Analysis

Universit´e Paris 1- Panth´eon Sorbonne UFR 27 Math´ematiques et Informatique

Analysis B. De Meyer

Master M1-MAEF and QEM-1 2007-2008

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Contents Chapter 1. Metric Spaces 1. Basic definitions 1.1. Distance 1.2. Balls, Open and closed sets, Neighborhoods 1.3. Convergent sequences 1.4. Continuity 1.5. Subspaces and Product of metric spaces 2. Completeness 2.1. Complete spaces 2.2. Contraction and fixed points 2.3. Uniform continuity 2.4. Cauchy completion 3. Compactness 3.1. Compact precompact and sequentially compact sets 3.2. Properties of compact sets 4. Connected sets 4.1. Basic definitions 4.2. Path-connected sets 4.3. Connected component

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Chapter 2. Linear Normed Spaces 1. Finite dimensional spaces 1.1. Equivalence of norms 1.2. Main properties 2. Linear maps 2.1. Continuous linear maps 2.2. The dual space 3. Absolutely convergent series 3.1. Series in a normed space 4. The space (Cb (E), k.k∞ ) 4.1. Ascoli Arzela Theorem 5. Frechet Differential 5.1. Definition 5.2. Implicit function theorem

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CHAPTER 1

Metric Spaces 1. Basic definitions 1.1. Distance. Let E be a set. A metric (or distance) d on E is a map d : E × E → R+ satisfying: 1) ∀x, y ∈ E : d(x, y) = 0 ⇔ x = y. 2) ∀x, y ∈ E : d(x, y) = d(y, x).(symmetry) 3) ∀x, y, z ∈ E : d(x, y) ≤ d(x, z) + d(z, y). (Triangle inequality) A set E endowed with a metric d is called a metric space. Example 1.1.1. (1) The discrete metric d on E is defined by ∀x, y ∈ E : d(x, y) = 0 if x = y, otherwise d(x, y) = 1. (2) On R, d(x, y) := |x − y| is a metric, referred to as the usual metric. (3) (Euclidian metric)p For x = (x1 , x2 ) and y = (y1 , y2 ) in R2 , the euclidian distance d(x, y) is (x1 − y1 )2 + (x2 − y2 )2 . (4) (Manhattan metric) For x = (x1 , x2 ) and y = (y1 , y2 ) in R2 , let d(x, y) be defined as d(x, y) := |x1 − y1 | + |x2 − y2 |. d is a metric on R2 sometimes called the Manhattan metric. (5) If k.k is a norm on a real vector space F (i.e. k.k : F → R+ satisfying 1) ∀x ∈ F : kxk = 0 ⇔ x = 0. 2) ∀x ∈ F, ∀α ∈ R : kα · xk = |α| · kxk. 3) ∀x, y ∈ F : kx + yk ≤ kxk + kyk.) then d(x, y) := kx − yk is a distance on F . (6) On the vector space Rk , one often consider the following norms: qP Pk k k 2 kxk1 := i=1 |xi |; kxk2 := i=1 xi and kxk∞ := maxi=1 |xi |. In case k = 2, the metrics associated with k.k1 and k.k2 are respectivly the Manhattan and the Euclidian metric. (7) A function f : E → R is bounded if kf k∞ := sup{|f (x)| : x ∈ E} < ∞. Prove that the set B(E) of bounded functions on E is a vector space and that k.k∞ is a norm on B(E). 1.2. Balls, Open and closed sets, Neighborhoods. In a metric space (E, d) the open ball B(x, ) with center x ∈ E and radius  ∈ R+ is defined as B(x, ) := {y ∈ E : d(x, y) < }. A subset A of E is a neighborhood of x ∈ E if there exists  > 0 such that B(x, ) ⊂ A. In particular ∅ is an open set. A subset A of E is an open set iif it is a neighborhood of all its points: ∀x ∈ A : ∃ > 0 : B(x, ) ⊂ A. 5

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A subset A is closed iif Ac is an open set, where Ac denotes the complementary of A in E: Ac := E − A. Exercise 1.2.1. (1) Draw the open ball B(0, 1) when d is the Euclidian metric on R2 . Same exercise when d is the Manhattan metric. (2) If d is the discrete metric on E and x ∈ E, what are the neighborhoods of x. (3) Is X := {(x, 0) : x ∈ R} an open or a closed set of R2 endowed with the Euclidian metric? Find a set that is neither open nor closed in R2 . (4) In a metric space (E, d) prove that E (resp ∅ ) is simultaneously open and closed. (5) In a metric space (E, d), prove that B(x, ) is an open set. (6) Let {Ai , i ∈ I} be a family of open sets in a metric space. Prove that ∪i∈I Ai is still an open set. Assuming that I is a finite set, prove that ∩i∈I Ai is also open. Note that ∩i∈I Ai may fail to be an open set when I is not finite. Find a counter example for this. What can be said about the union and the intersection of a family of closed sets. The interior A◦ of a set A is defined as the union of the open subsets of A. As it follows from the last exercise, A◦ is an open subset of A. Clearly A is an open set iif A◦ = A In the same way, the closure Cl(A) is defined as the intersection of the closed sets that contains A. Obviously: A◦ ⊂ A ⊂ Cl(A) = ((Ac )◦ )c . Furthermore A is closed iif A = Cl(A). Two metrics d and d0 on a set E are topologically equivalent if both metric spaces (E, d) and (E, d0 ) have the same open sets. Exercise 1.2.2. d(x,y) is a metric on (1) If (E, d) is a metric space, prove that d0 (x, y) := 1+d(x,y) 0 E equivalent to d that is bounded: ∀x, y ∈ E : d (x, y) ≤ 1. (2) Let d and d0 be two proportionally equivalent metrics on E (i.e. with the property that ∃α, β > 0 : ∀x, y ∈ E : αd(x, y) ≤ d0 (x, y) ≤ βd(x, y).) Prove that d and d0 are topologically equivalent.

1.3. Convergent sequences. A sequence {xn }n∈N in E is a map x. : N → E. A sequence {x0k }k∈N is a sub-sequence of {xn }n∈N if there exists a strictly increasing map n. : N → N such that for all k ∈ N : x0k := xnk . The sequence {xn }n∈N penetrates a set A if ∃N : ∀n ≥ N : xn ∈ A. Clearly if {xn }n∈N penetrates a set A, so does any sub-sequence of {xn }n∈N . The sequence {xn }n∈N is converging to a ∈ E if it penetrates all the neighborhoods of a. In this case, a is called the limit of {xn }n∈N and we write a = limn→∞ xn n→∞ or xn −→ a. Clearly, if {xn }n∈N is converging to a, then it penetrates all the balls B(a, ) with  > 0. In other words: ∀ > 0 : ∃N : ∀n ≥ N : d(a, xn ) < . This later condition is thus necessary to have the convergence of {xn }n∈N to a. It is also sufficient, since if V is a neighborhood of a, there exists a ball B(a, ) ⊂ V with  > 0. Since B(a, ) is penetrated by {xn }n∈N , so is V .

1. BASIC DEFINITIONS

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Theorem 1.3.1. 1) In a metric space (E, d), a sequence {xn }n∈N has at most one limit. 2) All the subsequences of a converging sequence {xn }n∈N converge to the same limit. Proof: 1) If {xn }n∈N converges to a and b, then, ∀ > 0, it penetrates B(a, ) and B(b, ). Therefore, there is an N such that xN ∈ B(a, )∩B(b, ). As a consequence: d(a, b) ≤ d(a, xN ) + d(b, xN ) ≤ 2. Since this holds for all  > 0 it must be the case that d(a, b) = 0 or equivalently that a = b. 2) If {xn }n∈N converges to a, it penetrates all the neighborhoods of a, and so does any sub-sequence {x0n }n∈N of {xn }n∈N . Therefore {x0n }n∈N also converges to a. Exercise 1.3.2. 1 (1) Prove that the sequence {xn }n∈N defined by xn = n+1 converges to 0 in R for the metric d(x, y) = |x − y|. (2) Analyze the convergence of the same sequence for the discrete metric on R. More generally, characterize the convergent sequences for the discrete metric. (3) A point a is called an accumulation point of a sequence {xn }n∈N if there is a subsequence {x0n }n∈N of {xn }n∈N converging to a. Prove that if b is not an accumulation point of {xn }n∈N , then there exists an open ball B(b, ) (with  > 0) such that {n ∈ N : xn ∈ B(b, )} is a finite set (i.e. the sequence {xn }n∈N penetrates the complementary of B(b, ). In R with the usual metric, find a sequence {xn }n∈N having no accumulation point. Find a non converging sequence {yn }n∈N having a single accumulation point. If the sequence {xn }n∈N is bounded from above in R, give the definition of lim sup xn . Prove that lim sup xn is an accumulation point of the sequence. Theorem 1.3.3. A set G is closed iif it contains the limit a of all the converging sequences {xn }n∈N with values in G. Proof: Necessary condition: Let {xn }n∈N be a sequence with values in the closed set G converging to a. If a were not in G, then a ∈ Gc which is by definition an open set. Gc is thus a neighborhood of a, and as such Gc must be penetrated by {xn }n∈N . This contradicts the fact the {xn }n∈N takes its values in G. Sufficient condition: Assume that G contains the limit a of all the converging sequences {xn }n∈N with values in G and let b ∈ Gc . If, for all n, there were a point xn ∈ G ∩ B(b, n1 ), the sequence {xn }n∈N with values in G would converge to b, and b would thus belong to G, which is impossible since b ∈ Gc . Therefore, there exists an n such that G ∩ B(b, n1 ) = ∅. This is equivalent to B(b, n1 ) ⊂ Gc . Gc is thus a neighborhood of b. Since this holds for all b in Gc , Gc results to be an open set and G is thus closed. A point a is adherent to a set A if any neighborhood of a has a non empty intersection with A. The adherence A of a set A is the set of all adherent point

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to A. Clearly A ⊂ A. A set A is dense in a metric space (E, d) iif E = A. A metric space is separable iif it contains a countable dense subset. R with the usual metric is a separable space since Q is dense in R. R with the discrete metric is not separable. Theorem 1.3.4. 1) Cl(A) = A. 2) A = A. c

Proof: 1) We first prove that A ⊂ Cl(A) or equivalently that Cl(A)c ⊂ A . Indeed, if a ∈ Cl(A)c , then Cl(A)c is an open set that contains a. Since A ⊂ Cl(A), Cl(A)c is thus a neighborhood of a that has an empty intersection with A. This indicates c that a 6∈ A and thus a ∈ A . c Conversely, if a ∈ A , then there exists a neighborhood V of a that does not meet A. There exists therefore an open set O such that a ∈ O ⊂ V (O could be taken to be an open ball). Since O has an empty intersection with A, we have A ⊂ Oc . Since Oc is a closed set that contains A, we have Cl(A) ⊂ Oc . Since c x ∈ O, we thus infer that x 6∈ Cl(A). Therefore A ⊂ Cl(A)c and thus Cl(A) ⊂ A. 2) With claim 1), we get A = Cl(Cl(A)) = Cl(A) = A. Exercise 1.3.5. (1) Prove that in a metric space, a point a ∈ A iif there exists a A valued sequence converging to a. 1.4. Continuity. Let (E, dE ) and (F, dF ) be two metric spaces. A function f : E → F is continuous at a point x ∈ E iif for all sequence {xn }n∈N in E converging to x, the sequence {f (xn )}n∈N converges to f (x) in F . The function f is continuous if it is continuous at any point x ∈ E. The reciprocal image f −1 (A) by f of a set A ⊂ F is define as f −1 (A) := {x ∈ E : f (x) ∈ A}. Theorem 1.4.1. The following claims are equivalent: 1) f : E → F is continuous at a point x ∈ E. 2) ∀ > 0 : ∃δ > 0 : ∀y ∈ E : dE (x, y) < δ =⇒ dF (f (x), f (y)) < . 3) The reciprocal image f −1 (V ) of any neighborhood V of f (x) is a neighborhood of x. Proof: 1) =⇒ 2): If 2) does not hold, there exists  > 0 such that for all δ > 0, there exists yδ such that dE (x, yδ ) < δ and dF (f (x), f (yδ )) ≥ . The sequence {y n1 }n∈N converges then to x but the sequence {f (y n1 )}n∈N does not penetrate the ball B(f (x), ) and so, does not converge to f (x). Therefore claim 1) does not hold either. So, if claim 1) holds, claim 2) must be satisfied. 2) =⇒ 3): If V is a neighborhood of f (x), it contains a ball B(f (x), ) for  > 0. Claim 2) indicates then that there is a δ > 0 such that for all y ∈ B(x, δ) : f (y) ∈ B(f (x), ) ⊂ V . Therefore B(x, δ) ⊂ f −1 (V ). f −1 (V ) is thus a neighborhood of x. 3) =⇒ 1): Assume that xn → x. According to claim 3), for all neighborhood V of f (x), f −1(V ) is a neighborhood of x. The sequence {xn }n∈N must then penetrate f −1(V ). Therefore {f (xn )}n∈N penetrates V . Since this holds for all neighborhood V of f (x), we conclude that f (xn ) → f (x).

1. BASIC DEFINITIONS

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Theorem 1.4.2. The following claims are equivalent: 1) f : E → F is continuous. 2) For all open set O in F , f −1 (O) is an open set in E. 3) For all closed set C in F , f −1 (C) is a closed set in E. Proof: 1) =⇒ 2) If f is continuous, if O is an open set in F , and if x ∈ f −1 (O), then f (x) ∈ O. O is thus a neighborhood of f (x). and 3) in the last theorem indicates that f −1 (O) is a neighborhood of x. Since this holds for all x in f −1 (O), the set f −1 (O) must be open. 2) =⇒ 1): Let x be a point in E. If V is a neighborhood of f (x), there exists an  > 0 such that O := B(f (x), ) ⊂ V . O is an open set that contains f (x) and thus, according to claim 2), f −1 (O) is an open set that contains x. f −1 (O) is thus a neighborhood of x that is contained in f −1 (V ): f −1 (V ) is thus also a neighborhood of x. Since this holds for all neighborhood V of f (x), we conclude with claim 3) in last theorem that f is continuous at x. The point x being arbitrary in E, f is thus continuous on E. The equivalence 2) ⇔ 3) is obvious since f −1 (Oc ) = f −1 (O)c . Theorem 1.4.3. If f : (E, dE ) → (F, dF ) is continuous at x ∈ E, if g : (F, dF ) → (G, dG ) is continuous at f (x), then the function h : F → G : y → h(y) := g(f (y)) is continuous at x. Exercise 1.4.4. (1) Prove the last theorem. (2) A function f : (E, dE ) → (F, dF ) is Lipschitz with coefficient α iif ∀x, y ∈ E : dF (f (x), f (y)) ≤ α · dE (x, y). Prove that a Lipschitz function is continuous. (3) Let a be a point of a metric space (E,d) and let g : E → R be defined by g(x) := d(a, x). Prove that g is Lipschitz with coefficient 1 when R is endowed with the usual metric. Conclude that B[a, ] := {y ∈ E : d(a, y) ≤ } is a closed set. Prove that B[a, ] = B(a, ), when  > 0. (4) Let A be a non empty set in a metric space (E,d) and let gA : E → R be defined by gA (x) := inf a∈A d(a, x). Prove that gA is Lipschitz with coefficient 1 when R is endowed with the usual metric. Prove that A = {x ∈ E : gA (x) = 0}. (5) Prove that the following maps are continuous: g1 : (R2 , k.k∞ ) → (R, |.|) : (x, y) → g1 (x, y) := x + y g2 : (R2 , k.k∞ ) → (R, |.|) : (x, y) → g2 (x, y) := x − y g3 : (R2 , k.k∞ ) → (R, |.|) : (x, y) → g3 (x, y) := x · y With theoem 3.2.3, conclude that the sum, the difference and the product of two continuous functions are still continuous functions. 1.5. Subspaces and Product of metric spaces. Exercise 1.5.1. (subspaces) (1) Let E be a subset of a metric space (F, dF ). The restriction dE of dF at E × E is defined as dE : E × E → R+ : (x, y) → dE (x, y) := dF (x, y). (E, dE ) is clearly a metric space. If a ∈ E, we may define the open balls BF (a, ) and BE (a, ) with center a and radius  in the metric spaces

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(F, dF ) and (E, dE ). We clearly have BE (a, ) = BF (a, ) ∩ E. Prove that O is an open set of (E, dE ) iif there is an open set O0 of (F, dF ) such that O = O0 ∩ E. (2) Prove that G is an closed set of (E, dE ) iif there is an closed set G0 of (F, dF ) such that G = G0 ∩ E. (3) Find an example of an E-valued sequence {xn }n∈N converging in (F, dF ) that is not converging in (E, dE ). Let (E i , di )i=1,...,m be a family of metric spaces. The cartesian product E := E 1 × · · · × E m can be endowed with the following metric d: if x := (x1 , . . . , xm ) ∈ E and y := (y 1 , . . . , y m ) ∈ E, d(x, y) := maxi=1,...,m di (xi , y i ). The space (E, d) will be referred to as the product space. Notice that if Rk is considered as the product of k copies of R with the usual topology, the corresponding metric d coincides with the metric induced by the norm k.k∞ . We will denote pri the projection pri : E → E i : x := (x1 , . . . , xm ) → pri (x) := xi .

Theorem 1.5.2. 1) A sequence {xn }n∈N is converging to a in (E, d) iif, for all i, the sequence {pri (xn )}n∈N is converging to pri (a) in (E i , di ). 2) A function f : (F, dF ) → (E, d) is continuous iif, for all i, the function f i : F → E i : y → f i (y) := pri (f (y)) is continuous. Exercise 1.5.3. (1) Prove the last theorem. (2) Let (F, dF ) be a metric space, and let E be the product (F × F ) endowed with the metric d as above. Prove that h : E → R : (x, y) → h(x, y) := dF (x, y) is Lipschitz from (E, d) to R with the usual metric. Prove that if f, g are continuous maps from (G, dG ) to (F, dF ), then the set A := {x ∈ G : f (x) = g(x)} is a closed set. (Hint: consider s : G → R : x → dF (f (x), g(x)). Prove that s is continuous and express A in terms of s.) Exercise 1.5.4. (Topology) (1) The notion of pointwise convergence for functions that will be introduced in a later section can not be described as the convergence in a metric space. The aim of this exercise is to define notion of convergence in a broader context than metric spaces. Let E be a set of point. A family O of subsets of E is called a topology if a) ∅ and E are in O. b) Any (infinite) union of elements in O is still in O. c) Any finite intersection of elements in O is still in O. The space (E, O) is then called a topological space. Prove that the family U of opens sets in a metric space (E, d) is a topology on E. (2) In a topological space (E, O), O will thus be interpreted as the class of open sets. Closed sets are thus defined as the complementary of an open set. A neighborhood V of a point x ∈ E is a subset V of E such that there exists O ∈ O satisfying x ∈ O ⊂ V . With this definition, prove that a subset U of E belongs to O iif it is a neighborhood of all of its points.

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(3) With the above definition of neighborhood, the notion of converging sequence in a topological space (E, O) is the same as in a metric space. The notions of interior, closure and adherence of a subset A of E are also defined in the same way. On R define O1 as the class of all subsets of R and O2 := {∅, R}. O1 and O2 are two topologies on R. What are the convergent sequences for these topologies? If A := [0, 1[ find A◦ Cl(A) and A for both topologies. 2. Completeness 2.1. Complete spaces. A Cauchy sequence in a metric space (E, d) is a sequence {xn }n∈N such that ∀η > 0 : ∃N : ∀k, l ≥ N : d(xk , xl ) < η. Theorem 2.1.1. A convergent sequence is a Cauchy sequence. Proof: Let {xn }n∈N be a sequence converging to a. For all η > 0, the number  := η/2 is strictly positive. According to the definition of a convergent sequence, there exists thus a N such that ∀n ≥ N : d(xn , a) < . Therefore, ∀k, l ≥ N , we have with the Triangle inequality: d(xk , xl ) ≤ d(xk , a) + d(xl , a) <  +  ≤ η. Exercise 2.1.2. (1) Prove that if a Cauchy sequence has an accumulation point a (see exercise 1.3.2-(3)), it converges to a. (2) A sequence {xn }n∈N in a metric space (E, d) is bounded if it is valued in a ball B(a, r) of finite radius r. Prove that a Cauchy sequence is bounded. As shows the following example, there are non converging Cauchy sequences: Consider E := R − {0} as a subspace of R with the usual metric d. The sequence 1 { n+1 }n∈N is a sequence valued in E. As as sequence in (R, d) it converges to 0. It is thus a Cauchy sequence. However, as a sequence in (E, dE ), it has no limit in E and is thus not converging. A metric space (E, d) is complete iif it has the property that all Cauchy sequence in (E, d) is converging. Theorem 2.1.3. R with the usual metric is a complete space. Proof: First observe that a Cauchy sequence {xn }n∈N in R is bounded. In the axiomatic defining R, the completeness axiom states that any set A in R bounded from above has a supremum (least upper bound): sup(A). This axiom allows the definition of lim sup xn for any bounded from above sequence {xn }n∈N . As seen in exercise 1.3.2-(3); lim sup xn is an accumulation point of the sequence. With exercise 2.1.2-(1), we conclude that {xn }n∈N converges to lim sup xn . Theorem 2.1.4. The product (E, d) space of finitely many complete metric spaces (E 1 , d1 ), . . . , (E k , dk ) is a complete space. Proof: Let {xn }n∈N be a Cauchy sequence in E. It results from the definition of the product metric d that di (pri (xk ), pri (xl )) ≤ d(xk , xl ). Since {xn }n∈N is a Cauchy sequence, we infer that, ∀i, {pri (xn )}n∈N is a Cauchy sequence in (E i , di ). Since this space is complete, the sequence {pri (xn )}n∈N has thus a limit ai in E i . Let a

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be the point a = (a1 , . . . , ak ) ∈ E. Since, ∀i, {pri (xn )}n∈N converges to pri (a) = ai , we conclude with theorem 1.5.2-1) that {xn }n∈N is converging in (E, d) to a. A Banach space is a complete normed vector space. Since (Rk , k.k∞ ) is the product space of k copies of the complete metric space (R, |.|), the previous theorem indicates that (Rk , k.k∞ ) is complete and thus Theorem 2.1.5. (Rk , k.k∞ ) is a Banach space. Theorem 2.1.6. (B(E), k.k∞ ) (see example 1.1.1-(7)) is a Banach space. Proof: Let {fn }n∈N be a Cauchy sequence in (B(E), k.k∞ ). Since, ∀x ∈ E, ∀k, l, |fk (x) − fl (x)| ≤ kfk − fl k∞ , we get that ∀x ∈ E, {fn (x)}n∈N is a Cauchy Sequence in the complete space (R, |.|). The sequence {fn (x)}n∈N is thus converging to a limit that will be denoted f (x). For all  > 0, there exists N such that:∀k, l ≥ N : kfk − fl k∞ ≤ . Therefore ∀k, l ≥ N, ∀x ∈ E, |fk (x) − fl (x)| ≤ . Since f (x) = liml→∞ fl (x), we also get that ∀k ≥ N : ∀x ∈ E : |fk (x) − f (x)| ≤ . Therefore ∀k ≥ N : kfk − fl k∞ ≤ . Theorem 2.1.7. If F is a closed set in a complete metric space (E, d), then the subspace (F, d) is also complete. Proof: Let {xn }n∈N be a Cauchy sequence in F . It is in particular a Cauchy sequence in the complete space (E, d) and has thus a limit a in E. Since a is the limit of a sequence {xn }n∈N valued in the closed set F , a must thus belong to F , and the sequence {xn }n∈N is thus converging in (F, d). If (E, d) is a metric space, we will denote C(E) the space of continuous functions from (E, d) to (R, |.|). It results from exercise 1.4.4-(5) that C(E) is a vector space. We also set Cb (E) := C(E) ∩ B(E): Cb (E) is the space of bounded continuous functions on E. Theorem 2.1.8. (Cb (E), k.k∞ ) is a Banach space. Proof: According to the previous theorem, we just have to prove that Cb (E) is a closed set in (B(E), k.k∞ ). With theorem 1.3.3, we just have to prove that the limit f in (B(E), k.k∞ ) of any sequence {fn }n∈N in Cb (E) is continuous. Let x be a point of E and let  be strictly positive. Since {fn }n∈N converges to f , we get: ∃N : ∀n ≥ N : kf − fn k∞ ≤ /3. Since fN is continuous at x, there exists a δ > 0 such that ∀y ∈ B(x, δ) : |fN (x) − fN (y)| ≤ /3. Therefore ∀y ∈ B(x, δ) : |f (x)−f (y)| ≤ |f (x)−fN (x)|+|fN (x)−fN (y)|+|fN (y)−f (y)| ≤ /3+/3+/3 = . Since  > 0 is arbitrary in the above reasoning, we have proved that ∀ > 0 : ∃δ > 0 : ∀y ∈ B(x, δ) : |f (x) − f (y)| ≤ . and f is thus continuous at x. Since x is arbitrary in E, we conclude as announced that f is continuous. 2.2. Contraction and fixed points. A contraction f on a metric space (E, d) is a map f : E → E that satisfy for some α < 1: ∀x, y ∈ E : d(f (x), f (y)) ≤ α · d(x, y). A point x∗ ∈ E is called a fixed point for a map f : E → E iif x∗ = f (x∗ ).

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Theorem 2.2.1. A contraction f on a complete metric space has a unique fixed point x∗ . Furthermore, ∀x0 ∈ E, the sequence obtained recursively by the formula xn+1 := f (xn ) converges to x∗ : ∀n : d(x∗ , xn ) ≤

αn d(x0 , f (x0 )). 1−α

Proof: For a given x0 , let us consider the corresponding sequence {xn }n∈N . Since (1)

d(xn+1 , xn ) ≤ α · d(xn , xn−1 ) ≤ · · · ≤ αn · d(x1 , x0 ),

we get with the triangle inequality ∀n, m, m ≥ n : d(xm , xn ) (2)

≤ d(xm , xm−1 ) + d(xm−1 , xm−2 ) + · · · + d(xn+1 , xn ) ≤ d(x1 , x0 ) · (αm−1 + αm−2 + · · · αn ) αn ≤ d(x1 , x0 ) · 1−α

Since αn → 0, we get at once that {xn }n∈N is a Cauchy sequence. Since (E, d) is complete, this sequence converges thus to a limit x∗ . Since inequality (2) holds αn as announced. for all m > n, we get letting m go to ∞: d(x∗ , xn ) ≤ d(x1 , x0 ) · 1−α ∗ We next prove that x is a fixed point for f : inequality (1) indicates that limn→∞ d(f (xn ), xn ) = 0. Since f is continuous, this limit is just d(f (x∗ ), x∗ ). Therefore f (x∗ ) = x∗ as announced. Let y be a fixed point. We prove now that it must coincide with x∗ . Indeed, we have then d(x∗ , y) = d(f (x∗ ), f (y)) ≤ α · d(x∗ , y). Since d(x∗ , y) ≥ 0 and α < 1, this inequality can only be satisfied if d(x∗ , y) = 0, which entails x∗ = y. The previous theorem is quite powerful. It is used for instance to prove that differential equation have a solution: Let F : R → R be a Lipschitz function with Lipschitz constant α. Let y0 ∈ R. The next exercise aim to prove the existence of a continuous solution y to the following differential problem:  d y(t) = F (y(t)) ∀t > 0 : dt y(0) = y0 Exercise 2.2.2. (1) For a function f : [0, ∞[→ R, define []f [] as []f [] := sup{|e−2αt f (t)| : t ∈ [0, ∞[}. Let V be the set of continuous function f : [0, ∞[→ R satisfying []f [] < ∞. Prove that (V, [].[]) is a Banach space. (2) For f ∈ V let us define the function T (f ) : [0, ∞[→ R by Z t ∀t ∈ [0, ∞[: T (f )(t) := y0 + F (f (s))ds. 0

Prove that T (0) ∈ V. Using the fact that F is α-Lipschitz and the definition of [].[], prove that 2αt ∀f, g ∈ V, ∀t ∈ [0, ∞[: |T (f )(t) − T (g)(t)| ≤ e 2 []f − g[]. Conclude that ∀f ∈ V : T (f ) ∈ V. Prove that T is a contraction on V (3) Prove that the fixed point of T is the unique solution of the above differential problem.

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2.3. Uniform continuity. Let (E, dE ) and (F, dF ) be two metric spaces. As it results from the definition of continuity, if {xn }n∈N is a converging sequence in (E, dE ), the image {f (xn )}n∈N of this sequence by a continuous function f : E → F is still convergent. It is thus a natural question to rise wether the image of a Cauchy sequence by a continuous function is still a Cauchy sequence. The answer is no as indicates the following example: The function f : (]0, 1], |.|) → (R, |.|) : x → f (x) := 1 sin( x1 ) is continuous and the sequence { n+1 }n∈N is clearly a Cauchy sequence. Its image by f is the sequence {sin(n + 1)}n∈N that keeps oscillating between − 1 and 1 and is thus neither convergent nor a Cauchy sequence. To preserve the Cauchy property, the function f must satisfy a stronger requirement than continuity: uniform continuity. A function f : (E, dE ) → (F, dF ) is uniformly continuous iif ∀ > 0 : ∃δ > 0 : ∀x, y ∈ E : dE (x, y) ≤ δ =⇒ dF (f (x), f (y)) < . It is useful to compare this definition with that of continuity: The function f is continuous iif ∀x ∈ E, ∀ > 0 : ∃δ > 0 : ∀y ∈ E : dE (x, y) ≤ δ =⇒ dF (f (x), f (y)) < . In the definition of continuity, the δ that is claimed to exist depends on  and x. In the definition of uniform continuity however, δ only depends on : it is a “uniform” δ, the same for all x. Observe that a uniformly continuous function is in particular continuous. Exercise 2.3.1. (1) Prove that a Lipschitz function f : (E, dE ) → (F, dF ) is uniformly continuous. (2) Prove that the image of a Cauchy sequence by a uniformly continuous function is afresh a Cauchy sequence. Theorem 2.3.2. Let A be a subset of a metric space (E, d) and the A be its adherence. If f is a uniformly continuous function from the subspace (A, d) to a complete metric space (F, dF ), then there exists a unique continuous function f : (A, d) → (F, dF ) such that ∀x ∈ A : f (x) = f (x). f is called the extension by continuity of f . It is furthermore uniformly continuous. Proof: Let us first prove the uniqueness of such an extension: If g and h are two continuous function from A to F that coincide with f on A, then the set Q := {x ∈ A|g(x) = h(x)} contains A. Q is a closed set in (A, d) as it results from exercise 1.5.3-(2). Therefore A ⊂ Q, and thus g and h coincide on A. Let us next prove the existence of such an extension: Let a be a point of A. According to exercise 1.3.5, there exists a A-valued sequence {xn }n∈N that converges to a. It is in particular a Cauchy sequence, and so is its image by f . Since (F, dF ) is a complete space, we conclude that {f (xn )}n∈N is convergent. Now, observe that if {yn }n∈N is another A-valued sequence that converges to a, then limn→∞ f (xn ) = limn→∞ f (yn ). Indeed, the sequence {xn }n∈N defined by z2n := xn and z2n+1 := yn is still a sequence converging to a. Therefore {f (zn )}n∈N is converging, and all its subsequences— in particular {f (xn )}n∈N and {f (yn )}n∈N — have the same limit. Let us then denote f (a) the common limit of the image by f of the A-valued sequences converging to a. We now argue that f coincide with f on A. Indeed, if a ∈ A, the constant sequence {xn }n∈N , where ∀n : xn := a, is A-valued and converges to a. Therefore f (a) = limn→∞ f (xn ) = f (a).

2. COMPLETENESS

15

We now prove that f is uniformly continuous: Let  be a strictly positive number. Since f is uniformly continuous, there exists δ > 0 such that ∀x, y ∈ A satisfying d(x, y) < δ, we have dF (f (x), f (y)) < . Let then a, a0 be two points of A satisfying d(a, a0 ) < δ. Let {xn }n∈N and 0 {xn }n∈N be two A-valued sequences converging respectively to a and a0 . Since limn→∞ d(xn , x0 n) = d(a, a0 ) < δ, we infer that there exists N such that ∀n ≥ N : d(xn , x0 n) < δ. Therefore ∀n ≥ N : dF (f (xn ), f (x0n )) ≤ . Since {f (xn )}n∈N and {f (x0n )}n∈N converge respectively to f (a) and f (a0 ), we conclude that dF (f (a), f (a0 )) = lim dF (f (xn ), f (x0n )) ≤ . n→∞

We thus have proved that ∀ > 0 : ∃δ > 0 : ∀a, a0 ∈ A : d(a, a0 ) < δ =⇒ dF (f (a), f (a0 )) ≤  f is thus uniformly continuous, as announced. This last theorem is quite important in mathematics. Here are two examples of application: Example 2.3.3. (1) How is the function 2x defined on R? An integer power of a number has an easy definition in terms of repeated product of that number. So 2x is well defined for x ∈ Z. This definition can be extended to x ∈ Q as follows: if x = nk , 2x is the unique number y such that y n = 2k . So 2x is well defined as a function from Q to R. The function 2x : R → R is then defined as the extention by continuity of the previous one. (2) The definition of the Riemann integral is another example of extension Rb by continuity: The integral a f (x)dx is first defined on the space of step Rb functions f . The map f → a f (x)dx is then proved to be uniformly continuous with respect to the uniform norm. This allows to extend the definition of the integral to the adherence of the set of step functions. 2.4. Cauchy completion. Theorem 2.4.1. Any metric space (E, d) is a dense subspace of a complete metric space (F, d). The space (F, d) is referred to as the Cauchy completion of (E, d). Proof: Let a0 be a point in E. For a ∈ E, let us define ga : E → R : x → d(a, x) and ha := ga −ga0 . ga and ha are thus continuous functions. The triangle inequality indicates that ∀x ∈ E : d(x, a) ≤ d(x, b) + d(a, b) and d(x, b) ≤ d(x, a) + d(a, b). This leads to |d(x, a) − d(x, b)| ≤ d(a, b), and thus ∀x ∈ E : |ga (x) − gb (x)| ≤ d(a, b). Since when x = b, |ga (x) − gb (x)| = d(a, b), we conclude that d(a, b) = kga − gb k∞ . In particular, with b = a0 , we conclude that ha ∈ Cb (E). Let us then consider the map H : E → Cb (E) : a → H(a) := ha . H is in fact isometric between (E, d) and (Cb (E), k.k∞ ). Indeed: ∀a, b ∈ E : kH(a) − H(b)k∞ = kha − hb k∞ = kga − gb k∞ = d(a, b). Let K denote K := {H(a) : a ∈ E}. Since H is isometric, it is a one to one map between E and K: If a, b ∈ E are such that H(a) = H(b), then d(a, b) = kH(a) − H(b)k∞ = 0 and thus a = b. Therefore (E, d) can be identified with (K, k.k∞ ).

16

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Let K be the adherence of K in Cb (E). As a closed subspace of the complete space (Cb (E), k.k∞ ), (K, k.k∞ ) is itself a complete space (theorem 2.1.7). The space (E, d) identified with (K, k.k∞ ) is thus a dense subspace of the complete space (K, k.k∞ ), and the theorem is proved. Exercise 2.4.2. The space L1 ([0, 1]) R1 (1) On C([0, 1]) define k.k1 by kf k1 := 0 |f (x)|dx. Prove that k.k1 is a norm on C([0, 1]). (2) Prove that (C([0, 1]), k.k1 ) is not a complete space. R1 (3) Prove that the map I : C([0, 1]) → R : f → I(f ) := 0 f (x)dx is uniformly continuous from (C([0, 1]), k.k1 ) to (R, |.|). (4) Let (L1 ([0, 1]), k.k1 ) denote the completion of (C([0, 1]), k.k1 ). Prove that the map I can be extended by continuity to (L1 ([0, 1]), k.k1 ). This define the Lebesgue integral on L1 ([0, 1]). Remark: The space L1 ([0, 1]) is introduced above in an abstract way, but any element in L1 ([0, 1]) can be identified with an (equivalence class) of measurable functions since from any Cauchy sequence of continuous function, one can extract a subsequence that converges pointwise on a set of Lebesgue measure 1. 3. Compactness 3.1. Compact precompact and sequentially compact sets. A subset K of a metric space (E, d) is compact iif any open covering of K (i.e. a family {Oi }i∈I of open sets such that K ⊂ ∪i∈I Oi ) contains a finite subcovering (i.e. a finite set J ⊂ I such that K ⊂ ∪i∈J Oi ). Exercise 3.1.1. (1) Prove that K is compact iif for all family {Fi }i∈I of closed sets such that ∩i∈I Fi ∩K = ∅ there is a finite sub-family J ⊂ I such that ∩i∈J Fi ∩K = ∅. A subset K of E is sequentially compact iif all K-valued sequence has an accumulation point in K. Theorem 3.1.2. In a metric space (E, d), a compact set is sequentially compact. Proof: Let K be a compact set and let {xn }n∈N be a K-valued sequence. We want to prove that {xn }n∈N has an accumulation point in K. Assume on the contrary that no point x in K is an accumulation point of {xn }n∈N . Then, according to exercise 1.3.2-(3), ∀x ∈ K, there exists (x) > 0 and N (x) ∈ N such that ∀n > N (x) : xn 6∈ B(x, (x)). Since x ∈ B(x, (x)), we get that K ⊂ ∪x∈K B(b, ). Therefore {B(b, )}x∈K is an open covering of K. There exists thus a finite set J ∈ K such that K ⊂ ∪x∈J B(b, ). But this is impossible: Indeed, if we set M := max{N (x), x ∈ J}, then M < ∞ and ∀n > M : xn 6∈ ∪x∈J B(x, (x)) ⊃ K, which contradicts the fact that {xn }n∈N is K-valued. Theorem 3.1.3. If K is a sequentially compact set in a metric space (E, d), then the metric subspace (K, d) is complete.

3. COMPACTNESS

17

Proof: Any Cauchy sequence in (K, d) has an accumulation point in K. It is therefore converging to this accumulation point according to exercise 2.1.2-(1). A subset K of a metric space (E, d) is precompact iif, for all  > 0, K has a finite covering by open balls of radius . (i.e. ∀ > 0 : ∃N ∈ N : ∃x1 , . . . , xN ∈ E : K ⊂ ∪N i=1 B(xi , ). ) Exercise 3.1.4. (1) Prove that K is precompact iif ∀ > 0 : ∃N ∈ N : ∃x1 , . . . , xN ∈ K : K ⊂ ∪N i=1 B(xi , ). In other words, the centers of the balls can be chosen in K itself. (2) Prove that [0, 1] is a precompact set of (R, |.|) Theorem 3.1.5. A sequentially compact set is precompact Proof: Let indeed K be a sequentially compact set. Would there be no finite covering of K by open balls of radius  > 0, we could construct recursively a sequence {xn }n∈N as follows. Pick x0 ∈ K. Once x0 , . . . , xn have been chosen, observe that K − ∪nj=0 B(xj , ) is a non empty set. Select then xn+1 in this set. Clearly ∀m 6= n : d(xn , xm ) ≥ . Therefore, the sequence {xn }n∈N could’nt have a converging sub-sequence, which would contradict the sequential compactness of K.

Theorem 3.1.6. Let f be continuous function from a compact set K to R. Then the set {f (x) : x ∈ K} is bounded and f reaches its supremum and infimum on K: ∃a∗ , a∗ ∈ K : f (a∗ ) = sup{f (x) : x ∈ K} and f (a∗ ) = inf{f (x) : x ∈ K}. Proof: Let us first prove that the set {f (x) : x ∈ K} is bounded from above. If it were not the case, then for all n ∈ N, there would exist xn ∈ K such that f (xn ) ≥ n. But the sequence {xn }n∈N would contain a convergent subsequence {x0n }n∈N with limit x in K. Due to the continuity of f at x, the sequence {f (x0n )}n∈N would thus converge to f (x). This is a contradiction since {f (x0n )}n∈N is an unbounded sequence. Therefore {f (x) : x ∈ K} must be bounded from above. Let then α denote the least upper bound of this set: α := sup{f (x) : x ∈ K}. Then ∀n ∈ N : α − n1 is not an upper bound of this set: ∃xn ∈ K : α ≥ f (xn ) > α − n1 . The sequence {f (xn )}n∈N converges thus to α. Since K is compact the sequence {xn }n∈N contains a convergent subsequence {x0n }n∈N with limit a∗ in K. Due to the continuity of f at a∗ , the sequence {f (x0n )}n∈N would thus converge to f (a∗ ). Since {f (x0n )}n∈N is a subsequence of a sequence {f (xn )}n∈N that converges to α, we also have that limn→∞ f (x0n ) = α and thus α = f (a∗ ). The prove is similar for the infimum. Theorem 3.1.7. Let K be a subset of a metric space (E, d). Then the following claims are equivalent: (1) K is a compact set. (2) K is a sequentially compact set. (3) K is precompact and the subspace (K, d) is complete. Proof: (1) =⇒ (2) was poven in theorem 3.1.2. Theorems 3.1.3 and 3.1.5 indicate that (2) =⇒ (3). A: Let us now prove that (3) =⇒ (2). Let K satisfy (3). For all m ≥ 1 m there exists a finite covering Cm := {B1m , . . . , BN } of K by open balls of radius m

18

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1 m+1 .

Let then {x0n }n∈N be a K-valued sequence. We will prove that it has an accumulation point in K. To do so, we will extract recursively subsequences of {x0n }n∈N . Once selected the subsequence {xm−1 }n∈N , the subsequence {xm n n }n∈N is selected as follows: Since m−1 ∀n : xn belongs to K, there must exists at least one ball Bjmm in Cm that is visited infinitely many times by {xm−1 }n∈N . In other words Am := {n ∈ N : n > n m } is not a finite set. Let then nm (k) denotes the k-th smallest 0 and xm−1 ∈ B n jm m−1 element in Am . We then define xm k := xnm (k) . m Next define ym := x0 . We claim that {yn }n∈N is a subsequence of {x0n }n∈N : m−1 m−2 0 indeed ym := xm 0 = xnm (0) = xnm−1 (nm (0)) = · · · = xn(m) , where n(m) := n1 (n2 (· · · (nm (0)) · · · )). So, we just have to prove that n(m) is strictly increasing in m. Observing that nm+1 (0) > 0 since 0 6∈ Am+1 , and that k → nq (k) is strictly increasing, we get successively: nm (nm+1 (0)) > nm (0), nm−1 (nm (nm+1 (0))) > nm−1 (nm (0)), · · · and thus n(m + 1) = n1 (n2 (· · · (nm (nm+1 (0))) · · · )) > n1 (n2 (· · · (nm (0)) · · · )) = n(m). n(m) is thus strictly increasing in m. We next claim that {yn }n∈N is a Cauchy sequence in (K, d). Indeed, for all  > 0, there exists N such that N2+1 < . For all n, m ≥ N , yn and ym are elements N of the subsequence {xN k }k∈N . Therefore, they both belong to the ball BjN of radius 1 2 N +1 . We infer therefore that d(ym , yn ) ≤ N +1 <  and {yn }n∈N is, as announced, a Cauchy sequence in (K, d). Since (K, d) is complete, we conclude that {yn }n∈N converges to a point y in K which is thus an accumulation point of {x0n }n∈N . B: We next prove that (2) =⇒ (1): Let K satisfy (2). Possibly by replacing d d by the topologically equivalent distance δ := 1+d , we may assume without loss of generality that d is a bounded metric on E. Let {Oi }i∈I be an open covering of K. We must prove that it contains a finite subcovering. For all i ∈ I, we may assume that E 6= Oi , otherwise {Oi } would be the required finite subcovering. We may thus define gi : E → R : x → gi (x) := inf{d(x, a) : a 6∈ Oi } which is a finite Lipschitz function with constant 1 as it follows from exercise 1.4.4-(4). Since Oi is open, this exercise also indicates that Oi = {x ∈ E : gi (x) > 0}. Let us then define g(x) := supi∈I gi (x) (since d is bounded, g(x) is a finite number). g is also a Lipschitz function with constant 1: Indeed, ∀x, y ∈ E : ∀ > 0 : ∃i ∈ I : g(x) ≤ gi (x) +  ≤ gi (y) + d(x, y) +  ≤ g(y) + d(x, y) + . Since this holds for all  > 0, we get: ∀x, y ∈ E : g(x) ≤ g(y) + d(x, y). Interchanging x and y, we obtain ∀x, y ∈ E : g(y) ≤ g(x) + d(x, y), and so: ∀x, y ∈ E : |g(x) − g(y)| ≤ d(x, y). Let next α denote α := inf{g(x) : x ∈ K}. It results from theorem 3.1.6 there exists a∗ ∈ K such that g(a∗ ) = α. But {Oi }i∈I is a covering of K. Therefore, there exists i ∈ I such that a∗ ∈ Oi . So α = g(a∗ ) ≥ gi (a∗ ) > 0. Since (2) ⇒ (3), K is precompact. According to exercise 3.1.4-(1), there exists thus a finite covering {B(x1 , α2 ), · · · , B(xN , α2 )} of K by balls of radius α2 such that x1 , . . . , xN ∈ K. From the definition of α, we get thus ∀k : g(xk ) ≥ α > α2 , and from the definition of g, we thus conclude that ∀k : ∃ik ∈ I : gik (xk ) > α2 . So, since gik is

3. COMPACTNESS

19

Lipschitz of constant 1, we get ∀x ∈ B(xk , α2 ): α α − = 0. 2 2 Therefore, ∀k : B(xk , α2 ) ⊂ Oik . Since {B(x1 , α2 ), · · · , B(xN , α2 )} covers K, so does {Oi1 , . . . , OiN } and we thus have found a finite subcovering in {Oi }i∈I , as announced. gik (x) ≥ gik (xk ) − d(x, xk ) >

3.2. Properties of compact sets. Theorem 3.2.1. Let f be a continuous map from (E, dE ) to (F, dF ). If K is a compact set in (E, dE ) then f (K) is a compact set in (F, dF ). Proof: If {Oi }i∈I is an open covering of f (K), then {f −1 (Oi )}i∈I is a family of open set in (E, d) as it follows from theorem 1.4.2. It is an open covering of K since ∀x ∈ K there exists i ∈ I such that f (x) ∈ Oi and thus x ∈ f −1 (Oi ). Since K is compact, there exists a finite set J ⊂ I such that K ⊂ ∪i∈J f −1 (Oi ). Since f (f −1 (Oi )) ⊂ Oi , we conclude that f (K) ⊂ ∪i∈J Oi and {Oi }i∈J is thus a finite subcovering of K. Theorem 3.2.2. If K in a compact set in a metric space (E, d), then K is a closed bounded set. Proof: Let us first prove that K is a closed set. According to theorem 1.3.3, we just have to prove that the limit in (E, d) of a K-valued converging sequence {xn }n∈N necessarily belongs to K. Since K is compact, the sequence {xn }n∈N has an accumulation point b ∈ K. Since this sequence is further converging, we clearly have b = limn→∞ xn ∈ K as announced. We next prove that K is bounded, i.e. contained in an open ball with finite radius. Let a be a point in E, and as in exercise 1.4.4-(3), let us consider the Lipschitz function g : x → g(x) := d(a, x). From theorem 3.1.6, the set g(K) is bounded from above: ∃M : ∀x ∈ K : g(x) ≤ M < M + 1. This implies that K ⊂ B(a, M + 1). Theorem 3.2.3. In a metric space (E, d), a closed subset A of a compact set K is a compact set. Proof: Let C := {Oi }i∈I be an open covering of A. Clearly C 0 := C ∪ {Ac } is an open covering of K. C 0 contains thus a finite covering of K. There exists thus a finite set J ⊂ I such that K ⊂ ∪i∈J Oi ∪Ac . Since A ⊂ K, we get then A ⊂ ∪i∈J Oi . So {Oi }i∈I is the required finite subcovering of A. Theorem 3.2.4. Let (E i , di )i=1,...,m be a family of metric spaces and, for all i, let Ki be a compact subset of Ei . Then the set K := K1 × · · · Km is compact in the product space (E, d) introduced in section 1.5. Proof: Let {xn }n∈N be a K-valued sequence. The sequence {pr1 (xn )}n∈N is thus K1 -valued. Since K1 is compact, there exists a subsequence {x1n }n∈N of {xn }n∈N such that {pr1 (x1n )}n∈N converges to x1 ∈ K1 . The sequence {pr2 (x1n )}n∈N is K2 -valued, there exists thus a subsequence {x2n }n∈N of {x1n }n∈N such that {pr2 (x2n )}n∈N converges to x2 ∈ E2 .

20

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Repeating this procedure m times, we finally obtain a subsequence {xm n }n∈N i m such that ∀i, {pri (xm )} converges to x . This indicates that {x } converges n∈N n∈N n n in the product space (E, d) to x = (x1 , · · · , xm ) ∈ K. The sequence {xn }n∈N has thus an accumulation point x in K as we wanted to prove. Exercise 3.2.5. (1) Theorem 3.2.2 indicates that, in any metric space, compact sets are closed and bounded. The next theorem states that in (Rn , k.k∞ ), any closed and bounded set is compact. The aim of this exercise is to prove that in other metric space, there are bounded closed sets that are not compact. Let fn be defined as  1 if x ≥ n fn : R → R : x → fn (x) := 0 otherwise Let A be the set A := {fn : n ∈ N}. It is clearly a bounded subset of (B(R), k.k∞ ). Observe that if n 6= m, then kfn − fm k∞ = 1. Conclude therefore that a A-valued Cauchy sequence must be constant after some rank (i.e. it must penetrate a singleton {f } ⊂ A.) Use this fact to prove that A is closed and not compact. Theorem 3.2.6. In (Rn , k.k∞ ), a set K is compact iif it is bounded and closed. Proof: Theorem 3.2.2 indicates that a compact set is necessarily bounded and closed. We next prove that this is also a sufficient condition for compactness: Let then K be a closed bounded set in (Rn , k.k∞ ). Since K is bounded, it is contains in a closed ball B[a, r]. We just have to prove that the set B[a, r] is compact. Indeed, theorem 3.2.3 indicates that a closed subset of a compact set is itself compact. Since (Rn , k.k∞ ) is the product space of n copies of (R, |.|), and B[a, r] = [a1 − r, a1 + r] × · · · × [an − r, an + r], we just have to prove that bounded closed intervals are compact in (R, |.|) to get with theorem 3.2.4 the compactness of B[a, r]. Since a bounded closed interval I can always be covered by finitely many open intervals of length 2 > 0 (which are the open balls of radius ), I is clearly a precompact set. As a closed subset of the complete space (R, |.|), (I, |.|) is itself a complete space according to theorem 2.1.7. We thus conclude with theorem 3.1.7 that I is compact in (R, |.|). One important feature of compact sets is that a ”local property” on this set is often a ”uniform property”. The next theorem is an illustration of this. Theorem 3.2.7. Let K be a compact set in (E, d). Then a continuous function f : (K, d) → (F, dF ) is uniformly continuous. Proof: Let f be a continuous function on a compact set K. For a given  > 0, for all x ∈ K, there exists δ(x) > 0 such that  (3) ∀y ∈ B(x, δ(x)) : dF (f (x), f (y)) ≤ . 3 Let us then consider the open covering {B(x, δ(x) 3 )}x∈K of K. There exists thus a δ(x) finite set J ⊂ K such that K ⊂ ∪x∈J B(x, 3 ). Let δ denote δ := minx∈J δ(x).

4. CONNECTED SETS

Then η :=

δ 3

21

> 0. For all x, y ∈ K such that d(x, y) ≤ η there exists x0 , y 0 ∈ J such 0

0

that x ∈ B(x0 , δ(x3 ) ) and y ∈ B(y 0 , δ(y3 ) ). With (3), we have thus dF (f (x), f (x0 )) ≤   0 3 and dF (f (y), f (y )) ≤ 3 . Next, the triangle identity gives us δ(x0 ) + δ + δ(y 0 ) ≤ max(δ(x0 ), δ(y 0 )). 3 Therefore either x0 ∈ B(y 0 , δ(y 0 )) or y 0 ∈ B(x0 , δ(x0 )). In any case, with inequality (3), we get: dF (f (x0 ), f (y 0 )) ≤ 3 . So we have proved that ∀ > 0 : ∃η > 0 : ∀x, y ∈ K such that d(x, y) ≤ η, we have: ++ = . dF (f (x), f (y)) ≤ dF (f (x), f (x0 ))+dF (f (x0 ), f (y 0 ))+dF (f (y 0 ), f (y)) ≤ 3 f is thus uniformly continuous on K. d(x0 , y 0 ) ≤ d(x0 , x) + d(x, y) + d(y, y 0 ) ≤

Theorem 3.2.8. Let f be a continuous one-to-one map from a compact (E, dE ) to (F, dF ). Then the inverse map of f is continuous from (F, dF ) to (E, dE ) Proof: Let g denote the inverse map of f : for all x ∈ E, we have thus g(f (x)) = x. To prove that g is continuous, we just have to prove that for all closed set C in (E, dE ), the set g −1 (C) is closed in (F, dF ) (see theorem 1.4.2). But g −1 (C) = {y ∈ F : g(y) ∈ C} = {f (x) : x ∈ C} = f (C). Next, C is a compact set as a closed subset of the compact set E (see theorem 3.2.3). Therefore g −1 (C) = f (C) is also compact (theorem 3.2.1) and thus closed (theorem 3.2.2), as required. 4. Connected sets 4.1. Basic definitions. Let E1 , E2 be two non-empty disjoint closed sets in a metric space (E, d) such that E = E1 ∪ E2 . Since E1 , E2 are also open sets, a sequence in E will converge iif it penetrates one of these sets Ei and if the corresponding Ei -valued subsequence converges in (Ei , d). In the same way, a function f : (E, d) → (F, dF ) will be continuous iif the restrictions fi of f on Ei are continuous from (Ei , d) to (F, dF ). Therefore, analyzing the properties of the space (E, d) comes essentially to analyzing the properties of its two pieces (E1 , d), (E2 , d). A space (E, d) that can not be separated in two such pieces is called connected. More precisely, a metric space (E, d) is connected iif the only sets that are both open and closed in (E, d) are E and ∅. Theorem 4.1.1. ([0, 1], |.|) is a connected space. Proof: Assume that [0, 1] = E1 ∪ E2 where E1 , E2 are two non-empty disjoint closed sets in ([0, 1], |.|). We will prove that this leads to a contradiction: Indeed, the point 1 must then belong to one of the Ei ’s— say 1 ∈ E2 . Define then α as α := sup E1 . Since E1 is closed, we get α ∈ E1 . Since E1 and E2 are disjoint, we get thus that α 6∈ E2 and thus α < 1. It also follows the definition of α that no point x > α belongs to E1 , and thus ]α, 1] ⊂ E2 . Since E2 is closed, α must then belong to E2 . This is the contradiction: α ∈ E1 ∩ E2 , and E1 and E2 can therefore not be disjoint.

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1. METRIC SPACES

A subset A in a metric space (E, d) is connected iif the subspace (A, d) is connected. Theorem 4.1.2. A subset A in a metric space (E, d) is connected iif for all closed sets F1 , F2 in (E, d), A ⊂ F1 ∪F2 and A∩F1 ∩F2 = ∅ imply either A∩F1 = ∅ or A ∩ F2 = ∅. Proof: Indeed, if A is connected and if F1 , F2 are closed sets in (E, d) such that A ⊂ F1 ∪ F2 and A ∩ F1 ∩ F2 = ∅, then Ei := A ∩ Fi , i = 1, 2 are closed disjoint set in (A, d) such that A = E1 ∪ E2 . Therefore Ei must either be A or ∅. Since E1 and E2 are further disjoint, one of these sets must be empty. Conversely, if A satisfies the above property, and if E1 , E2 are disjoint closed sets in (A, d) such that A = E1 ∪ E2 , there exist two closed sets F1 , F2 in (E, d) such that Ei = A ∩ Fi . Clearly A ⊂ F1 ∪ F2 and A ∩ F1 ∩ F2 = E1 ∩ E2 = ∅. Therefore E1 or E2 must be empty. A is thus connected since it can not be equal to the union of two disjoint non empty closed set in (A, d). Theorem 4.1.3. If A is a connected set in the metric space (E, d), A is also connected. Proof: Let E1 , E2 be two closed sets in (E, d) satisfying A ⊂ E1 ∪ E2 and A ∩ E1 ∩ E2 = ∅. Then A ⊂ E1 ∪ E2 and A ∩ E1 ∩ E2 = ∅. Since A is connected, this implies that A ∩ Ei = ∅ for i = 1 or 2. Assume that A ∩ E2 = ∅, then A ⊂ E1 . Since E1 is closed, we thus also have A ⊂ E1 , and thus A ∩ E2 = ∅. A is thus connected, since, for any pair of closed sets E1 , E2 satisfying A ⊂ E1 ∪ E2 and A ∩ E1 ∩ E2 = ∅, either A ∩ E1 = ∅ or A ∩ E2 = ∅. Theorem 4.1.4. The image of a connected set A by a continuous function is a connected set. Proof: Let f : (E, d) → (F, dF ) be a continuous function and let A be a connected set in (E, d). Then f (A) is connected in (F, dF ). Indeed, if G1 , G2 are two closed sets in (F, df ) satisfying f (A) ⊂ G1 ∪ G2 and f (A) ∩ G1 ∩ G2 = ∅, then A ⊂ f −1 (G1 ∪ G2 ) = f −1 (G1 ) ∪ f −1 (G2 ) and f −1 (G1 ) ∩ f −1 (G2 ) = ∅. Since f is continuous, f −1 (G1 ) and f −1 (G2 ) are closed sets. Since A is connected, we thus infer that either f −1 (G1 ) = ∅ or f −1 (G2 ) = ∅. Therefore either G1 ∩ f (A) = ∅ or G2 ∩ f (A) = ∅, and f (A) is thus connected. 4.2. Path-connected sets. A path joining two points a, b in a metric space (E, d) is a continuous function f : ([0, 1], |.|) → (E, d) such that a = f (0) and b = f (1). A metric space (E, d) is path connected iif all pair (a, b) of points in E can be joined by a path. A subset A of (E, d) is path connected iif the subspace (A, d) is path connected. Theorem 4.2.1. A path connected space (E, d) is connected. Proof: Indeed, if E1 , E2 are non empty closed sets in (E, d) satisfying E = E1 ∪ E2 , there are two points e1 ∈ E1 and e2 ∈ E2 that can be joined by a path f . f ([0, 1]) is thus a connected set. Furthermore f ([0, 1]) ⊂ E1 ∪ E2 and, for i = 1, 2: ei ∈ f ([0, 1]) ∩ Ei 6= ∅. Therefore, since f ([0, 1]) is connected, we infer

4. CONNECTED SETS

23

that f ([0, 1]) ∩ E1 ∩ E2 6= ∅, and thus E1 ∩ E2 6= ∅. So E can not be written as the union of two non empty disjoint closed sets, and (E, d) is thus connected. The next theorem indicates that the connected sets in (R, |.|) are the intervals (open, closed, semi open, bounded or unbounded) Theorem 4.2.2. If A is a connected set in (R, |.|) that contains x and y, with x < y then [x, y] ⊂ A. In particular, in (R, |.|), A is connected iif A is path-connected. Proof: Let z ∈]x, y[, and assume that z 6∈ A. Then E1 :=]−∞, z] and E2 := [z, ∞[ are closed sets that satisfy A ⊂ E1 ∪E2 and A∩E1 ∩E2 = ∅. However x ∈ A∩E1 6= ∅ and y ∈ A ∩ E2 6= ∅. A could therefore not be connected. Therefore z must belong to A. In other words: ]x, y[⊂ A. Since x, y ∈ A, this also implies [x, y] ⊂ A as announced. We next prove the second claim. According to the previous theorem, we just have to prove that a connected set A is path-connected. But this follows from the previous claim since if x, y ∈ A, the continuous function f : [0, 1] → R : t → f (t) := tx + (1 − t)y is an [x, y]-valued path joining x and y. f is thus also A-valued. In general metric spaces however there are connected sets that are not pathconnected. Exercise 4.2.3. (1) Let I := {(0, y) : y ∈ [−1, 1]}, J := {(x, sin( x1 )) : x > 0} and A := I ∪ J. a) Prove that I and J are connected. b) Prove that I ⊂ J. c) Prove that A is connected. (Hint: Let E1 , E2 be two closed sets in (R2 , k.k∞ ) satisfying A ⊂ E1 ∪ E2 , A ∩ E1 ∩ E2 = ∅, A ∩ E1 6= ∅ and A ∩ E2 6= ∅. Using a), prove J must then be included in one of the Ei , i = 1, 2—say E1 . With b) conclude that I ⊂ E1 and thus A ⊂ E1 . Find then a contradiction.) e) Prove that there is no path joining (0, 0) and (1, sin(1)). (Hint: let f (t) := (f1 (t), f2 (t)) be such a path. Let α := sup{t ∈ [0, 1] : f (t) ∈ I}. Prove that f (α) ∈ I and that ∀t > α : f (t) ∈ J. Let {βn }n∈N be a decreasing sequence in ]0, 1] that converges to 0 and such that {sin( β1n )}n∈N does not converge. Prove that there exists a decreasing sequence {tn }n∈N that converges to α such that f1 (tn ) = βn , for all n. Conclude that f2 can not be continuous at α. Theorem 4.2.4. Let f be a continuous function from a connected space (E, d) to (R, |.|). If there exist a, b ∈ E such that f (a) > 0 and f (b) < 0 then there exists c ∈ E such that f (c) = 0. Proof: f (E) is a connected set according to theorem 4.1.4. According to theorem 4.2.2, 0 ∈ [f (a), f (b)] ⊂ f (E). Therefore, there exists c ∈ E such that f (c) = 0. 4.3. Connected component. Let x be a point in a metric space (E, d). The connected component of x, denoted connect(x) is defined as the union of all connected sets A that contains x. Notice that {x} is a connected set that contains x, so that the above definition always makes sense.

24

1. METRIC SPACES

Theorem 4.3.1. For all x ∈ E, connect(x) is a closed connected set. Proof: Set K := connex(x). If E1 , E2 are two closed sets such that K ⊂ E1 ∪ E2 and K ∩ E1 ∩ E2 = ∅, then x belongs to E1 or E2 . Assume that x ∈ E1 . Let next A be a connected set that contains x. Since A ⊂ K, we also have A ⊂ E1 ∪ E2 and A ∩ E1 ∩ E2 = ∅. Therefore either A ∩ E1 = ∅ or A ∩ E2 = ∅. Since x ∈ A ∩ E1 , we thus conclude that A ∩ E2 = ∅, and thus A ⊂ E1 . Since this last inclusion holds for all connected set A that contains x, it also holds for their union K and so K ⊂ E1 . This also implies that K ∩ E2 = ∅. K is thus connected, as announced. Finally, since K is also a connected set that contains x, as it results from theorem 4.1.3, we get K ⊂ connect(x) = K, and thus K is closed. Theorem 4.3.2. Let x and y be two points of (E, d). Then either connect(x) = connect(y) or connect(x) ∩ connect(y) = ∅. In other words, the connected components form a partition of E. Proof: Indeed, assume that connect(x) ∩ connect(y) 6= ∅. Then there exists z ∈ connect(x) ∩ connect(y). In particular connect(x) is a connected set that contains z. Therefore x ∈ connect(x) ⊂ connect(z). Therefore connect(z) is a connected set that contains x, and so connect(z) ⊂ connect(x). Hence connect(x) = connect(z). A similar argument indicates that connect(y) = connect(z), and thus connect(x) = connect(y) as announced.

CHAPTER 2

Linear Normed Spaces 1. Finite dimensional spaces 1.1. Equivalence of norms. Let (E, ν) be a normed linear space of dimension n. Let {e1 , . . . , en }Pbe a basis of E. Then the map T : Rn → E : x := n (x1 , . . . , xn ) → T (x) := k=1 xk ek is a bijection between Rn and E (one to one since {e1 , . . . , en } is a linearly independent family and onto since the spann of {e1 , . . . , en } is E. For x ∈ Rn , let µ(x) denote µ(x) := ν(T (x)). Exercise 1.1.1. (1) Prove that µ is a norm on Rn . Prove that T is isometric between (Rn , µ) and (E, ν). This exercise indicates that analyzing the topological properties of a finite dimensional vector space turns out to be equivalent to analyzing (Rn , µ). Theorem 1.1.2. Any norm µ on Rn is equivalent to k.k∞ . In other words: ∃α, β > 0 : ∀x ∈ Rn : αkxk∞ ≤ µ(x) ≤ βkxk∞ . Proof: Indeed, let uk be the k-th element of the canonical basis of Rn . Then, if x = (x1 , . . . , xn ) we get with the triangle inequality: n n X X |xk |µ(ek ) ≤ βkxk∞ , xk ek ) ≤ µ(x) = µ( k=1

k=1

Pn

where β := k=1 µ(ek ). This is already half of our claim. It implies in particular that µ : (Rn , k.k∞ ) → (R, |.|) is Lipschitz with constant β. Indeed, for all x, y ∈ Rn , the triangle inequality for µ indicates that |µ(x) − µ(y)| ≤ µ(x − y) ≤ βkx − yk∞ . Let next B denote B := {x ∈ Rn : kxk∞ = 1}. B is a closed bounded subset of (Rn , k.k∞ ). B is therefore compact, as it results from theorem 3.2.6. According to theorem 3.1.6, the continuous function µ reaches its infimum on B: If α := inf{µ(x) : x ∈ B} ≥ 0, then, there exists a point x∗ in B such that µ(x∗ ) = α. Since kx∗ k∞ = 1, we infer that x∗ 6= 0 and so α = µ(x∗ ) > 0. So, if y ∈ Rn is such that y 6= 0, we get that x := y/kyk∞ ∈ B. Therefore µ(y) = kyk∞ µ(x) ≥ αkyk∞ . Since this inequality holds obviously for y = 0, the theorem is proved. 1.2. Main properties. The previous discussion indicates that the topological properties proved in the first chapter for (Rn , k.k∞ ) also apply to finite dimensional vector spaces: Corollary 1.2.1. A finite dimensional normed vector space is a Banach space. 25

26

2. LINEAR NORMED SPACES

Corollary 1.2.2. A set K in a finite dimensional normed vector space is compact iif K is bounded and closed. Corollary 1.2.3. An open set U in finite dimensional normed vector space is connected iif U is path connected Proof: We already know the U connected implies U path-connected (see theorem 4.2.2). We just have to prove the reciprocal implication. It results from our previous discussion that we just have to prove this implication on the normed vector space (Rn , k.k∞ ). To do so, observe that ∀x ∈ Rn , ∀ > 0, there exist y ∈ Qn and δ ∈ Q such that δ > 0 and x ∈ B(y, δ) ⊂ B(x, ). Therefore, since U is open, any point x of U is contained in a ball B(y, δ) ⊂ U , with center in Qn and radius in Q. Since there are only countably many such balls, we conclude that U is a countable union of open balls Bn : U = ∪n∈N Bn . We next will define a map j : N → N inductively: we set j(0) := 0, and when j(0), . . . , j(n) have been defined, we define j(n + 1) as the smallest integer j that differs form j(0), . . . , j(n) and such that Bj ∩ (∪nk=0 Bj(k) ) 6= ∅. Note that there must exists such a j, for otherwise we would have A ∩ B = ∅ with A = ∪k∈N−C Uk , B := ∪k∈C Uk where C := {j(k) : k = 0, . . . , n}. As union of open balls A and B are open sets and clearly U = A ∪ B. But this is impossible since U is connected: it can’t be the union of two non empty disjoint open sets. We next claim that j is onto N or in other words the set D := {j(k) : k ∈ N} coincides with N. Indeed, assume that D 6= N. Since, for all n j(n) differs from j(0), . . . , j(n − 1), D must contains infinitely many elements. For all l ∈ N − D, there must therefore exists m0 such that ∀m ≥ m0 : j(m) > l. Therefore ∀m ≥ m0 : Bl ∩ (∪m−1 k=0 Bj(k) ) = ∅ for otherwise we would have had j(m) ≤ l as it results from the definition of j. Therefore Bl ∩ (∪∞ k=0 Bj(k) ) = ∅, and thus A ∩ B = ∅ with A = ∪k∈N−D Uk , B := ∪k∈D Uk . Since U = A ∪ B, B 6= ∅ 6= A and both A and B are open sets, this is impossible, and so assuming D 6= N leads to a contradiction. We will next prove by induction that, for all n, Kn = ∪nk=0 Bj(k) is pathconnected. This will conclude the proof of the path-connectedness of U . Indeed, U = ∪∞ k=0 Bj(k) . Therefore, if x, y ∈ U there exists n such that x, y ∈ Kn . There exists a continuous path from x to y that remains in Kn and thus in U . Let us prove that Kn is path-connected: Since K0 is a ball, it is clearly pathconnected. Next assume that Kn−1 is path-connected. Then Kn = Kn−1 ∪ Bj(n) , and, by definition of j(n), Kn−1 ∩ Bj(n) 6= ∅. Let z ∈ Kn−1 ∩ Bj(n) . The for all x ∈ Kn , either x ∈ Kn−1 , and so there is a continuous path from x to z that remains in Kn−1 ⊂ Kn or x ∈ Bj(n) , and so there is a continuous path from x to z that remains in Bj(n) ⊂ Kn . So for all x, y ∈ Kn there exist two Kn -valued continuous paths, one from x to z and the other from z to y. Concatenating these paths yields a Kn -valued continuous path from x to y and Kn is thus path-connected as announced.

2. Linear maps 2.1. Continuous linear maps. Let (E, k.kE ) and (F, k.kF ) be two real normed spaces. We will denote L(E, F ) the vector space of the linear maps from E to F and Lc (E, F ) the subspace of continuous linear maps from (E, k.kE ) to (F, k.kF ).

2. LINEAR MAPS

27

Let B denotes B := {x ∈ E : kxkE ≤ 1} and, for f ∈ L(E, F ), let []f [] be defined as []f [] := supx∈B kf (x)kF . Theorem 2.1.1. Let f be in L(E, F ). Then f ∈ Lc (E, F ) iif []f [] < ∞. In this case f is Lipschitz with constant []f []. Proof: Let f ∈ L(E, F ) be such that []f [] < ∞. Then ∀x ∈ E, x 6= 0, we get y := x/kxkE ∈ B and thus kf (x)kF = kf (kxkE y)kF = kf (y)kF kxkE ≤ []f []kxkE . The inequality kf (x)kF ≤ []f []kxkE holds also when x = 0. Therefore ∀x, y ∈ E : kf (x) − f (y)kF = kf (x − y)kF ≤ []f []kx − ykE and the map f is thus Lipschitz with constant []f [] < ∞ and hence continuous. Conversely, if f is continuous, it is in particular continuous at 0. Would []f [] be infinite, there would be a sequence {xn }n∈N taking values in B such that kf (xn )kF ≥ n. The sequence {yn }n∈N defined by yn := xn /n converges thus to 0 in E since kyn kE ≤ 1/n. The continuity of f at 0 would then imply the convergence of {f (yn )}n∈N to f (0) = 0. But this is impossible since, for all n: kf (yn )kF = kf (xn )kF /n ≥ 1. Corollary 2.1.2. If (E, k.kE ) is a finite dimensional normed space then L(E, F ) = Lc (E, F ). Proof: As mentioned in the section 1.1, (E, k.kE ) is isometric to (Rn , ν) for an appropriate norm ν. Since this norm is equivalent to k.k∞ , we just have to prove that any linear map f on (Rn , k.k∞ ) is continuous. If {u1 , . . . , un } is the canonical basis of Rn , then forall x = (x1 , . . . , xn ) ∈ Rn : kf (x)kF = k

n X k=1

xk f (uk )kF ≤

n X

|xk |kf (uk )kF ≤ kxk∞ (

k=1

n X

kf (uk )kF ).

k=1

Pn This implies in particular that []f [] ≤ ( k=1 kf (uk )kF ) < ∞. Theorem 2.1.3. If (F, k.kF ) is a Banach space, so is (Lc (E, F ), [].[]). Exercise 2.1.4. Proof of theorem 2.1.3 (1) Prove that [].[] is a norm on Lc (E, F ). (2) Let B denotes B := {x ∈ E : kxkE ≤ 1}. Let (F, k.kF ) be a Banach space. Let B(B, F ) denote the set of functions g : B → F such that kgk∞ < ∞ where kgk∞ := sup{kg(x)kF : x ∈ B}. Prove that (B(B, F ), k.k∞ ) is a Banach space. (Hint: look at theorem 2.1.6). (3) Let Cb (B, F ) denotes the set of continuous functions g in B(B, F ). Prove that (Cb (B, F ), k.k∞ ) is a Banach space. (Hint: theorem 2.1.8). (4) For f ∈ Lc (E, F ), let fB denote the restriction of f to B. Prove that the map R : f ∈ Lc (E, F ) → R(f ) := fB ∈ Cb (B, F ) is isometric from (Lc (E, F ), [].[]) to (Cb (B, F ), k.k∞ ). (5) Let K be the set of fB where f ∈ Lc (E, F ). Prove that (K, k.k∞ ) is a Banach space isometric to (Lc (E, F ), [].[]).

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2. LINEAR NORMED SPACES

2.2. The dual space. If (E, k.kE ) is a normed space, the dual space, denoted E ∗ , is the space E ∗ := Lc (E, R) endowed with the norm [].[]. We just will consider in this course the dual space E ∗ of an Hilbert space E: A scalar product h., .i on a real vector space E is a map h., .i : E × E → R that satisfies: 1) ∀x ∈ E, x 6= 0 : hx, xi > 0 2) ∀x, y ∈ E : hx, yi = hy, xi. 3) ∀x, y, z ∈ E, ∀α, β ∈ R : hαx + βy, zi = αhx, zi + βhy, zi. Exercise 2.2.1. (1) Let x, y be two points of E. Using the fact that the quadratic map λ → hx + λy, x + λyi is positive for all λ ∈ R, prove the Cauchy Schwartz inequality: p p ∀x, y ∈ E : hx, yi ≤ hx, xi hy, yi p (2) Prove that kxk := hx, xi is a norm on E. This norm is said to be derived from the scalar product. An Hilbert space is a vector space E endowed with a scalar product that is complete for the corresponding norm. An Hilbert space is thus a particular case of Banach space. Theorem 2.2.2 (Riez representation theorem). Let (E, h.; .i) be an Hilbert space. Then for all f ∈ E ∗ there exists a point af ∈ E such that ∀x ∈ E : f (x) = haf , xi. The map f → af is a linear isometric bijection between (E, k.kE ) and (E ∗ , [].[]). Exercise 2.2.3. Proof of theorem 2.1.3 (1) If f = 0, the first claim holds with af = 0. Let f 6= 0 be in E ∗ . Let K be the kernel of f : K := {x ∈ E : f (x) = 0}. Prove that K is a closed vector space. f (x) Let a belong to E − K. ∀x ∈ E : x = z + ff (x) (a) a where z := (x − f (a) a). Prove that z ∈ K. (2) Prove that in a Hilbert space, the parallelogram relation holds: ∀u, v ∈ E : ku + vk2E + ku − vk2E = 2(kuk2E + kvk2E ) (3) Let α be α := inf{kx − akE : x ∈ K}. Let {xn }n∈N be a sequence of points in K such that kxn − akE converges to α. With u := xn − a and v := xm − a in the parallelogram relation, prove that kxn − xm k2E = 2ka − xn k2E + 2ka − xm k2E − 4ka −

xn + xm 2 kE 2

Prove that {xn }n∈N is a Cauchy sequence. m (Hint: prove that ka − xn +x kE ≥ α.) 2 (4) Let x∗ be the limit of {xn }n∈N . Prove that x∗ ∈ K and ka − x∗ kE = α. Let y ∗ be y ∗ := a − x∗ . Prove that for all x ∈ K : hy ∗ , xi = 0. Prove that hy ∗ , ai = α > 0. ∗ Prove that the first claim of the theorem holds for f with af := hyf ∗(a) ,ai y . (5) Use Cauchy Schwartz inequality to prove that []f [] = kaf kE . .

4. THE SPACE (Cb (E), k.k∞ )

29

3. Absolutely convergent series P∞3.1. Series in a normed space. Let (E, k.kE ) be a normed space. A series be convergent to the sum A iif the sequence k=0 ak where ∀k : ak ∈ E is said to P n {sn }n∈N converges to A, where s := n k=0 ak . P∞ A series a is said to be absolutely convergent iif the series of real numbers k=0 k P∞ ka k is convergent in (R, |.|). k E k=0 Theorem 3.1.1. In a Banach space (E, k.kE ), an absolutely convergent series is convergent. Exercise 3.1.2. (1) Prove this theorem. (2) Let {an }n∈N be a sequence of real numbers, and, for x ∈ R consider P∞ |an+1 | k the series k=0 ak x . Let γ denote γ := lim supn→∞ |an | . Prove P ∞ k that for all r < γ1 , the series k=0 ak x is absolutely convergent on (Cb ([−r, r]), k.k∞ ). Let g(x) denote the sum of this series.P Such a function is called analytic. ∞ k−1 (3) Consider next the derivative series: . Prove that this is k=0 ak kx also an absolutely convergent series on (Cb ([−r, r]), k.k∞ ). d g(x) = h(x) on Let h(x) denote the sum of this series. Prove that dx [−r, r]. Conclude that, ∀n, g has a continuous derivative of order n on [−r, r]. P∞ xk (4) Define the function ex as the sum of the series k=0 k! . Prove that this series converges uniformly on any bounded interval of R. Prove that d x x dx e = e . (5) Let (E, k.kE ) be a Banach space. Let I denote the identity map on E: ∀x ∈ E : I(x) := x. Prove that I ∈ Lc (E, E). If A, B ∈ Lc (E, E), the map A · B is defined by A · B : E → E : x → A · B(x) := A(B(x)). Prove that A · B ∈ Lc (E, E) and []A · B[] ≤ []A[] · []B[]. If A ∈ Lc (E, E) and n ∈ N, the map An is defined inductively by A0 := I and An := An−1 · A. ProveP that []An [] ≤ []A[]n . ∞ Consider the series g(x) = k=0 ak xk of points (2) and (3)Pabove. Prove ∞ c k that if A ∈ L (E, E) satisfies []A[] < γ1 then the series k=0 ak A is c absolutely convergent in L (E, E). Its sum is denoted g(A). We may in this way define the map eA for any A ∈ Lc (E, E). (6) A map C ∈ L(E, E) is said to be invertible if there exists B ∈ L(E, E) such that A · B = I = B · A. B is the called the inverse of C and denoted C −1 := B. Prove that C as at most one inverse. Prove that if A ∈ Lc (E, E) is such that []A[] < 1Pthen (I − A) has an ∞ k inverse in Lc (E, E) (Hint: consider the sequence k=0 A .) . 4. The space (Cb (E), k.k∞ ) 4.1. Ascoli Arzela Theorem. Let (E, d) be a metric space. We already know that (Cb (E), k.k∞ ) is a Banach space. The question we address in this section is what are the compact subset A of (Cb (E), k.k∞ )? According to theorems 3.1.7 and 3.2.2, if A is compact, then A must be both closed and precompact. These two conditions are also sufficient for A to be compact,

30

2. LINEAR NORMED SPACES

since a closed subset of the complete space (Cb (E), k.k∞ ) is complete (see theorem 2.1.7). It is in general relatively easy to determine wether a set is closed or not, so the main problem is to determine wether a set is precompact or not. Ascoli Arzela theorem completely solves this problem when E is a compact set. A set A of functions f : E → R is equi-continuous at x ∈ E iif ∀ > 0 : ∃δ > 0 : ∀y ∈ E : d(x, y) < δ =⇒ ∀f ∈ A : |f (x) − f (y)| < . A set A of functions f : E → R is equi-continuous iif A is equi-continuous at all x ∈ E. A set A of functions f : E → R is uniformly equi-continuous iif ∀ > 0 : ∃δ > 0 : ∀x, y ∈ E : d(x, y) < δ =⇒ ∀f ∈ A : |f (x) − f (y)| < .

Exercise 4.1.1. (1) Prove that a finite set of continuous functions is equi-continuous. (2) Adapt the proof of theorem 3.2.7 to prove that on a compact set E, an equicontinuous set A of functions f : E → R is uniformly equi-continuous. (3) Prove that, if all fonctions in A are Lipschitz with the same constant, then A is uniformly equi-continuous. Theorem 4.1.2 (Ascoli Arzela). If (E, d) is a compact metric space and if A is a set of functions f : E → R, then A is precompact iif the two following conditions are satisfied: 1) A is equi-continuous. 2) There exists a compact set K ⊂ R such that ∀x ∈ E : ∀f ∈ A : f (x) ∈ K. Proof: Necessary condition: Let A be a precompact subset of Cb (E). Let us first prove that 1) holds: For  > 0, there exists a finite covering of A by open balls {B(f1 , /3), . . . , B(fn , /3)} in (Cb (E), k.k∞ ). The finite set {f1 , . . . , fn } is equi-continuous (see exercise 4.1.1-1).) Therefore, ∀x ∈ E, there exists δ > 0 such that ∀y ∈ B(x, δ) : ∀k : |fk (x) − fk (y)| ≤ /3. Hence, if y ∈ B(x, δ) and f ∈ A, there exists k such that f ∈ B(fk , /3) and so |f (x) − f (y)| ≤ |f (x) − fk (y)| + |fk (x) − fk (y)| + |fk (x) − f (y)| ≤ : A is thus equi-continuous at x. This being true for all x, 1) is proved. We next prove that 2) holds: Let S be the set S := {f (x) : x ∈ E, f ∈ A}. Let us prove that S is precompact: for all  > 0, there exists a finite covering of A by open balls {B(f1 , /2), . . . , B(fn , /2)} in (Cb (E), k.k∞ ). Therefore, all the points f (x) ∈ S belong to some ball B(fk (x), /2) in R. Since fk is continuous, fk (E) is compact and thus precompact: there exists a finite family {ak1 , . . . , akmk } of real numbers such that {B(ak1 , /2), . . . , B(akmk , /2)} is a finite covering of fk (E) by open balls of radius /2. Therefore the point f (x) ∈ S belongs to some B(akj , ): {B(akj , ) : 1 ≤ k ≤ n, 1 ≤ j ≤ mk } is a finite covering of S by balls of radius . S is thus precompact and so is its adherence S (Prove it!). S, as a closed subset of the complete space (R, |.|), is also complete, and is thus a compact set that contains S: Therefore 2) is proved with K := S. Sufficient condition: Let A satisfy 1) and 2). According to exercise 4.1.12), A is uniformly equi-continuous. Therefore, ∀ > 0 : ∃δ > 0 : ∀x, y ∈ E such

5. FRECHET DIFFERENTIAL

31

that d(x, y) < δ, ∀f ∈ A : |f (x) − f (y)| < /3. Let then {B(x1 , δ), . . . , B(xn , δ)} be a covering of the compact set E by open balls of radius δ. Let also {B(y1 , /6), . . . , B(ym , /6)} be a covering of K by open balls in R of radius /6. Let φ be a map from {1, . . . , n} to {1, . . . , m}. We will say that a function f ∈ A is of type φ iif ∀k = 1, . . . , n : f (xk ) ∈ B(yφ(k) , /6). Let then Φ be the set of φ such that there exists a function f ∈ A of type φ. For all φ ∈ Φ let fφ be one function of type φ in A. Observe that Φ is a finite set since there are at most mn functions from {1, . . . , n} to {1, . . . , m}. We claim next that {B(fφ , ) : φ ∈ Φ} is a finite covering of A by balls of radius : Indeed, for all f in A, we have that ∀k : f (xk ) ∈ K. So, f (xk ) must belong to some B(yl , /6), let φ∗ (k) be the minimum l such that f (xk ) ∈ B(yl , /6). f is then clearly of type φ∗ and thus φ∗ ∈ Φ. For all x ∈ E, x must belong to some B(xj , δ). Therefore: |f (x) − fφ∗ (x)| ≤ |f (x) − f (xj )| + |f (xj ) − fφ∗ (xj )| + |fφ∗ (xj ) − fφ∗ (x)| Observe that |f (x) − f (xj )| < /3 and |fφ∗ (xj ) − fφ∗ (x)| < /3, since d(x, xj ) < δ and f, fφ∗ ∈ A. Next, since f and fφ∗ are of type φ∗ , we have f (xj ), fφ∗ (xj ) ∈ B(yφ∗ (j) , /6). Therefore |f (xj ) − fφ∗ (xj )| ≤ |f (xj ) − yφ∗ (j) | + |yφ∗ (j) − fφ∗ (xj )| < /3. Thus, ∀x ∈ E : |f (x) − fφ∗ (x)| <  and so kf − fφ∗ k∞ < : Indeed, the continuous function x → |f (x) − fφ∗ (x)| reaches its maximum at a point x∗ on the compact set E and so kf − fφ∗ k∞ = |f (x∗ ) − fφ∗ (x∗ )| < . In other words, f ∈ B(fφ∗ , ) and so as announced {B(fφ , ) : φ ∈ Φ} is a finite covering of A. Terefore 1) and 2) implies that A is a precompact set. Exercise 4.1.3. 1 (1) Let fn be the function fn : R → [0, 1] : x → fn (x) := 1+(x−n) 2 . Let A be the set A := {fn , n ∈ N}. Prove that A is uniformly equi-continuous. Prove that A is not precompact. (Hint: The sequence {fn }n∈N would have an accumulation point in (Cb (R), k.k∞ ).) This example indicates that one can’t dispense with the hypothesis that (E, d) is compact in Ascoli Arzela theorem. (2) Prove that the set X of Lipschitz functions f : [0, 1] → R of Lipschitz constant 1, such that f (0) = f (1) = 0 is compact in (Cb ([0, 1]), k.k∞ ). R1 (3) Prove that the optimization problem: maxf ∈X 0 (x − 1/2)f (x)dx has a solution. (4) Generalize the Ascoli Arzela theorem to the space (Cb (E, F ), k.k∞ ) introduced in exercise 2.2.3-3), when (E, d) is a compact metric space and (F, k.kF ) is a Banach space. Check the the proof can be adapted word by word in this setting. 5. Frechet Differential 5.1. Definition. Let f be a continuous function from R to R. The equation of the straight line that interpolates f at two points a 6= b ∈ R is y = f (a) + f (b)−f (a) (x − a). If b is getting closer and closer to a, the interpolating lines will b−a ”converge” to the tangent line at the point a. The slope of this line should thus be (a) limb→a f (b)−f which is the derivative f 0 (a) of f at a. b−a

32

2. LINEAR NORMED SPACES

Exercise 5.1.1. (1) Let f be a function from the metric space (E, dE ) to the metric space (F, dF ). The function f has a limit α at a ∈ E iif for all sequence {xn }n∈N in E − {a} that converges to a, the image sequence {f (xn )}n∈N converges to α. In this case we write limx→a f (x) = α. Prove that f is continuous at a iif f (a) = limx→a f (x). (2) Prove that, if the derivative f 0 (a) exists, f 0 (a) is the unique number α (a)−αh such that limh→0 f (a+h)−f = 0. |h| So the notion of derivative is intimately related with that of tangent line. To generalize these concepts to a function f : R2 → R, or more generally f : E → R, where (E, k.kE ) is a vector space, the notion of tangent line must be replaced by that of tangent hyperplane. The general equation of an hyperplane interpolating f at a ∈ E is y = f (a)+L(x−a) where L ∈ L(E, R). So, inspired by exercise 5.1.1-(2), the equation of the tangent hyperplane at a if it exists, should be y = f (a)+L(x−a) (a)−L(h) = 0. where L ∈ L(E, R) is such that limh→0 f (a+h)−f khk Exercise 5.1.2. (1) Prove that there is at most one such L This leads us to the following definition: Let f be a function from an open set U of a normed space E, k.kE to a normed space (F, k.kF ). F is Frechet differentiable at a point a ∈ U iif there exists L ∈ Lc (E, F ) such that kf (a + h) − f (a) − L(h)kF = 0. khkE In this case, L is unique. It is called the Frechet differential of f at a and denoted Df [a] (so L(x) = Df [a](x).) lim

h→0

Theorem 5.1.3. If f : (E, k.kE ) → (F, k.kF ) is differentiable at a point a ∈ E and g : (F, k.kF ) → (G, k.kG ) is differentiable at f (a), then the map h : x → h(x) := g(f (x)) is differentiable at a and, ∀x ∈ E : Df [a](x) = Dg[f (a)](Df [a](x)). Exercise 5.1.4. (1) Prove this theorem. (2) Prove that if f is differentiable at a then it is continuous at a. (Hint: use the fact that Df [a] ∈ Lc (E, F ).) (3) Let S : R2 → R : (x, y) → S(x, y) := x + y and P : R2 → R : (x, y) → P (x, y) := xy. What are DS[(x, y)] and DP [(x, y)]? (4) Prove that if f, g : E → F are differentiable at a then (f + g) and (f.g) are differentiable at a. Furthermore D(f + g)[a] = Df [a] + Dg[a] and D(f.g)[a] = g(a).Df [a] + f (a).Dg[a], ∂ The directional derivative − → f (a) of f : E → F at a point a ∈ E in the direction ∂b b ∈ E is defined as the derivative at 0 of the function g : R → F : t → g(t) := f (a+tx)−f (a) ∂ f (a + tx). In other words: − → f (a) := limt→0 t ∂b n If f is a map from R to R, and ei denotes the i-th element of the canonical basis ∂ of Rn , the partial derivative ∂x f (x1 , . . . , xn ) is defined as the directional derivative i ∂ of f in the direction ei : ∂xi f = −∂→ f . The gradient of f at x is then defined as the ∂ei

∂ vector ∇f (x) := ( ∂x f (x), . . . , ∂x∂n f (x)). Theorem 5.1.3 has as corollary: 1

5. FRECHET DIFFERENTIAL

33

Corollary 5.1.5. If a function f : E → F is Frechet differentiable at a ∈ E ∂ ∂ then for all b ∈ E, the directional derivative − → f (a) exists and − → f (a) = Df [a](b). ∂b

∂b

Exercise 5.1.6. (1) The Mean Value theorem claims that if g : [a, b] → R has a continuous derivative g 0 , then there exists a point c ∈]a, b[ such that g(b) − g(a) = g 0 (c)(b − a). Prove this theorem. (Hint: The function h(x) := g(x) − x−a x−b b−a g(b) − a−b g(a) is such that h(a) = h(b) = 0. Prove that |h(x)| must reach its maximal value on [a, b] at some point c ∈]a, b[. At that point h0 (c) = 0.) (2) Use the Mean Value theorem to prove that if f : E → R is such that the map a → Df [a] is continuous from (E, k.kE ) to (Lc (E, R), [].[]), then ∀a, b ∈ E, there exists θ ∈]0, 1[ such that, with c := θa + (1 − θ)b: f (b) − f (a) := Df [c](b − a). (3) Let a, b be two points of a normed space (E, k.kE ). The line segment [a, b] is then {ta + (1 − t)b : t ∈ [0, 1]}. If a map f : E → F is differentiable at all the points c ∈ [a, b] and if there is a constant k such that ∀c ∈ [a, b] : []Df [c][] ≤ k, prove that kf (b) − f (a)kF ≤ kkb − akE . The next corollary is a particular case of the previous one: Corollary 5.1.7. If f : Rn → R is differentiable at a point a ∈ Rn , then ∀i, ∂ ∂ f (a) exists and ∂x f (a) = Df [a](ei ). Therefore, the partial derivative ∂x i i ∀b ∈ Rn : Df [a](b) =

n X

bi

i=1

where hu, vi :=

Pn

i=1

∂ f (a) = h∇f (a), bi, ∂xi

ui vi .

This corollary indicates how to compute the differential of a function on Rn when it exists, just by computing the partial derivatives. However, as indicates the following exercise, the existence of the partial derivatives does not imply that f is Frechet differentiable. Exercise 5.1.8. (1) Consider the function f : R2 → R defined by ( 0 if x = y = 0 f (x, y) := √ xy else 2 2 x +y

Prove Prove Prove Prove

that f is continuous at (0, 0). ∂ ∂ that ∂x f (0, 0) = ∂y f (0, 0) = 0. that f is not differentiable at (0, 0). ∂ that ∂x f (x, y) is not continuous at (0, 0).

Theorem 5.1.9. If f : Rn → R is such that all the partial derivatives of f are continuous on a neighborhood of a then f is Frechet differentiable at a. Proof: Let B(a, r) be a ball on which the partial derivatives of f are continuous. Let h ∈ Rn be such that khk∞ < r. Let {e1 , . . . , en } be the canonical basis of Pk−1 Pn Pk Rn . Then f (a + h) − f (a) = k=1 f (a + i=1 hi ei ) − f (a + i=1 hi ei ). Let gk Pk−1 be the function gk : x ∈ R → gk (x) := f (a + i=1 hi ei + xek ). Then gk0 (x) =

34

2. LINEAR NORMED SPACES

Pk−1 + i=1 hi ei + xek ). In particular, gk0 (x) is continuous on ] − r, r[. Since |hk | < r, we infer with the Mean Value theorem that there exists θk ∈]0, 1[ such that gk (hk ) − gk (0) = gk0 (θk hk )hk . Therefore ∂ ∂xk f (a

f (a + h) − f (a) =

n k−1 X X ∂ f (a + hi ei + θk hk ek )hk ∂xk i=1

k=1

Pn

We next prove that L(h) := k=1 ∂x∂ k f (a)hk is the differential of f :  Pk−1 Pn  |f (a + h) − f (a) − L(h)| =| k=1 ∂x∂ k f (a + i=1 hi ei + θk hk ek ) − ∂x∂ k f (a) ·hk | Pn Pk−1 ≤khk∞ k=1 | ∂x∂ k f (a + i=1 hi ei + θk hk ek ) − ∂x∂ k f (a)|. The continuity of ∂x∂ k f implies then clearly that limh→0 the theorem is proved.

|f (a+h)−f (a)−L(h)| khk∞

= 0 and

A map f : E → F is C 1 (i.e. continuously differentiable) on a open set U ⊂ E iif the map a ∈ U → Df [a] is continuous from (U, k.kE ) to (Lc (E, F ), [].[]). 5.2. Implicit function theorem. Let us consider a map g : R2 → R and the set G := {(x, y) ∈ R2 : g(x, y) = 0}. If g is not ”pathological”, G will be a ”curve” in R2 : for instance, if g(x, y) := x2 + y 2 − 1, G is the circle of radius 1 with center (0, 0). Let (x0 , y0 ) be a point of G. (Think of the point (0, 1) in the circle.) Then for an x close to x0 , g(x, y0 ) should be close to 0 and there should be a y close to y0 such that g(x, y) = 0: Assume that this y is unique. Then by setting f (x) := y, we are defining a function f on a neighborhood U of x0 such that f (x0 ) = y0 and ∀x ∈ U : g(x, f (x)) = 0. In other words, the intersection of G and a small ball around (x0 , y0 ) coincide with the graph of f . The implicit function theorem specifies conditions under which such a function f exists. Theorem 5.2.1 (Implicit function theorem). Let g : R2 → R be a C 1 -mapping ∂ on a neighborhood of a point (x0 , y0 ) such that g(x0 , y0 ) = 0 and ∂y g(x0 , y0 ) 6= 0. Then there exist a neighborhood U of x0 , a neighborhood V of y0 , a C 1 function f : U → V such that, for all x ∈ U , f (x) is the unique root y ∈ V of the equation g(x, y) = 0. ∂

g(x,f (x))

Furthermore, for all x ∈ U , f 0 (x) = − ∂x . ∂ g(x,f (x)) ∂y

Proof: For two numbers u, v > 0 , let us define Uu := [x0 − u, x0 + u] and Vv := [y0 − v, y0 + v]. Let Hu,v be the set of continuous functions f from Uu to Vv such that f (x0 ) = y0 . As a closed subspace of (Cb (Uu ), k.k∞ ), (Hu,v , k.k∞ ) is a complete space. y) For x ∈ Uu and y˜ ∈ Vv , let us define Tx (˜ y ) := y˜ − ∂ g(x,˜ . Observe that the g(x ,y ) ∂y

0

0

fixed points of Tx are the roots y of g(x, y) = 0. For f ∈ Hu,v , let us define T (f ) as the map x ∈ Uu → Tx (f (x)). Since g is differentiable at (x0 , y0 ), there exists δ1 > 0 such that (4)

∀(x, y) ∈ B((x0 , y0 ), δ1 ) : |η(x, y)| ≤

∂ | ∂y g(x0 , y0 )|

2

k(x − x0 , y − y0 )k∞ ,

where η(x, y) := g(x, y) − g(x0 , y0 ) − Dg[(x0 , y0 )]((x − x0 , y − y0 )).

5. FRECHET DIFFERENTIAL

35

∂ Since ∂y g(x, y) is continuous at (x0 , y0 ), there exists δ2 > 0 such that ∀(x, y) ∈ B((x0 , y0 ), δ2 ) :

|

(5)

∂ g(x0 , y0 )| | ∂y ∂ ∂ g(x, y) − g(x0 , y0 )| ≤ . ∂y ∂y 2 ∂

g(x ,y )

∂y

0

0 0 | + 1)u. So, let u, v be two numbers in ]0, min(δ1 , δ2 )[ such that v ≥ (2| ∂x ∂ g(x ,y ) 0

For all (x, y) in Uu × Vv , we get then |Tx (y) − y0 |

g(x,y)−g(x0 ,y0 ) | ∂ ∂y g(x0 ,y0 ) η(x,y)+Dg[(x0 ,y0 )]((x−x0 ,y−y0 ) |(y − y0 ) − | ∂ ∂y g(x0 ,y0 ) Dg[(x0 ,y0 )]((x−x0 ,y−y0 ) |η(x,y)| |(y − y0 ) − | + | ∂ g(x ,y )| ∂ 0 0 ∂y g(x0 ,y0 ) ∂x ∂ g(x ,y ) |η(x,y)| 0 0 (x − x )| + | ∂x 0 ∂ ∂ | ∂x g(x0 ,y0 )| ∂y g(x0 ,y0 ) ∂ 1 ∂x g(x0 ,y0 ) | ∂ g(x ,y ) |u + 2 (u + v), 0 0 ∂y

= |(y − y0 ) − = ≤ = ≤

since k(x − x0 , y − y0 )k∞ ≤ max(u, v) ≤ u + v and η satisfies (4). Therefore |Tx (y) − y0 | ≤ v and thus Tx (y) ∈ Vv . If y 0 , y 00 ∈ Vv and x ∈ Uu then, we get with the mean value theorem (see exercise 5.1.6-(2)) that ∃y 000 ∈]y 0 , y 00 [ such that Tx (y 00 )−Tx (y 0 ) = (y 00 −y 0 )+

g(x, y 0 ) − g(x, y 00 ) = ∂ ∂y g(x0 , y0 )

∂ ∂ 000 ∂y g(x0 , y0 ) − ∂y g(x, y ) 00 (y −y 0 ). ∂ g(x , y ) 0 0 ∂y

Thus with inequality (5), we have: |Tx (y 00 ) − Tx (y 0 )| ≤ 21 |y 00 − y 0 |. Therefore, ∀x ∈ Uu , Tx is a contraction on Vv . Tx has thus a unique fixed point f (x) in Vv , which is the unique root y of the equation g(x, y) = 0. Furthermore, if h ∈ Hu,v , then T (h) is clearly continuous and for all x ∈ Uu : T (h)(x) := Tx (h(x)) ∈ Vv . Since furthermore T (h)(x0 ) = Tx0 (h(x0 )) = Tx0 (y0 ) = y0 , we conclude that T (h) ∈ Hu,v . Since ∀h1 , h2 ∈ Hu,v , we get ∀x ∈ Uu : |T (h1 )(x)−T (h2 )(x)| = |Tx (h1 (x))−Tx (h2 (x))| ≤

1 1 |h1 (x)−h2 (x)| ≤ kh1 −h2 k∞ , 2 2

Therefore, kT (h1 ) − T (h2 )k∞ ≤ 21 kh1 − h2 k∞ . So T is a contraction on Hu,v . It has thus a unique fixed point f ∗ . We conclude then that ∀x ∈ Uu : f ∗ (x) = T (f ∗ )(x) = Tx (f ∗ (x)). f ∗ (x) is thus a fixed point of Tx and must then coincide with f (x). The function f is thus continuous since so is f ∗ . We next prove that f is differentiable at any point x ∈ Uu◦ : First observe that ∂ ∂ (5) indicated that ∂y g(x, y) 6= 0, ∀(x, y) ∈ Uu ×Vv . In particular | ∂y g(x, f (x))| > 0. Since g is differentiable at (x, f (x)), for all  > 0, there exists δ3 > 0 such that ∀(h, ∆) ∈ R2 : k(h, ∆)k∞ < δ3 =⇒ |g(x + h, f (x) + ∆) − g(x, f (x)) − Dg[(x, f (x)](h, ∆)| ≤  · |

∂ g(x, f (x))| · k(h, ∆)k∞ . ∂y

Since f is continuous at x, there exists δ4 > 0 such that ∀h ∈ R : |h| < δ4 =⇒ |f (x + h) − f (x)| < δ3 .

36

2. LINEAR NORMED SPACES

Therefore, if δ5 := min(δ3 , δ4 ), we get ∀h ∈ R with |h| < δ5 : |g(x + h, f (x + h)) − g(x, f (x)) − Dg[(x, f (x)](h, f (x + h) − f (x))| ≤ · · · ∂  · | ∂y g(x, f (x))| · k(h, f (x + h) − f (x))k∞ . Since: 0 = g(x + h, f (x + h)) = g(x, f (x)), we get ∂ ∂ | ∂x g(x, f (x))h + ∂y g(x, f (x))(f (x + h) − f (x))| ≤ · · · ∂  · | ∂y g(x, f (x))| · k(h, f (x + h) − f (x))k∞ . ∂

g(x,f (x))

, we have: So, with φ(x) := − ∂x ∂ g(x,f (x)) ∂y

|f (x + h) − f (x) − φ(x)h| ≤  · k(h, f (x + h) − f (x))k∞ . Therefore |f (x + h) − f (x)| ≤ |φ(x)| · |h| +  · (|h| + |f (x + h) − f (x)|), which indicates that |f (x + h) − f (x)| ≤

|φ(x)| +  |h|, 1−

and thus |f (x + h) − f (x) − φ(x)h| ≤  · k(h, f (x + h) − f (x))k∞ ≤  max(1,

|φ(x)| +  ) · |h|. 1−

So, all together, for all η > 0, there exists  > 0 such that |φ(x)| +  ) < η. 1− For this  > 0, we have proved that there exists δ5 > 0 such that ∀h ∈ R:  max(1,

|h| < δ5 =⇒ |f (x + h) − f (x) − φ(x)h| ≤ η|h|. Therefore limh→0 |f (x+h)−f|h|(x)−φ(x)h| = 0, and we thus have proved that f is differentiable at x and that f 0 (x) = φ(x). Since f is continuous, so is φ, and f is thus C 1 on Uu◦ . The theorem is thus proved. The previous theorem can be generalized as follows: Theorem 5.2.2 (General implicit function theorem). Let g : Rn × Rk → Rk be a C -mapping on a neighborhood of a point (x0 , y0 ) such that g(x0 , y0 ) = 0. If the linear map h → Dg[(x0 , y0 )](0, h) from Rk to Rk is invertible, then there exist a neighborhood U of x0 , a neighborhood V of y0 , a C 1 function f : U → V such that, for all x ∈ U , f (x) is the unique root y ∈ V of the equation g(x, y) = 0. 1

∂

g(x,f (x))

Furthermore, for all x ∈ U , f 0 (x) = − ∂x . ∂ g(x,f (x)) ∂y

Exercise 5.2.3. ∂ (1) In the one dimensional implicit function theorem, the hypothesis ∂y g(x0 , y0 ) 6= 2 2 0 is fundamental: consider the case g(x, y) := x + y , x0 = −1, y0 = 0, ∂ where ∂y g(x0 , y0 ) = 0. Prove that it is impossible to find a neighborhood U of x0 and a function f : U → R such that ∀x ∈ U : g(x, f (x)) = 0 (2) Adapt the proof of the dimensional implicit function theorem to prove the general one.

5. FRECHET DIFFERENTIAL

37

(3) Prove the next corollary, which is a particular case of the general implicit function theorem. (Hint consider the map g(x, y) = x − h(y).) Corollary 5.2.4 (Inverse function theorem). Let h : Rk → Rk be a C 1 mapping on a neighborhood of y0 ∈ Rk . If Dh[y0 ] is invertible, then there exists an open neighborhood W of h(y0 ), and a C 1 function f : W → Rk such that ∀x ∈ W : h(f (x)) = x. In particular: V := f (W ) is an open neighborhood of y0 and h is a bijection between V and W .