379 / ,t BANACH SPACES AND WEAK AND VEAK TOPOLOGIES THESIS Presented to the Graduate Council of the University of North Texas in Partial Fulfillment of the Requirements For the Degree of MASTER OF ARTS By Andrew F. Kirk, B.S. Denton, Texas August, 1989 * Kirk, Andrew F., Banach Spaces and Weak and Weak Topologies. Master of Arts (Mathematics), August, 1989, 68 pp., bibliography, 3 titles. This paper examines several questions regarding Banach spaces, completeness and compactness of Banach spaces, dual * spaces and weak and weak topologies. Examples of completeness and isometries are given using the co and spaces. The Hahn-Banach extension theorem is presented, along with some applications. General theory about finite and infinite dimensional normed linear spaces is the bulk of the second chapter. A proof of the uniform boundedness principle is also given. Chapter three talks in detail about dual spaces and * weak and weak topologies. An embedding proof and proofs * involving weak and weak compactness are also given. The Cauchy-Bunyakowski-Schwarz inequality and Alaoglu's theorem are also proven. -I- --40,600 014i ."a = ow TABLE OF CONTENTS Chapter Page I. INTRODUCTION: LINEAR OPERATORS AND THE HAHN-BANACH THEOREM. .. a.a..... .. 1 II. FINITE AND INFINITE DIMENSIONAL BANACH SPACES . .18 III. DUAL SPACES AND VEAK AND VEAK TOPOLOGIES . .33 BIBLIOGRAPHY ..a................*....0. .a.. ........ 68 iii CHAPTER I INTRODUCTION: LINEAR OPERATORS AND THE HAHN-BANACH THEOREM The purpose of this thesis is to examine certain * questions concerning Banach spaces and weak and weak topologies. A basic knowledge of analysis, topology and algebra is assumed. The first chapter deals with supporting definitions of normed linear spaces and linear operators and includes an equivalence theorem concerning continuity and a proof of the Hahn-Banach theorem. Examples of complete normed linear spaces are also given. Chapter two discusses some general properties of finite and infinite dimensional Banach spaces. Theorems on completeness and compactness are given, as is a proof of the uniform boundedness principle. Chapter three deals primarily with dual spaces and weak * and weak topologies. Examples of isometric dual spaces are given, along with a proof showing the equivalence between reflexive spaces and weak compactness. Proofs of Alaoglu's theorem and the Cauchy-Bunyakowski-Schwarz inequality are also given. In this paper, we use R and M to denote the real and natural numbers, respectively. The zero element of a vector 1 2 space is denoted by 0, and the end of a proof is given by *. Also, the phrase "if and only if" is replaced by "if f." Further, all vector spaces are assumed to be over the real field R. Let X be a vector space (or linear space). A sublinear functional p on X is a real valued function with domain X satisfying: (i) p(x + y) p(x) + p(y), and (ii) p(ax) = ap(x), f or all x,y E X, a > 0. A seminorm p on X is a sublinear functional which is absolutely homogeneous; i.e., p(ax) = |alp(x) for all x E X, a E R. A norm 11-11 on X is a seminorm so that /jx jj= 0 iff x = 0. Since a norm is a seminorm, the only property the norm lacks to be a metric is established in the following proposition: Proposition 1.1: Let p be a seminorm on X. Then p(x) > 0 for all x E X. Proof: First we claim that p(0) = 0; for p(0) = p(0.x) for any x. But p(O-x) = j0jp(x) = 0. Now we suppose p(x) < 0 for some x E X, and choose x to be one such element. Thus I-1p(x) < 0 and hence p(-x) < 0. By adding the appropriate inequalities, and using the triangle inequality, we get p(-x + x) p(-x) + p(x) < 0. But p(-x + x) = p(0) < 0, and this contradicts our claim. A normed linear space is a pair (X,1- 1I) where X is a 3 linear space and 11-|1 is a norm on X. An abstract normed linear space is given by X, instead of the pair (X,I.I).Now, jj-j| defines a metric, d, on X where d(x,y) x - y . The topology on a normed linear space is inherited from the norm; i.e., a sequence < x, > -4 x E X iff || x. - x |- 0. Suppose X and Y are normed linear spaces. A linear function f is a function from X into Y so that f(x + y) = f(x) + f(y) and f(ax) = af(x), for all x,y E X, a E R. A linear functional is a linear function into R. We say a linear function is continuous if < f(x.) > -+ f(x) whenever < xn > -4 x in X. Other equivalences on continuity are given in the following theorem: Theorem 1.2: Suppose X and Y are normed linear spaces and T : X -+ Y is a linear function. Then the following are equivalent: (i) T : X - Y is continuous, (ii) sup{ fl T(x) j1: x E X, 1 xl || 1 } <w, (iii) There exists I > 0 so that jjT(x) I K-1l x jjfor all x E X, (iv) T is continuous at 0 in X, (v) There exists xo E X so that T is continuous at xo, (vi) T is uniformly continuous. Proof: (i) * (ii). Suppose T : X -4Y is continuous and suppose { |jT(x) jj : x E X, x | 1 } is unbounded. Then for each n E MN, choose xn E X, jjxn |j 1, so that T(x.) 11> n. Since 1J x (1, <- x> 0-. Since T is 4 continuous, we have T(j xn) - T(O) = 0. Thus f11T(xn) - 0. Hence, |11T(xn) ||1is1T(x,) 1| -4 0, and | eventually less than n for sufficiently large n, contradicting our supposition, making sup{ |1 T(x) | : x E X, x | 1 } <co (ii) * (vi). Suppose a = sup{ | T(x) : x E X, jj x || 1 }. Let e > 0, and choose 0 <8< 1 so that 6 < "/a+; i.e., aS < E. Let x,y E X so that x - yjj < 6. Thus,< | -(x - y) |< < 1, and 1 by supposition, we have |jT( 5(x - y)) |j a. Therefore, 11 T(x) - T(y) || aS < e, and T is uniformly continuous. (vi) * (i). The proof follows immediately from the definition. (i) * (iii). Suppose T : X - Y is continuous. Suppose further that if K > 0, then there exists an x E X so that jjT(x) jj > K-11 x 11. For each n E IN, choose x E X so that 11 T(x.) j|j> n-11 x. jj. Now, < -x> -40, and since T is continuous, T( 1 x) -4 T(O) = 0. Thus, as bef ore, this implies that njx -fI T(x )11 j-j 0, and hence 1j T(x) |1 must eventually be smaller than n-|j xn || for sufficiently large n, contradicting our supposition. (iii) * (iv). Suppose there exists a K > 0 so that T(x) 11 K-11 xl11 for all x. Let < x. > be a sequence in 11 n=1 X that converges to 0. Thus jjx,,j 0. Also, A " 9 - I I---- - - 'WIMMINIMI, loom 11is m- 5 jjT(xn) I1< K-11 xn jjfor all n, and as n - w, the right hand side tends to zero. Therefore, jjT(x.) f| -4 0, and T is continuous at 0. (iv) * (v). The proof follows immediately from the definition. (v) * (i). Suppose T : X - Y is continuous at some x in X. Let x E X and < xn >* be a sequence in X 0 n=1 converging to x. Let y. = x. - x. If it is true that jj T(y.) |1- 0 , then 11 T(x. - x) 1j -4 0 and hence T(xn) -4 T(x), proving that T is continuous at every x E X. Since xn -4 x, we see that 1| y. |11-+ 0, or y -4 0. Pick any sequence < zn > in X converging to x0 . Now we have < Zn + yn > -4 x0 + 0 = x; i.e., < Zn + yn > is a sequence converging to x0 . By the continuity of T, < T(zn + yJ) > -4 T(x0 ) or 11 T(zn + yJ) - T(x.) |11-4 0. Since Zn - x0 , we have (as before) 11 T(Zn) - T(xO) jj -4 0. Now, let E > 0, and choose N E I so that z+y,) - T(x.) | and T(z.) - T(xO) I<< for n > N. Adding appropriate sides of the previous two inequalities and using the triangle inequality, we arrive at 11 T(y) fj 1 E for n > N. This shows that |j T(y) -4 0, finishing the proof. * The statement that a sequence < xn > in a normed n=1 linear space is Cauchy means if e > 0, then there exist 6 N E M so that if n,m > N, then jjXn X-Xm 1< c. A normed linear space X is said to be complete if every Cauchy sequence from X converges and converges to a point in X. A complete normed linear space is called a Banach space. Two examples of Banach spaces are given in the following: Example 1.3: Let c0 { c = < c > : < c, > is a real sequence converging toO in R }. For c E c, define 11 c jjW = sup { cn : n is a positive integer }. Then (cOI-IM) is a complete normed linear space. Proof: Let < c > be a Cauchy sequence in c. Put cn = < cn.>CD , and fix i E M.