I'm looking at the following example from C* algebras by Murphy, and I'm totally lost on the notation they're using.
"Let $X$ be a locally compact Hausdorff space. Since $C_0(X) \subset C_b(X)$ is an essential ideal, there exists a unique injective $*$ homomorphism $\varphi: C_b(X) \longrightarrow M(C_0(X))$ extending the inclusion $C_0(X) \rightarrow M(C_0(X))$. We wish to show that $\varphi$ is onto, and to do this, it suffices to show that if $g \in M(C_0(X))^{+}$, then $\varphi(f)=g$ for some $f \in C_b(X)$ as $M(C_0(X))^{+}$ linearly spans the $C^{*}$ algebra $M(C_0(X))$.
Let $g \in M(C_0(X))^{+}$. If $(u_{\lambda})_{\lambda \in \Lambda}$ is an approximatte unit for $C_0(X)$, then for each $x \in X$, the net of real numbers $(gu_{\lambda}(x))_{\lambda \in \Lambda}$ is increasing and bounded by $||g||$, so it converges by MCT to some $h(x)$. Define the map $h: X \longrightarrow \mathbb{R}$ by $x \mapsto h(x)$. Then $h$ is bounded, and $hf=gf$."
How is $gu_{\lambda}(x) \in \mathbb{R}$? $g$ is some ordered pair $(L,R)$ of bounded linear maps from $C_0(X)$ into itself, so I dont understand what $gu_{\lambda}(x)$ even means. This also doesn't make any sense in the line below where they say that $hf=gf$.
Are they assuming that $g$ is an element of $C_0(X)^{+}$ identified in $M(C_0(X))^{+}$? If so, why do they get to assume that, when $g$ was assumed to simply be a positive element of $M(C_0(X))$?
Best Answer
By definition, $g$ is a multiplier of $C_0(X)$; that's the whole point of multipliers: that they multiply into the algebra. How the multipliers are constructed is immaterial.
Which means that $gu_\lambda\in C_0(X)$. The question is why the function $gu_\lambda$ is real-valued. And this is because it is a product of positive elements. Such a product is not necessarily selfadjoint in general; but it always has non-negative real spectrum. Combined with the fact that we are in a function algebra, where the spectrum is the closure of the image, we get that $gu_\lambda$ is real-valued.