ALMOST EVERYWHERE CONVERGENCE OF FEJÉR MEANS OF SOME SUBSEQUENCES OF FOURIER SERIES FOR INTEGRABLE FUNCTIONS WITH RESPECT TO THE KACZMARZ SYSTEM

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ADV MATH SCI JOURAL Advances in Mathematics: Scientic Journal 4 (205), no., 6577 ISS 857-8365 UDC: 57.58.

1 ADV MATH SCI JOURAL Advances in Mathematics: Scientic Journal 4 (205), no., 6577 ISS UDC: ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS OF SOME SUBSEQUECES OF FOURIER SERIES FOR ITEGRABLE FUCTIOS WITH RESPECT TO THE KACMAR SSTEM ACIMA MEMI Abstract. The rst result of this work is about almost everywhere convergence of Fejér means for some subsequences of partial sums of Fourier series of integrable functions with respect to the Kaczmarz system. The second result establishes almost everywhere convergence of subsequences of Fejér means for some specic families of subsequences.. Introduction For the Walsh-Paley system it is known that the sequence of Fejér means converges almost everywhere for every integrable function, see [3]. The question on the degree to which elements from the partial sums of Fourier series can be removed so that means of the remaining subsequence still converge almost everywhere was considered in [7]. In [2, Theorem ] G.Gát proves that the mean sequence of every lacunary sequence of partial sums of Fourier series for integrable functions with respect to the Walsh-Paley system is almost everywhere convergent to the function itself. The main point was about the level of condensation of the elements of the subsequence since this same question was mentioned for the trigonometric system, see []. amely, [2, Theorem ] gives the convergence result for less dense sequences than the sequence considered in []. The methods used in this work provide some answers on a.e. convergence with respect to the Kaczmarz system for some specic subsequences, where the question is more about structure and homogeneity rather than density of elements. In the sequel it would be interesting to study convergence of means for larger families of subsequences of Fourier series. Let 2 denote the discrete cyclic group 2 = f0; g, where the group operation is addition modulo 2. If jej denotes the measure of the subset E 2, then we have jf0gj = jfgj = : Mathematics Subject Classication. 42C0. Key words and phrases. Walsh-kaczmarz system, partial sums of Fourier series, weakly bounded operator. 65

2 66. MEMI The dyadic group G is obtained from G = Q 2, see[5], where topology and measure are obtained from the product. In this paper the notation jej is used for both of the probability measure for subsets of the dyadic group G and for the discrete measure of nite subsets of natural numbers. Let x = (x n ) n0 2 G. The sets I n (x) := fy 2 G : y 0 = x 0 ; :::; y n = x n g, n and I 0 (x) := G are dyadic intervals of G. Let I n = I n (0), and e n := ( in ) i. It is easily seen that (I n ) n is a decreasing sequence of subgroups. P Since every nonnegative integer i can be written in the form i = i k2 k, we dene the sequence (z i ) i0 by z i = i k e k : It is easily seen that for each positive integer n, the set fz i ; i < 2 n g is a set of representatives of I n -cosets. The Walsh-Paley system is dened as the set of Walsh-Paley functions: where n = P! n (x) = n k 2 k and r k (x) = ( n (x) := r jnj (x) for n ; 0 (x) := and jnj = max k. For every natural number n = n k 6=0 jnj P (r k (x)) nk ; n 2 ; x 2 G; ) xk. The n-th Walsh-Kaczmarz function is (r k (x)) nk ; n k 2 k, we dene the corresponding numbers jnj s n (s) = n k 2 k ; n (s) = n k 2 k : k=s We denote the Dirichlet kernel function by: D n := n n! k ; Dn := k ; n : For every integrable function f the maximal function f is dened by f (x) = sup jd 2 n f(x)j; x 2 G : n0 In [4] it was noticed that the Dirichlet kernel function with respect to the Kaczmarz system can be written in the form D n (x) = D 2 jnj(x) + r jnj(x)d n 2 jnj( jnj (x)) : The notation C is used for independent positive constant which may vary in dierent contexts.

3 ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS Main results Lemma 2.. For every positive integers s; n; k such that s < 2 k < 2 n, and every integrable function f we have (k(! s (D 2 n k n )) fk kfk : Proof. Let s; k and n be xed. Put F = (! s (D 2 n k n )) f. Since this convolution only depends on the mean values S 2 nf of the function f, we get We have It follows and we get F (x) = 2 n (! s (D 2 n k n )) (S 2 nf(z i ) In (z i ))(x) : (! s (D 2 n k n )) In (z i )(x) = 2 n k k(! s (D 2 n proving the lemma. k n )) In(zi) k = 2 k kf k 2 n I n (x+z i )\f n (t)2i n! s (t)dt k g = 2 k! s (x + z i ) fn(t)2in kg(x + z i ) : f n(x+zi)2in 2 n js 2 nf(z i )j kfk ; dx = 2 n ; kg Lemma 2.2. Let k and s be positive integers such that s < 2 k. Let ("(n)) n be a sequence where every term is either 0 or : Then, the operator L k dened on L by L k f(x) = sup is of weak type (L ; L ). jfn : k + n ; "(n) = gj j n=k+ "(n)(r n! s ((D 2 n k n )) f(x)j ; Proof. Let > kfk. There exists some 0 depending on f and such that jfl k f(x) > gj 2jf sup 0 jfn : k + n ; "(n) = gj j n=k+ "(n)(r n! s ((D 2 n k n )) f(x)j > gj : Then, we choose F = (! s (D 2 0 k 0 )) f. otice that for every n : k < n < 0, r n D 2 n (! s ((D 2 0 k 0 ))(x) = 2 n+ 0 k r n (x) = 2 n+ 0 k r n (x)! s (x) I n (x)\f 0 (t)2i 0 I n(x)\f0 (t)2i 0 r n (t)! s (t)dt kg dt kg = 2 n k r n (x)! s (x) fn (t)2i n k g(x) = r n (x)! s (x)(d 2 n k n )(x) :

4 68. MEMI Since = jfn : k + n ; "(n) = gj j jfn : k + n ; "(n) = gj j n=k+ n=k+ sup "(n)j(d 2 n+ D 2 n) F (x)j 2F (x) : n "(n)(r n D 2 n F )(x)j "(n)(d 2 n+ D 2 n) F (x)j Applying Lemma 2. we have jf sup 0 jfn : k + n ; "(n) = gj j n=k+ = jf sup 0 jfn : k + n ; "(n) = gj j "(n)(r n! s ((D 2 n n=k+ k n )) f(x)j > gj "(n)(r n D 2 n F )(x)j > gj jf2f (x) > gj C kf k C kfk ; where the constant C > 0 is independent on 0 ; k; f or. Theorem 2.. Let f 2 L and (k n ) n be a xed sequence of positive integers. Suppose that ((n)) n is an increasing sequence of positive integers satisfying (n + ) = 2 k n (n), for every n m, where m is some xed positive integer. Then, almost everywhere on G. n= D (n) f! f; Proof. Following Gát in [2], it suces to prove that the operator is of weak type (L ; L ). Putting D f in the form (n) since then it suces to estimate sup j n= D (n) fj; D (n) f = D 2 j(n)j f + r j(n)j(d (n) 2 j(n)j j(n)j ) f ; sup j sup j n= n= D 2 j(n)j fj f ; r j(n)j (D (n) 2 j(n)j j(n)j ) fj :

5 We have ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS sup j sup j m n= + sup j >m sup j m + sup j >m + sup j >m n= r j(n)j (D (n) 2 j(n)j j(n)j ) fj r j(n)j (D (n) 2 j(n)j j(n)j ) fj + n= n= r j(n)j (D (n) 2 j(n)j j(n)j ) fj r j(n)j (D (n) 2 j(n)j j(n)j ) fj + m n= n=m+ r j(n)j (D (n) 2 j(n)j j(n)j ) fj + r j(n)j (D (n) 2 j(n)j j(n)j ) fj : Since the rst two terms are obviously bounded, we only estimate the third term. For every n > m we have (n) = 2 kn+:::+km+ (m); j(n)j = k n + : : : + k m+ + j(m)j; and (n) 2 j(n)j = 2 k n+:::+k m+ ((m) 2 j(m)j ) : Suppose that (m) = P j(m)j m s 2 s, where m s 2 f0; g, for every s. Then we have (n) 2 j(n)j = which gives from [6, Theorem ] D (n) 2 j(n)j = j(m)j j(m)j m s D 2 s+j(n)j The result is true if we prove that the operator sup j >m n=m+ (r j(n)j (D 2 s+j(n)j j(m)j m s 2 s+kn+:::+km+ ; j(m)j j(m)j k=s+ j(m)j k=s+ r m k k+j(n)j r mk k+j(n)j j(m)j : j(m)j j(n)j)) fj is of weak type (L ; L ) for every s. This is true and follows from Lemma 2.2 since for every k : s + k j(m)j, we have r k+j(n)j j(m)j j(n)j = r j(m)j k : Lemma 2.3. Let m; n be natural numbers where jnj > jmj. Suppose that m = 2 s n (s), where s jnj. Then,

6 70. MEMI S2 jmjd n 2 jnj( jnj (x)) = D m 2 jmj( jmj (x)); n = n (s) ; S 2 jmjd n 2 jnj( jnj (x)) = D m 2 jmj( jmj (x)) +! m 2 jmj( jmj (x)); n > n (s). Proof. Let n = Pjnj n k 2 k and m = jmj P m k = n k+s. From the proof of [6, Theorem ] it can be seen that D n 2 jnj = m k 2 k. Then, for every k 2 f0; : : : ; jmjg we have i=k+ Hence, it suces to prove the following statements: () For every k s, if n k 6= 0 we have i=k+ i ) jnj ] = (D 2 k s i : jmj i=k s+ r m i i ) jmj : (2) For every k s satisfying the property n i = 0 for each i 2 fk + ; : : : ; s g if this set is not empty, we have i=k+ i ) jnj ] =! m 2 jmj jmj : (3) If k s 2 is so that there exists some i 2 fk + ; : : : ; s g where n i =, then In this way since if n (s) 6= 0 we have: S 2 jmj[( and if n (s) = 0 : Also we have S 2 jmj[( k=s s i=k+ i=k+ i ) jnj ] = 0: S 2 jmj[d n 2 jnj jnj ] = S 2 jmj[( i=k+ S 2 jmj[( +( k=s s i ) jnj ] =! m 2 jmj jmj ; s i ) jnj ] = ( = ( i=k+ k=s jmj i ) jnj ] = 0 : m k D 2 k s jmj i=k s+ jmj i=k+ i=k+ i=k+ i ) jnj i ) jnj ] ; r m i i ) jmj r m i i ) jmj = D m 2 jmj jmj ;

7 ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS... 7 and then, the result will follow. ow we prove the statements (),(2) and (3). Let x 2 G be xed and arbitrary. () Let k s and n k 6= 0 : Then we have i=k+ i ) jnj ](x) = 2 jmj+k = 2 jmj+k jnj I jmj (x)\f jnj (t)2i k g i=k+ i=k+ = 2 jmj+k jnj i=k+ = 2 jmj+k jnj jmj = (D 2 k s i=k+ jmj i=k s+ i ( jnj (t))dt i ( jnj (x)) fjmj (t)2i k s g(x) i s ( jmj(x)) fjmj (t)2i k s g(x) s r mi i ( jmj (x)) fjmj (t)2i k sg(x) r m i i )( jmj (x)) : (2) Let k s ; and n k = ; n i = 0 for each i 2 fk + ; : : : ; s g. In this case i ( jnj (t)) = i ( jnj (x)) ; whenever i k + and t 2 I jmj (x). Therefore, i=k+ i ) jnj ](x) = 2 jmj+k = 2 jmj+k = = i=s jmj I jmj (x)\f jnj (t)2i k g i=k+ i=k+ i ( jnj (x)) i ( jnj (x)) i ( jnj (t))dt I jmj (x)\f jnj (t)2i k g r mi i ( jmj (x)) =! m 2 jmj( jmj (x)) :

8 72. MEMI (3) Suppose that k s 2 is so that there exists some j 2 fk + ; : : : ; s g where n j = : Then we have: i=k+ This gives that i ) jnj ](x) = 2 jmj+k i ( jnj (t))dt I jmj (x)\f jnj (t)2i kg i=k+ = 2 jmj+k [I jmj (x)\f jnj (t)2i k g]+e jnj j i=k+ = 2 jmj+k I jmj (x)\f jnj (t)2i k g i=k+ = 2 jmj+k i I jmj (x)\f jnj (t)2i k g i=k+ = 2 jmj+k i=k+ I jmj (x)\f jnj (t)2i k g i=k+ i ) jnj ](x) = 0 : Lemma 2.4. Let m; n 2 be such that jmj < jnj. Then: i ( jnj (t))dt i ( jnj (t + e jnj j ))dt ( jnj (t)) i (e j )dt i ( jnj (t))dt : D 2m jnj+ (x) D m jnj (x) = r jnj (x)d m jnj (x); 8x 2 G : Proof. Let m = P jmj m i2 i : Then we have We get and D m = D 2m = 2m = jmj jmj jmj+ i= m i D 2 i m i D 2 i+ m i 2 i : jmj k=i+ jmj k=i+ r mk k ; r m k k+ : Suppose that x jnj = for some x 2 G. Then, for every k 0, D 2 k+ jnj+ (x) = 0 : Therefore, D 2m jnj+ (x) = 0; which gives D 2m jnj+ (x) D m jnj (x) = D m jnj (x) = r jnj (x)d m jnj (x) : For x jnj = 0, we have D 2 k+ jnj+ (x) = 2D 2 k jnj (x); 8k 0 :

9 ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS Besides, for every i it holds r i+ jnj+ (x) = r i jnj (x), which implies: We easily get D 2m jnj+ (x) = 2( jmj m i D 2 i jmj k=i+ r m k k ) jnj(x) = 2D m jnj (x) : D 2m jnj+ (x) D m jnj (x) = D m jnj (x) = r jnj (x)d m jnj (x) : Theorem 2.2. Let f be an integrable function. Let ((n)) n be an increasing sequence of numbers satisfying the following conditions : () If fj(k)j; k 2 g \ [m; n] = fm; ng; where m and n are natural numbers then the number jfk:j(k)j=ngj jfk:j(k)j=mgj denoted by b m is a natural number. Moreover, if k is so that j(k)j = m, then there are exactly b m elements from the set fk 0 : j(k 0 )j = ng, satisfying k = 2 jmj jnj k 0(jnj jmj), (2) jfk : j(k)j = ngj n, for some xed > 0. (3) For every k 2 ; Then, almost everywhere. k n:j(n)j=k jfn : j(n)j j()gj ((n)) i = O(k): n:j(n)j D n f! f;! ; Proof. Following the rst steps in Theorem 2. it suces to prove that the operator sup jfn : j(n)j j( )jgj j n:j(n)j is of weak type (L ; L ). From the property (2) we can easily deduce that because, jfn : j(n)j j( )j l r j(n)j (D (n) l j( )j + ; 2 j(n)j j(n)j ) fj gj < jfn : j(n)j j( )gj: Let > kfk. There exists a natural number such that jfsup j jfn : j(n)j j()gj n:j(n)j r j(n)j (D (n) 2 j(n)j j(n)j ) fj > gj < 2jf sup j r j(n)j (D jfn : j(n)j j( )gj (n) 2 j(n)j j(n)j ) fj > gj: n:j(n)j

10 74. MEMI Of course, the number depends on f and. For every s 2 f0; : : : ; [log 2 ]g; we dene the function F s = 2s Writing D (n) 2 j(n)j in the form D (n) 2 j(n)j = n:j(n)j=j([ 2s ])j j(n)j (D (n) 2 j(n)j j(n)j ) f: ((n)) i D 2 i j(n)j k=i+ r ((n)) k k ; we obtain that n:j(n)j=j([ 2s ])j D (n) 2 j(n)j j(n)j = j([ 2s ]))j j(n)j k=i+ n:j(n)j=j([ 2s ])j (r ((n)) k k j(n)j ) : ((n)) i (D 2 i j(n)j ) Applying Lemma 2. and the property (3) mentioned in this theorem, we get kf s k 2s j([ 2s ]))j n:j(n)j=j([ 2s ])j ((n)) i kfk = O(2 s j([ 2 s ])j )kfk j([ 2 s ])j kfk : Let n; k be so that j(n)j = j([ 2 s ])j, and (k) = 2 j(k)j j(n)j ((n)) (j(n)j j(k)j). We have (2.) r j(k)j D 2 j(k)j (D (n) 2 j(n)j j([ 2s ])j ) = S 2 j(k)j+ S 2 j(k)j(d (n) 2 j(n)j j([ 2s ])j ) : Then for ((n)) j(n)j j(k)j = 0 ; we have ((n)) (j(n)j j(k)j ) = ((n)) (j(n)j j(k)j) = 2 j(n)j j(k)j 2(k) ; and from Lemma 2.3 we have: (2.2) S 2 j(k)j+(d (n) 2 j(n)j j([ 2s ])j ) = = D 2(k) 2 j(k)j+ j(k)j+ +! 2(k) 2 j(k)j+ j(k)j+ ; where = 0; (n) = ((n)) (j(n)j j(k)j ), ; (n) > ((n)) (j(n)j j(k)j ). In a similar way if ((n)) j(n)j j(k)j = ; we have: ((n)) (j(n)j j(k)j ) = ((n)) (j(n)j j(k)j) +2 j(n)j j(k)j = 2 j(n)j j(k)j (2(k)+) ;

11 ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS and from Lemma 2.3 we have: S 2 j(k)j+(d (n) 2 j(n)j j([ 2s ])j ) = = D 2(k) 2 j(k)j+ + j(k)j+ +! 2(k) 2 j(k)j+ + j(k)j+ (2.3) = D 2(k) 2 j(k)j+ j(k)j+ +! 2(k) 2 j(k)j+ j(k)j+ + +! 2(k) 2 j(k)j+ + j(k)j+ : Then by Lemma 2.4 and equalities (2.), (2.2) and (2.3) we get (r j(k)j D 2 j(k)j) (D (n) 2 j(n)j j([ 2s ])j ) = = (D 2(k) 2 j(k)j+ j(k)j+ ) (D (k) 2 j(k)j j(k)j ) +! 2(k) 2 j(k)j+ + 2 j(k)j+ + 2! 2(k) 2 j(k)j+ j(k)j+ 3! (k) 2 j(k)j j(k)j = r j(k)j (D (k) 2 j(k)j j(k)j ) (2.4) +! 2(k) 2 j(k)j+ + 2 j(k)j+ + 2! 2(k) 2 j(k)j+ j(k)j+ 3! (k) 2 j(k)j j(k)j ; where 2 = ((n)) j(n)j j(k)j ; and 0; (n) = ((n)) (j(n)j j(k)j), 3 = ; (n) > ((n)) (j(n)j j(k)j). It follows that for every l < j([ 2 s ])j, where s 2 f0; : : : ; [log 2 ]g is xed, r l D 2 lf s = 2s jfn : j(n)j = j([ 2 s ])jgj r j(n)j (D jfn : j(n)j = lgj (n) 2 j(n)j j(n)j ) f n:j(n)j=l n:j(n)j=l + 2s jfn : j(n)j = j([ 2 s ])jgj jfn : j(n)j = lgj! 2(n) 2 j(n)j+ + 2 j(n)j+ + 2! 2(n) 2 j(n)j+ j(n)j+ 3! (n) 2 j(n)j j(n)j ; where, 2 and 3 are dened as previously. We get that for every : = = jfn : j(n)j j( )gj n:j(n)j r j(n)j (D (n) 2 s+ < 2 s, 2 j(n)j j(n)j ) f r j(n)j (D jfn : j(n)j j( )gj (n) 2 j(n)j j(n)j ) f l= n:j(n)j=l jfn : j(n)j = lgj jfn : j(n)j j( )gj 2 s jfn : j(n)j = j([ l= 2 s ])jgj r ld 2 lf s (! jfn : j(n)j j( )gj 2(n) 2 j(n)j+ + 2 j(n)j+ l= n:j(n)j=l + 2! 2(n) 2 j(n)j+ j(n)j+ 3! (n) 2 j(n)j j(n)j ) f :

12 76. MEMI It is easily seen that for every x 2 G, sup : 2s+ < 2s j jfn : j(n)j j()gj l= n:j(n)j=l (! 2(n) 2 j(n)j+ + 2 j(n)j+ + 2! 2(n) 2 j(n)j+ j(n)j+ 3! (n) 2 j(n)j j(n)j ) f(x)j 3kfk ; from which we get: k sup j jfn : j(n)j j( )gj l= n:j(n)j=l (! 2(n) 2 j(n)j+ + 2 j(n)j+ + 2! 2(n) 2 j(n)j+ j(n)j+ 3! (n) 2 j(n)j j(n)j ) fk k sup j (! jfn : j(n)j j( )gj 2(n) 2 j(n)j+ + 2 j(n)j+ l= n:j(n)j=l On the other side the term can be written as ow we have j + 2! 2(n) 2 j(n)j+ j(n)j+ 3! (n) 2 j(n)j j(n)j ) fk 3kfk : j l= jfn : j(n)j = j([ 2 s ])jgj = j jfn : j(n)j = lgj jfn : j(n)j = j([ 2 s ])jgj r ld 2 lf s j kc k jfn : j(n)j = j([ 2 s ])jgj l:jfn:j(n)j=lgj=k r l D 2 lf s j kc l:jfn:j(n)j=lgjk jfn : j(n)j = j([ 2 s ])jgj C F s CF s : r l D 2 lf s j jf sup j r j(n)j (D jfn : j(n)j j( )gj (n) 2 j(n)j j(n)j ) fj > gj n:j(n)j jf sup jfn : j(n)j j()gj j l= n:j(n)j=l (! 2(n) 2 j(n)j+ + 2 j(n)j+ + 2! 2(n) 2 j(n)j+ j(n)j+ 3! (n) 2 j(n)j j(n)j ) fj > 2 gj + [log 2 ] j l= jf sup : 2s+ < 2s jfn : j(n)j j( )gj 2 s jfn : j(n)j = lgj jfn : j(n)j = j([ 2 s ])jgj r ld 2 lf s j > gj 2

13 6 kfk ALMOST EVERWHERE COVERGECE OF FEJÉR MEAS C 6 kfk [log 2 ] + C jff s > 2 gj 6kfk [log 2 ] + C [log 2 ] j([ 2 s ])j kfk C kfk kf s k + [log 2 ] 2 s + C kfk : We provide an example that shows that the set of increasing sequences ((n)) n mentioned in Theorem 2.2 is not empty. Example 2.. Let for example =. For every l : 4 l < 6, we dene the set 2 fk : j(k)j = lg = f2 l ; 2 l + 2 l 4 g: Then if l is so that 2 2n l < 2 2n+2, we put j(k)j = l if and only if k 2 f2 l 22n 2 s; 2 l 22n 2 s + g; for some s being such that j(s)j = 2 2n 2. The condition () in the theorem is obviously satised. Recursively on n we get for 2 2n l < 2 2n+2, jfk : j(k)j = lgj = 2jfs : j(s)j = 2 2n 2 gj = 2 2 n l : We also use induction on n in order to prove the property (3), if we suppose that 2 2n 2 k:j(k)j=2 2n 2 ((k)) i 2 2n 2 ; we get for 2 2n l < 2 2n+2 l k:j(k)j=l ((k)) i = 2 2 2n 2 2 2n + 2 n < l : k:j(k)j=2 2n 2 ((k)) i + jfk : j(k)j = 2 2n 2 gj References [] E.S. Belinsky: Summability of Fourier series with the method of lacunary arithmetical means at the Lebesgue points, Proc. Am. Math. Soc. 25(2) (997), [2] G. Gát: Almost everywhere convergence of Fejér and logarithmic means of subsequnces of partial sums of the Walsh-Fourier series of integrable functions, J. of Approx. Theory 62 (200), [3].J. Fine: Cesàro summability of Walsh-Fourier series, Proc. at. Acad. Sci. U.S.A. 4 (955), [4] U. Goginava, K. agy: Maximal operators of Fejér means of Walsh-Kaczmarz Fourier series, J. Funct. Space. Appl., 8(2) (200), [5] F. Schipp, W.R. Wade, P. Simon, J. Pl: Walsh series: an introduction to dyadic harmonic analysis, Adam Hilger, Bristol and ew ork, 990. [6] W.S. oung: Mean convergence of generalized Walsh-Fourier series. Trans. Amer. Math. Soc., 28 (976), [7]. alcwasser: Sur la sommabilité des séries de Fourier, Stud. Math. 6 (936), Department of Mathematics Faculty of atural Sciences and Mathematics University of Sarajevo Bosnia and Herzegovina address: nacima.o@gmail.com

Summation of Walsh-Fourier Series, Convergence and Divergence

Bulletin of TICMI Vol. 18, No. 1, 2014, 65 74 Summation of Walsh-Fourier Series, Convergence and Divergence György Gát College of Nyíregyháza, Institute of Mathematics and Computer Science (Received December