Researcher, 1(1), 2009, Uniformly Convex Spaces. Usman, M.A, Hammed, F.A. And Olayiwola, M.

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1 Researcher, (, 9, Uniformly Convex Spaces Usman, M.A, Hammed, F.A. And Olayiwola, M.O Department Of Mathematical Sciences Olabisi Onabanjo University Ago-Iwoye, Nigeria ABSTARACT: This paper depicts the class OF Banach Spaces with normal structure and that they are generally referred to as uniformly convex spaces. A review of properties of a uniformly convex spaces is also examined with a view to show that all non-expansive mapping have a fixed point on this space. With the definition of uniformly convex spaces in mind, we also proved that some spaces are uniformly spaces. [Researcher. 9;(:74-85]. (ISSN: Keywords: Banach Spaces, Convex spaces, Uniformly Convex Space. INTRODUCTION The importance of Uniformly convex spaces in Applied Mathematics and Functional Analysis, it has developed into area of independent research, where several areas of Mathematics such as Homology theory, Degree theory and Differential Geometry have come to play a very significant role. [,,4] Classes of Banach spaces with normal structure are those generally refer to as Uniformly convex spaces. In this paper, we review properties of the space and show that all nonexpansive maps have a fixed-point on this space. [] Let x be a Banach Space. A Branch space x is said to be Uniformly convex if for ε > there exist a ( ε > such that if x, y x with x, y and x-y ε, then ( x + y. THEOREM (. Let x L P (µ denote the space of measurable function f such that f are integrable, endowed with the norm. x / f / μ f P LP( μ Then for <P<+, the space is uniformly convex for the proof of the above theorem, we need the following basic lemma. Lemma (. + Let XLp, then for p q>, such p q and for each pair f, g x, the following inequalities hold. (i For <p ( f g q + + ( f + g q ( f p + g q And (ii For p < p f + g p + f g p f pp + g q ( ( ( 74

2 Researcher, (, 9, We now apply lemma (. to prove theorem (. of theorem (. Choose f gε X Lp, such that f, g and for any ε >, we have f g ε. Two cases arise: Case: < p In this case (. yield q f + g q + ( f g q ( q ( f p + g q q Thus, ( f + g q / ( f + g f g / q g q ( ε > p Or ( ε δ > So that by choosing q ( f + g δ We obtain Case : p < ( ε q and so x Lp ( p is uniformly convex. As in Case, we use (ii of Lemma (.. to show that x Lp ( p < is uniformly convex, completing the proof of the theorem. Since Lemma (.. is also valid for p, p <, the following theorem is also true. Theorem (. For p <, the space p of all infinite (real or complex sequence, (x,x,x, / x i / p < + such that i is uniformly convex. A s a special case of theorem (., we have the following. Corollary (. Every Hibert space H is Uniformly convex. Although theorem (. and (. provide large classes of space which are Uniformly convex, a few well nown spaces are nown not to be Uniformly convex.. The Space λ is not Uniformly convex To see this ε and choose π (,,,,..., Υ λ.. π Υ λ > ε However, ( π Υ and there is no ε ( Χ Υ δ such that, showing that λ is not uniformly convex.. The space λ is not uniformly convex Consider U (,,,,,..., and V (,,,,,..., Both U and V ελ.tae ε, then U, V and U - V > ε ( U V and so λ. However, is not 75

3 Researcher, (, 9, uniformly convex.. Consider c(a,b the space of real-valued continuous function on the compact interval (a,b with Sup norm. Then C (a,b is not uniformly convex. To see this, choose two function f(t, g(t defined as follows: F(t for all t ε ( a, b And b t g ( t for each tε ( a, b b a ε Tae clearly f(t, g(t ε c( a,b. f g and f - g ε. ( f + g Also, and so c(a, b is Not uniformly convex. The following propositions are the consequences of the definition of uniform convexity. Proposition (. Suppose x is Uniformly convex Banach space, then for any >, ε > arbitrary ( ε δ > such that x + y δ Vectors x,y x with x, y, there exists a ε > be given and let z x /, z y / ε ε Let and suppose we set d. ε z z and z x y / d ε Now, by uniform convexity, we have z + δ ( ε z ( ε x + y δ That is 4 ( ε x + y δ Which implies, Proposition (. Let x be a uniformly convex Banach space. Then for any ε >, ε > and ε (, if x, yεχ such that x, x - y > ε, then exist a; δ ( ε > such that αx + ( α y ( ε minα,-α d ε (, Without loss of generality, we may tae Now, αx + ( α y α( + y + ( α y α ( x + y + ( α y...(. But by proposition (., we have that there exists aδ >, such that ( x + y δ ( ε Substitute this into (., to have 76

4 Researcher, (, 9, Since y ( ε - αδ ( ε α αε,, haveα min α,- α Put by the choice of Thus, we have, ( ε αx + - α δ min( α, α We now discuss a characteristic of some Banach space, which is related to uniform convexity.. STRICTLY CONVEX BANACH SPACES Definition (. A Banach space X is said to be strictly convex (or strictly rotund if for any pair of vecors x, y x, the equation x + yx+y, implies that there exists a λ such that x λ x ( or y λx. The following Lemma on uniform convexity will be useful in the sequence. Lemma (. Let X be a uniformly convex Banach space. λ +, x - λy, then ( x + λy x If Suppose < x < y. (The proof for the case < x < y is similar. x λ and set a x, b λy.the a - b x - λy >, Tae y Let ε > such that x - λy ε Observe that, x y a x, b λ y x y δ ( ε > δ So by proposition (.., there exist a x ( ε a + b < - x < x, Such that x ( x + y < x. That is Completing the proof of the lemma, we now prove the following theorem. Theorem (. Every uniformly convex space is strictly convex Suppose x is uniformly convex Let x, yεχ be non-zero vectors such that x+yx+y We need to show that there exist aλ > such that x λy. We consider two possible cases. Case I: x y Case II: < y < x, (The other case < x < / y is treated Similarly, 77

5 Researcher, (, 9, of Case I x x y; then as If x, then (II holds with λ, so, suppose y and x+y < x + y (Since x y. That is, x + y < x + y. Contradicting (I. Thus, x y is not possible and this proves case I. of Case Suppose xy x + y and that x λ y. Since x is uniform convex, lemma (. yields. ( x + λ y x...(. < x < y < x, let λ x + y x + y For y And x λ y λ( x + y λ x + y λx + λy - x + x We have x + λy + ( λ + x...(. λ( x + y x + λ y + λ x - x That is, λ( x + y...(. The inequalities (b and (c gives λ ( x + y x + λy + ( λ - x x λ ( x + λy x From which we obtain (since y Contracting (.. This completes the proof of the theorem. Theorem (.. gives a large class of strict convex Banach spaces. However, it can be shown easily that λ, L,λ and c(a,b are NOT strictly convex. For example, to see that λ is not strictly convex.. THE MODLUS OF CONVEXITY Definition (. Let x be a Banach space, the modulus of convexity of X is the function δ x : (, (, δx( ε Inf ( x + y : x, yε BC(, x - y ε { } defined by. We now give an important characteristic of the modulus of convexity in the following proposition. Proposition (. The modulus of convexity of a Banach space x is a non-decreasing convex function. The proof that δx( ε is non-decreasing is a trivial consequence of definition (. and it is therefore omitted. To proof convexity, suppose for any two vectors U, Vε X, we denote by N(U,V the set of all pairs x,,yε X with x, y ε B (, such that for some real scalars α, B we have x - y αu x + y BV x, y : x, y ε B and x - y U,x y BV α + That is N(U,V { } 78

6 Researcher, (, 9, For (,, define δ ( U V, r inf x + y..., x, yεn { U,V, x - y r}...(.4, It is easy to see that δ ( U, V, r for in (.4 { Since x, for all xεn( U, V } Moreover, of r, for given any λ, λ in (, and ε >, we can choose ( x y εn( U, V such that ( for K, x, y λ...(.5 and δ ( U, V, λ ( + ε x...(.6 + y δ U, V,ε The choice of (x, y is possible because of the definition Now, for λ x (, x λx + λ x Let y λy + λ y And x x ( x, x ( We have λ + λ Cas B in (.5 as infimum. Similarly Also, ( x, εn( U, V y implies that exist constants α, β (, such that x y α U and x y β V From (.6, we have x y λx + ( y x + λy ( y y λ( x y + ( y( x y λαu + ( y α U, from...(.7 λα + ( y α U If we set q λα + ( λ α, q some real number, We have x y λx + y x + λy y y λ( x + y + ( λ( x + y λβv + ( λ β V...(.8 λβ + ( λ β V γ λβ + λ β So that for some real number We have So that we have x y λα + λα λα λ α U ( λ α U + ( λ α U U + ( λ α U + ( λ α U.. (.9 79

7 Researcher, (, 9, x + y λβ β λ x + y + λ + y Now maing use of (..9, we get x y λε ( λ ε But then (.7 and (.8 give δ ( u, v, λε + λ ε x + y - λ x y ( + + λ λ / x λ + ( λ β v λβ + ( λ β v v + ( λ β v ( x...(. ( λ ( x + y + ( λ( x + y + y x + y x ( u, v, λε ( λ δ ( u, v, λε + ε / + + y λ From (. λδ ( u, v, λε + ( λ δ ( u, v, ε + ε Now since ε is arbitrary, we infer that δ u, v, λε + λ ε λδ u, v, ε + λ δ u, v ε ( ( (,.(. Thus, (u, v,ε is convex. Now from the definitions of N(u, v and δ (u, v, ε each pair (x, y B ( x B ( ε. Belong to some N(u, v, so that we have δ x ε Inf δ u, v, ε : U, Vε εχ, u, v and as δ u, v, ε is convex, { So is δ x( ε For the next proposition, we need the following lemma. Lemma (. Suppose f : (, (, is non-decreasing convex function and u < x < y then f ( u f ( v f ( v f ( x v - u y - x We can choose αβ ε, such that V αu + ( - α y and U βv + ( β y...(. f ( v f ( u f αu + ( - α y ( v - u αu + ( - α y u from (. αf ( u + ( α f ( y f ( u as if is αu + ( - α y u ( α f ( y ( α f ( u ( α ( y - u f ( y f ( u β ( f ( y f ( u y u β ( y u βf ( y βf ( u from (. y - x 8

8 Researcher, (, 9, f ( y βf ( u + βf ( y f ( y y x That is, f ( v f ( u f ( y βf ( u + ( β + ( β f ( y...(. v u y x But since f is convex we get from (. Hence, (. leads to f ( v f ( u f ( y f ( x v u y x We now prove the following proposition Proposition (. Let f : (, (, be a non-decreasing convex function with f(. Then for each x>, f is continuous and has Lipschitz constant (-x - As any point in the domain of f, r ε (,, say is finite and f is non-decreasing then at each x>, the right derivative f r(x of f exists (See Rocfella. Thus, f or v<x<, we have by definition f r (v+ lim f ( x f ( v lim f f ( x by x v x v x v x Lemma (. f f ( v v ( α f ( y ( α f ( u ( α ( y u f ( y f ( u β ( f ( y f ( u y u β ( y u ( f ( y βf ( u β from above y x f ( y f ( u + f ( y - f ( y...(. y - x That is, f ( v f ( u f ( y βf ( u( β f ( y...(.4 v u y x But since f is convex, we get from (. f ( x f βu + ( β y βf ( u + ( β f ( y Hence, (.. leads to f ( v f ( u f ( y f ( x v u y x We now prove the following proposition Proposition (4. Let f : (, (, be a non-decreasing convex function with f(. Then for each x>, f is continuous and has Lipschitz constant (-x - 8

9 Researcher, (, 9, f ( v f f r( v That is v - v Replacing v by x, we have f r( v - x Now, by the mean value theorem, for all x,y ε (, we have f x f y f x y ε x, y ( ε for some x - y/ / from (A - x That is f ( x f ( y ( x x y So that the function f is Lipschitz constant (-x -I on each interval (,x. The following proposition is also of interested Proposition (5. x ε > The Banach space x is uniformly convex if and only if δ for all ε ε, Suppose x is uniformly convex, then if x, yεx with x, y and x - y ε we have ( x + y δ, for δ > δx ε Inf - + y, x - y ε, x, y. By definition Now, ( x ( δ δ > Let δ x( ε > for rε (, and choose x, yεx such that x, y and x - y ε δx This implies that for all x, y ε B ( { and x - y } > ( ε Inf ( x + y : x, yε B ( with x - y ε,- ( x + y δx( ε This implies that there exists a δ > such that ( x + y δ, δ > Which implies uniform convexity 4. NORMAL STRUCTURE AND REFE LEXIVITY OF BANACH SPACES In this section, we deal with some other geometric properties which are important in studying the fixed point theory of non-expansive mapping. Let C be a bounded convex subset of a Banach spacex. The diameter d of c is define by D Sup{ zi-zj, zj, zica} A point Zo Χ is said to be a diameter point of C If Sup { z z : z } d And a point z { x is called } a non-diamental point of c If Sup z z : z C < Sup {zi - zj : zi, zj C} Definition. A bounded convex subset of Banach space is said to have normal structure if for each 8

10 Researcher, (, 9, convex subset x of C, consisting of more than one point A, contains a non-diamental point that is there exis ts a z A. z : Sup zi - zj : zi, zj A z z C < Such that Sup { } { } Geometrically, C is said to have normal structure if for every convex, C subset A of C there exist a ball whose radius is less than diameter of A centered at a point of A which contains A. In the following, we exhibit large classes of spaces with normal structure. Proposition (6. Every uniformly convex set in X, containing as least two different point Z, Z. z ( Ζ + Suppose is the diameter of C and Ζ for any Ζ C, proposition (. given us, from z - z ; Z Z Z - Z {( z Z + ( Z Z } {( Z Z } { δ ( Z Z } Since z - z The above inequality implies that C is contained in the ball of radius, say, r, less than centered at Z, that is, Z B Z, r C Br Z. r Sup z z : z C Where { } Propos ition (7. Every convex and impact subset of a Banach space normal structure. We proof this by this contradiction that is, we shall assume that a compact convex subset of C of a Banach space X does not have normal structure. Then we shall generate a sequence, which will contradict the hypothesis of compactness. Suppose C does not have normal structure, then we may assume that all point of C are diamental for C. let Z be the diameter of C, we shall construct a sequence Z, Z,. Of point of C that Z Z d i, j,,...i j i j To do this, we choose Z C arbitrary and assume that Z,Z,.Z n have already been chosen. By the convexity of C ( Z Z n n is a point of C and thus by assumption, is diamental for C. by the compactness of C, the Sup is achieved in C so that we can find a point Z, Z n+ C such that Z n+ Z Z n d n Consequently, Zn+-Z j d for j.., Which means that the sequences /{Z n } n has no convergent subsequence, and thus has no cluster value in C. this contradicts the compactness of C and completes the proof. We observe tha t if a convex set C has normal structure, then so does every convex subset of C. in particular, if the whole space X has normal structure, then every convex subset of X has normal structure. This follows from the definition. Some Banach space does not have the following. 8

11 Researcher, (, 9, Example. The Space X C(, with Sup norm does not have normal structure.. The space X L (,π does not have normal structure.. The space X I does not have normal structure. Is called a retracting mapping. Definition (.4. X is a retract of Y, if XCY and there exist a continuous mapping α. Lemma (.4. Let M be a closed convex subset of a Hilbert space. If T is a non-expansive mapping of M into H, then I-T is a restriction to M of a monotone operator. Let r be the metric retraction of H into H, for x, y H. (I-Trx (I-TR, x-y x-y (Trx-Try, x-y x - y Trx - Try x - y I-Tr is monotone The retraction to M I Tr is I T Lemma (.4.** If T is monotone and X and Y are normal of H such that Tx Y For any Y in H and T> Let y t X +ty With y- y t then (Ty t y, y so that (Ty, y y, y t y Tx t Let +, then Ty t Ty ( So that y Tx, y Then Y Tx (.4. Conjectures ( Let M be a be a convex subset of a normal linear space L let T be a non-expansive mapping on M into L, Then for <t<, the mapping St ti(- tt is non-expansive and the same set of fixed points is T. If TMCM, then St <Cm In fact, by ransnoselei, we have the following conjectives. (. Consider a mapping T : M L, Where T is non-e xpensive, L is a normed linear space MCL and convex, then if G ( I + T, G is non-expansive and the exists X εm such that GX X ( Also, X n + λ xm + λ T xn λε (, is true in a uniformly convex space. Conclusion We have examined in this paper project, the non-expansive mapping and the fixed point theorem. We have also been able to show that classes Banach of spaces with normal structures are examples of uniformly convex spaces. Thus, the Lp space ( < p < are classical examples of uniformly convex spaces. 84

12 Researcher, (, 9, Therefore, is application a non-expensive operator will have solution on these spaces. REFERENCES ( Kreyszig Erwin (978: Introductory Functional Analysis, John Wiley ( Kufuer A. et al: (977: Function Space. Publishing House of the Czechozbra Academy of Sciences. Pragne. ( Royden H. L (968: Real Analysis, Second Edition, Macmillan International Edition, (4 Smart D. R. (974: Functional Analysis, Printed by Cambridge Press, Syndics of Cambridge, University Press. (5 Walter Rudi (976: Principles of Mathematical Analysis, International Student Edition (Third Edition. Published by MCGRAW-HILL International Boo Company. /7/8 85