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21

MÖBIUS DISJOINTNESS FOR A CLASS OF EXPONENTIAL FUNCTIONS

WEICHEN GU AND FEI WEI

Abstract. A vast class of exponential functions is showed to be deterministic. This class in-cludes functions whose exponents are polynomial-like or “piece-wise” close to polynomials afterdifferentiation. Many of these functions are indeed disjoint from the Möbius function. As aconsequence, we show that Sarnak’s Disjointness Conjecture for the Möbius function (from de-terministic sequences) is equivalent to the disjointness in average over short intervals.

1. Introduction

Let µ(n) be the Möbius function, that is, µ(n) is 0 when n is not square free (i.e., divisible bya nontrivial square), and is (−1)r when n is a product of r distinct primes. Many problems innumber theory can be reformulated in terms of properties of the Möbius function. For example,the Prime Number Theorem is known to be equivalent to

∑n≤N µ(n) = o(N). The Riemann

hypothesis holds if and only if∑

n≤N µ(n) = o(N12+ǫ) for every ǫ > 0.

For a truly random sequence an of −1 and 1, the normalized average (∑N

n=1 an)/N12 obeys

Gaussian law in distribution as N tends to infinity, which implies that∑

n≤N an = o(N12+ǫ).

Under the Riemann hypothesis, the Möbius function shares this property as an indication thatcertain randomness may exist in the values of the Möbius function. It is widely believed that thisrandomness predicts significant cancellations in the summation of µ(n)ξ(n) for any “reasonable”sequence ξ(n). This rather vague principal is known as an instance of the “Möbius randomnessprincipal” (see e.g., [IK04, Sectinon 13.1]). In [Sar09], Sarnak made this principal precise byidentifying the notion of “reasonable” and proposed the following conjecture.

Conjecture 1.1 (Sarnak’s Möbius Disjointness Conjecture (SMDC)). Let ξ(n) be a determin-istic sequence. Then

limN→∞

1

N

N∑

n=1

µ(n)ξ(n) = 0.

Here, we recall the definition of deterministic sequences. Functions from N to C are calledarithmetic functions or sequences. An arithmetic function f(n) is said to be disjoint from

another one g(n) if∑N

n=1 f(n)g(n) = o(N). Let (X , T ) be a topological dynamical system,that is X is a compact Hausdorff space and T : X → X a continuous map. We say n 7→F (T nx0) an arithmetic function realized in the topological dynamical system (X , T ), and (X , T )a realization of this arithmetic function. Functions realized in topological dynamical systems ofzero topological entropy are called deterministic.

1

http://arxiv.org/abs/2011.08699v2

2

A lot of progress has been made on SMDC, while its general form remains open. We simplyrefer to survey papers [FKPL18], [KPL20] for the progress in this area, and discuss only theresults that are more related to this paper. The goal of this article is to show that a classof exponential functions are deterministic and verify the conjecture of Sarnak for them. Asan application, we show that Sarnak’s Disjointness Conjecture for the Möbius function (fromdeterministic sequences) is equivalent to the disjointness in average over short intervals.

1.1. Notation. We use e(f(n)) to denote the exponential function exp(2πıf(n)) when f is areal-valued arithmetic function, where ı is the imaginary unit. When we say the exponentialfunction of f(n), we mean e(f(n)). Sometimes we call e(f(n)) an f(n) phase.

We use 1S to denote the indicator of a predicate S, that is 1S = 1 when S is true and 1S = 0when S is false. We also denote 1A(n) = 1n∈A for any subset A of N. For any finite set C, |C|denotes the cardinality of C.

We use {x} and ⌊x⌋ to denote the fractional part and the integer part of a real number x,respectively. We use ‖x‖R/Z to denote the distance between x and the set Z, i.e., ‖x‖R/Z =min({x}, 1 − {x}). For ease of notation, we drop the subscript and write simply ‖x‖.

The difference operator △ is defined on the set of all arithmetic functions, mapping f(·) tof(·+1)−f(·). The k-th difference operator △k is defined by the composition of △ with k times.

For two arithmetic functions f(n) and g(n), f = o(g) means limn→∞ f(n)/g(n) = 0; f ≪ gmeans that there is an absolute constant c such that |f | ≤ c|g|; f = g+O(h) means f − g ≪ h.

1.2. A class of deterministic sequences. It is known that polynomial and bracket poly-nomial phases are deterministic sequences (see e.g., [GT12b]). In this paper, we first concernweather f(n) phase is deterministic when f(n) (or the k-th difference △kf(n)) is “piece-wise”close to polynomials.

Theorem 1.2. Let w be a positive integer. Suppose that p1(y), . . . , pw(y) are polynomials inR[y], and N = S1 ∪ S2 ∪ · · · ∪ Sw is a partition of N with each 1Sv(n) deterministic. Let

g(n) =

w∑

v=1

1Sv(n)pv(n). (1.1)

Then for any real-valued arithmetic function f(n) satisfying

limn→∞

‖△kf(n) − g(n)‖ = 0 (1.2)

for some k ∈ N, e(f(n)) is deterministic.Now, we explain briefly the main idea to prove the above result. The major tool we use

is anqie entropy (of arithmetic functions), which was introduced by Ge in [Ge16]. We referreaders to Section 2.1 for knowledge on anqie entropy. To prove Theorem 1.2, we first constructa sequence of arithmetic functions {fN (n)}∞N=0 with finite ranges that uniformly converges tof(n). By the lower semi-continuity of anqie entropy (see Proposition 2.2), it suffices to showthat the anqie entropy of fN is zero for N large enough. Note that the anqie entropy of fN (withfinite range) is given through the cardinality of different J-blocks occurring in it (see formula

3

(2.1)). So we focus on estimating this cardinality. Our method is to built a one-to-one map fromthe set of J-blocks occurring in the sequence fN(n) to the set of pieces of R

k cut by hyperplanes(for some k depending on J). So the cardinality of the first set is bounded by the cardinalityof the latter one. And we prove that the second cardinality has polynomial growth (see Lemma2.6). We refer readers to Section 3 for more details.

We also consider whether the exponential function of any concatenation of polynomials isdeterministic. Before stating the next result, we first introduce the definition of concatenationof arithmetic functions.

Definition 1.3. Let 0 = N0 < N1 < · · · be a sequence of natural numbers with limi→∞(Ni+1 −Ni) = ∞, and {fi(n)}∞i=0 be a sequence of arithmetic functions. We say that f(n) is theconcatenation of {fi(n)}∞i=0 with respect to the sequence {Ni}∞i=0 if f(n) = fi(n) whenNi ≤ n < Ni+1 for i = 0, 1, . . ..

For the exponential functions of concatenations, we obtain the following result.

Theorem 1.4. Let g(n) be defined as in Theorem 1.2. Suppose that {fi(n)}∞i=0 is a sequence ofreal-valued arithmetic functions such that

limn→∞

supi∈N

‖△kfi(n) − g(n)‖ = 0 (1.3)

holds for some k ∈ N. Then for any concatenation f(n) of {fi(n)}∞i=0, e(f(n)) is deterministic.1.3. Disjointness of Möbius from e(f(n)) with the k-th difference of f(n) tendingto zero. The disjointness of Möbius from exponential functions is important and has beenextensively studied in number theory. For example, the disjointness of µ from e(nα), for anyα ∈ R, is closely related to the estimate of exponential sums in prime variables, from which onecan deduce Vinogradov’s three primes theorem. In this part, we are interested in the followingproblem.

Problem 1. Let f(n) be a real-valued arithmetic function such that

limn→∞

‖△kf(n)‖ = 0 (1.4)

for some natural number k. Is µ(n) disjoint from e(f(n))?

As we have seen that e(f(n)) satisfying condition (1.4) is a deterministic sequence by Theorem1.2, so SMDC implies a positive answer to the above problem. In the following we investigateProblem 1 without assuming SMDC. For the case k = 0, it is obvious that µ(n) is disjoint fromsuch e(f(n)) by the Prime Number Theorem. For the case k = 1, we have the following result.

Proposition 1.5. Suppose that f(n) is a real-valued arithmetic function satisfying

limn→∞

‖△f(n) − c‖ = 0for some constant c ∈ R. Then

limN→∞

1

N

∑

1≤n≤N

µ(n)e(f(n)) = 0.

4

For k = 2, we show the following result.

Proposition 1.6. Suppose that f(n) is a real-valued arithmetic function satisfying that the set{e(△f(n)) : n = 0, 1, . . .} has finitely many limit points, and

limn→∞

‖△2f(n) − c‖ = 0

for some constant c ∈ R. Then

limN→∞

1

N

∑

1≤n≤N

µ(n)e(f(n)) = 0.

We remark that Proposition 1.5 is a special case of Proposition 1.6. They are both con-sequences of the following result on the disjointness of Möbius from concatenations of certainpolynomial phases.

Theorem 1.7. Let {Ni}∞i=0 be an increasing sequence of natural numbers with N0 = 0 andlimi→∞(Ni+1 − Ni) = ∞. Suppose that α0, α1, . . . are real numbers such that the sequence{e(αi)}∞i=0 has finitely many limit points. Suppose P (x) ∈ R[x]. Then

limm→∞

1

Nm

m−1∑

i=0

∣∣∣∣∣∑

Ni≤n

5

Proposition 1.9. Let k be a given positive integer. Let q ≥ 1 and a ≥ 0 be given integers.Denote Dk by the set of all polynomials in R[y] of degrees less than k. Assume that the followingestimate holds,

limh→∞

lim supX→∞

1

Xh

∫ 2X

X

supp(y)∈Dk

∣∣∣∣∣∑

x≤n

6

h = Xθ, the case that k = 1 and θ > 0.55 was obtained by Matomäki-Teräväinen in [MT19];the case that k = 2 and θ > 5/8 was previously established by Zhan in [Zhan91] and extendedto θ > 3/5 in [MT19]; the case that k ≥ 2 and θ > 2/3 was obtained by Matomäki-Shao in[MS19] (also see [Huang16] for related results on Λ(n) instead of µ(n)).

If we take f(n) ≡ 0 in equation (1.8), we obtain the following result.Corollary 1.13. SMDC holds if and only if for any deterministic sequence ξ(n),

limh→∞

lim supN→∞

1

N

N∑

n=1

∣∣∣∣∣1

h

∑

n≤m

7

The next one is about the lower semi-continuity of anqie entropy.

Proposition 2.2. If {fN (n)}∞N=0 is a sequence of bounded arithmetic functions converging tof(n) uniformly with respect to n ∈ N, then lim infN→∞ Æ(fN) ≥ Æ(f).

The above two propositions are claimed in the survey paper [Ge16, Section 4]. We referreaders to [Wei18] for proofs. Arithmetic functions of zero anqie entropy can be realized intopological dynamical systems of zero topological entropy [Ge16, Section 3]. Based on this fact,we have the following result.

Proposition 2.3. A sequence {f(n)}∞n=0 is deterministic if and only if Æ(f) = 0.2.2. Anqie entropy of arithmetic functions with finite ranges. In this subsection, weshow some results on computing anqie entropy of functions with finite ranges, which will beused in the proofs of Theorems 1.2 and 1.4. Let us first recall some basic concepts in symbolicdynamical systems. For a finite set A, a block over A is a finite sequence of symbols from A. AJ-block is a block of length J (J ≥ 1). For any given (finite or infinite) sequence x = (x0, x1, . . .)of symbols from A, we say that a block w occurs in x or x contains w if there are natural numbersi, j with i ≤ j such that (xi, . . . , xj) = w. A concatenation of two blocks w1 = (a1, . . . , ak) andw2 = (b1, . . . , bl) over A is the block w1w2 = (a1, . . . , ak, b1, . . . , bl).

Now suppose that f : N → C has finite range. Let BJ(f) denote the set of all J-blocksoccurring in f , i.e.,

BJ (f) = {(f(n), f(n+ 1), . . . , f(n+ J − 1)) : n ≥ 0}.A J-block of the form

(f(lJ), f(lJ + 1), . . . , f(lJ + J − 1))for some l ∈ N is called a regular J-block in f . Denote the set of all regular J-blocks in f byBrJ(f). A J-block, which occurs infinitely many times in the sequence {f(n)}∞n=0, is called aneffective J-block in f . Denote the set of all such blocks in f by BeJ(f). A J-block (a0, a1, . . . , aJ−1)is called regularly effective in f if there are infinitely many natural numbers l such that

(a0, a1, . . . , aJ−1) = (f(lJ), ..., f(lJ + J − 1)).The set of all regularly effective blocks in f is denoted by Be,rJ (f).

For a function f taking finitely many values, as we mentioned previously,

Æ(f) = limJ→∞

log |BJ(f)|J

. (2.2)

In the following, we show that Æ(f) also can be computed through the cardinality of BrJ(f),BeJ(f) or Be,rJ (f).Theorem 2.4. Let f(n) be an arithmetic function with finite range. Then

Æ(f) = limJ→∞

log |BrJ(f)|J

= limJ→∞

log |Be,rJ (f)|J

= limJ→∞

log |BeJ(f)|J

. (2.3)

8

Proof. We first show that the first equality in equation (2.3) holds. On one hand, BrJ(f) ⊆ BJ(f),so by formula (2.2),

Æ(f) ≥ lim supJ→∞

log |BrJ(f)|J

.

On the other hand, given J ≥ 1, for any l ≥ 1 and any (lJ)-block w occurring in f , there is aconcatenation of certain l + 1 successive regular J-blocks in f containing w. Thus |BlJ(f)| ≤J |BrJ(f)|l+1. This implies that

Æ(f) = liml→∞

log |BlJ(f)|lJ

≤ log |BrJ(f)|J

.

We then have

Æ(f) ≤ lim infJ→∞

log |BrJ(f)|J

.

So limJ→∞log |Br

J(f)|

Jexists and equals Æ(f).

Next we show that the second equality in equation (2.3) holds, i.e.,

limJ→∞

log |BrJ(f)|J

= limJ→∞

log |Be,rJ (f)|J

. (2.4)

Since Be,rJ (f) ⊆ BrJ(f),

limJ→∞

log |BrJ(f)|J

≥ lim supJ→∞

log |Be,rJ (f)|J

. (2.5)

So we only need to show that

limJ→∞

log |BrJ(f)|J

≤ lim infJ→∞

log |Be,rJ (f)|J

. (2.6)

In fact, for any given J ≥ 1, since the set BrJ(f) \Be,rJ (f) is finite, there is an integer lJ ≥ 1 suchthat all regular J-blocks in the set

{(f(nJ), ..., f(nJ + J − 1)) : n ≥ lJ}are regularly effective J-blocks in f . Then for each l > lJ , there is at most one regular (lJ)-blockin f which is not a concatenation of regularly effective J-blocks in f . This implies |BrlJ(f)| ≤|Be,rJ (f)|l + 1. Therefore

Æ(f) = liml→∞

log |BrlJ(f)|lJ

≤ liml→∞

log(|Be,rJ (f)|l + 1)lJ

=log |Be,rJ (f)|

J

holds. Letting J → ∞, we obtain formula (2.6).At last, note that

Be,rJ (f) ⊆ BeJ(f) ⊆ BJ(f),then

limJ→∞

log |Be,rJ (f)|J

≤ lim infJ→∞

log |BeJ (f)|J

, lim supJ→∞

log |BeJ(f)|J

≤ limJ→∞

log |BJ(f)|J

.

9

From formula (2.2) and the second equality in (2.3), we have

Æ(f) = limJ→∞

log |BJ(f)|J

= limJ→∞

log |Be,rJ (f)|J

.

Then limJ→∞

log |BeJ(f)|

Jexists and equals Æ(f). So the third equality in equation (2.3) holds. �

To estimate the cardinality of the set of J-blocks occurring in certain sequences, we introducethe following notion.

Definition 2.5. Let k,m ≥ 1 be integers and c1, . . . , cm ∈ R be constants. Suppose that forj = 1, . . . , m, Fj is a non-zero linear function of x1, ..., xk and Hj is the hyperplane in R

k givenby Fj(x1, ..., xk) = cj. Denote by

H+j = {(x1, ..., xk) ∈ Rk : Fj(x1, ..., xk) > cj},H−j = {(x1, ..., xk) ∈ Rk : Fj(x1, ..., xk) < cj}.

A non-empty subset P of Rk of the following form

P = P1 ∩ P2 ∩ · · · ∩ Pm,where each Pj ∈ {H+j , H−j , Hj}, is called a piece of Rk cut by H1, ..., Hm.

In the following lemma we give an upper bound for the cardinality of pieces of Rk cut byhyperplanes.

Lemma 2.6. Let k,m ≥ 1 be integers. Suppose that H1, ..., Hm are hyperplanes in Rk. LetC(H1, . . . , Hm, k) denote the cardinality of pieces of R

k cut by H1, ..., Hm and W (m, k) denotethe maximal value of C(H1, . . . , Hm, k) when H1, . . . , Hm go through all the possible hyperplanes.Then

W (m, k) ≤k∑

j=0

2j(m

j

).

In particular, W (m, k) ≤ (k + 1)2kmk.Proof. Notice that W (1, k) = 3 for any k ≥ 1. If we have

W (m, k) ≤W (m− 1, k) + 2W (m− 1, k − 1), (2.7)then one can easily draw the conclusion by induction on m.

Now we show formula (2.7) holds for m ≥ 2. Let ψ be the map from the set of all pieces ofRk cut by H1, ..., Hm, denoted by P, onto the set of all pieces of Rk cut by H2, ..., Hm, denotedby P̃ , given by

ψ : P1 ∩ P2 ∩ · · · ∩ Pm 7→ P2 ∩ · · · ∩ Pm.Then for any piece D ∈ P̃ , |ψ−1(D)| ≤ 3. Now we show that if |ψ−1(D)| ≥ 2, then H1 ∩ Dis nonempty. In fact, the only case we need to consider is when both H+1 ∩ D and H−1 ∩ Dare nonempty. In this case, suppose p1 ∈ H+1 ∩ D and p2 ∈ H−1 ∩ D, then there is a pointp0 ∈ H1 ∩ D by the convexity of D. Hence, |P| − |P̃| = C(H1, . . . , Hm, k) − C(H2, . . . , Hm, k)

10

does not exceed the cardinality of pieces of the form H1 ∩ P2 ∩ · · · ∩ Pm in P times 2. In thefollowing, we estimate this cardinality.

For each j = 2, . . . , m, H1 ∩ Hj is a hyperplane in H1, or empty, or equal to H1. DenoteG1, ..., Gn to be the ones which are hyperplanes in H1. Then n ≤ m − 1. We claim that ifH1 ∩ P2 ∩ · · · ∩ Pm = (H1 ∩ P2) ∩ · · · ∩ (H1 ∩ Pm) 6= ∅, then (H1 ∩ P2) ∩ · · · ∩ (H1 ∩ Pm) is apiece of H1 cut by G1, . . . , Gn. In fact, for j = 2, . . . , m, there are at most three cases. WhenH1 ∩ Hj = ∅, then H1 ∩ H+j = H1 or H1 ∩ H−j = H1 and then Pj = H+j or H−j , respectively.When H1 ∩Hj = H1, then Pj = Hj and H1 ∩Pj = H1. When H1 ∩Hj = Glj is a hyperplane inH1, then {H1 ∩H+j , H1 ∩H−j } = {G+lj , G

−lj}. Hence (H1 ∩ P2) ∩ · · · ∩ (H1 ∩ Pm) is of the form

T1 ∩ · · · ∩ Tn, where each Tl ∈ {Gl, G+l , G−l }.So the cardinality of pieces of the formH1∩P2∩· · ·∩Pm in P does not exceed C(G1, . . . , Gn, k−

1) which is at most W (n, k − 1) ≤ W (m − 1, k − 1). Therefore the inequality (2.7) holds andthe proof is completed. �

3. Proofs of Theorems 1.2 and 1.4

Recall that ‖x‖ = infm∈Z |x −m| = min{{x}, 1 − {x}}. Then ‖ · ‖ defines a metric on R/Zand the topology induced by it on R/Z is equivalent to the Euclid topology on the unit circle.The following lemma will be used in this section.

Lemma 3.1. Let f, g be real-valued arithmetic functions with

limn→∞

‖△kf(n) − g(n)‖ = 0 (3.1)for some k ∈ N. Then, for any ε > 0 and positive integer m ≥ 1, there is some L ∈ N suchthat, whenever n > L, the following holds for any j with 0 ≤ j ≤ m− 1,

‖f(n+ j) − Yn(n+ j)‖ ≤ ε,where Yn(n+ j) is defined to be f(n+ j) when 0 ≤ j ≤ k − 1 and to be the value determined bythe following linear equations when k ≤ j ≤ m− 1,

△kYn(n+ j) = g(n+ j) , j = 0, 1, ..., m− k − 1 . (3.2)Proof. When k = 0, the claim is trivial. In the following, we assume k ≥ 1. We use inductionon m to prove the lemma. For m ≤ k, since Yn(n+ j) = f(n+ j) for j = 0, 1, . . . , m− 1, chooseL = 0. Then we obtain the claim in the lemma. Assume inductively that the claim holds forsome m0 ≥ k. In the following we shall prove the claim holds for m0 + 1 case. By condition(3.1) and Proposition A.1,

limn→∞

‖k∑

l=0

(−1)k−l(k

l

)f(n+ l) − g(n)‖ = 0.

Then for any ǫ > 0, there is an L1 > 0 such that whenever n > L1,

‖k∑

l=0

(−1)k−l(k

l

)f(n+m0 − k + l) − g(n+m0 − k)‖ < ǫ/2,

11

i.e.,

‖f(n+m0) −(g(n+m0 − k) −

k−1∑

l=0

(−1)k−l(k

l

)f(n+m0 − k + l)

)‖ < ǫ/2. (3.3)

By the induction hypothesis, there is an L0 ∈ N, whenever n > L0,‖f(n+ j) − Yn(n+ j)‖ < ǫ/2k+1, j = 0, 1, . . . , m0 − 1. (3.4)

Let L = max{L0, L1}. Then by equations (3.3) and (3.4), whenever n > L,

‖f(n+m0) −(g(n+m0 − k) −

k−1∑

l=0

(−1)k−l(k

l

)Yn(n+m0 − k + l)

)‖ < ǫ. (3.5)

Define Yn(n +m0) to be the value determined in the following equation,

k∑

l=0

(−1)k−l(k

l

)Yn(n+m0 − k + l) = g(n+m0 − k).

Then by equation (3.5),

‖f(n+m0) − Yn(n+m0)‖ < ǫ.Combing with equation (3.4), we obtain the claim for m = m0+1, completing the induction. �

Theorem 1.2 (restated). Let w be a positive integer. Suppose that p1(y), . . . , pw(y) are poly-nomials in R[y], and N = S1 ∪ S2 ∪ · · · ∪ Sw is a partition of N with each 1Sv(n) deterministic.Let

g(n) =

w∑

v=1

1Sv(n)pv(n). (3.6)

Then for any real-valued arithmetic function f(n) satisfying

limn→∞

‖△kf(n) − g(n)‖ = 0 (3.7)

for some k ∈ N, e(f(n)) is deterministic.We use the following strategy to prove Æ(f(n)) = 0. Firstly, we construct a sequence

{gN(n)}∞N=0 of arithmetic functions with finite ranges to approach f(n) with respect to ‖ · ‖R/Z.Then it suffices to prove Æ(gN) = 0 for N large enough by Proposition 2.2. Secondly, we de-compose Be,r2m(gN), the set of all 2m-regularly effective blocks occurring in gN , into some subsetsAν,m according to the “zero entropy partition” given in g(n). Thirdly, for each ν, we constructa family of hyperplanes Hν,0, . . . , Hν,2m−1 in R

q based on the approximation of f(n) to polyno-mials after differentiation. Then there is a correspondence between the set of pieces of Rq cutby Hν,0, ..., Hν,2m−1 and Aν,m. Precisely, each element in Aν,m uniquely determines a piece of Rqcut by Hν,0, ..., Hν,2m−1. So |Aν,m|, moreover |Be,r2m(gN)|, is bounded by W (2m, q), which has thepolynomial growth rate with respect to 2m by Lemma 2.6.

12

Proof of Theorem 1.2. Given N ≥ 1, by Lemma 3.1, for each integer m ≥ 1, there is a sufficientlylarge Lm ∈ N with 2m|Lm, such that whenever n ≥ Lm, we have

‖f(n+ j) − Yn(n+ j)‖ ≤ 1/N, j = 0, . . . , 2m − 1, (3.8)where Yn(n+ j) is defined to be f(n+ j) when 0 ≤ j ≤ k − 1 and the value determined by thefollowing linear equations when k ≤ j ≤ 2m − 1,

△kYn(n+ j) = g(n+ j), j = 0, ..., 2m − k − 1. (3.9)Moreover, we may further assume that the sequence {Lm}∞m=0 (L0 = 0) chosen above satisfiesLm+1 > Lm for each m ≥ 1. Let dm = (Lm+1 −Lm)/2m. Then the following is a partition of N,

N =

∞⋃

m=0

dm−1⋃

a=0

{Lm + a2m, Lm + a2m + 1, ..., Lm + a2m + 2m − 1} .

We define YL0+a(L0 + a) = f(L0 + a) for a = 0, 1, . . . , L1 − 1 and define the arithmetic functiongN as follows: for a = 0, . . . , dm − 1 and j = 0, . . . , 2m − 1,

gN(Lm + a2m + j) =

t

N(3.10)

when {YLm+a2m(Lm + a2m + j)} ∈ [ tN , t+1N ) for some integer t with 0 ≤ t ≤ N − 1. By formula(3.8),

‖f(Lm + a2m + j) − gN(Lm + a2m + j)‖ < 2/N.Then supn∈N ‖f(n) − gN(n)‖ < 2/N . Hence limN→∞ supn∈N |e(f(n)) − e(gN (n))| = 0. Toprove e(f(n)) deterministic (that is Æ

(e(f(n))

)= 0 by Proposition 2.3), it suffices to prove

Æ(gN(n)) = 0 for N large enough by Propositions 2.1 and 2.2.In the remaining part of this proof, we shall prove Æ(gN) = 0 for any given N ≥ 1. Let η be

the function defined as η(n) = v if n ∈ Sv, v = 1. . . . , w. Then the anqie entropy of η is zero.Note that η(n) has finite range. By formula (2.2),

limm→∞

log |B2m(η)|2m

= 0,

where B2m(η) is the set of all 2m-blocks occurring in η. For any 2m-block ν in B2m(η), denote byAν,m = {(gN(n2m), ..., gN(n2m+2m−1)) : n ∈ N, n2m ≥ Lm, (η(n2m), ..., η(n2m+2m−1)) = ν}.Recall that Be,r2m(gN) denotes the set of all 2m-regularly effective blocks occurring in gN . Then

Be,r2m(gN) ⊆⋃

ν∈B2m (η)

Aν,m. (3.11)

Let d = max(deg(p1), ..., deg(pw)) + 1, where we define deg(pi) as −1 when pi = 0. In thefollowing, we estimate the cardinality of Aν,m for each given m with 2m > max(k + 1, d) andν = (ν0, ..., ν2m−1) in B2m(η). Denote by q = k + wd. If q = 0 (i.e., k = 0 and d = 0), then bythe definition, gN(n) = 0 when n ≥ L1. So it is easy to see that Æ(gN) = 0. Then Æ(e(f)) = 0.In the following, we may assume that q ≥ 1.

13

We first define linear functions Fν,0, . . . , Fν,2m−1 from Rq to R. Define Fν,j(x0, x1, . . . , xq−1) =

xj for j = 0, . . . , k − 1. Assume inductively that we have defined the linear function Fν,j0 forsome j0 with k − 1 ≤ j0 ≤ 2m − 2. Then we define Fν,j0+1(x0, x1, . . . , xq−1) to be the functionsatisfying

k∑

l=0

(−1)k−l(k

l

)Fν,j0+1−k+l(x0, x1, . . . , xq−1) = Pνj0+1−k(j0 + 1 − k), (3.12)

where

Pv(j0 + 1 − k) =d−1∑

r=0

xk+(v−1)d+r∏

0≤s≤d−1,s 6=r

(j0 + 1 − k) − sr − s , 1 ≤ v ≤ w, (3.13)

when d ≥ 2; Pv(j0 + 1− k) = 0 when d = 0; Pv(j0 + 1− k) = xk+(v−1)d when d = 1. By equation(3.12), it follows from the inductive assumption that Fν,j0+1(x0, x1, . . . , xq−1) is a linear functionof x0, x1, . . . , xq−1.

Next, given n with (η(n2m), η(n2m + 1), ..., η(n2m + 2m − 1)) = ν and n2m ≥ Lm. Supposethat Lm0 ≤ n2m < Lm0+1 for some m0 ≥ m and n2m = n02m0 + u with 0 ≤ u ≤ 2m0 − 2m.Taking yj = {Yn02m0 (n2m + j)} for 0 ≤ j ≤ k−1 and yk+(v−1)d+r = {pv(n2m +r)} for 1 ≤ v ≤ w,0 ≤ r ≤ d− 1. In the following, we show that

{Fν,j(y0, y1, . . . , yq−1)} = {Yn02m0 (n2m + j)}, j = 0, . . . , 2m − 1. (3.14)By the definition,

Fν,j(y0, y1, . . . , yq−1) = {Yn02m0 (n2m + j)}, j = 0, . . . , k − 1. (3.15)Plugging (y0, y1, . . . , yq−1) into equation (3.13), we have for j ≥ k,

Pv(j − k) =d−1∑

r=0

{pv(n2m + r)}∏

0≤s≤d−1,s 6=r

(j − k) − sr − s .

By Lemma A.4, {Pv(j − k)} = {pv(n2m + j − k)}. Then by equation (3.12),

{k∑

l=0

(−1)k−l(k

l

)Fν,j−k+l(y0, y1, . . . , yq−1)} = {pνj−k(n2m + j − k)} , j = k, . . . , 2m − 1. (3.16)

Using the condition (3.9), we have

k∑

l=0

(−1)k−l(k

l

)Yn02m0 (n2

m + j−k+ l) = g(n2m + j−k) = pνj−k(n2m + j−k), j = k, . . . , 2m−1.

Comparing with equation (3.16) and by (3.15), we conclude that equation (3.14) holds.Note that |j − k| ≤ 2m. Then by equation (3.13), |Pv(j − k)| ≤ (d + 1)2md when d = 0, or

d ≥ 1, xk+(v−1)d+r ∈ [0, 1), r = 0, . . . , d− 1. By equation (3.12) and Proposition A.5, we have|Fν,j(y0, y1, . . . , yq−1)| ≤ (k+1)jk(d+1)2md < (k+1)(d+1)2mk+md , j = 0, 1, . . . , 2m−1. (3.17)

14

Based on the linear functions Fν,0, . . . , Fν,2m−1 constructed above, we define a family of hyper-planes F = {HM,t,j : M, t, j ∈ Z, −(k + 1)(d+ 1)2m(k+d) − 1 ≤M ≤ (k + 1)(d+ 1)2m(k+d), 0 ≤t ≤ N − 1, 0 ≤ j ≤ 2m − 1}, where

HM,t,j = {(x0, x1, ..., xq−1) ∈ Rq : Fν,j(x0, x1, ..., xq−1) = M +t

N}.

Then |F| ≤ 2(k + 2)(d + 1)2m(k+d+1)N . By Lemma 2.6, there are at most W (|F|, q) pieces ofRq cut by the hyperplanes in F , where

W (|F|, q) ≤ (q + 1)2q(2(k + 2)(d+ 1)2m(k+d+1)N

)q. (3.18)

Now, we are ready to estimate |Aν,m|. Let (gN(n2m), ..., gN(n2m + 2m − 1)) ∈ Aν,m withn2m ≥ Lm. Then (η(n2m), ..., η(n2m + 2m − 1)) = ν. Suppose that Lm0 ≤ n2m < Lm0+1 forsome m0 ≥ m and n2m = n02m0 + u with 0 ≤ u ≤ 2m0 − 2m. Set

yj = {Yn02m0 (n2m + j)}for 0 ≤ j ≤ k − 1 and

yk+(v−1)d+r = {pv(n2m + r)}for 1 ≤ v ≤ w, 0 ≤ r ≤ d − 1. By formula (3.17), there are integers M0,M1, ...,M2m−1 ∈[−(k + 1)(d+ 1)2m(k+d) − 1, (k + 1)(d+ 1)2m(k+d)] and t0, t1, . . . , t2m−1 ∈ [0, N − 1] such that

Mj +tjN

≤ Fν,j(y0, y1, . . . , yq−1) < Mj +tj + 1

N, j = 0, . . . , 2m − 1. (3.19)

Note that any two pieces of Rq cut by hyperplanes in F are disjoint. Let P be the unique piececontaining the point (y0, y1, . . . , yq−1). Then it is not hard to check that formula (3.19) holdsfor each (x0, x1, . . . , xq−1) ∈ P . By equation (3.14), {Fν,j(y0, y1, . . . , yq−1)} = {Yn02m0 (n2m +j)} ∈ [ tj

N,tj+1

N). Then by formula (3.10), gN(n2

m + j) =tjN

, j = 0, . . . , 2m − 1. Moreover,from the above analysis, we conclude that if n, n′ ∈ N with (η(n2m), . . . , η(n2m + 2m − 1)) =(η(n′2m), . . . , η(n′2m + 2m − 1)) = ν, such that the corresponding points (y0, y1, . . . , yq−1) and(y′0, y

′1, . . . , y

′q−1) belong to the same piece of R

k, then (gN(n2m), . . . , gN(n2

m + 2m − 1)) =(gN(n

′2m), . . . , gN(n′2m + 2m − 1)). Hence

|Aν,m| ≤W (|F|, q).So by equation (3.11) and formula (3.18),

|Be,r2m(gN)| ≤∑

ν∈B2m (η)

|Aν,m| ≤ (q + 1)2q(2(k + 2)(d+ 1)2m(k+d+1)N

)q|B2m(η)|. (3.20)

Note that N, k, d, q are parameters independent of m. By Theorem 2.4,

Æ(gN) = limm→∞

log |Be,r2m(gN)|2m

≤ limm→∞

log |B2m(η)|2m

= Æ(η) = 0.

So we complete the proof of this theorem. �

Next, we prove Theorem 1.4, which discusses about the anqie entropy of exponential functionsof concatenations.

15

Theorem 1.4 (restated). Let g(n) be defined as in Theorem 1.2. Suppose that {fi(n)}∞i=0 isa sequence of real-valued arithmetic functions such that

limn→∞

supi∈N

‖△kfi(n) − g(n)‖ = 0 (3.21)

holds for some k ∈ N. Then for any concatenation f(n) of {fi(n)}∞i=0, e(f(n)) is deterministic..Proof. Given N ≥ 1. By condition (3.21) and Lemma 3.1, for each integer m ≥ 1, there is asufficiently large Lm ∈ N with 2m|Lm, such that for any i ∈ N, whenever n ≥ Lm, we have

‖fi(n + j) − Yn,i(n+ j)‖ ≤ 1/N, j = 0, 1, . . . , 2m − 1, (3.22)where Yn,i(n+ j) equals fi(n+ j) when 0 ≤ j ≤ k− 1, and is determined by the following linearequations when k ≤ j ≤ 2m − 1,

△kYn,i(n+ j) = g(n+ j), j = 0, 1, ..., 2m − k − 1. (3.23)Moreover, we may further assume that the sequence {Lm}∞m=0 (L0 = 0) chosen above satisfiesLm+1 > Lm for each m ≥ 1. Let dm = (Lm+1 −Lm)/2m. We define YL0+a,i(L0 + a) = fi(L0 + a)for i ∈ N and a = 0, 1, . . . , L1,i − 1, and a sequence {gN,i}∞i=0 of arithmetic functions with finiteranges as follows: for a = 0, 1, . . . , dm − 1 and j = 0, 1, . . . , 2m − 1, define

gN,i(Lm + a2m + j) =

t

N

when {YLm+a2m,i(Lm + a2m + j)} ∈ [ tN , t+1N ) for some integer t with 0 ≤ t ≤ N − 1. By formula(3.22),

‖fi(Lm + a2m + j) − gN,i(Lm + a2m + j)‖ < 2/N.Then

‖fi(n) − gN,i(n)‖ < 2/N, for any n, i ∈ N. (3.24)Denote by

Cm = {(gN,i(n2m), ..., gN,i(n2m + 2m − 1)) : i ∈ N, n ∈ N, n2m ≥ Lm}.Recall that

g(n) =

w∑

v=1

1Sv(n)pv(n), (3.25)

where each 1Sv(n) is deterministic (i.e., Æ(1Sv) = 0 by Proposition 2.3). Define the arithmeticfunction η by η(n) = v if n ∈ Sv, v = 1, . . . , w. Then Æ(η) = 0. Recall that B2m(η) denotes theset of all 2m-blocks occurring in η. Given ν ∈ B2m(η), denote by Ãν,m the set{(gN,i(n2m), ..., gN,i(n2m+2m−1)) : i ∈ N, n ∈ N, n2m ≥ Lm, (η(n2m), ..., η(n2m+2m−1)) = ν} .Then

Cm ⊆⋃

ν∈B2m (η)

Ãν,m. (3.26)

16

Let d = max(deg(p1), ..., deg(pw)) + 1 and q = k + wd. It follows, from a similar argument tothe proof of formula (3.20) in Theorem 1.2, that

|Cm| ≤∑

ν∈B2m (η)

|Ãν,m| ≤ (q + 1)2q(2(k + 2)(d+ 1)2m(k+d+1)N)q|B2m(η)| (3.27)

for any m with 2m > max(k + 1, d). Suppose that f(n) is the concatenation of {fi(n)}∞i=0 withrespect {Ni}∞i=0. Let gN(n) be the concatenation of {gN,i}∞i=0 with respect to {Ni}∞i=0, i.e.,

gN(n) = gN,i(n) if Ni ≤ n < Ni+1 .By formula (3.24),

‖gN(n) − f(n)‖ < 2/N, for any n ∈ N.This implies that limN→∞ supn∈N |e(gN(n)) − e(f(n))| = 0. So by Propositions 2.1 and 2.2, toprove e(f(n)) deterministic, it suffices to prove Æ(gN) = 0 for N large enough.

In the following, to show Æ(gN) = 0, we estimate |Be,r2m(gN)| for any given m with 2m >max(k+ 1, d), where |Be,r2m(gN)| is the cardinality of all regularly effective 2m-blocks occurring ingN . Let im be large enough such that Ni+1−Ni > 2m whenever i > im. Let (gN(n2m), gN(n2m +1), . . . , gN(n2

m + 2m−1)) be a 2m-block in gN with n2m > max{Lm, Nim}. It is easy to see thatthere are two cases about this block: one case is that there is an in such that (gN(n2

m), gN(n2m+

1), . . . , gN(n2m +2m−1)) = (gN,in(n2m), gN,in(n2m +1), . . . , gN,in(n2m +2m−1)); the other case

is that there are two integers in ≥ 0 and jn ≥ 1 such that gN(n2m + j) = gN,in(n2m + j) whenj = 0, 1, . . . , jn − 1 and gN(n2m + j) = gN,in+1(n2m + j) when j = jn, jn + 1, . . . , 2m − 1. So|Be,r2m(gN)| is less than or equal to 2m × |Cm|2. Note that N, k, d, q are parameters independentof m. Then by formula (3.27),

Æ(gN) = limm→∞

log |Be,r2m(gN)|2m

≤ limm→∞

log(|B2m(η)|2)2m

= 2Æ(η) = 0.

Now, we complete the proof of the theorem. �

Using a similar idea to the proof of Theorem 1.2, we obtain the following proposition thatgives many characteristic functions with zero anqie entropy.

Proposition 3.2. Let p1(y), p2(y) ∈ R[y]. Suppose thatS = {n ∈ N : {p1(n)} < {p2(n)}}.

Then 1S(n), the characteristic function defined on S, is a deterministic sequence.

Proof. Assume that the degrees of p1(n) and p2(n) are both less than k. Then △kp1(n) =△kp2(n) = 0. In the following, for any given integer J ≥ k + 1, we estimate |BJ(1S)|, thecardinality of the set of all J-blocks occurring in 1S.

Firstly, we define linear functions F0, F1, . . . , FJ−1 : Rk → R. Define Fj(x0, x1, . . . , xk−1) to

be xj when j = 0, 1, . . . , k − 1. Assume inductively that we have defined Fj0(x0, x1, . . . , xk−1)

17

for some j0 ≥ k − 1. Then define Fj0+1(x0, x1, . . . , xk−1) to be the linear function satisfyingk∑

l=0

(−1)k−l(k

l

)Fj0+1−k+l(x0, x1, . . . , xk−1) = 0, (3.28)

equivalently,

Fj0+1(x0, x1, . . . , xk−1) = −k−1∑

l=0

(−1)k−l(k

l

)Fj0+1−k+l(x0, x1, . . . , xk−1).

By Proposition A.1, it is not hard to see that for i = 1, 2 and any n ∈ N,{Fj({pi(n)}, {pi(n + 1)}, . . . , {pi(n+ k − 1)})} = {pi(n + j)}, j = 0, 1, . . . , J − 1. (3.29)

By Proposition A.5,

|Fj({pi(n)}, {pi(n + 1)}, . . . , {pi(n+ k − 1)})| < (k + 1)Jk, j = 0, 1, . . . , J − 1. (3.30)Secondly, we define a family of hyperplanes F = {HL,j, HM,j, HN,j : j, L,M,N ∈ Z, 0 ≤ j ≤

J − 1, −2(k + 1)Jk − 1 ≤ L ≤ 2(k + 1)Jk, −(k + 1)Jk − 1 ≤ M,N ≤ (k + 1)Jk} in R2k, whereHL,j = {(x0, x1, . . . , x2k−1) ∈ R2k : Fj(x0, x1, . . . , xk−1) − Fj(xk, xk+1, . . . , x2k−1) = L},

HM,j = {(x0, x1, . . . , x2k−1) ∈ R2k : Fj(x0, x1, . . . , xk−1) = M},and

HN,j = {(x0, x1, . . . , x2k−1) ∈ R2k : Fj(xk, xk+1, . . . , x2k−1) = N}.Then

|F| ≤ 8(k + 2)Jk+1.By Lemma 2.6, there are at most W (|F|, 2k) pieces of R2k cut by the hyperplanes in F , where

W (|F|, 2k) ≤ 82k(2k + 1)22k(k + 2)2kJ2k(k+1).Let P be the set of all pieces of R2k cut by the hyperplanes in F . Recall that BJ(1S)

denotes the set of all J-blocks occurring in 1S(n). Denote by |BJ(1S)| = CJ . Suppose thatBJ(1S) = {B1, . . . , BCJ}. Let nm = min{n ∈ N : (1S(n), 1S(n + 1), . . . , 1S(n + J − 1)) = Bm},for m = 1, . . . , CJ . Then Bm = (1S(nm), 1S(nm + 1), . . . , 1S(nm + J − 1)). Define the map

ψ : BJ (1S) → Pby ψ(Bm) = Pm, where Pm is the unique piece in P containing the point ({p1(nm)}, . . . , {p1(nm+k−1)}, {p2(nm)}, . . . , {p2(nm+k−1)}), form = 1, . . . , CJ . Since any two pieces in P are disjoint,ψ is well-defined.

In the following, we show that ψ is injective. Given n ∈ N, let(y0, . . . , yk−1, yk, . . . , y2k−1) = ({p1(n)}, . . . , {p1(n+ k − 1)}, {p2(n)}, . . . , {p2(n + k − 1)}).

Suppose that (y0, . . . , yk−1, yk, . . . , y2k−1) ∈ P , a piece of R2k cut by the hyperplanes in F . Thenby (3.30), there are integers M0, . . . ,MJ−1, N0, . . . , NJ−1 ∈ [−(k + 1)Jk − 1, (k + 1)Jk] andL0, . . . , LJ−1 ∈ [−2(k + 1)Jk − 1, 2(k + 1)Jk], such that

Lj ≤ Fj(y0, . . . , yk−1) − Fj(yk, . . . , y2k−1) < Lj + 1, (3.31)

18

and

Mj ≤ Fj(y0, . . . , yk−1) < Mj + 1 , Nj ≤ Fj(yk, . . . , y2k−1) < Nj + 1. (3.32)Moreover, the above inequalities also hold for each point in P . From formulas (3.31) and (3.32),it is not hard to see that for any given j with 0 ≤ j ≤ J − 1,

{Fj(x0, x1, . . . , xk−1)} < {Fj(xk, xk+1, . . . , x2k−1)}holds for all (x0, x1, . . . , x2k−1) ∈ P (when Lj = Mj −Nj − 1) or

{Fj(x0, x1, . . . , xk−1)} ≥ {Fj(xk, xk+1, . . . , x2k−1)}holds for all (x0, x1, . . . , x2k−1) ∈ P (when Lj = Mj−Nj). Then by equation (3.29), we concludethat if ({p1(n)}, . . . , {p1(n + k − 1)}, {p2(n)}, . . . , {p2(n + k − 1)}), ({p1(n′)}, . . . , {p1(n′ + k −1)}, {p2(n′)}, . . . , {p2(n′ + k − 1)}) ∈ P , then (1S(n), . . . , 1S(n + J − 1)) = (1S(n′), . . . , 1S(n′ +J − 1)). So ψ is injective. Hence |BJ(1S)| = |ψ(BJ(1S))| ≤ W (|F|, 2k) ≤ 82k(2k + 1)22k(k +2)2kJ2k(k+1). Then by formula (2.2),

Æ(1S) = limJ→∞

log |BJ(1S)|J

= 0.

So 1S(n) is deterministic by Proposition 2.3. �

As an application of the above proposition, we give the following example that satisfies thecondition in Theorem 1.2 with g(n) = △2f(n) 6= 0.Example 3.3. Let f(n) =

√3n{

√2n}. Then

△2f(n) =

2√

3(√

2 − 1), n ∈ S1,2√

3(√

2 − 2), n ∈ S2,2√

3(√

2 − 1) +√

3n, n ∈ S3,2√

3(√

2 − 2) −√

3n, n ∈ S4,where S1 = {n ∈ N : {

√2(n + 2)} > {

√2(n + 1)} > {

√2n}}, S2 = {n ∈ N : {

√2(n + 2)} <

{√

2(n + 1)} < {√

2n}}, S3 = {n ∈ N : {√

2(n + 2)} > {√

2(n + 1)}, {√

2(n + 1)} < {√

2n}},S4 = {n ∈ N : {

√2(n+ 2)} < {

√2(n+ 1)}, {

√2(n+ 1)} > {

√2n}}.

4. The Möbius disjointness of e(f(n)) with the k-th difference of f(n) tendingto zero

In this section, we shall study the disjointness of the Möbius function from exponential func-tions of arithmetic functions with the k-th differences tending to a constant (Propositions 1.5,1.6, 1.9, and Theorems 1.7, 1.8). We first show the following property that arithmetic func-tions with k-th differences tending to zero can be approximated by certain concatenations ofpolynomials of degrees less than k.

Lemma 4.1. Suppose that f(n) is a real-valued arithmetic function such that

limn→∞

‖△kf(n)‖ = 0,

19

for some integer k ≥ 1. Then for any integer N ≥ 1, there is an increasing sequence {Ni}∞i=0 ofnatural numbers with N0 = 0 and limi→∞(Ni+1 − Ni) = ∞, and a sequence {pi(y)}∞i=0 in R[y]of degrees less than k, such that

‖f(n) − gN(n)‖ ≤ 1/N, for any n ∈ N, (4.1)where gN is the concatenation of {pi(n)}∞i=0 with respect to {Ni}∞i=0.Proof. By Lemma 3.1, for each integer m ≥ 1 and N ≥ 1, there is a sufficiently large Lm ∈ Nwith 2m|Lm such that, whenever n ≥ Lm, we have

‖f(n+ j) − Yn(n+ j)‖ ≤ 1/N, j = 0, 1, . . . , 2m − 1, (4.2)where Yn(n+ j) is defined to be f(n+ j) when 0 ≤ j ≤ k − 1 and the value determined by thefollowing linear equations when k ≤ j ≤ 2m − 1,

△kYn(n + j) = 0, j = 0, 1, . . . , 2m − k − 1.It is not hard to check that

Yn(n+ j) =k−1∑

l=0

f(n+ l)k−1∏

t=0,t6=l

j − tl − t , j = 0, 1, . . . , 2

m − 1. (4.3)

We may further assume that the sequence {Lm}∞m=0 (L0 = 0) chosen above satisfies Lm+1 > Lmfor each m. Let dm = (Lm+1 − Lm)/2m. Then the following is a partition of N.

N =

∞⋃

m=0

dm−1⋃

q=0

{Lm + q2m, Lm + q2m + 1, ..., Lm + q2m + 2m − 1}.

Choose the sequence {Ni}∞i=0 with N0 < N1 < N2 < · · · such that{N0, N1, · · ·} = {Lm + q2m : m ∈ N, 0 ≤ q ≤ dm − 1}.

Define

pi(n) =

k−1∑

l=0

f(Ni + l)

k−1∏

t=0,t6=l

n−Ni − tl − t . (4.4)

It is a polynomial of degree less than k. Let

gN(n) = pi(n), when Ni ≤ n ≤ Ni+1 − 1.By formula (4.2) and equation (4.3), we obtain formula (4.1). �

Lemma 4.2. Let {Ni}∞i=0 be an increasing sequence of natural numbers with N0 = 0 andlimi→∞(Ni+1 − Ni) = ∞. Let {pi(y)}∞i=0 be a sequence in R[y] with degrees less than k forsome positive integer k. Suppose that f(n) is the concatenation of {pi(n)}∞i=0 with respect to{Ni}∞i=0. Let q be a positive integer and 0 ≤ a ≤ q − 1. Then

limN→∞

1

N

∑

1≤n≤Nn≡a(mod q)

µ(n)e(f(n)) = 0 (4.5)

20

if and only if

limm→∞

1

Nm

m−1∑

i=0

∑

Ni≤n≤Ni+1−1n≡a(mod q)

µ(n)e(pi(n)) = 0. (4.6)

Proof. It is obvious that (4.5) ⇒ (4.6). We now show (4.6) ⇒ (4.5). In this process, we needto use a classical result (see [Dav37] for k = 2 and [Hua65, Chapter 6, Theorem 10] for k > 2)stated as follows,

supp(y)∈R[y]

deg(p(y)) 0, there is a positive integer M such that whenever m ≥ M and N ≥ NM ,we have

m−1∑

i=0

∑

Ni≤n≤Ni+1−1n≡a(mod q)

µ(n)e(pi(n)) < (ǫ/2)Nm, (4.7)

and

supi∈N

∣∣∣∣∣∑

1≤n≤Nn≡a(mod q)

µ(n)e(pi(n))

∣∣∣∣∣ < (ǫ/4)N. (4.8)

Let N ≥ NM . Choose an appropriate l ≥M with Nl ≤ N ≤ Nl+1 − 1. Then∣∣∣∣∣N∑

n=1n≡a(mod q)

µ(n)e(f(n))

∣∣∣∣∣ ≤∣∣∣∣∣

l−1∑

i=0

∑

Ni≤n

21

So

limN→∞

1

N

∑

1≤n≤Nn≡a(mod q)

µ(n)e(f(n)) = 0

completing the proof. �

Next, we shall prove Propositions 1.5, 1.6. They are both consequences of Theorem 1.7 be-low. Before proving them, we need some preparations. The following Dirichlet’s approximationtheorem is classical and well-known (see e.g., [Tit86, Section 8.2]), which is proved via the factthat if there are m+ 1 points contained in m regions, then there must be at least two points liein the same region.

Lemma 4.3. Given L real numbers θ1, . . . , θL and a positive integer q, then we can find aninteger t ∈ [1, qL], and integers a1, . . . , aL such that |tθj − aj | ≤ 1/q, j = 1, 2, . . . , L.

The following “asymptotical periodicity” of the concatenation of certain linear phases will beused in the proof of Theorem 1.7.

Lemma 4.4. Let {Ni}∞i=0 be an increasing sequence of integers with N0 = 0 and limi→∞(Ni+1−Ni) = ∞. Suppose that α0, α1, . . . are real numbers such that the sequence {e(αi)}∞i=0 has finitelymany limit points. Assume that f(n) = e(nαi)e(βi) when Ni ≤ n < Ni+1 for i = 0, 1, 2, . . .,where β0, β1, . . . are real numbers. Then there is a δ with 0 < δ < 1 and a sequence {nj}∞j=0 ofpositive integers with limj→∞ nj = ∞ such that

limj→∞

nδj∑

l=1

lim supN→∞

1

N

N−1∑

n=0

|f(n+ lnj) − f(n)|2 = 0.

Proof. Suppose that the limit points of {e(αi)}∞i=0 are e(θ1), . . . , e(θL) for some integer L ≥ 1.Let qj = (j + 6)

2π2, j ≥ 0. By Lemma 4.3, we can find an integer nj with 1 ≤ nj ≤ qLj suchthat |θs − as,jnj | ≤

1njqj

, where as,j is some integer for s = 1, . . . , L. It is not hard to check that

the sequence {nj}∞j=0 can be chosen to satisfy limj→∞ nj = ∞. Moreover, there is an i0 suchthat when i ≥ i0 we can choose an s ∈ {1, 2, . . . , L} satisfying ‖αi − θs‖ < 1njqj . Then for i ≥ i0,‖njαi‖ < 2qj and |e(njαi) − 1| = 2| sin(πnjαi)| ≪

1qj

. So, for any given nj ,

n12Lj∑

l=1

lim supN→∞

1

N

(m−1∑

i=0

Ni+1−1∑

n=Ni

|f(n+ lnj) − f(n)|2 +N∑

n=Nm

|f(n+ lnj) − f(n)|2)

=

n12Lj∑

l=1

lim supN→∞

1

N

(m−1∑

i=0

Ni+1−lnj∑

n=Ni

|f(n+ lnj) − f(n)|2 +N−lnj∑

n=Nm

|f(n+ lnj) − f(n)|2)

=

n12Lj∑

l=1

lim supN→∞

1

N

(m−1∑

i=0

Ni+1−lnj∑

n=Ni

|e(lnjαi) − 1|2 +N−lnj∑

n=Nm

|e(lnjαm) − 1|2)

22

=

n12Lj∑

l=1

lim supm→∞

1

N

(m−1∑

i=i0

Ni+1−lnj∑

n=Ni

|e(lnjαi) − 1|2 +N−lnj∑

n=Nm

|e(lnjαm) − 1|2)

≪n

12Lj∑

l=1

l2

q2j≪ 1

q12j

→ 0, j → ∞.

The claimed result follows by choosing δ = 12L

. �

To prove Theorem 1.7, we also need the following result of the second author [Wei21, Lemma4.1] on the self-correlation of µ(n)e(P (n)) in short arithmetic progressions.

Lemma 4.5. Let s ≥ 1 and h ≥ 3 be integers. Suppose that P (x) ∈ R[x] of degree d ≥ 0.

lim supN→∞

1

N

N∑

n=1

∣∣∣∣∣1

h

h∑

l=1

µ(n+ ls)e(P (n+ ls)

)∣∣∣∣∣

2

≪ sϕ(s)

log log h

log h, (4.9)

where ϕ is the Euler totient function and the implied constant depends on d at most.

Proof of Theorem 1.7. Choose {βi}∞i=0 as a sequence of real numbers such that∣∣∣∣∣

∑

Ni≤n

23

N > M0,

1

nδj0

nδj0∑

l=1

1

N

N∑

n=1

|f(n+ lnj0) − f(n)|2 < ǫ2/9

and

1

N

N∑

n=1

∣∣∣1

nδj0

nδj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0))∣∣∣2

< ǫ2/9.

Then by the Cauchy-Schwarz inequality,

1

nδj0

nδj0∑

l=1

1

N

N∑

n=1

|f(n+ lnj0) − f(n)| < ε/3, (4.13)

and

1

N

N∑

n=1

∣∣∣∣∣1

nδj0

nδj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0))

∣∣∣∣∣ < ǫ/3. (4.14)

Observe that

1

N

N∑

n=1

µ(n)e(P (n)

)f(n) =

1

N

N∑

n=1

µ(n+ lnj0)e(P (n+ lnj0)

)f(n+ lnj0) +O(

lnj0N

).

Write hj0 = nδj0 . Then there is a positive integer M1 such that whenever N > M1,

∣∣∣∣∣1

N

N∑

n=1

µ(n)e(P (n)

)f(n) − 1

N

N∑

n=1

1

hj0

hj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0)

)f(n+ lnj0)

∣∣∣∣∣ < ǫ/3. (4.15)

Let M2 = max{M0,M1}. Then by formulas (4.13), (4.14) and (4.15), for N > M2, we have∣∣∣∣∣

1

N

N∑

n=1

µ(n)e(P (n)

)f(n)

∣∣∣∣∣ <∣∣∣∣∣

1

N

N∑

n=1

1

hj0

hj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0)

)f(n+ lnj0)

∣∣∣∣∣+ ǫ/3

≤∣∣∣∣∣

1

N

N∑

n=1

1

hj0

hj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0)

)(f(n+ lnj0) − f(n)

)∣∣∣∣∣

+

∣∣∣∣∣1

N

N∑

n=1

1

hj0

hj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0)

)f(n)

∣∣∣∣∣+ ǫ/3

≤ 1N

N∑

n=1

1

hj0

hj0∑

l=1

|f(n+ lnj0) − f(n)|

+1

N

N∑

n=1

∣∣∣∣∣1

hj0

hj0∑

l=1

µ(n+ lnj0)e(P (n+ lnj0)

)∣∣∣∣∣+ ε/3

24

0, by Lemma 4.1 and equation (4.4), there isan increasing sequence {Ni}∞i=0 with N0 = 0 and limi→∞(Ni+1 −Ni) = ∞, and a function gǫ(n)defined by gǫ(n) = n(f(Ni + 1) − f(Ni)) + (Ni + 1)f(Ni) − f(Ni + 1)Ni when Ni ≤ n < Ni+1for i = 0, 1, . . ., such that

supn∈N

|e(f(n)) − e(f0(n) + gǫ(n))| < ǫ. (4.16)

Since the set {e(f(Ni + 1) − f(Ni)

): i = 0, 1, . . .} has finitely many limit points,

limm→∞

1

Nm

m−1∑

i=0

∣∣∣∣∣∑

Ni≤n

25

Lemma 4.6. ([MRTTZ20, Theorems 1.3 and 1.8]) Let k be a given positive integer, and let5/8 < τ < 1 and 0 < θ < 1 be fixed. Denote by Dk the set of all polynomials in R[y] of degreesless than k. Let f : N → C be a multiplicative function with |f(n)| ≤ 1 for any n ∈ N. Supposethat X ≥ 1, Xθ ≥ H ≥ exp((logX)τ), and η > 0 are such that

∫ 2X

X

supp(y)∈Dk

∣∣∣∣∣∑

x≤n 0. For ε > 0, we have for X large enough,

M(µ;X,Q) ≥ (1/3 − ǫ) log logX +O(1).By the above proposition and Lemma 4.6, we have the following result which states that µ(n)

does not correlate with polynomial phases in short intervals on average.

Lemma 4.8. Let k be a given positive integer, and let 5/8 < τ < 1 and 0 < θ < 1 be fixed.Suppose that X ≥ 1 and Xθ ≥ H ≥ exp((logX)τ ). Denote by Dk the set of all polynomialsin R[y] of degrees less than k. Then for any η > 0, there is an X0 > 0 (at most depends onk, η, θ, τ) such that whenever X > X0,

∫ 2X

X

supp(y)∈Dk

∣∣∣∣∣∑

x≤n

26

Let Sj = {lhm + j : l = 0, 1, · · ·, ⌊Nmhm ⌋}, for 0 ≤ j ≤ hm − 1. Then, by (4.18), there is a j0such that

∑

x∈Sj0

supp(y)∈Dk

∣∣∣∣∣∑

x≤n

27

For any positive integer M with MC ≥ 1, choose L0 = 0 and Lm = 2m⌊exp(

log1τ (MC2mk)

)+1⌋

for m = 1, 2, . . .. Then by the above inequality, we have for n ≥ Lm and j = 0, 1, . . . , 2m − 1,∥∥∥f(n+ j) −

k−1∑

l=0

f(n+ l)

k−1∏

t=0,t6=l

j − tl − t

∥∥∥ ≤ 1M. (4.21)

Let dm = (Lm+1 − Lm)/2m. Setting 0 = N0 < N1 < N2 < · · · with {N0, N1, . . .} being the setof {Lm + t2m : m ∈ N, 0 ≤ t ≤ dm − 1}. Assume Lm = Nkm . Then km+1 = km + dm. Note thatk0 = 0. Then for m ≥ 1,

exp(

log1τ (MC2mk)

)≪ km ≪ m exp

(log

1τ (MC2mk)

).

Choose an appropriate ǫ > 0 with τ − ǫ > 5/8. Then for m large enough, we have 2m ≥exp((log km+1)

τ−ǫ). By the choice of Ni, this leads to Ni+1 − Ni > exp((log i)τ−ǫ) for i largeenough. Define

pM(n) =

k−1∑

l=0

f(Ni + l)

k−1∏

t=0,t6=l

n−Ni − tl − t

when Ni ≤ n < Ni+1, i = 0, 1, . . .. Hence by Theorem 4.9,

lims→∞

1

Ns

s−1∑

i=0

∑

Ni≤n

28

Suppose that {fi(n)}∞i=0 is a sequence of real-valued arithmetic functions such thatlimn→∞

supi∈N

‖△kfi(n) − g(n)‖ = 0 (5.1)

holds for some k ≥ 1. Let {Ni}∞i=0 be an increasing sequence of natural numbers with N0 = 0and limi→∞(Ni+1 −Ni) = ∞. Then SMDC implies that, for any deterministic sequence ξ(n),

limm→∞

1

Nm

m−1∑

i=0

∣∣∣∣∣∑

Ni≤n

29

limj→∞ hj = ∞, such that

lim supX→∞

1

X

∫ 2X

X

supf∈D

∣∣∣∣∣∑

x≤n 2δhj .

Given hj , choose Xj large enough with δXj > 16hj and Xj > 4Xj−1 satisfying

∫ 2Xj

Xj

supf∈D

∣∣∣∣∣∑

x≤n 2δXjhj .

By the pigeonhole principal, there is a yj ∈ [0, hj), such that⌊2Xj/hj⌋+1∑

lj=⌊Xj/hj⌋−1

(supf∈D

∣∣∣∣∣

(lj+1)hj+yj−1∑

n=ljhj+yj

w(n)e(f(n))

∣∣∣∣∣

)>

3

2δXj.

Furthermore, for each lj with ⌊Xj/hj⌋ − 1 ≤ lj ≤ ⌊2Xj/hj⌋ + 1, we can find glj(n) ∈ D suchthat

⌊2Xj/hj⌋+1∑

lj=⌊Xj/hj⌋−1

∣∣∣∣∣

(lj+1)hj+yj−1∑

n=ljhj+yj

w(n)e(glj(n))

∣∣∣∣∣ > δXj . (5.4)

Now we construct {Ni}∞i=0 and {fi(n)}∞i=0 in the following way. Choose N0 = 0 and Ni =⌊XJ+1/hJ+1⌋hJ+1+thJ+1+yJ+1 when i =

∑Jj=1⌊Xj/hj⌋−J+1+t, t = 0, 1, . . . , ⌊XJ+1/hJ+1⌋−2,

where J = 0, 1, . . .. Then limi→∞(Ni+1 − Ni) = ∞. For J = 0, 1, . . ., we choose fi(n) =g⌊XJ+1/hJ+1⌋+t(n) when i =

∑Jj=1⌊Xj/hj⌋−J + 1 + t, t = 0, 1, . . . , ⌊XJ+1/hJ+1⌋− 3, and fi(n) =

g⌊X1/h1⌋(n) otherwise. Then by equation (5.2), there is an m0 and J0 with m0 =∑J0−1

j=1 ⌊Xj/hj⌋−J0 + 2, such that

1

Nm0+⌊XJ0/hJ0⌋−2

(m0+⌊XJ0/hJ0⌋−3∑

i=0

∣∣∣∣∣∑

Ni≤n(δ/2)XJ02XJ0−hJ0

> δ4. This contradicts the right side of formula (5.5). Hence

equation (5.3) holds.Next, we show that (ii)⇒(i). Given ǫ with 0 < ǫ < 1. Let h be a fixed sufficiently large

positive integer. By equation (5.3) and a dyadic subdivision, there is an m0 ≥ 1 such thatwhenever m > m0, we have Nm −Nm−1 > 2ǫh and

(⌊Nm/h⌋+1)h−1∑

x=0

supf∈D

∣∣∣∣∣∑

x≤n

30

Let Sj = {lh + j : l = 0, 1, · · ·, ⌊Nm/h⌋}, for 0 ≤ j ≤ h − 1. Then, by (5.6), there is a j0 suchthat

∑

x∈Sj0

supf∈D

∣∣∣∣∣∑

x≤n

31

By the above formula, for N > max(X0, h/ǫ), we have

1

N

N∑

n=1

µ(n)ξ(n) =1

Nh

h−1∑

i=0

N∑

n=1

µ(n+ i)ξ(n+ i) +O(h

N)

=1

Nh

N∑

n=1

∑

n≤m k ≥ 0. Suppose that f(0), . . . , f(J − 1) satisfy∑kl=0(−1)k−l

(kl

)f(n+ l) = 0 for n = 0, . . . , J − 1 − k. Then

f(n) =k−1∑

j=0

f(j)∏

0≤i≤k−1i 6=j

n− ij − i

for n = 0, . . . , J − 1.Proof. Suppose g(n) =

∑k−1j=0 f(j)

∏0≤i≤k−1

i 6=j

n−ij−i

. Then g(n) is a polynomial of degree at most

k − 1 and satisfies g(0) = f(0), . . . , g(k − 1) = f(k − 1). By Proposition A.1,

△kg(n) =k∑

l=0

(−1)k−l(k

l

)g(n+ l) = 0, n ≥ 0.

For any given x0, . . . , xk−1, the solution (xk, . . . , xJ−1) that satisfies∑k

l=0(−1)k−l(kl

)xn+l = 0 for

n = 0, . . . , J − k − 1 is unique. Then (f(k), . . . , f(J − 1)) = (g(k), . . . , g(J − 1)). Hence f(n) isof the form as claimed in this proposition. �

The following simple fact is used in this paper.

32

Proposition A.3. For any n ∈ N, k ≥ 1 and 0 ≤ j ≤ k − 1, ∏0≤i≤k−1i 6=j

n−ij−i

is an integer.

Proof. For 0 ≤ n ≤ k − 1 and n 6= j, ∏0≤i≤k−1i 6=j

n−ij−i

= 0; for n = j,∏

0≤i≤k−1i 6=j

n−ij−i

= 1; for n ≥ k,∏

0≤i≤k−1i 6=j

n− ij − i =

∏

0≤i≤j−1

n− ij − i

∏

j+1≤i≤k−1

n− ij − i

= (−1)k−j−1 n!(n− k)!j!(k − 1 − j)!(n− j)

= (−1)k−j−1(n− j − 1n− k

)(n

n− j

)

is an integer. �

The next one gives a variant version of Proposition A.2.

Lemma A.4. Given integers J, k with J > k ≥ 0. Suppose that x0, . . . , xJ−1 ∈ R. Then thefollowing two statements are equivalent.

(i) For n = 0, . . . , J − 1 − k,

{k∑

l=0

(−1)k−l(k

l

){xn+l}} = 0. (A.2)

(ii) For n = 0, . . . , J − 1,

{xn} = {k−1∑

j=0

{xj}∏

0≤i≤k−1i 6=j

n− ij − i }, (A.3)

where {·} denotes the fractional part function.Proof. (i) ⇒ (ii). We first show that there are integers c0, c1, . . . , cJ−1 such that x0 + c0, x1 +c1, . . . , xJ−1 + cJ−1 satisfy the following linear equations

k∑

l=0

(−1)k−l(k

l

)(xn+l + cn+l) = 0, n = 0, . . . , J − 1 − k. (A.4)

Let c0 = · · · = ck−1 = 0. Letting n = 0 in equation (A.4), by equation (A.2) we have that thesolution ck is an integer. Repeating the above process with n = 1, . . . , J − 1 − k, we obtainsuccessively solutions ck+1, . . . , cJ−1, which are all integers. By equation (A.4) and PropositionA.2,

xn + cn =k−1∑

j=0

(xj + cj)∏

0≤i≤k−1i 6=j

n− ij − i , n = 0, . . . , J − 1.

By Proposition A.3, we have equation (A.3).

33

(ii)⇒ (i). Assume that x0, x1, . . . , xJ−1 satisfy equation (A.3). Let

g(n) =k−1∑

j=0

{xj}∏

0≤i≤k−1i 6=j

n− ij − i , n = 0, . . . , J − 1.

Then g(n) is a polynomial of degree at most k − 1. By Proposition A.1,

△kg(n) =k∑

l=0

(−1)k−l(k

l

)g(n+ l) = 0, n ≥ 0.

By (ii), {g(0)} = {x0}, {g(1)} = {x1}, . . . , {g(J − 1)} = {xJ−1}. Then

{k∑

l=0

(−1)k−l(k

l

){xn+l}} = {

k∑

l=0

(−1)k−l(k

l

){g(n+ l)}} = 0, n = 0, . . . , J − 1 − k.

�

The following gives an estimate of f(n) through the initial values and the upper bound of△kf(n).Proposition A.5. Given n ∈ N and a real number c > 0. Suppose integers J > k ≥ 0. Iff(n), f(n+ 1), . . . , f(n+ J − 1) satisfy:(a) |△kf(n+ j)| ≤ c, j = 0, 1, ..., J − 1 − k;(b) f(n+ j) ∈ [0, c], j = 0, 1, ..., k − 1.

Then we have

|f(n+ j)| ≤ (k + 1)jkc, j = k, k + 1, ..., J − 1.Proof. By (b), for j = 0, 1, ..., k − 2,

|△f(n+ j)| ≤ c,and by induction,

|△mf(n+ j)| ≤ 2m−1c, j = 0, 1, ..., k −m− 1, 1 ≤ m ≤ k − 1. (A.5)We first claim that, when 1 ≤ m ≤ k and k −m ≤ j ≤ J − 1 −m, we have

|△mf(n+ j)| ≤k−m−1∑

i=0

(j − (k −m) + ij − (k −m)

)2m−1+ic+

(j

k −m

)c, (A.6)

where(00

)= 1. In the following we shall prove formula (A.6). When m = k, the inequality (A.6)

holds by (a). Assume inductively that formula (A.6) holds when m = m0 + 1 ≤ k. We shallprove that formula (A.6) holds when m = m0 and k −m0 ≤ j ≤ J − 1 −m0. For j = k −m0,we have

|△m0f(n+ k −m0)| ≤ |△m0+1f(n+ k −m0 − 1)| + |△m0f(n+ k −m0 − 1)|.

34

Then by the inductive hypothesis on the m0 + 1 case and formula (A.5),

|△m0f(n+ k −m0)| ≤2m0−1c +k−m0−2∑

i=0

(k −m0 − 1 − (k −m0 − 1) + ik −m0 − 1 − (k −m0 − 1)

)2m0+ic

+

(k −m0 − 1k −m0 − 1

)c

=

k−m0−1∑

i=0

(k −m0 − (k −m0) + ik −m0 − (k −m0)

)2m0−1+ic+

(k −m0k −m0

)c.

Note that in the above formula the coefficients before 2m0+ic and 2m0−1+ic are all 1. So formula(A.6) holds for m = m0 and j = k −m0. Now assume inductively that formula (A.6) holds form = m0 and some j = j0 ≥ k −m0. When j = j0 + 1, we have

|△m0f(n+ j0 + 1)| ≤|△m0f(n+ j0)| + |△m0+1f(n+ j0)|

≤k−m0−1∑

i=0

(j0 − (k −m0) + ij0 − (k −m0)

)2m0−1+ic+

(j0

k −m0

)c

+

k−m0−2∑

i=0

(j0 − (k −m0 − 1) + ij0 − (k −m0 − 1)

)2m0+ic+

(j0

k −m0 − 1

)c

=k−m0−1∑

i=0

(j0 − (k −m0) + ij0 − (k −m0)

)2m0−1+ic+

(j0

k −m0

)c

+

k−m0−1∑

i=1

(j0 − (k −m0) + ij0 − (k −m0 − 1)

)2m0−1+ic +

(j0

k −m0 − 1

)c

=

k−m0−1∑

i=1

(j0 − (k −m0) + i + 1j0 − (k −m0) + 1

)2m0−1+ic+ 2m0−1c+

(j0 + 1

k −m0

)c

=

k−m0−1∑

i=0

(j0 + 1 − (k −m0) + ij0 + 1 − (k −m0)

)2m0−1+ic+

(j0 + 1

k −m0

)c.

Then formula (A.6) holds by the above induction process. In particular, taking m = 1 in formula(A.6), for k − 1 ≤ j ≤ J − 2,

|△f(n+ j)| ≤k−2∑

i=0

(j − k + i+ 1j − k + 1

)2ic+

(j

k − 1

)c.

Hence for k ≤ j ≤ J − 1, by the above inequality,

|f(n+ j)| ≤|f(n+ k − 1)| +j−1∑

l=k−1

|△f(n+ l)|

35

≤c+j−1∑

l=k−1

(k−2∑

i=0

(l − k + i+ 1l − k + 1

)2i +

(l

k − 1

))c

=c+

k−2∑

i=0

(j − 1 − k + i+ 2

i+ 1

)2ic+

(j

k

)c.

Since(jk

)≤ jk and

(j−k+i+1

j−k

)2i ≤ ji+1 when 0 ≤ i ≤ k − 2,

|f(n+ j)| ≤ (k + 1)jkc, k ≤ j ≤ J − 1.This completes the proof. �

The following proposition shows that f can be approached by polynomials if the k-th differenceof f is small. This proposition will be used in the proof of Theorem 1.8 in Section 5.

Proposition A.6. Given j ≥ k ≥ 1. There are constants ak, ..., aj such that for each arithmeticfunction f and each n ∈ N,

j∑

l=k

al · △kf(n+ l − k) = f(n+ j) −k−1∑

m=0

f(n+m)k−1∏

i=0,i 6=m

j − im− i . (A.7)

Proof. Let

g(l) =

(j − l + k − 1

k − 1

)=

(j − l + 1)(j − l + 2) · · · (j − l + k − 1)(k − 1)!

be a polynomial of l with degree k − 1. Choose al = g(l) when k ≤ l ≤ j + k − 1. So al = 0when l = j + 1, ..., j + k − 1. By equation (A.1), the left side of equation (A.7) is

j∑

l=k

al

k∑

t=0

(−1)k−t(k

t

)f(n+ l − k + t). (A.8)

In the following, we consider the coefficients of f(n+m) in the above formula for m = 0, . . . , j.When m = j, the coefficient of f(n + m) in (A.8) is aj = 1. For the case j ≥ 2k, whenk ≤ m ≤ j − k, the coefficient of f(n+m) in (A.8) is

k∑

t=0

am−t+k(−1)k−t(k

t

)= (−1)k

k∑

t=0

g(m+ t)(−1)k−t(k

t

)= (−1)k△kg(m) = 0.

Notice that al = 0 for l = j+1, ..., j+k−1. When j−k < m ≤ j−1, the coefficient of f(n+m)in (A.8) is

k∑

t=m+k−j

am−t+k(−1)k−t(k

t

)=

k∑

t=0

am−t+k(−1)k−t(k

t

)= (−1)k△kg(m) = 0. (A.9)

36

For the case j ≤ 2k− 1, when k ≤ m ≤ j− 1, by a similar argument to equation (A.9), we havethat the coefficient of f(n+m) in (A.8) is 0. Hence there are constants c0, ..., ck−1 such that

j∑

l=k

al · △kf(n+ l − k) = f(n+ j) −k−1∑

m=0

cmf(n+m) (A.10)

holds for each n ∈ N. To compute cm, let fm be the polynomial of degree k − 1 such thatfm(m) = 1, fm(t) = 0 for 0 ≤ t ≤ k − 1 and t 6= m. So

fm(t) =

k−1∏

i=0,i 6=m

t− im− i .

Note that △kfm = 0 and equation (A.10) holds for any arithmetic function f . Let f = fm andn = 0 in equation (A.10), then

cm = fm(j) =k−1∏

i=0,i 6=m

j − im− i .

Hence equation (A.7) holds with al = g(l) for l = k, . . . , j. �

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Sin. (3) 7, 259-272, 1991.

Department of Mathematics, University of New Hampshire, Durham, NH 03824, USA –and–

Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190,China

Email address : guweichen14@mails.ucas.ac.cn

Yau Mathematical Sciences Center, Tsinghua University, Beijing 100084, China

Email address : weif@mail.tsinghua.edu.cn

1. Introduction1.1. Notation1.2. A class of deterministic sequences1.3. Disjointness of Möbius from e(f(n)) with the k-th difference of f(n) tending to zero1.4. Applications

2. Preliminaries on anqie entropy of arithmetic functions2.1. Some properties of anqie entropy2.2. Anqie entropy of arithmetic functions with finite ranges

3. Proofs of Theorems 1.2 and 1.44. The Möbius disjointness of e(f(n)) with the k-th difference of f(n) tending to zero5. Proofs of Theorem 1.11 and Corollary 1.13Appendix A. Some properties of the difference operatorReferences