Andrica-Piticari. 86KB Jun 04 2011 12:08:21 AM

Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics - ICTAMI 2004, Thessaloniki, Greece

AN EXTENSIONS OF THE RIEMANN-LEBESGUE
LEMMA AND SOME APLICATIONS
by
Dorin Andrica,
Mihai Piticari

Abstract. We show that a similar relation as (1.2) holds for all continuous and bounded
functions g : [0, ∞) → R of finite Cesaro mean. A result concerning the asymptotic behavior
in (2.10) when a = 0, b = T and f ∈ C [0, T ] is given in Theorem 3.1. Some concrete
applications and examples are presented in the last section of the paper.
Keywords: Riemann-Lebesgue Lemma, T - periodic function.
Mathematics Subject Classification (2000): 26A42, 42A16
1

1. Introduction
The well-known Riemann-Lebesgue Lemma (see for instance [6, pp.22,
Problem 1.4.11(a) and (b)] or [9, Problem9.48]) states that if f : [a, b] → R is
a Riemann integrable function on [a, b] then

lim ∫ f ( x) sin nxdx = lim ∫ f ( x) cos nxdx = 0
b

b

n →∞ a

n →∞ a

(1.1)

That implies a classical result in Fourier analysis, i.e. if f : R → R is a
2π -periodic with Fourier coefficients an, bn, then under suitable conditions
S n ( x) → f ( x) , where
S n (x) =

1
a0 +
2

∑ (ak cos kx + bk sin kx) .
n

k =1

The following result is well-known and represents a nice generalization of the
previous one. It is also called the Riemann-Lebesque Lemma: Let f: [a,b] → R
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Dorin Andrica, Mihai Piticari - An extensions of the Riemann-Lebesgue
lemma and some aplications
be a continuous function, where 0 ≤ a < b. Suppose the function g:[0, ∞ ) → R
to be continuous and T-periodic. Then
f ( x) g (nx)dx = ∫ g ( x)dx ∫ f ( x)dx
a
n →∞ ∫a
T 0

lim

1

b

T

b

(1.2)

(see [5] for the special case a = 0, b = T).
There are other extensions and generalization of the important result
contained in (1.1). We mention here Knyazyuk A.V.[8] which gives necessary
and sufficient conditions for the existence of the limit.
h ( x, g (λ
λ →∞ ∫a

lim

b

(1.3)

x)dx

for a given function g : [0, ∞ ) → R and all continuous functions h : [a,b] ×
g([0, ∞ )) → R. The following multidimensional generalization of RiemannLebesgue Lemma is given in the recent paper Canada, A., Urena, A.J. [2,
Lemma 3.1] and it is used to study the asymptotic behavior of the solvability
set for pendulum-type equations with linear damping and homogeneous
Dirichlet conditions: Let g : R → R be a continuous and T-periodic function
with zero mean value and let u1 be given functions satisfying the following
property are real numbers sucs that
N


meas  x ∈ [0, π ] : ∑ ρ i u 'i ( x) = 0 > 0
i =1



then ρ1 = ... = ρ N = 0 . Let B ⊂ C 1[0, π ] be such that the set {b': b ∈ B} is
uniformly bounded in C[0, π ] . Then for any given function r ∈ L1[0, π ] we have
u

∫

π

|| ρ ||→∞ 0

lim

N

g  ∑ ρ i u i ( x) + b( x) r ( x)dx = 0
 i =1


(1.4)

uniformly with respect to b ∈ B. Other multidimensional version of RiemannLebesgue Lemma is mention in Canada, A.,Ruiz,D.[3, Lemma 2.1] and it is
applied to the study of periodic perturbations of a class of resonant problems.
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Dorin Andrica, Mihai Piticari - An extensions of the Riemann-Lebesgue
lemma and some aplications
Some other extensions and generalizations of Riemann-Lebesgue Lemma were
obtained by Bleistein, N., Handelsman, R. A., Lew, J. S. [1], Kantor, P. A. [7]
and Stadije, W. [10].
In this paper we extend the relation (1.2). Our main result is contained
in Theorem 2.1 and it shows that a similar relation as (1.2) holds for all
continuous and bounded function g : [0, ∞ ) → R of finite Cesaro mean. Some
applications are also given.

2. The main results

Let us begin whit some auxiliary results which will help us to derive our
main result.

Lemma 2.1. Let ω : [0, ∞ ) → R be a continuous function such that

ω ( x)
= 0. If (c n ) n≥1 is a sequence of non-negative real numbers such that
lim
x →∞
x
ω(cn )
 cn 
= 0.
  is bounded, then lim
n→∞
n
 n  n≥1
Proof. We consider the following cases.
Case 1. There exists lim c n and it is + ∞ . In that case we have
n →∞

ω (c n )
n

=

ω (c n ) c n
cn

⋅

n

≤M

ω (c n )
cn

,

(2.1)

ω ( x)

c

= 0, from (2.1) it follows
where M = sup  n : n ≥ 1 . Since lim
x →∞
x

n
ω (c n )
lim
= 0 hence the desired conclusion.
n →∞
n
Case 2. The sequence (c n ) n≥1 is bounded. Consider A > 0 such that
cn ≤ A for all positive integers n ≥ 1 , and define K= sup ω ( x) . It is clear that

ω (c n )
n

≤

ω(c )

K
for all n ≥ 1 , i.e. lim n =0.
n→∞
n
n
28

x∈[ 0, A]

Dorin Andrica, Mihai Piticari - An extensions of the Riemann-Lebesgue
lemma and some aplications
Case 3. The sequence (c n )n≥1 is unbounded and lim c n does not exist.

ω ( x)

For ε > 0 there exists δ > 0 such that

x

<

ε

M

n →∞

for any x > δ , where


c
M = sup  n : n ≥ 1 . Consider the sets

n

Aδ = {n ∈ N*: cn ≤ δ } and Bδ = {n ∈ N* : cn > δ }

If one of these sets is finite, then we immediately derive the desired
ω(cn )
= 0, it
result. Assume that Aδ and Bδ are both infinite. Since lim
n∈Aδ
n
ω (c n )
< ε for all n ∈ Aδ
follows that there exists N1 (ε ) with the property
n
with n ≥ N1 (ε ) . For n ∈ Bδ , we have

ω (c n ) ω (c n ) c n
ε
=
⋅
<
M =ε ,
n
cn
n
M
i.e

(2.2)

ω(c )
ω (c n )
< ε for any n ≥ N1 (ε ) . Finally lim n =0.
n→∞
n
n

Lemma 2.2 Let f :R → R be a continuous non-constant periodic function.
If F is an antiderivative of f, then

1 T
F(x) =  ∫ f (t )dt  x + g ( x) ,

T o

(2.2)

where T > 0 is the period of f and g : R → R is a T-periodic function.

Proof. Using the relation f(t + T) = f(t) for any t ∈ R it follows F(x + t) T

1 T
F(x) = ∫ f (t )dt for all x ∈ R. Considering the function h(x) =  ∫ f (t )dt  x ,
o
o

T

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Dorin Andrica, Mihai Piticari - An extensions of the Riemann-Lebesgue
lemma and some aplications

we have h(x + T) – h(x) =

∫o

T

f (t )dt , i.e. F(x + T) - h(x + T)=F(x) - h(x). That is

the function defined by g(x) = F(x) - h(x), x ∈ R, is periodic of period T.
Moreover the formula (2.2) holds.
Lemma 2.3 Consider 0 ≤ a < b to be real numbers and let f:[a,b] → R be

a function of class C1. Let g : [0, ∞ ) → R be a continuous function such that
1 x
lim ∫ g (t )dt = L (finite). Then the following relation holds:
x →∞ x 0
lim ∫ f ( x) g (nx)dx = L ∫ f ( x)dx
b

b

n →∞ a

(2.3)

a

Proof. If G(x)=

∫0 g (t )dt , then
x

define the function ω ( x) = G ( x) − Lx ,

x ≥ 0 . It is clear that ω is differentiable and it satisfies lim

x→∞

ω ( x)
x

= 0.

We have

∫
a

1
1
f ( x) g (nx)dx = (G (nb) f (b) − f (a )G (na )) − ∫ f ' ( x)G (nx)dx =
n
na
b

1
1
= (G (nb( f (b) − f (a )G (na )) − ∫ f ' ( x)( Lnx + ω (nx))dx =
n
na
b

b

1
1
= (G (nb) f (b) − f (a )G (na )) − L ∫ f ' ( x) xdx − ∫ ω (nx) f ' ( x)dx =
n
nb
a
b

a

1
(G (nb) f (b) − f (a )G (na )) − L(bf (b) − af (a )) +
n

+ L ∫ f ( x)dx −
b

a

1
ω (nx) f ' ( x)dx
n ∫a
b

Let us show that

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Dorin Andrica, Mihai Piticari - An extensions of the Riemann-Lebesgue
lemma and some aplications
1 b
ω (nx) f ' ( x)dx = 0
n→∞ n ∫a

(2.4)

lim

Indeed, from the relations
1 b
1 b
ω (nx) f ' ( x)dx ≤ ∫ | ω (nx) || f ' ( x) | dx =
∫
a
n a
n
| ω (c n ) | b
| f ' ( x) | dx,
=
n ∫a

where

na < c n < nb, by applying Lemma 2.1 it follows lim

n →∞

ω (c n )
n

= 0 , i.e.

the equality (2.4) holds.
Moreover we have
1
G (na )

 G (nb)
(G (nb) f (b) − G (na ) f (a )) = lim  b
f (a)  =
f (b) − a
n →∞ n
n →∞ 
na
nb

lim

= L(bf (b) − af (a ))

Using (2.4) and (2.5) the relation (2.3) follows.
Theorem 2.1 Let 0 ≤ a < b, be real numbers and let f : [a,b] → R be a
continuous function. Consider the function g : [0,∞) → R to be continuous and
1 x
bounded such that lim ∫ g (t )dt = L (finite). Then
x →∞ x 0

lim ∫

b

n →∞ a

f ( x) g (nx)dx = L ∫ f ( x)dx
b

a

(2.6)

Proof. Without loss of generality we can assume that 0 ≤ g(x) ≤ 1for
any x ∈ [0,∞). Applying the well-known Weierstrass approximation Theorem,
it follows that there exists a sequence ( f m ) m≥1 of polynomials, f m :[a,b] → R,

which converges uniformly to

f . Consider ε > 0 and define ε ' =

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Dorin Andrica, Mihai Piticari - An extensions of the Riemann-Lebesgue
lemma and some aplications

ε
. Then we can find a positive integer M( ε ) such that
(b − a)( L + 2)
| f m ( x) − f ( x) |< ε ’ for any m ≥ M (ε ) and for all x ∈ [a, b] . If m ≥ I (ε ) , then
we have

∫

b

f m ( x) g (nx)dx − ∫

b

∫

f ( xg (nx)dx ≤

b

f m ( x) − f ( x) g (nx)dx ≤ ε ' (b − a) ,

for all positive n . Therefore, for m ≥ M (ε ) and n ∈ N *
a

a

∫a

a

f m ( x) g (nx)dx − ε ' (b − a) ≤ ∫ f ( x) g (nx)dx ≤

b

b

≤ ε ' (b − a) + ∫ f m ( x) g (nx)dx
a

b

(2.7)

a

from Lemma 2.3. it follows
lim

n →∞

∫

b

a

f m ( x) g (nx)dx = L ∫

b

f m ( x)dx.

a

Hence, for ε > 0 one can find a positive integer N (ε , m ) such that for all
n ≥ N (ε , m ) we have
L ∫ f m ( x) − ε ' (b − a ) < ∫ f m ( x) g (nx)dx < L ∫ f m ( x)dx + ε ' (b − a ).
b

b

b

a

a

a

Using the last inequalities and (2.7) we get
L∫

b

a

f m ( x)dx − 2ε ' (b − a) < ∫

f ( x) g (nx)dx <

b

< L ∫ f m ( x)dx + 2ε ' (b − a ).
a

b

(2.8)

a

But for all m ≥ M( ε ) we have f ( x) - ε
x ∈ [a,b]. Thus

'

< f m ( x) < f ( x ) + ε

L ∫ f ( x)dx − Lε ' (b − a) < ∫ f m ( x)dx