Real Analysis with Economic Applications - Chapter J doc

In particular, we carry our earlier work on the Hahn-Banach type extension andseparation theorems to the realm of normed linear spaces, and talk about a fewfundamental results of infinit

Chapter J

Normed Linear Spaces

This chapter introduces a very important subclass of metric linear spaces, namely, theclass of normed linear spaces We begin with an informal discussion that motivatesthe investigation of such spaces We then formalize parts of that discussion, introduceBanach spaces, and go through a number of examples and preliminary results Thefirst hints of how productive mathematical analysis can be within the context ofnormed linear spaces are found in our final excursion to fixed point theory Here weprove the fixed point theorems of Glicksberg, Fan, Krasnoselski˘ı and Schauder, andprovide a few applications to game theory and functional equations We then turn tothe basic theory of continuous linear functionals defined on normed linear spaces, andsketch an introduction to classical linear functional analysis Our treatment is guided

by geometric considerations for the most part, and dovetails with that of Chapter

G In particular, we carry our earlier work on the Hahn-Banach type extension andseparation theorems to the realm of normed linear spaces, and talk about a fewfundamental results of infinite dimensional convex analysis, such as the ExtremePoint Theorem, Krein-Milman Theorem, etc In this chapter we also bring to theconclusion our work on the classification of the differences between the finite andinfinite dimensional linear spaces Finally, in order to give at least a glimpse of thepowerful Banach space methods, we establish here the famous Uniform BoundednessPrinciple as a corollary of our earlier geometric findings, and go through some of itsapplications

The present treatment of normed linear spaces is roughly at the same level withthat of the classic real analysis texts by Kolmogorov and Fomin (1970) and Roy-den (1994) For a more detailed introduction to Banach space theory, we shouldrecommend Kreyzig (1978), Maddox (1988) and/or the first chapter of Megginson(1998).1

1 My coverage of linear functional analysis here is directed towards particular applications, and is thus incomplete even at an introductory level A more leisurely introduction would cover the open mapping and closed graph theorems, and would certainly spend some time on Hilbert spaces I will not use these two theorems in this book, and talk about Hilbert spaces only in passing Moreover, my treatment completely ignores operator theory, which is an integral counterpart of linear functional analysis In this regard all I can do here is to direct your attention to the beautiful expositions of Megginson (1998) and Schechter (2002).

1 Normed Linear Spaces

The Separating Hyperplane Theorem attests to the fact that our program of marryingthe metric and linear analyses of Chapters C—G has been a successful one However,there are still a number of shortcomings we need to deal with For instance, although

we came close in Example I.9, we have so far been unable to establish the existence

of a nonzero continuous linear functional on an arbitrarily given metric linear space.And this is for a good reason: There are Fréchet spaces on which the only continuouslinear functional is the zero functional.2 In view of Propositions I.6 and I.7, thismeans that all hyperplanes in a metric linear space may be dense, which is a trulypathologic situation In such a space separation of convex sets is a dubious concept

at best Moreover, by the Separating Hyperplane Theorem, the only nonempty openconvex subset of such a space is the entire space itself (Why?) Let us step back for

a moment and ask: What is the source of these problems?

How would we separate two distinct points x and y in Rn by using the SeparatingHyperplane Theorem? The answer is easy: Just take an 0 < ε < d2(x, y),and separatethe open convex sets Nε,Rn(x) and Nε,Rn(y) by a closed hyperplane So why doesn’tthis argument work in an arbitrary metric linear space? Because an ε-neighborhood,while necessarily nonempty and open, need not be convex in an arbitrary metriclinear space (even if this space is finite dimensional) This, in turn, does not let usinvoke the Separating Hyperplane Theorem to find a closed hyperplane (and hence anonzero continuous functional) at this level of generality.3

So the problem lies in the fact that the ε-neighborhoods in a metric linear spacemay not be convex Or better, in an arbitrary metric linear space X, an ε-neighborhood

of 0 may not contain an open convex set O with 0 ∈ O Why don’t we then restrictattention to those metric linear spaces for which, for any ε > 0, there exists an openconvex subset of Nε,X(0) that includes 0? Indeed, such a space — called a locallyconvex metric linear space — is free of the difficulties we mentioned above After

2 I don’t know a simple example that would illustrate this If you’re familiar with measurable functions, you may be able to take comfort in the following example: The set of all measurable real functions on [0, 1], metrized via the map (f, g) → U 1

0 min{1, |f(t) − g(t)|}dt, is a Fréchet space on which there is no nonzero continuous linear functional (Even the linear functional f → f(0) is not continuous on this space Why?)

3 The metric linear space discussed in the previous footnote provides a case in point A simpler example is obtained by metrizing R2 with the metric d1/2(x, y) := s

on the former is continuous (Why?) The absence of “convex neighborhoods” yields a shortage of continuous linear functionals only in the case of infinite dimensional spaces.

all, any two distinct points of a locally convex metric linear space can be separated

by a closed hyperplane, and hence there are plenty of continuous functionals defined

on such a space Yet, even locally convex metric linear spaces lack structure that weneed for a variety of problems that we will explore in this chapter We want — youwill see why in due course — to work with metric linear spaces whose ε-neighborhoodsare not only convex but also behave well with respect to dilations and contractions.Let X be a metric linear space Take the open unit ball N1,X(0), and stretch

it in your mind to dilate it at a one-to-two ratio What would you get, intuitively?Well, there seems to be two obvious candidates: 2N1,X(0)and N2,X(0).Our Euclideanintuition suggests that these two sets should be the same, so perhaps there is no roomfor choice Let’s see Since the metric d on X is translation invariant, x ∈ N1,X(0)implies

d(2x, 0) = d(x,−x) ≤ d(x, 0) + d(0, −x) = 2d(x, 0) < 2,

so we have 2N1,X(0) ⊆ N2,X(0) indeed Conversely, take any x ∈ N2,X(0) Is x ∈2N1,X(0)?That is to say, is 12x∈ N1,X(0)? No, not necessarily After all, in a metriclinear space it is possible that d(x, 0) = d(12x, 0) (In R∞, for instance, the distancebetween 0 and (2, 2, ) and that between 0 and (1, 1, ) are both equal to 1.)This is another anomaly that we wish to avoid Since the distance between x and

0 is d(x, 0), and 12x is the midpoint of the line segment between 0 and x, it makessense to require the distance between 0 and 1

In particular, this guarantees that 2N1,X(0) = N2,X(0)

Of course, exactly the same reasoning would also justify the requirement d(1nx, 0)

= n1d(x, 0)for every n ∈ N This, in turn, implies d(mx, 0) = md(x, 0) for all m ∈ N,and it follows that d(rx, 0) = rd(x, 0) for all r ∈ Q++ (Why?) Consequently, sinced(·, 0) is a continuous map on X, we find that d(λx, 0) = λd(x, 0) for all λ > 0.(Yes?) Finally, if λ < 0, then, by translation invariance, d(λx, 0) = d(0, −λx) =d(−λx, 0) = −λd(x, 0) In conclusion, the geometric anomaly that we wish to avoidpoint towards the following homogeneity property:

d(λx, 0) =|λ| d(x, 0) for all x ∈ X (2)Metric linear spaces for which (2) holds provides a remarkably rich playground

In particular, all ε-neighborhoods in such a space are convex (and hence any suchspace is locally convex) Thus, the difficulty that worried us at the beginning of ourdiscussion does not arise in such metric linear spaces.4 In fact, we gain a lot more

by positing (2) It turns out that the metric linear spaces with (2) provides an idealenvironment to carry out a powerful convex analysis jointly with a functional analysis

4 More is true: (1) alone guarantees that Nε,X(0) (and hence Nε,X(x) = x + Nε,X(0) for all

x ∈ X) is a convex set for any ε > 0 After all, every open midpoint convex subset of a metric linear space is convex (Proof?)

of linear and nonlinear operators By the time you are done with this chapter andthe next, this point will have become abundantly clear.

Exercise 1 LetX be a metric linear space Show that everyε-neighborhood inX is convex iffd(x,·)is quasiconvex for each x∈ X.

Exercise 2.H As noted above, a metric linear space X is called locally convex if, for every ε > 0, there exists an open and convex set Oε in Nε,X(0) with 0 ∈ Oε

At least one of theε-neighborhoods in a locally convex metric linear space is convex True or false?

Exercise 3 LetX be a metric linear space with (1).Show that there exists a nonzero continuous linear functional on X

Exercise 4 Let X be a metric linear space with (1) Show that X is bounded iff

X ={0}

Exercise 5 In a metric linear space X with (1), would we necessarily have (2)?

Let us leave the geometric considerations we outlined above aside for a moment, andinstead focus on the following fundamental definition We will return to the discussionabove in about five pages

Dhilqlwlrq Let X be a linear space A function · : X → R+ that satisfies thefollowing properties is called a norm on X: For all x, y ∈ X,

(i) x = 0 if and only if x = 0,

(ii) (Absolute Homogeneity) λx = |λ| x for all λ ∈ R,

(iii) (Subadditivity) x + y ≤ x + y

If · is a norm on X, then we say that (X, · ) is a normed linear space If ·only satisfies the requirements (ii) and (iii), then (X, · ) is called a seminormedlinear space.5

Recall that the basic viewpoint of vector calculus is to regard a “vector” x in alinear space as a directed line segment that begins at zero and ends at x This allows

5 The idea of normed linear space was around since the 1906 dissertation of Fréchet, and a precursory analysis of it can be traced in the works of Eduard Helly and Hans Hahn prior to 1922 The modern definition was given first by Stefan Banach and Norbert Wiener, independently, in

1922 Banach then undertook a comprehensive analysis of such spaces which culminated in his ground breaking 1932 treatise (See Dieudonné (1981).)

one to think of the “length” (or the “magnitude”) of a vector in a natural way Forinstance, we think of the magnitude of a positive real number x as the length of theinterval (0, x], and that of −x as the length of [−x, 0) Indeed, it is easily verified thatthe absolute value function defines a norm on R Similarly, it is conventional to think

of the length of a vector in Rn as the distance between this vector and the origin,and as you would expect, x → d2(x, 0)defines a norm on Rn Just as the notion of ametric generalizes the geometric notion of “distance,” therefore, the notion of a normgeneralizes that of “length” or “magnitude” of a vector

This interpretation also motivates the properties that a “norm” must satisfy First,

a norm must be nonnegative, because “length” is an inherently nonnegative notion.Second, a nonzero vector should be assigned positive length, and hence property (i).Third, the norm of a vector −x should equal the norm of x, because multiplying avector by −1 should change only the direction of the vector, not its length Fourth,doubling a vector should double the norm of that vector, simply because the intuitivenotion of “length” behaves in this manner Property (ii) of a norm is a generaliza-tion of the latter two requirements Finally, our Euclidean intuition about “length”suggests that the norm of a vector that corresponds to one side of a triangle shouldnot exceed the sum of the norms of the vectors that form the other two sides of thattriangle This requirement is formalized as property (iii) in the formal definition of

a norm You may think that there may be other properties that an abstract notion

of “length” should satisfy But, as you will see in this chapter, the framework based

on the properties (i)-(iii) alone turns out to be rich enough to allow for a yieldinginvestigation of the properties of linearity and continuity in conjunction, along with

a satisfactory geometric analysis

When the norm under consideration is apparent from the context, it is customary

to dispense with the notation (X, · ), and refer to the set X itself as a normed linearspace We shall frequently adopt this convention in what follows That is, when wesay that X is a normed linear space, you should understand that X is a linear spacewith a norm · lurking in the background When we need to deal with two normedlinear spaces X and Y, the norms of these spaces will be denoted by · X and · Y ,respectively

Let X be a normed linear space For future reference, let’s put on record an mediate, yet important, consequence of the absolute homogeneity and subadditivityproperties of the norm on X Take any x, y ∈ X Note first that, by subadditivity,

im-x = im-x− y + y ≤ x − y + y so that

x − y ≤ x − y

Moreover, if we change the roles of x and y in this inequality, we get y − x ≤

y− x = x − y , where the last equality follows from the absolute homogeneity of

· Thus:

This inequality will be useful to us on several occasions, so please keep it in mind as

a companion to the subadditivity property

Exercise 6.H (Convexity of Norms) For any normed linear spaceX,show that

x + αy − x

α ≤ x + βyβ − x for all x, y ∈ Xand β ≥ α > 0

Exercise 7.H (Rotund Spaces) A normed linear spaceXis called rotund if x + y <

x + y for all linearly independent x, y ∈ X Show that X is rotund iff



1

2x +12y

 < 1

For any n ∈ N, we can norm Rn in a variety of ways to obtain a finite dimensionalnormed linear space For any 1 ≤ p ≤ ∞, the p-norm · p on Rn is defined as

x p :=

 nSi=1|xi|p

1 p

if p is finite, and x p := max{|xi| : i = 1, , n} if p = ∞ Using Minkowski’sInequality 1, it is readily checked that any · p is a norm on Rn

for any 1 ≤ p ≤ ∞.(Obviously, x p = dp(x, 0)for all x ∈ Rn

and 1 ≤ p ≤ ∞.)Recall that, for any 1 ≤ p, q ≤ ∞, Rn,p

and Rn,qare “identical” metric linear spaces(in the sense that they are linearly homeomorphic) It would thus be reasonable toexpect that (Rn, · p) and (Rn, · q) are “identical” in some formal sense as well.This is indeed the case, as we will discuss later in Section 4.2

Here are some examples of infinite dimensional normed linear spaces

E{dpsoh 1 [1] Let 1 ≤ p < ∞ The space p becomes a normed linear space whenendowed with the norm · p : p

→ R+ defined by

(xm) p :=

∞Si=1|xi|p

1 p

= dp((xm), 0)

(The subadditivity of · p is equivalent to Minkowski’s Inequality 2.) Similarly, wemake ∞a normed linear space by endowing it with the norm · ∞ : ∞→ R+definedby

(xm) ∞:= sup{|xm| : m ∈ N} = d∞((xm) , 0)

It is easily checked that · ∞ is indeed a norm on ∞

In what follows, when we consider p as a normed linear space, the underlyingnorm is always · p, 1≤ p ≤ ∞

[2] For any nonempty set T, the linear space B(T ) of all bounded real functions

on T is normed by · ∞ : B(T )→ R+,where

f ∞:= sup{|f(t)| : t ∈ T } = d∞(f, 0) For obvious reasons, · ∞ is called the sup-norm Of course, B(N) and ∞ are thesame normed linear spaces

[3]Let X be a normed linear space By a normed linear subspace Y of X, wemean a linear subspace of X whose norm is the restriction of the norm of X to Y(that is, y Y := y for each y ∈ Y ) Throughout the remainder of this book, werefer to a normed linear subspace simply as a subspace If Y is a subspace of X and

Y = X,then Y is called a proper subspace of X

As you would surely expect, given any metric space T, we view CB(T ) as asubspace of B(T ) Consequently, when we talk about the norm of an f ∈ CB(T )(or of f ∈ C(T ) when T is compact), what we have in mind is the sup-norm of thisfunction, that is, f ∞ Similarly, we norm c0, c0 and c by the sup-norm, and hence

c0 is subspace of c0, c0 is a subspace of c, and c is a subspace of ∞.6

[4] For any interval I, let CB1(I) denote the linear space of all bounded andcontinuously differentiable real maps on I whose derivatives are bounded functions

on I (Obviously, CB1([a, b]) = C1[a, b].) The standard norm on CB1(I) is denoted

by · ∞,∞, where

f ∞,∞:= sup{|f(t)| : t ∈ I} + sup {|f (t)| : t ∈ I} (We leave it as an exercise to check that (CB1(I), · ∞,∞)is a normed linear space.)From now on whenever we consider a space like C1[a, b],or more generally, CB1(I),

as a normed linear space, we will have the norm · ∞,∞ in mind

[5] Let X be the linear space of all continuous real functions on [0, 1] Then

· ∈ RX

+ defined by

f :=

] 1 0

[7] (Finite Product Spaces) Take any n ∈ N and let X1, , Xn be normed linearspaces We norm the product linear space X := XnXi by using the map · ∈ RX

+defined by (x1, x2, , xn) := x1 X

1 +· · · + xn X n (Is · a norm on X?)

6 Reminder c 0 is the linear space of all real sequences all but finitely many terms of which are zero; c0 is the linear space of all real sequences that converge to 0; and c is the linear space of all convergent real sequences.

[8] (Countably Infinite Product Spaces) Let Xi be a normed linear space, i =

1, 2, , and let X := X∞Xi We make X a normed linear space by means of theso-called product norm · ∈ RX

+ which is defined by(x1, x2, ) := S∞

(Why is · a norm on X?)

Exercise 8 Show that · ∞,∞ is a norm onCB1(I)for any intervalI

Exercise 9.H Define ϕ : ∞ → R+ by ϕ((xm)) := lim sup|xm| Show that ϕis a seminorm on ∞.Is ϕa norm on ∞? Compute ϕ−1(0)

Exercise 10 Determine all p in R+ such that p is a rotund normed linear space (Exercise 7).

Exercise 11 LetX denote the set of all Lipschitz continuous functions on[0, 1],and

a function such that, for allλ∈ R andx, x , y ∈ X,

(i) φ(x, x)≥ 0 andφ(x, x) = 0 iff x = 0,

Any metric linear space X that has the property (2) can be considered as a normedlinear space under the norm · d ∈ RX

+ with

x d:= d(x, 0)

(Right?) Therefore the geometric motivation we gave in Section 1.1 to study thosemetric linear spaces with (2) motivates our concentration on normed linear spaces.

In fact, not only that the metric of any such space arises from a norm, but it is alsotrue that there is an obvious way of “making” a normed linear space X into a metriclinear space that satisfies (2) Indeed, any given norm · on X readily induces adistance function d · on X in the following manner:

It is obvious that d · is translation invariant and satisfies (2)

This observation shows that there is a natural way of viewing a normed linearspace as a metric linear space For this reason, somewhat loosely speaking, one oftensays that a normed linear space “is” a metric linear space Throughout the rest ofthis book, you should always keep this viewpoint in mind

It is important to note that one cannot use (5) to derive a distance functionfrom a norm in the absence of (2) That is, if (X, d) is a metric space, the function

Rd ∈ RX

+ defined by Rd(x) := d(x, 0), is not necessarily a norm (even a seminorm).For instance, if d is the discrete metric, then Rd(x) = d(12x, 0) = Rd(12x)for all x = 0,which shows that Rdfails to be absolutely homogeneous For another example, noticethat Rd is not a seminorm on R∞ (Example I.1.[3]) In fact, since the metric of R∞fails to satisfy (2), it cannot possibly be induced by a norm on R∞.Therefore, we saythat a normed linear space “is” a metric linear space, but not conversely.8

Let X be a normed linear space Now that we agree in viewing X as a metricspace, namely (X, d · ), let us also agree to use any notion that makes sense in thecontext of a metric space also for X, with that notion being defined for (X, d · ).Forinstance, whenever we talk about the ε-neighborhood of a point x in X, we mean

Similarly, the closed unit ball of X, which we denote henceforth as BX, takes theform:

BX :={x ∈ X : x ≤ 1}

Continuing in the same vein, we declare a subset of X open iff this set is open in(X, d · ) By clX(S) we mean the closure of S ⊆ X in (X, d · ), and similarly forintX(S) and bdX(S) Or, we say that (xm) ∈ X∞ converges to x ∈ X (we writethis again as xm → x, of course) iff d · (xm, x) → 0, that is, xm− x → 0 Anytopological property, along with boundedness and completeness, of a subset of X is,again, defined relative to the metric d · By the same token, we view a real function

ϕ on X as continuous, if, for any x ∈ X and ε > 0, there exists a δ > 0 (which maydepend on both ε and x) such that

|ϕ(x) − ϕ(y)| < ε for any y ∈ X with x − y < δ

Similarly, when we talk about the continuity of a function Φ which maps X intoanother normed linear space Y, what we mean is: For all x ∈ X and ε > 0, thereexists a δ > 0 such that

Φ(x)− Φ(y) Y < ε for any y ∈ X with x − y < δ

By Proposition D.1, then, a map Φ ∈ YX is continuous iff, for any x ∈ X and(xm)∈ X∞ with xm

− x → 0, we have Φ(xm)− Φ(x) Y → 0

As an immediate example, let us ask if the norm of a normed linear space rendersitself continuous A moment’s reflection shows that not only is this the case, but themetric induced by a norm qualify that norm as nonexpansive

Proposition 1 The norm of any normed linear space X is a nonexpansive map onX

Proof Apply (3)

With these definitions in mind, the findings of Chapters C—G and I apply readily

to normed linear spaces However, thanks to the absolute homogeneity axiom, there

is considerably more structure in a normed linear space than that contained in anarbitrary metric linear space This point will be clear as we proceed

X\{0} byΦ(x) := x1 x.Show thatΦ is continuous.

Exercise 14 Let X be a normed linear space, and(xm)∈ X∞ a Cauchy sequence Show that( xm )is a convergent real sequence.

Exercise 15 Prove: A subsetSof a normed linear spaceX is bounded iffsup{ x :

x∈ S} < ∞

Exercise 16 (a) Show that any ε-neighborhood of any point in a normed linear space is a convex set (So, every normed linear space “is” a locally convex metric linear space.)

(b) Show that, while its metric is not induced by a norm, any ε-neighborhood of a point in R∞ is convex.

Exercise 17 For any normed linear spaceX,prove that the closure of the unit open ball equals the closed unit ball, that is,

clX({x ∈ X : x < 1}) = {x ∈ X : x ≤ 1}

Also show that the analogous result does not hold for an arbitrary locally convex metric linear space by establishing that

clR∞({x ∈ R∞ : ρ(x, 0) < 1}) ⊂ {x ∈ R∞ : ρ(x, 0)≤ 1},

whereρ is the product metric.

Exercise 18 For any normed linear space X,and any(x, ε)∈ X × R++,prove:

clX(Nε,X(x)) = x +clX(Nε,X(0)) and clX(Nε,X(0)) =−clX(Nε,X(0))

Are these equations true in an arbitrary metric linear space?

Exercise 19 A metric linear spaceX is said to be normable if there is a norm ·

(a) Prove: IfX is normable, then there is anε > 0such that, for every open convex subsetO of X that contains 0, we have Nε,X(0)⊆ λO for someλ > 0.9

(b) Show thatR∞ is a locally convex metric linear space which is not normable.

In Section I.2.4 we have seen that a concave (or convex) function whose domain is

a subset of a metric linear space must be continuous on the interior of its domain,provided that this map is locally bounded at a point in the interior of its domain

We now revisit this result, but this time in the context of normed linear spaces This

is our first main illustration of what sort of things one can do within the context ofnormed linear spaces, but not within that of metric linear spaces

Let X be a normed linear space, and O a nonempty open and convex subset of X.Take any concave ϕ : O → R, and assume that ϕ is locally bounded at some x ∈ Xfrom below.10 Of course, everything that we established in Section I.2.4 readily applies

9 Provided that X is locally convex, the converse of this statement also holds This is a special case of a classic theorem — it is called the Kolmogorov Normability Criterion — which was proved in

1934 by Andrei Kolmogorov, the founder of modern probability theory.

10 Just so that all is clear, in the language of normed linear spaces, the latter requirement means that there is an (ε, K) ∈ R ++ × R such that f(y) ≥ K for all y ∈ O with x − y < ε.

to ϕ (given that X “is” a metric linear space) In particular, by Proposition I.8, ϕ

is continuous Thanks to the norm structure of X (that is, thanks to the fact thatthe metric of X is induced by a norm), we can actually say much more here It turnsout that ϕ is locally Lipschitz continuous in the sense that, for any x ∈ O, thereexist a (δ, K) ∈ R2

++ such that |ϕ(y) − ϕ(z)| ≤ K y − z for all y, z ∈ Nδ,X(x)∩ O.The proof of this fact is an easy adaptation of the proof of Proposition A.14

Proposition 2 Let O be a nonempty open and convex subset of a normed linearspace X, and ϕ : O → R a concave function If ϕ is locally bounded at some x0 ∈ Ofrom below, then ϕ is locally Lipschitz continuous

Proof Assume that ϕ is locally bounded at some point in O from below ByProposition I.8, then, it is continuous Pick any x ∈ X Since it is continuous,

ϕ is locally bounded at x, so there exists an ε > 0 such that Nε,X(x) ⊆ O, and

α := inf ϕ(Nε,X(x)) and β := sup ϕ(Nε,X(x)) are real numbers To focus on thenontrivial case, we assume α < β

Let 0 < δ < ε Fix two distinct y, z ∈ Nδ,X(x) arbitrarily, and let

w := y + δ y−z1 (y− z) and λ := δ+ y−zy−z

It is readily verified that y = λw + (1 − λ)z and w − y = δ.11 We also wish toguarantee that w ∈ Nε,X(x), and for this we need to choose δ a bit more carefully.Notice that w − x ≤ w − y + y − x < 2δ, so if we choose, say, δ := ε2, we have

w∈ Nε,X(x)

Now we are all set The concavity of ϕ implies ϕ(y) ≥ λ(ϕ(w) − ϕ(z)) + ϕ(z), soϕ(z)− ϕ(y) ≤ λ(ϕ(z) − ϕ(w)) ≤ λ(β − α) = δ+ y−zy−z (β− α) < β−αδ y− z Interchanging the roles of z and y in this argument completes the proof

This result is not really a generalization of Proposition A.14, because that tion talks about the Lipschitz continuity of a concave function on any compact subset

proposi-of its domain, as opposed to its local Lipschitz continuity Yet, it is not difficult touse Proposition 2 to obtain generalizations of Proposition A.14 proper Two suchgeneralizations are reported in the following exercises

Exercise 20.H LetO be a nonempty open and convex subset of a normed linear space

X, and take any concaveϕ∈ RO Show that ifϕ is locally bounded from below at

a given point, then ϕ|S is Lipschitz continuous for any compact subsetS ofO

11 I chose w and λ as above in order to guarantee that (i) y = λw + (1 − λ)z, and (ii) w − y = δ After all, (i) says λ(y −w) = (1−λ)(z−y), so (ii) requires λδ = (1−λ) y − z , that is, λ = δ+ yy−z−z (Notice that I would lack the means to find such w and λ if X was known only to be a metric linear space.)

Exercise 21.H For any given n ∈ N, let O be a nonempty open and convex subset

of Rn Show that if ϕ ∈ RO is concave, then ϕ|S is Lipschitz continuous for any

2 Banach Spaces

A normed linear space X is called a Banach space if it is complete (that is, if (X, d · )

is a complete metric space) Clearly, every Fréchet space with (2) is a Banach space.This observation supplies us with many examples of Banach spaces For example, itfollows that Rn,p and p

are Banach spaces for any n ∈ N and 1 ≤ p ≤ ∞ (ExamplesI.1.[1] and[4]) Similarly, for any metric space T, both B(T ) and CB(T ) are Banachspaces (Example I.1.[6]) Here are some other examples

E{dpsoh 2.[1] CB1(I) is a Banach space for any interval I (Example 1.[4])

[2] The metric linear space considered in Example 1.[5] is not Banach (ExerciseC.42)

[3] Recall that a metric subspace of a complete metric space X is complete iff it

is closed in X (Proposition C.7) Thus: Every closed subspace of a Banach space isitself a Banach space

[4]The product of finitely (or countably infinitely) many Banach spaces is Banach.This follows from Theorem C.4

Exercise 22.H Show that c0 is a closed subspace of ∞ Thus c0 is a Banach space.

Exercise 23 Isc0 a Banach space?

Exercise 24 LetX be a Banach space, and ∞(X) := {(xm)∈ X∞ : sup{ xm :

m ∈ N} < ∞} We make ∞(X) a linear space by defining the operations of vector addition and scalar multiplication pointwise, and norm this space by · ∞ :

Banach space.

subspace ofX We define the binary relation∼ on X by x∼ y iff x = y + Y

(a) Show that∼ is an equivalence relation.

(b) Let[x]∼be the equivalence class of xrelative to∼, andX/∼ :={[x]∼: x∈ X}

(Section A.1.3) We define the operations of vector addition and scalar multiplication

on X/∼ as follows:

[x]∼+ [y]∼ := [x + y]∼ and λ[x]∼ := [λx]∼

Show thatX/∼ is a linear space under these operations.

(c) Define · : X/∼→ R+ by [x]∼ := d · (x, Y ), and show that(X/∼, · )is a normed linear space (Would this conclusion still hold ifY was not closed?)

(d ) Show that if X is a Banach space, so is(X/∼, · )

One major advantage of normed linear spaces is that they provide a suitable ground for developing a useful theory of infinite series In particular, there is a naturalway of defining the convergence and absolute convergence of an infinite series in anormed linear space

play-The following definitions are obvious generalizations of the ones we have given inSection A.3.4 for infinite series of real numbers By an infinite series in a normedlinear space X, we mean a sequence in X of the form (Sm

xi)for some (xm)∈ X∞

We say that this series is convergent (in X) if limSm

xi ∈ X (that is, when thereexists an x ∈ X with Sm

xi both for the sequence (Sm

xi) and for the vector limSm

xi (when the latterexists, of course) Thus we often talk about the convergence (or absolute convergence)

of the infinite series S∞

xi, but this should be understood as the convergence (orabsolute convergence) of the sequence (Sm

xi is a convergent series in a normed linear space, then(xm)must be convergent.

There is a tight connection between the notions of convergence and absolute vergence of infinite series in Banach spaces In fact, we may use these concepts toobtain a useful characterization of the notion of completeness (that is “Banachness”)

con-of a normed linear space This result allows us to think about completeness withoutdealing with Cauchy sequences

Proposition 3 (Banach) Let X be a normed linear space Then, X is Banach if,and only if, every absolutely convergent series in X is convergent

Proof Let X be a Banach space, and take any (xm)∈ X∞ with S∞

xi <∞.For any k, l ∈ N with k > l,

d ·

 kSi=1

xi,lSi=1

xi−lSi=1

xi must be convergent

Conversely, assume that every absolutely convergent series in X is convergent.Let (xm) be any Cauchy sequence in X We wish to show that (xm) converges in X.The “trick” is to view the limit of this sequence as an absolutely convergent series.First find positive integers m1 < m2 <· · · such that

(xm i+1− xm i) converges Since

xmk+1 = xm1+

kSi=1(xmi+1

− xmi), k = 1, 2, ,

this implies that the subsequence (xm1, xm2, ) converges in X Thus, since a Cauchysequence with a convergent subsequence must be convergent (Proposition C.6), wemay conclude that (xm) converges in X

Exercise 28.H Let (xm) be a sequence in a normed linear space such that S∞

xi

is absolutely convergent Show thatS∞

xσ(i) is also absolutely convergent for any bijective self-mapσ on N

Exercise 29 Let(xm)be a sequence in a normed linear space such that S∞

(xi+1

−

xi)is absolutely convergent Does(xm)have to be Cauchy? Must it be convergent?

∗ Exercise 30 (Schauder Bases) Let X be a Banach space A set S ⊆ X is said to

be a Schauder basis forX if, for every x∈ X,there exists a unique real sequence

(αm(x)) and a unique(xm)∈ S∞ such thatx =S∞

αi(x)xi.Prove:

(a) IfX is finite dimensional, a subset ofX is a basis forX iff it is a Schauder basis forX;

(b) IfX is infinite dimensional, no basis for X is a Schauder basis;

(c) Each p, 1≤ p < ∞,has a Schauder basis;

(d ) ∞ does not have a Schauder basis;

(e) IfX has a countable Schauder basis, then it is separable.

Remark 1 One major reason why an n-dimensional linear space is a well-behavedentity is that there exists a set {x1, , xn

} in such a space X such that, for each

x ∈ X, there exists a unique (α1, , αn) in Rn with x = Sn

αixi (Corollary F.2).(For instance, this is the reason why every linear functional on a finite dimensionalnormed linear space is continuous.) The Banach spaces with Schauder bases are those

in which this property is satisfied in terms of a sequence of elements in this space.Such spaces are of interest precisely because many finite dimensional arguments havenatural extensions to them through this property Unfortunately, the analogue ofTheorem F.1 is false for Schauder bases, as Exercise 30.(d) attests.12

The cardinality of any basis for an infinite dimensional normed linear space is, bydefinition, infinite Can this cardinality be countably infinite? Yes, of course Takecountably infinitely many linearly independent vectors in any given infinite dimen-sional space, and consider the subspace spanned by these vectors The resultingnormed linear space has a countably infinite basis Surprisingly, however, the space

we just created is sure to be incomplete Put differently, the answer to the questionabove is negative in the case of Banach spaces: A basis for an infinite dimensionalBanach space must be uncountable

This curious fact gives us a good reason to think of an infinite dimensional Banachspace as a “large” normed linear space In particular, any p is, intuitively, “muchlarger” than any of its subspaces that are spanned by countably infinitely manyvectors After all, the result noted above says that any basis for p is uncountable,

1≤ p ≤ ∞

The rest of this section is devoted to the derivation of this interesting property ofBanach spaces.13 We first prove an elementary geometric result about finite dimen-sional normed linear spaces

Lemma 1 Let X be a finite dimensional normed linear space and Y a closedproper subspace of X For every (y, α) ∈ Y × R++, there exists an x ∈ X such that

d · (x, Y ) = x− y = α

Proof This result is proved in exactly the same way it would have been proved

in R2 (Figure 1) Take any z ∈ X\Y Since Y is closed, there exists a w ∈ Y suchthat β := d · (z, Y ) = z− w > 0 (Why? Recall Example D.5.) Then, for any(y, α)∈ Y × R++,the vector x := αβ(z− w) + y satisfies d · (x, Y ) = x− y = α

∗ ∗ ∗ ∗ FIGURE J.1 ABOUT HERE ∗ ∗ ∗ ∗

12 A famous problem of linear analysis (the so-called basis problem) was to determine if at least all separable Banach spaces have Schauder bases After remaining open for over 40 years, this problem was settled in 1973 in the negative by Per Enflo who constructed a closed subspace of C[0, 1] with

no Schauder basis For a detailed account of the theory of Schauder bases, an excellent reference (for the more advanced reader) is Chapter 4 of Megginson (1998).

13 For another elementary proof of this fact, see Tsing (1984).

Proposition 4 No infinite dimensional Banach space has a countable basis.

Proof Let X be a Banach space with dim(X) = ∞, and suppose {x1, x2, }constitutes a basis for X For each m ∈ N, define Xm := span{x1, , xm

− y → 0 Given that {x1, x2, } is a basis for

X, we have y ∈ XM for some M ∈ N Clearly,



This contradiction completes the proof

Exercise 31.H Let X be a normed linear space and Y a closed proper subspace of

X

(a) (F Riesz’ Lemma) Show that, for every0 < α < 1, there exists an x∈ X such thatd · (x, Y )≥ α and x = 1

(b) (A Special Case of Theorem I.2 ) Use part (b) to prove thatclX(N1,X(0))is not

a compact subset of X unlessdim(X) < ∞

3 Fixed Point Theory IV

Our coverage of fixed point theory so far leaves something to be desired Tarski’sFixed Point Theorem (Section B.3.1) and the Banach Fixed Point Theorem (SectionC.6.2) work in very general environments, but they require quite a bit from theself-maps they work with By contrast, the corresponding demands of the BrouwerFixed Point Theorem (Section D.8.3) and Kakutani’s Fixed Point Theorem (SectionE.5.1) are relatively modest, but these theorems apply only within Euclidean spaces

This prompts the present section, our final excursion to fixed point theory Here weexamine a few different senses in which one can extend these results at least to thecontext of Banach spaces At the end of this section, you will begin to see what sort

of amazing things can be accomplished within the context of normed linear spaces,

as you will then possess some serious ammunition for solving existence problems in avariety of contexts

The following important result is a special case of the fixed point theorems proved,independently, by Irwing Glicksberg and Ky Fan in 1952

The Glicksberg-Fan Fixed Point Theorem Let S be a nonempty compact andconvex subset of a normed linear space X, and Γ a convex-valued self-correspondence

on S that has a closed graph Then, Γ has a fixed point, that is, there exists an x ∈ Swith x ∈ Γ(x)

Tm := N1

m ,X(0) for each m ∈ N,and define Γm : S(m)⇒ S(m) by

of Γ — we leave working out the details as an exercise

Let’s vary m now Since each S(m) lies in a metric linear space which is linearlyhomeomorphic to a Euclidean space (Theorem I.1), we may apply Kakutani’s Fixed

14 The idea of the proof is analogous to the way we have deduced Kakutani’s Fixed Point Theorem from Brouwer’s Fixed Point Theorem in Section E.5.3 Once again, “approximation” is the name

of the game We approximate Γ by means of a sequence of self-correspondences defined on finite dimensional subsets of S, and use Kakutani’s Fixed Point Theorem to obtain a fixed point for each term of the sequence The limit of any convergent subsequence of the resulting sequence of fixed points is then shown to be a fixed point of Γ.

Point Theorem to find an xm in S(m) such that xm ∈ Γm(xm), m = 1, 2, (Yes?)The sequence (xm)lies in the compact set S, so it must have a convergent subsequence(Theorem C.2) Let’s denote this subsequence again by (xm) and write x := lim xm.

We wish to show that x ∈ Γ(x) Since xm ∈ Γ(xm)+ clX(Tm), there is a ym ∈ Γ(xm)with xm

− ym

∈ clX(Tm), m = 1, 2, The sequence (ym) lies in S, so it has aconvergent subsequence as well Denoting this subsequence again by (ym), we findlim(xm− ym) = 0.But then

x = lim xm = lim ym+ lim(xm− ym) = lim ym,whereas it follows from the closed graph property of Γ that lim ym

∈ Γ(x)

This result generalizes Kakutani’s Fixed Point Theorem since the latter is proved

in the context of Euclidean spaces, and in that context a set is closed and bounded

iff it is compact However, as the examples given in Exercises D.73 and D.74 strate, in general the compactness requirement cannot be relaxed to being closed andbounded in the Glicksberg-Fan Fixed Point Theorem

demon-Exercise 32.H LetS be a nonempty subset of a metric linear spaceX IfΓ : S ⇒ X

is upper hemicontinuous, andΓ(x) is closed and convex for everyx ∈ S, then does the set of all fixed points ofΓ have to be closed? Convex?

Exercise 33 (Fan) LetX be a normed linear space, andAand B nonempty convex and compact subsets ofX.Supposef ∈ XAis a continuous map withf (A) ⊆ A+B

Show that there is anx∈ A such thatf (x) ∈ x + B

∗ Exercise 34.H Show that “normed linear space” can be replaced with “locally convex metric linear space” in the statement of the Glicksberg-Fan Fixed Point Theorem.

In Section E.6 we have introduced the theory of strategic games through games inwhich the action spaces of the players are subsets of a Euclidean space Some in-teresting economic games, however, do not satisfy this constraint For instance, if

we wished to model a game that is repeatedly played infinitely many times, then weneed to specify the action spaces of the players as a subset of an infinite dimensionalsequence space (as in dynamic programming) Or, if a player were allowed to makeher decisions by randomizing between some or all of her actions, this would meanthat her effective action space is in fact the set of all probability distributions definedover her original action space Provided that the player has infinitely many (non-randomized) action possibilities, her (effective) action space would then lie again in

an infinite dimensional linear space

All in all, it is of interest if we can extend the basic results we obtained in SectionE.6 to the case of strategic games in which the action spaces of the players are

subsets of an arbitrary normed (or metric) linear space As you should expect, theGlicksberg-Fan Fixed Point Theorem provides immediate help in this regard.

E{dpsoh 3 (A Generalization of Nash’s Existence Theorem) If each Si is a empty compact subset of a normed linear space Xi, then we say that the strategicgame G := ({Si, πi}i=1, ,m)is a compact game (Section E.6.2) If, in addition, each

non-πi ∈ RS is continuous (where S := XmSi lies in the product normed linear space

X := XmXi), we say that G is a continuous compact game If, instead, each Xi

is convex, and

πi(λxi+ (1− λ)yi, x−i)≥ min{πi(xi, x−i), πi(yi, x−i)}for any 0 < λ < 1, x−i ∈ X−i and i = 1, , n, then G is called a convex andcompact game Finally, a compact game which is both convex and continuous iscalled a regular game

Nash’s Existence Theorem says that every Euclidean regular game has a Nashequilibrium If you go back and check our proof of this important theorem, you willnotice that the only place we really needed the hypothesis of the game being Euclideanwas when we wanted to invoke Kakutani’s Fixed Point Theorem Consequently,

by replacing Kakutani’s Fixed Point Theorem with the Glicksberg-Fan Fixed PointTheorem in its proof, we may extend the coverage of Nash’s Existence Theorem

to games that are not Euclidean Put precisely: Every regular game has a Nashequilibrium

Exercise 35 Prove that every regular game has a Nash equilibrium.

Exercise 36.H LetG := ({Si, πi}i=1, ,m)be a convex and compact game Prove: If, for eachi, πi is upper semicontinuous, and, for eachiandxi ∈ Xi, the mapπi(xi,·)

is lower semicontinuous (onS−i), thenG has a Nash equilibrium.

We now turn to extending the Brouwer Fixed Point Theorem to the realm of normedlinear spaces Two major results of this sort were obtained by Julius Schauder in

1927 and 1930.15 The first of these theorems is, in fact, an immediate corollary ofthe Glicksberg-Fan Fixed Point Theorem

The Schauder Fixed Point Theorem 1 Every continuous self-map on a empty compact and convex subset of a normed linear space has a fixed point.16

non-15 Julius Schauder (1899-1943) was a student of Banach and made invaluable contributions to the theory of differential equations (most of which were in collaboration with Jean Leray) It would be fair to consider him as one of the founders of modern nonlinear analysis.

16 This result is, in fact, valid also in metric linear spaces In fact, Schauder’s original statement

That is, every nonempty compact and convex subset of a normed linear space hasthe fixed point property This is great news, to be sure Unfortunately, in applicationsone has to work all too often with either a noncompact set or a set the compactness

of which is difficult to establish It is thus worthwhile to think about to what extent

we can extend this result to the context of continuous self-maps on a noncompact set

It is obvious that we can’t just relax the compactness requirement in the SchauderFixed Point Theorem 1, as easy examples would demonstrate.17 In fact, we alreadyknow that compactness cannot be relaxed in the theorem above even to being closedand bounded (Recall Exercises D.73 and D.74) These are the bad news The goodnews is that, even when the domain of our self-map is known only to be closed,bounded and convex, we are in fact guaranteed a fixed point provided that the range

of our map is suitably well-behaved, namely, it is a relatively compact subset of aBanach space.18 This is the content of the Schauder’s 1930 theorem

The Schauder Fixed Point Theorem 2 Let S be a nonempty, closed, boundedand convex subset of a Banach space X, and Φ a continuous self-map on S such that

clX(Φ(S)) is compact Then, Φ has a fixed point

This result requires a bit more work The idea is to show that the restriction of Φ

to coX(Φ(S)) is a self-map, and then to apply the Schauder Fixed Point Theorem 1

to this map To this end, we need to first prove the following 1930 result of StanislawMazur which is quite important in its own right

Mazur’s Compactness Theorem.Let S be a relatively compact subset of a Banachspace X Then, coX(S) is compact

Proof Since coX(S) =coX(clX(S)) — yes? — we may assume that S is compact.Now recall that a metric space is compact iff it is complete and totally bounded(Theorem C.3) Since X is complete and coX(S) is a closed subset of X, coX(S)

is a complete metric space Therefore, all we need to do is to show that coX(S) isalso totally bounded Take any ε > 0 Since S is totally bounded, we can find a

was in the context of an arbitrary metric linear space His proof, however, contained a flaw, and established the result only for locally convex metric linear spaces Whether Schauder’s original claim was true remained as an open problem (known as the Schauder Conjecture) until Robert Cauty settled it in the affirmative in 2001.

Quiz Why isn’t the proof we gave for the Glicksberg-Fan Fixed Point Theorem valid in the case

of an arbitrary metric linear space?

17 Much more can be said here In 1955 Victor Klee proved that a nonempty convex subset of

a Banach space has the fixed point property if and only if it is compact In fact, it is even true that on every convex and noncompact subset of a Banach space there exists a Lipschitz continuous self-map without a fixed point (Lin and Sternfeld (1985)).

18 Reminder A subset of a metric space X is called relatively compact, if the closure of that set in

X is compact (Section D.6.4).

k ∈ N and x1, , xk ∈ S such that S ⊆Vk

Nε ,X(xi),that is, for any u ∈ S, there is aω(u)∈ {1, , k} such that

u − xω(u) < ε

4.Now take any z ∈ coX(S) Clearly, there

is a y ∈ co(S) with z − y < ε4 Moreover, y =Sl

λiyifor some l ∈ N, y1, , yl

∈ Sand (λ1, , λl)∈ Rl+ with Sl

coX(S) ⊆V

{Nε ,X(w) : w ∈ W }where W := co({x1, , xk

}) But W is compact (why?), so it is totally bounded, that

is, we can find finitely many w1, , wm

∈ W such that W ⊆Vm

Nε

2 ,X(wi).(Yes?) Itfollows that coX(S)⊆Vm

Nε,X(wi).Since ε > 0 was arbitrary in this discussion, wemay conclude that coX(S)is totally bounded

Warning The closed convex hull of a compact subset of a normed linear space maywell fail to be compact in general That is, “Banachness” is essential for the validity

of Mazur’s Compactness Theorem (Curiously, this is not so for the Schauder FixedPoint Theorem 2; see Exercise 38 below.) For instance, while S := {0, e1,12e2,13e3, },where e1 := (1, 0, 0, ), e2 := (0, 1, 0, ),etc., is a compact subset of c0 (the subspace

of ∞ which consists of all real sequences with finitely many nonzero terms), coc 0(S)

is not compact in c0 Indeed, the sequence (x1, x2, ), where xk := Sk 1

2 i+1ei foreach k = 1, 2, , lies entirely within coc 0(S), but it does not have a subsequence thatconverges in c0

Proof of the Schauder Fixed Point Theorem 2 Let T := coX(Φ(S)), which is

a nonempty and convex subset of X Moreover, it is compact in X, by Mazur’sCompactness Theorem And, since Φ is continuous, Φ(S) ⊆ S and S is convex, wehave Φ(T ) ⊆ T.19

Thus Φ|T must have a fixed point by the Schauder Fixed PointTheorem 1

19 What I have in mind is this:

Φ(co X (Φ(S))) ⊆ cl X (Φ(co(Φ(S))))

⊆ cl X (Φ(S))

⊆ cl X (co(Φ(S)))

= co X (Φ(S)).

Exercise 37 (The Leray-Schauder Fixed Point Theorem) LetXbe a Banach space

all x∈ X, thenΦhas a fixed point.

The following exercise derives a generalization of the Schauder Fixed Point Theorem

2 It turns out that by means of a different technique of proof we could prove that result without assuming that X is complete andS is closed.

∗ Exercise 38.H (A Generalization of the Schauder Fixed Point Theorem 2 ) LetS be

a nonempty bounded and convex subset of a normed linear spaceX, and let Φbe a continuous self-map onS such thatclX(Φ(S))is compact.

(a) Pick anym∈ N.Since clX(Φ(S)) is totally bounded, there exist akm ∈ N and

sup{ Φ(x) − pm(x) : x∈ S} ≤ m1 and dim(span(pm(S))) <∞

(b) Show thatpm◦Φis a continuous self-map onco({y1, , yk m})and apply er’s Fixed Point Theorem to find ay(m) ∈ co({y1, , yk m}) withpm(Φ(y(m))) =y(m)

must have a fixed point.

(d ) Describe the “idea” behind the proof outlined above.

Like all fundamental fixed point theorems, Schauder’s theorems too yield quite a fewfixed point theorems that are of interest on their own To keep our exposition self-contained, we examine next four such offsprings of these two theorems The first ofthese is a generalization of Proposition D.10 — it was proved by Erich Rothe in 1937.The proof of this result is analogous to how we obtained Proposition D.10 from theBrouwer Fixed Point Theorem, and is thus left as an exercise

Rothe’s Theorem Let X be a Banach space and BX the closed unit ball of X

If Φ : BX → X is a continuous function such that clX(Φ(BX)) is compact andΦ(bdX(BX))⊆ BX,then Φ has a fixed point

Exercise 39 Prove Rothe’s Theorem.

fixed point.

Exercise 41.H In the statement of Rothe’s Theorem BX can be replaced with any nonempty, closed, bounded and convex subset of X Prove! (Try making use of the Minkowski functionals.)

Our next result was proved first by Andrei Markov in 1936.20 It deals with theproblem of finding a common fixed point of a given family F of continuous self-maps

on a given set Any such family F is called commuting if f ◦ g = g ◦ f for all

f, g∈ F

Markov’s Fixed Point Theorem Let S be a nonempty compact and convexsubset of a normed linear space X If F is a commuting family of affine self-maps on

S,then there exists an x ∈ S such that x = f(x) for all f ∈ F

The proof is outlined in the following exercise

Fix(f ) stand for the set of all fixed points off ∈ F

(a) Show thatFix(f )is a nonempty compact and convex subset ofX for anyf ∈ F

(b) Show that{Fix(f) : f ∈ F} has the finite intersection property (Example C.8),

0 everywhere, then it becomes identical to the Schauder Fixed Point Theorem 2

20 In 1938 Kakutani proved a generalization of this result with a direct argument For this reason this result is often referred to as the Markov-Kakutani Fixed Point Theorem.

Proof of Krasnoselski˘ı’s Theorem Let f := idS− g.21 The following two tions about f are essential for the proof.

observa-Claim 1 f is an embedding from S into X

Proof of Claim 1 f is obviously a continuous map from S into X Moreover, forany x, y ∈ S,

f (x)− f(y) = x− y + (g(y) − g(x))

≥ x− y − g(x) − g(y)

≥ (1 + K) x − y ,where 0 < K < 1 is the contraction coefficient of g This observation proves that f isinjective and f−1 is a continuous function on f (S) (Why?)

(Recall Exercise D.20.) But Claim 2 guarantees that clf (S)(h(S)) equals the closure

of h(S) in X, so since the latter is compact (by hypothesis) and f−1 is continuous,

we may conclude that clX(Φ(S)) is compact (Why?) By the Schauder Fixed PointTheorem 2, therefore, there is a z ∈ S such that z = Φ(z) Then, z is a fixed point of

21 The idea of the proof stems from the fact that we are after a fixed point of f −1 ◦ h (Indeed, if

z = f −1 (h(z)), then z−g(z) = f(z) = h(z).) As you will see shortly, the hypotheses of Krasnoselski˘ı’s Theorem enables one to apply the Schauder Fixed Point Theorem 2 to this map.

Exercise 43 (The Sadovski˘ı Fixed Point Theorem) Let X be a Banach space and denote by B X the class of all bounded subsets ofX A mapζ :BX → R+ is called a measure of noncompactness on X if it satisfies the following properties: For all

A, B ∈ BX:

(i) ζ(A) = 0iff Ais totally bounded,

(ii)ζ(A) = ζ(clX(A))and ζ(A) = ζ(co(A)),

(iii) ζ(A∪ B) = max{ζ(A), ζ(B)}.

(For a concrete example of a measure of noncompactness, see Exercise C.46.) LetS be a nonempty, closed, bounded and convex subset of X, and f a continuous self-map on S Suppose that, for some a measure of noncompactness ζ on X, f is

ζ-condensing, that is,ζ(f (A)) < f (A)for all boundedA⊆ S which is not totally bounded Show that f must then have a fixed point by filling the gaps left in the following argument.

(a) Take anyx ∈ S and let Kbe the class of all closed and convex subsets A of S

to (g + h)(S) ⊆ S in the statement of Krasnoselski˘ı’s Theorem Prove this by using Sadovski˘ı’s Fixed Point Theorem in conjunction with the Kuratowski measure

of noncompactness (Exercise C.46).

One of the most widely used methods in establishing that a given functional equationhas a solution is to convert the problem into a fixed point problem, and see if thelatter can be solved by using a suitable fixed point argument You’re of course familiarwith this technique from our earlier work on integral/differential equations (SectionC.7) and Bellman’s Functional Equation (Section E.4.3) The examples consideredbelow illustrate how Schauder’s theorems commingles with this method of proof

E{dpsoh 4 Take any θ ∈ C([0, 1]2) and ϕ ∈ CB([0, 1] × R), and consider thefollowing equation:

f (x) =

] 1 0θ(x, t)ϕ(t, f (t))dt for all 0 ≤ x ≤ 1 (6)

This equation is called the Hammerstein integral equation Question: Does thereexist an f ∈ C[0, 1] that satisfies (6)? The answer is yes, as we demonstrate next

Define the operator Φ : C[0, 1] → R[0,1] by

Φ(f )(x) :=

] 1 0

It follows from the Riemann integration theory (Section A.4.3) that Φ is a self-map

on C[0, 1] (Why?) Moreover, Φ is continuous To prove this, take any f ∈ C[0, 1]and any sequence (fm) in C[0, 1] such that fm → f (relative to the sup-norm, ofcourse) It is easy to see that

Φ(fm)− Φ(f) ∞≤ α max{|ϕ(t, fm(t))− ϕ(t, f(t))| : 0 ≤ t ≤ 1},

where α := max{|θ(x, y)| : 0 ≤ x, y ≤ 1} (Verify!) It will thus follow that Φ iscontinuous if we can show that

limm→∞max{|ϕ(t, fm(t))− ϕ(t, f(t))| : 0 ≤ t ≤ 1} = 0

If this equation was false, we could then find a γ > 0 and a (tm)∈ [0, 1]∞ such that

|ϕ(t, fm(tm))− ϕ(t, f(tm))| ≥ γ, m = 1, 2, (8)Clearly, (tm)has a subsequence that converges in [0, 1] Let us denote this subsequenceagain by (tm)— relabelling if necessary — and set t∗ := lim tm.Since fm− f ∞→ 0,

we have

|fm(tm)− f(t∗)| ≤ |fm(tm)− f(tm)| + |f(tm)− f(t∗)| → 0,

that is, fm(tm)→ f(t∗).(Yes?) But then, since ϕ and f are continuous, lim ϕ(t∗, fm(tm))

= ϕ(t∗, f (t∗)) = lim ϕ(t∗, f (tm)),contradicting (8)

Okay, let’s see what we got here We now know that Φ is a continuous self-map

on C[0, 1], and that the integral equation at hand has a solution iff this map has

a fixed point We wish to find a fixed point of Φ by using one of the theorems ofSchauder, but since C[0, 1] is not bounded, we can’t do this right away We firstneed to find a suitable subset of C[0, 1] on which Φ acts still as a self-map To get

a feeling for what will do the job, let’s examine the range of Φ a bit more closely If

β := sup{|ϕ(t, z)| : (t, z) ∈ [0, 1] × R}, then it follows readily from the definition of Φthat Φ(f (x)) ≤ αβ for any 0 ≤ x ≤ 1 and f ∈ C[0, 1] But then, Φ(f) ∞≤ αβ for all

f ∈ C[0, 1] Put differently, Φ(C[0, 1]) ⊆ S, where S := {f ∈ C[0, 1] : f ∞ ≤ αβ}.Aha! Then, S is a nonempty, closed, bounded and convex subset of C[0, 1], and Φ|S

is a self-map on S Therefore, thanks to the Schauder Fixed Point Theorem 2, ourtask reduces to showing that Φ(S) is a relatively compact set (Why?)

To show that the closure of Φ(S) is compact in C[0, 1], observe first that Φ(S)

is obviously bounded — it is contained in S — so, by the Arzelà-Ascoli Theorem,

it is enough to show that Φ(S) is equicontinuous (Right?) To this end, fix any(ε, x)∈ R++× [0, 1] It is easy to see that, for any 0 ≤ y ≤ 1 and f ∈ C[0, 1],

|Φ(f)(x) − Φ(f)(y)| ≤ β

] 1

0 |θ(x, t) − θ(y, t)| dt