6. Banach algebras of continuous functions

6. Banach algebras of continuous functions In this section we apply several Banach space and Banach algebra techniques to prove several key result concerning the commutative Banach algebras of the form: C K (X), for a compact Hausdorff space, C0K (Ω) and CbK (Ω), for a locally compact space Ω. A first useful result is the following. Lemma 6.1 (Urysohn type density). Let Ω be a topological space, let C ⊂ CbR (Ω) be a linear subspace, which contains the constant function 1. Assume (u) for any two closed sets A, B ⊂ Ω, with A ∩ B = ∅, there exists a function h ∈ C, such that h A = 0, h B = 1, and h(Ω) ∈ [0, 1], for all Ω ∈ Ω. Then C is dense in CbR (Ω), in the norm topology. Proof. The key step in the proof will be the following: Claim: For any f ∈ CbR (Ω), there exists g ∈ C, such that kg − f k ≤

2 kf k. 3

To prove this claim we define α = inf f (p) and β = sup f (x), p∈Ω

p∈Ω

so that f (p) ⊂ [α, β], and kf k = max{|α|, |β|}. Define the sets α + 2β 2α + β and B = f −1 ,β . A = f −1 α, 3 3 so that both A and B are closed, and A ∩ B = ∅. Use the hypothesis, to find a function h ∈ C, such that h A = 0, h B = 1, and h(p) ∈ [0, 1], for all p ∈ Ω. Define the function g ∈ C by 1 g = α1 + (β − α)k . 3 Let us examine the difference g − f . Start with some arbitrary point p ∈ Ω. There are three cases to examine: α Case I: p ∈ A. In this case we have h(p) = 0, so we get g(p) = . By the 3 2α + β construction of A we also have α ≤ f (p) ≤ , so we get 3 2α α+β ≤ f (p) − g(p) ≤ . 3 3 β Case II: p ∈ B. In this case we have h(p) = 1, so we get g(p) = . We also 3 2β + α have ≤ f (p) ≤ β, so we get 3 α+β 2β ≤ f (p) − g(p) ≤ . 3 3 79

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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Case III: p ∈ Ω r (A ∪ B). In this case we have 0 ≤ h(p) ≤ 1, so we get α β 2α + β α + 2β ≤ g(p) ≤ , and < f (p) < . In particular we get 3 3 3 3 2α + β β 2α f (p) − g(p) > − = ; 3 3 3 α + 2β α 2β f (p) − g(p) < − = . 3 3 3 2α α+β 2β Since ≤ ≤ , we see that in all three cases we have 3 3 3 2α 2β ≤ f (p) − g(p) ≤ , 3 3 so we get 2β 2α ≤ inf f (p) − g(p) ≤ sup f (p) − g(p) ≤ , p∈Ω 3 3 p∈Ω so we indeed get the desired inequality kg − f k ≤

2 kf k. 3

Having proven the Claim, we now prove the density of C in CbR (Ω). Start with some f ∈ CbR (Ω), and we construct recursively two sequences (gn )n≥1 ⊂ C and (fn )n≥1 ⊂ CbR (Ω), as follows. Set f1 = f . Apply the Claim to find g1 ∈ C such that 2 kf1 k. 3 Once f1 , f2 , . . . , fn and g1 , g2 , . . . , gn have been constructed, we set kg1 − f k ≤

fn+1 = gn − fn , and we choose gn+1 ∈ C such that kgn+1 − fn+1 k ≤

2 kfn+1 k. 3

It is clear, by construction, that n−1 2 kfn k ≤ kf k, ∀ n ≥ 1. 3 Consider the sequence (sn )n≥1 ⊂ C of partial sums, defined by sn = g1 + g2 + · · · + gn , ∀ n ≥ 1. Using the equalities gn = fn − fn+1 , ∀ n ≥ 1, we get sn − f = g1 + g2 + · · · + gn − f1 = fn+1 , so we have ksn − f k ≤

n 2 kf k, ∀ n ≥ 1, 3

which clearly give f = limn→∞ sn , so f indeed belongs to the closure C. We are now in position to prove the following

§6. Banach algebras of continuous functions

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Theorem 6.1 (Tietze Extension Theorem). Let Ω be a normal topological space, let T ⊂ Ω be a closed subset. Let f : T → [0, 1] be a continuous function. (Here Y is equipped with the induced topology.) There there exists a continuous function g : Ω → [0, 1] such that g T = f . Proof. We consider the real Banach spaces (in fact algebras) C R (Ω) and To avoid any confusion, the norms on these Banach spaces will be denoted by k · kΩ and k · kT . If we define the restriction map R : CbR (Ω) 3 g 7−→ g T ∈ CbR (T ), CbR (T ).

then R is obviously linear and continuous. We define the subspace C = R CbR (Ω) ⊂ CbR (T ). Claim: For every f ∈ C, there exists some g ∈ CbR (Ω) such that f = Rg, and inf f (q) ≤ g(p) ≤ supq∈T f (q), ∀ p ∈ Ω.

q∈T

To prove this fact, we start first with some arbitrary g0 ∈ CbR (Ω), such that f = Rg0 = g0 Y . Put α = inf f (q) and β = sup f (q), q∈T

q∈T

so that kf kT = max |α|, |β| . Define the function θ : R → [α, β] by   α if t < α t if α ≤ t ≤ β θ(t) =  β if t > β Then obviously θ is continuous, and the composition g = θ ◦ g0 : Ω → [α, β] will still satisfy g T = f , and we will clearly have α ≤ g(p) ≤ β, ∀ p ∈ Ω. Having proven the Claim, we are going to prove that C is closed. We do this by showing that C is a Banach space, in the norm k · kT . By the Claim, we know that for every f ∈ C = Ran R, there exists some g ∈ CbR (Ω) with Rg = f and kgk = kf k. The fact, that C is a Banach space, then follows from Proposition 2.1. Let us remark now that obviously C contains the constant function 1 = R1. Using Urysohn Lemma (applied to T ) it is clear that C satifies the condition (u) in the above lemma. Using the Lemma 6.1, it follows that C = CbR (T ), i.e. R is surjective. To finish the proof, start with some arbitrary continuous function f : Y → [0, 1]. Use surjectivity of R, combined with the Claim, to find g ∈ CbR (Ω), such that Rg = f , and inf f (q) ≤ g(p) ≤ sup f (q), ∀ p ∈ Ω. q∈T

q∈T

This clearly forces g to take values in [0, 1].

Next we concentrate on the Banach algebras of the form C K (X), where X is a compact Hausdorff space. (When K = C, it will be ommited from the notation.) Exercise 1 ♦ . Define the sequence (Pn )n≥1 of polynomials, by P1 (t) = 0, and 1 Pn+1 (t) = t − Pn (t)2 + Pn (t), ∀ n ≥ 1. 2

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Prove that lim

n→∞

√ max Pn (t) − t = 0.

t∈[0,1]

√ Hint: Define the functions fn , f : [0, 1] → R by fn (t) = Pn (t) and f (t) = t. Prove that, for every t ∈ [0, 1], the sequence fn (t) n≥1 is incresing, bounded, and limn→∞ fn (t) = f (t). Then apply Dini’s Theorem.

Theorem 6.2 (Stone-Weierstrass). Let X be a compact Hausdorff space. Let A ⊂ C R (X) be a subalgebra, which contains the constant function 1. Assume A separates the points of X, i.e. for any p, q ∈ X, with p 6= q, there exists f ∈ A such that f (p) 6= f (q). Then A is dense in C R (X), in the norm topology. Proof. Let C denote the closure of A. Remark that C is again a sub-algebra, it contains the constant function 1, and it still separates the points of X. The proof will eventually use the Urysohn density Lemma. Before we get to that point, we need several preparations. Step 1. If f ∈ C, then |f | ∈ C. To prove this fact, we define g = f 2 ∈ C, and we set h = kgk−1 g, so that h ∈ C, and h(p) ∈ [0, 1], for all p ∈ K. Let Pn (t), n ≥ 1 be the polynominals defined in the above exercise. The functions hn = Pn ◦ h, n ≥ 1 are clearly all in C. By the above Exercise, we clearly get p lim max |hn (p) − h(p)| = 0, n→∞ p∈X √ √ which means that limn→∞ hn = h, in the norm topology. In particular, h belongs to C. Obviously we have √ h = kf k−1 · |f |, so |f | indeed belongs to C. Step 2: Given two functions f, g ∈ C, the continuous functions max{f, g} and min{f, g} both belong to C. This follows immediately from Step 1, and the equalities 1 1 max{f, g} = f + g + |f − g| and min{f, g} = f + g − |f − g| . 2 2 Step 3: For any two points p, q ∈ X, p 6= q, there exists h ∈ C, such that h(p) = 0, h(q) = 1, and h(s) ∈ [0, 1], ∀ s ∈ X. Use the assumption on A, to find first a function f ∈ A, such that f (p) 6= f (q). Put α = f (p) and β = f (q), and define 1 g= f − α1 . β−α The function g still belongs to A, but now we have g(p) = 0 and g(q) = 1. Define the function h = min{g 2 , 1}. By Step 3, h ∈ C, and it clearly satisfies the required properties. Step 4: Given a closed subset A ⊂ X, and a point p ∈ X r A, there exists a function h ∈ C, such that h(p) = 0, h A = 1, and h(q) ∈ [0, 1], ∀ q ∈ X. For every q ∈ A, we use Step 3 to find a function hq ∈ C, such that hq (p) = 0, hq (q) = 1, and hq (s) ∈ [0, 1], ∀ s ∈ X, and we define the open set Dq = {s ∈ X : hq (s) > 0}.


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Using the compactness of A, we find points q1 , . . . , qn ∈ A, such that A ⊂ Dq1 ∪ · · · ∪ Dqn . Define the function f = hq1 + · · · + hqn ∈ C, so that f (p) = 0, f (q) > 0, for all q ∈ A, and f (s) ≥ 0, ∀ s ∈ X. If we define m = min f (q), q∈A

then the function g = m f again belongs to C, and it satisfies g(p) = 0, g(q) ≥ 1, ∀ q ∈ A, and g(s) ≥ 0, ∀ s ∈ X. Finally, the function −1

h = min{g, 1} will satisfy the required properties. Step 5: Given closed sets A, B ⊂ X with A ∩ B = ∅, there exists h ∈ C, such that h A = 1, h B = 0, and h(q) ∈ [0, 1], ∀ q ∈ X. Use Step 4, to find for every p ∈ B, a function hp ∈ C, such that hp B = 1, hp (p) = 0, and hp (s) ∈ [0, 1], ∀ s ∈ X. Put gp = 1 − hp , so that gp (p) = 1, gp B = 0, and gp (s) ∈ [0, 1], ∀ s ∈ X. We the proceed as above. For each p ∈ A we define the open set Dp = {s ∈ X : gp (s) > 0}. Using the compactness of A, we find points p1 , . . . , pn ∈ A, such that A ⊂ Dp1 ∪ · · · ∪ Dpn . Define the function f = gp1 + · · · + gpn ∈ C, so that f B = 0, f (q) > 0, for all q ∈ A, and f (s) ≥ 0, ∀ s ∈ K. If we define m = min f (q), q∈A

then the function g = m f again belongs to C, and it satisfies g B = 0, g(q) ≥ 1, ∀ q ∈ A, and g(s) ≥ 0, ∀ s ∈ X. Finally, the function −1

h = min{g, 1} will satisfy the required properties. We now apply the Urysohn density Lemma, to conclude that C is dense in C R (X). Since C is already closed, this forces C = C R (X), i.e. A is dense in C R (X). Corollary 6.1 (Complex version of Stone-Weierstrass Theorem). Let X be a compact Hausdorff space. Let A ⊂ C(X) be a ?-subalgebra (i.e. f ∈ A ⇒ f ∈ A), which contains the constant function 1. Assume A separates the points of X. Then A is dense in C(X), in the norm topology. Proof. Consider the subalgebra AR = {f ∈ A : f = f }. It is clear that A = AR + iAR , and AR is a unital sub-algebra of C R (X), which separates the points of X. Using the real version, we know that AR is dense in C R (X). Then A is clearly dense in C(X).

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Example 6.1. Consider the unit disk D = {λ ∈ C : |λ| < 1}, and let D denote its closure. Consider the algebra A ⊂ C(D) consisting of all polynomial functions. Notice that, although A is unital and separates the points of D, it does not have the property f ∈ A ⇒ f ∈ A. In fact, one way to see that this property fails is by inspecting the closure of A in C(D). This closure is denoted by A(D) and is called the disk algebra. The main feature of A(D) is the following: Exercise 2*. Prove that A(D) = f : D → C : f continuous, and f D holomorphic . Exercise 3. Consider the unit circle T = {ζ ∈ C : |ζ| = 1}. (i) Prove that the restriction map Φ : A(D) 3 f 7−→ f T ∈ C(T) is an isometric unital algebra homomorphism. In particular, the space A = Ran Φ is a closed subalgebra of C(T), which contains the unit (the constant function 1). (ii) Consider the function z : T → C defined by z(ζ) = ζ, ∀ ζ ∈ T. Prove that SpecA (z) = D. Note that by the results from Section 5, we know that SpecC(T) (z) = Ran z = T, so one has the inclusion SpecA (z) ) SpecC(T) (z). Hints: (i). Use the Maximum Modulus Principle, combined with the preceding exercise. (ii). Use the fact that Φ implements an isomorphism, so SpecA (z) = SpecA(D) ¯ (Z), where Z ∈ A(D) is the function Z(ζ) = ζ, ζ ∈ D.

We continue now with a discussion on the topological duals of the Banach algebras C(X). Notations. Let X be a compact Hausdorff space, and let K be one of the fields R or C. We define the space MK (X) = C K (X)∗ = {φ : C K (X) → K : φ K-linear continuous}. The unit ball will be denoted by MK (X)1 . When K = C, the superscript C will be omitted from the notation. Remarks 6.1. Let X be a compact Hausdorff space. The space M(K) = C(X)∗ carries a natural involution, defined as follows. For φ ∈ M(X), we define the map φ? : C(X) → C by φ? (f ) = φ(f ), ∀ f ∈ C(X). For every φ ∈ M(X), the map φ? : C(X) → C is again linear, continuous, and has kφ? k = kφk. The map φ? will be called the adjoint of φ. We used the term involution, because the map M(X) 3 φ 7−→ φ? ∈ M(X) has the following properties: • (φ? )? = φ, ∀ φ ∈ M(X);


85

• (φ + ψ)? = φ? + ψ ? , ∀ φ, ψ ∈ M(X); • (λφ)? = λφ? , ∀ φ, ∈ M(X), λ ∈ C. If we define the space of self-adjoint maps Msa (X) = {φ ∈ M(X) : φ? = φ}, then is clear that, for any φ ∈ Msa (X), the restriction φ C R (K) is real-valued. In fact, for φ ∈ M(X), one has φ? = φ ⇐⇒ φ R is real-valued. C (X)

Moreover, one has a map (1)

Msa (X) 3 φ 7−→ φ C R (X) ∈ MR (X),

which is an isomorphism of R-vector spaces. The inverse of this map is defined as follows. Start with some φ ∈ MR (X), i.e. φ : C R (X) → R is R-linear and continuous, and we define φˆ : C(X) → C by ˆ ) = φ(Re f ) + iφ(Im f ), ∀ f ∈ C(X). φ(f It turns out that φˆ is again linear, continuous, and self-adjoint. Moreover, the correspondence MR (X) 3 φ 7−→ φˆ ∈ Msa (X) is the inverse of (1). Proposition 6.1. Let K be a compact Hausdorff space. Then the map Msa (X) 3 φ 7−→ φ C R (X) ∈ MR (X) is isometric. Moreover, when the two spaces are equipped with the w∗ topology, this map is a homeomorphism. sa Proof. To prove the first statement, fix φ ∈ M (X). It is obvious that kφ C R (X) k ≤ kφk. To prove the other inequality, fix for the moment ε > 0, and choose f ∈ C(X) such that kf k ≤ 1, and

|φ(f )| ≥ kφk − ε. Choose a complex number λ with |λ| = 1, such that |φ(f )| = λφ(f ) = φ(λf ). If we write λf = g + ih, with g, h ∈ C R (X), then using the fact that φ is self-adoint, we will have |φ(f )| = φ(g). Since kgk ≤ kλf k = kf k ≤ 1, we will get |φ(f )| ≤ kφ R k, C (X)

so our choice of f will give kφk − ε ≤ kφ C R (X) k. Since this holds for all ε > 0, we get kφk ≤ kφ C R (X) k. The w∗ continuity (both ways) is obvious.

Convention. From now on, we will identify the space M (X) with M (X). R

sa

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Remark 6.2. Let X be a compact Hausdorff space. We denote the character space of the Banach algebra C K (X) simply by EK (X). (When K = C, we ommit is from the notation.) The following results were proven in Section 5. • One has the inclusion EK (X) ⊂ MK (X)1 . • Every character in EK (X) is of the form p , for some p ∈ X, where p : C K (X) → K is defined by p (f ) = f (p), ∀ f ∈ C K (X). • When we equip EK (X) with the w∗ topology, the correspondence E : X 3 p 7−→ p ∈ EK (X) is a homeomorphism. Here is an interesting application of the above results to topology. Theorem 6.3 (Urysohn Metrizatbility Theorem). Let X be a compact Hausdorff space. The following are equivalent: (i) (ii) (iiiR ) (iiiC )

X is metrizable; X is second countable, i.e. the topology has a countable base; the Banach space C R (X) is separable; the Banach space C(X) is separable.

Proof. (i) ⇒ (ii). We already know this fact. (See the section on metric spaces). (ii) ⇒ (iiiR ). Assume X is second countable. Fix a countable base {Dn : n ∈ N} for the topology. Consider the countable set ∆ = {(m, n) ∈ N2 : Dm ∩ Dn = ∅}. Claim: For any two points p, q ∈ X, with p 6= q, there exists a pair (m, n) ∈ ∆ with p ∈ Dm and q ∈ Dn . Indeed, since X is Hausdorff, there exist open sets U0 , V0 ⊂ K with p ∈ U0 , q ∈ V0 , and U0 ∩ V0 = ∅. Since X is (locally) compact, there exist open sets U, V ⊂ K, such that p ∈ U ⊂ U ⊂ U0 and q ∈ V ⊂ V ⊂ V0 . Finally, since {Dn : n ∈ N} is a basis for the topology, there exist m, n ∈ N such that p ∈ Dm ⊂ U and q ∈ Dn ⊂ V . Then clearly we have Dm ⊂ U ⊂ U0 , and Dn ⊂ V ⊂ V0 , which forces Dm ∩Dn = ∅. Having proven the Claim, for every pair (m, n) ∈ ∆ we choose (use Urysohn Lemma) a continuous function hmn : X → [0, 1] such that hmn Dm = 0 and hmn Dn = 1, and we define the countable family F = {hmn : (m, n) ∈ ∆}. Using the Claim, we know that F separates the points of X. We set P = {h ∈ C R (X) : h is a finite product of functions in F}. Notice that P is still countable, it also separates the points of K, but also has the property: f, g ∈ P ⇒ f g ∈ P. If we define A = Span({1} ∪ P),


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then A ⊂ C R (X) satisfies the hypothesis of the Stone-Weierstrass Theorem, hence A is dense in C R (X). Notice that if we define AQ = SpanQ ({1} ∪ P), i.e. the set of linear combinations of elements in {1} ∪ P with rational coefficients, then clearly AQ is dense in A, and so AQ is dense in C R (X). But now we are done, since AQ is obviously countable. (iiiR ) ⇒ (iiiC ). Assume C R (K) is separable. Let S ⊂ C R (X) be a countable dense set. Then the set S + iS = {f + ig : f, g ∈ S} is clearly countable, and dense in C(X). (iiiC ) ⇒ (i). Assume C(X) is separable. By the results from Section 4, it follows that, when equipped with the w∗ topology, the compact space M(X)1 = C(X)∗ 1 is metrizable. Then the compact subset E(X) ⊂ M(X)1 is also metrizable. Since X is homeomorphic to E(X), it follows that X itself is metrizable. Definition. Let X be a compact Hausdorff space, and let K be one of the fields R or C. A K-linear map φ : C K (X) → K is said to be positive, if it has the property f ∈ C R (X), f ≥ 0 =⇒ φ(f ) ≥ 0. Proposition 6.2 (Automatic continuity for positive linear maps). Let X be a compact Hausdorff space, and let K be one of the fields R or C. Any positive K-linear map φ : C K (X) → K is continuous. Moreover, one has the equality kφk = φ(1). Proof. In the case when K = C, it suffices to prove that φ C R (X) is continuous. Therefore, it suffices to prove the statement for K = R. Start with some arbitrary f ∈ C R (X), and define the function f± ∈ C R (X) by f+ = max{f, 0} and f− = max{−f, 0}, so that f± ≥ 0, f = f+ − f− , and kf k = max{kf+ k, kf− k}. On the one hand, by positivity, we have the inequalities φ(f± ) ≥ 0, so we get −φ(f− ) ≤ φ(f+ ) − φ(f− ) ≤ φ(f+ ), which give (2)

|φ(f )| = |φ(f+ ) − φ(f− )| ≤ max{φ(f+ ), φ(f− )}.

On the other hand, we have kf± k · 1 − f± ≥ 0, so by positivity we get kf± k · φ(1) ≥ φ(f± ). Using this in (2) gives |φ(f )| ≤ φ(1) · max{kf+ k, kf− k} = φ(1) · kf k. Since this holds for all f ∈ C R (X), the continuity of φ follows, together with the estimate kφk ≤ φ(1). Since φ(1) ≤ kφk · k1k = kφk, the desired norm equality follows.

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Notations. Let X be a compact Hausdorff space. We define K MK + (X) = {φ : C (X) → K : φ K-linear, positive}; K K K MK + (X)1 = {φ ∈ M+ (X) : kφk ≤ 1} = M+ (X) ∩ M (X)1 .

When K = C, the superscript C will be ommitted. Remarks 6.3. Let X be a compact Hausdorff space. We have the inclusion M+ (X) ⊂ Msa (X). Indeed, if we start with φ ∈ M+ (X), then using the fact that every real-valued continuous function f ∈ C(X) is a difference of non-negative continuous functions f = f+ −f− , it follows that φ(f ) = φ(f+ )−φ(f− ) is a difference of two non-negative (hence real) numbers, so φ(f ) ∈ R. This implies φ? = φ. ∗ R ∗ The set MR + (X) is w -closed in M (X), and the set M+ (X) is w -closed in R M(X). This follows from the fact that, for each f ∈ C (X), the set K AK f = {f ∈ M (X) : φ(f ) ≥ 0}

is w∗ -closed, being the preimage of a closed set [0, ∞) ⊂ K, under the w∗ -continuous map MK (X) 3 φ 7−→ φ(f ) ∈ K. Then everything is a consequence of the equality \ MK AK + (X) = f. f ∈C R (X) f ≥0 ∗ In particular, the sets MR + (X)1 and M+ (X)1 are w -compact. R The sets M+ (X)1 and M+ (X)1 are convex. Using the identification MR (X) ' Msa (X), we have the following hierarchies:

MR M+ (X) + (X) ' ∩ ∩ MR (X) ' Msa (X) ∩ M(X)

MR + (X)1 ∩ MR (X)1

M+ (X)1 ∩ ' Msa (X)1 ∩ M(X)1 '

with ' isometric and w∗ -homeomorphism. Proposition 6.3. Let X be a compact Hausdorff space. (i) The set conv E(X) ∪ {0} is w∗ -dense in M+ (X)1 . (ii) The set conv E(X) ∪ −E(X) is w∗ -dense in Msa (X)1 . (iii) One has the equality Msa (X)1 = conv M+ (X)1 ∪ −M+ (X)1 . (Here “conv” denotes the convex cover.) Proof. Properties (i) and (ii) will be proven simulatenously. Define the sets C1 = conv E(X) ∪ {0} ; D1 = M+ (X)1 ; C2 = conv E(X) ∪ −E(X) ; D2 = Msa (X)1 . Fix for the moment k ∈ {1, 2}. With the notations above, we must show that Ck is w∗ -dense in Dk . It is obvious that one has the inclusion Ck ⊂ Dk . Dk is w∗ -closed, so it contains the w∗ -closure Ck of Ck , so all we have to prove is the inclusion Dk ⊂ Ck . We prove this inclusion by contradiction. Assume there exists


89

some φ ∈ Dk rCk . Since Ck is convex, by Corollary 4.2, there exists some f ∈ C(X) and a real number α, such that Re φ(f ) < α ≤ Re σ(f ), ∀ σ ∈ Ck . In particular, if we take h = −Re f , and β = −α, we get (3)

φ(h) > β ≥ σ(h), ∀ σ ∈ Ck .

Since 0 ∈ Ck (this is obvious for k = 1; for k = 2 this follows from the inclusion C1 ⊂ C2 ), we have β ≥ 0. Since E(X) ⊂ Ck , we also get β ≥ p (h) = h(p), ∀ p ∈ X, which means that β1 − h ≥ 0. At this point we break the proof in two cases. case I: k = 1. In this case, since φ is positive, we will have φ(β1 − h) ≥ 0, which gives φ(h) ≤ φ(β1) = βφ(1) = βkφk. Since we also have kφk ≤ 1, this gives φ(h) ≤ β, which clearly contradicts (3). case II: k = 2. In this case, the set Ck also contains −E(X), so we also have β ≥ −p (h) = −h(p), ∀ p ∈ X, thus proving that β1 + h ≥ 0. So now we have β1 ± h ≥ 0, and the fact that h is real-valued gives the inequality khk ≤ β. But then, the fact that kφk ≤ 1 will give |φ(h)| ≤ kφk · khk ≤ β, again contradicting (3). (iii). Denote the set conv M+ (X)1 ∪ −M+ (X)1 simply by C. Claim: One has the equality: (4)

C = {tφ − (1 − t)ψ : φ, ψ ∈ M+ (X)1 , t ∈ [0, 1]}. In particular, the set C is w∗ -compact.

Denote the set on the right hand side of (4) simply by C0 . The inclusion C ⊃ C0 is clear. To prove the inclusion C ⊂ C0 , we only need to prove that C0 is convex and it contains M+ (X)1 ∪ −M+ (X)1 . The second property is clear. The convexity of C0 is also clear, being a consequence of the convexity of ±M+ (X)1 . The w∗ -compactness of C is then a consequence of the compatness of the product space M+ (X)1 × M+ (X)1 × [0, 1], and of the fact that C is the range of the continuous map M+ (X)1 × M+ (X)1 × [0, 1] 3 (φ, ψ, t) 7−→ tφ − (1 − t)ψ ∈ Msa (X). Having proven the Claim, we now proceed with the equality Msa (X)1 = C. Since E(X) ⊂ M+ (X)1 , it is obvious that C ⊃ conv(E(X) ∪ −E(X)) = C2 . Since C is w∗ -closed, using part (ii), it follows that C ⊃ C2 = Msa (X)1 . The other inclusion C ⊂ Msa (X)1 is trivial, since Msa (X)1 is convex, and it contains ±M+ (X)1 . Corollary 6.2. Let X be a compact Hausdorff space, and let φ ∈ Msa (X). Then there exist φ1 , φ2 ∈ M+ (X), such that φ = φ1 − φ2 , and kφk = kφ1 k + kφ2 k.

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Proof. If φ ∈ M+ (X) ∪ −M+ (X), there is nothing to prove. Assume φ 6∈ φ M+ (X)∪−M+ (X), in particular φ 6= 0. We define ψ = , so that ψ ∈ Msa (X)1 . kφk Find ψ1 , ψ2 ∈ M+ (X)1 and t ∈ [0, 1], such that ψ = tψ1 − (1 − t)ψ2 . Since ψ 6∈ M+ (X) ∪ −M+ (X), it follows that 0 < t < 1. Notice that 1 = kψk = ktψ1 − (1 − t)ψ2 k ≤ tkψ1 k + (1 − t)kψ2 k. If kψ1 k < 1, or kψ2 k < 1, then this would imply tkψ1 k + (1 − t)kψ2 k < 1, which is impossible by the above estimate. This argument proves that we must have kψ1 k = kψ2 k = 1. If we define φ1 = tkφkψ1 and φ2 = (1 − t)kφkψ2 , then kφ1 k = tkφk and kφ2 k = (1 − t)kφk, so we indeed have kφ1 k + kφ2 k = kφk. Obviously φ1 and φ2 are positive, and φ1 − φ2 = kφk · tψ1 − (1 − t)ψ2 = kφk · ψ = φ. The results for the Banach spaces of the form C(X), with X compact Hausdorff space, can be generalized, with suitable modifications, to the situation when X is replaced with a locally compact space. The following result in fact reduces the analysis to the compact case. Theorem 6.4. Let Ω be a locally compact space, and let Ωβ be the Stone-Cech compactification of Ω. Then the restriction map R : C K (Ωβ ) 3 f 7−→ f Ω ∈ CbK (Ω) is an isometric unital algebra isomorphism. Proof. The fact that R is a unital algebra isomorphism is clear. We show that R is bijective, by exhibiting an inverse for it. For every h ∈ CbK (Ω), we consider the compact set Kh = {z ∈ K : |z| ≤ khk}, so that we can regard h as a continuous map Ω → Kh . We know from the functoriality of the Stone-Cech compactification that there exists a unique continuous map hβ : Ωβ → Khβ , with hβ Ω = h. Since Kh is compact, we have Khβ = Kh . In particular, this gives the inequality (5)

|hβ (x)| ≤ khk, ∀ x ∈ Ωβ .

Define the map T : CbK (Ω) 3 h 7−→ hβ ∈ C K (Ωβ ), and let us show that T is an inverse for R. The equality R ◦ T = Id is trivial, by construction. To prove the equality T ◦ R = Id, we start with some f ∈ CbK (Ω), and we consider h = Rf . Then T h = hβ , and since hβ Ω = h = f Ω , the denisty of Ω in Ωβ clearly forces f = hβ = T h = T (Rf ). The fact that R is isometric is now clear, because on the one hand we clearly have kRf k ≤ kf k, ∀ f ∈ C K (Ωβ ), and on the other hand, by (5), we also have kT hk ≤ khk, ∀ h ∈ CbK (Ω).


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We conclude with a couple of generalizations of the various results in this section, in which the algebras of the form C(X) - with X compact - are replaced with algebras of the form C0 (Ω) with Ω locally compact. The following result is a generalization of Proposition 6.2. Proposition 6.4. Let Ω be a locally compact space, and let φ : C0R (Ω) → R be a positive linear map. Then φ is continuous, and one has the equality (6)

kφk = sup{φ(f ) : f ∈ C0R (Ω), 0 ≤ f ≤ 1}.

Proof. Let us denote the right hand side of (6) by M . First we show that R M < ∞. If M = ∞, there exists a sequence (fn )∞ n=1 ⊂ C0 (Ω), such that 0 ≤ fn ≤ 1 and φ(fn ) ≥ 4n , ∀ n ≥ 1.

P∞ P∞ P∞ Consider then the function f = n=1 21n fn . Since n=1 21n fn ≤ n=1 21n = 1, it follows that f ∈ C0R (Ω). Notice however that, since we obviously have 21n fn ≤ f , by the positivity of φ, we get 1 1 φ(f ) ≥ φ n fn = n φ(fn ) ≥ 2n , ∀ n ≥ 1, 2 2 which is clearly impossible. Let us show now that φ is continuous, by proving the inequality (7)

|φ(f )| ≤ M, ∀ f ∈ C0R (Ω), with kf k ≤ 1.

Start with some arbitrary function f ∈ C0R (Ω). The functions g ± = |f |±f ∈ C0R (Ω), clearly satisfy g ≥ 0, so we get φ(|f | ± f ) ≥ 0, so we get φ(|f |) ≥ ±φ(f ). This gives |φ(f )| ≤ φ(|f |), and since 0 ≤ |f | ≤ 1, we immediately get (7). The inequality (7) proves the inequality kφk ≤ M . Since we obviously have M ≤ kφk, we get in fact the equality (6). Corollary 6.3. Let Ω be a locally compact space, which is non-compact, and let Ωα be the Alexandrov compactification of Ω. Using the identification C0R (Ω) = f ∈ C R (Ωα ) : f (∞) = 0 , every positive linear map φ : C0R (Ω) → R can be uniquely extended to a positive linear map ψ : C0R (Ω) → R, such that kψk = kφk. Proof. For every g ∈ C R (Ωα ), we know that there exists a unique λ ∈ R and f ∈ C0R (Ω), such that g = λ1 + f (namely λ = g(∞) and f = g − λ1). We then define ψ(g) = λkφk + φ(f ). Notice that ψ(1) = kφk. It is obvious that ψ : C R (Ωα ) → R is linear, and ψ C R (Ω) = φ. Let us show that ψ is positive. 0

Start with some g ∈ C R (Ωα ) with g ≥ 0, and let us prove that ψ(g) ≥ 0. Write g = λ1 + f with λ ∈ R and f ∈ C0R (Ω). We know that λ = g(∞) ≥ 0. If λ = 0, there is nothing to prove. If λ > 0, we define the function h = λ−1 f ∈ C0R (Ω), so that g = λ(1 + h). The positivity of g forces 1 + h ≥ 0, which means if we consider the function h− = max{−h, 0} ∈ C0R (Ω), then we have 0 ≤ h− ≤ 1, as well as h− + h ≥ 0. Using the above result, this will then give kφk + φ(h) ≥ φ(h− ) + φ(h) = φ(h− + h) ≥ 0, which means that ψ(1 + h) ≥ 0. Consequently we also get ψ(g) = ψ(λ(1 + h)) = λψ(h) ≥ 0.

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Having shown the positivity of ψ, we know that kψk = ψ(1) = kφk. R α To prove uniqueness, start with another positive linear map ξ : C (Ω ) → R, such that kξk = kφk, with ξ C R (Ω) = φ. Since ξ is positive, this forces ξ(1) = kξk = 0 kφk = ψ(1). But then we have

ξ(λ1 + f ) = λkφk + φ(f ) = ψ(λ1 + f ), ∀ λ ∈ R, f ∈ C0R (Ω), which proves that ξ = ψ.

Remark 6.4. Let Ω be a locally compact space, which is not compact. Consider the vector space CcR (Ω) of all real-valued continuous functions on Ω, with compact support. We know that CcR (Ω) is dense in C0R (Ω). The notion of positivity makes sense also for linear maps CcR (Ω) → R. When we deal only with this (smaller) subspace, positivity does not imply continuity (see the Exercise below). In fact, if φ : CcR (Ω) → R is a positive linear map, then the following are equivalent: (i) φ is continuous; (ii) sup φ(f ) : f ∈ CcR (Ω), 0 ≤ f ≤ 1 < ∞. The implication (i) ⇒ (ii) is trivial. To prove the implication (ii) ⇒ (i) we follow the exact same steps as in the proof of the equality (6) in Proposition 6.4. Denote the quantity in (ii) by M , and using the inequality |φ(f )| ≤ φ(|f |), we immediately get |φ(f )| ≤ M, ∀ f ∈ CcR (Ω), with kf k ≤ 1. Remark also that if φ is as above, then we have in fact the equality kφk = sup φ(f ) : f ∈ CcR (Ω), 0 ≤ f ≤ 1 . Exercise 4. Prove that, for every function f ∈ CcR (R), the sum φ(f ) =

∞ X

f (n)

n=1

has only finitely many non-zero terms. Prove that the map φ : CcR (R) → R, defined by the above formula, is linear, positive, but not continuous. The following is a generalization of Corollary 6.2. Proposition 6.5. Let Ω be a locally compact space, and let φ : C0R (Ω) → R be a linear continuous map. Then there exist positive linear maps φ1 , φ2 : C0R (Ω) → R, such that φ = φ1 − φ2 , and kφk = kφ1 k + kφ2 k. Proof. If Ω is compact there is nothing to prove (this is Corollary 5.3). Assume Ω is non-compact. Use Hahn-Banach Theorem to find a linear continuous map ψ : C R (Ωα ) → R, with kψk = 1 and ψ C R (Ω) = φ. Apply Corollary 5.3 to 0

find two positive linear maps ψ1 , ψ2 : C R (Ωα ) → R such that ψ = ψ1 − ψ2 and kψk = kψ1 k + kψ2 k. Define the positive linear maps φk = ψk C R (Ω) , k = 1, 2. We 0 clearly have φ = φ1 − φ2 , and kφ1 k + kφ2 k ≤ kψ1 k + kψ2 k = kψk = kφk = kφ1 − φ2 k ≤ kφ1 k + kφ2 k, which forces kφk = kφ1 k + kφ2 k.


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Proposition 6.6 (Stone-Weierstrass Theorems for locally compact spaces). Let Ω be a locally compact space, which is non-compact, and let A ⊂ C0K (Ω) be a subalgebra, with the following separation properties: (i) for any two points ω1 , ω2 ∈ Ω, with ω1 6= ω2 , there exists f ∈ A such that f (ω1 ) 6= f (ω2 ); (ii) for any ω ∈ Ω, there exists f ∈ A with f (ω) 6= 0. A. If K = R, then A is dense in C0R (Ω). B. If K = C, and if A is a ?-subalgebra (i.e. f ∈ A ⇒ f ∈ A), then A is dense in C0 (Ω). Proof. Work in the Banach algebra C K (Ωα ), where Ωα = Ω ∪ {∞} is the Alexandrov compactification, and identify C0K (Ω) = f ∈ C K (Ωα ) : f (∞) = 0 . Claim 1: Using the above identification, A separates the points of Ωα . Indeed if one starts with two different points in Ωα , then either they both belong to Ω, in which case we use (i), or one of them is in Ω and the other is ∞, in which case we use (ii). Claim 2: In either case the space B = K1 + A. is dense in C K (Ωα ). First of all, B is obviously a subalgebra of C K (Ωα ), which contains the constant function 1. Moreover, in case B, it follows that B is a ?-subalgebra. The Claim then follows from the Stone-Weierstrass Theorem and Claim 1 (due to inclusion A ⊂ B, it follows that B also separates the points of Ωα ). We can now finish the proof. Start with some arbitrary function f ∈ C0K (Ω), and let us find a sequence in A, which converges to f . Use Claim 2 to find a sequence (gn )∞ n=1 ⊂ B, such that f = limn→∞ gn , in the norm topology. For each n ≥ 1 we write gn = αn 1 + fn , with αn ∈ K and fn ∈ A. We are going to prove that limn→∞ fn = f . Since we have kfn − f k = kgn − f − αn 1k ≤ kgn − f k + kαn 1k = kgn − f k + |αn |, ∀ n ≥ 1, the proof will be finished once we show that limn→∞ αn = 0. But this is quite clear, since one has has the equalities αn = gn (∞), ∀ n ≥ 1, and then the fact that (gn )∞ n=1 converges uniformly to f forces limn→∞ αn = limn→∞ gn (∞) = f (∞) = 0.