# Stone-Weierstrass theorem (complex version)

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. 1.

$\forall x,y\in X,x\neq y,\exists f\in\mathcal{A}:f(x)\neq f(y)\;$, i.e. $\mathcal{A}$ separates points

2. 2.

$1\in\mathcal{A}\;$, i.e. $\mathcal{A}$ contains all constant functions

3. 3.

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

Then $\mathcal{A}$ is dense in $C(X)$.

$\,$

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

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

 $\displaystyle\mathcal{R}:=\{\mathrm{Re}(f):f\in\mathcal{A}\}$

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

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,

 $\displaystyle\|f-(g_{n}+ih_{n})\|_{\infty}\leq\|\mathrm{Re}(f)-g_{n}\|_{\infty% }+\|\mathrm{Im}(f)-h_{n}\|_{\infty}\longrightarrow 0$

Of course, the sequence $\{g_{n}+ih_{n}\}$ is in $\mathcal{A}$. Hence, $\mathcal{A}$ is dense in $C(X)$. $\square$

Title Stone-Weierstrass theorem (complex version) StoneWeierstrassTheoremcomplexVersion 2013-03-22 18:02:31 2013-03-22 18:02:31 asteroid (17536) asteroid (17536) 6 asteroid (17536) Theorem msc 46J10