Real Analysis with Economic Applications - Chapter D ppsx

Chapter D Con tinuit y I This c hapter pro vides a basic in troduction t o the theory o f functions in general, and to that of contin uous maps between t wo metric spaces in particular. Many of the results that you have seen in your earlier studie s in terms of real functions on R are derived h ere in the context of m e tric s p aces . Exa mp le s in clud e th e I nter me d iate Value T heorem, Wei erstrass’ T heorem, and the basic results on uniform convergence (such as tho se about the in tercha ngeability of limits and Dini’s Theorem). We also in troduce and lay out a basic analysis of a f ew c on cep ts that may be new to you, like stronger notions of con tin uit y ( e.g. uniform, Lipschitz and H ölder con tinuit y), w eak er notions of c ontinuity (e.g. upper and low er semicontin uity), homeomorphisms, and isometries. This chap ter contain s at least four topics that ar e often not covered in stan- dar d course s on real analysis, but that never t heless s ee good pla ying time in vario us branches of econom ic theory. In particular, and as ap plications of th e main body of the c ha pter, we study Caristi’s famous generalization of the Banach Fixed Poin t Theorem , the ch a ra cte r ization of additive contin uous maps on Euclidean space s , and de Finetti’s theorem on t he representation of additive preorders. We also revisit the problem of representing a preference relation by a utility func tion , a n d discuss the t wo o f the best known results of utility theory, namely the Debreu and Rader Utility Repres entation Theorems. This complet es ou r coverage o f ord inal u tility theory ; we will be able to t ak e up issues related to cardinal utilit y only in the second part of the text. The pace o f the chapter is leisurely for the most part, a nd o ur treatment i s fairly elementary. Tow ards the end, however, w e study t wo topics that ma y be considered relatively advanced. (These may be omitted in the first reading.) Fi rst, we discuss Marshall Stone’s important gen er alization of th e Weierstrass Approximation Theo- rem. We pro v e this landmark result, and consider a few of its applications, such as the proof of t he sepa rab ility of the set of a ll con t inuo us real fu n ctions de finedonacom- pact metric space. Second, w e explore the problem of extending a given continuous function defined on a subset of a metric space to the en tire space. The fundamental result in this regard is the Tietze Exten s ion Theorem . We prove this theorem , and then supplemen t it with further extension results, this time for functions that are either uniformly or Lipsc hitz continuou s. 1 The finalsectionofthechaptercontains our next major trip into fixed poin t theory. It con tains a preliminary discu ss ion of 1 This concludes our introduction to the classical theory o f functions. We barely touch upon approximation theory h ere, and omit matters related to differentiation altogether, other than one isolated instance. For those of you who wish to get a more complete introduction to the basic theory of real functions, a standard recommendation would be Rudin (1976) or Marsden and Hoffman (1993) at the entry level, and Apostol (1974), or Royden (1986), or Haaser and Sullivan (1991) 151 the fixed poin t propert y and retracts, and t hen goes on t o discuss the B rouw er Fixed Poin t Theorem and so me of its a pplication s. 1 Co ntinuity of Functio n s 1.1 De finitions and Examples Recall that a fun ction f ∈ R [0,1] is continuous if, for any x ∈ [0, 1], th e im ages of points nearby x under f are c lose to f(x). In conceptual term s, where we t hink f as transform ing inputs into outpu ts, this pro perty can be thought of as en suring that a small perturbation in the input entail only a small perturbation in t he output. It is easy to generalize the de finition of continu ity so that it applie s to function s defined on arbitrary metric spaces. If (X, d X ) and (Y,d Y ) are tw o metric spaces, and f ∈ Y X is any f unction, then, for any x ∈ X, the statement “the images of poin ts nearby x under f are close to f(x)” can be form aliz ed as follo ws : Howev e r s m a ll an ε > 0 one p icks, i f y is a poin t in X which is sufficiently c lose to x (closer than som e δ > 0), then the d istance bet ween f(x) and f(y) is boun d to be smaller than ε. Here goes the f or ma l d efinition. Dhilqlwlrq. Let (X, d X ) and (Y,d Y ) be two metric s paces. We say t hat the map f : X → Y is contin uous at x ∈ X if, for an y ε > 0, there exists a δ > 0 (which ma y d epend on both ε and x) suc h that d X (x, y) < δ implie s d Y (f(x),f(y)) < ε for each y ∈ X, that is, d Y (f(x),f(y)) < ε for all y ∈ N δ,X (x). Put differently, f is cont in uous at x if, for any ε > 0, th ere exists a δ > 0 such that f(N δ,X (x)) ⊆ N ε,Y (f(x)). If f is not continuous at x, then it is said to be discon tin uous at x. For any none mpty S ⊆ X, we say that f is con tin uous on S, if it is continuous at ever y x ∈ S. In turn, f is said to be con tin uous, if it is c ontinuous on X. Let’s look at some s imple examp le s. Consider th e se lf-map f on R ++ defined b y f(t):= 1 t . Of course you already know th a t f is co ntinuous, bu t if only for practice, at the intermediate level. I should go on record again, howev er, by confessing that my personal favorites are Körner (2003) at the introductory (but pleasantly challenging) level, and Carothers (2000) at the intermediate level. 152 let us give a rigorous proof an yway. Fix any x>0 an d notice tha t, for an y y>0, we have |f(x) −f(y)| =    1 x − 1 y    = |x−y| xy . Now take any ε > 0. We wish to find a δ > 0 such that |x−y| xy < ε for any y ∈ R ++ with |x −y| < δ.Sinceδ is allowed to d e pend both on x and ε, this is not difficult. Indeed, we h a ve |x−y| xy < δ x(x−δ) for an y δ ∈ (0,x) an d y>0 with |x −y| < δ. But δ x(x−δ) < ε if δ < εx 2 1+εx , so by choosing any suc h δ > 0 (which is necessarily smalle r than x), we find f(N δ,R + (x)) ⊆ N ε,R + (f(x)). (Notice that δ depends on both ε and x.)Sincex>0 is arbitrary in this observation, we m a y conclude that f is contin u ou s . Consider next the function ϕ :  ∞ → R + defined by ϕ((x m )) := sup{|x m | : m ∈ N}. For any bounded real s equences (x m ) an d (y m ), we have |ϕ((x m )) −ϕ((y m ))| =     sup m∈N |x m | − sup m∈N |y m |     ≤ sup m∈N |x m − y m | = d ∞ ((x m ), (y m )). (Why the first inequ ality?) Th is implie s that, for any ε > 0 an d any (x m ) ∈  ∞ , we have ϕ(N ε, ∞ ((x m ))) ⊆ N ε,R + (ϕ((x m ))). Thus: ϕ is continuous . A s im ilar argument would s how that the f u nctio n L ∈ R C[0,1] defined by L(f):=  1 0 f(t)dt is also continuous. Indeed, a n immed iate application o f Propositio n A.12 yie lds |L(f) − L(g)| ≤ d ∞ (f,h) for all f,g ∈ C[0, 1]. Amoment’sreflection shows that this is more than enough to conclude that L is continuous. 2 To give a s imple example of a discon tinuous function, consider f := 1 R ++ , the indicator function of R ++ in R (Example A.5.[3]). This function is discon tinuous at 0, because, for any δ > 0, we have f(N δ,R (0)) = {0, 1}, while N 1 2 ,R (f(0)) = (− 1 2 , 1 2 ). Th us there is no δ > 0 for which f (N δ,R (0)) ⊆ N 1 2 ,R (f(0)),whencef is not continuous at 0. In w ord s, the image of a point arbitrarily close to 0 under f is not necessarily arbitrarily close to f(0); th is is the source of d iscontinuit y of f at 0. As less trivial exa mples of discontinuo us functions on R, c onsider the maps 1 Q and g(t):=  t, if t ∈ Q −t, if t ∈ R\Q . You can check that 1 Q is discontin u ous at ev ery point in R while g is d iscontinuous at every poin t in R but at 0 . 2 Quiz. Define the s elf-map L on C[0, 1] by L(f)(x):=  x 0 f(t)dt, and show that L is continuous. 153 It is crucial to understand that the contin uity of a function that maps a metric space to another depends intrinsically on the inv olv ed metrics. Suppose we are given two metric spaces (X, d X ) and (Y, d Y ),andf ∈ Y X is con tinuous. No w suppose we were to endow X and/or Y with distance functions o th er tha n d X and/or d Y . Would f still be continuous in this new setting? Th e answer is no, not neces sarily ! For instance, co nsider f := 1 R ++ whichwehavejustseentobediscontinuousat0. This conclusion is valid c on d ition al on th e fact that we use (implicitly ) the stan d a rd metric d 1 on th e d om ain o f f. Suppose that we instead use the discrete metric on R (Exam ple C.1.[1]). In this case, d en oting the resultin g metric spac e by X, we would have f (N 1 2 ,X (0)) = {f(0)} = {0} ⊆ N ε,R (f(0)) for any ε > 0, and hence we would conc lude that f is c ontinuous at 0. The m oral o f the story is that con tin uit y of a function is conditional on the distance functions used to metrize the domain and codomain of the function. Notation. Some authors p refer t o write f :(X, d X ) → (Y, d Y ) to make it clear that the contin u ity proper ties of f depend both on d X and d Y . Since it l eads to somewhat cumbersome notation, we shall mo stly refrain from d o ing this, bu t it is advisable that you view the no tatio n f : X → Y (or f ∈ Y X ) as f :(X, d X ) → (Y, d Y ) throughout this chapter. After a while, this w ill become automatic a nyway. Notation. The symbols X, Y and Z are used in this chapter only to denote arbitrary metric spaces. Ge ne ric ally speaking, we deno te the metric on X simply by d, whereas the metrics on Y and Z are denoted more explicitly as d Y and d Z . E{dpsoh 1. [1] The iden tity function id X on a metric s pace X is continu ou s , for we have id X (N ε,X (x)) = N ε,X (x)=N ε,X (id X (x)) for all x ∈ X and ε > 0. Similarly, it is easily c hecked that a constant function on any metric space is continuous. [2] Fo r any given m etric space Y, if X is a discrete space, then any f ∈ Y X must becontinuous,becauseinthiscasewehavef(N 1 2 ,X (x)) = {f(x)} ⊆ N ε,Y (f(x)) for an y ε > 0. Th us: Any function defined on a discrete space is con tinuo us. [3] Let S be any nonempty subset of a metric space X. The distance bet ween S and a poin t x ∈ X is defined as d(x, S):=inf{d(x, z):z ∈ S}. Thus the function f ∈ R X + defined b y f(x):=d(x, S) mea sures the distance of an y given point in X from the set S in terms of the metric d. For self consistency, it is desirable t hat this function be con tinuous. This is indeed the case. Because, for each x, y ∈ X, the t rian gle inequality yields f(x)=d(x, S) ≤ inf{d(x, y)+d(y, z):z ∈ S} = d(x, y)+f(y), 154 and similarly, f(y) ≤ d(y, x)+f(x). Thus |f(x) − f(y)| ≤ d (x, y) for all x, y ∈ X, and it follows that f is continuo us . [4] Given t wo metric spaces X an d Y, if f ∈ Y X is continuous and S is a metric subspace of X,thenf| S is a contin uou s function. The conv erse is, of course, fa lse. For instance, while the indicator function of R ++ in R is discontinu ous at 0, the restriction of this function to R ++ is triv ially con tinuous on this metri c subspace. 3 [5] (O n the Continuity of C oncave Functions) In S ection A.4.5 w e sho wed that an y conca ve real fu n ction definedonanopenintervalmustbecontinuous.Thisfact generalizes to the case of real func tions definedonaEuclideanspace: For every n ∈ N, any concave (or convex ) function definedonanopensubsetof R n is continuous. 4  Exercise 1. Let X be any metric space, and ϕ ∈ R X . (a) Show that if ϕ is con tin uous, t hen the sets {x : ϕ(x) ≥ α} and {x : ϕ(x) ≤ α} are closed in X for any real number α. Also show that the continuit y of ϕ is necessary for this c onclusion t o hold. (b)Provethatif ϕ is continuous, and ϕ(x) > 0 for some x ∈ X, then there exists an open s ubset O of X such that ϕ(y) > 0 for all y ∈ O. Exercise 2. H Let A an d B be two nonempty closed subsets of a metric space X with A ∩ B = ∅. Define ϕ ∈ R X + and ψ ∈ R X by ϕ(x):=d(x, A) and ψ(x):= d(x, A) −d(x, B), respectively. Prove: (a) A = {x ∈ X : ϕ(x)=0}, so we have d(x, A) > 0 for all x ∈ X\A. (b) ψ is continuous, so {x ∈ X : ψ(x) < 0} and {x ∈ X : ψ(x) > 0} are open. (c) There exist disjoint open sets O and U in X such that A ⊆ O and B ⊆ U. (Compare with Exercise C.11.) Exercise 3. Let X be a metric space, and for any n ∈ N, define the map ρ : X n × X n → R + by ρ((x 1 , , x n ), (y 1 , ,y n )) :=  n  i=1 d(x i ,y i ) 2  1 2 . Now metrize X n × X n b y the product metric, and show that ρ is a continuous function on the resulting metric sp ace. Exercise 4. H Let ((X m ,d m )) be a sequence of metric spaces, and let X stand for the product of all (X i ,d i )s. Is the function f : X → X i defined b y f(x 1 ,x 2 , ):=x i contin uous? 3 It ma y be worthwhile to pursue this matter a little further. While we have seen earlier that a real function on R can well be discontinuous everywhere, it turns out that any such function is continuous on some dens e subspace S of R. That is, and this i s the famous Blumberg’s Theorem, for any f ∈ R R there exists a dense subset S of R such that f| S is a continuous member of R S . 4 The proof of this claim is a bit harder than that of Proposition A.14, so I don’t wan t to get into it here. But don’t despair, a substantially more general result will be proved later (in Section I.2.4). 155 E{dpsoh 2. [1] (Composition of c ontinu ou s functions) Fo r an y metric spaces, X, Y and Z, let f : X → Y and g : f(X) → Z be continuous function s . Then, w e claim , h := g ◦ f is a c ontinuous function on X. (He r e we obviously consider f(X) as a subspace of Y. So, a special case of this claim is the case in which g is c ontinuous on the entire Y.)Toseethis,takeanyx ∈ X and ε > 0. Since g is continu ou s at f(x), we can find a δ  > 0 such that g(N δ  ,f(X) (f(x)) ⊆ N ε,Z (g(f(x)) = N ε,Z (h(x)). But since f is continu o u s at x, there exists a δ > 0 with f(N δ,X (x)) ⊆ N δ  ,f(X) (f(x)) so that h(N δ,X (x)) = g(f(N δ,X (x)) ⊆ g(N δ  ,f(X) (f(x))). Combining these two observation s , w e find h(N δ,X (x)) ⊆ N ε,Z (h(x)), as we sough t. In words: The com position of two continuous functions is continuous. [2] For any given n ∈ N, take any metric spaces (X i ,d i ), i =1, , n, and let (X, ρ) be the p roduct of these spaces. (As usual, we abbreviate a point like (x 1 , , x n ) in X by x.) We define the ith projec tion map π i : X → X i by π i (x 1 , , x n ):=x i . It is easily seen that π i is a contin uous function. Indeed, w e h ave d i (π i (x), π i (y)) = d i (x i ,y i ) ≤ n  j=1 d j (x j ,y j )=ρ(x, y), for a ny x, y ∈ X an d i =1, , n. [3] For any given m ∈ N, ta ke any metric spaces Y i ,i=1, , m, and let Y be the product of these spaces. No w take an y other m etric space X, an d consider an y maps f i : X → Y i ,i=1, , m. Let us now d efine the function f : X → Y by f(x):=(f 1 (x), , f m (x)). (Here each f i is referred to as a componen t map of f.) Now if f is contin uous, then, b y observations [1] and [2] above, f i = π i ◦ f is contin u ous. Conversely, if each f i is con tin uous, then f mus t be continuou s as well. (Proof?) A special case of this observation is the following fact, which you must have seen before in some calculus course: For any m, n ∈ N, if Φ : R n → R m is defined as Φ(x):=(ϕ 1 (x), , ϕ m (x)) for some real maps ϕ i on R n ,i=1, , m, then Φ is continuous iff each ϕ i is continuous. [4] Fix any n ∈ N and any metric space X. Let ϕ i ∈ R X be a con tinuous map , i =1, , n, and pick any continuous F : R n → R. We wish to sho w that the map ψ ∈ R X defined by ψ(x):=F (ϕ 1 (x), ,ϕ n (x)) is con tinuous. To t his end, define ϕ : X → R n by ϕ(x):=(ϕ 1 (x), , ϕ n (x)), and ob serv e that ψ = F ◦ϕ. Applying the observations [1] and [3 ] abov e, therefore, we find that ψ is continuous. 5 5 It does not matter which of the m etrics d p weuseheretometrizeR n .Why?(Hint. Think about the strong equivalence of metrics.) 156 The follow in g situation obtains as a special case of this observation. If X is the product of the metric spaces (X 1 ,d 1 ), ,(X n ,d n ), and F : R n → R and φ i : X i → R are continuous, i =1, , n, then ψ ∈ R X defined by ψ(x 1 , ,x n ):= F (φ 1 (x 1 ), , φ n (x n )) is a con tinuous function. (Pr oof. By the findings of [1] and [2],themapϕ i := φ i ◦ π i is con tin uous (for eac h i). Now apply what we ha ve found in the previous paragraph.)  Exercise 5. Given any n ∈ N,letX be a metric space, and ϕ i ∈ R X a con- tinuous map, i =1, , n. Show that |ϕ 1 | ,  n ϕ i ,  n ϕ i , max{ϕ 1 , , ϕ n } and min{ϕ 1 , , ϕ n } are continuous r eal functions on X. Exercise 6. For any n ∈ N, afunctionϕ : R n → R is called a (multivariate) polynomial if there exist r eal numbers α i 1 , ,i n such that ϕ(t 1 , , t n )=  α i 1 , ,i n n  j=1 t i j j for all (t 1 , , t n ) ∈ R n , wher e the sum runs through a finite set of n-tuples of indices (i 1 , ,i n ) ∈ N n . Prov e that any polynomial is continuous. 1.2 Uniform Continuit y The notion of contin uity is an inheren tly local one. If f ∈ Y X is continu o u s , we know that, for any x ∈ X, “the images of points nearby x und er f are close to f(x),” but we do not know if the word “nearby” in this statemen t depends on x or not. A global property would a llow us to say something like this: “Giv e m e any ε > 0, and I can give you a δ > 0 such that, for any point x ∈ X, theimagesofpointsatmost δ-away from x under f ar e at most ε-aw ay from f(x).” This property says something about the behavior of f on its entire domain , not only in certain neighbor hoods of the points in its doma in. It i s called u niform continuity. Dhilqlwlrq. Let X an d Y be t wo m etric spaces. We sa y t hat a function f ∈ Y X is uniformly continuous if, for all ε > 0, there exists a δ > 0 (whic h m ay depend on ε)suchthatf(N δ,X (x)) ⊆ N ε,Y (f(x)) for all x ∈ X. Obvious ly, a uniformly continuous function is continuous. On the other hand, a contin uous function need not be uniformly continuous. Fo r instance, consider the contin uous fu nction f : R ++ → R ++ defined by f(x):= 1 x . Intuitively, yo u might sense that this function is not uniformly contin uous. It has a relatively peculiar behavior near 0; it is continuous , b ut the nature o f i ts cont inuity at 1 an d a t 0.0001 seems quite different. In a manner o f speaking, closer we a re to 0, the harder it gets to verify that f is con tinuous (in the sense that, in our ε-δ definition, for a given 157 ε > 0, w e nee d to choose smaller δ > 0). If f w as uniformly contin uous, this would not be the c a se. To demonstrate that x → 1 x is not uniformly continuous on R ++ formally, choose ε =1, and ask you r s elf if we can find a δ > 0 such that f(N δ,R ++ (x)) ⊆ N 1,R (f(x)) for all x>0. The q uestion is if there exists a δ > 0 su ch that, for an y x>0, we have    1 x − 1 y    < 1 (i.e. |x −y| <xy) whenev er y>0 satisfies |x − y| < δ. It is plain that the answ e r is no! Fo r instance, if we choose y = x + δ 2 , then we need to have δ 2 <x(x + δ 2 ) for all x>0. Obvious ly, n o δ > 0 is equal to this task, no matter ho w small. (If δ was allowed to depend on x, there would be no problem, of cou rse. After all, x → 1 x is co ntinuous on R ++ .) Warning. A continu ou s map from a metric space X into another metric space Y remains c ontin uou s if we remetrize X b y a metric equivalent to d X , and similarly for Y. This is not true for uniform continuity. Remetrizing the domain o f a uniformly continuous map f with an equivalent metric m a y render f not uniformly contin uous. (Can you give an examp le to illustrate this?) Remet rization with strongly equivalen t metrics, how ever, leaves uniformly contin uous maps uniformly continuous. (W h y?) Exercise 7. H (a) Show that the real map x → 1 x is uniformly continuous on [a, ∞) for any a>0. (b) Show that the real map x → x 2 is not uniformly continuous on R. (c)Isthemapf :  2 →  1 defined by f((x m )) := ( x m m ) uniformly c ontinuous? Exercise 8. H Let ϕ an d ψ be uniformly continuous real functions on a metric space X such that max{|ϕ| , |ψ|} <Kfor some K>0. Show that ϕψ is uniformly continuous. What if the boundedness condition did not hold? Why should you care about uniform continuity? There are plenty of reasons for this, andweshallencountermanyofthemlater. Inthemeantime,chewonthefollowing exercise. Exercise 9. Let X and Y be metric spaces and f ∈ Y X . Sho w that if (x m ) ∈ X ∞ is Cauch y and f is uniformly contin u ous, then (f(x m )) ∈ Y ∞ is C auch y. Would this be true if f was only known to be continuous? 1.3 Other Contin uit y Concepts Theordinarycontinuityanduniformcontinuityarethemostcommonlyusedconti- n uity properties i n practice. Ho wever, som etimes one needs to work with other kinds of co ntinuit y conditions that d ema n d more regularity from a function. For an y α > 0, a function f ∈ Y X is said to be α-Hölder contin u o u s , if there exis ts a K>0 such that d Y (f(x),f(y)) ≤ Kd(x, y) α for a ll x, y ∈ X. 158 (Recall that we denote the metric of X by d. ) It is called Hölde r contin uous if it is α-Hölder con tinuous for some α > 0, and Lipsc h itz contin uous if it is 1-Hölder contin uous, th at is, if th e r e exists a K>0 su ch that d Y (f(x),f(y)) ≤ Kd(x, y) for a ll x, y ∈ X. (The smallest such K is called the Lipsc hitz constant of f). O n the other hand, as y ou’d s urely guess, it is called a con t rac tion (or a con tractiv e m ap)ifthere exists a 0 <K<1 suc h tha t d Y (f(x),f(y)) ≤ Kd(x, y) for a ll x, y ∈ X, and nonexpansiv e if d Y (f(x),f(y)) ≤ d(x, y) for all x, y ∈ X. (The latter t wo definitions generalize the corresponding ones giv e n in Section C.6.1 which ap plied only to s e lf-m aps.) We hav e already seen s om e examples of n o nexpansive and Lipsch itz continuous functions. For instance, we hav e shown in S ection 1.1 that the functions ϕ ∈ R  ∞ + and L ∈ R C[0,1] + defined b y ϕ((x m )) := sup{|x m | : m ∈ N} and L(f):=  1 0 f(t)dt are nonexpansive. Similarly, we have seen that the map x → d(x, S) on any m etric space X (with S being a nonempty set in X) is nonexpansive. (Quiz. Is any o f these maps a c ontraction?) Finally, we also kno w that t he restriction of any concav e function o n R to a compac t in terval is Lipsch itz continu ous (Proposition A.14), but it does not have to be nonexpansive. It is often easy t o check whethe r a s e lf-map f on R is Lips chitz continuous or not. Indeed, any such f is Lipschitz continuous if its derivative is bounded, it is none xpansive if sup{|f  (t)| : t ∈ R} ≤ 1, and it is a contraction if sup{|f  (t)| : t ∈ R} ≤ K<1 for some real number K. These o bservations are straightforward consequences of th e Mean Va lue Theorem (Exercise A.56). For future reference, let us explicitly state the logical connections bet ween all of the continuity properties we introduced so far. contraction pro perty =⇒ nonexpansiv eness =⇒ Lipschitz con tin uit y =⇒ Hölder con tin uit y (1) =⇒ unifo r m con tin uit y =⇒ continuity 159 The con verse of any on e of thes e implicatio ns is false. As an example let us sho w that Hölder continuity does not imply Lipschitz contin uit y. Consider the function f ∈ R [0,1] + defined by f(x):= √ x. Th is function is 1 2 -Hölder continuous, because |f(x) − f(y)| =   √ x − √ y   ≤  |x −y| for all 0 ≤ x, y ≤ 1. (The proof of the claimed inequality is elementary.) On the other hand, f is not Lipschitz contin uous, because, for an y K>0, we have |f(x) −f(0)| = √ x>Kxfor an y 0 <x< 1 K 2 . Exercise 10. Prove (1) and pro vide examples to show that the con verse of any of the implications in (1) is false in general. Exercise 11. Let X be a metric space, ϕ ∈ R X , α > 0 and λ ∈ R. (a) Show that if ϕ and ψ are α-Hölder continuous, then so is λϕ + ψ. (b) Prove or disprove: If ϕ and ψ are nonexpansive, then so is λϕ + ψ. (c) Prove or disprove: If ϕ and ψ are Hölder continuous, then so is λϕ + ψ. Exercise 12. For any 0 < α < β ≤ 1, sh ow t ha t if f ∈ R [0,1] is β-Hölder con tin uous, then it is als o α-Hölder continuous. Exercise 13. Let Y be a metric space and α > 1. Show that F ∈ Y R is α-Hölder contin uous iff it is a constant function. 1.4 R emark s on the Differen tiabilit y of R eal Fu nctions We ha v e noted earlier that a monotonic function on R can ha ve at most countably many d iscon tinuity points (Exercise B.8). In f act, one can also say qu ite a bit about the differentiabilit y of su ch a fu n ctio n. Let us agree t o say that a s e t S in R is null if, for all ε > 0, there exis t countably m a ny interva ls such that (i) S is contained in the union of th ese intervals, an d (ii) the s um of th e lengths of these intervals is at most ε. For inst an ce, Q (or any countable s et) is null. 6 Clearly, one should intuitiv ely think of null s ets as being very “small” ( although, and this is important, suc h sets need not be countable). We therefore say that a property holds almost everywhere if it holds on R\S fo r some null subset S of R. For instance, w e can say that a monotonic function on R is con tin uous almost everywhere (but, again, Exercise B.8 says somet hing stronge r than this). 7 One of the main results of the theory of real functions concerns the differentiability of m onotonic functions; it establishes that any such real function on R is differentiable almost everywhere. This is: 6 Quiz. Show that a countable union of null sets is null. 7 There is a lot of stuff here that I don’t want to get into right now. All I expect you to do is to get an intuitive “feeling” for the idea that if something is tr ue almost everywher e,thenitistrue everywhere but on a negligibly small set. 160 [...]... any real numbers a and b with a < b, show that C[0, 1] and C[a, b] are isometric Exercise 18 Let d and D be two metrics on a nonempty set X (a) Show that d and D are equivalent iff idX is an homeomorphism (b) (Carothers) (X, d) and (X, D) may be homeomorphic, even if d and D are not equivalent For instance, let X := {0, 1, 1 , 1 , }, d := d1 , and define D on X 2 as 2 3 follows: D( x, 1) := x and D( x,... is satisfied by {Nε(x),X (x) : x ∈ S} = X, and we are done (Lesson: Compactness is your friend!) The proof we sketched for Corollary 2 above used the local-to-global method in disguise The property of interest there was the boundedness of ϕ ∈ C(X) Say that an open set O of X satisfies the property Λ if ϕ|O is bounded The local-to-global method says that, given that X is compact, all we need is to show... B.4 and C.2.3, and see how one may improve the utility representation results we have obtained there by using the real analysis technology introduced thus far Indeed, while fundamental, the results obtained in Section B.4 (and those in Section C.2.3) do need improvement They are of limited use even in the simple setting of the classical consumer theory where one usually takes Rn as the grand commodity... choice theory and related fields In later chapters, when we are equipped with more powerful methods, we will revisit this result and obtain a substantial generalization Theorem 2 (de Finetti) Let n ∈ N, and take any complete, continuous and strictly is additive, that is, increasing preorder on Rn If + x if and only if y x+z y+z for any x, y ∈ Rn and z ∈ R such that x + z, y + z ∈ Rn , then it admits a positive... bounded from below, then Φ has a fixed point in X It is easy to see that this theorem generalizes the Banach Fixed Point Theorem If Φ ∈ X X is a contraction with the contraction coefficient K ∈ (0, 1), then the hypoth1 esis of Caristi’s theorem is satisfied for ϕ ∈ RX defined by ϕ(x) := 1−K d( x, Φ(x)) + Indeed, ϕ is continuous (why?), and we have 1 d( x, Φ(x)) − d( Φ(x), Φ2 (x)) 1−K 1 > (d( x, Φ(x)) − Kd(x,... Homeomorphism Theorem Please go back and reexamine Exercise 16.(a), if you have any doubts about this A second corollary of Proposition 3 concerns the real functions Corollary 2 Any continuous real function ϕ defined on a compact metric space X is bounded 168 3.2 The Local-to-Global Method Since any compact set is bounded (Proposition C.5), Corollary 2 is really an immediate consequence of Proposition 3... Prove Corollary 3 by using Debreu’s Utility Representation Theorem 5.3 Cauchy’s Functional Equations: Additivity on Rn In this application, our objective is to understand the nature of additive functions defined on an arbitrary Euclidean space, and to see how continuity interacts with the property of additivity Let’s begin with clarifying what we mean by the latter property 182 D For any given n ∈ N,... irrational, and discontinuous everywhere These examples suggest that one can think of upper semicontinuity as allowing for upward jumps, and lower semicontinuity for downward jumps ∗ ∗ ∗ ∗ FIGURE D. 3 ABOUT HERE ∗ ∗ ∗ ∗ The characterization of semicontinuity in terms of ϕ• and ϕ• also allows us to generalize the principal definition to cover the extended real- valued functions D Let X be a metric space and ϕ :... closed subset S of Y, the set f −1 (S) is closed in X; (d) For any x ∈ X and (xm ) ∈ X ∞, xm → x implies f (xm ) → f(x).11 8 The modern proof of this result is based on a result called the Rising Sun Lemma (due to Frigyes Riesz) An easily accessible account is given in Riesz and Nagy (1990), pp 5-9 , but this result can be found in essentially any graduate textbook on real analysis 9 As explained by... is an interval and fi is a continuous real function on I, i = 1, , n By Example 2.[3] and Proposition 2, we may now conclude: Every path in Rn is connected (Compare with Example C.5.) and Exercise 23 A metric space X is called path-connected if, for any x, y ∈ X, there exists a continuous function F ∈ X [0,1] with F (0) = x and F (1) = y Show that every path-connected space is connected.15 Okay, now . homeomorphism. (b)(Carothers) (X, d) and (X, D) may be homeomorphic, even if d and D are not equivalent. For instance, let X := {0, 1, 1 2 , 1 3 , } ,d: = d 1 , and de fine D on X 2 as follows: D( x, 1) := x and D( x, 0) :=. ric ally speaking, we deno te the metric on X simply by d, whereas the metrics on Y and Z are denoted more explicitly as d Y and d Z . E{dpsoh 1. [1] The iden tity function id X on a metric s pace. olv ed metrics. Suppose we are given two metric spaces (X, d X ) and (Y, d Y ),andf ∈ Y X is con tinuous. No w suppose we were to endow X and/or Y with distance functions o th er tha n d X and/or