I'm a little embarassed that I can't answer this myself, so hopefully it will get answered very quickly.

Let $X$ be locally compact, Hausdorff. Consider $\text{C}_\text{b}(X)$ the $C^*$-algebra of bounded, complex-valued, continuous functions on it under the sup norm. The Gelfand transform takes $\text{C}_\text{b}(X)$ isomorphically to $\text{C}_0(Y)$ where $Y$ is the maximal ideal space of $\text{C}_\text{b}(X)$. Since $\text{C}_\text{b}(X)$ is unital, $Y$ must be compact.

It is well known that $Y$ is the Stone–Čech compactification of $X$. If $X = \mathbb{R}$ then $Y$ is the one-point-compactification. In general, the maximal ideal space is in one-to-one correspondence with the characters (nonzero multiplicative linear functionals) on $\text{C}_\text{b}(X)$.

**Question:** Say $X = \mathbb{R}$ and the maximal ideal space of $\text{C}_0(\mathbb{R})$ is made up of the point evaluations at points on $\mathbb{R}$. These point evaluations are also characters on $\text{C}_\text{b}(X)$, but its maximal ideal space is the one-point-compactification of $\mathbb{R}$. What character on $\text{C}_\text{b}(\mathbb{R})$ represents this single missing point of the maximal ideal space?

The big issue I'm having is that for a non-compact space like $\mathbb{R}$, we have an obvious problem that you could take a sequence of point evaluations $\chi_n(f) = f(n)$ where $n \to \infty$, which clearly goes to the zero functional if your $f \in \text{C}_0(\mathbb{R})$. This doesn't happen in $\text{C}_\text{b}(\mathbb{R})$ because the maximal ideal space doesn't need to add the zero functional in order to be compact, but that would imply that $\chi_n$ is converging to some point evaluation "at infinity" (which fits with the idea that $\mathbb{R}$ gets compactified by adding a point "at infinity"). But $\chi_n(f) = f(n)$ as $n\to\infty$ still doesn't make sense for functions $f \in \text{C}_\text{b}(\mathbb{R})$ (take $\sin(x)$ for example).