Every Hausdorff-compactification of the reals corresponds to a C*-subalgebra of bounded continuous functions on $\mathbb{R}$

Question

Let $\mathcal A$ denote the C*-algebra of complex valued, bounded continuous functions on the reals $\mathrm{C}_\mathrm{b}(\mathbb{R})$. Furthermore, let $\mathcal A_c$ denote the C*-subalgebra of $\mathcal A$ such that for every $f \in \mathcal A_c$ the limits $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty}$ exist and coincide.

On the one side, it is possible to identify the C*-subalgebra $\mathcal A_c$ with the continuous functions on the one-point compactification on $\mathbb{R}$. On the other side, one can see $\mathcal A$ as continuous functions on the Stone-Cech compactification of $\mathbb{R}$.

Let $K$ be another (Hausdorff-) compactification of the reals, i.e. $K$ is compact and we have a continuous embedding $\varphi \colon \mathbb{R} \to K$ such that $\overline{\varphi(\mathbb{R})} = K$. I want to show, that I can identify the C*-Algebra $\mathcal B := \mathrm{C}(K)$ with a sub-algebra of $\mathcal A$ which contains $\mathcal A_c$.

Consider some $f \in \mathcal B$. Since $\varphi(\mathbb{R})$ is dense in $K$, I can identify $f$ with its restriction $f_{|\mathbb{R}}$, which is bounded because $K$ is compact. This should show that we can consider $\mathcal B$ a subalgebra of $\mathcal A$, correct?

What about the other "inclusion". Given $f \in \mathcal A_c$ how would one "extend" this function to $K$. Since $K$ is Hausdorff, it is clear that once a function in $\mathcal A_c$ exhibits an extension, this extension will be unique.

Do you know how to prove the existence of a continuous extension to $K$?

Answer 1

Response to comment by OP: How does local compactness of $\mathbb R$ imply $\mathbb R$ is open in K? (Too long for a comment).

Lemma. Let $X$ be a dense subspace of $Y$ and let $U$ be an open subset of the space $X.$ If $Cl_X(U)$ is closed in $Y$ then $U$ is open in Y....And we also have $Cl_Y(U)=Cl_X(U)$.

Proof. Let $U=V\cap X$ where $V$ is an open subset of $Y.$ Since X is dense in $Y $ and $V$ is open in $Y$ we have $Cl_Y(V)=Cl_Y(V\cap X).$ Hence $Cl_Y(V)=Cl_Y(U).$ Now since $Cl_X(U)$ is closed in $Y,$ we have $$X\supset Cl_X(U)=Cl_Y(Cl_X(U))\supset Cl_Y(U)=Cl_Y(V)\supset V.$$ So $X\supset V,$ implying $U=V\cap X=V,$ so $U$ is open in $Y.$

And we also have $X\supset Cl_Y(V)=Cl_Y(U)$ so $Cl_X(U)=X\cap Cl_Y(U)=Cl_Y(U).$

Corollary. If $Y$ is Hausdorff and $X$ is dense in $Y,$ and if $U$ is an open subset of the space $X$ such that $Cl_X(U)$ is compact then $U$ is open in $Y$. Because $Cl_X(U)$ must be closed in $Y$ because $Y$ is Hausdorff and $Cl_X(U)$ is compact.

Now let $X$ be a locally compact Tychonoff space and $id_X:X\to Y$ be a compactification of $X$. For any $p\in X$ let $p\in U_p\subset X$ where $U_p$ is open in $X$ and $Cl_X(U_p)$ is compact. By the corollary, $U_p$ is open in $Y.$ So $X=\cup_{p\in X}U_p$ is open in $Y.$

BTW. The converse also holds: If $c:X\to K$ is a compactification of the Tychonoff space $X,$ such that $c(X)$ is open in $Y,$ then $X$ is locally compact.

Answer 2

Given $f \in \mathcal A_c$, let $L(f) = \lim_{x \to \infty} f(x)$. Then you extend $f$ to $\tilde{f} \in C(K)$ by $\tilde{f}(\varphi(x)) = f(x)$ while $\tilde{f}(k) = L(f)$ for $k \in K \backslash \varphi(\mathbb R)$. It is not hard to show that this is continuous.