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10. Uniformly Convex Spaces Researcher, 1(1), 2009, http://www.sciencepub.net, [email protected] Uniformly Convex Spaces Usman, M .A, Hammed, F.A. And Olayiwola, M.O Department Of Mathematical Sciences Olabisi Onabanjo University Ago-Iwoye, Nigeria [email protected] 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. 2009;1(1):74-85]. (ISSN: 1553-9865). 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. [1,3,4] 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 non- expansive maps have a fixed-point on this space. [2] Let x be a Banach Space. A Branch space x is said to be Uniformly convex if for ε > 0 there exist a ∂ = (ε ) > 0 such that if x, y £ x with //x//=1, //y//=1 and //x-y// ≥ ε , then 1 ()x+ y // ≤ 1 − ∂ // 2 . THEOREM (1.0) Let x = LP (µ) denote the space of measurable function f such that //f// are integrable, endowed with the norm. ()∂x// f ∂μ 1 // f // = P Then for 1<P<+ ∞ , the space LP(μ) is uniformly convex for the proof of the above theorem, we need the following basic lemma. Lemma (1.0) 1+ 1 =1 Let X=Lp, then for p q>0, such p q and for each pair f, g £ x, the following inequalities hold. (i) For 1<p ≤ 2 // 1 ()f+ g// q + // 1 () f + g// q ≤ // 2−1 ( //f // p+ // g // q) 2 2 And (ii) For 2 ≤p < ∞ − p //(f+ g // p + //( f − g // p ≤ // 2( //f // pp+ // g // q) 74 Researcher, 1(1), 2009, http://www.sciencepub.net, [email protected] We now apply lemma (1.0) to prove theorem (1.0) Proof of theorem (1.0) Choose f1 gε X= Lp, such that //f //≤ 1, // g //≤ 1 and for any ε > 0 , we have //f− g // ≥ ε . Two cases arise: Case1: 1<p ≤ 2 In this case (1.0) yield // 1 ()f+ g// q + // 1 () f − g// q ≤ 2 − ( q − 1)( // f // p+ // g //)q−1 // 2 2 ≤2 −)()(q − 1 2( q − 1) = 1 f− g (ε )q // 1 ()f+ g// q ≤ 1 − / /q ≤ 1 − 2 g 2 Thus, q ()ε 1 >0 // 1 ()f+ g // ≤ 1 − // p Or 2 2 ()ε δ =1 − // // 1 > 0 So that by choosing 2 q // 1 ()f+ g // ≤ 1 − δ We obtain 2 and so x= Lp( kp ≤ 2) is uniformly convex. Case 2: 2 ≤p < ∞ As in Case 1, we use (ii) of Lemma (3.1.1) to show that x= Lp( kp < ∞) is uniformly convex, completing the proof of the theorem. Since Lemma (3.1.1) is also valid for 1p, 1 ≤ p < ∞ , the following theorem is also true. Theorem (2.0) 1 ≤p < ∞ For , the space 1p of all infinite (real or complex) sequence, (x1,x2,x3,……) ∞ ∑ //xi p < +∞ such that i=1 is uniformly convex. A s a special case of theorem (1.0), we have the following. Corollary (1.0) Every Hibert space H is Uniformly convex. Although theorem (1.0) and (3.0) provide large classes of space which are Uniformly convex, a few well known spaces are known not to be Uniformly convex. 1. The Space λ is not Uniformly convex To see this ε =1 and choose π (1,0,0,0,..........) , //Υ //λ1 //π − Υ //λ1 = 2 > ε .. However, // 1 ()π − Υ// ≤ 1 // 1 (Χ − Υ// ≤ 1 − δ 2 and there is no ε = 0 such that 2 , showing that λ1 is not uniformly convex. 2. The space λ∞ is not uniformly convex Consider U = ()1,1,0,0,0,............, and V = (1,1,0,0,0,..........) , Both U and V ελ∞ . Take ε =1, then // 1 (UV−λ)//= 1 and so //U //∞ = 1, //V // ∞ and //U - V//∞ = 2 > ε . However, 2 ∞ is not 75 Researcher, 1(1), 2009, http://www.sciencepub.net, [email protected] uniformly convex. 3. 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) = 1 for all tε () a, b And b− t g() t = for each tε () a, b b− a 1 ε = Take 2 clearly f(t), g(t)ε c() a,b .//f //− // g //= 1 and //f - g// ⇒ ε . // 1 ()f+ g // = 1 Also, 2 and so c(a, b) is Not uniformly convex. The following propositions are the consequences of the definition of uniform convexity. Proposition (1.0) Suppose x is Uniformly convex Banach space, then for any ∂ >0, ε > 0 arbitrary ⎧ (ε )⎫ δ > 0 such that // 1 x +y // ≤ 1 − δ ∂ 2 ()⎨ ⎬ Vectors x,y£x with //x// ≤ ∂ , //y//≥ , there exists a ⎩ ∂ ⎭ Proof ε = ε Let ε > 0 be given and let z1 = x/ ∂ , z2 = y / ∂ and suppose we set d . 1 ε //z 1 //≤ 1 and //z1−z 2 // =//x − y // ≥ / d = ε 2 0 Now, by uniform convexity, we have =// 1 z + z // ≤ 1 − δ() ε 2 1 2 ()ε // 1 ∂()x + y // ≤ 1δ That is 2 4 ⎧ (ε )⎫ // 1 x+ y // ≤ 1 − δ ∂ 2 ()⎨ ⎬ Which implies, ⎩ ∂ ⎭ Proposition (2.0) Let x be a uniformly convex Banach space. Then for any ε >0,ε > 0 and ∂ε()0,1 if x, yεΧ such that // x//≤ ∂ , // x - y// > ε , then exist a; δ = ∂(ε )> 0 such that ∂ //αx+ 1 − α y // − 2()ε 1 minα , 1- α // d () ∂ ) Proof ε (0, 1 ) Without loss of generality, we may take 2 Now, //αx+() 1 −α y // = //α )( +y + (1− 2α ) //y // ≤ 2α // 1 x+ y // + 1 − 2α //y //..............................(1.0) 2 ()() But by proposition (1.0), we have that there exists aδ > 0 , such that // 1 ()x+ y // ≤ // −δ( ε ) // ∂ 2 Substitute this into (1.0), to have 76 Researcher, 1(1), 2009, http://www.sciencepub.net, [email protected] ⎧ (ε ) ⎫ Since //y// ≤ ∂ =⎨ ∂ - 2αδ ∂⎬ ⎩ ∂ ⎭ ⎛ ()ε ⎞ =⎜1 − 2α ∂ ∂⎟ ⎝ ∂ ⎠ αε (0, 1 ), have α≥ min( α ,1- α ) Put by the choice of 2 Thus, we have, (ε ) //α x+() 1- α // ≤ //1 − 2δ min()α ,1− α // ∂ ∂ We now discuss a characteristic of some Banach space, which is related to uniform convexity. 2.0 STRICTLY CONVEX BANACH SPACES Definition (1.0) 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 + y//=//x+//y//, implies that there exists a λ ≥ 0 such that x=λ x ()or y = λ x . The following Lemma on uniform convexity will be useful in the sequence. Lemma (1.0) Let X be a uniformly convex Banach space. λ + 0, // x -λ y//≥ 0, then // 1 ()x+λ y // ≤ //x // If 2 Proof Suppose 0< // x // < //y // . (The proof for the case 0< // x // < // y // is similar) . // x // λ = and set a= x, b = λ y. The // a - b // = // x -λ y //> 0, Take // y // Let ε > 0 such that //x -λ y // ≥ ε Observe that, // x //// y // // a //= //x//, // b // =λ // y // = = // x // // y // δ (ε ) > 0 δ = So by proposition (3.1.1), there exist a // x // ∂()ε // 1 ()a+ b // < 1- // x //< // x //, Such that 2 // x // // 1 ()x+ y // < // x //. That is 2 Completing the proof of the lemma, we now prove the following theorem. Theorem (1.2) Every uniformly convex space is strictly convex Proof Suppose x is uniformly convex Let x, yεΧ be non-zero vectors such that //x+y//=//x//+//y// We need to show that there exist aλ > 0 such that x= λ y . We consider two possible cases. Case I: // x //=// y // Case II: 0< // y// < // x //, (The other case 0< // x // < / y // is treated) Similarly, 77 Researcher, 1(1), 2009, http://www.sciencepub.net, [email protected] Proof of Case I // x // x≠ y; then as =1 If x=, then (II) holds with λ =1, so, suppose // y // and // x+y //<2 //x // =+ // y // (Since // x // y //). That is, // x + y // <1 // x // + //y //. Contradicting (I). Thus, x ≠ y is not possible and this proves case I. Proof of Case 2 Suppose // x=y // = // x // + // y // and that x ≠ λ y. Since x is uniform convex, lemma (2.0) yields. // 1 x+λ y // < // x // .................................(1.1) 2 () // x // 0< //y // < // x//, let λ = //x+ y // = // x// + //y// For // y // And x ≠ λ y λ()// x //+ // y // =λ //x + y//= //λ x+ λ y - x+ x // We have ≤//x +λ y // +() λ + 1 //x //................(1.2) λ()// x //+ // y // ≤ //x // + λ // y//+ λ // x// - // x // That is, =λ()// x // + // y // ................(1.3) The inequalities (b) and (c) gives λ()// x //+ // y // = // x+ λ y //+ (λ -1) //x // // x // λ = 1 x+λ y = // x // 2 () From which we obtain (since // y // Contracting (1.1). This completes the proof of the theorem. Theorem (3.2.2) gives a large class of strict convex Banach spaces.
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