Stone–Weierstrass Theorem

Show that the trigonometric polynomials on the circle $S^1$ , i.e. functions of the form $T(\theta) = \sum_{k=-n}^{n} c_k e^{ik\theta}$ , satisfy the hypotheses of the Stone–Weierstrass theorem: they form a subalgebra closed under products, contain the constants, and separate points.

e^{ik\theta} e^{im\theta} = e^{i(k+m)\theta}, \qquad \cos A \cos B = \tfrac12(\cos(A-B) + \cos(A+B))

\theta_1 \ne \theta_2 \pmod{2\pi} \ \Longrightarrow \ \cos\theta_1 \ne \cos\theta_2 \ \text{or} \ \sin\theta_1 \ne \sin\theta_2

Degree

n = 8

: uniform error

\lVert S_8 - f\rVert_\infty \approx 0.0792

— a trigonometric polynomial approximating the triangle wave

f(x) = |x|

, with no algebraic polynomial in sight.

Show proof

Subalgebra: the set $\mathcal{T}$ of finite linear combinations of the $e^{ik\theta}$ is a linear space, and the product of two elements stays in $\mathcal{T}$ because $e^{ik\theta} e^{im\theta} = e^{i(k+m)\theta}$ . In the real form $\sum_{k=0}^n (a_k \cos k\theta + b_k \sin k\theta)$ the product-to-sum identities (e.g. $\cos A \cos B = \tfrac12(\cos(A-B) + \cos(A+B))$ ) play the same role. Hence $\mathcal{T}$ is a subalgebra closed under products.
Contains constants: the $k = 0$ term $c_0 e^{i0\theta} = c_0$ produces every constant, so $\mathcal{T}$ contains all constant functions.
Separates points: let $\theta_1 \ne \theta_2 \pmod{2\pi}$ . If both $\cos$ and $\sin$ agreed there, then $e^{i\theta_1} = \cos\theta_1 + i\sin\theta_1 = \cos\theta_2 + i\sin\theta_2 = e^{i\theta_2}$ , forcing $\theta_1 = \theta_2 \pmod{2\pi}$ — a contradiction. So at least one of $u(\theta) = \cos\theta \in \mathcal{T}$ , $u(\theta) = \sin\theta \in \mathcal{T}$ takes different values at $\theta_1$ and $\theta_2$ : $\mathcal{T}$ separates points.
Numerical check (degree $n = 8$ ): by the coefficient identity $a_k = \frac{2}{\pi}\frac{(-1)^k - 1}{k^2}$ , the even $k$ vanish and only odd $k$ contribute $-\frac{4}{\pi k^2}$ , so $S_8(x) = \frac{\pi}{2} - \frac{4}{\pi}\sum_{\text{odd } k \le 8} \frac{\cos(kx)}{k^2}$ is a trigonometric polynomial. The uniform error on the 400-point grid is $0.0792$ , matching the caption, falling to $0.0159$ at $n = 40$ .

Let $\mathcal{A} \subset C(K, \mathbb{R})$ be a uniformly closed algebra. Show that $f \in \mathcal{A}$ implies $|f| \in \mathcal{A}$ , and deduce that $\max(f, g)$ and $\min(f, g)$ also belong to the (uniformly closed) $\mathcal{A}$ .

|f| = p_k \circ f \ \text{의 극한}, \quad p_k \to |t| \ \text{on } [-M, M], \quad M = \|f\|_\infty

\max(f, g) = \frac{f + g + |f - g|}{2}, \qquad \min(f, g) = \frac{f + g - |f - g|}{2}

Degree

n = 8

:

\lVert S_8 - f\rVert_\infty \approx 0.0792

— polynomial limits capture the absolute value of a continuous function uniformly, as used in the proof.

Show proof

Set $M := \lVert f\rVert_\infty$ . The absolute-value function $t \mapsto |t|$ is continuous on $[-M, M]$ , so by the Weierstrass approximation theorem (the preceding scene) there are polynomials $p_k$ with $p_k \to |t|$ uniformly on $[-M, M]$ .
Each composition $p_k \circ f$ is a polynomial in $f$ , hence a linear combination of products of elements of $\mathcal{A}$ — an element of $\mathcal{A}$ since $\mathcal{A}$ is an algebra. Because $f(K) \subset [-M, M]$ , uniform convergence of $p_k \to |t|$ on $[-M, M]$ implies uniform convergence of $p_k \circ f \to |f|$ on $K$ .
Since $\mathcal{A}$ is uniformly closed, the uniform limit $|f|$ lies in $\mathcal{A}$ . If $f - g \in \mathcal{A}$ then $|f - g| \in \mathcal{A}$ , so the identities $\max(f,g) = \frac{f+g+|f-g|}{2}$ and $\min(f,g) = \frac{f+g-|f-g|}{2}$ give $\max(f,g), \min(f,g) \in \mathcal{A}$ — a uniformly closed algebra is a lattice.
Note: the algebra hypothesis is essential for the $p_k \circ f$ step — it is what makes powers of $f$ available. The lattice property is what lets the density argument below carve approximants with $\min$ and $\max$ .

Complete the density argument of the Stone–Weierstrass theorem: for a subalgebra $\mathcal{A} \subset C(K, \mathbb{R})$ ( $K$ compact) that separates points and contains the constants, the uniform closure $\overline{\mathcal{A}}$ is uniformly dense in $C(K, \mathbb{R})$ .

\begin{pmatrix} u(x) & 1 \\ u(y) & 1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} f(x) \\ f(y) \end{pmatrix}, \quad \det = u(x) - u(y) \ne 0

g_x = \min_j g_{xy_j} \le f + \varepsilon, \quad h = \max_i g_{x_i}, \quad f - \varepsilon < h < f + \varepsilon

Degree

n = 25

:

\lVert S_{25} - f\rVert_\infty \approx 0.0245

— the lattice-refined approximant lands inside

\varepsilon

, shown numerically.

Show proof

The uniform closure $\overline{\mathcal{A}}$ is a uniformly closed algebra (closed under limits by uniform continuity of multiplication), still separating points and containing constants. It suffices to prove density for $\overline{\mathcal{A}}$ : applying the argument below to $\overline{\mathcal{A}}$ shows $\overline{\mathcal{A}}$ is dense in $C(K, \mathbb{R})$ , i.e. $\mathcal{A}$ itself is dense.
(a) Two-point interpolation: fix $f \in C(K, \mathbb{R})$ and $\varepsilon > 0$ . For distinct $x, y \in K$ , the separation hypothesis gives $u \in \mathcal{A}$ with $u(x) \ne u(y)$ . The system $\begin{pmatrix} u(x) & 1 \\ u(y) & 1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} f(x) \\ f(y) \end{pmatrix}$ has determinant $u(x) - u(y) \ne 0$ , so a unique solution $(a, b)$ exists, and $g_{xy} := au + b \in \mathcal{A}$ satisfies $g_{xy}(x) = f(x)$ , $g_{xy}(y) = f(y)$ (the $b$ term lies in $\mathcal{A}$ because $\mathcal{A}$ contains the constants). For $x = y$ take the constant function $f(x)$ .
(b) Pressing from above: fix $x$ . For each $y \in K$ , the set $U_y := \{z : g_{xy}(z) < f(z) + \varepsilon\}$ is open by continuity of $g_{xy}$ and $f$ , and $y \in U_y$ since $g_{xy}(y) = f(y)$ . So $\{U_y : y \in K\}$ is an open cover of the compact $K$ , and a finite subcover $U_{y_1}, \dots, U_{y_m}$ exists. By the lattice property from the previous problem, $g_x := \min_j g_{xy_j} \in \mathcal{A}$ ; since $g_{xy_j} < f + \varepsilon$ for every $j$ , we get $g_x \le f + \varepsilon$ on all of $K$ . Also $g_{xy_j}(x) = f(x)$ for every $j$ , so $g_x(x) = f(x)$ .
(c) Lifting from below: for each $x \in K$ , $V_x := \{z : g_x(z) > f(z) - \varepsilon\}$ is open and $x \in V_x$ since $g_x(x) = f(x)$ . The $\{V_x\}$ form an open cover of $K$ , so a finite subcover $V_{x_1}, \dots, V_{x_l}$ exists. With $h := \max_i g_{x_i} \in \mathcal{A}$ (lattice property), every $z \in K$ lies in some $V_{x_i}$ , giving $h(z) \ge g_{x_i}(z) > f(z) - \varepsilon$ , while $h \le f + \varepsilon$ by (b). Hence $f - \varepsilon < h < f + \varepsilon$ on $K$ .
Since $\varepsilon > 0$ was arbitrary, $f \in \overline{\mathcal{A}}$ ; since $f \in C(K, \mathbb{R})$ was arbitrary, $\overline{\mathcal{A}} = C(K, \mathbb{R})$ — $\mathcal{A}$ is uniformly dense. Numerical check: for $f(x) = |x|$ and the trigonometric-polynomial algebra $\mathcal{T}$ , at degree $n = 25$ we get $\lVert S_{25} - f\rVert_\infty \approx 0.0245$ , matching the caption, and the error tends to zero as the degree grows ( $0.0159$ at $n = 40$ ).

Stone–Weierstrass Theorem

Examples & Exercises