$C(K)^{*}$ in the weak*-topology It is clear that for a compact Hausdorff space $K$ the space of continuous complex (or real) functions $C(K)$ is a Banach space with the "sup" norm. What's not clear to me is whether there exists a homeomorphism between $K$ and the dual space $C(K)^{*}$ equipped with the weak*-topology. My guess is it would have to be linear and well defined in some special way.  
Rudin's $\textit{Functional Analysis}$ suggests choosing a point $p \in K$ and defining $\Lambda_{p} \in C(K)^{*}$ by $\Lambda_{p}(f)= f(p)$ (I think this justifies the existence of a weak*-topology on $C(K)^{*}$). Then the claim is that $p \mapsto \Lambda_{p}$ is the homeomorphism we want. Then again, this map is not onto, so calling it a homeomorphism doesn't seem to make sense at this point.
 A: The most natural map from a compact (nice, topological) space $K$ to $C^o(K)$ is $x\to \delta_x$, meaning $x\to (f \to f(x))$. This is a continuous linear functional when $C^o(K)$ is given the sup-norm topology (and convenient that $K$ is compact, so there's no issue of uniform-ness...).
But there are many other continuous linear functionals, unless $K$ is discrete, hence finite (or weird...)
EDIT: and/but re-reading the question, I don't understand the goal of the questioner. That is, I have the suspicion that the literal question is not the underlying question... ?
EDIT-EDIT: after clarification: to show that this map is a homeomorphism _to_its_image_, first, given continuous $f$ and $\epsilon>0$ and $x\in K$, there is a neighborhood $N$ of $x$ such that for $y\in N$ $|f(x)-f(y)|<\epsilon$, which is the continuity of $x\to \delta_x$. The map is injective because $K$ is presumably Hausdorff, so Urysohn's lemma gives functions separating points. To prove the map is open-to-its-image is to show that $\delta_x$ close to $\delta_y$ implies $x$ is close to $y$ in $K$. The contrapositive is that $x$ far from $y$ in $K$ implies $\delta_x$ is far from $\delta_y$. Again use Urysohn's lemma...
A: $K$ is certainly not homeomorphic to the dual space $C(K)^*$ because the latter is never compact. $K$ is homeomorphic to the unit ball of $C(K)^*$ endowed with the weak* topology.
