()


CUBO A Mathematical Journal
Vol.17, No¯ 01, (85–97). March 2015

Measure of noncompactness on Lp(RN) and applications

A. Aghajani

School of Mathematics,

Iran University of Science and Technology,

Narmak, Tehran 16844-13114, Iran.

aghajani@iust.ac.ir

D. O
,

Regan

School of Mathematics, Statistics and Applied Mathematics,

National University of Ireland, Galway, Ireland.

Nonlinear Analysis and Applied Mathematics (NAAM),

Department of Mathematics, King Abdulaziz University, Jeddah, Saudi Arabia.

donal.oregan@nuigalway.ie

A. Shole Haghighi

School of Mathematics,

Iran University of Science and Technology,

Narmak, Tehran 16844-13114, Iran.

a.sholehaghighi@kiau.ac.ir

ABSTRACT

In this paper we define a new measure of noncompactness on Lp(RN) (1 ≤ p < ∞) and
study its properties. As an application we study the existence of solutions for a class

of nonlinear functional integral equations using Darbo’s fixed point theorem associated

with this new measure of noncompactness.

RESUMEN

En este art́ıculo definimos una nueva medida de no-compacidad sobre Lp(RN) (1 ≤

p < ∞) y estudiamos sus propiedades. Como aplicación, estudiamos la existencia de
soluciones para una clase de ecuaciones integrales funcionales no lineales usando el

teorema de punto fijo de Darbo asociado a esta nueva medida de no-compacidad.

Keywords and Phrases: Measure of noncompactness, Darbo’s fixed point theorem, Fixed point.

2010 AMS Mathematics Subject Classification: 47H08, 47H10.



86 A. Aghajani, D. O
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1 Introduction

Measures of noncompactness and Darbo’s fixed point theorem play major roles in fixed point theory

and their applications. Measures of noncompactness were introduced by Kuratowski [19]. In 1955,

Darbo presented a fixed point theorem [12], using this notion. This result was used to establish

the existence and behavior of solutions in C[a,b], BC(R+) and BC(R+ × R+) to many classes of

integral equations; see [1, 2, 3, 4, 6, 9, 10, 16, 17] and the references cited therein. When one

seeks solutions in unbounded domains there are particular difficulties. The aim of this paper is to

construct a regular measure of noncompactness on the space Lp(RN) (1 ≤ p < ∞) and investigate
the existence of solutions of a particular nonlinear functional integral equation.

Let R+ = [0, + ∞) and (E,‖.‖) be a Banach space. The symbols X and ConvX stand for the
closure and closed convex hull of a subset X of E, respectively. Now ME denotes the family of all

nonempty and bounded subsets of E and NE denotes the family of all nonempty and relatively

compact subsets.

Definition 1.1. A mapping µ : ME −→ R+ is said to be a measure of noncompactness in E if it
satisfies the following conditions:

1◦ The family kerµ = {X ∈ ME : µ(X) = 0} is nonempty and kerµ ⊆ NE.

2◦ X ⊂ Y =⇒ µ(X) ≤ µ(Y).

3◦ µ(X) = µ(X).

4◦ µ(ConvX) = µ(X).

5◦ µ(λX + (1 − λ)Y) ≤ λµ(X) + (1 − λ)µ(Y) for λ ∈ [0,1].

6◦ If {Xn} is a sequence of closed sets from ME such that Xn+1 ⊂ Xn for n = 1,2, · · · and if

lim
n→∞

µ(Xn) = 0 then X∞ = ∩
∞

n=1Xn 6= ∅.

We say that a measure of noncompactness is regular [7] if it additionally satisfies the following

conditions:

7◦ µ(X ∪ Y) = max{µ(X),µ(Y)}.

8◦ µ(X + Y) ≤ µ(X) + µ(Y).

9◦ µ(λX) = |λ|µ(X) for λ ∈ R.

10◦ kerµ = NE.

The Kuratowski and Hausdorff measures of noncompactness have all the above properties (see

[5, 7]).

The following Darbo’s fixed point theorem will be needed in section 3.



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Measure of noncompactness on Lp(RN) and applications 87

Theorem 1.2. [12] Let Ω be a nonempty, bounded, closed and convex subset of a Banach space E

and let F : Ω −→ Ω be a continuous mapping such that there exists a constant k ∈ [0,1) with the
property

µ(FX) ≤ kµ(X) (1)

for any nonempty subset X of Ω. Then F has a fixed point in the set Ω.

Integral equations of Urysohn type in the space of Lebesgue integrable functions on bounded

and unbounded intervals and the concept of weak measure of noncompactness on L1(R+) was

studied in [8, 13, 14].

In Section 2, we define a new measure of noncompactness on Lp(RN) and study its properties.

In Section 3, using the obtained results in Section 2, we investigate the problem of existence of

solutions for a class of nonlinear integral equations.

2 Main results

Let Lp(U) (U ⊆ RN) denote the space of Lebesgue integrable functions on U with the standard

norm

‖x‖Lp(U) =
(

∫

U

|x(t)|pdt
)

1
p

.

Before introducing the new measures of noncompactness on Lp(RN), we need to characterize the

compact subsets of Lp(RN).

Theorem 2.1. [11, 18] Let F be a bounded set in Lp(RN) with 1 ≤ p < ∞. The closure of F in
Lp(RN) is compact if and only if

lim
h−→0

‖τhf − f‖Lp(RN) = 0 uniformly in f ∈ F, (2)

where τhf(x) = f(x+h) for all x,h ∈ R
N. Also for ǫ > 0 there is a bounded and measurable subset

Ω ⊂ RN such that

‖f‖Lp(RN\Ω) < ǫ for all f ∈ F. (3)

Now, we are ready to define a new measure of noncompactness on Lp(RN).

Theorem 2.2. Suppose 1 ≤ p < ∞ and X is a bounded subset of Lp(RN). For x ∈ X and ǫ > 0
let

ωT (x,ǫ) = sup{‖τhx − x‖Lp(BT ) : ‖h‖RN < ǫ},

ωT (X,ǫ) = sup{ωT(x,ǫ) : x ∈ X},

ωT (X) = lim
ǫ→0

ωT (X,ǫ),

ω(X) = lim
T→∞

ωT (X),

d(X) = lim
T→∞

sup{‖x‖Lp(RN\BT ) : x ∈ X},



88 A. Aghajani, D. O
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where BT = {a ∈ R
N : ‖a‖RN ≤ T}. Then ω0 : MLp(RN) −→ R given by

ω0(X) = ω(X) + d(X) (4)

defines a measure of noncompactness on Lp(RN).

Proof. First we show that 1◦ holds. Take X ∈ MLp(RN) such that ω0(X) = 0. Let η > 0 be

arbitrary. Since ω0(X) = 0, then limT→∞ limǫ→0 ω
T (X,ǫ) = 0 and thus, there exist δ > 0 and

T > 0 such that ωT (X,δ) < η implies that ‖τhx − x‖Lp(BT ) < η for all x ∈ X and h ∈ R
N such

that ‖h‖RN < δ. Since η > 0 was arbitrary, we get

lim
h→0

‖τhx − x‖Lp(RN) = lim
h→0

lim
T→∞

‖τhx − x‖Lp(BT ) = 0

uniformly in x ∈ X. Again, keeping in mind that ω0(X) = 0 we have

lim
T→∞

sup{‖x‖Lp(RN\BT ) : x ∈ X} = 0

and so for ε > 0 there exists T > 0 such that

‖x‖Lp(RN\BT ) < ǫ for all x ∈ X.

Thus, from Theorem 2.1 we infer that the closure of X in Lp(RN) is compact and kerω0 ⊆ NE.

The proof of 2◦ is clear. Now, suppose that X ∈ MLp(RN) and (xn) ⊂ X such that xn → x ∈ X in
Lp(RN) . From the definition of ωT (X,ǫ) we have

‖τhxn − xn‖Lp(BT ) ≤ ω
T (X,ǫ)

for any n ∈ N, T > 0 and ‖h‖RN < ǫ. Letting n → ∞ we get ‖τhx − x‖Lp(BT ) ≤ ω
T (X,ǫ) for any

‖h‖RN < ǫ and T > 0 , hence

lim
T→∞

lim
ǫ→0

ωT (X,ǫ) ≤ lim
T→∞

lim
ǫ→0

ωT (X,ǫ),

implies that

ω(X) ≤ ω(X). (5)

Similarly, we can show that d(X) ≤ d(X) so from (5) and 2◦ we get ω0(X) = ω0(X), so ω0 satisfies

condition 3◦ of Definition 1.1. The proof of conditions 4◦ and 5◦ can be carried out similarly by

using the inequality ‖λx + (1 − λ)y‖Lp(BT ) ≤ λ‖x‖Lp(BT ) + (1 − λ)‖y‖Lp(BT ).

To prove 6◦, suppose that {Xn} is a sequence of closed and nonempty sets from ME such that

Xn+1 ⊂ Xn for n = 1,2, · · · , and lim
n→∞

ω0(Xn) = 0. Now for any n ∈ N take an xn ∈ Xn and

set F = {xn}. We claim F is a compact set in L
p(RN). To prove the claim, we need to check

conditions (2) and (3) of Theorem 2.1. Let ε > 0 be fixed. Since lim
n→∞

ω0(Xn) = 0 there exists

k ∈ N such that ω0(Xk) < ε. Hence, we can find δ1 > 0 and T1 > 0 such that

ωT1(Xk,δ1) < ε,



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Measure of noncompactness on Lp(RN) and applications 89

and

sup{‖x‖Lp(RN\BT1 )
: x ∈ Xk} < ε.

Thus, for all n ≥ k and ‖h‖RN < δ1 we get

‖τhxn − xn‖Lp(RN) ≤ ‖τhxn − xn‖Lp(BT1)
+ ‖τhxn − xn‖Lp(RN\BT1)

≤ ‖τhxn − xn‖Lp(BT1)
+ 2‖xn‖Lp(RN\BT1 )

< 3ε

and

‖xn‖Lp(RN\BT1 )
< ε. (6)

The set {x1,x2, . . . ,xk−1} is compact, hence there exists δ2 > 0 such that

‖τhxn − xn‖Lp(RN) < ε (7)

for all n = 1,2, . . . ,k and ‖h‖RN < δ2, and there exists T2 > 0 such that

‖xn‖Lp(RN\T2) < ε (8)

for all n = 1,2, . . . ,k. Therefore by (6) and (7) we obtain

‖τhxn − xn‖Lp(RN) < 3ε

for all n ∈ N and ‖h‖ < min{δ1,δ2}, and from (6), (8) we get

‖xn‖Lp(RN\BT ) < ε (9)

for all n ∈ N, where T = max{T1,T2}. Thus all the hypotheses of Theorem 2.1 are satisfied and so

the claim is proved.

Hence there exist a subsequence {xnj} and x0 ∈ L
p(RN) such that xnj → x0, and since xn ∈ Xn,

Xn+1 ⊂ Xn and Xn is closed for all n ∈ N we get

x0 ∈
∞
⋂

n=1

Xn = X∞,

and this finishes the proof of the theorem. ✷

Now, we study the regularity of ω0.

Theorem 2.3. The measure of noncompactness ω0 defined in Theorem 2.1 is regular.

Proof. Suppose that X,Y ∈ MLp(RN). Since for all ε > 0, λ > 0 and T > 0 we have

ωT (X ∪ Y,ε) ≤ max{ωT (X,ε),ωT (Y,ε)} ,

ωT (X + Y,ε) ≤ ωT (X,ε) + ωT (Y,ε),

ωT (λX,ε) ≤ λωT (X,ε)



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and

sup
x∈X∪Y

‖x‖Lp(RN\BT ) ≤ max{sup
x∈X

‖x‖Lp(RN\BT ),sup
x∈Y

‖x‖Lp(RN\BT )},

sup
x∈X+Y

‖x‖Lp(RN\BT ) ≤ sup
x∈X

‖x‖Lp(RN\BT ) + sup
x∈Y

‖x‖Lp(RN\BT ),

sup
x∈λX

‖x‖Lp(RN\BT ) ≤ λ sup
x∈X

‖x‖Lp(RN\BT ),

then the hypotheses 7◦, 8◦ and 9◦ hold. To show that 10◦ holds, suppose that X ∈ NLp(RN). Thus,

the closure of X in Lp(RN) is compact and hence from Theorem 2.1, for any ǫ > 0 there exists

T > 0 such that ‖x‖Lp(RN\BT ) < ǫ for all x ∈ X and also limh−→0 ‖τhx − x‖Lp(RN) = 0 uniformly

in x ∈ X. From the first conclusion, there exists δ > 0 such that ‖τhx − x‖Lp(RN) < ǫ for any

‖h‖RN < δ. Then for all x ∈ X we have

ωT (x,δ) = sup{‖τhx − x‖Lp(BT ) : ‖h‖RN < δ} ≤ ǫ.

Therefore,

ωT (X,δ) = sup{‖ω(x,δ) : x ∈ X} ≤ ǫ,

which proves

lim
T→∞

lim
δ→0

ω(X,δ) = 0 (10)

and

lim
T→∞

sup{‖x‖Lp(RN\BT ) : x ∈ X} = 0. (11)

Now from (10) and (11) condition 10◦ holds. ✷

Theorem 2.4. Let Q = {x ∈ Lp(RN) : ‖x‖Lp(RN) ≤ 1}. Then ω0(Q) = 3

Proof. Indeed, we have

‖τhx − x‖Lp(RN) ≤ ‖τhx‖Lp(RN) + ‖x‖Lp(RN) ≤ 2

and

‖x‖Lp(RN\BT ) ≤ ‖x‖Lp(RN) ≤ 1

for all x ∈ Q, h ∈ RN and T > 0. Also for any ǫ > 0, T > 0 and x ∈ Q we have

ωT (x,ǫ) = sup{‖τhx − x‖Lp(BT ) : ‖h‖ < ǫ} ≤ 2.

Therefore we obtain ω0(Q) ≤ 3. Now we prove that ω0(Q) ≥ 3. For any k ∈ N there exists

Ek ⊂ R
N such that m(Ek) =

1
2k

(m is the Lebesgue measure on RN), diam(Ek) ≤
1
k
, Ek ∩Bk = ∅

and Ek ⊂ B2k. Define fk : R
N −→ R by

fk(x) =

{
(2k)

1
p x ∈ Ek

0 otherwise.
(12)

It is easy to verify that ‖fk‖Lp(RN) = 1, ‖τ1
k
fk − fk‖Lp(B2k) = 2 and ‖fk‖Lp(RN\Bk) = 1 for all

k ∈ N. Thus, we get ω0(Q) ≥ 3. This completes the proof. ✷



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Measure of noncompactness on Lp(RN) and applications 91

3 Application

In this section we show the applicability of our results.

Definition 3.1. We say that a function f : Rn ×Rm −→ R satisfies the Carathéodory conditions if
the function f(.,u) is measurable for any u ∈ Rm and the function f(x,.) is continuous for almost

all x ∈ Rn.

Theorem 3.2. Assume that the following conditions are satisfied:

(i) f : RN × R −→ R satisfies the Carathéodory conditions, and there exists a constant k ∈ [0,1)
and a ∈ Lp(RN) such that

|f(x,u) − f(y,v)| ≤ |a(x) − a(y)| + k|u − v|, (13)

for any u,v ∈ R and almost all x,y ∈ RN.

(ii) f(.,0) ∈ Lp(RN).

(iii) k : RN × RN −→ R satisfies the Carathéodory conditions and there exist g1,g2 ∈ Lp(RN)
and g ∈ Lq(RN) ( 1

p
+ 1

q
= 1) such that |k(x,y)| ≤ g(y)g1(x) for all x,y ∈ R

N and

|k(x1,y) − k(x2,y)| ≤ g(y)|g2(x1) − g2(x2)|. (14)

(iv) The operator Q acts continuously from the space Lp(RN) into itself and there exists a non-

decreasing function ψ : R+ −→ R+ such that

‖Qu‖Lp(RN) ≤ ψ(‖u‖Lp(RN)) (15)

for any u ∈ Lp(RN).

(v) There exists a positive solution r0 to the inequality

kr + ψ(r)‖K‖1 + ‖f(.,0)‖Lp(RN) ≤ r (16)

where

(Ku)(t) =

∫

RN

k(x,y)u(y)dy

and

‖K‖1 = sup{‖Ku‖Lp(RN) : ‖u‖Lp(RN) ≤ 1}.

Then the functional integral equation

u(x) = f(x,u(x)) +

∫

RN

k(x,y)(Qu)(y)dy (17)

has at least one solution in the space Lp(RN).



92 A. Aghajani, D. O
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Remark 3.3. The linear Fredholm integral operator K : Lp(RN) → Lp(RN) is a continuous
operator and ‖K‖1 < ∞.

Proof. First of all we define the operator F : Lp(RN) → Lp(RN) by

F(u)(x) = f(x,u(x)) +

∫

RN

k(x,y)(Qu)(y)dy. (18)

Now Fu is measurable for any u ∈ Lp(RN). Now we prove that Fu ∈ Lp(RN) for any u ∈ Lp(RN).

Using conditions (i)-(iv), we have the following inequality

|F(u)(x)| ≤ |f(x,u) − f(x,0)| + |f(x,0)| + |

∫

RN

k(x,y)(Qu)(y)ds|

a.e. x ∈ RN. Thus

‖Fu‖Lp(RN) ≤ k‖u‖Lp(RN) + ‖f(.,0)‖Lp(RN) + ‖K‖1ψ(‖u‖Lp(RN)). (19)

Hence F(u) ∈ Lp(RN) and F is well-defined and also from (19) we have F(Br0) ⊆ Br0, where r0
is the constant appearing in assumption (v). Also, F is continuous in Lp(RN), because f(t, .), K

and Q are continuous for a.e. x ∈ RN. Now we show that for any nonempty set X ⊂ Br0 we have

ω0(F(X)) ≤ kω0(X).

To do so, we fix arbitrary T > 0 and ε > 0. Let us choose u ∈ X and for x,h ∈ BT with ‖h‖RN ≤ ǫ,

we have

|(Fu)(x) − (Fu)(x + h)| ≤
∣

∣

∣
f(x,u(x)) +

∫

RN

k(x,y)(Qu)(y)dy

− f(x + h,u(x + h)) +

∫

RN

k(x + h,y)(Qu)(y)dy
∣

∣

∣

≤ |f(x,u(x)) − f(x + h,u(x))| + |f(x + h,u(x)) − f(x + h,u(x + h))|

+ |

∫

RN

k(x,y)(Qu)(y)dy −

∫

RN

k(x + h,y)(Qu)(y)dy|

≤ |a(x) − a(x + h)| + k|u(x) − u(x + h)| +

∫

RN

|k(x,y) − k(x + h,y)||Qu(y)|dy.



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Measure of noncompactness on Lp(RN) and applications 93

Therefore
(

∫

BT

|(Fu)(x + h) − (Fu)(x)|pdt
)

1
p

≤
(

∫

BT

|a(x) − a(x + h)|pdt
)

1
p

+ k
(

∫

BT

|u(x) − u(x + h)|pdt
)

1
p

+
(

∫

BT

∣

∣

∫

RN

|k(x,y) − k(x + h,y)||Qu(y)|dy
∣

∣

p
dx

)
1
p

≤
(

∫

BT

|a(x) − a(x + h)|pdx
)

1
p

+ k
(

∫

BT

|u(x) − u(x + h)|pdx
)

1
p

+
(

∫

BT

(

∫

RN

|k(x,y) − k(x + h,y)|qdy
)

p
q

dx
)

1
p

‖Qu‖Lp(RN)

≤
(

∫

BT

|a(x) − a(x + h)|pdx
)

1
p

+ k
(

∫

BT

|u(x) − u(x + h)|pdx
)

1
p

+
(

∫

BT

(

∫

RN

|g2(x) − g2(x + h)|
q
|g(y)|qdy

)

p
q

dx
)

1
p

‖Qu‖Lp(RN)

≤ ‖τha − a‖Lp(BT ) + k‖τhu − u‖Lp(BT )

+
(

∫

BT

|g2(x) − g2(x + h)|
pdx

)
1
p

‖g‖Lq(RN)‖Qu‖Lp(RN)

≤ ωT (a,ǫ) + kωT (u,ǫ) + ‖Qu‖Lp(RN)‖g‖Lq(RN)ω
T
(g2,ǫ).

Thus we obtain

ωT (FX,ǫ) ≤ ωT (a,ǫ) + kωT (X,ǫ) + ψ(r0)‖g‖Lq(RN)ω
T (g2,ǫ).

Also we have ωT (a,ǫ),ωT (g2,ǫ) → 0 as ǫ → 0. Then we obtain

ω(FX) ≤ kω(X). (20)

Next, let us fix an arbitrary number T > 0. Then, taking into account our hypotheses, for an

arbitrary function u ∈ X we have

(

∫

RN\BT

|(Fu)(x)|pdx
)

1
p

≤
(

∫

RN\BT

∣

∣

∣
f(x,u(x)) +

∫

RN

k(x,y)Qu(y)dy
∣

∣

∣

p

dt
)

1
p

≤
(

∫

RN\BT

|f(x,u(x)) − f(x,0)|pdx
)

1
p

+
(

∫

(RN\BT

|f(t,0)|pdx
)

1
p

+
(

∫

RN\BT

∣

∣

∣

∫

RN

k(x,y)Qu(y)dy
∣

∣

∣

p

dx
)

1
p

≤ k
(

∫

RN\BT

|u(x)|pdx
)

1
p

+
(

∫

RN\BT

|f(x,0)|pdx
)

1
p

+
(

∫

RN\BT

(

∫
∞

0

|k(x,y)|qdy
)

p
q

dx
)

1
p

‖Qu‖Lp(RN)

≤ k‖u‖LP(RN\BT ) + ‖f(.,0)‖LP(RN\BT ) + ‖g1‖LP(RN\BT )‖g‖Lq(RN)ψ(‖u‖Lp(RN)).

Also we have

‖f(.,0)‖LP(RN\BT ),‖g1‖LP(RN\BT ) −→ 0



94 A. Aghajani, D. O
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as T → ∞ and hence we deduce that

d(FX) ≤ kd(X). (21)

Consequently from (20) and (21) we infer

ω0(FX) ≤ kω0(X). (22)

From (22) and Theorem 1.2 we obtain that the operator F has a fixed-point u in Br0 and thus the

functional integral equation (17) has at least one solution in Lp(RN). ✷

In the example below we will use the following well known result.

Theorem 3.4. [15] Let Ω ⊆ Rn be a measure spaces and suppose k : Ω × Ω −→ R is an Ω × Ω-
measurable function for which there is constant C > 0 such that

∫

Ω

|k(x,y)|dx ≤ C for a.e. y ∈ Ω

and ∫

Ω

|k(x,y)|dy ≤ C for a.e. x ∈ Ω.

If K : Lp(Ω) −→ Lp(Ω) is defined by

(Kf)(x) =

∫

Ω

k(x,y)f(y)dy, (23)

then K is a bounded and continuous operator and ‖K‖1 ≤ C.

Example 3.5. Consider the integral equation

u(x) =
cosu(x)

‖x‖ + 2
+

∫

R3

e−(|x2|+|y2|+|y3|+1)

(|x1| + 3)
2(|y1| + 2)

2(1 + |x3|
2)
e−|u(y)|u(y)dy, (24)

where x = (x1,x2,x3) ∈ R
3 and ‖x‖ is the Euclidean norm. We study the solvability of the integral

equation (24) on the space Lp(RN) for p > 3. Let f(x,u) =
cosu

‖x‖ + 2
and note it satisfies hypothesis

(i) with a(x) =
1

‖x‖ + 2
and k =

1

2
. Indeed, we have

|f(x,u) − f(y,v)| = |
cosu

‖x‖ + 2
−

cosv

‖y‖ + 2
|

≤ |
1

‖x‖ + 2
−

1

‖y‖ + 2
|| cosu| +

1

‖y‖ + 2
| cosu − cosv|

≤ |
1

‖x‖ + 2
−

1

‖y‖ + 2
| +

1

2
|u − v|

= |a(x) − a(y)| + k|u − v|.



CUBO
17, 1 (2015)

Measure of noncompactness on Lp(RN) and applications 95

Also, it is easily seen that f(.,0) satisfies assumption (ii) and

‖f(.,0)‖
p

Lp(R3)
=

∫

R3

|
1

‖x‖ + 2
|pdx

=

∫2π

0

∫π

0

∫
∞

0

r2 sinϕ

(r + 2)p
drdϕdθ

≤ 4π

∫
∞

0

1

(r + 2)p−2
dr

=
4π

(p − 3)2p−3

for all p > 3. Thus , we have ‖f(.,0)‖Lp(R3) ≤ (
4π

p − 3
)

1

p . Moreover, taking

k(x,y) =
e−(|x2|+|y2|+|y3|+1)

(|x1| + 3)2(|y1| + 2)2(1 + |x3|2)
,

g1(x) = g2(x) =
e−|x2|

(|x1| + 3)
2(1 + |x3|

2)
and g(x) =

e−(|x2|+|x3|)

(|x1| + 2)
2
, we see that g1,g2,g ∈ L

p(R3) for

all 1 ≤ p < ∞ and k satisfies hypothesis (iii). Also, we have
∫

R3

|k(x,y)|dx =

∫
∞

−∞

∫
∞

−∞

∫
∞

−∞

e−(|x2|+|y2|+|y3|+1)

(|x1| + 3)
2(|y1| + 2)

2(1 + x2
3
)
dx1dx2dx3 ≤

π

3e
,

∫

R3

|k(x,y)|dy =

∫
∞

−∞

∫
∞

−∞

∫
∞

−∞

e−(|x2|+|y2|+|y3|+1)

(|x1| + 3)
2(|y1| + 2)

2(1 + |x3|
2)
dy1dy2dy3 ≤

4

9e

and thus from Theorem 3.2, ||K||1 ≤
π

3e
. Furthermore, Q(u)(x) = e−|u(x)|u(x) satisfies hypothesis

(iv) with ψ(t) = t. Finally, the inequality from assumption (v), has the form

kr + ψ(r)‖K‖1 + ‖f(.,0)‖Lp(R3) =
1

2
r +

π

3e
r + (

4π

p − 3
)

1

p = (
1

2
+
π

3e
)r + (

4π

p − 3
)

1

p ≤ r

Thus, for the number r0 we can take r0 = (
4π

p − 3
)

1

p ×
6e

3e − 2π
. Consequently, all the assumptions

of Theorem 3.2 are satisfied and thus equation (24) has at least one solution in the space Lp(R3)

if p > 3.

Received: December 2014. Accepted: January 2015.

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Measure of noncompactness on Lp(RN) and applications 97

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	Introduction
	Main results
	Application