Mathematics

The Stone-Weierstrass Theorem: Approximating Continuous Functions

We state and prove the Stone-Weierstrass theorem — the powerful generalization of Weierstrass's classical approximation theorem that characterizes when subalgebras of continuous functions are dense — and explore its wide-ranging applications.

Analysis Topology

The Theorem

Stone-Weierstrass Theorem (Real Version)

Let $X$ be a compact Hausdorff space and let $\mathcal{A} \subseteq C(X, \mathbb{R})$ be a subalgebra that:

Separates points: for any $x \neq y$ in $X$ , there exists $f \in \mathcal{A}$ with $f(x) \neq f(y)$ .
Vanishes nowhere: for every $x \in X$ , there exists $f \in \mathcal{A}$ with $f(x) \neq 0$ .

Then $\mathcal{A}$ is dense in $C(X, \mathbb{R})$ with respect to the supremum norm.

In short: every continuous function on a compact space can be uniformly approximated by functions in any "sufficiently rich" subalgebra.

The Classical Weierstrass Theorem

The original theorem (Weierstrass, 1885) is a special case:

Weierstrass Approximation Theorem

Every continuous function $f: [a, b] \to \mathbb{R}$ can be uniformly approximated by polynomials. That is, for every $\varepsilon > 0$ there exists a polynomial $p$ with:

$\sup_{x \in [a,b]} |f(x) - p(x)| < \varepsilon$

This follows from Stone-Weierstrass by taking $X = [a,b]$ and $\mathcal{A} = \mathbb{R}[x]$ (the polynomials), which separate points (the identity $x$ does) and vanish nowhere (the constant $1$ does).

Why It Matters

The Stone-Weierstrass theorem is one of the most widely used results in analysis. Its power lies in reducing questions about all continuous functions to questions about a specific class:

To prove a property holds for all continuous functions, prove it for the subalgebra and show it is preserved under uniform limits.
To construct approximation schemes, verify the Stone-Weierstrass hypotheses.

Key Lemma: Lattice Closure

The proof hinges on showing that the closure $\bar{\mathcal{A}}$ is a lattice: it is closed under $\max$ and $\min$ .

Lemma 1. If $f \in \bar{\mathcal{A}}$ , then $|f| \in \bar{\mathcal{A}}$ .

Proof. By the classical Weierstrass theorem, the function $|t|$ on $[-\|f\|, \|f\|]$ can be uniformly approximated by polynomials $p_n(t)$ with $p_n(0) = 0$ . Then $p_n \circ f \in \bar{\mathcal{A}}$ (since $\bar{\mathcal{A}}$ is an algebra) and $p_n \circ f \to |f|$ uniformly. $\square$

Corollary. Since $\max(f, g) = \frac{f+g+|f-g|}{2}$ and $\min(f, g) = \frac{f+g-|f-g|}{2}$ , the closure $\bar{\mathcal{A}}$ is closed under $\max$ and $\min$ .

Proof of the Stone-Weierstrass Theorem

Proof.

Let $f \in C(X, \mathbb{R})$ and $\varepsilon > 0$ . We must find $g \in \bar{\mathcal{A}}$ with $\|f - g\|_\infty < \varepsilon$ .

Step 1 — Interpolation at two points. For any $x, y \in X$ , there exists $h_{xy} \in \bar{\mathcal{A}}$ with $h_{xy}(x) = f(x)$ and $h_{xy}(y) = f(y)$ .

Proof: Since $\mathcal{A}$ separates points and vanishes nowhere, for any two points $x \neq y$ and any values $\alpha, \beta$ , we can find $h \in \mathcal{A}$ with $h(x) = \alpha$ and $h(y) = \beta$ . (Use the separation and non-vanishing properties to build appropriate linear combinations.)

Step 2 — Approximation from below at each point. Fix $x \in X$ . For each $y \in X$ , let $h_{xy}$ be as above. Since $h_{xy}(y) = f(y)$ , continuity gives an open set $U_y \ni y$ with $h_{xy}(t) > f(t) - \varepsilon$ for $t \in U_y$ .

By compactness, finitely many $U_{y_1}, \ldots, U_{y_k}$ cover $X$ . Define:

$g_x = \max(h_{xy_1}, \ldots, h_{xy_k}) \in \bar{\mathcal{A}}$

Then $g_x(t) > f(t) - \varepsilon$ for all $t \in X$ and $g_x(x) = f(x)$ .

Step 3 — Approximation from above. Since $g_x(x) = f(x)$ , continuity gives an open set $V_x \ni x$ with $g_x(t) < f(t) + \varepsilon$ for $t \in V_x$ .

By compactness, finitely many $V_{x_1}, \ldots, V_{x_m}$ cover $X$ . Define:

$g = \min(g_{x_1}, \ldots, g_{x_m}) \in \bar{\mathcal{A}}$

Then for all $t \in X$ :

$g(t) > f(t) - \varepsilon$ (since each $g_{x_i}(t) > f(t) - \varepsilon$ ).
$g(t) < f(t) + \varepsilon$ (since for some $i$ , $t \in V_{x_i}$ and $g_{x_i}(t) < f(t) + \varepsilon$ ).

Therefore $\|f - g\|_\infty < \varepsilon$ and $g \in \bar{\mathcal{A}}$ . $\square$

The Complex Version

Stone-Weierstrass (Complex Version). Let $X$ be compact Hausdorff. A self-adjoint subalgebra $\mathcal{A} \subseteq C(X, \mathbb{C})$ that separates points and vanishes nowhere is dense in $C(X, \mathbb{C})$ .

Here "self-adjoint" means: $f \in \mathcal{A} \implies \bar{f} \in \mathcal{A}$ .

Without the self-adjoint condition, the theorem fails: on the unit circle $S^1$ , the algebra generated by $z = e^{i\theta}$ (without $\bar{z}$ ) consists of boundary values of holomorphic functions and is not dense in $C(S^1, \mathbb{C})$ .

Applications

Trigonometric Approximation

The trigonometric polynomials $\sum_{k=-N}^{N} c_k e^{ik\theta}$ form a self-adjoint subalgebra of $C(S^1, \mathbb{C})$ that separates points and vanishes nowhere. By Stone-Weierstrass, they are dense — justifying Fourier analysis.

Polynomial Approximation on Products

On $[0,1]^n$ , the polynomials in $n$ variables separate points and form an algebra. Stone-Weierstrass gives uniform approximation by multivariate polynomials.

Müntz-Szász Theorem

Which subsets $S \subseteq \{0, 1, 2, \ldots\}$ have the property that $\operatorname{span}\{x^n : n \in S\}$ is dense in $C[0,1]$ ? The Müntz-Szász theorem answers: if and only if $0 \in S$ and $\sum_{n \in S, n > 0} 1/n = \infty$ . Stone-Weierstrass provides the framework.

Korovkin's Theorem

If a sequence of positive linear operators satisfies $T_n(1) \to 1$ , $T_n(x) \to x$ , $T_n(x^2) \to x^2$ uniformly on $[a,b]$ , then $T_n(f) \to f$ uniformly for all $f \in C[a,b]$ . This powerful result, used to prove the Bernstein polynomial approximation, connects to Stone-Weierstrass.

Bernstein's Constructive Proof

While Stone-Weierstrass is an existence result, Bernstein (1912) gave a constructive proof of Weierstrass's theorem using the Bernstein polynomials:

$B_n(f)(x) = \sum_{k=0}^{n} f\!\left(\frac{k}{n}\right) \binom{n}{k} x^k (1-x)^{n-k}$

One can verify $B_n(f) \to f$ uniformly on $[0,1]$ using the law of large numbers (or direct estimation). This is an explicit approximation, unlike the abstract Stone-Weierstrass argument.

Failure Without Hypotheses

Without point separation: The constant functions $\mathcal{A} = \mathbb{R}$ on $[0,1]$ form an algebra but do not separate points. The closure is just $\mathbb{R}$ , not all of $C[0,1]$ .

Without non-vanishing: The algebra $\{f \in C[0,1] : f(0) = 0\}$ separates points but vanishes at $0$ . Its closure is itself — it misses all functions with $f(0) \neq 0$ .

Without compactness: On $\mathbb{R}$ , polynomials separate points and vanish nowhere, but they are not dense in the bounded continuous functions (a polynomial that is close to a bounded function on all of $\mathbb{R}$ must itself be bounded, hence constant).

Summary

\begin{aligned} &\mathcal{A} \subseteq C(X, \mathbb{R}), \;\; X \text{ compact Hausdorff} \\[6pt] &\mathcal{A} \text{ subalgebra, separates points, vanishes nowhere} \\[6pt] &\Downarrow \\[6pt] &\overline{\mathcal{A}} = C(X, \mathbb{R}) \quad \text{(uniform closure)} \\[8pt] &\text{Special case: } \overline{\mathbb{R}[x]} = C[a,b] \quad \text{(Weierstrass)} \end{aligned}

References

Rudin, W., Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976.
Folland, G. B., Real Analysis, 2nd edition, Wiley, 1999.
Conway, J. B., A Course in Functional Analysis, 2nd edition, Springer, 1990.
Wikipedia — Stone-Weierstrass theorem
Wikipedia — Weierstrass approximation theorem
MIT OpenCourseWare — Real Analysis

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