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 , 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 . 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 ()yx 1// −≤+ ∂ // 2 . THEOREM (1.0)

Let x = LP (µ) denote the space of measurable function f such that //f// are integrable, endowed with the . ()fx // ∂∂ μ 1 // f // = P Then for 1

Lemma (1.0) 11 =+ 1 Let X=Lp, then for p q>0, such qp and for each pair f, g £ x, the following inequalities hold. (i) For 1

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We now apply lemma (1.0) to prove theorem (1.0)

Proof of theorem (1.0) Choose 1 ε = LpXgf , such that ≤ gf ≤ 1////,1//// and for any ε > 0 , we have gf //// ≥− ε . Two cases arise: Case1: p ≤< 21 In this case (1.0) yield // 1 ()//// 1 () −−≤−++ ( + gpfqqgfqgf )q−1 //////////)1(2// 2 2 ()()(qq ) =−−−≤ 11212

− gf (ε )q // 1 qgf −≤+ q 1//1// −≤ 2 () Thus, g 2 q ()ε 1 >0 // 1 ()gf 1// −≤+ // p Or 2 2 ()ε δ −= //1 // 1 > 0 So that by choosing 2 q // 1 ()gf 1// −≤+ δ We obtain 2 and so = (kpLpx ≤ 2) is uniformly convex. Case 2: 2 p <≤ ∞ As in Case 1, we use (ii) of Lemma (3.1.1) to show that = (kpLpx ∞< ) 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,……) ∞ ∑ i // px < +∞ 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 ( ) Υ λ1 π − Υ λ1 >= ε..2////////,.,...... 0,0,0,1 However, // 1 ()π ≤Υ− 1// // 1 ( 1// −≤Υ−Χ 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 = ()...,,...... 0,0,0,1,1 and V = ( ),.,...... 0,0,0,1,1 Both U and V ελ∞ . Take ε =1, then // 1 ( −λVU ) =1// and so U V ////,1//// ∞=∞ //Uand 2V//- >=∞ ε . However, 2 ∞ is not

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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 ε (),bat And − tb ()tg = for ε (),each bat − ab 1 ε = Take 2 clearly f(t), g(t)ε () − gf =1////////.ba,c //fand g//- ⇒ ε . // 1 ()gf =+ 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 that //such 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 1 ∂= ,/ 2 = yxz /z ∂ and suppose we set d . 1 ε z 1 ≤1//// //zand z21 // ///// dyx =≥−=− ε 2 0 Now, by uniform convexity, we have // 1 zz 1// −≤+= ()εδ 2 21 ()ε // 1 ()yx ≤+∂ 1// δ That is 2 4 ⎧ (ε )⎫ // 1 yx 1// −≤+ δ ∂ 2 ()⎨ ⎬ Which implies, ⎩ ∂ ⎭ Proposition (2.0) Let x be a uniformly convex Banach space. Then for any ε ε 0,0 and ∂>> ε()0,1 x,if y εΧ such that x,x//// y//-// >∂≤ ε , then exist a; δ ∂= (ε )> 0 such that ∂ αα yx −−+ 2//1// ()ε 1 ,min αα //-1 d () ∂ ) Proof ε ( ,0 1 ) Without loss of generality, we may take 2 Now, α ()α yx ////1// α ()y ++=−+ ( − α ) y ////21 ≤ α //2 1 yx −++ α y )0.1(...... ////21// 2 ()() But by proposition (1.0), we have that there exists aδ > 0 , such that // 1 ()yx (εδ )////// ∂−≤+ 2 Substitute this into (1.0), to have

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⎧ (ε ) ⎫ //y//Since ∂≤ ⎨∂= 2- αδ ∂⎬ ⎩ ∂ ⎭ ⎛ ()ε ⎞ ⎜ 21 α∂−= ∂⎟ ⎝ ∂ ⎠

αε ( ,0 1 ), have ≥ ( -,1min ααα ) Put by the choice of 2 Thus, we have, (ε ) ()αα −≤+ δ ()αα //1,min21////-1x// ∂− ∂ 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 xyor x x λx () == λxyor . The following Lemma on uniform convexity will be useful in the sequence. Lemma (1.0) Let X be a uniformly convex Banach space. λ + λ ≥ then //0,y//-x//,0 1 ()λ ≤+ x //////yx If 2 Proof Suppose ////0 << y .////x (The proof for the case 0 < // x // < // // isy similar) . x //// λ = and x,aset == λy.b The // b-a // = // x λ //- > 0,y Take y //// Let ε > 0 such that λy //-//x ≥ ε Observe that, y //// x //// y //// y // // // b// //x//, //a // //a = //x//, b// // λ // y // == = x //// y //// y //// δ (ε ) > 0 δ = So by proposition (3.1.1), there exist a x //// ∂()ε // 1 ()ba // <+ -1 //// < // x //,x Such that 2 // x // // 1 ()yx // <+ // 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 that = λyxsuch . We consider two possible cases. Case I: // x //=// y // Case II: << // x //,y////0 (The caseother 0 < // x // < / // treated)isy Similarly,

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Proof of Case I // x // ≠ y;x 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 λ <+ ////yx x // ...... (1.1) 2 () // x // y << x//,//////0 let λ = yx +=+ //y//x//////// For y //// And x ≠ λ y λ()x //// //// λ //xy +=+ = λ + λ x-yx//y// + x // We have )2.1...... (...... ////1//// yx ()λλ +++≤ x )2.1...... (...... ////1//// λ()x //// //// //x //y +≤+ λ + λ -x////y//// // x // That is, )3.1...... (...... y //// x //// λ() x //// += //// )3.1...... (...... y The inequalities (b) and (c) gives λ() x //// // // =+ + λ //x//y + (λ ) x ////1-y x //// λ = 1 λ =+ x ////yx 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. However, it can be shown easily that L11 ,, λλ ∞ and c(a,b) are NOT strictly convex. For example, to see that λ∞ is not strictly convex.

3.0 THE MODLUS OF CONVEXITY Definition (2.0) Let x be a Banach space, the modulus of convexity of X is the function δ ()→ ( 1,00,2:x ) ()εδ Inf { //1x 1 ()+−= yx x,:// y ε ( ) y//-x//,0BC ≥ ε} defined by 2 . We now give an important characteristic of the modulus of convexity in the following proposition. Proposition (3.0) The modulus of convexity of a Banach space x is a non-decreasing . Proof The proof that δx()ε is non-decreasing is a trivial consequence of definition (2.0) and it is therefore omitted. To proof convexity, suppose for any two vectors U, Vε X, we denote by α , B N(U,V) the set of all pairs x, ,yε X with x, yε B1 (0), such that for some real scalars 11 we have = αUy-x =+ BVyx That is N(U,V) = {}()x,:yx, ε 1 ()0By xand = α + = BVyxU,y-

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For rε (),2,0 define δ rVU 1inf,, 1 +−= yx εNyx,,...... //// //-x//,VU, ≥ ...... ry )4.1...... ( (){ 2 ( ) } It is easy to see that δ ( rVU ) = 0,, for (1.4)in {}Since x //// = 1, xallfor ε ()VU,N Moreover, of r, for given any λ ,λ21 in (0,2) and ε > 0 , we can choose εNy VU,x such that =1,2Kfor ()()k k ( )

kk //y,x// λ≥ k ...... )5.1...... ( and δ V,U, λ ε 1 1 +−≥+ y )6.1.....(...... //x// ()k 2 ( 2 k k ) δ ( V,U, ε ) The choice of (xk, yk) is possible because of the definition in (1.5) as infimum. Now, for λ = ()0,1x λ −+= λ x1xx Let 13 ()2 λ 1 −+= λ yyy And 13 ()2 λ λ ≤−+≤ 1//1//x////x x BCas 0x, We have // 3 1 () 121 ( ) Similarly y ε VU,N,x α , β Also, ()(kk ) implies that exist constants kk (k = 2,1 ) such that

x k =− α kk Uy xand =− β Vy k kk From (1.6), we have

1xy1x x y33 =− λ 1 + ()+− λ 12 − (1xy1x − )yyy 2 xy1yx λ()()()xy1yx −−+−= y 11 22 λα1 ()−+= α 2UyU ,1 from ...... 1.7)...... (

//y1// λα1 ()−+= α 2 //y1// U

If we set q= λα1 ()−+ λ α 2 ,1 q some real number, x − y = λ −+ xy1x + λ − 1 − yyy We have 33 1 () 12 ( ) 2

= λ()()()+ y11 −+ λ x1x + y22

λβ1V ()1−+= βλ 2V .(1.8)......

1 ()−+= βλλβ 2 //1// V

So that for some real number γ = λβ1 + (1 − λ)β 2

We have

= λα1 + ( − λ)α 2 U ////1

////1// 1 ()−+= αλλα 2 U ////1//

////1//// //x// y33 //x// =− λα1 U ()−+ αλ 2 U ////1//// ////1//// = αλ U −+ αλ U ////1//// 1 ()2 …………………………….. (1.9) So that we have

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////1 = λβ1 + ( − λ)β 2 v ////1

////1// //yx// 33 //yx// 1 ()−+=+− βλλβ 2 v ////1// //1// v −+= //1// βλβ v 1 ()2 = λ + y11 + ( − λ) + y22 ...... //x//1//x// )10.1..(...... Now making use of (3.3.9), we get − ≥ λε 1//yx// − λ ε 33 1 ()2 But then (1.7) and (1.8) give 1v,u, ελλεδ 1 1 +−≤−+ y //x// ()1 ()2 2 33 -1 = -1 1 λ λ +−++ //yx//1//yx//// 2 11 () 22 //yx//2/1 −= λ //yx//2/1 1 λ +−+−+ //yx//1 11 2 () 22 1 λ 1 1 λ 11//yx// 1 +−−++−= //yx// ()2 11 ()()2 22 8 ≤ ()v,u, λελ 1 ()(−++ 2 + ελεδλ //2/v,u,//1// 2 ………….(1.11) From (1.0) = λδ ()()λε1 −+ λ δ ( ε 2 ) + ε 2v,u,1v,u, Now since ε is arbitrary, we infer that =δ ()vu λε1 ()1,, −+ λ ε 2 ≤ λδ ( vu ε1 ) + ( − λ)δ ( vu ,,1,, ε 2 ) Thus, (u, v,ε ) is convex. Now from the definitions of N(u, v) and δ (u, v,ε ) each pair (x, y) ε B ()0 x B (0 ). Belong to some N(u, v), so that we have δ x ε = δ vuInf ε U,:,, ε ,V ≠Χ v0,u ≠ 0 and as δ vu ,, ε is convex, (){ ( ) ε ( ) So is δ x ()ε For the next proposition, we need the following lemma. Lemma (2.0) Suppose ()(→ 1,00,2:f ) is non-decreasing convex function and ≤ u < ≤< 2yx0 then ()− ()vfuf ()− ()xfvf ≤ u-v x-y Proof αβ 1,0 We can choose ε () such that = α + ()-1uV α y and β ()1−+= β yvU ...... )11.1...(...... ()− (ufvf ) α + ( -1uf// α )y = ()u-v ()-1u αα −+ uy from (1.11) α () (−+ α )()− (ufyf1uf ) ≤ as if is ()-1u αα −+ uy ()()()α 1yf1 −−− α (uf ) ()()1 − α u-y ()− ()ufyf β ( ( ) − (ufyf )) = = − uy β ()− uy β ()− β ()ufyf (1.11) from from (1.11) x-y

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()− β ()+ β ()− ()yfyfufyf = − xy That is, ()vf − ()uf ()− // β ()ufyf ≤ (1 )()()−+−+ ββ yf )12.1.....(...... //1 − uv − xy But since f is convex we get from (1.11)

Hence, (1.12) leads to ()vf − ()uf ()− fyf (x) ≤ − uv − xy We now prove the following proposition Proposition (3.0) Let ()()→ 1,00,2:f be a non-decreasing convex function with f(0) = 0. Then for each x>2, f is continuous an d has L ipschitz constant (2-x)-1 Proof As any point in the domain of f, rε ( 2,0 ), say is finite and f is non-decreasing then at each x>2, the right derivative f1 r(x) of f exists (See Rockfella). Thus, for v

()− ()ufyf β ( (yf ) − (uf )) = − uy β ()y − u β ()()yf − β ()uf above from from above − xy ()− ()+ ()yfufyf = ()yf- ...... (1.13) x-y That is, ()vf − ()uf ()− β ( )( − β ) (yfufyf )//1// ≤ )14.1.....(...... − uv − xy But since f is convex, we get from (1.12) ()= β (−+ β ) //1// ≤ β ( ) + ( − β ) (yfufyufxf )//1 Hence, (3.3.12) leads to ()vf − ()uf ()− fyf (x) ≤ − uv − xy We now prove the following proposition Proposition (4.0) Let ()()→ 1,00,2:f be a non-decreasing convex function with f(0) = 0. Then for each x>2, f is continuous an d has L ipschitz constant (2-x)-1

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()2 − ()vff 1 1 ()vrf ≤= ≤ That is 2 − v v-2 Replacing v by x, we have 1 = f 1 ()vr ≤ x-2 Now, by the mean value theorem, for all x,yε ()2,0 we have // () ()≤− ()ε − yxfyfxf //////// for some ε ( , yx ) 1 ≤ //x - / fromy/ (A) x-2 −1 That is // () () (−≤− ) ) − yxxyfxf ////2// So that the function f is Lipschitz constant (2-x)-I on each interval (0,x). The following proposition is also of interested Proposition (5.0) ε 2,0 The Banach space x is uniformly convex if and only if δ (ε ) > 0x for all ε () Proof Suppose x is uniformly convex, then if ε with //x/Xyx, ≤ //y//1,/ ≤1 /and y//-/x ≥ ε we have // 1 ()yx ,1// δδ >−≤+ 0for 2 Now , δ ()ε = Infx ( //x-1 + //xy//, ≥ ε //x//,y//- ≤ //y//1, ≤1. ( δ ) δ >=−−≤ 0//11// Let δx(ε )()> 0 ε 2,0rfor and x,choose y εx such that y //- //xand 1 //y //1,x //// //y //1,x ≤≤ 1 //xand y //- ≥ ε By definition ()εδ Infx { //1 1 ( +−= ) xyx yε B,:// (0 ) //xand }>≥ 0y//- 2 This implies that for all yx ε B, ()0 with ≥ ε //-,1y//-x// 1 ( + ) xyx (εδ ) 2 This implies that there exists a δ > 0 such that //1 1 ()yx ,1// δδ >−≤+− 0 2 Which implies uniform convexity

4.0 NORMAL STRUCTURE AND REFELEXIVITY 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 ://z// =∈− dzz If Sup {}0 ∗ And a point z ∈ x is called a non-diamental point of c ∗ If Sup { ://// Czzz }<−∈− Sup {}zi,:zj//-//zi ∈Czj Definition 3.0 A bounded convex subset of Banach space is said to have normal structure if for each

82 Researcher, 1(1), 2009, http://www.sciencepub.net, [email protected] convex subset x of C, consisting of more than one point A, contains a non-diamental point ∈ Az that is there exists a 0 . Azj zi,:zj//-//zi Sup ://// Czzz <∈− Sup zi,:zj//-//zi ∈ Azj Such that Sup {}0 { } 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 (6.0) Every uniformly convex set in X, containing as least two different point Z1, Z2. z 1 ( Ζ+Ζ= ) Suppose ∂ is the diameter of C and 0 2 21 for any Ζ∈C , proposition (1.2) given us, from

{ − δ ( − Z 21 //Z//1 )} 1 //Z;//z-z// Z 21 // =∂≤−∂≤ 0 //Z-//Z ≤ 0 1 ZZZz 1 −=−+− ZZ 2{}()()1 2 2{}( 2 ) z-z = z-z Since 0 The above inequality implies that C is contained in the ball of radius, say, r, less than

∈ ()0 , ⊆→ r (ZBCrZBZ 0 ) ∂ centered at Z0 , that is, . = ://// ∈− CzzzSupr Where {}0 Proposition (7.0) Every convex and impact subset of a Banach space normal structure. Proof We proof this by this contradic tion 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 Z0 be the diameter of C, we shall construct a sequence Z, Z2,…………. Of point of C that j...i1,2,...... j i, //// dZZ ()i, ==− ≠ j...i1,2,...... j ji ∈CZ To do this, we choose 1 arbitrary and assume that Z2,Z3,………………….Zn have already been chosen. By the convexity of C 1 ()Z + ...... + Z n 1 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, +1n ∈CZ such that // − ZZ + + Z //...... n+ 11 n = d n Consequently, //Zn+1-Zj// = d for j =1..,… Which means that the sequences /{Zn} n ≥ 1 has no convergent subsequence, and thus has no cluster value in C. this contradicts the compactness of C and completes the proof. We observe that 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.

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Example 1. The Space X = C(0,1) with Sup norm does not have normal structure. 2. The space X = L1 (0,2π ) does not have normal structure. 3. The space X = I1 does not have normal structure. Is called a retracting mapping. Definition (3.4.10) X is a retract of Y, if XCY and there exist a continuous mapping α . Lemma (3.4.10) Let M be a closed convex subset of a . If T is a non-expansive mapping of M into H, then I-T is a restriction to M of a monotone operator. Proof Let r be the metric retraction of H into H, for ∈ Hyx, . (I-Tr)x – (I-TR), x-y) =//x-y//2 – (Trx-Try, x-y) ≥ 2 − //x-//Trxy//-x// y//-Try// ≥ 0 I-Tr is monotone The retraction to M I – Tr is I – T Lemma (3.4.10)** If T is monotone and X0 and Y0 are normal of H such that Tx = Y0 Proof For any Y in H and T>0 Let yt = X0 +ty0

With y- yt then (Tyt – y0, y) ≥ 0 so that

≥ ()Tx00 y,y (Tyt – y0, y) t → + ,0 then Ty → Ty Let t 0 Tx y,y So that ()00 Then Y0 = Tx

(3.4.11) Conjectures (1) Let M be a b e a convex subset of a normal linear space L let T be a non-expansive mapping on M into L, Then for 0

G 1 ()+= TI ∗ ∗∗ 2 , G is non-expansive and the exists εMX such that GX = X X λ += 1 − λ T Also, +1n xm ()xn λε()1,0 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 ()1 p ∞<< are classical examples of uniformly convex spaces.

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Therefore, is application a non-expensive operator will have solution on these spaces.

REFERENCES (1) Kreyszig Erwin (1978): Introductory Functional Analysis, John Wiley (2) Kufuer A. et al: (1977): Function Space. Publishing House of the Czechozbrak Academy of Sciences. Pragne. (3) Royden H. L (1968): Real Analysis, Second Edition, Macmillan International Edition, (4) Smart D. R. (1974): Functional Analysis, Printed by Cambridge Press, Syndics of Cambridge, University Press. (5) Walter Rudi (1976): Principles of Mathematical Analysis, International Student Edition (Third Edition). Published by MCGRAW-HILL International Book Company.

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