Approximation of integrable functions by general linear matrix operators of their Fourier series

Vishnu Narayan Mishra; Włodzimierz Łenski; Bogdan Szal

doi:10.1515/dema-2022-0009

Artikel Open Access

Approximation of integrable functions by general linear matrix operators of their Fourier series

Vishnu Narayan Mishra , Włodzimierz Łenski und Bogdan Szal

Veröffentlicht/Copyright: 20. Mai 2022

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Aus der Zeitschrift Demonstratio Mathematica Band 55 Heft 1

Abstract

The pointwise estimates of the deviation T n , A f ( ⋅ ) − f ( ⋅ ) in terms of pointwise moduli of continuity based on the points of differentiability of indefinite integral of f , with application of the rth differences of the entries of A , are proved. The similar results in case of the Lebesgue points are considered, too. Analogical results on norm approximation with remarks and corollaries are also given.

Keywords: rate of approximation; summability of Fourier series

MSC 2010: 42A24

1 Introduction

Let L 2 π / r p ( 1 ≤ p ≤ ∞ ) [ L 2 π / r ∞ ] be the class of all 2 π / r –periodic real-valued functions, integrable in the Lebesgue sense with pth power [essentially bounded] over Q r = [ − π / r , π / r ] with the norm

‖ f ‖ L 2 π / r p ≔ ‖ f ( ⋅ ) ‖ L 2 π / r p = ∫ Q r ∣ f ( t ) ∣ p d t 1 / p when 1 ≤ p < ∞ , ess sup t ∈ Q r ∣ f ( t ) ∣ when p = ∞ ,

where r ∈ N . It is clear that L 2 π / r p ⊂ L 2 π / 1 p = L 2 π p with 1 ≤ p ≤ ∞ and for f ∈ L 2 π / r p with 1 ≤ p < ∞

‖ f ‖ L 2 π p = r 1 / p ‖ f ‖ L 2 π / r p .

We will consider for f ∈ L 2 π 1 the trigonometric Fourier series.

S f ( x ) ≔ a 0 ( f ) 2 + ∑ ν = 1 ∞ ( a ν ( f ) cos ν x + b ν ( f ) sin ν x )

with the partial sums S k f .

If A ≔ ( a n , k ) 0 ≤ n , k < ∞ = ( a n , k ) is an infinite matrix of real numbers such that

a n , k ≥ 0 when k , n = 0 , 1 , 2 , … , lim n → ∞ a n , k = 0 for every k = 0 , 1 , 2 , …

and

∑ k = 0 ∞ a n , k = 1 for every n = 0 , 1 , 2 , …

or A 0 ≔ ( a n , k ∘ ) 0 ≤ k ≤ n < ∞ = ( a n , k ∘ ) , where

a n , k ∘ = a n , k when k ≤ n and a n , k ∘ = 0 when k > n ,

then

T n , A f ( x ) ≔ ∑ k = 0 ∞ a n , k S k f ( x ) ( n = 0 , 1 , 2 , … )

T n , A 0 f ( x ) ≔ ∑ k = 0 n a n , k S k f ( x ) ( n = 0 , 1 , 2 , … ) ,

respectively.

As a measure of approximation of f ( x ) by the above quantities we use the pointwise moduli of continuity of f in the space L 2 π / r 1 defined by the formulas

w x f ( δ ) = sup 0 < t ≤ δ 1 t ∫ 0 t φ x ( u ) d u , w ¯ x f ( δ ) = sup 0 < t ≤ δ 1 δ ∫ 0 t φ x ( u ) d u , W ¯ x f ( δ ) = 1 δ ∫ 0 δ ∣ φ x ( u ) ∣ d u ,

where

φ x ( t ) ≔ f ( x + t ) + f ( x − t ) − 2 f ( x ) .

It is clear that w ¯ x f ( δ ) ≤ W ¯ x f ( δ ) , w ¯ x f ( δ ) ≤ w x f ( δ ) , and

(1) ‖ w ¯ ⋅ f ( δ ) ‖ L 2 π / r p ≤ ‖ W ¯ ⋅ f ( δ ) ‖ L 2 π / r p ≤ ω f ( δ ) L 2 π / r p ,

for any δ > 0 , where

ω f ( δ ) L 2 π / r p = sup 0 < t ≤ δ ‖ φ ⋅ ( t ) ‖ L 2 π / r p

is the usual integral modulus of continuity.

We will use the notations

A n , r = ∑ k = 0 ∞ ∣ a n , k − a n , k + r ∣ , A n , r ∘ = ∑ k = 0 n ∣ a n , k ∘ − a n , k + r ∘ ∣ , A ¯ n , r = ∑ k = 0 ∞ ( k + 1 ) ∣ a n , k − a n , k + r ∣ , A ¯ n , r ∘ = ∑ k = 0 n ( k + 1 ) ∣ a n , k ∘ − a n , k + r ∘ ∣ ,

for r ∈ N .

The deviations T n , A f ( ⋅ ) − f ( ⋅ ) and T n , A 0 f ( ⋅ ) − f ( ⋅ ) were estimated by many authors like e.g. [1,2, 3,4].

In our theorems, we will prove the pointwise estimates of deviations T n , A f ( ⋅ ) − f ( ⋅ ) and T n , A 0 f ( ⋅ ) − f ( ⋅ ) in terms of moduli w ⋅ f , w ¯ ⋅ f , and W ¯ ⋅ f . Consequently, we will also give some results on norm approximation and some remarks. Finally, we will derive some corollaries. Thus, we will use the weaker conditions than those in other papers and we will prove our statements for essentially wider set of points. These results will be presented in Section 2. In Sections 3 and 4, we give some informations on applications and prove auxiliary results, respectively. In Section 5, we prove our main results and formulate Conclusion. Finally, in the end of paper there are references listed and numbered in the order that they appear in the text.

We will write I 1 ≪ I 2 if there exists a positive constant K , sometimes depended on some parameters, such that I 1 ≤ K I 2 .

2 Statement of the results

We start with our main results on the degrees of pointwise summability.

Theorem 1

Let f ∈ L 2 π / r 1 , where r ∈ N . If the entries of the matrix A satisfy the condition

(2) ∑ k = 0 ∞ ( k + 1 ) a n , k ≪ n + 1 ,

then

(3) ∣ T n , A f ( x ) − f ( x ) ∣ ≪ A ¯ n , r ∑ k = 0 n w x f π k + 1 ,

for every natural n and all real x .

If in Theorem 1 we put the matrix A 0 instead of the matrix A , we obtain the following theorem:

Theorem 2

If f ∈ L 2 π / r 1 , where r ∈ N , then

(4) ∣ T n , A 0 f ( x ) − f ( x ) ∣ ≪ A ¯ n , r ∘ ∑ k = 0 n w ¯ x f π k + 1 ,

for every natural n and all real x .

If in Theorems 1 and 2 we assume additional conditions on quantities A ¯ n , r or A ¯ n , r ∘ , we will obtain the following remark:

Remark 1

If we suppose that

(5) A ¯ n , r = O 1 n + 1 or A ¯ n , r ∘ = O 1 n + 1 ,

then under the assumptions of Theorems 1 or 2

∣ T n , A f ( x ) − f ( x ) ∣ ≪ 1 n + 1 ∑ k = 0 n w x f π k + 1

∣ T n , A 0 f ( x ) − f ( x ) ∣ ≪ 1 n + 1 ∑ k = 0 n w ¯ x f π k + 1

hold, respectively. Additionally, if w x f ( δ ) = o x ( 1 ) as δ → 0 + at point x , then

T n , A f ( x ) − f ( x ) = o x ( 1 ) or T n , A 0 f ( x ) − f ( x ) = o x ( 1 ) ,

as n → ∞ , respectively.

We also note that the relations

w ¯ x f ( δ ) = o x ( 1 ) or w x f ( δ ) = o x ( 1 )

as δ → 0 + is more general than the condition

W ¯ x f ( δ ) = o x ( 1 ) as δ → 0 + .

We can check this for the function

f ( u ) = − 1 2 sin 1 u for − 1 ≤ u < 0 , 1 2 sin 1 u for 0 < u ≤ 1 , 0 for u = 0 and ∣ u ∣ > 1 .

In [5] it is proved that

lim λ → 0 + 1 λ ∫ 0 λ φ 0 ( u ) d u = 0 and lim λ → 0 + 1 λ ∫ 0 λ ∣ φ 0 ( u ) ∣ d u ≥ 2 π 2 ,

where φ 0 ( u ) = sin 1 u .

For the function W ¯ x f we can prove the next theorem.

Theorem 3

Let f ∈ L 2 π / r 1 , where r ∈ N . Then

(6) ∣ T n , A 0 f ( x ) − f ( x ) ∣ ≪ A n , r ∘ ∑ k = 0 n W ¯ x f π k + 1 ,

for every natural n and all real x.

Moreover, if the entries of matrix A satisfy condition (2), then

(7) ∣ T n , A f ( x ) − f ( x ) ∣ ≪ A n , r ∑ k = 0 n W ¯ x f π k + 1 ,

for every natural n and all real x .

Finally, we formulate the results on the estimates of L p norm of the deviations considered above.

Theorem 4

Let f ∈ L 2 π / r p , where r ∈ N and 1 ≤ p ≤ ∞ . Then

∥ T n , A 0 f ( ⋅ ) − f ( ⋅ ) ∥ L 2 π / r p ≪ A n , r ∘ ∑ k = 0 n ω f π k + 1 L 2 π / r p ,

for every natural n .

Moreover, if the entries of matrix A satisfy condition (2), then

∥ T n , A f ( ⋅ ) − f ( ⋅ ) ∥ L 2 π / r p ≪ A n , r ∑ k = 0 n ω f π k + 1 L 2 π / r p .

Let C n α = ( α + 1 ) ( α + 2 ) … ( α + n ) n ! for n ∈ N and α ∈ R . Suppose A 0 = A 0 α = ( a n , k ∘ ) , where a n , k ∘ = C n − k α − 1 C n α when k ≤ n and a n , k ∘ = 0 when k > n . In special case C n 0 = 1 and C 1 α = α + 1 for n ∈ N and α ∈ R . If α = 2 , then for r = 2 [6].

A n , r ∘ = ∑ k = 0 n ∣ a n , k ∘ − a n , k + r ∘ ∣ = ∑ k = 0 n − 2 ∣ a n , k ∘ − a n , k + 2 ∘ ∣ + a n , n − 1 ∘ + a n , n ∘ = 1 C n 2 ∑ k = 0 n − 2 ∣ C n − k 1 − C n − k − 2 1 ∣ + C 1 1 + C 0 1 = 1 C n 2 ∑ k = 0 n − 2 ∣ C n − k 1 − C n − k − 1 1 + C n − k − 1 1 − C n − k − 2 1 ∣ + 2 + 1 = 1 C n 2 ∑ k = 0 n − 2 ∣ C n − k 0 − C n − k − 1 0 ∣ + 3 = 3 C n 2 = O 1 ( n + 1 ) 2 ,

whence

A ¯ n , r ∘ = ∑ k = 0 n ( k + 1 ) ∣ a n , k ∘ − a n , k + r ∘ ∣ ≤ ( n + 1 ) A n , r ∘ = O 1 n + 1 .

Therefore, if f ∈ L 2 π / r 1 , where r ∈ N , then using Remark 1 we obtain

∣ T n , A 0 2 f ( x ) − f ( x ) ∣ ≪ 1 n + 1 ∑ k = 0 n w ¯ x f π k + 1 .

Remark 2

Let f ∈ L 2 π / r 1 , where r ∈ N . We know that T n , A 0 1 f ( x ) − f ( x ) does not tend to 0 at points of differentiability of indefinite integral of f . We can note that condition (5) does not hold when A 0 = A 0 1 , but in this case we can assume the following condition:

A n , r ∘ = O 1 n + 1 .

Then

∣ T n , A 0 f ( x ) − f ( x ) ∣ ≪ 1 n + 1 ∑ k = 0 n W ¯ x f π k + 1 .

Moreover, if the entries of matrix A satisfy condition (2) and

A n , r = O 1 n + 1 ,

then

∣ T n , A f ( x ) − f ( x ) ∣ ≪ 1 n + 1 ∑ k = 0 n W ¯ x f π k + 1 .

Remark 3

We note that in case A 0 2 = C n − k 1 C n 2 the relation (4) with (5) leads us to the result comparable with [7] on approximation of f by the Cesàro means ( C , 2 ) of its Fourier series. The same remark holds in the case of Theorem 3 with A 0 1 = 1 n + 1 (c.f. [3]).

3 Applications

The examined approximations of functions (signals) have wide applications in signal analysis in general and in signal processing. The obtained results are general in nature and can be applied to various problems of Mathematical Analysis, Mathematical Physics, and different Engineering branches. The estimates presented in the theorems allow us to determine the orders of deviation of a very wide class of linear means of partial sums of Fourier series from the examined functions depending on their differential properties. On the basis of these results, the considered deviations can be investigated with the use of other norms and approximation measures.

4 Auxiliary results

We begin this section by some notations from [8] and [6]. Let for r = ± 1 , ± 2 , …

D k , r ( t ) = sin ( 2 k + r ) t 2 2 sin r t 2 , D ˜ k , r ( t ) = cos ( 2 k + r ) t 2 2 sin r t 2 .

Then

S k f ( x ) = a 0 ( f ) 2 + ∑ ν = 1 k ( a ν ( f ) cos ν x + b ν ( f ) sin ν x ) = 1 π ∫ − π π f ( t + x ) D k , 1 ( t ) d t ,

and

S k f ( x ) − f ( x ) = 1 π ∫ 0 π φ x ( t ) D k , 1 ( t ) d t .

In the proofs we need the following known results.

Lemma 1

[6] If 0 < ∣ t ∣ ≤ π , then

∣ D k , 1 ( t ) ∣ ≤ π 2 ∣ t ∣ , ∣ D ˜ k , 1 ( t ) ∣ ≤ π 2 ∣ t ∣

and, for any real t , we have

∣ D k , 1 ( t ) ∣ ≤ k + 1 2 .

Lemma 2

[8,9] Let r ∈ N , l ∈ Z , and ( a n ) ⊂ C . If t ≠ 2 l π r , then for every m ≥ n

∑ k = n m a k sin k t = − ∑ k = n m ( a k − a k + r ) D ˜ k , r ( t ) + ∑ k = m + 1 m + r a k D ˜ k , − r ( t ) − ∑ k = n n + r − 1 a k D ˜ k , − r ( t ) , ∑ k = n m a k cos k t = ∑ k = n m ( a k − a k + r ) D k , r ( t ) − ∑ k = m + 1 m + r a k D k , − r ( t ) + ∑ k = n n + r − 1 a k D k , − r ( t ) .

5 Proofs of the results

We will use in the proofs, without references, the following inequalities:

(8) w ¯ x f π r ( n + 1 ) = sup 0 < t ≤ π r ( n + 1 ) r ( n + 1 ) π ∫ 0 t φ x ( u ) d u ∑ l = 0 n 2 ( l + 1 ) ( n + 1 ) ( n + 2 ) ≤ 2 r n + 1 ∑ l = 0 n ( l + 1 ) ( n + 1 ) π ( n + 2 ) sup 0 < t ≤ π r ( n + 1 ) ∫ 0 t φ x ( u ) d u ≤ 2 n + 1 ∑ l = 0 n r ( l + 1 ) π sup 0 < t ≤ π r ( l + 1 ) ∫ 0 t φ x ( u ) d u = 2 1 n + 1 ∑ l = 0 n w ¯ x f π r ( l + 1 )

and

(9) w x f π r ( n + 1 ) ≤ 1 n + 1 ∑ l = 0 n w x f π r ( l + 1 ) ,

when f ∈ L 2 π / r 1 , where r ∈ N .

We also note that for r ∈ N

(10) ∑ k = 0 r − 1 ( k + 1 ) a n , k ≤ ∑ k = 0 ∞ ( k + 1 ) a n , k − ∑ k = r ∞ ( k + 1 ) a n , k + r ∑ k = r ∞ a n , k = ∑ k = 0 ∞ ( k + 1 ) a n , k − ∑ k = r ∞ ( k − r + 1 ) a n , k = ∑ k = 0 ∞ ( k + 1 ) a n , k − ∑ l = 0 ∞ ( l + 1 ) a n , l + r = ∑ k = 0 ∞ ( k + 1 ) ( a n , k − a n , k + r ) ≤ A ¯ n , r .

5.1 Proof of Theorem 1

It is clear that

T n , A f ( x ) − f ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ 2 m π r 2 m π r + π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] − 1 ∫ 2 m π r + π r 2 ( m + 1 ) π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = I 1 ( x ) + I 2 ( x )

for an odd r , and

T n , A f ( x ) − f ( x ) = 1 π ∑ m = 0 [ r / 2 ] − 1 ∫ 2 m π r 2 m π r + π r + ∫ 2 m π r + π r 2 ( m + 1 ) π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = I 1 ′ ( x ) + I 2 ( x )

for an even r , whence

∣ T n , A f ( x ) − f ( x ) ∣ ≤ ∣ I 1 ( x ) ∣ + ∣ I 1 ′ ( x ) ∣ + ∣ I 2 ( x ) ∣ .

We will consider in detail the term ∣ I 1 ( x ) ∣ . It is clear that

I 1 ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ 2 m π r 2 m π r + π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = 1 π ∑ m = 0 [ r / 2 ] ∫ 2 m π r 2 m π r + π r ( n + 1 ) + ∫ 2 m π r + π r ( n + 1 ) 2 m π r + π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = J 1 ( x ) + J 2 ( x ) .

Using the integration by parts we obtain

∣ J 1 ( x ) ∣ = 1 π ∑ m = 0 [ r / 2 ] ∫ 0 π r ( n + 1 ) φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 t + 2 m π r d t ≤ ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ a n , k ∫ 0 π r ( n + 1 ) d d t ∫ 0 t φ x ( u ) d u D k , 1 t + 2 m π r d t ≤ ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ a n , k D k , 1 π r ( n + 1 ) + 2 m π r ∫ 0 π r ( n + 1 ) φ x ( u ) d u + ∫ 0 π r ( n + 1 ) ∫ 0 t φ x ( u ) d u d d t D k , 1 t + 2 m π r d t .

Using Lemma 1 we have for t ∈ ( 0 , π ]

(11) d d t D k , 1 ( t ) = 2 k + 1 2 cos ( 2 k + 1 ) t 2 sin t 2 − 1 2 cos t 2 sin ( 2 k + 1 ) t 2 2 sin 2 t 2 = 2 k + 1 2 D ˜ k , 1 ( t ) − cos t 2 2 sin t 2 D k , 1 ( t ) ≤ π ( 2 k + 1 ) 2 t .

Therefore,

(12) d d t D k , 1 t + 2 m π r ≤ π ( 2 k + 1 ) 2 t + 2 m π r ≤ π ( 2 k + 1 ) 2 t

and by (2) we obtain

∣ J 1 ( x ) ∣ ≪ 1 π ∑ m = 0 [ r / 2 ] ( n + 1 ) ∫ 0 π r ( n + 1 ) φ x ( u ) d u + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ a n , k ∫ 0 π r ( n + 1 ) ∫ 0 t φ x ( u ) d u π ( 2 k + 1 ) 2 t d t ≤ r 2 ( n + 1 ) π ∫ 0 π r ( n + 1 ) φ x ( u ) d u + ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ a n , k ( k + 1 ) ∫ 0 π r ( n + 1 ) 1 t ∫ 0 t φ x ( u ) d u d t ≪ 1 2 w ¯ x f π r ( n + 1 ) + ∑ m = 0 [ r / 2 ] ( n + 1 ) ∫ 0 π r ( n + 1 ) w x f ( t ) d t ≪ w x f π r ( n + 1 ) .

Furthermore, by Lemma 2,

J 2 ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) 2 sin t 2 ∑ k = 0 ∞ a n , k sin ( 2 k + 1 ) t 2 d t = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) cos t 2 2 sin t 2 − ∑ k = 0 ∞ ( a n , k − a n , k + r ) D ˜ k , r ( t ) − ∑ k = 0 r − 1 a n , k D ˜ k , − r ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) sin t 2 2 sin t 2 ∑ k = 0 ∞ ( a n , k − a n , k + r ) D k , r ( t ) + ∑ k = 0 r − 1 a n , k D k , − r ( t ) d t = − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ( a n , k − a n , k + r ) ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) cos t 2 2 sin t 2 D ˜ k , r ( t ) d t + − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) cos t 2 2 sin t 2 D ˜ k , − r ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ( a n , k − a n , k + r ) ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) sin t 2 2 sin t 2 D k , r ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) sin t 2 2 sin t 2 D k , − r ( t ) d t = − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ( a n , k − a n , k + r ) Q 1 ( x ) + − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k Q 2 ( x ) + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ( a n , k − a n , k + r ) Q 3 ( x ) + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k Q 4 ( x ) .

By standard calculations we have

Q 1 ( x ) Q 2 ( x ) = ± ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) cos t 2 cos ( 2 k ± r ) t 2 4 sin r t 2 sin t 2 d t = ± ∫ π r ( n + 1 ) π r d d t ∫ 0 t φ x ( u ) d u cos t + 2 m π r 2 cos ( 2 k ± r ) 2 t + 2 m π r 4 sin r t + 2 m π r 2 sin t + 2 m π r 2 d t = ± ∫ 0 t φ x ( u ) d u cos t + 2 m π r 2 cos ( 2 k ± r ) 2 t + 2 m π r 4 sin r t + 2 m π r 2 sin t + 2 m π r 2 π r ( n + 1 ) π r ∓ ∫ π r ( n + 1 ) π r ∫ 0 t φ x ( u ) d u d d t cos t + 2 m π r 2 cos ( 2 k ± r ) 2 t + 2 m π r 4 sin r t + 2 m π r 2 sin t + 2 m π r 2 d t

= ± ∫ 0 π r φ x ( u ) d u cos ( 2 m + 1 ) π 2 r cos ( 2 k ± r ) ( 2 m + 1 ) π 2 r 4 sin ( 2 m + 1 ) π 2 sin ( 2 m + 1 ) π 2 r ∓ ∫ 0 π r ( n + 1 ) φ x ( u ) d u cos π 2 r ( n + 1 ) + 2 m π r cos ( 2 k ± r ) 2 π r ( n + 1 ) + 2 m π r 4 sin π 2 r ( n + 1 ) + 2 m π 2 sin π 2 r ( n + 1 ) + m π r ∓ ∫ π r ( n + 1 ) π r ∫ 0 t φ x ( u ) d u d d t cos t + 2 m π r 2 cos ( 2 k ± r ) 2 t + 2 m π r 4 sin r t + 2 m π r 2 sin t + 2 m π r 2 d t .

Since

(13) d d t cos t 2 cos ( 2 k ± r ) t 2 4 sin r t 2 sin t 2 = 1 4 d d t cos ( 2 k ± r + 1 ) t 2 + cos ( 2 k ± r − 1 ) t 2 cos ( r − 1 ) t 2 − cos ( r + 1 ) t 2 = − ( 2 k ± r + 1 ) 2 sin ( 2 k ± r + 1 ) t 2 − ( 2 k ± r − 1 ) 2 sin ( 2 k ± r − 1 ) t 2 cos ( r − 1 ) t 2 − cos ( r + 1 ) t 2 16 sin 2 r t 2 sin 2 t 2 + cos ( 2 k ± r + 1 ) t 2 + cos ( 2 k ± r − 1 ) t 2 − ( r − 1 ) 2 sin ( r − 1 ) t 2 + ( r + 1 ) 2 sin ( r + 1 ) t 2 16 sin 2 r t 2 sin 2 t 2 = − ( 2 k ± r ) 2 sin ( 2 k ± r + 1 ) t 2 + sin ( 2 k ± r − 1 ) t 2 − 1 2 sin ( 2 k ± r + 1 ) t 2 − sin ( 2 k ± r − 1 ) t 2 8 sin r t 2 sin t 2 + 2 cos t 2 cos ( 2 k ± r ) t 2 r 2 sin ( r + 1 ) t 2 − sin ( r − 1 ) t 2 + 1 2 sin ( r + 1 ) t 2 + sin ( r − 1 ) t 2 16 sin 2 r t 2 sin 2 t 2 = − ( 2 k ± r ) 2 2 sin ( 2 k ± r ) t 2 cos t 2 − 1 2 2 cos ( 2 k ± r ) t 2 sin t 2 8 sin r t 2 sin t 2 + 2 cos t 2 cos ( 2 k ± r ) t 2 r 2 2 cos r t 2 sin t 2 + 1 2 2 sin r t 2 cos t 2 16 sin 2 r t 2 sin 2 t 2 = − ( 2 k ± r ) sin ( 2 k ± r ) t 2 cos t 2 − cos ( 2 k ± r ) t 2 sin t 2 8 sin r t 2 sin t 2 + cos t 2 cos ( 2 k ± r ) t 2 r cos r t 2 sin t 2 + sin r t 2 cos t 2 8 sin 2 r t 2 sin 2 t 2

taking into account the estimates

(14) sin t 2 ≥ ∣ t ∣ π for t ∈ [ 0 , π ] , sin r t 2 ≥ r t π − 2 m for t ∈ 2 m π r , 2 m π r + π r ,

where m ∈ { 0 , … , [ r / 2 ] } , we can see that

d d t cos t 2 cos ( 2 k ± r ) t 2 8 sin r t 2 sin t 2 ≤ 2 k + r + 1 8 sin r t 2 sin t 2 + 2 r 8 sin 2 r t 2 sin t 2 ≪ k + 1 r t π − 2 m t + 1 r t π − 2 m 2 t ,

when π r ( n + 1 ) + 2 m π r ≤ t ≤ π r + 2 m π r . Therefore,

(15) d d t cos t + 2 m π / r 2 cos ( 2 k ± r ) ( t + 2 m π / r ) 2 4 sin r ( t + 2 m π / r ) 2 sin t + 2 m π / r 2 ≪ k + 1 t t + 2 m π r + 1 t 2 t + 2 m π r ≪ k + 1 t 2 + 1 t 3 ,

when π r ( n + 1 ) ≤ t ≤ π r .

Hence,

∣ Q 1 ( x ) ∣ ∣ Q 2 ( x ) ∣ ≤ ∫ 0 π r φ x ( u ) d u cos ( 2 m + 1 ) π 2 r cos ( 2 k ± r ) ( 2 m + 1 ) π 2 r 4 sin ( 2 m + 1 ) π 2 sin ( 2 m + 1 ) π 2 r + ∫ 0 π r ( n + 1 ) φ x ( u ) d u cos π 2 r ( n + 1 ) + 2 m π r cos ( 2 k ± r ) 2 π r ( n + 1 ) + 2 m π r 4 sin π 2 ( n + 1 ) + m π sin π 2 r ( n + 1 ) + m π r + ∫ π r ( n + 1 ) π r ∫ 0 t φ x ( u ) d u d d t cos t + 2 m π r 2 cos ( 2 k ± r ) 2 t + 2 m π r 4 sin r t + 2 m π r 2 sin t + 2 m π r 2 d t ≪ ∫ 0 π r φ x ( u ) d u + ∫ 0 π r ( n + 1 ) φ x ( u ) d u ( n + 1 ) 2 + ( k + 1 ) ∫ π r ( n + 1 ) π r ∫ 0 t φ x ( u ) d u d t t 2 + ∫ π r ( n + 1 ) π r ∫ 0 t φ x ( u ) d u d t t 3 ≤ π r w ¯ x f π r + ( n + 1 ) w ¯ x f π r ( n + 1 ) + ( k + 1 ) ∫ π r ( n + 1 ) π r sup c ∈ [ 0 , t ] ∫ 0 c φ x ( u ) d u d t t 2 + ∫ π r ( n + 1 ) π r sup c ∈ [ 0 , t ] ∫ 0 c φ x ( u ) d u d t t 3 .

Using (8)

(16) ∣ Q 1 ( x ) ∣ ∣ Q 2 ( x ) ∣ ≪ w ¯ x f π r + ∑ l = 0 n w ¯ x f π r ( l + 1 ) + ( k + 1 ) ∫ 1 n + 1 w ¯ x f π r t d t t + ∫ 1 n + 1 w ¯ x f π r t d t t 2 ≪ w ¯ x f π r + ∑ l = 0 n w ¯ x f π l + 1 + ( k + 1 ) ∑ l = 1 n ∫ l l + 1 w ¯ x f π r l d t l + ∑ l = 1 n ∫ l l + 1 w ¯ x f π r l d t ≪ ( k + 1 ) ∑ l = 0 n w ¯ x f π r ( l + 1 ) .

Similarly, we can show

(17) ∣ Q 3 ( x ) ∣ ∣ Q 4 ( x ) ∣ ≪ ( k + 1 ) ∑ l = 0 n w ¯ x f π r ( l + 1 ) .

If i = 1 , 3 , then

1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ∣ a n , k − a n , k + r ∣ ∣ Q i ( x ) ∣ ≪ ∑ k = 0 ∞ ∣ a n , k − a n , k + r ∣ ∣ Q i ( x ) ∣ ≪ A ¯ n , r ∑ l = 0 n w ¯ x f π r ( l + 1 )

and if i = 2 , 4 , then using (10)

1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∣ Q i ( x ) ∣ ≪ ∑ k = 0 r − 1 a n , k ( k + 1 ) ∑ l = 0 n w ¯ x f π r ( l + 1 ) ≤ A ¯ n , r ∑ l = 0 n w ¯ x f π r ( l + 1 ) .

Finally,

∣ J 2 ( x ) ∣ ≪ A ¯ n , r ∑ l = 0 n w ¯ x f π r ( l + 1 )

and using (9) we obtain

∣ I 1 ( x ) ∣ ≪ A ¯ n , r ∑ l = 0 n w x f π r ( l + 1 ) .

Similarly, we can obtain the estimate

∣ I 1 ′ ( x ) ∣ ≪ A ¯ n , r ∑ l = 0 n w x f π r ( l + 1 ) .

It is clear that

I 2 ( x ) = 1 π ∑ m = 0 [ r / 2 ] − 1 ∫ 2 m π r + π r 2 ( m + 1 ) π r − π r ( n + 1 ) + ∫ 2 ( m + 1 ) π r − π r ( n + 1 ) 2 ( m + 1 ) π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = 1 π ∑ m = 0 [ r / 2 ] − 1 ∫ π r ( n + 1 ) π r + ∫ 0 π r ( n + 1 ) φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 2 ( m + 1 ) π r − t d t .

Using (11) we obtain

(18) d d t D k , 1 2 ( m + 1 ) π r − t ≤ π ( 2 k + 1 ) 2 2 ( m + 1 ) π r − t ≤ π ( 2 k + 1 ) 2 t .

Moreover, applying equality (13) and the inequalities: (14),

sin r t 2 ≥ 2 ( m + 1 ) − r t π for t ∈ 2 m π r + π r , 2 ( m + 1 ) π r ,

where m ∈ 0 , 1 , … , r 2 − 1 , we obtain

d d t cos t 2 cos ( 2 k ± r ) t 2 8 sin r t 2 sin t 2 ≤ 2 k + r + 1 8 sin r t 2 sin t 2 + 2 r 8 sin 2 r t 2 sin t 2 ≪ k + 1 2 ( m + 1 ) − r t π t + 1 2 ( m + 1 ) − r t π 2 t ,

when 2 ( m + 1 ) π r − π r ( n + 1 ) ≤ t ≤ 2 ( m + 1 ) π r − π r .

Therefore,

(19) d d t cos 2 ( m + 1 ) π / r − t 2 cos ( 2 k ± r ) ( 2 ( m + 1 ) π / r − t ) 2 4 sin r ( 2 ( m + 1 ) π / r − t ) 2 sin 2 ( m + 1 ) π / r − t 2 ≪ k + 1 t 2 ( m + 1 ) π r − t + 1 t 2 2 ( m + 1 ) π r − t ≪ k + 1 t 2 + 1 t 3 ,

when π r ( n + 1 ) ≤ t ≤ π r .

Using (18), (19) instead of (12), (15) we can estimate the quantity ∣ I 2 ( x ) ∣ similar to the quantity ∣ I 1 ( x ) ∣ .

Collecting the partial estimates we obtain (3).

5.2 Proof of Theorem 2

As usual

T n , A 0 f ( x ) − f ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ 2 m π r 2 m π r + π r φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] − 1 ∫ 2 m π r + π r 2 ( m + 1 ) π r φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t = I 1 ∘ ( x ) + I 2 ∘ ( x )

for an odd r , and

T n , A 0 f ( x ) − f ( x ) = 1 π ∑ m = 0 [ r / 2 ] − 1 ∫ 2 m π r 2 m π r + π r + ∫ 2 m π r + π r 2 ( m + 1 ) π r φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t = I 1 ′ ∘ ( x ) + I 2 ∘ ( x )

for an even r , whence

∣ T n , A f ( x ) − f ( x ) ∣ ≤ ∣ I 1 ∘ ( x ) ∣ + ∣ I 1 ′ ∘ ( x ) ∣ + ∣ I 2 ∘ ( x ) ∣ .

We will consider in detail the term ∣ I 1 ∘ ( x ) ∣ . It is clear that

I 1 ∘ ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ 2 m π r 2 m π r + π r ( n + 1 ) + ∫ 2 m π r + π r ( n + 1 ) 2 m π r + π r φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t = J 1 ∘ ( x ) + J 2 ∘ ( x )

and

J 1 ∘ ( x ) = 1 π ∫ 0 π r ( n + 1 ) φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t + 1 π ∑ m = 1 [ r / 2 ] ∫ 2 m π r 2 m π r + π r ( n + 1 ) φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t = ∑ 0 ∘ ( x ) + ∑ m ∘ ( x ) .

We know that D k , 1 ( t ) decreases with respect to t ∈ 0 , π n + 1 for k = 0 , 1 , 2 , … , n . The second mean value theorem for integrals with some c ∈ 0 , π n + 1 (cf. [10]) and Lemma 1 give

∑ 0 ∘ ( x ) ≤ 1 π ∑ k = 0 n a n , k ∘ D k , 1 ( 0 ) ∫ 0 c φ x ( t ) d t + D k , 1 π r ( n + 1 ) ∫ c π r ( n + 1 ) φ x ( t ) d t ≤ 1 π ∑ k = 0 n a n , k ∘ ( k + 1 ) ∫ 0 c φ x ( u ) d u + ∫ 0 π r ( n + 1 ) φ x ( u ) d u − ∫ 0 c φ x ( u ) d u ≤ 3 n + 1 π sup c ∈ 0 , π r ( n + 1 ) ∫ 0 c φ x ( t ) d t = 3 r w ¯ x f π r ( n + 1 ) .

Analogously as in the proof of Theorem 1 using (12)

∑ m ∘ ( x ) ≤ 1 π ∑ m = 1 [ r / 2 ] ∑ k = 0 n a n , k ∘ ( k + 1 ) ∫ 0 π r ( n + 1 ) φ x ( u ) d u + 1 π ∑ m = 1 [ r / 2 ] ∑ k = 0 n a n , k ∘ ∫ 0 π r ( n + 1 ) ∫ 0 t φ x ( u ) d u d d t D k , 1 t + 2 m π r d t ≤ 1 π ∑ m = 1 [ r / 2 ] ( n + 1 ) ∫ 0 π r ( n + 1 ) φ x ( u ) d u + 1 π ∑ m = 1 [ r / 2 ] ∑ k = 0 n a n , k ∘ ∫ 0 π r ( n + 1 ) ∫ 0 t φ x ( u ) d u π ( 2 k + 1 ) 2 t + 2 m π r d t ≤ r 2 ( n + 1 ) π ∫ 0 π r ( n + 1 ) φ x ( u ) d u + r 2 m π 2 ∑ m = 1 [ r / 2 ] ∑ k = 0 n a n , k ∘ ( k + 1 ) ∫ 0 π r ( n + 1 ) ∫ 0 t φ x ( u ) d u d t ≤ 1 2 w ¯ x f π r ( n + 1 ) + ∑ m = 1 [ r / 2 ] r ( n + 1 ) 2 m π ∫ 0 π r ( n + 1 ) ∫ 0 t φ x ( u ) d u d t ≪ w ¯ x f π r ( n + 1 ) ,

whence

∣ J 1 ∘ ( x ) ∣ ≪ w ¯ x f π r ( n + 1 ) .

Furthermore, by Lemma 2, similar to the proof of Theorem 1

J 2 ∘ ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) ∑ k = 0 n a n , k ∘ D k , 1 ( t ) d t = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) 2 sin t 2 ∑ k = 0 n a n , k ∘ sin ( 2 k + 1 ) t 2 d t = − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 n ( a n , k ∘ − a n , k + r ∘ ) ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) cos t 2 2 sin t 2 D ˜ k , r ( t ) d t + − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∘ ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) cos t 2 2 sin t 2 D ˜ k , − r ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 n ( a n , k ∘ − a n , k + r ∘ ) ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) sin t 2 2 sin t 2 D k , r ( t ) d t + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∘ ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) sin t 2 2 sin t 2 D k , − r ( t ) d t = − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 n ( a n , k ∘ − a n , k + r ∘ ) Q 1 ( x ) + − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∘ Q 2 ( x ) + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 n ( a n , k ∘ − a n , k + r ∘ ) Q 3 ( x ) + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k ∘ Q 4 ( x ) .

Thus using (16), (17), and (10)

∣ J 2 ∘ ( x ) ∣ ≪ ∑ k = 0 n ∣ a n , k ∘ − a n , k + r ∘ ∣ ( k + 1 ) ∑ l = 0 n w ¯ x f π r ( l + 1 ) + ∑ k = 0 r − 1 a n , k ∘ ( k + 1 ) ∑ l = 0 n w ¯ x f π r ( l + 1 ) ≪ A ¯ n , r ∘ ∑ l = 0 n w ¯ x f π r ( l + 1 )

and therefore by (8)

∣ I 1 ∘ ( x ) ∣ ≪ A ¯ n , r ∘ ∑ l = 0 n w ¯ x f π r ( l + 1 ) .

Analogously, we can obtain the estimates of ∣ I 1 ′ ∘ ( x ) ∣ and ∣ I 2 ∘ ( x ) ∣ .

Collecting the partial estimates we obtain that (4) holds.

5.3 Proof of Theorems 3

Similar to the proof of Theorem 1 we have

∣ T n , A f ( x ) − f ( x ) ∣ ≤ ∣ I 1 ( x ) ∣ + ∣ I 1 ′ ( x ) ∣ + ∣ I 2 ( x ) ∣

and

I 1 ( x ) = J 1 ( x ) + J 2 ( x ) .

By (2),

∣ J 1 ( x ) ∣ ≤ 1 π ∑ m = 0 [ r / 2 ] ∫ 2 m π r 2 m π r + π r ( n + 1 ) ∣ φ x ( t ) ∣ ∑ k = 0 ∞ a n , k ∣ D k , 1 ( t ) ∣ d t ≤ 1 π ∑ k = 0 ∞ a n , k ( k + 1 ) ∫ 0 π r ( n + 1 ) ∣ φ x ( t ) ∣ d t ≪ W ¯ x f π r ( n + 1 ) .

Next, by Lemma 2, we obtain

J 2 ( x ) = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) ∑ k = 0 ∞ a n , k D k , 1 ( t ) d t = 1 π ∑ m = 0 [ r / 2 ] ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r φ x ( t ) 2 sin t 2 ∑ k = 0 ∞ a n , k sin ( 2 k + 1 ) t 2 d t = − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ( a n , k − a n , k + r ) Q 1 ( x ) + − 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k Q 2 ( x ) + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 ∞ ( a n , k − a n , k + r ) Q 3 ( x ) + 1 π ∑ m = 0 [ r / 2 ] ∑ k = 0 r − 1 a n , k Q 3 ( x ) .

Furthermore,

∣ Q 1 ( x ) ∣ ≤ ∫ π r ( n + 1 ) + 2 m π r π r + 2 m π r ∣ φ x ( t ) ∣ cos t 2 cos ( 2 k ± r ) t 2 4 sin r t 2 sin t 2 d t ≪ ∫ π r ( n + 1 ) π r ∣ φ x ( t ) ∣ d t t 2 ≪ ∑ l = 0 n W ¯ x f π r ( l + 1 )

and analogously

∣ Q i ( x ) ∣ ≪ ∑ l = 0 n W ¯ x f π r ( l + 1 ) when i = 2 , 3 , 4 .

Therefore,

∣ J 2 ( x ) ∣ ≤ A n , r ∑ l = 0 n W ¯ x f π r ( l + 1 )

and

∣ I 1 ( x ) ∣ ≪ A n , r ∑ l = 0 n W ¯ x f π r ( l + 1 ) .

Analogously we can obtain the estimates of ∣ I 1 ′ ( x ) ∣ and ∣ I 2 ( x ) ∣ .

Collecting the partial estimates we obtain that (7) holds. The proof of estimate (6) is analogous to the above. This ends our proof.

5.4 Proof of Theorem 4

Taking into account estimates (7) and (6) from Theorem 3 and using property (1) the desired result follows immediately.

6 Conclusion

We proved very general estimates of the deviations T n , A f ( x ) − f ( x ) for any matrices A defining the regular methods of summability. Taking into account the particular cases of A we obtain the best orders of pointwise approximation in different sets of points x of full measure as well as the best orders of normwise approximation. An application of r -differences of the entries of A in the characteristics of the measures of approximation improve and generalize the earlier known results of such type. It could be very interesting to examine the conjugate functions and their Fourier series as well as the strong means.

Funding information: Not applicable.
Author contributions: This study was carried out in collaboration with equal responsibility. All authors read and approved the final manuscript.
Conflict of interest: The authors declare that they have no competing interests.
Data availability statement: Not applicable.

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Received: 2021-09-26

Revised: 2022-01-23

Accepted: 2022-03-29

Published Online: 2022-05-20

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https://doi.org/10.1515/dema-2022-0009

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rate of approximation; summability of Fourier series

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