BS III c©Gabriel Nagy

Banach Spaces III:

Banach Spaces of Continuous Functions

Notes from the Functional Analysis Course (Fall 07 - Spring 08)

Why do we call this area of mathematics Functional Analysis, after all? This section ismeant to justify this terminology, especially in the context of Banach space theory. As itturns out (see Remark 1 below), every Banach space can be isometrically realized as a closedsubspace in the Banach space C(Ω) of all continuous functions on some compact Hausdorffspace Ω. This trivial observation has many important consequences, one of them beingsomehow philosophical: if we want to prove some general result about all Banach spaces, itmay suffice to prove the result for all C(Ω). A perfect illustration of this point of view isMazur’s Theorem (see Theorem ?? below). It is then imperative to study the Banach spacesC(Ω) in detail, and the goal of this section is to give a comprehensive treatment. As weshall see, this study will also incorporate Measure Theory, so in a certain sense FunctionalAnalysis on C(Ω) sits on top of Real Analysis.

Convention. Throughout this note K will be one of the fields R or C, and all vectorspaces are over K.

A. The spaces Cb(Ω), C(Ω), Cc(Ω), and C0(Ω)

Notations. Let K be one of the fields C or R. For a topological space Ω, we denote byCK(Ω) the set of all continuous functions f : Ω → K, and by Cb

K(Ω) the space of all functionsf ∈ CK(Ω) that are bounded. These two spaces, as well as the other ones discussed below,where introduced in LCVS IV. Using the standard convention, when K = C, it will beremoved from the notations. Of course, if Ω is compact, then Cb

K(Ω) = CK(Ω). On CbK(Ω)

we have the norm (denoted in LCVS IV by p∞), denoted by ‖ . ‖∞, and defined by

‖f‖∞ = supω∈Ω

|f(ω)|, f ∈ CbK(Ω).

It is obvious that CbK(Ω) is a linear subspace in the Banach space `∞K (Ω) (which consists of

all bounded functions f : Ω → K) with the norm coming from the ‖ . ‖∞ norm on `∞K (Ω),and furthermore (see LCVS IV) we know that Cb

K(Ω) is norm-closed in `∞K (Ω). In particular,Cb

K(Ω) is a Banach space. Of course, the same is true about CK(Ω), when Ω is compact.For a locally compact Hausdorff space Ω we introduced in LCVS IV the subspace

Cc,K(Ω) = f ∈ CK(Ω) : supptopf compact in Ω ⊂ CbK(Ω),

where for a continuous function f ∈ CK(Ω) we defined its topological support to be the set

supptopf = ω ∈ Ω : f(ω).

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The norm closure (or equivalently the completion) of Cc,K(Ω) in CbK(Ω) is the space denoted

by C0,K(Ω). Of course, if Ω is compact, we have C0,K(Ω) = CbK(Ω) = CK(Ω). If Ω is

non-compact, we have a strict inclusion C0,K(Ω) ( CbK(Ω), since for a continuous function

f : Ω → K, the condition that f is in C0,K(Ω) is equivalent to the following:

• for every ε > 0, there exists a compact subset Kε ⊂ Ω, such that supω∈ΩrKε|f(ω)| < ε.

By our usual convention, in the complex case, the subscript K = C will be omitted from thenotations.

Comment. For an arbitrary set S, the Banach space `∞K (S) can also be understood as aBanach space of continuous functions, as follows. Equip S with the discrete topology, so Sin fact becomes a locally compact Hausdorff space, and then we clearly have `∞K (S) = Cb

K(S).Furthermore, `∞K (S) can also be identified as the Banach space CK(βS), for some compactHausdorff space βS, whose construction will be detailed in Exercise 1 below. The wholepoint in these observations is that, ultimately (see also the Comment that follows Exercise1) one only needs to consider two special cases of Banach spaces: either Cb

K(Ω), with Ωlocally compact Hausdorff, or CK(Ω), with Ω compact Hausdorff.

Exercise 1*. Suppose Ω is a locally compact Hausdorff space. Consider the spaceB = f : Ω → [0, 1] : f continuous ⊂ Cb

R(Ω), and the space TΩ = [0, 1]B, of all functionsθ : B → [0, 1], equipped with the product topology. According to Tihonov’s Theorem, TΩ isa compact Hausdorff space. Define the map b : Ω → TΩ by

b(ω) =(f(ω))f∈B, ω ∈ Ω.

Define the space βΩ = b(Ω), the closure of b(Ω) in TΩ. By construction, βΩ is a compactHausdorff space. Prove the following:

(i) The map b : Ω → b(Ω) is a homeomorphism, when we equip b(Ω) with the inducedtopology (from TΩ). Furthermore, when we equip the compact set βΩ with the inducedtopology, b(Ω) is open in βΩ.

(ii) For every f ∈ CbK(Ω), there exists a unique continuous function g ∈ CK(βΩ), such that

f = g b.

(iii) Prove that the correspondence CK(βΩ) 3 g 7−→ g b ∈ CK(Ω) establishes a linearisometric isomorphism Φ : CK(βΩ)

∼−→ CbK(Ω).

(iv) Prove that Φ is also a ring homomorphism, i.e. Φ(gh) = Φ(g)Φ(h), ∀ g, h ∈ CK(βΩ).

The compact Hausdorff space βΩ is called the Stone-Cech compactification of Ω.

Comment. The compactification operation Ω 7−→ βΩ can be extended to cover arbitrarytopological spaces (but with far fewer features), using a different approach – the GelfandNaimark Representation Theorem (see BA II). As it turns out, and entire class of Banachspaces, which include all spaces of the form X = Cb

K(Ω), with Ω arbitrary, can be isometrically(and multiplicatively) identified with CK(ΓX ) for a compact Hausdorff space ΓX – the so-called Gelfand spectrum of X . Therefore we will not loose much if, from now on, we restrict

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(with only one exception, in sub-section B) to topological spaces Ω that are already locallycompact Hausdorff.

We conclude this sub-section with the result mentioned at the beginning.

Remark 1. For any Banach space X , there exists a compact Haudorff space Ω and anisometric linear map T : X → CK(Ω). For instance we can take Ω = (X ∗)1, the unit ball inthe topological dual space X ∗. On the one hand, by Alaoglu’s Theorem, we know that Ω iscompact when equipped with the (induced) w∗-topology (from X ∗). On the other hand, weknow (see BS I) that for every x ∈ X , there exists φ ∈ Ω, such that |φ(x)| = ‖x‖. Of course,by the norm inequality this means that

‖x‖ = supφ∈Ω

|φ(x)|, (1)

so the operator T : X → CK(Ω) defined by

(Tx)(φ) = φ(x), x ∈ X , φ ∈ Ω

is indeed linear and isometric.

Exercise 2.♥ Let Ω be a locally compact Hausdorff space. For every ω ∈ Ω, let evω :C0,K(Ω) 3 f 7−→ f(ω) ∈ K be the evaluation map.

(i) Prove that, for every ω ∈ Ω, the map evω defines a linear continuous functional onC0,K(Ω), and furthermore ‖evω‖ = 1, ∀ω ∈ Ω.

(ii) Prove that the mapev : Ω 3 ω 7−→ evω ∈ (C0,K(Ω)∗)1

is continuous, when the unit ball (CK(Ω)∗)1 in the topological dual space is equippedwith the w∗-topology. Conclude that, if Ω is compact, then ev(Ω) is w∗-compact.

B. Dense subspaces

In this section we discuss various density criteria for subspaces in CbK(Ω). We begin

with a useful Urysohn-type density result (Lemma 1 below), which has several interestingapplications.

Lemma 1. Let Ω be a topological space. Assume X ⊂ CbR(Ω) is a linear sub-space with

the following properties

(i) for any two non-empty disjoint closed sets A,B ⊂ Ω, there exists f ∈ X with

• f∣∣A

= 0;

• f∣∣B

= 1;

• 0 ≤ f ≤ 1.

(ii) the constant function 1 belongs to X .

Then X is dense in CbR(Ω) in the norm topology.

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Proof. The key step in proving this result is contained in the following

Claim: For every g ∈ CbR(Ω), there exists f ∈ X , such that ‖g − f‖∞ ≤ 2

3‖g‖∞.

To prove this, fix g, and define the numbers α = infω∈Ω g(ω), β = supω∈Ω g(ω) and δ =(β − α)/3, and the sets A = g−1([α, α + δ]) and B = g−1([β − δ, β]). Clearly these two setsare closed, disjoint, and non-empty, so there exists f0 ∈ X with f

∣∣A

= 0, f0

∣∣B

= 1, and0 ≤ f0 ≤ 1. Consider then the function f = (α+ δ)1 + δf0, and let us check that it satisfiesthe desired property. Let us now notice that, for each ω ∈ Ω, one has the inequality

|g(ω)− f(ω)| ≤ δ. (2)

This can be immediately seen by inspecting the following three cases:

(a) ω ∈ A. In this case α ≤ g(ω) ≤ α+ δ, and f(ω) = α+ δ.

(b) ω ∈ B. In this case α+ 2δ =≤ g(ω) ≤ α+ 3δ = β, and f(ω) = α+ 2δ.

(c) ω ∈ Ωr(A∪B). In this case α+δ ≤ g(ω) ≤ β−δ = α+2δ, and α+δ ≤ f(ω) ≤ α+2δ.

Using (2) it follows that

‖g − f‖∞ ≤ δ =β − α

3≤ |α|+ |β|

3≤ 2

3max|α|, |β|,

and the Claim follows, since max|α|, |β| = ‖g‖∞.Having proved the Claim, we now prove that X is dense in Cb

R(Ω). Fix some g ∈ CbR(Ω),

and use the Claim to construct recursively a sequence (fn)∞n=1 ⊂ X , with f1 = 0, such that∥∥g − (f1 + · · ·+ fn+1)‖∞ ≤ 23‖g − (f1 + · · ·+ fn)‖∞, ∀n ∈ N. (3)

Of course, by (3) we get

‖g − (f1 + · · ·+ fn)‖∞ ≤(

23

)n−1‖g‖∞, ∀n ∈ N,

which proves that g =∑∞

n=1 fn, so g indeed belongs to the closure of X .

One nice application of the above result is the following.

Theorem 1 (Tietze). Suppose Ω is a normal1 topological space, and T ⊂ Ω is a non-empty closed subset. For any interval [α, β] ⊂ R, and any continuous function g : T → [α, β],there exists a continuous function f : Ω → [α, β], such that f

∣∣T

= g.

Proof. Consider the restriction operator

R : CbR(Ω) 3 f 7−→ f

∣∣T∈ Cb

R(T ),

1 This means that Ω for any two non-empty disjoint closed sets A,B ⊂ Ω, there exist disjoint open setsU, V ⊂ Ω with U ⊃ A and V ⊃ B, Accoring to Urysohn’s Lemma, for A and B as above, there exists acontinuous function h : Ω → [0, 1] with h

∣∣A

= 0 and h∣∣B

= 1.

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and let X = RangeR. It is pretty obvious that R is continuous (in fact a contraction), sinceit clearly satisfies the inequality

‖Rf‖∞ ≤ ‖f‖∞, ∀ f ∈ CbR(Ω).

Claim 1: For every g ∈ X , there exists f ∈ CbR(Ω) with Rf = g, such that ‖f‖∞ = ‖g‖∞.

Indeed, if one starts with some arbitrary g ∈ X , and we put α = inft∈T g(t) and β =supt∈T g(t) (so that α ≤ g ≤ β, and ‖g‖∞ = max|α|, |β|), and if we choose an arbitraryf0 ∈ Cb

R(Ω) with f0

∣∣T

= g, then the function f(ω) = maxminf0(ω), α, β is clearly

continuous and satisfies α ≤ f ≤ β, as well as f∣∣T

= g.

Claim 2: X = CbR(Ω).

By Claim 1 and Proposition 2 from BS II, we know that X is closed in CbR(T ), so it suffices

to show that X is dense in CbR(Ω). In turn, to prove this, it suffices to check that X satisfies

the density criterion from Lemma 1. Obviously X contains the constant function 1 = R1.To check the first condition from Lemma 1, we start with two disjoint non-empty closed setsA,B ⊂ T . Of course, on T we use the induced topology from Ω, but since T is closed inΩ, it follows that A and B are closed in Ω as well. By Urysohn’s Lemma, there exists acontinuous function h : Ω → [0, 1] with h

∣∣A

= 0 and h∣∣B

= 1, so its restriction F = Rh ∈ Xwill clearly satisfy condition (i) from Lemma 1.

Having proved Claim 2, we now know that R is surjective. As we did in Claim 1, if we startwith a continuous function g : T → [α, β], there exists a continuous function f0 : Ω → [α, β],such that f0

∣∣T

= g, and then the function f(ω) = maxminf0(ω), α, β will clearly do thejob.

Another application of Lemma 1 is contained in the following.

Exercise 3.♥ Prove that, for a compact Hausdorff space, the following are equivalent:

(i) Ω is metrizable;

(ii) Ω is second countable;

(iii) CR(Ω) is separable;

(iii’) C(Ω) is separable.

(Hint: Clearly (iii) ⇔ (iii′) and (i) ⇒ (ii). For (ii) ⇒ (iii), start with a countable base Bfor the topology of Ω, and consider the collections B∪ = B ∈ Ω : B finite union of sets in Band D = (A,B) ∈ B∪ × B∪ : A ∩ B = ∅, which are both countable. For every pair(A,B) ∈ D, choose some continuous function fA,B : Ω → [0, 1], such that f

∣∣A

= 0 and

f∣∣B

= 1. Allow for A = ∅ and B = Ω, in which case f∅,Ω = 1. Prove that the separablelinear subspace spanfA,B : (A,B) ∈ D is dense in CR(Ω). For the implication (iii) ⇒ (i),use Exercise 2 and Exercise ?? from BS II. The map ev : Ω → (CR(Ω)∗,w∗) which is injectiveand continuous, makes Ω homeomorphic to a w∗-compact subset in ((CR(Ω)∗)1,w

∗), whichis metrizable.) The above statement is known as Urysohn Meyrizability Theorem.

Other useful density criteria, this time in CK(Ω), with Ω compact Hausdorff space, areoffered in Theorems 2 and 3 below. As a preparation, we first prove a useful technical result.

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Lemma 2 (Dini’s Uniform Convergence Theorem). Let Ω be a compact space, and letf ∈ CK(Ω). If (fλ)λ∈Λ is a net in CK(Ω), such that

(i) fλ(ω) → f(ω), ∀ω ∈ Ω,

(ii) whenever λ µ, it follows that |fλ(ω)− f(ω)| ≤ |fµ(ω)− f(ω)|, ∀ω ∈ Ω,

then fλ → f , in the norm topology.

Proof. Without any loss of generality we can assume that f = 0. First of all, let us noticethat the net

(‖fλ‖

)λ∈Λ

⊂ [0,∞) is decreasing, i.e.

(∗) whenever λ µ, it follows that ‖fλ‖ ≤ ‖fµ‖.

Indeed, if we choose ωλ ∈ Ω, such that |fλ(ωλ)| = ‖fλ‖, then ‖fλ‖ ≤ |fµ(ωλ)| ≤ ‖fµ‖.Using (∗) we see that in order to prove that fλ → 0 in norm, which is equivalent to the

condition ‖fλ‖ → 0, it suffices to show that infλ∈Λ ‖fλ‖ = 0. Let t denote this infimum,and let us define, for every λ ∈ Λ the set Tλ = ω ∈ Ω : |fλ(ω)| ≥ t. By definition, since‖fλ‖ ≥ t, it follows that Tλ is non-empty and obviously closed in Ω. By condition (ii) it istrivial that, whenever λ µ, it follows that Tλ ⊂ Tµ. By directedness, it follows that for anyλ1, . . . , λn ∈ Λ, the intersection Tλ1 ∩ · · · ∩ Tλn is non-empty, so by compactness, it followsthat

⋂λ∈Λ Tλ 6= ∅. But now if we pick some ω ∈

⋂λ∈Λ Tλ, it follows that |fλ(ω)| ≥ t, ∀λ,

which by (i) clearly forces t = 0.

Corollary 1. For any real number a ≥ 0, there exists a sequence of polynomial functions(pn)∞n=1, such that limn→∞ pn(t) =

√t, uniformly on [0, a].

Proof. We can assume, without any loss of generality, that a = 1. (Indeed, if we prove thestatement for a = 1, and we define the polynomials qn(t) =

√apn(t/a), then qn(t) →

√t,

uniformly on [0, a].)Define p1(t) = 0 and pn+1(t) = pn(t) + 1

2(t − pn(t)2). For a fixed t ∈ [0, 1], an easy

inductive argument shows that

0 ≤ pn(t) ≤ pn+1(t) ≤√t, ∀n ∈ N,

and limn→∞ pn(t) =√t, so if we consider the function f(t) =

√t, t ∈ [0, 1], then

(i) limn→∞ pn(t) = f(t),

(ii) 0 ≤ f(t)− pn+1(t) ≤ f(t)− pn(t), ∀n ∈ N,

for every t ∈ [0, 1]. By Dini’s Theorem, pn → f in norm in CR([0, 1]).

The Theorem we are about to prove employs the algebraic structure of CK(Ω). Recall that– for an arbitrary commutative field K – a (not necessarily commutative) ring A is called aK-algebra, if it is endowed with a scalar multiplication operation K×A 3 (λ, a) 7−→ λa ∈ A,which together with the addition operation A × A 3 (a, b) 7−→ a + b ∈ A, turns A into aK-vector space, and furthermore, the multiplication operation A×A 3 (a, b) 7−→ ab ∈ A isK-bilinear. If A has a multiplicative unit, we say that A is unital. Given a K-algebra A, a

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subset B ⊂ A is referred to as a K-subalgebra of A, it B is simultaneously a linear sub-spaceand a sub-ring. (In particular B it itself a K-algebra, with the operations from A.)

When equipped with the point-wise operations, the Banach spaces CbK(Ω) are unital

K-algebras.

Theorem 2 (Stone-Weierstrass). Suppose Ω is a compact Hausdorff space, and A ⊂CK(Ω) is a K-subalgebra. Assume

• A contains the constant function 1;

• A is involutive2, i.e. f ∈ A ⇒ f ∈ A.

Then the following conditions are equivalent;

(i) A is dense in CK(Ω) in the norm topology;

(ii) A separates the points of Ω, i.e. for any two points ω, ω′ ∈ Ω, ω 6= ω′, there existsf ∈ A, such that f(ω) 6= f(ω′).

Proof. The implication (i) ⇒ (ii) is quite trivial. If ω 6= ω′, then by Urysohn’s Lemma3

there exists g ∈ CR(Ω) with g(ω) = 0 and g(ω′) = 1. If A is dense, there exists f ∈ A with‖g − f‖ ≤ 1

3, and this will clearly force |f(ω)| ≤ 1

3and |f(ω′)| ≥ 2

3.

To prove the implication (ii) ⇒ (i) we first treat the real case K = R. Assume Aseparates the points, and we let X denote the closure of A. It is obvious that X is nowa closed linear subspace, which separates the points of Ω, and also contains the constantfunction 1.

Claim 1: X is an R-subalgebra in CR(Ω).

All we need to do is to prove the implication f, g ∈ X ⇒ fg ∈ X . But this is quite clear,because if we choose two sequences (fn)∞n=1, (gn)∞n=1 ⊂ A, such that fn → f and gn → g,uniformly on Ω, then clearly fngn → fg, again uniformly on Ω, and since fngn ∈ A, ∀n, itfollows that fg indeed belongs to the closure of A.

Claim 2: If f ∈ X has Range f ⊂ [0,∞), then√f ∈ X .

Indeed, if we take a = ‖f‖, then Range f ⊂ [0, a], so if we take (using Corollary 1) asequence (pn)∞n=1 of polynomial functions, such that pn(t) →

√t, uniformly for all t ∈ [0, a],

then clearly the compositions fn = pn f converge uniformly to√f . To this end, all we

need to do is to observe that, since X is a subalgebra that contains 1, for every polynomialp, the composition p f again belongs to X . This argument shows that

√f is a norm limit

of a sequence in X , so√f itself belongs to X , because X is closed.

Claim 3: For any two functions, g, h ∈ X , the functions maxg, h and ming, h bothbelong to X .

This follows from Claim 2, applied to the function f = (g− h)2 ∈ X , which implies that thefunction

√f = |g − h| also belongs to X . The desired conclusion is them immediate from

the identities maxg, h = 12

(g + h+ |g − h|

)and ming, h = 1

2

(g + h− |g − h|

).

2 This stipulation is meaningful only if K = C.3 Every compact Hausdorff space is normal.

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Having proven Claim 3, we now wish to show that X satisfies the satisfies the conditionsfrom Lemma 1. This will be done in several steps.

Claim 4: For any two points ω0, ω1 ∈ Ω, ω 6= ω′, there exist f ∈ X , such that f(ω0) = 0 ≤f(ω) ≤ 1 = f(ω1), ∀ω ∈ Ω.

Since X separates the points, there exists h ∈ X with h(ω0) 6= h(ω1). If we put αk = h(ωk),k = 0, 1, then the function (α1 − α0)

−1(h − α01) ∈ X will now satisfy g(ωk) = k, k = 0, 1.Finally, the desired function f can be constructed as f = maxming, 0, 1, which belongsto X , by Claim 3.

Claim 5: Given some point ω0 ∈ Ω, and some closed subset B ⊂ Ω, such that B 63 ω, thereexists some f ∈ X , such that f

∣∣B

= 1, and f(ω0) = 0 ≤ f(ω) ≤ 1, ∀ω ∈ Ω.

For every point p ∈ B, choose (using Claim 4) some function hp ∈ X , such that hp(ω0) =0 ≤ hp(ω) ≤ 1 = hp(p), ∀ω ∈ Ω, and let Dp = ω ∈ Ω : hp(ω) > 1

2. Of course,

one has B ⊂⋃

p∈ADp, so by compactness (B is closed in Ω, hence compact), there existp1, . . . , pn ∈ B, such that B ⊂ Dp1 ∪ · · · ∪Dpn . The new function h = hp1 + · · · + hpn ∈ Xwill satisfy 0 = h(ω0) ≤ h(ω), ∀ω ∈ Ω, but also h(p) > 1

2, ∀ p ∈ B. The desired function can

then be constructed as f = min2h, 1, which again by Claim 3 belongs to X .

Claim 6: Given two non-empty disjoint closed sets A,B ⊂ Ω, there exists some f ∈ X ,such that f

∣∣A

= 0, f∣∣B

= 1, and 0 ≤ f(ω) ≤ 1, ∀ω ∈ Ω.

For p ∈ A, choose (using Claim 5) some function hp ∈ X , such that hp

∣∣B

= 1, and hp(p) =

0 ≤ hp(ω) ≤ 1, ∀ω ∈ Ω, and let Ep = ω ∈ Ω : hp(ω) < 12. As before, there exist

p1, . . . , pn ∈ A, such that A ⊂ Ep1 ∪ · · · ∪Epn . The new function h = hp1hp2 · · ·hpn ∈ X willsatisfy h

∣∣B

= 1, 0 ≤ h(ω) ≤ 1, ∀ω ∈ Ω, but also h(p) ≤ 12, ∀ p ∈ A. The desired function

can then be constructed as f = 2 maxh, 12 − 1, which again by Claim 3 belongs to X .

Now we are done, because Claim 6 shows that X satisfies condition (i) from Lemma 1.Since X also contains 1, it follows that X is dense in CR(Ω), and so will be A (which is densein X , by construction).

We now treat the complex case K = C. Consider the space AR = f ∈ A : f = f.Using the fact that A is involutive, it is easy to see that AR is an R-subalgebra in CR(Ω),which contains 1, and furthermore satisfies

A = AR + iAR, (4)

so AR separates the points. Using the real case, it follows that AR is dense in CR(Ω), andthen using (4) it follows that A is dense in C(Ω).

Examples 1-3. For a compact subset Ω ⊂ C, let us call an f ∈ C(Ω) a polyno-mial function in z and z, if there exists N ∈ N and (αmn)N

m,n=0 ⊂ C, such that f(z) =∑Nm,n=0 αmnz

mzn, ∀ z ∈ Ω. We say that f is an “honest” polynomial function, if if there

exists N ∈ N and (αm)Nm=0 ⊂ C, such that f(z) =

∑Nm=0 αmz

m, ∀ z ∈ Ω.

1. The space Polz,z(Ω) ⊂ C(Ω), consisting of all polynomial functions in z and z, clearlysatisfies the hypothesis of the Stone-Weierstrass Theorem, and separates the points ofthe Ω. Therefore Polz,z(Ω) is norm dense in C(Ω).

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2. If Ω ⊂ R, the space Pol(Ω) ⊂ C(Ω), consisting of all “honest” polynomial functionsalso satisfies the hypothesis of the Stone-Weierstrass Theorem, and separates the pointsof the Ω. Therefore Pol(Ω) is norm dense in C(Ω).

3. Consider the open unit disk D = z ∈ C : |z| < 1, its closure D = z ∈ C : |z| ≤ 1and its boundary T = z ∈ C : |z| = 1 = ∂ D. The algebra Pol(D), of “honest”polynomial functions on D is no longer dense in C(D). It norm closure in C(D),denoted from now on by A(D), is referred to as the disk algebra. One easy way to seewhy we have a strict inclusion A(D) ( C(D) is by noticing that, by the MaximumModulus Principle (“honest” polynomials are analytic), one has:

maxz∈D

|f(z)| = maxz∈T

|f(z)|, (5)

for all f ∈ Pol(D). Passing to norm closure, and using the continuity of the restrictionmap R : C(D) 3 f 7−→ f

∣∣T ∈ C(T), it follows that R : A(D) → C(T) will be

isometric, so (5) will also be satisfied for all f ∈ A(D). But clearly there are functionsf ∈ C(D) that fail to satisfy (5). Since R : A(D) → C(T) is isometric, the subalgebraA(T) = R

(A(D)

)⊂ C(T) is closed, but again we have a strict inclusion A(T) ( C(T),

as indicated in Exercise ??. Of course, by continuity, the subalgebra A(T) is the normclosure – in C(T) – of the algebra of all “honest” polynomial functions on T.

Exercise 4. Suppose Ω is a compact Hausforff space, and T ⊂ Ω is a closed (hencecompact) set. Consider the restriction map R : CK(Ω) 3 f 7−→ f

∣∣T∈ CK(T ). Use the

following steps to prove that R is surjective, without using Tietze’s Theorem.

(i) Show R is an algebra homomorphism, which is unital and involutive.

(ii) Show that, for every f ∈ CK(Ω), there exists g ∈ CK(Ω) such that Rf = Rg and‖Rf‖ = ‖g‖.

(iii) Using step (ii) conclude (see BS II C) that R has closed range. Using step (i) andStone-Weierstrass Theorem, conclude that RangeR is dense in CK(T ).

Exercise 5*. Use the notations from Example 3.

(i) Prove that, for any integer n ≥ 0 and any f ∈ A(T) one has∮

T f(z)zn dz = 0. Use thisto conclude that A(T) ( C(T).

(ii) Prove that A(D) = f ∈ C(D) : f∣∣D is holomorphic .

(iii) Prove that for g ∈ A(T), the unique function f ∈ A(D), that satisfies f∣∣T = g, is

defined on D by the Cauchy Formula: f(z) = 12πi

∮T g(ζ)(ζ − z)−1dζ, ∀ z ∈ D.

The Exercise below explains what happens with arbitrary unital involutive subalgebrasin CK(Ω).

Exercise 6.♥ Suppose Ω is a compact Hausdorff space, and A ⊂ CK(Ω) is an involutivesubalgebra, which contains the constant function 1. Consider the following equivalencerelation on Ω: ω1 ∼ ω2 ⇔ f(ω1) = f(ω2), ∀ f ∈ A. Let Ω/∼ denote the quotient space, andlet π : Ω → Ω/ ∼ denote the quotient map. Equip Ω/∼ with the quotient topology, i.e. thestrongest topology that makes π continuous.

9

(i) Prove that Ω/∼ is a compact Hausdorff space.

(ii) Consider the map T : CK(Ω/∼) 3 g 7−→ g π ∈ CK(Ω). Prove that T is an isometric,unital (i.e. T1 = 1), involutive (i.e. T g = Tg) algebra homomorphism.

(iii) Prove that the norm-closure A – of A in CK(Ω) – coincides with RangeT . Therefore,A is naturally (algebraically and isometrically) identified with CK(Ω/∼).

The Stone-Weierstrass Theorem has a version applicable to non-compact spaces Ω, whichgoes as follows.

Theorem 3. Suppose Ω is a locally compact Hausdorff space, and A ⊂ C0,K(Ω) is aK-subalgebra, which is involutive. Then the following conditions are equivalent;

(i) A is dense in C0,K(Ω) in the norm topology;

(ii) A separates the points of Ω, as well as any point in Ω from ∞, i.e. for any ω ∈ Ω,there exists f ∈ A, such that f(ω) 6= 0.

Sketch of Proof. If Ω is already compact, there is nothing to prove. Assume Ω is non-compact, and consider the Alexandrov compactification Ω = Ω ∪ ∞. We identify

C0,K(Ω) = f ∈ CK(Ω) : f(∞) = 0,

so that we identify CK(Ω) = C0,K(Ω) + K1. Using this identification we then consider thesubspace A = A+ K1 ⊂ CK(Ω. It is pretty obvious that A is an involutive subalgebra thatcontains the constant function 1. Condition (ii) above is equivalent to the condition that Aseparates the points in Ω, so by Theorem 2, we conclude that condition (ii) is equivalent to:

(i’) A is dense in CK(Ω),

so everything reduces to the proof of the equivalence (i) ⇔ (i′), which is pretty obvious.

Exercise 7. Fill in the details of the proof of Theorem 3.

C. The Arzela-Ascoli Precompactness Theorem

In this sub-section we discuss an important result concerning compactness in CK(Ω) inthe norm topology. Recall that, a metric space X is said to be precompact, if its completionis compact.

Remarks 2-3. Precompactness can be detected in two ways.

2. If Y is a complete metric space, and X ⊂ Y , then X is precompact (with the metriccoming from Y ), if and only if its closure X in Y is compact. Equivalently, X iscontained in some compact subset of Y .

3. Intrinsically, for a metric space (X, d), the following conditions are equivalent:

(i) (X, d) is precompact;

(ii) every sequence in X has a Cauchy sub-sequence;

10

(iii) for every r > 0, there exist points x1, . . . , xn ∈ X, such that⋃n

j=1Br(xj) = X,where Br(xj) denotes the ball x ∈ X : d(xj, x) < r;

(iv) for every ε > 0, all ε-discrete4 subsets of X are finite.

The following result should be regarded as the “linear version” of Theorem 3 below.

Lemma 3. Assume X is a normed vector space, S ⊂ X is a compact subset (in thenorm topology), and B ⊂ X ∗ is a norm bounded subset. Consider then the restriction map

R : B 3 φ 7−→ φ∣∣S ∈ CK(S)

(i) R is continuous, when B is equipped with the induced w∗-topology, and CK(S) isequipped with the norm topology.

(ii) The set R(B) is precompact – in norm – in CK(S).

Proof. (i). Let M = supφ∈B ‖φ‖. Start with some net (φλ)λ∈Λ ⊂ B and some φ ∈ B, such

that φλw∗−→ φ, and let us prove that (Rφλ) → (Rφ) in norm in CK(S). Fix some ε > 0, and

let us indicate how λε ∈ Λ can be constructed, so that

‖Rφλ −Rφ‖∞ ≤ ε, ∀λ λε. (6)

Using the compactness of S, we can find an integer n ≥ 1 and points x1, . . . , xn ∈ S, so that

(∗) for every x ∈ S, there exists j ∈ 1, . . . , n, such that ‖x− xj‖ ≤ ε3M

.

Since φλw∗−→ φ, we know that, φλ(xj) → φ(xj) for every j ∈ 1, . . . , n, so (using directed-

ness) there exists λε ∈ Λ, such that∣∣φλ(xj)− φ(xj)∣∣ ≤ ε

3, ∀λ λε, j ∈ 1, . . . , n. (7)

To check (6) we fix for the moment λ λε as well as some x ∈ S. By (∗) there existsj ∈ 1, . . . , n, such that ‖x − xj‖ ≤ ε

3M. Since ‖φλ‖, ‖φ‖ ≤ M , using the norm inequality

we have the inequalities∣∣φλ(x)− φλ(xj)∣∣ =

∣∣φλ(x− xj)∣∣ ≤ ‖φλ‖ · ‖x− xj‖ ≤

ε

3,∣∣φ(x)− φ(xj)

∣∣ =∣∣φ(x− xj)

∣∣ ≤ ‖φ‖ · ‖x− xj‖ ≤ε

3,

which by (7) give the estimate∣∣φλ(x)− φ(x)∣∣ ≤ ∣∣φλ(x)− φλ(xj)

∣∣ +∣∣φλ(xj)− φ(xj)

∣∣ +∣∣φ(x)− φ(xj)

∣∣ ≤ ε.

Since x ∈ S is arbitrary, we now have supx∈S∣∣(φλ−φ)(x)

∣∣ ≤ ε, which is precisely the desiredinequality (6).

(ii). It we now consider (with M defined as above) the ball B = φ ∈ X ∗ : ‖φ‖ ≤ M,then clearly R(B) ⊂ R(B), so it suffices to show that R(B) is compact. But this is immediatefrom part (i) applied to B and the fact that B is w∗-compact, by Alaoglu’s Theorem.

4 Recall that S ⊂ X is said to be ε-discrete, if whenever x, y ∈ S are two different points, it follows thatd(x, y) ≥ ε.

11

Theorem 4 (Arzela-Ascoli). Let Ω be a compact Hausdorff space. For a subset A ⊂CK(Ω), the following are equivalent.

(i) A is precompact (in norm).

(ii) The two conditions below are satisfied:

(a) A is point-wise bounded, i.e. supf∈A |f(ω)| <∞, ∀ω ∈ Ω;

(b) A is equi-continuous, i.e. for every ω ∈ Ω and every ε > 0, there exists aneighborhood Vε of ω, such that

∣∣f(ω′)− f(ω)∣∣ < ε, ∀ω′ ∈ Vε.

Proof. Before we proceed with the proof, let us point out that the equi-continuity condition(b) has the following equivalent formulation:

(b’) if (ωλ)λ∈Λ ⊂ Ω and ω ∈ Ω are such that ωλ → ω, then limλ

[supf∈A |f(ωλ)−f(ω)|

]= 0.

(i) ⇒ (ii). Assume A is precompact, and let us show is satisfies conditions (a) and (b).Since these two features pass to subsets, we can assume that A is in fact compact. Firstof all, since A is bounded in norm, condition (a) is trivially satisfied. To prove that A isequicontinuous, we use Lemma 3, with X = CK(Ω), S = A, and B = (X ∗)1. By part (i) ofLemma 3, it follows that:

(∗) if φλw∗−→ φ in (X ∗)1, then limλ

[supf∈A |φλ(f)− φ(f)

]= 0.

The desired condition (b’) now follows immediately from (∗) applied the the evaluation mapsφλ = evωλ

and φ = evω. (See Exercise 2.)(ii) ⇒ (i). Assume A is point-wise bounded and equi-continuous, and let us prove

that A is pre-compact. Let us argue by contradiction, assuming the existence of an infinitesubset D ⊂ A, which is ε-discrete for some ε > 0. Consider, for every ω ∈ Ω, the D-tupleΦ(ω) =

(f(ω)

)f∈D ∈

∏D K.

Claim 1: For every ω ∈ Ω, the D-tuple Φ(ω) is bounded, hence it defines an element in`∞K (D). Moreover, when we equip `∞K (D) with the norm topology, the map

Φ : Ω 3 ω 7−→ Φ(ω) ∈ `∞K (D)

is continuous.

The fact that Φ(ω) is bounded, for every ω, is immediate from the point-wise boundednesscondition (a). The continuity of Φ is immediate from the equicontinuity condition (b’).

Using Claim 1, it follows that the set S = Φ(Ω) is compact in `∞K (D) in the norm topology.Define for every f ∈ D, the linear continuous functional γf ∈ `∞K (D)∗ by: γf (x) = xf ,∀x = (xf )f∈D ∈ `∞K (D).

Claim 2: For any two distinct functions f, g ∈ D, there exists x ∈ S, such that∣∣γf (x)− γg(x)∣∣ ≥ ε.

Indeed, since D is ε-discrete, we know (by compactness of Ω) that there exists ω ∈ Ω with|f(ω)− g(ω)| = ‖f − g‖CK(Ω) ≥ ε, and the Claim immediately follows by taking x = Φ(ω).

12

We are now at the point of contradiction. On the one hand, if we apply Lemma 3 withX = `∞K (D), the compact set S = Φ(Ω), and the bounded set B = γf : f ∈ D ⊂ `∞K (D)∗,it follows that the set

R(B) =γf

∣∣S : f ∈ D

⊂ CK(S)

is pre-compact in CK(S). On the other hand, by Claim 2, it is obvious that R(B) is infiniteand ε-discrete, which is impossible.

Corollary 2 (Mazur’s Theorem) If X is a Banach space, and C ⊂ X is compact (in thenorm topology), then the norm closure conv C is also compact in norm.

Proof. Since X can be isometrically presented as a closed subspace in some CK(Ω), we mayvery well assume that X = CK(Ω) for some compact Hausdorff space Ω. Of course, it sufficesto show that the set A = conv C is precompact in norm. Since C is compact in CK(Ω), it isbounded and equi-continuous. Of course, the convex hull A = conv C is also bounded, sowe only need to show that A is equi-continuous. Fix some ω ∈ Ω and some ε > 0. By theequi-continuity of C, there exists some neighborhood Vε of ω, such that∣∣f(ω′)− f(ω)| ≤ ε, ∀ω′ ∈ Vε. (8)

for all f ∈ C. We claim that (8) holds for all f ∈ A. Indeed, if we start with somef ∈ A, we can write it as f = t1f1 + · · · + tnfn, with f1, . . . , fn ∈ C and t1, . . . , tn ≥ 0 witht1 + · · · + tn = 1, and then we have

∣∣fj(ω′) − fj(ω)| ≤ ε, ∀ω′ ∈ Vε, j ∈ 1, . . . , n, which

yields:

∣∣f(ω′)− f(ω)∣∣ =

∣∣ n∑j=1

tj[fj(ω′)− fj(ω)]

∣∣ ≤ n∑j=1

tj∣∣fj(ω

′)− fj(ω)∣∣ ≤ n∑

j=1

tjε = ε, ∀ω′ ∈ Vε.

Now that (8) holds for all f ∈ A, it means that A is equi-continuous, thus pre-compact.

D. The Dual Banach Spaces CK(Ω)∗

This sub-section relies heavily on Measure Theory. Until about halfway through ourdiscussion we are going to review the key definitions and results we are going to need. Sincemost results can be found in many Real Analysis textbooks, we will list them as Exercises8-12. (For those who are not familiar with these topics from earlier readings, these exercisesare arranged so one can re-discover them.)

Definitions and Notations. Suppose Ω is a topological Hausdorff space. Denote byTΩ the collection of all open sets in Ω and by KΩ the collection of all compact sets in Ω. Acontent on Ω is a map θ : KΩ → [0,∞), with the following properties:

(i) whenever K,L ∈ KΩ are such that K ⊂ L, it follows that θ(K) ≤ θ(L);

(ii) θ(K ∪ L) ≤ θ(K) + θ(L), ∀K,L ∈ KΩ;

(ii’) if K,L ∈ KΩ are disjoint, then θ(K ∪ L) = θ(K) + θ(L).

The content θ is said to be regular, if

13

(iii) θ(K) = infθ(L) : L ∈ KΩ, Int(L) ⊃ K

, ∀K ∈ KΩ.

Example 4. Fix some integer N ≥ 1. Call a subset B ⊂ RN a “bounded box” ifB = J1 × · · · × JN , where J1, . . . , JN ⊂ R are bounded intervals. In this case, we definevolN(B) =

∏nk=1 |Jk|, where for a bounded interval J we denote by |J | its length. (In the

case N = 1, “bounded boxes” are simply bounded intervals, and vol1 is the length function.)If we define, for a compact set K ⊂ RN the quantity

θN(K) = inf n∑

j=1

volN(Bj) : B1, . . . , Bn “bounded boxes” such thatn⋃

j=1

Bj ⊃ K,

this gives rise to a regular content θ on RN , which is referred to as the Jordan content.

Definitions and Notations. Let Ω be a topological Hausdorff space. We denote byBor(Ω) the σ-algebra on Ω generated by TΩ. (The sets in Bor(Ω) are called Borel sets.)

A positive Radon measure on Ω is an “honest” measure5 µ on Bor(Ω), with the followingproperties:

(r1) µ(K) <∞, ∀K ∈ KΩ;

(r2) µ(B) = infµ(D) : D ∈ TΩ, D ⊃ B

, ∀B ∈ Bor(Ω);

(r3) µ(D) = supµ(K) : K ∈ KΩ, K ⊂ D

, ∀D ∈ TΩ.

If Ω is a locally compact Hausdorff space (and from now on we are going to limit ourdiscussion to this type only), then a Radon measure µ on Ω also satisfies another version of(iii), namely:

(r′3) if B ∈ Bor(Ω) is µ-σ-finite, i.e. one can write B =⋃∞

n=1Bn with Bn ∈ Bor(Ω), suchthat µ(Bn) <∞, ∀n, then µ(B) = sup

µ(K) : K ∈ KΩ, K ⊂ B

.

One can easily show that, if Ω is a locally compact Hausdorff space, and µ is a positive Radonmeasure on Ω, then the restriction µ

∣∣KΩ

is a regular content. Remarkably, the converse isalso true, namely any regular content is the restriction of a positive Radon measure to KΩ.The proof of this fact is outlined in Exercises 8-11 below.

Exercises 8-9. Fix a locally compact Hausdorff space Ω and a regular content θ.

8. Define the map θ′ : TΩ → [0,∞] by

θ′(D) = supθ(K) : K ∈ KΩ, K ⊂ D, D ∈ TΩ.

Prove that

(i) if D,E ∈ TΩ are such that D ⊂ E, then θ′(D) ≤ θ′(E);

(ii) θ′( ⋃∞

n=1Dn

)≤

∑∞n=1 θ

′(Dn), ∀ (Dn)∞n=1 ⊂ TΩ;

(ii’) if (Dn)∞n=1 ⊂ TΩ consists of disjoint sets, then θ′( ⋃∞

n=1Dn

)=

∑∞n=1 θ

′(Dn).

5 The term “honest” measure is reserved for measures that take values in [0,∞].

14

(Hint: For (ii) use the following observation. If F is closed and F ⊂ A1 ∪ A2, withA1, A2 ∈ TΩ, then there exist closed subset F1, F2, such that F = F1∪F2 and Fj ⊂ Aj,j = 1, 2.)

9. Define the map θ′′ : P(Ω) → [0,∞] by

θ′′(A) = infθ′(D) : D ∈ TΩ, D ⊃ A, A ∈ P(Ω).

Prove that

(i) θ′′(∅) = 0;

(ii) if A,B ∈ P(Ω) are such that A ⊂ B, then θ′′(A) ≤ θ′′(B);

(ii) θ′′( ⋃∞

n=1An

)≤

∑∞n=1 θ

′′(An), ∀ (An)∞n=1 ⊂ P(Ω).

Comment. A map η : P(Ω) → [0,∞], satisfying conditions (i)-(iii) from Exercise 8, iscalled an outer measure on Ω. Given an outer measure η, one declares a set M ∈ P(Ω),η-measurable, if it satisfies the condition: η(A) = η(A ∩M) + η(ArM), ∀A ∈ P(Ω). If wedefine then the collection M(η) = M ∈ P(Ω) : M η-measurable , then Caratheodory’sTheorem states that M(η) is a σ-algebra on Ω, and η

∣∣M(η)

is an “honest” measure on

M(η).Exercises 10-11. Keep the assumptions and notations as in Exercises 8-9.

10. Prove that all open sets D ⊂ Ω are θ′′-measurable. In particular, one has the inclusionBor(Ω) ⊂ M(θ′′) and the map µθ = θ′′

∣∣Bor(Ω)

: Bor(Ω) → [0,∞] is an “honest” measure

on Bor(Ω).

11. Prove that µθ has the following additional properties

(i) µθ(D) = θ′(D), ∀D ∈ TΩ;

(ii) µθ(K) = θ(K), ∀K ∈ KΩ.

Conclude that µθ is a positive Radon measure on Ω.

The measure µθ constructed above is referred to as the Radon measure extension of thecontent θ.

Example 4 (continued). If we apply the above construction to the Jordan content θN ,the resulting positive Radon measure on RN is precisely the Lebesgue (volume) measure.In this case, θ′′N is referred to as the outer Lebesgue measure, and the sets in the σ-algebraM(θ′′N) are referred to as Lebesgue measurable.

Comment. Although the collection M(θ′′) is of little importance to us here (all wewanted is that fact that M(θ′′) contains all Borel sets), it is nevertheless significant, in thesense that it provides the largest σ-algebra to which θ can be extended to a measure. Wewill inspect this σ-algebra in BS IV. On the one hand, it is pretty clear property (r2) in thedefinition of a Radon measure also holds for B ∈ M(θ′′). On the other hand, “most” θ′′-measurable sets can be nicely characterized in a manner that is reminiscent of condition (r′3)

15

in the definition of Radon measures. Specifically, given6 A ∈ P(Ω), which is θ′′-σ-finite, i.e. Acan be written as A =

⋃∞n=1An with θ′′(An) <∞, ∀n, then A is θ′′-measurable, if and only if

there exists sequences (Kn)∞n=1 ⊂ KΩ and (Dn)∞n=1 ⊂ TΩ, such that⋃∞

n=1Kn ⊂ A ⊂⋂∞

n=1Dn,and θ′′

([ ⋂∞n=1Dn

]r

[⋃∞n=1Kn

])= 0. In particular, for such a set A one has the equality

θ′′(A) = supθ′′(K) : K ∈ KΩ, K ⊂ A.In the case of the Lebesgue (outer) measure θ′′N , all A ∈ P(RN) are θ′′N -σ-finite.

Remark 4. In “nice” cases, given a locally compact Hausdorff space Ω, in order tocheck that an arbitrary “honest” measure µ on Bor(Ω) is a positive Radon measure, verylittle is needed. More specifically, if Ω is second countable7, then a sufficient condition for µbeing a positive Radon measure is just (r1): µ(K) < ∞, ∀K ∈ KΩ. This follows from theobservation that, if Ω is second countable, then:

(a) every D ∈ TΩ is σ-compact, i.e. a countable union of compact sets, and

(b) for every K ∈ KΩ, there exists as sequence (Kn)∞n=1 ⊂ KΩ, with Kn+1 ⊂ Int(Kn), ∀n,such that K =

⋂∞n=1Kn.

It should be noticed also that, by (a) it follows that all B ∈ Bor(Ω) are µ-σ-finite, andproperty (r3) from the definition of positive Radon measures can be improved as:

(r′′3) µ(B) = supµ(K) : K ∈ KΩ, K ⊂ D

, ∀D ∈ Bor(Ω).

Remark 5. If µ is a finite8 positive Radon measure on Ω, then any other “honest”measure ν on Bor(Ω), satisfying

ν(B) ≤ µ(B), ∀B ∈ Bor(Ω),

is also a positive Radon measure. Indeed, if µ is finite, the conditions (r2) and (r3) can alsobe re stated as:

(r2) infµ(D rB) : D ∈ TΩ, D ⊃ B = 0, ∀B ∈ Bor(Ω),

(r3) infµ(D rK) : K ∈ KΩ, K ⊂ D = 0, ∀D ∈ TΩ.

and clearly ν will satisfy them as well.

In preparation for Theorem 5 below, we introduce the following terminology.

Definitions and Notations. Fix a locally compact Hausdorff space Ω, and let X beeither one of the vector spaces Cc(Ω), C0(Ω), C(Ω), or Cb(Ω). (Here we use K = C.) A(C-)linear functional φ : X → C is said to be positive, if whenever f ∈ X , f ≥ 0, it followsthat φ(f) ≥ 0.

Of course, positivity can also be defined in the real case, replacing X with its real coun-terpart XR – one of the spaces Cc,R(Ω), C0,R(Ω), CR(Ω), or Cb

R(Ω) – and using R-linearfunctionals φ : XR → R.

Of course, if a function f ∈ X is non-negative, then f ∈ XR. Moreover, any f ∈ XRis a difference of two non-negative functions in XR. Therefore, for a C-linear functionalφ : X → C, the following are equivalent:

6 The reader should be aware that neither A nor the An’s are assumed to be θ′′-measurable (yet).7 According to Urysohn’s Metrization Theorem, this condition is equivalent to the fact that Ω is metrizable.8 With a bit of work, one can replace the finiteness assumption with σ-finiteness.

16

(i) φ is positive, as a C-linear functional on X ;

(ii) the restriction φR = φ∣∣XR

is real valued, and positive as an R-linear functional φR :XR → R.

It should be pointed out that, by the obvious equality

X = XR + iXR, (9)

a C-linear functional φ : X → C is completely determined by its restriction φR, that is, thecorrespondence φ 7−→ φR is injective. In connection with the equivalence (i) ⇔ (ii), it isconvenient to isolate the first feature in (ii) using the following terminology. We say that aC-linear functional φ : X → C is hermitian, if its restriction φR = φ

∣∣XR

is real valued.The notion of hermitian functionals is an analogue of “real within complex.” Specifically,

one can define for every C-linear functional φ : X → C the adjoint functional φ? : X → C by

φ?(f) = φ(f), which is again C-linear, and then the C-linear functionals φRe, φIm : X → C,defined by φRe = 1

2(φ+ φ?) and φIm = 1

2i(φ− φ?) will be hermitian, and will satisfy

φ = φRe + iφIm and φ? = φRe − iφIm. (10)

Exercises 12-13. Suppose Ω is a locally compact Hausdorff space, let X be either oneof the spaces Cc(Ω), C0(Ω), or Cb(Ω), and let XR be its real counterpart.

12.♥ Using the notations as above, prove the following statements.

(i) The correspondence φ 7−→ φ? is an involution, i.e.

• (φ+ ψ)? = φ? + ψ?;

• (αφ)? = αφ?, α ∈ C;

• (φ?)? = φ.

(ii) For a C-linear functional φ, the following are equivalent

• φ is hermitian;

• φ = φ?;

• φ = φRe;

• φIm = 0.

(iii) For any R-linear functional ψ : XR → R, there exists a unique C-linear hermitianfunctional φ : X → C, such that φ

∣∣XR

= ψ.

13.♥ Equip both X and XR with the norm topology defined by ‖ . ‖∞, and consider the(complex) topological dual space X ∗ and the (real) topological dual space X ∗

R.

(i) Prove that, gor a C-linear functional φ : X → C, the following are equivalent:

• φ is continuous;

• φ? is continuous;

• both φRe and φIm are continuous.

17

Moreover, if this is the case, then ‖φ‖ = ‖φ?‖.(ii) For a hermitian C-linear functional φ : X → C, prove that the following are

equivalent:

• φ is continuous;

• the restriction φR = φ∣∣XR

: XR → R is continuous.

Moreover, if this is the case, then ‖φ‖ = ‖φR‖.(iii) Prove that the space X ∗

h = φ ∈ X ∗ : φ hermitian has the following properties:

• X ∗h is a real linear subspace in X ∗;

• the restriction map X ∗h 3 φ 7−→ φR ∈ X ∗

R is an isometric R-linear isomor-phism; in particular X ∗

h is closed in X ∗, hence a real Banach space;

• X ∗h is also w∗-closed in X ∗, and the isomorphism X ∗

h∼−→ X ∗

R is also a homeo-morphism, when both spaces are equipped with the (induced) w∗-topologies.

(Hint: For (ii) the only non-trivial inequality is ‖φ‖ ≤ ‖φR, for which we must showthat, whenever f ∈ X has ‖f‖∞ ≤ 1, it follows that |φ(f)| ≤ ‖φR‖. For such anf , choose a scalar α ∈ C with |α| = 1 such that |φ(f)| = αφ(f), and show that thefunction g = Re(αf) ∈ XR satisfies |φ(f)| = φR(g) ≤ ‖φR.)

Comment. Additional notions and results from Measure and Integration Theory willbe reviewed in BS IV. The reader is urged to browse sub-section A from BS IV, beforecontinuing with the remainder of this note.

Using the notion of positive Radon measures, one has the following fundamental result.

Theorem 5 (Riesz). If Ω is a locally compact Hausdorff space, and φ : Cc,K(Ω) → K isa positive linear functional, then there exists a unique Radon measure µφ on Ω, such that:

φ(f) =

∫Ω

f dµφ, ∀ f ∈ Cc,K(Ω). (11)

Sketch of Proof. Using Exercise 13, it suffices to prove the Theorem in the real case K = R.The proof is divided into several steps, as follows.

Step 1: Define, for every K ∈ KΩ, the quantity

θ(K) = infφ(f) : f ∈ Cc,R(Ω), f ≥ χK

.

(Recall that χK stands for the indicator function of K.) One can show that θ is a regularcontent on Ω.

Step 2 (Mean Value Property): One can show that for f ∈ Cc,R(Ω), and any K ∈ KΩ,such that K ⊃ supptopf , one has the inequalities[

minω∈K

f(ω)]· θ(K) ≤ φ(f) ≤

[maxω∈K

f(ω)]· θ(K). (12)

Step 3: Let µφ be the positive Radon measure extension of the regular content θ. UsingStep 2, one can show that µφ is the unique positive Radon measure on Ω, with the followinginterpolation property:

18

(∗) if K ∈ KΩ, D ∈ TΩ, and f ∈ Cc,R(Ω) are such that χK ≤ f ≤ χD, then ν(K) ≤ φ(f) ≤µ(D).

Using Steps 2 and 3, one can show that (11) holds. Using Step 3, one proves the unique-ness part.

Exercise 14. Fill in the details in the above proof.

Having reviewed all necessary Measure Theory results, the exposition will now continuein comprehensive fashion (complete proofs will be supplied). It is interesting to point outthat, if one considers positive linear functionals on spaces larger than Cc,K(Ω), continuity isautomatic, and one has the following version of Theorem 5.

Theorem 6. If Ω is a locally compact Hausdorff space, and φ : C0,K(Ω) → K is a positiveK-linear functional, then:

(i) φ is continuous;

(ii) there exists a unique finite positive Radon measure on Ω, such that

φ(f) =

∫Ω

f dµφ, ∀ f ∈ C0,R(Ω). (13)

In this case one has the equalities

‖φ‖ = µφ(Ω) = supφ(f) : f ∈ C0,R(Ω), f ≥ 0, ‖f‖∞ ≤ 1

. (14)

Moreover, if Ω is compact, then ‖φ‖ = φ(1).

Proof. Using Exercise ??, we can assume K = R. Let M denote the supremum consideredin (14). The key step for the proof of (i) is contained in the following

Claim 1: M <∞.

To prove this statement we argue by contradiction, assuming the existence of a sequence(fn)∞n=1 ⊂ C0,R(Ω), consisting of non-negative functions with ‖fn‖∞ ≤ 1, ∀n, such thatlimn→∞ φ(fn) = ∞. Without loss of generality (pass to a subsequence, if necessary) wecan assume that φ(fn) ≥ 4n, ∀n. Of course, since ‖fn‖∞ ≤ 1, the series

∑∞n=1 2−nfn is

convergent in C0,R(Ω) to some g. Clearly, g ≥ 2−nfn ≥ 0, so by positivity and linearity weget φ(g) ≥ 2−nφ(fn) ≥ 2n, ∀n, which is clearly impossible.

Claim 2: ‖φ‖ = M .

If f ∈ C0,R(Ω) has norm ‖f‖∞ ≤ 1, then by the obvious inequalities −|f | ≤ f ≤ |f |, andby positivity, we have

−M ≤ −φ(|f |) ≤ φ(f) ≤ φ(|f |) ≤M,

which yield |φ(f)| ≤ M . This argument shows that φ is indeed continuous, and ‖φ‖ ≤ M .The inequality M ≤ ‖φ‖ is trivial, by construction.

Having proved (i), we now concentrate on the second statement. Since the restrictionφ∣∣Cc,R(Ω)

clearly defines a linear, positive, and norm-continuous functional on Cc,R(Ω), by

Theorem 5 there exists a unique positive Radon measure µφ on Ω, such that

φ(f) =

∫Ω

f dµφ, ∀f ∈ Cc,R(Ω). (15)

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Claim 3: µφ(Ω) ≤ ‖φ‖, so in particular µφ is finite..

Since µφ(Ω) = supµ(K) : K ∈ KΩ, it suffices to prove that:

µφ(K) ≤ ‖φ‖, ∀K ∈ KΩ. (16)

But this is quite obvious, since for any compact K, there exists some continuous functionf ∈ Cc,R(Ω), such that f

∣∣K

= 1 and 0 ≤ f(ω) ≤ 1, ∀ω ∈ Ω. On the one hand, this functionhas norm 1, so φ(f) ≤ ‖φ‖. On the other hand, since χK ≤ f , we have

µφ(K) =

∫Ω

χK dµφ ≤∫

Ω

f dµφ = φ(f) ≤ ‖φ‖,

and (16) now follows.Let us observe now that, since µφ is finite, we can integrate every bounded continuous

function, so in particular we can consider the map

ψ : CbR(Ω) 3 f 7−→

∫Ω

f dµφ ∈ R,

which has the following properties:

(a) ψ is linear;

(b) |ψ(f)| =∣∣ ∫

Ωf dµ

∣∣ ≤ ∫Ω|f | dµ ≤ ‖f‖∞ · µ(Ω), ∀ f ∈ Cb

R(Ω);

(a) ψ(f) = φ(f), ∀ f ∈ Cc,R(Ω).

On the one hand, by (b) it follows that ψ is continuous, and has norm ‖ψ‖ ≤ µφ(Ω), whichby Claim 3 and (c) forces ‖ψ‖ = ‖φ‖ = µφ(Ω). On the other hand, since both ψ andφ are norm continuous on C0,R(Ω) in which Cc,R(Ω) is norm dense, by (c) it follows thatφ = ψ

∣∣C0,R(Ω)

, thus proving (13). The uniqueness of µφ is pretty obvious from Theorem 5.

Finally, in the compact case, the equality ‖φ‖ = µφ(Ω), combined with (13), clearly gives‖φ‖ = φ(1).

Corollary 3 (of the proof). Every positive K-linear functional φ : C0,K(Ω) → K admitsa continuous positive K-linear extension ψ : Cb

K(Ω) → K with ψ(1) = ‖ψ‖ = ‖φ‖.

Proof. Use the functional ψ introduced in the proof of Theorem 6.

Up to this point we dealt exclusively with positive functionals on C0,R(Ω). In order totreat the case of general linear continuous functionals (Theorems 7 and 8 below), we firsttake a look at the geometry of the (positive and full) unit ball in C0,K(Ω)∗.

Notations. Let Ω be a locally compact space. Denote by C0,K(Ω)∗+ the set of positiveK-linear functionals C0,K(Ω) → K. (By Theorem 6, such functionals are automaticallycontinuous.) We denote by C0,K(Ω)∗h the space of all hermitian functionals in C0,K(Ω)∗.(Strictly speaking, this notation means something only when K = C, in which case C0(Ω)∗his a real Banach space, norm- and w∗-closed in C0(Ω)∗, which can be identified with thereal dual Banach space C0,R(Ω)∗. For consistency reasons, if K = R, we define C0,R(Ω)∗h =C0,R(Ω)∗.)

20

If A is any one of the sets C0,K(Ω)∗, C0,K(Ω)∗+, or C0(Ω)∗h, we define

(A)1 = φ ∈ A : ‖φ‖ ≤ 1.

In Exercise 2, for every ω ∈ Ω, we defined the evaluation map evΩ : C0,K(Ω) 3 f 7−→ f(ω) ∈K, and the set ev(Ω) = evω : ω ∈ Ω ⊂ (C0,K(Ω)∗)1. It is trivial to see that evω is positive,of norm 1 (see Exercise 2), so we actually have the inclusion ev(Ω) ⊂ (C0,K(Ω)∗+)1.

Lemma 4. Suppose Ω is a locally compact Hausdorff space. With the notations as above,one has the following equalities:

(i) conv[bal ev(Ω)]w∗

= (C0,K(Ω)∗)1;

(i’) conv[ev(Ω) ∪ (−1)ev(Ω)]w∗

= (C0,K(Ω)∗h)1;

(ii) conv[ev(Ω) ∪ 0]w∗

= (C0,K(Ω)∗+)1;

(ii’) if Ω is compact, then conv ev(Ω)w∗

= φ ∈ C0,K(Ω)∗+ : ‖φ‖ = 1;

(iii) (C0,K(Ω)∗h)1 = tφ1 − (1− t)φ2 : t ∈ [0, 1], φ1, φ2 ∈ (C0,K(Ω)∗+)1.

Proof. (i). Using the dual pairing between C0,K(Ω) and C0,K(Ω)∗, by the Absolute Bi-polarTheorem (see DT I), we know that

conv[bal ev(Ω)] = [ev(Ω)].

Of course, the right-hand side is already convex and w∗-compact (by Alaoglu), hence w∗-closed, so we have (C0,K(Ω)∗)1 = [(C0,K(Ω)∗)

1 ], so in order to prove the equality (i) it sufficesto prove the equality

ev(Ω) = (C0,K(Ω)∗)

1 . (17)

The right-hand side of (17) is

(C0,K(Ω)∗)

1 = f ∈ C0,K(Ω) : |φ(f)| ≤ 1, ∀φ ∈ (C0,K(Ω)∗)1,

so (see BS I) it is given simply by

(C0,K(Ω)∗)

1 = f ∈ C0,K(Ω) : ‖f‖∞ ≤ 1(= (C0,K(Ω))1).

But then the desired equality is clear, since the left-hand side of (17) is

ev(Ω) = f ∈ C0,K(Ω) : |evω(f)| ≤ 1, ∀ω ∈ Ω =

= f ∈ C0,K(Ω) : |f(ω)| ≤ 1, ∀ω ∈ Ω = (C0,K(Ω))1.

Since the other equalities concern hermitian functionals, using Exercise ??, we can assumefor the rest of the proof that K = R, so in (i’), (ii) and (iii) the space C0,K(Ω)∗h is replacedwith C0,R(Ω)∗.

(i’). This equality follows from (i) – with K = R – and from the observation that in thereal case we have conv(balM) = conv(M∪ (−1)M), ∀M ⊂ C0,R(Ω)∗.

21

(ii). To prove this equality we argue as in (i), first noticing that (C0,R(Ω)∗+)1 is convexand w∗-compact, hence w∗-closed in C0,R(Ω)∗. Instead of the Absolute Bi-Polar Theorem,we use the Bi-Polar Theorem, so the equality (ii) will follows from the equality

ev(Ω) = (C0,R(Ω)∗+)1. (18)

Of course, since ev(Ω) ⊂ (C0,R(Ω)∗+)1, we have the inclusion ev(Ω) ⊃ (C0,R(Ω)∗+)1, so weonly need to prove that in (18) the left-hand side is contained in the right-hand side. Startwith some f ∈ ev(Ω), which means that

f(ω) = evω(f) ≤ 1, ∀ω ∈ Ω. (19)

We must show that f ∈ (C0,R(Ω)∗+)1, that is, φ(f) ≤ 1, ∀φ ∈ (C0,R(Ω)∗+)1. By Theorem 6this condition is equivalent to the fact that∫

Ω

f µ ≤ 1,

for every finite positive Radon measure µ with µ(Ω) ≤ 1, and this is now obvious, since by(19) we have

∫Ωf dµ ≤

∫Ω

1 dµ = µ(Ω) ≤ 1.(ii’). Let S denote the right-hand side. Since for every φ ∈ CR(Ω)∗+, one has the equality

‖φ‖ = φ(1), it is pretty obvious that S is convex and w∗-compact. Furthermore, since

ev(Ω) is clearly contained in S, we have the inclusion conv ev(Ω)w∗

⊂ S. Argue now bycontradiction, assuming that this inclusion is inclusion is strict, so there exist φ ∈ S, which

does not belong to to closure conv ev(Ω)w∗

. Using the Hahn-Banach Separation Theorem,this means that there exists a linear w∗-continuous functional η : CR(Ω)∗ → R, and someα ∈ R, such that

η(φ) > α ≥ η(ψ), ∀ψ ∈ ev(Ω).

Of course, every linear w∗-continuous functional is represented as η(ψ) = ψ(f), for somef ∈ CR(Ω), so the above inequalities read.

φ(f) > α ≥ f(ω), ∀ω ∈ Ω.

Now we reached a contradiction, since the second inequalities yield f ≤ α1, so by positivitywe have φ(f) ≤ φ(α1) = αφ(1) = α, which contradicts the first inequality.

(iii). Let C denote the right-hand side. On the one hand, since both (±1)(C0,R(Ω)∗+)1 areconvex and w∗-compact, it follows that C is also convex and w∗-compact, and furthermoreone has the equality

C = conv[(C0,R(Ω)∗+)1 ∪ (−1)(C0,K(Ω)∗+)1].

On the other hand, since (±1)ev(Ω) ⊂ (±)(C0,R(Ω)∗+)1 ⊂ (C0,R(Ω)∗)1, we have the inclusions

conv[ev(Ω) ∪ (−1)ev(Ω)]w∗

⊂ C ⊂ (C0,R(Ω)∗)1,

and then the desired equality C = (C0,R(Ω)∗)1 follows from (i’).

22

Corollary 1. With the notations and assumptions as above, for every φ ∈ C0,K(Ω)∗h,there exist φ1, φ2 ∈ C0,K(Ω)∗+, such that φ = φ1 − φ2 and ‖φ1‖+ ‖φ2‖ = ‖φ‖.

Proof. Without loss of generality we can assume ‖φ‖ = 1. By Lemma 4, there exist ψ1, ψ2 ∈(C0,K(Ω)∗+)1 and t ∈ [0, 1], such that φ = tψ1 − (1 − t)ψ2. We then define the functionalsφ1 = tψ1 and φ2 = (1− t)ψ2, which are clearly positive, and satisfy:

(a) ‖φ1‖ ≤ t and ‖φ2‖ ≤ 1− t;

(b) φ = φ1 − φ2.

Of course by (a) and (b) we have ‖φ‖ ≤ ‖φ1‖+‖φ2‖ ≤ t+(1−t) = 1, and since ‖φ‖ = 1, thiswould force equalities in (a), so we now have the equalities ‖φ1‖ = t and ‖φ2‖ = 1− t.

Comment. During the proof of Theorem 7 below, it will be shown that the abovedecomposition is unique.

Theorem 7. Suppose Ω is a locally compact Hausdorff space. For any hermitianφ ∈ C0,K(Ω)∗ there exist two unique finite positive Radon measures µ+

φ , µ−φ on Ω, withthe following properties:

(i) φ(f) =∫

Ωf dµ+

φ −∫

Ωf dµ−φ , ∀ f ∈ C0,K(Ω);

(ii) ‖φ‖ = µ+φ (Ω) + µ−φ (Ω).

Moreover, if one considers the signed measure

µφ = µ+φ − µ−φ , (20)

then (20) represents the Hahn-Jordan decomposition of µφ. (For explanations, see the proof.)

Proof. As before, we can assume that K = R, so “hermitian” is replaced by “arbitraryfunctional in C0,R(Ω)∗. By Lemma 4, there exists positive functionals φ+, φ−, such thatφ = φ+ − φ−, and ‖φ+‖+ ‖φ−‖ = ‖φ‖.

We now use Theorem 6 to produce two finite positive Radon measures µ±φ on Ω, whichrepresent φ±, i.e. such that

φ±(f) =

∫Ω

f dµ±φ , ∀ f ∈ C0,R(Ω),

which also satisfy µ±φ (Ω) = ‖φ±‖, so now the desired properties (i) and (ii) from the statementof the Theorem are satisfied, except for uniqueness.

For the uniqueness part, we fix µ±φ as above, and we consider the signed measure µφ =

µ+φ − µ−φ . Suppose now, we have another pair (µ+, µ−) of positive Radon measures that

satisfy (i) and (ii), and let us prove that µ± = µ±φ . Along the way, we are also going to provethe other statements made in the Theorem, concerning the Hahn-Jordan decomposition.

Claim 1: With µ± as above, the signed measure µ = µ+−µ− coincides with µφ, i.e. onehas the equality

µ+(B)− µ−(B) = µ+φ (B)− µ−φ (B), ∀B ∈ Bor(Ω).

23

Indeed, since both pairs (µ+, µ−) and (µ+φ , µ

−φ ) satisfy (i), it follows that

∫Ωf dµ+

φ +∫Ωt dµ− =

∫Ωf dµ+ +

∫Ωt dµ−φ , which can also be written as:∫

Ω

f d(µ+φ + µ−) =

∫Ω

f d(µ+ + µ−φ ), ∀ f ∈ C0,R(Ω).

Since µ+φ +µ− and µ+ +µ−φ are obviously finite positive Radon measures, by Theorem 5 they

must be equal, and the Claim follows.

The uniqueness part, as well as the remaining statement in the Theorem will now be aconsequence of the feature below.

Claim 2: With µ, µ+ and µ− as in Claim 1, the decomposition µ = µ+ − µ− is theHahn-Jordan decomposition of µ.

Let µ = ν+ − ν− be the Hahn-Jordan decomposition of µ, so that we must prove theequalities µ± = ν±. By the definition of the Hahn-Jordan decomposition, there exist disjointsets Ω± ∈ Bor(Ω), such that Ω+ ∪ Ω− = Ω, and ν±(B) = ±µ(B ∩ Ω±), ∀B ∈ Bor(Ω). Notethat, for every B ∈ Bor(Ω), one has the inequalities

ν+(B) = µ(B ∩ Ω+) = µ+(B ∩ Ω+)− µ−(B ∩ Ω+) ≤ µ+(B ∩ Ω+) ≤ µ+(B),

ν−(B) = −µ(B ∩ Ω−) = µ−(B ∩ Ω−)− µ+(B ∩ Ω−) ≤ µ−(B ∩ Ω−) ≤ µ−(B),

so by Remark 5, the measures ν+ and ν− are also finite positive Radon measures. Of course,the differences η+ = µ+ − ν+ and η− = µ− − ν− are also “honest” measures on Bor(Ω),satisfying

η±(B) ≤ µ±(B), ∀B ∈ Bor(Ω),

so again by Remark 5, it follows that η± are positive Radon measures. We now see that inorder to prove the equalities µ± = ν±, it suffices to prove that η± = 0, which by positivityis the same as η±(Ω) = 0, or equivalently, µ±(Ω) = ν±(Ω). Since ν±(Ω) ≤ µ±(Ω), and weassume µ+(Ω) + µ−(Ω) = ‖φ‖, the desired conclusion will follow from the inequality

‖φ‖ ≤ ν+(Ω) + ν−(Ω). (21)

To prove this inequality, we first argue as in Claim 1 (the equality µ+ − µ− = µ = ν+ − ν−

yields µ+ + ν− = µ− + ν+) so that φ can be represented as φ(f) =∫

Ωf dν+ −

∫Ωf dν−, so

for every f ∈ C0,R(Ω), we have the inequalities

|φ(f)| ≤∣∣∣∣ ∫

Ω

f dν+

∣∣∣∣ +

∣∣∣∣ ∫Ω

f dν−∣∣∣∣ ≤ ∫

Ω

|f | dν+ +

∫Ω

|f | dν− ≤ ‖f‖∞ ·[ν+(Ω) + ν−(Ω)

],

and the desired inequality (21) follows immediately.

The above Theorem, as well as the one below, can be stated better, by adopting followinglanguage from Measure Theory.

Definitions. Assume K is either R or C.

A. Suppose A is a σ-algebra on a set X. It can be shown that for a map µ : A → K, thefollowing are equivalent:

24

• µ is σ-additive, i.e. whenever (An)∞n=1 ⊂ A is a sequence consisting of disjointsets, it follows that

µ( ∞⋃

n=1

An

)=

∞∑n=1

µ(An); (22)

• µ can be written as a finite K-linear combination

µ =n∑

j=1

αkµk (23)

with µ1, . . . , µn “honest” finite measures on A.

A map satisfying these equivalent condition is called a K-valued measure on A. Inpractice, the number of terms in (23) can be reduced as follows.

(i) In the real case, µ can be written as a difference µ = µ1 − µ2 of finite “honest”measures. For instance, one can use the measures µ+ and µ−, that appear in theHahn-Jordan decomposition of µ.

(ii) In the complex case, µ can be written as µ = µ1 + iµ2, with µ1, µ2 R-valued, soin the sum (23) one can use just four terms.

B. For any K-valued measure µ on A one construct its associated variation measure |µ|,defined by |µ|(A) = sup

∑∞n=1 |µ(An)| : (An)∞n=1 ⊂ A,

⋃∞n=1An = A

, ∀A ∈ A.

(One can show that |µ| is always a finite “honest” measure. In the real case, one canuse the Hahn-Jordan decomposition µ = µ+ − µ−, to get |µ| = µ+ + µ−.)

C. Given a K-valued measure µ, although its presentation (23) is not unique, one canstill define un-ambiguously the integral

∫Ωf dµ, for arbitrary A-measurable and |µ|-

integrable function f : X → K, by∫

Ωf dµ =

∑nk=1 αk

∫Ωf dµk. This generalized

integral satisfies the inequality ∣∣∣∣ ∫Ω

f dµ

∣∣∣∣ ≤ ∫Ω

|f | d|µ|. (24)

D. If we restrict our attention to the case when A = Bor(Ω) – the Borel σ-algebra on alocally compact Hausdorff space Ω – one can show that for a K-valued measure µ onΩ, the following are equivalent:

• one can represent µ as a K-linear combination (23) with µ1, . . . , µn finite positiveRadon measures on Ω;

• the variation measure |µ| is a (automatically finite) Radon measure on Ω.

Such a µ is then referred to as a K-valued Radon measure on Ω. Exactly as discussedin Definition A, one has the following special cases.

(i) In the real case, µ is an R-valued Radon measure, if and only if measures µ+ andµ−, that appear in the Hahn-Jordan decomposition of µ, are Radon measures.

25

(ii) In the complex case, µ is a C-valued Radon measure, if and only if it can bewritten as µ = µ1 + iµ2, with µ1, µ2 R-valued Radon measures, so in the sum (23)one can use just four terms.

Using this terminology, Theorem 7 has the following extension and re-formluation.

Theorem 8. Assume Ω is a locally compact Hausdorff space. For any φ ∈ C0,K(Ω)∗

there exists a unique K-valued Radon measure µφ on Ω, such that

φ(f) =

∫Ω

f dµφ, ∀ f ∈ C0,K(Ω). (25)

Moreover, one has the equality ‖φ‖ = |µφ|(Ω).

Proof. The case when φ is hermitian (which includes the case K = R, with φ arbitrary) isalready contained in Theorem 7, so we will assume K = C. By Exercise ?? we know thereexists two unique functionals φRe, φIm ∈ C0(Ω)∗h, such that φ = φRe + iφIm, so by Theorem7, there exist R-valued Radon measures µRe

φ and µImφ , such that φRe(f) =

∫Ωf dµRe

φ andφIm(f) =

∫Ωf dµIm

φ , so if we define the C-valued Radon measure µφ = µReφ + iµIm

φ , then(25) is satisfied. For the uniqueness, one notices that if a C-valued Radon measure satisfies(25) then the R-valued measures ν1 = Reµ and ν2 = Imµ will give rise to two hermitianfunctionals ψk(f) =

∫Ωf dνk, k = 1, 2, that satisfy φ = ψ1 + iψ2. This will force ψ1 = φRe

and ψ2 = φIm, which by Theorem 7 yields ν1 = µReφ and ν2 = µIm

φ , therefore forcing µ = µφ.For the norm equality, we first notice that by (24) we have the inequality

|φ(f)| ≤∫

Ω

|f | d|µφ| ≤ ‖f‖∞ · |µφ|(Ω), ∀ f ∈ C0(Ω),

which yields the inequality ‖φ‖ ≤ |µφ|(Ω). For the other inequality we start off by provingthe following.

Claim: For any disjoint N-tuple (Kj)Nj=1 ⊂ KΩ, one has: ‖φ‖ ≥

∑Nj=1 |µφ(Kj)|.

It suffices to show that, for every ε > 0, there exists some f ∈ C0(Ω) with ‖f‖∞ ≤ 1,such that

|φ(f)|+ ε ≥ |µφ(K1)|+ · · ·+ |µφ(KN)|. (26)

Since K1, . . . , KN are disjoint, there exist9 disjoint open sets A1, . . . , AN , such that Aj ⊃ Kj,∀ j. Replace the A’s with smaller open sets Dj, such that we still have Dj ⊃ Kj, but also|µφ|(Dj r Kj) ≤ ε/N , ∀ j. For each j, by Urysohn’s Lemma for locally compact spaces,there exists hj ∈ Cc,R(Ω) such that hj

∣∣Kj

= 1, hj

∣∣ΩrDj

= 0, and 0 ≤ hj(ω) ≤ 1, ∀ω ∈ Ω.

Fix α1, . . . , αN ∈ C with |α1| = · · · = |αN | = 1, such that |µ(Kj)| = αjµ(Kj), ∀ j, and let

us consider the functions f =∑n

j=1 αjhj and g =∑N

j=1 αjχKj. On the one hand, it is clear

thatf(ω) = g(ω) = 0, ∀ω ∈ Ω r [D1 ∪ · · · ∪DN ], (27)

9 The case N = 2 is trivial, for instance using Urysohn Lemma for locally compact spaces. For N > 2use induction: find A and AN+1 open disjoint such that A ⊃ K1 ∪ · · · ∪KN and AN+1 ⊃ KN+1, and workin A to find A1, . . . , AN using the inductive hypothesis.

26

On the other hand, if ω ∈ Dj, then clearly hk(ω) = 0 = χKk(ω), ∀ k 6= j and also χKj

(ω) ≤hj(ω) ≤ χDj

(ω). This proves the inequalities

|fχDj| ≤ χDj

and |(f − g)χDj| ≤ χDjrKj

, ∀ j. (28)

The first inequality, combined with (27), clearly forces ‖f‖∞ ≤ 1. Using the second inequalityin (28), combined with (24) yields the inequality∣∣∣∣ ∫

Ω

(f − g)χDjdµφ

∣∣∣∣ ≤ ∫Ω

χDjrKjd|µφ| ≤ |µφ|(Dj rKj) ≤ ε/N, ∀ j.

Since by (27) we have∫

Ω(f − g) dµφ =

∑Nj=1

∫Ω(f − g)χDj

dµφ, we obtain∣∣∣∣[ ∫Ω

f dµφ

]−

[ ∫Ω

g dµφ

]∣∣∣∣ =

∣∣∣∣ N∑j=1

∫Ω

(f − g)χDjdµφ

∣∣∣∣ ≤ ε.

Of course, we have∫

Ωfdµφ = φ(f) and

∫Ωg dµφ =

∑Nj=1

∫ΩαjχKj

dµφ =∑N

j=1 αjµφ(Kj) =∑Nj=1 |µφ(Kj)|, so the preceding inequality yields∣∣∣∣φ(f)−

N∑j=1

|µφ(Kj)|∣∣∣∣ ≤ ε,

from which the desired inequality (26) easily follows.To finish the proof of the Theorem, we use the definition of the variation measure, so we

will be done once we prove the following.

Claim 2: For any disjoint sequence (Bn)∞n=1 ⊂ Bor(Ω), one has the inequality ‖φ‖ ≥∑∞n=1 |µφ(Bn)|.Of course, it suffices to prove the above inequality for finite sums/sequences. But given

Borel sets B1, . . . , BN , by the finiteness of the positive Radon measure |µφ| (see the discussionfrom Remark 5), for each j ∈ 1, . . . , N there exists a sequence (Kj

n)∞n=1 of compact subsetsof Bj, such that limn→∞ |µφ|(Bj rKj) = 0, which will force limn→∞ µφ(K

jn) = µ(Bj). This

argument shows that, in the statement of the Claim, in addition to the finite sum/sequencereduction, we can also assume that all the B’s are compact. But then everything is trivialby Claim 1.

Conclusion. Given a locally compact Hausdorff space Ω, one can consider the space

MK(Ω) = µ : K-valued Radon measure on Ω,

Of course MK(Ω) is a K-vector space, when equipped with point-wise addition and scalarmultiplication: (µ + ν)(B) = µ(B) + ν(B), (αµ)(B) = αµ(B), B ∈ Bor(Ω). Moreover,MK(Ω) becomes a Banach space, with the norm ‖µ‖ = |µ|(Ω), and the correspondence

Φ : MK(Ω) 3 µ 7−→ φµ ∈ C0,K(Ω)∗, (29)

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given by φµ(f) =∫

Ωf dµ, f ∈ C0,K(Ω), defines an isometric linear isomorphism. The

correspondence (29) is referred to as the Riesz correspondence. The Banach space MK(Ω),also carries another topology wmeas, referred to as the weak measure topology, which is definedas the pull-back of the w∗-topology via isomorphism (29). In other words,

µλwmeas−−−→ µ⇔ φµλ

w∗−→ φµ ⇔

∫Ω

f dµλ →∫

Ω

f dµ, ∀ f ∈ C(Ω).

(In order to remind us of this definition, the topology wmeas may be denoted simply by w∗.)Within MK(Ω) one has the following w∗-closed convex subsets:

M+(Ω) = µ ∈MK(Ω) : µ positive Radon measure ;M+(Ω)1 = µ ∈M+(Ω) : µ(Ω) ≤ 1;Prob(Ω) = µ ∈M+(Ω) : µ(Ω) = 1.

The set Prob(Ω) is referred to as the set of probability Radon measures on Ω. Using (29), forµ ∈ M+(Ω), one has the equality µ(Ω) = |µ|(Ω) = ‖µ‖, so the sets M+(Ω)1 and Prob(Ω)are in fact w∗-compact

Remark 6.. With this terminology, Lemma 4 can be restated by using the so-calledpoint-mass measures δω ∈ Prob(Ω), ω ∈ Ω, defined as

δω =

1 if B 3 ω0 ifB 63 ω B ∈ Bor(Ω),

which correspond under (29) to evaluation maps Φ(δω) = evω. The convex hull

Probdisc(Ω) = convδω : ω ∈ Ω,

is referred to as the space of discrete probability measures. With these definitions Lemma

4 (iv) reads: Prob(Ω) = Probdisc(Ω)w∗

, that is, every probability Radon measure on Ω is aw∗-limit of a net of discrete probability measures.

E*. Banach Spaces of Affine Functions

This optional section is devoted to the study of certain subspaces of CK(Ω) that appearnaturally when Ω carries additional structure. With one exception, all results here arepresented without proof, as a sequence of Exercises.

Definitions and Notations. Throughout this subsection we fix a locally convex topo-logical vector space X , and we assume Ω ⊂ X is some non-empty compact convex set. It isunderstood that on Ω one uses the induced topology.

Given a vector space Y , we say that a function f : Ω → Y is affine, if f(tx1 +(1− t)x2) =tf(x1) + (1− t)f(x2), ∀x1, x2 ∈ Ω, t ∈ [0, 1].

We specialize to Y = K, and we define the following spaces:

(a) CaffK (Ω) = f ∈ CK(Ω) : f affine ;

(b) if Ω 3 0, we also define Caff•,K(Ω) = f ∈ Caff

K (Ω) : f(0) = 0;

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As usual, when K = C, the subscript will be omitted from the notation.

Exercise 15. Prove that all spaces defined above are norm-closed in CK(Ω).

Exercise 16. Assume Ω 3 0. Prove that every f ∈ CaffK (Ω) can be uniquely represented

as f = α1+f0, with α ∈ K and f0 ∈ Caff•,K(Ω). Here 1 denotes the constant function 1, which

is obviously affine.Convention. From now on it will be understood that the spaces defined above are

equipped with the norm ‖ . ‖∞, coming from CK(Ω), so by the previous Exercises, they areBanach spaces.

Theorem-Definition 9. For any µ ∈ Prob(Ω), there exists a unique point xµ ∈ Ω, suchthat ∫

Ω

f dµ = f(xµ), ∀ fCaffK (Ω).

The point xµ is referred to as the µ-barycenter of Ω.

Proof. For the existence we argue as follows. Let φ ∈ CK(Ω)∗ be the linear continuousfunctional given by the Riesz correspondence, i.e.

φ(f) =

∫Ω

f dµ, f ∈ CK(Ω).

Using Lemma 4, there exists a net (φλ)λ∈Λ ⊂ conv[ev(Ω)], such that φλw∗−→ φ, i.e.

φλ(g) → φ(g), ∀ g ∈ CK(Ω). (30)

Write, for each λ ∈ Λ, the functional φλ as a convex combination of evaluation functionals

φλ =

Nλ∑k=1

t(k)λ ev

x(k)λ,

with x(1)λ , . . . , x

(Nλ)λ ∈ Ω, and t

(1)λ , . . . , t

(Nλ)λ ∈ [0, 1] satisfying

∑Nλ

k=1 t(k) = 1, so that

φλ(g) =

Nλ∑k=1

t(k)λ g(x

(k)λ ), ∀ g ∈ CK(Ω).

If we specialize to affine functions f , we have φλ(f) =∑Nλ

k=1 t(k)λ f(x

(k)λ ) = f

( ∑Nλ

k=1 t(k)λ x

(k)λ

),

so if we define for each λ the point yλ = t(k)λ x

(k)λ ∈ Ω, then

φλ(f) = f(yλ), ∀ f ∈ CaffK (Ω).

Since Ω is compact, the net (yλ)λ∈Λ ⊂ Ω has a sub-net (yγ(σ))σ∈Σ, which converges to somepoint x ∈ Ω, and by continuity and (30) we get:

f(x) = limσf(yγ(σ)) = lim

σφγ(σ)(f) = φ(f), ∀ f ∈ Caff

K (Ω).

For the uniqueness part, we start with two points x, x′ ∈ Ω, such that

f(x) = φ(f) = f(x′), ∀ f ∈ CaffK (Ω), (31)

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and we must show that x = x′. But this is trivial, since for every η ∈ X ∗, the restriction η∣∣Ω

is obviously an affine continuous function, so (31) implies

η(x) = η(x′), ∀ η ∈ X ∗,

which clearly forces x = x′, since X ∗ separates to points in X .

Remark 7. As seen in the above proof, the µ-barycenter uniquely determined by theidentity

η(xµ) =

∫Ω

η(x) dµ(x), ∀ η ∈ X ∗.

The point xµ is sometimes denoted by∫

Ωx dµ(x). This notation is dangerous, since it requires

some development of Integration Theory for X -valued functions. (In a very elementaryframework, we will discuss this in BS VI.)

Comment. Using the results from sub-section D, Theorem 9, combined with the Hahn-Banach Theorem, which shows that any linear continuous functional in Caff

K (Ω)∗ can beextended to a linear continuous functional in CK(Ω)∗, one can show that, for any φ ∈ Caff

K (Ω)∗,there exist x1, . . . , xn ∈ Ω and α1, . . . , αn ∈ K, such that

φ(f) =n∑

k=1

αkf(xk), ∀ f ∈ CaffK (Ω).

Comment. In what follows we shall see that the above presentation can be somehowsimplified, in the case when Ω 3 0 and K = R. (By an obvious translation argument this canbe adapted to the general case.) As pointed out in Exercise ??, if Ω 3 0, every f ∈ Caff

R (Ω)can be uniquely presented as f = α1 + f0, with α ∈ R and f0 ∈ Caff

•,R(Ω). The coefficient

α is nothing else but the value f(0) = ev0(f). In particular, for any φ ∈ CaffR (Ω), we have

φ(f) = f(0)φ(1) + φ(f0), so φ is completely determined by the value φ(1) ∈ R and by itsrestriction φ0 = φ

∣∣Caff•,R(Ω)

, so we must focus on the dual Banach space Caff•,R(Ω)∗.

Exercise 17*. Assume Ω 3 0, and consider the space Z = spanRΩ.

(i) Prove that every f ∈ Caff•,R(Ω) can be uniquely extended to a linear functional F : Z →

R. Therefore, the space Caff•,R(Ω) can be represented as:

Caff•,R(Ω) = F ∈ Z ′ : F

∣∣Ω

continuous .

(Warning! In the right-hand side, F only belongs to the algebraic dual of Z, so it isnot assumed that F is continuous on Z, relative to the induced topology from X .)

(ii) Prove that for every φ ∈ Caff•,R(Ω)∗. there exist two pairs (α1, x1), (α2, x2) ∈ [0,∞)×Ω,

such thatφ(f) = α1f(x1)− α2f(x2), ∀ f ∈ Caff

•,R(Ω). (32)

(iii) Prove that, if (α′1, x′1), (α

′2, x

′2) ∈ [0,∞) × Ω are two other pairs satisfying (32), then

α′1x′1 − α′2x

′2 = α1x1 − α2x2 (in Z).

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(iv) Show that there exists a unique linear isomorphism map Φ : Z → Caff•,R(Ω)∗, such that

(Φx)(f) = f(x), ∀ f ∈ Caff•,R(Ω), x ∈ Ω.

(v) Let B = conv[Ω ∪ (−1)Ω]. Use the notations from LCVS III C. Prove that, if weequip Z with the B-norm ‖ . ‖B (i.e. the Minkowski functional on Z, associated withB, which balanced and absorbing in Z), then Φ is isometric.

(vi) Prove that the restriction map

R : Caff•,R(B) 3 f 7−→ f

∣∣Ω∈ Caff

•,R(Ω)

is an isometric linear isomorphism.

Comment. The isomorphism Φ : Z ∼−→ Caff•,R(Ω)∗ has a complex counterpart. Specifically,

in the complex case, every φ ∈ Caff•,C(Ω)∗ can be uniquely presented as a formal combination

x1 + ix2, x1, x2 ∈ Z. (This is analogous to the hermitian decomposition.) This allows one oidentify the complex dual Banach space Caff

•,C(Ω)∗ with the complexification of Z. The detailsare left to the reader.

We conclude with two “affine” Stone-Weierstrass Theorems.

Exercise 18. Let Ω be compact convex with Ω 3 0, and let A ⊂ Caff•,K(Ω) be a linear

subspace, which is involutive10, i.e. f ∈ A ⇒ f ∈ A. Prove that the following are equivalent:

(i) A is dense in Caff•,K(Ω);

(ii) A separates the points in Ω.

(Hint: For the implication (ii) ⇒ (i), analyze first the real case. Assuming K = R, showthat if φ ∈ Caff

•,R(Ω)∗ is such that φ∣∣A = 0, then φ = 0. Use Exercise ??, by which there

exist – after a suitable rescaling – α ∈ K and x1, x2 ∈ Ω, such that φ(f) = α[f(x1)− f(x2)],∀ f ∈ Caff

•,R(Ω). Conclude that, since φ∣∣A = 0, then either α = 0, or x1 = x2. In the

complex case consider AR = Re f : f ∈ A and use the same arguments as in the proof ofStone-Weierstrass Theorem.)

Exercise 19. Let Ω be compact convex, and let A ⊂ CaffK (Ω) be a linear subspace, which

is involutive, and contains the constant function 1. Prove that the following are equivalent:

(i) A is dense in CaffK (Ω);

(ii) A separates the points in Ω.

(Hint: Since the statement does not account for the position of Ω in X , one can assumethat Ω 3 0. Consider then the space A0 = A ∩ Caff

•,K(Ω) = f ∈ A : f(0) = 0 and showthat any f ∈ A can be uniquely presented as f = α1+f0, with f0 ∈ A0. For the implication(ii) ⇒ (i), use the preceding Exercise to show that A0 is dense in Caff

•,K(Ω).)

10 This is meaningful only if K = C. It is pretty obvious that, if f is affine, then so is f .

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