View
15
Download
3
Category
Preview:
Citation preview
Chapter 1
Metric and Normed Spaces
There are things which seem incredibleto most men who have not studied mathematics.
Archimedes of Syracuse (287-212 BC)
1.1 Metric Spaces
13.10.08
Definition 1.1.1. Let X be a set. A function
d : X×X →R+ := [0,∞) = {x∈R : x≥ 0}is called adistanceor metric on X, if for all x,y,z∈ X(D1) d(x,y) = 0⇔ x = y; (Positivity)
(D2) d(x,y) = d(y,x); (Symmetry)
(D3) d(x,z) ≤ d(x,y)+d(y,z); (△-inequality)In this case(X,d) is called ametric space.
Remark 1.1.2. For a metric space(X,d) we often write simply X if it is clearwhich metric we mean. If we consider more metric spaces at onetime, e.g. X andY, we will denote the corresponding distances by dX and dY, respectively.
1.1.1 Sequences in Metric Spaces
Definition 1.1.3. Let (X,d) be a metric space,(xn)n a sequence in X and x∈ X.We say that(xn)n converges to x in(X,d) [limnxn = x in (X,d) or xn → x in (X,d)as n→ ∞] if d(xn,x) → 0 in R as n→ ∞. We say that a sequence(xn)n in X isconvergentin (X,d), if there exists x∈ X such thatlimnxn = x in (X,d).
1
2 CHAPTER 1. METRIC AND NORMED SPACES
Lemma 1.1.4. (Unicity of the limit)Let X be a metric space,(xn)n a sequence in X and x,y ∈ X. If limnxn = x andlimnxn = y, then x= y.
Proof. d(x,y)≤ d(x,xn)+d(xn,y)→ 0 asn→∞. Henced(x,y) = 0⇒ x= y.
Exercise 1.1.5. (A bounded distance)ExerciseLet (X,d) be a metric space,(xn)n be a sequence in X and x0 ∈ X. Defined1(x,y) := min{1,d(x,y)} for x,y∈ X. Showthat (X,d1) is a metric space andthat xn → x0 in (X,d) if and only if xn → x0 in (X,d1).
Definition 1.1.6. Let X be a metric space and let(xn)n be a sequence in X. Apoint x∈ X is called anaccumulation point of (xn)n if for all ε > 0 and all n∈Nthere exists m≥ n such that xm ∈ B(x,ε). It is immediately clear that x is anaccumulation point of(xn)n if and only if there exists a subsequence(xnk)k whichconverges to x.
1.1.2 Topological Notions
14.10.08
Definition 1.1.7. Let X be a metric space.
• For x0 ∈ X and r> 0 the set
BX(x0, r) := {x∈ X : dX(x,x0) < r}
is called theopen ball with center x0 and radius r. If it is clear in whichmetric space we consider the open ball, then we write simply B(x0, r).
• A set O⊂ X is calledopen, if for all x ∈ O there exists r> 0, such that
B(x, r) ⊂ O.
• A set A⊂ X is calledclosed, if Ac := X \A is open.
• Theinterior of a set S⊂ X is denoted byInt(S) or S◦ and is defined as theunion of all open sets contained in S, that is,
Int(S) :=⋃
O⊂S openO.
1.1 METRIC SPACES 3
• The closure of a set S⊂ X (in X) is denoted byS and is defined as theintersection of all closed sets A containing S, that is,
S:=⋂
A⊃S closedA.
• Theboundary of a set S⊂ X is denoted by∂S and is defined by
∂S:= S\ Int(S).
Exercise 1.1.8. Exercise
• Let X be a metric space, x0 ∈ X and r> 0. Show that the so called openball B(x0, r) is in fact open.
• Let X be a metric space, x0 ∈ X and r> 0. Showthat
B(x0, r) := {y∈ X : dX(x,y) ≤ r}
is closed and that in general
B(x0, r) 6= B(x0, r).
Lemma 1.1.9.Let X be a metric space. Then an arbitrary union of open sets Oα(α ∈ I) is open.
Proof. Let O :=⋃
α∈I Oα . Forx∈ O there existsα0 ∈ I such thatx∈ Oα0. SinceOα0 is open, there existsr > 0 such thatB(x, r) ⊂ Oα0 ⊂ O. By definitionO isopen.
Exercise 1.1.10. Let X be a metric space.Showthat an arbitrary intersection Exerciseof closed sets Aα (α ∈ I) is closed.
Lemma 1.1.11.Let X be a metric space and S⊂ X. Then the following hold.
1. The interiorInt(S) of S in X is open.
2. The closureS of S in X is closed.
3. The finite intersection of open sets is open.
4. The finite union of closed sets is closed.
5. The boundary∂S of S is closed.
4 CHAPTER 1. METRIC AND NORMED SPACES
6. One hasInt(S) ⊂ S⊂ S.
7. One has that S is open if and only if S= Int(S).
8. One has thatS= X \ Int(X \S).
9. One has that S is closed if and only if S= S.
10. One has thatInt(Int(S)) = Int(S) andS= S.
11. One has that X and/0 are open and closed.
Proof.
1. Int(S) is the union of open sets and hence by Lemma 1.1.9 open.
2. S is the intersection of closed sets and hence by Exercise 1.1.10 closed.
3. Let O :=⋂n
k=1Ok with open setsOk and letx ∈ O. Thenx ∈ Ok for allk = 1, . . . ,n and hence there existsrk > 0 such thatB(x, rk) ⊂ Ok. Let r :=min{rk : k = 1, . . . ,n} > 0. ThenB(x, r) ⊂ Ok for all k = 1, . . . ,n. HenceB(x, r) ⊂ O and by definitionO is open.
4. LetA :=⋃n
k=1Ak whereAk is closed. ThenO :=⋂n
k=1X \Ak is open as thefinite intersection of open sets. HenceA = X \O is closed.
5. ∂S= S∩(X \ Int(S)) is the intersection of two closed sets and hence closed.
6. This follows immediately from the definition of the interior and the closure.
7. If S is open, thenS is a member of the union in the definition of Int(S) andhenceS⊂ Int(S). The converse inclusion follows from the statement above.If S= Int(S) thenS is open since the interior ofS is open.
8. Let A := X \ Int(X \S). ThenA is closed sinceO := Int(X \S) is open.S= X \ (X \S) ⊂ X \ Int(X \S) = A. HenceS⊂ A. On the other hand,U := X \S⊂ X\Sis open and henceU ⊂ Int(X\S). ThereforeS= X\U ⊃X \ Int(X \S) = A. Both cases together imply thatS= A.
9. By definition we get thatS is closed if and only ifX \S is open. By theabove,X \S is open if and only ifX \S= Int(X \S). This equality holds ifand only ifS= X \ Int(X \S) = S.
10. This follows immediately from the definition of the interior and the closureby using that Int(S) is open andS is closed.
1.1 METRIC SPACES 5
11. This follows directly by definition.
Lemma 1.1.12.Let X be a metric space. A set A⊂ X is closed, if and only if thelimit of every convergent sequence(xn)n ⊂ A in X is in A.
Proof. Let A ⊂ X be closed and letxn ∈ A be such that limnxn = x. We have toshow thatx∈ A. Assume thatx 6∈ A, thenx∈ O := X \A. SinceO is open thereexistsr > 0, such thatB(x, r) ⊂ O. Sincexn → x there existsn0 ∈ N such thatd(xn,x) < r for all n≥ n0. Hencexn ∈ B(x, r) ⊂ O = X \A, a contradiction.Assume now that the limit of every convergent sequence(xn)n ⊂ A is in A. Weshow thatX \A is open. For this letx∈ X \A. If for all n∈N there existsxn ∈ Asuch thatd(xn,x) ≤ 1/n, thenxn → x and hencex∈ A, a contradiction. Thereforethere existsn∈N such thatd(y,x) ≥ 1/n for all y∈ A. HenceB(x,1/n) ⊂ X \Aand thereforeX \A is open, that is,A is closed.
Definition 1.1.13. Let X be a metric space. Aneighbourhoodof a point x∈ X isa set U⊂ X such that there exists r> 0 with B(x, r) ⊂U.
Exercise 1.1.14. Let X be a metric space and x∈ X. Show that a set U⊂ X Exerciseis a neighbourhood of x if and only if x∈ Int(U). In particular, every open setcontaining x is a neighbourhood of x.
1.1.3 Separable Metric Spaces
15.10.08
Definition 1.1.15.
• A set C is calledcountable if there exists an injective mappingϕ : C→N.• A set D⊂ X in a metric space X is calleddense, if D = X.
• A metric space X is calledseparableif there is a countable dense subset.
Example 1.1.16.
1. Let X:=R and d: X×X →R+ be given by d(x,y) := |x−y|. Then(X,d)is a metric space. Let D:=Q be the set of all rational numbers, then D is acountable and dense subset of X. Hence X is a separable metricspace.
6 CHAPTER 1. METRIC AND NORMED SPACES
2. Let X:=R and d: X×X →R+ be given by d(x,y) := 1 if x 6= y and0 else.Then(X,d) is a metric space. Let O⊂ X be an arbitrary set. Then O isopen. In fact, for x∈ O and r := 0.5 the open ball B(x, r) is just{x} andhence B(x, r) ⊂ O. This shows that every set in X is open and hence everyset in X is closed. Therefore the only set D⊂ X for whichD = X is X itself.But X is not countable and therefore(X,d) is not separable.
Exercise 1.1.17. Determinethe convergent sequences in(R,d) where d(x,y) :=Exercise1 if x 6= y and0 else.
1.1.4 Complete Metric Spaces
Definition 1.1.18.Let X be a metric space.
• A sequence(xn)n in X is called aCauchy sequence, if for all ε > 0 thereexists N∈N such that d(xn,xm) ≤ ε for all n,m≥ N.
• A metric space X is calledcomplete, if every Cauchy sequence in X isconvergent.
Exercise 1.1.19. Let X be a metric space,(xn)n a convergent sequence in X.ExerciseShow that (xn)n is a Cauchy sequence. That is, every convergent sequence is aCauchy sequence.
Lemma 1.1.20.Let X be a metric space and(xn)n be a Cauchy sequence. Then(xn)n is convergent if and only if(xn)n has a convergent subsequence.
Proof. Assume that(xn)n is convergent. Then there existsx0∈X such thatxn→ x0as n → ∞, i.e. d(xn,x0) → 0 asn → ∞. Let (xnk)k be a subsequence. Thend(xnk,x0) → 0 ask→ ∞ and hencexnk → x0 ask→ ∞.Assume that(xn)n is a Cauchy sequence and the subsequence(xnk)k converges.We have to show that(xn)n is convergent. For this letx0 := limk xnk andε > 0.Since(xn)n is a Cauchy sequence there existsN such thatd(xn,xm) ≤ ε/2 forall n,m≥ N. Moreover, there existsk ≥ N such thatd(xnk,x0) ≤ ε/2. Henced(xn,x0) ≤ d(xn,xnk)+d(xnk,x0) ≤ ε for all n≥ N.
1.1.5 Compact Metric Spaces
Definition 1.1.21.A metric space X is calledcompact, if every sequence in X hasa convergent subsequence.
1.1 METRIC SPACES 7
Lemma 1.1.22.Every compact metric space X is complete.
Proof. This follows immediately from Lemma 1.1.20.
Definition 1.1.23. A metric space X is calledprecompact, if for all ε > 0 thereare finitely many points x1, . . . ,xn ∈ X such that
X =n⋃
k=1
B(xk,ε).
Theorem 1.1.24.Let X be a metric space. Then X is compact if and only if X isprecompact and complete.
Proof. Let X be a compact metric space. Then by Lemma 1.1.22X is complete.Assume thatX is not precompact. Then there existsε > 0 such thatX can not becovered by finitely manyε-balls. By induction there existsxk ∈X such thatxk+1 6∈⋃k
j=1B(x j ,ε). Thusd(xn,xk) ≥ ε for all k > n. By assumption there exists anaccumulation pointx0 of (xk)k. Then there arek > n such thatd(xk,x0) < ε/2 andd(xn,x0) < ε/2. Henceε ≤ d(xn,xk) ≤ d(xn,x0)+d(x0,xk) < ε, a contradiction.Assume now thatX is precompact and complete. Let(xn)n be a sequence inX.We have to show that(xn)n has a convergent subsequence. SinceX is precompactthere exists forε > 0 a y ∈ X such thatxn ∈ B(y,ε) for infinitely many n’s. Byinduction we get for allp ∈ N pointsyp ∈ X and infinite setsJp ⊂ N such thatJp+1⊂ Jp andxn∈B(yp,1/p) for all n∈ Jp. Forp∈Nwe defineϕ(p) to be thep-th element ofJp. Thenϕ :N→N is strictly monotone increasing andϕ(m) ∈ Jpfor all m≥ p. Hence forn,m≥ p we get that
d(xϕ(m),xϕ(n)) ≤ d(xϕ(m),yp)+d(yp,xϕ(n)) ≤ 2/p.It follows that (xϕ(m))m is a Cauchy sequence and by completeness convergent.
Definition 1.1.25. Let (X,d) be a metric space. A set K⊂ X is calledcompact,if K is a compact metric space with respect to the metric dK := d|K×K, that is,if every sequence in K has a convergent subsequence in X with limit in K. A setR⊂ X is calledrelatively compact if R is compact.Exercise 1.1.26. Let X be a metric space.Showthat R⊂X is relatively compact Exerciseif every sequence in R has a convergent subsequence in X.
Definition 1.1.27.We say that a metric space X has theHeine-Borel12 propertyif for every family{Oα : α ∈ I} of open sets with X⊂
⋃
α∈I Oα there are finitelymanyα1, . . . ,αn such that X⊂
⋃nk=1Oαk.
1Heinrich Eduard Heine (1821-1881)2Félix Edouard JustińEmile Borel (1871-1956)
8 CHAPTER 1. METRIC AND NORMED SPACES
Theorem 1.1.28. Let(X,d) be a metric space. Then the following are equivalent.20.10.08
1. X is compact.
2. X has the Heine-Borel property.
Proof. (1) ⇒ (2): Let {Oα : α ∈ I} be a family of open sets withX ⊂⋃
α∈I Oα .
• First we show that there existsε > 0 such that for allx∈X there existsα ∈ Isuch thatB(x,ε) ⊂ Oα . If not, we choose for alln∈N pointsxn ∈ X suchthatB(xn,1/n) 6⊂ Oα for all α ∈ I . Let x be an accumulation point of(xn)n.Then there existsα0 ∈ I such thatx ∈ Oα0. Moreover, there existsn ∈ Nsuch thatB(x,2/n) ⊂ Oα0. Choosem≥ n such thatd(xm,x) ≤ 1/n. ThenB(xm,1/m)⊂ B(xm,1/n)⊂ B(x,2/n)⊂ Oα0, which is a contradiction to theconstruction ofxm.
• SinceX is precompact, there arey1, . . . ,yn ∈ X such thatX ⊂⋃n
k=1B(yk,ε).By the above there exists for allk∈ {1, . . . ,n} anαk ∈ I such thatB(yk,ε)⊂Oαk. HenceX ⊂
⋃nk=1Oαk.
(2) ⇒ (1): Let (xn)n be a sequence inX. Assume that(xn)n has no accumulationpoint. Then for ally ∈ X there existsεy > 0 such thatxn ∈ B(y,εy) for only afinite number ofn’s. By assumption there existy1, . . . ,ym ∈ X such thatX ⊂⋃m
k=1B(yk,εyk). Hencexn ∈ X for only finitely manyn’s, a contradiction.
Theorem 1.1.29.Every compact metric space X is separable.
Proof. Forn∈N there existxn1, . . . ,xnmn ∈ X such thatX ⊂⋃mnk=1 B(xnk,1/n). Thenthe setD :=
{
xnk : n∈N,k = 1, . . . ,mn} is dense inX.Theorem 1.1.30.A set K⊂R is compact if and only if K is closed and bounded.Proof. This is a well known fact from Analysis I.
1.1.6 Continuous Functions
Definition 1.1.31(Continuous functions).Let X,Y be metric spaces, f: X →Y a mapping.
1. The mapping f is calledcontinuous inx∈X if limnxn = x in X implies thatlimn f (xn) = f (x) in Y for each sequence(xn)n in X.
2. The mapping f is calledcontinuous if f is continuous in each point x∈ X.
1.1 METRIC SPACES 9
3. The family of all continuous functions from X into Y is denoted by C(X,Y).
4. The mapping f is calledLipschitz3-continuous if there exists L∈ R suchthat dY( f (x), f (y)) ≤ L · dX(x,y) for all x,y ∈ X. The smallest such L iscalled theLipschitz constant for f .
5. The mapping f is called acontraction, if f is Lipschitz continuous withLipschitz constant L∈ [0,1].
6. The mapping f is called astrict contraction if f is Lipschitz continuouswith Lipschitz constant L∈ [0,1).
Exercise 1.1.32. Let X and Y be a metric spaces and f: X → Y be Lipschitz Exercisecontinuous.Showthat f is continuous, i.e. every Lipschitz continuous function iscontinuous.
Exercise 1.1.33. Showthat f : X →Y is continuous in x∈ X if and only if Exercise∀ε > 0∃δ > 0 such that dX(x,y) ≤ δ implies dY( f (x), f (y)) ≤ ε.
Theorem 1.1.34(Characterization of Continuity). Let X and Y be metric spacesand let f be a mapping from X into Y.
1. Let x∈ X. Then the following assertions are equivalent:
(a) f is continuous in x.
(b) f−1(V) is a neighbourhood of x for each neighbourhood V of f(x).
2. The following assertions are equivalent:
(a) f is continuous.
(b) f−1(O) is open in X for all open sets O in Y.
Proof.
1. (a) ⇒ (b): Assume thatf is continuous inx and letV be a neighbourhoodof f (x). Then there existsε > 0 such thatB( f (x),ε) ⊂V. By continuity off in x there existsδ > 0 such thatdX(x,y)≤ δ implies thatdY( f (x), f (y))≤ε/2. HenceBX(x,δ ) ⊂ f−1(BY( f (x),ε)) ⊂ f−1(V), that is, f−1(V) is aneighbourhood ofx.(b) ⇒ (a): Let ε > 0. ThenV := B( f (x),ε) is a neighbourhood off (x)and henceU := f−1(V) is a neighbourhood ofx. In Particular, there existsδ > 0 such thatB(x,2δ ) ⊂ f−1(V). It follows that wheneverdX(x,y) ≤ δthen f (y) ∈V = BY( f (x),ε), i.e. dY( f (x), f (y)) < ε ≤ ε.
3Rudolf Otto Sigismund Lipschitz (1832-1903)
10 CHAPTER 1. METRIC AND NORMED SPACES
2. (a) ⇒ (b): Assume thatf is continuous and letO ⊂ Y be open. Then forall x ∈ f−1(O) the setO is a neighbourhood off (x) and hencef−1(O) isa neighbourhood ofx. This shows thatf−1(O) = Int( f−1(O)) and henceopen.(b) ⇒ (a): Let x∈ X be fixed and letV be a neighbourhood off (x). Thenthere existsε > 0 such thatO := B(x,ε) ⊂ V. Hence f−1(O) is open andcontainsx, this shows thatf−1(O) is a neighbourhood ofx. Consequently,f−1(V) ⊃ f−1(O) is a neighbourhood ofx. By the previous statementf iscontinuous inx and sincex∈ X was arbitrary, the proof is complete.
Exercise 1.1.35.ExerciseLet X,Y,Z be metric spaces.Showthat if f : X →Y and g:Y → Z are continuous,then g◦ f : X → Z is continuous, where g◦ f : X → Z is given by x7→ g( f (x)).
Theorem 1.1.36. Let (X1,d1) and(X2,d2) be metric spaces and f: X1 → X2 be22.10.08continuous. If X1 is compact, then f(X1) is compact.
Proof. Let (yn)n be a sequence inf (X1). Then there existxn ∈ X1 such thatyn =f (xn). SinceX1 is compact there exists a convergent subsequence(xnk)k, sayx0 =limk xnk. Henceynk = f (xnk)→ f (x0), that is,(yn)n has a convergent subsequence.
Theorem 1.1.37.Let X be a compact metric space and f: X → R continuous.Then there exist a,b∈ X such that
f (a) ≤ f (x) ≤ f (b) ∀x∈ X.
Proof. By Theorem 1.1.36f (X) is a compact setR and hence bounded and closedand hence there existm,M ∈ f (X) such thatm≤ y ≤ M for all y ∈ f (X). For aandb such thatf (a) = mand f (b) = M we get that
f (a) ≤ f (x) ≤ f (b) ∀x∈ X.
Remark 1.1.38.This shows that f is a bounded function which has a maximumand a minimum.
Definition 1.1.39. Let X be a metric space and A,B ⊂ X. Then we define thedistance from x∈ X to A by
d(x,A) := inf {d(x,y) : y∈ A}
1.1 METRIC SPACES 11
and the distance between A and B by
d(A,B) := inf {d(x,y) : x∈ A,y∈ B} = inf {d(x,B) : x∈ A} .
Exercise 1.1.40. Let X be a metric space and A⊂X. Showthat d(·,A) : X →R+ Exerciseis continuous.
Theorem 1.1.41.Let X be a metric space, A⊂ X compact and B⊂ X closed. IfA and B are disjoint, then d(A,B) > 0.
Proof. By Exercise 1.1.40 the functionx 7→ d(x,B) : A→R+ is continuous. SinceB is closed we get thatd(x,B) > 0 for all x∈ A. By Theorem 1.1.37 it follows that
d(A,B) = infx∈A
d(x,B) = minx∈A
d(x,B) > 0.
1.1.7 The Banach Fixed Point Theorem
Definition 1.1.42.Let X be a metric space and letϕ be a mapping from X into X.The iterations ofϕ are defined recursively by
ϕ1 := ϕ,ϕn+1 := ϕn◦ϕ for n∈N.
Theorem 1.1.43(Banach’s4 fixed point theorem). ImportantLet (X,d) be a complete metric space and letϕ : X → X be such that
d(ϕn(x),ϕn(y)) ≤ qnd(x,y) ∀x,y∈ M, n∈N,where qn ≥ 0 are such that∑nqn < ∞. Thenϕ has a uniquefixed point, that is,there exists a uniqueξ ∈ X such thatϕ(ξ ) = ξ .
Proof.
• Existence:Let x0 ∈ X be arbitrary. We setxn := ϕn(x0). Then
d(xn,xn−1) = d(ϕn−1(ϕ(x0)),ϕn−1(x0)) ≤ qn−1d(ϕ(x0),x0)4Stefan Banach (1892-1945)
12 CHAPTER 1. METRIC AND NORMED SPACES
for all n∈N, n≥ 2. Hence for alln,m∈N we get by the△-inequalityd(xn+m,xn) ≤ d(xn+m,xn+m−1)+ · · ·+d(xn+1,xn)
≤ (qn+m−1+ · · ·+qn)d(ϕ(x0),x0)
≤ d(ϕ(x0),x0) ·∞
∑k=n
qk.
Hence(xn)n is a Cauchy sequence inX. By completeness we get that thereexistsξ ∈ X such thatxn → ξ as n → ∞. Using thatϕ is continuous, itfollows that
ϕ(ξ ) = limn
ϕ(xn) = limn
xn+1 = ξ ,
that is,ξ is a fixed point ofϕ.
• Unicity:Let η ∈ X be an fixed point ofϕ, i.e. ϕ(η) = η. Thenϕn(η) = η for alln∈N. Hence
d(ξ ,η) = d(ϕn(ξ ),ϕn(η)) ≤ qnd(ξ ,η) → 0 asn→ ∞.
We get thatd(ξ ,η) = 0 and henceη = ξ .
Corollary 1.1.44. Let X be a complete metric space and letϕ : X → X be a strictImportantcontraction. Thenϕ has a unique fixed point.
Proof. Let L ∈ [0,1) be the Lipschitz-constant ofϕ and letqn := Ln. Then
d(ϕn(x),ϕn(y)) ≤ L ·d(ϕn−1(x),ϕn−1(y)) ≤ ·· · ≤ Lnd(x,y).
With qn := Ln all assumptions of Theorem 1.1.43 are fulfilled and henceϕ has aunique fixed point.
1.1.8 Product of Metric Spaces
Definition 1.1.45. Let (X1,d1), . . . ,(Xn,dn), N ∋ n ≥ 2, be metric spaces. Thenthe product X := ∏nk=1Xk is a metric space with the metric d
1 : X ×X → R+given by d1(x,y) := ∑nk=1dk(xk,yk) where x= (x1, . . . ,xn) and y= (y1, . . . ,yn).
Remark 1.1.46.In the following we will always use the metric d1 from Definition1.1.45 for a product of metric spaces unless something else is stated.
1.2 NORMED SPACES 13
Exercise 1.1.47.Exercise
1. Let(X,d) be a metric space.Showthat d : X×X →R is continuous.2. Let X and Y be metric spaces and let(xn)n be a sequence in X×Y. Show
that xn = (xn1,xn2) converges to x
0 = (x01,x02) in X ×Y if and only if(xn1)n
converges to x01 in X and(xn2)n converges to x
02 in Y .
1.2 Normed Spaces
The most important examples of metric spaces are normed spaces and their sub-sets. For a vector spaceE we consider a distance which is in some sense compat-ible with the structure of the vector space.
Definition 1.2.1(Norm).Let E be a vector space overK=R or C. A norm on E is a function‖ · ‖ : E →R+ such that for all x,y∈ E, λ ∈K(N1) ‖x‖ = 0 ⇔ x = 0; (Positivity)
(N2) ‖λx‖ = |λ | · ‖x‖; (Homogeneity)
(N3) ‖x+y‖ ≤ ‖x‖+‖y‖. (Subadditivity)
In this case we call(E,‖ · ‖) a normed vector space.
Remark 1.2.2. For a normed vector space(E,‖ · ‖) we often write simply E if 27.10.08it is clear which norm we mean. If we consider more normed vector spaces at thesame time, e.g. E and F, we denote the corresponding norms by‖ · ‖E and‖ · ‖F ,respectively.
1.2.1 Normed Vector Spaces as Metric Spaces
Definition 1.2.3. Let E be a normed vector space overK. Then we define thecorresponding distance dE : E×E →R+ on E by
dE(x,y) := ‖x−y‖E.
Remark 1.2.4. In this sense every normedK-vector space is also a metric space.Consequently, we will use all notions of metric spaces also in normed vectorspaces - for example convergence of sequences, continuous functions, open andclosed sets, etc. and in particular also Banach’s Fixed Point Theorem.
14 CHAPTER 1. METRIC AND NORMED SPACES
Exercise 1.2.5. Exercise
• Let (E,‖ · ‖E) be a normedK-vector space.Showthat (E,dE) is in fact ametric space, where dE is the corresponding distance on E.
• Let E be aK-vector space and let d be a distance on E such that d istranslation invariant and homogeneous, that is, d(x+z,y+z) = d(x,y) forall x,y,z∈ E and d(λx,λy) = |λ |d(x,y) for all x,y∈ E andλ ∈K. Showthat there is a norm‖ · ‖E on E such that d(x,y) = ‖x−y‖E.
Definition 1.2.6. Let E be a vector space and‖·‖1 and‖·‖2 be norms on E. Thenthese two norms are calledequivalent if there are constants c1,c2 > 0 such that
c1‖x‖1 ≤ ‖x‖2 ≤ c2‖x‖1
for all x ∈ E.
Exercise 1.2.7.Exercise
• Let E :=R2, ‖x‖1 := |x1|+ |x2| and‖x‖∞ := max{|x1|, |x2|}. Showthat thenorms‖ · ‖1 and‖ · ‖∞ are equivalent.
• Let E be aK-vector space,‖ · ‖1 and‖ · ‖2 be two equivalent norms on Eand let xn,x∈ E. Showthat xn → x in (E,‖ · ‖1) iff xn → x in (E,‖ · ‖2).This shows that equivalent norms yield the same notion of convergence.
Definition 1.2.8. Let E be a normed vector space.
• A subset S⊂ E is calledbounded, if supx∈S‖x‖ < ∞.
• A sequence(xn)n in E is calledbounded, if supn∈N ‖xn‖ < ∞.Lemma 1.2.9(Properties in normed vector spaces).Let E be a normed vector space, xn,yn,x,y∈ E, λn,λ ∈K.
1. Each convergent sequence in E is bounded.
2. Every Cauchy sequence in E is bounded.
3. If limnxn = x in E, limnyn = y in E andlimn λn = λ inK thenlim
n(xn+yn) = x+y and lim
nλnxn = λx.
That is, the mappings E×E → E, (x,y) 7→ x+y andK×E → E, (λ ,x) 7→λx are continuous(→ see 1.1.8 Product of Metric Spaces).
1.2 NORMED SPACES 15
4. One has,|‖x‖−‖y‖| ≤ ‖x−y‖ for all x,y∈ E.
5. In particular, that mapping‖ · ‖ : E →R+, x 7→ ‖x‖ is continuous.Proof.
1. Let (xn)n be a convergent sequence. Letx := limnxn. Then there existsN ∈N such that‖x−xn‖ ≤ 1 for all n≥ N. Then
supn‖xn‖ ≤ max{‖x1‖,‖x2‖, . . . ,‖xN−1‖,‖x‖+1} < ∞.
2. Let (xn)n be a Cauchy sequence inE. Then forε := 1 there existsn0 suchthat‖xn−xm‖ ≤ 1 for all n,m≥ n0. Hence‖xm‖ ≤ 1+‖xn0‖ for all m≥ n0.Therefore
supn‖xn‖ ≤ max{‖x1‖, . . . ,‖xn0−1,‖xn0‖+1‖} .
3. • ‖(xn+yn)− (x+y)‖ ≤ ‖xn−x‖+‖yn−y‖→ 0.• Let M := supn‖xn‖ < ∞. Then
‖λnxn−λx‖ = ‖(λn−λ )xn +λ (xn−x)‖≤ |λn−λ |M +λ‖xn−x‖→ 0.
4. One has
‖x‖ = ‖x−y+y‖ ≤ ‖x−y‖+‖y‖ ⇒ ‖x‖−‖y‖ ≤ ‖x−y‖.
By replacingx andy we get that‖y‖−‖x‖≤‖x−y‖ and hence the assertion.
5. Let (xn)n be a sequence inE converging tox. Then we have to show that(‖xn‖)n converges to‖x‖ in R. This follows from
|‖xn‖−‖x‖| ≤ ‖xn−x‖ → 0.
Definition 1.2.10(Banach Space).A normed vector space which is complete is called aBanach space.
16 CHAPTER 1. METRIC AND NORMED SPACES
1.2.2 Examples of Normed Vector Spaces
1.2.2.1 The Sequence Spacesℓp
Definition 1.2.11(The spacesℓp).For p∈ [1,∞) we letℓp be the set of allK-valued sequences x= (xn)n such that
‖x‖p :=(
∞
∑n=1
|xn|p)
1p
< ∞,
that is, ℓp ={
(xn)n ⊂K : ‖x‖p < ∞}. The setℓ∞ is the set of all boundedK-valued sequences and we set for x∈ ℓ∞
‖x‖∞ := supn|xn|.
Lemma 1.2.12.For p∈ [1,∞] the setℓp is a vector space with respect to scalarmultiplication and componentwise addition, that is, for x= (xn)n,y = (yn)n ∈ ℓp,λ ∈K
(x+y)n := xn +yn ∀n∈N, (λx)n := λxn ∀n∈N.Proof. Let p∈ [1,∞). Then forx,y∈ ℓp
|xn+yn|p ≤ (|xn|+ |yn|)p ≤ (2max{|xn|, |yn|})p ≤ 2pmax{|xn|p, |yn|p}≤ 2p(|xn|p+ |yn|p) .
It follows that‖x+y‖p < ∞ and hencex+y∈ ℓp. Moreover, forx∈ ℓp andλ ∈Kthe productλx is in ℓp since‖λx‖p = |λ |‖x‖p < ∞. Henceℓp is aK-vector space.Let p = ∞ andx,y∈ ℓ∞. Since the sum of two bounded sequences is bounded itfollows that that the sumx+y is in ℓ∞. Sinceλx is also a bounded sequence forλ ∈K andx∈ ℓ∞ it follows thatℓ∞ is aK-vector space.Remark 1.2.13. We will show that‖ ·‖p is a norm onℓp for p∈ [1,∞]. It is clear29.10.08that‖x‖p = 0 iff x = 0 and that‖λx‖p = |λ |‖x‖p for λ ∈K and x∈ ℓp. To provethe△-inequality we need some more preparation.Definition 1.2.14.Let p∈ [1,∞]. A number q∈ [1,∞] with 1/p+1/q= 1 is calledthe conjugate index to p and is denoted by p′. Note that in this context1/∞ isdefined to be zero.
Lemma 1.2.15(Hölder’s5 inequality - sequences). Let p∈ [1,∞] and q∈ [1,∞]be the conjugate index to p. For x∈ ℓp and y∈ ℓq we have
∞
∑n=1
|xnyn| ≤ ‖x‖p‖y‖q.
5Otto Ludwig Hölder (1859-1937)
1.2 NORMED SPACES 17
Proof. Let p ∈ (1,∞), thenq ∈ (1,∞). We may assume thatx,y 6= 0, otherwisethe inequality is trivial. Moreover, we can assume that‖x‖p = ‖y‖q = 1 (for thisreplacexn by xn/‖x‖p andyn by yn/‖y‖q). Hence it remains to show that
∞
∑n=1
|xn| · |yn| ≤ 1.
Let θ := 1/p ∈ (0,1). Then 1− θ = 1/q. Since log :(0,∞) → R is a concavefunction we get that fora,b > 0, θ log(a)+(1−θ) log(b) ≤ log(θa+(1−θ)b).Applying exp to both sides we get that
aθ b1−θ ≤ θa+(1−θ)b.
It follows that
|xn| · |yn| = (|xn|p)θ (|yn|q)1−θ ≤ θ |xn|p+(1−θ)|yn|q. (1.2.1)
Summing up and using that‖x‖p = 1 = ‖y‖q gives∞
∑n=1
|xn| · |yn| ≤ θ∞
∑n=1
|xn|p+(1−θ)∞
∑n=1
|yn|q = θ +(1−θ) = 1.
The easy casesp = 1,q = ∞ andp = ∞,q = 1 are left as an exercise. Exercise
Lemma 1.2.16(ℓp is a normed vector space).For p∈ [1,∞] the spaces(ℓp,‖ · ‖p) are normedK-vector spaces.Proof. By Lemma 1.2.12 and Remark 1.2.13 it remains to show that the normsatisfies the△-inequality. Letp ∈ (1,∞), q the conjugate index andx,y ∈ ℓp.Then(p−1)q = p and hence(|xn +yn|p−1)n ∈ ℓq. From Hölder’s inequality weget
‖x+y‖pp =∞
∑n=1
|xn+yn| · |xn+yn|p−1
≤∞
∑n=1
|xn| · |xn+yn|p−1+∞
∑n=1
|yn| · |xn+yn|p−1
≤ ‖x‖p(
∞
∑n=1
|xn +yn|p)1/q
+‖y‖p(
∞
∑n=1
|xn+yn|p)1/q
= (‖x‖p+‖y‖p)‖x+y‖p/qp
Hence the assertion follows sincep− p/q = 1.The easy casesp = 1,q = ∞ andp = ∞,q = 1 are left as an exercise. Exercise
18 CHAPTER 1. METRIC AND NORMED SPACES
Exercise 1.2.17. Exercise
1. Showthat ℓ1 andℓ∞ are normed vector spaces.
2. Let p∈ [1,∞] and let(xn)n with xn = (xkn)k be a convergent sequence inℓpwith limit x0 = (xk0)k ∈ ℓp. Show that for fixed k∈ N the sequence(xkn)nconverges inK to xk0.
Lemma 1.2.18.For p∈ [1,∞] the space(ℓp,‖ · ‖p) is a Banach space.
Proof. Let p∈ [1,∞) and let(xn)n be a Cauchy sequence inℓp, that is,xn = (xkn)k ∈ℓp for all n∈N. Then, for fixedk∈N, the sequence(xkn)n is a Cauchy sequenceinK and hence convergent toxk0 ∈K. Thenx0 = (xk0)k is a candidate for the limitof the sequence(xn)n. It remains two things to show: The first is thatx0 ∈ ℓp andthe second is thatxn → x0 in ℓp. Forε > 0 there existsN such that
‖xn−xm‖p ≤ ε ∀m,n≥ N.
In particular, for allM ∈N we get thatM
∑k=1
|xkn−xkm|p ≤ ‖xn−xm‖pp ≤ ε p ∀m,n≥ N.
Taking the limit forn→ ∞ we get that for allM ∈N andm≥ NM
∑k=1
|xk0−xkm|p ≤ ε p.
SinceM ∈N was arbitrary it follows that∞
∑k=1
|xk0−xkm|p ≤ ε p ∀m≥ N.
Consequently, we get thatx0−xN ∈ ℓp and hencex0 = (x0−xN)+xN ∈ ℓp. More-over, we have seen that‖x0−xm‖p≤ ε for all m≥N and sinceε > 0 was arbitrary,it follows that‖xm−x0‖p → 0 asm→ ∞.The casep = ∞ is left as an exercise (see Sheet 2, Exercise 1).Exercise
Definition 1.2.19. By c we denote theK-vector space consisting of allK-valuedconvergent sequences, that is,
c :={
(xn)n ⊂K : limn
xn exists inK} .
1.2 NORMED SPACES 19
The subspace ofc consisting of all sequences converging to zero is denoted by
c0 :={
(xn)n ⊂K : limn
xn = 0}
and the subspace ofc0 consisting of all sequences(xn)n from c0 such that xn = 0except for a finite number of n’s is denoted by
c00 :={
(xn)n ⊂K : xn = 0 except for a finite number of n′s} .Lemma 1.2.20.The vector spacec00 is dense inℓp for p∈ [1,∞).
Proof. Let x∈ ℓp andε > 0. Then there existsN such that∞
∑n=N
|xn|p ≤ ε p.
Let (yn)n ∈ c00 be given byyn := xn for n < N andyn := 0 for n≥ N. Then
‖y−x‖p =(
∞
∑n=N
|yn−xn|p)1/p
=
(
∞
∑n=N
|xn|p)1/p
≤ ε.
We have seen that for everyε > 0 and allx ∈ ℓp there existsy ∈ c00 such that‖x−y‖p ≤ ε. This means thatc00 is dense inℓp.
Exercise 1.2.21. Showthatc00 is not dense inℓ∞. Exercise
Lemma 1.2.22. The Banach space(ℓp,‖ · ‖p) is separable for p∈ [1,∞). 03.11.08
Proof. It is sufficient to show that(c00,‖ · ‖p) is separable. In fact, ifD ⊂ c00 is acountable and dense set in(c00,‖·‖p), thenD is dense inℓp. SinceK is separable,there exists a countable and dense setD0 inK. Then
D := {(xn)n ⊂ c00 : xn ∈ D0∪{0}}
is dense inℓp. In fact, for ε > 0 andx = (xn)n ∈ c00 there existsy = (yn)n ⊂ Dsuch that|xn−yn|p ≤ 2−nε p for all n∈N. Then
‖x−y‖pp =∞
∑n=1
|xn−yn|p ≤ ε p∞
∑n=1
2−n = ε p.
Hence‖x−y‖p≤ ε and sinceε > 0 was arbitrary, we see thatD is dense inℓp.
Summary 1.2.23.We have proved that(ℓp,‖ · ‖p) is a separable Banach space Importantfor p∈ [1,∞) and a Banach space for p= ∞. From Exercise 1 on sheet 2 we knowthat ℓ∞ is not separable.
20 CHAPTER 1. METRIC AND NORMED SPACES
1.2.2.2 Function Spaces
Definition 1.2.24(Space of Bounded Functions).Let Ω be a set. AK-valued function f defined onΩ is called bounded onΩ if‖ f‖∞ := supx∈Ω | f (x)| < ∞. The family of all bounded functions fromΩ intoK isdenoted by
Fb(Ω) := F b(Ω,K) := { f : Ω →K bounded} .
and‖ · ‖∞ is called thesup-norm.Lemma 1.2.25.Let Ω be a set. Then the familyF b(Ω) equipped with the sup-norm‖ · ‖∞ : F b(Ω) →R+ is a Banach space with respect to pointwise additionand scalar multiplication, that is, for f,g∈ F b(Ω), λ ∈K
( f +g)(x) := f (x)+g(x) ∀x∈ Ω, (λ f )(x) := λ f (x) ∀x∈ Ω.
Proof. Let ( fn)n be a Cauchy sequence inF b(Ω). Then forx ∈ Ω we get that| fn(x)− fm(x)| ≤ ‖ fn− fm‖∞. Hence( fn(x))n is a Cauchy sequence inK andf (x) := limn fn(x) exists. We will show thatf ∈ F b(Ω) and fn → f in F b(Ω).Let ε > 0. Then there existsn0 such that| fn(x)− fm(x)| ≤ ‖ fn− fm‖∞ ≤ ε for alln,m≥ n0 and allx∈ Ω. Taking the limit asm→ ∞ we get
| fn(x)− f (x)| ≤ ε ∀n≥ n0, x∈ Ω. (1.2.2)
In particular,| f (x)| ≤ | f (x)− fn0(x)|+ | fn0(x)| ≤ ε +‖ fn0‖∞ for all x∈ Ω. Hencef ∈ F b(Ω). Moreover, from (1.2.2) we get that‖ f − fn‖∞ ≤ ε for all n≥ n0 andsinceε > 0 was arbitrary we get that limn fn = f in F b(Ω).
Remark 1.2.26.Sinceℓ∞ = F b(N) we proved again thatℓ∞ is a Banach space.Note that on both vector space we consider the same norm, namely the sup-norm.
Definition 1.2.27.LetΩ be a metric space. ByC(Ω) we denote the vector space ofall continuous functions fromΩ intoK. The subspace Cb(Ω) is given by Cb(Ω) :=C(Ω)∩F b(Ω) and is equipped with the sup-norm‖ · ‖∞ defined above.Lemma 1.2.28.Let Ω be a metric space. Then Cb(Ω) is a Banach space withrespect to the sup-norm‖ · ‖∞.Proof. Let ( fn)n be a Cauchy sequence inCb(Ω). Then( fn)n is also a Cauchysequence in the Banach spaceF b(Ω) and hence there existsf ∈ F b(Ω) suchthat( fn)n converges uniformly tof , that is,‖ fn− f‖∞ → 0 asn→ ∞. It remainsto show thatf is continuous. For this letx ∈ Ω and (xn)n be a sequence inΩconverging tox and letε > 0. There there existsN such that
‖ fm− f‖∞ ≤ ε/2 ∀m≥ N.
1.2 NORMED SPACES 21
From this we get that
| f (xn)− f (x)| ≤ | f (xn)− fN(xn)|+ | fN(xn)− fN(x)|+ | fN(x)− f (x)|≤ ε + | fN(xn)− fN(x)|.
Since fN is continuous it follows that
limsupn
| f (xn)− f (x)| ≤ ε
and sinceε > 0 was arbitrary we get thatf (xn) → f (x) asn→ ∞.
Remark 1.2.29.Let K be a compact metric space. By Theorem 1.1.37 applied tothe function K→R, x 7→ | f (x)| we get that C(K) = Cb(K).1.2.3 Product of Banach Spaces
Definition 1.2.30. Let (X1,‖ · ‖X1), . . . ,(Xn,‖ · ‖Xn) be Banach spaces over thesame fieldK. Then the product X:= ∏nk=1Xk is a Banach space with the norm‖ · ‖X : X×X given by
‖u‖X :=(
n
∑k=1
‖uk‖2Xk
)1/2
.
Exercise 1.2.31. ExerciseLet (X1,‖ · ‖X1), . . . ,(Xn,‖ · ‖Xn) be Banach spaces over the same fieldK.
1. Checkthat X := ∏nk=1Xk is a vector space overK.2. Checkthat‖ · ‖X given in Definition 1.2.30 is a norm on X.
3. Checkthat (X,‖ · ‖X) is a Banach space.
1.2.4 An Application of Banach’s Fixed Point Theorem
In this subsection we show how we can prove existence and uniqueness of classi-cal solutions to differential equations with help of Banach’s Fixed Point Theorem.
Theorem 1.2.32.Let τ > 0 and f : [0,τ]×R→ R a continuous function suchthat
| f (t,x)− f (t,y)| ≤ L · |x−y|
22 CHAPTER 1. METRIC AND NORMED SPACES
for all t ∈ [0,τ] and all x,y∈R with L∈ [0,∞). Then for each u0 ∈R there existsa unique continuously differentiable function u: [0,τ] →R such that
{
u′(t) = f (t,u(t)) (t ∈ [0,τ]),u(0) = u0.
Proof. Let T : C([0,τ])→C([0,τ]) be given by
(Tu)(t) := u0+∫ t
0f (s,u(s)) ds.
We will show by induction that
|(Tnu)(t)− (Tnv)(t)| ≤ Ln
n!tn‖u−v‖∞ for all t ∈ [0,τ]. (1.2.3)
Forn = 1 we get
|(Tu)(t)− (Tv)(t)| =∣
∣
∣
∣
∫ t
0f (s,u(s))− f (s,v(s)) ds
∣
∣
∣
∣
≤ L∫ t
0|u(s)−v(s)| ds≤ Lt‖u−v‖∞.
We show that the assertion hold forn+1 assuming that it holds forn.
|(Tn+1u)(t)− (Tn+1v)(t)| =∣
∣
∣
∣
∫ t
0f (s, [Tnu](s))− f (s, [Tnv](s)) ds
∣
∣
∣
∣
≤ L∫ t
0|(Tnu)(s)− (Tnv)(s)| ds
≤ Ln+1
n!
∫ t
0sn ds· ‖u−v‖∞
=Ln+1
(n+1)!tn+1 ‖u−v‖∞
Hence (1.2.3) is proved. Taking the supremum over allt ∈ [0,τ] we get
‖Tnu−Tnv‖∞ ≤Lnτn
n!‖u−v‖∞
for all u,v∈C([0,τ]). By the Banach Fixed Point Theorem there exists a uniqueu∈C([0,τ]) such thatTu= u. Hence
u(t) = u0+∫ t
0f (s,u(s)) ds ∀t ∈ [0,τ].
This shows the existence. To show uniqueness letv be a solution of the differentialequation above. ThenTv= v and hence (by uniqueness of the fixed point)v = u.Hence uniqueness is proved.
1.2 NORMED SPACES 23
1.2.5 * The Theorem of Arzela-Ascoli
Lemma 1.2.33(Diagonal Sequences). Let (Xp,dp) be a compact metric space 05.11.08for each p∈N and let(xpn)n be a sequence in Xp. Then there existsϕ :N→Nstrictly monotone increasing, such that(xpϕ(n))n is convergent for each p∈N.Proof. By induction there exist infinite setsJp⊂N such that the sequence(xpn)n∈Jpis convergent. Moreover, we can chooseJp such thatJp+1 ⊂ Jp for all p∈N. Letϕ(p) be thep-th element ofJp. Thenϕ is strictly monotone increasing. LetJ := {ϕ(p) : p∈N}. ThenJ \ Jp is finite and hence(xpϕ(n))n is convergent foreachp∈N.Definition 1.2.34(Infinite Products of Metric Spaces).Let (Xn,dn) be a metric space for each n∈ N. Then the infinite product X:=∏n∈NXn is a metric space with respect to the metric d given by
d(x,y) :=∞
∑n=1
2−nmin{1,dn(xn,yn)}
where x= (xn)n and y= (yn)n.
Exercise 1.2.35. Exercise
1. Let X := ∏n∈NXn be the product of metric spaces(Xn,dn) and let d bethe distance given in Definition 1.2.34.Show that a sequence(uk)k in Xconverges if and only if(ukn)k converges in Xn for each n∈N.
2. Let X:= ∏n∈NXn be the product of compact metric spaces(Xn,dn) and letd be the distance given in Definition 1.2.34.Showthe(X,d) is a compactmetric space.Hint: Use Lemma 1.2.33
Our aim is now to characterize the compact sets in the Banach spaceC(K)whereK is a compact metric space. It is immediately clear that everycompactset is closed and bounded (see Exercise 1.2.38), but the converse is not true, ingeneral.
Definition 1.2.36. Let X be a metric space and H a subset of C(X). We say thatH is equicontinuous in x ∈ X if for eachε > 0 there exists aδ > 0 such that| f (x)− f (y)| ≤ ε for all y ∈ X with d(x,y) ≤ δ and all f ∈ H. We say that H isequicontinous if H is equicontinuous in each x∈ X. Moreover, we say that H isuniformly equicontinuous if for all ε > 0 there existsδ > 0 such that
| f (x)− f (y)| ≤ ε ∀x,y∈ X with d(x,y) ≤ δ and all f ∈ H.Finally, we say H ispointwise boundedif supf∈H | f (x)| < ∞ for all x ∈ X.
24 CHAPTER 1. METRIC AND NORMED SPACES
Remark 1.2.37. That H is uniformly equicontinuous means that the family His equicontinuous with aδ > 0 independent of x. If the set H consists only ofone function f , then H is equicontinuous means that f is continuous and H isuniformly equicontinuous means that f is uniformly continuous.
Exercise 1.2.38.Exercise
1. Let X be a metric space and K⊂ X be a compact set.Showthat K is closedand bounded.
2. Let K be a metric space and let H be a subset of C(K). Showthat if H isbounded then H is pointwise bounded.
Lemma 1.2.39.Let K be a compact metric space and let H be an equicontinuousfamily of C(K). Then H is uniformly equicontinuous.
Proof. Let ε > 0. Then for eachx∈K there existsδx > 0 such that| f (x)− f (y)| ≤ε/2 wheneverd(x,y) ≤ 2δx and f ∈ H. From the Heine-Borel-Property ofK weget that
K ⊂m⋃
j=1
B(x j ,δx j )
for somex1, . . . ,xm ∈ K. Let δ := min{
δx j : j = 1, . . . ,m}
> 0. Now letx,y∈ Kwith d(x,y) ≤ δ . Then there existsj ∈ {1, . . . ,m} such thatx∈ B(x j ,δx j ). Henced(y,x j) ≤ d(y,x)+d(x,x j) ≤ δ +δx j ≤ 2δx j . Hence
| f (x)− f (y)| ≤ | f (x)− f (x j)|+ | f (x j)− f (y)| ≤ ε/2+ ε/2 = ε
for all f ∈ H.
Theorem 1.2.40.Let K be a compact metric space and( fn)n be a equicontinuoussequence in C(K), that is, H:= { fn : n∈N} is equicontinuous. Let D be a densesubset of K. Iflimn fn(x) exists for all x∈ D then( fn)n is convergent in C(K).
Proof. SinceC(K) is a Banach space it is sufficient to show that( fn)n is a Cauchysequence inC(K). Let ε > 0. Then by Lemma 1.2.39 there existsδ > 0 such that
| fn(x)− fn(y)| ≤ ε/5 ∀x,y∈ K with d(x,y) ≤ δ
and for alln ∈ N. SinceK is compact there existx1, . . . ,xm ∈ K such thatK =⋃m
j=1B(x j ,δ ). SinceB(x j ,δ )∩D 6= /0 we choosed j ∈B(x j ,δ )∩D. By assumptionthere existsn0 ∈N such that
| fn(d j)− fm(d j)| ≤ ε/5 ∀n,m≥ n0, ∀ j = 1, . . . ,m.
1.2 NORMED SPACES 25
Let n,m≥ n0 andx∈ K. Then there existsj ∈ {1, . . . ,m} such thatx∈ B(x j ,δ ).Hence
| fn(x)− fm(x)| ≤ | fn(x)− fn(x j)|+ | fn(x j)− fn(d j)|+ | fn(d j)− fm(d j)|+| fm(d j)− fm(x j)|+ | fm(x j)− fm(x)| ≤ ε.
Sincex∈ K was arbitrary, we have proved that‖ fn− fm‖∞ ≤ ε for all n,m≥ n0,that is,( fn)n is a Cauchy sequence inC(K) and hence convergent.
Theorem 1.2.41(Arzela-Ascoli). Important
10.11.2008Let K be a compact metric space. A subset H of C(K) is relatively compact if andonly if H is (uniformly) equicontinuous and pointwise bounded.
Proof. Let H be equicontinuous and pointwise bounded. SinceK is separable(see Theorem 1.1.29) there exists a countable and dense setD =
{
dp : p∈N}in K. Let ( fn)n be a sequence inH. For eachp ∈ N the sequence( fn(dp))n isbounded inK and hence relatively compact. Lemma 1.2.33 gives us a strictlymonotone increasing functionϕ :N→N such that( fϕ(n)(dp))n is convergent foreachp∈N. It follows from Theorem 1.2.40 that( fϕ(n))n is convergent inC(K).
Assume now thatH is relatively compact. Forε > 0 there existf1, . . . , fm∈ Hsuch thatH ⊂ ⋃mj=1B( f j ,ε/3). For eachj ∈ {1, . . . ,m} there existsδ j such that
| f j(x)− f j(y)| ≤ ε/3 ∀x,y∈ K with d(x,y) ≤ δ j
Let δ := min{
δ j : j = 1, . . . ,m}
. Let x,y ∈ K with d(x,y) ≤ δ and let f ∈ H.Then there existsj ∈ {1, . . . ,m} such that‖ f − f j‖∞ ≤ ε/3. Hence
| f (x)− f (y)| ≤ | f (x)− f j(x)|+ | f j(x)− f j(y)|+ | f j(y)− f (y)| ≤ ε.
We have shown thatH is uniformly equicontinuous and since every compact setis bounded, we get thatH is bounded and in particular pointwise bounded.
26 CHAPTER 1. METRIC AND NORMED SPACES
Chapter 2
Measures and Integration
References for this Chapter are the book of R. G. Bartle [1] and the book ofH. Bauer [2].
2.1 Measurable Spaces and Measures
Definition 2.1.1. For a setΩ thepower setof Ω is denoted byP(Ω) and definedas the set of all subsets ofΩ, that is,
P(Ω) := {ω : ω ⊂ Ω} .
A mappingµ : P(Ω)→ [0,∞] is called anouter measureonΩ if
(O1) µ( /0) = 0,
(O2) µ(A) ≤ ∑∞k=1 µ(Ak) whenever A⊂⋃∞
k=1Ak.
Remark 2.1.2. If µ is an outer measure onΩ and A⊂ B⊂ Ω, then
µ(A) ≤ µ(B).
Definition 2.1.3. Let µ be an outer measure onΩ and B⊂ Ω. Thenµ restrictedto B (writtenµB) is the outer measure defined by
µB(C) := µ(B∩C) ∀C ⊂ Ω.
Definition 2.1.4. Let µ be an outer measure onΩ. A set A⊂ Ω is called µ-measurableif for all B ⊂ Ω
µ(B) = µ(B∩A)+ µ(B∩Ac).
This means, that a set A isµ-measurable iff A divides every set B nicely!
27
28 CHAPTER 2. MEASURES AND INTEGRATION
Exercise 2.1.5. Let µ be an outer measure onΩ. Showthat the following hold. Exercise
1. Every set A⊂ Ω with µ(A) = 0 is µ-measurable.
2. If A⊂ Ω is µ-measurable, then Ac is alsoµ-measurable.
3. If A⊂ Ω is µ-measurable and B⊂ Ω, then A isµB-measurable.
Lemma 2.1.6. Let Ω be a set andµ an outer measure onΩ. If A1 and A2 areµ-measurable, then the union A1∪A2 is alsoµ-measurable.
Proof. Let A := A1∪A2. Sinceµ(B)≤ µ(B∩A)+µ(B∩Ac) for every setB⊂Ω itsuffices to show the inverse inequality in order to show thatA is µ-measurable. Wehave forB⊂ Ω thatB∩(A1∪A2) = B∩(A1∪(A2\A1))⊂ (B∩A1)∪(B∩A2∩Ac1)and hence
µ(B) = µ(B∩A1)+ µ(B∩Ac1)= µ(B∩A1)+ µ((B∩Ac1)∩A2)+ µ((B∩Ac1)∩Ac2)≥ µ(B∩ (A1∪A2))+ µ(B∩ (A1∪A2)c).
HenceA1∪A2 is µ-measurable.
Exercise 2.1.7. Let Ω be a set andµ an outer measure onΩ. Show that forExerciseµ-measurable sets A1,A2 ⊂ Ω the intersection A1∩A2 is alsoµ-measurable.Hint: Use Exercise 2.1.5 part 2 and Lemma 2.1.6.
Remark 2.1.8. Using induction, Lemma 2.1.6 and Exercise 2.1.7 we get that thefinite union and finite intersection ofµ-measurable sets areµ-measurable.
Theorem 2.1.9(Properties of Measureable Sets).Let Ω be a set,µ an outer measure onΩ and let (Ak)k be a sequence ofµ-measurable sets. Then the following hold.
1. If the sets Ak are disjoint (i.e. Aj ∩Ak = /0 for j 6= k), then
µ
(
∞⋃
k=1
Ak
)
=∞
∑k=1
µ(Ak).
This property is calledσ -additivity .
2. If A1 ⊂ ·· · ⊂ Ak ⊂ Ak+1 ⊂ . . . , then
limk→∞
µ(Ak) = supk
µ(Ak) = µ
(
∞⋃
k=1
Ak
)
.
2.1 MEASURABLE SPACES AND MEASURES 29
3. If A1 ⊃ ·· · ⊃ Ak ⊃ Ak+1 ⊃ . . . andµ(A1) < ∞, then
limk→∞
µ(Ak) = infk
µ(Ak) = µ
(
∞⋂
k=1
Ak
)
.
4. The sets U:=⋃∞
k=1Ak and I :=⋂∞
k=1Ak are µ-measurable.
Proof.
1. Assume that theµ-measurable setsAk are disjoint and letB j :=⋃ j
k=1Ak forj ∈N. Using thatA j+1 is µ-measurable we get that
µ(B j+1) = µ(B j+1∩A j+1)+ µ(B j+1∩Acj+1) = µ(A j+1)+ µ(B j).
By induction, using thatµ(⋃1
k=1Ak) = ∑1k=1 µ(Ak) we get that
µ
(
j⋃
k=1
Ak
)
=j
∑k=1
µ(Ak).
It follows that
j
∑k=1
µ(Ak) ≤ µ(
∞⋃
k=1
Ak
)
⇒∞
∑k=1
µ(Ak) = limj
j
∑k=1
µ(Ak) ≤ µ(
∞⋃
k=1
Ak
)
.
Hence, using (O2), we get the proposed equality.
2. For j ∈N we letB j := A j \A j−1 whereA0 := /0. Then, by Remark 2.1.8,the setsB j areµ-measurable and disjoint and hence using (1) we get that
Ak =k⋃
j=1
B j ⇒ µ(Ak) =k
∑j=1
µ(B j)
Taking the limit ask→ ∞ gives
limk→∞
µ(Ak) =∞
∑j=1
µ(B j) = µ
(
∞⋃
j=1
B j
)
= µ
(
∞⋃
j=1
A j
)
.
3. We show that (2) implies (3). In fact, 12.11.08
30 CHAPTER 2. MEASURES AND INTEGRATION
µ(A1)− limk
µ(Ak) = limk
µ(A1∩Ack) = µ(
∞⋃
k=1
(A1∩Ack))
= µ
(
A1∩∞⋃
k=1
Ack
)
= µ
(
A1∩(
∞⋂
k=1
Ak
)c)
≥ µ(A1)−µ(
∞⋂
k=1
Ak
)
.
Consequently, we get that limk µ(Ak) ≤ µ (⋂∞
k=1Ak) ≤ limk µ(Ak). Hencewe get equality which proves (3).
4. Recall, that by Exercise 2.1.7 forB ⊂ Ω arbitrary, everyµ-measurable setis alsoµB measurable. Moreover,B j :=
⋃ jk=1Ak areµ-measurable sets and
for B⊂ Ω with µ(B) < ∞ we get
µ (B∩U)+ µ (B∩Uc) = µB(U)+ µB(Uc)
= µB
(
∞⋃
k=1
Bk
)
+ µB
(
∞⋂
k=1
Bck
)
= limk
µB(Bk)+ limk
µB(Bck) = µ(B).
If B⊂ Ω is such thatµ(B) = ∞, then it is obvious that
µ(B∩U)+ µ(B∩Uc) = ∞ = µ(B).
ThusU is µ-measurable. ThatI is µ-measurable follows from the equality
I =∞⋂
k=1
Ak =
(
∞⋃
k=1
Ack
)c
whereAck areµ-measurable sets.
Definition 2.1.10(σ -Algebra).Let Ω be a set. A subsetA ⊂ P(Ω) is called aσ -algebraor σ -field on Ω if thefollowing properties hold.
(S1) /0∈ A .
(S2) A∈ A ⇒ Ac ∈ A . (Stable under complementing)
(S3) Ak ∈ A ⇒⋃
k Ak ∈ A . (Stable under countable unions)
2.1 MEASURABLE SPACES AND MEASURES 31
Definition 2.1.11.LetΩ be a set and letA be aσ -algebra onΩ. Then the orderedpair (Ω,A ) is called ameasurable spaceand every element inA is called anA -measurableset or simplymeasurable setif it is clear which measurable spacewe mean.
Definition 2.1.12.Let (Ω,A ) be a measurable space. A mappingµ : A → [0,∞]is called ameasureon (Ω,A ) if the following holds.
(M1) µ( /0) = 0;
(M2) µ(⋃∞
n=1An) = ∑∞n=1 µ(An) if An ∈ A and An∩Am = /0 for n 6= m.
In this case the triple(Ω,A ,µ) is called ameasure space. A measure space(Ω,A ,µ) is calledσ -finite, if there exists a sequence(An)n⊂A such thatµ(An) <∞ and
⋃
nAn = Ω.
Remark 2.1.13. We have proved in Theorem 2.1.9 that the collection of allµ-measurable sets is aσ -algebra which we will denote byσµ . Moreover, it followsfrom the same Theorem that ifµ is an outer measure onΩ thenµ|σµ : σµ → [0,∞]is a measure on(Ω,σµ) and a set A⊂ Ω is by definitionµ-measurable iff A isσµ -measurable.
Example 2.1.14.LetΩ be a set. ThenS (Ω) := { /0,Ω} andP(Ω) areσ -algebrason Ω, whereS (Ω) is the smallest andP(Ω) is the biggest.
Exercise 2.1.15. Let Ω be a set and let(Aα)α∈I be a family ofσ -algebras on ExerciseΩ, I 6= /0. ShowthatA :=⋂α∈I Aα is aσ -algebra onΩ.
Lemma 2.1.16.Let Ω be a set andA a σ -algebra onΩ. If (An)n is a sequencein A , then A:=
⋂
nAn ∈ A .
Proof. Let Bn := Acn ∈A (S2) andB :=⋃
nBn ∈A (S3). ThenA = Bc ∈A (S2).
Definition 2.1.17. Let Ω be a set andD ⊂ P(Ω). Then the smallestσ -algebracontainingD is called theσ -algebra generated byD and is denoted byσ(D),that is,σ(D) :=
⋂
A where the intersection is taken over allσ -algebrasA ⊃D .
Remark 2.1.18.That σ(D) is in fact aσ -algebra follows from Exercise 2.1.15and from the fact thatP(Ω) is a σ -algebra containingD .
Definition 2.1.19.TheBorel σ -algebra on a metric space X is denoted byB(X) Importantand is theσ -algebra generated by the open sets of X. A set A⊂X is called aBorelset if A ∈ B(X).
32 CHAPTER 2. MEASURES AND INTEGRATION
Example 2.1.20(Examples of Measures).
1. LetΩ be a set. For a set A⊂ Ω we define
µ(A) :={
♯A if A has finitely many elements∞ else.
Thenµ is a measure on(Ω,P(Ω)), the so calledcounting measureonΩ.
2. Letµ be a measure on(Ω,A ) and letΩ′ ⊂Ω be measurable. LetA ′ be theintersection-σ -algebra given byA ′ := {A∩Ω′ : A∈ A }. Thenµ ′ : A ′ →[0,∞] given byµ ′(A) := µ(A) for A∈A ′ is a measure on(Ω′,A ′). We callµ ′ the restriction ofµ to (Ω′,A ′).
Exercise 2.1.21. Let Ω ⊂RN be a set,B(Ω) the Borelσ -algebra onΩ and letExercisex0 ∈ Ω. For A∈ B(Ω) we define
δx0(A) :={
1 if x0 ∈ A,0 else.
Showthat δx0 is a measure on(Ω,B(Ω)). We callδx0 theDirac1 measurein thepoint x0.
Theorem 2.1.22(Properties in Measure Spaces). Compare with Theorem 2.1.9ImportantLet (Ω,A ,µ) be a measure space.
1. If (Ak)k is an increasing sequence inA , then
µ
(
⋃
k
Ak
)
= supk
µ(Ak) = limk
µ(Ak).
2. If (Ak)k is a decreasing sequence inA and if µ(A1) < ∞, then
µ
(
⋂
k
Ak
)
= infk
µ(Ak) = limk
µ(Ak).
Proof. The proof is left to the reader.Exercise
Definition 2.1.23. Compare with Definition 2.1.3Let (Ω,A ,µ) be a measure space and let B∈A . Thenµ restricted to B (writtenµB) is the measure defined by
µB(C) := µ(B∩C) ∀C∈ A .
2.2 THE LEBESGUEMEASURE 33
Exercise 2.1.24. Let (Ω,A ,µ) be a measure space and let B∈ A . ShowthatExercise(Ω,A ,µB) is a measure space.
Exercise 2.1.25. Letµ1, . . . ,µn be measures on a fixed measurable space(Ω,A )Exerciseand letλ1, . . . ,λn ∈R+. Showthatµ := ∑nk=1 λkµk is a measure on(Ω,A ) where
µ(A) :=n
∑k=1
λkµk(A), A∈ A .
Definition 2.1.26.Let (Ω,A ,µ) be a measure space. Aµ-null-set (or null-set) isa set N∈ A with µ(N) = 0. We say that a property holdsµ-almost everywhere(µ-a.e.) (or almost everywhere) if there exists a null-set N such that the propertyholds onΩ\N.
Exercise 2.1.27. Showthat the countable union of null-sets is a null-set. Exercise
2.2 The Lebesgue Measure
Definition 2.2.1(Cells). 17.11.08Let ak,bk ∈R, ak ≤ bk for k = 1, . . . ,N. Then a set Q⊂RN of then form
• Q = (a1,b1]×·· ·× (aN,bN] is called aleft-open cell,
• Q = (a1,b1)×·· ·× (aN,bN) is called anopen cell.
For such cells Q we define the natural volume by|Q| := ∏Nk=1(bk−ak).
Definition 2.2.2. Let λ ⋆, λ ⋆o : P(RN) → [0,∞] be given byλ ⋆(A) := inf
{
∑k∈N |Qk| : Qk is a left-open cell and A⊂ ⋃k∈NQk}
λ ⋆o(A) := inf
{
∑k∈N |Qk| : Qk is an open cell and A⊂ ⋃k∈NQk}
Lemma 2.2.3.For all A ⊂RN we haveλ ⋆(A) = λ ⋆o (A).Proof. Assume thatλ ⋆(A) andλ ⋆o(A) are finite.
For ε > 0 there exist left-open cellsQk such thatA ⊂⋃
k Qk and∑k |Qk| ≤λ ⋆(A)+ ε. For eachk there exists an open cellUk ⊃ Qk such that|Uk| ≤ |Qk|+
1Paul Adrien Maurice Dirac (1902-1984)
34 CHAPTER 2. MEASURES AND INTEGRATION
ε2−k. ThenA⊂ ⋃kUk and∑k |Uk| ≤ ∑k |Qk|+ ε ≤ λ ⋆(A)+2ε. Sinceε > 0 wasarbitrary we get thatλ ⋆o (A) ≤ λ ⋆(A).
Forε > 0 there exist open cellsUk such thatA⊂⋃
kUk and∑k |Uk| ≤ λ ⋆o(A)+ε. Now letQk ⊃Uk be the left-open cell such that|Qk|= |Uk|. ThenA⊂
⋃
k Qk and∑k |Qk| ≤ λ ⋆o (A)+ ε. This shows thatλ ⋆(A) ≤ λ ⋆o(A) which finishes the proof.
The easy casesλ ⋆(A) = ∞ andλ ⋆o (A) = ∞ are left as an exercise.Exercise
Theorem 2.2.4.The mappingλ ⋆ : P(RN) → [0,∞] is an outer measure, calledthe (N-dimensional)outer Lebesgue measure, also noted by byλ ⋆N to specify thedimension N.
Proof. It is clear that the property (O1) is fullfilled. To show the property (O2) letA⊂⋃kAk. If λ ⋆(Ak) = ∞ for somek, then it is obvious thatλ ⋆(A)≤∑∞k=1 λ ⋆(Ak).Hence we assume thatλ ⋆(Ak) < ∞. Let ε > 0 be given. Then there exist left-opencellsQkl such thatAk ⊂
⋃
l∈NQkl and∑l∈N |Qkl| ≤ λ ⋆(Ak)+2−kε.
ThenA⊂⋃k∈NAk ⊂ ⋃k,l∈NQkl andλ ⋆(A) ≤ ∑
k∈N ∑l∈N |Qkl| ≤ ∑k∈Nλ ⋆(Ak)+2−kε = ε + ∑k∈Nλ ⋆(Ak).Sinceε > 0 was arbitrary, we get (O2), that is,
λ ⋆(A) ≤ ∑k∈Nλ ⋆(Ak).
Lemma 2.2.5.Let Q be a left-open cell inRN. Thenλ ⋆(Q) = |Q|.Proof. For Q1 := Q andQk := /0 we have thatQ ⊂
⋃
k Qk and henceλ ⋆(Q) ≤∑k |Qk|= |Q|. To get the other inequality letU be an open cell such thatU ⊂Qand|Q| ≤ |U |+ε. LetUk be open cells such thatQ⊂
⋃
kUk and∑k |Uk| ≤ λ ⋆(Q)+ε.ThenU is compact and contained in
⋃
kUk. By the Heine-Borel property weget a finite number of open cells such thatU ⊂ ⋃rk=1Uk. (a) Assume that|U | ≤∑rk=1 |Uk|. Then
|Q| ≤ |U |+ ε ≤r
∑k=1
|Uk|+ ε ≤∞
∑k=1
|Uk|+ ε ≤ λ ⋆(Q)+2ε.
2.2 THE LEBESGUEMEASURE 35
Sinceε > 0 was arbitrary we get that|Q|= λ ⋆(Q). Now we prove our assumption(a). For this we let(ak1,b
k1)×·· ·× (akN,bkN) = Qk for k = 1, . . . , r and
Xj :={
akj : k = 1, . . . , r}
∪{
bkj : k = 1, . . . , r}
for j = 1, . . . ,N.
Then the(N−1)-dimensional hyperplanesH j(c) :={
x∈RN : x j = c} with c∈Xj for j = 1, . . . ,N divide the open cellsUk into distinct open cellsC1, . . . ,Cn. LetV1, . . . ,Vm be the open cells in which the hyperplanes divide the cellU . Then
|U |=m
∑k=1
|Vk| ≤n
∑k=1
|Ck| ≤r
∑k=1
|Uk|.
Definition 2.2.6. The restrictionλ := λ ⋆|σλ⋆ of the outer Lebesgue measureλ ⋆ totheσ -algebraσλ ⋆ of λ ⋆-measurable sets is called the (N-dimensional)Lebesguemeasure. Thatλ is in fact a measure on(RN,σλ ⋆) follows from Remark 2.1.13.Further, we call a set A⊂RN Lebesgue measurableif A ∈ σλ ⋆ .Exercise 2.2.7. ExerciseShowthat every left-open cell Q is Lebesgue measurable.
Lemma 2.2.8. Every open set inRN is a countable union of disjoint left-open 19.11.08cubes.
Proof. Let Ω ⊂RN be an open set. Fork∈N we letSk := (0,2−k]N, Ω0 := /0 andCk :=
{
v∈ 2−kZN : v+Sk ⊂ Ω\Ωk−1} , Ωk := Ωk−1∪ ⋃v∈Ck
(v+Sk).
Then, by construction,Ω =⋃
k∈N⋃v∈Ck Sk + v is a countable union of disjointleft-open cubes. In fact, the inclusion ”⊃” is clear. On the other hand, ifx ∈ Ωthen there existsk ∈N such that dist(x,Ωc) > 21−k√N. For j = 1, . . . ,N we letmj ∈ Z andv j ∈ 2−kZ be given by
2−kmj < x j ≤ 2−k(mj +1), v j := 2−kmj .
Thenx∈ (v+Sk) ⊂ Ωk.
Theorem 2.2.9.Every Borel inRN set is Lebesgue measurable. ImportantProof. By Exercise 2.2.7 the left-open cells are in theσ -algebraσλ ⋆ . Using theσ -algebra property we get that every countable union of left-open cells is inσλ ⋆.By Lemma 2.2.8 every open set is inσλ ⋆ and since the Borelσ -algebra is thesmallestσ -algebra containing all open sets we get thatB(RN) ⊂ σλ ⋆ .
36 CHAPTER 2. MEASURES AND INTEGRATION
Lemma 2.2.10.For A⊂RN, x∈R andα > 0 we have thatλ ⋆(x+αA) = αNλ ⋆(A).
Proof. Forε > 0 there exist left-open cellsQk such thatA⊂⋃
k Qk and∑k |Qk| ≤λ ⋆(A)+ε. LetCk be the left-open cells given byCk := x+αQk. Then(x+αA)⊂⋃
kCk and hence
λ ⋆(x+αA) ≤ ∑k
|Ck| = ∑k
αN|Qk| ≤ αNλ ⋆(A)+αNε.
Henceλ ⋆(x+αA) ≤ αNλ ⋆(A).LetUk be left-open cells such thatx+αA⊂
⋃
kUk and∑k |Uk| ≤ λ ⋆(x+αA)+ε. Define the left-open cellsQk by Qk := (−x+Uk)/α. ThenA⊂
⋃
k Qk and
λ ⋆(A) ≤ ∑k
|Qk| = ∑k
α−N|Uk| ≤ α−Nλ ⋆(x+αA)+ εα−N.
Henceλ ⋆(A) ≤ α−Nλ ⋆(x+αA).
Exercise 2.2.11. Let A⊂ RN, x∈ RN and α > 0. Show that A is Lebesgue-Exercisemeasurable iff x+αA is Lebesgue-measurable.
Theorem 2.2.12(Approximation by Open Sets).For M ⊂RN we have that
λ ⋆(M) = inf {λ (O) : O open and M⊂ O} . (2.2.1)
Proof. Let I denote the right hand side of (2.2.1). It is clear thatλ ⋆(M)≤ λ (O) =λ ⋆(O) for every open setO containingM. Henceλ ⋆(M) ≤ I . On the otherhand, forε > 0 there exist open cellsUk such thatM ⊂ O :=
⋃
kUk and∑k |Uk| ≤λ ⋆(M)+ε (even ifλ ⋆(M) = ∞). HenceI ≤ λ ⋆(O) = λ (O) = ∑k λ (Uk)≤ λ ⋆(M)+ε. This gives thatI ≤ λ ⋆(M) and hence the desired equality.
Exercise 2.2.13. Let T :RN →RN be linear.Exercise1. Showthat T :RN →RN is continous.2. Let T be invertible.Showthat T−1 :RN →RN is linear.3. Let T be invertible.Showthat T−1 :RN →RN is continuous.
Lemma 2.2.14.Let T : RN → RN be linear and invertible and S:= (0,1]N ⊂RN. Then TS is a Borel set. Moreover, forρ := λ (TS) we get thatλ ⋆(T(M)) =ρλ ⋆(M) for all M ⊂RN.
2.2 THE LEBESGUEMEASURE 37
Proof. The setS is the union of countably many compact sets, in fact,
S=∞⋃
k=1
[1/k,1]N ⇒ T(S) =∞⋃
k=1
T([1/k,1]N).
SinceT is continuous we get thatT(S) is a countable union of compact sets andhenceT(S) ∈ B(RN) ⊂ σλ ⋆ . So the definition ofρ makes sense. LetΩ ⊂RN bean open set. By Lemma 2.2.8Ω is a countable union of disjoint left-open cubes:
Ω = ˙⋃
k(vk + tkS), vk ∈RN, tk ∈R.
SinceT is one to one we get thatT(Ω) = ˙⋃
kTvk + tkT(S) and
λ (Tvk + tkT(S)) = λ (tkT(S)) = tNk λ (T(S)) = tNk ρλ (S)
= ρλ (tkS) = ρλ (vk + tkS).
We conclude thatλ (T(Ω)) = ∑k λ (Tvk + tkT(S)) = ρ ∑k λ (vk + tkS) = ρλ (Ω).Let M ⊂RN be arbitrary. Then there exists an open setΩ ⊃ M such thatλ ⋆(Ω)≤λ ⋆(M)+ ε. Hence
λ ⋆(T(M)) ≤ λ ⋆(T(Ω)) = ρλ ⋆(Ω) ≤ ρλ ⋆(M)+ρε.
Sinceε > 0 was arbitrary we get thatλ ⋆(T(M)) ≤ ρλ ⋆(M). LetV be an open setsuch thatV ⊃T(M) andλ ⋆(V)≤ λ ⋆(T(M))+ε. Thenρλ ⋆(M)≤ ρλ ⋆(T−1(V)) =λ ⋆(T[T−1(V)]) = λ ⋆(V) ≤ λ ⋆(T(M))+ ε. Sinceε > 0 was arbitrary, we get thedesired equality.
Lemma 2.2.15. Every open setΩ ⊂RN is a countable union of compact sets. 24.11.08Proof. For n ∈ N we let Kn := {x∈ Ω : dist(x,Ωc) ≥ 1/n and‖x‖ ≤ n}. ThenKn is a closed (sinced(·,Ωc) and‖ · ‖ are continuous) and bounded and hencecompact. Moreover,Ω =
⋃
nKn. In fact, ”⊃” is clear. To see ”⊂” let x∈ Ω. Thendist(x,Ωc) > 0 and hence there existsn0 ∈ N such that dist(x,Ωc) ≥ 1/n0 and‖x‖ ≤ n0, that is,x∈ Kn0.
Remark 2.2.16.Every compact set is the complement of an open set and hencea Borel set. Moreover, we get by Lemma 2.2.15 that the Borelσ -algebra is thesmallestσ -algebra containing all compact sets.
Theorem 2.2.17(Application of Linear Mappings). ImportantLet T :RN →RN be linear and let M⊂RN. Thenλ ⋆(T(M)) = |det(T)|λ ⋆(M).Moreover, if M is Lebesgue-measurable then T(M) is Lebesgue-measurable andλ (T(M)) = |det(T)|λ (M).
Recommended