What is the Gelfand-Naimark representation of functions that don't vanish at infinity?

Question

The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:X\rightarrow\mathbb{C}$ 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 $f:Y\rightarrow\mathbb{C}$, 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?

Answer 1

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 $\Delta:Y\to\beta Y$ such that $\Delta(Y)$ is dense in $\beta Y$. The important property is that $f\mapsto f\circ\Delta:C(\beta Y)\to C_b(Y)$ 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.

Answer 2

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 $A=C_0(X)\oplus\Bbb C\Bbb1$ follows, the second ist $A=C_b(X)$. The first case will give you the one-point compactification of $X$, ie $\sigma(C_0(X)\oplus\Bbb C\Bbb1)\cong X\cup\{\infty\}$ (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 $f:X\to\Bbb C$ extends to a function $\beta X\to \Bbb C$ 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): $$\text{unitisations of $C_0(X)$}\leftrightarrow\text{compactifications of $X$}.$$ 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 $x\mapsto x$ and the function $x\mapsto 1-x$. 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 $x\mapsto x$ 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.

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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.