Stone–Weierstrass Theorem

Recall the blog "Beyond Sequences: A Topological Approach to Density Arguments", where we prove that

and discuss several applications. I would like to add one more important result to it: the Stone–Weierstrass theorem.

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Corollary. Let $f:X\to Y$$f: X \to Y$ be a continuous function. Then

$$f(\bar{A})\subseteq \overline{f(A)}.$$

Proof. Since $f(A)\subseteq \overline{f(A)}$$f(A) \subseteq \overline{f(A)}$, we have:

$$f(\bar{A})\subseteq \overline{f(A)}.◻$$

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Let $R,S$$R, S$ be two topological rings and let $f:R\to S$$f: R \to S$ be a continuous ring homomorphism with $f(R)\subseteq Z(S)$$f(R) \subseteq Z(S)$, the center of $S$$S$. Then $S$$S$ becomes a topological $R$$R$-algebra. The scalar multiplication is continuous because it is a composition of continuous maps:

$$R\times S\stackrel{(f,{\mathrm{id}}_S)}{\to }S\times S\stackrel{\times }{\to }S$$

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Proposition. Let $A$$A$ be a subalgebra of $S$$S$, then $\bar{A}$$\overline{A}$ is also a subalgebra.

Proof. By the corollary above,

$$+(A\times A)\subseteq A\subseteq \bar{A}\Rightarrow +(\overline{A\times A})=+(\bar{A}\times \bar{A})\subseteq \bar{A}.$$

So $\bar{A}$$\overline{A}$ is closed under addition. Similarly for multiplication.
For scalar multiplication $\cdot$$\cdot$, we have

$$\cdot (R\times A)\subseteq A\subseteq \bar{A}\Rightarrow \cdot (\overline{R\times A})=\cdot (R\times \bar{A})\subseteq \bar{A}.◻$$

The same argument works for topological groups.

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Now consider $C(X,\mathbb{R})$$C(X, \mathbb{R})$, where $X$$X$ is a compact space. Let $A$$A$ be a subalgebra (possibly a ring without identity). We want to know when

$$\bar{A}=C(X,\mathbb{R}).$$

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(1) Vanishing at a point

If $A\subseteq \ker ({\mathrm{ev}}_a)$$A \subseteq \ker (\mathrm{ev}_a)$, where ${\mathrm{ev}}_a:C(X,\mathbb{R})\to \mathbb{R}$$\mathrm{ev}_a: C(X, \mathbb{R}) \to \mathbb{R}$ is defined by ${\mathrm{ev}}_a(f)=f(a)$$\mathrm{ev}_a(f) = f(a)$, then $A$$A$ is not dense.

Why? Since ${\mathrm{ev}}_a$$\mathrm{ev}_a$ is continuous, its kernel is closed. So

$$\bar{A}\subseteq \ker ({\mathrm{ev}}_a)⊊C(X,\mathbb{R}).$$

If $A$$A$ is not contained in the kernel of any evaluation map, we say $A$$A$ is a non-vanishing subalgebra.

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(2) Failure to separate points

If $A\subseteq \ker ({\mathrm{ev}}_a-{\mathrm{ev}}_b)$$A \subseteq \mathrm{ker}(\mathrm{ev}_a - \mathrm{ev}_b)$, then again $A$$A$ is not dense.

Since ${\mathrm{ev}}_a-{\mathrm{ev}}_b$$\mathrm{ev}_a - \mathrm{ev}_b$ is continuous, its kernel is closed:

$$\bar{A}\subseteq \ker ({\mathrm{ev}}_a-{\mathrm{ev}}_b).$$

If $A$$A$ is not contained in any such kernel, we say $A$$A$ separates points.

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Lemma (Weierstrass Approximation Theorem).
The ring of polynomial functions on $[0,1]$$[0,1]$ is dense in $C[0,1]$$C[0,1]$. That is, for all $\epsilon >0$$\varepsilon > 0$ and any $f\in C[0,1]$$f \in C[0,1]$, there exists a polynomial $P$$P$ such that

$$\parallel f-P{\parallel }_{\mathrm{\infty }}<\epsilon .$$

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Corollary. For any $0<a<b$$0 < a < b$ and any $\epsilon >0$$\varepsilon > 0$, there exists a polynomial $p(t)$$p(t)$ such that $p(0)=0$$p(0) = 0$ and

$$p([a,b])\subseteq (1-\epsilon ,1+\epsilon ).$$

Proof. Define

$$\begin{array}{c}f(t)=\{\begin{array}{ll}t/a,& t\in [0,a]\\ 1,& t\in [a,b]\end{array}\end{array}$$

By the Weierstrass theorem, there exists a polynomial $q$$q$ such that

$$\parallel q-f{\parallel }_{\mathrm{\infty }}<\epsilon /2.$$

Let $p(t)=q(t)-q(0)$$p(t) = q(t) - q(0)$. Then

$$\underset{t\in [a,b]}{sup}|p(t)-1|\le \underset{t\in [a,b]}{sup}|q(t)-1|+q(0)<\epsilon .◻$$

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Proposition. Let $A$$A$ be a non-vanishing subalgebra of $C(X,\mathbb{R})$$C(X, \mathbb{R})$ where $X$$X$ is compact. Then $1\in \bar{A}$$1 \in \overline{A}$.

Proof. For each $x\in X$$x \in X$, there exists $f_x\in A$$f_x \in A$ such that $f_x(x)\ne 0$$f_x(x) \ne 0$. Let

$$U_x:=\{y\in X\mid f_x(y)\ne 0\}.$$

Then $\{U_x{\}}_{x\in X}$$\{ U_x \}_{x \in X}$ is an open cover. By compactness, there is a finite subcover $\{U_{x_1},\dots ,U_{x_n}\}$$\{U_{x_1}, \dots, U_{x_n}\}$. Let

$$f_1=f_{x_1}^2+\cdots +f_{x_n}^2,$$

then $f_1>0$$f_1 > 0$ on $X$$X$. Since $f_1$$f_1$ is continuous on a compact space, there exist constants $a,b>0$$a, b > 0$ such that

$$a\le f_1(x)\le b.$$

By the previous corollary, there exists a polynomial $p$$p$ such that

$$p(f_1(x))\in (1-\epsilon ,1+\epsilon ).$$

Let $f=p(f_1)$$f = p(f_1)$, then

$$\parallel f-1{\parallel }_{\mathrm{\infty }}<\epsilon ,$$

and since $p$$p$ is a polynomial and $f_1\in A$$f_1 \in A$, we have $f\in A$$f \in A$. So $1\in \bar{A}$$1 \in \overline{A}$. $◻$$\square$

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Proposition. Let $A$$A$ be a unital closed subalgebra of $C(X,\mathbb{R})$$C(X, \mathbb{R})$ where $X$$X$ is compact. Then:

1. $f\in A\Rightarrow |f|\in A$$f \in A \Rightarrow |f| \in A$.

2. $f_1,\dots ,f_n\in A\Rightarrow max\{f_1,\dots ,f_n\}\in A$$f_1, \dots, f_n \in A \Rightarrow \max\{f_1, \dots, f_n\} \in A$ and $min\{f_1,\dots ,f_n\}\in A$$\min\{f_1, \dots, f_n\} \in A$.

Proof.

(1) Since $f$$f$ is bounded, there exists a sequence of polynomials $p_n$$p_n$ on $[0,\parallel f{\parallel }_{\mathrm{\infty }}]$$[0, \|f\|_\infty]$ converging uniformly to $\sqrt{t}$$\sqrt{t}$. Then $p_n(f^2)\to |f|$$p_n(f^2) \to |f|$ uniformly.

(2) Note that

$$f\vee g=\frac{f+g+|f-g|}{2},f\wedge g=\frac{f+g-|f-g|}{2}.◻$$

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Stone–Weierstrass Theorem

Let $X$$X$ be a compact Hausdorff space. If a subalgebra $A\subseteq C(X,\mathbb{R})$$A \subseteq C(X, \mathbb{R})$ is non-vanishing and separates points, then $A$$A$ is dense in $C(X,\mathbb{R})$$C(X, \mathbb{R})$.

By previous propositions, this is equivalent to: if $A$$A$ is a unital algebra that separates points, then $\bar{A}=C(X,\mathbb{R})$$\overline{A} = C(X, \mathbb{R})$.

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Proof.

Let $f\in C(X,\mathbb{R})$$f \in C(X, \mathbb{R})$. For any $\epsilon >0$$\varepsilon > 0$, we need to find $f_{\epsilon }\in A$$f_\varepsilon \in A$ such that:

$$\parallel f-f_{\epsilon }{\parallel }_{\mathrm{\infty }}<\epsilon .$$

Pick distinct $a,b\in X$$a, b \in X$. Since $A$$A$ separates points, there exists $g\in A$$g \in A$ such that $g(a)\ne g(b)$$g(a) \ne g(b)$. Define

$$f_{a,b}(x):=f(a)+\frac{f(b)-f(a)}{g(b)-g(a)}\cdot (g(x)-g(a)).$$

Then $f_{a,b}\in A$$f_{a,b} \in A$, with

$$f_{a,b}(a)=f(a),f_{a,b}(b)=f(b).$$

Let

$$U_{a,b,\epsilon }:=\{x\in X\mid |f_{a,b}(x)-f(x)|<\epsilon \}.$$

This is open. Fixing $a$$a$, the collection $\{U_{a,b,\epsilon }{\}}_{b\in X}$$\{U_{a,b,\varepsilon}\}_{b \in X}$ is an open cover of $X$$X$. By compactness, there exists a finite subcover:

$$\{U_{a,b_1,\epsilon },\dots ,U_{a,b_n,\epsilon }\}.$$

Define

$$f_a^{\epsilon }:=min\{f_{a,b_1},\dots ,f_{a,b_n}\}.$$

Then

$$f_a^{\epsilon }(x)<f(x)+\epsilon \text{ for all }x\in X,f_a^{\epsilon }(a)=f(a).$$

Now define the sets

$$V_{b,\epsilon }:=\{x\in X\mid f_a^{\epsilon }(x)>f(x)-\epsilon \},$$

which form another open cover. Compactness gives a finite subcover:

$$\{V_{b_1,\epsilon },\dots ,V_{b_m,\epsilon }\}.$$

Define

$$f_{\epsilon }:=max\{f_{b_1}^{\epsilon },\dots ,f_{b_m}^{\epsilon }\}.$$

Then

$$f(x)-\epsilon <f_{\epsilon }(x)<f(x)+\epsilon \text{for all }x\in X.$$

So

$$\parallel f-f_{\epsilon }{\parallel }_{\mathrm{\infty }}<\epsilon .◻$$

Example. Let consider the family of $p(e^x)$$p(e^x)$, where $p$$p$ is polynomial on $[0,1]$$[0,1]$, easy to see that it is a untial algebra and separate points, hence it is dense in $C[0,1]$$C[0,1]$. Hence, if $\mathrm{\forall }n\in \mathbb{N},{\int }_0^{1}f(x)e^{nx}\mathrm{d}x=0$$\forall n\in\mathbb{N},\int_0^1 f(x) e^{nx}\mathrm{d}x=0$ then $f=0$$f=0$. Since ${\int }_0^1(-)\mathrm{d}x$$\int_0^1(-)\mathrm{d}x$ is a continuous function on $C[0,1]$$C[0,1]$.