30.6 $BC(X)$
Definition 30.6.1 (Algebra of Bounded Continuous Functions).label Let $X$ be a topological space, then $BC(X; \complex)$ is the space of bounded continuous functions on $X$.
Equipped with pointwise operations and the uniform norm, $BC(X; \complex)$ forms a commutative Banach algebra.
Theorem 30.6.2.label Let $X$ be a completely regular space and $\beta X$ be its Stone-Čech compactification. For each $f \in BC(X; \complex)$, let $\beta f \in BC(\beta X; \complex)$ be its unique extension to $\beta X$, then the mapping
is a homeomorphism. Under the identification $\beta X = \Omega(BC(X; \complex))$, $\Gamma_{BC(X; \complex)}= \beta$.
Proof, [Theorem I.6.4, Zhu93]. Let $\phi \in BC(X; \complex)^{*} \setminus \ol{E(X)}$, then there exists $\seqf{f_k}\subset BC(X; \complex)$ and $\eps > 0$ such that for every $x \in X$,
In which case, if $\phi \in \Omega(BC(X; \complex))$, then
Since $f(x) > 0$ for all $x \in X$, $f \in G(BC(X; \complex))$. As $\phi \in \Omega(BC(X; \complex))$, the above contradicts (3) of Proposition 29.7.2. Thus $\Omega(BC(X; \complex)) \setminus \ol{E(X)}$ is empty, and $E(X)$ is dense in $\Omega(BC(X; \complex))$.
By (U) of the Stone-Čech compactification, $E|_{X}$ extends uniquely to a surjective continuous map. Since $BC(\beta X; \complex)$ and $BC(X; \complex)$ are in bijection via the restriction map, $E$ is injective. Therefore $E$ is bijective, and $E$ is a homeomorphism by Proposition 5.16.4.$\square$
Post a Comment