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 $K$?**