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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 R2. The periodicity defined by lattices is the most useful one.

In the Euclidean plane R2, besides the standard lattice Z2 and the rect- angular domain [−12,12)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 latticeHZ2 are given by

H =

√3 0

−1 2

!

and

H =n(x1, x2)∈R2:−1≤x2,

3

2 x1±12x2 <1o.

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

(2)

It is more convenient to use the homogeneous coordinates (t1, t2, t3) that satis- fiest1+t2+t3 = 0. As it is pointed out in [21], using homogeneous coordinates reveals symmetry in various formulas. If we set

t1 :=−x22 +3x2 1, t2 :=x2, t3:=−x22

3x1

2 , the hexagon ΩH becomes

Ω =n(t1, t2, t3)∈R3 :−1≤t1, t2,−t3<1, t1+t2+t3 = 0o,

which is the intersection of the plane t1+t2+t3= 0 with the cube [−1,1]3. We use bold letterst for homogeneous coordinates and we set

R3H =nt= (t1, t2, t3)∈R3 :t1+t2+t3 = 0o, Z3H :=Z3∩R3H. A function f :R2 →Cis called H-periodic if

f(x+Hk) =f(x)

for all k∈Z2 and x∈R2.If we define ts (mod 3) as t1s1t2s2t3s3 (mod 3)

for t = (t1, t2, t3), s = (s1, s2, s3)∈ R3H, it follows that the function f is H-periodic if and only if f(t) =f(t+s) whenever s0 (mod 3),and

Z

f(t+s)dt=Z

f(t)dt s∈R3H

forH-periodic integrable functionf [21].

L2(Ω) becomes a Hilbert space with respect to the inner product hf, giH := |Ω|1 Z

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

φj(t) :=e2πi3 hj,ti (t∈R3H),

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

nφj:j∈Z3H

o

becomes an orthonormal basis ofL2(Ω) [3] (see also [11]). For every natural number n,we define a subset of Z3H by

Hn:=nj= (j1, j2, j3)∈Z3H :−n≤j1, j2, j3no. The subspace

Hn:= spanφj:j∈Hn (n∈N)

has dimension #Hn = 3n2+ 3n+ 1, and its members are called hexagonal trigonometric polynomials of degree n.

(3)

The hexagonal Fourier series of anH-periodic function fL1(Ω) is

(1) f(t)∼ X

j∈Z3H

fbjφj(t), where

fbj = |Ω|1 Z

f(t)φj(t)dt (j∈Z3H).

The nth hexagonal partial sum of the series (1) is defined by Sn(f) (t) := X

j∈Hn

fbjφj(t) (n∈N). It is clear that

Sn(f) (t) = |Ω|1 Z

f(tu)Dn(u)du, whereDn is the Dirichlet kernel, defined by

Dn(t) := X

j∈Hn

φj(t).

It is known that the Dirichlet kernel can be expressed as (2) Dn(t) = Θn(t)−Θn−1(t) (n≥1), where

(3) Θn(t) := sin(n+1)(t31−t2)πsin(n+1)(t32−t3)πsin(n+1)(t33−t1)π

sin(t1−t32)πsin(t2−t33)πsin(t3−t31)π

fort= (t1, t2, t3)∈R3H [11].

The degree of approximation ofH-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 CH(Ω) be the Banach space of complex valued H-periodic continuous functions defined on R3H,whose norm is the uniform norm:

kfkC

H() := sup

n|f(t)|:t∈Ωo.

The modulus of continuity of the functionfCH(Ω) is defined by ωH(f, δ) := sup

0<ktk≤δ

kf −f(·+t)kC

H(Ω), where

ktk:= max{|t1|,|t2|,|t3|}

(4)

fort= (t1, t2, t3)∈R3H. ωH(f,·) is a nonnegative and nondecreasing function, and satisfies

(4) ωH(f, λδ)≤(1 +λ)ωH(f, δ) forλ >0 [21].

A functionfCH(Ω) is said to belong to the H¨older spaceHα(Ω) (0< α≤1) if

Λα(f) := sup

t6=s

|f(t)−f(s)|

kt−skα <∞.

Hα(Ω) becomes a Banach space with respect to the H¨older norm kfkHα() :=kfkC

H() + Λα(f).

LetA= (an,k) (n, k= 0,1, . . .) be a lower triangular infinite matrix of real numbers. TheA-transform of the sequence (Sn(f)) of partial sums the series (1) is defined by

Tn(A)(f) (t) :=

n

X

k=0

an,kSk(f) (t) (n∈N).

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

(5) an,k ≥0 (n= 0,1, . . . ,0≤kn), (6) an,k+1an,k (n= 0,1, . . . ,0≤kn−1), and

(7)

n

X

k=0

an,k = 1 (n= 0,1, . . .). Also we use the notations

An,k:=

n

X

ν=k

an,ν (0≤kn), An(u) :=An,n−[u] ,an(u) :=an,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 thatxcy holds for quantitiesx and y.

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

(8) fTn(A)(f)

CH() .log (n+ 1) (

ωH(f, an,n) +Xn

k=1

ωH(f,1k)

k An,n−k )

holds.

(5)

Corollary 2. If fHα(Ω) (0< α≤1) and A = (an,k) (n, k= 0,1, . . .) as in Theorem 1, then

(9) fTn(A)(f)C

H() .log (n+ 1) (

aαn,n+

n

X

k=1 An,n−k

k1+α

) .

Theorem 3. Let 0 ≤ β < α ≤ 1, fHα(Ω) and let A = (an,k) (n, k= 0,1, . . .) be a lower triangular infinite matrix of real numbers which satisfies (5), (6) and (7). Then

(10)

fTn(A)(f)Hβ().log (n+ 1)

n

X

k=1 An,n−k

k

!αβ

aα−βn,n +

n

X

k=1 An,n−k

k1+α

!1−βα

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

3. PROOFS OF MAIN RESULTS

Proof of Theorem 1. It is clear that

f(t)−Tn(A)(f) (t)|Ω|1 Z

|f(t)−f(tu)|

n

X

k=0

an,kDk(u)

du

. |Ω|1 Z

ωH(f,kuk)

n

X

k=0

an,kDk(u)

du.

If we set Θ−1(u) := 0,by (2) we get Z

ωH(f,kuk)

n

X

k=0

an,kDk(u)

du=

=Z

ωH(f,kuk)

n

X

k=0

an,kΘk(u)−Θk−1(u)

du.

The function

tωH(f,ktk)

n

X

k=0

an,kk(t)−Θk−1(t))

is symmetric with respect to variablest1, t2 and t3, where t= (t1, t2, t3)∈Ω. Hence it is sufficient to estimate the integral over the triangle

∆ :=nt= (t1, t2, t3)∈R3H : 0≤t1, t2,−t3 ≤1o

={(t1, t2) :t1≥0, t2 ≥0, t1+t2 ≤1},

(6)

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

Z

ωH(f,ktk)

n

X

k=0

an,kk(t)−Θk−1(t))

dt=

=Z

ωH(f, t1+t2)

n

X

k=0

an,k

sin(k+1)(t31−t2)πsin(k+1)(t32−t3)πsin(k+1)(t33−t1)π sin(t1−t32)πsin(t2−t33)πsin(t3−t31)π

sin

k(t1−t2)π

3 sink(t2−t3 3)πsink(t3−t3 1)π sin(t1−t32)πsin(t2−t33)πsin(t3−t31)π

dt.

If we use the change of variables

(11) s1 := t1−t3 3 = 2t13+t2, s2 := t2−t3 3 = t1+2t3 2, the integral becomes

3Z

e

ωH(f, s1+s2)

n

X

k=0

an,k

sin((k+1)(s1−s2)π) sin((k+1)s2π) sin((k+1)(−s1π)) sin((s1−s2)π) sin(s2π) sin(−s1π)

sin(k(s1−s2)π) sin(ks2π) sin(k(−s1π)) sin((s1−s2)π) sin(s2π) sin(−s1π)

ds1ds2, where∆ is the image of ∆ in the plane, that ise

∆ :=e {(s1, s2) : 0≤s1 ≤2s2, 0≤s2 ≤2s1, s1+s2 ≤1}.

Since the integrated function is symmetric with respect tos1ands2,estimating the integral over the triangle

:=n(s1, s2)∈∆ :e s1s2

o={(s1, s2) :s1s2 ≤2s1, s1+s2 ≤1}, which is the half of∆e,will be sufficient.The change of variables

(12) s1 := u1−u2 2, s2 := u1+u2 2 transforms the triangle ∆ to the triangle

Γ :=(u1, u2) : 0≤u2u31, 0≤u1≤1 . Thus we have to estimate the integral

In:=Z

Γ

ωH(f, u1)

n

X

k=0

an,kDk(u1, u2)

du1du2, where

Dk(u1, u2) = sin((k+1)(u2)π) sin (k+1)u1+2u2π

sin (k+1) u1−u2 2π

sin((u2)π) sin u1+2u2π

sin u1−u2 2π

sin(k(u2)π) sin k

u1+u2 2 π

sin k u1−u2 2π

sin((u2)π) sin u1+2u2π

sin u1−u2 2π .

(7)

By elementary trigonometric identities, we obtain

(13) Dk(u1, u2) =Dk,1(u1, u2) +Dk,2 (u1, u2) +Dk,3 (u1, u2), where

Dk,1(u1, u2) := 2 cos(k+12)u2πsin(12u2π)sin (k+1)u1+2u2π

sin (k+1)u1−u2 2π

sin(u2π) sin u1+2u2π

sin u1−u2 2π , Dk,2(u1, u2) := 2 cos(k+12)u1+u2 2πsin(ku2π) sin

1 2

u1+u2 2 π

sin (k+1)u1−u2 2π

sin(u2π) sin u1+2u2π

sin u1−u2 2π and

Dk,3 (u1, u2) := 2 cos(k+12)u1−u2 2πsin(ku2π) sin k

u1+u2 2 π

sin 12u1−u2 2π

sin(u2π) sin u1+2u2π

sin u1−u2 2π . We partition the triangle Γ as Γ = Γ1∪Γ2∪Γ3,where

Γ1 :={(u1, u2)∈Γ :u1an,n},

Γ2 :=(u1, u2)∈Γ :u1an,n, u2an,n3 , Γ3 :=(u1, u2)∈Γ :u1an,n, u2an,n3 . Hence In=In,1+In,2+In,3,where

In,j :=Z

Γj

ωH(f, u1)

n

X

k=0

an,kDk(u1, u2)

du1du2 (j= 1,2,3). We need the inequalities

(14)

sinnt

sint

n, (n∈N), and

(15) sintπ2t, 0≤tπ2

to estimate integrals In,1, In,2 and In,3.

We divide Γ1 into three parts to estimateIn,1 as follows:

Γ01:=n(u1, u2)∈Γ1 :u1n+11 o,

Γ001 :=n(u1, u2)∈Γ1 :u1n+11 , u23(n+1)1 o, Γ0001 :=n(u1, u2)∈Γ1 :u1n+11 , u23(n+1)1 o. Hence we have

In,1 =

Z

Γ01

+Z

Γ001

+Z

Γ0001

ωH(f, u1)

n

X

k=0

an,kDk(u1, u2)

du1du2.

(8)

By (14),

Z

Γ01

ωH(f, u1)

n

X

k=0

an,kDk(u1, u2)

du1du2 .

. Z

Γ01

ωH(f, u1)

n

X

k=0

(k+ 1)2an,k

!

du1du2

≤(n+ 1)2

1 3(n+1)

Z

0

1 n+1

Z

3u2

ωH(f, u1)du1du2

ωHf,n+11 ωH(f, an,n). By (15),

Z

Γ001

ωH(f, u1)

n

X

k=0

an,kDk,1(u1, u2)

du1du2 .

.

1 3(n+1)

Z

0 an,n

Z

1 n+1

ωH(f,u1)

u21 du1du2ωH(f, an,n)

1 3(n+1)

Z

0 an,n

Z

1 n+1

1

u21du1du2ωH(f, an,n). (14) and (15) gives forj= 1,2,

Z

Γ001

ωH(f, u1)

n

X

k=0

an,kDk,j (u1, u2)

du1du2.

. (n+ 1)

1 3(n+1)

Z

0 an,n

Z

1 n+1

ωH(f,u1)

u1 du1du2≤(n+ 1)ωH(f, an,n)

1 3(n+1)

Z

0 an,n

Z

1 n+1

1

u1du1du2

≤ log (n+ 1)ωH(f, an,n). Since

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

(16) Dk(u1, u2) =Hk,1(u1, u2) +Hk,2(u1, u2) +Hk,3(u1, u2), where

Hk,1(u1, u2) := 12 cos((2k+1)u2π) sin u1+2u2π

sin u1−u2 2π, Hk,2(u1, u2) :=−12 cos (2k+1)

u1+u2 2 π

sin(u2π) sin u1−u2 2π,

(9)

Hk,3(u1, u2) := 12 cos (2k+1)u1

−u2 2 π

sin(u2π) sin u1+2u2π. By the method used in [12, p. 179] we get

(17)

n

X

k=0

an,kcos (2k+ 1)t

.An1t+sin1tan1t (0< t < π), and

(18)

n

X

k=0

an,kcos (2k+ 1)t

.An1t 0< tπ2. By aim of (18) and (15) we obtain

(19)

n

X

k=0

an,kHk,1(u1, u2)

. u12 1

Anπu1

2

and (20)

n

X

k=0

an,kHk,3(u1, u2)

. u11u2Anπu3

1

,

where both of for u1 and u2 are away from the origin. Also, for such u1 and u2,it follows from (17),(15) and from the fact

sin u12π.sin(u1+u2 2 that

(21)

n

X

k=0

an,kHk,2(u1, u2)

. u11u2Anπu3

1

.

Using (19) and the inequality (22) ωH(f, δ2)

δ2

≤2ωH(f, δ1)

δ1 (δ1 < δ2),

which is obtained from (4), and considering that the function An is nonde- creasing yield

Z

Γ0001

ωH(f, u1)

n

X

k=0

an,kHk,1(u1, u2)

du1du2.

(10)

.

an,n 3

Z

1 3(n+1)

an,n

Z

3u2

ωH(f,u1)

u21 Anπu1

2

du1du2

≤2

an,n 3

Z

1 3(n+1)

an,n

Z

3u2

ωH(f,3u2)

3u1u2 Anπu1

2

du1du2

= 23

an,n 3

Z

1 3(n+1)

ωH(f,3u2)

u2 loga3un,n2Anπu1

2

du2

≤log ((n+ 1)an,n)

an,n

Z3

1 3(n+1)

ωH(f,3u2)

u2 Anπu1

2

du2

≤log (n+ 1)

an,n

Z3

1 3(n+1)

ωH(f,3u2)

u2 Anπu1

2

du2 = log (n+ 1)

3 π(n+1)

Z

3 π

ωH(f,πt3)

t An(t)dt

= log (n+ 1)

n

X

k=1

3 π(k+1)

Z

3 πk

ωH(f,πt3)

t An(t)dt

≤log (n+ 1)Xn

k=1 ωH(f,k1)

k An(k+ 1). Forj = 2,3,by (20) and (21)

Z

Γ0001

ωH(f, u1)

n

X

k=0

an,kHk,j(u1, u2)

du1du2 .

.

an,n

Z

1 n+1

u1

Z3

1 3(n+1)

ωH(f,u1)

u1u2 Anπu3

1

du2du1

=

an,n

Z

1 n+1

ωH(f,u1)

u1 log ((n+ 1)u1)Anπu3

1

du1

≤log ((n+ 1)an,n)

an,n

Z

1 n+1

ωH(f,u1)

u1 Anπu3

1

du1 =

(11)

= log ((n+ 1)an,n)

3 π(n+1)

Z

3 π

ωH(f,πt3)

t An(t)dt

≤log (n+ 1)

n

X

k=1

ωH(f,k1)

k An(k+ 1). Thus we get the estimate

(23) In,1 .log (n+ 1) (

ωH(f, an,n) +

n

X

k=1

ωH(f,k1)

k An(k+ 1) )

. We divide Γ2 into two domains as

Γ02:=n(u1, u2)∈Γ2 :u23(n+1)an,n o, Γ002 :=n(u1, u2)∈Γ2 :u23(n+1)an,n o to estimate In,2.By (15) and (22),

Z

Γ02

ωH(f, u1)

n

X

k=0

an,kDk,1 (u1, u2)

du1du2 .

.

an,n 3(n+1)

Z

0 1

Z

an,n

ωH(f,u1)

u21 du1du2≤2ωH(f,aan,nn,n)

an,n 3(n+1)

Z

0 1

Z

an,n

1

u1du1du2

= 2ωH(f,aan,nn,n)logan,n1 3(n+1)an,n ≤log (n+ 1)ωH(f, an,n). (14),(15) and (22) yield

Z

Γ02

ωH(f, u1)

n

X

k=0

an,kDk,j(u1, u2)

du1du2 .

. (n+ 1)

an,n 3(n+1)

Z

0 1

Z

an,n

ωH(f,u1)

u1 du1du2≤2 (n+ 1)ωH(f,aan,nn,n)

an,n 3(n+1)

Z

0 1

Z

an,n

du1du2

ωH(f, an,n)

forj= 2,3.Since (15) implies|Hk,1(u1, u2)|. u12

1 for (u1, u2)∈Γ002,we get Z

Γ002

ωH(f, u1)

n

X

k=0

an,kHk,1(u1, u2)

du1du2 .

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• Linear algebra including solution of systems of linear equations, matrix manipulation, eigenvalues and eigenvectors, and elementary vector space concepts such as basis and

Adrian Iftene, Faculty of Computer Science, “Alexandru Ioan Cuza” University of Iași Elena Irimia, Research Institute for Artificial Intelligence “Mihai Drăgănescu”, Romanian

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