Let $X$ be a locally compact, non-compact Hausdorff space. Let $X^+$ denote the one point compactification of $X.$ Let $C_0 (X)$ denote the space of all complex valued continuous functions on $X$ vanishing at infinity and $C_0 (X)^{+}$ be it's unitization. Then $C_0(X)^{+}$ is $\ast$– isomorphic to $C_0 \left (X^{+} \right )$ as $C^{\ast}$-algebras.
Could anyone give me some suggestion as to how to draw such an isomorphism? Any help would be a boon for me at this stage.
Thanks a bunch.
EDIT $:$ I am thinking along the following lines $:$
Consider any sequence $\{x_n\}_{n \geq 1}$ in $X$ converging to the point $\infty \in X^{+}.$ Now let $\varepsilon \gt 0.$ Since $f \in C_0 (X)$ we may consider a compact subset $K \subseteq X$ such that $|f(x)| \lt \varepsilon$ for all $x \in X \setminus K.$ Now $X^{+},$ being the one point compactification of $X,$ any open neighborhood of $\infty$ in $X^{+}$ is the complement of a compact subset of $X$ in $X^{+}.$ Since $\{x_n\}_{n \geq 1}$ is a sequence in $X$ converging to $\infty$ there exists $k \in \mathbb N$ such that $x_n \in X \setminus K$ for all $n \geq k.$ But that will imply $|f(x_n)| \lt \varepsilon$ for all $n \geq k.$ This shows that $f(x_n) \to 0.$ This will allow us to extend our $f \in C_0(X)$ to a function $\overline {f} \in C_0 \left (X^{+} \right )$ in the following way $:$ $$\overline {f} (x) = \begin{cases} f(x) & x \in X \\ 0 & x = \infty \end{cases}$$ Also since $X^{+}$ is compact, $C_0 \left (X^{+} \right )$ is unital. Let $1$ be it's identity. Now define a map $\varphi : C_0(X)^{+} \longrightarrow C_0 \left (X^{+} \right )$ by $$\varphi ((f, \lambda)) = \overline {f} + \lambda\ 1,\ f \in C_0 (X),\ \lambda \in \mathbb C.$$ I think it will give us the required isometric $\ast$-isomorphism.
Best Answer
Your argument is fine, minus a few minor details.
First, you cannot use sequences, you have to use nets. But it doesn't change your argument. The argument can be simplified a bit by noting that by definition the neighbourhoods of $\infty $ are the complements of compact sets. So taking the $\varepsilon$ and $K$ as you have them, you need $|f(\infty)|<\varepsilon$ if you want to extend $f$ as continuous. That is, the only possible continuous extension of $f$ is to define $f(\infty)=0$.
The isomorphism is fine, and you only have to check it is bijective, which is trivial.