Real Analysis with Economic Applications - Chapter I ppt

Chapter I
M etric Linear Spaces
In Chapters C - G we have laid out a foundat ion for s t udying a number o f issues that
arise in metric s paces ( such as cont inuit y and completeness) and others that arise
in linear sp aces (such as linear e xtension s a n d c o nvexity). However, sa ve for a few
exceptions in the context of Euclidean spaces, we ha ve so far studied such mat ters in
isolation from each other. This should really be remedie d , for in most app lications one
works with a space structure t ha t allows for th e sim ulta neo u s con sid er ation of me tric
and linear properties. In this chapte r, therefore, we bring our earlier metric and linear
analyses together, and explore a framework that is g eneral enough to encompa ss such
situations. I n p artic ula r, this setup w ill allow us to t alk a bout thin gs like continuous
linear functions, complete linear spaces, closed convex sets, and so on.
We begin the chapter b y discussing in which sense one ma y think of a metric
structure to be imposed on a linear s p ace as “compatible” with the inheren t algebraic
structure of that space. This leads us to the no tion of m etric linear space. After
goin g through several exam ples of suc h linear spa ces, we derive som e elementar y
properties pertaining to them, and exam in e w h en a l in ear functional defined on suc h
a space w ould be con tinuous. We discuss the basic properties of contin uous linear
functionals here in some d etail, and high light the significance of them from a geometric
viewpoint. We the n consider several ch aracterizations of finite dimensional metric
linear spaces, and finally, prov ide a pr imer on convex analysis on infinite dimensional
metric linear s p ac es. The most i mportant results w ith in this frame work will clarify
the basic connection between the notions of “ openness” and “algebraic openness,”
and sharpen the separation theorems w e have obtained in Chapter G.
A c ommon theme that keeps coming up t h rou g hout the chapter pertains t o t h e
cruc ia l differences between the finite and infinite dimensional m etric linear spaces.
In particular, y o u will s e e that t h e infinite dime nsional case con ta ins a good deal of
surprises, suc h as the p resence of discon tinuous linear functionals, linear subspaces
that are not closed, and open neighborhoods the closures of which are n e cessarily
unbounded. 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 fa miliarity with general
topology here, and topological considerations arise in economic applications mo stly through metric
spaces, I chose to depart from the standa rd treatments. If you wish to see the results I present here in
their n atural (topological-algebraic) habitat, you should try outlets like Holmes (1975), Megginson
(1998) and/or Aliprantis and Border ( 1999).
413
1 Metric Linear Spaces
Take a linear space X. Giv en the objectives of this chapter, w e wish to endow X
with a metric structure that is compatible with the algebraic/geometric structure of
X. We th us n eed to choose a distance fun ction on X which goes well — w hatever this
means — with the linear structure of X.
Let’s think of t he things we can do i n X. We can shift around the vecto r s in X (b y
using the v ector addition operation) and, for any giv en vector in X, we can construct
the ray th at originates from the origin 0 and that goes through this vector (by using
the scalar m u ltiplication operation). Let’s focu s on shifting, th at is, translating ,
vectors first. What w e mean b y this precisely is t hat, for a ny giv e n x, z ∈ X, the
vector x + z is well-de fined in X, and geometrica lly speakin g , we think of this vecto r
as th e “translation of x by z” (or the “tran slation of z by x”). Now tak e another
vector y ∈ X and translate it by z as well to obtain the v ector y + z. Here is a good
question: How s ho uld the d istan ce between x and y relate to that between x + z and
y +z? If w e wish to model a space whose g eometry is homogeneo us —atermwhichwe
will formalize later on — it makes good sense to require these two distances be equ al.
After all, if posited globally, this sort of a propert y would ensure that the distance
bet ween any t wo giv en points be preserved when these vectors are translated (in the
same way) an ywhere else in the space.
2
This is , for instance, prec ise ly the ca se for
the geometry of the plane (or of any other Euclidean space).

Dhilqlwlrq. Let X be a linear space. A metric d ∈ R
X×X
+
is called translation
invariant if
d(x + z,y + z)=d(x, y) for all x, y , z ∈ X. (1)
Observe tha t (1) is a statem ent th at ties the metric d with the group st ructu re
of the lin ear space X. Translation in variance is thus a concept whic h is partly met-
ric a nd partly algebraic — it connects in a particular w ay the distance function o n a
linear space with the operation of v ector addition. In fact, this connection is tigh ter
than what ma y seem at first. If you endow a linear space with a translation invari-
an t m etric, then, per force, you make the vector addition o peration of your space
continuous!
Proposition 1. Let X be a linear space which is also a metric space. If the metric
d of X is t ran slatio n invariant, th e n th e map (x, y) → x + y is a c ontinuous fu n ction
from X × X into X.
2
I have so far tried t o be careful in referring to t he elements of a metric space as “point s,” 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 us e these two terms interchangeably.
414
Proof. Su ppose d is translation invariant, and take any (x
m
), (y
m
) ∈ X
∞
and
x, y ∈ X with x
m

→ x and y
m
→ y. Then, for any m =1, 2, ,
d(x
m
+ y
m
,x+ y)=d(x
m
+ y
m
− (x + y
m
),x+ y − (x + y
m
)) (by (1))
= d(x
m
− x, y − y
m
)
≤ d(x
m
− x, 0)+d(0,y− y
m
) (by the triangle inequality)
= d(x
m
,x)+d(y
m

,y) (b y (1) and symmetry of d),
so d(x
m
+ y
m
,x+ y) → 0,thatis,x
m
+ y
m
→ x + y. 
Exercise 1.LetX be a lin ear space which is also a metric space. Show that if the
metric
d of X is translation invariant, then, for any k ∈ Z, the self-map f on X,
defined by f(x):=kx, is continuous.
Unfortunately, t ranslation invariance alone does not yield a sufficiently rich frame-
work, we need a somewhat tighter link between th e metric and linear structures in
general. For instance, metriz ing a linear spac e by means of a translation invariant
distance function does not guaran tee the con tin uit y o f the scalar multiplication op-
eration. As an extreme examp le, consider m etrizing a given nontrivial m etric space
X by the d iscre te metric (which is, obviously, translation invariant). Then the map
λ → λx (from R into X) is not contin uous for a n y x ∈ X\{0}. (Proof. If x = 0, then
(
1
m
x) is a sequenc e which is not eventually const ant, so en d owing X with the discrete
metric d yields d

1
m
x, 0


=1for each m.)
To connect th e metric and lin ear structures in a more s a tisfactory manner, ther e-
fore, we n eed to do better than asking for the translation invariance property.
3
In
particular, we should certainly make sure that a map lik e λ → λx (in X
R
)andamap
like x → λx (in X
X
) are decla red c ontinuous by t h e metric we impose on X. Well,
then wh y d on’t w e co ncen trate on the case where the m etric at hand render the map
(λ,x) → λx (in X
R×X
) continuous? We kn ow that translation in variance gives us th e
continuit y of the vector addition operation. This wo uld, in turn, render the scalar
multiplication operation on X contin uous. And putting these together, we arriv e at
a c lass of spaces which are both metric and linear, and in which the metric and linear
structures of the space are naturally compatible.
Dhilqlwlrq. Let X be a linear space which is also a m etric space. If th e metric d
of X is translation in variant, and, for all con ver gent (λ
m
) ∈ R
∞
and (x
m
) ∈ X
∞
, we

have
lim λ
m
x
m
=(limλ
m
)(limx
m
) ,
3
To be fair, I should sa y that a reasonably complete algebraic theory can be developed using only
the translation invariance property, but this theory would lack geometric co mpetence, so I will not
pursue it here. (But see Exercises 15 and 16.)
415
then X is called a metric linear space. If, in addition, (X, d) is comp lete, the n, we
say tha t X is a Fr éc het space. A metric l inear spa ce X is said to be nontrivial
if X = {0}, finite dimensio n al if dim(X) < ∞, and infinite dimensional if
dim(X)=∞.
Put succinctly, a metric linear space is a linear space endowed with a translation
invariant distance function that renders the scalar m ultiplication operation contin u-
ous.
4
So, it follows fro m P roposition 1 that X is a metric linear spa ce iff it is both a
linear and a metric space such that
(i) the distance between any tw o points are preserv ed under the iden tical t ransla-
tions of these poin ts,
(ii) the scalar multiplicat io n map (λ,x) → λx is a continuous function from R×X
in to X, and
(iii) the vector addition map (x, y) → x + y is a continuous function from X × X

in to X.
Let’s look at some exam p le s.
E{dpsoh 1. [1] R
n
is a Fréchet space, for any n ∈ N. T his is easily proved by using
the fact that a seque nce converges in a Euclidean space i ff each of its coordinate
sequences conv erges in R. (Similarly, R
n,p
is a Fréchet space, for an y n ∈ N and
1 ≤ p ≤∞.)
[2] If we metrize R b y the discrete metric, we do not obtain a metric linear space
even tho u gh this metric is translation invariant. By cont rast, if w e m e trize R by
d ∈ R
R×R
+
with d(a, b):=|a
3
− b
3
| , then we guarantee that the scalar mu ltip lication
operation on R is continuous, but do not make R a metric linear space because the
metric d is not translation invarian t.
[3] C onsider the linear space R
∞
of all real sequences wh ic h i s m etrized by m eans
of the product metric:
ρ((x
1
,x
2

, ), (y
1
,y
2
, )) :=
∞
S
i=1
1
2
i
min{1, |x
i
− y
i
|}.
4
Some authors assume that the vector addition operation is continuous instead of positing the
translation invariance of the metric when defining a metric linear space. My definition is thus
a bit more demanding than the u sual. Yet, insofar as the topological properties are concerned,
this is only a matter of convention, since, by a well-known result of Shizuo Kakutani (and as an
immediate c orollary of the famous Birkho ff-Kakutani Metrization Theorem), every metric linear
space is homeomorphic to a translation inva riant metric linear space. (For a proof, see Rolewicz
(1985), pp. 2-4.)
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 typ e of spaces in the next chapter.)
416
(Recall Section C.8.2.) This metric is obviously translation invariant. To see that it
renders the scalar mu ltip lica tion operation on R

∞
continuous, take any (λ
m
) ∈ R
∞
and an y sequ ence (x
m
) in R
∞
such that λ
m
→ λ and x
m
→ x for some (λ,x) ∈ R×R
∞
.
(Of course, (x
m
) is a sequ e nc e of sequences.) By Proposition C.8, x
m
i
→ x
i
(as
m →∞), so we have λ
m
x
m
i
→ λx

i
(as m →∞), for each i ∈ N. App lyin g Proposition
C.8 again, we find lim λ
m
x
m
= λx. Comb ining t h is obs ervation with Th e ore m C.4,
therefore, w e may conclude: R
∞
is a Fréch et space.
[4] 
p
is a metric linear space for any 1 ≤ p<∞. Fix any such p. That d
p
is translation invariant is obvious . To see the con tinuity of scalar m u ltiplication
operation, take any (λ
m
) ∈ R
∞
and any sequence (x
m
) in 
p
suc h that λ
m
→ λ and
d
p
(x
m

,x) → 0 for some (λ,x) ∈ R × 
p
. By the triangle inequality, we ha ve
d
p
(λ
m
x
m
, λx) ≤ d
p
(λ
m
x
m
, λx
m
)+d
p
(λx
m
, λx)
=

∞
S
i=1
|λ
m
− λ|

p
|x
m
i
|
p

1
p
+

∞
S
i=1
|λ|
p
|x
m
i
− x
i
|
p

1
p
= |λ
m
− λ| d
p

(x
m
, 0)+|λ| d
p
(x
m
,x).
But since (x
m
) is con ve r ge nt, (d
p
(x
m
, 0)) is a bounded real sequence (f o r, d
p
(x
m
, 0) ≤
d
p
(x
m
,x)+d
p
(x, 0) → d
p
(x, 0)). Therefore, the inequalit y above ensures that d
p
(λ
m

x
m
,
λx) → 0, as wa s to be pro ve d.
Com bining this observation with Example C.11.[4], we ma y conclude that 
2
is a
Fréchet space. Moreover, t he argumen t given in that example can be g eneralized to
an y 
p
space — but w e om it 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 Pr oposition D.7, both of th ese s pa ces a re Fréc het spaces. 
Exercise 2. Supply the missing arguments in Examples 1.[5]-[6].
Exercise 3. Show that the metric space in troduced in Exer cise C.42 i s a metric linear
space (under the point wise definedoperations)whichisnotFréchet.
Exercise 4. Show that
C
1
[0, 1] is a Fréchet space.
Exercise 5. Let
X := {(x
m
) ∈ R

∞
:sup{|x
m
|
1
m
: m ∈ N} < ∞}, and define
d ∈ R
X×X
+
by d((x
m
), (y
m
)) := sup{|x
m
− y
m
|
1
m
: m ∈ N}. Is X a metric linear
space relative to
d and the usual addition and scalar multiplication operations?
Exercise 6. (Product Spaces )Let(X
1
,X
2
, ) be a sequence of metric linear spaces,
and

X := X
∞
X
i
. We endow X with the product metric (S ection C.8.2), and mak e it
a linear s pace by defining the operations o f scalar multiplication and vector addition
point wise. Show that
X is a metric linear space, and it is a Fréc h et space whenever
each
X
i
is complete.
417
Exercise 7. Let X be an y linear space, and ϕ aseminormonX (Section G.2.1).
Define
d
ϕ
∈ R
X×X
+
by d
ϕ
(x, y):=ϕ(x − y). Is d
ϕ
necessarily a distance function?
Show that
X would become a metric linear space when endo wed with d
ϕ
, provided
that

ϕ
−1
(0) = {0}.
Th e following proposition collects some basic facts about met ric linear spaces that
we will need later o n. It also provides a good illustration of the interplay bet ween
algebraic and me tric consid erations, which is ch arac te ristic of metric linear spaces.
Proposition 2 . For any subsets A and B of a metric l inear space X, the follo wing
are true:
(a) cl
X
(A + x)=cl
X
(A)+x for all x ∈ X;
(b) If A is open, then so is A + B;
(c) If A is comp act 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
(x
m
) be a sequence in A + B suc h that x
m
→ x for so me x ∈ X. By definition, there
exist a sequ ence (y
m
) in A and a sequence (z
m
) in B such that x
m
= y
m

+z
m
for each
m. Since A is co m p act, Theorem C .2 im p lies that there exists a strictly increas ing
sequence (m
k
) in N such that (y
m
k
) converges to a vector, say y, in A. But t hen (z
m
k
)
converges to x −y by the continuity of vecto r addition, so, s in ce B is closed, it follows
that x − y ∈ B. Thus, x = y +(x − y) ∈ A + B, which proves t hat A + B is clo se d. 
Exercise 8.
H
Complete the proof of Proposition 2.
Parts (b) a nd ( c ) of Proposition 2 point to a s ign ificant di fference in the behavior
of sums of open sets and closed sets: While the sum of a ny two open sets is open,
thesumoftwoclosedsetsneednotbeclosed. Thisdifference is wo rth k eeping in
mind . It exists even in the case o f our beloved real line. Fo r instan c e, let A := N an d
B := {−2+
1
2
, −3+
1
3
, }. Then, both A and B are closed subsets of R, but A + B
is not closed in R, for

1
m
∈ A + B for each m =1, 2, , but lim
1
m
=0/∈ A + B. As
another (more geometric example), let A := {(x, y) ∈ R
2
: x =0and y ≥
1
x
} and
B := R×{0}. (Dra w a picture.) Then, while both A and B are closed in R
2
, A + B
is not a closed subset of R
2
. (Pr oof. A + B = R × R
++
.)
Themostcommonmethodofcreatinganewmetriclinearspacefromagiven
metric linear sp ac e is to loo k for a su b set of this space w h ich inhe rits th e me tric
linear structure of the or igina l space. T his leads us to t h e n otion of a metri c linear
subspace of a metric linear s pace X, which is defined as a subset of X which is
both a l inear and a metric subspace of X. (For instance, R
2
×{0} is a metric linear
subspace o f R
3
, and C[0, 1] is a metric linea r subs pace of B[0, 1].) Throughout this

418
chapter, by a su bsp ace of a metric l inear space X, we mean a m etric linear s ubspace
of X.
Exercise 9.
H
Find a subspace of R
∞
which is not closed. Is t his subspace dense in
R
∞
?
Exercise 10. Show that the c losure of any subspace (affine manifold) of a metric
linear space
X is a subspace (affine m anifold) of X.
Exercise 11. For any subsets A and B of a metric linear space X,prove:
(a)
cl
X
(A)+cl
X
(B) ⊆ cl
X
(A + B);
(b) cl
X
(cl
X
(A)+cl
X
(B)) = cl

X
(A + B);
(c) int
X
(A)+int
X
(B) ⊆ A +int
X
(B) ⊆ int
X
(A+B), pro vided th at int
X
(B) = ∅.
Exercise 12.
H
Show that every metric linear space is connected.
Exercise 13.
H
(Nikodem) L et X and Y be two metric linear spaces, S anonempty
con vex subset of
X, and Γ : S ⇒ Y a correspondence which has a closed graph.
Show that if
1
2
Γ(x)+
1
2
Γ(y) ⊆ Γ(
1
2

x +
1
2
y) for any x, y ∈ S,
then Γ is a convex correspondence (Exercise G.8).
2 Continuou s Linea r Operato r s a nd Func t io n a ls
The primary objects of analysis w ithin the context of metric linear spaces are those
linear operators L that map a metric linear space X (with metric d) to ano ther metric
linear space Y (with metric d
Y
) such that, fo r all x ∈ X and ε > 0, there exists a
δ > 0 with
d(x, y) < δ implie s d
Y
(L(x),L(y)) < ε.
For obvious re a sons, we call any such map a continuous lin e ar operato r,except
when Y = R, in which case w e refer to it as a con tin uous linear functional.
2.1 Examples of (Dis-)Contin u ous Linear Operators
E{dpsoh 2. [1] Any linear operator that ma ps R
n
to R
m
is a continuous linear
operator. This follow s from Examples D.2.[4] and F.6.[1].(Verify!)
[2] Let L ∈ R
B[0,1]
be defined b y L(f):=f(0).Lis obviously a linear function al
on C[0, 1]. It is also continuou s — 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].
419
[3] Let L ∈ R
C[0,1]
be defined b y
L(f):=
]
1
0
f(t)dt.
It follo ws from R ie mann integration t h eo ry t ha t L is a l inear fu nct ion al (Ex ercise A.
57). Moreover, L is a nonexpansive map : For any f,g ∈ C[0, 1], we have
|L(f) − L(g)| =




]
1
0
(f(t) − g(t))dt




≤ d
∞
(f,g)
b y Propositions A.12 and A .1 3. Conclu sion: L is a continuou s linear fun c tion al on
C[0, 1]. 

Most studen ts are startled when they hear the term “con tinuous linear functional.”
Aren’t linear f u n ction s always con tinuous? The answer is no, not necessarily. True,
any linear real function o n a Euclidean space is co ntinuous, but this is no t t he case
when th e linear function under consideration is definedonametriclinearspacewhich
is — guess what? — infin ite dim en siona l. H e re are a few examples that illustrate this.
E{dpsoh 3. [1] Let X be the linear space of all real sequences that are abs o l utely
summable , that is,
X :=

(x
1
,x
2
, ) ∈ R
∞
:
∞
S
i=1
|x
i
| < ∞

.
Let u s make this s pace a metric l inear s pace by usi ng the sup-metric d
∞
. (N otice t hat
X is a metric linear sp a ce that differs from 
1
only in its metric stru c tu re.) Define

L ∈ R
X
by
L(x
1
,x
2
, ):=
∞
S
i=1
x
i
,
which is ob v iou sly a linear functional on X. Is L continuous? No! Consider the
following sequence (x
m
) ∈ X
∞
:
x
m
:=

1
m
, ,
1
m
, 0, 0,


,m=1, 2, ,
where exactly m en tries of x
m
are nonzero. Clearly, d
∞
(x
m
, 0)=
1
m
→ 0 . Yet L(x
m
)=
1 for each m, so L(lim x
m
)=0=1=limL(x
m
). Conclusion: L is linear but it is not
continuous at 0.
[2] We play on this theme a bit more. Let
X := {(x
1
,x
2
, ) ∈ R
∞
:sup{|x
i
| : i ∈ N} < ∞},

420
and define L ∈ L(X, R) by
L(x
1
,x
2
, ):=
∞
S
i=1
δ
i
x
i
,
where 0 < δ < 1. It is easy to verify that if X wa s endowed with the sup-metric, t hen L
w ould be a continuous f unction. (You now know that this w ould not be true if δ =1.
Yes?) But what if w e endo wed X with the product metric? Then, interestingly,
L w ould not be continuous. Indeed, if x
1
:= (
1
δ
, 0, 0, ),x
2
:= (0,
1
δ
2
, 0, ), etc.,

then x
m
→ (0, 0, ) with respect to t he product metric (P r oposition C.8 ), and yet
L(x
m
)=1for eac h 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 C
1
[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 neither D nor L is continuous.
Wh ile it is true th at line arity of a map does n ot nec essa rily guara ntee its continu-
ity, it still brings quite a bit of d isc ip line into the picture. Ind e e d, linearit y spreads
even the t inniest b it o f con tin uit y a function may ha ve o nto t he entire dom ain o f that
function . Put differently, there is no reason to distinguish between l ocal and g lobal
continuity conce p ts in the p resence of lin earit y. ( Th is is quite reminisc e nt of a d d itive
real functions on R; recall Exercise D.40.) We for malize this point next.

Proposition 3 . Let X and Y be t wo metric linear spaces. A linear o perator L from
X into Y is uniformly continu ou s if, and only if, it is continuous at 0.
Proof. Let L ∈ L(X, Y ) be con tin uous at 0,andtakeanyε > 0 and x ∈ X.
By con tinuity at the origin, there exist s a δ > 0 such that d
Y
(L(z), 0) < ε whenever
d(z,0) < δ. No w take an y y ∈ X with d(x, y) < δ. By translation invariance of d, we
have d(x − y,0) < δ, and hence, by translation invariance of d
Y
and linearity,
d
Y
(L(x),L(y)) = d
Y
(L(x) − L(y), 0)=d
Y
(L(x − y), 0) < ε.
Since y is arbitrary in X here, and δ does not depend on x, we ma y conclude that L
is uniformly con tinuous.
5

5
Another Proof. If L is continuous at 0, the n, 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 w ait, where did I use the translation
invariance of d and d
Y
?)
421
Even thou gh its p roof is easy, a first comer to the topic may be a bit su rp rised
by Proposition 3. Let us try to think through things more clearly here. The key
observation is that a ny metric linear spa ce Z is homogeneous in the sense that, fo r
an y giv en z
0
,z
1
∈ Z, we can map Z onto Z in such a way that (i) z
0
is mapped to z
1
,
and (ii) all topological properties of Z are left inta ct. M ore precisely, the translation
map τ : Z → Z defined b y τ(z):=z +(z

1
− z
0
) is an hom eomorphism, thanks to
the continuity o f vector ad dition o n Z. But then if we kno w that an L ∈ L(X, Y )
is continu ou s at a giv e n point, sa y 0, to sho w that L must then be contin uous at an
arbitrary point x
0
∈ X, all we need is to translate X so that x
0
“becomes” 0 (of X),
and trans late Y so that 0 (of Y )“becomes”y
0
:= L(x
0
). So define τ : X → X by
τ (x):=x − x
0
, and ρ : Y → Y by ρ(y):=y + y
0
. Since τ and ρ are c ontinuou s
everywhere, and L is continuous at 0, an d s in ce L = ρ ◦ L ◦ τ, it follows that L is
continuous at x
0
. Th u s linearity (in fac t, ad d itivity ) spreads contin u ity at a sin g le
point to t he entire doma in pr e cisely via translation maps that are always continuou s
in a metric linear sp ace . (The situation could be drastically different if the metric
and linear struc tu re s of the spa ce were no t compatible enough to yield t h e con tinuity
of vector addition.)
Often in real analys is, one gets a “clea rer” view of things upon suitably general-

izing the mathe matical structure at hand . The follo win g exercis e s aim to clarify the
origin of Proposition 3 further by means of such a generalizatio n.
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 contin uous map from
X × X into X. For any x
∗
∈ X, we define the self-map τ on X by τ(x):=x − x
∗
,
whic h is called a left translation.(Right translations are definedasmapsofthe
form
x →−x
∗
+ x, and coincide with left translations when (X, +) is Abelian.)
(a) Show that any left or right translation on
X is a homeomorphism.
(b)Showthatif
O is an open set that contains the identity element 0 of X, the n so
is
−O.
(c)ShowthatifO is an open set that contains the identity element 0 of X, then
thereexistsanopensubset
U of X with 0 ∈ U and U + U ⊆ O.
Exercise 16. Let (X, +
X
,d
X
) and (Y,+
Y

,d
Y
) be two metric groups, and c onsider a
map
h : X → Y such that h(x +
X
y)=h(x)+
Y
h(y) for all x,y ∈ X. (Recall that
such a map is called a homomorphism from
X into Y.)Provethath is continuous
iff it is continuous at
0. (How does this fact relate to Proposition 3?)
It is now time t o consider a non trivial example of a con tinuous linear fu nc tion al
definedonaninfinite dimensional metric linear space.
E{dpsoh 4. For any n ∈ N, Exam ples 2.[1] an d F.6.[1] show that an y con tinu ou s lin-
ear functional L on R
n
is of the f o r m x →
S
n
α
i
x
i
(for some real nu mbers α
1
, , α
n
).

422
We n ow wish t o determine the general structure of continuous linear functionals o n
R
∞
.
Tak e an y continuous L ∈ L(R
∞
, R). Define α
i
:= L(e
i
) for each i ∈ N, where
e
1
:= (1, 0, 0, ), e
2
:= (0, 1, 0, ), etc We claim that α
i
=0for all but finitely
man y i. To see this, note th a t
(x
m
) = lim
M→∞
M
S
i=1
x
i
e

i
for any (x
m
) ∈ R
∞
.
(Proof. Observ e that the mth term of k th term of the sequence of s equences (x
1
e
1
,
x
1
e
1
+ x
2
e
2
, ) is x
m
for any k ≥ m an d m ∈ N, and apply Proposition C.8.) So, by
continuity and linearity of L, for an y (x
m
) ∈ R
∞
,
L((x
m
)) = L


lim
M→∞
M
S
i=1
x
i
e
i

=lim
M→∞
L

M
S
i=1
x
i
e
i

=lim
M→∞
M
S
i=1
α
i

x
i
. (2)
Since L is real-valued, this can’t hold true for all (x
m
) ∈ R
∞
, unless α
i
=0for all
but finite ly m any i.
6
Thus S := {i ∈ N : α
i
=0} is finite, and (2) yields
L((x
m
)) =
S
i∈S
α
i
x
i
for any (x
m
) ∈ R
∞
. (3)
Since it is easy to ch e ck that this in de ed defines a continuous linear functional on

R
∞
, w e conclude: L is a co ntinu ous linear functional on R
∞
if, and only if, there
exists a finite subset S of N and real numbers α
i
,i∈ S, such tha t (3) holds.
7

Remark 1. The argument given in Ex ample 4 relies on the fact that (x
m
)=
lim
M→∞
S
M
x
i
e
i
for any real sequen ce (x
m
). This ma y perhaps tempt y ou to view
the set {e
i
: i ∈ N} as a basis for R
∞
. H oweve r, this is n ot the c ase , for w e may
not be able to express a real sequence as a linear combination of finitely man y e

i
s.
(Consider, for instance, the sequence (1, 1, ).) Instead, one sa ys that {e
i
: i ∈ N}
is a Schauder basis for R
∞
.Thedefining feature of this concept is expressing the
v ectors i n a metric lin ear s pace as a linear i nfinite ser ies. As opposed to the s tan dard
one, this basis concept depends on the metric in question since it in v olv es the notion
of “conv ergence” in its definition. (Mo re on this in Section J.2.2.) 
Exercise 17. Let S be a finite subset of [0, 1]. Show that, for any λ ∈ R
S
, the map
L ∈ R
C[0,1]
defined by L(f):=
S
x∈S
λ(x)f(x) is a continuous linear functional on
C[0, 1].
6
If there w as a subsequence (α
m
k
) in R\{0}, by defining (x
m
) as x
m
:=

1
α
m
if m = m
k
and
x
m
:= 0 otherwise, we would get
S
∞
α
i
x
i
=1+1+···= ∞, contradicting (2).
7
Corollary. There is no strictly positiv e continuous linear functional on R
∞
. (In fact, there is no
strictly positive linear functional on R
∞
, but showing this requires more work.)
423
Exercise 18. G ive an example of a discon tinuous linear functional on 
∞
.
Exercise 19. Prove that an upper semicontinuous linear functional on a metric lin ear
space is continuous.
Exercise 20.

H
Show that a linear functional L on a metric linear space X is continuou s
iff there exists a continuous seminorm
ϕ on X such that |L(x)| ≤ ϕ(x) for all x ∈ X.
Exercise 21.
H
Let L be a linear functional on a metric linear space X. Pro ve:
(a)
L is continuous iff it is bounded on some open neighborhood O of 0;
(b)IfL is continuous, then L(S) is a bounded set for an y bounded subset S of X.
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)LetX and Y be two metric
linear spaces
, and Γ : X ⇒ Y a linear correspondence.
(a) Show that
Γ is upper hemicontinuous iff it is upper hemicon tinuous at 0.
(b)ShowthatΓ is lower h emicon tinuous iff it is lower hemicontinuous at 0.
Exercise 23.
H
(Continuous Linear Selections)LetX and Y be two m etric linear
spaces
, and Γ : X ⇒ Y a linear correspondence.
(a) Show that if
Γ admits a con tin u ous linear selection, then it is contin uous.
(b)Prove:IfP ∈ L(Γ(X), Γ(X)) is continuous, idempotent (i.e. P ◦ P = P)and
null(P )=Γ(0), then P ◦ Γ is a continuous linear selection from Γ.

(c) In the special case where X and Y are Euclidean spaces, p ro ve that if Γ is upper
hemicontinuous, then it admits a con tin uous linear selection.
2.2 C o ntinuit y of Positive Linear Functionals
Is a positive linea r fun ct ion al de fined on a preordered metric linear space necessarily
contin uous? A very good question, to be sure. Monotonic real functions possess,
in general, r easonably strong con tin uity properties. (Any such function on R is,
for instance, continuous everywhere but coun tably man y points.) So, while a linear
functional need not be continuous in general, perhaps monotonic linear functionals
are.
The bad news is that the answ er is no! Indeed, all of the linear functionals
considere d in Exam p le 3 are positiv e (with R
∞
being partially o r dered by means of
the c oordinatew ise order ≥), but as we have seen th ere, these functionals may well
turn up discon tinuous, depending on ho w we c hoose to metrize their domain. The
good news is that this is not th e end of the story. It is in fac t possible to pinpoint the
source of the problem, and thus find out when it would not a r is e. Our next r es u lt,
which pro vides us w ith a rich class of continu ous linear functiona l s, d oes precisely
this.
424
Proposition 4. (Shaefer) Le t X be a p reorder ed metric linear space. If int
X
(X
+
) =
∅, then any positiv e linear functional on X is continuous.
8
Proof. Let L be a positive linear functional on X, and take an y sequence (y
m
) ∈

X
∞
such that y
m
→ 0. By Proposition 3, all we need to show is that L(y
m
) → 0.
No w assume int
X
(X
+
) = ∅, an d pick any x ∈ int
X
(X
+
). The crux of the argument is
to establish the following: For every α > 0, we can find an M
α
∈ N such that
αx 
X
y
m

X
−αx for all m ≥ M
α
. (4)
Why?Becauseifwecanprovethis,thenwemayusethepositivityofL to find, for
ever y k ∈ N, a positive integer M

k
with
1
k
L(x) ≥ L(y
m
) ≥−
1
k
L(x) for all m ≥ M
k
.
Letting k →∞here yields L(y
m
) → 0, as we seek.
9
Now fixanyα > 0. Clearly, αx ∈ int
X
(X
+
) — why? — so t here exist s a δ > 0 such
that N
δ,X
(αx) ⊆ X
+
. Notice that, for any y ∈ N
δ,X
(0), we have
y + αx ∈ N
δ,X

(0)+αx = N
δ,X
(αx) ⊆ X
+
,
so y 
X
−αx. But if y ∈ N
δ,X
(0), then −y ∈ N
δ,X
(0) by translation invariance, so the
same argument yields αx 
X
y. Thus, if we choose M
α
∈ N suc h that y
m
∈ N
δ,X
(0)
for all m ≥ M
α
, w e obtain (4). 
2.3 C losed vs. Dense Hyperplanes
Recall that we can identify a h y perplane in a linear space with a nonz ero linear
functional up to an additiv e constant (Corollary F.4). In this subsection we show
that the closed h yperplanes in a metric line ar space have the same relatio nsh ip with
the c ontinu ous nonzero linear functionals on that s pace. The crux of the a rgument
iscontainedinthefollowingfact.

Proposition 5. Let X be a metr i c linear space an d L ∈ L(X, R). Then L is cont in -
uous if, and only if, null(L) is a closed subspace of X.
10
Proof. Th e “only if” part is easy. To establish the “if” part, tak e 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 m etric lin ear space endowed w ith a
vector preorder. (In particular, no (direct) connection betw een 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, M
k
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).
425
to check that L is continuou s at 0. So, take any (x
m
) ∈ X
∞
with x
m
→ 0. We wish
to show that L(x
m
) → 0.
If L is the zero functional, then the claim is o b v iously true, so assume that it is
nonzero. Then, by P roposition 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 x
m
= y
m
+ λ
m
w for some (λ
m
,y
m
) ∈ R × Y, m =1, 2,
11

Clearly, for
each m, we have L(x
m
)=L(y
m
)+λ
m
L(w)=λ
m
L(w) (since y
m
∈ null(L)), while
L(w) =0(since w/∈ Y ). Thus, by the continu ity of scalar m u ltip licatio n, all we n eed
to do is to show that λ
m
→ 0.
Let’s first verify that (λ
m
) is a bounded real sequence. If this was not the case,
we could find a subsequence (λ
m
k
) of this sequence s uch that λ
m
k
=0for e ach k, and
1
λ
m
k

→ 0 (as k →∞). But i f w e let θ
k
:=
1
λ
m
k
, we may write w = θ
m
k
x
m
k
− θ
m
k
y
m
k
for e a ch k. Since t he m etric d of X is tra nslation invariant an d Y is a linear subspac e
of X, therefore,
d(θ
m
k
x
m
k
− θ
m
k

y
m
k
,Y)=d(θ
m
k
x
m
k
,Y + θ
m
k
y
m
k
)=d(θ
m
k
x
m
k
,Y) ≤ d(θ
m
k
x
m
k
, 0),
k =1, 2,
12

Then, letting k →∞and using the continuity of scalar multiplic ation ,
we get d(w, Y )=0, which is impossible, given that w/∈ Y and Y is closed (Exercise
D.2). We conclude that (λ
m
) is bounded.
Now let λ := lim sup λ
m
. Then λ ∈ R (because (λ
m
) is bounded) and there is
a subsequence (λ
m
k
) with λ
m
k
→ λ (as k →∞). Since y
m
= x
m
− λ
m
w for ea ch
m, and lim x
m
= 0, we have lim y
m
k
= λw by the continuity of scalar multip lica t ion
and v ector addition. But since y

m
k
∈ Y = L
−1
(0) for each k, and Y is closed,
lim y
m
k
∈ L
−1
(0), so, λL(w)=0. Sinc e w/∈ Y, we thus find λ =0. But notice that
thesameargumentwouldgothroughverbatim,ifweinsteadhadλ := lim inf λ
m
. We
ma y thus conclude that lim λ
m
=0. 
Hereistheresultwepromisedabove.
Proposition 6. A subset H of a met ric linear space X is a closed h y perplane in X
if, and only if,
H = {x ∈ X : L(x)=α}
for some α ∈ R and a con tin uous nonzero linear functional L on X.
11
False Proof.“Giventhatx
m
= y
m
+ λ
m
w for each m, and x

m
→ 0,wehave0 = lim y
m
+
(lim λ
m
)w. But since Y is closed, lim y
m
∈ Y, that is, L(lim y
m
)=0, and h ence applying L to
both sides of this equation, we obtain 0=(limλ
m
)L(w). Since w/∈ Y ,wehaveL(w) =0, so it
follows that lim λ
m
=0. But then, since L is linear and y
m
∈ Y for each m, we have L(x
m
)=
L(y
m
)+λ
m
L(w)=λ
m
L(w) for each m, so, l etting m →∞, we get lim L(x
m
)=0as we s o ught.”

Unfortunately, things are a bit more complicated than this. Please find what’s wrong with this
argument before proceeding further.
12
Reminder. For an y nonempty subset S of a metric space X, and x ∈ X, d(x, S):=inf{d(x, z):
z ∈ S}.
426
Proof. The “if” part o f t h e claim follows readily from Proposition D.1 and C orol-
lary F.4. Conversely, if H is a close d h yper plan e in X, then by Cor ollary F.4 t here
exist an α ∈ R and a nonzero L ∈ L(X, R) such that H = L
−1
(α). Take any x
∗
∈ H
and d efine Y := H − x
∗
. B y Proposition 2 .(c), Y is cl osed subspace of X. But it is
evide nt that Y = null(L) — yes? — so by Proposition 5, L is continuou s . 
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 con tinuous iff null(L) is a closed
subspace of
Y.
Proposition 6 entails that an y hyperplane that is induced b y a discontinu ou s linea r
functional canno t be closed. Let us inquire into t he natu re of such hy perplanes a bit
further. In Exercise 10, we have noted that t he closure o f a subspace Z of a metric
linear space X is itse lf a subsp ac e. Th u s the closu re of a ⊇-maximal proper subspace
Z of X is either Z or X. Put d ifferently, any ⊇-maximal proper linear subspa ce of

X is either closed or (exclusiv e) dense. S ince any hyperplane H canbewrittenas
H = Z +x
∗
for so m e ⊇-maximal proper linea r subspac e Z of X and x
∗
∈ X, an d since
cl
X
(H)=cl
X
(Z + x
∗
)=cl
X
(Z)+x
∗
(why ?), this observa tion gives us the following
insight.
Proposition 7. A h y perplane i n a metric linear s pace is either clos ed o r (exclusiv e)
dense.
How can a h yperplane be d en se in the gran d space that i t lies in? This seem in gly
paradoxic al situation is just anoth e r illustration of h ow 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 in
infinite dimensional spaces. Since a linear functional on such a space n eed n ot be
continuo us , a h yperplane in it is not necessarily closed.
Actually, this is not as crazy as it m ight first seem. After all, thanks to the
Weierstrass Appro ximation Theorem, we know that every continuous real function
on [0, 1] is a uniform limit of a sequence of polyn om ials defined o n [0, 1].Butthis
means that P[0, 1] is not a closed subset of C[0, 1], wh ile, of c ou rse , it is a linear

subspace of C[0, 1]. Thus,inaninfinite dimensional metric linear space, a proper
subspace which is not even ⊇-maximal can be dense!
To su m up, one i mportant an alyt ic les son we learn here is that a linear fu nc tional
is in general not cont inuous, whereas the geom etric lesson is that a hyperplane is
in general not closed (i n w hich case it is dense). While the latter findin g may defy
our geomet ric intuition (which is, unfortunately, finite dimensional), this is just how
things are in infinite dimensional metric linear spaces.
427
E{dpsoh 5. Consider the metric linear space X and the discontinuous L ∈ L(X, R)
definedinExample3.[1]. Propositions 5 an d 7 together say t h at null(L) must be
dense in X. Let us verify this fact directly. Take any y ∈ X, and define t he sequence
(x
m
) ∈ X
∞
by
x
m
:=

y
1
−
1
m
∞
S
i=1
y
i

, , y
m
−
1
m
∞
S
i=1
y
i
,y
m+1
,y
m+2
,

,m=1, 2,
(Each x
m
is well-defined, because, by definition of X, we have
S
∞
y
i
∈ R.) Clearly,
for eac h m ∈ N,
L(x
m
)=
m

S
i=1
y
i
−
∞
S
i=1
y
i
+
∞
S
i=m+1
y
i
=0
so that x
m
∈ null(L). But we hav e
d
∞
(x
m
,y)=sup{|x
m
i
− y
i
| : i ∈ N} =

1
m
∞
S
i=1
y
i
→ 0
since
S
∞
y
i
is finite. This sho ws that, for an y vector in X, we can find a sequence in
null(L) that con verges to that vector, that is, null(L) is dense in X. 
E{dpsoh 6. In the p revious ex ample we i nferred t he dens eness o f a hyperplane from
the d isc ont inu ity of its defining linear functional. In this example, we shall construct
a discontinuous linear functional on the good old C[0, 1] b y using the denseness of a
hy perplane that this functional induces. Define f
i
∈ P[0, 1] by f
i
(t):=t
i
for eac h
i ∈ N, and recall that B := {f
i
: i ∈ N} is a basis for P[0, 1]. Now extend B to a
basis A for C[0, 1] (whichcanbedoneasintheproofofTheoremF.1). Evidently,
A\B = ∅. Pick an y g ∈ A\B, let 1

{g}
stand for the indicator function of {g} in A,
and define L : C[0, 1] → R by
L(f):=λ
1
(f)1
{g}
(h
1,f
)+···+ λ
m
f
(f)1
{g}
(h
m
f
,f
),
where the n umbers m
f
∈ N and λ
i
(f) ∈ R\{0}, and the basis vectors h
i,f
∈ A,
i =1, ,m
f
, are uniquely d e fined throu gh the e q uation f =
S

m
f
λ
i
(f)h
i,f
. (R ecall
Corollary F.2.) It is readily checked that L is linear. Yet, L is not cont inu ou s. Why?
BecausewehaveP[0, 1] ⊆ null(L) (y es?), and hence by the Weierstrass Approxima-
tion Theorem, null(L) must be dense in C[0, 1]. By Propositions 5 and 7, therefore,
L cannot be contin uous. 
2.4 Digression: On the Contin u it y of Conca ve Functions
We have noted earlier t hat a concav e real function de finedonanopensubsetofa
Euclidean space must be continuous. This need not be true for a concave function
defined on an open s ub s et of a metric line ar s p ac e. Afte r all, we now know t h at a
428
linear fu nctional (which is obviously concave) may well be di scontin uous i f its do main
is an infinite dimensional metric linear space. This said, w e also know that linearity
spreads the minimal amount of continu it y a function m a y have o nto i ts en tire do -
main. Put differently, a linear function al is either co ntinuous or it is discontinuou s
everywhere (Proposition 3). Rem arkably, concav ity m atches the s trength o f linearit y
on this score. That is to say, any concave (or convex) function wh o se domain i s an
open and c onvex s ubset of a m etric l inear space is either contin uous or discontin uous
ever ywh ere.
In fact, w e can say somethin g a bit stronger than this. Giv en a met ric s pace X
and a poin t x ∈ X, let us agree to call a function ϕ ∈ R
X
locally bounded at x
from below if there ex ists a real number a andanopensubsetU of X such that
x ∈ U an d ϕ(U) ≥ a. If ϕ is locally bou nd e d at x f rom below for every x ∈ X, th en

we say that it is locally bounded from below. Finally, if both ϕ and −ϕ are locally
bounded from below, then w e simp ly say that ϕ is locally bounded.
Evide ntly, an y con tinuous real func tion on a met ric space is locally bounde d .
(Why?) While the conv erse is of course false in general,
13
it is true for concave
function s . That is, if suc h a function is locally bound e d at a given po int in the
in terior of its domain, it must be continuous ev ery wh ere.
Proposition 8. Let O be a n onem pt y open and convex subset of a metric linear
space X,andϕ : O → R a c on cave function. If ϕ is locally bounded at some x
0
∈ O
from below, then it is co ntinuous. There fore, ϕ is continuous if, and only if, i t is
continuous at some x
0
∈ O.
In the context of Euclidean s paces, we ma y sharpen this result significantly. In-
deed,aswehaveassertedafewtimesearlier,anyconcavemaponanopensubsetof
a E uc lidean space i s con tinuous. As we sho w next, this is because any suc h map is
locally bounded.
Corollary 1. Given any n ∈ N, let O be a nonempty open a nd co nv ex su bset of R
n
.
If ϕ : O → R is concave, then it is con tin uous.
Proof. It is without loss of generality to assume that 0 ∈ O.
14
Then, since O
is open, w e can find a small enough α > 0 su ch that αe
i
∈ O and −αe

i
∈ O
for each i =1, , n. (H ere {e
1
, , e
n
} is the standard basis for R
n
.)Define S :=
{αe
1
, , αe
n
, −αe
1
, , −αe
n
}, and let T := co(S). Clearly, if x ∈ T, then there exists
a λ ∈ [0, 1]
S
such that x =
S
y∈S
λ(y)y and
S
y∈S
λ(y)=1. So, if ϕ is con cave,
ϕ(x) ≥
S
y∈S

λ(y)ϕ(y) ≥ min ϕ(S)
13
In a big way! An everywhere (uniformly) bounded function may be contin uous nowhere! (Think
of 1
Q
on R.)
14
Why? First follow the argumen t , and then go back and rethink how it would modify if we didn’t
assume 0 ∈ O.
429
for any x ∈ T, and h ence ϕ(T) ≥ min ϕ(S). Since 0 ∈ int
R
n
(T ) — yes? — it follo ws
that ϕ mu st be locally bou n ded at 0 in this case. App lying Proposition 8 completes
the proof. 
It re m a ins to p rove Proposition 8 . Th e involve d argument is a good illustration
of how convex and m e tric analyses intertwine in the c ase of metric linea r spaces. You
should go through it carefully.
Of course, the second assertion of Proposition 8 is a n immediate consequence of
its first assertion, so we need to focus only on the l atter. We d iv ide th e p roof in to
two observations (that are stated for O and ϕ of Proposition 8).
Observation 1. Suppose that ϕ is locally bound e d at some x
0
∈ O fr om below.
Then ϕ is locally bounded at any x ∈ O from below.
Observation 2. Suppose that ϕ is locally bounde d at s ome x ∈ O from below . Then
ϕ is contin uous at x.
Befor e w e move to prov e thes e facts, let’s note that it is enough to establish
Observation 1 under the additional hypotheses 0 ∈ O, x

0
= 0 and ϕ(0)=0.To
see this, suppose we were able to prove the assertion with these assumptions. Then,
giv en any nonempty open and convex set O ⊆ X with x
0
∈ O, and conca ve ϕ ∈ R
O
,
we would let U := O −x
0
and define ψ ∈ R
U
by ψ(x):=ϕ(x+x
0
) − ϕ(x
0
). Clearly, ψ
is concave, 0 ∈ U, and ψ(0)=0, while ψ is locally bounded at y − x
0
from belo w iff
ϕ is locally bounded at y from below (for an y y ∈ O). C o n sequently, applying what
w e have established to ψ would readily yield Observation 1 for ϕ. (Agreed?)
ProofofObservation1. Let 0 ∈ O and ϕ(0)=0, and suppose that ϕ is locally
bounded at 0 from below. Then there exists a real num ber a andanopensubsetU
of O such that 0 ∈ U and ϕ(U) ≥ a. Fix an arbitrary x ∈ O. Since O is open and
scalar mu ltip lication on X is continuous, we can find an α > 1 such that αx ∈ O.
Let
V :=

1 −

1
α

U +
1
α
αx
(Figure 1). Clearly, x ∈ V and V is an open subset of O (Pr oposition 2). Moreov e r,
for an y y ∈ V, we have y =

1 −
1
α

z +
1
α
αx for some z ∈ U, so, by concavity,
ϕ(y) ≥

1 −
1
α

ϕ(z)+
1
α
ϕ(αx) ≥

1 −

1
α

a +
1
α
ϕ(αx).
Thus, wher e b :=

1 −
1
α

a +
1
α
ϕ(αx), we have ϕ(V ) ≥ b, which p roves that ϕ is
locally bounded at x from belo w.
15

15
Quiz. Show that, under the givens of Proposition 8, if ϕ is locally bounded a t x
0
from below,
then it must be locally bounded. (Prov ing this is easier than proving Observation 2, so give it a try
before studying t he proof o f that claim.)
430
∗∗∗∗FIGU R E I.1 ABOUT HERE ∗∗∗∗
It remains t o p rove O b servation 2. Just as we could assume x
0

= 0 while prov ing
Observation 1, we c an take x in O bse rvation 2 to be 0 without loss of generalit y.
(Why?)
ProofofObservation2. Let 0 ∈ O and ϕ(0)=0, and suppose that ϕ is locally
bounded at 0 from belo w . Fix a n arbitrarily small ε > 0. We wish to find an open
subset U of O such that 0 ∈ U and ϕ(U) ⊆ (−ε, ε). (Right?) We know th at t here
exists a δ > 0 and an a<0 suc h that N
δ,X
(0) ⊆ O and ϕ(N
δ,X
(0)) ≥ a. (Wh y ?)
Of course, λN
δ,X
(0) is an open subset of O that contains 0, for a ny 0 < λ ≤ 1.
(Why?) So a g ood starting poin t is to check if we can make ϕ(λN
δ,X
(0)) small
enough to be conta ined within (−ε, ε) by choosing λ > 0 sufficien tly small. (This
is the main idea behind the proof.) Let’s see. For any y ∈ λN
δ,X
(0), we have
ϕ(y)=ϕ

(1 − λ)0 + λ

1
λ
y

≥ λϕ


1
λ
y

≥ λa. Moreo ver,
0=ϕ(0)=ϕ

1
1+λ
y +
λ
1+λ

−
1
λ
y

≥
1
1+λ
ϕ (y)+
λ
1+λ
ϕ

−
1
λ

y

and h ence −ϕ(y) ≥ λϕ(−
1
λ
y). But, by translation invariance,
1
λ
y ∈ N
δ,X
(0) implies
−
1
λ
y ∈ N
δ,X
(0), so we ha ve ϕ(−
1
λ
y) ≥ a, that is, −ϕ(y) ≥ λa. Thus, letting b := −a>
0, we have λb ≥ ϕ(y) ≥−λb for each y ∈ λN
δ,X
(0). So,ifwetakeany0 < λ <
ε
b
and
set U := λN
δ,X
(0), we find ϕ(U) ⊆ (−ε, ε). Since ε > 0 was a rbitrarily chosen above,
this comp lete s the proof. 

We will return to the investigation of con cave functions Ch apters J and K. For
now , we conclu de with some exercises that w ill extend the presen t analysis to the
context of con vex corresponden ces (Exerc ise G.8).
∗
Exercise 26.
H
(Borwein) Let X and Y be two metric linear spaces, O anonempty
open and convex subset of
X,andΓ : O ⇒ Y a convex correspondence. Prove:
(a)If
Γ(x
0
) is a bounded subset of Y for some x
0
∈ X, then Γ(x) is a bounded
subset of
Y for any x ∈ X;
(b)If
Γ is lower hemicon tinuous at some x
0
∈ X, then it is low er hemicontinu ous;
(c)If
Γ is lo wer hemicontinuous at some x
0
∈ X, and Γ(x
0
) is a bounded subset of
Y, then Γ is contin u ous.
3 F inite Dimensio nal Metric Lin e a r Spa ces
Recall that two isomorphic linear spaces cannot be d istinguished from each other in

terms of th eir linear algebraic properties (Section F.2.3). Sim ilarly, two ho me omor-
phic me tric spaces e njoy the s ame s tructure insofar as their topological properties are
431
concer ne d (Se ction D.1.6). It is thus plain how to ide ntify tho se m et ric linear sp a ces
that share the same linear algebraic and topological properties.
Dhilqlwlrq. Let X and Y be tw o metr ic linear spaces and L ∈ L(X, Y ). If L is a
continuo us bijection with a continuous inverse, then it i s called a l inear homeomor-
phism. If such a linear operator L exists, we t hen s ay that X and Y are linearly
homeomorphic. Similar ly, if L is an isometry, then it is called a linear isometry,
andifsuchanoperatorexists,wesaythatX and Y are linearly isome tr ic .
We know that two linear spaces with the sam e finite dimension a re necessarily
isomorphic (Proposition F.5). Moreove r, we can construct an explicit linear isomor-
phism between an y two such linear spaces by mapping the basis elements of one
space to the basis elemen ts of the other. As we sho w next, in the case of metric
linear spaces with the same finite dimension, an y such linear isomorphism is, in fact,
a linear hom eomorphism. We thu s have the following sharpening of Corollary F.3.
Theorem 1. (Tychonoff) E ver y non triv ial finite dimensional m etric linear s pace is
linearly homeomorphic to R
n
, for some n ∈ N.
We shall use the follo wing technical observation to prove this importan t result.
Lem ma 1. Given any n ∈ N,takeanyλ
i
∈ R
N
and α
i
∈ R, i =1, , n. Let Z be a
metric linear spa ce an d {z
1

, , z
n
} a linearly independent subset of Z. Then,
lim
m→∞
n
S
i=1
λ
i
(m)z
i
=
n
S
i=1
α
i
z
i
if and only if lim
m→∞
λ
i
(m)=α
i
, i =1, ,n.
This is hardly an incre dib le assertion, but its p roof require s a small dose of razzle-
dazzle. Let u s then postpone its proof un til t he e n d of this section, an d rather try to
seewhatonecandowiththisresult. Hereisthemainconclusionwewishtoderive

from it.
Proposition 9. Ev ery lin ear isom orph ism from one finite dimensional linear space
on to another is a homeom orph ism.
It is quite easy to prove this fact by using Lemma 1. We giv e the proof in the form
of an exercise (which replicates a chunk of the analysis presented in Section F.2.3).
432
Exercise 27 . Let X and Y be two finite dimensional metric linear spaces, and
L ∈ L(X, Y ) abijection.
(a)UseLemma1toshowthat
L is continuous.
(b)Let
A be a basis for X. Sho w that L(A) is a basis for Y.
(c)Define ϕ ∈ X
Y
by ϕ(z):=
S
x∈A
λ
z
(L(x))x, where λ
z
is the unique map in
R
L(A)
with z =
S
y∈L(A)
λ
z
(y)y. Sh ow that L

−1
= ϕ.
(d)UseLemma1toshowthatL
−1
is continuous.
Exercise 28. E v ery closed and bounded subset of a finite dimensional metric linear
space is compact. True or false?
Theorem 1 is, of course, a trivial consequence of Proposition 9. Suppose X is
a nontrivial metric linear space with n := di m(X) ∈ N. Then, by Corollary F.3,
there is a linear isomorph is m from X onto R
n
. ButProposition9saysthatthis
linear isomorphism is a homeomo rphism, s o X and R
n
are linearly homeomo rph ic,
as asserted b y Theorem 1. As a n imm ediate corollary, we find that R
n,p
and R
n,q
are
linearly homeomorphic, for any n ∈ N and 1 ≤ p, q ≤∞. It is in t his sense that we
think of these Fréchet spa ces as “identic al.”
The main lesson here is that the linear a nd topological properties of any giv en
nontriv ial finite dimensional metric linear space are e xactly th e same as those of a
Euclidean space. The following three corollaries feast on this observation.
Corollary 2. Any finite dim ensional met r ic linea r sp ac e X is a Fréchet space.
Proof. If X = {0}, there is nothing to prove, so assume that X is nontrivial,
and take an y Cauchy sequence (x
m
) ∈ X

∞
. By Theorem 1, there exists a linear
homeom orph ism L ∈ L(X, R
n
), where n := dim (X). By Proposition 3, L is unifor m ly
continuous, so (L(x
m
)) is a Cauc hy sequence in R
n
(Exerc ise D.9). Since R
n
is
comp le te, t h ere is a y ∈ R
n
with L(x
m
) → y.Butthen,sinceL
−1
is c ontinu ou s ,
x
m
= L
−1
(L(x
m
)) → L
−1
(y) ∈ X, that is, (x
m
) converges in X. 

The follow ing observation i s an im mediate consequ e n ce of Co rollary 2 and Pr opo-
sition C.7.
Corollary 3. An y finite dimen sion a l sub s p ace of a metric linear space X is closed in
X.
Exercise 29 . Show that every affine manifold in a finite dimensional metric linear
space is closed.
Exercise 30.
H
Show that cone(S) is closed for any finite subset S o f a m etric linear
space.
433
Anoth e r application of Theorem 1 giv es us the follow ing result which y o u might
have already anticipated .
Corollary 4. Any linear functional definedonafinite dimensional metric linear space
is continuous.
Again, why should you expect such a result to be true? Your argumen t should
go along the follow in g lines: “I know the precise structu re of an arbitrary linear
functional on R
n
(Example F.6) — any such functional is contin uous. Thus the claim
advanced in Corollary 4 holds in R
n
(for an y n). But, insofar as continuity and
linearity properties are concerned, any giv en nontrivial finite dimensional metric linear
space can be id entified with R
n
(for som e n)—wemayregardtheformerspaceasifit
is obtained from R
n
b y relabelling its vecto rs. Thus, the claim of Corollary 4 should

hold in an y finite dimension al metric linear spac e.”
Exercise 31. Prove Corollary 4.
Finally, we note that one should be careful in identifying two linearly homeomor-
phic m etric linear spaces. True, these spaces are indistinguishable from each other
insofar a s their linear a lgebraic and topological structures a re concerned, but they
ma y behave quite differe ntly with regar d to some ot her p roperties. The following
exercise illustrates this point.
Exercise 32. AsubsetS of a metric space X is said to satisfy the Unique Nearest
Point Property if for every point
x outside S there is a unique point in S whic h is
closest to
x.
(a) Show that every nonempty closed and convex subset of R
2
has the Unique Nearest
Point Property.
16
(b)Letp ∈ {1, ∞}, and show that a nonempty closed and conv ex subset of R
n,p
need
not have the Unique Nearest Point Property
. Conclude that having a u nique nearest
point is a property which need not be preserved under a linear homeomorphism. (Of
course, this property would be preserved under any linear isometry.)
All this is g ood, but remember that we ha ve left a big gap in our investigation
of th e finite dimensional m etric linear spaces above. While our proof of Theorem 1
depends vitally on Lemma 1, we have not ye t established the latter result. To wrap
things up, then, w e prove Lemma 1 no w .
16
A classical result of convex analysis maintains that the converse of this also holds: A closed

subset S of R
n
is convex if, and only if, for every vector x in R
n
, there exists a unique nearest v ector
in S with respect to d
2
. (This is Motzkin’s Theorem.)
434
ProofofLemma1. The “ if” part o f the claim is easy. If λ
i
(m) → α
i
for eac h i,
then by the continuity of scalar multiplication, we have λ
i
(m)z
i
→ α
i
z
i
for each i, so
b y the continuity of vector add ition,
S
n
λ
i
(m)z
i

→
S
n
α
i
z
i
(as m →∞).
We will prov e the “only if” part by an inductiv e argument. In what follow s w e
assume
S
n
λ
i
(m)z
i
→
S
n
α
i
z
i
, and denote the set {1, , n} by N.
Case 1. Assume: There is a j
1
∈ N such that λ
i
(·)=0for all i ∈ N \{j
1

}.
17
Let j
1
=1fo r ease o f reference. If (λ
1
(m)) is un bounded, then there exists a
subsequence (λ
1
(m
k
)) of this s e qu e nce with λ
1
(m
k
) =0for each k, and
1
λ
1
(m
k
)
→ 0
(as k →∞). Th us,
z
1
=
1
λ
1

(m
k
)

λ
1
(m
k
)z
1

→ 0
n
S
i=1
α
i
z
i
= 0
b y the continuity of scalar multiplic ation . But then z
1
= 0, which contradicts
{z
1
, , z
n
} being linearly independent. Therefore, (λ
1
(m)) must be a bounded se-

quence, so if β
1
∈ {lim inf λ
1
(m), lim sup λ
1
(m)}, then β
1
∈ R. Clear ly, there exists
a subsequence (λ
1
(m
k
)) of (λ
1
(m)) with λ
1
(m
k
) → β
1
(as k →∞), and hence
λ
1
(m
k
)z
1
→ β
1

z
1
b y the con tin uity of scalar multiplication. Thus, under o ur case
hypothesis, β
1
z
1
=
S
n
α
i
z
i
. By linear independence, then, β
1
= α
1
and α
i
=0fo r
all i ∈ N with i>1.
18
Thus: lim λ
i
(m)=α
i
for eac h i ∈ N. 
Case 2. Assume: There are j
1

,j
2
∈ N such th at λ
i
(·)=0for all i ∈ N\{j
1
,j
2
}.
Let j
1
=1and j
2
=2for ease of r eference. If (λ
1
(m)) is un bounded, then there
exists a subsequence (λ
1
(m
k
)) of this sequence with λ
1
(m
k
) =0for each k, and
1
λ
1
(m
k

)
→ 0 (as k →∞). Thus
z
1
+
λ
2
(m
k
)
λ
1
(m
k
)
z
2
=
1
λ
1
(m
k
)

λ
1
(m
k
)z

1
+ λ
2
(m
k
)z
2

→ 0
by the contin uit y o f scalar multiplication. But then, since the metric of Z is trans-
lation invaria nt, −
λ
2
(m
k
)
λ
1
(m
k
)
z
2
→ z
1
, which is impossible i n view of w hat we h ave es-
tablishedinCase1. Thus(λ
1
(m)) is bounded. Besides, the analogous argumen t
shows tha t (λ

2
(m)) is bounded as well. Now let β
i
∈ {lim inf λ
i
(m), lim sup λ
i
(m)},
i =1, 2. Clearly, f or each i =1, 2, there exists a subsequence (λ
i
(m
k
)) of (λ
i
(m)) with
λ
i
(m
k
) → β
i
(as k →∞), and hence λ
1
(m
k
)z
1
+ λ
2
(m

k
)z
2
→ β
1
z
1
+ β
2
z
2
, which
implies β
1
z
1
+ β
2
z
2
=
S
n
α
i
z
i
. By linear independence, then, β
i
= α

i
,i=1, 2, and
α
i
=0for all i ∈ N with i>2. Thus: λ
i
(m)=α
i
for all i ∈ N. 
Con tin uing this way inductiv ely yields the proof. 
17
Observ e that pro ving the assertion with t his assumption establishes the following: For a ny real
sequence (θ
m
), we have θ
m
z
i
→ z
j
iff i = j and θ
m
→ 1. (Why?) We will use this observation in
Case 2.
18
Since β
1
may equal either lim inf λ
1
(m) or lim sup λ

1
(m) here, it follows that lim λ
1
(m)=α
1
.
435
4 Com pact Sets in Metric Lin ear Spaces
We ha ve m otivated the notion of com pactness in C hapter C a s s ome sort of a “finite-
ness” property whic h is su itab le for in finite sets. So, in an intuitive sense, comp ac t
sets are “not very large.” On the o ther hand, w e ha ve argued in Chapter G that
there is reason to think of algebraically open sets in infinite dim ensional l inear spaces
as “very large.” (Recall Example G.8.) A s we shall see in the next section, the same
argument ap p lies to open subsets of a ny infinite dimensional metric linear space as
well. But ho w can we rec on cile these intuitive point s of view? How should we think
about a compact set (which is supposedly “small”) t hat contains a nonempty open
set (which is supposedly “large” )?
Observe firstthattheissuedoesnotariseinthefinite dimensional case. For in-
stance, there is good reason to “view” the closed unit ball {x ∈ R
n
: d
2
(x, 0) ≤ 1}
(which is ob viou s ly compact) as a “small” set. Even though this set contains an open
set, the “size” in terpretation of open sets applies o nly to infinite dimensional spaces,
so we are fine here. Things are more subtle in the infinite dimensional case, howev er.
As we show below, a comp act s ub set of an infinite dimensional metric linear space X
can never conta in an open subset of this space. So, somewhat unexpectedly, the po-
tential c onflict between o ur i ntuitive vi ewpoin ts does not arise in infinite dim ensional
spaces either.

This is a prett y startling obser vation about whic h you should pau se an d reflect. On
one hand it s how s again that we should alwa y s keep ou r finite dimensional intuitions
in check when dealing with infinite dimensional metric linear spaces. On the other
hand, it tells us that our heuristic way of viewing compact sets as “sma ll” and infinite
dime nsion a l open sets as “large” holds water in the context of metric linear spaces:
The closure o f a n onempty open subset o f an infinite dimensional metric
linear space cannot be bounded (and is thus not c ompact).
Let us no w move on to the formal dev elopment. We begin with an observation
that might ring a bell (if yo u reca ll Section C.4).
Lem ma 2. If S is a n onempty compact subset of a metric l inear space X,then,
given a ny θ > 0 an d ε > 0, there e xist finitely man y vectors x
1
, , x
k
in S such that
S ⊆
k
V
i=1

x
i
+ θN
ε,X
(0)

.
Exercise 33 .
H
Pro ve Lemma 2.

Hereisthemainresultofthissection.
436
Theorem 2. (Riesz) Let X be a metric linear space. If there is a compact subset S
of X with in t
X
(S) = ∅,thenX is finite dimensional.
That is, the interior of an y comp act set in an infinite dimensional metric lin-
ear s pace must be empty. This situation is, of course, m arkedly different than the
cor r espon ding scenario in finite dimensional linear spaces. For, b y Theorem C.1,
the closure of ev ery bounded set (e.g., an y ε-neigh borhood of 0)iscompactinany
Euclidean space. Thus, by Theorem 1, t his is the case in any finite dimensional metric
linear space. Consequen tly, Theorem 2 p rovides us with the follo wing propert y that
distingu ish e s infinite and finite dimensional metric linear spaces.
Corollary 5. (Riesz) A m etric linear s pace X is finite dimensional if, and only if,
the closure of every bounded subset of X is comp ac t in X.
AmetricspaceX is said to b e locally compact if every point of th e space has an
ε-neighborhood the closure of which is compact in X. O bv iously, every compact met-
ric space is locally compact, but not con ve rsely. (For instance, R is locally comp act,
while, of cou rs e, it is not compa ct.) Thus, Corollary 5 is sometimes paraphrased as
follows:
A metric linear space is locally compact iff it is finite dimensio nal.
19
Exercise 34 . Derive Corollary 5 from Theorem 2.
An immediate application of T heorem 2 (or Corollary 5) sho ws that t he closure of
theopenunitballofaninfinite dimensional metric linear space cannot be compact.
While this fact may conflict with our “Euclidean ” intuition , it is not all that m y ste-
rious. Perhaps a s imp le illu stration may con v inc e you. Take any 1 ≤ p ≤∞, and
let O denote the open unit ball in 
p
, that is, O := {(x

m
) ∈ 
p
:
S
∞
|x
i
|
p
< 1}. The
closure of this set equals the closed unit ball B

p
:= {(x
m
) ∈ 
p
: d
p
((x
m
), 0) ≤ 1}.
But an easy computation sho ws that
d
p
(e
k
, e
l

)=

2
1
p
, if 1 ≤ p ≤∞
1, if p = ∞
,
where e
1
:= (1, 0, 0, ), e
2
:= (0, 1, 0, ), e tc It follows that (e
1
, e
2
, ) is a sequenc e
in B

p
without a con v ergent subsequence. Thus: The closure of O is not (seque ntially )
compac t.
19
This importan t result was proved in 1918 by Frigyes Riesz (1880-1956). It is thus often referred
to as Frigyes Riesz’s Theorem. (Riesz’s younger brother, Marcel Riesz, was also a well-known
mathematician.) Frigy es Riesz h as contributed to the foundations of functional analysis and operator
theory in a substantial way — his contributions have accen tuated the development of quantum
mechanics early in the 20th century. There are at least tw o major results in functional analysis
that carries his name: the Riesz-Fisc her Theorem and the Riesz Representation Theorem.
437