8 Show that the open unit ball inRnis an open subset ofRn.
9 SupposeG1is a nonempty subset ofRmandG2is a nonempty subset ofRn. Prove thatG1×G2is an open subset ofRm×Rnif and only ifG1is an open subset ofRmandG2is an open subset ofRn.
[One direction of this result was already proved(see5.36); both directions are stated here to make the result look prettier and to be comparable to the next exercise, where neither direction has been proved.]
10 SupposeF1is a nonempty subset ofRmand F2is a nonempty subset ofRn. Prove thatF1×F2is a closed subset ofRm×Rnif and only ifF1is a closed subset ofRmandF2is a closed subset ofRn.
11 SupposeEis a subset ofRm×Rnand
A={x∈Rm:(x,y)∈Efor somey∈Rn}.
(a) Prove that ifEis an open subset ofRm×Rn, thenAis an open subset ofRm.
(b) Prove or give a counterexample: IfEis a closed subset ofRm×Rn, then Ais a closed subset ofRm.
12 (a) Prove thatlimn→∞λn(Bn) =0.
(b) Find the value ofnthat maximizesλn(Bn).
13 For readers familiar with the gamma functionΓ: Prove that λn(Bn) = π
n/2
Γ(n2+1) for every positive integern.
14 Define f: R2→Rby
f(x,y) =
xy(x2−y2)
x2+y2 if(x,y)̸= (0, 0), 0 if(x,y) = (0, 0). (a) Prove thatD1(D2f)andD2(D1f)exist everywhere onR2. (b) Show that D1(D2f)(0, 0)̸= D2(D1f)(0, 0).
(c) Explain why (b) does not violate5.48.
Banach Spaces
We begin this chapter with a quick review of the essentials of metric spaces. Then we extend our results on measurable functions and integration to complex-valued functions. After that, we rapidly review the framework of vector spaces, which allows us to consider natural collections of measurable functions that are closed under addition and scalar multiplication.
Normed vector spaces and Banach spaces, which are introduced in the third section of this chapter, play a hugely important role in modern analysis. Most interest focuses on linear maps on these vector spaces. Key results about linear maps that we develop in this chapter include the Hahn–Banach Theorem, the Open Mapping Theorem, the Closed Graph Theorem, and the Principle of Uniform Boundedness.
Market square in Lviv, a city that has had several names and has been in several countries because of changing international borders. From 1772 until 1918, the
city was in Austria and was called Lemberg. Between World War I and World War II, the city was in Poland and was called Lwów. During this time,
mathematicians in Lwów, particularly Stefan Banach (1892–1945) and his colleagues, developed the basic results of modern functional analysis, using tools of analysis to study infinite-dimensional vector spaces. Since World War II
ended, Lviv has been in Ukraine, which was part of the Soviet Union until Ukraine became an independent country in 1991.
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146
6A Metric Spaces
Open Sets, Closed Sets, and Continuity
Much of analysis takes place in the context of a metric space, which is a set with a notion of distance that satisfies certain properties. The properties we would like a distance function to have are captured in the next definition, where you should think ofd(f,g)as measuring the distance between f andg.
Specifically, we would like the distance between two elements of our metric space to be a nonnegative number that is0if and only if the two elements are the same. We would like the distance between two elements not to depend on the order in which we list them. Finally, we would like a triangle inequality (the last bullet point below), which states that the distance between two elements is less than or equal to the sum of the distances obtained when we insert an intermediate element.
Now we are ready for the formal definition.
6.1 Definition metric space
Ametricon a nonempty setVis a functiond:V×V→[0,∞)such that
• d(f,f) =0for all f ∈V;
• if f,g∈Vandd(f,g) =0, then f =g;
• d(f,g) =d(g,f)for all f,g∈V;
• d(f,h)≤d(f,g) +d(g,h)for all f,g,h∈V.
Ametric spaceis a pair(V,d), whereVis a nonempty set anddis a metric onV.
6.2 Example metric spaces
• SupposeVis a nonempty set. DefinedonV×Vby settingd(f,g)to be1if f ̸=gand to be0if f =g. Thendis a metric onV.
• DefinedonR×Rbyd(x,y) =|x−y|. Thendis a metric onR.
• Forn∈Z+, definedonRn×Rnby
d (x1, . . . ,xn),(y1, . . . ,yn)=max{|x1−y1|, . . . ,|xn−yn|}. Thendis a metric onRn.
• DefinedonC([0, 1])×C([0, 1])byd(f,g) =sup{|f(t)−g(t)|:t∈[0, 1]}; hereC([0, 1])is the set of continuous real-valued functions on[0, 1]. Thendis a metric onC([0, 1]).
• Definedonℓ1×ℓ1byd (a1,a2, . . .),(b1,b2, . . .)=∑∞k=1|ak−bk|; hereℓ1 is the set of sequences(a1,a2, . . .)of real numbers such that∑∞k=1|ak| < ∞.
Thendis a metric onℓ1.
This book often uses symbols such as f,g,has generic elements of a generic metric space because many of the important metric spaces in analysis are sets of functions; for example, see the fourth bullet point of Example6.2.
The material in this section is proba- bly review for most readers of this book.
Thus more details than usual are left to the reader to verify. Verifying those details and doing the exercises is the best way to solidify your understanding of these concepts. You should be able to transfer familiar definitions and proofs from the
context ofRorRnto the context of a metric space.
We will need to use a metric space’s topological features, which we introduce now.
6.3 Definition open ball;B(f,r); closed ball;B(f,r) Suppose(V,d)is a metric space, f ∈V, andr>0.
• Theopen ballcentered atf with radiusris denotedB(f,r)and is defined by B(f,r) ={g∈V:d(f,g)<r}.
• Theclosed ballcentered atf with radiusris denotedB(f,r)and is defined by
B(f,r) ={g∈V:d(f,g)≤r}.
Abusing terminology, many books (including this one) include phrases such as supposeVis a metric spacewithout mentioning the metricd. When that happens, you should assume that a metricdlurks nearby, even if it is not explicitly named.
Our next definition declares a subset of a metric space to be open if every element in the subset is the center of an open ball that is contained in the subset.
6.4 Definition open
A subsetGof a metric spaceVis calledopenif for every f ∈ G, there exists r>0such thatB(f,r)⊂G.
6.5 open balls are open
SupposeVis a metric space, f ∈V, andr>0. ThenB(f,r)is an open subset ofV.
Proof Supposeg ∈ B(f,r). We need to show that an open ball centered atgis contained inB(f,r). To do this, note that ifh∈B g,r−d(f,g), then
d(f,h)≤d(f,g) +d(g,h)<d(f,g) + r−d(f,g)=r,
which implies thath∈ B(f,r). ThusB g,r−d(f,g)⊂ B(f,r), which implies thatB(f,r)is open.
Closed sets are defined in terms of open sets.
6.6 Definition closed
A subset of a metric spaceVis calledclosedif its complement inVis open.
For example, each closed ballB(f,r)in a metric space is closed, as you are asked to prove in Exercise3.
Now we define the closure of a subset of a metric space.
6.7 Definition closure; E
SupposeVis a metric space andE⊂V. TheclosureofE, denotedE, is defined by
E={g∈V:B(g,ε)∩E̸=∅for everyε>0}.
Limits in a metric space are defined by reducing to the context of real numbers, where limits have already been defined.
6.8 Definition limit in metric space;limk→∞fk
Suppose(V,d)is a metric space, f1,f2, . . .is a sequence inV, andf ∈V. Then
klim→∞fk= f means lim
k→∞d(fk,f) =0.
In other words, a sequence f1,f2, . . .inVconverges to f ∈Vif for everyε>0, there existsn∈Z+such that
d(fk,f)<εfor all integersk≥n.
The next result states that the closure of a set is the collection of all limits of elements of the set. Also, a set is closed if and only if it equals its closure. The proof of the next result is left as an exercise that provides good practice in using these concepts.
6.9 closure
SupposeVis a metric space andE⊂V. Then
(a) E={g∈V: there exist f1,f2, . . . inEsuch that lim
k→∞fk =g}; (b) Eis the intersection of all closed subsets ofVthat containE;
(c) Eis a closed subset ofV;
(d) Eis closed if and only ifE=E;
(e) Eis closed if and only ifEcontains the limit of every convergent sequence of elements ofE.
The definition of continuity that follows uses the same pattern as the definition for a function from a subset ofRtoR.
6.10 Definition continuous
Suppose(V,dV)and(W,dW)are metric spaces andT:V→Wis a function.
• Forf ∈V, the functionTis calledcontinuousatf if for everyε>0, there existsδ>0such that
dW T(f),T(g)<ε for allg∈VwithdV(f,g)<δ.
• The functionTis calledcontinuousifTis continuous atf for every f ∈V. The next result gives equivalent conditions for continuity. Recall thatT−1(E) is called the inverse image ofEand is defined to be{f ∈ V : T(f)∈ E}. Thus the equivalence of (a) and (c) below could be restated as saying that a function is continuous if and only if the inverse image of every open set is open. The equivalence of (a) and (d) below could be restated as saying that a function is continuous if and only if the inverse image of every closed set is closed.
6.11 equivalent conditions for continuity
SupposeVandWare metric spaces andT: V →W is a function. Then the following are equivalent:
(a) Tis continuous.
(b) lim
k→∞fk= f inVimplies lim
k→∞T(fk) =T(f)inW.
(c) T−1(G)is an open subset ofVfor every open setG⊂W.
(d) T−1(F)is a closed subset ofVfor every closed setF⊂W.
Proof We first prove that (b) implies (d). Suppose (b) holds. SupposeFis a closed subset ofW. We need to prove thatT−1(F)is closed. To do this, suppose f1,f2, . . . is a sequence inT−1(F)andlimk→∞fk= f for somef ∈V. Because (b) holds, we know thatlimk→∞T(fk) =T(f). Because fk ∈T−1(F)for eachk∈Z+, we know thatT(fk)∈ Ffor eachk∈ Z+. BecauseFis closed, this implies thatT(f)∈ F.
Thus f ∈T−1(F), which implies thatT−1(F)is closed [by6.9(e)], completing the proof that (b) implies (d).
The proof that (c) and (d) are equivalent follows from the equation T−1(W\E) =V\T−1(E)
for everyE⊂Wand the fact that a set is open if and only if its complement (in the appropriate metric space) is closed.
The proof of the remaining parts of this result are left as an exercise that should help strengthen your understanding of these concepts.
Cauchy Sequences and Completeness
The next definition is useful for showing (in some metric spaces) that a sequence has a limit, even when we do not have a good candidate for that limit.
6.12 Definition Cauchy sequence
A sequence f1,f2, . . .in a metric space(V,d)is called aCauchy sequenceif for everyε>0, there existsn ∈Z+such thatd(fj,fk)<εfor all integersj≥n andk≥n.
6.13 every convergent sequence is a Cauchy sequence
Every convergent sequence in a metric space is a Cauchy sequence.
Proof Supposelimk→∞fk = f in a metric space(V,d). Supposeε > 0. Then there existsn∈Z+such thatd(fk,f)< ε2for allk≥n. Ifj,k∈Z+are such that j≥nandk≥n, then
d(fj,fk)≤d(fj,f) +d(f,fk)< ε2+2ε =ε.
Thus f1,f2, . . .is a Cauchy sequence, completing the proof.
Metric spaces that satisfy the converse of the result above have a special name.
6.14 Definition complete metric space
A metric spaceVis calledcompleteif every Cauchy sequence inVconverges to some element ofV.
6.15 Example
• All five of the metric spaces in Example6.2are complete, as you should verify.
• The metric spaceQ, with metric defined byd(x,y) =|x−y|, is not complete.
To see this, fork∈Z+let xk = 1
101!+ 1
102! +· · ·+ 1 10k!. Ifj<k, then
|xk−xj|= 1
10(j+1)! +· · ·+ 1
10k! < 2 10(j+1)!.
Thusx1,x2, . . .is a Cauchy sequence inQ. However,x1,x2, . . .does not con- verge to an element ofQbecause the limit of this sequence would have a decimal expansion0.110001000000000000000001 . . .that is neither a terminating deci- mal nor a repeating decimal. ThusQis not a complete metric space.
Entrance to the École Polytechnique, Paris, where Augustin-Louis Cauchy (1789–1857)was a student and a faculty member. Cauchy wrote almost 800 mathematics papers and the highly influential textbookCours d’Analyse (published
in 1821), which greatly influenced the development of analysis.
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Every nonempty subset of a metric space is a metric space. Specifically, suppose (V,d)is a metric space andUis a nonempty subset ofV. Then restrictingd to U×Ugives a metric onU. Unless stated otherwise, you should assume that the metric on a subset is this restricted metric that the subset inherits from the bigger set.
Combining the two bullet points in the result below shows that a subset of a complete metric space is complete if and only if it is closed.
6.16 connection between complete and closed
(a) A complete subset of a metric space is closed.
(b) A closed subset of a complete metric space is complete.
Proof We begin with a proof of (a). SupposeUis a complete subset of a metric space V. Suppose f1,f2, . . . is a sequence inU that converges to some g ∈ V. Then f1,f2, . . .is a Cauchy sequence inU(by6.13). Hence by the completeness ofU, the sequence f1,f2, . . . converges to some element ofU, which must beg (see Exercise7). Henceg∈U. Now6.9(e) implies thatUis a closed subset ofV, completing the proof of (a).
To prove (b), supposeUis a closed subset of a complete metric spaceV. To show thatUis complete, suppose f1,f2, . . .is a Cauchy sequence inU. Thenf1,f2, . . .is also a Cauchy sequence inV. By the completeness ofV, this sequence converges to somef ∈V. BecauseUis closed, this implies that f ∈U(see6.9). Thus the Cauchy sequence f1,f2, . . .converges to an element ofU, showing thatUis complete. Hence (b) has been proved.