Theorem - Let $X$ be a compact space and $C(X)$ the algebra of continuous functions $X \longrightarrow \mathbb{C}$ endowed with the sup norm $\| \cdot \|_{\infty}$ . Let $\mathcal{A}$ be a subalgebra of $C(X)$ for which the following conditions hold:

1. $\forall x, y \in X, x \ne y, \exists f \in \mathcal{A} : f(x) \neq f(y)\;$ , i.e. $\mathcal{A}$ separates points
2. $1 \in \mathcal{A}\;$ , i.e. $\mathcal{A}$ contains all constant functions
3. If $f \in \mathcal{A}$ then $\overline{f} \in \mathcal{A}\;$ , i.e. $\mathcal{A}$ is a self-adjoint subalgebra of $C(X)$

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Proof: The proof follows easily from the real version of this theorem (see the parent entry).

Let $\mathcal{R}$ be the set of the real parts of elements $f \in \mathcal{A}$ , i.e.

We can see that $\mathcal{R} \subseteq \mathcal{A}$ . In fact, $\mathrm{Re}(f)= \frac{f + \overline{f}}{2}$ and by condition 3 this element belongs to $\mathcal{A}$ .

Moreover, $\mathcal{R}$ is a subalgebra of $\mathcal{A}$ . In fact, since $\mathcal{A}$ is an algebra, the product of two elements $\mathrm{Re}(f)$ , $\mathrm{Re}(g)$ of $\mathcal{R}$ gives an element of $\mathcal{A}$ . But since $\mathrm{Re}(f).\mathrm{Re}(g)$ is a real valued function, it must belong to $\mathcal{R}$ . The same can be said about sums and products by real scalars.

Let us now see that $\mathcal{R}$ separates points. Since $\mathcal{A}$ separates points, for every $x \neq y$ in $X$ there is a function $f \in \mathcal{A}$ such that $f(x) \neq f(y)$ . But this implies that $\mathrm{Re}(f(x)) \neq \mathrm{Re}(f(y))$ or $\mathrm{Im}(f(x)) \neq \mathrm{Im}(f(y))$ , hence there is a function in $\mathcal{R}$ that separates $x$ and $y$ .

Of course, $\mathcal{R}$ contains the constant function $1$ .

Hence, we can apply the real version of the Stone-Weierstrass theorem to conclude that every real valued function in $X$ can be uniformly approximated by elements of $\mathcal{R}$ .

Let us now see that $\mathcal{A}$ is dense in $C(X)$ . Let $f \in C(X)$ . By the previous observation, both $\mathrm{Re}(f)$ and $\mathrm{Im}(f)$ are the uniform limits of sequences $\{g_n\}$ and $\{h_n\}$ in $\mathcal{R}$ . Hence,