Metric, Normed, and Topological Spaces


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2 Chapter 13 Metric, Normed, ad Topological Spaces A metric space is a set X that has a otio of the distace d(x, y) betwee every pair of poits x, y X. A fudametal example is R with the absolutevalue metric d(x, y) = x y, ad early all of the cocepts we discuss below for metric spaces are atural geeralizatios of the correspodig cocepts for R. A special type of metric space that is particularly importat i aalysis is a ormed space, which is a vector space whose metric is derived from a orm. O the other had, every metric space is a special type of topological space, which is a set with the otio of a ope set but ot ecessarily a distace. The cocepts of metric, ormed, ad topological spaces clarify our previous discussio of the aalysis of real fuctios, ad they provide the foudatio for wideragig developmets i aalysis. The aim of this chapter is to itroduce these spaces ad give some examples, but their theory is too extesive to describe here i ay detail Metric spaces A metric o a set is a fuctio that satisfies the miimal properties we might expect of a distace. Defiitio A metric d o a set X is a fuctio d : X X R such that for all x, y, z X: (1) d(x, y) 0 ad d(x, y) = 0 if ad oly if x = y (positivity); (2) d(x, y) = d(y, x) (symmetry); (3) d(x, y) d(x, z) + d(z, y) (triagle iequality). A metric space (X, d) is a set X with a metric d defied o X. 271
3 Metric, Normed, ad Topological Spaces I geeral, may differet metrics ca be defied o the same set X, but if the metric o X is clear from the cotext, we refer to X as a metric space. Subspaces of a metric space are subsets whose metric is obtaied by restrictig the metric o the whole space. Defiitio Let (X, d) be a metric space. A metric subspace (A, d A ) of (X, d) cosists of a subset A X whose metric d A : A A R is is the restrictio of d to A; that is, d A (x, y) = d(x, y) for all x, y A. We ca ofte formulate itrisic properties of a subset A X of a metric space X i terms of properties of the correspodig metric subspace (A, d A ). Whe it is clear that we are discussig metric spaces, we refer to a metric subspace as a subspace, but metric subspaces should ot be cofused with other types of subspaces (for example, vector subspaces of a vector space) Examples. I the followig examples of metric spaces, the verificatio of the properties of a metric is mostly straightforward ad is left as a exercise. Example A rather trivial example of a metric o ay set X is the discrete metric { 0 if x = y, d(x, y) = 1 if x y. This metric is evertheless useful i illustratig the defiitios ad providig couterexamples. Example Defie d : R R R by d(x, y) = x y. The d is a metric o R. The atural umbers N ad the ratioal umbers Q with the absolutevalue metric are metric subspaces of R, as is ay other subset A R. Example Defie d : R 2 R 2 R by d(x, y) = x 1 y 1 + x 2 y 2 x = (x 1, x 2 ), y = (y 1, y 2 ). The d is a metric o R 2, called the l 1 metric. (Here, l 1 is proouced elloe. ) For example, writig z = (z 1, z 2 ), we have d(x, y) = x 1 z 1 + z 1 y 1 + x 2 z 2 + z 2 y 2 x 1 z 1 + z 1 y 1 + x 2 z 2 + z 2 y 2 d(x, z) + d(z, y), so d satisfies the triagle iequality. This metric is sometimes referred to iformally as the taxicab metric, sice it s the distace oe would travel by taxi o a rectagular grid of streets. Example Defie d : R 2 R 2 R by d(x, y) = (x 1 y 1 ) 2 + (x 2 y 2 ) 2 x = (x 1, x 2 ), y = (y 1, y 2 ). The d is a metric o R 2, called the Euclidea, or l 2, metric. It correspods to the usual otio of distace betwee poits i the plae. The triagle iequality is geometrically obvious but a aalytical proof is otrivial (see Theorem below).
4 13.1. Metric spaces Figure 1. The graph of a fuctio f C([0, 1]) is i blue. A fuctio whose distace from f with respect to the suporm is less tha 0.1 has a graph that lies iside the dotted red lies y = f(x) ± 0.1 e.g., the gree graph. Example Defie d : R 2 R 2 R by d(x, y) = max ( x 1 y 1, x 2 y 2 ) x = (x 1, x 2 ), y = (y 1, y 2 ). The d is a metric o R 2, called the l, or maximum, metric. Example Defie d : R 2 R 2 R for x = (x 1, x 2 ), y = (y 1, y 2 ) as follows: if (x 1, x 2 ) k(y 1, y 2 ) for k R, the d(x, y) = x 21 + x22 + y1 2 + y2 2 ; ad if (x 1, x 2 ) = k(y 1, y 2 ) for some k R, the d(x, y) = (x 1 y 1 ) 2 + (x 2 y 2 ) 2. That is, d(x, y) is the sum of the Euclidea distaces of x ad y from the origi, uless x ad y lie o the same lie through the origi, i which case it is the Euclidea distace from x to y. The d defies a metric o R 2. I Eglad, d is sometimes called the British Rail metric, because all the trai lies radiate from Lodo (located at 0). To take a trai from tow x to tow y, oe has to take a trai from x to 0 ad the take a trai from 0 to y, uless x ad y are o the same lie, whe oe ca take a direct trai. Example Let C(K) deote the set of cotiuous fuctios f : K R, where K R is compact; for example, K = [a, b] is a closed, bouded iterval. If f, g C(K) defie d(f, g) = sup f(x) g(x) = f g, x K f = sup f(x). x K
5 Metric, Normed, ad Topological Spaces x x 1 Figure 2. The uit balls B 1 (0) o R 2 for differet metrics: they are the iterior of a diamod (l 1 orm), a circle (l 2 orm), or a square (l orm). The l ball of radius 1/2 is also idicated by the dashed lie. The fuctio d : C(K) C(K) R is welldefied, sice a cotiuous fuctio o a compact set is bouded, ad d is a metric o C(K). Two fuctios are close with respect to this metric if their values are close at every poit x K. (See Figure 1.) We refer to f as the suporm of f. Sectio 13.6 has further discussio Ope ad closed balls. A ball i a metric space is aalogous to a iterval i R. Defiitio Let (X, d) be a metric space. The ope ball B r (x) of radius r > 0 ad ceter x X is the set of poits whose distace from x is less tha r, B r (x) = {y X : d(x, y) < r}. The closed ball Br (x) of radius r > 0 ad ceter x X as the set of poits whose distace from x is less tha or equal to r, B r (x) = {y X : d(x, y) r}. The term ball is used to deote a solid ball, rather tha the sphere of poits whose distace from the ceter x is equal to r. Example Cosider R with its stadard absolutevalue metric, defied i Example The the ope ball B r (x) = {y R : x y < r} is the ope iterval of radius r cetered at x, ad the closed ball Br (x) = {y R : x y r} is the closed iterval of radius r cetered at x.
6 13.1. Metric spaces 275 Example For R 2 with the Euclidea metric defied i Example 13.6, the ball B r (x) is a ope disc of radius r cetered at x. For the l 1 metric i Example 13.5, the ball is a diamod of diameter 2r, ad for the l metric i Example 13.7, it is a square of side 2r. The uit ball B 1 (0) for each of these metrics is illustrated i Figure 2. Example Cosider the space C(K) of cotiuous fuctios f : K R o a compact set K R with the suporm metric defied i Example The ball B r (f) cosists of all cotiuous fuctios g : K R whose values are withi r of the values of f at every x K. For example, for the fuctio f show i Figure 1 with r = 0.1, the ope ball B r (f) cosists of all cotiuous fuctios g whose graphs lie betwee the red lies. Oe has to be a little careful with the otio of balls i a geeral metric space, because they do t always behave the way their ame suggests. Example Let X be a set with the discrete metric give i Example The B r (x) = {x} cosists of a sigle poit if 0 r < 1 ad B r (x) = X is the whole space if r 1. (See also Example ) A aother example, what are the ope balls for the metric i Example 13.8? A set i a metric space is bouded if it is cotaied i a ball of fiite radius. Defiitio Let (X, d) be a metric space. A set A X is bouded if there exist x X ad 0 R < such that d(x, y) R for all y A, meaig that A B R (x). Ulike R, or a vector space, a geeral metric space has o distiguished origi, but the ceter poit of the ball is ot importat i this defiitio of a bouded set. The triagle iequality implies that d(y, z) < R + d(x, y) if d(x, z) < R, so B R (x) B R (y) for R = R + d(x, y). Thus, if Defiitio holds for some x X, the it holds for every x X. We ca say equivaletly that A X is bouded if the metric subspace (A, d A ) is bouded. Example Let X be a set with the discrete metric give i Example The X is bouded sice X = B r (x) if r > 1 ad x X. Example A subset A R is bouded with respect to the absolutevalue metric if A ( R, R) for some 0 < R <. Example Let C(K) be the space of cotiuous fuctios f : K R o a compact set defied i Example The set F C(K) of all cotiuous fuctios f : K R such that f(x) 1 for every x K is bouded, sice d(f, 0) = f 1 for all f F. The set of costat fuctios {f : f(x) = c for all x K} is t bouded, sice f = c may be arbitrarily large. We defie the diameter of a set i a aalogous way to Defiitio 3.5 for subsets of R.
7 Metric, Normed, ad Topological Spaces Defiitio Let (X, d) be a metric space ad A X. The diameter 0 diam A of A is diam A = sup {d(x, y) : x, y A}. It follows from the defiitios that A is bouded if ad oly if diam A <. The otios of a upper boud, lower boud, supremum, ad ifimum i R deped o its order properties. Ulike properties of R based o the absolute value, they do ot geeralize to a arbitrary metric space, which is t equipped with a order relatio Normed spaces I geeral, there are o algebraic operatios defied o a metric space, oly a distace fuctio. Most of the spaces that arise i aalysis are vector, or liear, spaces, ad the metrics o them are usually derived from a orm, which gives the legth of a vector. We assume that the reader is familiar with the basic theory of vector spaces, ad we cosider oly real vector spaces. Defiitio A ormed vector space (X, ) is a vector space X together with a fuctio : X R, called a orm o X, such that for all x, y X ad k R: (1) 0 x < ad x = 0 if ad oly if x = 0; (2) kx = k x ; (3) x + y x + y. The properties i Defiitio are atural oes to require of a legth. The legth of x is 0 if ad oly if x is the 0vector; multiplyig a vector by a scalar k multiplies its legth by k ; ad the legth of the hypoteeuse x + y is less tha or equal to the sum of the legths of the sides x, y. Because of this last iterpretatio, property (3) is called the triagle iequality. We also refer to a ormed vector space as a ormed space for short. Propositio If (X, ) is a ormed vector space, the d : X X R defied by d(x, y) = x y is a metric o X. Proof. The metricproperties of d follow directly from the properties of a orm i Defiitio The positivity is immediate. Also, we have d(x, y) = x y = (x y) = y x = d(y, x), d(x, y) = x z + z y x z + z y = d(x, z) + d(y, z), which proves the symmetry of d ad the triagle iequality. If X is a ormed vector space, the we always use the metric associated with its orm, uless stated specifically otherwise. A metric associated with a orm has the additioal properties that for all x, y, z X ad k R d(x + z, y + z) = d(x, y), d(kx, ky) = k d(x, y),
8 13.2. Normed spaces 277 which are called traslatio ivariace ad homogeeity, respectively. These properties imply that the ope balls B r (x) i a ormed space are rescaled, traslated versios of the uit ball B 1 (0). Example The set of real umbers R with the absolutevalue orm is a oedimesioal ormed vector space. Example The discrete metric i Example 13.3 o R, ad the metric i Example 13.8 o R 2 are ot derived from a orm. (Why?) Example The space R 2 with ay of the orms defied for x = (x 1, x 2 ) by x 1 = x 1 + x 2, x 2 = x x2 2, x = max ( x 1, x 2 ) is a ormed vector space. The correspodig metrics are the taxicab metric i Example 13.5, the Euclidea metric i Example 13.6, ad the maximum metric i Example 13.7, respectively. The orms i Example are special cases of a fudametal family of l p  orms o R. All of the l p orms reduce to the absolutevalue orm if = 1, but they are differet if 2. Defiitio For 1 p <, the l p orm p : R R is defied for x = (x 1, x 2,..., x ) R by x p = ( x 1 p + x 2 p + + x p ) 1/p. The l 2 orm is called the Euclidea orm. For p =, the l orm : R R is defied by x = max ( x 1, x 2,..., x ). The otatio for the l orm is explaied by the fact that x = lim p x p. Moreover, cosistet with its ame, the l p orm is a orm. Theorem Let 1 p. The space R with the l p orm is a ormed vector space. Proof. The space R is a dimesioal vector space, so we just eed to verify the properties of the orm. The positivity ad homogeeity of the l p orm follow immediately from its defiitio. We verify the triagle iequality here oly for the cases p = 1,. Let x = (x 1, x 2,..., x ) ad y = (y 1, y 2,..., y ) be poits i R. For p = 1, we have x + y 1 = x 1 + y 1 + x 2 + y x + y x 1 + y 1 + x 2 + y x + y x 1 + y 1.
9 Metric, Normed, ad Topological Spaces For p =, we have x + y = max ( x 1 + y 1, x 2 + y 2,..., x + y ) max ( x 1 + y 1, x 2 + y 2,..., x + y ) max ( x 1, x 2,..., x ) + max ( y 1, y 2,..., y ) x + y. The proof of the triagle iequality for 1 < p < is more difficult ad is give i Sectio We ca use Defiitio to defie x p for ay 0 < p. However, if 0 < p < 1, the p does t satisfy the triagle iequality, so it is ot a orm. This explais the restrictio 1 p. Although the l p orms are umerically differet for differet values of p, they are equivalet i the followig sese (see Corollary 13.29). Defiitio Let X be a vector space. Two orms a, b o X are equivalet if there exist strictly positive costats M m > 0 such that m x a x b M x a for all x X. Geometrically, two orms are equivalet if ad oly if a ope ball with respect to either oe of the orms cotais a ope ball with respect to the other. Equivalet orms defie the same ope sets, coverget sequeces, ad cotiuous fuctios, so there are o topological differeces betwee them. The ext theorem shows that every l p orm is equivalet to the l orm. (See Figure 2.) Theorem Suppose that 1 p <. The, for every x R, x x p 1/p x. Proof. Let x = (x 1, x 2,..., x ) R. The for each 1 i, we have which implies that x i ( x 1 p + x 2 p + + x p ) 1/p = x p, x = max { x i : 1 i } x p. O the other had, sice x i x for every 1 i, we have which proves the result x p ( x p ) 1/p = 1/p x, As a immediate cosequece, we get the equivalece of the l p orms. Corollary The l p ad l q orms o R are equivalet for every 1 p, q. Proof. We have 1/q x q x x p 1/p x 1/p x q. With more work, oe ca prove that that x q x p for 1 p q, meaig that the uit ball with respect to the l q orm cotais the uit ball with respect to the l p orm.
10 13.3. Ope ad closed sets Ope ad closed sets There are atural defiitios of ope ad closed sets i a metric space, aalogous to the defiitios i R. May of the properties of such sets i R carry over immediately to geeral metric spaces. Defiitio Let X be a metric space. A set G X is ope if for every x G there exists r > 0 such that B r (x) G. A subset F X is closed if F c = X \ F is ope. We ca rephrase this defiitio more geometrically i terms of eighborhoods. Defiitio Let X be a metric space. A set U X is a eighborhood of x X if B r (x) U for some r > 0. Defiitio the states that a subset of a metric space is ope if ad oly if every poit i the set has a eighborhood that is cotaied i the set. I particular, a set is ope if ad oly if it is a eighborhood of every poit i the set. Example If d is the discrete metric o a set X i Example 13.3, the every subset A X is ope, sice for every x A we have B 1/2 (x) = {x} A. Every set is also closed, sice its complemet is ope. Example Cosider R 2 with the Euclidea orm (or ay other l p orm). If f : R R is a cotiuous fuctio, the E = { (x 1, x 2 ) R 2 : x 2 < f(x 1 ) } is a ope subset of R 2. If f is discotiuous, the E eed t be ope. We leave the proofs as a exercise. Example If (X, d) is a metric space ad A X, the B A is ope i the metric subspace (A, d A ) if ad oly if B = A G where G is a ope subset of X. This is cosistet with our previous defiitio of relatively ope sets i A R. Ope sets with respect to oe metric o a set eed ot be ope with respect to aother metric. For example, every subset of R with the discrete metric is ope, but this is ot true of R with the absolutevalue metric. Cosistet with our termiology, ope balls are ope ad closed balls are closed. Propositio Let X be a metric space. If x X ad r > 0, the the ope ball B r (x) is ope ad the closed ball Br (x) is closed. Proof. Suppose that y B r (x) where r > 0, ad let ɛ = r d(x, y) > 0. The triagle iequality implies that B ɛ (y) B r (x), which proves that B r (x) is ope. Similarly, if y B r (x) c ad ɛ = d(x, y) r > 0, the the triagle iequality implies that B ɛ (y) B r (x) c, which proves that B r (x) c is ope ad B r (x) is closed. The ext theorem summarizes three basic properties of ope sets. Theorem Let X be a metric space. (1) The empty set ad the whole set X are ope. (2) A arbitrary uio of ope sets is ope.
11 Metric, Normed, ad Topological Spaces (3) A fiite itersectio of ope sets is ope. Proof. Property (1) follows immediately from Defiitio (The empty set satisfies the defiitio vacuously: sice it has o poits, every poit has a eighborhood that is cotaied i the set.) To prove (2), let {G i X : i I} be a arbitrary collectio of ope sets. If x i I G i, the x G i for some i I. Sice G i is ope, there exists r > 0 such that B r (x) G i, ad the B r (x) i I G i. Thus, the uio G i is ope. The prove (3), let {G 1, G 2,..., G } be a fiite collectio of ope sets. If x G i, the x G i for every 1 i. Sice G i is ope, there exists r i > 0 such that B ri (x) G i. Let r = mi(r 1, r 2,..., r ) > 0. The B r (x) B ri (x) G i for every 1 i, which implies that B r (x) G i. Thus, the fiite itersectio G i is ope. The previous proof fails if we cosider the itersectio of ifiitely may ope sets {G i : i I} because we may have if{r i : i I} = 0 eve though r i > 0 for every i I. The properties of closed sets follow by takig complemets of the correspodig properties of ope sets ad usig De Morga s laws, exactly as i the proof of Propositio Theorem Let X be a metric space. (1) The empty set ad the whole set X are closed. (2) A arbitrary itersectio of closed sets is closed. (3) A fiite uio of closed sets is closed. The followig relatioships of poits to sets are etirely aalogous to the oes i Defiitio 5.22 for R. Defiitio Let X be a metric space ad A X. (1) A poit x A is a iterior poit of A if B r (x) A for some r > 0. (2) A poit x A is a isolated poit of A if B r (x) A = {x} for some r > 0, meaig that x is the oly poit of A that belogs to B r (x). (3) A poit x X is a boudary poit of A if, for every r > 0, the ball B r (x) cotais a poit i A ad a poit ot i A.
12 13.3. Ope ad closed sets 281 (4) A poit x X is a accumulatio poit of A if, for every r > 0, the ball B r (x) cotais a poit y A such that y x. A set is ope if ad oly if every poit is a iterior poit ad closed if ad oly if every accumulatio poit belogs to the set. We defie the iterior, boudary, ad closure of a set as follows. Defiitio Let A be a subset of a metric space. The iterior A of A is the set of iterior poits of A. The boudary A of A is the set of boudary poits. The closure of A is Ā = A A. It follows that x Ā if ad oly if the ball B r(x) cotais some poit i A for every r > 0. The ext propositio gives equivalet topological defiitios. Propositio Let X be a metric space ad A X. The iterior of A is the largest ope set cotaied i A, A = {G A : G is ope i X}, the closure of A is the smallest closed set that cotais A, Ā = {F A : F is closed i X}, ad the boudary of A is their settheoretic differece, A = Ā \ A. Proof. Let A 1 deote the set of iterior poits of A, as i Defiitio 13.38, ad A 2 = {G A : G is ope}. If x A 1, the there is a ope eighborhood G A of x, so G A 2 ad x A 2. It follows that A 1 A 2. To get the opposite iclusio, ote that A 2 is ope by Theorem Thus, if x A 2, the A 2 A is a eighborhood of x, so x A 1 ad A 2 A 1. Therefore A 1 = A 2, which proves the result for the iterior. Next, Defiitio ad the previous result imply that (Ā)c = (A c ) = {G A c : G is ope}. Usig De Morga s laws, ad writig G c = F, we get that Ā = {G A c : G is ope} c = {F A : F is closed}, which proves the result for the closure. Fially, if x A, the x Ā = A A, ad o eighborhood of x is cotaied i A, so x / A. It follows that x Ā \ A ad A Ā \ A. Coversely, if x Ā \ A, the every eighborhood of x cotais poits i A, sice x Ā, ad every eighborhood cotais poits i A c, sice x / A. It follows that x A ad Ā \ A A, which completes the proof. It follows from Theorem 13.36, ad Theorem that the iterior A is ope, the closure Ā is closed, ad the boudary A is closed. Furthermore, A is ope if ad oly if A = A, ad A is closed if ad oly if A = Ā. Let us illustrate these defiitios with some examples, whose verificatio we leave as a exercise.
13 Metric, Normed, ad Topological Spaces Example Cosider R with the absolutevalue metric. If I = (a, b) ad J = [a, b], the I = J = (a, b), Ī = J = [a, b], ad I = J = {a, b}. Note that I = I, meaig that I is ope, ad J = J, meaig that J is closed. If A = {1/ : N}, the A = ad Ā = A = A {0}. Thus, A is either ope (A A ) or closed (A Ā). If Q is the set of ratioal umbers, the Q = ad Q = Q = R. Thus, Q is either ope or closed. Sice Q = R, we say that Q is dese i R. Example Let A be the uit ope ball i R 2 with the Euclidea metric, A = { (x, y) R 2 : x 2 + y 2 < 1 }. The A = A, the closure of A is the closed uit ball ad the boudary of A is the uit circle Ā = { (x, y) R 2 : x 2 + y 2 1 }, A = { (x, y) R 2 : x 2 + y 2 = 1 }, Example Let A be the uit ope ball with the xaxis deleted i R 2 with the Euclidea metric, A = { (x, y) R 2 : x 2 + y 2 < 1, y 0 }. The A = A, the closure of A is the closed uit ball Ā = { (x, y) R 2 : x 2 + y 2 1 }, ad the boudary of A cosists of the uit circle ad the xaxis, A = { (x, y) R 2 : x 2 + y 2 = 1 } { (x, 0) R 2 : x 1 }. Example Suppose that X is a set cotaiig at least two elemets with the discrete metric defied i Example If x X, the the uit ope ball is B 1 (x) = {x}, ad it is equal to its closure B r (x) = {x}. O the other had, the closed uit ball is B 1 (x) = X. Thus, i a geeral metric space, the closure of a ope ball of radius r > 0 eed ot be the closed ball of radius r Completeess, compactess, ad cotiuity A sequece (x ) i a set X is a fuctio f : N X, where x = f() is the th term i the sequece. Defiitio Let (X, d) be a metric space. A sequece (x ) i X coverges to x X, writte x x as or lim x = x, if for every ɛ > 0 there exists N N such that > N implies that d(x, x) < ɛ. That is, x x i X if d(x, x) 0 i R. Equivaletly, x x as if for every eighborhood U of x there exists N N such that x U for all > N.
14 13.4. Completeess, compactess, ad cotiuity 283 Example If d is the discrete metric o a set X, the a sequece (x ) coverges i (X, d) if ad oly if it is evetually costat. That is, there exists x X ad N N such that x = x for all > N; ad, i that case, the sequece coverges to x. Example For R with its stadard absolutevalue metric, Defiitio is the defiitio of the covergece of a real sequece. As for subsets of R, we ca give a sequetial characterizatio of closed sets i a metric space. Theorem A subset F X of a metric space X is closed if ad oly if the limit of every coverget sequece i F belogs to F. Proof. First suppose that F is closed, meaig that F c is ope. If (x ) be a sequece i F ad x F c, the there is a eighborhood U F c of x which cotais o terms i the sequece, so (x ) caot coverge to x. Thus, the limit of every coverget sequece i F belogs to F. Coversely, suppose that F is ot closed. The F c is ot ope, ad there exists a poit x F c such that every eighborhood of x cotais poits i F. Choose x F such that x B 1/ (x). The (x ) is a sequece i F whose limit x does ot belog to F, which proves the result. We defie the completeess of metric spaces i terms of Cauchy sequeces. Defiitio Let (X, d) be a metric space. A sequece (x ) i X is a Cauchy sequece for every ɛ > 0 there exists N N such that m, > N implies that d(x m, x ) < ɛ. Every coverget sequece is Cauchy: if x x the give ɛ > 0 there exists N such that d(x, x) < ɛ/2 for all > N, ad the for all m, > N we have d(x m, x ) d(x m, x) + d(x, x) < ɛ. Complete spaces are oes i which the coverse is also true. Defiitio A metric space is complete if every Cauchy sequece coverges. Example If d is the discrete metric o a set X, the (X, d) is a complete metric space sice every Cauchy sequece is evetually costat. Example The space (R, ) is complete, but the metric subspace (Q, ) is ot complete. I a complete space, we have the followig simple criterio for the completeess of a subspace. Propositio A subspace (A, d A ) of a complete metric space (X, d) is complete if ad oly if A is closed i X. Proof. If A is a closed subspace of a complete space X ad (x ) is a Cauchy sequece i A, the (x ) is a Cauchy sequece i X, so it coverges to x X. Sice A is closed, x A, which shows that A is complete.
15 Metric, Normed, ad Topological Spaces Coversely, if A is ot closed, the by Propositio there is a coverget sequece i A whose limit does ot belog to A. Sice it coverges, the sequece is Cauchy, but it does t have a limit i A, so A is ot complete. The most importat complete metric spaces i aalysis are the complete ormed spaces, or Baach spaces. Defiitio A Baach space is a complete ormed vector space. For example, R with the absolutevalue orm is a oedimesioal Baach space. Furthermore, it follows from the completeess of R that every fiitedimesioal ormed vector space over R is complete. We prove this for the l p orms give i Defiitio Theorem Let 1 p. The vector space R with the l p orm is a Baach space. Proof. Suppose that (x k ) k=1 is a sequece of poits x k = (x 1,k, x 2,k,..., x,k ) i R that is Cauchy with respect to the l p orm. From Theorem 13.28, x i,j x i,k x j x k p, so each coordiate sequece (x i,k ) k=1 is Cauchy i R. The completeess of R implies that x i,k x i as k for some x i R. Let x = (x 1, x 2,..., x ). The, from Theorem agai, x k x p C max { x i,k x i : i = 1, 2,..., }, where C = 1/p if 1 p < or C = 1 if p =. Give ɛ > 0, choose N i N such that x i,k x i < ɛ/c for all k > N i, ad let N = max{n 1, N 2,..., N }. The k > N implies that x k x p < ɛ, which proves that x k x with respect to the l p orm. Thus, (R, p ) is complete. The BolzaoWeierstrass property provides a sequetial defiitio of compactess i a geeral metric space. Defiitio A subset K X of a metric space X is sequetially compact, or compact for short, if every sequece i K has a coverget subsequece whose limit belogs to K. Explicitly, this coditio meas that if (x ) is a sequece of poits x K the there is a subsequece (x k ) such that x k x as k where x K. Compactess is a itrisic property of a subset: K X is compact if ad oly if the metric subspace (K, d K ) is compact. Although this defiitio is similar to the oe for compact sets i R, there is a sigificat differece betwee compact sets i a geeral metric space ad i R. Every compact subset of a metric space is closed ad bouded, as i R, but it is ot always true that a closed, bouded set is compact. First, as the followig example illustrates, a set must be complete, ot just closed, to be compact. (A closed subset of R is complete because R is complete.)
16 13.4. Completeess, compactess, ad cotiuity 285 Example Cosider the metric space Q with the absolute value orm. The set [0, 2] Q is a closed, bouded subspace, but it is ot compact sice a sequece of ratioal umbers that coverges i R to a irratioal umber such as 2 has o coverget subsequece i Q. Secod, completeess ad boudedess is ot eough, i geeral, to imply compactess. Example Cosider N, or ay other ifiite set, with the discrete metric, { 0 if m =, d(m, ) = 1 if m. The N is complete ad bouded with respect to this metric. However, it is ot compact sice x = is a sequece with o coverget subsequece, as is clear from Example The correct geeralizatio to a arbitrary metric space of the characterizatio of compact sets i R as closed ad bouded replaces closed with complete ad bouded with totally bouded, which is defied as follows. Defiitio Let (X, d) be a metric space. A subset A X is totally bouded if for every ɛ > 0 there exists a fiite set {x 1, x 2,..., x } of poits i X such that A B ɛ (x i ). The proof of the followig result is the completely aalogous to the proof of the BolzaoWeierstrass theorem i Theorem 3.57 for R. Theorem A subset K X of a metric space X is sequetially compact if ad oly if it is is complete ad totally bouded. The defiitio of the cotiuity of fuctios betwee metric spaces parallels the defiitios for real fuctios. Defiitio Let (X, d X ) ad (Y, d Y ) be metric spaces. A fuctio f : X Y is cotiuous at c X if for every ɛ > 0 there exists δ > 0 such that d X (x, c) < δ implies that d Y (f(x), f(c)) < ɛ. The fuctio is cotiuous o X if it is cotiuous at every poit of X. Example A fuctio f : R 2 R, where R 2 is equipped with the Euclidea orm ad R with the absolute value orm, is cotiuous at c R 2 if x c < δ implies that f(x) f(c) < ɛ Explicitly, if x = (x 1, x 2 ), c = (c 1, c 2 ) ad this coditio reads: f(x) = (f 1 (x 1, x 2 ), f 2 (x 1, x 2 )), (x1 c 1 ) 2 + (x 2 c 2 ) 2 < δ implies that f(x 1, x 2 ) f 1 (c 1, c 2 ) < ɛ.
17 Metric, Normed, ad Topological Spaces Example A fuctio f : R R 2, where R 2 is equipped with the Euclidea orm ad R with the absolute value orm, is cotiuous at c R 2 if x c < δ implies that f(x) f(c) < ɛ Explicitly, if f(x) = (f 1 (x), f 2 (x)), where f 1, f 2 : R R, this coditio reads: x c < δ implies that [f 1 (x) f 1 (c)] 2 + [f 2 (x) f 2 (c)] 2 < ɛ. The previous examples geeralize i a atural way to defie the cotiuity of a mcompoet vectorvalued fuctio of variables, f : R R m. The defiitio looks complicated if it is writte out explicitly, but it is much clearer if it is expressed i terms or metrics or orms. Example Defie F : C([0, 1]) R by F (f) = f(0), where C([0, 1]) is the space of cotiuous fuctios f : [0, 1] R equipped with the suporm described i Example 13.9, ad R has the absolute value orm. That is, F evaluates a fuctio f(x) at x = 0. Thus F is a fuctio actig o fuctios, ad its values are scalars; such a fuctio, which maps fuctios to scalars, is called a fuctioal. The F is cotiuous, sice f g < ɛ implies that f(0) g(0) < ɛ. (That is, we take δ = ɛ). We also have a sequetial characterizatio of cotiuity i a metric space. Theorem Let X ad Y be metric spaces. A fuctio f : X Y is cotiuous at c X if ad oly if f(x ) f(c) as for every sequece (x ) i X such that x c as, We defie uiform cotiuity similarly to the real case. Defiitio Let (X, d X ) ad (Y, d Y ) be metric spaces. A fuctio f : X Y is uiformly cotiuous o X if for every ɛ > 0 there exists δ > 0 such that d X (x, y) < δ implies that d Y (f(x), f(y)) < ɛ. The proofs of the followig theorems are idetically to the proofs we gave for fuctios f : A R R. First, a fuctio o a metric space is cotiuous if ad oly if the iverse images of ope sets are ope. Theorem A fuctio f : X Y betwee metric spaces X ad Y is cotiuous o X if ad oly if f 1 (V ) is ope i X for every ope set V i Y. Secod, the cotiuous image of a compact set is compact. Theorem Let f : K Y be a cotiuous fuctio from a compact metric space K to a metric space Y. The f(k) is a compact subspace of Y. Third, a cotiuous fuctio o a compact set is uiformly cotiuous. Theorem If f : K Y is a cotiuous fuctio o a compact set K, the f is uiformly cotiuous.
18 13.5. Topological spaces Topological spaces A collectio of subsets of a set X with the properties of the ope sets i a metric space give i Theorem is called a topology o X, ad a set with such a collectio of ope sets is called a topological space. Defiitio Let X be a set. A collectio T P(X) of subsets of X is a topology o X if it satisfies the followig coditios. (1) The empty set ad the whole set X belog to T. (2) The uio of a arbitrary collectio of sets i T belogs to T. (3) The itersectio of a fiite collectio of sets i T belogs to T. A set G X is ope with respect to T if G T, ad a set F X is closed with respect to T if F c T. A topological space (X, T ) is a set X together with a topology T o X. We ca put differet topologies o a set with two or more elemets. If the topology o X is clear from the cotext, the we simply refer to X as a topological space ad we do t specify the topology whe we refer to ope or closed sets. Every metric space with the ope sets i Defiitio is a topological space; the resultig collectio of ope sets is called the metric topology of the metric space. There are, however, topological spaces whose topology is ot derived from ay metric o the space. Example Let X be ay set. The T = P(X) is a topology o X, called the discrete topology. I this topology, every set is both ope ad closed. This topology is the metric topology associated with the discrete metric o X i Example Example Let X be ay set. The T = {, X} is a topology o X, called the trivial topology. The empty set ad the whole set are both ope ad closed, ad o other subsets of X are either ope or closed. If X has at least two elemets, the this topology is differet from the discrete topology i the previous example, ad it is ot derived from a metric. To see this, suppose that x, y X ad x y. If d : X X R is a metric o X, the d(x, y) = r > 0 ad B r (x) is a oempty ope set i the metric topology that does t cotai y, so B r (x) / T. The previous example illustrates a separatio property of metric topologies that eed ot be satisfied by ometric topologies. Defiitio A topological space (X, T ) is Hausdorff if for every x, y X with x y there exist ope sets U, V T such that x U, y V ad U V =. That is, a topological space is Hausdorff if distict poits have disjoit eighborhoods. I that case, we also say that the topology is Hausdorff. Nearly all topological spaces that arise i aalysis are Hausdorff, icludig, i particular, metric spaces. Propositio Every metric topology is Hausdorff.
19 Metric, Normed, ad Topological Spaces Proof. Let (X, d) be a metric space. If x, y X ad x y, the d(x, y) = r > 0, ad B r/2 (x), B r/2 (y) are disjoit ope eighborhoods of x, y. Compact sets are defied topologically as sets with the HeieBorel property. Defiitio Let X be a topological space. A set K X is compact if every ope cover of K has a fiite subcover. That is, if {G i : i I} is a collectio of ope sets such that K i I G i, the there is a fiite subcollectio {G i1, G i2,..., G i } such that K G ik. k=1 The HeieBorel ad BolzaoWeierstrass properties are equivalet i every metric space. Theorem A metric space is compact if ad oly if it sequetially compact. We wo t prove this result here, but we remark that it does ot hold for geeral topological spaces. Fially, we give the topological defiitios of covergece, cotiuity, ad coectedess which are essetially the same as the correspodig statemets for R. We also show that cotiuous maps preserve compactess ad coectedess. The defiitio of the covergece of a sequece is idetical to the statemet i Propositio 5.9 for R. Defiitio Let X be a topological space. A sequece (x ) i X coverges to x X if for every eighborhood U of x there exists N N such that x U for every > N. The followig defiitio of cotiuity i a topological space correspods to Defiitio 7.2 for R (with the relative absolutevalue topology o the domai A of f) ad Theorem Defiitio Let f : X Y be a fuctio betwee topological spaces X, Y. The f is cotiuous at x X if for every eighborhood V Y of f(x), there exists a eighborhood U X of x such that f(u) V. The fuctio f is cotiuous o X if f 1 (V ) is ope i X for every ope set V Y. These defiitios are equivalet to the correspodig ɛδ defiitios i a metric space, but they make sese i a geeral topological space because they refer oly to eighborhoods ad ope sets. We illustrate them with two simple examples. Example If X is a set with the discrete topology i Example 13.71, the a sequece coverges to x X if a oly if its terms are evetually equal to x, sice {x} is a eighborhood of x. Every fuctio f : X Y is cotiuous with respect to the discrete topology o X, sice every subset of X is ope. O the other had, if Y has the discrete topology, the f : X Y is cotiuous if ad oly if f 1 ({y}) is ope i X for every y Y.
20 13.6. * Fuctio spaces 289 Example Let X be a set with the trivial topology i Example The every sequece coverges to every poit x X, sice the oly eighborhood of x is X itself. As this example illustrates, ohausdorff topologies have the upleasat feature that limits eed ot be uique, which is oe reaso why they rarely arise i aalysis. If Y has the trivial topology, the every fuctio X Y is cotiuous, sice f 1 ( ) = ad f 1 (Y ) = X are ope i X. O the other had, if X has the trivial topology ad Y is Hausdorff, the the oly cotiuous fuctios f : X Y are the costat fuctios. Our last defiitio of a coected topological space correspods to Defiitio 5.58 for coected sets of real umbers (with the relative topology). Defiitio A topological space X is discoected if there exist oempty, disjoit ope sets U, V such that X = U V. A topological space is coected if it is ot discoected. The followig proof that cotiuous fuctios map compact sets to compact sets ad coected sets is the same as the proofs give i Theorem 7.35 ad Theorem 7.32 for sets of real umbers. Note that a cotiuous fuctio maps compact or coected sets i the opposite directio to ope or closed sets, whose iverse image is ope or closed. Theorem Suppose that f : X Y is a cotiuous map betwee topological spaces X ad Y. The f(x) is compact if X is compact, ad f(x) is coected if X is coected. Proof. For the first part, suppose that X is compact. If {V i : i I} is a ope cover of f(x), the sice f is cotiuous {f 1 (V i ) : i I} is a ope cover of X, ad sice X is compact there is a fiite subcover { f 1 (V i1 ), f 1 (V i2 ),..., f 1 (V i ) }. It follows that {V i1, V i2,..., V i } is a fiite subcover of the origial ope cover of f(x), which proves that f(x) is compact. For the secod part, suppose that f(x) is discoected. The there exist oempty, disjoit ope sets U, V i Y such that U V f(x). Sice f is cotiuous, f 1 (U), f 1 (V ) are ope, oempty, disjoit sets such that X = f 1 (U) f 1 (V ), so X is discoected. It follows that f(x) is coected if X is coected * Fuctio spaces There are may fuctio spaces, ad their study is a cetral topic i aalysis. We discuss oly oe importat example here: the space of cotiuous fuctios o a compact set equipped with the sup orm. We repeat its defiitio from Example 13.9.
21 Metric, Normed, ad Topological Spaces Defiitio Let K R be a compact set. The space C(K) cosists of the cotiuous fuctios f : K R. Additio ad scalar multiplicatio of fuctios is defied poitwise i the usual way: if f, g C(K) ad k R, the (f + g)(x) = f(x) + g(x), (kf)(x) = k (f(x)). The suporm of a fuctio f C(K) is defied by f = sup f(x). x K Sice a cotiuous fuctio o a compact set attais its maximum ad miimum value, for f C(K) we ca also write f = max x K f(x). Thus, the suporm o C(K) is aalogous to the l orm o R. I fact, if K = {1, 2,..., } is a fiite set with the discrete topology, the it is idetical to the l orm. Our previous results o cotiuous fuctios o a compact set ca be formulated cocisely i terms of this space. The followig characterizatio of uiform covergece i terms of the suporm is easily see to be equivalet to Defiitio 9.8. Defiitio A sequece (f ) of fuctios f : K R coverges uiformly o K to a fuctio f : K R if lim f f = 0. Similarly, we ca rephrase Defiitio 9.12 for a uiformly Cauchy sequece i terms of the suporm. Defiitio A sequece (f ) of fuctios f : K R is uiformly Cauchy o K if for every ɛ > 0 there exists N N such that m, > N implies that f m f < ɛ. Thus, the uiform covergece of a sequece of fuctios is defied i exactly the same way as the covergece of a sequece of real umbers with the absolute value replaced by the suporm. Moreover, like R, the space C(K) is complete. Theorem The space C(K) with the suporm is a Baach space. Proof. From Theorem 7.15, the sum of cotiuous fuctios ad the scalar multiple of a cotiuous fuctio are cotiuous, so C(K) is closed uder additio ad scalar multiplicatio. The algebraic vectorspace properties for C(K) follow immediately from those of R. From Theorem 7.37, a cotiuous fuctio o a compact set is bouded, so is welldefied o C(K). The suporm is clearly oegative, ad f = 0 implies that f(x) = 0 for every x K, meaig that f = 0 is the zero fuctio.
22 13.6. * Fuctio spaces 291 We also have for all f, g C(K) ad k R that kf = sup kf(x) = k sup f(x) = k f, x K x K f + g = sup f(x) + g(x) x K sup { f(x) + g(x) } x K sup f(x) + sup g(x) x K x K f + g, which verifies the properties of a orm. Fially, Theorem 9.13 implies that a uiformly Cauchy sequece coverges uiformly so C(K) is complete. For compariso with the suporm, we cosider a differet orm o C([a, b]) called the oeorm, which is aalogous to the l 1 orm o R. Defiitio If f : [a, b] R is a Riema itegrable fuctio, the the oeorm of f is f 1 = b a f(x) dx. Theorem The space C([a, b]) of cotiuous fuctios f : [a, b] R with the oeorm 1 is a ormed space. Proof. As show i Theorem 13.86, C([a, b]) is a vector space. Every cotiuous fuctio is Riema itegrable o a compact iterval, so 1 : C([a, b]) R is welldefied, ad we just have to verify that it satisfies the properties of a orm. Sice f 0, we have f 1 = b f 0. Furthermore, sice f is cotiuous, a Propositio shows that f 1 = 0 implies that f = 0, which verifies the positivity. If k R, the kf 1 = b a kf = k b a f = k f 1, which verifies the homogeeity. Fially, the triagle iequality is satisfied sice f + g 1 = b a f + g b a f + g = b a f + b a g = f 1 + g 1. Although C([a, b]) equipped with the oeorm 1 is a ormed space, it is ot complete, ad therefore it is ot a Baach space. The followig example gives a ocoverget Cauchy sequece i this space. Example Defie the cotiuous fuctios f : [0, 1] R by 0 if 0 x 1/2, f (x) = (x 1/2) if 1/2 < x < 1/2 + 1/, 1 if 1/2 + 1/ x 1.
23 Metric, Normed, ad Topological Spaces If > m, we have f f m 1 = 1/2+1/m 1/2 f f m 1 m, sice f f 1. Thus, f f m 1 < ɛ for all m, > 1/ɛ, so (f ) is a Cauchy sequece with respect to the oeorm. We claim that if f f 1 0 as where f C([0, 1]), the f would have to be { 0 if 0 x 1/2, f(x) = 1 if 1/2 < x 1, which is discotiuous at 1/2, so (f ) does ot have a limit i (C([0, 1]), 1 ). To prove the claim, ote that if f f 1 0, the 1/2 f = 0 sice 0 1/2 0 f = 1/2 0 f f 1 0 f f 0, ad Propositio implies that f(x) = 0 for 0 x 1/2. Similarly, for every 0 < ɛ < 1/2, we get that 1 f 1 = 0, so f(x) = 1 for 1/2 < x 1. 1/2+ɛ The sequece (f ) is ot uiformly Cauchy sice f f m 1 as for every m N, so this example does ot cotradict the completeess of (C([0, 1]), ). The l orm ad the l 1 orm o the fiitedimesioal space R are equivalet, but the suporm ad the oeorm o C([a, b]) are ot. I oe directio, we have b f (b a) sup f, a [a,b] so f 1 (b a) f, ad f f 0 implies that f f 1 0. As the followig example shows, the coverse is ot true. There is o costat M such that f M f 1 for all f C([a, b]), ad f f 1 0 does ot imply that f f 0. Example For N, defie the cotiuous fuctio f : [0, 1] R by { 1 x if 0 x 1/, f (x) = 0 if 1/ < x 1. The f = 1 for every N, but 1/ f 1 = (1 x) dx = [x 12 ] 1/ x2 so f 1 0 as. 0 0 = 1 2, Thus, ulike the fiitedimesioal vector space R, a ifiitedimesioal vector space such as C([a, b]) has may iequivalet orms ad may iequivalet otios of covergece. The icompleteess of C([a, b]) with respect to the oeorm suggests that we use the larger space R([a, b]) of Riema itegrable fuctios o [a, b], which
24 13.7. * The Mikowski iequality 293 icludes some discotiuous fuctios. A slight complicatio arises from the fact that if f is Riema itegrable ad b f = 0, the it does ot follows that f = 0, a so f 1 = 0 does ot imply that f = 0. Thus, 1 is ot, strictly speakig, a orm o R([a, b]). We ca, however, get a ormed space of equivalece classes of Riema itegrable fuctios, by defiig f, g R([a, b]) to be equivalet if b f g = 0. a For istace, the fuctio i Example is equivalet to the zerofuctio. A much more fudametal defect of the space of (equivalece classes of) Riema itegrable fuctios with the oeorm is that it is still ot complete. To get a space that is complete with respect to the oeorm, we have to use the space L 1 ([a, b]) of (equivalece classes of) Lebesgue itegrable fuctios o [a, b]. This is aother reaso for the superiority of the Lebesgue itegral over the Riema itegral: it leads fuctio spaces that are complete with respect to itegral orms. The iclusio of the smaller icomplete space C([a, b]) of cotiuous fuctios i the larger complete space L 1 ([a, b]) of Lebesgue itegrable fuctios is aalogous to the iclusio of the icomplete space Q of ratioal umbers i the complete space R of real umbers * The Mikowski iequality Iequalities are essetial to aalysis. Their proofs, however, may require cosiderable igeuity, ad there are ofte may differet ways to prove the same iequality. I this sectio, we complete the proof that the l p spaces are ormed spaces by provig the triagle iequality give i Defiitio This iequality is called the Mikowski iequality, ad it s oe of the most importat iequalities i mathematics. The simplest case is for the Euclidea orm with p = 2. We begi by provig the followig fudametal CauchySchwartz iequality. Theorem (CauchySchwartz iequality). If (x 1, x 2,..., x ), (y 1, y 2,..., y ) are poits i R, the ( ) 1/2 ( ) 1/2 x i y i x 2 i yi 2. Proof. Sice x i y i x i y i, it is sufficiet to prove the iequality for x i, y i 0. Furthermore, the iequality is obvious if x = 0 or y = 0, so we assume that at least oe x i ad oe y i is ozero. For every α, β R, we have 0 (αx i βy i ) 2. Expadig the square o the righthad side ad rearragig the terms, we get that 2αβ x i y i α 2 x 2 i + β 2 yi 2.
25 Metric, Normed, ad Topological Spaces We choose α, β > 0 to balace the terms o the righthad side, ( ) 1/2 ( ) 1/2 α = yi 2, β = x 2 i. The divisio of the resultig iequality by 2αβ proves the theorem. The Mikowski iequality for p = 2 is a immediate cosequece of the Cauchy Schwartz iequality. Corollary (Mikowski iequality). If (x 1, x 2,..., x ) ad (y 1, y 2,..., y ) are poits i R, the [ ] 1/2 ( ) 1/2 ( ) 1/2 (x i + y i ) 2 x 2 i + yi 2. Proof. Expadig the square i the followig equatio ad usig the Cauchy Schwartz iequality, we get (x i + y i ) 2 = x 2 i + 2 x i y i + ( x 2 i + 2 x 2 i ( ) 1/2 ( x 2 i + y 2 i ) 1/2 ( ) 1/2 yi 2 + y 2 i ) 1/2 Takig the square root of this iequality, we obtai the result. To prove the Mikowski iequality for geeral 1 < p <, we first defie the Hölder cojugate p of p ad prove Youg s iequality. Defiitio If 1 < p <, the the Hölder cojugate 1 < p < of p is the umber such that 1 p + 1 p = 1. If p = 1, the p = ; ad if p = the p = 1. The Hölder cojugate of 1 < p < is give explicitly by p = p p 1. Note that if 1 < p < 2, the 2 < p < ; ad if 2 < p <, the 1 < p < 2. The umber 2 is its ow Hölder cojugate. Furthermore, if p is the Hölder cojugate of p, the p is the Hölder cojugate of p. Theorem (Youg s iequality). Suppose that 1 < p < ad 1 < p < is its Hölder cojugate. If a, b 0 are oegative real umbers, the ab ap p + bp p. 2. y 2 i
26 13.7. * The Mikowski iequality 295 Moreover, there is equality if ad oly if a p = b p. Proof. There are several ways to prove this iequality. We give a proof based o calculus. The result is trivial if a = 0 or b = 0, so suppose that a, b > 0. We write a p a p ). p + bp p ( ab = b p 1 p b p + 1 p a The defiitio of p implies that p /p = p 1, so that a p ( a ) p ( a = b p b p /p = b p 1 Therefore, we have p + bp p a p ) p b p 1 ab = b p f(t), f(t) = tp p + 1 p t, t = a b p 1. The derivative of f is f (t) = t p 1 1. Thus, for p > 1, we have f (t) < 0 if 0 < t < 1, ad Theorem 8.36 implies that f(t) is strictly decreasig; moreover, f (t) > 0 if 1 < t <, so f(t) is strictly icreasig. It follows that f has a strict global miimum o (0, ) at t = 1. Sice f(1) = 1 p + 1 p 1 = 0, we coclude that f(t) 0 for all 0 < t <, with equality if ad oly if t = 1. Furthermore, t = 1 if ad oly if a = b p 1 or a p = b p. It follows that p + bp p ab 0 a p for all a, b 0, with equality if ad oly a p = b p, which proves the result. For p = 2, Youg s iequality reduces to the more easily proved iequality i Propositio 2.8. Before cotiuig, we give a scalig argumet which explais the appearace of the Hölder cojugate i Youg s iequality. Suppose we look for a iequality of the form ab Ma p + Na q for all a, b 0 for some expoets p, q ad some costats M, N. Ay iequality that holds for all positive real umbers must remai true uder rescaligs. Rescalig a λa, b µb i the iequality (where λ, µ > 0) ad dividig by λµ, we fid that it becomes ab λp 1 µ Map + µq 1 λ Nbq. We take µ = λ p 1 to make the first scalig factor equal to oe, ad the the iequality becomes ab Ma p + λ r Nb q, r = (p 1)(q 1) 1. If the expoet r of λ is ozero, the we ca violate the iequality by takig λ sufficietly small (if r > 0) or sufficietly large (if r < 0), sice it is clearly
27 Metric, Normed, ad Topological Spaces impossible to boud ab by a p for all b R. Thus, the iequality ca oly hold if r = 0, which implies that q = p. This argumet does ot, of course, prove the iequality, but it shows that the oly possible expoets for which a iequality of this form ca hold must satisfy q = p. Theorem proves that such a iequality does i fact hold i that case provided 1 < p <. Next, we use Youg s iequality to deduce Hólder s iequality, which is a geeralizatio of the CauchySchwartz iequality for p 2. Theorem (Hölder s iequality). Suppose that 1 < p < ad 1 < p < is its Hölder cojugate. If (x 1, x 2,..., x ) ad (y 1, y 2,..., y ) are poits i R, the ( ) 1/p ( ) 1/p x i y i x i p y i p. Proof. We assume without loss of geerality that x i, y i are oegative ad x, y 0. Let α, β > 0. The applyig Youg s iequality i Theorem with a = αx i, b = βy i ad summig over i, we get The, choosig αβ α = ( x i y i αp p y p i x p i + βp p ) 1/p, β = ( x p i y p i. ) 1/p to balace the terms o the righthad side, dividig by αβ, ad usig the fact that 1/p + 1/p = 1, we get Hölder s iequality. Mikowski s iequality follows from Hölder s iequality. Theorem (Mikowski s iequality). Suppose that 1 < p < ad 1 < p < is its Hölder cojugate. If (x 1, x 2,..., x ) ad (y 1, y 2,..., y ) are poits i R, the ( ) 1/p ( ) 1/p ( ) 1/p x i + y i p x i p + y i p. Proof. We assume without loss of geerality that x i, y i are oegative ad x, y 0. We split the sum o the lefthad side as follows: x i + y i p = x i + y i x i + y i p 1 x i x i + y i p 1 + y i x i + y i p 1. By Hölder s iequality, we have ( ) 1/p ( ) 1/p x i x i + y i p 1 x i p x i + y i (p 1)p,
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