PlanetMath: Stone-Weierstrass theorem (complex version)

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Stone-Weierstrass theorem (complex version)

(Theorem)

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

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

Then $\mathcal{A}$ is dense in

$\,$

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.

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

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

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 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 and .

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

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

Let us now see that $\mathcal{A}$ is dense in . 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,

$\displaystyle \Vert f - (g_n+ih_n)\Vert _{\infty} \leq \Vert\mathrm{Re}(f)-g_n\Vert _{\infty} + \Vert\mathrm{Im}(f)-h_n\Vert _{\infty} \longrightarrow 0$

Of course, the sequence $\{g_n + i h_n\}$ is in $\mathcal{A}$ . Hence, $\mathcal{A}$ is dense in . $\square$

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Cross-references: sequences, limits, observation, Stone-Weierstrass theorem, implies, scalars, sums, product, function, imaginary parts, clear, real parts, real, dense in, constant functions, contains, points, subalgebra, sup norm, continuous functions, algebra, compact

This is version 3 of Stone-Weierstrass theorem (complex version), born on 2008-05-03, modified 2008-05-04.
Object id is 10564, canonical name is StoneWeierstrassTheoremComplexVersion.
Accessed 98 times total.

Classification:

AMS MSC:

46J10 (Functional analysis :: Commutative Banach algebras and commutative topological algebras :: Banach algebras of continuous functions, function algebras)

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