Real Analysis with Economic Applications - Chapter I ppt

In particular, you will see that the infinite dimensional case contains a good deal ofsurprises, such as the presence of discontinuous linear functionals, linear subspacesthat are not cl

Chapter I

Metric Linear Spaces

In Chapters C-G we have laid out a foundation for studying a number of issues thatarise in metric spaces (such as continuity and completeness) and others that arise

in linear spaces (such as linear extensions and convexity) However, save for a fewexceptions in the context of Euclidean spaces, we have so far studied such matters inisolation from each other This should really be remedied, for in most applications oneworks with a space structure that allows for the simultaneous consideration of metricand linear properties In this chapter, therefore, we bring our earlier metric and linearanalyses together, and explore a framework that is general enough to encompass suchsituations In particular, this setup will allow us to talk about things like continuouslinear functions, complete linear spaces, closed convex sets, and so on

We begin the chapter by discussing in which sense one may think of a metricstructure to be imposed on a linear space as “compatible” with the inherent algebraicstructure of that space This leads us to the notion of metric linear space Aftergoing through several examples of such linear spaces, we derive some elementaryproperties pertaining to them, and examine when a linear functional defined on such

a space would be continuous We discuss the basic properties of continuous linearfunctionals here in some detail, and highlight the significance of them from a geometricviewpoint We then consider several characterizations of finite dimensional metriclinear spaces, and finally, provide a primer on convex analysis on infinite dimensionalmetric linear spaces The most important results within this framework will clarifythe basic connection between the notions of “openness” and “algebraic openness,”and sharpen the separation theorems we have obtained in Chapter G

A common theme that keeps coming up throughout the chapter pertains to thecrucial differences between the finite and infinite dimensional metric linear spaces

In particular, you will see that the infinite dimensional case contains a good deal ofsurprises, such as the presence of discontinuous linear functionals, linear subspacesthat are not closed, and open neighborhoods the closures of which are necessarilyunbounded So, hopefully, it should be fun!1

1 In textbooks on functional analysis, metric linear spaces are covered, at best, as special cases of either topological groups or topological linear spaces Since I do not presume familiarity with general topology here, and topological considerations arise in economic applications mostly through metric spaces, I chose to depart from the standard treatments If you wish to see the results I present here in their natural (topological-algebraic) habitat, you should try outlets like Holmes (1975), Megginson (1998) and/or Aliprantis and Border (1999).

1 Metric Linear Spaces

Take a linear space X Given the objectives of this chapter, we wish to endow Xwith a metric structure that is compatible with the algebraic/geometric structure of

X.We thus need to choose a distance function on X which goes well — whatever thismeans — with the linear structure of X

Let’s think of the things we can do in X We can shift around the vectors in X (byusing the vector addition operation) and, for any given vector in X, we can constructthe ray that originates from the origin 0 and that goes through this vector (by usingthe scalar multiplication operation) Let’s focus on shifting, that is, translating,vectors first What we mean by this precisely is that, for any given x, z ∈ X, thevector x + z is well-defined in X, and geometrically speaking, we think of this vector

as the “translation of x by z” (or the “translation of z by x”) Now take anothervector y ∈ X and translate it by z as well to obtain the vector y + z Here is a goodquestion: How should the distance between x and y relate to that between x + z and

y + z?If we wish to model a space whose geometry is homogeneous — a term which wewill formalize later on — it makes good sense to require these two distances be equal.After all, if posited globally, this sort of a property would ensure that the distancebetween any two given points be preserved when these vectors are translated (in thesame way) anywhere else in the space.2 This is, for instance, precisely the case forthe geometry of the plane (or of any other Euclidean space)

Dhilqlwlrq Let X be a linear space A metric d ∈ RX×X+ is called translationinvariant if

d(x + z, y + z) = d(x, y) for all x, y, z ∈ X (1)

Observe that (1) is a statement that ties the metric d with the group structure

of the linear space X Translation invariance is thus a concept which is partly ric and partly algebraic — it connects in a particular way the distance function on alinear space with the operation of vector addition In fact, this connection is tighterthan what may seem at first If you endow a linear space with a translation invari-ant metric, then, per force, you make the vector addition operation of your spacecontinuous!

met-Proposition 1 Let X be a linear space which is also a metric space If the metric

d of X is translation invariant, then the map (x, y) → x + y is a continuous functionfrom X × X into X

2 I have so far tried to be careful in referring to the elements of a metric space as “points,” and those of a linear space as “vectors.” The spaces we will work with from now on will be both metric and linear, so I will use these two terms interchangeably.

Proof Suppose d is translation invariant, and take any (xm), (ym) ∈ X∞ and

≤ d(xm− x, 0) + d(0, y − ym) (by the triangle inequality)

= d(xm, x) + d(ym, y) (by (1) and symmetry of d),

frame-X by the discrete metric (which is, obviously, translation invariant) Then the map

λ→ λx (from R into X) is not continuous for any x ∈ X\{0} (Proof If x = 0, then(m1x)is a sequence which is not eventually constant, so endowing X with the discretemetric d yields d1

mx, 0

= 1for each m.)

To connect the metric and linear structures in a more satisfactory manner, fore, we need to do better than asking for the translation invariance property.3 Inparticular, we should certainly make sure that a map like λ → λx (in XR) and a maplike x → λx (in XX) are declared continuous by the metric we impose on X Well,then why don’t we concentrate on the case where the metric at hand render the map(λ, x)→ λx (in XR×X) continuous? We know that translation invariance gives us thecontinuity of the vector addition operation This would, in turn, render the scalarmultiplication operation on X continuous And putting these together, we arrive at

there-a clthere-ass of spthere-aces which there-are both metric there-and linethere-ar, there-and in which the metric there-and linethere-arstructures of the space are naturally compatible

Dhilqlwlrq Let X be a linear space which is also a metric space If the metric d

of X is translation invariant, and, for all convergent (λm)∈ R∞ and (xm)∈ X∞, wehave

lim λmxm = (lim λm) (lim xm) ,

3 To be fair, I should say that a reasonably complete algebraic theory can be developed using only the translation invariance property, but this theory would lack geometric competence, so I will not pursue it here (But see Exercises 15 and 16.)

then X is called a metric linear space If, in addition, (X, d) is complete, then, wesay that X is a Fréchet space A metric linear space X is said to be nontrivial

if X = {0}, finite dimensional if dim(X) < ∞, and infinite dimensional ifdim(X) = ∞

Put succinctly, a metric linear space is a linear space endowed with a translationinvariant distance function that renders the scalar multiplication operation continu-ous.4 So, it follows from Proposition 1 that X is a metric linear space iff it is both alinear and a metric space such that

(i) the distance between any two points are preserved under the identical tions of these points,

transla-(ii) the scalar multiplication map (λ, x) → λx is a continuous function from R×Xinto X, and

(iii) the vector addition map (x, y) → x + y is a continuous function from X × Xinto X

Let’s look at some examples

E{dpsoh 1.[1] Rn is a Fréchet space, for any n ∈ N This is easily proved by usingthe fact that a sequence converges in a Euclidean space iff each of its coordinatesequences converges in R (Similarly, Rn,p

is a Fréchet space, for any n ∈ N and

[3] Consider the linear space R∞of all real sequences which is metrized by means

of the product metric:

Warning What I call here a Fréchet space is referred to as an F -space in some texts which reserve the term “Fréchet space” for locally convex and complete metric linear spaces (I will talk about the latter type of spaces in the next chapter.)

(Recall Section C.8.2.) This metric is obviously translation invariant To see that itrenders the scalar multiplication operation on R∞ continuous, take any (λm) ∈ R∞and any sequence (xm)in R∞such that λm → λ and xm

→ x for some (λ, x) ∈ R×R∞.(Of course, (xm) is a sequence of sequences.) By Proposition C.8, xmi → xi (as

m→ ∞), so we have λmxm

i → λxi (as m → ∞), for each i ∈ N Applying PropositionC.8 again, we find lim λmxm = λx Combining this observation with Theorem C.4,therefore, we may conclude: R∞ is a Fréchet space

[4] p is a metric linear space for any 1 ≤ p < ∞ Fix any such p That dp

is translation invariant is obvious To see the continuity of scalar multiplicationoperation, take any (λm)∈ R∞ and any sequence (xm) in p such that λm → λ and

dp(xm, x)→ 0 for some (λ, x) ∈ R × p By the triangle inequality, we have

dp(λmxm, λx) ≤ dp(λmxm, λxm) + dp(λxm, λx)

=

∞Si=1|λm− λ|p|xmi |p

1 p

+

∞Si=1|λ|p|xmi − xi|p

1 p

= |λm− λ| dp(xm, 0) +|λ| dp(xm, x)

But since (xm)is convergent, (dp(xm, 0))is a bounded real sequence (for, dp(xm, 0)≤

dp(xm, x)+dp(x, 0)→ dp(x, 0)) Therefore, the inequality above ensures that dp(λmxm,λx)→ 0, as was to be proved

Combining this observation with Example C.11.[4], we may conclude that 2 is aFréchet space Moreover, the argument given in that example can be generalized toany p space — but we omit doing this in this text — so: p is a Fréchet space for any

1≤ p < ∞

[5] ∞ is a Fréchet space

[6]For any metric space T, both B(T ) and CB(T ) are metric linear spaces In fact,

by Example C.11.[5] and Proposition D.7, both of these spaces are Fréchet spaces

Exercise 2 Supply the missing arguments in Examples 1.[5]-[6].

Exercise 3 Show that the metric space introduced in Exercise C.42 is a metric linear space (under the pointwise defined operations) which is not Fréchet.

Exercise 4 Show thatC1[0, 1]is a Fréchet space.

Exercise 5 Let X := {(xm) ∈ R∞ : sup{|xm|m1 : m ∈ N} < ∞}, and define

d ∈ RX×X+ by d((xm), (ym)) := sup{|xm− ym|m1 : m ∈ N} Is X a metric linear space relative to dand the usual addition and scalar multiplication operations? Exercise 6 (Product Spaces) Let(X1, X2, ) be a sequence of metric linear spaces, andX := X∞Xi.We endowXwith the product metric (Section C.8.2), and make it

a linear space by defining the operations of scalar multiplication and vector addition pointwise Show that X is a metric linear space, and it is a Fréchet space whenever each Xi is complete.

Exercise 7 Let X be any linear space, and ϕ a seminorm on X (Section G.2.1) Define dϕ

∈ RX×X+ by dϕ(x, y) := ϕ(x− y) Is dϕ necessarily a distance function? Show thatX would become a metric linear space when endowed with dϕ, provided thatϕ−1(0) ={0}.

The following proposition collects some basic facts about metric linear spaces that

we will need later on It also provides a good illustration of the interplay betweenalgebraic and metric considerations, which is characteristic of metric linear spaces

Proposition 2 For any subsets A and B of a metric linear space X, the followingare true:

(a) clX(A + x) = clX(A) + x for all x ∈ X;

(b) If A is open, then so is A + B;

(c) If A is compact and B is closed, then A + B is closed;

(d) If both A and B are compact, so is A + B

Proof We only prove part (c) here, leaving the other three claims as exercises Let(xm)be a sequence in A + B such that xm

→ x for some x ∈ X By definition, thereexist a sequence (ym)in A and a sequence (zm)in B such that xm = ym+ zm for each

m Since A is compact, Theorem C.2 implies that there exists a strictly increasingsequence (mk)in N such that (ym k)converges to a vector, say y, in A But then (zm k)converges to x − y by the continuity of vector addition, so, since B is closed, it followsthat x − y ∈ B Thus, x = y + (x − y) ∈ A + B, which proves that A + B is closed

Exercise 8.H Complete the proof of Proposition 2.

Parts (b) and (c) of Proposition 2 point to a significant difference in the behavior

of sums of open sets and closed sets: While the sum of any two open sets is open,the sum of two closed sets need not be closed This difference is worth keeping inmind It exists even in the case of our beloved real line For instance, let A := N and

B :={−2 + 12,−3 +13, } Then, both A and B are closed subsets of R, but A + B

is not closed in R, for m1 ∈ A + B for each m = 1, 2, , but limm1 = 0 /∈ A + B Asanother (more geometric example), let A := {(x, y) ∈ R2 : x = 0 and y ≥ 1x} and

B := R×{0} (Draw a picture.) Then, while both A and B are closed in R2, A + B

is not a closed subset of R2 (Proof A + B = R × R++.)

The most common method of creating a new metric linear space from a givenmetric linear space is to look for a subset of this space which inherits the metriclinear structure of the original space This leads us to the notion of a metric linearsubspace of a metric linear space X, which is defined as a subset of X which isboth a linear and a metric subspace of X (For instance, R2

× {0} is a metric linearsubspace of R3, and C[0, 1] is a metric linear subspace of B[0, 1].) Throughout this

chapter, by a subspace of a metric linear space X, we mean a metric linear subspace

(c)intX(A) +intX(B)⊆ A + intX(B)⊆ intX(A+B),provided thatintX(B) =∅

Exercise 12.H Show that every metric linear space is connected.

Exercise 13.H (Nikodem) Let X and Y be two metric linear spaces, S a nonempty convex subset of X, and Γ : S ⇒ Y a correspondence which has a closed graph Show that if

1

2Γ(x) + 12Γ(y)⊆ Γ(12x +12y) for any x, y ∈ S,

thenΓ is a convex correspondence (Exercise G.8).

The primary objects of analysis within the context of metric linear spaces are thoselinear operators L that map a metric linear space X (with metric d) to another metriclinear space Y (with metric dY) such that, for all x ∈ X and ε > 0, there exists a

δ > 0 with

d(x, y) < δ implies dY(L(x), L(y)) < ε

For obvious reasons, we call any such map a continuous linear operator, exceptwhen Y = R, in which case we refer to it as a continuous linear functional

2.1 Examples of (Dis-)Continuous Linear Operators

E{dpsoh 2 [1] Any linear operator that maps Rn to Rm is a continuous linearoperator This follows from Examples D.2.[4] and F.6.[1] (Verify!)

[2] Let L ∈ RB[0,1] be defined by L(f ) := f (0) L is obviously a linear functional

on C[0, 1] It is also continuous — in fact, it is a nonexpansive map — because

|L(f) − L(g)| = |f(0) − g(0)| ≤ d∞(f, g) for any f, g ∈ B[0, 1]

[3] Let L ∈ RC[0,1] be defined by

L(f ) :=

] 1 0

is — guess what? — infinite dimensional Here are a few examples that illustrate this

E{dpsoh 3 [1] Let X be the linear space of all real sequences that are absolutelysummable, that is,

X :=

(x1, x2, )∈ R∞ :S∞

i=1|xi| < ∞

Let us make this space a metric linear space by using the sup-metric d∞.(Notice that

X is a metric linear space that differs from 1 only in its metric structure.) Define

where exactly m entries of xmare nonzero Clearly, d∞(xm, 0) = m1 → 0 Yet L(xm) =

1for each m, so L(lim xm) = 0 = 1 = lim L(xm).Conclusion: L is linear but it is notcontinuous at 0

[2] We play on this theme a bit more Let

X :={(x1, x2, )∈ R∞ : sup{|xi| : i ∈ N} < ∞},

L would not be continuous Indeed, if x1 := (1δ, 0, 0, ), x2 := (0,δ12, 0, ), etc.,then xm

→ (0, 0, ) with respect to the product metric (Proposition C.8), and yetL(xm) = 1 for each m while L(0, 0, ) = 0

Exercise 14.H Let X denote the linear space of all continuously differentiable real functions on [0, 1], and make this space a metric linear space by using the sup- metric d∞ (Thus X is a (dense) subspace of C[0, 1]; it is not equal to C1[0, 1].) Define L ∈ L(X, R) and D ∈ L(X, C[0, 1]) by L(f ) := f (0) and D(f ) := f ,

respectively Show that neitherD norLis continuous.

While it is true that linearity of a map does not necessarily guarantee its ity, it still brings quite a bit of discipline into the picture Indeed, linearity spreadseven the tinniest bit of continuity a function may have onto the entire domain of thatfunction Put differently, there is no reason to distinguish between local and globalcontinuity concepts in the presence of linearity (This is quite reminiscent of additivereal functions on R; recall Exercise D.40.) We formalize this point next

continu-Proposition 3 Let X and Y be two metric linear spaces A linear operator L from

X into Y is uniformly continuous if, and only if, it is continuous at 0

Proof Let L ∈ L(X, Y ) be continuous at 0, and take any ε > 0 and x ∈ X

By continuity at the origin, there exists a δ > 0 such that dY(L(z), 0) < ε wheneverd(z, 0) < δ.Now take any y ∈ X with d(x, y) < δ By translation invariance of d, wehave d(x − y, 0) < δ, and hence, by translation invariance of dY and linearity,

dY(L(x), L(y)) = dY(L(x)− L(y), 0) = dY(L(x− y), 0) < ε

Since y is arbitrary in X here, and δ does not depend on x, we may conclude that L

is uniformly continuous.5

5 Another Proof If L is continuous at 0, then, for any ε > 0, there exists a δ > 0 such that L(N δ,X (0)) ⊆ N ε,Y (L(0)) = N ε,Y (0), and thus, for any x ∈ X,

L(N δ,X (x)) = L(x + N δ,X (0)) = L(x) + L(N δ,X (0)) ⊆ L(x) + N ε,Y (0) = N ε,Y (L(x)), and we are done (See, all I need here is the additivity of L But wait, where did I use the translation invariance of d and d Y ?)

Even though its proof is easy, a first comer to the topic may be a bit surprised

by Proposition 3 Let us try to think through things more clearly here The keyobservation is that any metric linear space Z is homogeneous in the sense that, forany given z0, z1 ∈ Z, we can map Z onto Z in such a way that (i) z0 is mapped to z1,and (ii) all topological properties of Z are left intact More precisely, the translationmap τ : Z → Z defined by τ(z) := z + (z1

− z0) is an homeomorphism, thanks tothe continuity of vector addition on Z But then if we know that an L ∈ L(X, Y )

is continuous at a given point, say 0, to show that L must then be continuous at anarbitrary point x0

∈ X, all we need is to translate X so that x0 “becomes” 0 (of X),and translate Y so that 0 (of Y ) “becomes” y0 := L(x0) So define τ : X → X by

τ (x) := x− x0, and ρ : Y → Y by ρ(y) := y + y0 Since τ and ρ are continuouseverywhere, and L is continuous at 0, and since L = ρ ◦ L ◦ τ, it follows that L iscontinuous at x0 Thus linearity (in fact, additivity) spreads continuity at a singlepoint to the entire domain precisely via translation maps that are always continuous

in a metric linear space (The situation could be drastically different if the metricand linear structures of the space were not compatible enough to yield the continuity

of vector addition.)

Often in real analysis, one gets a “clearer” view of things upon suitably izing the mathematical structure at hand The following exercises aim to clarify theorigin of Proposition 3 further by means of such a generalization

general-Exercise 15 We say that (X, +, d) is a metric group if (X, d) is a metric space and (X, +) is a group such that the binary relation + is a continuous map from

X× X intoX For anyx∗ ∈ X,we define the self-mapτ on X by τ (x) := x− x∗,

which is called a left translation (Right translations are defined as maps of the formx→ −x∗+ x, and coincide with left translations when(X, +) is Abelian.) (a) Show that any left or right translation onX is a homeomorphism.

(b) Show that if O is an open set that contains the identity element0of X,then so

iff it is continuous at0 (How does this fact relate to Proposition 3?)

It is now time to consider a nontrivial example of a continuous linear functionaldefined on an infinite dimensional metric linear space

E{dpsoh 4 For any n ∈ N, Examples 2.[1]and F.6.[1]show that any continuous ear functional L on Rn

lin-is of the form x →Sn

αixi (for some real numbers α1, , αn)

We now wish to determine the general structure of continuous linear functionals on

R∞

Take any continuous L ∈ L(R∞, R) Define αi := L(ei) for each i ∈ N, where

e1 := (1, 0, 0, ), e2 := (0, 1, 0, ), etc We claim that αi = 0 for all but finitelymany i To see this, note that

(xm) = lim

M →∞

MSi=1

xiei for any (xm)∈ R∞

(Proof Observe that the mth term of kth term of the sequence of sequences (x1e1,

x1e1+ x2e2, ) is xm for any k ≥ m and m ∈ N, and apply Proposition C.8.) So, bycontinuity and linearity of L, for any (xm)∈ R∞,

L((xm)) = L

lim

M →∞

MSi=1

Since it is easy to check that this indeed defines a continuous linear functional on

R∞, we conclude: L is a continuous linear functional on R∞ if, and only if, thereexists a finite subset S of N and real numbers αi, i∈ S, such that (3) holds.7

Remark 1 The argument given in Example 4 relies on the fact that (xm) =limM →∞SM

xiei for any real sequence (xm) This may perhaps tempt you to viewthe set {ei : i ∈ N} as a basis for R∞ However, this is not the case, for we maynot be able to express a real sequence as a linear combination of finitely many eis.(Consider, for instance, the sequence (1, 1, ).) Instead, one says that {ei : i ∈ N}

is a Schauder basis for R∞ The defining feature of this concept is expressing thevectors in a metric linear space as a linear infinite series As opposed to the standardone, this basis concept depends on the metric in question since it involves the notion

of “convergence” in its definition (More on this in Section J.2.2.)

Exercise 17 Let S be a finite subset of [0, 1] Show that, for anyλ ∈ RS, the map

Exercise 18 Give an example of a discontinuous linear functional on ∞.

Exercise 19 Prove that an upper semicontinuous linear functional on a metric linear space is continuous.

Exercise 20.HShow that a linear functionalLon a metric linear spaceXis continuous

iff there exists a continuous seminormϕonXsuch that|L(x)| ≤ ϕ(x)for allx∈ X

Exercise 21.H LetLbe a linear functional on a metric linear space X.Prove: (a) Lis continuous iff it is bounded on some open neighborhood O of 0;

(b) IfL is continuous, thenL(S) is a bounded set for any bounded subsetS ofX

The following two exercises further develop the theory of linear correspondences sketched in Exercise F.31 They presume familiarity with the definitions and results

of that exercise.

Exercise 22.H (Continuous Linear Correspondences) Let X and Y be two metric linear spaces,and Γ : X ⇒ Y a linear correspondence.

(a) Show thatΓ is upper hemicontinuous iff it is upper hemicontinuous at0

(b) Show thatΓ is lower hemicontinuous iff it is lower hemicontinuous at0

Exercise 23.H (Continuous Linear Selections) Let X and Y be two metric linear spaces,and Γ : X ⇒ Y a linear correspondence.

(a) Show that ifΓ admits a continuous linear selection, then it is continuous

(b) Prove: If P ∈ L(Γ(X), Γ(X)) is continuous, idempotent (i.e. P ◦ P = P) and

null(P ) = Γ(0), thenP ◦ Γis a continuous linear selection from Γ

(c) In the special case whereXand Y are Euclidean spaces, prove that ifΓis upper hemicontinuous, then it admits a continuous linear selection.

2.2 Continuity of Positive Linear Functionals

Is a positive linear functional defined on a preordered metric linear space necessarilycontinuous? A very good question, to be sure Monotonic real functions possess,

in general, reasonably strong continuity properties (Any such function on R is,for instance, continuous everywhere but countably many points.) So, while a linearfunctional need not be continuous in general, perhaps monotonic linear functionalsare

The bad news is that the answer is no! Indeed, all of the linear functionalsconsidered in Example 3 are positive (with R∞ being partially ordered by means ofthe coordinatewise order ≥), but as we have seen there, these functionals may wellturn up discontinuous, depending on how we choose to metrize their domain Thegood news is that this is not the end of the story It is in fact possible to pinpoint thesource of the problem, and thus find out when it would not arise Our next result,which provides us with a rich class of continuous linear functionals, does preciselythis

Proposition 4 (Shaefer) Let X be a preordered metric linear space If intX(X+) =

∅, then any positive linear functional on X is continuous.8

Proof Let L be a positive linear functional on X, and take any sequence (ym)∈

2.3 Closed vs Dense Hyperplanes

Recall that we can identify a hyperplane in a linear space with a nonzero linearfunctional up to an additive constant (Corollary F.4) In this subsection we showthat the closed hyperplanes in a metric linear space have the same relationship withthe continuous nonzero linear functionals on that space The crux of the argument

is contained in the following fact

Proposition 5 Let X be a metric linear space and L ∈ L(X, R) Then L is uous if, and only if, null(L) is a closed subspace of X.10

contin-Proof The “only if” part is easy To establish the “if” part, take any L ∈ L(X, R)with Y := null(L) being a closed subspace of X In view of Proposition 3, it is enough

8 Reminder We denote the vector preorder of X as X , and the positive cone induced by X

as X+ By the way, a preordered metric linear space is a metric linear space endowed with a vector preorder (In particular, no (direct) connection between the metric of the space and its vector preorder is postulated — these relate to each other only through being consistent with the operations

of vector addition and scalar multiplication.)

9 Yes, Mk depends on k here, but no matter! For any ε > 0, there exists a k ∈ N such that

1

k L(x) 

 < ε, and hence |L(y m )| < ε for all m ≥ M, for some M ∈ N.

10 Reminder null(L) := L −1 (0).

to check that L is continuous at 0 So, take any (xm)∈ X∞ with xm → 0 We wish

to show that L(xm)→ 0

If L is the zero functional, then the claim is obviously true, so assume that it isnonzero Then, by Proposition F.6, null(L) is a ⊇-maximal proper subspace of X.(Why?) So, for an arbitrarily fixed w ∈ X\Y, we have span(Y + w) = X, and hence

we may write xm = ym+ λmwfor some (λm, ym)∈ R × Y, m = 1, 2, 11 Clearly, foreach m, we have L(xm) = L(ym) + λmL(w) = λmL(w) (since ym ∈ null(L)), whileL(w) = 0(since w /∈ Y ) Thus, by the continuity of scalar multiplication, all we need

to do is to show that λm → 0

Let’s first verify that (λm) is a bounded real sequence If this was not the case,

we could find a subsequence (λm k)of this sequence such that λm k = 0for each k, and1

λmk → 0 (as k → ∞) But if we let θk := λ1

Then, letting k → ∞ and using the continuity of scalar multiplication,

we get d(w, Y ) = 0, which is impossible, given that w /∈ Y and Y is closed (ExerciseD.2) We conclude that (λm) is bounded

Now let λ := lim sup λm Then λ ∈ R (because (λm) is bounded) and there is

a subsequence (λm k) with λm k → λ (as k → ∞) Since ym = xm

− λmw for each

m, and lim xm = 0, we have lim ymk = λw by the continuity of scalar multiplicationand vector addition But since ym k ∈ Y = L−1(0) for each k, and Y is closed,lim ym k ∈ L−1(0), so, λL(w) = 0 Since w /∈ Y, we thus find λ = 0 But notice thatthe same argument would go through verbatim, if we instead had λ := lim inf λm Wemay thus conclude that lim λm= 0

Here is the result we promised above

Proposition 6 A subset H of a metric linear space X is a closed hyperplane in X

if, and only if,

H ={x ∈ X : L(x) = α}

for some α ∈ R and a continuous nonzero linear functional L on X

11 False Proof “Given that xm = ym+ λ m w for each m, and xm → 0, we have 0 = lim ym+ (lim λ m )w But since Y is closed, lim ym ∈ Y, that is, L(lim ym) = 0, and hence applying L to both sides of this equation, we obtain 0 = (lim λm)L(w) Since w / ∈ Y , we have L(w) = 0, so it follows that lim λm = 0 But then, since L is linear and y m ∈ Y for each m, we have L(x m ) = L(y m ) + λmL(w) = λmL(w) for each m, so, letting m → ∞, we get lim L(x m ) = 0 as we sought.” Unfortunately, things are a bit more complicated than this Please find what’s wrong with this argument before proceeding further.

12 Reminder For any nonempty subset S of a metric space X, and x ∈ X, d(x, S) := inf{d(x, z) :

z ∈ S}.

Proof The “if” part of the claim follows readily from Proposition D.1 and lary F.4 Conversely, if H is a closed hyperplane in X, then by Corollary F.4 thereexist an α ∈ R and a nonzero L ∈ L(X, R) such that H = L−1(α) Take any x∗ ∈ Hand define Y := H − x∗ By Proposition 2.(c), Y is closed subspace of X But it isevident that Y = null(L) — yes? — so by Proposition 5, L is continuous.

Corol-Exercise 24 Show that any open (closed) halfspace induced by a closed hyperplane

of a metric linear space is open (closed).

Exercise 25 Let X and Y be two metric linear spaces and L ∈ L(X, Y ) Prove

or disprove: If Y is finite dimensional, then L is continuous iff null(L) is a closed subspace ofY

Proposition 6 entails that any hyperplane that is induced by a discontinuous linearfunctional cannot be closed Let us inquire into the nature of such hyperplanes a bitfurther In Exercise 10, we have noted that the closure of a subspace Z of a metriclinear space X is itself a subspace Thus the closure of a ⊇-maximal proper subspace

Z of X is either Z or X Put differently, any ⊇-maximal proper linear subspace of

X is either closed or (exclusive) dense Since any hyperplane H can be written as

H = Z +x∗ for some ⊇-maximal proper linear subspace Z of X and x∗ ∈ X, and since

clX(H) = clX(Z + x∗) =clX(Z) + x∗ (why?), this observation gives us the followinginsight

Proposition 7 A hyperplane in a metric linear space is either closed or (exclusive)dense

How can a hyperplane be dense in the grand space that it lies in? This seeminglyparadoxical situation is just another illustration of how our finite dimensional intu-ition can go astray in the realm of infinite dimensional spaces Indeed, there cannot

be a dense hyperplane in a Euclidean space (Why?) But all such bets are off ininfinite dimensional spaces Since a linear functional on such a space need not becontinuous, a hyperplane in it is not necessarily closed

Actually, this is not as crazy as it might first seem After all, thanks to theWeierstrass Approximation Theorem, we know that every continuous real function

on [0, 1] is a uniform limit of a sequence of polynomials defined on [0, 1] But thismeans that P[0, 1] is not a closed subset of C[0, 1], while, of course, it is a linearsubspace of C[0, 1] Thus, in an infinite dimensional metric linear space, a propersubspace which is not even ⊇-maximal can be dense!

To sum up, one important analytic lesson we learn here is that a linear functional

is in general not continuous, whereas the geometric lesson is that a hyperplane is

in general not closed (in which case it is dense) While the latter finding may defyour geometric intuition (which is, unfortunately, finite dimensional), this is just howthings are in infinite dimensional metric linear spaces

E{dpsoh 5 Consider the metric linear space X and the discontinuous L ∈ L(X, R)defined in Example 3.[1] Propositions 5 and 7 together say that null(L) must bedense in X Let us verify this fact directly Take any y ∈ X, and define the sequence(xm)∈ X∞ by

xm :=



y1 −m1

∞Si=1

yi, , ym− m1

∞Si=1

yi, ym+1, ym+2,

, m = 1, 2,

(Each xm is well-defined, because, by definition of X, we have S∞

yi ∈ R.) Clearly,for each m ∈ N,

L(xm) =

mSi=1

yi−S∞i=1

yi+ S∞i=m+1

yi = 0

so that xm∈ null(L) But we have

d∞(xm, y) = sup{|xmi − yi| : i ∈ N} = m1

∞Si=1

yi → 0sinceS∞

yi is finite This shows that, for any vector in X, we can find a sequence innull(L) that converges to that vector, that is, null(L) is dense in X

E{dpsoh 6 In the previous example we inferred the denseness of a hyperplane fromthe discontinuity of its defining linear functional In this example, we shall construct

a discontinuous linear functional on the good old C[0, 1] by using the denseness of ahyperplane that this functional induces Define fi

L cannot be continuous

2.4 Digression: On the Continuity of Concave Functions

We have noted earlier that a concave real function defined on an open subset of aEuclidean space must be continuous This need not be true for a concave functiondefined on an open subset of a metric linear space After all, we now know that a

linear functional (which is obviously concave) may well be discontinuous if its domain

is an infinite dimensional metric linear space This said, we also know that linearityspreads the minimal amount of continuity a function may have onto its entire do-main Put differently, a linear functional is either continuous or it is discontinuouseverywhere (Proposition 3) Remarkably, concavity matches the strength of linearity

on this score That is to say, any concave (or convex) function whose domain is anopen and convex subset of a metric linear space is either continuous or discontinuouseverywhere

In fact, we can say something a bit stronger than this Given a metric space Xand a point x ∈ X, let us agree to call a function ϕ ∈ RX locally bounded at xfrom below if there exists a real number a and an open subset U of X such that

x∈ U and ϕ(U) ≥ a If ϕ is locally bounded at x from below for every x ∈ X, then

we say that it is locally bounded from below Finally, if both ϕ and −ϕ are locallybounded from below, then we simply say that ϕ is locally bounded

Evidently, any continuous real function on a metric space is locally bounded.(Why?) While the converse is of course false in general,13 it is true for concavefunctions That is, if such a function is locally bounded at a given point in theinterior of its domain, it must be continuous everywhere

Proposition 8 Let O be a nonempty open and convex subset of a metric linearspace X, and ϕ : O → R a concave function If ϕ is locally bounded at some x0 ∈ Ofrom below, then it is continuous Therefore, ϕ is continuous if, and only if, it iscontinuous at some x0 ∈ O

In the context of Euclidean spaces, we may sharpen this result significantly deed, as we have asserted a few times earlier, any concave map on an open subset of

In-a EuclideIn-an spIn-ace is continuous As we show next, this is becIn-ause In-any such mIn-ap islocally bounded

Corollary 1.Given any n ∈ N, let O be a nonempty open and convex subset of Rn

If ϕ : O → R is concave, then it is continuous

Proof It is without loss of generality to assume that 0 ∈ O.14 Then, since O

is open, we can find a small enough α > 0 such that αei

∈ O and −αei

∈ Ofor each i = 1, , n (Here {e1, , en} is the standard basis for Rn.) Define S :={αe1, , αen,−αe1, ,−αen

}, and let T := co(S) Clearly, if x ∈ T, then there exists

a λ ∈ [0, 1]S such that x =S

y∈Sλ(y)y and S

y∈Sλ(y) = 1 So, if ϕ is concave,ϕ(x) ≥ S

y∈Sλ(y)ϕ(y)≥ min ϕ(S)

13 In a big way! An everywhere (uniformly) bounded function may be continuous nowhere! (Think

of 1Q on R.)

14 Why? First follow the argument, and then go back and rethink how it would modify if we didn’t assume 0 ∈ O.

for any x ∈ T, and hence ϕ(T ) ≥ min ϕ(S) Since 0 ∈ intR n(T ) — yes? — it followsthat ϕ must be locally bounded at 0 in this case Applying Proposition 8 completesthe proof.

It remains to prove Proposition 8 The involved argument is a good illustration

of how convex and metric analyses intertwine in the case of metric linear spaces Youshould go through it carefully

Of course, the second assertion of Proposition 8 is an immediate consequence ofits first assertion, so we need to focus only on the latter We divide the proof intotwo observations (that are stated for O and ϕ of Proposition 8)

Observation 1 Suppose that ϕ is locally bounded at some x0 ∈ O from below.Then ϕ is locally bounded at any x ∈ O from below

Observation 2 Suppose that ϕ is locally bounded at some x ∈ O from below Then

ϕis continuous at x

Before we move to prove these facts, let’s note that it is enough to establishObservation 1 under the additional hypotheses 0 ∈ O, x0 = 0 and ϕ(0) = 0 Tosee this, suppose we were able to prove the assertion with these assumptions Then,given any nonempty open and convex set O ⊆ X with x0 ∈ O, and concave ϕ ∈ RO,

we would let U := O − x0 and define ψ ∈ RU by ψ(x) := ϕ(x + x0)− ϕ(x0).Clearly, ψ

is concave, 0 ∈ U, and ψ(0) = 0, while ψ is locally bounded at y − x0 from below iff

ϕ is locally bounded at y from below (for any y ∈ O) Consequently, applying what

we have established to ψ would readily yield Observation 1 for ϕ (Agreed?)

Proof of Observation 1 Let 0 ∈ O and ϕ(0) = 0, and suppose that ϕ is locallybounded at 0 from below Then there exists a real number a and an open subset U

of O such that 0 ∈ U and ϕ(U) ≥ a Fix an arbitrary x ∈ O Since O is open andscalar multiplication on X is continuous, we can find an α > 1 such that αx ∈ O.Let

V :=

1−α1



U +α1αx(Figure 1) Clearly, x ∈ V and V is an open subset of O (Proposition 2) Moreover,for any y ∈ V, we have y =