Stone-Weierstrass theorem (complex version)

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.

$\forall x,y\in X,x\neq y,\exists f\in\mathcal{A}:f(x)\neq f(y)\;$ , 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 (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)
Canonical name	StoneWeierstrassTheoremcomplexVersion
Date of creation	2013-03-22 18:02:31
Last modified on	2013-03-22 18:02:31
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	6
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46J10