 2 n2 1 subtract (46) from (45), and establish with (44) the left-hand side of the desired inequality.

Lemma 4.4 Using the conditions and notations from Lemma 4.3 and Theorem 2.8 we obtain the following estimates: (i) 2 xα bα bα log + ,α= 1,...,A, α n n

(ii) xα+1 + xα 2(2α +1), α = 1,...,A − 1, (iii) 0 < xα+1 − xα 3, α = 1,...,A − 1, (iv) −bα/n xα − α 2α2b/n, α = 1,...,A.

∗ Proof Recall the deﬁnition of Qα in (7). For α = 1 we obtain

∗ ∗ n − b b + bQα n, 123 A scaling property of Farey fractions 405 and for α 2, n − b(α −1) n + b b∗ + bQ∗ . α α α

This gives n n α xα α ,α= 1,...,A, (47) n + b n − αb n n (α +1) xα+ (α +1) ,α= 1,...,A − 1. (48) n + b 1 n − αb Regarding b n/3, we have for α = 1,...,A: nb n n αb < √ < , (49) 2 6 2

n < 2. (50) n − αb

This implies (ii), xα + xα+1 2(2α +1), α = 1,...,A − 1. We use (47), (49) and (50) to derive (i): 2 xα n b b bα b α log log =−log 1 + − − − , α n + b n n n n ∞ j xα 1 1 αb log log = α 1 − αb/n j n j=1 α α 2 α 3 b + 1 b + b +··· n 2 n n α α 2 α α 2 = b + 1 b n < b + b . n 2 n n − αb n n

We show (iii) and take care for α(α+1) A2 n/(2b): n αn 1 1 + α xα+ − xα (α +1) − = 1 + αb + 1 n − αb n + b b + n n − αb 1 α(α+1) 3 n/(2b) 3 1 n < 1 + + b < + b = + < 3. 2 n − αb 2 n − αb 2 2 n − αb

Finally, (iv) easily follows from (47).

The following function g plays a crucial role in the present paper, especially as a limit function in Theorem 4.6 below.

123 406 M. Kunik

Lemma 4.5 Deﬁne the continuous, piecewise smooth and odd function g : R → R by ⎧ ⎪ 6 1 ⎪ − xfor0 x 1, ⎪ π 2 ⎨⎪ 2 x ( ) = 6 1 6 1 ϕ(k) g x ⎪ − + − dt for x > 1, (51) ⎪ π 2 2 π 2 t k ⎪ kt ⎩⎪ 1 −g(−x) for x < 0.

Then we have for k ∈ N and k < x k + 1,

k 6 ϕ(m) x g(x) = g(k) + (x −k) − · log , (52) π 2 m k m=1 and especially

k− 6 1 1 ϕ(m) m g(k) = k − + · log for all k ∈ N. (53) π 2 2 m k m=1

Proof Since g is continuous and piecewise smooth, (52) can be obtained by forming the derivatives on each interval (k, k +1), whereas (53) results from (52) by Abelian summation.

/ ∈ Fext Theorem 4.6 (scaling property of fn near Farey fractions) Assume that a b n and put a x a g , (n, x) = b f + − f , x ∈ R. a b n b bn n b

Then for n →∞the sequence of functions ga,b(n, ·) converges uniformly on each interval [−x∗, x∗],x∗ > 0 ﬁxed, towards the limit function g√ from Lemma 4.5. More precisely, if n, b, A ∈ N such that n 3b and A n/(2b) − 1/4, then we obtain for all x ∈ R with |x| An/(n +b),

+ 2 log n 2 b |g , (n, x) − g(x)| A + cA . (54) a b n n

Here c > 0 is an absolute constant.

∈ N / ∈ Fext Proof For given n and given reduced fraction a b n there holds the symmetry relation

ga,b(n, x) =−gb−a,b(n, −x), x ∈ R. 123 A scaling property of Farey fractions 407

Thus it is sufﬁcient to prove the theorem under the restriction that x 0, and inequality (54) results from the statement sn 6 2 b x − − cA g , (n, x) − g(x) n π 2 n a b (55) s 6 b An x n − + cA2 , 0 x , n π 2 n n + b = ϕ( )/ see Lemma 4.1, inequality (40), and recall that sn kn k k. We use the conditions and notations from Lemma 2.5 and recall a n x+(q) = nb ξ+(q) − = , q 0, b b∗+bq = ( ∗), = ,..., . x j x+ Q j j 1 A

( ∗ ) = /( + ) Note that x+ Q A x A An n b ,see(48) in the proof of Lemma 4.4 with α + = ∗ > 1 A, and that Q A 0 due to Lemma 2.6.Forx 0 we deﬁne the functions

sn g , (n, x) = x − g , (n, x), a b n a b (56) 6 g(x) = x − g(x) π 2 and will show that

2 b |g , (n, x) − g(x)| cA for all x ∈[0, x ] (57) a b n A with a sufficiently large constant c > 0 independent of n, a, b, A, x, such that (57) implies (55). ∗ =( − ∗)/ From the definition Q1 n b b it follows that x1 1. We also obtain from the definition of x1 that 1 b 1 x1 < 2, > 1 − . (58) x1 n

The following equations result from the deﬁnitions of fn, g, ga,b,from(56), (58) and Theorem 2.4(b): x x 1 b g(x) = , ga,b(n, x) = + , 0 x 1, (59) 2 2 x1 n 1 x 1 b g(x) = + log x, ga,b(n, x) = + , 1 x x1. (60) 2 2 x1 n From (58), (59) and (60) we conclude that

b |g , (n, x) − g(x)| , 0 x x . a b n 1 123 408 M. Kunik

Thus we may assume in the sequel that x is a fixed real number which satisfies x1 < x x A with A > 1. Due to Theorem 2.8, there is exactly one index α ∈{1,...,A−1} with xα < x xα+1, (61) and it is sufficient to prove (57)forA = α + 1 with the index α in (61). From x we also obtain an index qx ∈ N which satisfies the two inequalities

∗ < ∗ , ( + ) ( ). Qα+1 qx Qα x+ qx 1 x x+ qx (62)

We supplement the deﬁnition of the sums Sα,n(a/b) in Lemma 4.3 by a 1 S ,n = . (63) 0 b 2b

We obtain for the index α in (61) that α−1 ( , ) = a + 1 b , ga,b n xα b Sβ,n ∗ ∗ (64) b 2 b +bQα β=0 and from Lemma 4.4,(47), (50),

α α 1 b 1 b b . ∗ ∗ (65) 2 b +bQα 2 n − αb n

Due to Lemmas 4.3 and 4.4, there is an absolute constant c1 > 0 such that for β = 1,...,α− 1,

β bβ a ϕ(k) xβ+1 bβ −c1 bSβ,n − log c1 . n b k xβ n k=1

Summation with respect to β and (63), (64), (65) yields with an absolute constant c2 > 0

β 2 α−1 2 α b 1 ϕ(k) xβ+1 α b − c2 ga,b(n, xα) − − log c2 . (66) n 2 k xβ n β=1 k=1

Abelian summation gives

α−1 β α−1 α−1 ϕ(k) xβ+1 ϕ(k) ϕ(k) log = log xα − log xk. (67) k xβ k k β=1 k=1 k=1 k=1 123 A scaling property of Farey fractions 409

Now we use (56), (67) and (53) in Lemma 4.5 and get

α−1 β ϕ(k) xβ+ log 1 k xβ β=1 k=1 (68) α−1 α−1 1 xα ϕ(k) ϕ(k) x = g(α) − + log − log k . 2 α k k k k=1 k=1

| ( )|= Since g is bounded, supy>0 g y c3, we obtain from Lemma 4.4(iv)

2c3 2 |g(xα) − g(α)| c |xα − α| α b, (69) 3 n and from (68), (69) and Lemma 4.4(i) with some constant c4 > 0 we obtain

β 2 α−1 2 α b ϕ(k) xβ+1 1 α b − c4 log − g(xα) − c4 . (70) n k xβ 2 n β=1 k=1

The sum of inequalities (66), (70) gives with c5 = c2 + c4,

α2b |g , (n, xα) − g(xα)| c . (71) a b 5 n

Due to (61), (62) there holds

xα x+(qx +1) x x+(qx ) xα+1. (72)

We deﬁne = 1 , = 1 , 1 b 2 b / ∈Fext∩[ξ ( ∗ ),ξ ( + )) / ∈Fext∩[ξ ( + ),ξ ( )] a b n + Qα + qx 1 a b n + qx 1 + qx and regard that the sum 2 also contains the inverse denominator of the boundary fraction ξ+(qx ).Wehave

b ga,b(n, x+(qx +1)) − ga,b(n, xα) = b + · (x+(qx +1) − xα), (73) 1 2n b 3 b 0 · (x+(qx +1) − xα) , 2n 2 n because x+(qx +1) − xα xα+1 − xα 3 from Lemma 4.4(iii). 123 410 M. Kunik

For the estimation of 1 we use the same method like in the proof of Lemma 4.3, see (41)–(46),

∗ − Qα 1 α 1 + 1 + ( ∗+ ) + − (α + ) + = + ba b bq b n 1 b 1 a /b ∈Fα q qx 1 α Q∗ −1 ϕ(k) α 1 α + (74) k b∗+bq n + 1 − b(α +1) k=1 q=qx +1 α ! ϕ(k) 1 x+(q +1) x+(q +1) − xα α log x + x + . k b xα n n + 1 − b(α +1) k=1

For the lower bound of 1 we have the estimation

∗ ∗ Qα−1 Qα−1 1 1 − a b 1 ∗+ 2 ( ∗+ )2 + b = + b bq + b = + b bq a /b ∈Fα q qx 1 a /b ∈Fα q qx 1 α ! ϕ(k) 1 x+(q +1) x+(q +1) − xα log x + x k b xα 2n k=1 ! α ϕ( ) ( + ) − 1 1 k x+ qx 1 xα b 2 2 − + + x+(qx +1) − xα . 2 2 k n n2 k=1

This can be simpliﬁed if we note that

α ϕ( ) α + 1 + 1 k 1 = A , 2 2 k 2 2 k=1 then α 1 ϕ(k) x+(qx +1) 1 log b k xα k=1 (75) ( + ) − x+ qx 1 xα bA 2 2 − − x+(qx +1) − xα . 2n 2n2

We put x+(q +1) − xα λ = x , y = α + λ, xα+1 − xα and regard that 0 λ 1, see (72). Then

x+(q +1) xα+ xα x = λ 1 + (1 − λ ) y α + 1 α 123 A scaling property of Farey fractions 411 with the quantity α + λ = λ 1 ∈[ , ]. α + λ 0 1

Using α in (61), we obtain from (47) and (48) that

n x+(q +1) n x . (76) n + b y n − α b

Exactly as in the proof of Lemma 4.4(i) we conclude from (76) that 2 x+(qx +1) αb αb log + . (77) y n n

Recall that A = α + 1. Another consequence of (76)is

2b 2 |x+(q +1) − y| A . (78) x n

= | ( )| Using that c3 supy>0 g y we obtain from (78) that

b 2 |g(x+(q +1)) − g(y)| 2c A . (79) x 3 n

For (74), (75) we note the identity

x+(q +1) y x+(q +1) xα log x = log + log x − log xα α y α with the main term log (y/α) and the two error terms log ((x+(qx +1))/y),log(xα/α) on the right-hand side. We use (73), (74), (75), (77) and Lemma 4.4(i), and conclude for an appropriate constant c6 > 0 that

A2 b −c g , (n, x+(q +1)) − g , (n, xα) 6 n a b x a b α (80) y ϕ(k) A2 b − log c . α k 6 n k=1

Regarding y = α + λ ∈[α, α + 1] and Lemma 4.5 we have

α y ϕ(k) log = g(y) − g(α). (81) α k k=1 123 412 M. Kunik

From (69), (79), (80), (81) we get an absolute constant c7 > 0 such that

A2 b −c g , (n, x+(q +1)) − g(x+(q +1)) 7 n a b x x A2 b − g , (n, xα) + g(xα) c , a b 7 n and regarding (71) with c8 = c5 + c7,

2 A b g , (n, x+(q +1)) − g(x+(q +1)) c . (82) a b x x 8 n

Finally we use the two simple inequalities ga,b(n, x) − ga,b(n, x+(qx +1)) b 2, (83)

2 A b g(x) − g(x+(q +1)) c x+(q ) − x+(q +1) c (84) x 3 x x 3 n and the following estimates for b 2 with an absolute constant c9 > 0:

b bα b + 2 + ( ∗+ ) + − (α + ) ba b bqx b n 1 b 1 a /b ∈Fα α ϕ( ) α b · + k + b ∗ 1 (85) b +bqx k n + 1 − b(α +1) k=1 bxα+ bα bA2 1 A + c . n n + 1 − b(α +1) 9 n

With c = c3 + c8 + c9 and (82)–(85) we obtain inequality (57).

The plot in Fig. 5 illustrates the good agreement between the limit function g,see Fig. 4, and the function g2,5(1000, ·) in the range x = 5000·(ξ − 2/5) ∈[0, 10].

5 Asymptotic behaviour of the limit function g

In this section we need some input from the analytic theory of numbers and the theory of the Riemann zeta function, for details we refer to the standard textbooks [3,9,11,15].

Lemma 5.1 For Re s > 0,

∞ g(x) 1 6 1 6 1 ζ(s) G(s) = dx = − + · − (86) xs+1 s π 2 2 π 2 s − 1 sζ(s +1) 1 123 A scaling property of Farey fractions 413

limit function g

0.1

0.08

0.06

0.04

0.02

0 g(x) −0.02

−0.04

−0.06

−0.08

−0.1

−30 −20 −10 0 10 20 30 x axis Fig. 4 The limit function g(x) in the range |x| 30

g (1000, ⋅ ) and g 2,5

0.1

0.08

0.06

0.04 (1000, x) and g(x)

2,5 0.02 g

−0.02 1 2 3 4 5 6 7 8 9 10 x axis

Fig. 5 Comparison between g2,5(1000, ·) and g deﬁnes a holomorphic function G, which has an analytic continuation on the axis Re s = 0. The Mellin integral converges absolutely for Re s > 0. Regard a limit for s → 1 on the right-hand side in (5.1).

123 414 M. Kunik

Proof Due to (40), Lemma 4.1 and (51) we have uniformly in x 2 log x g (x) = O for x →∞. x

This implies the rough estimate g(x) = O(log2 x) for x →∞, such that the Mellin integral in (86) is absolutely convergent for Re s > 0. The Mellin integral deﬁnes a holomorphic function in the half plane Re s > 0. Representation (86) is valid for Re s > 1 because ∞ ϕ(k)/k 1 ζ(s) k x dx = , xs+1 s ζ(s +1) 1 ∞ x 1/t · ϕ(k)/kdt 1 ζ(s) 1 k t dx = . xs+1 s2 ζ(s +1) 1

Hence it is also valid for Re s > 0 by analytic continuation. Due to Hadamard, 1/ζ(s +1) has an analytic continuation on the axis Re s = 0. From

1 ζ(0) =− , lim sζ(s +1) = 1 2 s→0

2 2 we obtain that lims→0[6/π − 1/2 + 6/π · 1/(s −1) − ζ(s)/(sζ(s +1))]=0.

Lemma 5.2 There is an absolute constant C > 0 with ζ(s) C (87) s for all s ∈ C with Re s 0 and |s| 2.

Proof From Ingham [9, Chapter II, Theorem 9] we use (6) for δ = 1/2, " |ζ(σ +it)| C1 |t|, (88) which is valid for all σ 1/2 and all t ∈ R with |t| 1. Here C1 denotes a sufﬁciently large absolute constant. We also note, with an absolute constant C2 > 0, the trivial estimate

|ζ(σ +it)| C2 (89) for all σ 3/2 and all t ∈ R. Dueto(88), (89) it is sufﬁcient to prove (87) in the domain

1 0 Re s , |Im s| 1. (90) 2 123 A scaling property of Farey fractions 415

For values of s in this domain we use the functional equation of the Riemann zeta function in the form

− / ((1−s)/2) ζ(s) = π s 1 2 ζ(1−s). (91) (s/2)

From the textbook [1, Kapitel 1.4, Corollary 1.4.4] we derive that (( − )/ ) | | 1/2−σ s− / 1 s 2 t 1 π 1 2 = 1 + O (92) (s/2) 2π |t| with σ = Re s, t = Im s.From(88), (91) and (92) we obtain (87) in the region (90) and hence for all s with Re s 0 and |s| 2. Lemma 5.3 The holomorphic function G in Lemma 5.1 is a member of the Hardy 2 = 2( > ). ∈ 2 space H0 H Re s 0 Here any member h H0 is equipped with the norm ∞ 1/2 = 1 | (σ + )|2 . h H 2 sup h it dt 0 2π σ>0 −∞

Proof From Ingham [9, Chapter II, Theorem 10] we obtain the following estimate with an absolute constant C > 0: 1 C log7(2 +|Im s|), Re s 0, (93) |ζ(s +1)| see also Ingham’s general discussion before Theorem 9 concerning the use of the big-O symbols for the correct interpretation of his result. Using (93) and Lemma 5.2 we obtain that 7 ζ(s) log (2 +|Im s|) CC s2 ζ(s +1) |s| for all s ∈ C with Re s 0 and |s| 2 . Finally we use the analytical continuation of G from Lemma 5.1 and conclude

log7(2 +|Im s|) |G(s)| C |s|+1 for all s ∈ C with Re s 0 and a new constant C > 0. A good introduction to the theory of Hardy spaces is presented in the textbooks [7,12, 16]. Theorem 5.4 (integral representation of the limit function g) For x = eu > 1, i.e. for u > 0, we have 6 1 1 g(x) = g (x) + − · 0 π 2 2 x 123 416 M. Kunik with ∞ 1 1 6 1 6 1 g (x) = g (eu) = eitu − − · 0 0 2π it − t2 π 2 2 π 2 t2 + it −∞ πit−1/2 ((1−it)/2) ζ(1−it) + · · dt. t2 (it/2) ζ(1+it)

The integral above is absolutely convergent.

∈ 2 = 2( > ) Proof We also have G0 H0 H Re s 0 , from Lemma 5.3 for 6 1 1 G (s) = G(s) − − , 0 π 2 2 s + 1 or from (86) in Lemma 5.1, regarding a limit for s → 1, 1 6 1 6 1 ζ(s) G (s) = − + − . (94) 0 s(s +1) π 2 2 π 2 s(s −1) s2 ζ(s +1)

Using the functional equation (91)forζ(s) in the nominator of the last fraction in (94), we obtain that G0(s) decays faster then G(s) along Re s = σ 0. Especially on the boundary Re s = 0 there holds the following representation for t ∈ R\{0}, regarding a limit for t → 0: 1 6 1 6 1 πit−1/2 ((1−it)/2) ζ(1−it) G (it) = − − + , 0 it − t2 π 2 2 π 2 t2 + it t2 (it/2) ζ(1+it) and from (92) the boundary function G0(i ·) satisﬁes

−3/2 G0(it) = O(|t| ) for |t|→∞

1 2 with G0(i ·) ∈ (L ∩ L )(R). Thus we can apply Mellin’s inversion formula and obtain for x = eu > 1,

∞ 1 g (x) = g (eu) = eituG (it) dt, 0 0 2π 0 −∞ 6 1 1 g(x) = g (x) + − . 0 π 2 2 x

Corollary 5.5 limx→∞ g(x) = limx→−∞ g(x) = 0.

Proof Apply the Riemann–Lebesgue lemma of Fourier analysis on the absolutely convergent integral in Theorem 5.4. 123 A scaling property of Farey fractions 417

6 Conclusions and outlook

In Sect. 3 we have introduced 1-periodic polygonal functions fn which carry important number theoretical information about Farey sequences, see especially Corollary 3.10. The main result Theorem 4.6 was obtained by using only elementary properties of Farey fractions, namely from Theorem 2.8. The limit function g plays a major role in Theorem 4.6, such that we have studied its asymptotic behaviour in Sect. 5. An appropriate complex integration technique may reﬁne Theorem 5.4 and can afford quantitative remainder term estimates for Corollary 5.5 when this is needed for later purposes. Theorem 4.6 describes the behaviour of the Farey sequence and the function fn with large order n in the vicinity of any ﬁxed fraction. It would be an interesting future task to describe the local behaviour of fn for almost all irrational numbers in terms of appropriate distribution functions for the Farey sequences. Distribution functions for the Farey sequences were already studied in [2], see also the references therein.

Acknowledgments The author is thankful to Professor Jörn Steuding for encouragement and support in work.

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