JAMES’ WEAK COMPACTNESS THEOREMpeople.maths.ox.ac.uk/sojmark/jwct.pdffunctional on Xattains its...

JAMES’ WEAK COMPACTNESS THEOREM

ANDREAS SØJMARK

Abstract. The purpose of this note is a careful study of James’ Weak Compact-ness Theorem. Strikingly, this theorem completely characterizes weak compactnessof weakly closed sets in a Banach space by the property that all bounded linearfunctionals attain their supremum on the set.

To fully appreciate this statement, we first present a simple proof in a familiarfinite dimensional setting. Subsequently, we examine the subtleties of compactnessin the weak topologies on infinite dimensionsal normed spaces. Building on this, wethen provide detailed proofs of both a separable and non-separable version of James’Weak Compactness Theorem, following R.E. Megginson (1998).

Using only James’ Weak Compactness Theorem, we procede to derive alternativeproofs for the celebrated James’ Theorem and the equally famous Krein-ŠmulianWeak Compactness Theorem as well as another interesting compactness result byŠmulian. Furthermore, we present some insightful applications to measure theoryand sequence spaces.

Robert C. James (1918-2004) — Photo by Paul R. Halmos

Acknowledgements. The contents of this note constituted my thesis for the degree of BSc inMathematics at the University of Copenhagen. I would like to extend my warmest thanks tomy thesis advisor Dr Job J. Kuit for wonderful guidance and inspiration in the preparation ofthe note. The proofs of the theorems in Section 7 follow the book by R.E. Megginson (1998).

Date: The present state of the note was completed on 5th August 2014.1

JAMES’ WEAK COMPACTNESS THEOREM 2

Contents

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1 Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4

2. Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3. The Finite Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4. Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5. Weak Topologies & Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135.1 The Dual Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .155.2 The Double Dual & Reflexivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.3 Boundedness & Comparison of Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17

6. Weak Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.1 The Eberlein-Šmulian Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7. James’ Weak Compactness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.1 The Separable Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.2 Extension to the Non-Separable Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347.3 James’ Weak Compactness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

8. Corollaries, Counterexamples, & Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498.1 The Assumption of Weak Closedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508.2 The Assumption of Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.3 The Rôle of Reflexivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538.4 James Weak* Compactness Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53

Appendix A. List of Analysis Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Appendix B. List of Topology Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57


1. Introduction

The subject of this note is a thorough study of James’ Weak Compactness Theoremby Robert C. James (1964, 1972). Informally, the theorem states that: A weakly closedsubset A of a Banach space X is weakly compact if and only if every continuous linearfunctional on X attains its supremum on A.

This theorem is fundamental to a wide range of topics in Functional Analysis andstands as one of the deepest results in the study of weak topologies. Moreover, ithas important applications in vector space optimization and has recently found uses inmathematical finance (see e.g. Cascales et al. 2010). As a testimony to its fame, theEncyclopedia of Optimization (Floudas & Pardalos 2009, p. 2091) refers to the theoremas “one of the most beautiful results in functional analysis”.

At the same time, however, the proof has a redoutable reputation for being highlyinaccessible and it is thus omitted in most textbooks on the subject. The plan for thepresent exposition is therefore to provide an illuminating and detailed account of thetheorem and its proof. By striving for clarity and precision, as opposed to concision,we aim to equip the reader with a complete and comprehensive appreciation of the keyaspects of the theorem. As we make no claims of originality, this is indeed the truejustification for the existence of this note.

Before proceeding to introduce the basic notation in Section 1.1 below, we first providethe reader with a brief outline of the progression of the paper:

In Sections 2 and 3 we motivate our interest in James’ Weak Compactness Theoremand explore the theorem in a finite dimensional setting. In Sections 4 and 5 we thenpresent a comprehensive introduction to topological vector spaces in order to fully ap-preciate the theory underlying James’ Weak Compactness Theorem. The experiencedreader may wish to skip these sections and return to them as needed. With the the-oretical underpinnings now in place, we devote Section 6 to an investigation of weakcompactness in normed spaces. After this, we embark on the heart and soul of this ex-position as we present a detailed proof of James’ Weak Compactness Theorem in Section7. We finish Section 7 with three simple, yet insightful, applications of the theorem. InSection 8 we then derive a series of well-established theorems as corollaries of James’Weak Compactness Theorem. Moreover, we provide a series of examples to emphasizethat the assumptions of the theorem cannot be relaxed. In addition to this, we discussthe influence of reflexivity on the proof of James’ Weak Compactness Theorem and howthe ability to apply the theorem will in many cases automatically imply reflexivity. Fi-nally, we finish this notes by proving a weak* analogue of James’ Weak CompactnessTheorem.

As a final remark, it should be noted that many of our arguments will rely on theHahn-Banach Theorem and hence the Axiom of Choice is implicitly assumed. We shallnot pursue this issue any further.

1.1. Preliminaries

Definition 1.1. Let X be a vector space over F, where F denotes either the field ofreal numbers R or the field of complex numbers C. Suppose ‖·‖ : X → R+ is a norm onX. Then the ordered pair (X, ‖·‖) is called a normed space. If, in addition, the inducedmetric d(x, y) := ‖x− y‖ is complete, then (X, ‖·‖) is said to be a Banach space.

When the context is clear, we usually write only X, when we are in fact referring to(X, ‖·‖). If a norm ‖·‖ induces a complete metric, then it is sometimes called a Banachnorm. The topology on X induced by the metric d(x, y) := ‖x− y‖ is referred to asthe norm topology on X. Whenever a topological property of X is mentioned withoutfurther qualification, this norm topology is in mind.

All standard properties of normed spaces are assumed to be well-known to the reader.Nevertheless, we shall always mention explicitly any property that we employ. Thesesame principles apply to the properties of linear operators between normed spaces. As arule of thumb, the material in Chapter 5 of Folland (1999) is considered well-known (thisbook is used for the first course in Functional Analysis at the University of Copenhagen).


Definition 1.2. Suppose (X, ‖·‖) is a normed space. Then we define the closed unitball of X as x ∈ X : ‖x‖ ≤ 1 and denote it by BX .

The notation BX will be used extremely often. It has the advantage that the closedunit balls BX and BY of two different normed spaces (X, ‖·‖X) and (Y, ‖·‖Y ) are easilydistinguished without any notational clutter.

Throughout the text, the notions of linear span and convex hull will appear repeat-edly. For this reason, we introduce the definitions here and remind the reader of somebasic facts.

Definition 1.3. Let X be a vector space over of F and let A be a given subset ofX. Then the smallest convex set that includes A is called the convex hull of A and isdenoted by co(A). Analogously, the smallest closed convex set that includes A is calledthe closed convex hull of A and is denoted by co(A).

It is easily seen that co(A) is simply the set of all convex combinations of elementsin A. That is, the set of all sums

∑nj=1 tjxj where x1, . . . , xn ∈ A, t1, . . . , tn ≥ 0, and∑n

j=1 tj = 1. Moreover, if X is a normed space, then co(A) = co(A).

Definition 1.4. Let X be a vector space over of F and let A be a given subset of X.Then the smallest subspace of X that includes A is called the linear span of A and isdenoted by span(A). Analogously, the smallest closed subspace of X that includes A iscalled the closed linear span of A and is denoted by span(A).

Similarly to the case of the convex hull, one easily realizes that span(A) is simply theset of all linear combinations

∑nj=1 λjxj of elements x1, . . . , xn ∈ A where λ1, . . . , λn ∈ F.

In addition, we also have span(A) = span(A) whenever X is a normed space.In the event that F = C, we will often be interested in the restriction of multiplication

by scalars to R. This motivates the following definition.

Definition 1.5. Let X be a vector space over F. If we restrict the multiplication byscalars map m : F×X → X to R×X, then we obtain a new vector space over R, which wedenote by XR. Suppose f is a real-valued function on X such that f(x+y) = f(x)+f(y)for all x, y ∈ X and f(λx) = λf(x) for any λ ∈ R. Then f is a linear functional on XRin the usual sense. When this is the case, we say that f is a real-linear functional onX. Analogously, we sometimes say that f is a complex-linear functional on X if it is alinear functional on X over the field of complex numbers.

Note that if F = R, then XR is simply the original vector space. We also emphasizethat if X is a normed space or a Banach space, then so is XR with respect to the samenorm. Some useful results about the intimate relationship between real- and complex-linear functionals are given below. It is these properties that make the study of XRfruitful.

Theorem 1.6. If f is a linear functional on X, then the real part u := Re(f) is areal-linear functional on X with f(x) = u(x)− iu(ix) for all x ∈ X. Conversely, if u isreal-linear functional on X, then f(x) = u(x)− iu(ix) uniquely defines a complex-linearfunctional f on X such that u = Re(f). When X is a normed space, f is continuous ifand only if u is continuous, in which case ‖f‖ = ‖u‖.

Proof. See Proposition 5.5 in Folland (1999).

Basically, there are two main reasons for our interest in XR and the appertainingreal-linear functionals. First of all, the behavior of the real-linear functionals on X issufficient for the characterization of weak compactness in James’ Weak CompactnessTheorem. Secondly, temporary restriction to XR will simplify our arguments in section7, where we prove the theorem.

On a completely different note, the next result will play an important role in some ofour applications of James’ Weak Compactness Theorem. Although the proof is straight-forward, we include it here for completeness (and in lack of a good reference).


Theorem 1.7. Let X be a topological space and suppose that f : X → R is a continuousreal-valued function. Then sup f(x) : x ∈ A = sup

f(x) : x ∈ A

for any subset A of

X.

Proof. Given a subset A of X, let s := supf(x) : x ∈ A

. If A = ∅, then the claim is

trivial, so we may assume that A 6= ∅. Clearly, we have sup f(x) : x ∈ A ≤ s sinceA ⊆ A. Hence s is an upper bound of f(x) : x ∈ A. It remains to show that s is thesmallest upper bound.

Under the assumption that s < ∞ (as is sometimes required for the supremum toexist), we need to show that, for every ε > 0, there exists x ∈ A such that s− ε < f(x).To this end, let ε > 0 be given. Since s is the supremum of

f(x) : x ∈ A

, there

exists a′ ∈ A such that s− ε/2 < f(a′). Then, by continuity of f , there exists an openneighborhood U of a′ such that |f(x)− f(a′)| < ε/2 whenever x ∈ U . In particular, wehave f(a′) − ε/2 < f(x) whenever x ∈ U . Now observe that U ∩ A 6= ∅ by the factthat a′ ∈ A. Hence there exists a ∈ A such that f(a′) − ε/2 < f(a). But then f(a) >f(a′)−ε/2 > (s−ε/2)−ε/2 = s−ε as desired. This proves that s = sup f(x) : x ∈ Ain the cases where s <∞.

The case where s = ∞ proceeds similarly. First of all, fix an N ∈ N. Then thereexists an a′ ∈ A such that f(a′) > N + 1/2. Moreover, by continuity of f , there isan open neighborhood U of a′ such that f(a′) − 1/2 < f(x) whenever x ∈ U . ButU ∩A 6= ∅ since a′ ∈ A, so this implies that f(a′)− 1/2 < f(a) for some a ∈ A. Hencef(a) > f(a′)− 1/2 > (N + 1/2)− 1/2 = N for some a ∈ A. Since N ∈ N was arbitrary,this proves that sup f(x) : x ∈ A = ∞.

Finally, we remind the reader that the notions of a bounded linear functional (not tobe confused with a bounded image) and a continuous linear functional are equivalent.In line with tradition, we shall prefer the term bounded, except when we explicitly wishto stress the continuity (as we e.g. do in Sections 2 and 3 below).


2. Motivation

We begin our jorney by considering some motivating theorems from analysis andtopology. First and foremost, we present a slight generalization of the Extreme ValueTheorem, which is one of the staples of analysis and ordinary calculus.

Theorem 2.1. (Extreme Value Theorem) Let A be a compact subset of a topologicalspace X. Then any continuous real-valued function f : A→ R attains its supremum andinfimum on A. Analogously, any continuous complex-valued function f : A→ C attainsthe supremum and infimum of its absolute value |f | on A.

Proof. For the first statement, observe that f(A) is compact in R by continuity of f .Hence f(A) is closed and bounded in R by the Heine-Borel Theorem. It follows thatsup f(a) | a ∈ A = sup (f(A)) ∈ f(A) = f(A) and likewise for the infimum. Thesecond statement follows from the first by noting that |f | = |·| f is a continuousreal-valued function.

Perhaps even more interestingly, it turns out that whenever X is metrizable, theconverse of the Extreme Value Theorem is also true. This paves the way for a completecharacterization of compactness in metric spaces by means of continuous functions.

Theorem 2.2. Let X be a metrizable topological space and let A be subset of X. Ifevery continuous real-valued function f : A→ R is bounded on A, then A is compact inX.

Proof. We proceed by contraposition and show that if A is not compact, then there existsan unbounded continuous function f : A → R. Since A inherits the metrizability of X,A is in particular a Normal space. Hence the Tietze Extension Theorem (Theorem B.1)applies to A, so it is enough to construct an unbounded continuous function f : S → Ron a closed subset S in A. Now, since A is not compact, it also not sequentially compactby virtue of being metrizable. Therefore, there is an infinite sequence (an)n∈N in A withno convergent subsequence. Let (ank

)k∈N be a subsequence where no two terms agree.Then S := ank

| k ∈ N is a countably infinite discrete subset of A. Since it has nolimit points in A (if it had a limit point, then by T1-ness we could choose a subsequenceconverging to it), it is closed in A. Now let f : S → R be given by f(ank

) = k, which iswell-defined since no two terms in S are equal. Then f is clearly unbounded. Also, f isautomatically continuous because S is discrete. Thus, f extends to a continuous functionon all of A by the Tietze Extension Theorem (Theorem B.1). Since f is unbounded onS ⊆ A, it is also unbounded on A.

By combining Theorems 2.1 and 2.2, we obtain, as promised, a highly instructivecharacterization of compactness in terms of continuous real-valued functions.

Theorem 2.3. Let A be a subset of a metrizable topological space X. Then the followingstatements are equivalent.

(1) A is compact(2) Every real-valued continuous function f : A → R attains its supremum and

infimum on A(3) Every real-valued continuous function f : A→ R is bounded on A

Proof. The first implication (1) ⇒ (2) follows from Theorem 2.1. The second implication(2) ⇒ (3) is trivial. Finally, (3) ⇒ (1) follows from Theorem 2.2.

It is an interesting question whether a similar result to Theorem 2.3 can be obtainedfor continuous linear functionals on normed spaces. A priori, there is no reason tosuspect that Theorem 2.3 should generalize to this situation since, on a given normedspace, the set of all continuous linear functionals is not even dense in the set of allcontinuous scalar-valued functions.

In finite dimensions, however, the answer turns out to be a fairly straightforward“yes” if only we demand that A is closed. This is the topic of Section 3 below. Ininfinite dimensions, on the other hand, we need to look for a coarser topology with a


natural connection to the continuous linear functionals: This will be the so-called weaktopology, which we introduce in Section 5. The weak topology is not only intricatelylinked to the set of continuous linear functionals, it also shares many useful propertieswith the metric topologies.

As such, it is worth keeping in mind that it was precisely metrizability that broughtus the nice characterization of compactness in Theorem 2.3. In particular, we used theequivalence of compactness and sequential compactness in Theorem 2.2, which is alsotrue in the weak topology by virtue of the remarkable Eberlein-Šmulian Theorem (seeTheorem 6.5).


3. The Finite Dimensional Case

In the cases where X is a finite dimensional Banach space, the proof of James’ WeakCompactness Theorem is immensely simplified. The main reason for this simplificationis the simple fact that we have the Heine-Borel Theorem at our disposal. Somethingwhich we generally do not have in infinite dimensions.

In addition, the norm topology agrees with the weak topology precisely for finitedimensional Banach spaces (see Theorem 5.22), so James’ Weak Compactness Theorembecomes a statement about standard compactness. Thus, the finite dimensional versionof James Weak Compactness Theorem is essentially the best we can do if we want torestrict attention to linear functionals in Theorem 2.3. Specifically, we need to restrictourselves to normed spaces (rather than metric spaces) and we need to require that Ais closed.

Theorem 3.1. (Finite Dimensional James’ Weak Compactness Theorem) LetX be a finite dimensional Banach space. Suppose that A is a nonempty closed (equiva-lently, weakly closed) subset of X. Then the following statements are equivalent.

(1) A is compact (equivalently, weakly compact)(2) Every continuous linear functional f on X attains the supremum of its absolute

value |f | on A.(3) Every continuous real-linear functional f on X attains the supremum of its

absolute value |f | on A.(4) Every continuous real-linear functional f on X attains its supremum on A.

Proof. Observe first that the implications (1) ⇒ (2), (1) ⇒ (3), and (1) ⇒ (4) all followfrom Theorem 2.1. We finish the proof by showing the implications (2) ⇒ (1), (3) ⇒ (1),and (4) ⇒ (3). For this, recall that all linear functionals on X are continuous by thefinite dimensionality.

We begin by proving (2) ⇒ (1). Since A is assumed closed, it follows from the finitedimensionality of X together with the Heine-Borel Theorem that it suffices to show thatA is bounded. Moreover, since any two norms on a finite dimensional Banach space areequivalent, it suffices to prove that A is bounded in the `2-norm. To this end, fix a basise1, . . . , en for X and write the elements of X as x = λ1e1 + · · ·+ λnen where λi ∈ F.Then the scalar-valued functions πi : X → F defined by πi(λ1e1 + · · ·+ λnen) = λi arelinear functionals for each i = 1, . . . , n, so it follows from (2) that each |πi| attains itssupremum on A. But then it holds for every x ∈ A that

‖x‖22 =

∥∥∥∥ n∑i=1

λiei

∥∥∥∥22

=

n∑i=1

|λi|2 ≤n∑

i=1

(sup |πi(x)| : x ∈ A)2 .

Hence A is bounded and therefore A is compact.Now consider the implication (3) ⇒ (1). If F = R, then the real-linear functionals on

X are linear functionals in the normal sense. Hence the proof is analogous to the above.If F = C, we appeal to Theorem 1.6 and write the complex-linear functionals in terms ofreal-linear functionals. Specifically, we may write πi(λ1e1 + · · ·+λnen) = Reλi + iImλi,where λ1e1 + · · · + λnen 7→ Reλi and λ1e1 + · · · + λnen 7→ Imλi are the real-linearfunctionals Reπi and Imπi on X. It follows that

‖x‖22 =

∥∥∥∥ n∑i=1

λiei

∥∥∥∥22

=

n∑i=1

|λi|2 =

n∑i=1

(Reλi) 2 + (Imλi) 2

≤n∑

i=1

(sup |(Reπi) (x)| : x ∈ A)2 + (sup |(Imπi) (x)| : x ∈ A)2 ,

so A is bounded in the `2-norm. Since A is closed and bounded, we conclude that A iscompact.

Finally, we prove the implication (4) ⇒ (3). Let f be an arbitrary real-linear func-tional on X and recall that inf f(x) : x ∈ A = − sup −f(x) : x ∈ A. Since −f is alsoa linear functional on X, both the infimum and supremum of f are thus attained.


We conclude thatsup |f(x)| : x ∈ A = max

sup f(x) : x ∈ A , |inf f(x) : x ∈ A|

is attained for every real linear functional f on X. This finishes the proof.

Recall the basic fact that every finite dimensional normed space over C or R is com-plete (see Theorem 4.7 for a stronger result). Hence there are no restrictions involved inthe requirement that X should be a Banach space in the above theorem. On the otherhand, as we shall see in Example 3.2 below, the assumption that the subset A shouldbe closed in X is paramount for the implications (2), (3), (4) ⇒ (1) to hold. Observethat this assumption was not needed in Theorem 2.3, when we were not restricting ourattention to linear functionals.

Example 3.2. Let X be a finite dimensional Banach space. We show that there aresubsets in X on which every continuous linear functional attains its supremum despitethe sets not being compact.

To see this, let K be any closed and bounded subset of X with nonempty interior.Then sup

∣∣f(x)∣∣ : x ∈ K

is attained for every continuous linear functional on X by theExtreme Value Theorem (Theorem 2.1). We claim that the supremum is always attainedon the boundary of K. To see this, fix an arbitrary continuous linear functional on X.Notice first that the only constant linear functional is the zero functional, which triviallyattains its supremum everywhere on K. Hence we may assume that f is not constant.

Suppose that the supremum is attained at some x0 ∈ K. Then x0 ∈ B(x0, ε) ⊆ Kfor some ε > 0. Since f is not the zero functional, and its supremum is attained at x0, wemust have f(x0) 6= 0. Take λ := 1+ ε

2‖x0‖ > 1. Then λx0 ∈ B(x0, ε) since ‖λx0 − x0‖ =

|1− λ| ‖x0‖ = ε2‖x0‖ ‖x0‖ = ε

2 . By linearity, we have |f(λx0)| = |λ| |f(x0)| > |f(x0)|,which is clearly a contradiction. Hence we conclude that the supremum is attained onthe boundary ∂K.

Now let C ( K be a proper closed subset of the interior of K. Then ∂K ⊆ K \ C.Hence every continuous linear functional on X still attains its supremum on A := K \Cby the previous result. Yet A is not compact precisely because it fails to be closed. Thisclearly stresses the necessity of the assumption that A should be closed in Theorem 3.1above.


4. Topological Vector Spaces

We do not always have the luxury of a norm, and even when we do, the topology thatit induces may not be the most practical for our purposes. Moreover, it is quintessentialfor a full understanding of normed spaces that we know exactly what properties drivewhich behavior. Indeed, for many of the key properties of normed spaces, the heart ofthe matter turns out to be the continuity of the vector space operations rather than anypecularities of the norm. This leads the way for a natural generalization of normed spacesto vector spaces equipped with a topology that makes addition and scalar multiplicationcontinuous. It is precisely this link between the vector space structure of X and itstopology that allows us to join the forces of linear algebra and analysis.

Definition 4.1. Let X be a vector space over F. If T is a topology on X such that(x, y) 7→ xy is continuous from X ×X into X and (λ, x) 7→ λx is continuous from F×Xinto X, then T is a vector topology for X and (X,T) is called a Topological Vector Space(abbreviated TVS).

In order to generalize key results from normed spaces, such as the Hahn-BanachTheorems, we also need the existence of convex neighborhoods around every point inX. This motivates the following definition.

Definition 4.2. Let (X,T) be a Topological Vector Space. If T has a basis consistingonly of convex sets, then T is a locally convex topology and (X,T) is called a LocallyConvex Space (abbreviated LCS).

Observe that a TVS is, in partiuclar, a topological group since the inversion mapx 7→ −x is just multiplication by −1, which is assumed continuous. It is a testimony tothe sensibility of the above definitions that every normed space is a LCS.

Theorem 4.3. Let X be a vector space equipped with a norm topology T‖·‖. Then(X,T‖·‖) is a LCS.

Proof. The vector space operations are clearly continuous and the collection of openballs constitutes a basis of convex sets.

We shall need the notion of a balanced set as we go along. Hence we include a formaldefinition here.

Definition 4.4. Let A be any subset of a vector space X. If αA ⊆ A whenever |α| ≤ 1,then we say that A is balanced.

It is immediate from the definition that any open or closed ball in the norm topologycentered at 0 is balanced. As the following theorem shows, this property generalizes toarbitrary topological vector spaces.

Theorem 4.5. Let (X,T) be a TVS. Then every open neighborhood of 0 contains abalanced open neighborhood of 0.

Proof. Let U ∈ T be an open neighborhood of 0. By continuity of scalar multiplicationthere exist an open ball B(0, δ) in F and an open neighborhood V of 0 in X such thatλV = λv : v ∈ V ⊆ U whenever λ ∈ B(0, δ). Now define a new open neighborhoodV ′ :=

⋃λ∈F:|λ|≤δ λV . By construction, V ′ ⊆ U and 0 ∈ V ′. Let α be an arbitrary

scalar with |α| ≤ 1 and observe that if y = αλv for some v ∈ V and some |λ| ≤ δ, then|αλ| = |α| |λ| ≤ δ implies that y ∈ V ′. Hence

αV ′ =⋃

λ∈F:|λ|≤δ

(αλ)V ⊆⋃

λ∈F:|λ|≤δ

λV = V ′.

for any α ∈ F with |α| ≤ 1. Consequently, V ′ is a balanced open neighborhood of 0contained in U .

An extremely useful idea in the theory of metric spaces (and hence normed spaces)is the notion of boundedness. A subset of a metric space is said to be bounded if it isincluded in an open ball. In a normed space we thus have that a subset A is bounded if


and only if there exists M ∈ R such that A ⊆ B(0, r) = rB(0, 1) for every r > M . Thefollowing definition generalizes this idea to topological vector spaces.

Definition 4.6. A subset A of a TVS (X,T) is said to be bounded if, for each openneighborhood U ∈ T of 0, there exists sU > 0 such that A ⊆ tU for every t > sU .

By the above remarks, it is clear that a subset of a normed space is bounded accordingto the metric space definition if and only if it is bounded according to the presentdefintion. In what follows, the term “bounded” shall therefore always refer to Definition4.4 unless we specifically state otherwise.

We finish this section by showing that the finite dimensional Hausdorff topologicalvector spaces are very well-known objects. In fact, they are simply copies of the euclideanvector spaces.

Theorem 4.7. Suppose (X,T) is a finite dimensional Hausdorff TVS. Then (X,T) islinearly isomorphic and homeomorphic with Fn in the euclidean topology.

Proof. Let v1, . . . , vn be a basis for X and let e1, . . . , en be the standard basis forFn. Based on this, we define a linear functional T : (Fn, ‖·‖2) → (X,T) by T (ei) := visuch that

T (λ1, . . . , λn) = T (λ1e1 + · · ·+ λnen) = λ1v1 + · · ·+ λnvn

for each (λ1, . . . , λn) ∈ Fn. It is immediate that T is a linear isomorphism from Fn ontoX. Hence it only remains to show that T and T−1 are continuous with respect to anyHausdorff vector topology T on X and the euclidean topology T‖·‖2

on Fn.That T is continuous follows easily from the T-continuity of scalar multiplication and

vector addition. To see that T−1 is continuous, consider the unit sphere SFn of Fn. Bythe Heine-Borel property of Fn, SFn is compact. Hence T (SFn) is compact in X by thecontinuity of T . Since X is Hausdorff, T (SFn) is in particular closed in X. Notice that0 = T (0) /∈ T (SFn) since 0 /∈ SFn . Hence X \T (SFn) is an open neighborhood of 0, so byTheorem 4.5 there is a balanced open neighborhood U of 0 contained in X \T (SFn). Byconstruction U ∩ T (SFn) = ∅, so T−1(U)∩SFn = ∅. Moreover, by linearity of T−1 andbalancedness of U , we have αT−1(U) = T−1(αU) ⊆ T−1(U) whenever |α| ≤ 1. HenceT−1(U) is balanced.

Now observe that ‖v‖2 < 1 for every v ∈ T−1(U). To see this, simply note that if‖v‖2 ≥ 1 for some v ∈ T−1(U), then 1/ ‖v‖2 ≤ 1 implies that v/ ‖v‖2 ∈ T−1(U) by thebalancedness of T−1(U). This clearly contradicts T−1(U) ∩ SFn = ∅.

Finally, let πi T−1 : X → F denote the coordinate maps of T−1 and observe thateach πi T−1 is linear. Since ‖v‖2 < 1 for every v ∈ T−1(U), we have

∣∣πi(T−1(x))∣∣ ≤∥∥T−1(x)

∥∥2≤ 1 for every x ∈ U . Now let ε > 0 be given. By linearity the previous result

implies that∣∣(πi T−1

)(x)∣∣ ≤ ε whenever x ∈ εU . Hence∣∣(πi T−1

)(x)−

(πi T−1

)(y)∣∣ = ∣∣(πi T−1

)(x− y)

∣∣ < ε

whenever y ∈ x + εU . Furthermore, note that x + εU is open since (X,T) is a TVS.It follows that each coordinate map of T−1 is continuous and therefore T−1 itself iscontinuous.

It is an immediate consequence of Theorem 4.7 that the unique Hausdorff vector spacetopology on a finite dimensional vector space is induced by a Banach norm. Furthermore,it follows from Theorems 4.3 and 4.7 that any two norms on a finite dimensional vectorspace induce the same topology.


5. Weak Topologies and Dual Spaces

The stronger the topology, the easier it is for functions to be continuous. On the otherhand, too many open sets makes it diffucult for sets to be compact because of finer opencovers. If both continuity and compactness is of interest, we thus face a trade-off in ourchoice of topology. Given a family of functions F, Theorem 5.1 below suggests a naturalchoice of topology in this situation.

Theorem 5.1. Let X be a set and let F be a family of maps f : X → Yf , where eachYf is equipped with a topology Tf . Then there exists a unique weakest topology TF forX such that each f ∈ F is continuous.

Proof. Observe that S :=f−1(U) : f ∈ F, U ∈ Tf

is a subbasis for a topology on X

since the union of the elements in S equals X. Let TF be the topology generated by S.Then each f ∈ F is continuous since S ⊆ TF. Now suppose TX is another topology forX such that each f ∈ F is continuous. Then by defintion of continuity, S ⊆ TX . But,by construction, TF equals the intersection of all topologies for X containing S. Hencewe have TF ⊆ TX as desired.

The topology TF in Theorem 5.1 is sometimes called the initial or weak topology onX with respect to F. We shall say that F is a topologizing family for X and denote TF

by σ(X,F).In functional analysis, we are mainly interested in the cases where X is a vector space

and F is a vector space of linear functionals. This leads us to introduce the algebraicdual space X# := f : X → F | f is linear consisting of all the linear functionals on X.It is easy to see that X# is a vector space. For the remainder of this section we shallconsider some important properties of the weak topologies induced by subspaces of X#.

Theorem 5.2. Let X be a vector space over F and consider a subspace X ′ of thealgebraic dual space X#. Let ε > 0 be given. Then the collection of all “open balls”

Bε(x,A) := y ∈ X : |f(y − x)| < ε ∀f ∈ A , x ∈ X,A ⊆ X ′

where A ⊆ X ′ is finite, constitutes a basis for σ(X,X ′). The subcollection of all setsBε(x, f) := y ∈ X : |f(y − x)| < ε , x ∈ X, f ∈ X ′

forms a subbasis for σ(X,X ′).

Proof. Let S :=f−1(U) : f ∈ X ′, U ∈ TF

be the subbasis for σ(X,X ′) from Theorem

5.1 and let B be the basis consisting of all finite intersections of elements in S. Firstobserve that each Bε(x,A) is open with respect to σ(X,X ′) since

Bε(x,A) = y ∈ X : |f(y)− f(x)| < ε ∀f ∈ A=

⋂f∈A

y ∈ X : |f(y)− f(x)| < ε

=⋂f∈A

f−1 (B(f(x), ε)) ,

which is a finite intersection of elements from S. Now let x ∈ X and consider any basiselement O ∈ B with x ∈ O. Then O is of the form f−1

1 (U1) ∩ · · · ∩ f−1n (Un) for some

fi ∈ X ′ and Ui ∈ TF. For each i = 1, . . . , n choose an open ball B(fi(x), δi) ⊆ Ui.Let δ := min δ1, . . . , δn and define ϕi := ε

δfi, which is again an element of X ′ inthe natural vector space structure on X#.Then y ∈ Bε(x, ϕ1, . . . , ϕn) implies thatεδ |fi(y)− fi(x)| = |ϕi(y − x)| < ε for all i = 1, . . . n or equivalently |fi(y)− fi(x)| < δfor all i = 1, . . . , n. Hence fi(y) ∈ B(fi(x), δi) ⊆ Ui for each i = 1, . . . , n. It follows thatBε(x, ϕ1, . . . , ϕn) is contained in O. Thus, we conclude that the collection of “openballs” Bε(x,A) is indeed a basis for σ(X,X ′). The second statement follows from thefact that each Bε(x,A) equals the finite intersection

⋂f∈A Bε(x, f).

Theorem 5.3. Let X be a set and let F be a topologizing family of functions for X.Suppose that (xα)α∈A is a net in X and x ∈ X. Then xα → x with respect to σ(X,F)if and only if f(xα) → f(x) for every f ∈ F.


Proof. Suppose first that xα → x with respect to σ(X,F). By definition of σ(X,F),each f ∈ F is continuous. Hence f(xα) → f(x) for every f ∈ F as desired. For theconverse, assume instead that f(xα) → f(x) for every f ∈ F. Then, for each f ∈ F, itholds for every neighborhood U of f(x) that there is an αf,U ∈ A such that f(xα) ∈ Uwhenever α αf,U . It follows that for every ϕ−1(V ) ∈

f−1(U) : f ∈ F, U ∈ σ(X,F)

with x ∈ ϕ−1(V ), there is an αϕ,V ∈ A such that xα ∈ ϕ−1(V ) whenever α αϕ,V . Nowrecall from the proof of Theorem 5.1 that

f−1(U) : f ∈ F, U ∈ σ(X,F)

is a subbasis

for σ(X,F). Hence xα → x with respect to σ(X,F) by Theorem B.2.

Theorem 5.4. Let X be a vector space and suppose that X ′ is subspace of the algebraicdual X#. Then σ(X,X ′) is locally convex, and therefore (X,σ(X,X ′)) is a LCS.

Proof. Let (xα), (yα) and (λα) be nets in X and F, respectively. Assume that xα → x,yα → y, and λα → λ for some x, y ∈ X and λ ∈ F. By continuity of multiplication andaddition in F, we have

f(λαxα + yα) = λαf(xα) + f(yα) → λf(x) + f(y) = f(λx+ y)

for every f ∈ X ′. Hence λαxα + yα → λx + y with respect to σ(X,X ′) by Theorem5.3. Therefore, multiplication by scalars and vector addition are continuous with respectσ(X,X ′), so it is a vector topology. Furthermore, B(0, r) : r > 0 is a basis for F, soS :=

f−1(B(0, r)) : f ∈ X ′, r > 0

is a subbasis for σ(X,X ′) by the proof of Theorem

5.1. Since each f−1(B(0, r)) is convex, the basis generated by S consists of convex sets.In conclusion, σ(X,X ′) is locally convex.

Theorem 5.5. Let X be a vector space and suppose that X ′ is subspace of the algebraicdual X#. Then a subset A of X is bounded in (X,σ(X,X ′)) if and only if f(A) isbounded in F for every f ∈ X ′.

Proof. Suppose first thatA ⊆ X is bounded with respect to σ(X,X ′). Then f−1(B(0, 1))is an open neighborhood of 0 in X, so according to Defintion 4.6 there exists sU > 0such that A ⊆ tf−1(B(0, 1)) = f−1(tB(0, 1)) = f−1(B(0, t)) for every t > sU . Hencef(A) ⊆ f(f−1(B(0, t))) ⊆ B(0, t) for every t > sU . This implies that f(A) is boundedin F.

For the converse, assume that f(A) is bounded in F for every f ∈ X ′. Let U ⊆X be an open neighborhood of 0. By definition of σ(X,X ′) there is a basis elementf−11 (B(0, r1)) ∩ · · · ∩ f−1

n (B(0, rn)) ⊆ U . Since each fi(A) is bounded and each B(0, ri)is an open neigborhood of 0, there exists si > 0 such that fi(A) ⊆ tB(0, ri) whenevert > si for each i = 1, . . . , n. In particular, letting sU := max s1, . . . , sn we have fi(A) ⊆tB(0, ri) whenever t > sU for each i = 1, . . . , n. It follows that A ⊆ f−1

i (fi(A)) ⊆tf−1

i (B(0, ri)) whenever t > sU for each i = 1, . . . , n and therefore

A ⊆n⋂

i=1

tf−1i

(B(0, ri)

)= t

( n⋂i=1

f−1i

(B(0, ri)

))⊆ tU

for every t > sU . This proves that A is bounded in (X,σ(X,X ′)).

Theorem 5.6. Let X be a vector space and suppose that X ′ is subspace of the algebraicdual X#. If X ′ is infinite dimensional, then every nonempty open set U ∈ σ(X,X ′) isunbounded with respect to σ(X,X ′).

Proof. Fix U ∈ σ(X,X ′) with U 6= ∅. Clearly x + U is bounded if and only U isbounded, so we may assume that 0 ∈ U . Then there is a basis element f−1

1 (B(0, r1)) ∩· · · ∩ f−1

n (B(0, rn)) ⊆ U . Now consider the subspace⋂n

j=1 ker fj of X and notice that⋂nj=1 ker fj ⊆ f−1

1 (B(0, r1)) ∩ · · · ∩ f−1n (B(0, rn)) ⊆ U . Hence it suffices to show that⋂n

j=1 ker fj is unbounded. Since X ′ is infinite dimensional, we can choose f ∈ X ′

which is not a linear combination of f1, . . . , fn. Therefore,⋂n

j=1 ker fj * ker f byLemma 7.2, so there exists x0 ∈

⋂nj=1 ker fj \ ker f . Now, for any k ∈ N we have

kx0 ∈⋂n

j=1 ker fj and f(kx0) = kf(x0) 6= 0. It follows that for any N ∈ N we canchoose f(kx0) ∈ f

(⋂nj=1 ker fj

)with k > N/

∣∣f(x0)∣∣ so that∣∣f(kx0)∣∣ = k

∣∣f(x0)∣∣ > N .


Hence f(⋂n

j=1 ker fj

)is unbounded in F and therefore

⋂nj=1 ker fj is unbounded in

(X,σ(X,X ′)) by Theorem 5.5.

A final property of σ(X,X ′) is worth mentioning. By construction, every memberof X ′ is continuous with respect to σ(X,X ′). It is not difficult to show, however, thatthese are in fact all the continuous linear functionals. It follows that the continuouslinear functionals with respect to σ(X,X ′) are precisely the members of X ′.

To verify this claim, it suffices to show that every continuous linear functional on Xcan be obtained as a linear combination of finitely many members of X ′. To this end,we shall need the result that if f, f1, . . . , fn are linear functionals on a vector space X,then f ∈ span f1, . . . , fn if and only if

⋂nj=1 ker fj ⊆ ker f . This result is presented as

Lemma 7.2 in Section 7, where a full proof is given.First of all, observe that if f is any continuous linear functional on X, then the preim-

age f−1(B(0, r)) is an open neighborhood of 0 in X for some fixed r > 0. Hence there isa basis element f−1

1 (B(0, r1)) ∩ · · · ∩ f−1n (B(0, rn)) around 0 contained in f−1(B(0, r)),

where f1, . . . , fn ∈ X ′. Now, if x ∈⋂n

j=1 ker fj , then fj(kx) = kfj(x) = 0 for eachj = 1, . . . , n, so kx ∈ f−1

1 (B(0, r1)) ∩ · · · ∩ f−1n (B(0, rn)) ⊆ f−1(B(0, r)) for any k ∈ N.

Thus, if f(x) 6= 0 we would have |f(kx)| = k |f(x)| > r whenever k > r/ |f(x)|, whichclearly contradicts kx ∈ f−1(B(0, r)) for all k ∈ N. Consequently, x ∈ ker f and there-fore

⋂nj=1 ker fj ⊆ ker f . The claim now follows from the aforementioned result.

5.1. The Dual Space

Given a normed space X, the set of bounded linear functionals f : X → F constitutesa particularly important subspace of X#. This leads us to the following defintion.

Definition 5.7. Let X be a normed space. Then the dual space of X is the normedspace B(X,F), where B(X,F) is the set of all bounded linear functionals on X equippedwith the operator norm. It is denoted by X∗.

We remind the reader that X∗ is always a Banach space even if X is not complete.Since the topologizing family X∗ appears so frequently in functional analysis, we reservea special name for the corresponding topology σ(X,X∗).

Definition 5.8. Let X be a normed space. The topology σ(X,X∗) on X induced bythe topologizing family X∗ is called the weak topology of X.

When the context is clear, we shall denote the weak topology by Tw and use thequalifications w, weak, or weakly to refer to its topological properties.

Theorem 5.9. Let X be a normed space equipped with the weak topology. Then (X,Tw)is a completely regular LCS.

Proof. First observe that (X,σ(X,X∗)) is a LCS by Theorem 5.4. Next, notice that X∗

is a separating family of functions by Corollary A.4. Hence Theorem B.3 applies. Con-sequently, it follows from the complete regularity of F that (X,σ(X,X∗)) is completelyregular.

Theorem 5.10. Let X be normed space. Suppose V ⊆ X is a linear subspace of X.Then the weak topology σ(V, V ∗) on V equals the subspace topology that V inherits as asubspace of (X,σ(X,X∗)). In particular, if A ⊆ V is any subset of V , then the subspacetopology on A induced by σ(X,X∗) agrees with the subspace topology on A induced byσ(V, V ∗).

Proof. Recall from Theorem 5.2 that, given ε > 0, the collection BX of “open balls”BXε (x, S) := y ∈ X : |x∗(y − x)| < ε ∀x∗ ∈ S for all x ∈ X and all finite subsets S ⊆

X∗ constitutes a basis for σ(X,X∗). As a linear subspace, V is itself a normed spaceand hence has a dual space V ∗. The collection BV of all “open balls” BV

ε (x, S) for x ∈ Vand S a finite subset of V ∗ is a basis for σ(V, V ∗).

Now consider the basis BVX for the subspace topology on V induced by σ(X,X∗) and

fix an arbitrary basis element V ∩ BXε (x, S) . Let ι : V → X be the inclusion map and


define v∗x∗ := x∗ ι for each x∗ ∈ S. Clearly v∗x∗ ∈ V ∗ for each x∗ ∈ S. Notice thatx, y ∈ V implies y − x ∈ V and hence

BVε (x, v∗x∗ : x∗ ∈ S) = y ∈ V : |v∗x∗(y − x)| < ε for all v∗x∗with x∗ ∈ S

= y ∈ V : |(x∗ ι) (y − x)| < ε for all x∗ ∈ S= y ∈ V : |x∗(y − x)| < ε for all x∗ ∈ S= V ∩ y ∈ X : |x∗(y − x)| < ε for all x∗ ∈ S= V ∩ BX

ε (x, S).

It follows that BVX ⊆ BV .

Conversely, fix an element BVε (x, S) of BV . Then each v∗ ∈ S has a Hahn-Banach

extension x∗v∗ ∈ X∗ with x∗v∗ ι = v∗. Again it follows from y − x ∈ V thatBVε (x, S) = y ∈ V : |v∗(y − x)| < ε for all v∗ ∈ S

= y ∈ V : |(x∗v∗ ι)(y − x)| < ε for all v∗ ∈ S= y ∈ V : |x∗v∗(y − x)| < ε for all x∗v∗with v∗ ∈ S= V ∩ y ∈ X : |x∗v∗(y − x)| < ε for all x∗v∗with v∗ ∈ S= V ∩ BX

ε (x, x∗v∗ : v∗ ∈ S).Hence we also have BV ⊆ BV

X . Thus BV = BVX , which proves the first statement.

For the second statement, note that U ⊆ A is open in the subspace topology inducedby σ(V, V ∗) if and only if U = A∩OV for some OV ∈ σ(V, V ∗). By the first statement,σ(V, V ∗) equals the subspace topology on V induced by σ(X,X∗), from whence it followsthat OV ∈ σ(V, V ∗) if and only if OV = V ∩ OX for some OX ∈ σ(X,X∗). Therefore,U = A∩OV for some OV ∈ σ(V, V ∗) if and only if U = A∩V ∩OX = A∩OX for someOX ∈ σ(X,X∗). But U = A ∩ OX for some OX ∈ σ(X,X∗) if and only if U is open inthe subspace topology induced from σ(X,X∗). This prove the second statement.

Theorem 5.11. Let X be a normed space. Suppose (xα) is a net in X and x ∈ X.Then xα

w−→ x if and only if x∗(xα) → x∗(x) for every x∗ ∈ X∗.

Proof. This follows immediately from Theorem 5.3 by considering F = X∗.

Since (X,Tw) is Hausdorff, the weak limit of a sequence is unique. Moreover, normconvergence is easily seen to imply weak convergence.

5.2. The Double Dual and Reflexivity

Given a normed space X, the dual space (X∗)∗ of the dual space X∗ is also ofsignificant interest. We denote it by X∗∗ and refer to it as the double dual of X.

Definition 5.12. Let X be a normed space and define a linear map J : X → X∗∗ byx 7→ J(x) where (J(x)) (x∗) = x∗(x) for every x∗ ∈ X∗. Then we say that J is thecanonical embedding from X into X∗∗.

To see that J is well-defined, simply note that each J(x) : X∗ → F is linear by thevector space structure on X∗ and bounded because of the following inequality:

‖J(x)‖X∗∗ = sup|x∗(x)| : x∗ ∈ BX∗

≤ sup

‖x∗‖X∗ ‖x‖X : x∗ ∈ BX∗

≤ ‖x‖X .

The next theorem shows that J is in fact an isometric embedding of X in X∗∗.

Theorem 5.13. Let X be a normed space. Then the canonical embedding J : X → X∗∗

has the following properties:(1) J : X → X∗∗ is linear and injective.(2) X is isometrically isomorphic to J(X) ⊆ X∗∗.(3) J(X) is closed in X∗∗ if and only if X is a Banach space.

Proof. First note that J is linear by the linearity of x∗. Hence (1) and (2) bothfollow if we can show that ‖J(x)‖X∗∗ = ‖x‖X for all x ∈ X, since this automati-cally implies injectivity of J . We have already seen that ‖J(x)‖X∗∗ ≤ ‖x‖X in theabove computation. To see the reverse inequality, observe that by Corollary A.3 (with


Y = 0) there exists x∗0 ∈ X∗ such that ‖x∗0‖X∗ = 1 and x∗0(x) = ‖x‖X . Hence‖J(x)‖X∗∗ = sup

|x∗(x)| : x∗ ∈ BX∗

≥ |x∗0(x)| = ‖x‖X as desired. In order to prove

(3), recall that a subspace of a complete metric space is closed if and only if it is com-plete. Hence J(X) is closed in the Banach space X∗∗ if and only if it is complete. Byvirtue of (2), J(X) is complete if and only if X is complete. This proves (3).

Definition 5.14. A normed space X is said to be reflexive if J(X) = X∗∗.

It is easy to see that every finite dimensional normed space is reflexive. Indeed, theinjectivity of J together with the fact that dimX = dimX∗ = dimX∗∗ forces J to besurjective.

Definition 5.15. Let X be a normed space. Then the topology σ(X∗, J(X)) on X∗

induced by the topologizing family J(X) is called the weak* topology of X∗.

The weak* topology is often denoted by σ(X∗, X) in view of the isometric isomor-phism J(X) ∼= X. When the context is clear, we shall denote it simply by Tw∗ and usethe qualifications w∗, weak* or weakly* to refer to is topological properties.

Theorem 5.16. Let X∗ be the dual space of a normed space X. Then (X∗,Tw∗) is acompletely regular LCS.

Proof. It follows from Theorem 5.4 that (X∗, σ(X∗, J(X))) is a LCS. Moreover, we ob-serve that if x∗ 6= y∗, then x∗(x0) 6= y∗(x0) for some x0 ∈ X, so (J(x)) (x0) 6= (J(y)) (x0).Hence J(X) ⊆ X∗∗ is a separating family of functions on X∗. Therefore, Theorem B.3applies and we conclude from the complete regularity of F that (X∗, σ(X∗, J(X))) iscompletely regular.

Theorem 5.17. Let X∗ be the dual of a normed X. Suppose (x∗α) is a net in X∗ andx∗ ∈ X∗. Then x∗α

w∗

−→ x∗ if and only if x∗α(x) → x∗(x) for every x ∈ X.

Proof. This follows immediately from Theorem 5.3 by considering F = J(X).

By the Hausdorffness of (X∗,Tw∗), the weak* limit of a convergent sequence in X∗

is unique. Moreover, one easily sees that norm convergence implies weak* convergence.

5.3. Boundedness & Comparison of Topologies

Theorem 5.18. Let X be a normed space. Then σ(X,X∗) ⊆ T‖·‖Xon X. On the dual

space X∗, we have σ(X∗, J(X)) ⊆ σ(X∗, X∗∗) ⊆ T‖·‖X∗ .

Proof. For the first statement, simply note that all elements of X∗ are continuous withrespect to T‖·‖ by definition. Hence σ(X,X∗) ⊆ T‖·‖ since σ(X,X∗) is the weakesttopology with respect to which the elements of X∗ are continuous. For the secondstatement, first observe that every member of X∗∗ is continuous with respect to T‖·‖X∗ .By construction, the weak topology σ(X∗, X∗∗) on X∗ is weakest topology such thateach x∗∗ ∈ X∗∗ is continuous. Hence σ(X∗, X∗∗) ⊆ T‖·‖X∗ . Finally, every memberof J(X) ⊆ X∗∗ is continuous with respect to σ(X∗, X∗∗). However, the weak* topol-ogy σ(X∗, J(X)) on X∗ is the weakest topology for which every element in J(X) iscontinuous. Hence σ(X∗, J(X)) ⊆ σ(X∗, X∗∗). This completes the proof.

As an immediate consequence of Theorem 5.18, we emphasize that compactness inthe norm topology implies weak compactness.

In accordance with Definition 4.6, we shall say that a subset A of a normed space Xis weakly bounded (respectively, weakly* bounded) if, for each weakly open (respectively,weakly* open) neighborhood U of 0, there exists sU > 0 such that A ⊆ tU whenevert > sU .

Theorem 5.19. Let X be a normed space. Then a subset A ⊆ X is norm bounded ifand only if it is weakly bounded.


Proof. Suppose that A is bounded in the norm topology. Let U be an arbitrary weaklyopen neighborhood of 0. Then U is also open in the norm topology by Theorem 5.18.Hence there exists sU > 0 such that A ⊆ tU for every t > sU by the assumption that Ais norm bounded. Since U was an arbitrary weakly open neighborhood of 0, A is weaklybounded.

Assume conversely that A is weakly bounded. If A = ∅ the statement is trivial, sowe assume A 6= ∅. Then J(A) is a nonempty family of bounded linear functionals onthe Banach space X∗. Now notice that x∗(A) is bounded in F for each x∗ ∈ X∗ byTheorem 5.5 because A is weakly bounded. Hence

sup |(J(x))(x∗)| : J(x) ∈ J(A) = sup |x∗(x)| : x ∈ A <∞for every x∗ ∈ X∗. Therefore, the Uniform Boundedness Principle (Theorem A.5)applied to J(A) implies that

sup ‖J(x)‖ : J(x) ∈ J(A) <∞.

Since ‖x‖ = ‖J(x)‖ by Theorem 5.13, it follows thatsup ‖x‖ : x ∈ A = sup ‖J(x)‖ : J(x) ∈ J(A) <∞.

This proves that A is norm bounded.

Corollary 5.20. Let X be a normed space. Then a subset A ⊆ X is norm bounded (orequivalently, weakly bounded) if and only if the image x∗(A) is bounded in F for everyx∗ ∈ X∗.

Proof. According to Theorem 5.5, A is weakly bounded if and only if x∗(A) is boundedin F for every x∗ ∈ X∗. To finish the proof, simply note that A is weakly bounded ifand only if it is norm bounded by Theorem 5.19.

The analogue of Theorem 5.19 for the weak* topology requires X to be complete.The reason being that we need to apply the Uniform Boundedness Principle to linearfunctionals on X. This was not an issue in Theorem 5.19, where we applied the Uni-form Boundedness Principle to the linear functionals on X∗, since X∗ is automaticallycomplete even if X is not.

Theorem 5.21. Let X be a Banach space. Then a subset of A ⊆ X∗ is norm boundedif and only if it is weakly* bounded.

Proof. As in Theorem 5.19, the “only if” part follows from the fact every weakly* openneighborhood of 0 is also an open neighborhood of 0 in the norm topology by Theorem5.18. For the “if” part, assume that A is weakly* bounded. If A = ∅ the statementis trivial, so we assume A 6= ∅. Then A is a nonempty family of linear functionals onthe Banach space X. Since J(X) is the topologizing family for the weak* topology, itfollows from Theorem 5.5 that

(J(x)) (A) = (J(x)) (x∗) : x∗ ∈ A = x∗(x) : x∗ ∈ Ais bounded in F for every J(x) ∈ J(X). In other words, x∗(x) : x∗ ∈ A is bounded inF for every x ∈ X. Hence

sup |x∗(x)| : x∗ ∈ A <∞for every x ∈ X, so the Uniform Boundedness Principle (Theorem A.5) applied to Aensures that

sup ‖x∗‖ : x∗ ∈ A <∞.

This is precisely the statement that A ⊆ X∗ is norm bounded.

It follows from the preceding results that the study of bounded sets is essentially thesame in the weak topologies vis-à-vis the norm topology. As the next two theoremsshow, when X is finite dimensional, then the topological properties are, in fact, also thesame. This, however, is far from the truth when X is infinite dimensional.

Theorem 5.22. Let X be a normed space. Then σ(X,X∗) = T‖·‖ if and only if X isfinite dimensional.


Proof. Suppose that X is finite dimensional. Recall from Theorems 4.3 and 5.9 that(X,T‖·‖) and (X,σ(X,X∗)) are both Hausdorff topological vector spaces. Hence itfollows from Theorem 4.7 that (X,T‖·‖) and (X,σ(X,X∗)) are linearly isomorphic andhomeomorphic to Fn in the euclidean topology. In particular, the two topologies agree.For the converse, suppose instead that X is infinite dimensional. By Theorem 5.18, it isalways the case that σ(X,X∗) ⊆ T‖·‖. We claim that T‖·‖ is strictly finer than σ(X,X∗).Since X is infinite dimensional, the dual space X∗ is also infinite dimensional. Henceevery nonempty U ∈ σ(X,X∗) is unbounded with respect to σ(X,X∗) by Theorem 5.6.According to Theorem 5.19, this is equivalent to every nonempty U ∈ σ(X,X∗) beingunbounded in the norm topology. Since, for example, the open balls of finite radii arebounded in the norm topology, it follows that they cannot be open in the weak topology.This proves the claim.

Theorem 5.23. Let X∗ be the dual of a normed space. Then σ(X∗, X∗∗) = σ(X∗, J(X))if and only if X is reflexive. And σ(X∗, J(X)) = T‖·‖ if and only if X is finite dimen-sional.

Proof. By the remarks following Theorem 5.6 we have σ(X∗, X∗∗) = σ(X∗, J(X)) ifand only if X∗∗ = J(X). Since X∗∗ = J(X) is precisely the definition of reflexivity,this proves the first statement. Now consider the second statement. By Theorems 4.3,5.9, and 5.16 all three topologies are Hausdorff vector topologies. Hence they agreewhenever X is finite dimensional by Theorem 4.7. This proves the “if” part. Conversly,suppose that X is infinite dimensional. Then σ(X∗, X∗∗) is strictly coarser than T‖·‖by Theorem 5.22 and therefore σ(X∗, J(X)) is also strictly coarser than T‖·‖ sinceσ(X∗, J(X)) ⊆ σ(X∗, X∗∗) by Theorem 5.18.

It follows from Theorem 5.21 that any weakly closed subset of X is also closed. It alsofollows, however, that if X is infinite dimensional, then there are closed sets which arenot weakly closed. Perhaps surprisingly, the next theorem shows that this discrepancynever arises for convex sets.

Theorem 5.24. (Mazur’s Theorem) Let X be a normed space. Then the closureand weak closure of any convex subset of X are the same. In particular, a convex subsetof X is closed if and only if it is weakly closed.

Proof. Let C be an arbitrary convex subset of X. By Theorem 5.18 we have Tw ⊆ T‖·‖,so the limit points of C in the norm topology are also weak limit points of C. HenceC

‖·‖ ⊆ Cw. We claim that C‖·‖ is convex. To see this, fix x, y ∈ C

‖·‖ and t ∈ (0, 1). Let(xn), (yn) be sequences in C converging to x and y, respectively. Then txn+(1−t)yn ∈ C

for all n ∈ N and txn + (1− t)yn converges to tx+ (1− t)y. Hence tx+ (1− t)y ∈ C‖·‖

and so C‖·‖ is convex as claimed.Now suppose that x0 ∈ C

w \ C‖·‖. Then d(x0, C‖·‖

) > 0 since C‖·‖ is closed. Seeingthat C‖·‖ is also convex, it follows from the Hahn Banach Separation Theorem (TheoremA.8) that there exists t ∈ R such that Ref(x0) > t > Ref(x) for every x ∈ C

‖·‖. Butthen (Ref)−1

((t,∞)) ∩ C‖·‖

= ∅ and x0 ∈ (Ref)−1((t,∞)). Since Ref : X → R is

weakly continuous (see the proof of Lemma 7.6) and (t,∞) is open in R, we concludethat (Ref)−1

((t,∞)) is a weakly open neighborhood of x0, which does not intersectC

‖·‖. This clearly contradicts x0 ∈ Cw. Hence Cw

= C‖·‖ as desired.

It is an immediate consequence of Mazur’s Theorem that the closed unit ball BX isalso weakly closed. This result should by no means be taken for granted. For example,it can be shown that BX is in fact the weak closure of the unit sphere SX whenever X isinfinite dimensional. Observe also that any linear subspace of a normed space is convexand therefore closed if and only if it is weakly closed.


6. Weak Compactness

Compactness is one of the most important topological properties in analysis becauseof its intimate interplay with continuous functions. The Extreme Value Theorem fromSection 2 is arguably the most famous example of this. Another important example isthe Heine-Cantor Theorem on uniform continuity.

In finite dimensional normed spaces, the Heine-Borel theorem gives a particularlysimple characterization of compactness as equivalent to being closed and bounded. This,however, is by no means true for infinite dimensional normed spaces, where even theclosed unit ball fails to be compact. Indeed, it turns out that the finite dimensionalnormed spaces are chacterized precisely by the property that every closed and boundedsubset is compact. We shall refer to this property as the Heine-Borel property.

Theorem 6.1. Let X be a normed space. Then the following statements are equivalent:(1) X is finite dimensional.(2) X has the Heine-Borel property.(3) The closed unit ball BX is compact.(4) The unit sphere SX is compact.

Proof. Suppose that dimX = n. Then there is a linear homeomorphism T : X → Fn byTheorem 4.7. Let A be any closed and bounded subset of X. Clearly T (A) is then alsoclosed in Fn. Moreover, since A is bounded we have A ⊆ tT−1 (B(0, 1)) = T−1(B(0, t))for some t > 0. Hence T (A) ⊆ B(0, t), so T (A) is bounded in Fn. It follows that T (A)is compact in Fn by the Heine-Borel Theorem and therefore A must be compact in X.This proves that (1) implies (2). That (2) implies (3) is trivial and the implication(3) ⇒ (4) follows from the fact that SX is a closed subset of BX .

In order to prove the implication (4) ⇒ (1), we show that SX fails to be compactwhenever X is infinite dimensional. Since compactness implies sequential compactnessin metrizable spaces, it suffices to show that there is a sequence (xn) in SX with no con-vergent subsequence. Specifically, we use induction to construct a linearly independentsequence (xn) with ‖xn‖ = 1 for every n ∈ N and ‖xn − xk‖ ≥ 1 whenever n 6= k. Tobegin the induction, choose an arbitrary x1 ∈ SX . Now assume that m linearly indepen-dent elements x1, . . . , xm ∈ SX have been selected such that ‖xn − xk‖ ≥ 1 whenevern 6= k. Then Em := span x1, . . . , xm ( X is a subspace of X with dimEm = m. ByTheorem 4.7, the finite dimensionality of Em implies completeness. Hence Em is closedsince every complete subspace of a metric space is closed. Fix ym ∈ X \ Em and letYm := span ym, Em. Then Ym is a normed space of dimension m + 1. Since Em is aclosed subspace of Ym and ym ∈ Ym \Em, it follows from Corollary A.3 that there existsa bounded linear functional f ∈ Y ∗

m such that f(ym) = d(ym, Em) > 0, Em ⊆ ker f ,and ‖f‖ = 1. Because Ym is finite dimensional, the unit sphere SYm

of Ym is compactby the implication (1) ⇒ (4) proved above. Hence it follows from the Extreme ValueTheorem (Theorem 2.1) that ‖f‖ = sup |f(y)| : y ∈ SYm

is attained. Consequently, wecan choose xm+1 ∈ SYm

⊆ SX such that |f(xm+1)| = ‖f‖ = 1. Note that xm+1 /∈ Em

since |f(xm+1)| = 1 and Em ⊆ ker f . Hence xm+1 is linearly independent of x1, . . . , xm.Finally, observe that for every n < m+ 1 we have xn ∈ Em ⊆ ker f and therefore

‖xm+1 − xn‖ ≥ |f(xm+1 + xn)|‖f‖

= |f(xm+1) + f(xn)| = |f(xm+1)| = 1.

This completes the induction and thus finishes the proof.

One of the main motivations for the introduction of the weak and weak* topologiesin Section 5 was the greater abundance of compact sets. Hence we may hope to recoupthe Heine-Borel property in these topologies. For the dual of a Banach space, this isexactly what happens in the weak* topology.

Theorem 6.2. Let X∗ be the dual of a Banach space. Then (X∗,Tw∗) has the Heine-Borel property in the sense that every weakly* closed and weakly* bounded subset isweakly* compact.


Proof. First note that the closed unit ball BX∗ is weak* compact by the Banach-AlaogluTheorem (Theorem A.6). Since (X∗,Tw∗) is a TVS by Theorem 5.16, the maps x 7→ λxand x 7→ λ−1x are both continuous with respect to the weak* topology. Hence BX∗ ishomeomorphic to tBX∗ for every t > 0 and therefore tBX∗ is weakly* compact for everyt > 0. Now suppose that A ⊆ X∗ is weakly* closed and weakly* bounded. BecauseX is Banach, Theorem 5.21 ensures that A is also norm bounded. Thus, there existst > 0 such that A ⊆ tBX∗ ⊆ tBX∗ since BX∗ is an open neighborhood of 0 in thenorm topology. But then A is a weakly* closed subset of the weakly* compact set tBX∗ ,whereby A itself is weakly* compact.

In the weak topology on normed spaces, it turns out that we need the further restric-tion of reflexivity in order to have the Heine-Borel property. This is proved in Theorem6.4 as a corollary of the next result.Theorem 6.3. A normed space X is reflexive if and only if the closed unit ball BX isweakly compact.Proof. Suppose that X is reflexive. Then J(X) = X∗∗ by definition of reflexivity. Since‖J(x)‖X∗∗ = ‖x‖X . by Theorem 5.13 we thus have J(BX) = BX∗∗ . Now recall thatBX∗∗ is weakly* compact in X∗∗ by the Banach-Alaoglu Theorem (Theorem A.6). Itfollows that BX is weakly compact in X if we can show that J is a homeomorphism withrespect to the weak toplogy σ(X,X∗) on X and the weak* topology σ(X∗∗, JX∗(X∗))on X∗∗. To this end, let (xα) be an arbitrary net in X, let x ∈ X and observe that

xαw−→ x in X ⇔ x∗(xα) → x∗(x) in F for every x∗ ∈ X∗ [by Theorem 5.11]

⇔ (J(xα)) (x∗) → (J(x)) (x∗) in F for every x∗ ∈ X∗

⇔ J(xα)w∗

−−→ J(x) in X∗∗ [by Theorem 5.17].Hence J : X → X∗∗ is weak-weak* continuous and J−1 : X∗∗ → X is weak*-weakcontinuous. That is, J is a homeomorphism and therefore BX is weakly compact.

Conversely, assume that BX is weakly compact. Analogously to the above, J is ahomeomorphism from X in the weak toplogy onto J(X) as a subspace of X∗∗ in theweak* topology. Hence J(BX) is weakly* compact in X∗∗. In particular, J(BX) isweakly* closed in X∗∗ since the weak* topology is Hausdorff by Theorem 5.16. Further-more, the fact that J is an isometry by Theorem 5.13, implies that J(BX) ⊆ BX∗∗ . Now,according to Goldstine’s Theorem (Theorem A.5), J(BX) is weakly* dense in BX∗∗ .But J(BX) is closed in the weak* subspace topology on BX∗∗ by the preceding remarks.Hence J(BX) = BX∗∗ . We claim that this implies surjectivity of J : X → X∗∗. To seethis, let x∗∗0 ∈ X∗∗ be arbitrary and notice that x∗∗0 / ‖x∗∗0 ‖ ∈ BX∗∗ . Thus, it followsfrom J(BX) = BX∗∗ that there exists x0 ∈ BX such that J(x0) = x∗∗0 / ‖x∗∗0 ‖. But thenJ (‖x∗∗0 ‖x0) = ‖x∗∗0 ‖ J(x0) = x∗∗0 , which proves that J is surjective. Consequently, X isreflexive.

Theorem 6.4. Let X be a normed space and suppose that X is reflexive. Then (X,Tw)has the Heine-Borel property in the sense that every weakly closed and weakly boundedsubset is weakly compact.Proof. Assume that X is reflexive. Then BX is weakly compact by Theorem 6.3. SinceX is a TVS by Theorem 5.9, it follows that BX and tBX are homeomorphic for anyt > 0. In particular, tBX is weakly compact for every t > 0. Now suppose that A ⊆ Xis weakly closed and weakly bounded. Then A is also norm bounded by Theorem 5.19.Since the open unit ball BX is an open neighborhood of 0 in the norm topology, itfollows that there exists t > 0 such that A ⊆ tBX ⊆ tBX . But then A is a weakly closedsubset of the weakly compact set tBX . This immediately implies that A itself is weaklycompact.

In fact, the converse of Theorem 6.4 is also true. To see this, simply note that BX isweakly closed by Mazur’s Theorem. Hence the Heine-Borel property would imply thatBX is compact and therefore X would be reflexive by Theorem 6.2. It follows that theweak analogue of the Heine-Borel property actually characterizes reflexivity.


6.1. The Eberlein-Šmulian Theorem

Another striking aspect of the weak topology is its similarity to metric spaces. Whilethe weak topology is not metrizable when X is infinite dimensional, it nevertheless sharessome surprisingly nice properties with metric spaces. One of the most beautiful results inthis respect is undoubtedly the Eberlein-Šmulian Theorem, which we state here withoutproof.

Theorem 6.5. (The Eberlein-Šmulian Theorem) Suppose A is a subset of a normedspace X. Then the following statements are equivalent:

(1) The set A is weakly compact.(2) The set A is weakly countably compact.(3) The set A is weakly limit point compact.(4) The set A is weakly sequentially compact.

The same is true if “weakly” is replaced by “relatively weakly”.

Proof. See Theorem 2.8.6 in Megginson (1998).

Recall that compact metric spaces are always complete. By means of the Eberlein-Šmulian Theorem we obtain the same result for the weak topology.

Theorem 6.6. Every weakly compact subset of a normed space is complete.

Proof. Let X be a normed space and suppose that A is a weakly compact subset ofX. We need to show that every cauchy sequence in A is convergent. To this end, fixan arbitrary cauchy sequence (xn) in A. Observe that A is weakly limit point compactby the Eberlein-Šmulian Theorem. Hence the infinite subset xn : n ∈ N in A has aweak limit point x ∈ A. Now fix an arbitrary x∗ ∈ X∗ and let ε > 0 be given. Since(xn) is cauchy, there exists N ∈ N such that ‖xn − xm‖ < ε/ (2 ‖x∗‖) whenever n,m ≥N . Hence |x∗(xn)− x∗(xm)| ≤ ‖x∗‖ ‖xn − xm‖ < ε/2 for all n,m ≥ N . Moreover,according to Theorem 5.2, y ∈ X : |x∗(y − x)| < ε/2 is a basic neighborhood of x inthe weak topology. Hence there exist k ≥ N such that xk ∈ y ∈ X : |x∗(y − x)| < ε/2.That is, |x∗(xk)− x∗(x)| < ε/2 for some k ≥ N . It follows that

|x∗(xn)− x∗(x)| ≤ |x∗(xn)− x∗(xk)|+ |x∗(xk)− x∗(x)| < ε

2+ε

2= ε.

Hence x∗(xn) converges to x∗(x) in F. Since x∗ ∈ X∗ was arbitrary, we have x∗(xn) →x∗(x) in F for every x∗ ∈ X∗. Thus, we conclude that xn

w−→ x in X by Theorem 5.11.We can, however, do even better than this. As such, we claim that xn → x in norm.

To see this, simply observe that if we choose N such that ‖xn − xm‖ < ε/2 for alln,m > N in the above, then the inequality |x∗(xn)− x∗(x)| < ε holds for all x∗ ∈ X∗

with ‖x∗‖ ≤ 1 and n > N . Letting J : X → X∗∗ denote the canonical embedding, wethus have

‖J(xn − x)‖X∗∗ = sup|x∗(xn − x)| : x∗ ∈ BX∗

< ε

for all n > N . Hence it follows from Theorem 5.13 that ‖xn − x‖ = ‖J(xn − x)‖X∗∗ < εfor all n > N . Thus xn converges to x in norm, which proves that every cauchy sequencein A is convergent. Consequently, A is complete.

We finish the section with an important theorem that relates weak compactness toseparability. Specifically, we show that weak compactness is completely determined bythe separable closed subspaces of the given normed space. This result will prove itselfimmensely helpful to our plan of attack in Section 7.

Theorem 6.7. Let A be a subset of a normed space X. Then A is weakly compact ifand only if A ∩ Y is weakly compact in Y whenever Y is a separable closed subspace ofX.

Proof. Assume A is weakly compact and let Y be any closed subspace of X. Then Yis also weakly closed in X by Theorem 5.24, so A ∩ Y is a weakly closed subset of theweakly compact set A in X. Hence A∩Y is weakly compact in X and therefore compactin Y with respect to the subspace topology. By Theorem 5.10, the subspace topology


on Y induced by σ(X,X∗) is the same as the weak topology σ(Y, Y ∗) on Y . Thus A∩Yis weakly compact in Y .

Conversely, suppose that A ∩ Y is weakly compact in Y for every closed separablesubspace Y of X. Let (xn) be an arbitrary sequence in A. Then span (xn : n ∈ N)is a closed and separable subspace of X. Indeed, the closed linear span restrictedto the subfield Q if F = R, or Q [i] if F = C, is a countable dense subset. HenceA ∩ span (xn : n ∈ N) is weakly compact in span (xn : n ∈ N) and therefore weaklysequentially compact in span (xn : n ∈ N) by the Eberlein-Šmulian Theorem. It fol-lows that (xn) ⊆ A ∩ span (xn : n ∈ N) has a weakly convergent subsequence (withrespect to the weak topology on span (xn : n ∈ N), that is) whose limit belongs toA ∩ span (xn : n ∈ N). In particular, the limit must belong to A. By Theorem 5.10,the subsequence is also convergent with respect to the weak topology on X. Since (xn)was arbitrary, it follows that A is weakly sequentially compact in X and therefore weaklycompact in X by the Eberlein-Šmulian Theorem.

It should be noted that the first three equivalences in the relative version of theEberlein-Šmulian Theorem in fact follow from James’ Weak Compactness Theorem. Tosee this, it suffices to show that relative weak countable compactness implies relativeweak limit point compactness since the remaining implications then follow easily fromstandard results in topology. As such, suppose that A is relatively weakly countablycompact as a subset of a normed space X. Then A

w is weakly countably compactand hence it is easily seen that x∗(Aw

) is also weakly countably compact in F for anyx∗ ∈ X∗ by weak continuity of x∗. By metrizability of F we thus have that x∗(Aw

)

is weakly compact in F. But then supx∗(x) : x ∈ Aw is attained for any x∗ ∈ X∗.

Since Aw is weakly closed, this implies that Aw is weakly compact in X by James’ WeakCompactness Theorem. In conclusion, A is relatively weakly compact.

The previous paragraph can in some sense be summarized as saying that James’ WeakCompactness Thereom “almost” implies the Eberlein-Šmulian Theorem. This lends ussome reassurance that the Eberlein-Šmulian Theorem is not too big a result to utilize inour proof of James’ Weak Compactness Theorem (as we do in Theorem 7.11 via Theorem6.7). We leave it to the reader to consider if more can be said about the relationshipbetween these two grand theorems.


7. James’ Weak Compactness Theorem

As emphasized in the introduction, Section 7 is truly the heart and soul of thisexposition. For the next many pages we shall dig deeper and deeper into the proofof James’ Weak Compactness Theorem. In doing so, we shall follow the lead of R.C.James’ simplified proof from 1972 as expounded in Megginson (1998).

Our strategy is to first prove a version of James’ Weak Compactness Theorem forseparable subsets of the closed unit ball. This is done in Theorem 7.5 but requires atechnical lemma in the form of Lemma 7.4. After this, we prove what is essentially astronger version of Lemma 7.4, namely Lemma 7.10. Together with Theorem 7.5 (andTheorem 6.7 above) this allows us to lose the separability requirement, thereby provinga version of James’ Weak Compactness Theorem for real Banach spaces and balancedsubsets of the closed unit ball in Theorem 7.11. Finally, the full version of the theoremis proved in Theorem 7.12 with the help of Theorem 7.11.

We begin by proving the so-called Helly’s Theorem, which plays an important role inTheorem 7.5 below. In order to prove Helly’s Theorem, we need two well-known lemmas.The first lemma is the following standard corollary of the Hahn-Banach Theorem, whichwe state here without proof.

Lemma 7.1. Let Y be a closed subspace of a normed space X. If x ∈ X \ Y then thereexists a bounded linear functional f : X → F such that f(x) = d(x, Y ) with Y ⊆ ker fand ‖f‖ = 1.

Proof. See Corollary A.3 in Appendix A.

The second lemma that we need is the following result from linear algebra.

Lemma 7.2. Let f1, . . . , fn be a finite family of linear functionals on a vector spaceX. If f is another linear functional on X, then f ∈ span (f1, . . . , fn) if and only if⋂n

j=1 ker fj ⊆ ker f .

Proof. Suppose f ∈ span (f1, . . . , fn). Then f =∑n

j=1 λjfj so if fj(x) = 0 for allj = 1, . . . , n we also have f(x) = 0. Hence

⋂ni=1 ker fi ⊆ ker f .

For the converse, assume that⋂n

i=1 ker fi ⊆ ker f . Define a linear operator T : X →Fn by T (x) := (f1(x), . . . , fn(x)). The linearity of T follows immedaitely from thelinearity of each fj . Moreover, T (x) = 0 if and only if fj(x) = 0 for each j = 1, . . . , n, sokerT =

⋂nj=1 ker fj . Now define a linear functional S : T (X) → F by S(T (x)) := f(x),

where T (X) ⊆ Fn. Linearity follows from

S(αT (x) + βT (y)

)= S

(T (αx+ βy)

)= f(αx+ βy).

= αf(x) + βf(y) = αS(T (x)

)+ βS

(T (y)

)To see that S is well-defined, notice that T (x) = T (y) implies f(x) = f(y). Indeed,T (x) − T (y) = 0 implies fj(x − y) = fj(x) − fj(y) = 0 for each j = 1, . . . , n so thatx− y ∈

⋂ni=1 ker fi ⊆ ker f and hence f(x)− f(y) = f(x− y) = 0.

Observe that S can be extendend to a linear functional S on all of Fn such thatS = S

∣∣T (X). To see this, first note that T (X) is a linear subspace of Fn by the linearity

of T . Now fix a basis v1, . . . , vk for T (X) and extend it to a basis v1, . . . , vk, wk+1, . . . , wn

for Fn. Then the linear functional S : Fn → F defined by

S(α1v1 + · · ·+ αkvk + αk+1wk+1 + · · ·+ αnwn) := S(α1v1 + · · ·+ αkvk).

clearly does the job. Finally, let e1, . . . , en be the standard basis for Fn and notice that

f(x) = S((f1(x), . . . , fn(x)

))= S

( n∑j=1

fj(x)ej

)=

n∑j=1

S(ej)fj(x).

for all x ∈ X. Setting λj := S(ej), we have f =∑n

j=1 λjfj , so f ∈ span (f1, . . . , fn).This finishes the proof.


Theorem 7.3. (Helly’s Theorem) Let X be a normed space. Let f1, . . . , fn be afinite family of bounded linear functionals on X and let λ1, . . . , λn be a correspondingcollection of scalars. Then the following two statements are equivalent:

(1) There is an x0 ∈ X such that fi(x0) = λi for each i = 1, . . . , n(2) There is a nonnegative real number M ∈ R+ such that

|α1λ1 + · · ·+ αnλn| ≤M ‖α1f1 + · · ·+ αnfn‖for each linear combination α1f1 + · · ·+ αnfn in span (f1, . . . , fn).

If (2) is satisfied, then given any ε > 0 we can choose x0 in (1) such that ‖x0‖ ≤M +ε.

Proof. Suppose statement (1) is true. Then there exists x0 ∈ X such that fi(x0) = λifor each i = 1, . . . , n. Hence an arbitrary element α1f1+· · ·+αnfn in span (f1, . . . , fn)obeys the following inequality:

|α1λ1 + · · ·+ αnλn| = |α1f1(x0) + · · ·+ αnfn(x0)|= |(α1f1 + · · ·+ αnfn) (x0)|≤ ‖x0‖ ‖α1f1 + · · ·+ αnfn‖ .

In turn, statement (2) is satisfied with M := ‖x0‖. Note also that ‖x0‖ ≤M + ε for anyε > 0.

For the converse, assume that statement (2) is true. If λ1 = · · · = λn = 0 then 0 ∈ Xis the desired x0. Hence we may assume that at least one λj is nonzero. In particular,0 < |λj | ≤ M ‖fj‖ by (2), so at least one of the linear functionals fj are nonzero. Byrearranging the indices if necessary, we can thus assume that f1, . . . , fm is a maximallinearly independent subcollection of f1, . . . , fn. It follows that for each k = m+1, . . . , nthere are scarlars α1,k, . . . , αm,k such that fk =

∑mj=1 αj,kfj .

Suppose we could prove the existence of an x0 ∈ X such that fk(x0) = λk for eachof the linearly independent f1, . . . , fm and such that ‖x0‖ ≤M + ε. Then the fact thatfk =

∑mj=1 αj,kfj for each k = m+1, . . . , n and the fact that the inequality in (2) holds

for the full collection f1, . . . , fn would imply that

|fk(x0)− λk| =∣∣∣∣ m∑j=1

αj,kλj + (−1)λk

∣∣∣∣ ≤M

∥∥∥∥ m∑j=1

αj,kfj + (−1)fk

∥∥∥∥ =M ‖fk − fk‖ = 0

for each k = m+ 1, . . . , n. Hence fk(x0) = λk would also hold for k = m+ 1, . . . , n. Inconclusion, we need only find x0 ∈ X such that fk(x0) = λk for each k = 1, . . . ,m and‖x0‖ ≤M + ε for any ε > 0.

Define a linear operator T : X → Fm by T (x) := (f1(x), . . . , fm(x)). The linearity ofT is an immediate consequence of the linearity of each fj . If m = 1, then T is triviallya surjection. We claim that the same is true when m ≥ 2. By Lemma 7.2, the linearindependence of f1, . . . , fm implies that

⋂j 6=k ker fj * ker fk for each k = 1, . . . ,m. It

follows that for each k = 1, . . . ,m, we can choose yk ∈⋂

j 6=k ker fj \ ker fk such thatfk(yk) = 1 and fj(yk) = 0 whenever j 6= k. In turn, T (y1) = e1, . . . , T (ym) = em, whereei denotes the i’th standard basis vector of Fm. We conclude that T is a surjection fromX onto Fm.

By the surjectivity of T , there exists y0 ∈ X such that (f1(y0), . . . , fm(y0)) = T (y0) =(λ1, . . . , λm). As at least one λj is nonzero, we have y0 /∈

⋂mj=1 ker fj . Therefore,

Lemma 7.1 ensures the existence of a bounded linear functional f on X such thatf(y0) = d

(y0,⋂m

j=1 ker fj

),⋂m

j=1 ker fj ⊆ ker f , and ‖f‖ = 1. Appealing to Lemma 7.2once more, it follows from

⋂mj=1 ker fj ⊆ ker f that f ∈ span (f1, . . . , fm). In other

words, there are scalars β1, . . . , βm such that f =∑m

j=1 βjfj . We deduce that

d(y0,

m⋂j=1

ker fj

)= f(y0) =

m∑j=1

βjfj(y0) =

m∑j=1

βjλj

≤∣∣∣∣ m∑j=1

βjλj

∣∣∣∣ ≤M

∥∥∥∥ m∑j=1

βjfj

∥∥∥∥ =M ‖f‖ =M.


Thus, for any ε > 0, there exists z0 ∈⋂m

j=1 ker fj with ‖y0 − z0‖ ≤M+ε as the distancewould otherwise be larger than M . Since z0 ∈

⋂mj=1 ker fj we have fk(y0−z0) = fk(y0) =

λk for each k = 1, . . . ,m. Setting x0 := y0 − z0 completes the proof.

7.1. The Separable Case

In Theorem 7.5 below we shall prove that if every x∗ ∈ X∗ attains its supremumon a separable and weakly closed subset A of BX , then that subset is weakly compact.To prove this, we proceed by contraposition. Hence, we shall need a bounded linearfunctional z∗ ∈ X∗ that does not attain its supremum.

The point of Lemma 7.4 is precisely to ensure the existence of such a linear functional,in the shape of

∑∞j=1 βjy

∗j , under just the right circumstances. That these circumstances

are in fact satisfied when A is not weakly compact and that z∗ :=∑∞

j=1 βjy∗j really does

not attain its supremum are then the topics of Theorem 7.5.

Lemma 7.4. Let X be a normed space. Let 0 < θ < 1 be given and suppose that(i) A is a nonempty subset of the closed unit ball BX .(ii) (βn)n∈N is a sequence of strictly positive reals with

∑n βn = 1.

(iii) (x∗n)n∈N is a sequence in BX∗ such that sup |x∗(x)| : x ∈ A ≥ θ for everyx∗ ∈ co (x∗n : n ∈ N).

Then there exists a real number α with θ ≤ α ≤ 1 and a sequence (y∗n)n∈N in BX∗ suchthat

(1) y∗n ∈ co(x∗j : j ≥ n

)⊆ BX∗ for all n ∈ N.

(2) sup∣∣∣∑∞

j=1 βjy∗j (x)

∣∣∣ : x ∈ A= α.

(3) sup∣∣∣∑n

j=1 βjy∗j (x)

∣∣∣ : x ∈ A< α(1− θ

∑∞j=n+1 βj) for every n ∈ N.

Proof. The proof proceeds in the following manner: First, we employ an inductive argu-ment to construct a sequence (y∗n) satisfying (1) along with a sequence (αn) whose limitwe claim to be the desired α. Then we prove that the inequality θ ≤ α ≤ 1 is indeedsatisfied by this limit. Finally, we show that these choices of (y∗n) and α satisfy (2) and(3). For the sake of clarity, we split the proof into four steps.

Step 1. First of all, we introduce the seminorm x∗ A := sup |x∗(x)| : x ∈ A definedfor every x∗ ∈ X∗. Observe that the supremum is always finite since

x∗ A = sup |x∗(x)| : x ∈ A ≤ sup|x∗(x)| : x ∈ BX

= ‖x∗‖ <∞,

where we have used that A ⊆ BX by (i) and ‖x∗‖ <∞ for every x∗ ∈ X∗ by definitionof X∗. To see that · A : X∗ → R+ indeed defines a seminorm on X∗, we observe that

λx∗ A = sup |λx∗(x)| : x ∈ A = |λ| sup |x∗(x)| : x ∈ A = |λ| x∗ A

andx∗ + y∗ A = sup |(x∗ + y∗)(x)| : x ∈ A = sup |x∗(x) + y∗(x)| : x ∈ A

≤ sup |x∗(x)|+ |y∗(x)| : x ∈ A≤ sup |x∗(x)| : x ∈ A+ sup |y∗(x)| : x ∈ A= x∗ A + y∗ A

Step 2. Because the sequence (βn) in (ii) sums to 1, we can choose a sequence ofpositive reals (εn)n∈N converging to 0 such that

0 <∞∑k=1

βkεk∑∞j=k+1 βj

∑∞j=k βj

< 1− θ.

To see this, note that εk can be chosen smaller than (1− θ)∑∞

j=k+1 βj∑∞

j=k βj > 0.Step 3. We are now ready for the construction of the sequences (y∗n) and (αn). Our

plan is to use induction to prove the existence of a sequence (y∗n)n∈N in BX∗ such thatfor each n ∈ N we have

y∗n ∈ co(x∗j : j ≥ n

)⊆ BX∗ (7.1)


andn−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n

A

< αn(1 + εn) (7.2)

where

αn := inf

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗

A

: y∗ ∈ co(x∗j : j ≥ n

). (7.3)

Note that (7.3) is well-defined since X∗ is a vector space and x∗ A < ∞ for everyx∗ ∈ X∗.

Step 3.1. We begin the induction by noticing that

α1 = infy∗ A : y∗ ∈ co

(x∗j : j ≥ 1

)≥ θ > 0

by the assumption in (iii). Hence there exists a y∗1 in co(x∗j : j ≥ 1

)so that

y∗1 A < α1(1 + ε1).

Indeed, α1(1 + ε1) would otherwise be a larger lower bound than α1. This means that(7.1) and (7.2) are satisfied for n = 1.

Step 3.2. Now, let n ≥ 2 be given and suppose that y∗1 , . . . , y∗n−1 have been selectedin accordance with (7.1) and (7.2). Observe that if y∗ ∈ co

(x∗j : j ≥ n

), then

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗ =

n−2∑j=1

βjy∗j + βn−1y

∗n−1 +

( ∞∑j=n

βj

)y∗

=

n−2∑j=1

βjy∗j +

( ∞∑j=n−1

βj

)(βn−1∑∞j=n−1 βj

y∗n−1 +

∑∞j=n βj∑∞

j=n−1 βjy∗),

whereβn−1∑∞j=n−1 βj

y∗n−1 +

∑∞j=n βj∑∞

j=n−1 βjy∗ ∈ co

(x∗j : j ≥ n− 1

)since it is a convex combination of the elements y∗n−1 and y∗ in co

(x∗j : j ≥ n− 1

).

But then the set, over which αn is the infimum, is contained in the set, over which αn−1

is the infimum (see (7.3)). Hence αn−1 ≤ αn. Since α1 > 0 by Step 3, it follows thatαn > 0. Therefore, there exists y∗n in co

(x∗j : j ≥ n

)satisfying (7.2) by definition of

the infimum. This completes the induction.Step 4. For each n ∈ N, the set from (7.3), over which αn is the infimum, turns out

to be bounded above by 1. To see this, recall from Step 1 that · A is a seminorm withx∗ A ≤ ‖x∗‖. Hence it follows from y∗j , y

∗ ∈ BX∗ and∑

n βn = 1 thatn−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗

A

≤n−1∑j=1

βj y∗j A +

( ∞∑j=n

βj

)y∗ A

≤n−1∑j=1

βj∥∥y∗j∥∥+ ( ∞∑

j=n

βj

)‖y∗‖ ≤

n−1∑j=1

βj +

∞∑j=n

βj = 1.

By the result in Step 3.2 that αn−1 ≤ αn for each n ∈ N, we thus have

0 < θ ≤ α1 ≤ · · · ≤ αn ≤ · · · ≤ 1.

It follows that (αn) is increasing and bounded from above, so the sequence converges toits supremum. The limit α thus exists and satisfies θ ≤ α ≤ 1 as well as αn ≤ α for alln ∈ N.

Step 4.1. By construction of (αn) we have

αn ≤n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n

A

≤ αn(1 + εn)


for every n ∈ N. Recall from Step 2 that (εn) was chosen such that εn → 0 for n→ ∞.Letting n tend to infinity, it thus follows from the squeeze theorem that

α = limn→∞

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n

A

.

In order to evaluate this limit, we first observe that · A is Lipschitz continuous. Beinga seminorn, it satisfies x∗ A = x∗−y∗+y∗ A ≤ x∗−y∗ A+ y∗ A, which in turn impliesthat x∗ A − y∗ A ≤ x∗ − y∗ A. Analogously, we have y∗ A − x∗ A ≤ y∗ − x∗ A =x∗ − y∗ A. The Lipschitz continuity then follows from∣∣ x∗ A − y∗ A

∣∣ ≤ x∗ − y∗ A ≤ ‖x∗ − y∗‖ ,

where we have used that x∗ A ≤ ‖x∗‖ for every x∗ ∈ X∗ by Step 1. Secondly, we notethat

∑∞j=1 βjy

∗j is convergent in the norm on X∗. To see this, recall that X∗ is complete

and notice that the sequence of partial sums(∑n

j=1 βjy∗j

)is cauchy. Indeed, if n > m,

then ∥∥∥∥ n∑j=1

βjy∗j −

m∑j=1

βjy∗j

∥∥∥∥ ≤n∑

j=m+1

∥∥βjy∗j∥∥ =

n∑j=m+1

|βj |∥∥y∗j∥∥ ≤

n∑j=m+1

βj ,

which converges to 0 as m tends to infinity. Likewise,(∑∞

j=n βj)y∗n → 0 for n → ∞

since ‖y∗n‖ ≤ 1. It follows from the continuity of · A and the preceding remarks aboutconvergence that

α = limn→∞

n−1∑j=1

βjy∗j + lim

n→∞

( ∞∑j=n

βj

)y∗n

A

=

∞∑j=1

βjy∗j

A

.

This is precisely the statement in (2).Step 4.2. It only remains to show that (y∗n) and α, as constructed above, satisfy the

inequality in (3). To this end, fix an arbitrary n ∈ N. Observe that if n ≥ 2, then

n∑j=1

βjy∗j

A

=

n−1∑j=1

βjy∗j + βny

∗n

A

=βn∑∞j=n βj

( n−1∑j=1

βjy∗j

)+

∑∞j=n+1 βj∑∞j=n βj

( n−1∑j=1

βjy∗j

)+ βny

∗n

A

=βn∑∞j=n βj

( n−1∑j=1

βjy∗j

)+


( n−1∑j=1

βjy∗j

)+ βn

∑∞j=n βj∑∞j=n βj

y∗nA

=βn∑∞j=n βj

( n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n

)+


( n−1∑j=1

βjy∗j

)A

≤ βn∑∞j=n βj

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n

A

+


n−1∑j=1

βjy∗j

A

<βn∑∞j=n βj

αn (1 + εn) +


n−1∑j=1

βjy∗j

A

(7.4)

=

( ∞∑j=n+1

βj

)(βnαn (1 + εn)∑∞j=n+1 βj

∑∞j=n βj

+

∑n−1j=1 βjy

∗j A∑∞

j=n βj

),

where the strict inequality in (7.4) follows from our construction of (y∗n) as indicated informula (7.2) on p. 27.


It follows, by replacing n with n− k in the previous result, that if n− k ≥ 2, then

∑n−kj=1 βjy

∗j A∑∞

j=n−k+1 βj<

(βn−kαn−k (1 + εn−k)∑∞j=n−k+1 βj

∑∞j=n−k βj

+

∑n−k−1j=1 βjy

∗j A∑∞

j=n−k βj

)

Hence it holds for any n ≥ 2 that

n∑j=1

βjy∗j

A

<

( ∞∑j=n+1

βj


∑∞j=n βj

+

∑n−1j=1 βjy

∗j A∑∞

j=n βj

)

<

( ∞∑j=n+1

βj


∑∞j=n βj

+βn−1αn−1 (1 + εn−1)∑∞

j=n−1 βj∑∞

j=n−1 βj+

∑n−2j=1 βjy

∗j A∑∞

j=n−1 βj

)

<

( ∞∑j=n+1

βj


∑∞j=n βj

+βn−1αn−1 (1 + εn−1)∑∞

j=n βj∑∞

j=n−1 βj

+βn−2αn−2 (1 + εn−2)∑∞

j=n−1 βj∑∞

j=n−2 βj+

∑n−3j=1 βjy

∗j A∑∞

j=n−2 βj

)

< · · ·

<

( ∞∑j=n+1

βj

)( n∑k=2

(βkαk (1 + εk)∑∞j=k+1 βj

∑∞j=k βj

)+

β1y∗1 A∑∞

j=2 βj

)

<

( ∞∑j=n+1

βj

)( n∑k=2


∑∞j=k βj

)+

β1α1 (1 + ε1)∑∞j=2 βj

∑∞j=1 βj

)

=

( ∞∑j=n+1

βj

) n∑k=1


∑∞j=k βj

).

For n = 1 the above inequality reduces to

β1y∗1 A <

( ∞∑j=2

βj

)β1α1 (1 + ε1)∑∞j=2 βj

∑∞j=1 βj

= β1α1 (1 + ε1) ,

which is also true since β1y∗1 A = β1 y

∗1 A < β1α1 (1 + ε1) by (7.2). Now, recall from

Step 3.1 that αn ≤ α for all n ∈ N and recall from Step 2 that

0 <βkεk∑∞

j=k+1 βj∑∞

j=k βj< 1− θ.


We conclude that for every n ∈ N it holds thatn∑

j=1

βjy∗j

A

<

( ∞∑j=n+1

βj

) n∑k=1


∑∞j=k βj

)

≤ α

( ∞∑j=n+1

βj

) n∑k=1

(βk∑∞

j=k+1 βj∑∞

j=k βj+


∑∞j=k βj

)

< α

( ∞∑j=n+1

βj

) n∑k=1

(∑∞j=k βj −

∑∞j=k+1 βj∑∞

j=k+1 βj∑∞

j=k βj+ (1− θ)

)

= α

( ∞∑j=n+1

βj

)( n∑k=1

(1∑∞

j=k+1 βj− 1∑∞

j=k βj

)+ (1− θ)

)

= α

( ∞∑j=n+1

βj

)((1∑∞

j=n+1 βj− 1∑∞

j=1 βj

)+ (1− θ)

)

= α

( ∞∑j=n+1

βj

)(1∑∞

j=n+1 βj− 1

)+ (1− θ)

)

= α

( ∞∑j=n+1

βj

)(1∑∞

j=n+1 βj− θ

)

= α

(1− θ

∞∑j=n+1

βj

).

This is precisely the statement in (3). Hence the proof is complete.

As already explained, Lemma 7.4 plays an integral role in the proof of the nextthereom. Specifically, we shall prove that when a subset is not weakly compact, thenthe assumptions in Lemma 7.4 are fulfilled, and hence we get a linear functional of theform z∗ =

∑∞j=1 βjy

∗j , which fails to attain its supremum.

Strictly speaking, we only need the implication (1) ⇒ (2) from Theorem 7.5 for theensuing results in Section 7.2. Nevertheless, we prove all four equivalences. First of all,the extra work involved is negligible and the proof provides a valuable guide for howto proceed in the non-separable case. Secondly, it is well worth the effort to have acomplete proof of James’ Weak Compactness Thereom for separable subsets.

Theorem 7.5. Let X be a Banach space. Suppose that A is a nonempty, separable,weakly closed subset of the closed unit ball BX . Then the following statements areequivalent:

(1) The set A is not weakly compact.(2) There is a θ ∈ R with 0 < θ < 1 and a sequence (x∗n) in BX∗ such that

limn→∞ x∗n(x) = 0 for each x ∈ A and sup∣∣x∗(x)∣∣ : x ∈ A

≥ θ for every

x∗ ∈ co (x∗n : n ∈ N).(3) There is a θ ∈ R with 0 < θ < 1 such that if (βn) is a sequence of strictly positive

reals with∑

n βn = 1, then there is an α ∈ R with θ ≤ α ≤ 1 and a sequence(y∗n) in BX∗ such that(a) limn→∞ y∗n(x) = 0 for each x ∈ A.(b) sup

∣∣∣∑∞j=1 βjy

∗j (x)

∣∣∣ : x ∈ A= α.

(c) sup∣∣∣∑n

j=1 βjy∗j (x)

∣∣∣ : x ∈ A< α(1− θ

∑∞j=n+1 βj) ∀n ∈ N.

(4) There is a z∗ in X∗ such that sup∣∣z∗(x)∣∣ : x ∈ A

is not attained.

Proof. The plan for the proof is to show that each statement implies the one immediatelyafter it and then close the chain by showing that (4) implies (1). The hard part is proving(1) ⇒ (2). The implication (2) ⇒ (3) is also deep, but its proof is an easy consequenceof Lemma 7.4. The two remaining implications require little work.


(1) ⇒ (2): Assume A is not weakly compact. Let V := span(A) and notice that thisis again a normed space, so in particular it has a dual space V ∗. Now, let W denotethe vector space underlying V ∗ and equip W with the norm ‖·‖W : W → R+ given by‖v∗‖W := sup |v∗(x)| : x ∈ A instead of the usual operator norm. That ‖v∗‖W is aseminorm follows by the exact same arguments as in Step 1 of the proof of Lemma 7.4.To see that it is a norm, we need to show that if ‖v∗‖W = 0, then v∗ is the zero elementin W . So suppose ‖v∗‖W = 0, which means that v∗(x) = 0 for every x ∈ A, and observethat this implies v∗(x) = 0 for every x ∈ span(A). Since A is a subset of the normedspace X, we have span(A) = span(A), and hence v∗(x) = 0 for every x ∈ span(A) = Vby the continuity of v∗. Thus v∗ is the zero functional on V , which is precisely the zeroelement of W . We shall refer to the normed space (W, ‖v∗‖W ) simply as W .

Now, let f : A→W ∗ be the canonical map from A into W ∗ defined by (f(x)) (v∗) =v∗(x) for x ∈ A and v∗ ∈ W . That is, f(x) acts on v∗ ∈ W by evaluation in x. Thatf(x) is linear is immediate from the linearity of v∗. Let ‖·‖W∗ denote the operator normon W ∗. To see that f(x) is also bounded on W , and hence belongs to W ∗ as required,notice that |(f(x)) (v∗)| = |v∗(x)| ≤ sup |v∗(x)| : x ∈ A = ‖v∗‖W ≤ 1 for v∗ ∈ BW .Therefore,

‖f(x)‖W∗ = sup|(f(x))(v∗)| : v∗ ∈ BW

≤ 1

for every x ∈ A, so f(x) :W → F is bounded on W for each f(x) ∈ f(A). In particular,we have f(A) ⊆ BW∗ .

By Corollary A.4, V ∗ is a separating family of functions on V . It follows that ifx 6= y then there exists v∗x,y ∈ V ∗ such that v∗x,y(x) 6= v∗x,y(y) and hence (f(x)) (v∗x,y) 6=(f(y)) (v∗x,y), so f(x) 6= f(y). In conclusion, f is injective and therefore bijective ontof(A). In fact, f is a homeomorphism onto its image when A is considered as a subspaceof X in the weak topology and f(A) is considered as a subpace of W ∗ in the weak*topology. Since V = span(A) is a linear subspace of X with A ⊆ V , it follows fromTheorem 5.10 that the topology which A inherits as a subspace of (X,σ(X,X∗)), isequivalent to the topology that it inherits as a subspace of (V, σ(V, V ∗)). This impliesthat if (xα) is a net in A ⊆ V and x ∈ A ⊆ X, then

xαw−→ x in (V, σ(V, V ∗)) ⇔ v∗(xα) → v∗(x) in F for all v∗ ∈W [by Theorem 5.11]

⇔ (f(xα)) (v∗) → (f(x)) (v∗) in F for all v∗ ∈W

⇔ f(xα)w∗

−−→ f(x) in (W ∗, σ(W ∗,W )) [by Theorem 5.17].

Hence f : A → f(A) and f−1 : f(A) → A are both continuous with respect to thesetopologies. We conclude that A, as a subspace of X in the weak topology, is homeomor-phic to f(A) as a subspace of W ∗ in the weak* topology.

Recall that, by assumption, A is not weakly compact in X. Hence f(A) is not weakly*compact in W ∗ by the above homeomorphism result. Since f(A) ⊆ BW∗ and BW∗ isweakly* compact in W ∗ by the Banach-Alaoglu Theorem (Theorem A.6), it follows thatf(A) cannot be weakly* closed in W ∗. Thus f(A)

w∗

\ f(A) is nonempty. Now fix anelement F ∈ f(A)

w∗

\ f(A) ⊆W ∗.Observe that no v ∈ V satisfies F (v∗) = v∗(v) for every v∗ ∈W . To prove this claim,

suppose for contradcition that such a v0 ∈ V exists. By defintion of F ∈ f(A)w∗

thereis a net (f(xα)) in f(A) such that f(xα)

w∗

−−→ F . Hence v∗(xα) = (f(xα))(v∗) → F (v∗)

in F for every v∗ ∈W by Theorem 5.17. Then F (v∗) = v∗(v0) for every v∗ ∈W impliesthat v∗(xα) → F (v∗) = v∗(v0) in F for every v∗ ∈W . By Theorem 5.11 this implies thatxα

w−→ v0 in X, from which it follows that v0 ∈ Aw. On the other hand, the fact that

F /∈ f(A) forces v0 /∈ A. Indeed, v0 ∈ A would imply that F (v∗) = v∗(v0) = (f(v0))(v∗)

for all v∗, which in turn would imply that F = f(v0) ∈ f(A). We conclude thatv0 ∈ A

w \A, which contradicts the fact that A is weakly closed.Notice that if F = 0, then 0 ∈ V satisfies F (v∗) = 0 = v∗(0) for every v∗ ∈W , which

violates the above result. Hence F 6= 0 and in particular ‖F‖W∗ > 0.


Moreover, it follows from A ⊆ BX ∩ V = BV that

BV ∗ = v∗ ∈W : ‖v∗‖V ∗ ≤ 1 =v∗ ∈W : sup

|v∗(x)| : x ∈ BV

≤ 1

⊆ v∗ ∈W : sup |v∗(x)| : x ∈ A ≤ 1 = v∗ ∈W : ‖v∗‖W ≤ 1 = BW

so that

‖F‖V ∗∗ = sup|F (v∗)| : v∗ ∈ BV ∗

≤ sup

|F (v∗)| : v∗ ∈ BW

= ‖F‖W∗ <∞.

Thus we have F ∈ V ∗∗ and ‖F‖V ∗∗ ≤ ‖F‖W∗ .Let JV : V → V ∗∗ be the canonical embedding from V into V ∗∗. Note that V is itself

a Banach space as a closed (linear) subspace of the Banach space X. Hence JV (V ) isa closed subspace of V ∗∗ by Theorem 5.13. Since no v ∈ V satisfies F (v∗) = v∗(v) forevery v∗ ∈W we have F 6= JV (v) for each v ∈ V and hence F /∈ JV (V ). It follows thatdV ∗∗(F, JV (V ) > 0.

Now choose a real number ∆ such that

0 < ∆ < dV ∗∗ (F, JV (V )) .

By the separability of A, we can also choose a countable dense subset an : n ∈ N inA. Let n ∈ N be given and consider any set of n+1 scalars α1, . . . , αn+1. Then we have

∣∣∣∣α1∆+

n+1∑j=2

αj0

∣∣∣∣ = |α1|∆ ≤ |α1|∆

∥∥∥F − JV (−∑n+1

j=2αj

α1aj−1)

∥∥∥V ∗∗

dV ∗∗(F, JV (V ))

=∆

dV ∗∗(F, JV (V ))

∥∥∥∥α1F +

n+1∑j=2

αjJV (aj−1)

∥∥∥∥V ∗∗

.

Hence the collection of bounded linear functionals F, JV (a1), . . . , JV (an) on V ∗ and thecorresponding collection of scalars ∆, 0, . . . , 0 satisfy statement (2) in Helly’s Theoremwith M = ∆

dV ∗∗ (F,JV (V )) > 0. Helly’s Theorem thus assures the existence of a v∗0 ∈ V ∗

such that F (v∗0) = ∆, (JV (a1)) (v∗0) = 0, . . . , (JV (an)) (v

∗0) = 0 and such that

‖v∗0‖V ∗ ≤ ∆

dV ∗∗(F, JV (V ))+dV ∗∗(F, JV (V ))−∆

2 · dV ∗∗(F, JV (V ))=dV ∗∗(F, JV (V )) + ∆

2 · dV ∗∗(F, JV (V ))< 1.

Since n ∈ N was arbitrary, it follows that for each n ∈ N there exists v∗n ∈ V ∗ such that(i) ‖v∗n‖V ∗ < 1(ii) F (v∗n) = ∆(iii) v∗n(aj) = (JV (aj)) (v

∗n) = 0 if n ≥ j

For each n ∈ N, let x∗n ∈ X∗ be the Hahn-Banach extension of v∗n : V → F to all of Xwith ‖x∗n‖X = ‖v∗n‖V < 1 by (i). It is immediate from (iii) that limn→∞ x∗n(aj) = 0 forevery aj ∈ an : n ∈ N. Let ε > 0 be given. If a ∈ A \ an : n ∈ N then the density ofan : n ∈ N in A implies that there exists ajε ∈ an : n ∈ N such that ‖a− ajε‖X < ε.By limn→∞ x∗n(ajε) = 0 there is an Nε ∈ N such that |x∗n(ajε)| < ε for all n > Nε.Together with ‖x∗n‖X < 1 for all n ∈ N, the previous remarks imply that

|x∗n(a)| ≤ |x∗n(a− ajε)|+ |x∗n(ajε)| ≤ ‖x∗n‖X∗ ‖a− ajε‖X + |x∗n(ajε)|≤ sup

n∈N(‖x∗n‖X∗) ‖a− ajε‖X + |x∗n(ajε)| < 1 · ‖a− ajε‖X + |x∗n(ajε)|

< ε+ |x∗n(ajε)| < 2ε

for all n > Nε. Hence limn→∞ x∗n(a) = 0. In summary, we have thus constructed asequence (x∗n) in BX∗ with limn→∞ x∗n(x) = 0 for each x ∈ A.

It remains to show that there exists 0 < θ < 1 such that sup |x∗(x)| : x ∈ A ≥ θwhenever x∗ ∈ co (x∗n : n ∈ N). Suppose x∗ ∈ co (x∗n : n ∈ N) and let v∗ be therestriction of x∗ to V . Then it follows from (ii) that

∆ = F (v∗) ≤ ‖F‖W∗ ‖v∗‖W = ‖F‖W∗ sup |v∗(x)| : x ∈ A= ‖F‖W∗ sup |x∗(x)| : x ∈ A .


Letting θ := ∆/ ‖F‖W∗ we thus have sup |x∗(x)| : x ∈ A ≥ θ. Finally, observe that0 < ∆ = F (v∗1) ≤ ‖F‖V ∗∗ ‖v∗1‖V ∗ < ‖F‖V ∗∗ ≤ ‖F‖W∗ ,

so that 0 < θ < 1 as desired.(2) ⇒ (3): Assume statement (2) is true. That is, assume there exists θ ∈ (0, 1)

and a sequence (x∗n) in BX∗ such that limn→∞ x∗n(x) = 0 for each x ∈ A and such thatsup

∣∣x∗(x)∣∣ : x ∈ A≥ θ for every x∗ ∈ co (x∗n : n ∈ N). Recall that A is a nonempty

subset of BX and let (βn) be any sequence of positive reals with∑∞

n=1 βn = 1. ThenLemma 7.4 ensures the existence of an α with θ ≤ α ≤ 1 and a sequence (y∗n) in BX∗

such that parts (b) and (c) of statement (3) are satisfied.Now, Lemma 7.4 also ensures that y∗n ∈ co

(x∗j : j ≥ n

)for each n ∈ N. Together

with the assumption that limn→∞ x∗n(x) = 0 for all x ∈ A, this is sufficient for (y∗n) toalso satisfy part (a) of statement 3. To see this, fix an arbitrary x ∈ A and let ε > 0 begiven. Then limn→∞ x∗n(x) = 0 implies the existence of an Nε ∈ N such that

∣∣x∗j (x)∣∣ < ε

for all j > Nε. Observe that for each n > Nε we have y∗n =∑kn

i=1 ti,nx∗ji,n

for somex∗j1,n, . . . , x

∗jk,n

∈x∗j : j ≥ n > Nε

, where t1,n, . . . , tk,n ≥ 0 and

∑kn

i=1 ti,n = 1. Hence

|y∗n(x)| =∣∣∣∣ kn∑i=1

ti,nx∗ji,n(x)

∣∣∣∣ ≤ kn∑i=1

ti,n∣∣x∗ji,n(x)∣∣ < kn∑

i=1

ti,nε = ε

for all n > Nε. It follows that limn→∞ y∗n(x) = 0 for each x ∈ A as desired.(3) ⇒ (4): Fix a sequence (βn) of positive reals with

∑∞n=1 βn = 1 and let θ, α, and

(y∗n) have the properties listed in statement (3). Recall from Lemma 7.4 (Step 4.1) that∑∞j=1 βjy

∗j is convergent, whence z∗ :=

∑∞j=1 βjy

∗j is a well-defined element of X∗. We

proceed to show that this particular choice of z∗ ∈ X∗ satisfies statement (4) inasmuchas sup |z∗(x)| : x ∈ A is not attained.

Fix an arbitrary x0 ∈ A. Since 0 < θ ≤ α we have αθ > 0. By part (a) of statement(3), we have limn→∞ y∗n(x0) = 0. Thus there exists n ∈ N such that

∣∣y∗j (x0)∣∣ < αθ forall j > n. Together with parts (b) and (c) of statement (3), this in turn implies that

|z∗(x0)| =

∣∣∣∣ ∞∑j=1

βjy∗j (x0)

∣∣∣∣≤

∣∣∣∣ n∑j=1

βjy∗j (x0)

∣∣∣∣+ ∞∑j=n+1

βj∣∣y∗j (x0)∣∣

≤ sup

∣∣∣∣ n∑j=1

βjy∗j (x)

∣∣∣∣ : x ∈ A

+

∞∑j=n+1

βj∣∣y∗j (x0)∣∣

< sup

∣∣∣∣ n∑j=1

βjy∗j (x)

∣∣∣∣ : x ∈ A

+ αθ

∞∑j=n+1

βj

< α

(1− θ

∞∑j=n+1

βj

)+ αθ

∞∑j=n+1

βj (7.5)

= α

= sup |z∗(x)| : x ∈ A . (7.6)The inequality in (7.5) is precisely (c) and the equality in (7.6) is precisely (b). Sincex0 ∈ A was arbitrary, we conclude that sup |z∗(x)| : x ∈ A is not attained at anyx ∈ A. This completes the proof of (3) ⇒ (4).

(4) ⇒ (1): Assume there is a z∗ ∈ X∗ such that sup |z∗(x)| : x ∈ A is not attained.We need to show that this implies A is not weakly compact. To this end, supposefor contradiction that A is weakly compact. Then it follows from the Extreme ValueTheorem (Theorem 2.1) that every weakly continuous scalar-valued function attains thesupremum of its absolute value on A. Now recall that all elements of X∗ are continuouswith respect to the weak topology by its very definition. Hence sup |x∗(x)| : x ∈ A isattained for every x∗ ∈ X∗.


This, however, clearly contradicts the existence of a z∗ ∈ X∗ for which the supremumis not attained. Consequently, A cannot be weakly compact.

Observe that the equivalence of (1) and (4) in Theorem 7.5 is precisely the statementof James’ Weak Compactness Theorem for separable subsets of a Banach space. To seethis, one simply needs to realize that whenever the subset A is weakly compact or thesupremum of each linear functional is attained on A, then A is automatically bounded,and thus there is no loss of generality in assuming that A is contained in the closed unitball (see Theorem 7.12 for the details).

7.2. Extension to the Non-Separable Case

The goal of the present section is to obtain new versions of Lemma 7.4 and Theorem7.5, which do not require the subset A to be separable, but instead only require it to bebalanced. Thankfully, we will be able to reuse a lot of the results from Section 7.1.

Before endeavoring to remove the separability requirement, however, we begin bymaking a simple, yet clever, observation. As the next lemma shows, we lose nothingin terms of the topology by passing to the real Banach space XR induced by a Banachspace X (recall that XR was defined in Definition 1.5).

Lemma 7.6. Let X be a Banach space and let XR be the real Banach space obtainedby restricting multiplication by scalars to R×X. Then the weak topology σ

(XR, (XR)

∗)on XR is equivalent to the weak topology σ(X,X∗) on X.

Proof. If F = R, then (XR, ‖·‖) = (X, ‖·‖), so the statement is trivial. Hence we mayassume that F = C. Let (xα)α∈A be a net in X and let x ∈ X. We prove both inclusions.

Suppose xα → x in(XR, σ

(XR, (XR)

∗)). By construction, every u∗ ∈ (XR)∗ is

σ(XR, (XR)

∗) continuous, so u∗(xα) → u∗(x) in C. For each x∗ ∈ X∗ we have x∗(x) =u∗(x) − iu∗(ix) with u∗ = Re(x∗) ∈ (XR)

∗ by Theorem 1.6. Moreover, x 7→ u∗(ix)is clearly also a bounded real-linear functional and hence continuous with respect toσ(XR, (XR)

∗). Hence u∗(ixα) → u∗(ix) in C. It follows that

x∗(xα) = u∗(xα)− iu∗(ixα) → u∗(x)− iu∗(ix) = x∗(x)

in C for all x∗ ∈ X∗. We conclude that every x∗ ∈ X∗ is continuous with respect toσ(XR, (XR)

∗). By Theorem 5.1, σ (X,X∗) is the unique weakest topology for whichevery member of X∗ is continuous, whence σ (X,X∗) ⊆ σ

(XR, (XR)

∗).Conversely, suppose xα → x in (X,σ (X,X∗)). Fix an arbitrary u∗ ∈ (XR)

∗ and recallfrom Theorem 1.6 that there is a unique bounded complex-linear functional x∗ ∈ X∗

such that u∗ = Re(x∗). Now, by construction of σ (X,X∗) we have x∗(xα) → x∗(x) inC. Hence (x∗(xα))α∈A is a net in C converging to x∗(x), which means that for all r > 0there is an αr ∈ A such that x∗(xα) ∈ B(x∗(x), r) for every α αr in A. Moreover, wehave

|u∗(xα)− u∗(x)| = |u∗(xα − x)| = |Re(x∗(xα − x))|≤ |x∗(xα − x)| = |x∗(xα)− x∗(x)| ,

which implies that u∗(xα) ∈ B(u∗(x), r) whenever x∗(xα) ∈ B(x∗(x), r). It followsthat given r > 0, u∗(xα) ∈ B(u∗(x), r) for every α αr in A. Thus the net (u∗(xα))converges to u∗(x) in C, so u∗ is continuous with respect to σ (X,X∗). By appealing toTheorem 5.1 again, we conclude that σ

(XR, (XR)

∗) ⊆ σ (X,X∗).

Lemma 7.6 allows us to restrict attention to XR when working with the weak topologyon a normed space X. Therefore, we shall only consider real Banach spaces in theremainder of Section 7.2. By means of Lemma 7.6, our results will then be seamlesslytranslated to the case of complex Banach spaces in the final proof of James’ WeakCompactness Theorem in Section 7.3. Before we can get any further, however, we needto build some extra machinery.


Definition 7.7. Let X be a normed space over R. Suppose that (x∗n)n∈N is a boundedsequence in X∗. Then we define

L(x∗n) :=x∗ ∈ X∗ | x∗(x) ≤ lim sup

n→∞x∗n(x) for all x ∈ X

and

V(x∗n) :=(y∗n)n∈N ⊆ X∗ | y∗n ∈ co

(x∗j : j ≥ n

)for each n ∈ N

.

We emphasize that the field in question is R. Hence lim supn x∗n(x) is well-defined.

In Lemma 7.8 below, we list a series of useful properties for the sets in Defintion 7.7. Inparticular, we show in statement (6) of this lemma that L(x∗n) in fact satisfies

L(x∗n) =x∗ ∈ X∗ | lim inf

n→∞x∗n(x) ≤ x∗(x) ≤ lim sup

n→∞x∗n(x) for all x ∈ X

.

This suggests that L(x∗n) is in some sense a set of potential limit points for (x∗n). WhileV(x∗n) is primarily introduced for notational convenience, the set L(x∗n) is really atthe heart of our strategy to lose the separability requirement. We will have more to sayabout this right after the statement of Lemma 7.10. For now, we shall prove two lemmasabout the properties of L(x∗n) and V(x∗n), which will make our lives considerably easierin the proofs of Lemma 7.10 and Theorem 7.11.

Lemma 7.8. Let L(x∗n) and V(x∗n) be as in Definition 7.7 above. Then the followingstatements are true:

(1) Every bounded sequence (x∗n) in X∗ satisfies (x∗n) ∈ V(x∗n).(2) If (y∗n) ∈ V(x∗n), then (y∗n) ⊆ B(0, sup ‖x∗n‖ : n ∈ N).(3) If (y∗n) ∈ V(x∗n), then V(y∗n) ⊆ V(x∗n).(4) If (y∗n) ∈ V(x∗n), then L(y∗n) ⊆ L(x∗n).(5) If x∗ ∈ L(x∗n), then x∗ ∈ B(0, sup ‖x∗n‖ : n ∈ N).(6) If x∗ ∈ L(x∗n), then lim infn x

∗n(x) ≤ x∗(x) for all x ∈ X.

Proof. The first statement, (1), follows immediately from x∗n ∈ co(x∗j : j ≥ n

). The

second statement, (2), follows from the fact that y∗n ∈ co(x∗j : j ≥ n

)implies

‖y∗n‖ ≤ sup∥∥x∗j∥∥ : j ≥ n

≤ sup

∥∥x∗j∥∥ : j ∈ N

for each n ∈ N. Hence y∗n : n ∈ N is contained the closed ball with center 0 and radiussup ‖x∗n‖ : n ∈ N.

To see that statement (3) is true, fix a sequence (y∗n) ∈ V(x∗n) and notice thatco(x∗j : j ≥ N

)⊆ co

(x∗j : j ≥ n

)whenever N ≥ n. This implies that y∗N ∈ co

(x∗j : j ≥ n

)for every N ≥ n, so that

co(y∗j : j ≥ N

)⊆ co

(x∗j : j ≥ n

). Thus (z∗n) ∈ V (y∗n) implies

z∗n ∈ co(y∗j : j ≥ n

)⊆ co

(x∗j : j ≥ n

)for each n ∈ N, so that (z∗n) ∈ V (x∗n). This proves the desired inclusion V (y∗n) ⊆ V (x∗n).

For statement (4) we need to show that L (y∗n) ⊆ L (x∗n) whenever (y∗n) ∈ V(x∗n). Tothis end, fix a sequence (y∗n) ∈ V(x∗n) and suppose z∗ ∈ L(y∗n). Then by definition ofz∗ ∈ L(y∗n) we have

z∗(x) ≤ lim supn→∞

y∗n(x) = infn∈N

(supj≥n

y∗j (x))

for all x ∈ X. Now, for each n ∈ N it follows from (y∗n) ∈ V(x∗n) that y∗j ∈ co (x∗i : i ≥ n)for all j ≥ n. Hence y∗j (x) ≤ supi≥n x

∗i (x) for all j ≥ n and therefore supj≥n y

∗j (x) ≤

supi≥n x∗i (x). We conclude that

z∗(x) ≤ lim supn→∞

y∗n(x) = infn∈N

(supj≥n

y∗j (x))≤ inf

n∈N

(supj≥n

x∗j (x))= lim sup

n→∞x∗n(x)

for all x ∈ X. That is, z∗ ∈ L(x∗n). This proves L(y∗n) ⊆ L(x∗n).We now prove statement (5). First observe that, for each x ∈ X, x∗ ∈ L(x∗n) satisfies

x∗(x) ≤ lim supn→∞

x∗n(x) ≤ supn∈N

x∗n(x) ≤ supn∈N

‖x∗n‖X∗ ‖x‖X .


In particular, we also have

−x∗(x) = x∗(−x) ≤ supn∈N

‖x∗n‖X∗ ‖−x‖X = supn∈N

‖x∗n‖X∗ ‖x‖X

for all x ∈ X. Therefore, |x∗(x)| ≤ supn∈N ‖x∗n‖X∗ ‖x‖X for each x ∈ X. It follows that

‖x∗‖X∗ = sup|x∗(x)| : x ∈ BX

≤ sup

n∈N‖x∗n‖X∗ .

Finally, statement (6) follows easily from lim infn∈N αn = − lim supn∈N −αn. Indeed,if x∗ ∈ L(x∗n) then x∗(x) ≤ lim supn x

∗n(x) for all x ∈ X and therefore −x∗(x) =

x∗(−x) ≤ lim supn x∗n(−x) so that − lim supn x

∗n(−x) ≤ x∗(x). Hence

lim infn→∞

x∗n(x) = − lim supn→∞

−x∗n(x) = − lim supn→∞

x∗n(−x) ≤ x∗(x)

for all x ∈ X as desired.

In what follows, we need L(x∗n) to be nonempty. Luckily, the following lemma assuresus that this is always the case, as long as the sequence in question is bounded (which isprecisely what we required in the definition).

Lemma 7.9. Let X be a normed space over R. Suppose (x∗n) is a bounded sequence inX∗. Then L(x∗n) is nonempty.

Proof. Define a sublinear functional p : X → R on X by p(x) := lim supn→∞ x∗n(x).The sublinearity follows immediately from the linearity of x∗n together with the well-known properties of the limit superior. Now consider the trivial subspace 0 of X andlet z∗ denote the zero functional on 0. Then z∗(0) = 0 = p(0), so z∗(x) ≤ p(x)for every x ∈ 0. Hence the vector space version of the Hahn-Banach Theorem(Theorem A.1) yields the existence of linear (but not necessarily continuous) exten-sion x∗ such that x∗(x) ≤ p(x) for every x ∈ X. It follows that x∗(x) ≤ p(x) =lim supn x

∗n(x) ≤ supn∈N x

∗n(x) ≤ supn∈N ‖x∗n‖X∗ ‖x‖X for every x ∈ X and analogously

−x∗(x) = x∗(−x) ≤ supn∈N ‖x∗n‖X∗ ‖x‖X . Hence

|x∗(x)| ≤ supn∈N

‖x∗n‖X∗ ‖x‖X

for every x ∈ X, which together with the boundedness of (x∗n) implies that

‖x∗‖X∗ = sup|x∗(x)| : x ∈ BX

≤ sup

n∈N‖x∗n‖X∗ <∞.

Thus x∗ : X → R is bounded, so x∗ ∈ X∗ as desired. As we have already seen thatx∗(x) ≤ lim supn x

∗n(x) for all x ∈ X, it follows that x∗ ∈ L(x∗n).

We are now ready to embark upon the key technical lemma of this section, in theshape of Lemma 7.10. Before actually proving it, however, we shall take some time totruly appreciate the motivation behind our introduction of L(·). Hopefully, this will alsoprovide the reader with a firmer grasp of the overall strategy underlying this section.

Lemma 7.10. Let X be a normed space over R. Let 0 < θ < 1 be given and supposethat

(i) A is a nonempty balanced subset of the closed unit ball BX .(ii) (βn)n∈N is a sequence of strictly positive reals with

∑n βn = 1.

(iii) (x∗n)n∈N is a sequence in BX∗ such that sup |(x∗ − w∗) (x)| : x ∈ A ≥ θ forevery x∗ ∈ co (x∗n : n ∈ N) and w∗ ∈ L(x∗n).

Then there exists a real number α with θ ≤ α ≤ 2 and a sequence (y∗n)n∈N in BX∗ suchthat whenever w∗ ∈ L(y∗n) then it holds that

(1) sup∣∣∣∑∞

j=1 βj(y∗j − w∗) (x)∣∣∣ : x ∈ A

= α.

(2) sup∣∣∣∑n

j=1 βj(y∗j − w∗) (x)∣∣∣ : x ∈ A

< α(1− θ

∑∞j=n+1 βj) for n ∈ N.


The key observation here is that in the setting of Theorem 7.5 we had x∗n → 0 and y∗n → 0in the above (with no mention of w∗ and L(·), of course). The fact that x∗n → 0, however,was obtained directly from the separability of A (see the second to last paragraph of“(1) ⇒ (2)” in the proof of Theorem 7.5) and the fact that y∗n → 0 in turn followedfrom x∗n → 0 (see the last paragraph of “(2) ⇒ (3)” in the proof of Theorem 7.5). Thus,we have no reason to believe that (x∗n) and (y∗n) should converge to 0 now that we areno longer assuming A to be separable, and hence we will not be able to utilize this inTheorem 7.11.

For this reason, L(x∗n) and L(y∗n) are introduced precisely such that (x∗n − w∗) and(y∗n − w∗) are in a certain sense close to converging to 0. For example, if x∗n → 0 as inthe separable case, then w∗ ∈ L(x∗n) implies w∗ = 0, so x∗n −w∗ = x∗n → 0. Similarly, ify∗n → 0, then we also get y∗n − w∗ = y∗n → 0 for any w∗ ∈ L(y∗n). This should make itclear, in what ways Lemma 7.10 is simply a generalization of Lemma 7.4.

In Theorem 7.5, y∗n → 0 was used to squeeze the tail of∑∞

j=1 βjy∗j less than some

desired number (see the second paragraph of (3) ⇒ (4) in the proof of Theorem 7.5).Obviously, we don’t have quite the same level of control over the tail of

∑∞j=1 βj(y

∗j −w∗)

when (y∗n) is not convergent to 0. Nonetheless, as we shall see later in Theorem 7.11,w∗ ∈ L(y∗n) allows us to find certain terms of y∗n − w∗ that are as small as we like, andthis turns out to be sufficient for our purposes.

For now, however, it is time to return to the proof of Lemma 7.10, which is the onethat follows below.

Proof. As in Lemma 7.4, we introduce the seminorm x∗ A := sup |x∗(x)| : x ∈ Adefined for all x∗ ∈ X∗. Recall that this was shown to be continuous and satisfyx∗ A ≤ ‖x∗‖X∗ for every x∗ ∈ X∗. Precisely as in Step 2 of Lemma 7.4, we also fix a

convergent sequence (εn) of positive reals with limn→∞ εn = 0 such that

0 <

∞∑k=1


∑∞j=k βj

< 1− θ.

The remaining part of the proof proceeds in four steps.Step 1. We wish to construct a sequence of scalars (αj)j∈N in F, a sequence of linear

functionals (y∗j )j∈N in X∗, and two collections of sequences (0x∗j ), (1x

∗j ), (2x

∗j ), . . . and

(1z∗j ), (2z

∗j ), (3z

∗j ), . . . in X∗. By means of induction we show that these sequences can

be chosen such that(0x

∗j

)is in BX∗ and such that, for each n ∈ N, we have

(I) y∗n, (nz∗j ), and (nx∗j ) lie in BX∗

(II) (nz∗j ) ∈ V(n−1x

∗j )

(III) (nx∗j ) is a subsequence of (nz∗j )

(IV) y∗n ∈ co(

n−1x∗j : j ≥ n

)(V) θ ≤ αn ≤ 2

(VI) αn−1 ≤ αn if n ≥ 2

Before beginning the induction, first choose (0x∗j ) := (x∗j ), where (x∗j ) is the sequence

from assumption (iii). Then the condition that(0x

∗j

)should lie in BX∗ is satisfied.

In Step 2 below, we proceed to show that the above conditions are all satisfied forthe base case n = 1. The inductive step is then completed in Step 3.

Step 2. Let y∗ ∈ co(

0x∗j : j ≥ 1

)and (v∗j ) ∈ V(0x

∗j ). Then (0x

∗j ) ⊆ BX∗ implies

y∗ ∈ BX∗ . Also, by definition of (v∗j ) ∈ V(0x∗j ), we have v∗n ∈ co

(0x

∗j : j ≥ n

)⊆ BX∗

for each n ∈ N, so (v∗j ) ⊆ BX∗ . In particular, we thus have L(v∗j ) ⊆ BX∗ by statement(5) in Lemma 7.8. It follows that if w∗ ∈ L(v∗j ), then

y∗ − w∗A ≤ y∗ A + w∗

A ≤ ‖y∗‖+ ‖w∗‖ ≤ 2.

Moreover, (v∗j ) ∈ V(0x∗j ) implies that L(v∗j ) ⊆ L(0x

∗j ) by statement (4) of Lemma 7.8,

so if w∗ ∈ L(v∗j ), then w∗ ∈ L(0x∗j ). Thus, assumption (iii) ensures that

y∗ − w∗A = sup |(x∗ − w∗) (x)| : x ∈ A ≥ θ.


We conclude that the set

S1

(y∗, (v∗j )

):=y∗ − w∗

A : w∗ ∈ L(v∗j )

is bounded below by θ and above by 2. Also, L(v∗j ) is nonempty according to Lemma7.9, so S1

(y∗, (v∗j )

)is nonempty. In particular,

α1 := infsupS1

(y∗, (v∗j )

): y∗ ∈ co

(0x

∗j : j ≥ 1

), (v∗j ) ∈ V(0x

∗j )

is a well-defined number between θ and 2. Thus condition (V) is satisfied for n = 1.Step 2.1. Given our of choice of α1, and the definition of the infimum, we can choose

y∗1 from co(

0x∗j : j ≥ 1

)and (1z

∗j ) from V(0x

∗j ) such that

α1 ≤ supy∗1 − w∗

A : w∗ ∈ L(1z∗j )< α1(1 + ε1). (7.7)

This takes care of conditions (II) and (IV). Also, y∗1 ∈ BX∗ and (1z∗j ) ⊆ BX∗ by the

initial remarks in Step 1. All that remains is to choose an appropriate subsequence(1x

∗j ) of (1z∗j ) in BX∗ . Then conditions (I) and (II) would also be satisfied, which would

conclude the base case. To this end, notice that because of equation (7.7), and thedefinition of the supremum, we can choose w∗

1 from L(1z∗j ) such that

α1(1− ε1) < y∗1 − w∗1 A = sup |y∗1(x)− w∗

1(x)| : x ∈ A . (7.8)Since A is balanced by assumption (i), we have −A = A, so the absolute value inequation (7.8) can be dropped. Hence it follows from equation (7.8), and the defintionof the supremum, that there exists x1 ∈ A such that

α1(1− ε1) < y∗1(x1)− w∗1(x1). (7.9)

Clearly,∣∣lim infj

(1z

∗j (x1)

)∣∣ ≤ 1 by the fact that (1z∗j ) is contained in BX∗ and x1 ∈ A ⊆BX . Now choose (1x

∗j ) to be a subsequence of (1z

∗j ) such that (1x

∗j (x1)) converges to

lim infj(1z

∗j (x1)

). Then α1, y

∗1 , (1z

∗j ) and (1x

∗j ) fulfills all the requirements for the base

case of n = 1.Step 3. In Step 2 we chose α1, y

∗1 , (1z

∗j ) and (1x

∗j ) in a very particular way. We now

show that αn, y∗n, (nz

∗j ) and (nx

∗j ) can be chosen according to the same pattern for all

n ∈ N. The argument is based on strong induction and proceeds through a series ofintermediate claims.

Induction Hypothesis. Let m ≥ 2 be given. Then we assume that αn, y∗n, (nz

∗j ),

and (nx∗j ) have already been chosen such that they satisfy conditions (I) through (VI)

for each n = 1, . . . ,m−1. The particular way, in which they are chosen, will be detailedas we move along.

Step 3.1. We claim that if (v∗j ) ∈ V(m−1x∗j ), then (v∗j ) is contained in BX∗ . It suf-

fices to show that (m−1x∗j ) is contained in BX∗ since then v∗n ∈ co

(m−1x

∗j : j ≥ n

)⊆

BX∗ for each n ∈ N. That (m−1x∗j ) ⊆ BX∗ follows immediately from our induction

hypothesis with regard to condition (I).Step 3.2. With a view towards the construction of (αj), we claim that if y∗ ∈

co(

m−1x∗j : j ≥ m

)and (v∗j ) ∈ V(m−1x

∗j ), then

Sm

(y∗, (v∗j )

):=

m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)y∗ − w∗

A

: w∗ ∈ L(v∗j )

defines a nonempty subset of the closed interval [0, 2] ⊆ R for each pair y∗ and (v∗j ). Inparticular, the scalar

αm := infsupSm

(y∗, (v∗j )

): y∗ ∈ co

(m−1x

∗j : j ≥ m

), (v∗j ) ∈ V(m−1x

∗j )

is then a well-defined number between 0 and 2. To see that the claim is indeed true, fix anarbitrary y∗ ∈ co

(m−1x

∗j : j ≥ m

)and an arbitrary (v∗j ) ∈ V(m−1x

∗j ). Since (m−1x

∗j )

is in BX∗ by the induction hypothesis, we have y∗ ∈ co(

m−1x∗j : j ≥ m

)⊆ BX∗ . In

addition, the induction hypothesis also ensures that y∗1 , . . . , y∗m−1 ∈ BX∗ .Furthermore, (v∗j ) ∈ V(m−1x

∗j ) implies (v∗j ) ⊆ BX∗ by the claim in Step 3.1, so

L(v∗j ) ⊆ BX∗ by statement (5) of Lemma 7.8.


It follows from the above that

0 ≤m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)y∗ − w∗

A

≤m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)y∗

A

+ w∗A

≤ 1 + 1 = 2

for every w∗ ∈ L(v∗j ) since L(v∗j ) ⊆ BX∗ and∑m−1

j=1 βjy∗j +

(∑∞j=m βj

)y∗ belongs to

BX∗ as a convex combination of elements in BX∗ . Recalling that L(v∗j ) is nonempty byLemma 7.9, the claim follows.

Step 3.3. We claim that V(m−1x∗j ) ⊆ V(m−2x

∗j ) for every m ≥ 2. By Lemma 7.8(3)

it suffices to show that (m−1x∗n) ∈ V(m−2x

∗j ). To see that this is the case, simply observe

that if m ≥ 2, then (m−1x∗j ) is a subsequence of (m−1z

∗j ) by the induction hypothesis wih

regard to (III). Since (m−1z∗j ) ∈ V(m−2x

∗j ) by the induction hypotheses in relation to

(II), we thus have m−1x∗n ∈ co

(m−2x

∗j : j ≥ n

)for each n ∈ N. This, in turn, implies

that (m−1x∗n) itself belongs to V(m−2x

∗j ).

Step 3.4. We claim that if m ≥ 2 and x∗ ∈ co(

m−1x∗j : j ≥ m

), then

βm−1∑∞j=m−1 βj

y∗m−1 +

∑∞j=m βj∑∞

j=m−1 βjx∗ ∈ co

(m−2x

∗j : j ≥ m− 1

). (7.10)

To see that this is true, we first observe that (m−1x∗j ) ∈ V(m−1x

∗j ) ⊆ V(m−2x

∗j ) by

Lemma 7.8(1) and the claim in Step 3.3. Therefore, m−1x∗n ∈ co

(m−2x

∗j : j ≥ n

)for

each n ∈ N. In particular, m−1x∗k ∈ co

(m−2x

∗j : j ≥ m− 1

)for all k ≥ m. This

implies that

x∗ ∈ co (m−1x∗k : k ≥ m) ⊆ co

(m−2x

∗j : j ≥ m− 1

).

In addition to this, y∗m−1 ∈ co(

m−2x∗j : j ≥ m− 1

)by the induction hypothesis with

respect to (IV). Since every convex combination of elements in co(

m−2x∗j : j ≥ m− 1

)is again in co

(m−2x

∗j : j ≥ m− 1

), equation (7.10) is true.

Step 3.5. We claim that αm−1 ≤ αm as long as m ≥ 2. To see this, first recallthe definition of Sn

(y∗, (v∗j )

)from Step 3.2 and notice that when n = 1, this definition

agrees with the definition of S1

(y∗, (v∗j )

)in Step 2. Now observe that

Sm−1

(βm−1∑∞j=m−1 βj

y∗m−1 +

∑∞j=m βj∑∞

j=m−1 βjx∗, (v∗j )

)

=

m−2∑j=1

βjy∗j +

( ∞∑j=m−1

βj

)(βm−1∑∞m−1 βj

y∗m−1 +

∑∞m βj∑∞

m−1 βjx∗)− w∗

A

: w∗ ∈ L(v∗j )

=

m−2∑j=1

βjy∗j + βm−1y

∗m−1 +

( ∞∑j=m

βj

)x∗ − w∗

A

: w∗ ∈ L(v∗j )

=

m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)x∗ − w∗

A

: w∗ ∈ L(v∗j )

= Sm(x∗, (v∗j )).

Recall that the infimum over a subset is at least as large as the infimum over theoriginal set. Hence it follows from the claims in Steps 3.3 and 3.4, together with the


above computation, that

αm−1 = inf

supSm−1

(y∗, (v∗j )

): y∗ ∈ co

(m−2x

∗j : j ≥ m− 1

), (v∗j ) ∈ V(m−2x

∗j )

≤ inf

supSm−1

(y∗, (v∗j )

): y∗ ∈

βm−1∑∞j=m−1 βj

y∗m−1 +

∑∞j=m βj∑∞

j=m−1 βjx∗ :

x∗ ∈ co(

m−1x∗j : j ≥ m

), (v∗j ) ∈ V(m−1x

∗j )

= inf

supSm−1

(βm−1∑∞j=m−1 βj

y∗m−1 +

∑∞j=m βj∑∞

j=m−1 βjx∗, (v∗j )

):

x∗ ∈ co(

m−1x∗j : j ≥ m

), (v∗j ) ∈ V(m−1x

∗j )

= inf

supSm

(x∗, (v∗j )

): x∗ ∈ co

(m−1x

∗j : j ≥ m

), (v∗j ) ∈ V(m−1x

∗j )

= αm.

This proves the claim. In particular, condition (VI) is satisfied for n = m.Step 3.6. We claim that θ ≤ αm ≤ 2. Notice that, because of the claim in Step 3.5,

it suffices to show that α1 ≥ θ. This, however, was already proved in Step 2, when α1

was chosen. Hence condition (V) is satisfied for n = m.Step 3.7 Given our choice of αm, and by definition of the infimum, we can choose

y∗m in co(

m−1x∗j : j ≥ m

)and (mz

∗j ) in V(m−1x

∗j ) such that

αm ≤ supSm

(y∗m, (mz

∗j ))< αm(1 + εm).

Then conditions (II) and (IV) are satisfied for n = m. Note also that y∗m ∈ BX∗ since(m−1x

∗j ) ⊆ BX∗ by the induction hypothesis. Likewise, (mz∗j ) ⊆ BX∗ by the claim in

Step 3.1 because (mz∗j ) ∈ V(m−1x

∗j ). Hence the only unresolved issue is our choice of the

subsequence (mx∗j ) of (mz∗j ), which will automatically satisfy the remaining conditions

(I) and (III).To this end, we begin by writing out the expression behind Sm

(y∗m, (mz

∗j ))

in theabove inequality, which leaves us with

αm ≤ sup

m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)y∗m − w∗

A

: w∗ ∈ L(mz∗j )

< αm(1 + εm). (7.11)

Based on equation (7.11), and by definition of the supremum, we can thus choose w∗m

in L(mz∗j ) such that

αm(1− εm) <

m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)y∗m − w∗

mA

.

Now using the fact that A is balanced by assumption (i), we get that A = −A, andtherefore

m−1∑j=1

βjy∗j +

( ∞∑j=m

βj

)y∗m − w∗

mA

= supx∈A

∣∣∣∣m−1∑j=1

βjy∗j (x) +

( ∞∑j=m

βj

)y∗m(x)− w∗

m(x)

∣∣∣∣

= supx∈A

m−1∑j=1

βjy∗j (x) +

( ∞∑j=m

βj

)y∗m(x)− w∗

m(x)

Hence there is an xm ∈ A such that

αm(1− εm) <

m−1∑j=1

βjy∗j (xm) +

( ∞∑j=m

βj

)y∗m(xm)− w∗

m(xm). (7.12)


By the claim in Step 3.1, (mz∗j ) ∈ V(m−1x∗j ) implies (mz

∗j ) ⊆ BX∗ , so xm ∈ A ⊆ BX

ensures that∣∣lim infj

(mz

∗j (xm)

)∣∣ ≤ 1. Given this we may choose (mx∗j ) as a subse-

quence of (mz∗j ) such that (mx∗j (xm)) converges to lim infj

(mz

∗j (xm)

)in BF. Since our

particular choices of αm, y∗m, (mz

∗j ), and (mx

∗j ) satisfy all the conditions (I) through (V),

the induction is complete.Step 4. With the help of the sequences just constructed we shall now prove that (y∗j )

satisfies (1) and (2) from the statement of the lemma. First we need two intermediateresults, which are presented in Steps 4.1 and 4.2. Then we prove (1) in Step 4.3 and (2)in Step 4.4.

Step 4.1. We claim that L(y∗j ) ⊆⋂∞

n=0 L(nx∗j ) ⊆

⋂∞n=1 L(nz

∗j ). For the second

inclusion it is enough to show that L(nx∗j ) ⊆ L(nz∗j ) for every n ∈ N. To this end, fix

n ∈ N and observe that since (nx∗j ) is a subsequence of (nz∗j ) we have (nx

∗j ) ∈ V(nz

∗j ).

Hence L(nx∗j ) ⊆ L(nz∗j ) by Lemma 7.8(4).

For the first inclusion it suffices to show that lim supj(y∗j (x)

)≤ lim supj

(nx

∗j (x)

)for every x ∈ A and every n ∈ N. To prove that this is the case, fix any n ∈ N andlet k > n be arbitrary. Since (k−ix

∗j ) is a subsequence of (k−iz

∗j ) by (III), we have

co(

k−ix∗j : j ≥ k

)⊆ co

(k−iz

∗j : j ≥ k

). Moreover, since (k−iz

∗j ) ∈ V(k−i−1x

∗j ) by

(II), it holds that k−iz∗k, k−iz

∗k+1, k−iz

∗k+2, . . . ∈ co

(k−i−1x

∗j : j ≥ k

)and therefore

co(

k−iz∗j : j ≥ k

)⊆ co

(k−i−1x

∗j : j ≥ k

). It follows that

co(

k−1x∗j : j ≥ k

)⊆ co

(k−1z

∗j : j ≥ k

)⊆ co

(k−2x

∗j : j ≥ k

)⊆ · · ·

· · · ⊆ co(

n+1x∗j : j ≥ k

)⊆ co

(n+1z

∗j : j ≥ k

)⊆ co

(nx

∗j : j ≥ k

).

In particular, we have y∗k ∈ co(

k−1x∗j : j ≥ k

)⊆ co

(nx

∗j : j ≥ k

)for every k > n

by an appeal to (IV) and the above result. Hence supj≥k

(y∗j (x)

)≤ supj≥k

(nx

∗j (x)

)for

every k > n so that limk→∞ supj>k

(y∗j (x)

)≤ limk→∞ supj>k

(nx

∗j (x)

). Since n ∈ N

was arbitrary we have lim supj y∗j (x) ≤ lim supj(nx

∗j (x)) for every n ∈ N, which proves

the claim.Step 4.2. Suppose that w∗ ∈ L(y∗j ). Then we claim that

αn(1− εn) <

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n − w∗

A

< αn(1 + εn) (7.13)

for every n ∈ N. For the second inequality, simply note that if w∗ ∈ L(y∗j ), thenw∗ ∈ L(nz

∗j ) for every n ∈ N by the claim in Step 4.1. If n ≥ 2, then the conclusion

follows from (7.11) since w∗ ∈ L(nz∗j ). Analogously, the case where n = 1 follows

from (7.7) since w∗ ∈ L(1z∗j ). For the first inequality it suffices to show that (7.12)

and (7.9) remain true if w∗n is replaced by any w∗ ∈ L(y∗j ). To prove this, we show

that w∗(xn) ≤ w∗n(xn) for every n ∈ N. Now, if w∗ ∈ L(y∗j ), then w∗ ∈ L(nx

∗j )

for every n ∈ N by the claim in Step 4.1. Thus, it follows from Lemma 7.8(6) thatlim infj

(nx

∗j (xn)

)≤ w∗(xn) ≤ lim supj

(nx

∗j (xn)

). Recall from our choice of (nx

∗j )

in Step 3.7 (for n ≥ 2) and Step 3 (for n = 1) that (nx∗j (xn)) is convergent with

lim infj(nz

∗j (xn)

)as its limit. Hence

w∗(xn) = limj→∞

(nx∗j (xn)) = lim inf

j→∞

(nz

∗j (xn)

).

Since w∗n was chosen from L(nz

∗j ) in Step 3.7 (for n ≥ 2) and Step 3 (for n = 1), another

application of Lemma 7.8(6) yields that lim infj(nz

∗j (xn)

)≤ w∗

n(xn). From this weconclude that w∗(xn) ≤ w∗

n(xn) as desired.Step 4.3. We now show that statement (1) is satisfied with α := limj→∞ αj . First

of all, it follows from (VI) and (V) that (αj) is increasing, bounded below by θ, andbounded above by 2. Hence the sequence is convergent and θ ≤ limj αj ≤ 2. Now fixany w∗ ∈ L(y∗j ). Then (7.13) applies by the claim in Step 4.2.


Using the same arguments as in Step 4.1 of Lemma 7.4, we conclude from the squeezetheorem and the continuity of · A that

α = limn→∞

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n − w∗

A

=

∞∑j=1

βjy∗j − w∗

A

=

∞∑j=1

βj(y∗j − w∗)

A

.

This proves statement (1).Step 4.4. It only remains to show that statement (2) is satisfied for the same α as

in Step 4.3. The proof of this is completely analogous to the proof in Step 4.2 of Lemma7.4 with y∗j replaced by (y∗j − w∗). Indeed, by virtue of (7.13), the strict inequality in(7.4) remains true when y∗j replaced by (y∗j − w∗) inasmuch asn−1∑j=1

βj(y∗j −w∗)+

( ∞∑j=n

βj

)(y∗n −w∗)

A

=

n−1∑j=1

βjy∗j +

( ∞∑j=n

βj

)y∗n −w∗

A

< αn(1+ εn)

for every n ∈ N. All the other arguments in Step 4.2 of Lemma 7.4 rely only on theproperties of · A, (βj), and (εk), which satisfy all the same conditions here as they didin Lemma 7.4.

For the all-important Theorem 7.11, we shall need the concept of annihilators. Recallthat, given a normed space X, the annihilator of a subset A ⊆ X is denoted by A⊥ anddefined as

A⊥ := x∗ ∈ X∗ : x∗(x) = 0 for all x ∈ A .In relation to Theorem 7.11, there are two key observations to make. First, if we areonly interested in the behavior of a linear functional on a certain subset of X, then wemay freely add or deduct linear functionals in the annihilator of that subset. This ideamotivates the statement in (2) of Theorem 7.11. Secondly, if (x∗n) is a sequence in X∗,then L(x∗n) is contained in the annihilator of any subset of x ∈ X : x∗n(x) → 0. Thissecond observation will allow us to apply Lemma 7.10 in the proof of (2) ⇒ (3) below.We also remind the reader that the rôle of L(·) was already discussed after the statementof Lemma 7.10 above.

Theorem 7.11. Let X be a Banach space over R. Suppose that A is a nonempty,balanced, weakly closed subset of the closed unit ball BX . Then the following statementsare equivalent:

(1) The set A is not weakly compact.(2) There is a θ ∈ R with 0 < θ < 1, a subset A0 ⊆ A of A, and a sequence (x∗n)

in the closed unit ball BX∗ with limn→∞ x∗n(x) = 0 for each x ∈ A0 such thatsup

∣∣(x∗ − w∗) (x)∣∣ : x ∈ A

≥ θ for every x∗ ∈ co (x∗n : n ∈ N) and w∗ ∈ A⊥

0 .(3) There is a θ ∈ R with 0 < θ < 1 such that if (βn) is a sequence of strictly positive

reals with∑

n βn = 1, then there is an α ∈ R with θ ≤ α ≤ 2 and a sequence(y∗n) in BX∗ such that whenever w∗ ∈ L(y∗n) we have(a) sup

∣∣∣∑∞j=1 βj

(y∗j − w∗) (x)∣∣∣ : x ∈ A

= α.

(b) sup∣∣∣∑n

j=1 βj(y∗j − w∗) (x)∣∣∣ : x ∈ A

< α(1− θ

∑∞j=n+1 βj), n ∈ N.

(4) There is a z∗ in X∗ such that sup∣∣z∗(x)∣∣ : x ∈ A

is not attained.

Proof. We prove that (1) ⇒ (2) ⇒ (3) ⇒ (4) ⇒ (1). The first implication is proved byapplying the similar implication (1) ⇒ (2) from Theorem 7.5 to a particular separablesubset A0 of A. The second implication boils down to proving the inclusion L(x∗n) ⊆ A⊥

0

in order to apply Lemma 7.10. The third implication involves some clever computationsand an appropriate choice of (βn), but just as in Theorem 7.5 the idea is simply tocontrol the tail of z∗ =

∑∞j=1 βj

(y∗j − w∗) and squeeze the whole thing strictly less than

α. Finally, the proof that (4) implies (1) is completely analogous to the argument givenin Theorem 7.5.

(1) ⇒ (2): Suppose that (1) is true. Then it follows from Theorem 6.7 that thereexists a separable closed subspace Y of X such that A∩Y is not weakly compact. SettingA0 := A ∩ Y we have A0 ⊆ A ⊆ BX .


Moreover, 0 ∈ A by the assumption that A is balanced and 0 ∈ Y since it is asubspace, so 0 ∈ A∩Y . Hence A0 is nonempty. Also, A0 is separable by the separabilityof Y . Finally, Y is closed in the norm topology, so in particular it is weakly closed, andso A0 is weakly closed. Thus, A0 satisfies all the conditions on A in Theorem 7.5.

By the fact that A0 is not weakly compact, we therefore conclude from the implication(1) ⇒ (2) in Theorem 7.5 that there exists a real number θ ∈ (0, 1) and a sequence oflinear functionals (x∗n) ⊆ BX∗ such that limn→∞ x∗n(x) = 0 for each x ∈ A0 and suchthat sup

∣∣x∗(x)∣∣ : x ∈ A0

≥ θ for every x∗ ∈ co (x∗n : n ∈ N).

Now, if w∗ ∈ A⊥0 , then w∗(x) = 0 for every x ∈ A0. Together with A0 ⊆ A and the

above observation, this implies that

sup∣∣(x∗ − w∗) (x)

∣∣ : x ∈ A

= sup∣∣x∗(x)− w∗(x)

∣∣ : x ∈ A

≥ sup∣∣x∗(x)− w∗(x)

∣∣ : x ∈ A0

= sup

∣∣x∗(x)∣∣ : x ∈ A0

≥ θ

for every x∗ ∈ co (x∗n : n ∈ N) and w∗ ∈ A⊥0 . This proves the implication (1) ⇒ (2).

(2) ⇒ (3): Let (βn) be any sequence of strictly positive reals summing to 1. Then Aand (βn) satisfy assumptions (i) and (ii) in Lemma 7.10. Now observe that if (2) remainstrue when A⊥

0 is replaced by L(x∗n), then assumption (iii) in Lemma 7.10 would alsobe satisfied for the given θ ∈ (0, 1). Hence (3) would follow immediately from (2) andthe conclusion in Lemma 7.10. Consequently, it suffices to show that L(x∗n) is a subsetof A⊥

0 . To this end, notice that the sequence (x∗n) in (2) satifies limn→∞ x∗n(x) = 0for all x ∈ A0 ⊆ A. Now recall that if w∗ ∈ L(x∗n), then lim infn x

∗n(x) ≤ w∗(x) ≤

lim supn x∗n(x) for every x ∈ X ⊇ A0. Hence w∗(x) = limn→∞ x∗n(x) = 0 for each

x ∈ A0, so L(x∗n) ⊆ A⊥0 as desired.

(3) ⇒ (4): Assume that (3) is true. For the given θ ∈ (0, 1), choose a positive realnumber ∆ such that 0 < ∆ < θ2/2. Now define a sequence of positive reals (βn) by

βn :=2−∆

∆

(∆

2

)n

> 0

for each n ∈ N. Note that 0 < ∆2 < θ2/2

2 = θ2

4 < 14 , so

∑∞m=1 βn is a geometric series

with

∞∑n=1

βn =2−∆

∆

∞∑n=1

(∆

2

)n

=2−∆

∆

∆2

1− ∆2

=2−∆

∆

∆

2−∆= 1.

Let α and (y∗n) be as in (3) and define z∗ ∈ X∗ by z∗(x) :=∑∞

j=1 βj(y∗j − w∗) (x) for

an arbitrary w∗ ∈ L(y∗n). If we can show that sup |z∗(x)| : x ∈ A is not attained, thenthe proof will be complete. To this end, fix an arbitrary x0 ∈ A. Since w∗ ∈ L(y∗n) wehave lim infj y

∗j (x0) ≤ w∗(x0) so for every ε > 0 there are infinitely many n ∈ N with

y∗n(x0) < lim infn y∗n(x0) + ε ≤ w∗(x0) + ε. In particular, the fact that θ2 − 2∆ > 0

implies the existence of an n ∈ N with y∗n+1(x0) < w∗(x0) + (θ2 − 2∆). Since θ ≤ α itfollows that there is an n ∈ N such that

(y∗n+1 − w∗) (x0) < θ2 − 2∆ ≤ αθ − 2∆. (7.14)

Moreover, by Lemma 7.8(5) it also follows from w∗ ∈ L(y∗n) and (y∗n) ⊆ BX∗ that‖w∗‖ ≤ 1. Hence ∥∥y∗j − w∗∥∥ ≤

∥∥y∗j∥∥+ ‖w∗‖ ≤ 1 + 1 = 2. (7.15)


We conclude from (7.14), (7.15) and statement (b) in (3) that

z∗(x0) =

∞∑j=1

βj(y∗j − w∗) (x0)

=

n∑j=1

βj(y∗j − w∗) (x0) + βn+1

(y∗n+1 − w∗) (x0) + ∞∑

j=n+2

βj(y∗j − w∗) (x0)

<

n∑j=1

βj(y∗j − w∗) (x0) + βn+1 (αθ − 2∆) + 2

∞∑j=n+2

βj

≤ sup

∣∣∣∣ n∑j=1

βj(y∗j − w∗) (x)∣∣∣∣ : x ∈ A

+ βn+1 (αθ − 2∆) + 2

∞∑j=n+2

βj

< α

(1− θ

∞∑j=n+1

βj

)+ (αθ − 2∆)βn+1 + 2

∞∑j=n+2

βj

= α− αθ

∞∑j=n+1

βj + (αθ − 2∆)βn+1 + 2

∞∑j=n+2

βj (7.16)

Now observe that∞∑

j=n+2

βj =

∞∑j=n+2

2−∆

∆

(∆

2

)j

=∆

2

∞∑j=n+1

2−∆

∆

(∆

2

)j

=∆

2

∞∑j=n+1

βj < ∆

∞∑j=n+1

βj .

Therefore, it follows from (7.16) and statement (a) in (3) that

z∗(x0) < α− αθ

∞∑j=n+1

βj + (αθ − 2∆)βn+1 + 2∆

∞∑j=n+1

βj

= α− (αθ − 2∆)

∞∑j=n+1

βj + (αθ − 2∆)βn+1

= α− (αθ − 2∆)

∞∑j=n+2

βj

< α

= sup

∣∣∣∣ ∞∑j=1

βj(y∗j − w∗) (x)∣∣∣∣ : x ∈ A

= sup |z∗(x)| : x ∈ A .

Finally, recall that A is balanced by assumption. Hence x0 ∈ A implies that −x0 ∈A. In particular, since the above inequality holds for arbitrary x0 ∈ A, we also have−z∗(x0) = z∗(−x0) < sup |z∗(x)| : x ∈ A. From this we conclude that |z∗(x0)| <sup |z∗(x)| : x ∈ A whenever x0 ∈ A, so the supremum is not attained.

(4) ⇒ (1): Finally, assume that (4) is true. Then there is a weakly continuous linearfunctional z∗ ∈ X∗ which does not attain its supremum on A. If A was weakly compact,then this would be a clear contradiction of the Extreme Value Theorem. Hence A cannotbe weakly compact.

7.3. James’ Weak Compactness Theorem

At last, we are now fully equipped to give the proof of James’ Weak CompactnessTheorem in all its generality. Most of the hard work has already been done in theprevious sections. In short, our primary task is to allow for a final generalization of theequivalence of statements (1) and (4) in Theorem 7.11.


Theorem 7.12. (James’ Weak Compactness Theorem) Let X be a Banach spaceand suppose that A is a nonempty weakly closed subset of X. Then the following state-ments are equivalent.

(1) A is weakly compact in X(2) Every bounded linear functional x∗ on X attains the supremum of its absolute

value |x∗| on A.(3) Every bounded real-linear functional u∗ on X attains the supremum of its abso-

lute value |u∗| on A.(4) Every bounded real-linear functional u∗ on X attains its supremum on A.

Proof. We first prove that (1) implies both (2), (3), and (4). We then prove that (2)implies (1) and (3) implies (1) such that (1) ⇔ (2) and (1) ⇔ (3). Finally, we show that(4) implies (3) such that (1) ⇔ (4).

(1) ⇒ (2), (3), (4): First recall that all bounded linear functionals on X (i.e. themembers of the dual space) are weakly continuous by definition. Likewise, all boundedreal-linear functionals on X are weakly continuous by Lemma 7.6. Now assume that (1)is true. Then A is weakly compact. Thus, it follows from the Extreme Value Theoremthat every real-valued weakly continuous function attains its supremum and infimumon A. This proves (4). In fact, it also proves (3) since the supremum of |u∗| over A iseither the supremum of u∗ or the absloute value of the infimum of u∗, both of which areattained on A. Finally, the Extreme Value Theorem also asserts that every complex-valued weakly continuous function attains the supremum and infimum of its absolutevalue on A. This proves (2). We conclude that (1) implies (2), (3), and (4) as desired.

In order to apply Theorem 7.11 in the proof of the final three implications, we needA to be contained in the closed unit ball of X. Claims 1 and 2 below justify that thiscan be assumed without any loss of generality.

Claim 1. We claim that A is bounded whenever (2), (3), or (4) is true. To see this,first suppose that (2) holds and fix any x∗ ∈ X∗. Then |x∗(x0)| ≤ sup |x∗(x)| : x ∈ A <∞ for every x0 ∈ A. Hence x∗(A) is bounded in F for every x∗ ∈ X∗, and therefore Ais bounded in X by Corollary 5.20. To prove the claim for (3) and (4), fix an arbitraryu∗ ∈ (XR)

∗. If (3) is true, then |u∗(x0)| ≤ sup |u∗(x)| : x ∈ A < ∞ for every x0 ∈ A,so u∗(A) is bounded in F. Suppose instead that (4) is true. Since u∗ ∈ (XR)

∗ implies−u∗ ∈ (XR)

∗, we have inf u∗(x) : x ∈ A = − sup −u∗(x) : x ∈ A. In turn, both theinfimum and supremum of u∗ is attained, so u∗(A) is bounded in F. In sum, u∗(A)is bounded in F whenever (3) or (4) holds. Now observe that x 7→ u∗(ix) is also abounded real-linear functional on X. Hence u∗(iA) is also bounded in F. By Theorem1.6, each x∗ ∈ X∗ can be written as x∗(x) = u∗(x)−iu∗(ix) where u∗ = Re(x∗) ∈ (XR)

∗.Since |u∗(x)− iu∗(ix)| ≤ |u∗(x)| + |u∗(ix)|, it follows that x∗(A) is bounded for everyx∗ ∈ X∗. Another application of Corollary 5.20 thus proves that A is weakly boundedin X whenever (3) or (4) is true.

Claim 2. We claim that, when proving the implications (2) ⇒ (1), (3) ⇒ (1), and(4) ⇒ (3), it may be assumed that A ⊆ BX . To see this, notice that by virtue of Claim 1there is an r > 0 such that A ⊆ B(0, r) whenever (2), (3), or (4) is assumed. Now recallthat (X,Tw) is a TVS by Theorem 5.9. Hence the maps x 7→ rx and x 7→ r−1x are bothcontinuous in the weak topology. Consequently, x 7→ r−1x is a homeomorphism fromA onto r−1A and therefore A ⊆ B(0, r) and r−1A ⊆ B(0, 1) ⊆ BX are homeomorphicas subsets of X in the weak topology. In particular, A is weakly compact if and only ifr−1A ⊆ BX is weakly compact.

(2) ⇒ (1): Assume that (2) is true. In accordance with Claim 2 we may in additionassume that A ⊆ BX . Now define

E :=⋂

x∗∈X∗

x ∈ X : |x∗(x)| ≤ sup |x∗(y)| : y ∈ A

⊆ BX .

Notice that if x ∈ A, then |x∗(x)| ≤ sup |x∗(y)| : y ∈ A for every x∗ ∈ X∗. HenceA ⊆ E by construction. Also, E is weakly closed as an intersection of weakly closedsets. This follows from the observation that the individual sets are simply preimagesunder x∗ of the closed ball with center 0 and radius sup |x∗(y)| : y ∈ A.


Finally, E is also balanced. To see this, let λ be any scalar with |λ| ≤ 1 and supposethat y ∈ λE. Then y = λx for some x ∈ E, which implies that

|x∗(λx)| = |λ| |x∗(x)| ≤ |x∗(x)| ≤ sup |x∗(y)| : y ∈ A .

for all x∗ ∈ X∗. Hence y ∈ E and therefore λE ⊆ E. This proves that E is balanced.In conclusion, E is a nonempty, balanced, and weakly closed subset of BX , which

also satisfies A ⊆ E.Note that, by construction of E, we have sup |x∗(x)| : x ∈ E ≤ sup |x∗(x)| : x ∈ A

for every x∗ ∈ X∗. Furthermore, recall that sup |x∗(x)| : x ∈ A is attained for everyx∗ ∈ X∗ by our initial assumption. Hence A ⊆ E implies that sup |x∗(x)| : x ∈ E isalso attained for every x∗ ∈ X∗.

We now claim that

sup |x∗(x)| : x ∈ E = sup |Re(x∗)(x)| : x ∈ E (7.17)

for every x∗ ∈ X∗. To see this, first observe that |Re(x∗)(x)| ≤ |x∗(x)| trivially im-plies sup |x∗(x)| : x ∈ E ≥ sup |Re(x∗)(x)| : x ∈ E. In order to prove the reverseinequality, we fix an arbitrary x∗ ∈ X∗ and let x0 ∈ E be such that |x∗(x0)| =sup |x∗(x)| : x ∈ E. If x∗(x0) = 0, then the result is trivial, so we may assumethat x∗(x0) 6= 0. Thus λ := x∗(x0)/ |x∗(x0)| is a well-defined scalar with |λ| = 1.Since |x∗(x0)|2 = x∗(x0)x

∗(x0) we have |x∗(x0)| = λx∗(x0). In particular, λx∗(x0)must be real, so |x∗(x0)| = Re(λx∗)(x0) = Re(x∗)(λx0). Now notice that λx0 ∈ Eby the balancedness of E. Hence |x∗(x0)| ≤ sup |Re(x∗)(x)| : x ∈ E and thereforesup |x∗(x)| : x ∈ E ≤ sup |Re(x∗)(x)| : x ∈ E. As x∗ was arbitrary, this proves theclaim in (7.17).

By Theorem 1.6, every u∗ ∈ (XR)∗ is the real part Re(x∗) of some x∗ ∈ X∗. Hence

it follows from (7.17) that sup |u∗(x)| : x ∈ E is attained for every u∗ ∈ (XR)∗. Now

observe that Theorem 7.11 applies to XR and E since XR is a real Banach space (inthe original norm) and E is a nonempty, balanced, weakly closed subset of BX = BXR .Thus, we conclude from the equivalence of (1) and (4) in Theorem 7.11 that E is weaklycompact in XR. By Lemma 7.6, the weak topologies on XR and X coincide. Hence E isalso weakly compact in X. Since A is a weakly closed subset of E, we conclude that Ais weakly compact in X as well.

(3) ⇒ (1): Suppose that (3) holds. Then sup |u∗(x)| : x ∈ A is attained for everyu∗ ∈ X∗. Let E be as above and replace X,x∗ by XR, u

∗ in the appertaining argumentsup untill the claim in (7.17). Then we conclude that sup |u∗(x)| : x ∈ E is attainedfor every u∗ ∈ (XR)

∗. Hence E is weakly compact in XR by Theorem 7.11 and thereforeweakly compact in X by Lemma 7.6.

(4) ⇒ (3): For the final implication we assume that (4) is true. That is, we assumethat sup u∗(x) : x ∈ A is attained for every u∗ ∈ (XR)

∗. Now observe that −u∗ ∈(XR)

∗ whenever u∗ ∈ (XR)∗. Hence sup −u∗(x) : x ∈ A is also attained for every

u∗ ∈ (XR)∗. Since

sup |u∗(x)| : x ∈ A = maxsup u∗(x) : x ∈ A , sup −u∗(x) : x ∈ A

it follows that sup |u∗(x)| : x ∈ A is attained for every u∗ ∈ (XR)

∗ as desired.

7.4. Examples

We finish the section with some simple but insightful applications of James’ WeakCompactness Theorem.

Example 7.13. Consider the Banach space c0 := (αn) ∈ `∞ : limn xn = 0 equippedwith the `∞-norm. We claim that the closed unit Bc0 is not weakly compact. Tosee this, recall that c∗0 ∼= `1 and observe that (2−n) ∈ `1 with ‖(2−n)‖1 = 1 since∑∞

n=1 |2−n| = 1/21−1/2 = 1. It follows that if x∗ ∈ c∗0 is the bounded linear functional

represented by (2−n) ∈ `1, then sup|x∗(αn)| : (αn) ∈ Bc0

= ‖x∗‖ = 1.


Now notice that ‖(αn)‖∞ = 1 implies that |αn| ≤ 1 for all n ∈ N while (αn) ∈ c0implies that |αn| < 1 eventually. Hence

|x∗(αn)| =∣∣∣∣ ∞∑n=1

1

2nαn

∣∣∣∣ ≤ ∞∑n=1

1

2n|αn| <

∞∑n=1

1

2n= 1

for all (αn) ∈ Bc0 . Consequently, sup|x∗(αn)| : (αn) ∈ Bc0

is not attained and there-

fore it is an immediate consequence of James’ Weak Compactness Theorem that Bc0

cannot be weakly compact.

We shall now apply the full force of James’ Weak Compactness Theorem to providean alternative proof of what Diestel & Uhl (1977, p. 14) refers to as a key structuretheorem for countably additive vector measures on σ-algebras.

Example 7.14. Let (Ω,F ) be a measurable space and let (X, ‖·‖) be a real Banachspace. A map µ : F → X is said to be a vector measure if µ(E1 ∪E2) = µ(E1) + µ(E2)whenever E1 and E2 are disjoint members of F . Let (Ej) be any sequence of pairwisedisjoint members of F such that

⋃∞j=1Ej ∈ F . If µ

(⋃∞j=1Ej

)=∑∞

j=1 µ(Ej), wherethe infinite sum is convergent in the norm on X, then µ is said to be a countably additive(or σ-additive) vector measure.

It is proved in Corollary 7 of Diestel & Uhl (1977, p.14) that a countably additivevector measure µ has a relatively weakly compact range. This, of course, is a strongresult in relation to the characterization of the values µ(E) ∈ X for E ∈ F . In thefollwing, we present a short and non-technical proof using James’ Weak CompactnessTheorem.

Let µ(F ) denote the range of µ in X. Fix an arbitrary x∗ ∈ X∗ and observe thatx∗ µ : F → R defines a signed measure on (Ω,F ). Indeed, if (Ei) is a sequence ofpairwise disjoint memebers of F such that

⋃∞n=1En ∈ F , then

n∑j=1

x∗(µ(Ej)) = x∗( n∑

j=1

µ(Ej)

)→ x∗

( ∞∑j=1

µ(Ej)

)= x∗

(µ

( ∞⋃n=1

Ej

))by the continuity of x∗ since

∑∞j=1 µ(Ej) is convergent. Consequently,

x∗ µ( ∞⋃

n=1

Ej

)=

∞∑j=1

x∗ µ(Ej).

Because x∗ µ is a signed measure on (Ω,F ), there is a Hahn-Decomposition of Ω(Theorem A.9). That is, there exists P,N ∈ F such that Ω = P ∪N and P ∩N = ∅with x∗ µ(E) ≥ 0 for every E ⊆ P and x∗ µ(E) ≤ 0 for every E ⊆ N . It follows thatfor any E ∈ F , we have

x∗ µ(E) = x∗ µ((E ∩ P ) ∪ (E ∩N)

)= x∗ µ(E ∩ P ) + x∗ µ(E ∩N)

≤ x∗ µ(E ∩ P ) ≤ x∗ µ(P )

Hence sup x∗ µ(E) : E ∈ F is attained at P . By the weak continuity of x∗, itfollows that

supx∗(x) : x ∈ µ(F )

w

= sup x∗(x) : x ∈ µ(F )

= sup x∗(µ(E)) : E ∈ F= sup x∗ µ(E) : E ∈ F= x∗ µ(P ).

Since x∗ ∈ X∗ was arbitrary, we conclude that supx∗(x) : x ∈ µ(F )

w

is attained

for every x∗ ∈ X∗. Seeing as µ(F )w

is weakly closed, it follows from James’ WeakCompactness Theorem that µ(F )

wis weakly compact. This finishes the proof.


One of the very first encounters with compactness in analysis is the case of the closedintervals on the real line. If a real Lp-space is partially ordered by the relation f g iff f ≥ g a.e., then a similar situation arises, where the order intervals [ϕ,ψ] :=f ∈ Lp : ϕ f ψ turn out to be weakly compact. This is in fact intimately linkedto the result in Example 7.14, since a weakly compact order interval in a Banach lattice(of which the Lp-spaces are special cases) is always the range of a countably additivevector measure (Diestel & Uhl, 1977, p. 275).

Example 7.15. Let (Ω,F , µ) be a measure space and consider the real Lp-spaceLp(Ω,F , µ) with 1 ≤ p < ∞. If p = 1 it is assumed that µ is σ-finite (such thatthe dual of L1 is L∞). Recall that the Lp-spaces are Banach spaces. Define

[ϕ,ψ] := f ∈ Lp(Ω,F , µ) : ϕ ≤ f ≤ ψ µ-a.e.for ϕ,ψ ∈ Lp such that ϕ ≤ ψ µ-a.e. We claim that [ϕ,ψ] is weakly compact. First recallthat L∗

p∼= Lq where p and q are conjugate. Fix x∗ ∈ L∗

p and let g be the correspondingelement in Lq. Then x∗(f) =

∫fgdµ whenever f ∈ Lp. Now define h ∈ [ϕ,ψ] by

h(x) :=

ϕ(x), x ∈ g < 0ψ(x), x ∈ g ≥ 0

.

Then it holds for any f ∈ [ϕ,ψ] that

x∗(f) =

∫Ω

fgdµ =

∫g<0

fgdµ+

∫g≥0

fgdµ

≤∫g<0

ϕgdµ+

∫g≥0

ψgdµ

=

∫g<0

hgdµ+

∫g≥0

hgdµ =

∫Ω

hgdµ = x∗(h).

Hence sup x∗(f) : f ∈ [ϕ,ψ] is attained at h ∈ [ϕ,ψ]. Since x∗ ∈ L∗p was arbitrary, it

follows that sup x∗(f) : f ∈ [ϕ,ψ] is attained for every x∗ ∈ X∗.Moreover, we claim that [ϕ,ψ] is convex and closed in the Lp-norm. The convexity is

clear, but the closedness requires a little work. Let (fn) be a convergent sequence in [ϕ,ψ]with fn → f ∈ Lp(Ω,F , µ) in the Lp-norm. Then there exists a sequence of measurablesets (Fn) in F such that, for each n ∈ N, FC

n is a null set and ϕ(x) ≤ fn(x) ≤ ψ(x)for all x ∈ Fn. Moreover, there is a subsequence (fnk

) such that fnk→ f a.e. by a

standard result in Measure Theory. That is, there exists F ∈ F such that FC is anull set and fnk

(x) → f(x) for each x ∈ F . Now define A := F ∩ (⋂

n Fn) and observethat AC = FC ∪

(⋃n F

Cn

)is again a null set. Obersve that, for all x ∈ A, we have

ϕ(x) ≤ fnk(x) ≤ ψ(x) for each k ∈ N and fnk

(x) → f(x). Thus, by the SqueezeTheorem, we have ϕ(x) ≤ f(x) ≤ ψ(x) for every x ∈ A, where AC is a null set. In otherwords, the limit f of (fn) belongs to [ϕ,ψ], so [ϕ,ψ] is closed as claimed.

Since [ϕ,ψ] is closed and convex it is also weakly closed by Mazur’s Theorem (Theorem5.24). Consequently, it follows from James’ Weak Compactness Theorem that [ϕ,ψ] isweakly compact.

In the event that 1 < p <∞ the Lp spaces are reflexive and so the result in Example7.15 could have been obtained much less profoundly by simply appealing to the Heine-Borel Property of reflexive normed spaces. The space L1, on the other hand, is in generalnot reflexive, so in this case the Heine-Borel Theorem could not have been applied. HenceExample 7.15 is by no means a trivial result and it serves well to illustrate the force ofJames’ Weak Compactness Theorem.

Moreover, it is a famous result of Dunford that the relatively weakly compact subsetsof L1 are uniformly integrable (See Diestel & Uhl, 1977, p. 105). Hence the previ-ous example also serves to show that the ordered intervals in L1 are always uniformlyintegrable.


8. Corollaries, Counterexamples and Comments

We begin the section by considering the celebrated James’ Theorem on the reflexivityof Banach spaces (James, 1964), which can now be obtained as an easy corollary ofJames’ Weak Compactness Theorem.

Theorem 8.1. (James’ Theorem) Let X be a Banach space. Then X is reflexive ifand only if every bounded linear functional on X is norm-attaining.

Proof. Assume that every linear functional on X is norm-attaining. This is simply thestatement that ‖x∗‖ = sup

|x∗(x)| : x ∈ BX

is attained for every x∗ ∈ X∗. Now recall

that BX is weakly closed by Mazur’s Theorem (Theorem 5.24). Hence it follows fromJames’ Weak Compactness Theorem that BX is weakly compact. By Theorem 6.2 thisis equivalent to reflexivity of X.

Conversely, suppose that X is reflexive. Then BX is weakly compact by Theorem 6.2.Hence, James’ Weak Compactness Theorem ensures that ‖x∗‖ = sup

|x∗(x)| : x ∈ BX

is attained for every x∗ ∈ X∗. Consequently, every bounded linear functional on X isnorm-attaining.

Example 8.2. We claim that the Banach space (c0, ‖·‖∞) is not reflexive. To see this,simply recall from Example 7.13 that the bounded linear functional (2−n) ∈ `1 ∼= c∗0does not attain its norm. Hence c0 is not reflexive by James’ Theorem.

As another striking example of the power of James’ Weak Compactness Theorem,we shall now present an alternative and highly instructive proof of the famous Krein-Šmulian Weak Compactness Theorem.

Theorem 8.3. (The Krein-Šmulian Weak Compactness Theorem) Let X be aBanach space. Then the closed convex hull of any weakly compact subset of X is itselfweakly compact.

Proof. Fix any weakly compact subset A of X. Then A is in particular weakly closed,since the weak topology is Hausdorff by Theorem 5.9. Hence James’ Weak CompactnessTheorem ensures that sup |x∗(x)| : x ∈ A is attained for every x∗ ∈ X∗. Now fix anarbitrary x∗ ∈ X∗ and let x0 ∈ A be such that |x∗(x0)| = sup |x∗(x)| : x ∈ A. Observethat if x ∈ co(A), then x =

∑ni=1 λixi for some x1, . . . , xn ∈ A and λ1, . . . , λn ≥ 0 with∑n

i=1 λi = 1. Hence

|x∗(x)| =

∣∣∣∣x∗( n∑i=1

λixi

)∣∣∣∣ = ∣∣∣∣ n∑i=1

λix∗(xi)

∣∣∣∣ ≤ n∑i=1

λi |x∗(xi)|

≤n∑

i=1

λi max |x∗(xj)| : x1, . . . , xn ∈ A

= max |x∗(xj)| : x1, . . . , xn ∈ A≤ sup |x∗(y)| : y ∈ A .

It follows that sup |x∗(x)| : x ∈ co(A) ≤ sup |x∗(y)| : y ∈ A. Since X is normed, wehave co(A) = co(A) and therefore

sup |x∗(x)| : x ∈ co(A) ≤ sup |x∗(y)| : y ∈ A

by the continuity of x∗. Now recall that x0 ∈ A ⊆ co(A) satisfies

|x∗(x0)| = sup |x∗(y)| : y ∈ A .

Hence sup |x∗(x)| : x ∈ co(A) is attained at x0 ∈ co(A). Since x∗ ∈ X∗ was arbitrary,we conclude that sup |x∗(x)| : x ∈ co(A) is attained for every x∗ ∈ X∗. Finally, sinceco(A) is a closed and convex, it follows from Mazur’s Theorem (Theorem 5.24) that co(A)is also weakly closed. Thus, co(A) is weakly compact by James’ Weak CompactnessTheorem.


As a final application of James’ Weak Compactness Theorem, we provide an alterna-tive proof of a very nice theorem by Šmulian (1939). What is so nice about this particulartheorem, is that it presents a characterization of weak compactness for convex subsetsof a Banach space without any reference to the dual space.

Theorem 8.4. A convex subset C of a Banach space X is weakly compact if and only if⋂∞n=1 Cn is nonempty whenever C1, C2, . . . is a nested sequence (that is, C1 ⊇ C2 ⊇ · · · )

of nonempty convex subsets of C that are closed in C.

Proof. First suppose that C is a convex and weakly compact subset of X. Then C isweakly closed in X and therefore also closed in X. Now, for each n ∈ N, Cn is closed inC by assumption and therefore also closed in X. But then the convexity of Cn impliesthat Cn is weakly closed in X for every n ∈ N by Mazur’s Theorem (Theorem 5.24).

In other words, C is weakly compact and each Cn is a weakly closed subset of C.Since the sequence C1, C2, . . . is nested, it follows from Cantor’s Intersection Theorem(Theorem B.5) that

⋂∞n=1 Cn is nonempty.

For the converse implication, fix an arbitrary convex subset C of X and suppose that⋂∞n=1 Cn is nonempty whenever C1, C2, . . . is a nested sequence of nonempty convex

subsets of C that are closed in C.First we claim that C is weakly closed in X. Since C is convex it suffices to show

that C is closed in X by Mazur’s Theorem (Theorem 5.24). To see this, suppose thatx0 ∈ C

‖·‖. Then by definition we have B(x0,1n ) ∩ C 6= ∅ for every n ∈ N. Hence the

sets B(x0,1n )∩C form a nested sequence of nonempty convex sets that are closed in C.

In turn,⋂∞

n=1B(x0,1n ) ∩ C must be nonempty by our initial assumption. But clearly⋂∞

n=1B(x0,1n ) = x0, which implies that x0 ∈ C. Consequently, C‖·‖

= C and henceC is closed in X as claimed.

Next we claim that s := sup u∗(x) : x ∈ C is finite for any bounded real-linearfunctional u∗ on X (this is not clear a priori since we do not know if C is bounded).To see this, fix an arbitrary u∗ ∈ (XR)

∗ and notice that the preimages u∗−1([n,∞))are closed and convex. In turn, the sets u∗−1([n,∞)) ∩ C form a nested sequenceof convex sets that are closed in C. Hence, if each u∗−1([n,∞)) is nonempty, then⋂∞

n=1 u∗−1([n,∞))∩C is nonempty by our initial assumption and so there exists x0 ∈ C

with u∗(x0) = ∞, which is clearly a contradiction. Thus, s ≤ N for some N ∈ N.Now, for all n ∈ N there exists x ∈ C such that u∗(x) ≥ s− 1/n by definition of the

supremum s. Hence u∗−1([s− 1/n,∞))∩C is nonempty for every n ∈ N and these setsclearly form a nested sequence. Since each set u∗−1([s−1/n,∞))∩C is also convex andclosed in C, it follows from our initial assumption that

⋂∞n=1 u

∗−1([s− 1/n,∞)) ∩ C isnonempty. Thus, there exists x0 ∈ C such that s ≥ u∗(x0) ≥ s − 1/n for all n ∈ N,which implies u∗(x0) = s. That is, u∗ attains its supremum at x0 ∈ C.

Since C is weakly closed, and each bounded real-linear functional u∗ attains its supre-mum on C, James’ Weak Compactness Theorem ensures that C is weakly compact.

8.1. The Assumption of Weak Closedness

It is assumed in the statement of James’ Weak Compactness Theorem that the subsetin question is weakly closed. The below example shows that this assumption is indeednecessary for the theorem to hold.

Example 8.5. Consider again the Banach space (c0, ‖·‖∞). Let en = (0, . . . , 0, 1, 0, . . .)

denote the n’th standard unit vector in c0 and let A :=n−1en : n ∈ N

. We show that

every bounded linear functional on c0 attains its supremum on A despite the fact that Ais not weakly compact. Since A also fails to be weakly closed, this serves to illustrate theimportance of the assumption that A should be weakly closed in order to apply James’Weak Compactness Theorem.

First recall that x ∈ c0 is a weak limit point of A if and only if there is a net in A\xthat converges weakly to x. Now observe that

∥∥n−1en∥∥∞ =

∣∣n−1∣∣ ‖en‖∞ = n−1 → 0,

so(n−1en

)converges to 0 in the norm topology and hence also in the weak topology.


Thus, A is not weakly closed (and not closed either). By the Hausdorffness of the weaktopology, this in partiuclar implies that A is not weakly compact.

As in Example 7.13, recall that c∗0 ∼= `1. Now fix an arbitrary x∗ ∈ c∗0 represented by(αj) ∈ `1 and observe that

x∗(n−1en) =

∞∑j=1

αjn−1en =

αn

n.

We need to show that s := sup |αn| /n : n ∈ N is attained. If there are only finitelymany αn 6= 0, then the supremum is trivially attained. So assume there are infinitelymany αn 6= 0 and suppose, for contradiction, that the supremum is not attained. Bydefinition of the supremum, there exists k ∈ N such that |αk| /k > s

2 . If only finite terms|αk| /k satisfy this inequality, then the supremum would be attained at the maximum ofthese. Hence there are infinitely many k ∈ N such that |αk| > k s

2 . But then ‖(αj)‖1 =∑∞j=1 |αj | = ∞, which contradicts (αj) ∈ `1. It follows that sup |αn| /n : n ∈ N is

indeed attained for every x∗ ∈ c∗0.

As the next example shows, even closedness in the norm topology is a too weakrestriction on the subset.

Example 8.6. Given an infinite dimensional reflexive Banach space X, we claim thatnorm closedness of a subset A ⊆ X is not enough for the implication (2) ⇒ (1) in James’Weak Compactness Theorem to hold. To see this, we consider the unit sphere SX .

First of all, SX is clearly closed in the norm topology. On the other hand, SX is notweakly closed sinceX is infinite dimensional. To prove this, we show that 0 belongs to theweak closure of SX (in fact, the closure equals BX). Notice that by the proof of Theorem5.6, every weakly open neighborhood U of 0 contains a nontrivial subspace Y (equal tothe intersection of the kernels of a finite collection of bounded linear functionals). SinceY is nontrivial, there exists x0 ∈ Y ⊆ U and thus 1

‖x0‖x0 ∈ Y ⊆ U . Hence U ∩ SX

is nonempty for every weakly open neighborhood U of 0. This proves that the weakclosure of SX contains 0. Notice that, in particular, SX cannot be weakly compact bythe Hausdorffness of the weak topology.

Now observe that since X is reflexive, each x∗ ∈ X∗ attains it supremum on BX byJames’ Theorem. Fix x∗ ∈ X∗ and suppose that the supremum is attained at some x0with ‖x0‖ < 1. Then 1

‖x0‖ > 1, so∣∣∣∣x∗( 1

‖x0‖x0)

∣∣∣∣ = 1

‖x0‖|x∗(x0)| > |x∗(x0)| ,

where 1‖x0‖x0 ∈ SX ⊆ BX , which is a contradiction. Hence the supremum must be

attained at a point in SX . It follows that every bounded linear functional on X attainsits supremum on SX even though SX is not weakly compact.

8.2. The Assumption of Completeness

In the next example we pay a final visit to the Banach space (c0, ‖·‖∞) from Examples7.13, 8.2, and 9.1. Specifically, we provide an example where James Weak CompactnessTheorem fails in an incomplete, albeit dense, subspace of c0.

Example 8.7. Let c00 := (αn) ∈ `∞ : αn 6= 0 for only finitely many n ∈ N denote thesubset of c0 consisting of sequences with finite support. We first show that c00 is notcomplete in the `∞ norm. To see this, consider the sequence

(∑nj=1 j

−1ej)

where ejdenotes the j’th standard unit vector in c00. Assume that n > m and observe that∥∥∥∥ n∑

j=1

j−1ej −m∑j=1

j−1ej

∥∥∥∥∞

=∥∥(0, . . . , 0, 1

m+ 1, . . . ,

1

n, 0 . . .)

∥∥∞ =

1

m+ 1→ 0.

Hence(∑n

j=1 j−1ej

)is a cauchy sequence in c00. Suppose that the sequence has its

limit x = (ξj) in c00. Then there exists n ∈ N such that ξj = 0 for all j > n. But then∥∥(∑nj=1 j

−1ej)− x∥∥∞ ≥ 1

n+1 > 0 for all j > n, which is a contradiction.


Consequently, the limit must have infinitely many nonzero terms and therefore belongsto c0 \ c00. We conclude that c00 is not complete.

Next we claim that c∗00 ∼= `1. In order to prove this, we use that c00 is dense in c0 toconstruct an isometric isomorphism from c∗0 onto c∗00. The conclusion then follows fromthe well-known result that c∗0 ∼= `1. Define a linear map T : c∗0 → c∗00 by x∗ 7→ x∗ ι,where ι : c00 → c0 denotes the natural injection of c00 into c0. Trivially, the map is well-defined since x∗ι is a composition of continuous maps. Surjectivity follows immediatelyfrom the Hahn-Banach Theorem (since c00 is dense in c0 this can be proved directelywithout Hahn-Banach). To see that T is injective, notice that if x∗(x) = y∗(x) for allx ∈ c00, then x∗(x) = y∗(x) for all x ∈ c00 = c0 by continuity. Finally, Bc00 is dense inBc0 , so

∥∥x∗ ι∥∥c∗00

=∥∥x∗∥∥

c∗0. This proves the claim.

Now consider the subset K := 0 ∪n−1en : n ∈ N

of c00. We claim that K is

compact. As argued in Example 8.5, the sequence (n−1en) converges to 0 in norm.Let U be an open cover of K. Take an open neighborhood U0 of 0 in U . Then thereexists k ∈ N such that n−1en ∈ U0 for all n > k. Choosing neigborhoods U1, . . . , Uk of1−1e1, . . . , k

−1ek in turn yields a finite subcover U0, U1, . . . , Uk.What we are really interested in, however, is the closed convex hull of K as a subset

of c00. It is worth emphasising that if we take the closed convex hull of K in thecomplete space c0 (instead of c00), then it is immediately seen to be compact by Mazur’sCompactness Theorem (Theorem A.10).

Nevertheless, we claim that co(K) is not even weakly compact when taken in theincomplete subspace c00. To see this, first observe that

∑nj=1 2

−j < 1. Hence xn :=∑nj=1 2

−jj−1ej belongs to co(K) for each n ∈ N since 0 ∈ K. Because compact-ness always implies limit point compactness, it suffices to show that the infinite subsetxn : n ∈ N ⊆ co(K) has no weak limit point in co(K).

Suppose, for contradiction, that there exists a weak limit point x = (ξj) ∈ co(K). Byweak continuity, x∗(x) would then be a limit point of x∗(xn) : n ∈ N for each x∗ ∈ c∗00.Now take the j’th standard unit vector ej in `1 ∼= c∗00 and observe that ej(xn) = 0 whenn < j while ej(xn) = 2−jj−1ej for all n ≥ j. Thus, we must have ej(x) = 2−jj−1ej inorder for ej(x) to be a limit point of ej(xn) : n ∈ N. But then ξj = 2−jj−1ej for eachj ∈ N. Hence the weak limit point x = (ξj) has infinitely many nonzero terms. Thisis clearly a contradiction, since x is then not even an element in c00. We conclude thatco(K) cannot be weakly compact.

Finally, we claim that every x∗ ∈ c∗00 attains its supremum on co(K) despite thefact that co(K) is not weakly compact. By an argument analogous to the one given inthe proof of the Krein-Smulian Weak Compactness Theorem (Theorem 8.3), it sufficesto show that sup

∣∣x∗(x)∣∣ : x ∈ K

is attained for every x∗ ∈ c∗00. That this is thecase follows immediately from the compactness of K. Alternatively, one can proceedanalogously to Example 8.5 and reach the desired conclusion directly.

Since co(K) is closed and convex by definition, Mazur’s Theorem (Theorem 5.24)ensures that co(K) is weakly closed. Thus, the only assumption in James’ Weak Com-pactness Theorem, which is not satisfied, is the completeness of c00. Consequently, thisassumption cannot be dropped.

Notice that Example 8.7 also shows that the assumption of completeness is necessaryin the Krein-Smulian Weak Compactness Theorem. This provides some comfort thatwe did not lose out on anything by using James Weak Compactness Theorem to proveit in Theorem 8.3.

It is important to emphasize that the true problem in Example 8.7 is really one ofincompleteness. Recall from Theorem 6.6 that the weakly compact subsets of a normedspace are always complete. In that sense, c00 is simply too incomplete to contain itslimit points.

Another way of seeing this issue directly is by noting that the sequence(∑n

j=1 2−jj−1ej

)in co(K) is cauchy but its limit

∑∞j=1 2

−jj−1ej is not in co(K) for the very reason thatc00 is not complete. Indeed,

∑∞j=1 2

−jj−1ej would have belonged to the closed convexhull in the Banach space c0.


8.3. The Rôle of Reflexivity

Recall from Section 3 that James’ Weak Compactness was proved quite easily forfinite dimensional Banach spaces by a simple application of the Heine-Borel Theorem.As we have already hinted at in Section 3, it is really the ability to apply the Heine-Borel Theorem that is the key to a non-technical proof of the theorem rather than, forexample, finite dimensionality.

Now recall from the remarks after Theorem 6.4 that the reflexive normed spacesare precisely the spaces that have the Heine-Borel Property when equipped with theweak topology. As such, James’ Weak Compactness Theorem is not at all surprising forreflexive spaces. To see that this is indeed the case, let A be a weakly closed subset ofa reflexive Banach space X and suppose that x∗ attains its supremum on A for everyx∗ ∈ X∗. In order to show that A is weakly compact we only need to show that A isbounded because X has the Heine-Borel Property. To this end, observe that x∗(A) isbounded for each x∗ ∈ X∗ by the assumption that each x∗ ∈ X∗ attains its supremumon A (and hence also attains its infimum by linearity). By Corollary 5.30 this impliesthat A is bounded. Consequently, A is weakly compact by the Heine-Borel Theorem.

Thus, what complicates our proof in Section 7 is simply that if X is not reflexive, thenit is not enough for a set to be weakly closed and bounded in order to be weakly compact.The next theorem exhibits just how intimate the connection between reflexivity andJames’ Weak Compactness Theorem really is.

Theorem 8.8. Let X be a Banach space. Suppose that there is a subset A of X withnonempty interior such that every x∗ ∈ X∗ attains its supremum on A. Then X isreflexive.

Proof. Let A be as in the theorem. Then Aw is weakly closed and sup

x∗(x) : x ∈

Aw

= sup x∗(x) : x ∈ A is still attained for every x∗ ∈ X∗. Hence Aw is weaklycompact by James’ Weak Compactness Theorem. Since A has nonempty interior wecan choose r > 0 such that B(0, r) is contained in A

w. Now, B(0, r) is weakly closedby convexity and therefore B(0, r) is weakly compact by the weak compactness of Aw.Since B(0, r) is homeomorphic to BX in the weak topology, it follows that BX is weaklycompact. But then X is reflexive by Theorem 6.3. This completes the proof.

It follows from Theorem 8.8 that if James’ Weak Compactness Theorem can be appliedto a set with nonempty interior, then the Banach space in question is necessarily reflexive.Similarly, if James’ Weak Compactness Theorem can be applied to a subset of a non-reflexive Banach space, then that subset must have empty interior.

As an example of this, recall Example 7.15. There it was shown that the orderintervals [ϕ,ψ] of L1(Ω,F , µ) are weakly compact. Now, if F has infinitely manymembers, then L1(Ω,F , µ) is not reflexive and so the order intervals [ϕ,ψ] must haveempty interior according to Theorem 8.8. The reader should verify that this is indeedthe case.

8.4. James Weak* Compactness Theorem

Given the dual spaceX∗ of a Banach spaceX, it is interesting to consider the analogueof James’ Weak Compactness Theorem for the weak* topology on X∗. First recall fromSection 5.3 that the weak* topology is coarser than the weak topolgoy, so compactnessshould be more easily achieved. More importantly, however, recall from Theorem 6.2that (X∗,Tw∗) always has the Heine-Borel Property when X is Banach. Thus, in lightof the discussion in Section 8.3, the ease with which the following result is obtainedshould come as no surprise.

Theorem 8.9. (James’ Weak* Compactness Theorem) Let X be a Banach space.Suppose that A is a nonempty weakly* closed subset of X∗. Then the following state-ments are equivalent.

(1) A is weakly* compact in X∗.(2) Every J(x) ∈ J(X) ⊆ X∗∗ attains the supremum of its absolute value on A.


Since (J(x))(x∗) = x∗(x), the statement in (2) simply says that sup |x∗(x)| : x∗ ∈ Ais attained for every x ∈ X.

Proof. Suppose that A is weakly* compact. By definition of the weak* topology, everyJ(x) ∈ J(X) ⊆ X∗∗ is weakly* continuous. Hence (2) follows from the Extreme ValueTheorem.

For the converse, suppose that sup |x∗(x)| : x∗ ∈ A is attained for every x ∈ X.Then sup |x∗(x)| : x∗ ∈ A is finite for each x ∈ X, so the Uniform Boundedness Prin-ciple (Theorem A.5) ensures that sup ‖x∗‖ : x∗ ∈ A is finite. This means that A iscontained in the closed ball with center 0 and radius sup ‖x∗‖ : x∗ ∈ A. Hence A isbounded in the norm topology and therefore also weakly* bounded by Theorem 5.19.Now recall that A is weakly* closed by assumption. Since A is both weakly* boundedand weakly* closed, we conclude from Theorem 6.4 that A is weakly* compact.

It should be noted that the name “James’ Weak* Compactness Theorem” has beeninvented for our purposes only and does not appear in the literature. In fact, the theoremis not really mentioned anywhere, except as an exercise in Megginson (1998).


Appendix A. List of Analysis Theorems

Theorem A.1. (The Hahn-Banach Theorem for Vector Spaces). Let X be avector space over R. Suppose that p is a sublinear functional on X and f0 is a linearfunctional on a subspace Y of X such that f0(y) ≤ p(y) for every y ∈ Y . Then there is alinear functional f on all of X such that the restriction of f to Y is f0 and f(x) ≤ p(x)for every x ∈ X.


Theorem A.2. (The Hahn-Banach Theorem for Normed Spaces). Let X be anormed space over F. Suppose that f0 is a bounded linear functional on a subspace Y ofX. Then there is a bounded linear functional f on all of X such that the restriction off to Y is f0 and ‖f‖ = ‖f0‖.


Corollary A.3. Let Y is a closed subspace of a normed space X. If x ∈ X \ Y thenthere exists a bounded linear functional f : X → F such that f(x) = d(x, Y ), Y ⊆ ker f ,and ‖f‖ = 1.

Proof. See Corollary 1.9.7 in Megginson (1998).

Corollary A.4. Suppose x, y ∈ X are distinct elements of a normed space X. Thenthere is a bounded linear functional x∗ ∈ X∗ such that x∗(x) 6= x∗(y).

Proof. See Corollary 1.9.9 in Megginson (1998).

Theorem A.5. (The Uniform Boundedness Principle). Let F be a family ofbounded linear operators from a Banach space X into a normed space Y . If the supremumsup ‖T (x)‖ : T ∈ F is finite for each x ∈ X, then sup ‖T‖ : T ∈ F is also finite.


Theorem A.6. (The Banach-Alaoglu Theorem). Let X be a normed space. Thenthe closed unit ball BX∗ of X∗ is weakly* compact.


Theorem A.7. (Goldstine’s Theorem). Let X be a normed space and let J : X →X∗∗ be the canonical embedding of X in X∗∗. Then J(BX) is weakly* dense in BX∗∗ .


Theorem A.8. (Hahn Bahnach Separation Theorem) Let X be a normed space.If C is a nonempty convex subset of X and x0 ∈ X satisfies d(x0, C) > 0, then thereexists a bounded linear functional f on X such that Ref(x0) > sup Ref(x) : x ∈ C.In particular, there exists t ∈ R such that Ref(x0) > t > Ref(x) for every x ∈ C.

Proof. See Proposition 1.9.15 in Megginson (1998).

Theorem A.9. (The Hahn Decomposition Theorem) Let (Ω,F ) be a measurablespace. If µ is a signed measure on (Ω,F ), then there exists a positive set P ∈ F withµ(E) ≥ 0 for every E ⊆ P and a negative set N ∈ F with µ(E) ≤ 0 for every E ⊆ Nsuch that P ∪N = Ω and P ∩N = ∅.

Proof. See Theorem 3.3 in Folland (1999).

Theorem A.10. (Mazur’s Compactness Theorem) Let X be a Banach space. Thenthe closed convex hull of any compact subset of X is itself compact.



Appendix B. List of Topology Theorems

Theorem B.1. (The Tietze Extension Theorem) Let X be a normal (T4) spaceand suppose that A is a closed subset of X. Then any continuous map of A into R maybe extended to a continuous map of all of X into R.

Proof. See Theorem 35.1 in Munkres (2000).

Theorem B.2. Let (X,T) be a topological space and let S be a subbasis for T. Supposethat (xα)α∈A is a net in X and x ∈ X. Then xα → x if and only if, for every U ∈ Scontaining x, there is an αU ∈ A such that xα ∈ U whenever α αU .


Theorem B.3. Let X be a set and let F = f : X → (Yf ,Tf ) be a separating topol-ogizing family of functions for X. If each (Yf ,Tf ) is T0, T1, Hausdorff (T2), regular(T3), or completely regular (T3 1

2), then so is (X,σ(X,F)).


Theorem B.4. (The Heine-Borel Theorem) Let A be a subset of Rn. Then A iscompact if and only if it is closed and bounded in the euclidean metric.


Theorem B.5. Let X be a topological space. Then X is compact if and only if, for everycollection C of closed sets in X having the finite intersection property, the intersection⋂

C∈C C is nonempty. In particular, if we have a nested sequence C1 ⊇ C2 ⊇ · · · ofnonempty closed sets in a compact space X, then the collection (Cn) automatically hasthe finite intersection property, and hence the intersection

⋂n Cn is nonempty.



References

Cascales, B., Orihuela, J., and Galán, M. R., 2010. Compactness, optimality, and risk.In: Bailey, D. H. et al., eds. 2013. Computational and analytical mathematics. NewYork: Springer. Ch. 10.

Diestel, J. and Uhl, J. J., 1977. Vector Measures. New York: American Mathemat-ical Society.

Floudas, C. A., Pardalos, P. eds., 2009. Encyclopedia of Optimization. 2nd ed. NewYork: Springer.

Folland, G. B., 1999. Real Analysis: Modern Techniques and Their Applications. 2nded. New York: John Wiley & Sons.

James, R.C., 1964. Weakly Compact Sets. Trans. Amer. Math. Soc., 113, pp. 129-140.

James, R.C., 1972. Reflexivity and the sup of linear functionals. Ann. of Math., 66, pp.159-169.

Megginson, R. E., 1998. An Introduction to Banach Space Theory. New York: Springer-Verlag.

Munkres, J. R., 2000. Topology. 2nd ed. New York: Pearson.

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