Section V.8

An Application: The Stone-Weierstrass Theorem

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Problem V.8.1

Suppose that A\mathcal{A} is a subalgebra of C(X)C(X) that separates the points of XX and 1∈A1 \in \mathcal{A}. Show that if x1,…,xnx_1, \dots , x_n are distinct points in XX and α1,…,αn∈F\alpha_1, \dots , \alpha_n \in \mathbb {F}, there is an ff in A\mathcal{A} such that f(xj)=αjf(x_j) = \alpha_j for 1≤j≤n1 \leq j \leq n.

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Problem V.8.2

Give the details of the proof of Corollary 8.3: if XX is locally compact and A\mathcal{A} is a closed subalgebra of C0(X)C_0(X) such that

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(a)

for each xx in XX there is an ff in A\mathcal{A} such that f(x)≠0f(x) \neq 0;

(b)

A\mathcal{A} separates the points of XX;

(c)

fˉ∈A\bar{f} \in \mathcal{A} whenever f∈Af \in \mathcal{A};

then A=C0(X)\mathcal{A} = C_0(X).

Problem V.8.3

If XX is compact, show that for each xx in XX, δx\delta_x is an extreme point of P(X)P(X) and αδx\alpha \delta_x, ∣α∣=1\left|\alpha \right| = 1, is an extreme point of \ballM(X)\ball M(X).

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Problem V.8.4

Let XX be compact and let A\mathcal{A} be a closed subalgebra of C(X)C(X) such that 1∈A1 \in \mathcal{A} and A\mathcal{A} is closed under conjugation. Define an equivalence relation ∼\sim on XX by declaring x∼yx \sim y if and only if f(x)=f(y)f(x) = f(y) for all ff in A\mathcal{A}. Let X/∼X / \sim be the corresponding quotient space and let π:X→X/∼\pi : X \to X / \sim be the natural map. Give X/∼X / \sim the quotient topology.

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(a)

Show that if f∈Af \in \mathcal{A}, then there is a unique function π∗(f)\pi^*(f) in C(X/∼)C(X / \sim ) such that π∗(f)∘π=f\pi^*(f) \circ \pi = f.

(b)

Show that π∗:A→C(X/∼)\pi^* : \mathcal{A} \to C(X / \sim ) is an isometry.

(c)

Show that π∗\pi^* is surjective.

(d)

Show that A={f∈C(X):f(x)=f(y) whenever x∼y}\mathcal{A} = \left\{ f \in C(X) : f(x) = f(y) \text{ whenever } x \sim y \right\}.

Problem V.8.5

Let XX be completely regular and topologize C(X)C(X) as in Example IV.1.5. If A\mathcal{A} is a closed subalgebra of C(X)C(X) such that 1∈A1 \in \mathcal{A}, A\mathcal{A} separates the points of XX, and fˉ∈A\bar{f} \in \mathcal{A} whenever f∈Af \in \mathcal{A}, then A=C(X)\mathcal{A} = C(X).

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Remark.
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This exercise requires Exercise IV.4.7.

Problem V.8.6

Let X,YX, Y be compact spaces and show that if f∈C(X×Y)f \in C(X \times Y) and ε>0\varepsilon > 0, then there are functions g1,…,gng_1, \dots , g_n in C(X)C(X) and h1,…,hnh_1, \dots , h_n in C(Y)C(Y) such that ∣f(x,y)−∑k=1ngk(x)hk(y)∣<ε\left|f(x,y) - \sum_{k=1}^n g_k(x) h_k(y)\right| < \varepsilon for all (x,y)(x,y) in X×YX \times Y.

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Problem V.8.7

Let A\mathcal{A} be the uniformly closed subalgebra of Cb(R)C_b(\mathbb {R}) generated by sin⁡x\sin x and cos⁡x\cos x. Show that A={f∈Cb(R):f(t)=f(t+2π) for all t in R}\mathcal{A} = \left\{ f \in C_b(\mathbb {R}) : f(t) = f(t + 2\pi ) \text{ for all } t \text{ in } \mathbb {R} \right\}.

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Problem V.8.8

If KK is a compact subset of C\mathbb {C}, f∈C(K)f \in C(K), and ε>0\varepsilon > 0, show that there is a polynomial p(z,zˉ)p(z, \bar{z}) in zz and zˉ\bar{z} such that ∣f(z)−p(z,zˉ)∣<ε\left|f(z) - p(z,\bar{z})\right| < \varepsilon for all zz in KK.

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