LEC ANA2 11

Stone Weierstrass Theorem

Note

A list of all versions of S-W we’ve seen:

1. Theorem 6: S-W for $\mathbb{R}$-subalgebras
2. Corollary 10: S-W for $\mathbb{R}$-subalgebras as seen in Rudin (1976) 7.32.
3. Theorem 11: S-W for $\mathbb{C}$-subalgebras
4. Theorem 324.1: characterization of $\mathbb{R}$-subalgebras which separate points
5. Theorem 324.10: S-W for $\mathbb{R}$-subalgebras and locally compact spaces

Proposition 315.1(Dini).

Let $X$ be a compact metric space. Suppose $\{f_n\}\subseteq C(X)$ is monotone¹ and converges pointwise to $f\in C(X)$. Then, $\{f_n\}$ converges to $f$ uniformly.

Proof.

We may assume $f_n$ converges to $0$, and is monotone decreasing. Given $ϵ>0$, define

$$U_n:=\{x\in X:|f_n(x)|<ϵ\text{ for every }m\ge n\}=\{x\in X:|f_n(x)|<ϵ\}.$$

These open sets $\{U_n\}$ increase to $X$. By compactness of $X$, there exists $N$ such that $U_N=X$.□

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Note that $C(X,\mathbb{R})$, with its vector space ($\mathbb{R}$-module) and ring structure, is an $\mathbb{R}$-algebra (Definition 108.5).

Definition 315.2(Definition).

Let $X$ be compact. A subset $\mathcal{A}\subseteq C(X,\mathbb{R})$ is a subalgebra if

1. for every $f,g\in \mathcal{A}$, $fg\in \mathcal{A}$
2. $\forall f,g\in \mathcal{A}$ and $\forall \alpha ,\beta \in \mathbb{R}$, $\alpha f+\beta g\in \mathcal{A}$.

$\mathcal{A}$ separates points if for every distinct $x,y\in X$, there exists $f\in \mathcal{A}$ such that $f(x)\ne f(y)$. $\mathcal{A}$ is said to contain constants/be unital if $1\in \mathcal{A}$.

Lemma 315.3(Lemma).

There is a sequence $\{p_n\}$ of real valued polynomials with zero constant coefficient which converge uniformly to $f(x)=\sqrt{x}$ on $[0,1]$.

Proof.

Recursively define

$$\begin{array}{rlr}& p_1=0& \\ & p_{n+1}(t)=p_n(t)+1/2(t-p_n^2(t))& n\ge 1.\end{array}$$

We will first show $0\le p_n(t)\le \sqrt{t}$ for all $n$. This is true for $n=1$; assume it is for $n=k$.

$$\begin{array}{rl}0\le p_{k+1}(t)& =p_k(t)+1/2(t-p_k^2(t))\\ & \le p_k(t)+1/2(\sqrt{t}+p_k(t))(\sqrt{t}-p_k(t))\\ & \le p_k(t)+\sqrt{t}-p_k(t)=\sqrt{t}.\end{array}$$

It follows that $p_{n+1}(t)\ge p_n(t)$ for all $n$ and for each $t$, so $\{p_n\}$ is monotone increasing. Now, letting $n\to \infty$ on both sides of the definition of $p_{n+1}(t)$ (which we can do since monotone bounded sequences converge), we obtain ${\lim }_{n\to \infty }{p}_n(t)=\sqrt{t}$, for each $t\in [0,1]$. We are done by Proposition 1.□

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Definition 315.4(Definition).

A subset $S\subseteq C(X,\mathbb{R})$ is a lattice if for every $f,g\in S$, $\text{min}(f,g)$ and $\text{max}(f,g)$ are in $S$.

Lemma 315.5(Lemma).

Let $X$ be compact. A closed subalgebra $\mathcal{A}$ of $C(X,\mathbb{R})$ is a lattice.

Proof.

Let$f\in \mathcal{A}$, $M=\Vert f{\Vert }_{\infty }$. Note that $f^2/M^2\in \mathcal{A}$. Let $\{p_n\}$ be as in Lemma 3. $p_n\circ f^2/M^2$ is a polynomial in $f^2/M^2$, so $p_n\circ f^2/M^2\in \mathcal{A}$² for every $n$. $p_n\circ f^2/M^2$ converges uniformly to $|f|/M$. Since $\mathcal{A}$ is closed, $|f|/M\in \mathcal{A}$, so $|f|\in \mathcal{A}$.

$$\begin{array}{rl}\text{min}(f,g)& =\frac{1}{2}(f+g-|f-g|),\\ \text{max}(f,g)& =\frac{1}{2}(f+g+|f-g|).\end{array}$$ □

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Theorem 315.6(Stone-Weierstrass for$\mathbb{R}$, v1).

Let $X$ be a compact metric space. Let $\mathcal{A}\subseteq C(X,\mathbb{R})$ be a unital subalgebra which separates points. Then, $\mathcal{A}$ is dense in $C(X,\mathbb{R})$.

Proof.

Let $f\in C(X)$. For distinct $x,y$, let $s_{xy}\in \mathcal{A}$ be such that $s_{xy}(x)\ne s_{xy}(y)$. For distinct $x,y$, define

$$h_{xy}(z):=f(x)+(f(y)-f(x))\left(\frac{s_{xy}(z)-s_{xy}(x)}{s_{xy}(y)-s_{xy}(x)}\right).$$

Clearly, $h_{xy}\in C(X)$. Note that $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$.

Let $ϵ>0$. Fix $x\in X$. For $y\in X$, define

$$U_y:=\{z\in X:h_{xy}(z)<f(z)+ϵ\}.$$

Note that each $U_y$ is open³ and that $y\in U_y$ for all $y\in X$. Since $X$ is compact, $X={\bigcup }_{y\in X}{U}_y={\bigcup }_{i=1}^n{U}_{y_i}$. By Lemma 5, $\overline{\mathcal{A}}$ is a lattice, so

$$h_x:=\text{min}\{h_{x{y}_1},\dots ,h_{x{y}_n}\}\in \overline{\mathcal{A}}.$$

So, for each $x\in X$, we have $h_x\in \overline{\mathcal{A}}$ such that $h_x<f+ϵ$ and $h_x(x)=f(x)$.

Now vary $x$: for $x\in X$, define

$$V_x:=\{z\in X:h_x(z)>f(z)-ϵ\}.$$

Again, these are open sets, and $X={\bigcup }_{x\in X}{V}_x={\bigcup }_{i=1}^m{V}_{x_i}$.

$$h=\text{max}\{h_{x_1},\dots ,h_{x_m}\}\in \overline{\mathcal{A}}.$$

Finally, for all $z\in X$,

$$f(z)-ϵ<h(z)<f(z)+ϵ,$$

so $\Vert f-h{\Vert }_{\infty }<ϵ$. It follows that $\overline{\mathcal{A}}$ is dense in $C(X,\mathbb{R})$. But, $\overline{\mathcal{A}}$ is closed, so $\overline{\mathcal{A}}=C(X,\mathbb{R})$, so $\mathcal{A}$ is dense in $C(X)$.□

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Corollary 315.7(Corollary).

Let $K\subseteq {\mathbb{R}}^n$ be compact. Then polynomials in coordinates $x_1,\dots ,x_n$ are dense in $C(K,\mathbb{R})$.

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Exercise 315.8(Exercise).

Suppose $f\in C([0,1])$ such that

$${\int }_0^{1}f(t)t^{i}dt=0$$

for all $i⩾0$. Show that $f=0$.

Proof.

By Corollary 7, we have a sequence of polynomials $\{p_n\}$ converging uniformly to $f$. It follows that the sequence $\{p_n\cdot f\}$ converges uniformly to $f^2$. Let the degree of $p_n$ be $d_n$. By Theorem 155.7,

$$\begin{array}{rl}{\int }_0^1{f}^2(t)dt& =\underset{n\to \infty }{\lim }{\int }_0^1{p}_n(t)f(t)dt\\ & =\underset{n\to \infty }{\lim }\left[\sum _{i=0}^{d_n}{\int }_0^1{p}_{n,i}{t}^{i}f(t)dt\right]\\ & =0.\end{array}$$

Since $f^2⩾0$, it follows that $f^2=0$, so $f=0$.□

Lemma 315.9(Lemma).

Let $X$ be a compact metric space. If $\mathcal{A}$ is a subalgebra of $C(X,\mathbb{R})$ which vanishes at no point of $X$, then $\overline{\mathcal{A}}$ is unital.

Proof.

We need to prove $1\in \overline{\mathcal{A}}$. Since $\mathcal{A}$ vanishes at no point of $X$, for each $x\in X$ there exists $f_x\in \mathcal{A}$ such that $f_x(x)=1$. Let $1>m>0$. As before, define

$$U_x:=\{y\in X:f_x(y)>m\}.$$

These are non-empty, since $x\in U_x$. Since $X$ is compact,

$$X=\bigcup _{x\in X}{U}_x=\bigcup _{i=1}^n{U}_{x_i}.$$

By Lemma 5, $f=\max \{f_{x_1},\dots ,f_{x_n}\}\in \overline{\mathcal{A}}$. Note that $f(t)>m$ for all $t\in X$. Define $g:[0,\Vert f{\Vert }_{\infty }]\to \mathbb{R}$ such that $g(0)=0$, $g{|}_{[m,\Vert f{\Vert }_{\infty }]}=1$, and $g$ is continuous. Then, $g\in C([0,\Vert f{\Vert }_{\infty }])$. By Corollary 7, there exist polynomials $\{p_n\}$ defined on $[0,\Vert f{\Vert }_{\infty }]$ which converge uniformly to $g$. Let $\{c_n\}$ be the sequence of constant coefficients of $\{p_n\}$. Then, $\{c_n\}\to 0$, and $\{q_n\}:=\{p_n-c_n\}$ is a sequence of polynomials with constant zero coefficient also converging uniformly to $g$. Each $q_n(f)\in \mathcal{A}$, and $\{q_n(f)\}\to g(f)=1$, so $1\in \overline{\mathcal{A}}$.□

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We have the following corollary of Theorem 6 and Lemma 9.

Corollary 315.10(Stone-Weierstrass over$\mathbb{R}$, v2).

Let $X$ be a compact metric space. Let $A\subseteq C(X,\mathbb{R})$ be a subalgebra which separates points and vanishes at no point of $X$. Then, $\mathcal{A}$ is dense in $C(X,\mathbb{R})$.

Proof.

By Lemma 9, $\overline{\mathcal{A}}$ is unital. By Theorem 6, $\overline{\mathcal{A}}$ is dense in $C(X,\mathbb{R})$. This implies $\overline{\mathcal{A}}=C(X,\mathbb{R})$, which implies $\mathcal{A}$ is dense in $C(X,\mathbb{R})$.□

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The Theorem 6 analog for complex algebras requires additional hypotheses. A complex algebra $\mathcal{A}$ is called self-adjoint if $f\in \mathcal{A}$ implies $\overline{f}\in \mathcal{A}$.

Theorem 315.11(Stone-Weierstrass for$\mathbb{C}$, v3).

Let $X$ be a compact metric space. Let $\mathcal{A}\subseteq C(X,\mathbb{C})$ be a unital self-adjoint subalgebra which separates points. Then $\mathcal{A}$ is dense in $C(X,\mathbb{C})$.

Proof.

Let $f\in \mathcal{A}$. $\text{Re}(f)=(f+\overline{f})/2\in \mathcal{A}$ and $\text{Im}(f)=(f-\overline{f})/2i\in \mathcal{A}$. Let ${\mathcal{A}}_{\mathbb{R}}\subseteq \mathcal{A}$ be the subalgebra of real valued functions. Since $\mathcal{A}$ separates points, ${\mathcal{A}}_{\mathbb{R}}$ separates points: if $x\ne y$, there exists $h\in \mathcal{A}$ such that $h(x)\ne h(y)$, so either $\text{Re}(h(x))\ne \text{Re}(h(y))$ or $\text{Im}(h(x))\ne \text{Im}(h(y))$. By Theorem 6, ${\mathcal{A}}_{\mathbb{R}}$ is dense in $C(X,\mathbb{R})$.

Note that $C(X,\mathbb{C})\cong C(X,\mathbb{R})+\sqrt{-1}C(X,\mathbb{R})$. So $\mathcal{A}\cong {\mathcal{A}}_{\mathbb{R}}+\sqrt{-1}{\mathcal{A}}_{\mathbb{R}}$. Given $f\in C(X,\mathbb{C})$, $\text{Re}(f)$ and $\text{Im}(f)$ can be uniformly approximated by functions in ${\mathcal{A}}_{\mathbb{R}}$: ${\phi }_n\to \text{Re}(f)$ and ${\psi }_n\to \text{Im}(f)$. Therefore, ${\phi }_n+\sqrt{-1}{\psi }_n\to \text{Re}(f)+\sqrt{-1}\text{Im}(f)=f$. It follows that $\overline{\mathcal{A}}=C(X,\mathbb{C})$ and $\mathcal{A}$ is dense.□

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Footnotes

1. $f_n(x)$ is a monotone sequence for every $x\in X$ ↩

2. We can say this because the polynomials $p_n$ have zero constant coefficient. If that were not the case, we would have had to require $\mathcal{A}$ to contain constants. ↩

3. $U_y=(h_{xy}-f{)}^{-1}(-\infty ,ϵ)$. ↩

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References

Rudin, W. (1976). Principles of Mathematical Analysis (3d ed). McGraw-Hill.