Krdytkyu

The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:Xrightarrowmathbb{C}$ that vanish at infinity, for some locally compact Hausdorff space $X$.

What happens when we apply this construction to a commutative C*-algebra that looks almost like $C_0(X)$, but we relax one of the requirements? For example, we can consider the C*-algebra $C_b(Y)$, the set of bounded continuous functions $f:Yrightarrowmathbb{C}$, with no requirement they vanish at infinity. Is there any relation between the the space $Y$ of the original C*-algebra and the space $X$ that we generate using Gelfand-Naimark?

Bonus followup: What if Y is not Hausdorff? Or not locally compact?

Since $C_b(Y)$ has a unit, if it is isometrically $^*$-isomorphic to $C_0(X)$ then $C_0(X)$ must also have a unit, so $X$ is compact. This is intuitively why $X$ will need to be some sort of compactification of $Y$. To be exact, we will get the Stone Cech compactification.

The Stone Cech compactification $beta Y$ of $Y$ comes with an embedding $Delta:Ytobeta Y$ such that $Delta(Y)$ is dense in $beta Y$. The important property is that $fmapsto fcircDelta:C(beta Y)to C_b(Y)$ is an isometric $^*$-isomorphism. Since $beta Y$ is by definition also compact, we have $C_0(beta Y)=C(beta Y)$, so $C_b(Y)$ is isometrically $^*$-isomorphic to $C_0(beta Y)$.

Do note that $Y$ is required to be completely regular for the Stone Cech compactification to be well-defined. I do not know about the case where $Y$ is not completely regular.

Here is a slight elaboration of the previous answer. Lets us look at a system $mathcal F$ of (bounded) continuous functions on $X$ that do not vanish at infinity and "include" these in our algebra, what space do we get with the Gelfand isomorphism?

To be more concrete, let $A$ be the $C^*$ sub-algebra of $C_b(X)$ generated by $C_0(X)$ and $mathcal F$, what is $sigma(A)$?

There are two such completions I can always write down. The simplest case is $mathcal F = {1}$ (the constant function $1$) from which $A=C_0(X)oplusBbb CBbb1$ follows, the second ist $A=C_b(X)$. The first case will give you the one-point compactification of $X$, ie $sigma(C_0(X)oplusBbb CBbb1)cong Xcup{infty}$ (either this is clear to you or its a very illustrative exercise). Meanwhile the second will return the Stone-Cech compactification, to see this note that any bounded function $f:XtoBbb C$ extends to a function $beta Xto Bbb C$ by the universal property of $beta X$ (note that $mathrm{im}(f)$ is contained in a compact set), furthermore the extension is unique by denseness of $X$ in $beta X$. From this $C_b(X)subset C(beta X)$ follows and the other direction is clear.

Both of the spaces we got back were compact, which is not a coincidence. Note that in both cases we added the constant functions, which results in the unit of $C(X)$ being added to the algebra. I believe that it is more or less clear that the Gelfand isomorphism results in a compact space iff the original algebra was unital, ie it gives a correspondence
$$text{unital commutative $C^*$ algebras} leftrightarrow text{compact locally compact Hausdorff spaces}$$
in particular we find via this correspondence another correspondence (for $X$ locally compact Hausdorff):
$$text{unitisations of $C_0(X)$}leftrightarrowtext{compactifications of $X$}.$$
So basically when you add functions at infinity you are compactifying $X$, provided you add enough functions so you can reconstruct $1$. Lets consider another example.

For $X=(0,1)$ you can choose to add the function $xmapsto x$ and the function $xmapsto 1-x$. If you do this you will get a unital algebra and indeed you will get all polynomials, so by Stone-Weierstrass your algebra will have all functions in $C([0,1])$. Since you also don't have more functions than that you see that adding these two functions results in taking the closure of $(0,1)$ in $Bbb R$.

Indeed if $X$ is a bounded open subset of $Bbb R^n$, the compactification you get by considering its closure will correspond to the unitsation you get by adding the $n$ coordinate functions $x_i$ to the algebra.

What if you don't add enough functions to recover identity? In our example $X=(0,1)$ if we just add $xmapsto x$ the space we get is $(0,1]$. Indeed this is what always will happen: You will get an open subspace of a compactification of $X$.

The general lesson is thus: If you add functions at infinity you are basically extending $X$ by points at infinity.

To address your follow up: Continuous functions from a non-Hausdorff space into $Bbb C$ cannot distinguish two non-separated points, ie if $x,y$ are such that every neighbourhood of $x$ intersects every neighbourhood of $y$ then $f(x)=f(y)$ for any continuous $f$. For that reason $C(X)=C(X/sim)$ if $X$ is not Hausdorff and $sim$ is the relation that glues all points that are not separated together. So basically continuous functions to $Bbb C$ do not actually the non-Hausdorff parts of your space.

For spaces that are not locally compact I believe $C_0(X)$ will be entirely inadequate. Notice for example that $C_0(V)=0$ if $V$ is an infinite dimensional Banach space (for me the model of non locally compact spaces). I have no idea what the space associated to $C_b(V)$ looks like.