Chapter I M etric Linear Spaces In Chapters C - G we have laid out a foundat ion for s t udying a number o f issues that arise in metric s paces ( such as cont inuit y and completeness) and others that arise in linear sp aces (such as linear e xtension s a n d c o nvexity). However, sa ve for a few exceptions in the context of Euclidean spaces, we ha ve so far studied such mat ters in isolation from each other. This should really be remedie d , for in most app lications one works with a space structure t ha t allows for th e sim ulta neo u s con sid er ation of me tric and linear properties. In this chapte r, therefore, we bring our earlier metric and linear analyses together, and explore a framework that is g eneral enough to encompa ss such situations. I n p artic ula r, this setup w ill allow us to t alk a bout thin gs like continuous linear functions, complete linear spaces, closed convex sets, and so on. We begin the chapter b y discussing in which sense one ma y think of a metric structure to be imposed on a linear s p ace as “compatible” with the inheren t algebraic structure of that space. This leads us to the no tion of m etric linear space. After goin g through several exam ples of suc h linear spa ces, we derive som e elementar y properties pertaining to them, and exam in e w h en a l in ear functional defined on suc h a space w ould be con tinuous. We discuss the basic properties of contin uous linear functionals here in some d etail, and high light the significance of them from a geometric viewpoint. We the n consider several ch aracterizations of finite dimensional metric linear spaces, and finally, prov ide a pr imer on convex analysis on infinite dimensional metric linear s p ac es. The most i mportant results w ith in this frame work will clarify the basic connection between the notions of “ openness” and “algebraic openness,” and sharpen the separation theorems w e have obtained in Chapter G. A c ommon theme that keeps coming up t h rou g hout the chapter pertains t o t h e cruc ia l differences between the finite and infinite dimensional m etric linear spaces. In particular, y o u will s e e that t h e infinite dime nsional case con ta ins a good deal of surprises, suc h as the p resence of discon tinuous linear functionals, linear subspaces that are not closed, and open neighborhoods the closures of which are n e cessarily unbounded. So, hopefully, it should be fun! 1 1 In textbooks on functional analysis, metric linear spaces are covered, at best, as special cases of either topological groups or topological linear spaces. Since I do not presume fa miliarity with general topology here, and topological considerations arise in economic applications mo stly through metric spaces, I chose to depart from the standa rd treatments. If you wish to see the results I present here in their n atural (topological-algebraic) habitat, you should try outlets like Holmes (1975), Megginson (1998) and/or Aliprantis and Border ( 1999). 413 1 Metric Linear Spaces Take a linear space X. Giv en the objectives of this chapter, w e wish to endow X with a metric structure that is compatible with the algebraic/geometric structure of X. We th us n eed to choose a distance fun ction on X which goes well — w hatever this means — with the linear structure of X. Let’s think of t he things we can do i n X. We can shift around the vecto r s in X (b y using the v ector addition operation) and, for any giv en vector in X, we can construct the ray th at originates from the origin 0 and that goes through this vector (by using the scalar m u ltiplication operation). Let’s focu s on shifting, th at is, translating , vectors first. What w e mean b y this precisely is t hat, for a ny giv e n x, z ∈ X, the vector x + z is well-de fined in X, and geometrica lly speakin g , we think of this vecto r as th e “translation of x by z” (or the “tran slation of z by x”). Now tak e another vector y ∈ X and translate it by z as well to obtain the v ector y + z. Here is a good question: How s ho uld the d istan ce between x and y relate to that between x + z and y +z? If w e wish to model a space whose g eometry is homogeneo us —atermwhichwe will formalize later on — it makes good sense to require these two distances be equ al. After all, if posited globally, this sort of a propert y would ensure that the distance bet ween any t wo giv en points be preserved when these vectors are translated (in the same way) an ywhere else in the space. 2 This is , for instance, prec ise ly the ca se for the geometry of the plane (or of any other Euclidean space). Dhilqlwlrq. Let X be a linear space. A metric d ∈ R X×X + is called translation invariant if d(x + z,y + z)=d(x, y) for all x, y , z ∈ X. (1) Observe tha t (1) is a statem ent th at ties the metric d with the group st ructu re of the lin ear space X. Translation in variance is thus a concept whic h is partly met- ric a nd partly algebraic — it connects in a particular w ay the distance function o n a linear space with the operation of v ector addition. In fact, this connection is tigh ter than what ma y seem at first. If you endow a linear space with a translation invari- an t m etric, then, per force, you make the vector addition o peration of your space continuous! Proposition 1. Let X be a linear space which is also a metric space. If the metric d of X is t ran slatio n invariant, th e n th e map (x, y) → x + y is a c ontinuous fu n ction from X × X into X. 2 I have so far tried t o be careful in referring to t he elements of a metric space as “point s,” and those of a linear space as “vectors.” The spaces we will work with from now on will be both metric and linear , so I will us e these two terms interchangeably. 414 Proof. Su ppose d is translation invariant, and take any (x m ), (y m ) ∈ X ∞ and x, y ∈ X with x m → x and y m → y. Then, for any m =1, 2, , d(x m + y m ,x+ y)=d(x m + y m − (x + y m ),x+ y − (x + y m )) (by (1)) = d(x m − x, y − y m ) ≤ d(x m − x, 0)+d(0,y− y m ) (by the triangle inequality) = d(x m ,x)+d(y m ,y) (b y (1) and symmetry of d), so d(x m + y m ,x+ y) → 0,thatis,x m + y m → x + y.  Exercise 1.LetX be a lin ear space which is also a metric space. Show that if the metric d of X is translation invariant, then, for any k ∈ Z, the self-map f on X, defined by f(x):=kx, is continuous. Unfortunately, t ranslation invariance alone does not yield a sufficiently rich frame- work, we need a somewhat tighter link between th e metric and linear structures in general. For instance, metriz ing a linear spac e by means of a translation invariant distance function does not guaran tee the con tin uit y o f the scalar multiplication op- eration. As an extreme examp le, consider m etrizing a given nontrivial m etric space X by the d iscre te metric (which is, obviously, translation invariant). Then the map λ → λx (from R into X) is not contin uous for a n y x ∈ X\{0}. (Proof. If x = 0, then ( 1 m x) is a sequenc e which is not eventually const ant, so en d owing X with the discrete metric d yields d  1 m x, 0  =1for each m.) To connect th e metric and lin ear structures in a more s a tisfactory manner, ther e- fore, we n eed to do better than asking for the translation invariance property. 3 In particular, we should certainly make sure that a map lik e λ → λx (in X R )andamap like x → λx (in X X ) are decla red c ontinuous by t h e metric we impose on X. Well, then wh y d on’t w e co ncen trate on the case where the m etric at hand render the map (λ,x) → λx (in X R×X ) continuous? We kn ow that translation in variance gives us th e continuit y of the vector addition operation. This wo uld, in turn, render the scalar multiplication operation on X contin uous. And putting these together, we arriv e at a c lass of spaces which are both metric and linear, and in which the metric and linear structures of the space are naturally compatible. Dhilqlwlrq. Let X be a linear space which is also a m etric space. If th e metric d of X is translation in variant, and, for all con ver gent (λ m ) ∈ R ∞ and (x m ) ∈ X ∞ , we have lim λ m x m =(limλ m )(limx m ) , 3 To be fair, I should sa y that a reasonably complete algebraic theory can be developed using only the translation invariance property, but this theory would lack geometric co mpetence, so I will not pursue it here. (But see Exercises 15 and 16.) 415 then X is called a metric linear space. If, in addition, (X, d) is comp lete, the n, we say tha t X is a Fr éc het space. A metric l inear spa ce X is said to be nontrivial if X = {0}, finite dimensio n al if dim(X) < ∞, and infinite dimensional if dim(X)=∞. Put succinctly, a metric linear space is a linear space endowed with a translation invariant distance function that renders the scalar m ultiplication operation contin u- ous. 4 So, it follows fro m P roposition 1 that X is a metric linear spa ce iff it is both a linear and a metric space such that (i) the distance between any tw o points are preserv ed under the iden tical t ransla- tions of these poin ts, (ii) the scalar multiplicat io n map (λ,x) → λx is a continuous function from R×X in to X, and (iii) the vector addition map (x, y) → x + y is a continuous function from X × X in to X. Let’s look at some exam p le s. E{dpsoh 1. [1] R n is a Fréchet space, for any n ∈ N. T his is easily proved by using the fact that a seque nce converges in a Euclidean space i ff each of its coordinate sequences conv erges in R. (Similarly, R n,p is a Fréchet space, for an y n ∈ N and 1 ≤ p ≤∞.) [2] If we metrize R b y the discrete metric, we do not obtain a metric linear space even tho u gh this metric is translation invariant. By cont rast, if w e m e trize R by d ∈ R R×R + with d(a, b):=|a 3 − b 3 | , then we guarantee that the scalar mu ltip lication operation on R is continuous, but do not make R a metric linear space because the metric d is not translation invarian t. [3] C onsider the linear space R ∞ of all real sequences wh ic h i s m etrized by m eans of the product metric: ρ((x 1 ,x 2 , ), (y 1 ,y 2 , )) := ∞ S i=1 1 2 i min{1, |x i − y i |}. 4 Some authors assume that the vector addition operation is continuous instead of positing the translation invariance of the metric when defining a metric linear space. My definition is thus a bit more demanding than the u sual. Yet, insofar as the topological properties are concerned, this is only a matter of convention, since, by a well-known result of Shizuo Kakutani (and as an immediate c orollary of the famous Birkho ff-Kakutani Metrization Theorem), every metric linear space is homeomorphic to a translation inva riant metric linear space. (For a proof, see Rolewicz (1985), pp. 2-4.) Warning. What I call here a Fréchet space is referred to as an F -space in some texts which reserve the term “Fréchet space” for locally convex and complete metric linear spaces. ( I will talk about the latter typ e of spaces in the next chapter.) 416 (Recall Section C.8.2.) This metric is obviously translation invariant. To see that it renders the scalar mu ltip lica tion operation on R ∞ continuous, take any (λ m ) ∈ R ∞ and an y sequ ence (x m ) in R ∞ such that λ m → λ and x m → x for some (λ,x) ∈ R×R ∞ . (Of course, (x m ) is a sequ e nc e of sequences.) By Proposition C.8, x m i → x i (as m →∞), so we have λ m x m i → λx i (as m →∞), for each i ∈ N. App lyin g Proposition C.8 again, we find lim λ m x m = λx. Comb ining t h is obs ervation with Th e ore m C.4, therefore, w e may conclude: R ∞ is a Fréch et space. [4]  p is a metric linear space for any 1 ≤ p<∞. Fix any such p. That d p is translation invariant is obvious . To see the con tinuity of scalar m u ltiplication operation, take any (λ m ) ∈ R ∞ and any sequence (x m ) in  p suc h that λ m → λ and d p (x m ,x) → 0 for some (λ,x) ∈ R ×  p . By the triangle inequality, we ha ve d p (λ m x m , λx) ≤ d p (λ m x m , λx m )+d p (λx m , λx) =  ∞ S i=1 |λ m − λ| p |x m i | p  1 p +  ∞ S i=1 |λ| p |x m i − x i | p  1 p = |λ m − λ| d p (x m , 0)+|λ| d p (x m ,x). But since (x m ) is con ve r ge nt, (d p (x m , 0)) is a bounded real sequence (f o r, d p (x m , 0) ≤ d p (x m ,x)+d p (x, 0) → d p (x, 0)). Therefore, the inequalit y above ensures that d p (λ m x m , λx) → 0, as wa s to be pro ve d. Com bining this observation with Example C.11.[4], we ma y conclude that  2 is a Fréchet space. Moreover, t he argumen t given in that example can be g eneralized to an y  p space — but w e om it doing this in this text — so:  p is a Fréchet space for any 1 ≤ p<∞. [5]  ∞ is a Fréchet space. [6] For any metric space T, both B(T ) and CB(T ) are metric linear spaces. In fact, by Example C.11.[5] and Pr oposition D.7, both of th ese s pa ces a re Fréc het spaces.  Exercise 2. Supply the missing arguments in Examples 1.[5]-[6]. Exercise 3. Show that the metric space in troduced in Exer cise C.42 i s a metric linear space (under the point wise definedoperations)whichisnotFréchet. Exercise 4. Show that C 1 [0, 1] is a Fréchet space. Exercise 5. Let X := {(x m ) ∈ R ∞ :sup{|x m | 1 m : m ∈ N} < ∞}, and define d ∈ R X×X + by d((x m ), (y m )) := sup{|x m − y m | 1 m : m ∈ N}. Is X a metric linear space relative to d and the usual addition and scalar multiplication operations? Exercise 6. (Product Spaces )Let(X 1 ,X 2 , ) be a sequence of metric linear spaces, and X := X ∞ X i . We endow X with the product metric (S ection C.8.2), and mak e it a linear s pace by defining the operations o f scalar multiplication and vector addition point wise. Show that X is a metric linear space, and it is a Fréc h et space whenever each X i is complete. 417 Exercise 7. Let X be an y linear space, and ϕ aseminormonX (Section G.2.1). Define d ϕ ∈ R X×X + by d ϕ (x, y):=ϕ(x − y). Is d ϕ necessarily a distance function? Show that X would become a metric linear space when endo wed with d ϕ , provided that ϕ −1 (0) = {0}. Th e following proposition collects some basic facts about met ric linear spaces that we will need later o n. It also provides a good illustration of the interplay bet ween algebraic and me tric consid erations, which is ch arac te ristic of metric linear spaces. Proposition 2 . For any subsets A and B of a metric l inear space X, the follo wing are true: (a) cl X (A + x)=cl X (A)+x for all x ∈ X; (b) If A is open, then so is A + B; (c) If A is comp act and B is closed, then A + B is closed ; (d) If both A and B are compact, so is A + B. Proof. We only prove part (c) here, leaving the other three claims as exercises. Let (x m ) be a sequence in A + B suc h that x m → x for so me x ∈ X. By definition, there exist a sequ ence (y m ) in A and a sequence (z m ) in B such that x m = y m +z m for each m. Since A is co m p act, Theorem C .2 im p lies that there exists a strictly increas ing sequence (m k ) in N such that (y m k ) converges to a vector, say y, in A. But t hen (z m k ) converges to x −y by the continuity of vecto r addition, so, s in ce B is closed, it follows that x − y ∈ B. Thus, x = y +(x − y) ∈ A + B, which proves t hat A + B is clo se d.  Exercise 8. H Complete the proof of Proposition 2. Parts (b) a nd ( c ) of Proposition 2 point to a s ign ificant di fference in the behavior of sums of open sets and closed sets: While the sum of a ny two open sets is open, thesumoftwoclosedsetsneednotbeclosed. Thisdifference is wo rth k eeping in mind . It exists even in the case o f our beloved real line. Fo r instan c e, let A := N an d B := {−2+ 1 2 , −3+ 1 3 , }. Then, both A and B are closed subsets of R, but A + B is not closed in R, for 1 m ∈ A + B for each m =1, 2, , but lim 1 m =0/∈ A + B. As another (more geometric example), let A := {(x, y) ∈ R 2 : x =0and y ≥ 1 x } and B := R×{0}. (Dra w a picture.) Then, while both A and B are closed in R 2 , A + B is not a closed subset of R 2 . (Pr oof. A + B = R × R ++ .) Themostcommonmethodofcreatinganewmetriclinearspacefromagiven metric linear sp ac e is to loo k for a su b set of this space w h ich inhe rits th e me tric linear structure of the or igina l space. T his leads us to t h e n otion of a metri c linear subspace of a metric linear s pace X, which is defined as a subset of X which is both a l inear and a metric subspace of X. (For instance, R 2 ×{0} is a metric linear subspace o f R 3 , and C[0, 1] is a metric linea r subs pace of B[0, 1].) Throughout this 418 chapter, by a su bsp ace of a metric l inear space X, we mean a m etric linear s ubspace of X. Exercise 9. H Find a subspace of R ∞ which is not closed. Is t his subspace dense in R ∞ ? Exercise 10. Show that the c losure of any subspace (affine manifold) of a metric linear space X is a subspace (affine m anifold) of X. Exercise 11. For any subsets A and B of a metric linear space X,prove: (a) cl X (A)+cl X (B) ⊆ cl X (A + B); (b) cl X (cl X (A)+cl X (B)) = cl X (A + B); (c) int X (A)+int X (B) ⊆ A +int X (B) ⊆ int X (A+B), pro vided th at int X (B) = ∅. Exercise 12. H Show that every metric linear space is connected. Exercise 13. H (Nikodem) L et X and Y be two metric linear spaces, S anonempty con vex subset of X, and Γ : S ⇒ Y a correspondence which has a closed graph. Show that if 1 2 Γ(x)+ 1 2 Γ(y) ⊆ Γ( 1 2 x + 1 2 y) for any x, y ∈ S, then Γ is a convex correspondence (Exercise G.8). 2 Continuou s Linea r Operato r s a nd Func t io n a ls The primary objects of analysis w ithin the context of metric linear spaces are those linear operators L that map a metric linear space X (with metric d) to ano ther metric linear space Y (with metric d Y ) such that, fo r all x ∈ X and ε > 0, there exists a δ > 0 with d(x, y) < δ implie s d Y (L(x),L(y)) < ε. For obvious re a sons, we call any such map a continuous lin e ar operato r,except when Y = R, in which case w e refer to it as a con tin uous linear functional. 2.1 Examples of (Dis-)Contin u ous Linear Operators E{dpsoh 2. [1] Any linear operator that ma ps R n to R m is a continuous linear operator. This follow s from Examples D.2.[4] and F.6.[1].(Verify!) [2] Let L ∈ R B[0,1] be defined b y L(f):=f(0).Lis obviously a linear function al on C[0, 1]. It is also continuou s — in fact, it is a nonexpansive map — because |L(f) − L(g)| = |f(0) − g(0)| ≤ d ∞ (f,g) for any f,g ∈ B[0, 1]. 419 [3] Let L ∈ R C[0,1] be defined b y L(f):= ] 1 0 f(t)dt. It follo ws from R ie mann integration t h eo ry t ha t L is a l inear fu nct ion al (Ex ercise A. 57). Moreover, L is a nonexpansive map : For any f,g ∈ C[0, 1], we have |L(f) − L(g)| =     ] 1 0 (f(t) − g(t))dt     ≤ d ∞ (f,g) b y Propositions A.12 and A .1 3. Conclu sion: L is a continuou s linear fun c tion al on C[0, 1].  Most studen ts are startled when they hear the term “con tinuous linear functional.” Aren’t linear f u n ction s always con tinuous? The answer is no, not necessarily. True, any linear real function o n a Euclidean space is co ntinuous, but this is no t t he case when th e linear function under consideration is definedonametriclinearspacewhich is — guess what? — infin ite dim en siona l. H e re are a few examples that illustrate this. E{dpsoh 3. [1] Let X be the linear space of all real sequences that are abs o l utely summable , that is, X :=  (x 1 ,x 2 , ) ∈ R ∞ : ∞ S i=1 |x i | < ∞  . Let u s make this s pace a metric l inear s pace by usi ng the sup-metric d ∞ . (N otice t hat X is a metric linear sp a ce that differs from  1 only in its metric stru c tu re.) Define L ∈ R X by L(x 1 ,x 2 , ):= ∞ S i=1 x i , which is ob v iou sly a linear functional on X. Is L continuous? No! Consider the following sequence (x m ) ∈ X ∞ : x m :=  1 m , , 1 m , 0, 0,  ,m=1, 2, , where exactly m en tries of x m are nonzero. Clearly, d ∞ (x m , 0)= 1 m → 0 . Yet L(x m )= 1 for each m, so L(lim x m )=0=1=limL(x m ). Conclusion: L is linear but it is not continuous at 0. [2] We play on this theme a bit more. Let X := {(x 1 ,x 2 , ) ∈ R ∞ :sup{|x i | : i ∈ N} < ∞}, 420 and define L ∈ L(X, R) by L(x 1 ,x 2 , ):= ∞ S i=1 δ i x i , where 0 < δ < 1. It is easy to verify that if X wa s endowed with the sup-metric, t hen L w ould be a continuous f unction. (You now know that this w ould not be true if δ =1. Yes?) But what if w e endo wed X with the product metric? Then, interestingly, L w ould not be continuous. Indeed, if x 1 := ( 1 δ , 0, 0, ),x 2 := (0, 1 δ 2 , 0, ), etc., then x m → (0, 0, ) with respect to t he product metric (P r oposition C.8 ), and yet L(x m )=1for eac h m while L(0, 0, )=0.  Exercise 14. H Let X denote the linear space of all continuously differentiable real functions on [0, 1], and make this space a metric linear space by using the sup- metric d ∞ . (Thus X is a (dense) subspace of C[0, 1]; it is not equal to C 1 [0, 1].) Define L ∈ L(X, R) and D ∈ L(X, C[0, 1]) by L(f):=f  (0) and D(f):=f  , respectively. Show that neither D nor L is continuous. Wh ile it is true th at line arity of a map does n ot nec essa rily guara ntee its continu- ity, it still brings quite a bit of d isc ip line into the picture. Ind e e d, linearit y spreads even the t inniest b it o f con tin uit y a function may ha ve o nto t he entire dom ain o f that function . Put differently, there is no reason to distinguish between l ocal and g lobal continuity conce p ts in the p resence of lin earit y. ( Th is is quite reminisc e nt of a d d itive real functions on R; recall Exercise D.40.) We for malize this point next. Proposition 3 . Let X and Y be t wo metric linear spaces. A linear o perator L from X into Y is uniformly continu ou s if, and only if, it is continuous at 0. Proof. Let L ∈ L(X, Y ) be con tin uous at 0,andtakeanyε > 0 and x ∈ X. By con tinuity at the origin, there exist s a δ > 0 such that d Y (L(z), 0) < ε whenever d(z,0) < δ. No w take an y y ∈ X with d(x, y) < δ. By translation invariance of d, we have d(x − y,0) < δ, and hence, by translation invariance of d Y and linearity, d Y (L(x),L(y)) = d Y (L(x) − L(y), 0)=d Y (L(x − y), 0) < ε. Since y is arbitrary in X here, and δ does not depend on x, we ma y conclude that L is uniformly con tinuous. 5  5 Another Proof. If L is continuous at 0, the n, for any ε > 0, there exists a δ > 0 such that L(N δ,X (0)) ⊆ N ε,Y (L(0)) = N ε,Y (0), and thus, for any x ∈ X, L(N δ,X (x)) = L(x + N δ,X (0)) = L(x)+L(N δ,X (0)) ⊆ L(x)+N ε,Y (0)=N ε,Y (L(x)), and we are done. (See, all I need here is the additivity of L. But w ait, where did I use the translation invariance of d and d Y ?) 421 Even thou gh its p roof is easy, a first comer to the topic may be a bit su rp rised by Proposition 3. Let us try to think through things more clearly here. The key observation is that a ny metric linear spa ce Z is homogeneous in the sense that, fo r an y giv en z 0 ,z 1 ∈ Z, we can map Z onto Z in such a way that (i) z 0 is mapped to z 1 , and (ii) all topological properties of Z are left inta ct. M ore precisely, the translation map τ : Z → Z defined b y τ(z):=z +(z 1 − z 0 ) is an hom eomorphism, thanks to the continuity o f vector ad dition o n Z. But then if we kno w that an L ∈ L(X, Y ) is continu ou s at a giv e n point, sa y 0, to sho w that L must then be contin uous at an arbitrary point x 0 ∈ X, all we need is to translate X so that x 0 “becomes” 0 (of X), and trans late Y so that 0 (of Y )“becomes”y 0 := L(x 0 ). So define τ : X → X by τ (x):=x − x 0 , and ρ : Y → Y by ρ(y):=y + y 0 . Since τ and ρ are c ontinuou s everywhere, and L is continuous at 0, an d s in ce L = ρ ◦ L ◦ τ, it follows that L is continuous at x 0 . Th u s linearity (in fac t, ad d itivity ) spreads contin u ity at a sin g le point to t he entire doma in pr e cisely via translation maps that are always continuou s in a metric linear sp ace . (The situation could be drastically different if the metric and linear struc tu re s of the spa ce were no t compatible enough to yield t h e con tinuity of vector addition.) Often in real analys is, one gets a “clea rer” view of things upon suitably general- izing the mathe matical structure at hand . The follo win g exercis e s aim to clarify the origin of Proposition 3 further by means of such a generalizatio n. Exercise 15. We say that (X, +,d) is a metric group if (X, d) is a metric space and (X, +) is a group such that the binary relation + is a contin uous map from X × X into X. For any x ∗ ∈ X, we define the self-map τ on X by τ(x):=x − x ∗ , whic h is called a left translation.(Right translations are definedasmapsofthe form x →−x ∗ + x, and coincide with left translations when (X, +) is Abelian.) (a) Show that any left or right translation on X is a homeomorphism. (b)Showthatif O is an open set that contains the identity element 0 of X, the n so is −O. (c)ShowthatifO is an open set that contains the identity element 0 of X, then thereexistsanopensubset U of X with 0 ∈ U and U + U ⊆ O. Exercise 16. Let (X, + X ,d X ) and (Y,+ Y ,d Y ) be two metric groups, and c onsider a map h : X → Y such that h(x + X y)=h(x)+ Y h(y) for all x,y ∈ X. (Recall that such a map is called a homomorphism from X into Y.)Provethath is continuous iff it is continuous at 0. (How does this fact relate to Proposition 3?) It is now time t o consider a non trivial example of a con tinuous linear fu nc tion al definedonaninfinite dimensional metric linear space. E{dpsoh 4. For any n ∈ N, Exam ples 2.[1] an d F.6.[1] show that an y con tinu ou s lin- ear functional L on R n is of the f o r m x → S n α i x i (for some real nu mbers α 1 , , α n ). 422 [...]... it lies in? This seemingly paradoxical situation is just another illustration of how our finite dimensional intuition can go astray in the realm of infinite dimensional spaces Indeed, there cannot be a dense hyperplane in a Euclidean space (Why?) But all such bets are off in infinite dimensional spaces Since a linear functional on such a space need not be continuous, a hyperplane in it is not necessarily... never contain an open subset of this space So, somewhat unexpectedly, the potential conflict between our intuitive viewpoints does not arise in infinite dimensional spaces either This is a pretty startling observation about which you should pause and reflect On one hand it shows again that we should always keep our finite dimensional intuitions in check when dealing with infinite dimensional metric linear spaces... function defined on an open subset of a metric linear space After all, we now know that a 428 linear functional (which is obviously concave) may well be discontinuous if its domain is an infinite dimensional metric linear space This said, we also know that linearity spreads the minimal amount of continuity a function may have onto its entire domain Put differently, a linear functional is either continuous... then, β i = i , i = 1, 2, and i = 0 for all i ∈ N with i > 2 Thus: i (m) = i for all i ∈ N Continuing this way inductively yields the proof 17 Observe that proving the assertion with this assumption establishes the following: For any real sequence (θm ), we have θm z i → z j i i = j and θ m → 1 (Why?) We will use this observation in Case 2 18 Since β 1 may equal either lim inf λ1 (m) or lim sup... which is not even ⊇-maximal can be dense! To sum up, one important analytic lesson we learn here is that a linear functional is in general not continuous, whereas the geometric lesson is that a hyperplane is in general not closed (in which case it is dense) While the latter finding may defy our geometric intuition (which is, unfortunately, finite dimensional), this is just how things are in infinite dimensional... Here is the main conclusion we wish to derive from it Proposition 9 Every linear isomorphism from one finite dimensional linear space onto another is a homeomorphism It is quite easy to prove this fact by using Lemma 1 We give the proof in the form of an exercise (which replicates a chunk of the analysis presented in Section F.2.3) 432 Exercise 27 Let X and Y be two finite dimensional metric linear... and in particular z, lies in intX (S) 443 The proof of the third assertion is analogous, and left as an exercise ∗ ∗ ∗ ∗ FIGURE I. 2 ABOUT HERE ∗ ∗ ∗ ∗ In words, the algebraic interior of a given convex subset of a metric linear space X is always larger than its interior More precisely, there are two possibilities in general Either the interior of the set is empty while its algebraic interior is nonempty,... identify those metric linear spaces that share the same linear algebraic and topological properties D Let X and Y be two metric linear spaces and L ∈ L(X, Y ) If L is a continuous bijection with a continuous inverse, then it is called a linear homeomorphism If such a linear operator L exists, we then say that X and Y are linearly homeomorphic Similarly, if L is an isometry, then it is called a linear isometry,... functional is continuous Thus the claim advanced in Corollary 4 holds in Rn (for any n) But, insofar as continuity and linearity properties are concerned, any given nontrivial finite dimensional metric linear space can be identified with Rn (for some n) — we may regard the former space as if it is obtained from Rn by relabelling its vectors Thus, the claim of Corollary 4 should hold in any finite dimensional... Show that if K is an invertible compact operator, then K −1 is a bounded linear operator i dim(X) < ∞ 5 Convex Analysis in Metric Linear Spaces In Chapter G we have investigated the basic algebraic structure of convex sets In concert with our general program, we will now study the metric linear properties of convex sets in an arbitrary metric linear space Our ultimate objective is to obtain a suitable . that i t lies in? This seem in gly paradoxic al situation is just anoth e r illustration of h ow our finite dimensional intu- ition can go astray in the realm of infinite dimensional spaces. Indeed,. line arity of a map does n ot nec essa rily guara ntee its continu- ity, it still brings quite a bit of d isc ip line into the picture. Ind e e d, linearit y spreads even the t inniest b it o f. Euclidean spaces, p ro ve that if Γ is upper hemicontinuous, then it admits a con tin uous linear selection. 2.2 C o ntinuit y of Positive Linear Functionals Is a positive linea r fun ct ion