Stone-Weierstrass theorem (complex version)

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

1. $\forall x,y\in X,x\ne y,\exists f\in 𝒜:f(x)\ne f(y)$, i.e. $𝒜$ separates points

2. $1\in 𝒜$, i.e. $𝒜$ contains all constant functions

3. If $f\in 𝒜$ then $\overline{f}\in 𝒜$, i.e. $𝒜$ is a self-adjoint (http://planetmath.org/InvolutaryRing) subalgebra of $C(X)$

Then $𝒜$ is dense in $C(X)$.

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Proof: The proof follows easily from the real version of this theorem (see the parent entry (http://planetmath.org/StoneWeierstrassTheorem)).

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

$ℛ:=\{\mathrm{Re}(f):f\in 𝒜\}$

It is clear that $ℛ$ contains (it is in fact equal) to the set of the imaginary parts of elements of $𝒜$. This can be seen just by multiplying any function $f\in 𝒜$ by $-i$.

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

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

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

Of course, $ℛ$ 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 $ℛ$.

Let us now see that $𝒜$ 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 $ℛ$. Hence,

${\parallel f-(g_n+i{h}_n)\parallel }_{\mathrm{\infty }}\le {\parallel \mathrm{Re}(f)-g_n\parallel }_{\mathrm{\infty }}+{\parallel \mathrm{Im}(f)-h_n\parallel }_{\mathrm{\infty }}⟶0$

Of course, the sequence $\{g_n+i{h}_n\}$ is in $𝒜$. Hence, $𝒜$ is dense in $C(X)$. $\mathrm{\square }$