Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis 22 (2006), 41–61 www.emis.de/journals ISSN 1786-0091 UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES WITH RESPECT TO THE WALSH–PALEY SYSTEM V. TEVZADZE Abstract. In the present paper we prove a number of statements dealing with the uniform convergence of Cesàro means of negative order of the Fourier– Walsh series. 1. Definitions and Notation Let r0 (x) be the function defined by £ ¢ 1 if x ∈ 0, 12 , r0 (x) = −1 if x ∈ £ 1 , 1¢ , r0 (x + 1) = r0 (x). 2 The Rademacher system is defined by rn (x) = r0 (2n x), n ≥ 1, and x ∈ [0, 1). Let ψ0 (x), ψ1 (x), ψ2 (x), . . . represent the Walsh functions, i.e. ψ0 (x) = 1, and if k = 2n1 + 2n2 + · · · + 2ns is a positive integer with n1 > n2 > · · · > ns , then ψk (x) = rn1 (x) · rn2 (x) · · · rns (x). Denote by Knα (t) the kernel of the method (C, α) and call it the Cesàro kernel: Knα (t) = Aα k = n 1 X α An−ν ψν (t), Aα n ν=0 (α + 1)(α + 2) · · · (α + k) k! (α 6= −k). It is well-known ([19, Ch. 3]), that n X Aα−1 (I) Aα = n n−k ; k=0 (II) (III) α α−1 Aα ; n − An−1 = An α α An ∼ n . 2000 Mathematics Subject Classification. 42C19. Key words and phrases. Walsh system, Cesàro summability, negative order, uniform convergence, modulus of continuity, modulus of variation. 41 42 V. TEVZADZE By C(0, 1) we denote the space of the continuous periodic functions with period 1 and norm kf kC = max |f (x)|. 0≤x≤1 Let f ∈ C(0, 1); the modulus of continuity of function f is called the function ω(δ, f ) = max |f (x) − f (y)|, |x−y|≤δ x,y∈[0,1] 0 ≤ δ ≤ 1. A modulus of continuity is called the nonnegative function ω of the nonnegative argument possessing the following properties: (1) (2) (3) (4) ω(0) = 0; ω(δ) is nondecreasing; ω(δ) is continuous on [0, 1]; ω(δ1 + δ2 ) ≤ ω(δ1 ) + ω(δ2 ) for 0 ≤ δ1 ≤ δ2 ≤ δ1 + δ2 ≤ 1. Given the modulus of continuity ω(δ), by H ω we denote a set of all those functions f ∈ C(0, 1) for each of which ω(δ, f ) = O(ω(δ)) as δ → 0. If, however, α ω(δ) = δ α (0 < α ≤ 1), then by Lip α we denote a class H δ . Let φ be an increasing continuous function on [0, ∞), and φ(0) = 0. By Vφ it is denoted the class of bounded on [0, 1] functions f for which Vφ (f ) = sup Π n X φ (|f (xk ) − f (xk−1 )|) < ∞, k=1 where Π = {0 ≤ x0 < x1 < x1 , · · · < xn ≤ 1} is an arbitrary partitioning of the segment [0, 1]. Let M (0, 1) denote a class of bounded functions on [0, 1]. The modulus of variation of the function f ∈ M (0, 1) is called the function of an entire argument υ(n, f ) defined as follows: υ(0, f ) = 0, and for n ≥ 1 υ(n, f ) = sup Πn n−1 X |f (t2k+1 ) − f (t2k )| , k=0 where Πn is an arbitrary system of n nonintersecting intervals (t2k , t2k+1 ) (k = 0, 1, 2, . . . , n − 1) of the segment [0, 1]. The notion of modulus of variation has been introduced by Z. Chanturia [3]. If υ(n) is nondecreasing convex upwards function and υ(0) = 0, then υ(n) is called the modulus of variation. Given the modulus of variation υ(n), by V [υ] we denote the class of those functions which satisfy the relation υ(n, f ) = O(υ(n)) as n → ∞. Next, let f ∈ C(0, 1), and σ(f ) be the Fourier–Walsh–Paley series of that function, i.e. σ(f ) ∼ ∞ X k=0 Z1 ck ψk (x), where ck = f (t)ψk (t) dt, k = 0, 1, 2, . . . . 0 UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 43 By σnα (f, x) we denote Cesàro means, or (C, α) means of the Fourier–Walsh– Paley series of the function f , i.e. σnα (f, x) = n 1 X α An−ν cν ψν (t). Aα n ν=0 2. Introduction N. Fine [5] has proved that for any summable function f the means σnα (x, f ) converge almost everywhere for all α > 0, and for any continuous function σ(f ) is uniformly (C, 1) summable to f . In [18], S. Yano has studied the points (C, α) convergence of the summable function f , namely: if lim f (x) = A, then σ(f ) is (C, α) summable to A at the x→x0 point x0 (α > 0). Moreover, Yano has shown that if f (x) satisfies the Lipschitz condition of order α (0 < α < 1), then for every β > α |σnβ (f, x) − f (x)| = O(n−α ). This result has been somewhat amplified by V. Kokilashvili [12], who found that n β kσn−1 (f, x) − f (x)kC ≤ 1X Eν (f ), n ν=1 β ≥ 1, En (f ) is the best approximation of f (x) in the metric C(0, 1) with the help of polynomials by the Walsh system. The problems of summability of Cesàro means of positive order for Walsh–Fourier series were studied in [8]–[7]. The questions dealing with the uniform convergence of Cesàro means of negative order were first studied by the author, and the obtained results without proof were published in [17]. Important results in this direction have been obtained by Goginava in [9], [10]. 3. Main Results The main results of the paper are presented in the form of the following propositions. Theorem 1. Let ω(δ, f ) be the modulus of continuity, and let υ(n, f ) be the modulus of variation of the function f ∈ C(0, 1). If ) ( µ ¶X n m X 1 υ(k, f ) 1 ,f + = 0, 0 < α < 1, lim min ω n→∞ 1≤m≤n n k 1−α k 2−α k=1 k=m+1 then σ(f ) is uniformly (C, −α) summable to f . Theorem 1 can be rewritten in the following equivalent form. Theorem 2. Let ω(δ, f ) be the modulus of continuity, and let υ(n, f ) be the modulus of variation of the function f . Then 44 V. TEVZADZE ( µ ¶ mX 0 (n) 1 1 −α |σn (f, x) − f (x)| ≤ c(α) ω ,f n k 1−α k=1 n X + k=m0 (n)+1 υ(k, f ) − υ(k − 1, f ) k 1−α ) + o(1), where m0 (n) = m0 is defined by the inequality µ ¶ υ(m0 , f ) υ(m0 + 1, f ) 1 0 ≤ω ,f ≤ . (1 ) m0 + 1 n m0 If ω( n1 , f ) < υ(n,f ) n , then we take m0 = n. Let ω(δ) be an arbitrary modulus of continuity and υ(n) be an arbitrary modulus of variation. Furthermore, let (20 ) τ (n) = ω µ ¶ mX 0 (n) 1 1 + 1−α n k k=1 n X k=m0 (n)+1 υ(k) − υ(k − 1) . k 1−α where m0 (n) is defined by the relation (10 ) and omitting in it the function f . Suppose lim τ (n) = lim τ (ni ) = τ0 > 0. Then the following theorem is valid. n→∞ i→∞ Theorem 3. In the class H ω ∩ V (υ) there exists a function f0 , such that (30 ) lim |f0 (0) − σ2−α ni (f0 , 0)| > 0. τ (2ni ) 4. Auxiliary Results We shall need the following Lemma 1 ([16]). Let Knα (t) = n 1 X α An−ν ψν (t). Aα n ν=0 Then the estimate |Kn−α (t)| ≤ c(α) 1 1 , 1−α t A−α n t ∈ (0, 1), 0 < α < 1, holds. Lemma 2 ([16]). For any α ∈ (0, 1) and p ≥ 2m the equality Ã2m −1 ! X −α Sgn An−ν ψν (t) = Sgn(ψ2m −1 (t)), t ∈ [0, 1), ν=0 is valid. Lemma 3 ([3]). If f ∈ C ∩ Vφ , where φ(u) is strictly increasing for u ∈ [0, ∞) and φ(0) = 0, then µ ¶ 1 −1 , n ≥ 1. υ(n, f ) ≤ c(f )nφ n UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 45 Lemma 4. If f ∈ C ∩ Vφ , where φ satisfies the conditions of Lemma 3, then the following two conditions µ ¶ ∞ X 1 1 −1 (1) < ∞ (0 < α < 1) φ −α k k k=1 and Z1 (2) 0 1 dτ < ∞ φα (τ ) (0 < α < 1) are equivalent. Proof. We have (3) m X k=2 1 k 1−α φ −1 µ ¶ Zm µ ¶ µ ¶ m−1 X 1 1 1 1 1 −1 −1 ≤ φ dt ≤ φ . k t1−α t k 1−α k k=1 1 Since Zm 1 1 t1−α φ µ ¶ Z1 1 1 dt = u1−α φ−1 (u) 2 du t u 1 m Z1 = −1 φ−1 Z (1) 1 u1+α φ−1 (u) du = 1 m 1 φ−1 ( m ) ¯φ−1 (1) 1 τ ¯¯ 1 =− + α φα (τ ) ¯φ−1 ( 1 ) α (4) τ φ0 (τ ) dτ φ1+α (τ ) m ¯φ−1 (1) 1 τ ¯¯ 1 =− + α φα (τ ) ¯φ−1 ( 1 ) α m 1 = φ−1 α µ 1 m ¶ φ−1 Z (1) τ φ0 (τ ) dτ φ1+α (τ ) 1 ) φ−1 ( m φ−1 Z (1) τ dτ φα (τ ) 1 ) φ−1 ( m 1 1 mα − φ−1 (1) + α α φ−1 Z (1) 1 dτ, φα (τ ) 1 φ−1 ( m ) therefore if the condition (1) is fulfilled, then the condition (2) is likewise fulfilled; and if the condition (2) is fulfilled, then Z1 0 1 dτ ≥ α φ (τ ) 1 φ−1 ( m ) Z 0 dτ ≥ φ−1 α φ (τ ) µ 1 m ¶ mα and by virtue of (3) and (4) the condition (1) is fulfilled. ¤ Lemma 5. ([15]) Let f ∈ C(0, 1), then for every α ∈ (0, 1) the estimation ¯ Z1 2k−1 −1 ¯ X ¯ 1 ¯¯ −α An−ν ψν (u)[f (x+̇t) − f (x)]dt¯¯ −α ¯ An ν=0 0 where 2k ≤ n < 2k+1 , holds. ≤ c(α) k−1 X r=0 µ 2 r−k ω 1 ,f 2r ¶ , 46 V. TEVZADZE 5. Proof of Main Results Proof of Theorem 1. Represent n ≥ 1 in the form n = 2m + p, 0 ≤ p < 2m . As is known (see [4, p. 393]),1 Z1 σn−α (f, x) Kn−α (t)[f (x+̇t) − f (x)] dt − f (x) = 0 = 1 A−α n Z1 2m−1 X−1 1 + −α An (5) = 1 A−α n + ν=0 0 1 + −α An A−α n−ν ψν (t)[f (x+̇t) − f (x)] dt Z1 m 2X −1 m−1 0 ν=2 m Z1 2X +p 0 A−α n−ν ψν (t)[f (x+̇t) − f (x)] dt ν=2m Z1 2m−1 X−1 1 + −α An A−α n−ν ψν (t)[f (x+̇t) − f (x)] dt ν=0 0 1 A−α n A−α n−ν ψν (t)[f (x+̇t) − f (x)] dt Z1 2m−1 X−1 A−α n−2m−1 −ν ψν (t)[f (x+̇t) − f (x)] dt ν=0 0 Z1 X p A−α n−2m −ν ψ2m (t)ψν (t)[f (x+̇t) − f (x)] dt 0 ν=0 = A1 + A2 + A3 . Estimate A1 . By Lemma 5, we have ¯ ¯ Z1 2m−1 X−1 ¯ 1 ¯ −α ¯ |A1 | = ¯ −α An−ν ψν (t)[f (x+̇t) − f (x)] dt¯¯ ≤ A n ≤ c(α) ν=0 0 m−1 X µ 2ν−m ω ν=0 1 ,f 2ν ¶ , whence (6) A1 = o(1) as n → ∞. 1 x+̇y means that x, y is written as dyadic fractions which are combined pairwise with respect to [mod 2]. UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 47 Now we proceed to estimation A2 ; first we estimate the sum 2m−1 X−1 A−α n−2m−1 −ν ψν (t). ν=0 We have ¯ m−1 ¯ ¯2 X−1 ¯ ¯ ¯ −α ¯ An−2m−1 −ν ψν (t)¯¯ ¯ ¯ ν=0 ¯ ¯ ¯ m−1 m−1 ¯n−2 ¯ n−2 X ¯ X ¯ −α −α ¯ =¯ An−2m−1 −ν ψν (t) − An−2m−1 −ν ψν (t)¯¯ ¯ ν=0 ¯ ν=2m−1 ¯ p ¯ ¯ q ¯ ¯X ¯ ¯X ¯ ¯ ¯ ¯ ¯ ≤¯ A−α A−α p−ν ψν (t)¯ + ¯ q−ν ψν (t)¯ , ¯ ¯ ¯ ¯ ν=0 ν=0 where p = n − 2m−1 , q = n − 2m . This by virtue of Lemma 1 implies that ¯ m−1 ¯ ¯2 ¯ ¯ X−1 −α ¯ 1 ¯ An−2m−1 −ν ψν (t)¯¯ ≤ c(α) 1−α . ¯ t ¯ ν=0 ¯ (7) For A2 we have 1 |A2 | = −α An ¯ 1 m−1 ¯ ¯Z 2 ¯ X−1 ¯ ¯ ¡ ¢ −α ¯ An−2m−1 −ν ψ2m−1 (t)ψν (t) f (x+̇t) − f (x) dt¯¯ ¯ ¯ ¯ ν=0 0 2j+1 ¯ 2m−1 −1 µ Z2m 1 ¯¯ X = −α ¯ An j=0 2j 2m 2j+2 2m 2m−1 −1 Z − 2j+1 2m X 2m−1 X−1 ¡ ¢ A−α n−2m−1 −ν ψν (t) f (x+̇t) − f (x) dt ν=0 ¶¯ ¯ ¡ ¢ ¯. A−α ψ (t) f (x +̇t) − f (x) dt m−1 n−2 −ν ν ¯ ν=0 Taking into account the fact that the functions ψν (t) (ν = 0, 1, . . . , 2m−1 − 1) ¢ £ j m−1 , 2j+1 − 1, we find that if are constant in the intervals 2m−1 m−1 , j = 0, 1, . . . , 2 £ 2j 2j+1 ¢ ¢ ¡ ¡ 2j ¢ 1 t ∈ 2m , 2m , then ψν t + 2m = ψν (t) = ψν 2m , and hence 48 V. TEVZADZE 2j+1 ¯ 2m−1 −1 Z2m 1 ¯¯ X |A2 | = −α ¯ A n − j=0 2m−1 X−1 2m−1 X−1 ν=0 2j 2m A−α n−2m−1 −ν ψν j=0 2j+1 ¯ 2m−1 −1 Z2m 1 ¯¯ X = −α ¯ An j=0 ¡ ¢ A−α n−2m−1 −ν ψν (t) f (x+̇t) − f (x) µ ¶µ µ µ ¶¶ ¶ ¯ ¯ 1 1 t+ m f x+̇ t + m − f (x) dt¯¯ 2 2 2m−1 X−1 µ A−α n−2m−1 −ν ψν ¶ 2m−1 ν=0 2j 2m j µ µ µ ¶¶¶ ¯ ¯ 1 × f (x+̇t) − f x+̇ t + m dt¯¯ 2 1 (8) ¯ 2m−1 −1 Z2m 2m−1 µ ¶ X−1 1 ¯¯ X j −α = −α ¯ An−2m−1 −ν ψν 2m−1 An ν=0 j=0 0 µ µ µ ¶¶ µ µ ¶¶¶ ¯ ¯ 2j 2j + 1 × f x+̇ t + m − f x+̇ t + dt¯¯ m 2 2 1 m−1 ¯ 2m−1 −1 ¶ µ Z 2 X−1 1 ¯ X −m j = −α ¯¯ 2 A−α ψ m−1 ν n−2 −ν 2m−1 An ν=0 j=0 0 µ µ ¶ µ ¶¶ ¯ ¯ t + 2j t + 2j + 1 × f x+̇ m − f x+̇ dt¯¯ m 2 2 1 2m−1 −1 ¯ 2m−1 −1 ¶ µ Z X 1 ¯¯ X −m j −α ≤ −α ¯ An−2m−1 −ν ψν 2 2m−1 An ν=0 j=1 0 µ µ ¶ µ ¶¶ ¯ ¯ t + 2j t + 2j + 1 × f x+̇ m − f x+̇ dt¯¯ m 2 2 1 m−1 ¯ Z 2 X−1 ¶ µ ¶¶ ¯ µ µ ¯ ¯ 1 t t+1 −f x +̇ dt¯¯ + −α 2−m ¯¯ A−α f x +̇ m−1 n−2 −ν m m 2 2 An ν=0 0 = (1) A2 + (2) A2 . It can be seen easily that µ (2) A2 α −m ≤ c(α)n 2 (9) µ ≤ c(α)ω ω 1 ,f 2m 1 ,f 2m ¶ 2m−1 X−1 A−α n−2m−1 −ν ν=0 ¶ α −m m(1−α) n 2 2 µ ≤ c(α)ω 1 ,f 2m (1) Estimate now A2 . Applying (7), we have (1) A2 ¶1−α 2m−1 j j=1 0 ¯µ µ ¶ µ ¶¶¯ ¯ ¯ t + 2j t + 2j + 1 ¯ ¯ dt × ¯ f x+̇ m − f x+̇ ¯ m 2 2 α −m ≤ c(α)n 2 Z1 2m−1 X−1 µ ¶ . UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES Z1 2m−1 X−1 j 1−α j=1 0 ¯ µ µ ¶ ¶¯ ¯ ¯ t + 2j + 1 t + 2j ¯f x+̇ ¯ dt. − f x+̇ ¯ ¯ 2m 2m 1 ≤ c(α) 49 Estimate the sum A= 2m−1 X−1 1 j 1−α j=1 ¯ µ ¶ µ ¶¯ ¯ ¯ ¯f x+̇ t + 2j − f x+̇ t + 2j + 1 ¯ . ¯ ¯ m m 2 2 It is evident that for every t ∈ [0, 1) there exists a y(t) ∈ [0, 1), such that q m x+̇ t+q 2m = y +̇ 2m , q = 1, 2, . . . , 2 − 1. Thus ¯ ¶ µ ¶¯ 2m−1 X−1 1 ¯ µ ¯ ¯f y +̇ 2j − f y +̇ 2j + 1 ¯ . A= ¯ ¯ 1−α m m j 2 2 j=1 Using the Abelian transformation and taking into account (see [19, p. 378]) that |(x + h) − x| ≤ h, x, h ∈ [0, 1) and (see [3, p. 536]) υ(n, f ) ≤ c(f )nω( n1 , f ), we obtain ¯ µ ¶ µ ¶¯ s X 1 ¯¯ 2j + 1 ¯¯ 2j A= − f y +̇ + f y +̇ + ¯ j 1−α ¯ 2m 2m j=1 + 2m−1 X−1 j=s+1 µ ≤ω + 1 ,f 2m − j 1−α ¶X s j=1 2m−1 X−2 µ j=s+1 + 1 j 1−α 2 1 (s + 1)1−α 1 ,f 2m µ ω 1 2m−1 ,f ¶X s 1 (j + 1)1−α ¶X ¶ µ ¶¯ j ¯ µ ¯ ¯ ¯f y +̇ 2k − f y +̇ 2k + 1 ¯ ¯ ¯ m m 2 2 k=1 ¯ ¶ µ ¶¯ X−1 ¯ µ ¯ ¯f y +̇ 2j − f y +̇ 2j + 1 ¯ ¯ ¯ m m 2 2 j=1 ¶ µ ¶¯ s ¯ µ X ¯ ¯ ¯f y +̇ 2j − f y +̇ + 2j + 1 ¯ ¯ ¯ m m 2 2 j=1 ¶X s j=1 Since υ(n, f ) ≤ cnω therefore − m−1 (2m−1 )1−α µ 1 j 1−α 1 1 ≤ω ¯ µ ¶ µ ¶¯ ¯ ¯ ¯f y +̇ + 2j − f y +̇ + 2j + 1 ¯ ¯ ¯ m m 2 2 1 j 1−α ¡1 n, f ¢ + m−1 2X υ(j, f ) υ(2m−1 , f ) + m−1 1−α . j 2−α (2 ) j=s+1 and υ(n,f ) n ↓ 0 due to the convexity of υ(n, f ) [3], m−1 2X υ(j, f ) + 1−α j j 2−α j=s+1 j=1 m−1 s 2X m−1 m−1 X υ(2 , f ) υ(2 , f ) 1 1 ≥ c 2m−1 j 1−α 2m−1 j 1−α j=1 j=s+1 1 s ≥c υ(2m−1 , f ) X 1 υ(2m−1 , f ) ≥ c 2m−1 j 1−α (2m−1 )1−α j=1 50 V. TEVZADZE and hence µ (1) A2 ≤ c(α) ω 1 ,f 2m ¶X s j=1 1 + j 1−α 2m−1 X−1 j=s+1 υ(j, f ) . j 2−α It can be easily seen that the last relation is valid for all s, 1 ≤ s ≤ n. Thus (1) finally, for A2 we obtain the estimate ! à µ ¶X s n X 1 1 υ(k, f ) (1) ,f . (10) A2 ≤ c(α) ω + n k 1−α k 2−α k=1 k=s+1 Analogous estimate is obtained for A3 , à µ ! ¶X s n X 1 1 υ(k, f ) (11) |A3 | ≤ c(α) ω ,f + . n k 1−α k 2−α k=1 k=s+1 Taking into account (6), (8), (9), (10) and (11), from (5) we get (12) |σn−α (x, f ) − f (x)| à µ ≤ c(α) min 1≤m≤n ω 1 ,f n ¶X m k=1 1 k 1−α n X υ(k, f ) + k 2−α ! + o(1), k=m+1 where o(1) is the value tending to zero as n → ∞. This implies that Theorem 1 is valid. ¤ From Theorem 1 we can obtain a number of corollaries. Corollary 1. If ω(δ, f ) = O(δ α ) (0 < α < 1), then σ(f ) is uniformly (C, −α) summable to f . Corollary 2. If f ∈ C ∩ V [υ], and ∞ X υ(k) < ∞, k 2−α 0 < α < 1, k=1 then σ(f ) is uniformly (C, −α) summable to f . Corollary 3. If f ∈ C ∩Vφ , where φ(u) is strictly increasing convex for u ∈ [0, ∞), φ(0) = 0, and µ ¶ ∞ X 1 1 −1 φ < ∞, 0 < α < 1, k 1−α k k=1 then σ(f ) is uniformly (C, −α) summable to f . Indeed, in the conditions of Corollary 3, by Lemma 3, the relation µ ¶ 1 −1 u(n, f ) ≤ c(f ) nφ , n ≥ 1, n is valid, and hence nφ−1 ( n1 ) υ(n, f ) 1 ≤ c(f ) = c(f ) 1−α φ−1 n2−α n2−α n from which, by virtue of Corollary 2, follows Corollary 3. µ ¶ 1 n UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 51 Corollary 4. If f ∈ C ∩ Vφ , where φ satisfies the conditions of Corollary 3, and Z1 (13) 0 1 φα (τ ) dτ < ∞, 0 < α < 1, then σ(f ) is uniformly (C, −α) summable to f . Corollary 4 follows directly from Corollary 3 by using Lemma 4. Corollary 5. If f ∈ C(0, 1), m = min f (t), M = max f (t), and the Banach 0≤t≤1 0≤t≤1 indicatrix2 N (y, f ) satisfies the condition ZM N α (y, f ) dy < ∞, 0 < α < 1, m then σ(f ) is uniformly (C, −α) summable to f . This corollary follows from Theorem 1 by virtue of the results of [1]. Corollary 6. Let f ∈ C ∩Vφ , where φ(u) is a strictly increasing on [0, ∞) function, φ(0) = 0, and (13) is fulfilled, then 1 ω( n ,f ) Z |σn−α (f, x) − f (x)| ≤ c(α) 0 Vφ (f ) dτ + o(1), φα (τ ) where Vφ (f ) is a full φ variation of the function f on [0, 1]. Corollary 6 follows from Theorem 1 by virtue of the results obtained in [3]. Proof of Theorem 2. Let m0 < n such that ! à µ ¶X m n X 1 1 υ(k, f ) min ω ,f + 1≤m≤n n k 1−α k 2−α k=1 k=m+1 (14) µ ¶X m0 n X 1 1 υ(k, f ) =ω ,f + , n k 1−α k 2−α k=1 Then ¶X µ m0 1 1 ,f + ω 1−α n k k=1 n X k=m0 +1 k=m0 +1 υ(k, f ) ≤ω k 2−α µ 1 ,f n ¶ mX 0 −1 k=1 1 k 1−α + n X υ(k, f ) k 2−α k=m0 and µ ω 1 ,f n ¶X m0 k=1 1 k 1−α + n X k=m0 +1 υ(k, f ) ≤ω k 2−α µ 1 ,f n ¶ mX 0 +2 k=1 1 k 1−α + n X k=m0 +1 υ(k, f ) . k 2−α The above inequalities imply that µ ¶ µ ¶ υ(m0 , f ) υ(m0 + 1, f ) 1 1 1 1 ≤ , ω ,f ≤ ω , f , 2−α 2−α n (m + 1) n (m + 1)1−α m1−α m 0 0 0 0 2The Banach indicatrix N (y, f ) is a number (finite or infinite) of solutions of the equation f (x) = y. 52 V. TEVZADZE whence υ(m0 + 1, f ) ≤ω m0 + 1 (15) µ 1 ,f n ¶ ≤ υ(m0 , f ) . m0 ) Because of the fact that υ(n,f is strictly decreasing (see [3, p. 544]) starting n from some n0 , then for n ≥ n0 the m0 (n) from the relation (15) is defined uniquely. If, however, the minimum in the left-hand side of (14) is attained for m0 = n, we have one relation µ ¶ 1 υ(m0 , f ) ω ,f ≤ . n m0 Using now the Abelian transformation, we get n X k=m0 +1 υ(k, f ) − υ(k − 1, f ) = k 1−α + (16) k=m0 +1 k=m0 +1 1 k 1−α υ(n, f ) υ(m0 , f ) − ≥ c(α) 1−α n (m0 + 1)1−α whence n X µ n−1 X υ(k, f ) ≤ c(α) k 2−α − 1 (k + 1)1−α n−1 X k=m0 +1 ¶ υ(k, f ) υ(k, f ) υ(m0 , f ) − , 2−α k (m0 + 1)1−α à n X υ(k, f ) − υ(k − 1, f ) υ(m0 , f ) + 1−α 1−α k (m 0 + 1) n=m +1 ! , 0 and since υ(m0 , f ) υ(m0 + 1, f ) ≤ (m0 + 1)α ≤ 2ω 1−α (m0 + 1) m0 + 1 µ 1 ,f n ¶ mα 0, taking into account (12) and (16), we obtain kσn−α (f ) ¶X ½ µ m 1 1 ,f − f kC ≤ c(α) ω 1−α n k k=1 + n X k=m0 +1 υ(k, f ) − υ(k − 1, f ) k 1−α Thus Theorem 2 is proved. ¾ + o(1), ¤ To prove Theorem 3 we will need some lemmas. Lemma 6. If υ(n) = o(n1−α ), υ(n) → ∞, as n → ∞, and υ(n) is convex, then there exists the sequence of natural numbers {ϕ(n)} possessing the following properties: (a) ϕ(n) = o(n) as n → ∞; n X υ(k) − υ(k − 1) = o(1) as n → ∞. (b) k 1−α k=ϕ(n)+1 Proof. Suppose µ ¶ υ(m) − υ(m − 1) 1 ϕ(n) = max m : ≥ m1−α n Because υ(n) is convex, υ(k) − υ(k − 1) ↓, and since ϕ(n) ↑ ∞. υ(k)−υ(k−1) k1−α ↓ 0, therefore UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 53 From the definition of ϕ(n) it follows that υ(ϕ(n)) − υ(ϕ(n) − 1) 1 υ(ϕ(n) + 1) − υ(ϕ(n)) 1 ≥ and < . ϕ1−α (n) n (ϕ(n) + 1)1−α n (17) Therefore ϕ1−α (n) υ(ϕ(n)) ≤ υ(ϕ(n)) − υ(ϕ(n) − 1) ≤ , n ϕ(n) whence υ(ϕ(n)) ϕ(n) ≤ 1−α = o(1). n ϕ (n) By virtue of (17) we have n X υ(k) − υ(k − 1) ≤ (υ(ϕ(n) + 1) − υ(ϕ(n))) k 1−α k=ϕ(n)+1 ≤ c(α) (ϕ(n) + 1) n 1−α n X 1 k 1−α k=ϕ(n)+1 nα ≤ c(α) ϕ1−α (n) = o(1). n1−α Thus the lemma is proved. ¤ Lemma 7. Let m K2−α m (t) 2 1 X −α = −α A m ψν (t), A2m ν=0 2 −ν 0 < α < 1. There exists a natural number N such that for i < m − N the estimate 2i Z2m iα |K2−α m (t)| dt ≥ c(α)2 2i−1 2m is valid for sufficiently large m. Proof. We have 2i 2i Z2m |K2−α m (t)|dt ≥ 2i−1 2m 1 A−α 2m m ¯ Z2m ¯ 2X ¯ −1 −α ¯ ¯ A2m −ν (t)ψν (t)¯¯dt ¯ 2i−1 2m ν=0 2i − (18) 1 A−α 2m Z2m |A−α 0 (t)ψ2m (t)|dt 2i−1 2m 2i = 1 A2−α m m ¯ Z2m ¯ 2X ¯ −1 −α ¯ 1 2i−1 ¯ ¯ = A1 − A2 . A m −ν (t)ψν (t) dt − 2 ¯ ¯ m A−α 2m 2 ν=0 2i−1 2m Here we present lower bound of A1 . By Lemma 2, we have 2i 1 A1 = −α A2 m m ¯ Z2m ¯ 2X ¯ −1 −α ¯ ¯ A2m −ν (t)ψν (t)¯¯dt ¯ 2i−1 2m ν=0 54 V. TEVZADZE 2i = 1 A−α 2m m Z2m µ 2X −1 ¶ m (t)ψ (t)ψ (t) dt A−α ν 2 −1 2m −ν ν=0 2i−1 2m k+1 m i m ¶ 2X −1 Z2 µ 2X −1 1 −α = −α A2m −ν (t)ψν (t)ψ2m −1 (t) dt A2m k=2i−1 ν=0 k 2m = 1 A−α 2m m 2X −1 k+1 A−α 2m −ν m i µ 2X −1 Z2 ν=0 k=2i−1 k 2m µ 2X 2 −1 −1 ³ k ´ ³ k ´¶ 1 1 X −α m A ψ ψ , m ν 2 −1 2 −ν m 2m 2m A−α 2m 2 ν=0 k=2i−1 m = ¶ ψν (t)ψ2m −1 (t)dt i from which it follows that since (see [2, p. 17]) µ ¶ ³ ν ´ k , ν, k = 0, 1, 2, . . . , ψk m = ψν 2 2m and ([4, p. 379]) µ ¶ ³ ν ´ µ 2m − 1 ¶ ν 2m − 1 +̇ ψk = ψ ψk , k 2m 2m 2m 2m ν, k = 1, 2, . . . , 2m − 1, we get i 2m −1 2 −1 ³ ν ´ µ 2m − 1 ¶ 1 1 X −α X A1 = −α m Am ψk m ψk 2 2m A2m 2 ν=0 2 −ν k=2i−1 i µ ¶ 2m −1 2 −1 1 1 X −α X ν 2m − 1 = −α m Am +̇ m ψk 2m 2 A2m 2 ν=0 2 −ν i−1 k=2 = 1 A−α 2m ¶ µ ¶¸ · µ 2 −1 ν 2m − 1 1 X −α ν 2m − 1 Am +̇ m − D2i−1 +̇ m . D 2i 2m ν=0 2 −ν 2m 2 2m 2 m Since (see [4]) 2m , D2m (t) = 0, (19) 0 ≤ t < 2−m , 2−m ≤ t < 1, where Dn (t) = n−1 X ψk (t), k=0 therefore for ν < 2m−1 + · · · + 2m−i+1 we will have µ ¶ µ ¶ ν 2m − 1 ν 2m − 1 D2i +̇ m = 0, D2i−1 +̇ m =0 2m 2 2m 2 and hence A1 = µ µ ¶ ν 2m − 1 i D +̇ A−α 2 2m −ν 2m 2m ν=pi +qi +1 µ ¶¶ ν 2m − 1 − D2i−1 +̇ m 2m 2 µ µ ¶ µ ¶¶ ¾ pX i +qi ν 2m − 1 ν 2m − 1 −α + A2m −ν D2i +̇ m − D2i−1 +̇ m , 2m 2 2m 2 ν=p 1 1 m 2 A−α m 2 ½ m 2X −1 i UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 55 where pi = 2m−1 + 2m−2 + · · · + 2m−i+1 = 2m − 2m−i+1 , qi = 2m−i+1 + 2m−i+2 + · · · + 21 + 20 = 2m−i − 1. α−1 α Taking again into account (19) and equalities Aα , we obtain k − Ak−1 = Ak m 2m X −2m−i 2X −1 1 2i−1 −α A−α A1 = −α m A2m −ν − 2m −ν A2m 2 ν=2m −2m−i+1 ν=2m −2m−i m−i m−i 2 X−1 2X i−1 1 2 = −α m A−α A−α 2m−i −ν − 2m−i+1 −ν A2m 2 ν=0 ν=0 (20) −1 2i−1 = −α m A2m 2 ¾ + A−α 2m−i ½ 2m−i X−1 −α−1 −α−1 (A−α−1 2m−i −ν A2m−i +1−ν + · · · + A2m−i+1 −ν ) ν=0 1 2i−1 = − −α m A2m 2 m−i 2m−i X−1 2X ν=0 A−α−1 2m−i +k−ν k=0 i−1 1 2 − −α m A−α 2m−i = B1 + B2 . A2m 2 Estimate now B1 . By virtue of (III) we have (21) mα ≥ c(α)2 2m−i X−1 2i−1 2m |B1 | ≥ c(α)(2m )α ν=0 m−i 1−α (2 (2m−i − ν)−α ) ≥ c0 (α)2iα . For B2 we have |B2 | ≤ c(α)2mα (22) 2i 2i−1 ≤ c(α) . 2m 2m(1−α) Similar estimate is obtained for A2 . That is, (23) A2 ≤ c(α) 2i 2m(1−α) . Taking into account (20), (21), (22) and (23), from (18) we obtain 2i Z2m iα |K2−α − c1 (α) m (t)| dt ≥ c0 (α)2 2i 2m(1−α) . 2i−1 2m Since 2−(m−i)(1−α) → 0 as (m − i) → ∞, there exists a natural number N such that 2i c0 (α) iα c1 (α) m(1−α) < 2 for i < m − N, 2 2 and therefore 2i Z2m |K2−α m (t)| dt ≥ c0 (α) iα 2 , 2 i < m − N. 2i−1 2m Thus the lemma is proved. ¤ 56 V. TEVZADZE Proof of Theorem 3. Let τ (n) be defined by the relation (20 ), where m0 (n) for n > n0 is defined uniquely by inequality (15) by omitting the function f . Next, let {ni } ⊂ N such that lim τ (2ni ) = τ0 > 0. i→∞ Without restriction of generality we can assume (see [3, p. 1545]) that υ(n) is convex, and υ(n) ≤ cnω( n1 ). There can take place two cases: (1) υ(n) = o(n1−α ); (2) υ(n) 6= o(n1−α ). Let us consider the case υ(n) = o(n1−α ). Suppose that υ(n) → ∞ because τ (n) → 0 as n → ∞, otherwise. By Lemma 6, there exists the sequence of natural numbers {ϕ(n)} with the properties indicated in the lemma. Note that if υ(n) = o(n1−α ) as n → ∞, then (24) ω µ ¶ mX 0 (n) 1 1 → 0, 1−α n k n → ∞. k=1 Indeed, ω µ ¶ mX 0 (n) 1 υ(m0 (n)) α 1 ≤ c(α) m0 (n) 1−α n k m0 (n) k=1 = c(α) υ(m0 (n)) = o(1), m1−α (n) 0 n → ∞. Therefore by virtue of the fact that lim τ (2ni ) = τ0 > 0, starting from some i0 ϕ(2ni ) > m0 (2ni ). Taking into account (24) and also the properties of the sequence {ϕ(n)}, we find that (2ni ) 2n i µ 1 ¶ m0X X 1 υ(k) − υ(k − 1) + lim ω i→∞ 2ni k 1−α k 1−α n k=1 µ = lim ω i→∞ 1 2ni ¶ m0X (2 ni ) + lim i→∞ ni 2 X + lim i→∞ k=ϕ(2ni )+1 ϕ(2ni ) X = lim i→∞ i )+1 ϕ(2ni ) 1 k 1−α k=1 (25) k=m0 (2 k=m0 (2ni )+1 X k=m0 (2ni )+1 υ(k) − υ(k − 1) k 1−α υ(k) − υ(k − 1) k 1−α υ(k) − υ(k − 1) = τ0 . k 1−α Suppose ξ(n) = 2q(n)+4 X k=2r(n)+3 υ(k) − υ(k − 1) , k 1−α where r(n) = [log2 m0 (n)], q(n) = [log2 ϕ(n)]. Now we construct the sequence of natural numbers {`k } and the sequence of functions {fk }. Let `1 ∈ {ni } such that `1 > ni0 , and `1 > n0 (see the definition of n0 and i0 ), `1 ) ≤ 1. 2`1 − ϕ(2`1 ) > N (N appears in Lemma 7) and 4ϕ(2 2` 1 UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES 57 The function ϕ1 (x) is defined as follows: `1 υ(r + 1) − υ(r) for x = 22r+1 ), . . . , 4ϕ(2`1 ) − 1, `1 +2 , r = m0 (2 0 for x = 2`1r+1 , r = m0 (2`1 ), . . . , 4ϕ(2`1 ), ϕ1 (x) = £ ¤ £ `1 ¤ `1 ) ) 0 for x ∈ 0, m02(2 ,1 ∪ 4ϕ(2 `1 `1 2 is linear and continuous for the rest x from [0, 1]. Assume f1 (x) = ϕ1 (x) Sgn K2−α `1 (x). Let the numbers `1 , `2 , . . . , `k−1 and periodic with period 1 functions f1 , f2 , . . . , fk−1 be already constructed. Then `k and fk can be constructed as follows: we choose `k in such a way that the following conditions be fulfilled: (1) `k > `k−1 ; (2) `k ∈ {ni }; `k `k −1 ) ) (3) 4ϕ(2 < m02(2 (ϕ(n) = o(n)); ` `k −1 ¡2 k1 ¢ (4) ω 2`k ≤ υ(4ϕ(2`k −1 )) − υ(4ϕ(2`k −1 ) − 1); (5) m0 (2`k ) > ϕ(2`k −1 ); ¡ ¢ `k−1 ) ≤ c1 (α)ξ(`k−1 ); (6) ω 21`k mα 0 (2 (7) k−1 P ω i=1 ¡ 1 2`i ¢ 1 ω( `i ϕ(2 ) R2`k 0 ) 1 τα dτ < c2 (α)ξ(2`k ) (the constants c1 (α) and c2 (α) will be chosen below). Let 2r+1 ` ` υ(r + 1) − υ(r) for x = 2`k +2 , r = m0 (2 k ), . . . , 4ϕ(2 k ) − 1, 0 for x = r , r = m (2`k ), . . . , 4ϕ(2`k ), 0 2`k +1 ϕk (x) = £ m0 (2`k ) ¤ £ 4ϕ(2`k ) ¤ 0 for x ∈ 0, 2`k ∪ ,1 2` k is linear and continuous for the rest x from [0, 1]. Assume fk (x) = ϕk (x) Sgn K2−α `k (x). Let now f0 (x) = ∞ X fk (x) k=1 Reasoning just as in [3, p. 548], we can show that f0 ∈ H ω ∩ V [υ]. It remains to prove the relation (30 ). Suppose ∞ X fi (x). Fk (x) = i=k+1 Taking into account the definition of the function fk (x) and Lemma 1, we obtain ¯ 1 ¯ ¯Z ¯ ¯ ¯ −α ¯ ¯ |σ2−α (F , 0) − F (0)| = F (t)K (t)dt k k k `k ¯ ¯ 2`k ¯ ¯ 0 58 V. TEVZADZE ≤ kFk kC 1 A−α 2` k m0 (2`k ) 2`k Z µ 1 dt ≤ c(α) ω t1−α 0 ¶ ¶α m0 (2`k ) 2`k +1 2`k ¶ µ 1 `k ≤ c0 (α)ω mα 0 (2 ). 2`k +1 1 µ 2`k α By virtue of the property (6) of the sequence {`k }, we get `k |σ2−α ), `k (Fk , 0)| ≤ c0 (α)c1 (α)ξ(2 (26) Further, using Lemma 7 and the definition of fk (x), we can write ¯ ¯ ¯ −α |σ2`k (fk , 0) − fk (0)| = ¯ ¯ 4ϕ(2`k ) 2`k ¯ ¯ Z ¯ fk (t)K2−α `k (t)dt¯ m0 (2`k ) 2` k `k )+2 2q(2 2` k 4ϕ(2`k ) 2`k Z Z ϕk (t)|K2−α `k (t)|dt ≥ = m0 (2`k ) 2`k (27) ` 2r(2 k )+1 2`k q(2`k )+1 ≥ ϕk (t)|K2−α `k (t)|dt X 2i+1 2`k Z ϕk (t)|K2−α `k (t)|dt i=r(2`k )+1 2i 2`k q(2`k )+1 X ≥ c(α) [υ(2i+2 + 1) − υ(2i+2 )]2iα i=r(2`k )+1 `k )+4 2q(2 X ≥ c3 (α) i=2r(2 `k )+3 υ(i) − υ(i − 1) = c3 (α)ξ(2`k ) i1−α +1 since `k )+4 2q(2 X `k )+3 i=2r(2 υ(i) − υ(i − 1) ≤ c(α) i1−α +1 `k )+4 2q(2 X q(2`k )+1 = c(α) `k )+3 i=2r(2 X i+3 2X i=r(2`k )+1 j=2i+2 υ(i + 1) − υ(i) i1−α υ(j + 1) − υ(j) j 1−α q(2`k )+1 X ≤ c(α) [υ(2i+2 + 1) − υ(2i+2 )]2iα . i=r(2`k )+1 Estimate now σ2−α `k (gk , 0), where gk (t) = k P i=1 fi (t). Since gk (t) ∈ H ω ∩ V , using Corollary 6, we find that ω( 1 Z2`k |σ2−α `k (gk , 0)| ≤ c(α) 0 ) V (gk ) dτ + o(1). τα UNIFORM (C, α) (−1 < α < 0) SUMMABILITY OF FOURIER SERIES It is easy to see that V (gk ) ≤ 2 k−1 X µ 1 2`i ω i=1 59 ¶ ϕ(2`i ), and therefore |σ2−α `k (gk , 0)| ≤ c1 (α) k−1 X i=1 µ ω 1 2`i ω( ¶ 1 Z2`k ϕ(2`i ) 0 ) 1 dτ + o(1). τα Taking into account the property (7) of the sequence {ϕ(n)}, we have (28) `k |σ2−α ) + o(1). `k (gk , 0)| ≤ c4 (α)c2 (α)ξ(2 It follows from (26), (27) and (28) that −α −α −α |σ2−α `k (f0 , 0) − f0 (0)| = |σ `k (fk , 0) + σ `k (gk , 0) + σ `k (Fk , 0)| 2 2 2 −α −α ≥ |σ2−α `k (fk , 0)| − |σ2`k (Fk , 0)| − |σ2`k (gk , 0)| ≥ c3 (α)ξ(2`k ) − x0 (α)c1 (α)ξ(2`k ) − c4 (α)c2 (α)ξ(2`k ) − o(1). Choosing now c1 (α) = c3 (α) 3c0 (α) , c2 (α) = c3 (α) 3c4 (α) , |σ2−α `k (f0 , 0) − f0 (0)| ≥ whence we get c3 (α)ξ(2`k ) − o(1), 3 |σ2−α c3 (α) o(1) `k (f0 , 0) − f0 (0)| ≥ − ` k ξ(2 ) 3 ξ(2`k ) and since ξ(2`k ) ∼ τ (2`k ) by virtue of (25), from the last relation we obtain (30 ). Consider now case 2), i.e. υ(n) 6= o(n1−α ). We define the function υ1 (x) as follows: υ(n) for x = n, n ∈ N, υ1 (x) = 0 for x = 0, is linear and continuous for the rest x from [0, ∞). Let ω1 (δ) = δυ1 µ ¶ 1 . δ It is easily seen that ω1 (δ) is the modulus of continuity, and since υ(n) ≤ cnω( n1 ), therefore ω1 (δ) ≤ cω(δ). By S. Stechkin’s lemma, there exists the convex modulus of continuity ω0 (δ) satisfying the condition c1 ω0 (δ) < ω1 (δ) < c2 ω0 (δ), and since υ1 (n) = υ(n), we have H ω0 ⊂ H ω ∩ V [υ]. By virtue of the fact that υ(n) 6= o(n1−α ), it follows that ω0 (δ) 6= o(δ α ) (0 < α < 1). Let us now show that in the class H ω0 there exists the function f0 whose Fourier–Walsh–Paley series fails to be summable by the method (C, −α) to f0 at the point x = 0. 60 V. TEVZADZE By the condition ω0 (δ) 6= o(δ α ), there exists the sequence {ni } ⊂ N, such that µ ¶ 1 ω 2ni α ≥ c > 0, i = 1, 2, . . . . 2ni We choose from {ni } the sequence with the following properties: ∞ ¡ ¢ ¡ ¢ P ω 21`i ≤ 2ω 21`k ; 1) i=k+1 ³ ´ ¡ ¢ ¡ ¢ 1 2) 2`k+1 ω 2`k+1 > 2 · 2`k ω 21`k since 2`k ω 21`k → ∞; ³ k−1 ¡ ¢ P ¡ 1 ¢ `i ´α ¡ 1 ¢ 3) ω 2`i 2 ω 2`k ≤ c0 (α)ω 21`k 2`k α (c0 (α) depends on our choice). Let i=1 ¡ ¢ `k +1 for x = 22r+1 − 1, ω 1 `k +2 , r = 0, 1, , . . . , 2 2`k r ` +1 ϕk (t) = 0 for x = `k +1 , r = 0, 1, , . . . , 2 k , 2 is linear and continuous for the rest x from [0, 1]. Suppose f0 (x) = ∞ X ϕk (t) Sgn K2−α `k (t). k=1 Reasoning analogously as in case 1), we can show that f0 is the unknown function. Thus Theorem 3 is proved. ¤ From Theorems 1 and 3 follows Theorem 4. 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