*J. Numer. Anal. Approx. Theory, vol. 49 (2020) no. 2, pp. 138–154*
**ictp.acad.ro/jnaat**

TRIGONOMETRIC APPROXIMATION ON THE HEXAGON

ALI GUVEN^{†}

*Dedicated to the memory of Dr. Figen Kiraz*

**Abstract.** The degree of trigonometric approximation of continuous functions,
which are periodic with respect to the hexagon lattice, is estimated in uniform
and H¨older norms. Approximating trigonometric polynomials are matrix means
of hexagonal Fourier series.

**2020 Mathematics Subject Classification.** 41A25, 41A63, 42B08.

**Keywords.** Hexagonal Fourier series, H¨older class, matrix mean, regular
hexagon.

1. INTRODUCTION

Approximation problems of functions of several variables defined on cubes
of the Euclidean space are usually studied by assuming that the functions
are periodic in each of their variables (see, for example [20, §§ 5.3, 6.3], [23,
vol. II, ch. XVII], [22, part 2], [15], [16] and [17]). But, we need other defini-
tions of periodicity to study approximation problems on non-tensor product
domains, for example on hexagonal domains of R^{2}. The periodicity defined
by lattices is the most useful one.

In the Euclidean plane R^{2}*,* besides the standard lattice Z^{2} and the rect-
angular domain [−^{1}_{2}*,*^{1}_{2})^{2}*,* the simplest lattice is the hexagon lattice and the
simplest spectral set is the regular hexagon. The hexagon lattice has impor-
tance, since it offers the densest packing of the plane with unit circles. In
this section we give basic information about hexagonal lattice and hexagonal
Fourier series. More detailed information can be found in [11] and [21].

The generator matrix and the spectral set of the hexagonal lattice*H*Z^{2} are
given by

*H* =

√3 0

−1 2

!

and

Ω*H* =^{n}(x1*, x*2)∈R^{2}:−1≤*x*2*,*

√3

2 *x*1±^{1}_{2}*x*2 *<*1^{o}*.*

†Department of Mathematics, Balikesir University, 10145 Balikesir, Turkey, e-mail:

It is more convenient to use the homogeneous coordinates (*t*1*, t*2*, t*3) that satis-
fies*t*_{1}+*t*_{2}+*t*_{3} = 0. As it is pointed out in [21], using homogeneous coordinates
reveals symmetry in various formulas. If we set

*t*1 :=−^{x}_{2}^{2} +^{√}^{3x}_{2} ^{1}*, t*2 :=*x*2*, t*3:=−^{x}_{2}^{2} −

√3x1

2 *,*
the hexagon Ω*H* becomes

Ω =^{n}(*t*_{1}*, t*_{2}*, t*_{3})∈R^{3} :−1≤*t*_{1}*, t*_{2}*,*−t_{3}*<*1*, t*_{1}+*t*_{2}+*t*_{3} = 0^{o}*,*

which is the intersection of the plane *t*_{1}+*t*_{2}+*t*_{3}= 0 with the cube [−1*,*1]^{3}*.*
We use bold letters**t** for homogeneous coordinates and we set

R^{3}* _{H}* =

^{n}

**t**= (

*t*

_{1}

*, t*

_{2}

*, t*

_{3})∈R

^{3}:

*t*

_{1}+

*t*

_{2}+

*t*

_{3}= 0

^{o}

*,*Z

^{3}

*:=Z*

_{H}^{3}∩R

^{3}

_{H}*.*A function

*f*:R

^{2}→Cis called

*H*-periodic if

*f*(*x*+*Hk*) =*f*(*x*)

for all *k*∈Z^{2} and *x*∈R^{2}*.*If we define **t**≡**s** (mod 3) as
*t*_{1}−*s*_{1} ≡*t*_{2}−*s*_{2}≡*t*_{3}−*s*_{3} (mod 3)

for **t** = (t1*, t*2*, t*3)*,* **s** = (s1*, s*2*, s*3)∈ R^{3}_{H}*,* it follows that the function *f* is
*H*-periodic if and only if *f*(**t**) =*f*(**t**+**s**) whenever **s**≡**0** (mod 3)*,*and

Z

Ω

*f*(**t**+**s**)*dt*=^{Z}

Ω

*f*(**t**)*dt* ^{}**s**∈R^{3}*H*

for*H*-periodic integrable function*f* [21]*.*

*L*^{2}(Ω) becomes a Hilbert space with respect to the inner product
hf, gi* _{H}* :=

_{|Ω|}

^{1}

^{Z}

Ω

*f*(**t**)*g*(**t**)*dt,*
where|Ω|denotes the area of Ω*.*The functions

*φ*** _{j}**(

**t**) :=

*e*

^{2πi}

^{3}

^{hj,ti}(

**t**∈R

^{3}

*H*)

*,*

where hj,**ti** is the usual Euclidean inner product of **j** and **t,** are *H*-periodic,
and by a theorem of B. Fuglede the set

n*φ*** _{j}**:

**j**∈Z

^{3}

*H*

o

becomes an orthonormal basis of*L*^{2}(Ω) [3] (see also [11])*.*
For every natural number *n,*we define a subset of Z^{3}* _{H}* by

H*n*:=^{n}**j**= (*j*_{1}*, j*_{2}*, j*_{3})∈Z^{3}*H* :−n≤*j*_{1}*, j*_{2}*, j*_{3} ≤*n*^{o}*.*
The subspace

H* _{n}*:= span

^{}

*φ*

**:**

_{j}**j**∈H

*n*(

*n*∈N)

has dimension #H*n* = 3*n*^{2}+ 3*n*+ 1*,* and its members are called hexagonal
trigonometric polynomials of degree *n*.

The hexagonal Fourier series of an*H*-periodic function *f* ∈*L*^{1}(Ω) is

(1) *f*(**t**)∼ ^{X}

**j∈**Z^{3}*H*

*f*b_{j}*φ*** _{j}**(

**t**)

*,*where

*f*b**j** = _{|Ω|}^{1} ^{Z}

Ω

*f*(**t**)*φ***j**(**t**)dt (**j**∈Z^{3}* _{H}*).

The *nth hexagonal partial sum of the series (1) is defined by*
*S** _{n}*(

*f*) (

**t**) :=

^{X}

**j∈**H*n*

*f*b_{j}*φ*** _{j}**(

**t**) (

*n*∈N)

*.*It is clear that

*S**n*(*f*) (**t**) = _{|Ω|}^{1} ^{Z}

Ω

*f*(**t**−**u**)*D**n*(**u**)*du,*
where*D**n* is the Dirichlet kernel, defined by

*D**n*(**t**) := ^{X}

**j∈**H*n*

*φ***j**(**t**)*.*

It is known that the Dirichlet kernel can be expressed as
(2) *D** _{n}*(

**t**) = Θ

*n*(

**t**)−Θ

*n−1*(

**t**) (

*n*≥1)

*,*where

(3) Θ*n*(**t**) := ^{sin}^{(n+1)(t}^{3}^{1}^{−t}^{2)}^{π}^{sin}^{(n+1)(t}^{3}^{2}^{−t}^{3)}^{π}^{sin}^{(n+1)(t}^{3}^{3}^{−t}^{1)}^{π}

sin^{(t}^{1}^{−t}_{3}^{2)}* ^{π}*sin

^{(t}

^{2}

^{−t}

_{3}

^{3)}

*sin*

^{π}^{(t}

^{3}

^{−t}

_{3}

^{1)}

^{π}for**t**= (*t*1*, t*2*, t*3)∈R^{3}*H* [11].

The degree of approximation of*H*-periodic continuous functions by Ces`aro,
Riesz and N¨orlund means of their hexagonal Fourier series was investigated by
us in [4], [5], [6], [7] and [8]. In this paper, we studied the degree of approxi-
mation by matrix means of hexagonal Fourier and we obtained generalizations
of previous results.

2. MAIN RESULTS

Let *C** _{H}*(Ω) be the Banach space of complex valued

*H*-periodic continuous functions defined on R

^{3}

_{H}*,*whose norm is the uniform norm:

kfk_{C}

*H*(^{Ω}) := sup

n|f(**t**)|:**t**∈Ω^{o}*.*

The modulus of continuity of the function*f* ∈*C**H*(Ω) is defined by
*ω** _{H}*(

*f, δ*) := sup

0<ktk≤δ

kf −*f*(·+**t**)k_{C}

*H*(Ω)*,*
where

ktk:= max{|t_{1}|*,*|t_{2}|*,*|t_{3}|}

for**t**= (*t*1*, t*2*, t*3)∈R^{3}_{H}*. ω** _{H}*(

*f,*·) is a nonnegative and nondecreasing function, and satisfies

(4) *ω**H*(*f, λδ*)≤(1 +*λ*)*ω**H*(*f, δ*)
for*λ >*0 [21]*.*

A function*f* ∈*C**H*(Ω) is said to belong to the H¨older space*H** ^{α}*(Ω) (0

*< α*≤1) if

Λ* ^{α}*(

*f*) := sup

**t6=s**

|f(t)−f(s)|

kt−sk^{α}*<*∞.

*H** ^{α}*(Ω) becomes a Banach space with respect to the H¨older norm
kfk

_{H}*(*

_{α}^{Ω}) :=kfk

_{C}*H*(^{Ω}) + Λ* ^{α}*(

*f*)

*.*

Let*A*= (*a** _{n,k}*) (

*n, k*= 0

*,*1

*, . . .*) be a lower triangular infinite matrix of real numbers. The

*A*-transform of the sequence (

*S*

*(*

_{n}*f*)) of partial sums the series (1) is defined by

*T*_{n}^{(A)}(*f*) (**t**) :=

*n*

X

*k=0*

*a*_{n,k}*S** _{k}*(

*f*) (

**t**) (

*n*∈N)

*.*

We shall assume that the lower triangular matrix *A* = (a*n,k*) satisfies the
conditions

(5) *a** _{n,k}* ≥0 (

*n*= 0

*,*1

*, . . . ,*0≤

*k*≤

*n*)

*,*(6)

*a*

*n,k+1*≥

*a*

*n,k*(

*n*= 0

*,*1

*, . . . ,*0≤

*k*≤

*n*−1)

*,*and

(7)

*n*

X

*k=0*

*a** _{n,k}* = 1 (

*n*= 0

*,*1

*, . . .*)

*.*Also we use the notations

*A*^{∗}* _{n,k}*:=

*n*

X

*ν=k*

*a** _{n,ν}* (0≤

*k*≤

*n*)

*, A*

^{∗}

*(*

_{n}*u*) :=

*A*

^{∗}

*,*

_{n,n−[u]}*a*

^{∗}

*(*

_{n}*u*) :=

*a*

*(*

_{n,n−[u]}*u >*0)

*,*where [

*u*] is the integer part of

*u.*

Hereafter, the relation *x* . *y* will mean that there exists an absolute con-
stant *c >*0 such that*x*≤*cy* holds for quantities*x* and *y*.

Theorem 1. *Let* *f* ∈*C**H*(Ω)*and let* *A*= (*a**n,k*) (*n, k*= 0*,*1*, . . .*) *be a lower*
*triangular infinite matrix of real numbers which satisfies*(5)*,*(6)*and*(7)*. Then*
*the estimate*

(8) ^{}^{}_{}*f*−*T*_{n}^{(A)}(f)^{}^{}_{}

*C**H*(^{Ω}) .log (n+ 1)
(

*ω**H*(f, a*n,n*) +^{X}^{n}

*k=1*

*ω**H*(^{f,}^{1}*k*)

*k* *A*^{∗}* _{n,n−k}*
)

*holds.*

Corollary 2. *If* *f* ∈ *H** ^{α}*(Ω) (0

*< α*≤1)

*and*

*A*= (

*a*

*) (*

_{n,k}*n, k*= 0

*,*1

*, . . .*)

*as in*

*Theorem 1, then*

(9) ^{}^{}*f* −*T*_{n}^{(A)}(*f*)^{}^{}_{C}

*H*(^{Ω}) .log (*n*+ 1)
(

*a*^{α}* _{n,n}*+

*n*

X

*k=1*
*A*^{∗}_{n,n−k}

*k*^{1+α}

)
*.*

Theorem 3. *Let* 0 ≤ *β < α* ≤ 1*,* *f* ∈ *H** ^{α}*(Ω)

*and let*

*A*= (

*a*

*n,k*) (

*n, k*= 0

*,*1

*, . . .*)

*be a lower triangular infinite matrix of real numbers which*

*satisfies*(5)

*,*(6)

*and*(7)

*. Then*

(10)

*f* −*T*_{n}^{(A)}(*f*)^{}^{}_{H}* _{β}*(

^{Ω}).log (

*n*+ 1)

*n*

X

*k=1*
*A*^{∗}_{n,n−k}

*k*

!_{α}* ^{β}*

*a*^{α−β}* _{n,n}* +

*n*

X

*k=1*
*A*^{∗}_{n,n−k}

*k*^{1+α}

!1−^{β}* _{α}*

*.*
Analogues of these results were obtained in [10], [13], [14] and [2] for matrix
means of trigonometric Fourier series of 2*π*-periodic continuous functions.

3. PROOFS OF MAIN RESULTS

*Proof of* *Theorem 1.* It is clear that

*f*(**t**)−*T*_{n}^{(A)}(*f*) (**t**)^{}^{} ≤ _{|Ω|}^{1}
Z

Ω

|f(**t**)−*f*(**t**−**u**)|

*n*

X

*k=0*

*a**n,k**D**k*(**u**)

*du*

. _{|Ω|}^{1}
Z

Ω

*ω** _{H}*(

*f,*kuk)

*n*

X

*k=0*

*a*_{n,k}*D** _{k}*(

**u**)

*du.*

If we set Θ−1(**u**) := 0*,*by (2) we get
Z

Ω

*ω**H*(f,kuk)

*n*

X

*k=0*

*a**n,k**D**k*(**u**)

*du*=

=^{Z}

Ω

*ω** _{H}*(

*f,*kuk)

*n*

X

*k=0*

*a*_{n,k}^{}Θ*k*(**u**)−Θ*k−1*(**u**)^{}

*du.*

The function

**t**→*ω**H*(f,ktk)

*n*

X

*k=0*

*a**n,k*(Θ*k*(**t**)−Θ*k−1*(**t**))

is symmetric with respect to variables*t*1*, t*2 and *t*3, where **t**= (*t*1*, t*2*, t*3)∈Ω*.*
Hence it is sufficient to estimate the integral over the triangle

∆ :=^{n}**t**= (*t*1*, t*2*, t*3)∈R^{3}*H* : 0≤*t*1*, t*2*,*−t_{3} ≤1^{o}

={(*t*1*, t*2) :*t*1≥0*, t*2 ≥0*, t*1+*t*2 ≤1}*,*

which is one of the six equilateral triangles in Ω*.*By considering the formula
(3), we obtain

Z

∆

*ω** _{H}*(

*f,*ktk)

*n*

X

*k=0*

*a** _{n,k}*(Θ

*k*(

**t**)−Θ

*k−1*(

**t**))

*dt*=

=^{Z}

∆

*ω** _{H}*(

*f, t*

_{1}+

*t*

_{2})

^{}

^{}

^{}

*n*

X

*k=0*

*a*_{n,k}

sin^{(k+1)(t}_{3}^{1}^{−t}^{2)}* ^{π}*sin

^{(k+1)(t}

_{3}

^{2}

^{−t}

^{3)}

*sin*

^{π}^{(k+1)(t}

_{3}

^{3}

^{−t}

^{1)}

*sin*

^{π}^{(t}

^{1}

^{−t}

_{3}

^{2)}

*sin*

^{π}^{(t}

^{2}

^{−t}

_{3}

^{3)}

*sin*

^{π}^{(t}

^{3}

^{−t}

_{3}

^{1)}

^{π}−^{sin}

*k(t*1−t2)*π*

3 sin^{k(t}^{2}^{−t}_{3} ^{3)}* ^{π}*sin

^{k(t}^{3}

^{−t}

_{3}

^{1)}

*sin*

^{π}^{(t}

^{1}

^{−t}

_{3}

^{2)}

*sin*

^{π}^{(t}

^{2}

^{−t}

_{3}

^{3)}

*sin*

^{π}^{(t}

^{3}

^{−t}

_{3}

^{1)}

^{π}

*dt.*

If we use the change of variables

(11) *s*_{1} := ^{t}^{1}^{−t}_{3} ^{3} = ^{2t}^{1}_{3}^{+t}^{2}*, s*_{2} := ^{t}^{2}^{−t}_{3} ^{3} = ^{t}^{1}^{+2t}_{3} ^{2}*,*
the integral becomes

3^{Z}

∆e

*ω**H*(f, s1+*s*2)^{}^{}^{}

*n*

X

*k=0*

*a**n,k*

sin((k+1)(s1−s2)π) sin((k+1)s2*π) sin((k+1)(−s*1*π))*
sin((s1−s2)π) sin(s2*π) sin(−s*1*π)*

− ^{sin(k(s}^{1}^{−s}^{2}^{)π) sin(ks}^{2}*π) sin(k(−s*_{1}*π))*
sin((s1−s_{2})π) sin(s2*π) sin(−s*_{1}*π)*

*ds*_{1}*ds*_{2}*,*
where∆ is the image of ∆ in the plane, that is^{e}

∆ :=e {(*s*_{1}*, s*_{2}) : 0≤*s*_{1} ≤2*s*_{2}*,* 0≤*s*_{2} ≤2*s*_{1}*, s*_{1}+*s*_{2} ≤1}*.*

Since the integrated function is symmetric with respect to*s*_{1}and*s*_{2}*,*estimating
the integral over the triangle

∆^{∗} :=^{n}(*s*1*, s*2)∈∆ :^{e} *s*1 ≤*s*2

o={(*s*1*, s*2) :*s*1 ≤*s*2 ≤2*s*1*, s*1+*s*2 ≤1}*,*
which is the half of∆^{e}*,*will be sufficient*.*The change of variables

(12) *s*_{1} := ^{u}^{1}^{−u}_{2} ^{2}*, s*_{2} := ^{u}^{1}^{+u}_{2} ^{2}
transforms the triangle ∆^{∗} to the triangle

Γ :=^{}(*u*1*, u*2) : 0≤*u*2 ≤ ^{u}_{3}^{1}*,* 0≤*u*1≤1 *.*
Thus we have to estimate the integral

*I**n*:=^{Z}

Γ

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a**n,k**D*_{k}^{∗}(*u*1*, u*2)

*du*1*du*2*,*
where

*D*^{∗}* _{k}*(

*u*

_{1}

*, u*

_{2}) = sin((k+1)(u2)π) sin (k+1)

^{u}^{1+}

_{2}

^{u}^{2}

*π*

sin (k+1) ^{u}^{1}^{−u}_{2} ^{2}*π*

sin((u2)π) sin ^{u}^{1+}_{2}^{u}^{2}*π*

sin ^{u}^{1}^{−u}_{2} ^{2}*π*

−^{sin(k(u}^{2}^{)π) sin} ^{k}

*u*1+*u*2
2 *π*

sin *k* ^{u}^{1}^{−u}_{2} ^{2}*π*

sin((u2)π) sin ^{u}^{1+}_{2}^{u}^{2}*π*

sin ^{u}^{1}^{−u}_{2} ^{2}*π* *.*

By elementary trigonometric identities, we obtain

(13) *D*^{∗}* _{k}*(

*u*1

*, u*2) =

*D*

^{∗}

*(*

_{k,1}*u*1

*, u*2) +

*D*

_{k,2}^{∗}(

*u*1

*, u*2) +

*D*

_{k,3}^{∗}(

*u*1

*, u*2)

*,*where

*D*^{∗}* _{k,1}*(

*u*

_{1}

*, u*

_{2}) := 2 cos

^{}(

*k*+

^{1}

_{2})

*u*

_{2}

*π*

^{}

^{sin}(

^{1}2

*u*2

*π*)

^{sin (k+1)}

^{u}^{1+}2

^{u}^{2}

*π*

sin (k+1)^{u}^{1}^{−u}_{2} ^{2}*π*

sin(u2*π) sin* ^{u}^{1+}_{2}^{u}^{2}*π*

sin ^{u}^{1}^{−u}_{2} ^{2}*π* *,*
*D*^{∗}* _{k,2}*(

*u*

_{1}

*, u*

_{2}) := 2 cos

^{}(

*k*+

^{1}

_{2})

^{u}^{1}

^{+u}

_{2}

^{2}

*π*

^{}

^{sin(ku}

^{2}

^{π) sin}1 2

*u*1+*u*2
2 *π*

sin (k+1)^{u}^{1}^{−u}_{2} ^{2}*π*

sin(u2*π) sin* ^{u}^{1+}_{2}^{u}^{2}*π*

sin ^{u}^{1}^{−u}_{2} ^{2}*π*
and

*D*_{k,3}^{∗} (*u*1*, u*2) := 2 cos^{}(*k*+^{1}_{2})^{u}^{1}^{−u}_{2} ^{2}*π*^{}^{sin(ku}^{2}^{π) sin}^{k}

*u*1+*u*2
2 *π*

sin ^{1}_{2}^{u}^{1}^{−u}_{2} ^{2}*π*

sin(u2*π) sin* ^{u}^{1+}_{2}^{u}^{2}*π*

sin ^{u}^{1}^{−u}_{2} ^{2}*π* *.*
We partition the triangle Γ as Γ = Γ1∪Γ2∪Γ3*,*where

Γ1 :={(*u*_{1}*, u*_{2})∈Γ :*u*_{1} ≤*a** _{n,n}*}

*,*

Γ2 :=^{}(u1*, u*2)∈Γ :*u*1 ≥*a**n,n**, u*2≤ ^{a}^{n,n}_{3} *,*
Γ3 :=^{}(*u*1*, u*2)∈Γ :*u*1 ≥*a**n,n**, u*2≥ ^{a}^{n,n}_{3} *.*
Hence *I**n*=*I**n,1*+*I**n,2*+*I**n,3**,*where

*I** _{n,j}* :=

^{Z}

Γ*j*

*ω** _{H}*(

*f, u*

_{1})

*n*

X

*k=0*

*a*_{n,k}*D*_{k}^{∗}(*u*_{1}*, u*_{2})

*du*_{1}*du*_{2} (*j*= 1*,*2*,*3)*.*
We need the inequalities

(14) ^{}^{}^{}

sin*nt*

sin*t*

≤*n,* (*n*∈N)*,*
and

(15) sin*t*≥ _{π}^{2}*t,* 0≤*t*≤ ^{π}_{2}^{}

to estimate integrals *I*_{n,1}*, I** _{n,2}* and

*I*

_{n,3}*.*

We divide Γ1 into three parts to estimate*I**n,1* as follows:

Γ^{0}1:=^{n}(*u*_{1}*, u*_{2})∈Γ1 :*u*_{1} ≤ _{n+1}^{1} ^{o}*,*

Γ^{00}1 :=^{n}(*u*_{1}*, u*_{2})∈Γ1 :*u*_{1} ≥ _{n+1}^{1} *, u*_{2} ≤ _{3(n+1)}^{1} ^{o}*,*
Γ^{000}_{1} :=^{n}(u1*, u*2)∈Γ1 :*u*1 ≥ _{n+1}^{1} *, u*2 ≥ _{3(n+1)}^{1} ^{o}*.*
Hence we have

*I**n,1* =

Z

Γ^{0}_{1}

+^{Z}

Γ^{00}_{1}

+^{Z}

Γ^{000}_{1}

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a**n,k**D*_{k}^{∗}(*u*1*, u*2)

*du*1*du*2*.*

By (14),

Z

Γ^{0}_{1}

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a**n,k**D*^{∗}* _{k}*(

*u*1

*, u*2)

*du*1*du*2 .

. Z

Γ^{0}_{1}

*ω** _{H}*(

*f, u*

_{1})

*n*

X

*k=0*

(*k*+ 1)^{2}*a*_{n,k}

!

*du*_{1}*du*_{2}

≤(n+ 1)^{2}

1 3(n+1)

Z

0

1
*n+1*

Z

3u2

*ω**H*(f, u1)*du*1*du*2

≤*ω*_{H}^{}*f,*_{n+1}^{1} ^{}≤*ω** _{H}*(

*f, a*

*)*

_{n,n}*.*By (15)

*,*

Z

Γ^{00}_{1}

*ω** _{H}*(

*f, u*

_{1})

*n*

X

*k=0*

*a*_{n,k}*D*^{∗}* _{k,1}*(

*u*

_{1}

*, u*

_{2})

*du*_{1}*du*_{2} .

.

1 3(n+1)

Z

0
*a**n,n*

Z

1
*n+1*

*ω**H*(f,u1)

*u*^{2}_{1} *du*_{1}*du*_{2} ≤*ω** _{H}*(

*f, a*

*)*

_{n,n}1 3(n+1)

Z

0
*a**n,n*

Z

1
*n+1*

1

*u*^{2}_{1}*du*_{1}*du*_{2} ≤*ω** _{H}*(

*f, a*

*)*

_{n,n}*.*(14) and (15) gives for

*j*= 1

*,*2

*,*

Z

Γ^{00}_{1}

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a*_{n,k}*D*_{k,j}^{∗} (*u*1*, u*2)

*du*1*du*2.

. (*n*+ 1)

1 3(n+1)

Z

0
*a**n,n*

Z

1
*n+1*

*ω**H*(f,u1)

*u*1 *du*_{1}*du*_{2}≤(*n*+ 1)*ω** _{H}*(

*f, a*

*)*

_{n,n}1 3(n+1)

Z

0
*a**n,n*

Z

1
*n+1*

1

*u*1*du*_{1}*du*_{2}

≤ log (*n*+ 1)*ω** _{H}*(

*f, a*

*n,n*)

*.*Since

sin 2*x*+ sin 2*y*+ sin 2*z*=−4 sin*x*sin*y*sin*z*
for*x*+*y*+*z*= 0*,*we also get the expression

(16) *D*^{∗}* _{k}*(

*u*

_{1}

*, u*

_{2}) =

*H*

*(*

_{k,1}*u*

_{1}

*, u*

_{2}) +

*H*

*(*

_{k,2}*u*

_{1}

*, u*

_{2}) +

*H*

*(*

_{k,3}*u*

_{1}

*, u*

_{2})

*,*where

*H** _{k,1}*(

*u*

_{1}

*, u*

_{2}) :=

^{1}

_{2}cos((2k+1)u2

*π)*sin

^{u}^{1+}

_{2}

^{u}^{2}

*π*

sin ^{u}^{1}^{−u}_{2} ^{2}*π**,*
*H** _{k,2}*(

*u*

_{1}

*, u*

_{2}) :=−

^{1}

_{2}

^{cos (2k+1)}

*u*1+*u*2
2 *π*

sin(u2*π) sin* ^{u}^{1}^{−u}_{2} ^{2}*π**,*

*H** _{k,3}*(

*u*

_{1}

*, u*

_{2}) :=

^{1}

_{2}

^{cos (2k+1)}

^{u}^{1}

−u2
2 *π*

sin(u2*π) sin* ^{u}^{1+}_{2}^{u}^{2}*π**.*
By the method used in [12, p. 179] we get

(17)

*n*

X

*k=0*

*a**n,k*cos (2k+ 1)*t*

.*A*^{∗}_{n}^{}^{1}_{t}^{}+_{sin}^{1}_{t}*a*^{∗}_{n}^{}^{1}_{t}^{} (0*< t < π),*
and

(18)

*n*

X

*k=0*

*a**n,k*cos (2*k*+ 1)*t*

.*A*^{∗}_{n}^{}^{1}_{t}^{} 0*< t*≤ ^{π}_{2}^{}*.*
By aim of (18) and (15) we obtain

(19)

*n*

X

*k=0*

*a*_{n,k}*H** _{k,1}*(

*u*

_{1}

*, u*

_{2})

. _{u}^{1}2
1

*A*^{∗}_{n}^{}_{πu}^{1}

2

and (20)

*n*

X

*k=0*

*a*_{n,k}*H** _{k,3}*(

*u*

_{1}

*, u*

_{2})

. _{u}_{1}^{1}_{u}_{2}*A*^{∗}_{n}^{}_{πu}^{3}

1

*,*

where both of for *u*1 and *u*2 are away from the origin. Also, for such *u*1 and
*u*_{2}*,*it follows from (17)*,*(15) and from the fact

sin ^{u}^{1}_{2}^{π}^{}.sin^{}^{(u}^{1}^{+u}_{2} ^{2}^{)π}^{}
that

(21)

*n*

X

*k=0*

*a*_{n,k}*H** _{k,2}*(

*u*

_{1}

*, u*

_{2})

. _{u}_{1}^{1}_{u}_{2}*A*^{∗}_{n}^{}_{πu}^{3}

1

*.*

Using (19) and the inequality
(22) *ω** _{H}*(

*f, δ*

_{2})

*δ*2

≤2*ω** _{H}*(

*f, δ*

_{1})

*δ*1 (*δ*_{1} *< δ*_{2})*,*

which is obtained from (4)*,* and considering that the function *A*^{∗}* _{n}* is nonde-
creasing yield

Z

Γ^{000}_{1}

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a**n,k**H**k,1*(*u*1*, u*2)

*du*1*du*2.

.

*an,n*
3

Z

1 3(n+1)

*a**n,n*

Z

3u2

*ω**H*(f,u1)

*u*^{2}_{1} *A*^{∗}_{n}^{}_{πu}^{1}

2

*du*1*du*2

≤2

*an,n*
3

Z

1 3(n+1)

*a**n,n*

Z

3u2

*ω**H*(f,3u2)

3u1*u*2 *A*^{∗}_{n}^{}_{πu}^{1}

2

*du*_{1}*du*_{2}

= ^{2}_{3}

*an,n*
3

Z

1 3(n+1)

*ω**H*(f,3u2)

*u*2 log^{}^{a}_{3u}^{n,n}_{2}^{}*A*^{∗}_{n}^{}_{πu}^{1}

2

*du*_{2}

≤log ((*n*+ 1)*a** _{n,n}*)

*an,n*

Z3

1 3(n+1)

*ω** _{H}*(f,3u2)

*u*2 *A*^{∗}_{n}^{}_{πu}^{1}

2

*du*_{2}

≤log (*n*+ 1)

*an,n*

Z3

1 3(n+1)

*ω** _{H}*(f,3u2)

*u*2 *A*^{∗}_{n}^{}_{πu}^{1}

2

*du*_{2} = log (*n*+ 1)

3
*π*(n+1)

Z

3
*π*

*ω**H*(f,_{πt}^{3})

*t* *A*^{∗}* _{n}*(

*t*)

*dt*

= log (*n*+ 1)

*n*

X

*k=1*

3
*π*(k+1)

Z

3
*π**k*

*ω**H*(^{f,}*πt*^{3})

*t* *A*^{∗}* _{n}*(

*t*)

*dt*

≤log (n+ 1)^{X}^{n}

*k=1*
*ω**H*(f,_{k}^{1})

*k* *A*^{∗}* _{n}*(k+ 1)

*.*For

*j*= 2

*,*3

*,*by (20) and (21)

Z

Γ^{000}_{1}

*ω** _{H}*(

*f, u*

_{1})

*n*

X

*k=0*

*a*_{n,k}*H** _{k,j}*(

*u*

_{1}

*, u*

_{2})

*du*_{1}*du*_{2} .

.

*a**n,n*

Z

1
*n+1*

*u*1

Z3

1 3(n+1)

*ω**H*(f,u1)

*u*1*u*2 *A*^{∗}_{n}^{}_{πu}^{3}

1

*du*2*du*1

=

*a**n,n*

Z

1
*n+1*

*ω**H*(f,u1)

*u*1 log ((n+ 1)*u*1)*A*^{∗}_{n}^{}_{πu}^{3}

1

*du*1

≤log ((*n*+ 1)*a** _{n,n}*)

*a**n,n*

Z

1
*n+1*

*ω**H*(f,u1)

*u*1 *A*^{∗}_{n}^{}_{πu}^{3}

1

*du*_{1} =

= log ((*n*+ 1)*a** _{n,n}*)

3
*π*(n+1)

Z

3
*π*

*ω**H*(^{f,}*πt*^{3})

*t* *A*^{∗}* _{n}*(

*t*)

*dt*

≤log (*n*+ 1)

*n*

X

*k=1*

*ω**H*(^{f,}*k*^{1})

*k* *A*^{∗}* _{n}*(

*k*+ 1)

*.*Thus we get the estimate

(23) *I** _{n,1}* .log (

*n*+ 1) (

*ω** _{H}*(

*f, a*

*) +*

_{n,n}*n*

X

*k=1*

*ω**H*(^{f,}*k*^{1})

*k* *A*^{∗}* _{n}*(

*k*+ 1) )

*.*
We divide Γ2 into two domains as

Γ^{0}_{2}:=^{n}(u1*, u*2)∈Γ2 :*u*2 ≤ _{3(n+1)}^{a}^{n,n}^{o}*,*
Γ^{00}2 :=^{n}(*u*1*, u*2)∈Γ2 :*u*2 ≥ _{3(n+1)}^{a}^{n,n}^{o}
to estimate *I**n,2**.*By (15) and (22)*,*

Z

Γ^{0}_{2}

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a*_{n,k}*D*_{k,1}^{∗} (*u*1*, u*2)

*du*1*du*2 .

.

*an,n*
3(n+1)

Z

0 1

Z

*a**n,n*

*ω**H*(f,u1)

*u*^{2}_{1} *du*1*du*2≤2^{ω}^{H}^{(f,a}_{a}_{n,n}^{n,n}^{)}

*an,n*
3(n+1)

Z

0 1

Z

*a**n,n*

1

*u*1*du*1*du*2

= 2^{ω}^{H}^{(f,a}_{a}_{n,n}^{n,n}^{)}log^{}_{a}_{n,n}^{1} ^{}_{3(n+1)}^{a}* ^{n,n}* ≤log (n+ 1)

*ω*

*H*(f, a

*n,n*)

*.*(14)

*,*(15) and (22) yield

Z

Γ^{0}_{2}

*ω** _{H}*(

*f, u*1)

*n*

X

*k=0*

*a*_{n,k}*D*^{∗}* _{k,j}*(

*u*1

*, u*2)

*du*1*du*2 .

. (*n*+ 1)

*an,n*
3(n+1)

Z

0 1

Z

*a**n,n*

*ω**H*(f,u1)

*u*1 *du*1*du*2≤2 (*n*+ 1)^{ω}^{H}^{(f,a}_{a}_{n,n}^{n,n}^{)}

*an,n*
3(n+1)

Z

0 1

Z

*a**n,n*

*du*1*du*2

≤ *ω**H*(f, a*n,n*)

for*j*= 2*,*3*.*Since (15) implies|H* _{k,1}*(

*u*

_{1}

*, u*

_{2})|.

_{u}^{1}2

1 for (*u*_{1}*, u*_{2})∈Γ^{00}_{2}*,*we get
Z

Γ^{00}_{2}

*ω**H*(*f, u*1)

*n*

X

*k=0*

*a**n,k**H**k,1*(*u*1*, u*2)

*du*1*du*2 .