Measure of Noncompactness in Banach Algebra and Application to the Solvability of Integral Equations in BC(R + )

Inf. Sci. Lett. 4, No. 2, 93-99 215 93 Information Sciences Letters An International Journal http://dx.doi.org/1.12785/isl/426 Measure of Noncompactness in Banach Algebra and Application to the Solvability of Integral Equations in BCR A. Aghajani and M. Aliaskari School of Mathematics, Iran University of Science and Technology, Narmak, Tehran 16846-13114, Iran Received: 2 Nov. 214, Revised: 25 Mar. 215, Accepted: 26 Mar. 215 Published online: 1 May 215 Abstract: In this paper an attempt is made to prove a fixed point theorem for the product of two operators each of which satisfies a special conditions in Banach algebra, using the technique of measure of noncompactness. Also we show that how it can be used to investigate the solvability of integral equations. Keywords: Measure of noncompactness, Fixed point. 1 Introduction It is has been witnessed that the differential and integral equations that appear in many physical problems are generally nonlinear and fixed point theory presents a strong tool for obtaining the solutions of such equations which otherwise are hard to solve by other ordinary procedures for example, see [1,2,3]. In this paper, we analyze solvability of a certain functional-integral equation which consist of many special cases of integral and functional-integral equations, which are applicable in various real world problems of engineering, economics, physics and similar fields see [4, 5]. Indeed, we are going to investigate the solvability of the integral equation xt= T xt ft,xt gt,s,xsds, 1 where t R and T is an operator acting from the Banach algebra BCR consisting of all functions x : R R which are continuous and bounded on R into itself and the functions f, g are continuous and satisfy certain conditions. Eq.1 includes many known integral equations as model cases. In the case T x 1 the equation 1 turns into xt= ft,xt gt, s, xsds, which has been investigated in [6]. What we are going to achieve in this paper, will extend the findings already obtained in [8,9,1,12,14]. The main tool used in our investigation is the technique associated with measure of noncompactness. For a discussion of existence of solution to the above-mentioned integral equation, the used measure of noncompactness must also satisfy an additional condition. Indeed, we will use a class of measures of noncompactness which satisfies a condition calledm. Such a condition will guarantee the solvability of operator equations in Banach algebra. It is worthwhile mentioning that the important measures of noncompactness in notable spaces satisfy condition m see [8,9,1,12,18]. This condition had first been used in for Hausdorff measure of noncompactness in Banach algebra CI consisting of real continuous functions defined on a closed and bounded interval I see [15]. 2 Preliminaries For this reason, suppose that E is a given Banach space which has the norm. and zero element θ. If the closed ball in E is centered at x and has radius r, we show it by Bx,r. In order to show Bθ,r, we write B r. If X is a subset of E, in that case, we can show the closure and the closed convex hull of X with the symbols X and ConvX respectively. Also X Y and λ X λ R are used to show the algebraic operation on sets. Moreover, by the Corresponding author e-mail: aghajani@iust.ac.ir c 215 NSP

94 A. Aghajani, M. Aliaskari: Measure of Noncompactness in Banach Algebra... symbol X we will denote the norm of a bounded set X, i.e, X =sup{ x : x X}. Furthermore M E is used to denote the family of all nonempty bounded subsets of E and N E denote its subfamily includes all relatively compact sets. Definition 1[11]. A mapping µ : M E R is said to be measure of noncompactness in E if it is satisfies the following conditions 1The family kerµ = {X M E : µx=} is nonempty and kerµ N E. 2X Y µx µy. 3µX=µX. 4µConvX = µx. 5µλ X1 λy λ µx1 λµy for λ [,1]. 6If X n is a nested sequence of closed sets from M E such that lim n µx n =, then the intersection set X = n=1 X n is nonempty. Observe that the intersection set X from axiom 6 is a member of the kerµ. In fact, since µx µx n for any n, we have that µx =. This yields that X kerµ see [13]. Now suppose Banach space E has the structure of Banach algebra. For given subsets X and Y of a Banach algebra E, let XY ={xy : x X,y Y}. Definition 2.We state that measure of noncompactness µ which has been defined in Banach algebra E satisfies the condition m, if for arbitrary sets X,Y M E, the following inequality is satisfied µxy X µy Y µx. Now we present an example of a measure of noncompactness in Banach algebra which satisfies condition m. Let us consider the Banach space BCR consisting of all functions x : R R which are continuous and bounded on R. This space is endowed with the standard norm x = sup{ xt : t R }. Obviously BCR has also the structure of Banach algebra with the standard multiplication of functions. In addition, fix a set X M BCR and numbers ε > and L >. For an arbitrary function x X let us denote by ω L x,ε the modulus of continuity of x on the interval [,L], i.e. ω L x,ε=sup{ xt xs : t,s [,L],[t s] ε}. In addition ω L X,ε=sup{ωx,ε : x X}, ω L X= lim ωx,ε, ε ω X= lim L ω L X,ε. Moreover, if t R is a fixed number, let us denote Xt={xt : x X}, diamxt= sup{ xt yt : x,y X}, cx=limsupdiamxt. With help of the above mappings we denote the following measures of noncompactness in BCR cf. [16,17] µ c X=ω XcX. 2 Theorem 1.The measure of noncompactness µ c defined by 2 satisfies conditionm on the family of all nonempty and bounded subsets X of Banach algebra BCR such that functions belonging to X are nonnegative onr. Proof.If x,y C[a,b], ε > then for t,s [a,b] such that t s ε, we get xtyt xsys xtyt xtys As a result xtys xsys xt yt ys ys xt xs x ωy,ε y ωx,ε. ωxy,ε x ωy,ε y ωx,ε. So ω X satisfies conditionm. Now, fix arbitrarily sets X;Y M BCR. Choose arbitrary functions z 1,z 2 XY. This means that there exist functions x 1,x 2 X and y 1,y 2 Y such that z 1 = x 1 y 1,z 2 = x 2 y 2. Next, for t R we get z 1 t z 2 t = x 1 ty 1 t x 2 ty 2 t x 1 ty 1 t x 1 ty 2 t x 1 ty 2 t x 2 ty 2 t = x 1 t y 1 t y 2 t y 2 t x 1 t x 2 t X diamyt Y diamxt. Hence we obtain diamxtyt X diamyt Y diamxt and consequently cxy X cy Y cx. So, that the measure of noncompactness µ c satisfies conditionm. In order to achieve the main purpose of this paper, the following theorem plays a crucial role Theorem 2[7]. Let Ω be a bounded, nonempty, convex and closed subset of a Banach space E. Then each continuous and compact map T : Ω Ω has at least one fixed point in the set Ω. Obviously the above formulated theorem constitutes the well know Schauder fixed point principle. c 215 NSP

Inf. Sci. Lett. 4, No. 2, 93-99 215 / www.naturalspublishing.com/journals.asp 95 3 Main result Now it is time to put forward the main theorem of this paper. Theorem 3.Assume that Ω is a nonempty, bounded, closed and convex subset of the Banach algebra E and the operators P and T continuously transform the set Ω into E such that PΩ and TΩ are bounded. Moreover, we assume that the operator S=P.T transform Ω into itself. If the operator P and T on the set Ω satisfy the following conditions { µpx ψ 1 µx, µtx ψ 2 µx, for any nonempty subset X of Ω, where µ is an arbitrary measure of noncompactness satisfying condition m and ψ 1,ψ 2 :R R are nondecreasing functions such that lim n ψ n 1 t=, lim n ψ n 2 t=, lim n PΩ ψ2 TΩ ψ 1 nt=, for any t, then S has at least fixed point in the set Ω. Proof.Let us take an arbitrary nonempty subset X of the set Ω. Then in view of the assumption that µ satisfies conditionm we obtain µsx µpx.tx PX µtx TX µpx PΩ µtx TΩ µpx PΩ ψ 2 µx TΩ ψ 1 µx = PΩ ψ 2 TΩ ψ 1 µx. 3 Now, letting ϕt = PΩ ψ 2 TΩ ψ 1 t, then from??, we have µsx ϕµx. Now, with regard to the fact that ψ 1 and ψ 2 are nondecreasing, we conclude ϕ is nondecreasing and in view of lim n ϕ n t =, we can apply main result in [6], to get the desired result. But for the convenience of the reader, we add the scheme of proof of aforementioned theorem. We define sequence Ω n as Ω = Ω, Ω n = ConvSΩ n 1 for n 1. Furthermore we assume µω n > for all n = 1,2,.... Keeping this condition in mind, we get µω n 1 =µconvsω n = µsω n ϕµω n ϕ 2 µω n 1... ϕ n µω. This showed that µω n as n. Now, we can use axiom 6 of definition of measure of noncompactness and conclude that Ω = n=1 Ω n is nonempty, convex and closed subset of the set Ω. Moreover it is noteworthy that Ω is compact. With regard to the above discussion Schauder fixed point principle guarantees the existence of a fixed point for the operator S. Remark.By letting { ψ 1 t=k 1, k 1 < 1 ψ 2 t=k 2, k 2 < 1 in Theorem 3, we obtain a special case of above theorem which has already been studied in[9, 1, 12, 18], where the application of that special case in the existence of solutions of many integral equation has been investigated. 4 Application In this section we use the main theorem of this paper to prove the solvability of integral equation xt=t xt ft,xt we define Fxt= ft,xt gt,s,xsds, t R, gt,s,xsds, t R where the operator T, F are defined on the Banach algebra BCR. Notice that F represented the so-called Volterra integral operator. Now, we formulate the assumptions under which the equation 1 will be investigated. We will assume the following hypotheses: IT is an operator acting continuously from Banach algebra BCR into itself which satisfies the following condition µ c TX ψ 1 µ c X for any nonempty subset X of Ω in which Ω is a nonempty, bounded, closed and convex subset of the Banach algebra BCR and ψ 1 : R R is a nondecreasing function such that lim n ψ n 1 t = for any t. IIThere exists a constant b such that Tx ψ 1 x b III f :R R is a continuous function. Moreover, t ft, is a member of the space BCR. IVThere exists an upper semicontinuous function ψ 2 : R R is nondecreasing function such that lim n ψ2 n t= for any t, we have that ft,x ft,y ψ 2 x y, t R, x,y R. Moreover, we assume that ψ 2 is superadditive i.e., for each t,s, R,ψ 2 tψ 2 s ψ 2 t s. c 215 NSP

96 A. Aghajani, M. Aliaskari: Measure of Noncompactness in Banach Algebra... Vg : R R R R is a continuous function and there exist continuous functions c,d : R R such that lim ct dsds= and gt,s,x ctds for t,s R such that s t, and for each x R. VIThe inequality ψ 1 r bψ 2 r q r has a positive solution r in which q is constant and defined as { } q=sup ft, ct dsds} : t. Moreover, the number r is such that ψ2 r qψ 1 ψ 1 r bψ 2 t<t for t R. The following lemma is necessary to prove the theorem 4. Lemma 1[6]. Let ϕ :R R be a nondecreasing and upper semicontinuous function. Then the following two conditions are equivalent 1lim n ϕ n t= for each t. 2ϕt< t for any t >. Theorem 4.Under the assumptions I to V I, the integral equation 1 has at least one solution in the space BCR. Proof.We define the operator A as follows Axt=T xtfxt. With regard to the above assumptions, the functions T x and Fx are continuous functions on R for any x BCR. For an arbitrary fixed function x BCR, we have So, we get Axt = T xt Fxt ψ 1 x b ft,xt ft, ft, gt,s,xs ds ψ 1 x bψ 2 xt ft, gt,s,xs ds ψ 1 x bψ 2 xt ft, ct dsds ψ 1 x bψ 2 xt q. Ax ψ 1 x bψ 2 xt q, in which b and q are constant, defined in assumptionsii, IV. So A maps the space BCR into itself. Moreover based of assumption IV, we conclude that A maps the ball B r into itself in which r is a constant appearing in assumption V I. Now we show that operator A is continuous on the ball B r. To do this, let us first observe that the continuity of the operator T on the ball B r is an easy consequence of the assumptions I, II, V I. Thus, it suffices to show that the operator F is continuous on B r. Fix an arbitrary ε > and x,y B r such that x y ε. So we can conclude Fxt Fyt ψ 2 xt yt where we denoted gt,s,xs gt,s,ys ds ψ 2 xt yt gt, s, xs ds gt, s, ys ds ψ 2 ε2kt, 4 kt=ct dsds. Further, in view of assumption V, we deduce that there exists a number L> such that 2kt = 2ct dsds ε, 5 for each t L. Thus, taking into account Lemma 1 and linking 5 and 4, for an arbitrary t L we get Fxt Fyt 2ε. 6 Now, we define the quantity ω L g,ε as follows ω L g,ε=sup{ gt,s,x gt,s,y : t,s [,L],x,y [ r,r ], x y ε}. Now with regard to the fact that the function gt,s,x is uniformly continuous on the set [,L] [,L] [ r,r ], so lim ε ωl g,ε=. By considering 4 for an arbitrary fixed t [, L], we conclude that L Fxt Fyt ψ 2 ε ω L g,εds = ψ 2 εlω L g,ε. 7 Combining 6 and 7, it is possible to conclude that the operator F is continuous on the ball B r. Now, let X be an arbitrary nonempty subset of the ball B r. Fix numbers ε > and L >. Next, choose t,s [,L] such that t s ε. Without loss of generality, we assume that c 215 NSP

Inf. Sci. Lett. 4, No. 2, 93-99 215 / www.naturalspublishing.com/journals.asp 97 s< t. Then, for x X we conclude Fxt Fxs ft,xt fs,xs gt,τ,xτdτ s gs, τ, xτdτ ft,xt fs,xt fs,xt fs,xs gt,τ,xτdτ gs,τ,xτdτ s ω L 1 f,εψ 2 xt xs gs, τ, xτdτ gs, τ, xτdτ gt,τ,xτ gs,τ,xτ dτ gs, τ, xτ dτ s ω1 L f,εψ 2 ω L x,ε ω1g,εdτ L cs dτdτ s ω1 L f,εψ 2 ω L x,ε Lω L 1g,εε sup{csdt : t,s [,L]} 8 where we denote ω1 L f,ε=sup{ ft,x fs,x : t,s [,L],x [ r,r ], t s <ε}, ω L 1g,ε=sup{ gt,t,x gs,t,x : t,s,t [,L],x [ r,r ], t s <ε}. Now with regard to the fact that f is uniformly continuous on the set [,L] [ r,r ] and g is uniformly continuous on the set [,L] [,L] [ r,r ], we can conclude ω1 L f,ε and ωl 1 g,ε as ε. Moreover, since c = ct and d = dt are continuous on R, the quantity sup{csdt : t,s [,L]} is finite. From 8, we conclude ω L FX lim ε ψ 2 ω L X,ε. Now with regard to the fact that ψ 2 is upper semicontinuous, so and so ω L FX ψ 2 ω L X, ω FX ψ 2 ω X. 9 Now we choose two arbitrary functions x,y X. Then for t R we have Fxt Fyt ft,xt ft,yt gt, s, xs ds gt, s, ys ds ψ 2 xt yt 2ct dsds ψ 2 xt yt 2kt. This estimate allows us to get the following one diamfxt ψ 2 diamxt 2kt. Now with regard to the upper semicontinuity of the functions ψ 2 we obtain cfx=limsup diamfxt ψ 2 limsup diamxt = ψ 2 cx. 1 So, combining 9 and 1, we can conclude µ c FX=ω FXcFX or, equivalently = ω FXlimsupdiamFXt ψ 2 ω Xt ψ 2 limsup diamxt ψ 2 ω Xtlimsup diamxt ψ 2 ω XtcX µ c FX ψ 2 µc X, moreover, by considering assumptioni we have µ c T X ψ 1 µc X, in which µ c is the defined measure of noncompactness on the space BCR. Also, we get TB r ψ 1 r b, FB r ψ 2 r q. So, according to assumptionv I, we have FBr ψ 1 T B r ψ 2 t< ψ2 r qψ 1 ψ 1 r bψ 2 t < t for all t R 11 Now, linking 11 and lemma 1 we get nt=. FBr ψ 1 TB r ψ 2 lim n Thus, all the conditions of Theorem 4 hold. Therefore Eq.1 has at least one solution in the space BCR. 5 Example Example 1.Consider the following functional integral equation t 2 xt= 1t 4 ln1 xt se t sinxs 1 cos xs ds t 2 55t 4 ln1 xt se t sinxs 3 cos xs ds, we define T xt= t2 1 t 4 ln1 xt se t sinxs 1 cosxs ds, Fxt= t2 55t 4 ln1 xt se t sinxs 3 cosxs ds. 12 c 215 NSP

98 A. Aghajani, M. Aliaskari: Measure of Noncompactness in Banach Algebra... Now, we show that all the conditions of Theorem 4 are satisfied for the functional integral equation 12. To do so, first we checked out whether condition IV and V are satisfied. Similar fashion, by putting ψ 1 t = 4 1 ln1 t condition I and II satisfied for the operator T, too. Moreover, we put ft,x = t2 ln1 x,gt,s,x = se t sinx 1t 6 1 cosx and ψ 2 t = 1 5 ln1 t. obviously, ψ 2 is nondecreasing and concave onr and ψ 2 t<t for all t >. In addition, for arbitrarily fixed x,y R such that x y and for t > we get ft,x ft,y = 1 t 2 5 22t 4 ln 1 x 1 y 1 5 ln 1 x y 1 y < 1 5 ln 1 x y < 1 4 ln 12 x y = ψ 2 x y. The case y x can be dealt with in the same way. Conditions III of Theorem 4 are clearly evident. In addition, pay close attention that the function g is continuous and maps the set R R R into R. Also, we have gt,s,x e t s for t,s R and x R. So, if we put ct= e t, and ds= s, then we can see that assumption V is satisfied. Indeed, we have lim ct dsds=. Now, let us calculate the constant q which appears in assumption VI. We obtain q=sup{ ft, ct dsds : t } = sup{2t 2 e t/2 : t }=2e 2. Just like the above way checked out condition II, we get b=2e 2 also see [6]. Furthermore, we can check that the inequality from assumptionv I takes the form 1 4 ln1rb 1 5 ln1rq < r. It is obvious that this inequality has a positive solution r, say r = 1. Moreover, we have ψ2 r qψ 1 ψ 1 r bψ 2 t < t for t R. Consequently, all the conditions of Theorem 4 are satisfied. Therefore the functional integral equation 12 has at least one solution in the space BCR. References [1] M. Kir and H.Kiziltunc, Fixed Point Theorems for Nonself Contraction Mappings in N-Normed Spaces, Inf. Sci. Lett. 3, No. 3, 111-115 214. [2] A. Aghajani, M. Aliaskari, Generalization of Darbo s fixed point theorem and application. International Journal of Nonlinear Analysis and Applications, 52, 86-95, 214. [3] A. Aghajani, M. Aliaskari, Commuting mappings and generalization of Darbos fixed point theorem, Math. Sci. Lett. In press. [4] C. Corduneanu, Integral Equations and Applications, Cambridge University Press, New York, 199. [5] K. Deimling, Nonlinear Functional Analysis, Springer- Verlag, Berlin, Heidelberg, New York, Tokyo, 1985. [6] A. Aghajani, J. Banaś, N, Sabzali, Some generalizations of Darbo fixed point theorem and applictions, Bull. Belg. Math. Soc. Simon Stevin 2 213, 345-358. [7] R. P. Agarwal, M. Meehan and D. O Regan, Fixed Point Theory and Applications, Cambridge University Press, Cambridge, 24. [8] J. Caballero, A. B. Mingarelli, K. Sadarangani, Existence of solutions of an integral equation of Chandrasekhar type in the theory of radiative transfer. Electronic Journal of Differential Equations, 26. Chicago [9] J. Banaś, S. Dudek, The Technique of Measures of Noncompactness in Banach Algebras and Its Applications to Integral Equations. In Abstract and Applied Analysis Vol. 213. Hindawi Publishing Corporation. [1] J. Banaś, M. Lecko, Fixed points of the product of operators in Banach algebra, Panamerican Mathematical Journal, 122, 1119, 22. [11] J. Banaś, K. Goebel, Measures of Noncompactness in Banach Spaces, Lecture Notes in Pure and Applied Mathematics, vol. 6, Marcel Dekker, New York, 198. [12] J. Banaś, L. Olszowy, On a class of measures of noncompactness in Banach algebras and their application to nonlinear integral equations, Zeitschrift fur Analysis und ihre Anwendungen, 284, 475-498, 29. [13] J. Banas, On measures of noncompactness in Banach spaces, Comment. Math. Univ. Carolinae 21 198 131-143. [14] X. Yu, C. Zhu, J.Wang. On a weakly singular quadratic integral equations of Volterra type in Banach algebras. Advances in Difference Equations, 2141, 13. [15] J. Banaś, M. Lecko, Fixed points of the product of operators in Banach algebra. Panamer. Math. J. 12 22, 11-19. [16] J. Banaś,K. Goebel, Measures of Noncompactness in Banach Spaces. Lect. Notes Pure Appl. Math. 6. New York: Marcel Dekker 198. [17] J. Banaś, Measure of noncompactness in the space of continuous tempered functions. Demonstratio Math. 14 1981, 127-133. [18] H. K. Pathak, Study on existence of solutions for some nonlinear functional-integral equations with applications. Mathematical Communications, 181213, 97-17. c 215 NSP

Inf. Sci. Lett. 4, No. 2, 93-99 215 / www.naturalspublishing.com/journals.asp 99 Asadollah Aghajani was born in 197. He graduated in 1993 and got his Ph.D. degree in 1999 in Pure Mathematics from Tarbiat Modares University Iran. Since 21 he is a Associate Professor of Mathematics at Iran University of Science and Technology IUST. His research interests are ordinary differential equations, partial differential equations, fixed point theory and applications and measure of noncompactness. Mahdi Aliaskari was born in 1988. He received his Bs.C. degree in mathematics 211, Ms.C. in mathematics 213 from IUST in the field of Pure Mathematics. He is a Ph.D. student at department of mathematics in IUST. His research interests include fixed point theory, measure of noncompactness and Geometry of Banach spaces and applications. c 215 NSP