UNIVERSITY OF CRAIOVA

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1 UNIVERSITY OF CRAIOVA DEPARTMENT OF MATHEMATICS Doctoral Thesis Qualitative Analysis on Some Classes of Nonlinear Problems Ph D Candidate: CEZAR LUPU Supervisor: Dr. VICENŢIU RĂDULESCU A thesis submitted in fulfilment of the requirements for the degree in PHILOSOPHY of MATHEMATICS CRAIOVA, ROMANIA 213

2 2 An overview of the Thesis This thesis contains two parts. The first one contains three chapters and it is dedicated to the theory of mean value theorems and their applications to linear integral operators. The second part contains three chapters and it is devoted to theory of variationalhemivariational inequalities and generalized variational-like inequalities. In Chapter 1 entitled The development of some mean value theorems and the Volterra operator, we present one of the main developments in the theory of mean value problems: Flett s mean value theorem. This result was proved by Flett [6] in 1958 and asserts that for a real valued function on closed interval [a, b], differentiable on (a, b), continuous on [a, b], and f (a) = f (b), there exists c (a, b) such that f (c) = f(c) f(a). c a This theorem as well as other of its extensions and generalizations will be used in the next two chapters in proving some mean value properties for some linear integral operators of Volterra type. On the other hand, in the second section of the chapter, we introduce the Volterra map, V : L 2 ([, 1]) L 2 ([, 1]) given by V f(x) := x f(t)dt, and we state and prove some of its basic properties. Chapter 2 entitled Mean value theorems for some linear integral operators is based on the paper Mean value theorems for some linear integral operators published in Electronic Journal of Differential Equations, 29 (29), no. 117, pp ([93]) and written in collaboration with Tudorel Lupu. The paper was cited in: Y. Lu, N. Yang, Symbol error probability of QAM with MRC diversity in two-wave with diffuse power fading channels, IEEE Communications Letters, 15 (211), (see [91]) D. Cakmak, A. Tiryaki, Mean value theorem for holomorphic functions, Eletronic Journal of Differential Equations, 212 (212), No. 34, 1 6. (see [86]) O. Hutnik, J. Molnarova, Flett s mean value theorem: a survery, preprint (September 213) at arxiv.org/pdf/ pdf. (see [61])

3 3 In this chapter, we study some mean value problems involving linear integral operators on the Banach space of continuous functions. The main tools used are are mean value theorems like Rolle and Flett in order to study existence of zeroes for linear integral operators like and T ϕ(t) := ϕ(t) SΨ(t) := tψ(t) ϕ(x)dx, xψ(x)dx, Rξ(t) := ξ(t) ξ (t) V ρ(t) := tρ(t) ρ (t) ξ(x)dx, xρ(x)dx, where ϕ, Ψ C([, 1]) and ξ, ρ C 1 ([, 1]). The main results of the chapter assert that the following mean value problems f(x)dx T g(c 1 ) = g(x)dx T f(c 1 ), T f(c 2 ) = Sf(c 2 ), (.1) f(x)dx Sg( c 4 ) = have solutions c 1, c 2, c 4 (, 1) for f, g C([, 1]), (1 x)f(x)dx T g(c 7 ) = (1 x)f(x)dx Sg( c 7 ) = have solutions c 7, c 7 (, 1) for f, g C([, 1]), f(x)dx Rg(c 3 ) = f(x)dx V g(c 4 ) = have solutions c 3, c 4 (, 1) for f, g C 1 ([, 1]), and g(x)dx Sf( c 4 ) (1 x)g(x)dx T f(c 7 ), g(1 x)(x)dx Sf( c 7 ) g(x)dx Rf(c 3 ), g(x)dx V f(c 4 ) (.2) (.3)

4 4 (1 x)f(x)dx Rg(c 8 ) = (1 x)f(x)dx V g( c 8 ) = (1 x)g(x)dx Rf(c 8 ), (1 x)g(x)dx V f( c 8 ) (.4) have solutions c 8, c 8 (, 1) for f, g C([, 1]), The main idea of the proof of the problem (.1) is based on the construction of the following auxiliary functions: and h 1 (t) = f(t) g(x)dx g(t) h 2 (t) = (t 1)f(t), f(x)dx, together with the facts: if h 1 C([, 1]) satisfies the condotion such that h 1 (c 1 ) = and c 4 h 1 ( c 4 ) = c1 h 1 (x)dx =, then there exist c 1, c 4 (, 1) c4 h 1 (x)dx, xh 1 (x)dx. if h 2 C([, 1]) satisfies the condition h 2 (1) =, then there exists c 2 (, 1) such that h 2 (c 2 ) = c2 h 2 (x)dx, In the case of problem (.2) the procedure is similar, but the auxiliary function we construct is given by h 8 (t) = f(t) (1 x)g(x)dx g(t) (1 x)f(x)dx, and by applying the same mean value results as above, but the condition is h 8 (x)dx = xh 8 (x)dx. Problems (.3) and (.4) are treated similarly by constructing the the same auxiliary functions but employing different mean value problems. Chapter 3 entitled Mean value problems of Flett type for a Volterra operator is mainly based on the paper Mean value problems of Flett type for a Volterra operator published in Electronic Journal of Differential Equations, vol. 29 (29), No. 53, pp.

5 (see [69]) However, section 3.2 contains a result about zeroes in the image of a Volterra map and it is based on the paper Zeroes of functions in the image of a Volterra operator published in Gazeta Matematică, A-series, 27 (29), with Radu Gologan. The papers have been cited in: O. Hutnik, J. Molnarova, Flett s mean value theorem: a survery, preprint (September 213) at arxiv.org/pdf/ pdf. (see [61]) In this chapter, we study mean value problems for pairs of integrals involving a Volterra map. The first problems we deal with is the following: f(x)dx V ϕ g(c) = g(x)dx V ϕ f(c) (.5) has a solution c (, 1), where f, g C([, 1]) and ϕ C 1 ([, 1]) satisfying certain conditions, where V ϕ f(t) := ϕ(x)f(x)dx is the Volterra map. Problem (.5) is treated in [88] under minimal assumptions for ϕ. In fact, it is shown that the mean value problem (.5) has a solution c (, 1) if ϕ : [, 1] R is nondecreasing, continuous at, and ϕ() =. On the other hand, in [96] it is assumed that ϕ C 1 ([, 1]), ϕ(x) and ϕ (x) for all x (, 1). The proof of this result is rather elementary, but tricky! The idea goes as it follows: 1 consider h : [, 1] R, given by h(x) = f(x) λg(x) where λ = f(x)dx 1. Now, problem g(x)dx (.5) reduces to prove that there exists c (, 1) such that provided that V ϕ h(c) = c ϕ(x)f(x)dx =, h(x)dx =. In this sense, we denote H : [, 1] R, given by H(t) = V ϕ h(t) = ϕ(x)h(x)dx. H(t) We first show that lim = and the argument continuues with integration by t + ϕ(t) parts in the Riemann-Stieltjes setting followed by the intermediate value property. In the next section of this chapter we prove more general results than (.5). The first main result is that the mean value problem V g f(c) = g(a) V f(c) (.6)

6 6 has a solution c (a, b) where f C null ([a, b]) and g C 1 ([a, b]), and g (x) for all x [a, b]. The main idea of the proof is to construct the following functions: and u(t) = a f(x)g(x)dx g(t) v(t) = g(t), t [a, b]. a f(x)dx, By an extension of Flett s mean value theorem, the conclusion follows immediately. On the other hand, the second main result states that the mean value problem V ϕ f(x ) g(x)dx V ϕ g(x ) ( f(x)dx = ϕ() V f(x ) has a solution x (, 1) for f, g C([, 1]). g(x)dx V g(x ) ) f(x)dx As in the proof of (.6) the idea is to consutruct the following auxiliary functions: (.7) and u(t) = (ϕ(t)v f(t) V ϕ f(t)) g(x)dx (ϕ(t)v g(t) V ϕ g(t)) v(t) = ϕ(t), t [, 1]. f(x)dx By the same extension of Flett s mean value theorem as in (.6) we obtain the conclusion. As a particular case, if ϕ() =, we obtain problem (.5). Chapter 4 entitled Variational, Variational-Hemivariational and Variationallike Inequalities. A brief review. introduces us to the theory of variational, variationalhemivariational inequalities and variational-like inequalities. Firstly, we present some basic, but important tools from Convex and Nonsmooth Analysis like: semicontinuous and convex functionals in Banach spaces, convex and Lipschitz functions, Clarke s generalized directional derivative, etc. Secondly, we review some important facts about monotone operators and related results that will be used in the next two chapters in proving existence of solutions for some variational-hemivariational and variational-like inequality problems. The variational inequality problem is the problem of finding u K, where K is a subset of a Banach space X which satisfies some properties of compactness and convexity, such that Au, v u, v K, (.8)

7 7 where, : X X R is the duality pairing. In general, a variational inequality can be formulated in any finite or infinite dimensional spaces. The ida of variational inequalities were extended and generalized to another class, which is known nowdays as the theory of hemivariational inequalities. The hemivariational inequalities have been introduced and inverstigated by Panagiotopoulos [27]. The problem is the following: Find u K such that given a nonlinear operator A : H H, where H is a real Hilbert space, we have Au, v u + f (x, u, v u)dω, v K. (.9) Ω Here f (x, u, v u) denotes the generalized directional derivative. Panagiotopoulos studied these kind of inequalities in order to formulate variational principles associated with energy functions which are not neither convex nor smooth. If f =, then problem (.9) becomes the classical variational inequality (.8). In 1989, Parida, Sahoo and Kumar proposed another type of variational inequality, namely variational-like inequality which is devoted to the following problem: have (.8). Find u K such that for two continuous maps A : K R n and η : K K R n, we Au, η(v, u), v K. (.1) If η(u, v) = v u, then we obtain again the classical variational inequality problem The proofs of the existence of solutions for these variational, variational-hemivariational and variational-like inequality problems are based on some topological (fixed point) principles for set-valued mappings due to Knaster-Kuratowski-Mazurkiewicz, Ky Fan, Kakutani, Tarafdar, Mosco or Ansari-Yao. For instance, one of the most used tools is the celebrated Knaster-Kuratowski-Mazurkiewicz (KKM) [15] topological principle obtained from the Sperner combinatorial lemma, and applied it to a simple proof of Brouwer s fixed point theorem, KKM principle. Let K be a nonempty subset of a Haussdorff topological vector space E and let G : K 2 E be a set-valued mapping satisfying the following conditions: (i) G has the KKM property: for any {x 1, x 2..., x n } K, we have co{x 1, x 2,..., x n }

8 8 n G(x i ), where co{x 1, x 2,..., x n } denotes the convex hull of {x 1, x 2,..., x n }; i=1 (ii) g(x) is closed in E for every x K; (iii) G(x ) is compact in E for some x K. Then G(x). x K In Chapter 5 entitled Variational-Hemivariational inequalities involving setvalued mappings is based on the paper On a class of variational-hemivariational inequalities involving set valued mappings published in Advances in Pure and Applied Mathematics, 1 (21), , in collaboration with Nicuşor Costea. (see [5]). The paper has been cited in: Y. L. Zhang, Y. R. He, On stably quasimonotone hemivariational inequalities, Nonlinear Analysis (TMA) 74 (211), (see [57]) G. Tang, N-J. Huang, Existence theorems of the variational-hemivariational inequalities, Journal of Global Optimization 56 (213), (see [55]) Y. L. Zhang, Y. R. He, The hemivariational inequalities for an upper semicontinuous set-valued mapping, Journal of Optimization Theory and Applications 156 (213), (see [58]) R. Wankeeree, P. Preeschasilp, Existence theorems of the hemivariational inequality governed by a multi-valued map perturbed with a nonlineaer term in Banach spaces, J. Global Optim. 57 (213), (see [56]) In this chapter, we establish some existence results for variational-hemivariational inequalities involving monotone set-valued mappings on bounded, closed and convex subsets in reflexive Banach spaces. We also derive several conditions for the existence of solutions in the case of unbounded subsets. The main goal of is to study the following problem: Find u K and u A(u) such that u, v u + ϕ(v) ϕ(u) + j (x, û(x); ˆv(x) û(x)) dx, v K, (.11) which is closely related to the following dual problem Ω

9 9 sup v, u v ϕ(v) ϕ(u) + j (x, û(x); ˆv(x) û(x)) dx, v K. (.12) v A(v) Ω Using the celebrated principle Knaster-Kuratowski-Mazurkiwievicz (KKM) and some other related topological results, we establish that if K is a non-empty, bounded, closed and convex subset of a real reflexive Banach space X, and A : K 2 X is a set-valued mapping which is monotone, lower hemicontinuous on K, and if T : X L p (Ω; R k ) is linear and compact, and j satisfies the condition: z C(1 + y p 1 ) almost everywhere x w, for all y R k and z j(x, y), then thevariational-hemivariational inequality (.11) has at least one solution. and The main idea of the proof is to define F, G : D(ϕ) 2 X as follows: G(v) = { F (v) = u K D(ϕ) : u A(u) such that u, v u + ϕ(v) ϕ(u)+ } + j (x, û(x); ˆv(x) û(x)) dx Ω { u K D(ϕ) : } sup v, u v ϕ(v) ϕ(u) + j (x, û(x); ˆv(x) û(x)) dx. v A(v) Ω The proof is divided into several steps, but the first goal is to show that the maps F and G are KKM followed by the fact that G is weakly compact for all v K D(ϕ). In the case when K is unbounded, let u n K n := {u K : u n} and u n A(u n ) be two sequences that satisfy u n, v u n + ϕ(v) ϕ(u n ) + j (x, û n (x); ˆv n (x) û n (x))dx (.13) Ω for all v K n and for every n 1. The second main result of the chapter asserts that under some hypothesis, the existence of solutions is guaranted for problem (.11). Chapter 6 entitled Variational-like inequality problems involving set-valued maps and generalized monotonicity is based on the paper Variational-like inequality problems involving set-valued maps and generalized monotonicity published in Journal of Optimization Theory and Applications, 155 (212), (see [51]) in collaboration with Nicuşor Costea and Daniel Alexandru Ion. The paper has been recently cited in:

10 1 S. Park, Recent applications of the Fan-KKM theorem, Publications of the Research Institute for Mathematical Sciences (RIMS-Kyoto), to appear. (see [54]) The aim of this chapter is to establish existence results for some variational-like inequality problems involving set-valued maps, in reflexive and non-reflexive Banach spaces. When the set K, in which we seek for solutions, is compact and convex, we do not impose any monotonicity assumptions on the set-valued map A, which appears in the formulation of the inequality problems. In the case when K is only bounded, closed, and convex, certain monotonicity assumptions are needed: We ask A to be relaxed η α monotone for generalized variational-like inequalities and relaxed η α quasimonotone for variationallike inequalities. We also provide sufficient conditions for the existence of solutions in the case when K is unbounded, closed, and convex. The main problems we study in this chapter are the following variational-like inequalities: Find u K D(ϕ) such that u A(u) : u, η(v, u) + ϕ(v) ϕ(u), v K, (.14) and Find u K such that u A(u) : u, η(v, u), v K, (.15) where K X is nonempty, closed, and convex, η : K K X, A X is a set-valued map and ϕ : X R {+ } is a proper convex and lower semicontinuous functional such that K ϕ := K D(ϕ). Here D(ϕ) stands for the domain of the functional ϕ, i.e. D(ϕ) = {u X : ϕ(u) < + }. One of the main results of this chapter states that for K nonempty, bounded, closed, and convex subset of the real reflexive Banach space X, and for A : K X relaxed η α monotone map, under certain conditions, problem (.14) admits at least one classical and strong solutions. The main idea of the proof is the use of the Mosco s alternative for the weak topology of X, and to define ξ, ζ : X X R, and ξ(v, u) = inf v A(v) v, η(v, u) + α(v u), ζ(v, u) = sup u, η(u, v). u A(u)

11 11 On the other hand, we also apply the celebrated KKM principle to derive classical and strong solutions to other variational-like inequality problems. More exactly, if we weaken even more the hypothesis on the set-valued mapping A : K X is relaxed η α quasimonotone, the existence of solutions for problem (.14) is still open. However, we prove that problem (.15) admits, under certain hypothesis, at least one classical solutions and at least one strong solution. The main idea is to define the map G : K X in the following way: G(v) = {u K : v, η(v, u) α(v u), v A(v)}. The proof continues with the use of the KKM principle in order to obtain at least one strong solution for problem (.15). Acknoledgements First of all, I would like to express my deepest respect to my advisor, Professor Vicenţiu Rădulescu who offered me a chance and accepted me as his Ph D student. His help and guidance during my thesis were crucial and his personality influenced me a lot in my research and probably in my future career as a mathematician although I must confess I had some moments of imaturity regarding research projects. I benefited his trust and support in all moments in my last years of mathematical activity. Furthermore, I am indebted to my collaborators: Nicuşor Costea, Radu Gologan, Daniel Ion, and Tudorel Lupu. This thesis would have been poorer without their support and fruitful cooperation. A special mention here goes to Nicuşor Costea who taught me some tricks related to variational-hemivariational inequalities. Also, I would like to thank my friends Grigore Danciu, Alin Gălăţan, Octavian Ganea, Vlad Matei, Roxana Tănase and Adrian Vladu for their valuable suggestions regarding the content of this thesis and for L A TEX help. Nevertheless, I have to express my gratitude to my professors who influenced and taught me mathematics during all these years. I will name them in alphabetical order: Alexander Borisov, Andrea Cianchi, George Dincă, Piotr Hajlasz, Radu Gologan, Chris Lennard, Constantin Niculescu, Yibiao Pan, Laurenţiu Panaitopol, Călin Popescu, Vicenţiu Rădulescu, Şerban Strătilă, and Victor Vuletescu. Also, I want to thank the

12 12 Department of Mathematics for their help and for the constructive mathematical and non-mathematical conversations we had during the years. A special mention goes here to: Magdalena Boureanu, Dumitru Buşneag, Mihai Mihăilescu, Constantin Niculescu, Ionel Rovenţa, Denisa Stancu. Many thanks to George Gîrleşteanu and Mădălin Ticu from the strategic grant POSDRU /6/1.5/S/14 for their unconditional help and patience on the burreaucracy regarding research projects Last but not least, I would like to thank my family for support and love in all these years. A special mention goes to my mother Silvia who carried out during college and Ph D and, of course, my aunt Lucia for treating me like a son over the last ten years. This thesis is dedicated to both of them. This research has been supported by the strategic grant POSDRU /6/1.5/S/14, cofinanced by the European Social Fund-Investing in People, within the Sectorial Operational Programme Human Resources Development Craiova, 213

13 CONTENTS Part I Mean Value Theorems and Linear Integral Operators The development of some Mean Value Theorems and the Volterra operator Flett s mean value theorem and its extensions and generalizations The Volterra operator Mean Value Theorems for some Linear Integral Operators Introduction and Preliminaries Main results Mean Value Problems of Flett type for a Volterra operator Introduction Zeroes in the image of a Volterra operator Main result Discussion and Some examples Part II Variational-Hemivariational and Variational-like Inequalities Variational, Variational-Hemivariational and Variational-like Inequalities. A brief review Some preliminaries of Convex and Nonsmooth analysis Multivalued mappings. Continuity and monotonicity Some inequality problems Variational-Hemivariational Inequalities Involving Set-Valued Mappings Introduction The abstract framework Existence results

14 Contents Variational-like inequality problems involving set-valued maps and generalized monotonicity Introduction The Studied Problems and Preliminary Results Existence Results The Case of Nonreflexive Banach Spaces The Case of Reflexive Banach Spaces Concluding Remarks

15 Part I MEAN VALUE THEOREMS AND LINEAR INTEGRAL OPERATORS

16 1. THE DEVELOPMENT OF SOME MEAN VALUE THEOREMS AND THE VOLTERRA OPERATOR 1.1 Flett s mean value theorem and its extensions and generalizations Mean value theorems of differential and integral calculus are an essential and powerful tools in Real Analysis. The simplest forms of the mean value theorems are due to Rolle, Fermat, Lagrange, Cauchy, and so on. For a continuous real-valued function f on a compact interval [a, b], the difference of the values of f at the end points of [a, b], provided that the derivative f (a) exists, can be estimated as f(b) f(a) f (a)(b a). In 1958, T. M. Flett [6] proved a surprising result which can be interpreted as a variant of Lagrange s mean value theorem with Rolle-type condition. Theorem (Flett, 1958) If f : [a, b] R is differentiable on (a, b), continuous on [a, b], and f (a) = f (b), then there exists c (a, b) such that f(c) f(a) c a = f (c). We will present two different proofs for this theorem. First proof is the original proof of Flett and the second can be found in [67] and [69]. Moreover, there is a nice geometric interpretation of Flett s theorem: if the curve y = f(x) has tangent at each point in (a, b) and if the tangents at (a, f(a)) and (b, f(b)) are parallel, then there exists a point (c, f(c)) such that the tangent at (c, f(c)) passes through the point (a, f(a)). First proof. [6] Without loss of generality, we may assume that f (a) = f (b) =, otherwise one can work with f(x) xf (a). Now, let us consider the following function Ψ(x) = f(x) f(a) x a Ψ(a) = f (a),a < x b,x = a

17 1. The development of some Mean Value Theorems and the Volterra operator 17 It is evident that Ψ is continuous in a x b and differentiable in a < x b, and moreover, we have Ψ f(x) f(a) (x) = (x a) 2 + f (x) x a, a < x b. Therefore, it is sufficient to prove that there exists c (a, b) such that Ψ (c) =. If Ψ(b) =, by Rolle s theorem we are done. If not, suppose that Ψ(b) > so that Ψ (b) = Ψ(b) b a <. Then, there exists x 1 with a < x 1 < b such that Ψ(x 1 ) > Ψ(b). Since Ψ is continuous in a x x 1 and Ψ(a) < Ψ(b) < Ψ(x 1 ), by the intermediate value property, there exists a point x 2 with a < x 2 < x 1 such that Ψ(x 2 ) = Ψ(b). Now, BY applying again Rolle s theorem on the interval (x 2, b) we have the conclusion. Similarly, one can argue in the case when Ψ(b) <. Second proof. [67, 69] Let us consider the folowing continuous function: Ψ(x) = f(x) f(a) x a f (a),a < x b,x = a If Ψ achieves an extremum at an interior point c (a, b), then by Fermat s theorem, Ψ (c) = and we are done. Now, let s assume that the only extremum points of Ψ are a and b. Without loss of generality, one can suppose that for all x [a, b] we have Ψ(a) Ψ(x) Ψ(b). We have and for any x [a, b], we obtain f(b) f(x) b x f(x) f(a) + (x a)ψ(b), f(b) f(x) (x a)ψ(b) b x = f(b) f(a). b a By passing to the limit when x b, x < b, we have f (b) Ψ(b) and by f (a) Ψ(b), we obtain Ψ(a) Ψ(b). Thus, it follows that Ψ is constant, so Ψ (x) = for all x (a, b) and the proof ends here. Ana analogue of Flett s theorem for twice differentiable functions is given by the following Theorem Let f : [a, b] R be a twice differentiable function such that f (a) = f (b). Then ther exists c (a, b) such that

18 1. The development of some Mean Value Theorems and the Volterra operator 18 f(c) f(a) = (c a)f (c) (c a)2 f (c). 2 Proof. From f (a) = f (b), by Flett s mean value theorem, there exists λ (a, b) such that (λ a)f (λ) = f (λ) f (a). Now, let us consider the function Ψ : [a, b] R, given by Ψ(x) = (x a)f (x) 2f(x) + xf (a). A simple computation of the first derivative shows that Ψ (x) = (x a)f (x) f (x) + f (a). Clearly Ψ (a) = Ψ (λ) =, so by the Flett s mean value theorem there exists c (a, λ) (a, b) such that (c a)ψ (c) = Ψ(c) Ψ(a), which is equivalent to (c a) 2 f (c) (c a)(f (c) f (a)) = (c a)f (c) 2f(c) + cf (a) + 2f(a) af (a) or f(c) f(a) = (c a)f (c) (c a)2 f (c). 2 Going further, in 1966, Trhan [99] removed the boundary condition on derivatives in Flett s theorem and obtained the following Theorem (Trahan, 1966)Let f : [a, b] R be differentiable function on (a, b), continuous on [a, b] and (f (a) m)(f (b) m) >, where m = f(b) f(a) b a. Then there exists c (a, b) such that f (c) = f(c) f(a). c a Proof. [99] First, we start with the following lemmata: Lemma If f : [a, b] R is differentianble on (a, b], continuous on [a, b] and f (b)(f(b) f(a)), then there exists c (a, b] such that f (c) =.

19 1. The development of some Mean Value Theorems and the Volterra operator 19 Proof. If f(a) = f(b), then the conclusion is clear by Rolle s theorem. If f (b) =, then let c = b. If f (b)(f(b) f(a)) <, then there is a maximum or minimum value at c (a, b) and f (c) =. The next lemma is obvious. Lemma If f : [a, b] R is differentiable on (a, b], continuous on [a, b], and f (b)(f(b) f(a)) <, then there exists c (a, b) such that f (c) =. Now, let us consider the function h : [a, b] R given by h(x) = f(x) f(a) x a f (a),a < x b,x = a Clearly, h is continuous on [a, b] and differentiable on (a, b]. Moreover, the derivative is given by h (x) = (x a)f (x) (f(x) f(a)) (x a) 2. By Lemma 1, providing that h (b)(h(b) h(a)), there exists c (a, b] such that h (c) = or equivalently (c a)f (c) = f(c) f(a). A complementary result with Flett s theorem is the following variant obtained by Meyers [95], Theorem (Meyers, 1977) If f : [a, b] R is differentiable on [a, b], and f (a) = f (b), then there exists c (a, b) such that f (c) = f(b) f(c). b c In 1999, Davitt, Powers, Riedel and Sahoo [71] extended Flett s mean value theorem by removing the hypothesis f (a) = f (b). They proved the following: Theorem (Davitt-Powers-Riedel-Sahoo, 1999) If f : [a, b] R is a differentiable function, then there exists c (a, b) such that f(c) f(a) = (c a)f (c) 1 2 f (b) f (a) (c a) 2. b a

20 1. The development of some Mean Value Theorems and the Volterra operator 2 Proof. [71] Construct the following auxilliary function Ψ : [a, b] R, Ψ(x) = f(x) 1 2 f (b) f (a) (x a) 2. b a Clearly, Ψ is differentiable on [a, b], and the derivative is given by Ψ (x) = f (x) f (b) f (a) (x a). b a It follows that Ψ (a) = Ψ (b) = f (a) and by Flett s mean value theorem, there exists c (a, b) such that which is equivalent to our conslusion. Ψ(c) Ψ(a) = (c a)ψ (c), In fact, thwe above theorem was used to extend Flett s mean value theorem for holomorphic functions. More details are given in [71]. In 1999, Pawlikowska [78] generalized Flett s mean value theorem answering a question of Zsolt Pales raised during the 35-th International Symposium on Functional Equations held in Graz in Theorem (Pawlikowska, 1999) Let f : [a, b] R be a n-times differentiable on [a, b), continuous on [a, b], and f (n) (a) = f (n) (b). Then there exists c (a, b) such that f(c) f(a) = n ( 1) k 1 k=1 k! (c a) k f (k) (c) + ( 1)n (n + 1)! f (n) (b) f (n) (a) (c a) n+1. b a The proof given bt Pawlikowska [78] follows the original idea of Flett [6] by considering the following auxilliary function: F f (x) = and using succesively Rolle s theorem. f (n) (x) f (n) (a) x a 1 n f (n) (a),a < x b,x = a How ever, Molnarova [97] gave a different proof for Pawlikowska s result by considering the following auxilliary function: k ϕ k (x) = xf (n k+1) ( 1) i+1 (a) + (k i)(x a) i f (n k+i) (x), k = 1, 2,..., n, i! i=

21 1. The development of some Mean Value Theorems and the Volterra operator 21 and by applying Flett s mean value theorem for all its derivatives. In the last decade, there have been many applications of Flett s mean value theorem in connection with Functional Equations, Integral Operators, divided differences, stability pooints, etc. For example, Riedel and Sablik [82] proved a mean value theorem similar with Flett s and Meyers mean value theorems. MOreover, they associatted it to a certain functional equation that it is solved in a general setting on abelian groups. The following is proved Theorem (Riedel-Sablik, 24) Le tf : [a, b] R be a real valued differentiable function on [a, b]. Then there exists c (a, b) such that ( 1 f (c) c a ) f(c) f(a) + 1 ( f (c) c a c b Proof. [82] Construct the following auxilliary function: ) f(c) f(b) c b = f (b) f (a). b a ϕ(x) = f(x) f(a) x a + f(x) f(b) x b,x (a, b) f (a) + f(a) f(b) a b,x = a f(a) f(b) a b + f (b),x = b Clearly ϕ is continuous on [a, b] and differentiable on (a, b). Moreover, the first derivative is given by ϕ (x) = f (x) f(x) f(a) x a (x a) 2 + f (x) f(x) f(b) x b (x b) 2. By the Lagrange s mean value theorem, the conclusion follows immediately. This thoerem is applied in solving the following functional-differential equation ( ) x + y 8f + 4f(x) + 4f(y) f (y)x f (y)y f (x)x + f (x)y =. 2 On the other hand, the mean value theorem of Flett plays an important role in proving some other mean value thoerems of the same type for generalized derivatives and divided differences. For distinct points t 1, t 2,..., t n in [a, b] we define the divided differences of a function f as follows [t; f] = f(t 1 ) and [t 1, t 2,..., t n ; f] := [t 1, t 2,... t n 1 ; f] [t 2, t 3,... t n ; f] t 1 t n.

22 1. The development of some Mean Value Theorems and the Volterra operator 22 For all positive integers n 2, the n-point divided difference of f can be expressed as [t 1, t 2,... t n ; f] = n i=1 f(t i ) n j i,j=1 (x i x j ). An example is the mean value theorem for n-point difference (see [73]) which states that if f : [a, b] R is (n 1)-times differentiable and t 1, t 2,... tn are n distinct points in the interval [a, b], then there exists c [min{t 1,... t n }, max{t 1,... t n }] such that [t 1, t 2,... t n ; f] = f (n 1) (c) (n 1)!. In 24, Ivan, Abel and Riedel [76] proved the following extensions of Flett s mean value theorem for divided differences Theorem If f : [a, b] R is continuous and the generalized derivatives satisfy D n f(a) = D n f(b), then there exists a system of knots a < t <... < t n < b (a, b) such that and [a, t,... t n ; f] =. Theorem If f : [a, b] R is continuous and possesses derivatives of order n at a and b such that f (n) (a) = f (n) (b), then there exists c (a, b) such that in any neighborhood of the point c there exist equidistant points c < c 1 <... < c n, c < c < c n such that [a, c,..., c n ; f] =. 1.2 The Volterra operator As it has already been highlighted in [64], the Volterra operator v defined on the space of square integrable functions L 2 ([, 1]) by V f(x) := x f(t)dt, plays an important role in the development of the operator theory in Hilbert spaces. Usually, the Volterra oeprator is defined by V f(x) = x K(x, t)f(t)dt, where K(x, t) is called the kernel of V. This operator is strongly related to the Volterra equations of first and second kind,

23 1. The development of some Mean Value Theorems and the Volterra operator 23 and g(x) = x K(x, t)f(t)dt, g(x) = f(x) + x K(x, t)f(t)dt. In our presentation, we shall prove some basic properties for the Volterra map when the kernel K(x, t) = 1. On the other hand, one can view Volterra operator as a special case of Fredholm operator when K(x, y) = for y > x. Regarding the second kind Volterra integral equations we have the following Theorem ([64]) The Volterra integral equation of second kind, g(x) = f(x) + x K(x, t)f(t)dt, with continuous kernel K for each right hand side f C([a, b]) has a unique solution g C([a, b]). In what follows we state and prove some well-known properties of the Volterra map. Theorem The Volterra operator V : L 2 ([, 1]) L 2 ([, 1]) defined by V f(x) = has the following properties: x f(t)dt, x [, 1] a) V is positive semidefinite for any f L 2 ([, 1]); b) V has an adjoint V given by V f(x) = x f(t)dt, f L 2 ([, 1]). c) V is compact and the norm is given by V = 2 π. d) I + V is invertible and (I + V ) 1 = 1. e) V has no eigenvalues and it is quasinilpotent. Proof. [63, 68, 64, 65] a) This is evident since V f; f = 1 ( 2 f(t)dt). 2 b) For the kernel K : [, 1] [, 1] R, K(x, t) = χ [,x] (t), for x, t [, 1], we define the map

24 1. The development of some Mean Value Theorems and the Volterra operator 24 and observe that V f(x) := K(x, t)f(t)dt V f(x) = K(x, t)f(t)dt = χ [,t] (x)f(t)dt = x f(t)dt. c) We first show that V is bounded. By the Cauchy-Schwarz s inequality, we have V f 2 L 2 = x f(t)dt 2 dx ( ( x ) 1/2 ( ) x 1/2 2 f(t) 2 dt dt) dx f(t) 2 dt = f 2 L 2. Suppose that {f n } n is a bounded sequence of functions in L 2. We have to prove that {V f n } n contains a uniformly convergent subsequence, and hence a L 2 subsequence. We shall use Arzela-Ascoli in this sense. First, we show that {V f} is uniformly bounded sequence of continuous functions. Indeed, this follows from the fact that V f n (x) = x f n (t)dt x f n L 2 f n L 2. Now, we show that V f n is a sequence of equicontinuous functions. Take ɛ >, and let δ = ɛ2 C 2, where f n L 2 C for all n. Then for x y < δ, we have V f n (x) V f n (y) = y x f n (t)dt x y f n L 2 ɛ. By the Arzela-Ascoli theorem, V f n contains a convergent subsequence and thus V is compact in the Hilbert space L 2 ([, 1]). eigenvalue of the map V V. We obtain the following equation: x ( y We need to compute the root of the greatest ) f(t)dt dy = λf(x), x [, 1]. By differentiating the above equation and taking into account that f(1) =, we obtain x f(t)dt = λf (x), x [, 1], and thus we have f () =. By differentiating one more time, we obtain f(x) = λf (x), x [, 1].

25 1. The development of some Mean Value Theorems and the Volterra operator 25 This differential equation has the solution f(x) = a cos x λ + b sin x λ, a, b C, x [, 1]. From f(1) = f () = it follows that cos 1 λ = and thus, the solutions are given by 4 λ k = (2k + 1) 2 π 2, k =, 1, 2,.... Finally, V 2 = sup λ k = 4 k π 2, so V = 2 π. d) Since V < 1 it follows that I V is invertible. Now, for f L 2 ([, 1]), we have (I + V )(f) 2 = (I + V )(f); (I + V )(f) = f 2 + (V + V )(f); f + V f 2 f 2. This implies that (I + V ) 1 1. Now, let us prove that (I + V ) 1 1. Assuem by contradiction that (I + V ) 1 < 1. Again, this implies that I (I + V ) 1 is invertible. We have (I (I + V ) 1 ) 1 (I + V ) 1 = [(I + V )(I (I + V ) 1 )] 1 = (I + V I) 1 = V 1, in contradiction with the fact that V is not invertible. e) Assume that V has an eigenvalue λ. Thus, we have V f = λf for f which is equivalent to Since f L 2 ([, 1]), we have: f(x) = 1 λ x f(t)dt. f(y) f(x) = x y y f(t)dt f(t) dt = x ( 1/2 f(t) χ [x,y] (t)dt f L 2 (χ [x,y] (t)) dt) 2 = = f L 2 y x. This implies that f is continuous, and thus, f is differentiable. By differentiating we obtain the equation: f(t) = λf (t), and the solution is given by f(t) = Ce t/λ. We obtain: Ce t/λ = 1 λ x Ce t/λ dt = Ce x/λ C,

26 1. The development of some Mean Value Theorems and the Volterra operator 26 so C =, and thus f = is the only solution, false. So, λ cannot be an eigenvalue for V. The only thing we need to show now is that V is quasinilpotent. In fact, we have to prove that ρ(v ) = lim n (V f(x))n 1/n =. By an easy induction combined with Fubini s theorem, one can prove that Now, we estimate the norm V n f 2 L 2 = (V n f)(x) = x f(t) x f(t) (x t)n 1 (n 1)! (x t)n 1 dt. (n 1)! ( dt dx 1 (n 1)! since (x t) n 1 for t [, x] and x 1. Now, we clearly have ( ) V n f 1/n 1 2. (n 1)! ) 2 f 2 L 2, On the other hand, by the Stirling approximation, we have the inequality (n 1)! ( 2π) 1/n (n 1) 1 1 2n e 1+ 1 n, and we finally obtain that ρ(v ) = lim n V n 1/n =.

27 2. MEAN VALUE THEOREMS FOR SOME LINEAR INTEGRAL OPERATORS 2.1 Introduction and Preliminaries Mean value theorems play a key role in analysis. The simplest form of the mean value theorem is the next basic result due to Rolle, namely Theorem Let f : [a, b] R be a continuous function on [a, b], differentiable on (a, b) and f(a) = f(b). Then there exists a point c (a, b) such that f (c) =. Rolle s theorem has also a geometric interpretation which states that if f(a) = f(b) then there exists a point in the interval (a, b) such that the tangent line to the graph of f is parallel to the x-axis. There is another geometric interpretation as pointed out in [67], namely the polar form of Rolle s theorem. As been noticed in [67], if we take into account the geometric interpretation of Rolle s theorem, one expects that it is possible to relate the slope of the chord connecting (a, f(a)) and (b, f(b)) with the value of the derivative at some interior point. There are also other mean value theorems like Lagrange, Cauchy, Darboux which are well-known and can be found in any undergraduate Real Analysis course. In 1958, Flett gave a variation of Lagrange s mean value theorem with a Rolle type condition, namely Theorem ([67, 69]). Let f : [a, b] R be a continuous function on [a, b], differentiable on (a, b) and f (a) = f (b). Then there exists a point c (a, b) such that f (c) = f(c) f(a). c a A detailed proof can be found in [69] and some applications are provided too. The same proof appears also in [67]. A slightly different proof which uses Rolle s theorem instead of Fermat s, can be found in [6] and [98]. There is a nice geometric interpretation for Theorem 2.1.2, namely: If the curve y = f(x) has a tangent at each point in [a, b], and if the tangents at (a, f(a)) and (b, f(b)) are parallel, then there is an intermediate point c (a, b) such that the tangent at (c, f(c)) passes through the point (a, f(a)). Later,

28 2. Mean Value Theorems for some Linear Integral Operators 28 Riedel and Sahoo [98] removed the boundary assumption on the derivatives and prove the following Theorem ([98]). Let f : [a, b] R be a differentiable function on [a, b]. Then there exists a point c (a, b) such that f(c) f(a) = (c a)f (c) 1 2 f (b) f (a) (c a) 2. b a The proof relies on Theorem applied to the auxiliary function α : [a, b] R defined by α(x) = f(x) 1 2 f (b) f (a) (x a) 2. b a We leave the details to the reader. We also point out that this theorem is used to extend Flett s mean value theorem for holomorphic functions. In this sense, one can consult [71]. On the other hand, there exists another result due to Flett as it is pointed out in [67] for the second derivative of a function. Theorem If f : [a, b] R is a twice differentiable function such that f (a) = f (b) then there exists c (a, b) such that f(c) f(a) = (c a)f (c) (c a)2 f (c). 2 There exists another version of Flett s theorem for the antiderivative: Theorem Let f : [, 1] R be a continuous function such that f(a) = f(b). Then there exists c (a, b) such that c a f(x)dx = (c a)f(c). Similar to Theorem there exists another mean value theorem due to Penner (problem 987 from the Mathematics Magazine, [67]) that we shall apply in the next section. Theorem Let f : [a, b] R be a differentiable function with f be continuous on [a, b] such that there exists λ (a, b) with f (λ) =. Then there exists c (a, b) such that f (c) = f(c) f(a). b a The proof of the above theorem can be found in [67, page 233]. The version of Theorem for antiderivative is the following theorem.

29 2. Mean Value Theorems for some Linear Integral Operators 29 Theorem Let f : [a, b] R be a continuous function such that there is λ (a, b) such that f(λ) =. Then there is c (a, b) such that c a f(x)dx = (b a)f(c). In 1966, Trahan [99] extended Theorem by removing the condition f (a) = f (b) to the following theorem. Theorem Let f : [a, b] R be a continuous function on [a, b], differentiable on (a, b) such that (f (a) m)(f (b) m) >, where m = f(b) f(a) b a. Then there exists a point c (a, b) such that f (c) = f(c) f(a). c a A proof for the above theorem can be found in [74]. At the 35-th International Symposium on Functional Equations held in Graz in 1997, Zsolt Pales raised a question regarding a generalization of Flett s theorem. An answer to his question was given by Pawlikowska in [78], namely: Theorem If f possesses a derivative of order n on the interval [a, b], then there exists a point c (a, b) such that n ( 1) k 1 f(c) f(a) = f (k) (c)(c a) k + ( 1)n k! (n + 1)! f (n) (b) f (n) (a) (c a) n+1. b a k=1 Other generalizations and several new mean value theorems in terms of divided differences are given in [76]. For further reading concerning mean value theorems we recommend [74]. 2.2 Main results In this section, we shall prove mean value value problems for some function mapping. Our main results are for continuous, real-valued functions defined on the interval the [, 1]. We also mention that the results can be easily extended to the interval [a, b]. Before proceeding into the main results of the paper, we state and prove a lemmata. We give more proofs to some lemmas, which we consider very instructive. We start with the first two lemmas involving the integrant factor e t.

30 2. Mean Value Theorems for some Linear Integral Operators 3 Lemma Let h 1 : [, 1] R be a continuous function with h 1(x) =. Then there exists c 1 (, 1) such that h 1 (c 1 ) = c1 h 1 (x)dx. Proof. [First proof] We assume by the way of contradiction, that h 1 (t) > h 1 (x)dx, t [, 1]. Now, we consider the auxiliary function ζ 1 : [, 1] R, given by A simple calculation gives ζ 1 (t) = e t h 1 (x)dx. ζ 1(t) = e t( h 1 (t) ) h 1 (t)dt >, and from our assumption we deduce that ζ 1 is strictly increasing. This means that ζ 1 () < ζ 1 (1) which is equivalent to < 1 e h 1(x)dx =, a contradiction. Second proof. Like in the previous proof, let us consider the differentiable function γ 1 = ζ 1 : [, 1] R, defined by A simple calculation yelds γ 1 (t) = e t h 1 (x)dx. γ 1(t) = e t( h 1 (t) ) h 1 (t)dt. More than that we have γ 1 () = γ 1 (1) =, so by applying Theorem 2.1.1, there exists c 1 (, 1) such that γ 1 (c 1) =, i.e. h 1 (c 1 ) = c1 h 1 (x)dx. Third proof. We shall use Theorem Indeed from the hypothesis h 1(x)dx =, by applying the first mean value theorem for integrals we obtain the existence of λ (, 1) such that = h 1 (x)dx = h 1 (λ). Now, by Theorem there exists c 1 (, 1) such that h 1 (c 1 ) = c1 h 1 (x)dx. Similarly with Lemma 2.2.1, we prove the following result.

31 2. Mean Value Theorems for some Linear Integral Operators 31 Lemma Let h 2 : [, 1] R be a continuous function with h 2 (1) =. Then there exists c 2 (, 1) such that h 2 (c 2 ) = c2 h 2 (x)dx. Proof. [First proof] Let us consider the following auxiliary function ζ 2 : [, 1] R, given by ζ 2 (t) = e t h 2 (x)dx. Suppose by the way of contradiction that h 2 (t) h 2(x)dx, t [, 1]. This means that, without loss of generality, we can assume that h 2 (t) > h 2 (x)dx, t [, 1]. (2.1) A simple calculation of derivatives of the function ζ 2 combined with the inequality above, gives us the following inequality ζ 2(t) = e t( h 2 (t) ) h 2 (t)dt >, so our function ζ 2 is strictly increasing for every t (, 1). This means that ζ 2 (1) > ζ 2 (). It follows immediately that 1 e h 2(x)dx >. On the other hand, taking into account our assumption, namely (2.1), we deduce in particular, that h 2 (1) > h 2(x)dx > which contradicts the hypothesis h 2 (1) =. Second proof. Let us consider the differentiable function γ 2 : [, 1] R, defined by A simple calculation yields γ 2(t) = e t( γ 2 (t) = te t h 2 (x)dx. ( h 2 (x)dx t h 2 (t) )) h 2 (x)dx. Taking into account that h 2 (1) =, it is clearly that γ 2 () = γ 2 (1), so by Flett s mean value theorem (Theorem 2.1.2) (see [6]), we deduce the existence of c 2 (, 1) such that which is equivalent to or ( c 2 e c c 2 2 h 2 (x)dx c 2 (h 2 (c 2 ) γ 2(c 2 ) = γ 2(c 2 ) γ 2 () c 2 h 2 (c 2 ) = c2 c2 )) c2 h 2 (x)dx = c 2 e c 2 h 2 (x)dx, h 2 (x)dx.

32 2. Mean Value Theorems for some Linear Integral Operators 32 Following the same idea from lemma 2.1 we state and prove the following result. Lemma Let h 3 : [, 1] R be a differentiable function with continuous derivative such that h 3(x)dx =. Then there exists c 3 (, 1) such that h 3 (c 3 ) = h 3(c 3 ) c3 h 3 (x)dx. Proof. As in the proof of Lemma 2.2.1, let us consider the differentiable function γ 3 = ζ 3 : [, 1] R, defined by A simple calculation yields γ 1 (t) = e h 3(t) γ 3(t) = e h 3(t) ( h 3 (t) h 3(t) h 3 (x)dx. ) h 3 (t)dt. More than that we have γ 3 () = γ 3 (1) =, so by applying Theorem 2.1.1, there exists c 3 (, 1) such that γ 3 (c 3) =, i.e. h 3 (c 3 ) = h 3(c 3 ) c3 h 3 (x)dx. In what will follow we prove other technical lemmas without the integrant factor e t. Lemma Let h 4 : [, 1] R be a continuous function with h 4(x)dx =. Then there exists c 4 (, 1) such that c4 xh 4 (x)dx =. Proof. [First proof] We assume by contradiction that xh 4(x)dx, for all t (, 1). Without loss of generality, let xh 4(x)dx >, for all t (, 1) and let H 4 (t) = h 4(x)dx. Integrating by parts, we obtain < xh 4 (x)dx = th 4 (t) H 4 (x)dx, t (, 1). Now, by passing to the limit when t 1, and taking into account that H 4 (1) =, we deduce that H 4 (x)dx. (2.2)

33 2. Mean Value Theorems for some Linear Integral Operators 33 Now, we consider the differentiable function, µ : [, 1] R defined by 1 t t µ(t) = H 4(x)dx, if t, if t =. It is easy to see µ (t) = (th 4 (t) H 4(x)dx)/t 2 >, so µ is increasing on the interval (, 1), so it is increasing on the interval [, 1] (by continuity argument). Because µ() =, it follows that H 4 (x)dx >, which is in contradiction to (2.2). So, there exists c 4 (, 1) such that c4 xh 4 (x)dx =. Second proof. We consider the differentiable function H : [, 1] R, defined by H(t) = t h 4 (x)dx xh 4 (x)dx with H (t) = h 4(x)dx. It is clear that H () = H (1) = h 4(x)dx =. Applying Flett s mean value theorem (see [6]), there exists c 4 (, 1) such that or H (c 4 ) = H(c 4) H() c 4 c4 c4 c 4 h 4 (x)dx = c 4 c4 h 4 (x)dx xh 4 (x)dx which is equivalent to c 4 xh 4(x)dx =. Third proof. Let H 4 (t) = xh 4(x)dx which is continuous on [, 1]. By L Hopital rule we derive that lim H t + 4 (t)/t =. Integrating by parts, we obtain h 4 (x)dx = xh 4 (x) dx = x H 4 (x) 1 x 1 + H 4 (x) 1 x 2 dx = H 4 (1) + H 4 (x)x 2 dx. Now, since h 4(x)dx =, by the equality above H 4 (x) cannot be positive or negative for all x (, 1). So, by the intermediate value property there exists c 4 (, 1) such that H 4 (c 4 ) = and thus the conclusion follows. Remark Using the same idea as in Lemmas and 2.2.2, we define the auxiliary function like in the first solution, we define the auxiliary function γ 4 = ζ 4 : [, 1] R, given by γ 4 (t) = e t xh 4 (x)dx,

34 2. Mean Value Theorems for some Linear Integral Operators 34 whose derivative is γ 4(t) = e t( th 4 (t) ) xh 4 (x)dx. Since γ 4 () = γ 4 (c 3 ) (by Lemma 2.2.4), by applying Rolle s theorem on the interval [, c 4 ], there exists c 4 (, c 4 ) such that γ ( c 4 ) =, i.e. c 4 h 4 ( c 4 ) = c4 xh 4 (x)dx. Remark As we have seen in the first remark, if we consider the differentiable function γ 4 : [, 1] R defined by whose derivative is γ 4 (t) = e h 4(t) γ 4 (t) = e h 4(t) ( th 4 (t) h 4(t) xh 4 (x)dx, ) xh 4 (x)dx. Now, it is clear that γ 4 () = γ 4 (c 4 ) = by Lemma So, by applying Rolle s theorem there exists c 4 (, 1) such that γ 4 ( c 4 ) = which is equivalent to c 4 h 4 ( c 4 ) = h 4( c 4 ) c4 xh 4 (x)dx. In what follows we prove two lemmas starting from the same hypothesis. Lemma Let h 5 : [, 1] R be a continuous function such that h 5 (x)dx = xh 5 (x)dx. Then, there exists c 5 (, 1) such that c 5 h 5(x)dx =. Proof. [First proof] Consider the differentiable function I : [, 1] R defined by We have I(t) = t I (t) = h 5 (x)dx h 5 (x)dx. xh 5 (x)dx. Moreover I() = I(1), so by Rolle s theorem, there exists c 5 (, 1) such that I (c 5 ) = c5 Second proof. Let H 5 : [, 1] R defined by H 5 (t) = h 5 (x)dx =. h 5 (x)dx.

35 2. Mean Value Theorems for some Linear Integral Operators 35 Integrating by part and using the hypothesis, we have H 5 (1) = h 5 (x)dx = xh 5 (x)dx = xh 5(x)dx = H 5 (1) H 5 (x)dx, and we get H 5(x)dx =. By the first mean value theorem for integrals we have the existence of c 5 (, 1) such that = which is equivalent to c 5 h 5(x)dx =. H 5 (x)dx = H 5 (c 5 ) Third proof. Let us rewrite the hypothesis in the following (x 1)h 5 (x)dx =. The answer is given by the following mean value theorem for integrals (see [84], page 193) Proposition. If Ω 1, Ω 2 : [a, b] R are two integrable functions and Ω 2 is monotone, then there exists c 5 (a, b) such that b a c5 b Ω 1 (x)ω 2 (x) = Ω 2 (a) a Ω 1 (x) + Ω 1 (b) Ω 2 (x)dx. c 5 Now, we consider a =, b = 1 and Ω 1 (x) = h 5 (x) and Ω 2 (x) = x 1 which is increasing. By the mean value theorem in Lemma 2.2.5, there is c 5 (, 1) such that equivalent to c 5 = (x 1)f(x)dx = c5 f(x)dx f(x)dx =. Lemma Let h 6 : [, 1] R be a continuous function such that h 6 (x)dx = xh 6 (x)dx. Then, there exists c 6 (, 1) such that c 6 xh 6(x)dx =. Proof. [First proof] Let us consider the function ϕ : [, 1] R given by H 6 (t) = h 6 (s)ds, where 1 s s h 6 (s) = 2 xh 6(x)dx, if s (, 1] h 6 ()/2, if s =

36 2. Mean Value Theorems for some Linear Integral Operators 36 Clearly the function h 6 is continuous and H 6 () =. Next, we compute H 6 (1) = lim ɛ,ɛ> ɛ = lim = ɛ,ɛ> 1 s ( 1 s ) ( s s xh 6 (x)dx + ) xh 6 (x)dx ds xh 6 (x)dx 1 ɛ+ lim ɛ ɛ h 6 (x)dx =. ( 1 s sh 6(s))ds By Rolle s theorem there exists c 6 (, 1) such that H 6 (c 6) = ; i.e. c 6 xh 6(x)dx =. Second proof. Consider the differentiable function H 6 ; [, 1] R defined by H 6 (t) = t h 6 (x)dx xh 6 (x)dx. It is obvious that H 6 (t) = h 6(x)dx. By Lemma there exists c 6 (, 1) such that H 6 (c 6) = c 5 h 6(x)dx =. On the other hand, since H 6 () = H 6 (c 5) =, by Theorem there exists c 6 (, c 5 ) such that H 6 (c 6) = H 6 (c 6 ) H 5 () c 6 which is equivalent to c 6 xh 6(x)dx =. Combining Lemmas and 2.2.6, one can easily derive the following result. Theorem Assume h 7 : [, 1] R is continuous such that Then there are c 7, c 7 (, 1) such that Proof. It is clear that h 7 (x)dx = h 7 (c 7 ) = c 7 h 7 ( c 7 ) = c7 c7 xh 7 (x)dx. h 7 (x)dx, xh 7 (x)dx. Let us define the auxiliary functions ζ 7, ζ 7 : [, 1] R given by ζ 7 (t) = e t h 7 (x)dx, ζ 7 (t) = e t xh 7 (x)dx. ζ 7(t) = e t( h 7 (t) ζ 7 (t) = e t( th 7 (t) ) h 7 (x)dx, ) xh 7 (x)dx.