Accepted June N¨orlund) logarithmic means of Fourier series

Vol. 20, No. 1, 2016, 48–56

Convergence of the Logarithmic Means of Two-Dimensional Trigonometric Fourier Series

Davit Ishkhnelidze^∗

Batumi Shota Rustaveli State University

(Received April 28, 2016; Revised June 20, 2016; Accepted June 30, 2016)

(N¨orlund) logarithmic means of Fourier series.

Keywords:Double Fourier series, logarithmic means, convergence in norm.

AMS Subject Classification: 42A24.

1. Main Results Let f ∈ (

T²)

, T² = [−π, π]² be a 2π-periodic functions with respect to each variable. The two-dimensional Fourier series off with respect to the trigonometric system is the series

s[f] =

+∞

∑

m,n=−∞

fb(m, n)e^imxe^iny,

where

fb(m, n) = 1 4π²

∫ _π

−π

∫ _π

−πf(x, y)e^−imxe^−inydxdy

are the Fourier coefficients of the functionf.

Let C(T²)be the space of continuous functions are 2π-periodic with respect to each variable with the norm

∥f∥_c= sup

x,y∈T²|f(x, y)|. Let f ∈C(T²).The expression

ω(δ, f)c= sup{

∥f(·+u,·+v)−f(·,·)∥_c:u²+v² ≤δ²}

∗Corresponding author. Email: [email protected]

Abstract. We discuss on some convergence and divergence properties of two-dimensional

is called the total modulus of continuity of the functionf. The partial modulus of continuity are defined by

ω1(δ, f)c= sup{∥f(·+u,·)−f(·,·)∥_c:|u| ≤δ},

ω2(δ, f)c= sup{∥f(·,·+v)−f(·,·)∥_c:|v| ≤δ}.

We also use the notion of a mixed modulus of continuity. They are defined as follows:

ω1,2(δ1, δ2, f)c= sup

{∥f(·+u,·+v)−f(·+u,·)−f(·,·+v) +f(·,·)∥c

:|u| ≤δ1,|v| ≤δ2

}

, f ∈C(T²).

The Riesz’s means of the Fourier series has been studied by a lot of authors.

We mention for instance the papers of Szasz [11] and Yabuta [12], devoted to the logarithmic means. Similar means with respect to the Walsh and Vilenkin systems were discussed by Simon [10], and Gat [5]. The Norlund logarithmic means has been studied in ([1-7],[10-12]).

In this paper we investigate the approximation properties of two-dimensional logarithmic means of double trigonometric Fourier series off defined as follows:

tn,m(f, x, y) = 1 l_nl_m

n−1

∑

i=1 m∑−1

j=1

si,j(f, x, y)

(n−i)(m−j), ln=

∑n

k=1

1 k,

whereS_M,N(f, x, y) is the partial sum of double Fourier series off defined by

s_M,N(f, x, y) =

∑M m=−M

∑N n=−N

fb(m, n)e^imxe^iny.

It is evident that

t_n,m(f, x, y)−f(x, y) =

∫ _π

−π

∫ _π

−π

[f(x+t, y+s)−f(x, y)]F_n(t)F_m(s)dtds, where

F_n(t) = 1 l_n

n−1

∑

k=1

D_k(t) n−k andD_k(t)is Dirichlet kernel.

For one dimensional trigonometric Fourier series Goginava and Tkebuchava [6]

proved that the following are true

Theorem A [6]. Letf ∈C(T) and ω(δ, f)_c=o

( 1 log(1/δ)

)

then

||t_n(f)−f||c→oasn→ ∞.

Theorem B [6]. There exists a functionf ∈C(T) such that ω(δ, f)c=O

( 1 log(1/δ)

)

andt_n(f,0) diverges.

It is well-known that the following statement is true [13].

Theorem C (Zhizhiashvili). Letf ∈C(T²), then

∥S_n,m(f)−f∥c≤c {

ω₁(1 n, f)

clog(n+ 1) +ω₂(1 m, f)

clog(m+ 1) +ω1,2

(1 n, 1

m, f)

clog(n+ 1) log(m+ 1) }

.

From (1) and (2)Let A=(a_mnjk) denote a positive rectangular matrix, i. e., a_mnjk=0 for j > m or k > n, a a_mnjk > 0 for each 0 ≤ j ≤ m,0 ≤ k ≤ n and

∑m j=0

∑n

k=0

a_mnjk = 1.

For any double sequence (S_jk), define t_mn =

∑m

j=0

∑n

k=0

a_mnjk·s_jk, m, n= 0, 1, 2, . . .

The sequence (S_jk) is said to be summable by A if t_mn tends to a finite limit as m, n→ ∞.

A double rectangular matrix A is said to be regular if it sums every bounded convergent double sequence (Sjk) to the same limit. Necessary and sufficient con- ditions for the matrix A to be regular are known (see, e.g. [9]):

m,nlim→∞

∑m

j=0

a_mnjk = 0 (k= 0,1, ...), (1)

m,nlim→∞

∑n

k=0

amnjk = 0 (j= 0,1, ...). (2)

Since

∥tn,m(f)−f∥_c≤ 1 l_nl_m

n∑−1

i=0 m∑−1

j=0

∥Si,j(f)−f∥_c (n−i)(m−j), from (1) and (2) we conclude that the following theorem is true.

Theorem 1.1 : Let f ∈C(T²) and ω(δ, f)c=o

(( 1 log(1/δ)

)₂) .

Then

∥tn,m(f)−f∥_C →0 as m, n→ ∞.

In the paper we investigate sharpness of Theorem 1.1. In particular, the following is true

Theorem 1.2 : There exist a function f ∈C(T²) such that ω(δ, f)c=O

(( 1 log(1/δ)

)₂) ,

andtn,n(f,0,0)diverges.

Proof : (of Theorem 1.2) We choose a monotonically increasing sequence of posi- tive integers{n_k;k≥1} such that

n₁ ≥2,

n²_k ≤n_k+1, (3)

k−1∑

l=1

2^2nl

n²_l < 2^2nk

n²_k , (4)

( n_k 2^2nk

)2k∑−1 i=0

(2²ⁿⁱ ni

)2

< 1

k. (5)

We construct a function f defined as follows. Set

f(x, y) =

∑∞ k=1

f_k(x)·f_k(y) n²_k ,

where

fk(x) = sin(

2^2nk+1 2

)x·1[⁶·γ_nk,6·m(nk)·γ_nk](x),

k= 1,2, . . . , x∈[−π, π], where 1A is the characteristic function of a set A and

m(n_nk) = max{

s:sγ_nk ≤γ_nk−1}

, γ_nk = π

6(2^2nk+ 1/2)..

First we prove that

ω(δ, f)C =O

(( 1 log(1/δ)

)₂)

. (6)

For every sufficiently small δ >0 there exists a positive integer ksuch that π

2^2nk+ 1/2 ≤δ < π 2^2nk−1+ 1/2. Since

|fnl(x+δ)−f(x)|=O(δ2^2nl), l= 1,2..., k−1, from (3) and (4) we get

|f(x+δ, y)−f(x, y)| ≤

k−1

∑

l=1

1

n²_l · |fnl(x+δ)−fnl(x)|+ 2

∑∞ l=k

1 n²_l

=O (

δ

k−1

∑

l=1

2^2nl n²_l

) +O

( 1 n²_k

)

=O (

δ2^2nk−1 n²_k−1

) +O

( 1 n²_k

)

=O

(( 1 log(1/δ

)2) .

Consequently,

ω₁(δ, f)_C =O

(( 1 log(1/δ)

)2)

. (7)

Analogously, we obtain

ω2(δ, f)C =O

(( 1 log(1/δ

)₂)

. (8)

Since

ω(δ, f)_C ≤ω₁(δ, f)_C+ω₂(δ, f)_C from (7) and (8) we get (6)

Next, we shall prove that t₂²nk,2^2nk(f,0,0) diverges.

It is clear that

t₂²nk,2^2nk(f,0,0)−f(0,0)=t₂²nk,2^2nk(f,0,0)

= ∫ _π

−π

∫ _π

−π

f(t, s)F₂²nk(t)F₂²nk(s)dtds

≥ c n²_k

(∫ _π

−π

fnk(t)F₂²nk(t)dt )₂

−

k−1

∑

i=1

c n²_i

(∫ _π

−π

fni(t)F₂²nk(t)dt )₂

−

∑∞ i=k+1

c n²_i

(∫ π

−π

fni(t)F₂²nk(t)dt )2

=I−II−III. (9)

Since (see [6])

l2²ⁿF2²ⁿ(x)

= sin(2²ⁿ+¹₂)x 2 sin^x₂

2²ⁿ∑−2 k=1

2

k(k+ 1)(k+ 2) sin²(

(k+ 1)^x₂) 2 sin²(x/2)

+ 1

2²ⁿ(2²ⁿ−1)×sin(2²ⁿ+¹₂) 2 sin (x/2)

sin²2²ⁿ⁻¹x 2 sin²(x/2) + 1

2²ⁿ

sin²(2²ⁿ+¹₂)x 4 sin²(x/2) − 3

4

sin(2²ⁿ+ ¹₂)x 2 sin (x/2)

−cos(2²ⁿ+¹₂)x 2 sin (x/2) (

∑n

k=1

sinkx k ),

we have

I = c n²_k

(∫ π

−π

f_nk(t)F₂²nk(t)dt )2

≥

≥ (

c n²_k

∫ _2nk^πm(nk)

+1/2

π 22nk+1/2

sin²(2^2nk+ 1/2)t 2 sin(t/2)

2²∑^nk−2

i=1

2

i(i+ 1)(i+ 2) ·sin²(i+ 1)₂^t 2 sin²(t/2) dt

− c n²_k

1 2^2nk(2^2nk−1)

∫ ^π·m(nk)

22nk+1/2

π 22nk+1/2

sin²(2^2nk+ 1/2)t 2 sin(t/2)

sin²2^nk−1t 2 sin²(t/2)dt

− c n²_k

1 2^2nk

∫ ^πm(nk)

22nk+1/2

π 22nk+1/2

sin(

2^2nk+ 1/2)

tsin²(2^2nk+ 1/2)t 4 sin²(t/2) dt

− c n²_k

∫ ^πm(nk)

22nk+1/2

π 22nk+1/2

sin²(

2^2nk+ 1/2) t 2 sin (t/2) dt

− c n²_k

∫ ^πm(nk)

22nk+1/2

π 22nk+1/2

sin(

2^2nk+ 1/2) tcos(

2^2nk+ 1/2) t 2 sin (t/2)

(₂∑2nk

i=1

sinit i

) dt

)2

=

= (I1−I2−I3−I4−I5)². (10) It is evident that

I2,I3, I4, I5= 0 (

1 n²_k ·

∫ ^πm(nk)

22nk+1/2

π 22nk+1/2

1 tdt

)

= 0 ( 1

n_k )

. (11)

Since (see [14])

sin(i+ 1)· t 2 ≥ 2

π i+ 1

2 t, i= 1,2, ...,2^nk−1−1,

for t∈I_nk, I_n=

2∪ⁿ⁻¹

m=1

[α_mn, β_mn],

where

αmn= π·(12m+ 1)

6·(2²ⁿ+ 1/2), βmn= π·(12m+ 5)

6·(2²ⁿ+ 1/2), m, n= 1,2, ....

and

sin(

2^2nk+ 1/2)

t≥1/2,

2²ⁿ

∑

k=1

sinkx k

≤c <∞, forI₁ we have

I₁ ≥ c n²_k

2^nk∑⁻¹−1 i=1

(i+ 1)² i(i+ 1) (i+ 2)

2∑^nk−1

m=1

∫ _βm,nk

α_m,nk

1

t dt≥c >0. (12) Combining (11) and (12) we conclude that

I ≥c >0. (13)

Now, we estimate II. Since [6]

∥t_n(f)−f∥_c≤c·ω(1/n, f)_clog (n+ 1) and

ω (

f_ni, 1 2^2nk

)

c

= 0 (2²ⁿⁱ

2^2nk )

, i= 1,2, ..., k−1, from (4) and (5) we get

II ≤C

k−1

∑

i=1

1

n²_i ∥t2^2nk(fni)−(fni)∥²_c ≤C

k−1

∑

i=1

(1 ni

ω (

fni, 1 2^2nk

) nk

)₂

(14)

≤ C·

k−1

∑

i=1

( 1 n_i

2²ⁿⁱ

2^2nkn_k)² ≤C( n_k 2^2nk)²

k−1

∑

i=1

(2²ⁿⁱ n_i )² ≤ c

k =o(1) as k→ ∞. It is obvious that

∥F_n∥_L=O (

1 logn·

n−1

∑

i=1

∥D_i∥₁ n−i

)

=O (

1 logn·

n−1

∑

i=1

log (i+ 1) n−i

)

=O(log (n+ 1)).

Then we have III =O

( _∞

∑

i=k+1

1

n²_i · ∥F₂²nk∥²₁ )

=O(

∑∞ i=k+1

1

n²_in²_k) (15)

=O

(( n_k n_k+1

)2)

=O(n²_k

n⁴_k) =O( 1

n²_k) =o(1) as k→ ∞. After substituting 13, (14) and (15) in (9) we obtain

klim→∞ t₂²nk,2^2nk(f,0,0)−f(0,0)>0.

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