379
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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 } (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 } 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. Let f > 0, and choose N E M so 1 i1 1 that jjcn - cm j < e for n,m > N. Hence, |cn. - cm. < e for I I for Therefore, < cn.> * all j E N, and in particular j = i. j n=1 is a Cauchy sequence in R, and hence, convergent; choose c E R so that< cj> -n- c. Further, let c =< c >. We claim that cn - c, and c E co. To show the first statement, fix n > N. Therefore, for any m > N, we have || - cm < c. Thus, sup{ |c. - cm.l : j E NE} < E I .1 for any m > N. By letting m tend to w, we arrive at sup{ lc . - c I : n > N } c. This shows that we can make all the coordinates of some c as close to c as we want simply by choosing n sufficiently large, thus showing cn - c, wherever c happens to lie. Now we will show that c is in fact in co. Let c > 0, and choose N,N E I so that jc<- c i for n I N, and lc << for 7 nj - c~~~~~il -Fon NadIn 2 '7 i > N . Using the triangle inequality on the former inequality, we get Icil Proof: Let E > 0 and let < xn > be a Cauchy sequence in 11. Choose N E M so thatI11 x - xmII1 < e for n,m > N. Put xn= < x.>.. Since |jx. - xmI < eforn,m N, Therefore, Ix. - xmI Fix n > N. Thus, { IEx -xml : m> N } is bounded above by E. By letting m approach w, we get 00 .Nx- xj c f or n > N. The above statement not only says that xn - x E V, but also shows that < x > - x. By dint of the fact that I is a vector space we conclude that x E e. If R is any set, then a partial order on R is a relation < on R x R that is transitive and reflexive; i.e., if x < y and y z, then x ( z, and x ( x for all x,y,z E R. A partially ordered system is a pair (R, ), where < is a 8 partial order on R. If (R, ) is a partially ordered system, then a subset S of R is linearly ordered if whenever x,y E S, either x < y or y < x. A linearly ordered subset S of a partially ordered system (R, ) is said to be maximal if whenever T is a linearly ordered set in (R,<) so that T 2 S, then S = T. The following lemma is equivalent to the Axiom of Choice: Lemma 1.5: (Hausdorff Maximal Principle) ' Every partially ordered system contains a maximal linearly ordered set. A proof of the the equivalency between lemma 1.5 and the axiom of choice can be found in Kelly (1, pp. 31 - 36). The following theorem is an application of the maximal principle and has applications further in this chapter and in chapter three (see theorems 1.9 and 3.18, lemma 3.16). Theorem 1.6: (iahn-Banach) Let X be a vector space and let p be a sublinear functional defined on X. Suppose S is a linear subspace of X and f : S - R is a linear function so that f (s) p(s) for all s E S. Then there exists a linear function f : X - R so that f(s) = f(s) for all s E S and f(x) C p(x) for all x E X. Proof: Suppose that the hypotheses are satisfied. We first show that we can extend by one dimension. Let x E X\S and consider the space V spanned by S U {x}. Suppose we have a linear functional f that satisfies f(s) = f (s) f or all s E S and f (v) p(v) f or all v E V. 9 Each v E V can be written as v = ax + s for some a E R, s E S. Thus, f(ax + s) < p(ax + s). In particular, the above statement is true for a = 1,-1; i.e., f(x + s) p(x + s) and f(-x + t) < p(-x + t), for all s,t E S. Thus, f(x) p(x + s) - f(s) and f (x) f(t) - p(-x + t) for all s,t E S. Is it true that f satisfies f(t) -p(-x + t) f(x) p(x + s) - f(s) for all s,t E S ? The answer is "yes," and to show this, we merely show that sup{f(t) - p(--x + t) t E S } < inf{ p(x + s) -- f(s) s E S }. We know that f(s + t) p(s + t) for all choices of s and t. Thus, since f is linear, f(s) + f(t) 1p(s + t + x - x) p(x + s) + p(-x + t) for all s,t E S. By rearranging the first and third terms of the above inequality, we arrive at the desired conclusion. Now we are able to choose any a in the interval [sup{ f(t) - p(-x + t) : t E S }, inf{ p(x + s) - f(s) s E S }] and define f(x) = a. Let v E V; recall that v = ax + s f or some unique a E R and s E S. Define f(v) = af (x) + f (s) . We assert that f extends f linearly to V. First, note that f is well defined as a result of the uniqueness of the representaion of v; 10 also, f(O) = f(Ox + 0) = Of(x) + f(0) = 0 + 0 = 0. To show the linearity of f, let u,v E V, u = apx +s, v = a2x + S2. Then f(u + v) = f((apx + s1) + (a 2x + S2)) = f((a+ a2)x + (s1+ S2)) = (ao+ a2)f(x) + f(s1+ S2) = alf(x) + f (s1) + a2f(x) + f (s2) - f(u) + f(v). Also, f (au) =f (aapxAA + as1)= aalf (x) + af (s1 ) = af (u) , and thus f is linear. Also, for any s E S, f(s) = f(Ox + s) = f(s), and f agrees with f on S. To show f(v) p(v) for all v E V, where v = ax + s, we consider two cases: (i) a > 0 and (ii) a < 0. (We took care of the case a = 0 above.) (i) a > 0. We know f(x) p(x + s) - f(s) for all s E S, and, in particular, for is.a Thus we have f(x) p(x + -s) - f (-s) and f(x) + f( s) ( p(x + 1s). Since a > 0, af (x) + af(as) ap(x + a s) and af(x) + f(s) p(ax + s) = p(v). This leads us to f(v) < p(v). If a < 0, we use the fact that f(x) > f(s) - p(-x + s) for all s E S. By substituting 1 -s for s, and noting that - a > 0, we argue similarly as above concluding that f(v) p(v) for all v E V. 11 The above three cases have shown f agrees with f everywhere on S, and f is dominated by p everywhere on V. Thus, we have extended f on S by one dimension. Consider { (S', g) S' is a linear subspace. of X containing S and g is a linear functional on S' satisfying f(s) = g(s) for all s E S and g(s') p(s') for all s E S' } =. Partially order Yby (S', g) < (T', h) iff g c h. This also partially orders the domain of 5. Invoking the Hausdorff Maximal Principle, we let eYbe a maximal linearly ordered subset of Y. Let f = U { g (S', g) E }. The linear ordering on e/ensures that f is actually a function. We claim that f is the desired linear functional. By definition of Y, f(s) = f(s) for all s E S. Let D = Domain(f) and x,y E D, a E R. Thus, x,y E S' for some (S', g) E QY. Hence, g(x + y) = g(x) + g(y) and since g C f, f(x + y) = f(x) + f(y) for any choice of x,y E D. Also, ax E T' for some (T', h) E /, and h(ax) = ah(x) implies that f(ax) = af(x) since h c f. Furthermore, f(x) = g(x) p(x) for x arbitrary in D, and thus f is dominated everywhere on D by p. Finally, to show D = X, note that if it is not, simply choose an element v E X\D and extend f by one dimension as before. Thus, Q/is not maximal, a 12 contradiction. Therefore, D = X and we have reached the desired conclusion. * If X is a linear space, then a linearly independent set B is an algebraic basis (or Hammel basis) if every element in X can be uniquely represented as a finite linear combination of members of B . When we speak of bases in linear spaces, they will be understood to be algebraic unless otherwise stated. Another application to the maximal principle is given in the following: Proposition 1.7: Suppose X is a vector space and x is a non-zero vector in X. Then there is a linear functional f : X -4 Rso that f(x) #0. Proof: Let x E X be as above. We wish to construct a basis for X containing x. Let 2= { B : B is a linearly independent and x E B }. Put 2= { Ba : a E r }. Note that 2 is partially ordered under normal set containment. Again invoking the maximal principle, we get a maximal linearly ordered subset of 2; call it /. Let A cLrso thatBa E Yiff a E A. Let B= UB . We claim that B is a basis for X. To show aEA a linear independence, we simpley suppose not and choose v E B that is a linear combination of finitely many members of B \ {v}, say { v,,*-, v }. But {v1,---, v , v} is contained in some Ba, a E A, and hence Ba is not linearly independent, contradicting the definition of 2. Thus, B is linearly independent. To show B spans X, we again suppose 13 not and choose v E X \ span(B). Thus, B U {v} is a linearly independent set properly containing B, contradicting the maximality of B. Thus we have shown that B is a basis for X containing x. Now each v E X can be written as v = aibi + ax where a, a,---,a. E R and each bi E B. For v defined as above, def ine f (v) = a, and note that f (x) # 0. We claim that f : X - R is the linear functional we want. First note that f is well defined since the unique representation of v as a linear combination of members of B yields a unique a above. Let u,v E X, E R with u = abi + ax, v = .. c. +OX, i=1 j=1 where a,# E R, ai,3 E R, and bjc3 E B ,1 < i < n,1 Now f(u) = a, f(v) = , and u + v = . abi +.flc + (a + f)x. i=1 j=1 Since the two summands can be written as a single summand, f(u + v) = a + l. Also, 7U = .7ajbj + (ya)x 1=1 and f (7u) = 7a = 7f (u). Thus, f has the desired properties of the conclusion to the theorem. * An immediate corollary to the above proposition is: Corollary 1.8: If X is a vector space and x is a non-zero vector in X, then there is a sublinear functional p : X -uRso that p(x) #0. Theorem 1.9: Suppose X is a normed linear space and x is a non-zero vector in X. Then there exists a continuous linear functional f : X - IR so that f (x) = jj x 11. 14 Proof: Let x E X satisfy the hypotheses. First we note that p(v) = jfv jjis a sublinear functional defined on X. Let S = { ax : a E RR }. Clearly, S is a linear subspace of X. Let f : S - R be defined by f(ax) = all x j1. Note that f (x) = |l x j|. Now f is linear since for if u = ax, V = (x E S, 7 E R, f(u + v) = f((a + #)x) = (a + P)l1 x =all xl|l +fllxlI1 = f(u) + f(v), and f (7u) = f(7ax) = 7|a| x || = 7f (u). To show f (s) < p(s) for all s E S, s = ax, we consider the two cases where a >0 and a < 0. If a 0, then f (s) = f(ax) = all xl1l=l|l axl11l=I11 sll| = p(s). If a < 0, then f(s) = f(ax) = all x 11i= - |1 ax 11 <|11 ax ||= ||1sl| = p(s). In either case, f(s) p(s). Invoking the Hahn-Banach theorem, we construct a linear functional f : X -R so that f agrees with f everywhere on S and is dominated by p everywhere on X. We claim that f satisfies the conclusion to the theorem. Since x E S, f(x) = f(x) = jjx |. Now the question of continuity must be settled. But continuity follows from theorem 1.2 since |f(v)| 1-| v 11.U If X and Y are linear spaces, then another linear space can be constructed by considering the space of all 15 continuous linear functions from X into Y, B(XY). This clearly forms a vector space. Further, if T : X - Y is a continuous linear function, we can define a norm on T (and hence on B(X,Y)) by 11T 11 = sup { 1 T(x) |: x E X, ix 11 1 }. The following theorem will show that if X is a vector space, then the space of all continuous linear functionals on X, B(X,R), is complete when R is endowed with the usual metric norm. Theorem 1.10: If Y is complete, then B(X,Y) is complete. Proof: Let < fn > be a Cauchy sequence in B(X,Y) and n=1 e> 0. Choose N E sothat fn-fm1'< j'forn,m> N. Thus, if n,m > N, then sup{ f| f.(x) - fm(x) 1 : x E X, lx1 1 } < Let x E X. Ifil1 xl|l 1, thenll1 fn(x) - fm(x)il1 <'j for n,m > N, and hence < fn(x) > is a Cauchy (and thus n=i convergent) sequence in Y. Now, fix x E X. Thus, If(x) - fm( X) ii1 || f1 - fm || for n,m > N and hence -| fn(x) - fm(X) 1 f. f- fml|- Now, || f.(x) - fm(X) Iiixli1l-1fn-fmlL- As n,m - 0, the right hand side tends to zero. Theref ore < f.(x) > a' is a cauchy and hence convergent sequence. We n=1 may now define f : X - Y by f(x) = lim fn(x). Note that f n 16 is linear by the linearity of fn. Since we are dealing with Hausdorff spaces, limits are unique, and f is well defined. Fixn > N; thus { fl f.(x) -fm(x) 11:11 x < 1, m > N } is bounded above by j. By letting m -w, jjf.(x) - f(x) <1 . for allil1 1 < 1 and all n > N. Now, sup{ |jfn(x) - f(x) 1|:11 xi |< 1 } for n > N. The above statement not only states' that fn - f is in B(X,Y), but also < fn > - f uniformly. Since B(X,Y) is a vector space, we conclude that f E B(X,Y). * CHAPTER BIBLIOGRAPHY 1. Kelley, John L., General Topology, New York, D. Van Nos- trand Company, 1955. 17 CHAPTER II FINITE AND INFINITE DIMENSIONAL BANACH SPACES In this chapter, we examine some general properties of finite and infinite dimensional normed linear spaces. Lemmas involving 4 and R" are given, as well as a proof that every finite dimensional normed linear space is a Banach space. Compactness proofs are given, along with a proof of the Uniform Boundedness principle. Lemma 2.1: Let = { x = Proof: Since normed linear spaces form metric spaces, compactness is equivalent to sequential compactness. We will show that every infinite sequence in S possesses a convergent subsequence. Let < ak > m be a sequence in S Put ak=< ak>. and k=1'n1 i=1 note that j1a || = .S |ak I = 1 for all k. Therefore IakI 1 for all k E IN, 1 < i 5 n. Thus ak E [-1,11 for all n E N, a compact set. We will now denote the sequence <&ak > *'by i k=1 < a(k,i) > * . For i = 1, < a(k,1) > is an infinite k=1 k=1 sequence in [-1,1], and hence has a convergent subsequence. Let < a(k1,1) >." be a subsequence of < a(k,1) > ' and I j1=1 k=1 18 19 a, E [-1,1] so that < a(k ,1i) > T-4a,. Note that k. > k. Now, < a(k,2) >* is an inf inite sequence in [-1,1], and has a convergent subsequence. Choose < a(k,2) > a subsequence of < a(k,2) >. and 2 2= 1 1 a2 E [-1,1] so that < a(k32,2) > T2a2- Suppose < a(k.,i) >.* and ai have been chosen so that ii Y=1 < a(k.,i) >. is a subsequence of < a(k,i) > and .1 k=1 1 < a(k .,i) > 1 -+ aj for 1 < i < n - 1. Now < a(k ,n) >. is a subsequence of Jn-1'= < a(k,n) > . As before choose < a(kn) >. a k=1 kn j=1 subsequence of < a(k ,n) >. * and an E [-1,1] so that n-I < a(k ,n) > - a. in Jn n Note that we now have k. > - --.> k. k. Consider the in ~~ i3 subsequence < ak. >. of < ak > W and a = < al,...1,an >. We k=1 3 n ri=I claim that < ak > a j n and a E S Now < a(k ,i) >. is a subsequence of n j = n < a(k.,n) >.'D for all 1 i < n, and thus ii Y=1 < a(k ,i) > - a n' n 1 for all 1 < i ( n. In order for < a. > - a, we need n Sak. - a T-0 an n 20 But ak. - -v II | iI .J |a(k n ,i) -ail T-0n n -3 1=1 iff ia(ki) - a| -- 0 for all 1 ( i < n if f < a(k ,i)4>a, which we have shown to be true. Thus < a. > T-- a. Now we n must show that a E S Since < a. > is a subsequence of Si.j njnJ=1 | n.- a < c. By the triangle inequality, In |akj 1< E+jj1a n for in N. Thus, <|| s. 1 > Ty-- 11 a ll. Jnn But < |1 ak. f >. is just the constant sequence 1 and must n n=I converge to 1. Therefore, 1a j = 1 implying that a E S . Thus, we have shown every infinite sequence from S possesses a convergent subsequence whose limit is in S . Therefore, S is compact. * Theorem 2.2: Any two n - dimensional normed linear spaces are linearly homeomorphic. Proof: We will show that any n - dimensional normed linear space X is linearly homeomorphic to (A, I|Il-), where im - , ,;,, '..- - .- - -,-, 40"s - 21 (Q, 1-|| ) is def ined as in the previous problem. Thus, if Y is any n - dimensional normed linear space, then Y is homeomorphic to X via function composition. Let { ei } be a basis for 4 where i1 ei =< 0, 0,...,1 ,..., > with a 1 in the ith coordinate position, and { xi } be a i=1 basis for X where 11 xij< = 1, 1 i n. Define the function f :A -+ X by f(x) = f(.aiej) = .aixi. Clearly, f defines a linear, one-to-one function onto X; to show that f is a homeomorphism, we must show that f and f' are continuous. Suppose that there exists positive constants M and m so that m <11 f(a) | < M for all a E S . Then for any x(Ef(< M and thus jJ f(x) 11 M-|l x 1J. Therefore f is continuous. Now let a E i , f(a) = y. Thus, m < jff( a) |1, and m.ll a |11 |jf(a) j1. Since f-1(y) = a, we have m-1l f~1(y) I| 1 |1 yjj. Therefore 11 f-(y) |1 -||1 y||, and f-1 is continuous. Let a E S a =.ae. Thus, f (a) = aixi Since 11 a l = 1, |1ail = 1. But || f(a) | = I . aix | . ai| = 1. Thus, | f(a) 1| < 1 for all a E S i, and f is continuous. We now wish to show that f-1 is continuous. Define 22 T : S- IR by T(a) = 11f(a) 11. We claim that T is continuous. Let < a, > - a in S. Since f is continuous, we n have that 1| f(a.) - f(a) j - 0. If E > 0, then I|f(a.) - f(a) j< | IIf(an )1 1 - 1I f(a) I < |T(a.) - T(a)j < e for n > N. Thus, T(a.) -4 T(a), and T is continuous. Since 0 % S4, f(a) # 0 E X for all a E S , since isomorphisms map zero to zero. Thus, jjf(a) jj> 0 for all a E SI. By the previous lemma, T is a continuous function on a compact set, and T attains its minimum value which is greater than zero. If m = inf{ T(a) a E S4 }, then 0 < m < || f(a) j for all a E S4, and thus f-1 is continuous. * Theorem 2.2 is very helpful in the respect that we may now use previously defined finite dimensional normed linear spaces instead of abstract spaces. In particular, we will show that every finite dimensional normed linear space is complete by showing R" is complete. Lemma 2.3: r" is complete under the usual norm. Proof: Recall for x = 0 < < 1, and choose K E 9 so that if xk = < Xk i >ni1j . Let k,1 >K, then II Xi )\2] /2 < 1 Xk -X 1 1 = l(xki - 1J 2 Now, 2 2 x i= 1i2 - Xi)2]1=1T I I/ 2 < iff (Xk.-- x1.) <1 2 Since each (xk. - x1 ) > 0, 1 < i < n, we have 2 (xk - X1 ) 2 2 -X11 < Cxk - x1 < C ff Ixk SIffI k -xii * Therefore, Ixk. - xj < for 1 ( i ( n, k,1 K. Thus, for i12 each 1 < i < n, < Xk. > ' is a Cauchy, and hence convergent, 1 k=1 sequence. Let xi = lim Xk., 1 < i n, and put x= < xi >. k 1 1 Clearly, x E In. Now, for kl > K, [.(Xk - x1 .)2]'/2< Fix k > K. Thus, { [I(xk -- xi)2] 1/2: 1 K } is bounded above by f. Therefore, sup{ [(xk. - xi )2]1/2 :1 K } < . 1=1 1 i2 Thus, [t(Xk - X)2]1/2 < e provided k > K. Therefore, Xk - x jj< c for k > K and thus < Xk > -4 x, proving that R' is complete. * Theorem 2.4: Every finite dimensional normed linear space is complete. Proof: Let X be an n - dimensional normed linear mm"mto 24 space, f : R" - X be a linear homeomorphism, and < xk >'be a Cauchy sequence in X. Since f -1 is a continuous linear function, f-1 is uniformly continuous. Let e > 0. Choose 6, > 0 so that if x,y E X and j x - y jj< , then jjf-(x) - f-1(y) < 1 1 f-(xk) - f ~(x1) |11< f f or k,1 > N. Theref ore, < f -1(xk) >k* is a Cauchy and hence k=1 convergent sequence in R". Choose r so that < f -1(xk) > - r E R . Since f is continuous and linear, f is uniformly continuous. Let b2 > 0 so that if s E R' and <2, then ||r - s < 11f (r) - f (s) 1 Let K E IN so that if k >.K then 11 f -1(xk) - r |1 < 82. Thus, 11f[f-'(x)] - f(r) |1 < e for k > K implying that 11 Xk - f(r) 1 Therefore, < Xk >kw is a convergent sequence, and X is k=1 complete. * Corollary 2.5: Let X be a finite dimensional normed linear space. Then,B={x EX: | xII1}and SX = { x E X :1|xI| =} are compact. Also, a finite dimensional subspace of an infinite dimensional normed linear space is closed. Proof: Let X be an n - dimensional normed linear space, and consider (es, | We have shown that S4is 25 compact; the proof is easily adapted to show that Biis also compact. Also note that scalar multiples of each are compact. Let f : 1. -+ X be a linear homeomorphism. Since f is a homeomorphism, both f and f~1 are closed functions. Therefore, f-1(BX) is a closed set. Choose K > 0 so that f-1(x) j K-11 x 11 for all x E Bx. Thus, jjf-(x) j| K for all x E BX which implies f-1(BX) L K-B . Since f'-(BX) is a closed subset of a compact space, f-1(BX) is compact. Since f is a homeomorphism, f[f'-(BX)] = BX is compact. Also, since SX is a closed subset of a compact space ( BX )' 5X is also compact. Finally, suppose that X is an infinite dimensional normed linear space. Note that (X, I-|I) defines a metric space. We have shown that every f inite dimensional normed linear space is complete; since complete subspaces of metric spaces are closed, our claim is verified. * Theorem 2.6: Let Y be a proper closed subspace of a normed linear space X, and 0 <( < 1. Then there exists a x E Sx so that 1j x, - y [1 for all y E Y. Proof: Define the distance function d : X - R by d(x) = inf{ 1 x - y 11 : y E Y }. We claim that d(x) > 0 for x 0 Y. Let x 0 Y and suppose- that d(x) = 0. For each n E IN, choose y E Y so that j|x-y j|j< . Clearly, < y > - x, and since Y is closed, this puts x in Y, a contradiction. 26 Let x 0 Y and consider d(x). Let ( E (0,1), and choose 0 < e < d(x)-( ~ . Let y E Y so that 1jx - yjj || d g(x -xy) = inf{ 11 (x - Y) - w |I: w E Y } = inf{ 1 1| (x - y) - x - y -w : w E Y } ||x-yI1 = IIx-yi-inf { ii x - (11x - y 1-w + y) 11: w E Y }- Now we claim that { j x - y I-w + y : w E Y } = Y. Clearly, we have{I x - y I|-w + y : w E Y }C Y. Let z E Y. Thus 1 - y) E Y and hence 11 x - y |11 1(z - y) + y E { IJ x - y 1-w + y : w E Y }. But the latter term is just z, proving our claim. Now, IIxY1i-inf { x - (|1 x - y I w + y) |: w E Y } E Y I -yId -Ix inf{| x -y | y Since jjx - y || d(x) + f, d(x) d(x) ||x-yJJ d(x) + C' But d(x) _ d(x) d(x) + d(x) d(x) d(x)(1 + ) d(x) d(x 27 Thus , |1 (x - y) - w 11 > for all w E Y. I1x-y- Theorem 2.7: If I is an infinite dimensional normed linear space, then B. is not compact in the norm topology. Proof: Let X be an infinite dimensional normed linear space and 0 # x E X. Now the span of x (denoted by span(x)) is a finite dimensional subspace of X and hence is closed. Put S1 = span(x). Let x, E SX so that 1 I x1 - s Is>- for all s E Si. Let 2 = span(S1 ,x1 ). Again, S2 is a finite dimensional subspace of X and is closed. Choose x 2 E SX so that 1 || x s >||2 1 for all s E S2. Note that 11 x 2 - X, |> since x, E S2- Suppose finite dimensional proper closed subspaces S1,...,s, of X and elements x1,... ,xn E S have been chosen so that ii 1 1xi - s 11| 2 for alls E S , 1 < i n. Notethat |xi - xi.. 11|| 2 for 2 < i < n. Let S,+1 = span(Sn,xn). Choose xn+ 1 E SX SO that ||x - s for all s E Sn+ * Consider the sequence < x.1 >* in B1 . For n=1 X any pair m,n E N, m > n, j|xm - xK jj since jjxm - s 28 for all s E Sn. Let< x. >. be any subsequence of < xa > *c. By the same argument jjx. - xj j1>I for all n=1 2 i,j E IN, and hence < x. > is not Cauchy, implying that it is not convergent. Therefore,< x > has no convergent n=1 subsequence, and BX is not compact. M . One might also note that the same argument shows that SX is not compact in the norm topology. As we have noted before, a linearly independent set B of a linear space X is a (algebraic) basis for X provided every element of X can be uniquely represented as a linear combination of finitely many members of B . If X is an infinite dimensional Banach space, then obviously no basis can be finite. This raises the question: "Can an infinite dimensional Banach space have a countable (algebraic) basis?" The reason we speak of a Banach space instead of an arbitrary linear space is due to the Baire-Category theorem. A proof of theorem 2.8 may be found in Royden (1, p. 159).: Theorem 2.8: (Baire-Category) No complete metric space can be expressed as a countable union of nowhere dense sets. (A subset N of a metric space M is dense if the closure of N is M, and is nowhere dense if the closure of N contains no open sets (open with respect to the metric).) The answer to the above question is a resounding "No!". In fact when we go from finite to infinite dimensional Banach spaces, we actually go from finite to uncountable dimensions. 29 Theorem 2.9: An infinite dimensional Banach space cannot have a countable basis. Proof: Let X be an infinite dimensional Banach space basis, say { xi . Let and suppose that X has a countable 1=1 X1 = span(x1), and for any i E M, i > 1, let xi =span(Xi..,xi).- Thus, X = .X., and note that Xj. 1 2 Xi. i=1 We now claim that a finite dimensional subspace of X cannot contain any open set of X, for suppose it did. Let Y be a finite dimensional subspace of X and suppose that 0 is an open set in X contained in Y. Let x E (0andconsider 6?- x = { 4- x : azE 0}. Note that 6?- x is open in Y. Since x e 0, x - x = 0 EO- x. Since 6?- x is open containing 0, There exists E > 0 so that e-BX = { x E X : 11 xi |1 Thus e-BX - Y and by taking scalar multiples of -BX we see that Y = X. Therefore, no finite dimensional subspace of X can contain any open set in X. Recall that the Baire Category Theorem states that no complete normed linear space can be written as a countable union of nowhere dense sets. Since each Xi is a finite dimensional subspace of X, each Xi is closed. By our claim, each Xi = XY cannot contain any non-empty open set, showing that each Xi is nowhere dense. Now, X = .UJX is a countable i=1 union of nowhere dense sets, a contradiction. 30 Another application of the Baire-Category theorem is given in the following theorem, known as the uniform- boundedness principle. Theorem 2.10: (Unif orm boundedness) Let each of X and Y be a Banach space and let Ta: X -4 Y be a bounded linear map for all a E P. If { Ta(x) : aEP} is bounded for each x E X, then sup{ |j Ta l: a E P } < M. Proof: For each a E P and each m E IN, let A = { x E X : 11Tla(x)t11 Since each A is the (continuous) inverse image of the closed set [-m, m], each A is closed. Let Am = nA Now, each Am is closed and Am = f{xEX : jTa(x) II m} aEP = { xE X :j Ta(x) || m, aE P }. Further, UAm = X, since for any x E X, m=1 sup{ Ta(x) : a E P } k f or some k E M. Since X is a Banach space, X cannot be written as the countable union of nowhere dense sets, and Ak (for some k) must contain some non-empty open set, say 0. Let x E a, and consider the open set 0&- x, containing 0. We will now show that 0 - x C A2 m. Let z E 0- x, and write z as aZ- x f or some aE 0. Thus, x,x + z E 0, and since 6 c Am, jjTa(x + z) 1 10 mmmm IN _11 momommummill 11,1111, - I N 31 I Ta(x) < m for all a E P. Now for any a E r, TlTa(z) = T (z +Tx) +Ta(~) Ix fl T (z + x) I| + j| Ta(X) II < 2m. Since OE O-xC A, let e>O so that -BXcA2m. We now have sup{ jj Ta(x) j1 :xE E-BX, a E 1 <2m. But sup{ |j Ta(x) jj x E E.BX, a E P } = sup{ |1Ta(x) 11:11 x 11 < E, a E r } = sup{ I1T.(x) II: II x < 1, a E }. By letting y =x, we have that sup{ Ta(x) 11: ?x |I 1, aE P } = sup{ Ta(Ey) j :I y 11 1, a E P } 2m. After factoring out and dividing both sides of the above in- equality by e, we see that 2m sup{ II Ta(y) 11 :11| y || 1, a E P } $ - how CHAPTER BIBLIOGRAPHY 1. Royden, H. L., Real Analysis, New York, MacMillan Pub- lishing Co., Inc., 1968. 32 . - --l "'I'll- ,.- -= 14Q"- Ai ---- . -- mummommoommpo - - - CHAPTER III DUAL SPACES AND * WEAK AND WEAK TOPOLOGIES In this chapter we examine some questions regarding the * weak and weak topologies on Banach spaces and dual spaces. We show that any Banach space can be isometrically embedded in its second dual, and we look at some examples of * isometric spaces. Theorems on weak and weak compactness are given, including a proof of Alaoglu's theorem. Also, some supporting lemma's on convex subsets are included. We begin by defining dual spaces. If X is a normed linear space, then the dual of X (or continuous dual) is the space of all continuous linear functionals on X. The dual * of a space is denoted by X , and from earlier results, we * have that X = B(X,R) is a Banach space. Analogously, we * define the second dual of X (i.e., the dual of X ) and ** denote it by X If (X,II.II) and (Y,I-I) are two normed linear spaces, then we say that X and Y are isometric if there exists a linear homeomorphism f : X -4 Y so that i f(x) I = |x jj for all x E X. Isometric spaces give us more to work with than homeomorphic spaces because "distance" is preserved. With this, we present a theorem that any normed linear space can be isometrically embedded in its second dual. ** Theorem 3.1: Let r; X -' X be the mapping defined by 33 34 * * n(x) (x) = x (x). The function n is an isometric isomorphism from X into X ; i.e., u(ax + fly) = aq(x) + fln(y) for all x,y E X, a,f E IR, and 1l 1 (x) 11 = 11 x |1 f or all x E X. Proof: Let x,y E X, a,f E R, and consider n(ax + fy). * * * Recall n(ax + fly) : X - R; choose x E X . Now, /(ax + fly)(x) = x(ax + fly), * and by the linearity of x , we get ax (x) + Ox (x) = a)(x ) + fl(x)(x * * This is true for all x in X , thus n(ax + fly) = aw(x) + #rn(y). ** To show117r(x) =xj11, recall: X-X . Let * x E X. Thus n(x) : X - R. By definition of the norm of a continuous function, * * j11(x) 11 =sup{ J7(x) (x)j : x EBX* } * * = sup{ Ix (x)l : x E BX* Since jx (x)| ix |1-1lxil, |1 r (x) |1 ||ix I|. To show equality, all we need to do is exhibit a continuous function f E BX* so that lf (x)I = |1 x ||. Recall that x E X has been fixed; if x = 0, then choose f : X -4R to be the function that is identically 0, which is certainly continuous and in BX*. Then lf(x)I = 101 = 0 = 1| 0 ||. If x is non-zero, we invoke theorem 1.9 and get a continuous linear functional f defined on X so that f(x) =1 x |. Further recall that theorem 1.9 produced f such that |f (x)| ( Jjx 11 f or all x E X. Theref ore, we have proven the existence of such a function, and have reached the desired 35 conclusion. * For the following examples, we will refer to the spaces co and P. Recall that co is the space of all real sequences that converge to zero, and for any 1 < p < P ={ x = CD P I x.] )/P. |1 x lip = ( n=1E For p = w, P is defined to be the set of all bounded sequences with the sup norm; i.e., for x = < x > Ee, D=I 1 x 11 = sup{ |xI }. Sn Note that P and c are infinite dimensional vector spaces and recall that no infinite dimensional Banach space can have a countable algebraic basis; i.e., there does not exist a countable set so that every element of the space can be expressed uniquely as a finite linear combination of that set. This motivates the following definition. Let X be an infinite dimensional normed linear space. A Schauder basis for X is a countable collection of members of X so that every element of X can be uniquely expressed as a countable linear combination (i.e., an infinite series) of members of the (Schauder) basis. Since members of P are convergent infinite series, it is inherent that P C c. We need only to establish the following proposition for c: Proposition 3.2: Let < x. > ' E c,, and for each n let n=1 en = < 0, 0, ... , 1, 0, ... >, the vector with a 1 in the 36 nth position and zeros elsewhere. Then < xn > can be n=1 uniquely expressed as Nx.e.. Remark: For future reference when referring to sequences, we will dispense with the indexing of the sequence when indexing is obvious. Proof: Let < xn > E co, and let E > 0. Choose N E IN so that IxI < e for all n > N. Now N-1 < xn > - Exnen n~ I < 0, 0,1 . xN' xN+1 .. > Since jxj < e for all n > N, II< 0, 0, ... , XN, xN+l,...1>I<< C N-1 which implies 11< x. > -E- 1nen1< c for all n > N. n= 1 Therefore, < x. > = Sx e. n=1 With this representation for elements in c and P, we consider the following: Example 3.3: The dual of co may be identified isometrically with 11. * ** Proof: Let x E (co) and x = E xnen E c. Then n=1 * * U * x = by the continuity of x , (x) x (Exnen),n=1 and (x) = xx (en) E R. Thus, the final sum converges n=1 provided x = < x > E c. We now claim that < x (e.) > C E 11. To prove this we must show n=1 U * E|x (en)j < w. Let n=1 y 1 = < sgn x (e1), ... , sgn x (en), 0, 0,. > where sgn(r) = 1 if r > 0, and sgn(r) = - 1 if r < 0. Note 37 ** that each yn E B . Now, .Eyx (el) < m f or all n E I. c 0 1 Also, .Xy x (ei) = . yn.x (ei) i=1 1 =1 j =tIx (ei) I Since x is continuous, sup{ Ix (z) I z E Bc } < . Thus, ao * sup{ IE zx (e.)j: z E BC n=10 * D = sup{ Ix ( E znee)|: z E Bc}< n=10 By considering { yn : n E N } as a subset of Bc we see that a * sup{ |.ty.x (ei) I: n E IN } j=1 GD * < sup{ E znx (en)I z E BC } n=1 0 But the right hand term of the above inequality is (after OD * taking the supremum) EIx (e)I < w, verifying our claim. * * * Define f (cO) *- 7 by f(x) = < x (e) > . Let n=1 * * *n= x ,y E (c0) , , E R. Then * ** * f(ax + y ) = < (ax + #y)(e.) > =< (ax * )(en) + (#y* )(en)* > = < (ax )(en) > + < (fly )(en) > = a< x (en) > + k< y (e) > = af(x) + #f(y) and we conclude that f is linear. To show that f is continuous, we will show f*1 * jjf(x ) || J|x | 38 * * for all x E (c) .But IIf(x ) | < II x H iff I< x* (en) > |j x|x | iff EIx (e.)II sup{ Ix (z)I z E bc}iff n=10 Go*OC E I|x (en) I 5 sup{ I E znx (e.)| : z E bcI n=1 n=10 which we have shown to be true for any x * * To show f is one-to-one, suppose f(x ) = f(y ). Thus * * * * < x (en) > = < y (en) > implying x(en) = y (en) for all * * * * n E N. Suppose x f y . Then x(x) y (x) for some xE Co o * * and Exnx (e) # Exy (en). But by observing that n=1 n=1 x (en) = y (en) for all n, we see a clear contradiction. Thus, f is one-to-one. We now must show that f is onto 11; let = y n=1ye E. * * c Recall that y E c.. Define y co -4R by y (a) =Eya. Note that y is well defined since < ans> -+ 0 and Ov E Iyj < ,we have Jay.< < yj| for sufficiently large n. n=1 Thus, for some N E N, E Iy an| < Eyj < EIyj < ca n=N n=N "=1 cv * and hence E jynaj < O. Clearly y is linear. We will show n=1 y is continuous by showing sup{ |y (z)j z E Bc } c< But * sup{ |y (z)1 : 11 z 1 } = sup{ IE ynznj : z jj 1 } n=1 39 < sup{ NIy||z :11 z i| 1 }. n=,t Since for any z E BC0 , |zni 1 for all n, the right hand side of the above inequality is less than or equal to Ejyj * * < w; thus y is continuous and in (co) . Finally, we claim * * * that f(y ) = y, for f(y ) = < y (en) > where y (e.)= ** 0 + 0 +-- + yn + 0 + -- =yn for all n. Therefore, < y (en) > = < y >, and f is onto . Showing that f is an requires verifying that f-1 is ** * continuous and || x j| | f (x ) || for all x (since we have * * * * previously shown that l f (x) || lix ii). Let x E (ce) * * Now, li x 11 1 f(x) j|iff sup{ Ix(z)i :IzEB } x (e)I C0 n=1 Let z E Bco. Thus * CD * = D=Ez~x1 (en)i jx (z)j n=1 I W1 IE z,x (e) C * SE Ix (e.) I n=1 since zJ| 5 1 for all n. Therefore f is an isometry. Example 3.4: The dual of 11 may be identified isometrically with fD. Proof: As in the previous example, if x E (i') and x = xe jEl, then x (x) = x x (en)'. We want to n=1 n=1 associate each x E (1i) with a bounded sequence, and we * claim that < x (es) > E f". Since e, E B11 , q 40 ** * Ix (e.)I sup{ |x (x)l :|11 x 1 1 } =|11 x j, a fixed constant. Thus * * sup{ Ix (e.)I n E IN } 11 x | < co for each x E (P) and < x(en) > E IF. * * * Define f (1') -4 f by f (x) = < x (e.) >. Note that * * * we have shown |jf(x) 11 1 x |for all x ; thus f is continuous provide it is linear. Analogous to the preceeding example, * *** f(ax + #ly) = af(x) + #if(y) * * * for all x ,y E ( ), a,fi E R; further, f is one-to-one. To show that f is onto f, let x = < x > E tD and note that E1 ix.Y1 1 I=x= x |11 1y.1 =||1 x|1 -(61y| n=I n=1 W Con=Im for all y E 2. Consider the function x defined by x(y) = xy. n=1 We claim that x E (11). The linearity of x can easily be * * seen, and since x (y) lix |I,| y j, x is continuous. Also analogous to the preceeding example, f(x*) = x and f is onto ex. Again, we need only show that jjx* || j f(x*) in order to complete the proof. But * * |lix 11 f (x) iff * * |ix 11 < x (en) > ||0 iff sup{ Ix (x)i :|11 x II<5 1 } 11 < x (en) > 11. Let x E B11. Thus 41 * w * Ix (x)I = xx (en)I In=1 E x1|x (e.)> E x n=1 sE|x,-||< x (en) n=1 > 0| 5|< x (e.) > |;|,,n since x E B11. By taking the supremum over the left hand side, we obtain the desired result. * A normed linear space is said to be reflexive if n is ** onto X . With this, we give an example of a non reflexive space. Example 3.5: co is not reflexive. ** ** First proof: We need to exhibit an x E (c) so ** ** * that n (x) # x for any choice of x E co. Define x : (co) ** * DO * -> R by x (xE) = Ex (e.). Note that x defined in this n=1 way is well def ined since the summation converges f or all x E (co) (cf. example 3.3). Clearly, x is linear, and we will show x (x )j |1 x* j to prove continuity. But * ** * jx (x )I 11 x 11 iff I E x (e)II sup{ x (x)| : x E BC } n=1 0 Nowfor anykEM, e E B . Thus Ix ( en)I sup{ x (x)| : x E BC n=1 0 for all k E M, and by taking the supremum over the right hand side, we obtain the desired result. We will now show i(x) # x for any choice of 42 00 x= Exnen E c. Now, n=1 t/(x)(x ) = x (x ) for all x E (co) iff ** ** * x (x) x (x ) for all x iff CD * * E xx (en) = Ex (en) f or all x if f n=1 n=1 CD * * * Ex (e.) [x. - 1] = 0 f or all x E (c0 ) n=1 Suppose SXx (en) [x - 1] = 0 for all x E (co) . Consider n=1 the sequence of functions xi defined by xi(x) = xi, i E IN, where x = Sxe. E c; clearly each xi is linear and n=1 continuous. Note: xi(e.) =1i1ff i = n. Thus Exi(en)[xn - 1] = 0 for n=1 all i E IN, and f or a given i, CD * E x i (en)x.- 1] = (x, - 1) = 0 implies xi = 1. Thus n=1 n=i xn = 1 for all n, contradicting the fact that x E co. ** Therefore, rn(x) j x for any choice of x E c, completing the proof. Second proof: Since (co) is isometric to I and (1) is isometric to 27, we see (co)*** is isometric to ?. Since 77(c0) is embedded in (c0) (i.e., in 4P), all we need to exhibit is a bounded sequence which does not converge to 0. The constant sequence 1 satisfies this, and in fact is the same sequence exhibited above. * For the next example, we will need to establish the following lemma, known as the Cauchy-Bunyakowski-Schwarz inequality. 43 Lemma 3.6: (C.B.S. inequality) If < x, > W3, < y > W n=1 n=1 are real sequences, then xy ( x,2*y2)1/2 for all natural numbers k. Proof: Proof by induction. The case where k = 1 is clearly true, and suppose the inequality is false for some natural number k, k being of least magnitude. Now (t xnyn)2 > tX n2 yn n=1 n=1 n=iy and kEi k-i k1 n= n=1 n=1y k-i k-i k-i But (kIxnyn) 2 xy - xyk (tX - n=1(t ) 2< n=1 n2 Xk2) (y2n = y ~Yk-2 Multiplying out both sides of the above inequality, we get (tx yn )2 - 2xkyktXnyn + (xy)2 < n=1 n=1 2 2 + (xyk)2, xn21 ty.2 - Xk tYn2 - yk txn2 n=1 n= 1 n=1 and, after cancelling (xy) 2 , (tx yn)2 - 2xkyktXfnyn n=1 n=1 <(tx n2 tyn2 - xk 2 tyn2 - yk2 txfn2. n=1 n=1 n=-1 n=1 By supposition ( txnyn)2 > tXn2 tyn2, and thus n=1 n=1 n=1 txn2 yn2 - 2xkyk xlyn n=1 n=1 n=1 - - - tKnyn <(2xkyk - Xk2 t yn2 - y2 txfn2 n=1 n=1 n=1 or equivalently 2xkykt xyfl > Xk 2 y, 2 + yk2 txn 2 . n=1 n=1 n=1 Expanding both sides we get 2 2 2 2 xkykxjyI + ..-- + xkykXkyk > Xk y1 +-- + xk2yk2 44 2 2 + yk2X1 2 + + Yk Xk , and cancelling again, we get 2 2 2 2 2 2 xkykx1y1 + ... + xkykxk.ylY.. > Xk y, + - + Xk Yk_1 2 2 2 + yk X1 + ... + Yk Xk12. If we can show that for all 1 < i 2 XkYkXiYi k 2y,2 + Yk2 X 2 , then the inequality in question is clearly false, contradicting our original supposition and proving the lemma. But 2 xkykxiyi xk2y,2 + yk 2 x,2 i 0 < xk2y,2 -- 2 xkykiyi + yk2X,2 - (xky ~~Yk- 2 which is true. Example 3.7: (13) is isometric to (12). * * aD Proof: As before, if x E (12) and x = Xen then n=1 *D * * Sxnx (en) Suppose vk # 0. Now = (x(e)2)'/2-vk tvx E B1 2. Thus, E -v x (en) xj n=I|Vk |k~n~ <| 1 and 45 00 1* * -1/2*C -v x (ex) = ( x(e.)2))/2. ; x(e.) =1lVk |k~nn= n=1 k n (x (en)2)1/2.X(en)2 n=1 1n=1 = (tx(en)2)1/2 1X implying that tx(en)2 x 12. By taking the supremum n=1 over k E I, we arrive at ao * * E x (e.)2X n=1 If Vk = 0, then clearly the inequality is satisfied. * * * We may now define f: (1) - 2 by f (x ) = < x (en) >. From the previous examples, we have that f is well defined, linear and one-to-one. Also, since Ex (en) 2 2, we n=1 *2< see that / (Ex (en)2)x = x 11f(xE) n=1x and f is continuous. To show that f is onto 12, let y = < y > E 1 2 . Jf k E IN, by lemma 3.6, we have that tIxnynj (txn2 tyn 1/2 =2( txn2)1/2( ty2)1/2 n=1 n=1 n=1 n==1n But n =1X n=1 n=1 n=1 = lix |-|11 y jj< 0. By taking the supremum over k E IN, we arrive at x y jj n=1 Ixy || Now, we may define x by x (x) = E xy for all n=1 x = Exe. E j2 since |x(x) E= xny| xE yn< * n=1 n=1 n= i Also analogous to the preceeding examples, x is linear, and 46 * * * since |x (x)| I || x | -|y |1, x is continuous and in (2)*. Further, we note that f(x) = y, and f is onto t. To show that f is an isometry, we will show || x || |f(x ) jj1 Now Sjj | x11 f(xf) | for all x E () .iff sup{ Ix (x)I x E B } (D x *E(e)2) f or all x E (2)*. C Let x = E xe E B 2 . Thus, n=1 Ix (x) = n=1E xnx (en)I n=1 1; ( E x g(en)2 n=1 n=1 D( nx2 n=1= 2(2n=1 OD* Ex* ( )1/2 n=1 By taking the supremum over Bp, we obtain the desired result. If X is a normed linear space with dual X , then the weak topology (or w-topology or X topology induced on X) is the weakest (in the sense of set containment) topology on X so that every element of X is continuous. The weak * * ** topology (or w -topology or X topology induced on X ) is the weakest topology on X so that every element of r(X) is continuous. We denote the weak and weak topologies on X by (X,w) and (X,w ), respectively. A typical w-neighborhood of X0 E X may be described as 47 * * **, *E(o) = { x E X :Ixi(x0 ) - x 1 (x)I < c } where E>0 and xi E X , 1 i of convergence. Proposition 3.8: Let (xa) be a net in X. Then * -* * * (Xa O in X iff x (x) X (xo) for allix E X Proof: Suppose (xa) -2- X0 E X. Thus, if x , then MX1 *).,x *;E)(Xo) is any weak neighborhood of 0 * * xa E W for sufficiently large a. Let x E X and Y= (x (x0 ) - e, x (x.) + c), E > 0. It is readily seen that * x (x ) E M,X*;C) for sufficiently large a, and x (xa) ~4x (xO). * * * * Suppose x (xa) -~ x (x0) for all x E X . Let = 11(x*,...,x*;E)(XO) be any weak neighborhood of x0 . Now, xi (xO) -4 xi (x,), < i < n. Choose # so that if a > , then lxi (xv) - xi (xO)I < c, 1 < i < n. Now we have x# EV whenever #> a, proving (xa) ~~'X0 . * * * * * If x E X , then a typical w -neighborhood of x may be described as )x-;)(xX) = { y E X : Ix(xi) - y*(xi)| < E } where c > 0 and { xi }. is a finite collection of members i=1 of X. Analogous to proposition 3.8 is the following, which we will not prove: Proposition 3.9: Let (x,) be a net in X . Then * a * * ** * ** * ** ** (x) ~-4x EXI iff x (xa) -. x (x )for all x E X . 48 As, * w * * * * Also, (xa) ~a x E X iff (x) )-4(x)(x)-(x) for all x E X. To prove the following compactness theorem due to Alaoglu, we will use the famous Tychanoff product theorem, whose proof can be found in Willard (1, p. 120). Theorem 3.10: (Tychanoff product) If { Sa : a E A } is a collection of compact Hausdorff spaces, then the product H S ={ fIf :A - U Sa where f(a) E Sa for all a E A} aEAa aEA is a compact Hausdorff space. Theorem 3.11: (Alaoglu) If X is a Banach space, then (BX*, w ) is a compact Hausdorff space. Proof: To prove the theorem, we will show that BX* is a closed subset of a compact Hausdorff space. Utilizing the Tychanoff product theorem, we will construct a compact product space H Sa of compact spaces Sa that is a superset aEA of BX*. Suppose we have a product space H S that is a aEAa superset of BX*. Then H Sa = { f j f : A - U Sa, f(a) E Sa for all a E A } aEA aEA For our indexing set A we will use X, and for x E X, define * * S= { x (x) : x E BX* }. We claim that S = [ - 11 x I|, IIx || ]. This will ensure that each Sx is compact; also note that separation of points is inherited from R. Let x E X and consider Sx. For a 49 * * * given x E BX*, jjx |1 1, and hence x (x)|I |1 x 1|. Thus, * - || x ( x (x) II x jj and we have Sx C [ -1||x |, |x | ] We now wish to show that Sx contains the. entire interval. Let a E [ - |1 x j1, jjx 1 ]. By theorem 1.9, we can get f E BX* so that f (x) =11 x . Suppose x 0 0. Clearly, the function a E BX*. Thus, af(x) = a, for x #j0,and a e Sx. If x = 0, then x (x) = 0 for all x E BX*. Thus, S x= { 0 } = [ - H1x fl, ||1x j11]. Now, H Sx is a compact Hausdorff space. XEX We must now show that H S, is a superset of BX*. Let xEX * x E BX*. Elements of H Sx are arbitrary functions xEX f X -+ Rsatisfying f(x) E S for all x E X. Let x E X. * * * Since x E BX*, lix 11 1. Thus, |x (x)j <|1 x |1and * - | x j1 x (x) x fl. * Therefore, x (x) E [ - |x j, jjx Jj] = Sx and BX* C 1IS. xEXx Finally, we wish to show that BX* is closed in H S . xEX To do this, we will show that ( Bx* )' c BX*. Let f be a limit point of BX*; we will show that f E BX*. We know basic neighborhoods of f in ( H S , w*) are of the form xEX { g E H S 1If(xi) - g(xi)| 0 and xi E X, I1 i < n. Denote the above open set by 50 (f; E;xl),...,)' We want f (x + y) = f (x) + f (y) and f (ax) = af (x) f or all x,y E X, a E R. To show f(x + y) = f(x) + f(y), we will show If(x + y) - f(x) - f(y) (f x yx+y) = { g EH SH : jf(x) - g(x)I < If(y) - g(y)I < ,and If(x + y) - g(x + y)I < } Since f is a limit point of BX*, g E f;x yBx+y) nBx* is non-empty. Choose g to be an element of this intersection. Therefore, If(x) - g(x)| + If (y) - g(y)I + jf (x + y) - g(x + Y)I < C. By using the triangle inequality, rearranging the terms, and using the linearity of g, we arrive at jf(x + y) - f(x) - f(y)| I E for every e > 0. To show that f (ax) = af (x), we will show that If(ax) - af (x)I < E, for all E > 0. Consider now cE) and &(f. . (f ; -g; ax) ( ; ;21ax+2) The intersection of these two open sets is also open and f;(;ax) 2 2 X)f; 51 { g E Il S : |f(ax) - g(ax)| I< xEX < a+2 |Wf(x) - g(x)|I 2 Since f is a limit point of BX*, choose g E BX* in the above open set. Thus, If(ax) - g(ax)< 2 and |f(x) - g(x)I < 21a+2' The latter inequality implies (jai + 1)|f(x) - g(x)j < ' and IaiIf(x) - g(x)I (jaj + 1)|f (x) - g(x)I j. Thus, Iaf (x) - ag(x) < . Adding inequalities, we get |af (x) - ag(x)| + If (ax) - g(ax) I< c. Again using the triangle inequality and the linearity of g, we arrive at If (ax) - af (x)j ( E for all e > 0. This shows that f is linear. We must now show that jjf || ( 1. But |1 f || 1 if f If(x)I I x 1| (making f continuous) f or all x E X. Consider 0f. Choose g E f n Bx*. Therefore, f(x) - g(x) E and If(x)I < E + Ig(x)I. Since g E BX*, jg(x) 11| x jjand If(x)| < e + jjx jj. But since f was chosen arbitrarily, as well as x, we can conclude that If(x)I I <1 x 11 for all x E X, and hence 11 f 11 < 1. Theref ore, f E BX*, and the theorem is proven. 52 An application of Alaoglu's theorem can be seen in theorem 3.18 of this chapter. Let X and Y be vector spaces. A bilinear mapping on X x Y is a function f X x Y - R satisfying: (i) f(xi + x2 ,Y) = f(X11,y) + f (X2IY) I (ii) f(x,y + y2) = f(xy 1) + f (xIy2) (iii) f(ax,y) = f(x,ay) = af(x,y), and (iv) f(0,y) = f(x,O) = 0. The function f is denoted by f(x,y) = < x,y >. If X and Y are vector spaces and <-,-> a bilinear map on X x Y which separates points in X x Y, then we refer to X x Y as a dual pair (X,Y). A net (xa) in X is said to converge to x0 in X in the topology of the dual pair if 1< xay > - < x0,y >1 a 140 f or all y E Y. The topology induced by this convergence is denoted by o(X,Y) (or the Y-topology induced on X). If y E Y, def ine py : X - R+ by py(x) = I< x,y >I. Note that py is a seminorm on X. Lemma 3.12: Let (X,Y) be a dual pair. A linear mapping f : X -4 R is (X,Y)-continuous if f there exists an M > 0 and finitely many y,,...,y E Y so that f (X) < 1( i=1 Yi~x.x)t Proof: Suppose for every M > 0 and every finite subset {y 1,. ..,y.} of Y there exists an x E X so that |f (x) > MtPy (x). We need to exhibit a net (zn) in X so that (za) a z0 (f or some z. E X) in the a(X,Y)-topology, 53 but f (za) does not converge to f (z.). Let Y be the set of all finite subsets of Y, directed by set containment. For each F E Y, choose xF E X so that If(xF)I > |Fj jpyi(xF) where F = {y,,...,yj and IFI denotes the cardinality of F. Further, since If(xF)I > 0 for all F E 5, we may scale up the right hand side of the above inequality by choosing an aF E R large enough so that tf(aFxF)I > IFItpY(aFxF) + 1. With this, we dispense with the aF's and simply choose xF E X so that If (xF)I > IFIY(xF) + 1. Define 1 ZF = Px. F=IFI%pY(xF) + 1 F We assert that (zF) -r . But (zF) -r 0 iff < zFy > - <0O,y >1 - 0 for all y E Y iff 1< zF'y >1 Fr 0 for all y E Y iff < 1=1-.xFyI r f or all y E Y if f F|9p,.(xF) + 1 1 1 1 FI>1 < 1 1< IGI IG=tp p (xF)xFy+ix, + 1M| IGI - IFI|F and convergence is verified. Finally, we show that f(zF) does not converge to 0. Let F E Y. Therefore, 54 If(zF)I =FI~t 1 F+ If(XF)I IF|pY (xF) + 1 (FF . py.(xF) + 1) = 1. IFIj=p (xF) + 1 11 1 Thus, f (zF) is bounded away from 0, and we conclude that f is not o(X,Y)-continuous. Let f X - R be a linear mapping. Suppose there exists an M > 0 and finitely many y1,...,yn E Y so that f (x) < Mtpy (x). Let (xa) be a net in X converging to x in 1=1 Ia the Y-topology on X. Thus I< x y > -i->a 0 for all y E Y. Let E>0 and choose #so that if a >g, then < xa - xy >1 < , 1 < i < n. Note: Since < -xy > = |< x,y >1, we have that f (-x) < M.%py.(x) and consequently If(x)I %Mp(x). Consider If(xa) - f(x)I a > . Now, f(xa) - (x)I I(xa - X)j K = En whenever a > #. Thuis f (xa) -a f (x) in the c(X,Y)-topology. U Lemma 3.13: Let X be a linear space. Suppose that f and { fU }' are linear functionals so that j=1 .n 1ker(fj) C ker(f). Then f is a linear combination of { fi n i=1 Proof: Define b : X - R" by O(x) = (f 1(x),. ..,f,(x)) and :(X) -4 R by 55 p(f 1(x),. .. ,fn(x)) = f(x). Clearly 0 is well defined, and to show v is well defined, suppose (f (x),... ,fn(x)) = (fl(y),....,f 1 (y)). Now (f1(x - y) ,...,f(x-y)) = 0, which implies fi(x - y) = 0, 1 i < n, and thus x -yE ker(fi) for all 1 i < n. By hypothesis, x-y Eker(f) and f(x-y) = f(x) -f(y) = 0. Thus so(f 1(x),... ,f1 (x)) = s(f 1(y),... and s is well defined on O(X). Suppose we can extend s to a linear function o defined on all of R" so that o(x) = (x) for all x E O(X). Then s can be represented as a 1 x n matrix, say (a, ... an). Now, f = soob = so's, and for any x E X, f(x) = V(b(x)) = s(f 1(x),.. ) = (al ... a-) (f 1(x ,...,f (x)) =9aif i (x) , i= 1 proving the lemma. We now need to extend s to all of R". Since 0 is a linear mapping from X - R" (i.e., a homomorphism), O(X) is a linear subspace (i.e., subgroup) of R" having dimension k, k 1 < k < n. Let { b }. be a basis for O(X), and extend this 1=1 basis to { bi }, a basis for R". Define so :R" .-4R' by W(.tfjb) = .l jW(b). It is easily seen that V is well defined, so let x = . 3ibi , y = .97ibi E n , /,7 E R. j=1 = 56 Thus, W(#x + yy) = o(f#.fibi + 7.7ibi) = V(t(fifi3 + 77 )bj) 1=1 =fi ( b i ) + 'j #ij (b ) = fo(x) + 7(y), and V is linear. Finally, let (f 1(x),...,f.(x)) E O(X). This element can be represented as t#j3ibi where #li E R, b f3=i'bi where Pi' =Pi if1 1 <( i andfi' =Oif k (x), ... ,f.(x)) = Pi(.I 'bi) i= 1 iljpo(bj) i= 1 == = p(f 1 (x),...f(x)), and so agrees with V everywhere on O(X), completing the proof . ** * Now we note that <-,-> X x X - R defined by ** * ** * < x ,x > E x (x) defines a bilinear functional since < x + y x >=(x + y)(x) ** * ** * =x (x)+y (x) ** * ** * = = X (x) + x(y) ** * ** * = < x ,x > + < x ,y >, and ** * ** * < ax ,x >=(MC )(x) ** * = ax (x ) WIN 11 57 = x (ax) ** * = 0 < A < 1 } is called the line segment joining x and y. A point x0 is said to be an internal point of a convex set K if f or each x E X, there exists an E = E(x) > 0 so that X0 + Sx E K whenever ISj < e. This is to say that every line passing through x0 contains an open interval around x0 that is contained in K. If the origin 0 is an internal point of a convex set K, then we define the support function p of K (with respect to 0) to be p(x) = inf{ A > 0 : x E AK}. Some basic properties of the support function are given in the following lemma: Lemma 3.15: If K is a convex set containing the origin as an internal point, the support function p satisfies the following: (i) p(Ax) = Ap(x) for all A > 0, (ii) p(x + y) S p(x) + p(y), and (iii) { x E X :p(x)<1< } IXC{x EX:p(x) 1}. Proof: (i) Suppose A > 0. Then p(Az) = inf{ 7 > 0 : Ax E 7K } = inf{ 7 > 0 : x EEK } = inf{ AS > 0 : x E 6K } = A-inf{ S > 0 : x E 3K } = Ap(x). If A = 0, then p(Ax) = p(O) = inf{ 7> 0 : 0 E 7K } = 0 = Ap(x). (ii) Suppose x E aK and y E #K. Further, we may 59 choose a and P so that a < p(x) + and /3 p(y) + 6, for a given E > 0. Now, -x and -y are in K, and since K is convex, a (\a \1 + ~ay) EK. X) -+-p Simplifying, we get a1(x + y) E K, i.e., x + y E (a + #)K and p(x + y) a + /. Thus, p(x + y) < p(x) + p(y) + E, and since e was an arbitrary positive number, we see that p(x + y) p(x) + p(y). (iii) Suppose p(x) < 1. Therefore, inf{ 7 > 0 : x E 7K } < 1. Choose p(x) < 6 < 1 so that x E 3K. Since 0 is an internal point of K, 3K C K, and hence x E K. Now if x E K, then x E 1-K and hence p(x) 1. U Note that the support function is a sublinear functional. Let X be a vector space. Two convex sets K, and K2 are said to be separated by a linear functional f if f sup{ f(x) : x E KI} inf{ f(x) : x E K2 }. Lemma 3.16: Let K, and K2 be disjoint convex subsets of a vector space X, and suppose x, is an internal point of K1. Then there exists a non-zero linear functional f that separates K1 and K 2 . Proof: Let x,, K, and K2 be as hypothesized. We claim that Ki - K 2 is convex. Let x = k1 - k 2, y = h,- h 2 E K, - K 2 for appropriate elements in each convex set, and 6 E [0, 1]. Now 60 Sx + (1 - 6)y = 6(k, - k2) + (1-b) (h, - h2 ) = 6k (1 1 + - 6)h1 - (6k 2 + (1 - 6)h2 ) which is clearly in K, - K2 by the convexity of each set, and K, - K2 is convex. Further, we claim that x0 = xI - x2 is internal to K, - K2 for any choice of x 2 E K2. Let x E X, x2 E K 2 and choose E > 0 so that if 161 < c, then xi + 6x E K1 . Thus, (x 1 -x 2) +6x = (x1 +Sx) - x2 E Ki - K 2 whenever 161 < e. Let K = KI - K2 - x0. As before, K is convex since it is the difference of two convex sets, and contains 0 as an internal point. Since Kin K2 = O ki - k2 f 0 for any choice of k, and k2. Thus, 0 % KI - K2 and - x0 f K. Let p be the support function for K. Since - x0 f K, p(- xO) 1. Let S be the linear subspace consisting of all multiples of x0. Define f : S RbyD- f (axo) = - a. Clearly f is linear. If a > 0, then f(axo)= - a 0 and f(axo) S p(axo) since p > 0. If a < 0, then f (axo) = - a > 0, and since p(- xO) 1, we have - a 5 - ap(- xO) = p(ax0 ). In either case, f is dominated by p, a sublinear functional. Appealing to the Hahn-Banach theorem, we extend f to a linear functional f on all of X that agrees with f on S and is dominated by p everywhere. Now, for any x E K, f(x) 5 p(x) < 1. To show that f satisfies the conclusion to the theorem, let z, E K, and z2 E K2. Now zi - z2 -X 0 E K and f(zI - Z XO) = f(z) - f(z2) - f(XO) < 1- Since f(xO) = f(xO) = -1, f(z1) - f(z2) + 1 5 1 and 61 f(z) f(z2). By taking the supremum over K, and the infimum over K2 , we obtain the desired result. * Suppose (X,Y) is a dual pair and N is c(X,Y)-open in X. Then for each x0 e N, there exists an E > 0 and finitely many y,.... ,yn E Y so that V = { x E X :.p(x 0 - x) < E } is contained in N. Note that V is convex and contains x0 as an internal point, which is proven in the following lemma: Lemma 3.17: Let (X,Y) be a dual pair, and suppose that K is a o(X,Y)-compact, convex subset of X. Further suppose that u E X \ K. Then there exists a 0 # y E Y so that < u,y > > sup{ < k,y > : k E K }. Proof: Suppose u and K are as above. By translating by - u we reduce the problem to u = 0. Since 0 % K (and K is closed), there exists an c > 0 and finitely many yI,...,yn E Y so that 0 E V = { x E X : .p(x) < } i=1 I and V is contained in X \ K. Let x,y E V, a E [0, 1], and consider ax + (1 - a)y. Now p(ax+ (1 - a)y) . (py.(ax) + py((' - a)y)) i.t p.(ax) + .S p ((1 - a)y) i=1 I1= = a.:+p1.(x)E( - a).* p (y) < ac + (1 - a) e =e and we conclude that V is convex. Further, let x E X and a = sup{ < x,y >| : 1 i < n }, and suppose 6< f na + 1 Now x E V iff p (Sx) ( c. But i=1 Yi 62 .t (x)1= 1 ,pY,(x) |Slna E 1 = 1 ii=1P and thus 0 is an internal point of V. Further note that if x E V, then - x E V (since p(x) = p(- x)), and by the preceeding theorem, there exists a non-zero linear functional f so that sup{ f(v) } < inf{ f(k) }. V K Call the previous inf imum 7, and thus f (v) , f (- v) 7 for all v E V, implying If (v)1 7. Let e > 0 and consider !V. If x = E -V, then if(x)I = If(7v)j = |-f(v)| I -'Y = E 7 '7Y7 7 making f continuous at 0 and hence continuous everywhere. Now, we may choose y E Y so that f = <-,y >. Since sup{ < v,y > } inf{ < k,y > }, VK we have < 0,y > < inf{ < k,y > } = 7. K We need only show If > 0. Since <-,y > j 0, there exists an x E X so that < x,y > # 0. Since 0 is internal to V, choose e > 0 so that if 0 < j< < E, then &x E V. Therefore, < 0,y > = 0 < k< x,y > = < Sx,y > < 7 and hence 7 is positive. By considering - y, we see that < 0,- y > > sup{ < k,- y > }, K and we have reached the desired conclusion. We conclude with a theorem which characterizes reflexive spaces in terms of weak topologies. Theorem 3.18: Let X be a Banach space. The following are equivalent: 63 (i) X is reflexive, (ii) (Bx,w) is compact, * (iii) X is reflexive, (iv) (BX*,w) is compact. Proof:(i) * (ii). Suppose X is reflexive, and let (xa) be a net in BX. Now, n(xa) is a net in BX**. By Alaoglu's * * theorem, we have that (Bx**,w ) is compact (thinking of X as a Banach space). We choose n(xa), a subnet of n(xa) ** and x0 E BX** so that ** Bx so that q (xa)) x . By hypothesis, choose x0 E * ** q(xo) = x0 . Now 77(x ) iw i (xo) . This is to say that * * * * (x) (x )= x (x ) (x) (x ) = x (xo) * * for all x E X . Consider the subnet (xa) of (xa). For (xa ) >X0, we need x (x ) x (xO) for all x E X , which we have shown to be true, and (BXw) is compact. (iii) 4 (iv). This follows from the preceding argument. (ii) * (i) Suppose (BX,w) is compact. The weak * topology on X is the X topology induced on X, which is the * ** X topology on y(X) in X . Therefore, since ** * * r;:(X,w) - (X ,w ) is continuous , n/(Bx) is w compact in 64 ** ** BX**. If v(X) is not all of X , then we choose x E BX \ r(BX); there is no problem in doing this, since any element in X \ 7(X) can be scaled down to BX**. Since linear subspaces are convex, and r(BX) is o(X,X )-compact, the conditions of the preceeding theorem are satisfied. Let * * x E X so that * ** * ** ** < x ,x > > sup{ < x ,y > : y E'n(Bx) * * =sup{< x ,x> x E BX} x Further, < X ,x > ||x |-|x | |x | ** ** since x E BX**. We now have || x | x |, a clear ** contradiction, and n is onto X (iv) * (iii). This follows from the preceding argument. (i) * (iii). Suppose X is reflexive, and let x EX . Now * * *** * * n (x)=x (for some x E X) iff * * ** *** ** ** (x )(x x (x for all x E X iff ** * *** ** ** ** x (x)=x (x )for all x E X iff * ** y(x)(x ) = x (n(x)) for all x E X (since n is onto X) iff *** x (x) = x (n(x)) for all x E X. * * *** * Define x X - R by x (x) = x (7(x)). Note that x is well defined, since both n7 and x are well defined, and linear. Now for a given x E X, *I*** Jx (x)I = lx (r7(x))l 65 and since x is continuous, Ix (M/x))*** I 5 K|| 1)1||= KI| x || for some K > 0, making x continuous, completing the proof. * (iii) 4 (i). Suppose X is reflexive and X is not reflexive. Further, suppose we can show there exists a ******* ** ** Oj x E X so that x (x ) =for all x Er(X), *** * and let x be one such. Also, choose x E X so that ** ** ** *** ** * * 77(x ) = x . Thus, V(x)(x) = x (x ) for all ** ** * * ** ** x E X and n (x )(x ) =0 for all x E r(X). Therefore, x (x*) = 0 for all x E n(X) and n(x)(x) = 0 * for all x E X. Now we have x (x) = 0 for all x E X making * * * x the zero function of X . Further, since r is linear, * ***** y (x ) = 0 = x , a contradiction of x being non-zero, finishing the proof. Now we will show the existence of such an x . Since X is closed and ni is an isometry, y(X) is closed and by ** * hypothesis, we choose x E X \ y(X). Let ** 'y=inf{llx -ij(x)| x E X}>0, ** and consider the subspace V spanned by y(X) and x . For ** v = ax + y(x) E V, define f : V - IR by f (v) = a7. Clearly ** f is linear, and f (x ) = 7, making f non-zero. Also, for any x E X, f(r7(x)) = 0. Now, 1|-|| is a sublinear functional ** and if v=ax + r(x) E V, a #0, then ** |lvII=ll ax +7r(x) || = Ial-1| x - (- -) (x) || > Ia.7 a7 = f(v) 66 and f is dominated by |1-11 on V. If a = 0, then clearly the inequality is satisfied. Now we may extend f to a linear *** ** functional x def ined on all of X that agrees with f on *** ** ** ** V so that x (x ) < f|x | everywhere on X . The last ** *** ** inequality further states that Ix (x.)I <11 x 1 for all ** ****** x E X , making x continuous and in X , completing the theorem. * -OM NI wwww" I 1 110 I-I'Mma-l" 6"'Illm Nmoommow CHAPTER BIBLIOGRAPHY 1. Willard, Stephen, General Topology, Reading, Massachu- setts,, Addison-Wesley Publishing Company, 1970. 67 BIBLIOGRAPHY Kelley, John L., General Topology, New York, D. Van Nostrand Company, 1955. Royden, H. L., Real Analysis, New York, MacMillan Publishing Co., Inc., 1968. Willard, Stephen, General Topology, Reading, Massachusetts, Addison-Wesley Publiching Company, 1970. 68