Multiple Fourier series and lattice point problems · problem on the Fourier series of the function u 0;a. On the other hand, in the case 1

Multiple Fourier series and lattice point problems

Shigehiko Kuratsubo and Eiichi Nakai

Abstract

For the multiple Fourier series of the periodization of some radial functionson Rd, we investigate the behavior of the spherical partial sum. We show theGibbs-Wilbraham phenomenon, the Pinsky phenomenon and the third phe-nomenon for the multiple Fourier series, involving the convergence propertiesof them. The third phenomenon is closely related to the lattice point prob-lems, which is a classical theme of the analytic number theory. We also provethat, for the case of two or three dimension, the convergence problem on theFourier series is equivalent to the lattice point problems in a sense. In partic-ular, the convergence problem at the origin in two dimension is equivalent toHardy’s conjecture on Gauss’s circle problem.

1 Introduction

It is well known as the Gibbs-Wilbraham phenomenon that, for the Fourier series of

piecewise continuous functions, in the neighborhood of each jump, the partial sums

overshoot the jump by approx 9% of the jump. This phenomenon can be seen not

only in one dimension but also in higher dimensions (see for example [4, 24, 42]).

In one dimension, it is also well known as the localization property that, if

the function is zero on an interval, then the Fourier series converges to zero there.

However, in higher dimensions this property is no longer valid. In 1993, Pinsky,

Stanton and Trapa [36] showed that, for the Fourier series of the indicator function

of a d-dimensional ball with d ≥ 3, the spherical partial sum diverges at the centerof the ball. That is, the Fourier series diverges at a smooth point - even a point of

2010 Mathematics Subject Classification. Primary 42B05Key words and phrases. multiple Fourier series, lattice point problem, Fourier transform, Hardy’s

identity, Gibbs-Wilbraham phenomenon, Pinsky phenomenon, spherical partial sum, radial func-tion.This work was supported by JSPS KAKENHI Grant Numbers JP22540166, JP24540159,

JP17K18731.

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local constancy - of the function, resulting from global rather than local properties.

This phenomenon is called the Pinsky phenomenon.

In 2010, the third phenomenon was discovered in [18, 22]. Namely, for the Fourier

series of the indicator function of a d-dimensional ball with d ≥ 5, the sphericalpartial sum diverges at all rational points, while it converges almost everywhere.

This third phenomenon was proved by using results on the lattice point problems.

The study of lattice point problems is a classical theme of analytic number

theory which is concerned with the number of integer points. It has a long history

and deep accumulations since G. Voronöı, G. H. Hardy, E. Landau, J. G. Van der

Corput, V. Jarńık and A. Walfisz, see [5, 15, 16]. For example, in R2 = {(x1, x2) :x1 and x2 are real numbers}, by A(s) we denote the number of lattice points, whereare points with integral co-ordinates, inside the circle

x21 + x22 = s,

see Figure 1. Then A(s) is the same as the area of the polygon in Figure 2, since the

polygon is the union of the unit squares whose centers are the lattice points inside

the circle. Let P (s) = A(s)− πs, where πs is the area of this circle. Gauss showedthat

P (s) = O(s1/2) as s→∞.

In the above O is Landau’s symbol, that is, f(s) = O(g(s)) as s→∞ means that

Figure 1: Gauss’s circle problem Figure 2: The union of the unit squares.

lim sups→∞ |f(s)|/g(s) < ∞ for the positive valued function g. Similarly, f(s) =

Multiple Fourier series and lattice point problems 3

o(g(s)) as s→∞ means that lims→∞ f(s)/g(s) = 0. In 1915 Hardy [8] proved that

P (s) 6= O(sθ) if θ ≤ 1/4.

In fact P (s) 6= o(s1/4 log1/4 s). The best bound on θ for P (s) = O(sθ) is a very fa-mous open problem known as Gauss’s circle problem. While Hardy’s conjecture([9])

is P (s) = O(s1/4+ε) for any ε > 0, the most sharp result up to now is P (s) =

O(s131/416(log s)18637/8320) by M. Huxley [12] in 2003, where 131/416 = 0.3149 . . . .

(Recently, Bourgain and Watt [1] gave θ = 517/1648 = 0.31371 . . . in arXiv, 2017.)

Note that A(s) is a special case of

∑m21+···+m2d≤s

exp

(2πi

d∑k=1

mkxk

), (1.1)

where m1, · · · ,md are integers and x1, · · · , xd are real numbers, that is, A(s) is thecase (x1, · · · , xd) = (0, · · · , 0) of (1.1) and d = 2. The sum (1.1) is related to theFourier series. Especially the research on the sum (1.1) by Czechoslovakian mathe-

matician B. Novák (1938–2003) is very important for the study of the convergence

problem of multiple Fourier series.

Recently, Taylor [43, 44] found that the Pinsky phenomenon arises even in two-

dimension. He treated the radial function

U(x) =

{1/√a2 − |x|2, |x| < a,

0, |x| ≥ a,x ∈ R2, a > 0,

Then U(x) is the fundamental solution to the wave equation on R × T2, evaluatedat t = a. Our aim in this paper which is motivated by Taylor [43, 44] is to study the

Gibbs-Wilbraham phenomenon, the Pinsky phenomenon and the third phenomenon

on the Fourier series of

Uβ,a(x) =

{(a2 − |x|2)β, |x| < a0, |x| ≥ a,

x ∈ Rd, a > 0, (1.2)

for β > −1, a > 0 and dimension d, involving the convergence properties of them.If β = 0, then Uβ,a(x) is the same as the indicator function of the ball centered at

the origin and of radius a.


By Rd, Zd and Td = Rd/Zd we denote the d-dimensional Euclidean space, integerlattice and torus, respectively. In this paper, however, we always identify Td with(−1/2, 1/2]d, that is, x ∈ Td means x ∈ (−1/2, 1/2]d and Td ⊂ Rd. Let Q be the setof all rational numbers, and let Qd = {(x1, · · · , xd) : x1, · · · , xd ∈ Q}.

For an integrable function F (x) on Rd, its Fourier transform F̂ (ξ) and its Fourierspherical partial integral σλ(F )(x) of order λ ≥ 0 are defined by

F̂ (ξ) =

∫RdF (x)e−2πiξx dx, ξ = (ξ1, · · · , ξd) ∈ Rd, (1.3)

σλ(F )(x) =

∫|ξ|


If β = 0, then u0,a is the periodization of the indicator function of the ball centered

at the origin and of radius a. See Figure 3 and also Figure 4. Let

Figure 3: uβ,a(x) (d = 2, β = 0, a = 3/4).

u0,a(x) =∑m∈Zd

U0,a(x+m), x ∈ Td, U0,a(x) =

1, |x| < 1,1/2, |x| = 1,0, |x| > 1,

x ∈ Rd.

If d = 2, then

Sλ(u0,a)(x) = πa2 + a

∑0


We first show an identity (Theorem 2.1) for the periodization of any integrable

radial function with compact support. Then it turns out that the difference between

the Fourier partial sum and the Fourier partial integral is closely related to lattice

point problems. Therefore, to study the convergence problem on the Fourier series

of uβ,a(x) we must also investigate the behavior of σλ(Uβ,a)(x) as λ→∞ and latticepoint problems.

In the case β = 0, the convergence problem on the Fourier series of the function

u0,a has been studied in detail by [18, 19, 20, 21, 22, 23, 24, 36, 34, 35]. Some of

results in these papers were proved by using Novák’s results in [28, 29, 30, 32]. His

results were very useful and sufficient for the affirmative results of the convergence

problem on the Fourier series of the function u0,a.

On the other hand, in the case −1 < β < 0, Novák’s results on the sum (1.1) arenot sufficient to study the convergence problem on the Fourier series of uβ,a. Our

results on the convergence of the Fourier series of uβ,a are obtained by using the

best estimates up to now on lattice point problems. Therefore, if the lattice point

problems will be improved in the future, then our results can be also improved. Ac-

tually, the convergence problem on the Fourier series and the lattice point problems

are equivalent in a sense as we will show in Section 7. In particular, the conver-

gence problem at the origin in two dimension is equivalent to Hardy’s conjecture on

Gauss’s circle problem, see Remark 7.1.

To state our main results, for a > 0, let

Ea = {x ∈ Td : x 6= 0 and |x−m| 6= a for all m ∈ Zd}, (1.11)

Ga = {x ∈ Td : x 6= 0 and |x−m| = a for some m ∈ Zd}. (1.12)

Then Td = {0} ∪Ga ∪ Ea. See Figures 4. If 0 < a < 1/2, then

Ea = {x ∈ Td : x 6= 0 and |x| 6= a},

Ga = {x ∈ Td : |x| = a},

because x ∈ Td means x ∈ (−1/2, 1/2]d and then {x ∈ Td : |x −m| = a} is emptyfor m 6= 0. For a > 0, let also

rd(a : x) =∑

m∈Zd, |x−m|=a

1, x ∈ Td. (1.13)


Then rd(a : x) = 0 for x ∈ Ea. If 0 < a < 1/2, then rd(a : 0) = 0 and rd(a : x) = 1for x ∈ Ga.

x1

x2

O

T2

R2

Ea

Ga

− 1 1

−1

1

x1

x2

O

T2

R2

−1 1

−1

1

Figure 4: Ea and Ga for a = 1/4 and a = 3/4

In the following, Theorem 1.1 and Corollary 1.2 deal with the behavior of Sλ(uβ,a)

at x = 0 including the Pinsky phenomenon, Theorem 1.3 deals with the point-

wise behaviors including the third phenomenon, Theorem 1.4 deals with the Gibbs-

Wilbraham phenomenon near Ga. Theorem 1.5 deals with the almost everywhere

convergence.

Our first result is on the behavior of Sλ(uβ,a)(0) which includes the Pinsky phe-

nomenon. Let Γ(s) be the Gamma function. For β > −1, a > 0 and the dimensiond, let

P[d]β,a =

Γ(β + 1)

Γ(d/2)a(d−3)/2+βπ(d−4)/2−β (1.14)

and

Lβ,a =Γ(β + 1)

2

(aπ

)βsin βπ2βπ2

, (1.15)where

(sin βπ

2

)/βπ

2is regarded as 1 if β = 0, that is, L0,a = 1/2.

Theorem 1.1. Let β > −1 and a > 0.


(i) If d ≥ 1 and β > (d− 3)/2, then

Sλ(uβ,a)(0) = uβ,a(0) + rd(a : 0) (Lβ,a + o(1))λ−β +O(λ

d−32−β)

as λ→∞. (1.16)

(ii) If d ≥ 2 and −1 < β ≤ (d−3)/2, then Sλ(uβ,a) reveals the Pinsky phenomenon.More precisely,

Sλ(uβ,a)(0) = uβ,a(0) +(σλ(Uβ,a)(0)− Uβ,a(0)

)+ rd(a : 0)(Lβ,a + o(1))λ

−β + o(λd−32−β) as λ→∞,

where

σλ(Uβ,a)(0)−Uβ,a(0) = −P [d]β,a cos(

2πaλ− d− 1 + 2β4

π

)λd−32−β +O(λ

d−52−β).

If 0 < a < 1/2, then rd(a : 0) = 0. Hence we have the following corollary.

Corollary 1.2. Let d ≥ 1 and 0 < a < 1/2.

(i) If β > d−32

, then

limλ→∞

Sλ(uβ,a)(0) = uβ,a(0).

(ii) If −1 < β ≤ d−32

, then

lim infλ→∞

Sλ(uβ,a)(0)− uβ,a(0)λd−32−β

= −P [d]β,a,

lim supλ→∞

Sλ(uβ,a)(0)− uβ,a(0)λd−32−β

= P[d]β,a.

For example, even if d = 2, we can see the Pinsky phenomena on the graphs of

Sλ(uβ,a) with β = −1/2 and a = 1/4 for λ ∈ N, as Taylor pointed out in [43, 44],see Figure 6. In this case P

[d]β,a = 4.

If d = 3, β = 0 and a = 1/4, then P[d]β,a = 2/π, see Figure 7. If d = 4, β = 1/2

and a = 1/4, then P[d]β,a = 1/8, see Figure 8.

Next, we state the pointwise behaviors of Sλ(uβ,a) on Td \{0} which includes thethird phenomenon. We define the number c(d) for the dimension d as the following:

c(d) = d− 2dd+ 1

− d+ 12

=d− 5

2+

2

d+ 1=d(d− 4)− 1

2(d+ 1), (1.17)


Figure 5: uβ,a(x1, x2) (d = 2, β = −1/2, a = 1/4).

Figure 6: Sλ(uβ,a)(x1, x2) (d = 2, β = −1/2, a = 1/4) for λ = 9, 10, 11, 12.


-0.4 -0.2 0.2 0.4

0.5

1.0

1.5

-0.4 -0.2 0.2 0.4

0.5

1.0

1.5

-0.4 -0.2 0.2 0.4

0.5

1.0

1.5

-0.4 -0.2 0.2 0.4

0.5

1.0

1.5

Figure 7: Sλ(uβ,a)(x, 0, 0) (d = 3, β = 0, a = 1/4) for λ = 46, 47, 48, 49.

-0.4 -0.2 0.2 0.4-0.1

0.1

0.2

0.3

0.4

-0.4 -0.2 0.2 0.4-0.1

0.1

0.2

0.3

0.4

-0.4 -0.2 0.2 0.4-0.1

0.1

0.2

0.3

0.4

-0.4 -0.2 0.2 0.4-0.1

0.1

0.2

0.3

0.4

Figure 8: Sλ(uβ,a)(x, 0, 0, 0) (d = 4, β = 1/2, a = 1/4) for λ = 43, 44, 45, 46.


that is,

c(1) = −1, c(2) = −5/6, c(3) = −1/2, c(4) = −1/10.

Theorem 1.3. Let β > −1 and a > 0.

(i) If 1 ≤ d ≤ 4, then,

(a) for β > −1,

limλ→∞

Sλ(uβ,a)(x)− uβ,a(x)λ−β

= rd(a : x)Lβ,a for all x ∈ Ga.

(b) for β > c(d),

limλ→∞

Sλ(uβ,a)(x) = uβ,a(x) uniformly on any compact set in Ea.

(ii) If d ≥ 5, then

limλ→∞

1

λd−52

∣∣∣∣Sλ(uβ,a)(x)− uβ,a(x)λ−β − rd(a : x)Lβ,a∣∣∣∣ = 0for all x ∈ (Ea ∪Ga) \Qd,

and

0 < lim supλ→∞

1

λd−52

∣∣∣∣Sλ(uβ,a)(x)− uβ,a(x)λ−β − rd(a : x)Lβ,a∣∣∣∣


-0.4 -0.2 0.2 0.4

-20

-10

10

20

Figure 9: uβ,a(x, 0, 0, 0, 0) (d = 5, β = −1/2, a = 1/4).

-0.4 -0.2 0.2 0.4

-20

-10

10

20

-0.4 -0.2 0.2 0.4

-20

-10

10

20

-0.4 -0.2 0.2 0.4

-20

-10

10

20

-0.4 -0.2 0.2 0.4

-20

-10

10

20

Figure 10: Sλ(uβ,a)(x, 0, 0, 0, 0) (d = 5, β = −1/2, a = 1/4) for λ = 100, . . . , 103.

0.25 0.30 0.35 0.40 0.45 0.50

-4

-2

2

4

0.25 0.30 0.35 0.40 0.45 0.50

-4

-2

2

4

0.25 0.30 0.35 0.40 0.45 0.50

-4

-2

2

4

0.25 0.30 0.35 0.40 0.45 0.50

-4

-2

2

4

Figure 11: Sλ(uβ,a)(x, 0, 0, 0, 0) (d = 5, β = −1/2, a = 1/4) for λ = 100, . . . , 103:Expansion of the part 0.2 ≤ x ≤ 0.5 in Figure 10.


Next, we state the Gibbs-Wilbraham phenomenon near Ga for 1 ≤ d ≤ 4.

Theorem 1.4. Let 1 ≤ d ≤ 4, c(d) < β ≤ 0 and 0 < a < 1/2. Then, on anyneighborhood of the set Ga, Sλ(uβ,a) reveals a phenomenon like the Gibbs-Wilbraham

phenomenon. More precisely, the following holds: For each x0 ∈ Ga, let {x±λ } ⊂ Td

be the sequences which satisfy limλ→∞

x±λ = x0 and |x±λ | = a∓ (2± β)/(4λ). Then

limλ→∞

Sλ(uβ,a)(x±λ )− uβ,a(x

±λ )

λ−β= G±β,a,

where

G±β,a = ∓Γ(β + 1)aβ(2± β)β

π2β

∫ ∞π

sin s

(s± β2π)β+1

ds. (1.18)

If d = 3, β = 0 and a = 1/4, then we can see both the Gibbs-Wilbraham

phenomena and the Pinsky phenomena on the graphs in Figure 7. If d = 4, β =

1/2 and a = 1/4, then we can see the Pinsky phenomena but not see the Gibbs-

Wilbraham phenomena on the graphs in Figure 8.

Remark 1.2. (i) The constant G+β,a is positive and G−β,a is negative. Especially

G±0,a = ∓1

π

∫ ∞π

sin s

sds = ∓1

2± 1π

∫ π0

sin s

sds = ±0.08949 · · · .

(ii) In the case 2 ≤ d ≤ 4 and −1 < β ≤ c(d), Theorem 1.4 is also an open problemby the same reason as Remark 1.1.

Finally, we state the almost everywhere convergence of Sλ(uβ,a) for d ≥ 4.

Theorem 1.5. Let d ≥ 4, β > −1/2 and a > 0. Then

limλ→∞

Sλ(uβ,a)(x) = uβ,a(x), a.e. x ∈ Td.

Remark 1.3. (i) From Theorems 1.3 and 1.5 we get Theorem in [18] as a corollary,

which is the case β = 0.

(ii) In the case d ≥ 4 and −1 < β ≤ −1/2, the almost everywhere convergence ofSλ(uβ,a) is an open problem.

We first prove in Section 2 a fundamental identity for the periodization f of any

integrable radial function F on Rd with compact support. This identity gives the


relation among Sλ(f)(x), σλ(F )(x) and the term related to lattice point problems.

In Section 3 we collect the results with respect to Bessel functions. In Section 4

we investigate the behavior of σλ(Uβ,a)(x). We show the uniform convergence of

σλ(Uβ,a)(x) on any compact subset of Ea, the Pinsky phenomenon at the origin,

and, the Gibbs-Wilbraham phenomenon near the spherical surface. In Section 5

we prove lemmas related to the lattice point problems. Then, in Section 6, using

the results in Sections 2–5, we prove the main results, in which it turns out that

the third phenomenon is closely related to the lattice point problems. Finally, in

Section 7 we give the relation between the convergence of the spherical partial sum

and lattice point problems.

2 Fundamental identity

Let Jν be the Bessel function of order ν. If ν > −1, then

Jν(s)

sν→ 1

2νΓ(ν + 1)as s→ 0.

For this reason, in this paper we always regard

Jν(s)

sν=

1

2νΓ(ν + 1)for s = 0. (2.1)

For ν > −1, let

Λν(t : s) =Jν(2πt

√s)

sν2

, t ≥ 0, s > 0. (2.2)

By (2.1) we regard

Λν(t : 0) =(πt)ν

Γ(ν + 1), t ≥ 0.

For any α > −1 we define Dα(s : x), Dα(s : x) and ∆α(s : x) as the following:

Dα(s : x) =

1

Γ(α + 1)

∑|m|2 0,

0, s = 0,

x ∈ Rd,

Dα(s : x) =

1

Γ(α + 1)

∫|ξ|2 0,

0, s = 0,

x ∈ Rd,


and

∆α(s : x) = Dα(s : x)−Dα(s : x), s ≥ 0, x ∈ Rd. (2.3)

We consider a function φ on [0,∞) and the periodization fφ of the functionFφ(x) = φ(|x|), x ∈ Rd, that is,

fφ(x) =∑m∈Zd

Fφ(x+m) =∑m∈Zd

φ(|x+m|), x ∈ Td. (2.4)

If the radial function Fφ is integrable on Rd, then the Fourier transform F̂φ is alsoradial and has the form

F̂φ(ξ) =

∫Rdφ(|x|)e−2πixξ dx = 2π

∫ ∞0

φ(t)J d

2−1(2πt|ξ|)

(t|ξ|) d2−1td−1 dt, (2.5)

that is,

F̂φ(ξ) = 2π

∫ ∞0

φ(t)Λ d2−1(t : |ξ|

2) td2 dt. (2.6)

For the equation (2.5), see [40, Theorem 3.3, page 155] if d ≥ 2. If d = 1, then (2.5)is also valid by the equation J− 1

2(s) =

√(2/πs) cos s. Let

Aφ(s) = 2π

∫ ∞0

φ(t)Λ d2−1(t : s) t

d2 dt, s ≥ 0. (2.7)

Then

F̂φ(ξ) = Aφ(|ξ|2). (2.8)

Moreover, if the support of φ is compact, then Aφ is infinitely differentiable and

A(j)φ (s) = 2π(−π)

j

∫ ∞0

φ(t)Λ d2−1+j(t : s) t

d2+j dt, s ≥ 0, j = 1, 2, · · · ,

since∂

∂sΛµ(t : s) = (−πt)Λµ+1(t : s), (2.9)

by the Bessel recurrence formula (see (2.15) below).

Now, we define d] as the smallest integer that is greater than (d− 1)/2, that is,

d] =

[d− 1

2

]+ 1, (2.10)

where [t] is the integer part of t ≥ 0. In this section we prove the following:


Theorem 2.1. Let φ be a function on [0,∞) with compact support. Suppose thatFφ is integrable on Rd. Let fφ be the periodization of the function Fφ. Then

Sλ(fφ)(x) = σλ(Fφ)(x) +

d]∑j=0

(−1)j∆j(λ2 : x)A(j)φ (λ2)

+ (−1)d]+1∑

m∈Zd\{0}

∫ λ20

Dd](s : x−m)A(d]+1)

φ (s) ds, (2.11)

for all x ∈ Td.

As φ in Theorem 2.1, take

φβ,a(t) =

{(a2 − t2)β, 0 ≤ t < a,0, t ≥ a,

(2.12)

with β > −1 and a > 0. Then Uβ,a(x) = φβ,a(|x|), x ∈ Rd, and the function uβ,ais the periodization of Uβ,a(x) defined by (1.9). In this case we denote Aφ by Aβ,a.

Then Aβ,a can be calculated explicitly by using (2.5), (2.8) and (2.19) below. Its

derivatives A(j)β,a are also calculated by (2.9). That is,

Aβ,a(s) =Γ(β + 1)

πβad2+βΛ d

2+β(a : s),

A(j)β,a(s) = (−1)

jΓ(β + 1)

πβ−jad2+β+jΛ d

2+β+j(a : s).

(2.13)

Then we have the following corollary:

Corollary 2.2. Let β > −1 and a > 0. Then

Sλ(uβ,a)(x) = σλ(Uβ,a)(x) +

d]∑j=0

(−1)j∆j(λ2 : x)A(j)β,a(λ2)

+ (−1)d]+1∑

m∈Zd\{0}

∫ λ20

Dd](s : x−m)A(d]+1)

β,a (s) ds, (2.14)

for all x ∈ Td.

To prove Theorem 2.1, first we state the properties of the Bessel functions, Dα

and Dα. The following are the recurrence formulas for Bessel functions.d

ds(sνJν(s)) = s

νJν−1(s),d

ds

(Jν(s)

sν

)= −Jν+1(s)

sν. (2.15)


The Bessel functions have also the following asymptotic behavior.

Jν(s) =sν

2νΓ(ν + 1)+O(sν+1) as s→ 0, (2.16)

Jν(s) =

√2

πscos

(s− 2ν + 1

4π

)+O(s−3/2) as s→∞. (2.17)

If α > −1, then

∫ t0

Dα(s : x) ds = Dα+1(t : x), t ≥ 0, x ∈ Rd, (2.18)

and

Dα(s : x) =1

Γ(α + 1)

∫|ξ|2 (d− 1)/2, for example, α = d], then

Dα(s : x) =∑m∈Zd

Dα(s : x−m) =sd2+α

πα

∑m∈Zd

J d2+α(2π

√s|x−m|)

(√s|x−m|) d2+α

, (2.20)

where the sum in (2.20) converges absolutely. In the above the equation (2.18)

follows from the definition. For the equalities (2.19) and (2.20), see (4.2) and (4.3)

in [24], respectively. By (2.15) and (2.19) we have

∂

∂sDα(s : x) = Dα−1(s : x), α > 0,

∂

∂sD0(s : x) = π

(√s

|x|

) d2−1

J d2−1(2π|x|

√s), α = 0.

(2.21)

Proof of Theorem 2.1. Combining (1.8) with (2.8) we have

f̂φ(m) = F̂φ(m) = Aφ(|m|2), m ∈ Zd.


Then

Sλ(fφ)(x)−D0(λ2 : x)Aφ(λ2)

=∑|m|


Using (2.21) and integration by parts again, we have∫ λ20

Dd](s : x)A(d]+1)

φ (s) ds

= Dd](λ2 : x)A(d])

φ (λ2)−

∫ λ20

Dd]−1(s : x)A(d])

φ (s) ds

= · · · · · ·

=

d]∑j=0

(−1)jDd]−j(λ2 : x)A(d]−j)φ (λ

2)

+ (−1)d]+1∫ λ20

π

(√s

|x|

) d2−1

J d2−1(2π|x|

√s)Aφ(s) ds

=

d]∑j=0

(−1)d]−jDj(λ2 : x)A(j)φ (λ2)

+ (−1)d]+1(2π)∫ λ0

Aφ(u2)J d

2−1(2π|x|u)

(|x|u) d2−1ud−1 du.

Here we note that, by (2.8) and (2.5),

σλ(Fφ)(x) =

∫|ξ|


Bessel functions:

Jν(s) =

√2

πs

(cos

(s− 2ν + 1

4π

)− 4ν

2 − 18s

sin

(s− 2ν + 1

4π

))+O(s−5/2)

as s→∞. (3.1)

Lemma 3.1 (Gradshteyn and Ryzhik [6], page 676, 6.561, 14). Let ν + 1 > −µ >−1/2. Then ∫ ∞

0

xµJν(x) dx = 2µΓ((1 + ν + µ)/2)

Γ((1 + ν − µ)/2).

Take

ν =1

2+ β and µ = −1

2− β

in Lemma 3.1, and we have the following corollary.

Corollary 3.2. Let β > −1. Then∫ ∞0

J 12+β(s)

s12+β

ds =1

2βΓ(1 + β)

√π

2.

Lemma 3.3. Let β > −1. Then∫ ∞u

J 12+β(s)

s12+β

ds =

√2

πu−β−1 cos

(u− β

2π

)+O(u−β−2) as u→∞.

Proof. Use (3.1) and two equations∫ ∞u

cos(s− θ)s1+β

ds = u−β−1 cos(u− θ + π

2

)+O(u−2−β),∫ ∞

u

sin(s− θ)s2+β

ds = O(u−2−β),

and we have the conclusion.

Lemma 3.4 (Gradshteyn and Ryzhik [6], page 683, 6.575, 1). Let ν+1 > µ > −1,A > 0 and B > 0. Then

∫ ∞0

Jν+1(As)Jµ(Bs)sµ−ν ds =

0 (A < B)(A2 −B2)ν−µBµ2ν−µAν+1Γ(ν − µ+ 1)

(A > B).


Take

ν + 1 = µ+ β + 1, A = 1 and B =t

a,

in Lemma 3.4, and we have the following corollary.

Corollary 3.5. Let µ > −1, a > 0, β > −1 and t > 0. Then

2βΓ(β + 1)a2β∫ ∞0

Jµ(tas)Jµ+β+1(s)(ta

)µsβ

ds =

{0 (t > a)

(a2 − t2)β (0 < t < a),

Lemma 3.6. Let ν > −1, µ > −1, A > 0 and B > 0. Then

Jν(Au)Jµ(Bu) =1

π√ABu

(cos

((A−B)u− ν − µ

2π

)+ cos

((A+B)u− ν + µ+ 1

2π

))+

1√AB

(1

A+

1

B

)O(u−2) as u→∞.

Proof. Use (2.17) and 2 cos θ cosφ = cos(θ + φ) + cos(θ − φ), and we have theconclusion.

Let

sign(r) =

1, r > 0,

0, r = 0,

−1, r < 0.

Lemma 3.7. For β > −1, let

ψβ(λ, r) = |r|β∫ ∞2π|r|λ

1

sβ+1cos

(s− sign(r)β + 1

2π

)ds, λ > 0, r ∈ R\{0}. (3.2)

Then ψβ has the following properties:

(i) For each r ∈ R \ {0},

ψβ(λ, r) = |r|−1O(λ−β−1) as λ→∞. (3.3)

(ii) If β > 0, then

ψβ(λ, r) = O(λ−β) as λ→∞ (3.4)

uniformly with respect to r ∈ R \ {0}.


(iii) If −1 < β ≤ 0, then, for each λ > 0, −ψβ(λ, r) has values(2 + β

4λ

)β ∫ ∞π

− sin s(s+ β

2π)β+1

ds at r =2 + β

4λ,

and (2− β

4λ

)β ∫ ∞π

sin s

(s− β2π)β+1

ds at r = −2− β4λ

.

Proof. (i) Using the equation∫ ∞v

cos(s− θ)sβ+1

ds = O(v−β−1), (3.5)

we have ψβ(λ, r) = |r|βO((|r|λ)−β−1) = |r|−1O(λ−β−1).(ii) Changing of variables, we have

ψβ(λ, r) =1

(2πλ)β

∫ ∞1

cos(2π|r|λu− sign(r)β+12π)

uβ+1du,

which shows ψβ(λ, r) = O(λ−β) uniformly with respect to r, since β > 0.

(iii) If r > 0, then

−ψβ(λ, r) = rβ∫ ∞2πrλ

− sin(s− β

2π)

sβ+1ds = rβ

∫ ∞2πrλ−β

2π

− sin s(s+ β

2π)β+1 ds,

and, if r < 0, then

−ψβ(λ, r) = (−r)β∫ ∞2π(−r)λ

sin(s+ β

2π)

sβ+1ds = (−r)β

∫ ∞2π(−r)λ+β

2π

sin s(s− β

2π)β+1 ds.

Then we have the conclusion.

Lemma 3.8. Let µ > −1, β > −1 and a > 0. If 0 < t < a or t > a, then∫ ∞2πaλ

Jµ(tau)Jµ+β+1(u)(ta

)µuβ

du

=1

πaβ

(at

)µ+ 12ψβ(λ, a− t) +

1

aβ+1

(at

)µ+ 12(

1 +a

t

)O(λ−β−1) as λ→∞,

where the term O(λ−β−1) is uniform with respect to t.


Proof. By Lemma 3.6 we have

Jµ(t

au)Jµ+β+1(u)

=1

πu

√a

t

(cos

(|a− t|a

u− sign(a− t)1 + β2

π

)+ cos

(a+ t

au− 2µ+ β + 1

2π

))+

√a

t

(1 +

a

t

)O(u−2).

Let

I1 =1

π

(at

)µ+ 12

∫ ∞2πaλ

cos(|a−t|au− sign(a− t)β+1

2π)

uβ+1du,

I2 =1

π

(at

)µ+ 12

∫ ∞2πaλ

cos(a+tau− 2µ+β+1

2π)

uβ+1du.

Then ∫ ∞2πaλ

Jµ(tau)Jµ+β+1(u)(ta

)µuβ

du = I1 + I2 +(at

)µ+ 12(

1 +a

t

)a−β−1O(λ−β−1).

Changing of variables and using (3.5), we have

I1 =1

π

(at

)µ+ 12

(|a− t|a

)β ∫ ∞2π|a−t|λ

cos(s− sign(a− t)β+1

2π)

sβ+1ds

=1

πaβ

(at

)µ+ 12ψβ(λ, a− t)

and

I2 =1

π

(at

)µ+ 12

(a+ t

a

)β ∫ ∞2π(a+t)λ

cos(s− 2µ+β+1

2π)

sβ+1ds

=(at

)µ+ 12

(a+ t

a

)β(a+ t)−β−1O(λ−β−1)

=1

aβ+1

(at

)µ+ 12

(a

a+ t

)O(λ−β−1).

Then we have the conclusion.

Lemma 3.9 (Gradshteyn and Ryzhik [6], page 683, 6.574, 2). Let ν+µ+1 > κ > 0

and b > 0. Then∫ ∞0

Jν(bs)Jµ(bs)

sκds =

bκ−1Γ(κ)Γ(µ+ν−κ+12

)

2κΓ(µ−ν+κ+12

)Γ(−µ+ν+κ+12

)Γ(µ+ν+κ+12

).


Especially, for µ > −1 and β > 0,∫ ∞0

Jµ+β+1(s)Jµ(s)

sβds = 0.

Lemma 3.10 (Gradshteyn and Ryzhik [6], page 660, 6.512, 3). Let ν > 0. Then∫ ∞0

Jν(s)Jν−1(s) ds =1

2.

Lemma 3.11. Let µ > −1 and β ≥ 0. Then

∫ ∞u

Jµ+β+1(s)Jµ(s)

sβds =

−1

πβsin

(βπ

2

)u−β +O(u−1−β), if β > 0,

O(u−1), if β = 0,

as u→∞.

Proof. By Lemma 3.6 we have

Jµ+β+1(s)Jµ(s) =1

πs

(cos

(β + 1

2π

)+ cos

(2s− 2µ+ 2 + β

2π

))+O(s−2). (3.6)

Using

1

π

∫ ∞u

1

sβ+1cos

(β + 1

2π

)ds =

−1

πβsin

(βπ

2

)u−β, β > 0,

0, β = 0,

and (3.5), we have the conclusion.

Corollary 3.12. Let µ > −1 and β > −1. Then

2βΓ(β + 1)a2β∫ 2πaλ0

Jµ+β+1(s)Jµ(s)

sβds

= Lβ,aλ−β +

{O(λ−β−1), β ≥ 0,O(1), −1 < β < 0,

as λ→∞. (3.7)

In the above Lβ,a is defined by (1.15), that is,

Lβ,a =Γ(β + 1)

2

(aπ

)βsin βπ2βπ2

,where

(sin βπ

2

)/βπ

2is regarded as 1 if β = 0.


Proof. Case 1. Let β ≥ 0. By Lemmas 3.9 and 3.10 we have

∫ ∞0

Jµ+β+1(s)Jµ(s)

sβds =

{0, if β > 0,12, if β = 0.

Then ∫ 2πaλ0

Jµ+β+1(s)Jµ(s)

sβds = −

∫ ∞2πaλ

Jµ+β+1(s)Jµ(s)

sβds+

{0, if β > 0,12, if β = 0.

By Lemma 3.11 we have (3.7).

Case 2. Let −1 < β < 0. By (3.6) we have∫ 2πaλ1

Jµ+β+1(s)Jµ(s)

sβds

=1

π

∫ 2πaλ1

1

sβ+1

{cos

(β + 1

2π

)+ cos

(2s− 2ν + 2 + β

2π

)}ds+O(λ−β−1)

=1

π

(− sin

(βπ

2

))(2πaλ)−β

−β+O(1).

Since ∫ 10

Jµ+β+1(s)Jµ(s)

sβds

is a constant, we have the conclusion.

The following lemma is an extension of [24, Lemma 3.3] which is for the case

β = 0.

Lemma 3.13. Let ν > 0, λ > 0, ω ≥ 0 and β > −1. Then∫ λ0

Jν+β+1(s)Jν(ωs)

ωνsβds =

∫ λ0

Jν+β(s)Jν−1(ωs)

ων−1sβds− Jν+β(λ)Jν(ωλ)

ωνλβ. (3.8)

In the above, if ω = 0, then by (2.1) we regard (3.8) as∫ λ0

Jν+β+1(s)

sβsν

2νΓ(ν + 1)ds =

∫ λ0

Jν+β(s)

sβsν−1

2ν−1Γ(ν)ds− Jν+β(λ)

λβλν

2νΓ(ν + 1).


Proof. Case 1. Let ω > 0. Then, using (2.15) and integration by parts, we have∫ λ0

Jν+β(s)Jν−1(ωs)

sβων+1 ds

=

∫ λ0

Jν+β(s)

sν+β((ωs)νJν−1(ωs)

)ω ds

=

∫ λ0

Jν+β(s)

sν+β∂

∂s

((ωs)νJν(ωs)

)ds

=

[Jν+β(s)

sν+β((ωs)νJν(ωs)

)]λ0

+

∫ λ0

Jν+β+1(s)

sν+β((ωs)νJν(ωs)

)ds

=Jν+β(λ)Jν(ωλ)

λβων +

∫ λ0

Jν+β+1(s)Jν(ωs)

sβων ds.

Case 2. Let ω = 0. Then,∫ λ0

Jν+β(s)

sβsν−1

2ν−1Γ(ν)ds

=

∫ λ0

Jν+β(s)

sβ+νs2ν−1

2ν−1Γ(ν)ds

=

[Jν+β(s)

sβ+νs2ν

2ν2ν−1Γ(ν)

]λ0

+

∫ λ0

Jν+β+1(s)

sβ+νs2ν

2ν2ν−1Γ(ν)ds

=Jν+β(λ)

λβλν

2νΓ(ν + 1)+

∫ λ0

Jν+β+1(s)

sβsν

2νΓ(ν + 1)ds.

Therefore we have the conclusion.

Using Lemma 3.13 several times, we have the following.

Corollary 3.14. Let d ≥ 3, β > −1, a > 0 and t ≥ 0, and let d] be as in (2.10).

(i) If t > 0, then

∫ 2πaλ0

J d2+β(s)J d

2−1(

tas)

( ta)d2−1sβ

ds

=

∫ 2πaλ0

J d2+β−d]+1(s)J d2−d]

( tas)

( ta)d2−1−d]+1sβ

ds−d]−1∑`=1

J d2+β−`(2πaλ)J d

2−`(2πtλ)

( ta)d2−`(2πaλ)β

,


where the first term of the right hand side in the above equation is equal to

∫ 2πaλ0

Jβ+ 12(s)J− 1

2( tas)

( ta)−

12 sβ

ds, if d is odd,

∫ 2πaλ0

Jβ+1(s)J0(tas)

sβds, if d is even.

(ii) If t = 0, then

∫ 2πaλ0

J d2+β(s)s

d2−1

2d2−1Γ(d

2)sβ

ds

=

∫ 2πaλ0

J d2+β−d]+1(s)s

d2−d]

2d2−d]Γ(d

2− d] + 1)sβ

ds−d]−1∑`=1

(πaλ)d2−`J d

2+β−`(2πaλ)

Γ(d2− `+ 1)(2πaλ)β

,

where the first term of the right hand side in the above equation is equal to

√2

π

∫ 2πaλ0

Jβ+ 12(s)

sβ+12

ds, if d is odd,

∫ 2πaλ0

Jβ+1(s)

sβds, if d is even.

4 Fourier inversion for the function Uβ,a(x)

Recall that Uβ,a(x) = φβ,a(|x|) with

φβ,a(t) =

{(a2 − t2)β, 0 ≤ t < a,0, t ≥ a.

Then φβ,a and Uβ,a have the following properties:

(i) φβ,a is continuous and bounded variation if and only if β > 0;

(ii) φ0,a is discontinuous and bounded variation and U0,a is the indicator function

of the ball in Rd centered at the origin with the radius a;

(iii) φβ,a is not bounded variation if and only if β < 0;

(iv) Uβ,a is integrable on Rd if and only if β > −1 for all dimensions d.


In this section, for β > −1, we investigate the behavior of the Fourier sphericalpartial integral σλ(Uβ,a)(x) as λ→∞.

For a > 0, let

Ẽa = {x ∈ Rd : x 6= 0 and |x| 6= a}, (4.1)

G̃a = {x ∈ Rd : |x| = a}. (4.2)

Then Rd = {0} ∪ Ẽa ∪ G̃a.

Theorem 4.1. Let d ≥ 1, a > 0 and β > −1. Then

σλ(Uβ,a)(x) = 2βΓ(β + 1)a2β

∫ 2πaλ0

J d2−1(

|x|as)J d

2+β(s)(

|x|a

) d2−1sβ

ds, (4.3)

for all x ∈ Rd and λ > 0. Moreover, σλ(Uβ,a) has the following properties:

(i) At x = 0,

σλ(Uβ,a)(0) = Uβ,a(0)− P [d]β,a cos(

2πaλ− d− 1 + 2β4

π

)λd−32−β

+O(λd−52−β) as λ→∞, (4.4)

where P[d]β,a is the constant defined by (1.14). Consequently,

(a) if β > (d− 3)/2, then

σλ(Uβ,a)(0) = Uβ,a(0) +O(λd−32−β) as λ→∞,

(b) if −1 < β ≤ (d− 3)/2, then σλ(Uβ,a)(x) reveals the Pinsky phenomenon,that is,

lim infλ→∞

σλ(Uβ,a)(0)− Uβ,a(0)λ(d−3)/2−β

= −P [d]β,a,

lim supλ→∞

σλ(Uβ,a)(0)− Uβ,a(0)λ(d−3)/2−β

= P[d]β,a.

(ii) For x ∈ G̃a,

limλ→∞

σλ(Uβ,a)(x)

λ−β= Lβ,a,


where Lβ,a is the constant defined by (1.15), and consequently,

limλ→∞

σλ(Uβ,a)(x) =

0, β > 0,1

2, β = 0,

∞, −1 < β < 0.

(iii) For x ∈ Ẽa,

σλ(Uβ,a)(x) = Uβ,a(x) +O(λ−β−1) as λ→∞,

where the last term O(λ−β−1) is uniform on any compact subset of Ẽa.

(iv) If β > 0, then

limλ→∞

σλ(Uβ,a)(x) = Uβ,a(x) for x ∈ Ẽa ∪ G̃a

where the convergence is uniform on any compact subset of Ẽa ∪ G̃a andσλ(Uβ,a) does not reveal the Gibbs-Wilbraham phenomenon.

(v) If −1 < β ≤ 0, then

limλ→∞

σλ(Uβ,a)(x) = Uβ,a(x) for x ∈ Ẽa,

where the convergence is uniform on any compact subset of Ẽa, and σλ(Uβ,a)

reveals a phenomenon like the Gibbs-Wilbraham phenomenon near G̃a. More

precisely, the following holds: For each x0 ∈ G̃a, let {x±λ } be the sequenceswhich satisfy lim

λ→∞x±λ = x0 and |x

±λ | = a∓ (2± β)/(4λ). Then

limλ→∞

σλ(Uβ,a)(x+λ )− Uβ,a(x

+λ )

λ−β= G+β,a,

and

limλ→∞

σλ(Uβ,a)(x−λ )− Uβ,a(x

−λ )

λ−β= G−β,a,

where G±β,a are the constants defined by (1.18).

Remark 4.1. The function Uβ,a is piecewise smooth in the sense of Pinsky [34] if

and only if β is a nonnegative integer. In this case the result in Theorem 4.1 (i) is

contained in [34, Theorem 1a].


In the following we first prove (4.3) in Subsection 4.1. Then, using (4.3), we

prove (i)–(v) of Theorem 4.1 in Subsections 4.2–4.6, respectively. Let

U[d]β,a,λ(t) = 2

βΓ(β + 1)a2β∫ 2πaλ0

J d2−1(

tas)J d

2+β(s)(

ta

) d2−1sβ

ds. (4.5)

Then (4.3) means that

σλ(Uβ,a)(x) = U[d]β,a,λ(|x|). (4.6)

4.1 Proof of (4.3)

By (2.19) the Fourier transform of Uβ,a(x) is expressed by the following:

Ûβ,a(ξ) = Γ(β + 1)Dβ(a2 : ξ) = Γ(β + 1)ad+2β

πβ

J d2+β(2πa|ξ|)

(a|ξ|) d2+β. (4.7)

Since Ûβ,a(ξ) is a radial function, using (2.5), we have

σλ(Uβ,a)(x) =

∫|ξ| −1, a > 0 and λ > 0. Then

U[1]β,a,λ(0) = a

2β − P [1]β,a cos(

2πaλ− β2π

)λ−β−1 +O(λ−β−2) as λ→∞.

U[2]β,a,λ(0) = a

2β − P [2]β,a cos(

2πaλ− 2β + 14

π

)λ−β−

12 +O(λ−β−

32 ) as λ→∞.


Lemma 4.3. Let d ≥ 3, β > −1, a > 0 and λ > 0, and let

P [d]β,a,λ(t) = −Γ(β + 1)aβ

(πλ)β

d]−1∑`=1

J d2−`+β(2πaλ)J d

2−`(2πtλ)(

ta

) d2−`

, t ≥ 0, (4.9)

where d] is defined by (2.10). Then

U[d]β,a,λ(t) =

{U

[1]β,a,λ(t) + P

[d]β,a,λ(t), if d is odd,

U[2]β,a,λ(t) + P

[d]β,a,λ(t), if d is even,

t ≥ 0. (4.10)

Moreover, if t 6= 0, then P [d]β,a,λ(t) = O(λ−β−1) as λ→∞, If t = 0, then

P [d]β,a,λ(0) = −P[d]β,a cos

(2πaλ− d− 1 + 2β

4π

)λd−32−β +O(λ

d−52−β) as λ→∞.

(4.11)

Remark 4.2. (i) In Lemma 4.2, if β = −1/2, then

U[2]−1/2,a,λ(0) =

1− cos (2πaλ)a

+O(λ−1) as λ→∞. (4.12)

since P[2]−1/2,a = a

−1. This shows the amplitude of Pinsky phenomenon which was

found by Taylor [44].

(ii) Recently, Grafakos and Teschl [7] showed some related results with Lemma 4.3.

Proof of Lemma 4.2. Let d = 1. By (4.5) with (2.1) and Corollary 3.2 we have

U[1]β,a,λ(0) = 2

βΓ(β + 1)a2β√

2

π

∫ 2πaλ0

J 12+β(s)

s12+β

ds

and

a2β = 2βΓ(β + 1)a2β√

2

π

∫ ∞0

J 12+β(s)

s12+β

ds,

respectively. Then by Lemma 3.3 we have

U[1]β,a,λ(0)− a

2β = −2βΓ(β + 1)a2β√

2

π

∫ ∞2πaλ

J 12+β(s)

s12+β

ds

= −P [1]β,a cos(

2πaλ− β2π

)λ−β−1 +O(λ−β−2).


Let d = 2. By (4.5) with (2.1) we have

U[2]β,a,λ(0) = 2


J1+β(s)

sβds

= 2βΓ(β + 1)a2β[−Jβ(s)

sβ

]2πaλ0

= a2β − a2βΓ(β + 1)Jβ(2πaλ)(πaλ)β

= a2β − P [2]β,a cos(

2πaλ− 2β + 14

π

)λ−β−1/2 +O(λ−β−3/2).

Then the proof is complete.

Proof of Lemma 4.3. The equation (4.10) follows from Collorary 3.14 and the defi-

nition of U[d]β,a,λ(t) (see (4.5)). If t 6= 0, then from (2.17) it follows that


2−`(2πtλ) = O(λ

−1), ` = 1, . . . , d] − 1.

Then we have P [d]β,a,λ(t) = O(λ−β−1). If t = 0, then by (2.1) we regard


2−`(2πtλ)(

ta

) d2−`

=(πaλ)

d2−`

Γ(d2− `+ 1)

J d2−`+β(2πaλ), ` = 1, . . . , d] − 1.

Then, using (2.17), we have

P [d]β,a,λ(0)

= −Γ(β + 1)aβ

(πλ)β

d]−1∑`=1

(πaλ)d2−`

Γ(d2− `+ 1)

J d2−`+β(2πaλ)

= −Γ(β + 1)aβ

(πλ)β(πaλ)

d2−1

Γ(d2)

J d2−1+β(2πaλ) +O(λ

d−52−β)

=

(−Γ(β + 1)

Γ(d2)

ad−32

+βπd−42−β

)λd−32−β cos

(2πaλ− d− 1 + 2β

4π

)+O(λ

d−52−β).

This is the conclusion.


4.3 Proof of Theorem 4.1 (ii)

For x ∈ G̃a, by (4.5) and (4.6) we have

σλ(Uβ,a)(x) = U[d]β,a,λ(a) = 2


J d2+β(s)J d

2−1(s)

sβds.

Then, using Corollary 3.12 with µ = d2− 1, we have

U[d]β,a,λ(a) = Lβ,aλ

−β +

{O(λ−β−1), β ≥ 0,O(1), −1 < β < 0,

as λ→∞, (4.13)

which shows the conclusion.

4.4 Proof of Theorem 4.1 (iii)

Let x ∈ Ẽa and |x| = t. Then 0 < t < a or t > a. In this case, by (4.5), (4.6) andCorollary 3.5, we have

σλ(Uβ,a)(x)− Uβ,a(x) = U [d]β,a,λ(t)− φβ,a(t)

= −2βΓ(β + 1)a2β∫ ∞2πaλ

J d2−1(

tas)J d

2+β(s)(

ta

) d2−1sβ

ds.

Using Lemma 3.8 with µ = d2− 1, we have

U[d]β,a,λ(t)− φβ,a(t)

= −2βΓ(β + 1)aβ

π

(at

) d−12ψβ(λ, a− t) +

(at

) d−12(

1 +a

t

)O(λ−β−1). (4.14)

Since ψβ(λ, a− t) = |a− t|−1O(λ−β−1) by Lemma 3.8 (i),

U[d]β,a,λ(t)− φβ,a(t) = O(λ

−β−1)

uniformly on any closed interval in (0, a) ∪ (a,∞). This shows the conclusion.

4.5 Proof of Theorem 4.1 (iv)

Let β > 0. By (4.13) we have

U[d]β,a,λ(a)− φβ,a(a) = U

[d]β,a,λ(a) = Lβ,aλ

−β +O(λ−β−1).


If 0 < t < a or t > a, then by (4.14) and Lemma 3.7 (ii) we have

U[d]β,a,λ(t)− φβ,a(t) =

(at

) d−12O(λ−β) +

(at

) d−12(

1 +a

t

)O(λ−β−1),

where the terms O(λ−β) and O(λ−β−1) are uniform with respect to t. Hence, U[d]β,a,λ(t)

converges to φβ,a(t) uniformly on any closed interval in (0,∞). That is, σλ(Uβ,a)(x)converges to Uβ,a(x) uniformly on any compact subset of Ẽa ∪ G̃a = Rd \ {0},and consequently, it does not reveal any phenomenon like the Gibbs-Wilbraham

phenomenon.

4.6 Proof of Theorem 4.1 (v)

Let −1 < β ≤ 0. The first assertion is already shown by (iii). Next we show theGibbs-Wilbraham phenomenon. Here, we recall that

G±β,a = ∓Γ(β + 1)aβ(2± β)β

π2β

∫ ∞π

sin s

(s± β2π)β+1

ds.

By (4.14) we have


+λ ) = U

[d]β,a,λ(|x

+λ |)− φβ,a(|x

+λ |)

= −2βΓ(β + 1)aβ

π

(a

|x+λ |

) d−12

ψβ(λ, a− |x+λ |) +(

a

|x+λ |

) d−12(

1 +a

|x+λ |

)O(λ−β−1).

Since a− |x+λ | =2+β4λ

, using Lemma 3.7 (iii), we have

−ψβ(λ, a− |x+λ |) =(

2 + β

4λ

)β ∫ ∞π

− sin s(s+ β

2π)β+1

ds.

Observing (a

|x+λ |

) d−12

=

(1− 2 + β

4λa

)− d−12

= 1 +O(λ−1),

we conclude that


+λ )

λ−β

= −2βΓ(β + 1)aβ

π

(2 + β

4

)β ∫ ∞π

sin s

(s+ β2π)β+1

ds(1 +O(λ−1)

)+O(λ−1)

→ G+β,a as λ→∞.


In a similar way we also have

σλ(Uβ,a)(x−λ )− Uβ,a(x

−λ )

λ−β→ G−β,a as λ→∞.

5 Results related to lattice point problems

The terms ∆j(s : x), j = 0, 1, · · · , are closely related to lattice point problems whichhave been studied by Landau, Jarńık, Szegö, Novák and others, see [5, 13, 15, 16,

26, 27, 46]. Recall that

∆α(s : x) = Dα(s : x)−Dα(s : x), α > −1, s ≥ 0, x ∈ Rd,

where

Dα(s : x) =

1

Γ(α + 1)

∑|m|2 0,

0, s = 0,

x ∈ Rd,

Dα(s : x) =

1

Γ(α + 1)

∫|ξ|2 0,

0, s = 0,

x ∈ Rd.

In this section we consider the behavior of ∆α(s : x) as s→∞. In Stein [39], theestimation of ∆ d−1

2(s : x) was treated. We shall prove the following four lemmas: In

the following f(s) = Ω(g(s)) means f(s) 6= o(g(s)).

Lemma 5.1. Let d ≥ 1. Then, as s→∞,

∆α(s : x) =

O(s

d2− dd+1 ), if α = 0,

O(sd2− dd+1

+ αd+1

+ε) for every ε > 0, if 0 < α ≤ d−12,

O(sd−14

+α2 ), if α > d−1

2,

(5.1)

uniformly with respected to x ∈ Td.

Lemma 5.2. Let d ≥ 5. If 0 ≤ α < (d− 4)/2, then, as s→∞,

∆α(s : x) =

{O(s

d2−1), Ω(s

d2−1) for x ∈ Td ∩Qd,

o(sd2−1) for x ∈ Td \Qd.

(5.2)

If (d− 4)/2 ≤ α ≤ (d− 1)/2, then, for every ε > 0, as s→∞,

∆α(s : x) =

{O(s

d−1+α3

+ε) for x ∈ Td ∩Qd,o(s

d−1+α3

+ε) for x ∈ Td \Qd.(5.3)


The following lemma gives more precise information on the estimate (5.2) for

x ∈ Td ∩Qd.

Lemma 5.3. Let d ≥ 5, α = 0 and β > −1. For k ∈ N = {1, 2, · · · }, let

`k = `k(d, β) =1

2a

(k +

d+ 2β + 1

4− 1

4

), (5.4)

mk = mk(d, β) =1

2a

(k +

d+ 2β + 1

4+

1

4

). (5.5)

Then, for large k ∈ N, there exists sk ∈ [`2k,m2k] such that

|∆0(sk : x)| ≥ C(x)sd2−1

k , x ∈ Td ∩Qd, (5.6)

where C(x) is a positive constant dependent on x, but independent of sk for large k.

Lemma 5.4. Let d ≥ 4 and 0 ≤ α ≤ (d− 1)/2. Then, for every ε > 0, as s→∞,

∆α(s : x) = O(sd4+ d−2

2(d−1)α+ε) for a.e.x ∈ Td. (5.7)

To prove above four lemmas we state known results (Theorems 5.5–5.9): See also

Jarńık [14] for Theorem 5.5. For α ≥ 0, let

Pα(s : x) = Dα(s : x)−πd2 s

d2+α

Γ(d2

+ α + 1)δ(x), x ∈ Rn, s ≥ 0, (5.8)

where δ(x) is the indicator function of Zd.

Theorem 5.5 (Landau [25, 27]). Let d ≥ 2. Then, for x ∈ Rd,

Pα(s : x) =

O(s

d2+α− d

d+1−2α ), if 0 ≤ α < d−12,

O(sd−12 log s), if α = d−1

2,

O(sd−14

+α2 ), if α > d−1

2.

(5.9)

Theorem 5.6 (Novák [32]). Let d ≥ 5, and let 0 ≤ α < (d− 4)/2. Then

Pα(s : x) =

O(s

d2−1), Ω(s

d2−1) for x ∈ Qd,

o(sd2−1) for x /∈ Qd,

O(sd4+α

2 logτ s) for a.e.x ∈ Rd,(5.10)

where τ = 3d if α = 0 and τ = 3d− 1 if α > 0.


Theorem 5.7 (Novák [30]). Let d ≥ 3. Then, for all x ∈ Qd, there exists a positiveconstant Kd(x) such that

∫ s0

|P0(t : x)|2 dt =

Kd(x)s

2 log s+O(s2 log1/2 s), if d = 3,

Kd(x)s3 +O(s5/2 log s), if d = 4,

Kd(x)s4 +O(s3 log2 s), if d = 5,

Kd(x)sd−1 +O(sd−2), if d ≥ 6.

(5.11)

Remark 5.1. In Theorem 5.7 the positive constant Kd(x) is given explicitly for

each x ∈ Qd, see [22]. For example, if d ≥ 4 and x = 0, then

Kd(0) =πd(2d + 8)ζ(d− 2)

12(d− 1)(2d − 1)ζ(d)Γ2(d/2),

where ζ is the Riemann’s zeta function, see [46].

Theorem 5.8 (Kuratsubo [17]). Let d ≥ 2 . Then, for every τ > 3/2,

P0(s : x) = O(sd4 logτ s) for a.e.x.

Remark 5.2. Theorems 5.5–5.8 valid for ∆α(s : x) instead of Pα(s : x), if x ∈ Td.Actually,

∆α(s : x)− Pα(s : x) =

{0, if x = 0,

−Dα(s : x) = O(sd−14

+α2 ), if x ∈ Td \ {0},

since (see (2.19) and (2.17))

Dα(s, x) =

πd2 s

d2+α

Γ(d2

+ α + 1), x = 0,

sd2+α

πα

J d2+α(2π

√s|x|)

(√s|x|) d2+α

= O(sd−14

+α2 ), x 6= 0.

Therefore,

|∆α(s : x)| = |Pα(s : x)|+O(sd−14

+α2 ), if x ∈ Td.

Remark 5.3. Let d = 1. Then

∆α(s : x) =

{O(1), if α = 0,

O(sα2 ), if α > 0,

(5.12)


uniformly with respect to x ∈ T. Actually, for all s > 0, choosing N ∈ N such thatN <

√s ≤ N + 1, we have by an elementary calculation

∆0(s : x) =sin(2π(N + 1

2)x)

sin πx− sin(2π

√sx)

πx= O(1).

For α > 0, from (2.19) and (2.20) it follows that

∆α(s : x) =s

12+α

πα

∑m∈Z, m6=0

J 12+α(2π

√s|x−m|)

(√s|x−m|) 12+α

.

By (2.17) we have

|J 12+α(2π

√s|x−m|)|

(√s|x−m|) 12+α

≤ C(√s|x−m|)1+α

,

for some positive constant C. Since |x −m| ≥ 1/2 for x ∈ T and m 6= 0, the sumconverges absolutely and ∆α(s : x) = O(s

α2 ).

The following is the Riesz’ convexity theorem. (see [40, page 285] and [3, page

13]).

Theorem 5.9. Let 0 ≤ α0 < α1 < ∞. For x ∈ Rd, let V0(s : x) and V1(s : x) betwo positive nondecreasing functions with respect to s > 0. Assume that

|∆αi(s : x)| ≤ Vi(s : x), i = 0, 1.

Then, for 0 ≤ θ ≤ 1,

|∆(1−θ)α0+θα1(s : x)| ≤ CV0(s : x)1−θV1(s : x)θ,

where C is a positive constant dependent on α1, α0 and θ, and independent of s and

x.

Proof of Lemma 5.1. The case d = 1 has been already proven in Remark 5.3. Then

we consider the case d ≥ 2. In general, for a function f : [0,∞)→ R, the differenceof f with h > 0 is defined by

δhf(s) = δ1hf(s) = f(s+ h)− f(s),


and

δk+1h f(s) = δkhf(s+ h)− δkhf(s), k = 1, 2, · · · .

Then

δkhf(s) =

∫ s+hs

ds1 · · ·∫ sk−1+hsk−1

f (k)(sk) dsk.

Let k = d] as in (2.10) and use this relation for f(s) = ∆k(s : x). Then, from

(2.18) and (2.21) we see that

δkh∆k(s : x)− hk∆0(s : x)

=

∫ s+hs

ds1 · · ·∫ sk−1+hsk−1

(∆0(sk : x)−∆0(s : x)) dsk

=

∫ s+hs

ds1 · · ·∫ sk−1+hsk−1

((D0(sk : x)−D0(s : x))− (D0(sk : x)−D0(s : x))

)dsk

=

∫ s+hs

ds1 · · ·∫ sk−1+hsk−1

∑s≤|m|2


where vd is the volume of the d-dimensional unit ball. Hence∣∣δkh∆k(s : x)− hk∆0(s : x)∣∣≤∫ s+hs

ds1 · · ·∫ sk−1+hsk−1

∣∣∣∣∣∣∑

s≤|m|2


Therefore, combining (5.13) and (5.14), we have

|∆0(s : x)| ≤ h−k(∣∣δkh∆k(s : x)− hk∆0(s : x)∣∣+ ∣∣δkh∆k(s : x)∣∣)

≤ C(sd2− dd+1 + hs

d2−1)

+ C( sh

) d−12, if kh ≤ s.

Then, setting h as hsd2−1 = (s/h)

d−12 , that is, h = s

1d+1 , we have that

|∆0(s : x)| ≤ Csd2− dd+1 , if s ≥ k

d+1d ,

where C is a positive constant independent of x ∈ Td and s ≥ k d+1d .If α > (d− 1)/2, then we have by (2.19) and (2.20) that

∆α(s : x) =sd2+α

πα

∑m∈Zd,m 6=0

J d2+α(2π

√s|x−m|)

(√s|x−m|) d2+α

.

Moreover, since |x−m| ≥ 1/2 for x ∈ Td and m 6= 0, the sum converges absolutelyand

|∆α(s : x)| ≤ Csd−14

+α2 ,

where C is a positive constant independent of x ∈ Td and s > 0. Therefore, we have

∆α(s : x) =

{O(s

d2− dd+1 ), if α = 0,

O(sd−14

+α2 ), if α > d−1

2,

uniformly with respect to x ∈ Td.Applying Theorem 5.9 as

α0 = 0, α1 =d− 1

2and α = θα1,

we have

∆α(s : x) = O(sd2− dd+1

+ αd+1

+ε) for every ε > 0, if 0 < α ≤ d− 12

,

since

(1− θ)(d

2− dd+ 1

)+ θ

(d− 1

4+α12

)=d

2− dd+ 1

+α

d+ 1.

Then the proof is complete.


Remark 5.4. We have the following comparison between Lemma 5.1 and Landau’s

estimate (5.9):

d

2− dd+ 1

+α

d+ 1<d

2+ α− d

d+ 1− 2α, if 0 < α <

d− 12

.

Proof of Lemma 5.2. If 0 ≤ α < (d−4)/2, then, by Theorem 5.6 and Remark 5.2 wehave (5.2) immediately. Next we show (5.3). By Theorems 5.5, 5.6 and Remark 5.2

we have

∆α(s : x) =

{O(s

d2−1) for x ∈ Td ∩Qd,

o(sd2−1) for x ∈ Td \Qd,

if α <d− 4

2,

and

∆α(s : x) = O(sd−12 log s) for x ∈ Td, if α = d− 1

2.

Applying Theorem 5.9 as

α0 =d− 4

2, α1 =

d− 12

and α = (1− θ)α0 + θα1,

we have

∆α(s : x) = O(sd−1+α

3+ε) for every ε > 0, if

d− 42≤ α ≤ d− 1

2,

since

(1− θ)(d

2− 1)

+ θ

(d− 1

2

)=d− 1 + α

3.

Proof of Lemma 5.3. By Theorem 5.7 and Remark 5.2 we have∫ s0

|∆0(t : x)|2 dt = Kd(x)sd−1 +O(sd−2 logτ s),

where τ = 2 if d = 5, τ = 0 if d ≥ 6. Using

m2k − `2k =1

(2a)2

(k +

d+ 2β + 1

4

),

m2(d−1)k − `

2(d−1)k =

d− 1(2a)2(d−1)

k2d−3 +O(k2d−4),

we have

1

m2k − `2k

∫ m2k`2k

|∆0(t : x)|2 dt =1

m2k − `2k

(∫ m2k0

|∆0(t : x)|2 dt−∫ `2k0

|∆0(t : x)|2 dt

)

= Kd(x)m

2(d−1)k − `

2(d−1)k

m2k − `2k+O(k2(d−2)−1 logτ k)

=d− 1

(2a)2(d−2)Kd(x) k

2d−4 +O(k2(d−2)−1 logτ k).


Hence we can take K̃d(x) such that

1

m2k − `2k

∫ m2k`2k

|∆0(t : x)|2 dt ≥ K̃d(x)2m2d−4k for large k.

Therefore, for large k, there exists sk ∈ [`2k,m2k] such that

|∆0(sk : x)| ≥ K̃d(x)md−2k ≥ K̃d(x)sd2−1

k .

Proof of Lemma 5.4. By Theorems 5.5, 5.8 and Remark 5.2 we have

∆α(s, x) =

{O(s

d4 logτ s), if α = 0,

O(sd−12 log s), if α = d−1

2,

for a.e.x ∈ Td.


α0 = 0, α1 =d− 1

2and α = θα1,

we have

∆α(s, x) = O(sd4+ d−2

2(d−1)α+ε) for every ε > 0, if 0 < α <d− 1

2,

since

(1− θ)d4

+ θd− 1

2=d

4+

d− 22(d− 1)

α.

Remark 5.5. We have the following comparison between Lemma 5.1 and Lemma 5.4:

d

4+

d− 22(d− 1)

α <d

2− dd+ 1

+α

d+ 1, if d ≥ 4 and 0 ≤ α < d− 1

2.

6 Proof of the main theorems

In this section we prove the pointwise convergence of the Fourier series of the func-

tion uβ,a(x) described in the main theorems (Theorems 1.1–1.5). First we state a

generalized Hardy’s identity (Theorem 6.1) and three lemmas (Lemmas 6.4–6.6).

Next, using the generalized Hardy’s identity and three lemmas, we will prove the

main theorems in Subsection 6.2. The proofs of Theorem 6.1 and Lemmas 6.4–6.6

are in Subsections 6.3, 6.4, 6.5 and 6.6, respectively.


6.1 Generalized Hardy’s identity and three lemmas

Recall that uβ,a(x) is the periodization of Uβ,a(x) = φβ,a(|x|) with

φβ,a(t) =

{(a2 − t2)β, 0 ≤ t < a,0, t ≥ a.

That is,

uβ,a(x) =∑m∈Zd

Uβ,a(x+m) =∑

|x+m| 1/2 and then

supx∈Td

∑m∈Zd\{0}

1

|x−m| d+12 +d]

d. By Lemma 3.7 (i) we have that

ψβ(λ, |a− |x−m||) = |a− |x−m||−1O(λ−β−1) as λ→∞.

Letting

Ma(x) = maxm∈Zd\{0}, |x−m|6=a

|a− |x−m||−1,

we have ∑m∈Zd\{0}, |x−m|6=a

(a

|x−m|

) d+12

+d] ∣∣ψβ(λ, |a− |x−m||)∣∣≤ C

∑m∈Zd\{0}

1

|x−m| d+12 +d]

Ma(x)λ−β−1


Hence,

Gβ,a(λ, x) = Ma(x)O(λ−β−1) as λ→∞. (6.6)

Next, recall that

rd(a : x) =∑

m∈Zd, |x−m|=a

1, x ∈ Td,

and let

r̃d(a : x) =∑

m∈Zd\{0}, |x−m|=a

1, x ∈ Td. (6.7)

Then

r̃d(a : 0) = rd(a : 0) and r̃d(a : x) = rd(a : x) = 0 for x ∈ Ea. (6.8)

Theorem 6.1 (Generalized Hardy’s identity). Let d ≥ 1, β > −1 and a > 0. Then

Sλ(uβ,a)(x) = uβ,a(x) +(σλ(Uβ,a)(x)− Uβ,a(x)

)+

(Gβ,a(λ : x) + r̃d(a : x)

(Lβ,a + o(1)

)λ−β

)+Kβ,a(λ2 : x) +O(λ−β−1) as λ→∞ (6.9)

for all x ∈ Td.

If 0 < a < 1/2, then r̃d(a : x) = 0 and uβ,a(x) = Uβ,a(x) in x ∈ Td. Combiningthese and (6.6), we have the following corollary of Theorem 6.1.

Corollary 6.2. Let d ≥ 1, β > −1 and 0 < a < 1/2. Then

Sλ(uβ,a)(x) = σλ(Uβ,a)(x) +Kβ,a(λ2 : x) +O(λ−β−1) as λ→∞ (6.10)

for all x ∈ Td.

If d = 2, β = 0 and a > 0, then U0,a(x) is the indicator function of the ball

{x ∈ R2 : |x| < a}. In this case, from Theorem 4.1 (i) (a), (ii) and (iii) it followsthat

limλ→∞

(σλ(U0,a)(x)− U0,a(x)

)=

L0,a =1

2, |x| = a,

0, |x| 6= a,x ∈ R2.

Then we conclude that

limλ→∞

(σλ(U0,a)(x)− U0,a(x)

)+ r̃2(a : x)L0,a =

1

2r2(a : x) =

1

2

∑|x+m|=a

1, x ∈ T2.


By (6.6) and Lemma 6.4 bellow we also have

G0,a(λ : x) +K0,a(λ2 : x)→ 0 as λ→∞, x ∈ T2.

Since u0,a(x) =∑|x+m| 0.

Then

limλ→∞

Sλ(u0,a)(x) =∑

|x+m| 0. By (1.8) and (4.7) together with (2.1)

we have û0,a(0) = πa2 and û0,a(m) = aJ1(2πa|m|)/|m| for m 6= 0. Then

Sλ(u0,a)(x) = πa2 + a

∑0


and Sλ(u− 12,a)(x) behaves about as nicely on the torus as does σλ(U− 1

2,a)(x) on the

Euclidean space. Now, from (6.12) we know that the set of these discrete values of

λ is

{λ > 0 : sin(2πaλ) = 0}.

For example, if a = 3/(4π), then λ = (2π/3)n for n = 1, 2, . . . , which are correspon-

dent with Figure 7A (n = 6) and Figure 7F (n = 7) in [43]. For the coefficient of

sin(2πaλ), we can calculate by Lemma 5.1 as

∆0(λ2 : x)

λ= O(λ−

13 ).

In the rest of this subsection we state three lemmas. Recall that

c(d) = d− 2dd+ 1

− d+ 12

=d− 5

2+

2

d+ 1=d(d− 4)− 1

2(d+ 1)=d− 3

2− d− 1d+ 1

.

Lemma 6.4. Let d ≥ 1, β > −1 and a > 0. Then

Kβ,a(λ2 : x) = O(λc(d)−β) uniformly on Td.

Consequently, if 1 ≤ d ≤ 4 and β > c(d), then

limλ→∞Kβ,a(λ2 : x) = 0 uniformly on Td,

and, if d ≥ 2, then

limλ→∞

|Kβ,a(λ2 : x)|λd−32−β

= 0 uniformly on Td.

Lemma 6.5. Let d ≥ 5, β > −1 and a > 0. Then

limλ→∞

|Kβ,a(λ2 : x)|λd−52−β

= 0, if x ∈ Td \Qd,

0 < lim supλ→∞

|Kβ,a(λ2 : x)|λd−52−β

−1/2 and a > 0. Then

limλ→∞Kβ,a(λ2 : x) = 0, a.e.x ∈ Td.

6.2 Proof of Theorems 1.1–1.5

In this section, using Theorem 6.1 and Lemmas 6.4–6.6, we prove the main theorems.


6.2.1 Proof of Theorem 1.1

For all β > −1, from (6.6) and Lemma 6.4 it follows that

Gβ,a(λ : 0) = O(λ−β−1) and Kβ,a(λ2 : 0) = O(λ

d−32−β),

respectively. Since r̃d(a, 0) = rd(a, 0) as in (6.8), by Theorem 6.1 we have

Sλ(uβ,a)(0) = uβ,a(0) +(σλ(Uβ,a)(0)− Uβ,a(0)

)+ rd(a : 0) (Lβ,a + o(1))λ

−β +O(λd−32−β). (6.13)

If β > (d− 3)/2, then, using Theorem 4.1 (i) (a), we have

σλ(Uβ,a)(0)− Uβ,a(0) = O(λd−32−β).

Combining this with (6.13), we have

Sλ(uβ,a)(0) = uβ,a(0) + rd(a : 0) (Lβ,a + o(1))λ−β +O(λ

d−32−β),

which shows (i).

If d ≥ 2 and −1 < β ≤ (d− 3)/2, then, using (4.4) in Theorem 4.1 (i), we have

σλ(Uβ,a)(0)− Uβ,a(0) = −P [d]β,a cos(

2πaλ− d− 1 + 2β4

π

)λd−32−β +O(λ

d−52−β),

which shows (ii).

6.2.2 Proof of Theorem 1.3

By Theorem 4.1 (ii) and (iii) we have

limλ→∞

σλ(Uβ,a)(x)− Uβ,a(x)λ−β

=

{Lβ,a, x ∈ G̃a,0, x ∈ Ẽa,

which shows

limλ→∞

σλ(Uβ,a)(x)− Uβ,a(x)λ−β

+ r̃d(a : x)Lβ,a = rd(a : x)Lβ,a, x ∈ Ea ∪Ga. (6.14)

Proof of (i) (a) Let 1 ≤ d ≤ 4. By (6.6) and Lemma 6.4 we have

Gβ,a(λ : x)

λ−β= O(λ−1),

Kβ,a(λ2 : x)λ−β

= O(λd−52

+ 2d+1 ) = o(1).


By Theorem 6.1 and (6.14) we have

Sλ(uβ,a)(x)− uβ,a(x)λ−β

=σλ(Uβ,a)(x)− Uβ,a(x)

λ−β+Gβ,a(λ : x)

λ−β+ r̃d(a : x)

(Lβ,a + o(1)

)+Kβ,a(λ2 : x)

λ−β+O(λ−1)

→ rd(a : x)Lβ,a as λ→∞,

which shows the conclusion.

Proof of (i) (b) Let β > c(d) = d−52

+ 2d+1

. Then by Lemma 6.4 we have

Kβ,a(λ2 : x) = O(λd−52

+ 2d+1−β) = o(1) uniformly on Td.

If x ∈ Ea, then r̃d(a : x) = 0 as (6.8). By Theorem 4.1 (iii) and (6.6) we have

limλ→∞

(σλ(Uβ,a)(x)− Uβ,a(x)) = 0, limλ→∞

Gβ,a(λ : x) = 0

uniformly on any compact set in Ea. Hence, by Theorem 6.1 we have

Sλ(uβ,a)(x)− uβ,a(x)

= (σλ(Uβ,a)(x)− Uβ,a(x)) +Gβ,a(λ : x) +Kβ,a(λ2 : x) +O(λ−β−1)

→ 0 as λ→∞

uniformly on any compact set in Ea, which shows the conclusion.

Proof of (ii) Let d ≥ 5 and x ∈ Ea ∪Ga. By Theorem 6.1 we have

1

λd−52

(Sλ(uβ,a)(x)− uβ,a(x)

λ−β− rd(a : x)Lβ,a

)=

1

λd−52

(σλ(Uβ,a)(x)− Uβ,a(x)

λ−β+ r̃d(a : x)

(Lβ,a + o(1)

)− rd(a : x)Lβ,a

)+Gβ,a(λ : x)

λd−52−β

+Kβ,a(λ2 : x)λd−52−β

+O(λ−d−52−1).

By (6.14), (6.6) and Lemma 6.5 we have

limλ→∞

1

λd−52

(Sλ(uβ,a)(x)− uβ,a(x)

λ−β− rd(a : x)Lβ,a

)= lim

λ→∞

Kβ,a(λ2 : x)λd−52−β

= 0


for all x ∈ (Ea ∪Ga) \Qd, and

0 < lim supλ→∞

1

λd−52

∣∣∣∣Sλ(uβ,a)(x)− uβ,a(x)λ−β − rd(a : x)Lβ,a∣∣∣∣

= lim supλ→∞

|Kβ,a(λ2 : x)|λd−52−β

0.

Hence, by (6.6) we haveGβ,a(λ : x

±λ )

λ−β= O(λ−1).

Since x±λ ∈ Ea, r̃d(a : x±λ ) = 0 as in (6.8). By Theorem 6.1 and Theorem 4.1 (v) we

have

Sλ(uβ,a)(x±λ )− uβ,a(x

±λ )

λ−β

=

(σλ(Uβ,a)(x

±λ )− Uβ,a(x

±λ )

λ−β

)+Gβ,a(λ : x

±λ )

λ−β+Kβ,a(λ2 : x±λ )

λ−β+O(λ−1)

→ G±β,a as λ→∞,

which is the conclusion.

6.2.4 Proofs of Theorem 1.5

Let d ≥ 4, β > −1/2 and a > 0. Then by lemma 6.6 we have

limλ→∞Kβ,a(λ2 : x) = 0, a.e.x ∈ Td.

Let x ∈ Ea. Then r̃d(a : x) = 0 as in (6.8). By Theorem 4.1 (iii) and (6.6) we have

σλ(Uβ,a)(x)− Uβ,a(x) = O(λ−β−1) and Gβ,a(λ : x) = O(λ−β−1).


Then by Theorem 6.1 we have, for a.e.x ∈ Ea,

Sλ(uβ,a)(x)− uβ,a(x)

=(σλ(Uβ,a)(x)− Uβ,a(x)

)+Gβ,a(λ : x) +Kβ,a(λ2 : x) +O(λ−β−1)

→ 0 as λ→∞,

which shows the conclusion, since the measure of Td \ Ea is zero.

6.3 Proof of Theorem 6.1

By Corollary 2.2 and (6.2) we have that, for all x ∈ Td,

Sλ(uβ,a)(x) = σλ(Uβ,a)(x) +Kβ,a(λ2 : x)

+ (−1)d]+1∑

m∈Zd\{0}

∫ λ20

Dd](s : x−m)A(d]+1)

β,a (s) ds.

Using (2.19) and (2.13) we have

Dd](s : x−m)A(d]+1)

β,a (s)

=sd2+d]

πd]

J d2+d]

(2π√s|x−m|)

(√s|x−m|) d2+d]

(−1)d]+1 Γ(β + 1)πβ−(d]+1)

ad2+β+d]+1

J d2+β+d]+1

(2πa√s)

sd2+β+d]+1

= (−1)d]+12βΓ(β + 1)a2βJ d

2+d]

(2π√s|x−m|)J d

2+β+d]+1

(2πa√s)

( |x−m|a

)d2+d](2πa

√s)β

πa√s.

Then, letting u = 2πa√s, we have

(−1)d]+1∫ λ20

Dd](s : x−m)A(d]+1)

β,a (s) ds

= 2βΓ(β + 1)a2β∫ 2πaλ0

J d2+d]

( |x−m|au) J d

2+β+d]+1

(u)

( |x−m|a

)d2+d] uβ

du. (6.15)

If |x−m| 6= a, then Corollary 3.5 shows that

2βΓ(β + 1)a2β∫ ∞0

J d2+d]

( |x−m|au) J d

2+β+d]+1

(u)

( |x−m|a

)d2+d] uβ

du

= Uβ,a(x−m) =

{0, if a < |x−m|,(a2 − |x−m|2)β, if 0 < |x−m| < a.

(6.16)


If m 6= 0 and |x−m| 6= a, then Lemma 3.8 shows that

2βΓ(β + 1)a2β∫ ∞2πaλ

J d2+d]

( |x−m|au) J d

2+β+d]+1

(u)

( |x−m|a

)d2+d] uβ

du

=2βΓ(β + 1)aβ

π

(a

|x−m|

) d+12

+d]

ψβ(λ, a− |x−m|) +1

|x−m| d+12 +d]O(λ−β−1),

(6.17)

since |x − m| ≥ 1/2 for m 6= 0 and x ∈ Td. Combining (6.15), (6.16) and (6.17),and observing (6.3)–(6.5), we have

(−1)d]+1∑

m∈Zd\{0}, |x−m|6=a

∫ λ20

Dd](s : x−m)A(d]+1)

β,a (s) ds

=

∑m∈Zd\{0}, |x−m|6=a

Uβ,a(x−m)

−Gβ,a(λ : x) +O(λ−β−1).= uβ,a(x)− Uβ,a(x)−Gβ,a(λ : x) +O(λ−β−1).

On the other hand, if m 6= 0 and |x−m| = a, then, using (6.15) and Corollary 3.12,we have

(−1)d]+1∫ λ20

Dd](s : x−m)A(d]+1)

β,a (s) ds

= 2βΓ(β + 1)a2β∫ 2πaλ0

J d2+d]

(u) J d2+β+d]+1

(u)

uβdu

= Lβ,aλ−β +

{O(λ−β−1), β ≥ 0,O(1), −1 < β < 0

= (Lβ,a + o(1))λ−β.

Therefore, we have the conclusion.

6.4 Proof of Lemma 6.4

To estimate

Kβ,a(λ2 : x) =d]∑j=0

(−1)j∆j(λ2 : x)A(j)β,a(λ2),


we combine the estimates of ∆j(λ2 : x) and A

(j)β,a(λ

2). Firstly, by (2.17) we see that

A(j)β,a(s) = (−1)

jΓ(β + 1)

πβ−jad2+β+j

J d2+β+j(2πa

√s)

s12( d2+β+j)

.

= (−1)jΓ(β + 1)πβ−j+1

ad2+β+j− 1

2

s12( d2+β+j+ 1

2)

cos

(2πa√s− d+ 2β + 2j + 1

4π

)+O(s−

12( d2+β+j+ 3

2)) as s→∞. (6.18)

That is, for some positive constant C,

|A(j)β,a(λ2)| ≤ Cλ−(

d2+β+j+ 1

2), λ ≥ 1, (6.19)

For the terms ∆j(s : x), we use Lemma 5.1, that is,

∆j(λ2 : x) =

O(λd−

2dd+1 ), if j = 0,

O(λd−2dd+1

+ 2jd+1

+ε) for every ε > 0, if 0 < j ≤ d−12,

O(λd−12

+j), if j > d−12.

If j = 0, then

|∆0(λ2 : x)Aβ,a(λ2)| ≤ Cλd−2dd+1λ−(

d2+β+ 1

2) = Cλ

d−52

+ 2(d+1)

−β. (6.20)

If 0 < j < d], then 0 < j ≤ (d− 1)/2 and then, for any small ε > 0,

|∆j(λ2 : x)A(j)β,a(λ2)| ≤ Cλd−

2dd+1

+ 2jd+1

+ελ−(d2+β+j+ 1

2) = Cλ

d−52

+2−j(d−1)

(d+1)−β+ε. (6.21)

If j = d], then j > (d− 1)/2 and then

|∆d](λ2 : x)A(d])

β,a (λ2)| ≤ Cλ

d−12

+d]λ−(d2+β+d]+

12) = Cλ−1−β. (6.22)

Comparing (6.20), (6.21) and (6.22), we have the conclusion.


By Lemma 5.2 we have that, as λ→∞,

∆j(λ2 : x) =

O(λd−2) for x ∈ Td ∩Qd and if 0 ≤ j < (d− 4)/2,o(λd−2) for x ∈ Td \Qd and if 0 ≤ j < (d− 4)/2,O(λ

2(d−1+j)3

+ε) for x ∈ Td and if (d− 4)/2 ≤ j ≤ (d− 1)/2.


Combining this and (6.19), we have the following estimates: If 0 ≤ j < (d − 4)/2and x ∈ Td \ Qd, then, for some decreasing function ϕ(λ) which satisfies ϕ(λ) → 0as λ→∞,

|∆j(λ2 : x)A(j)β,a(λ2)| ≤ Cλd−2ϕ(λ)λ−(

d2+β+j+ 1

2) = Cλ

d−52−β−jϕ(λ).

If 0 ≤ j < (d− 4)/2 and x ∈ Td ∩Qd, then

|∆j(λ2 : x)A(j)β,a(λ2)| ≤ Cλd−2λ−(

d2+β+j+ 1

2) = Cλ

d−52−β−j.

If (d− 4)/2 ≤ j ≤ (d− 1)/2, then, for small ε > 0,

|∆j(λ2 : x)A(j)β,a(λ2)| ≤ Cλ

2(d−1+j)3

+ελ−(d2+β+j+ 1

2) = Cλ

d−76−β− j

3+ε ≤ Cλ−

12−β+ε.

If j = d], then we have (6.22). Comparing these estimates, we have

|Kβ,a(λ2 : x)| ≤

{Cλ

d−52−βϕ(λ) for x ∈ Td \Qd,

Cλd−52−β for x ∈ Td ∩Qd,

which shows

limλ→∞

|Kβ,a(λ2 : x)|λd−52−β

= 0, if x ∈ Td \Qd,

lim supλ→∞

|Kβ,a(λ2 : x)|λd−52−β


If λ ∈ [`k,mk], then

kπ − 14π ≤ 2πaλ− d+ 2β + 1

4π ≤ kπ + 1

4π,

that is, ∣∣∣∣cos(2πaλ− d+ 2β + 14 π)∣∣∣∣ ≥ 1√2 .

Therefore, if x ∈ Td ∩Qd, then, for λk ∈ [`k,mk],

|∆0(λ2k : x)Aβ,a(λ2k)| ≥ Cβ,aC(x)λd−2k λ−( d

2+β+ 1

2)

k = Cβ,aC(x)λd−52−β

k ,

that is,

0 < lim supλ→∞

|∆0(λ2 : x)Aβ,a(λ2)|λd−52−β

.

The proof is complete.


If j = d], then we have (6.22). If 0 ≤ j ≤ (d − 1)/2, then, using Lemma 5.4 and(6.19), we have, for small ε > 0,

|∆j(λ2 : x)A(j)β,a(λ2)| ≤ Cλ

d2+ d−2d−1 j+ελ−(

d2+β+j+ 1

2) = Cλ−

12−β− j

d−1+ε ≤ Cλ−12−β+ε,

λ ≥ 1, a.e.x ∈ Td.

This shows the conclusion.

7 A relation between multiple Fourier series and

lattice point problems

Let d ≥ 1, β > −1 and 0 < a < 1/2. Then Corollary 6.2 has shown that

Sλ(uβ,a)(x) = σλ(Uβ,a)(x) +Kβ,a(λ2 : x) +O(λ−β−1) as λ→∞

for all x ∈ Td. The behavior of σλ(Uβ,a)(x) has been clarified by Theorem 4.1.Therefore, to clarify the behavior of Sλ(uβ,a)(x) we need to investigate the term

Kβ,a(λ2 : x) =d]∑j=0

(−1)j∆j(λ2 : x)A(j)β,a(λ2).


Since the estimates for the terms A(j)β,a(s) are gotten as (6.19), the convergence prob-

lem depends only on the terms ∆j(s : x), which are connected with the lattice point

problem. Especially, the estimate for ∆0(s : x) is very important and difficult.

First we state the following theorem, which will be proven later:

Theorem 7.1. Let d = 2 or 3, 0 < a < 1/2 and x0 ∈ Td. Then,

Sλ(uβ,a)(x0)− σλ(Uβ,a)(x0) = o(1) as λ→∞ for all β > −1,

if and only if

∆0(s : x0) = O(sd−14

+ε) as s→∞ for all ε > 0. (7.1)

For d = 1 the convergence property of Sλ(uβ,a)(x) and σλ(Uβ,a)(x) are well known

and the estimate (7.1) holds. Actually, we have ∆0(s : x) = O(1) for all x ∈ T, seeRemark 5.3. However, for d = 2 and d = 3, the estimate (7.1) is an open problem,

see Remarks 7.1 and 7.2, respectively. For d ≥ 4, see Remarks 7.3 and 7.4.

Remark 7.1. If d = 2, then the following fact is well known (see [30, Hauptsatz 3]

and Remark 5.2):

C2(x) t14 <

(1

t

∫ t0

|∆0(s : x)|2 ds) 1

2

< D2(x) t14 for all x ∈ T2, (7.2)

where C2(x) and D2(x) are positive constants depending on x ∈ T2. Therefore it isnatural to conjecture that ∆0(s : x) = O(s

14+ε) for all x ∈ T2. However,

“∆0(s : 0) = O(s14+ε) for all ε > 0”

is an open problem as famous Hardy’s conjecture on Gauss’s circle problem, since

∆0(s : 0) = D0(s : 0)−D0(s : 0) =∑|m|2


Remark 7.2. If d = 3 and for all x ∈ T3 then (see [30, Hauptsatz 3] and Remark 5.2)

C3(x) t12 <

(1

t

∫ t0

|∆0(s : x)|2 ds) 1

2

< D3(x) t12 log

12 t for all x ∈ T3, (7.3)

where C3(x) and D3(x) are positive constants depending on x ∈ T3. Therefore it isnatural to conjecture that ∆0(s : x) = O(s

12+ε) for all x ∈ T3. However, this problem

is also very hard. Up to now the best result on this problem is ∆0(s : 0) = O(s21/32+ε)

by D. R. Heath-Brown [11] in 1999.

For d ≥ 4 we have the following theorem. The proof will be given later:

Theorem 7.2. Let d ≥ 4, β > −1 and 0 < a < 1/2. For x0 ∈ Td, if (7.1) holds,then

Sλ(uβ,a)(x0)− σλ(Uβ,a)(x0) = o(1) as λ→∞.

Remark 7.3. If d ≥ 4 and x ∈ Td∩Qd, then (see [30, Hauptsatz 1] and Remark 5.2)(1

t

∫ t0

|∆0(s : x)|2 ds) 1

2

= Md(x) td2−1 +O(t

d2− 3

2+ε),

where Md(x) is a positive constant depending on x. Therefore, the estimate (7.1)

fails at x ∈ Td ∩Qd for d ≥ 4.

By Propositions 7.1 and 7.2 we have the following corollary immediately:

Corollary 7.3. Let d ≥ 2, β > −1 and 0 < a < 1/2. Assume that, for all ε > 0,

∆0(s : x) = O(sd−14

+ε) as s→∞ for a.e. x ∈ Td. (7.4)

Then

Sλ(uβ,a)(x)− σλ(Uβ,a)(x) = o(1) as λ→∞ for a.e. x ∈ Td.

Consequently,

limλ→∞

Sλ(uβ,a)(x) = uβ,a(x) for a.e. x ∈ Td.

Remark 7.4. If d = 2 or d = 3, then (7.4) is conjectured naturally by (7.2) and

(7.3). If d ≥ 4, then the following is known (see [31, Theorem 5] and Remark 5.2)(1

t

∫ t0

|∆0(s : x)|2 ds)1/2

= O(td−14 ) for a.e. x.

Therefore (7.4) is conjectured naturally for d ≥ 4 also.


Proof of Theorem 7.1. First observe that Theorem 6.1 implies that

Sλ(uβ,a)(x0) = σλ(Uβ,a)(x0) + o(1) as λ→∞,

if and only if

Kβ,a(λ2 : x0) =d]∑j=0

(−1)j∆j(λ2 : x0)A(j)β,a(λ2) = o(1). (7.5)

If j = d], then by (6.22) we have

∆d](λ2 : x0)A

(d])

β,a (λ2) = O(λ−1−β).

Note that d] = 1 if d = 2 and d] = 2 if d = 3. For the case d = 3 and j = 1, by

(6.21), we have, for small ε0 > 0,

∆1(λ2 : x0)A

(1)β,a(λ

2) = O(λ−1−β+ε0).

Hence

Kβ,a(λ2 : x0) =

{∆0(λ

2 : x0)Aβ,a(λ2) +O(λ−1−β), d = 2,

∆0(λ2 : x0)Aβ,a(λ

2) +O(λ−1−β+ε0), d = 3.

Since we can take ε0 > 0 as −1− β + ε0 < 0 in the case d = 3, we see that (7.5) isequivalent to

∆0(λ2 : x0)Aβ,a(λ

2) = o(1). (7.6)

(i) Assume that the estimate (7.1) holds. Then, for all β > −1, we can takeε > 0 as −1− β + ε < 0 and

|∆0(λ2 : x0)Aβ,a(λ2)| ≤ Cλd−12

+ελ−(d2+β+ 1

2) = Cλ−1−β+ε,

where we use (7.1) and (6.19). This shows (7.6).

(ii) Conversely, assume that the estimate (7.6) holds for all β > −1. Then by(6.18) we have that

∆0(λ2 : x)

λd2+β+ 1

2

cos

(2πaλ− d+ 2β + 1

4π

)= o(1)

for all β > −1. Now, for all ε > 0, take β(1) and β(2) such that −1 < β(1) <−1 + � = β(2) < β(1) + 1. Then

∆0(λ2 : x)

λd2+β(i)+ 1

2

cos

(2πaλ− d+ 2β(i) + 1

4π

)= o(1) (i = 1, 2). (7.7)


Let θ0 = (β(2)− β(1))π/4. Then 0 < θ0 < π/4 and

minθ∈R

(max {| cos θ|, | cos(θ − 2θ0)|}) = min−π≤θ≤π

(max {| sin θ|, | sin(θ − 2θ0)|}) ≥ sin θ0.

Therefore,

minλ>0

(max

{∣∣∣∣cos(2πaλ− d+ 2β(i) + 14 π)∣∣∣∣ : i = 1, 2}) ≥ sin θ0.

Combining this and (7.7), and observing that β(1) < β(2), we conclude that

∆0(λ2 : x0)

λd2+β(2)+ 1

2

=∆0(λ

2 : x0)

λd−12

+�= o(1).

This shows the conclusion.

Proof of Theorem 7.2. From Theorem 6.1 it is enough to prove that

Kβ,a(λ2 : x0) =d]∑j=0

(−1)j∆j(λ2 : x0)A(j)β,a(λ2) = o(1).

By the assumption and Lemma 5.1, for all ε > 0,

∆α(s : x0) =

{O(s

d−14

+ε), if α = 0,

O(sd−12

+ε), if α = d−12.


α0 = 0, α1 =d− 1

2and α = θα1,

we have

∆α(s : x0) = O(sd−14

+α2+ε), if 0 ≤ α ≤ d− 1

2,

since

(1− θ)d− 14

+ θd− 1

2=d− 1

4+α

2.

Take ε > 0 as −1− β + ε < 0. Then we have by (6.19) and (6.22)

|Kβ,a(λ2 : x0)| ≤d]−1∑j=0

|∆j(λ2 : x0)A(j)β,a(λ2)|+O(λ−1−β)

≤ Cd]−1∑j=0

λd−12

+j+ελ−(d2+β+j+ 1

2) +O(λ−1−β)

= O(λ−1−β+ε)

= o(1).

The proof is complete.


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Shigehiko KuratsuboDepartment of Mathematical SciencesHirosaki UniversityHirosaki 036-8561, JapanE-mail: [email protected]

Eiichi NakaiDepartment of Math

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Multiple Fourier series and lattice point problems · problem on the Fourier series of the function u 0;a. On the other hand, in the case 1