Bootcamp. Christoph Thiele. Summer 2012

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1 Bootcamp Christoph Thiele Summer 212 Contents 1 Metric spaces Eamples of metric spaces Cones Completeness Baire argument Separability Compactness Total boundedness Cauchy sequences, convergent sequences Eamples Mappings between metric spaces Contraction mapping principle Curves The space C(M, M ) Arzela Ascoli The Riemann integral for continuous functions Metric spaces 1.1 Eamples of metric spaces Definition 1 A metric space is a pair (M, d) consisting of a set M with a function d : M M R + with the following properties that hold for all, y, z M 1. Zero distance for identical element: d(, ) =. 2. Non-zero distance for non-identical elements: If y, then d(, y) > 3. Symmetry: d(, y) = d(y, ) 4. Triangle inequality: d(, z) d(, y) + d(y, z) 1

2 Sometimes we refer to M as thye metric space, when the distance function d is assumed clear from the contet. Note that any subset M of a metric space M is again a metric space, if the distance function is restricted to M M from M M. We list some eamples of metric spaces, by the above remarks any subset of these spaces is again a metric space. 1. D with distance d(, y) = y = 2. R with distance d(, y) = y = ( y) 2 ( y) 2 3. R n, i.e. the set of tuples = ( 1,..., n ) with i R with distance d(, y) = ni=1 ( i y i ) 2 4. R n with distance d(, y) = sup 1 i n i y i 5. l 2 (N), the set of sequecnes : N R with distance d(, y) = i=1 ( i y i ) 2 6. l (N), the set of bounded sequences, with distance d(, y) = sup i N i y i 7. l 1 (N), the set of absolutely summable sequences, with d(, y) = i=1 i y i 8. The set BV (D) of functions of bounded variation on D, i.e. the set of pairs (f, g) with f, g right continuous (say) monotone increasing bounded functions modulo the equivalence relation (f, g) (f, g ) if f + g = f + g with distance d((f, g), (h, k)) = inf[ sup (f +k )()) (f +k )() + sup(g +h )() (g +h )() ] where the infimum is taken over all representatives of ther respective the equivalence classes. 9. The space C([, 1]) of continuous functions f : [, 1] R with distance sup f() g() [,1] 1. The space C(R + ) of continuous functions f : R + R with distance sup f() g() R The space C (R + ) C(R + ) of functions satisfying with the same distance. lim sup f() = 2

3 We leave the verification of the metric space properties in most of these cases as an eercise, just elaborating on some specific eamples. We show the triangle inequality in the case of l 2 (N). Let i and y i two sequences with i 2 i < and i y 2 i <. We frist claim the inequality i y i i ( i 2 i )( i Note that this inequality is homogeneous, if we multiply i or y i by a positive real number, then the inequality remains true. Hence it suffices to prove this inequality under the assumption 2 i = yi 2 = 1 i i But then the inequality becomes equivalent to y 2 i ) 2 i i y i 2 = ( i 2 i ) + ( i y 2 i ) = i ( 2 i + y 2 i ) But then the inequality follwos by the comparison principle since for each i which follows from 2 i y i 2 i + y 2 i ( i y i ) 2 by adding 2 i y i on both sides. We thus conclude the Cauchy Schwarz inequality But then we have i 2 i + i i y i i ( i + y i ) 2 = i i yi ( i ( i 2 i + i 2 i )( i 2 i )( i y 2 i ) y 2 i + 2 i i y i yi 2 ) = ( 2 i + yi 2 ) 2 i i Upon setting i = a i b i and y i = b i c i this becomes the triangle inequality for the sequences a i, b i, c i We also show the triangle inequality for the space BV (D). For simplicity of notation we restrict attention to the space of functions therein which vanish at, which can be epressed using pairs of functions vanishing at. Then the distance is epressed as d((f, g), (h, k)) = inf[sup(f + k )() + sup(g + h )()] 3

4 where again the infimum is taken over all choices of representatives of the given class. (eercise: do the reduction of the general case to this case.) Let (f, g), (h, i), (j, k) be three pairs of functions. Assume that the first two are chosen appropriate representatives so that d((f, g), (h, i)) + ɛ sup (f + i)() + sup(g + h)() Assume that (j, k) is chosen so that for some equivalent pair h, i to h, i d((j, k), (h, i)) + ɛ sup (j + i )() + sup(k + h )() Then we have h + i = h + i and (f, g) is equivalent to (f + i + h, g + h + i ). Hence d((f, g), (j, k)) sup sup (f + i + h + k)() + sup(g + h + i + j)() (f + i)() + sup(g + h)() + sup(h + k)() + sup(i + j)() d((f, g), (h, i)) + ɛ + d((j, k), (h, i)) + ɛ Since ɛ was arbitrary, this proves the triangle inequality. Approimation in metric spaces We begin by noting some similarity of the metric space aioms to the aioms of an equivalence relation. Asd a matter of fact, we have the following lemma: Lemma 1 Assume a M is a set and d a function M M R + satisfying properties 1, 3, 4 of a metric space. Then the relation defined by y if d(, y) = is an equivalence relation on M. Proof: For all M we have d(, ) = by the first property and thus which imlies refleivity. Symmetry follows from symmetry of d. Transitivity: If y and y z then d(, z) d(, y) + d(y, z) = + =. hence z. One can use thsi equivalence relation to pass to the space of equivalence classes, which then becomes a metric space. Lemma 2 Assume a M is a set and d a function M M R + satisfying properties 1, 3, 4 of a metric space. Let M be the set of equivalence classes with respect to the equivalence relation y if d(, y) =. Then the function d from M M R + defined by d([], [y]) = d(, y) is well defined, and makes ( M, d) a metric space. 4

5 Proof: We need to prove that the distance function depends only on the equivalence class [] and [y] of, y. Let and y y. Then by the tirangle inequality d(, y) d(, ) + d(, y ) + d(y, y) = d(, y ) where we have used the similarity relation in the last identity. Similarly we have the reverse inequality. Thus d(, y) = d(, y ). Finally we show that M is a metric space with metric d. Properties 1, 3, 4 follow immediately form the corresponding properties of M. To verify Property 2, observe that if and y are not equivalent, then d(, y) and hence d(, y). One may make an attempt to take the the above insight further saying that the relation y if d(, y) ɛ is almost an equivalence relation. It is refleiv and symmetric, but fails to be transitive by a small error: if d(, y) and d(y, z) ɛ, then by the triangle inequality d(, z) 2ɛ, which is still small though not sufficiently small to guarantee equivalence. This almost equivalence relation is behind the idea of approimation in metric spaces. We define almost equivalence classes, called balls: Definition 2 (Ball) Let (M, d) be a metric space. For M and r D we define the ball with center and radius r to be The empty set is a ball with radius r =. B r () = {y M : d(, y) < r} Note that the balls encode the full information of the metric space: knowing which element is in which ball is enough to determine the function d. A dense set may be considered a sufficient set to pick a representtaive out of each almost equivalence class: Definition 3 (Dense) Let (M, d) be a metric space. A subset M of M is called dense in M, if every ball B r () in M contains at least one point of M. 1.2 Cones Given a metric space (M, d), we call M D the upper half space of the metric space. For each point M we can define a cone C() = {(y, r) M D : d(y, ) r} Thus the horizontal section of the cone at height r is the intersection of all balls of radius s about with s > r. Definition 4 A pre-cone S is a subset of the upper half space satisfying 1. For each r > there is < s r and a y M such that (y, s) S 5

6 2. For each pair (, r), (y, s) S we have the inequality d(, y) r + s By the triangle inequality, the cone C() is a pre-cone. Namely, for all (y, s), (z, t) S we have d(y, z) d(y, ) + d(, z) s + t Cones are pre-cones that are maimal in the sense: Definition 5 A pre-cone C is called a cone if for all (y, s) / C there is a (z, t) C with d(y, z) > s + t Lemma 3 For each pre-cone, there is a unique cone containing the pre-cone. Proof: Eistence: define C to be the set of all (, r) M D such that for all (y, s) S we have d(, y) r + s By definition of the pre cone we have S C. Then we have for all (, r), (, r ) in C and all (y, s) S d(, ) d(, y) + d(y, ) r + s + s + r Now there eists (y, s) S with arbitrarily small s, hence d(, ) r + r Hence C is a pre cone. If (y, r) / C then by definition there eists (y, s) S such that d(, y) > r + s Since S C, this proves that C is a cone. Uniqueness: Let C be another cone containing S. Since S C and for each (y, s) C we have d(, y) r + s for each (, r) S, the above argument shows that (y, s) C. Note that if (z, t) C, then since C C we have d(y, z) s + t for each (y, s) C, hence (z, t) C. Hence C = C. Lemma 4 A nested family of balls is a family of balls B r ( r ) parameterized by r in a subset A D such that B r ( r ) B s ( s ) whenever r s. Given such a family where A contains arbitrarily small positive elements, then the set is a pre cone. C = {(z, 2t) : z B t ( t )} 6

7 Proof: Let (y, s), (z, t) C and assume without loss of generality that s < t. Then we have: d(y, z) d(y, s ) + d( s, z) 2t + s 2t + 2s Definition 6 A tip of a pre-cone C is an element of the form (, ) C Lemma 5 There is at most one tip of any pre-cone. Proof: Let (, ) and (y, ) be two tips, then by the cone property we have which proves = y. 1.3 Completeness A pre-cone may or may not contain a tip. d(, y) = + = Definition 7 A metric space is called complete, if every pre-cone has a tip. A cone is the natural pendant for metric spaces of a Dedekind cut for D. This suggest we may seek an analoguous process in general metric spaces for the etensions from the metric space D to the metric space R +. Lemma 6 Let (M, d) be a metric space. The set M of cones of this metric space with distance d(c, C ) defined as the infimum of all numbers c such that d(, y) r + c + s for all (, r) C and (y, s) C is a metric space. Proof: We first show that the infimum is not and hecne in R +. Fi (, r ) C and (y, s ) C. Then for all (, r) C and (y, s) C we have d(, y) d(, ) + d(, y ) + d(y, y) r + r + d(, y ) + s + s This implies d(c, C ) r + d(, y ) + s which shows that this distance is finite. If C = C, then we have d(, y) r + + s for all (, r), (y, s) C. hence equal cones have zero distance. Assume d(, y) =. Fi (, r) C, then we have for all ɛ > and (y, s) C d(, y) r + ɛ + s 7

8 Since ɛ was arbitrary, we have d(, y) r + s. Since (y, s) C was arbitrary, we have (, r) C and hance C C. By symmetry we also have C C, and hence C = C. Symmetry of the distance d is clear by definiton. Now let C, C, C be cones. Then we have for every (, r) C, (y, s) C and (z, t) C d(, z) d(, y) + d(y, z) r + d(c, C ) + s + s + d(c, C ) + t Since C contains poitns with arbitrarily small s, we conclude d(, z) d(, y) + d(y, z) r + d(c, C ) + d(c, C ) + t By definition of the infimum, this implies d(c, C ) d(c, C ) + d(c, C ) Lemma 7 Let (M, d) be a metric space and ( M, d) be the space of cones of M. Then there is an isometric embedding from M to M sending to the cone C(). The range of this map is dense in M Proof: Let, y M. Then for (, r) C() and (y, s) C(y) we have d(, y ) d(, ) + d(, y) + d(y, y ) r + d(, y) + s hecne d(c(), C(y)) d(, y). On the other hand we have (, ) C() and (y, ) C(y) and d(, y) = + d(, y) + shows that d(c(), C(y)) d(, y). hence the above map is an isometry. It is therefore also injective, and thus an embedding. To see density, let C be a cone in M and let r >. By definition of the cone there is M with (, s) C and < s < r. Then for any (z, t) C we have d(z, ) t + s = t + s + and hence This proves density.. d(c, C()) s Lemma 8 With the above setup, the space ( M, d) is complete. Let ι be the embedding from M to M. Let C be a cone in M. Let S = C ι(m) D. By density of M for each r > there eists M with (C(), s) C with < s < r. Hence S is a pre-cone. Let S be the subset of M D of all points (, r) with (ι(), r) S. Since ι is an isometry, S is a pre-cone. Let C be the associated cone. We claim C = C(C) 8

9 Eercise 1 Let (M, d) be a complete metric space. Let M be any subset of M. Then the embedding map i : M M etends to the completion of M. Any cone in M which has arbitrarily low elements in M has a tip in the image of the completion. Eercise 2 Consider a complete metric space. Let B t ( t ) be a set of nested balls parameterized by t A where A D. Then there eists an element M which is contained in the balls B 2t ( t ). Eercise 3 Let (M, d) be a complete metric space and M be a subspace. If and only if M/M is the union of balls, then M is complete. 1.4 Baire argument Lemma 9 Let (M, d) be a nonempty complete metric space. Let M n be a sequence of subspacess such that each M n is a union of balls and dense in M. Then tyhe intersection of the spaces M n is not empty. Proof: Let B r ( ) be a ball contained in M. Assume we have already chosen B rn ( n ). Since A n+1 is dense, it has nonempty intersection with B rn ( n ). Let B n be a ball contained in A n+1 that has nonempty intersection with B rn ( n ). Then there eists a ball B rn+1 ( n+1 ) contained in B rn ( n ) B n. Choose this ball, we may assume that its double ball is contained in B rn ( n ) B n. This gives a nested sequence of balls. By completeness, the associated pre-cone has a tip, which is contained in B 2rn+1 ( n+1 ) which in turn is contained in B rn ( n ) which in turn is contained in A n. hecne the intersection of the A n is not empty. Definition 8 A set X M is called nowehre dense if its complement is dense and the union of balls. As a corollary of the above lemma, a complete metric space is not the union of nowhere dense sets. Definition 9 A point in a metric space M is called isolated, if there eists r > such that B r () = {}. Lemma 1 A complete metric space without isolated points is uncountable. Proof. For each M. the set M \ {} is dense. Namely, since is not isolated, the ball B r () contains some point of M \ {}. For balls centered at points other than, these balls trivially contain soem point in the set. The set M \ {} is alos the union of balls, since for every y the bal B d(,y) (y) is contained in M \{}. Assume to get a contradiction that M is countable. Then M is the union of countably many sets of the above type. But this contradicts the previous lemma. 9

10 1.5 Separability Definition 1 Let (M, d) be a metric space and let M be a subset of M. A set C of balls in M is called a cover of M, if M B A subset C C is called a subcover, if B C M B C B A cover is called countable if it consists of countably many or balls. Our use of the phrase countable set includes the possibility that the set is finite. A subset C of a cover C is called subcover if it is itself a cover of the subspace in question. Definition 11 (Separable) A metric space (M, d) is called separable, if every cover of M has a countable subcover. One could interpret this definition as follows: no matter how one chooses almost equivalence classes, countably many are enough. Lemma 11 Countable metric spaces are separable. Proof: Assume we are given an arbitrary cover. For each element in the separable space choose one ball of the cover that contains the element. This gives a countable subcover. Lemma 12 If a metric space has a countable dense set, then it is separable. Proof: Assume M has a countable dense subset M. Consider the set C of all balls B r () with r D and M. This is a countable set. Now let C be any cover of M. Let C be the set of all balls in C which are contained in some ball of C. We claim C is a cover of M. Namely, let M. then there eists B r (y) C which contains. Then B s () B r (y) for some s >. Then there eists z M contained in B s/2 () By the triangle inequality B s/2 (z) B r (y). Hence B s/2 (z) C and C is a cover. For each ball B r () in C choose one ball in C which contains the ball B r (). the set of chosen balls is countable and covers M. Lemma 13 If a metric space is separable, then it has a countable dense set. Proof. for each r > consider the set M of all balls of radius r. Clearly every point M is contained in the ball B r (), so these balls cover M. By assumption there is a countable subcover C r. Let X r be the set of centers of balls in C r and let X be the countable union of the X r. Then X is countable. We claim it is dense. Let B s (y) be any ball in M. Since C s is a cover, there is a ball B s (z) C s which contains y. But then z B s (y). This proves that X is dense.. 1

11 Eercise 4 Prove any number of the following: 1. l 2 (N) is separable. 2. BV (D) is not separable. Lemma 14 Any subset of a separable metric space is separable. Note that given any countable dense subset of M, it is not clear that this set has any point in common with M. hence construction of a countable dense subset of M has to be more careful. Proof: Since M is separable, we find a countable dense subset X. Consider the countable collection C of all balls with centers in X and radius in D. For each ball B r () in C such that intersects M, pick a point in B r (z) M. This gives a countable collection Y of points in M. We claim that this collection is dense in M. Pick z M and s >. Since C s/2 covers M, there is a ball in C s/2 which contains z. This ball contains a chosen point y Y as well, and we have by the triangle inequality y B s (z). This proves density of Y. Conversely we have Lemma 15 Let M be a metric subspace of a metric space M. Assuyme we have a cover of M by balls in M, then there eists a countable subcover. Note that this lemma is not immediate from the definiiton of separability, since the balls in the given cover may not have centers in M and thus their restrictions to M may not be balls in M. Proof: For each point y M there is a ball B r () in the cover containing y. Then there is a small ball B s (y) contained in B r (). The collection of all such small balls for y M covers M and their restrictions to M are a cover in the space M. Hence there is a finite subcover. For each ball B s (y) in this subcover pick one of the big balls B r () as above. This gives a countable subcover of the original cover. Definition 12 Let (M, d) and (M, d ) be two metric spaces. The Cartesian product of the two metric spaces is the set of all ordered pairs (, ) with M and M and distance d((, ), (y, y )) = ma{d(, y), d(, y )} Eercise 5 Prove that this is indeed a metric space. Eercise 6 The Cartesian product of two separable metric spaces is separable. 11

12 1.6 Compactness Definition 13 A metric space is called compact, if every cover of the space has a finite subcover. As in the case of separability we have the following two observations: Lemma 16 Finite sets are compact. Lemma 17 If a metric subspace M of a space M is compact and we are given any collection of balls in M that cover M, then there is a finite subcover. The analogy of the theories of separable and compact sets has an end when the countability of the set of radii D comes into play. For eample, compact spaces do not have finite dense sets (finite sets are never dense in any larger space). Also, a subset of a compact set need not be compact. On the other hand one has some immediate consequences from applying the compactness property to all balls centered at a given point: Definition 14 A metric space (M, d) is bounded if there eists a ball which contains the whole space M. This is clear if M is empty by using the empty ball. Lemma 18 If M is a metric space and M a subspace, and if M is contained in a ball with center in M, then M is bounded. Proof: We need to show there eists a ball with center in M which contains M. We may assume M is nonempty. Let B r () a ball with center in M which contains M. Let y be any point in M. Then B 2r (y) contains M by the triangle inequality. Lemma 19 A compact metric space (K, d) is bounded. Proof: We may assume K is not empty. Pick a point K and observe that M = r D B r (), since every point y in K has some finite distance d(, y) to and there eists r D with d(, y) < r and hence y B t (). Hence the collection of balls about covers the compact set, and hence there is a finite collection of balls about which cover the compact set. Since the balls are nested, it suffices to take the largest ball to cover the compact set. This proves boundedness. There is a dual observation using complements of balls Lemma 2 Every compact metric space (K, d) is complete. 12

13 Proof: Pick a cone B in K. For every (, r) / B there eists ɛ > and (y, s) B such that d(, y) r + s + ɛ Pick the ball B ɛ (r) and let C be the collection of all these balls. We have two cases: 1. C covers K. Then there is some finite subcover. Each ball in the cover comes with a point (y, s) B as above, pick s the minimal s. If s = s, we are done since then the cone has a tip. We claim s > is impossible. For assume to get a contradiction that s >. Pick (z, t) B with t < s. The point z is in the finite subcover. Let B be the ball in the cover containing z, it comes with a point (, y) / B and (y, s) B with Then r + s + ɛ d(, y) r + t + ɛ d(, z) + d(y, z) d(, y) r + s + ɛ This is a contradiction since t < s s. 2. Assume C does not cover K, and let z be a point not covered. Then there does not eist (y, s) B with d(z, y) r Hence the cone is of the form B(z) Lemma 21 Let (M, d) be a compact metric space and M be a subspace. If and only if M/M is the union of balls, then M is compact. Proof: Let C be a cover of M. Since M \ M is the union of balls, the set C of these balls forms a cover of M \ M. Hence C C covers M. Find a finite subcover. Remove all balls in C from this finite subcover, then it still covers M The only if part follows form only if for completeness. Lemma 22 Let K be a compact metric space and let K n be a nested sequence of compact subsets of K, i.,e., K n+1 K n. Then the intersection of the sets K n is not empty. Proof: the complement of K n is a union of open balls by the previous lemma. Take the collection of all such balls for all n. If these balls cover all of K, there is a finite subcover. This finite subcover is disjoint from some K n. Thsi is a contradiction. Hence these balsl do not cover K, hence the K n have nonempty intersection. The following is a version of the fact that finite subsets in a totally ordered space have a maimum and a minimum. 13

14 Lemma 23 Let (M, d) be a metric space and let K be a subspace. Let M, then there eists y and z in K such that d(, y) = inf d(, t) t K d(, y) = sup d(, t) t K Proof: There is a sequence t n of elements in K such that d(, t n ) is monotone increasing and sup d(, t n ) = sup d(, t) = S n t K Pick a subsequence that is Cauchy. Then this seqeunce has a limit t K. Then we have d(, t) d(, t n ) d(t, t n ) S ɛ ɛ for arbitrarily small ɛ. Hence d(, t) = S The proof of eistence of a minimum is similar. Lemma 24 The Cartesian product of two compact metric spaces is compact. Proof: It suffices to find finite subcovers of covers by balls in this metric. Suppose we are given such a cover. For each M this covers in particular the set {} M. Find a finite subcover. Pick the minimal radius r of this finite subcover. Then this finite subcover covers B r () M. Now these B r () cover all of M. Find a finite subcover. This produces a finite colelction of ifnite sets covering all of M M 1.7 Total boundedness The following definition is a property between separability and compactness. Definition 15 A subset M of a metric space M is called totally bounded, if for every r > there eists a cover of M by balls of radius r. Lemma 25 A totally bounded space is separable. Proof: For each r pick a finite cover C r of the space M. Let C be any cover of M. Let be a point in M. and B r (y) a ball in C containing. then there is a ball B s () contained in B r (y). Pick a ball B in C s/2 containing, then this ball is contained in B r (y). The collection of all B is countable since contained in the countable union of finite sets C r. Picking for each ball B a ball in C that contains B gives a countable subcover of C. Lemma 26 A compact metric space is totally bounded. 14

15 Proof: Apply the definiiton of compactness to the collection of balls of radius r. Lemma 27 Let (K, d) be a totally bounded complete metric space. Then K is compact. Proof: Denote by C r a finite cover of M by balls of radius r. Pick a ball B r ( ) in C 1 that is not covered by finitely many balls in C. Such ball eists, since if each of the finitely many balls in C 1 where covered by finitely many balls in C, then there was a finite subcover of C for M. Assume we have already selected the ball B rn ( n ). Pick one ball B rn+1 /2( n+1 ) in C rn/2 that intersects B rn ( n ) and is not covered by finitely many balls in C. Such ball eists since C r/2 is a finite cover of M and since B rn ( n ) is not covered by finitely many balls in C. The sequecne of balls B 4rn ( n ) is nested. Let be the tip of the associated precone. It is contained in each ball B 8rn ( n ). Let B r (y) be a ball in C that contains. Then this ball contains a ball B s (). Then this ball contains the selected ball B 8rn ( n ) provided 16r n s, which can be achieved for large enough n. This however contradicts the fact that B rn ( n ) is not covered by finitely many balls of C. Eercise 7 let M be a metric space and M be a subspace. If for each r the space M is covered by finitely many balls of radius r with center in M, then also for each r it is covered by finitely many balls of radius r with center in M. Eercise 8 Let M be a totally bounded metric space and assume there is a sequence f : N M such that for any two balls B r () and B s (y) with B 2r () B 2s () = at most one of them contains infinitely many point of the sequence. Then there is eactly one cone B with the property that for every r there are at most finitely many points of the sequence not in { : (, r) C} 1.8 Cauchy sequences, convergent sequences Minimal eamples of pre cones are called Cauchy sequences. Definition 16 Let (M, d) be a metric space. A sequence f : N M is called Cauchy, if for every r > there eists n N such that sup d(f(n), f(n + k)) r k Definition 17 Let (M, d) be a metric space. A sequence f : N M is said to converge to if for every r > there eists n N such that d(f(n), ) r 15

16 Eercise 9 If a sequence converges to a point, then it is Cauchy. Lemma 28 Let f : N M be a Cauchy sequence. Then there is eactly one cone C such that for every r the set of points in the seuquence that are not in { : (, r) C} is finite. If f converges, it converges to the tip of the cone. Eercise 1 A metric space is complete if and only if every cauchy sequecnes converges. Lemma 29 A metric space M is toatlly bounded, if and only if every sequence in the space has a subsequence which is Cauchy. Proof: First assume the space is totally bounded. Let f : N M be a sequence. define f = f. Assume we have defined f n for n. Define a subsequecne f n+1 of f n so that f n+1 (m) coincides with f n(m) for m n, and so that f n+1 (m) B 2 n() for some ball of radius 2 n. This is possible, since we can cover the space M by finitely many balls, so at least one of them needs to contain infinitely many values of the sequence f n. Then define g(n) = f n (n). This is a subsequecne of f, and it is Cauchy. Now assume every sequecne has a subsequence that is Cauchy. Proof: Fi r and pick a point M. Assume we have already picked n. If the collection of balls B r ( k ) with k n covers M, we are done having constructed a finite cover of M. If not, pick n+1 not covered by these balls. We claim this process cannot go on indefinitely. If it does, then we have a seuqence n which has no Cauchy subsequecne since any two elements in the seuqence have distance at least r. this is a contradiction. Eercise 11 A metric space is compact if and only if ever sequence has a convergent subsequence. Eercise 12 Let f : N M be a sequence in a compact space. and let M If every subsequence of f has a further subsequence which converges to, then f itself converges to. 1.9 Eamples Note that R n with Euclidean metric is complete. Lemma 3 Bounded complete subsets of R n with Euclidean metric are compact. Proof: We prove that every sequence in K has a subsequence which is Cauchy. let ( 1,..., n ) denote the sequence. Since the sequence is bounded, each entry is bounded. Choose a subsequence so that 1 is monmotone. Then choose a subsequence so that 2 is monotone, and so on. Hence we may assume all sequences are monotone. But then all coordinate sequences are Cauchy, and hence the full sequence is Cauchy. Since K is complete, it has a limit. 16

17 Eercise 13 The space l 2 (N) is separable and complete. Lemma 31 The ball B 1 () in l 2 (N) is not compact. Proof: Let f n be the element in l 2 (N) which satisfies f n (m) = if n m and f n (m) = 1 if n = m. Then Any two elements have distance > 1, hence this sequence has no subsequence that is Cauchy. Eercise 14 Let a : N R + be a sequence with lim sup n a n =. Let K be the set of all sequences f in l 2 (N) (with values in R) such that Prove that K is compact. a(n)f(n) 2 < < n= Eercise 15 Prove the converse of the above. If K l 2 (N) is compact, then there eists a sequence a n as above such that n= a(n)f(n) 2 < for all f K 1.1 Mappings between metric spaces We study mappings f : M M between two metric spaces (M, d) and (M, d ) that in one way or the other respect the metric. However canonical the definition of a metric space appeared, there are now a whole variety of classes of mappings to consider. Essentially, one would like to have an upper bound for the distance of the function values in terms of the original distance. A mapping f : M M is called: 1. An isometry, if d (f(), f(y)) = d(, y) for all, y 2. A contraction, if d (f(), f(y)) d(, y) for all, y 3. A Lipshitz map, if there eists L (Lipschitz constant) such that we have d (f(), f(y)) Ld(, y) for all, y 4. A strict contraction if it is Lipshitz with constant L < 1 5. A Hölder map with eponent < α if there eists a constant C such that d (f(), f(y)) Cd(, y) α for all, y The case α = 1 is that of a Lipschitz map. 6. A uniformly continuous map, if there eists a right continuous (conve?) monotone increasing function g : R + R + with g() = such that g(d (f(), f(y))) d(, y) for all, y. The function g is called the modulus of continuity of the function. In case of a Hölder function with eponent alpha this finction is of the form c 1/α for some small c. 17

18 7. A continuous map, if for every M there eists a function g as above such that for all y we have g (d (f(), f(y))) d(, y) for all, y. The function g is called the modulus of continuity of the function at. Eercise 16 Any isometry is injective. Isometries are therefore often called isometric embeddings, they are bijective onto their range, and they describe a complete metric isomorphism between M and its range. The eample of isometries also suggests why one does not look at maps corresponding to the revers inequalities in the above list. these maps are informally of an epanding nature, but they are injective in all these cases. Thus one may simply studey the inverse map on the range and obtain a map of the above list. Thus one does not epect much new insight from studying reverse inequalities. The class of contraction mappings plays a role in fied point theorems, we will show the classical fied point theorem for strict contractions. Contractions still do not have an interesting class of isomorphisms, if a bijective map is a contraction and its inverse is a contraction, then the map is already an isometry. Note that for a metric space (M, d), the metric space (M, λd) for λ > behaves very similarly, ecept distances are measured in another unit. If one tries to look at equivalence relations of metric spaces under this scaling, then the notion of contraction mapping makes no sense anymore. Then the class of Lipschitz maps is the immediate generalization of contraction mappings that is invariant under this scaling. Now Lipschitz maps allow for inetresting non-trivial isomorphisms. A bijective map is bi-lipschitz if both the map and its inverse are Lipschitz. There is a very rich class of bi-lipshchitz maps, studied etensively. It is the most natural and most simple notion of non-trivial isomorphisms of metric spaces if one wants to consider global properties of metric spaces and not just infinitesimally small features of the space. Hölder functions appear prominently in nature, for eample the Brownian path of a particle is almost surely Hölder continuous with eponent α for any α < 1/2 but not Hölder continuous with eponent α = 1/2. Lemma 32 Functions f : D D which satsify Hölder condition with α > 1 are constant. Proof: Let C be the the best constant in the Hölder estimate for any three points, y [, 1]. We need to show that C =. Assume to get a contradiction C >. Consider a pair of points such that C(1 ɛ) y α < f() f(y) Then clearly y. Let z be the midpoint of and y, then f() f(y) f() f(z) + (f(z) f(y) C z α + C z y α 18

19 C2( 1 2 )α y α This however is a contradiction if 1 ɛ > 2 1 α Since we could choose ɛ arbitrary and 2 1 α < 1 this is a contradiction. More generally this prove works in any metric space that has midpoints, we will discuss these in more detail later. On bounded spaces (upper bound on all distances) and without midpoints, the notion of Hölder α with α > 1 is less trivial. One major point distinguishing uniform continuity and continuity is the following lemma, which works only for uniformly continuous maps. Lemma 33 Let (M, d) be a metric space and M a subspace of M, and let ( M, d) be some complete metric space. If f : M M is a uniformly continuous map, then there is a unique etension of this map from M to M which is also uniformly continuous. Proof: Eercise: fi this proof: Let M. Since g has a right limit of at, for each dyadic r > there eists ɛ r such that 2g(ɛ r ) + ɛ r < r. Define S = {(f(y), r) : (y, g(r)) B() M } Since M is dense in M, this set is good. This set is a restrained set. namely, let (z, g(r)) in this set. Then z is of the form f(y) with (y, g(r)) B(), that is We have to show lim inf s lim inf s sup d(y, u) + ɛ g(r) (u,s) S sup d(z, v) + ɛ r (v,s) S Since M is complete, the associated dedekind cone is of the form B(y) for some y M. Define f() to be y. Then f thus defined is uniformly continuous (eercise) A point with f() = is called a fied point for a map. Lemma 34 (Contraction mapping principle) If f : M M is a strict contraction from a complete non-empty metric space to itself, then there eists a point M with f() = 19

20 Proof: Since M is non-empty, there eists a point in M. defined 1 already, then define n+1 = f( n ). We claim that for every n we have Assume we have d( n+1, n ) L n d( 1, ) For n = this is clear since L = 1. Assume this inequality holds for Note that by the strict contraction property we have d( n+2 ), n+1 ) = d(f( n+1 ), f( n )) = Ld( n+1 ), ( n )) L n+1 d( 1, ) Now we claim that there is a constant C such that for any n, m d( n, n+m ) C(2L n L n+m ) This is clear for m = as long as C is at least d(, 1 ). Assume we have proven this for m Then d( n, n+m+1 ) d( n, n+m ) + d( n+m, n+m+1 ) C(2L n L n+m ) + L n+m d(, 1 ) = C(2L n L n+m+1 ) + L n+m (d(, 1 ) C(L 1 1)) The last term is negative as long as C is larger than a number determined by L. This completes the induction. Now since f is a contraction we have d(f(), ) d(f(), f( n )) + d(f( n ), n ) + d( n, ) ɛ + CL n + ɛ Where ɛ and L n can both be made arbitrarile small by choosing n large enough. Henc ed(f(), ) = and hence f() =. Eercise 17 The fi point guaranteed to eists in the above lemma is in fact unique. The contraction mapping principle is a very ubiquitous method of choice to prove eistence for ODE or PDE. We discuss an eample Lemma 35 Let F be a Lipschitz map R R and let c R. Then there eists a solution to the integral equation in C[, ). f() = c + F (f(t)) dt 2

21 Prove. We first construct ɛ > such taht there is a solution to this equation in the metric space C([, ɛ]) with the supremum norm. Define a map on this space by T (f)() = c + F (f(t)) dt This is a contraction mapping with respect to the supremums First of all if f is continuosu, the F f is continuous since it is the composition of continuous maps.the primitive of a continuosu function is continuous. Hence T maps M to itself. We have, using that F is Lipshitz d(t (f), T (g)) = sup F (f(t)) dt F (g(t)) dt d(t (f), T (g)) = sup F (f(t)) F (g(t)) dt sup F (f(t)) F (g(t)) dt sup C f(t) g(t) dt sup Cd(f, g) dt ɛcd(f, g) Hence this is a contraction mapping provided ɛc < 1 Hence there is a a continuous function f : [, ɛ] R satisfying the equation. We shall by induction prove that for every n that there eists continuous f on [, nɛ, ] solving the equation. Assume this is shown for n. We shalls eek an etension of f on [, nɛ] to [, (n + 1)ɛ]. That satisfies for n f() = c + f() = c + nɛ F (f(t)) dt F (f(t)) dt + This can be done by solving on [, ɛ] the equation g() = c + nɛ F (f(t)) dt + nɛ F (f(t) dt F (g(t) dt and then setting f() = g( nɛ) Hiowever, to find a solution to the equation for g we simply apply the previous argument. Note that for c = 1 and F the identity map, this eistence proof can be used to prove eistence of the eponential function. 21

22 Lemma 36 Assume (K, d) is a compact metric space and (M, d ) is a metric space. Assume f : K M is continuous. Then the range of f is compact. Proof: Let C be an open cover of f(k). We produce a cover of K. Let K. Let B be a ball in C that contains f(). By continuity of f, there is a ball B in K such that f(b ) B. The collection of these balls B covers K. Pick a finite subcover. Consider all B corresponding to the chosen B. This is a finite collection. We claim it covers f(k). Let y f(k). The there is z such that f(z) = y. Let B be the chosen ball containing z. Then B contains y. Lemma 37 Let (K, d) be a compact space and (M, d ) be a metric space. Let f : K M be continuous. then f is uniformly continuous. Fi s >. define for each K r() to be the supremum of all numbers r 1 such d(, y) r implies d (f(), f(y)) δ. We claim that r is continuous, and first see how this will finish the proof. If continuous, then it attains its minimum on the compact set K. The minimum cannot be by continuity at that point. Hecne the minimu is larger than zero, which proves uniform continuity since s was arbitrary. Now we turn to the proof of continuity. Fi and note that in the vicinity oif we will use Contraction mapping principle A point with f() = is called a fied point for a map. Lemma 38 (Contraction mapping principle) If f : M M is a strict contraction from a complete non-empty metric space to itself, then there eists a point M with f() = Proof: Since M is non-empty, there eists a point in M. defined 1 already, then define n+1 = f( n ). We claim that for every n we have Assume we have d( n+1, n ) L n d( 1, ) For n = this is clear since L = 1. Assume this inequality holds for some n. Note that by the contraction property we have d( n+2, n+1 ) = d(f( n+1 ), f( n )) Ld( n+1 ), ( n )) L n+1 d( 1, ) Now we claim that there is a constant C such that for any n, m d( n, n+m ) C(2L n L n+m ) 22

23 This is clear for m = as long as C is at least d(, 1 ). Assume we have proven this for some m. Then d( n, n+m+1 ) d( n, n+m ) + d( n+m, n+m+1 ) C(2L n L n+m ) + L n+m d(, 1 ) = C(2L n L n+m+1 ) + L n+m (d(, 1 ) C(L 1 1)) The last term is negative as long as C is larger than a number determined by L < 1. This completes the induction. Now since f is a contraction we have d(f(), ) d(f(), f( n )) + d(f( n ), n ) + d( n, ) ɛl + 2CL n + ɛ Where ɛ and L n can both be made arbitrarily small by choosing n large enough. Hence d(f(), ) = and hence f() =. Lemma 39 The fi point guaranteed to eists in the above lemma is in unique. Proof: assume that there are two fied points and y. Then d(f(), f(y) = d(, y) because both points are fied points, but als d(f(), f(y) Ld(, y) because of the contraction property. Since L < 1, this is impossibly for d(, y). Hecne d(, y) = which implies = y. The contraction mapping principle is a very ubiquitous method of choice to prove eistence for ODE or PDE. We discuss an eample Lemma 4 Let F be a Lipschitz map R R and let c R. Then there eists a solution to the integral equation in C[, ). f() = c + F (f(t)) dt Prove. We first find small ɛ > and a solution to this equation in the metric space C([, ɛ]) with the uniform metric given by sup [,ɛ] f() g(). Define a map on this space by T (f)() = c + F (f(t)) dt We claim that this is a contraction mapping with respect to the uniform metric. First of all if f is continuous, then F f is continuous since it is the composition of continuous maps. The primitive of a continuous function is continuous. Hence T maps M to itself. We have, using that F is Lipschitz d(t (f), T (g)) = sup F (f(t)) dt F (g(t)) dt 23

24 d(t (f), T (g)) = sup sup sup F (f(t)) F (g(t)) dt F (f(t)) F (g(t)) dt sup C f(t) g(t) dt Cd(f, g) dt ɛcd(f, g) Hence this is a contraction mapping provided ɛc < 1. Hence choosing ɛ smaller than 1/C there is a a continuous function f : [, ɛ] R satisfying the equation. We shall by induction prove that for every n there eists a continuous f on [, nɛ, ] solving the equation, the case n = just having been established. Assume this is shown for some n. We shall seek an etension of f on [, nɛ] to [, (n + 1)ɛ]. That etension needs to satisfy for n f() = c + f() = c + nɛ F (f(t)) dt F (f(t)) dt + This can be done by solving on [, ɛ] the equation g() = c + nɛ F (f(t)) dt + nɛ F (f(t) dt F (g(t) dt using the previous arguments and then setting f() = g( nɛ). Note that for c = 1 and F the identity map, this eistence proof can be used to prove eistence of the eponential function. Eercise 18 (Newton method) Assume we have a function f : [, 1] R that satisfies for any, y [, 1] the Taylor epansion estimate for some constant C Define the iteration f(y) f() + f ()(y ) C 2 n+1 = n f( n) f ( n ) And prove that if is within ɛ of a point with f() =, then this iteration converges to. This is called the Newton iteration method to find a zero. 24

25 1.12 Curves We consider mappings f : [, 1] M with M a complete metric space. Definition 18 An mapping f : [, 1] M is called 1. geodesic, if it is an isometry or equivalent to an isometry, that is of a becomes an isometry when the distance function on the target space is multiplied by a nonzero constant. 2. a rectifiable curve, if it is Lipschitz 3. a curve if it is continuous (hence uniformly continuous) Sometimes one restricts attention to injective mappins which are then called nonselfintersecting curves. We will however not need injectivity here. We say two points and y can be connected by a curve f if there is a curve with f() = and f(1) = y. Eercise 19 Prove that a subset of [, 1] which contains the points and 1, these two points are connected by a curve if and only if it is the set [, 1] itself. Lemma 41 Let M be a metric space. Then the following two are equivalent: 1. For every, y M there is a geodesic f : [, 1] M such that f() = and f(1) = y. 2. For every, y M there is a midpoint of and y, that is a z M such that d(, z) = d(z, y) = d(, y)/2 Proof: If, y can be connected by a geodesic curve, then we pick such a geodesic curve and define z = f(1/2). If λ is such that d(f(s), f(t)) = λd(s, t) for this geodesic, then d(, z) = λ/2 = d(, z)/2 and similarly for d(z, y). Conversely, assume there is a midpoint for any two points in M. Assume without loss of generality that d(, y) = 1. Define a map f : [, 1] M as follows: Set f() = and f(1) = y. Assume we have already defined f on D n [, 1]. Let z D n+1 \ D n and let u D n such that u z ν n (u). Then define f(z) to be the midpoint of f(u) and f(ν n (u)). By induction we see that d(t, t + ν m (t)) = 2 m for D m. Let s < t be any two points in D m, then we have by the triangle inequality r D m [,1),r<s 1 = d(f(), f(1)) d(f(), f(s)) + d(f(s), f(t)) + d(f(t), f(1)) d(f(r), f(ν m (r)))+ r D m [,1),s r<t d(f(r), f(ν m (r)))+ t D m [,1) d(f(t), f(ν m (t))) 25

26 = r D m [,1),r<s 2 m r D m [,1),s r<t 2 m = (s ) + (t s) + (1 t) = 1 t D m [,1) Since both sides of the inequality are 1, all intermediate inequalities in this argument have to be equalitites and in particular d(f(s), f(t)) = t s. Hence f is an isometry and thus a geodesic. Definition 19 We define the length of a curve to be sup n D n, <1 d(f(), f(ν n ()) Eercise 2 Let f : [, 1] M be a curve. Let g : [, 1] D [, 1] be a monotone curve with g() = and g(1) = 1 (called a reparameterization). Then the length of f equals the length of f g. Lemma 42 A curve has finite length if and only if there is a curve g onto the range of f which is Lipschitz. Proof: Assume we have the map g such that h = f g is rectifiable, with Lipschitz constant L. Then we have for every n D n, <1 D n, <1 d(h(), h(ν n ()) L2 n L Conversely, assume the curve has finite length. Assume without loss of generality that this length is 1. Then we can define the arclength for every partial piece [, ] to be λ() = sup n d(f(t), f(ν n (t)) t D n, t< Then λ is obviously a monotone function [, 1] to [, 1). We define a function g[, 1] [, 1] by setting g(λ()) = f() To see that this is well defined, we need to verify that if λ() = λ(y) then f() = f(y). However, if λ() = λ(y), then by definiiton of λ in particular d(f(), f(y)) = because that term may appear in a sum that we take supremum over. Similarly we see that g is Lipschitz, since we have for y > Clearly g is onto the range of f. λ() λ(y) d(f(), f(y)) 2 m 26

27 Definition 2 A curve f : D [, 1] M is called of finite r-variation, if there eists a constant C such that for any partition < 1 <... < n of [, 1] (in D m say) we have n ( d( i, i+1 ) r ) 1/r C i=1 Eercise 21 A curve f is of bounded r variation if ane only if there is a curve g : [, 1] onto the range of f with which is Hölder with eponent 1/r. Definition 21 Let M be a complete metric space and M. then the set of all y such that there eists a curve f with f() = and f(1) = y is called the connected component M of. Definition 22 A metric space is called pathwise connected, if for any two points, y there eists a curve f with f() = and f(1) = y. In other words, it is pathwise connected if M = M for every M. Definition 23 A metric space is called locally pathwise connected, if for any point there eists a r > so that B r () M. Lemma 43 Assume M is locally pathwise connected. Then for every the set M is open. Proof: Let y M. Since M is locally pathwise connected there eists r such that B r (y) M y. It suffices to show that B r M. Let z B r (y). Since also z M y, there eist g : [, 1] M with g() = y and g(1) = z. Since y M there eists f : [, 1] M with g() = y and g(1) = z. Now define h : [, 1] M as follows 1. If t 1/2 then h(t) = f(2t) 2. If t 1/2 then h(t) = f(2t 1) This clearly defines a curve connecting with z. Lemma 44 Assume M is locally pathwise connected. Then for every the set M\M is open. Proof: If y M \ M, then also M y M \ M. We prove this by contradiction, assume z M y M. Then we can connect z both with and y, and then we can construct a curve from to y as in the proof of the previous lemma. Since M y is open for every y M \ M, then M \ M is open. Definition 24 A metric space is called connected if it is not the union of two nonempty disjoint open sets. Lemma 45 A locally pathwise connected and connected set is pathwise connected. Proof: For every, since both M and M \ M are open, one of them has to be empty. Since M, we have M \ M =. Eercise 22 Present and prove carefully an eample for a connected metric sapce which is not pathwise connected. 27

28 1.13 The space C(M, M ) Lemma 46 For two metric spaces (M, d) and (M, d ) consider the space ( M, d) of uniformly continuous functions f : M M with distance d(f, g) = sup d(f(), g()) M Then this space is a complete metric space. Note: this lemma works also with uniform continuity replaced by continuity. Proof: Suppose we have a Dedekind cone Bin M. Then for every M the set B = {(f(), r) : (f, r) B is a Dedekind cone in M. To see that, we need to show that for (y, r) B there eists ɛ with inf s sup d (y, y ) + ɛ r (y,s) B Given such y, pick f B such that y = f(). Choose ɛ as given by Dedekind cone property of B fro (f, r). Then inf s = inf s inf s sup d (f(), z) + ɛ (z,s) B sup d (f(), g()) + ɛ (g,s) B sup d(f, g) + ɛ r (g,s) B This proves that B is a Dedekind cone. Since M is complete, there is a y M such that B = B(y). This gives a function g : M M assigning to each this y. We claim that thsi function is uniformly continuous. For this we need to estimate d (g(), g( )). This is the distance of two Dedekind cones, which by definition this is inf s sup sup d(f(), f ( )) f B s f B s Fi s and poick an individual f s B s. then we estimate sup sup d(f(), f ( )) f B s f B s sup sup (d(f(), f s ()) + d(f s (), f s ( )) + d(f ( ), f ( ))) f B s f B s Now estimate sup by triangle inequality, giving three terms total. The first and last term are estimated alike, we discuss the first sup sup d(f(), f s ()) f B s f B s 28

29 = sup f B s d(f(), f s ()) = sup f B s d(f, f s ) This can be estimated by 2s since B is a Dedekind cone. Now the middle term can be estimated by the modulus G s of continuity of s sup sup d(f s (), f s ( )) f B s f B s G s (d(, )) This proves uniform continuity of g. Note that the modulus of continuity of the function g produced was assembled by the modulus of continuity of a sequence of functions f s. Thus its modulus of continuity may be worse than the modulus of continuity of either of these functions. This is why the corresponding sttement does nopt work for the class of Lipschitz functions, or the class of Hölder functions. However, a similar theorem holds for continuous functions (we were only working locally at some at a time, or say the class of contraction mappings since then all functions along the way can be estimated by the same modulus of continuity. Eercise 23 The space of Lipschitz functions from M to M is dense in the space C(M, M ) Eercise 24 The space of contractions from M to M is complete in C(M, M ) 1.14 Arzela Ascoli Lemma 47 Let M and M be compact spaces. Assume F C(M, M ). Assume that there is a modulus of continuity g such that all f F satisfy Then F is totally bounded. g(d (f(), f(y)) d(, y) Proof: Pick r >. Pick s = g(r/4). Cover M by finitely many balls B s ( i ), i = 1,..., n. Cover the compact set M by finitely many balls B r/4 (y j ), j = 1,... m. For each of the finitely many functions h : {1,..., n} {1,..., m} for which there eists a function f F with f( i ) B r/4 (y h(i) ) 29