http://jipam.vu.edu.au/
Volume 5, Issue 4, Article 92, 2004
ON RELATIONS OF COEFFICIENT CONDITIONS
LÁSZLÓ LEINDLER
B OLYAI I NSTITUTE
J OZSEF A TTILA U NIVERSITY
A RADI VERTANUK TERE 1
H-6720 S ZEGED
H UNGARY .
leindler@math.u-szeged.hu
Received 27 April, 2004; accepted 20 October, 2004
Communicated by A. Babenko
A BSTRACT . We analyze the relations of three coefficient conditions of different type implying one by one the absolute convergence of the Haar series. Furthermore we give a sharp condition which guaranties the equivalence of these coefficient conditions.
Key words and phrases: Haar series, Absolute convergence, Equivalence of coefficient conditions.
2000 Mathematics Subject Classification. 26D15, 40A30, 40G05.
1. I
NTRODUCTION
A known result of P.L. Ul’janov [4] asserts that the condition
(1.1) σ
1
:=
∞
X a
√ n n n =3
< ∞ ( a n
≥ 0) implies the absolute convergence of the Haar series, i.e.
∞
X
2 m
X b
( k ) m
χ
( k ) m
( x ) ≡
∞
X
| a n
χ n
( x ) | < ∞ n =0 m =0 k =1 almost everywhere in (0 , 1) . He also verified, among others, that if the sequence { a n
} is mono-
tone then the condition (1.1) is not only sufficient, but also necessary to the absolute conver-
gence of the Haar series.
In [1] we verified that if the condition
(1.2) σ
2
:=
∞
X
( 2 m +1
X m =1 n =2 m
+1 a
2 n
)
1
2
< ∞
ISSN (electronic): 1443-5756 c 2004 Victoria University. All rights reserved.
Partially supported by the Hungarian NFSR Grand # T042462.
085-04

2 L ÁSZLÓ L EINDLER holds then the Haar series is absolute ( C, α ) -summable for any α ≥ 0 , consequently the condi-
tion (1.2) also guarantees the absolute convergence of the Haar series.
Recently, in [3], we showed that if the sequence
{ a n
} is only locally quasi decreasing, i.e. if a n
≤ K a m for m ≤ n ≤ 2 m and for all m, and the Haar series is absolute ( C, α ≥ 0)
-summable almost everywhere, then (1.2) holds.
Here and in the sequel, K and K i will denote positive constants, not necessarily the same at each occurrence. Furthermore we shall say that a sequence { a n
} is quasi decreasing if
(0 ≤ ) a n
≤ K a m holds for any n ≥ m . This will be denoted by { a n
} ∈ QDS , and if the sequence { a n
} is a locally quasi decreasing, then we use the short notion { a n
} ∈ LQDS .
P.L. Ul’janov [5], implicitly, gave a further condition in the form
(1.3) σ
3
:=
∞
X
1 m =3 m (log m )
1
2
( ∞
X n = m a
2 n
)
1
2
< ∞ which also implies the absolute convergence of the Haar series.
These results propose the question: What is the relation among these conditions?
We shall show that the condition (1.3) claims more than (1.2), and (1.2) demands more than
(1.1); and in general, they cannot be reversed. In order to get an opposite implication, a certain
monotonicity condition on the sequence { a n
} is required.
2. R
ESULTS
We establish the following theorem.
Theorem 2.1. Suppose that a := { a n
} is a sequence of nonnegative numbers. Then the following assertions hold:
(2.1) and if a ∈ LQDS then
σ
1
≤ K σ
2
,
(2.2) σ
2
≤ K σ
1
.
Similarly
(2.3) σ
2
≤ K σ
3
, and if the sequence { A m
} defined by
A m
:=
( 2 m +1
X k =2 m
+1 a
2 k
)
1
2 belongs to QDS then
(2.4) σ
3
≤ K σ
2
.
Finally
(2.5) σ
1
≤ K σ
3
, and if the sequence { n a
2 n
} ∈ QDS then
(2.6) σ
3
≤ K σ
1
.
Corollary 2.2. If the sequence { n a 2 n
} ∈ QDS
then the conditions (1.1), (1.2) and (1.3) are
equivalent.
J. Inequal. Pure and Appl. Math., 5(4) Art. 92, 2004 http://jipam.vu.edu.au/

O N R ELATIONS OF C OEFFICIENT C ONDITIONS
Next we show that the assumption { n a claim only that the sequence { n α a 2 n
2 n
} ∈ QDS in a certain sense is sharp. Namely if we
} ∈ QDS with α < 1 ,
then already the implication (1.1)
⇒
(1.3), in general, does not hold.
Proposition 2.3. If (0 ≤ ) α < 1 then there exists a sequence { a n
} such that the sequence
{ n α a 2 n
} ∈ QDS , furthermore
σ
1
< ∞ but σ
3
= ∞ .
Finally we verify the following.
Proposition 2.4. The requirements
(2.7) { n a
2 n
} ∈ QDS and the following two assumptions jointly
(2.8) { A m
} ∈ QDS and { a n
} ∈ LQDS are equivalent.
Acknowledgement 1. I would like to sincerest thanks to the referee for his worthy sugges-
tions, exceptionally for the remark that the inequality (2.6) also follows from (2.2), (2.4) and
Proposition 2.4.
3. L EMMA
We require the following lemma being a special case of a theorem proved in [2, Satz] ap-
pended with the inequality (3.2) which was also verified, in the same paper, in the proof of the
"Hilfssatz" (see p. 217).
Lemma 3.1. The inequality (1.3) holds if and only if there exists a nondecreasing sequence
{ µ n
} of positive numbers with the properties
(3.1)
∞
X
1 n =1 n µ n
< ∞ and
∞
X a
2 n
µ n n =1
< ∞ .
Furthermore
(3.2)
∞
X n =3
1 n (log n )
1
2
( ∞
X a
2 k
)
1
2 k = n
≤ K
( ∞
X a
2 n
µ n
)
1
2 n =3
( ∞
X n =1
1 n µ n
)
1
2 also holds.
4. P ROOFS
Proof of Theorem 2.1. The inequality (2.1) can be verified by then Hölder inequality. Namely
σ
1
=
∞
X
2 m +1
X m =1 n =2 m
+1 a
√ n n
≤
∞
X
( 2 m +1
X m =1 n =2 m
+1 a
2 n
)
1
2 ( 2 m +1
X n =2 m
+1
1 n
)
1
2
≤ σ
2
.
To prove the inequality (2.2) we utilize the monotonicity assumption and thus we get that
σ
2
≤ K
∞
X
2 m/ 2 a
2 m +1 m =1
≤ K
1
∞
X
2 m +1
X m =1 n =2 m +1
1
√ n a n
= K
1
σ
1
.
J. Inequal. Pure and Appl. Math., 5(4) Art. 92, 2004 http://jipam.vu.edu.au/
3

4 L ÁSZLÓ L EINDLER
The inequality (2.3) also comes via the Hölder inequality. Let
R m
:=
∞
P n = m a 2 n
1
2
.
Then
σ
2
=
∞
X
2
ν +1 − 1
X
( 2 m +1
X
ν =0 m =2 ν n =2 m +1 a
2 n
)
1
2
≤
∞
X
2
ν/ 2
ν =0


2
2
ν +1
X
 n =2 2
ν
+1
 a
2 n


1
2
≤
∞
X
2
ν/ 2
ν =0


∞
X
 n =2
2
ν
+1

1
2 a
2 n


≤ R
3
+ K
∞
X
2
2
ν
X
1 n (log n )
1
2
R
2 2
ν
+1
ν =1 n =2 2
ν − 1
+1
≤ K
1
∞
X n (log n )
1
2 n =3
1
R n
= K
1
σ
3
.
In order to prove (2.4) first we define a nondecreasing sequence
{ µ n
} as follows. Let
µ n
:= max
1 ≤ k ≤ m
A
− 1 k for 2 m
< n ≤ 2 m +1
, m = 1 , 2 , . . . , furthermore let µ
1
= µ
2
= µ
3
.
It is clear by { A m
} ∈ QDS that
(4.1) A
− 1 m
≤ µ
2 m +1
≤ K A
− 1 m
( m ≥ 1) ,
holds. Hence we obtain by (1.2) and (4.1) that
(4.2)
∞
X
2 m +1
X m =1 n =2 m
+1 a
2 n
µ n
≤ K σ
2
< ∞ and
(4.3)
∞
X
1 n =1 n µ n
≤ K
∞
X n µ n n =3
1
= K
∞
X
2 m +1
X
1 m =1 n =2 m +1 n µ n
∞
X
1
≤ K
1
≤ K
1
µ
2 m +1 m =1
∞
X
A m
= K
1
σ
2
< ∞ .
m =1
Finally, using the inequality (3.2), the estimations (4.2) and (4.3) clearly imply the statement
(2.4).
The assertion (2.5) is an immediate consequence of (2.1) and (2.3).
The proof of the declaration (2.6) is analogous to that of (2.4). The assumption
{ n a 2 n
} ∈
QDS enables us to define again a nondecreasing sequence { µ n
} satisfying the inequalities in
J. Inequal. Pure and Appl. Math., 5(4) Art. 92, 2004 http://jipam.vu.edu.au/

O N R ELATIONS OF C OEFFICIENT C ONDITIONS
(3.1). We can clearly assume that all
a k for n ≥ 3
> 0 ,
otherwise (2.6) is trivial if
µ n
:= max
1 ≤ k ≤ n a k
1
√ k
, and µ
1
= µ
2
= µ
3
.
{ n a 2 n
} ∈ QDS . Let
The definition of µ n and the assumption { n a 2 n
} ∈ QDS certainly imply that
(4.4) a n
1
√ n
≤ µ n
≤ a n
K
√ n is valid. The definition of σ
1
given in (1.1) and (4.4) convey the estimations
∞
X a
2 n
µ n
≤ K
∞
X a
√ n n n =3 n =3
≤ K σ
1
< ∞ and
∞
X
1 n =1 n µ n
≤ K
∞
X n µ n n =3
1
= K
∞
X a
√ n n n =3
= K σ
1
< ∞ .
These estimations and (3.2) verify (2.6).
Herewith the whole theorem is proved.
Proof of Corollary 2.2. The inequalities (2.1), (2.3) and (2.6) proved in the theorem obviously
deliver the assertion of the corollary. The proof is ready.
Proof of Proposition 2.3. Setting
ν m
:= 2
2 m
α − 1
, ε m
:= 2
− m/ 2
ν 2 m +1 and a
2 n
:= ε
2 m n
− α if ν m
< n ≤ ν m +1
, m = 0 , 1 , . . .
Then however, with R n
:=
∞
∞
X a
√ n n n =3
=
∞
X
ε m
ν m +1
X n
−
1+ α
2
≤ m =0 m =0 n = ν m
+1
∞
X
ε m
1 − α
ν 2 m +1
=
∞
X m =0
2
− m/ 2
< ∞ ,
P k = n a 2 k
1
2
,
σ
3
=
∞
X n =3
1 n (log n )
1
2
R n
=
≥
∞
X
ν m +1
X
1 n (log n )
1
2
R n m =0 n = ν m
+1
1
∞
X
4 m =0
R
ν m +1
2 m/ 2
,
5
J. Inequal. Pure and Appl. Math., 5(4) Art. 92, 2004 http://jipam.vu.edu.au/

6 L ÁSZLÓ L EINDLER furthermore
R
2
ν m
≥
≥
∞
X
ν k +1
X a
2 n k = m n = ν k
+1
1
K
∞
X
ε
2 k
ν
1 − α k +1 k = m
=
=
∞
X
ε
2 k k = m
1
K
ν k +1
X k
− α n = ν k
+1
∞
X
2
− k ≥ k = m
1
K
2
− m
.
From the last two estimations we clearly get that σ
3
The proof is complete.
= ∞ , as stated.
Proof of Proposition 2.4. First we prove that the assumption (2.7) implies both properties claimed in (2.8). Namely by
{ n a
2 n
} ∈ QDS we get that if µ > m then
A
2 m
=
2 m +1
X n =2 m +1 a 2 n n n
≥
1
2 m +1
2 m
1
K a
2
2 m +1
2 m +1
≥
1
2 K 2 a
2
2 µ
2
µ
≥
1
2
µ +1
X
2 K 3 n =2 µ +1 a
2 n
=
1
2 K 3
A
2
µ
, i.e.
{ n a 2 n
} ∈ QDS
The implications
⇒ {
{ n a
A
2 n n
} ∈
} ∈
QDS
QDS holds.
⇒ { a n
} ∈ QDS ⇒ { a n
To prove the implication (2.8)
⇒
(2.7) we first prove by
{ a
} ∈ LQDS are trivial.
n
} ∈ LQDS that if µ > m then
2 m +1
X k =2 m
+1 a
2 k
≤ K 2 m a
2
2 m and
2
µ
X k =2
µ − 1
+1 thus by { A n
} ∈ QDS we obtain that a
2 k
≥ 2
µ − 1
1
K a
2
2 µ
,
2
µ a
2
2 µ
≤ K
1
2 m a
2
2 m holds, whence { n a 2 n
} ∈ QDS plainly follows.
The proof is ended.
R EFERENCES
[1] L. LEINDLER, Über die absolute Summierbarkeit der Orthogonalreihen, Acta Sci. Math. (Szeged),
22 (1961), 243–268.
[2] L. LEINDLER, Über einen Äquivalenzsatz, Publ. Math. Debrecen, 12 (1965), 213–218.
[3] L. LEINDLER, Refinement of some necessary conditions, Commentationes Mathematicae Prace
Matematyczne, (in press).
[4] P.L. UL’JANOV, Divergent Fourier series, Uspehi Mat. Nauk (in Russian), 16 (1961), 61–142.
[5] P.L. UL’JANOV, Some properties of series with respect to the Haar system, Mat. Zametki (in Russian), 1 (1967), 17–24.
J. Inequal. Pure and Appl. Math., 5(4) Art. 92, 2004 http://jipam.vu.edu.au/