chapter 1 metric and normed spaces - uni ulmchapter 1 metric and normed spaces there are things...

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

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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.


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.


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)


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)


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)


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.



• 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.


• 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).


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.


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)


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



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} .


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.


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.


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|


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.


1.2.5 * 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.


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.


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.


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.


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.

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


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).


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)


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.


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 ⋆ .


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.


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).

chapter 1 metric and normed spaces - uni ulmchapter 1 metric and normed spaces there are things...

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