Stone–Weierstrass Theorem

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.

Conjugate-closed (the complex Stone–Weierstrass hypothesis): $\overline{e^{ik\theta}} = e^{-ik\theta}$, and $-k$ is again an allowed exponent, so the conjugate lies in $\mathcal{T}$ — $\mathcal{T}$ is closed under conjugation. Also, membership of $\cos\theta, \sin\theta$ in $\mathcal{T}$ is derived, not assumed: $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} \in \mathcal{T}$ and $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} \in \mathcal{T}$, both being linear combinations of the exponentials.

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 $n = 40$.

Hypothesis check: $K$ is assumed compact (the Stone–Weierstrass setting), so the continuous $f$ is bounded on $K$ and $M := \lVert f\rVert_\infty < \infty$. The absolute-value function $t \mapsto |t|$ is continuous on $[-M, M]$, so by the Weierstrass approximation theorem (the preceding scene) there are polynomials $q_k$ with $q_k \to |t|$ uniformly on $[-M, M]$.

Now set $p_k(t) := q_k(t) - q_k(0)$. Then $p_k(0) = 0$, and since $q_k(0) \to |0| = 0$, we have $\lVert p_k - |t|\rVert \le \lVert q_k - |t|\rVert + |q_k(0)| \to 0$ — so $p_k \to |t|$ uniformly on $[-M, M]$ as well.

Since $p_k(0) = 0$, $p_k$ is a polynomial with zero constant term: $p_k(t) = \sum_{j \ge 1} a_{kj} t^j$. Hence the composition $p_k \circ f = \sum_{j \ge 1} a_{kj} f^j$ lies in $\mathcal{A}$ — $\mathcal{A}$, being an algebra (a subspace closed under products), contains all powers of $f$ and their linear combinations. (This removal of the constant term is essential: $\mathcal{A}$ is not assumed to contain $1$ or the constants.)

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 $\max$.

The uniform closure $\overline{\mathcal{A}}$ is a uniformly closed algebra, still separating points and containing constants. That it is an algebra: if $f_n \to f$ and $g_n \to g$ uniformly on $K$, the convergent sequences are uniformly bounded (some constant bounds all $\lVert f_n\rVert, \lVert g_n\rVert$), and $\lVert f_n g_n - fg\rVert \le \lVert f_n\rVert \lVert g_n - g\rVert + \lVert g\rVert \lVert f_n - f\rVert \to 0$ — hence $fg \in \overline{\mathcal{A}}$. (This boundedness step is needed because multiplication on all of $C(K)$ is not globally uniformly continuous.)

The lattice steps (b)–(c) below are carried out inside $\overline{\mathcal{A}}$, not $\mathcal{A}$ (the absolute-value lemma applied to the uniformly closed algebra $\overline{\mathcal{A}}$). Proving $\overline{\mathcal{A}}$ dense in $C(K, \mathbb{R})$ then gives density of $\mathcal{A}$.

(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 (run with $\varepsilon/2$ slack): fix $x$. For each $y \in K$, the set $U_y := \{z : g_{xy}(z) < f(z) + \varepsilon/2\}$ 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.

Quantifier care: $g_{xy_j} < f + \varepsilon/2$ holds only on $U_{y_j}$, not on all of $K$. But for each $z \in K$ there is some $j$ with $z \in U_{y_j}$, and for that $j$, $g_{xy_j}(z) < f(z) + \varepsilon/2$; hence $g_x(z) := \min_j g_{xy_j}(z) \le g_{xy_j}(z) < f(z) + \varepsilon/2$ holds pointwise for every $z$. By the lattice property applied to $\overline{\mathcal{A}}$, $g_x = \min_j g_{xy_j} \in \overline{\mathcal{A}}$. 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 $h := \max_i g_{x_i} \in \overline{\mathcal{A}}$ (lattice property in $\overline{\mathcal{A}}$), every $z \in K$ lies in some $V_{x_i}$, giving $h(z) \ge g_{x_i}(z) > f(z) - \varepsilon$, while the $\varepsilon/2$ slack in (b) gives $g_{x_i} \le f + \varepsilon/2$, so $h \le f + \varepsilon/2 < f + \varepsilon$. Hence $f - \varepsilon < h < f + \varepsilon$ on $K$, with $h \in \overline{\mathcal{A}}$.

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 $n = 40$).