6. Banach Algebras of Continuous Functions

Home , Disk algebra, Tietze extension theorem

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: K K K C (X), for a compact Hausdorﬀ space, C0 (Ω) and Cb (Ω), for a locally compact space Ω. A ﬁrst useful result is the following.

R Lemma 6.1 (Urysohn type density). Let Ω be a topological space, let C ⊂ Cb (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. R Then C is dense in Cb (Ω), in the norm topology. Proof. The key step in the proof will be the following: R Claim: For any f ∈ Cb (Ω), there exists g ∈ C, such that 2 kg − fk ≤ kfk. 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 kfk = max{|α|, |β|}. Define the sets 2α + β α + 2β A = f −1 α, and B = f −1 , β . 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 80 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 2α 2β ≤ inf f(p) − g(p) ≤ sup f(p) − g(p) ≤ , 3 p∈Ω p∈Ω 3 so we indeed get the desired inequality 2 kg − fk ≤ kfk. 3 R Having proven the Claim, we now prove the density of C in Cb (Ω). Start with R some f ∈ Cb (Ω), and we construct recursively two sequences (gn)n≥1 ⊂ C and R (fn)n≥1 ⊂ Cb (Ω), as follows. Set f1 = f. Apply the Claim to ﬁnd g1 ∈ C such that 2 kg − fk ≤ kf k. 1 3 1

Once f1, f2, . . . , fn and g1, g2, . . . , gn have been constructed, we set

fn+1 = gn − fn, and we choose gn+1 ∈ C such that 2 kg − f k ≤ kf k. n+1 n+1 3 n+1 It is clear, by construction, that 2n−1 kf k ≤ kfk, ∀ n ≥ 1. n 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 2n ks − fk ≤ kfk, ∀ n ≥ 1, n 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 81

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) CR(Ω) and R Cb (T ). 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 R R : Cb (Ω) 3 g 7−→ g T ∈ Cb (T ), then R is obviously linear and continuous. R R We deﬁne the subspace C = R Cb (Ω) ⊂ Cb (T ). R Claim: For every f ∈ C, there exists some g ∈ Cb (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 ∈ CR(Ω), such that f = b Rg0 = g0 Y . Put α = inf f(q) and β = sup f(q), q∈T q∈T so that kfkT = max |α|, |β| . Deﬁne the function θ : R → [α, β] by   α if t < α θ(t) = t if α ≤ t ≤ β  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 R for every f ∈ C = Ran R, there exists some g ∈ Cb (Ω) with Rg = f and kgk = kfk. 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) R in the above lemma. Using the Lemma 6.1, it follows that C = Cb (T ), i.e. R is surjective. To finish the proof, start with some arbitrary continuous function f : Y → [0, 1]. R Use surjectivity of R, combined with the Claim, to find g ∈ Cb (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 CK(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 P (t) = t − P (t)2 + P (t), ∀ n ≥ 1. n+1 2 n n 82 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Prove that √ lim max Pn(t) − t = 0. n→∞ 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 ⊂ CR(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 CR(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−1g, 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 = kfk−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{g2, 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}. §6. Banach algebras of continuous functions 83

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−1f again belongs to C, and it satisfies g(p) = 0, g(q) ≥ 1, ∀ q ∈ A, and g(s) ≥ 0, ∀ s ∈ X. Finally, the function 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 = 1, h = 0, and h(q) ∈ [0, 1], ∀ q ∈ X. A B Use Step 4, to find for every p ∈ B, a function hp ∈ C, such that hp = 1, B 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 −1 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 h = min{g, 1} will satisfy the required properties. We now apply the Urysohn density Lemma, to conclude that C is dense in CR(X). Since C is already closed, this forces C = CR(X), i.e. A is dense in CR(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, R and AR is a unital sub-algebra of C (X), which separates the points of X. Using R the real version, we know that AR is dense in C (X). Then A is clearly dense in C(X). 84 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

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( ) = f : → : f continuous, and f holomorphic . D D C D Exercise 3. Consider the unit circle T = {ζ ∈ C : |ζ| = 1}. (i) Prove that the restriction map

Φ: A( ) 3 f 7−→ f ∈ C( ) D T 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 Spec (z) = Ran z = , C(T) T so one has the inclusion Spec (z) Spec (z). A ) C(T) Hints: (i). Use the Maximum Modulus Principle, combined with the preceding exercise. (ii). Use the fact that Φ implements an isomorphism, so Spec (z) = Spec ¯ (Z), where A A(D) 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) = CK(X) = {φ : CK(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); §6. Banach algebras of continuous functions 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 is real-valued. In CR(K) fact, for φ ∈ M(X), one has φ? = φ ⇐⇒ φ is real-valued. CR(X) Moreover, one has a map (1) Msa(X) 3 φ 7−→ φ ∈ MR(X), CR(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 : CR(X) → R is R-linear and continuous, and we define φˆ : C(X) → C by φˆ(f) = φ(Re f) + iφ(Im f), ∀ f ∈ C(X). 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 ∈ MR(X) CR(X) is isometric. Moreover, when the two spaces are equipped with the w∗ topology, this map is a homeomorphism. Proof. To prove the first statement, fix φ ∈ Msa(X). It is obvious that kφ k ≤ kφk. To prove the other inequality, fix for the moment ε > 0, and CR(X) choose f ∈ C(X) such that kfk ≤ 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 ∈ CR(X), then using the fact that φ is self-adoint, we will have |φ(f)| = φ(g). Since kgk ≤ kλfk = kfk ≤ 1, we will get |φ(f)| ≤ kφ k, CR(X) so our choice of f will give kφk − ε ≤ kφ k. CR(X) Since this holds for all ε > 0, we get kφk ≤ kφ k. CR(X) ∗ The w continuity (both ways) is obvious. Convention. From now on, we will identify the space MR(X) with Msa(X). 86 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Remark 6.2. Let X be a compact Hausdorﬀ space. We denote the character space of the Banach algebra CK(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 : CK(X) → K is deﬁned by

p(f) = f(p), ∀ f ∈ CK(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 Haus- dorﬀ space. The following are equivalent: (i) X is metrizable; (ii) X is second countable, i.e. the topology has a countable base; R (iiiR) the Banach space C (X) is separable; (iiiC) 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 2 ∆ = {(m, n) ∈ N : 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 D = 0 and m hmn = 1, and we deﬁne the countable family Dn

F = {hmn :(m, n) ∈ ∆}. Using the Claim, we know that F separates the points of X. We set P = {h ∈ CR(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 ⇒ fg ∈ P. If we define A = Span({1} ∪ P), §6. Banach algebras of continuous functions 87 then A ⊂ CR(X) satisfies the hypothesis of the Stone-Weierstrass Theorem, hence A is dense in CR(X). Notice that if we define A = Span ({1} ∪ P), Q Q i.e. the set of linear combinations of elements in {1} ∪ P with rational coefficients, R then clearly AQ is dense in A, and so AQ is dense in C (X). But now we are done, since AQ is obviously countable. R R (iiiR) ⇒ (iiiC). Assume C (K) is separable. Let S ⊂ C (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 : CK(X) → K is said to be positive, if it has the property f ∈ CR(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 : CK(X) → K is continuous. Moreover, one has the equality kφk = φ(1).

Proof. In the case when = , it suffices to prove that φ is continuous. K C CR(X) Therefore, it suffices to prove the statement for K = R. Start with some arbitrary f ∈ CR(X), and define the function f± ∈ CR(X) by

f+ = max{f, 0} and f− = max{−f, 0}, so that f± ≥ 0, f = f+ − f−, and kfk = 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) · kfk. Since this holds for all f ∈ CR(X), the continuity of φ follows, together with the estimate kφk ≤ φ(1). Since φ(1) ≤ kφk · k1k = kφk, the desired norm equality follows. 88 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Notations. Let X be a compact Hausdorff space. We define K K M+(X) = {φ : C (X) → K : φ K-linear, positive}; K K K K M+(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 sa M+(X) ⊂ M (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 ∗ R ∗ The set M+(X) is w -closed in M (X), and the set M+(X) is w -closed in M(X). This follows from the fact that, for each f ∈ CR(X), the set

K K Af = {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

K \ K M+(X) = Af . f∈CR(X) f≥0

R ∗ In particular, the sets M+(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:

R R M+(X) ' M+(X) M+(X)1 ' M+(X)1 ∩ ∩ ∩ ∩ sa sa MR(X) ' M (X) MR(X)1 ' M (X)1 ∩ ∩ M(X) 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. ∗ sa (ii) The set conv E(X) ∪ −E(X) is w -dense in M (X)1. (iii) One has the equality sa M (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; sa C2 = conv E(X) ∪ −E(X) ; D2 = M (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 §6. Banach algebras of continuous functions 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 0 C 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

sa M+(X)1 × M+(X)1 × [0, 1] 3 (φ, ψ, t) 7−→ tφ − (1 − t)ψ ∈ M (X). sa Having proven the Claim, we now proceed with the equality M (X)1 = C. Since E(X) ⊂ M+(X)1, it is obvious that C ⊃ conv(E(X) ∪ −E(X)) = C2. Since C ∗ sa is w -closed, using part (ii), it follows that C ⊃ C2 = M (X)1. The other inclusion sa sa C ⊂ M (X)1 is trivial, since M (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 = kφ1k + kφ2k. 90 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

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) . + + kφk 1 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)ψ2k ≤ tkψ1k + (1 − t)kψ2k.

If kψ1k < 1, or kψ2k < 1, then this would imply tkψ1k + (1 − t)kψ2k < 1, which is impossible by the above estimate. This argument proves that we must have kψ1k = kψ2k = 1. If we deﬁne

φ1 = tkφkψ1 and φ2 = (1 − t)kφkψ2, then kφ1k = tkφk and kφ2k = (1 − t)kφk, so we indeed have kφ1k + kφ2k = 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

K β K R : C (Ω ) 3 f 7−→ f Ω ∈ Cb (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 ∈ K Cb (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 func- toriality of the Stone-Cech compactiﬁcation that there exists a unique continuous β β β β β map h :Ω → Kh , with h Ω = h. Since Kh is compact, we have Kh = Kh. In particular, this gives the inequality (5) |hβ(x)| ≤ khk, ∀ x ∈ Ωβ.

K β K β Deﬁne the map T : Cb (Ω) 3 h 7−→ h ∈ C (Ω ), and let us show that T is an inverse for R. The equality R ◦ T = Id is trivial, by construction. To prove the K equality T ◦ R = Id, we start with some f ∈ Cb (Ω), and we consider h = Rf. β β β Then T h = h , and since h Ω = h = f Ω, the denisty of Ω in Ω 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 kRfk ≤ kfk, ∀ f ∈ CK(Ωβ), and on the other hand, by (5), we also have K kT hk ≤ khk, ∀ h ∈ Cb (Ω). §6. Banach algebras of continuous functions 91

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.

R Proposition 6.4. Let Ω be a locally compact space, and let φ : C0 (Ω) → R be a positive linear map. Then φ is continuous, and one has the equality

R (6) kφk = sup{φ(f): f ∈ C0 (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 n 0 ≤ fn ≤ 1 and φ(fn) ≥ 4 , ∀ n ≥ 1. P∞ 1 P∞ 1 P∞ 1 Consider then the function f = n=1 2n fn. Since n=1 2n fn ≤ n=1 2n = 1, R 1 it follows that f ∈ C0 (Ω). Notice however that, since we obviously have 2n fn ≤ f, by the positivity of φ, we get 1 1 φ(f) ≥ φ f = φ(f ) ≥ 2n, ∀ n ≥ 1, 2n n 2n n which is clearly impossible. Let us show now that φ is continuous, by proving the inequality

R (7) |φ(f)| ≤ M, ∀ f ∈ C0 (Ω), with kfk ≤ 1. R ± R Start with some arbitrary function f ∈ C0 (Ω). The functions g = |f|±f ∈ C0 (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

R R α C0 (Ω) = f ∈ C (Ω ): f(∞) = 0 ,

R every positive linear map φ : C0 (Ω) → R can be uniquely extended to a positive R linear map ψ : C0 (Ω) → R, such that kψk = kφk. Proof. For every g ∈ CR(Ωα), we know that there exists a unique λ ∈ R R and f ∈ C0 (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 : CR(Ωα) → is linear, and ψ = φ. Let us show that ψ is positive. R R C0 (Ω) Start with some g ∈ CR(Ωα) with g ≥ 0, and let us prove that ψ(g) ≥ 0. Write R g = λ1 + f with λ ∈ R and f ∈ C0 (Ω). We know that λ = g(∞) ≥ 0. If λ = 0, −1 R there is nothing to prove. If λ > 0, we deﬁne the function h = λ f ∈ C0 (Ω), so that g = λ(1 + h). The positivity of g forces 1 + h ≥ 0, which means if we consider − R − the function h = max{−h, 0} ∈ C0 (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. 92 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Having shown the positivity of ψ, we know that kψk = ψ(1) = kφk.

To prove uniqueness, start with another positive linear map ξ : CR(Ωα) → R, such that kξk = kφk, with ξ = φ. Since ξ is positive, this forces ξ(1) = kξk = R C0 (Ω) kφk = ψ(1). But then we have

R ξ(λ1 + f) = λkφk + φ(f) = ψ(λ1 + f), ∀ λ ∈ R, f ∈ C0 (Ω), which proves that ξ = ψ. Remark 6.4. Let Ω be a locally compact space, which is not compact. Con- R sider the vector space Cc (Ω) of all real-valued continuous functions on Ω, with R R compact support. We know that Cc (Ω) is dense in C0 (Ω). The notion of posi- R tivity makes sense also for linear maps Cc (Ω) → R. When we deal only with this (smaller) subspace, positivity does not imply continuity (see the Exercise below). R In fact, if φ : Cc (Ω) → R is a positive linear map, then the following are equivalent: (i) φ is continuous; R (ii) sup φ(f): f ∈ Cc (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 R get |φ(f)| ≤ M, ∀ f ∈ Cc (Ω), with kfk ≤ 1. Remark also that if φ is as above, then we have in fact the equality

R kφk = sup φ(f): f ∈ Cc (Ω), 0 ≤ f ≤ 1 .

R Exercise 4. Prove that, for every function f ∈ Cc (R), the sum ∞ X φ(f) = f(n) n=1

R has only ﬁnitely many non-zero terms. Prove that the map φ : Cc (R) → R, deﬁned by the above formula, is linear, positive, but not continuous. The following is a generalization of Corollary 6.2.

R Proposition 6.5. Let Ω be a locally compact space, and let φ : C0 (Ω) → R be R a linear continuous map. Then there exist positive linear maps φ1, φ2 : C0 (Ω) → R, such that φ = φ1 − φ2, and kφk = kφ1k + kφ2k. Proof. If Ω is compact there is nothing to prove (this is Corollary 5.3). As- sume Ω is non-compact. Use Hahn-Banach Theorem to find a linear continuous map ψ : CR(Ωα) → , with kψk = 1 and ψ = φ. Apply Corollary 5.3 to R R C0 (Ω) α find two positive linear maps ψ1, ψ2 : CR(Ω ) → R such that ψ = ψ1 − ψ2 and kψk = kψ k + kψ k. Define the positive linear maps φ = ψ , k = 1, 2. We 1 2 k k R C0 (Ω) clearly have φ = φ1 − φ2, and

kφ1k + kφ2k ≤ kψ1k + kψ2k = kψk = kφk = kφ1 − φ2k ≤ kφ1k + kφ2k, which forces kφk = kφ1k + kφ2k. §6. Banach algebras of continuous functions 93

Proposition 6.6 (Stone-Weierstrass Theorems for locally compact spaces). K Let Ω be a locally compact space, which is non-compact, and let A ⊂ C0 (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. R A. If K = R, then A is dense in C0 (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 CK(Ωα), where Ωα = Ω ∪ {∞} is the Alexandrov compactification, and identify K K α C0 (Ω) = f ∈ C (Ω ): 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 CK(Ωα). First of all, B is obviously a subalgebra of CK(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). K We can now finish the proof. Start with some arbitrary function f ∈ C0 (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 = αn1 + fn, with αn ∈ K and fn ∈ A. We are going to prove that limn→∞ fn = f. Since we have

kfn − fk = kgn − f − αn1k ≤ kgn − fk + kαn1k = kgn − fk + |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.