Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis
28 (2012), 177–187
www.emis.de/journals
ISSN 1786-0091
ESTIMATES FOR THE INTEGRAL OF MAXIMAL
FUNCTIONS OF FEJÉR KERNEL
NACIMA MEMIĆ
Abstract. In the first part of this paper we prove an estimate for the
integral of the maximal function of Fejér kernel related to Walsh-Paley
system. In the second part we study local integrability of partial maximal
functions obtained from Walsh-Kaczmarz system.
1. Introduction
G. Gát [1] estimated the maximal function of Fejér kernel with respect to
Walsh-Paley system, where a specific form of the Fejér kernel Kn has been
used. Here we express Kn in a new form which enables us to obtain a sharper
estimate.
Many authors have compared the behaviors of Walsh-Paley and WalshKaczmarz systems. Parallel studies resulted in some analogies ([5], [8]) and
many contrasts ([7], [2]).
As G. Gát proved in [2] the maximal function of Fejér kernel with respect to
Walsh-Kaczmarz system is not integrable on any interval of the dyadic group.
However, U. Goginava [3] obtained that the same function is an element from
Lp for every p ∈ (0, 1). Besides, K. Nagy [4] established necessary and sufficient
conditions for the integrability of weight maximal functions related to the same
system.
Our attempt is to study local integrability for partial maximal functions,
namely those related to some subsequences of the set of natural numbers.
Let Z2 denote the discrete cyclic group Z2 = {0, 1}, where the group operation is addition modulo 2. If |E| denotes the measure of the subset E ⊂ Z2 ,
then |{0}| = |{1}| = 21 .
∞
Q
The dyadic group G is obtained from G =
Z2 , where topology and meai=0
sure are obtained from the product.
2010 Mathematics Subject Classification. 42C10.
Key words and phrases. Walsh-Kaczmarz system, Fejér kernels.
177
178
NACIMA MEMIĆ
Let x = (xn )n≥0 ∈ G. The sets In (x) := {y ∈ G : y0 = x0 , . . . , yn−1 = xn−1 },
n ≥ 1 and I0 (x) := G are dyadic intervals of G. Set en := (δin )i . The WalshPaley system is defined as the set of Walsh-Paley functions:
χn (x) =
∞
Y
(rk (x))nk , n ∈ N, x ∈ G,
k=0
where n =
∞
P
nk 2k and rk (x) = (−1)xk . The n-th Walsh-Kaczmarz function
k=0
is
|n|−1
κn (x) := r|n| (x)
Y
(r|n|−1−k (x))nk ,
k=0
for n ≥ 1, k0 (x) := 1 and |n| = max k, so that n ∼ 2|n| .
nk 6=0
We denote the Dirichlet and the Fejér kernel functions by:
n−1
X
Dn :=
χk ,
Dnκ
:=
n−1
X
κk ,
k=0
k=0
1X
Kn :=
Dk ,
n k=1
n
1X κ
:=
D ,
n k=1 k
n
Knκ
where n ≥ 1.
2. On Walsh-Paley system
Lemma 2.1. The Fejér kernel satisfies the following formula
nKn ==
N
−1
X
D2Aj
j=1
!
N
X
2Ai
χ2A1 χ2A2 . . . χ2Aj−1
i=j+1
+
N
X
(χ2A1 χ2A2 . . . χ2Aj−1 )2Aj K2Aj .
j=1
Proof. Suppose that the number n has the form n = 2A1 + 2A2 + · · · + 2AN ,
A1 > A2 > . . . > AN .
nKn =
n
X
k=1
2 1
X
A
Dk =
k=1
Dk +
1 +2A2
2AX
k=2A1 +1
Dk + · · · +
n
X
k=n−2AN +1
Dk .
ESTIMATES FOR THE INTEGRAL . . .
179
Notice that for l = 1, 2, . . . , N − 1,
A
2A1 +...+2
X l+1
Dk =
k=2A1 +...+2Al +1
Al+1
=
2X

X
X
2A1 +2A2 −1
χi +
i=0
k=1
D2A1 +...+2Al +k =
k=1
2A1 −1

Al+1
2X
χi + . . . +
i=2A1

Al+1
A1 −1
A2 −1
2X
2X
2X

=
χi + χ2A1
χi + · · · + χ
k=1
i=0
A
2A1 +...+2
X l −1
χi +
i=2A1 +...+2Al−1
Al −1
2X
2A1 +···+2Al−1
i=0
A
2A1 +...+2
X l +k−1
+
χi 
i=2A1 +...+2Al
χi + χ2A1 +···+2Al
i=0
h
i
Al+1
=2
D2A1 + χ2A1 D2A2 + · · · + χ2A1 χ2A2 . . . χ2Al−1 D2Al
Al+1
2X

k−1
X

χ
i
i=0
χ2A1 χ2A2 . . . χ2Al Dk .
k=1
Where we have considered that the product χ2A1 χ2A2 . . . χ2Aj−1 = 1, for
j = 1.
Since
Al+1
2X
χ2A1 χ2A2 . . . χ2Al Dk = χ2A1 χ2A2 . . . χ2Al 2Al+1 K2Al+1 ,
k=1
for l = 0 we get
A1
2P
Dk = 2A1 K2A1 .
k=1
Now, summing over l = 0, 1, . . . , N − 1, completes the proof of Lemma 2.1.
Theorem 2.2. The integral of the maximal function of Fejér kernel can be
estimated by
Z
sup |Kn (x)|dx ≤ C
G\Ik |n|≥A
A−k
,
2A−k
where k < A.
Proof. Using Lemma 2.1, we estimate sup |Kn (x)| if x ∈ Is2 +1 (es1 +es2 ), where
|n|≥A
s1 < k ≤ A and s1 < s2 . Let n = 2A1 + 2A2 + . . . + 2AN , A1 > A2 > . . . > AN ,
180
NACIMA MEMIĆ
and A1 ≥ A. We have
N −1
N
X
1X
Kn (x) =
D Aj (
2Ai )χ2A1 (x)χ2A2 (x) . . . χ2Aj−1 (x)
n j=1 2 i=j+1
1X
+
(χ2A1 (x)χ2A2 (x) . . . χ2Aj−1 (x))2Aj K2Aj (x)
n j=1
X
1
1 X
≤
4Aj +
2Aj 2s1
n
n A ≤s ,j≤N
N
(Aj ≤s1 ,j≤N −1)
j
2
C s1
(4 + 2min(s2 ,A1 ) 2s1 + 4s1 )
A
1
2
( s +s
C min(s2 ,A1 )+s1
C 2 22A1 1 ,
≤
≤ A1 2
2
C2s1 ,
≤
if s2 ≤ A1 ;
if s2 > A1 .
Hence, sup |Kn (x)| ≤ 2s1 , for every x ∈ Is2 +1 (es1 + es2 ), and if s2 ≤ A, then
|n|≥A
s2 ≤ |n| for all |n| ≥ A, which means that
Now,
Z
X Z
sup |Kn (x)|dx =
Is1 +1 (es1 ) |n|≥A
s2 ≥s1 +1
2s1 +s2
2|n|
2s1 +s2
.
2A
≤
sup |Kn (x)|dx
Is2 +1 (es1 +es2 ) |n|≥A
X
X
2s1 +s2
|I
|
+
C
2s1 |Is2 +1 |
s
+1
2
A
2
s1 +1≤s2 ≤A
s2 ≥A+1
!
X 2s1
X 2s1
A − s1
≤C
+
≤
C
.
A
s2
A−s1
2
2
2
s +1≤s ≤A
s ≥A+1
≤C
1
It follows that
Z
k−1 Z
X
sup |Kn (x)|dx =
G\Ik |n|≥A
s=0
≤C
2
2
sup |Kn (x)|dx ≤ C
Is+1 (es ) |n|≥A
k−1
X
A−s
s=0
2A−s
A−k
.
2A−k
3. On Walsh-Kaczmarz system
Lemma 3.1. For every x ∈ Is2 +1 (es1 + es2 ), s1 < s2 , the inequality
|AX
n |−1
1 s1
s2 −s1
i
i (τi (x)) ≤ C max(2 , 2
2
r
(x)K
)
sup
i
2
n An i=0
holds
ESTIMATES FOR THE INTEGRAL . . .
181
Proof. First, we take |An | ≤ s1 . We have
|An |−1
|An |−1
1 X i
1 X i 4|An |
4 ≤ |An | = 2|An | ≤ 2s1 .
|
2 ri (x)K2i (τi (x))| ≤
An i=0
An i=0
2
Now if s1 < |An |, we have
|An |−1
X
1 i
2 ri (x)K2i (τi (x)) =
An i=0
sX
min(|AX
n |−1,s2 )
1 −1
1 1
1
i
i−1
s1
s1 −1
i i−s1 −2 =
2
+
2
2
2
2
+
−
2
+
An i=0
2
2
i=s1 +1
s
1 4 1 − 1 1 4s1 4min(|An |,s2 +1)| − 4s1 +1 =
− −
+
An 6
2
2
3 · 2s1 +2
min(|An |,s2 )
1
s1 4
≤ C |An | max 4 ,
≤ C max(2s1 , 2s2 −s1 ).
2
2s1
Theorem 3.2. For every natural number k and every increasing sequence
(An )n of natural numbers, we have
Z
sup |KAκ n (x)|dx = +∞,
n
Ik (ej )
for all j ≤ k − 1.
Proof. Let j be arbitrary, but fixed. We may assume that k ≥ 4j.
Clearly,
Z
∞ Z
X
κ
sup |KAκ n (x)|dx.
sup |KAn (x)|dx =
Ik (ej )
n
s=k
Is (ej )\Is+1 (ej )
n
However, it suffices to consider the sum over s ≥ k being such that s − 1 =
|An | ≥ 16, for some n. Then, we have r|An | (x) = 1 if x ∈ Is (ej ) \ Is+1 (ej ).
Moreover,
|An |−1
X
i
2 ri (x)K2i (τi (x)) =
i=0
j−1
X
|An |−1
2 K2i (τi (x)) − 2 K2j (τj (x)) +
i
j
i=0
=
j−1
X
2i 2i−1 +
1
2
− 2j 2j−1 +
|An |
4
−4
1
4 −1 4
−
+
−
6
2
2j+2
2
|An |−1
|An |−4
4
4
≥ −4j+1 + j+2 ≥ j+2 .
2
2
=
2i K2i (τi (x))
i=j+1
i=0
j
X
j
j+1
1
2
|An |−1
+
X
i=j+1
2i 2i−j−2
182
NACIMA MEMIĆ
It follows immediately that
2|An |
≥ j+11 ,
2
for every x ∈ Is (ej ) \ Is+1 (ej ), and that
|KAκ n (x)|
Z
sup |KAκ n (x)|dx ≥
Is (ej )\Is+1 (ej )
n
2|An |
2s+j+12
=
1
.
8 · 2j+11
Since the set {|An | ≥ 16} is infinite, the result follows.
Lemma 3.3. Let (Nn )n be any sequence of nonnegative integers with Nn ≤ n.
Define the sets
En,Nn = {(xi )i ∈ G, xn−k = 0, ∀k ∈ {1, 2, . . . , Nn }}.
Let S denote the set of integers n such that Nn 6= 0, n − Nn > n − k − Nn−k ,
for every k ≤ n − 1 such that n − k ∈ S, where the least integer from S is the
least nonnegative integer n for which Nn 6= 0. Then
\
Y
Y
1
1
c
1 − Nn ≤ En,Nn ≤
1 − Nn +1
2
2
n∈S
n∈S
n∈N
Proof. ForTevery natural number i let ui denote the number of Ii+1 -cosets
T c inc
cluded in En,Nn , and zi denote the number of Ii+1 -cosets included in En,Nn
n
n
having 0 as i-th component.
Clearly, (ui )i and (zi )i are increasing sequences, ui+1 ≤ 2ui and zi ≤ u2i , for
every i.
Notice that if n + k − Nn+k ≤Tn − Nn , forTsome k ≥ 1, Nn , Nn+k 6= 0, then
c
c
c
c
En,N
⊂ En+k,N
En,N
En,N
. Therefore,
=
.
n
n
n
n+k
n∈S
n∈S
Also, notice that ui = zi + ui−1 , for every
S i ≥ 1, because un−1 represents the
number of In+1 -cosets not contained in En having 1 as n-th component.
n
Clearly, if i + 1 is not an element from S, then zi = ui−1 , which gives
ui = 2ui−1 .
We prove that for every i being such that (i + 1) ∈ S, we have
1
1
2 1 − Ni+1 ui−1 ≤ ui ≤ 2 1 − Ni+1 +1 ui−1 .
2
2
It can be seen that for Ni+1 ≥ 3, i + 1 ∈ S, we have
zi = ui−1 − zi+1−Ni+1 +
i−1
X
uk−1 − zk ,
k=i+2−Ni+1
because the number (uk−1 − zk ) represents the number of chains of zeros beginning at some k − Nk+1 , where (k + 1) ∈ {i + 3 − Ni+1 , . . . , i} ∩ S.
ESTIMATES FOR THE INTEGRAL . . .
183
In this case
i
X
i−1
X
zk =
k=i+1−Ni+1
uk ,
k=i+1−Ni+1
from which we get
i−1
X
i−1
X
zk + zi + ui−1 = ui−1 +
k=i+1−Ni+1
uk ,
k=i+1−Ni+1
and
i−1
X
ui = ui−1 +
i−1
X
(uk − zk ) =
k=i+1−Ni+1
uk .
k=i−Ni+1
Now, if Ni+1 = 2, i + 1 ∈ S, then zi = ui−1 − zi−1 .
The expression
i−1
X
uk ,
ui =
k=i−Ni+1
is easily verified in this case as
ui = 2ui−1 − zi−1 = ui−1 + ui−2 .
Therefore, we obtain in the general case that
i−2
X
ui − ui−1 ≥
k=i−Ni+1
It follows that
1
2 1−
1−
ui−1 =
2i−1−k
1
2Ni+1 −1
ui−1 .
1
ui−1 ≤ ui .
2Ni+1
In order to prove the second inequality for (i + 1) ∈ S, let j be the largest
integer such that j ≤ i − 1 and (j + 1) ∈ S.
If j ≤ i − Ni+1 , then for every k ∈ {i − Ni+1 , . . . , i − 2} ∩ S, we have
1
uk = 2i−1−k
ui−1 .
Hence,
i−2
X
ui − ui−1 =
uk =
k=i−Ni+1
i−1
=
We get
2
−2
2i−1
i−2
X
1
2i−1
2k ui−1
k=i−Ni+1
i−Ni+1
ui−1 =
ui = 2 1 −
1
2Ni+1
1−
ui−1 .
1
2Ni+1 −1
ui−1 .
184
NACIMA MEMIĆ
Suppose that j > i − Ni+1 . We have for j ≤ i − 2,
ui − ui−1 =
j−1
X
uk +
i−Ni+1 −1
u k = uj −
k=j
k=i−Ni+1
=
i−2
X
2i−1−j
X
ui−1 −
i−2
1 X
uk +
2i−1
k=j−Nj+1
i−Ni+1 −1
1
X
k=j−Nj+1
2k ui−1
k=j
i−Ni+1 −1
X
2i−1 − 2j
u
=
u
−
uk +
uk
i−1
i−1
2i−1
k=j−N
j+1
i−Ni+1 −1
X
2i−Ni+1 − 2j−Nj+1
2k
u
=
u
−
ui−1
i−1
i−1 i−1
i−1
2
2
k=j−Nj+1
1
1
2j
1
= 1 − Ni+1 −1 + i−1 Nj+1 ui−1 ≤ 1 − Ni+1 +1 ui−1 .
2
2 2
2
≤ ui−1 −
If j = i − 1, then in a similar way we obtain
i−Ni+1 −1
ui − ui−1 = ui−1 −
uk ≤ ui−1 −
k=i−1−Ni
=
X
1−
1
2Ni+1 −1
Therefore, we obtain
1
+ Ni
2
1−
ui−1 ≤
1
1−
1
2Ni+1
ui−1 .
2Ni+1 +1
ui−1 .
N
\
Y
un
c
=
En,Nn = lim
N →∞
2un−1
n=1
n∈N
(2i−Ni+1 − 2i−1−Ni )ui−1
ui ≤
Since
1
2i−1
Y
(n+1)∈S
un
,
2un−1
we have
Y
(n+1)∈S
2 1−
1
2Nn+1
2un−1
un−1
\
c
En,N
≤
≤
n
n∈N
2 1−
Y
(n+1)∈S
1
2Nn+1 +1
2un−1
un−1
.
Theorem 3.4. Let k be a natural number and (An )n an increasing sequence
of natural numbers. For every M > 0 define the sets (El,Nl ) such that Nl = 0
|An | )2
if l 6= |An | for every n, or if M > (An −2
for every n such that l = |An |.
An
If l = |An | for some n such that M ≤
Nl =
min
n:l=|An |,M ≤
(An −2|An | )2
An
(An −2|An | )2
,
An
log2
then
M An
− 1.
An − 2|An |
ESTIMATES FOR THE INTEGRAL . . .
Then define the sets SM as in Lemma 3.3. If
∞
X An − 2|An | X
X
1
RM :=
+
An
M +k
k=1
|An |∈SM
|An |∈SM +k
185
An − 2|An |
< +∞,
An
where in the sum every |An | ∈ S is taken at most only once with the largest
|An | )2
|An |
value of An −2
satisfying M ≤ (An −2
, then we have
An
An
Z
sup |KAκ n (x)|dx < +∞.
S
G\
k
j=0 Ik (ej )
n
Proof. We use a formula proved in [6]
Knκ (x)
|n|−1
|n|−1
1 X i
1 X i
1
= +
2 D2i (x) +
2 ri (x)K2i (τi (x))
n n i=0
n i=0
1
1
(n − 2|n| )D2|n| (x) + (n − 2|n| )r|n| (x)Kn−2|n| (τ|n| (x)),
n
n
where τn (x0 , . . . , xn , . . . ) = (xn−1 , xn−2 , . . . , x0 , xn , xn+1 , . . . ).
The first sum and the fourth term clearly represent integrable functions. We
only need to study
S the integrability of the second sum and the last term.
Let x ∈ G \ kj=0 Ik (ej ). Then, x ∈ Is2 +1 (es1 + es2 ), where s1 < s2 ≤ k − 1.
Using Lemma 3.1, we get
|An |−1
Z
X
1 i
sup
2 ri (x)K2i (τi (x)) dx =
Sk
G\ j=0 Ik (ej ) n An i=0
|An |−1
k−2 X
k−1 Z
X
X
1 i
2 ri (x)K2i (τi (x)) dx
=
An i=0
s1 =0 s2 =s1 +1 Is2 +1 (es1 +es2 )
!
2s1
k−1
k−2
X
X
X
2s2 −s1
2s1
+
≤
2s2 +1 s =2s +1 2s2 +1
s1 =0 s2 =s1 +1
2
1
k−2
k−2
X
X
k
1
2s1
k
+ s1 ≤ + k
≤ 3k.
≤
s
+1
1
2
2
2
2s1
s =0
s =0
+
1
1
An − 2|An |
FM = x ∈ G : sup
|KAn −2|An | (τ|An | (x))| ≥ M .
An
n
n
If x ∈ FM , then |KAn −2|An | (τ|An | (x))| ≥ AnM−2A|A
for some n, which implies
n|
Let
|An | )2
that M ≤ (An −2
An
This gives that
and x ∈ E|An |,sn , for sn ≥ [log2
FM ⊂
[
l∈SM
El,Nl .
M An
]
An −2|An |
− 1.
186
NACIMA MEMIĆ
By Lemma 3.3, we get
\
Y 1
c El,Nl ≥
1− N =
2 l
l∈S
l∈S
M
M
Y |An |∈SM
An − 2|An |
1−
M An
,
where in the last product every |An | is taken at most only once with the largest
|An |
(An −2|An | )2
value of An −2
satisfying
M
≤
.
An
An
Since the convergence of RM implies that RM 0 converges for every M 0 > M .
Then we can choose M sufficiently large so that


\
|An |
X
2
An − 2 
c .
El,N
≥ exp −
l
M
A
n
l
|An |∈SM
It follows

|FM | ≤ 1 − exp −
We obtain
Z
S
G\ kj=0 Ik (ej )
sup
n
2
M
X
|An |∈SM
|An |
An − 2
An

≤ C
M
X
|An |∈SM
An − 2|An |
.
An
An − 2|An |
|KAn −2|An | (x)|dx ≤
An
≤ M + (M + 1)|FM | +
∞
X
(M + k + 1 − (M + k))|FM +k |
k=1
≤ M + CRM < +∞.
Example 3.5. Consider the sequence (An = 2n + 2αn )αn∈N , for some fixed
rational number α such that 1 > α > 12 . Applying Theorem 3.4, we get
RM =
X
|An |∈SM
≤
X
n≥1
∼
An − 2|An | X 1
+
An
M +k
k=1
X
∞
∞
X
2αn
1
+
n
αn
2 +2
M +k
k=1
∞
X
1
k=1
(M + k)1+ 2α−1
1−α
< +∞.
|An |∈SM +k
X
n≥
log2 (M +k)
2α−1
An − 2|An |
An
2αn
2n + 2αn
ESTIMATES FOR THE INTEGRAL . . .
187
While, if we consider the sequence An = 2n + 2n−l−1 , n > 2l + 2, used by G.
Gát in [2], where l is a fixed positive integer, we have for arbitrary M > 0,
∞
X
X
X
2n−l−1
1
2n−l−1
RM =
+
2n + 2n−l−1 k=1 M + k
2n + 2n−l−1
n≥log M +2l+2
n≥log2 (M +k)+2l+2
2
>
X
n≥log2
−l−1
2
= +∞.
1 + 2−l−1
M +2l+2
References
[1] G. Gát. On the Fejér kernel functions with respect to the Walsh-Paley system. Acta
Acad. Paedagog. Agriensis, Sect. Mat. (N.S.), 24(v. 24):105–110, 1997.
[2] G. Gát. On the Fejér kernel functions with respect to the Walsh-Kaczmarz system. Acta
Math. Acad. Paedagog. Nyházi. (N.S.), 17(2):121–126 (electronic), 2001.
[3] U. Goginava. On the conjecture of Gát. Acta Math. Acad. Paedagog. Nyházi. (N.S.),
20(1):39–42 (electronic), 2004.
[4] K. Nagy. On the L1 norm of the weighted maximal function of Fejér kernels with respect
to the Walsh-Kaczmarz system. JIPAM. J. Inequal. Pure Appl. Math., 9(1):Article 16,
9, 2008.
[5] F. Schipp. Pointwise convergence of expansions with respect to certain product systems.
Anal. Math., 2(1):65–76, 1976.
[6] V. A. Skvorcov. On Fourier series with respect to the Walsh-Kaczmarz system. Anal.
Math., 7(2):141–150, 1981.
[7] A. A. Šneı̆der. On series of Walsh functions with monotonic coefficients. Izvestiya Akad.
Nauk SSSR. Ser. Mat., 12:179–192, 1948.
[8] W. S. Young. On the a.e. convergence of Walsh-Kaczmarz-Fourier series. Proc. Amer.
Math. Soc., 44:353–358, 1974.
Received May 9, 2012.
Department of Mathematics,
University of Sarajevo,
Zmaja od Bosne 33-35,
Sarajevo, Bosnia and Herzegovina
E-mail address: nacima.o@gmail.com