Questions about the C* subalgebra $h(\Gamma)$ of $l^{\infty}(\Gamma)$

Question

Given $(\Gamma,d)$ a metric space that is discrete. We know that $l^{\infty}(\Gamma)$ is a commutative $C^*$ algebra that is isometrically isomorphic to $C(\beta\Gamma)$. Let $h(\Gamma)$ be a subset of $l^{\infty}(\Gamma)$ containing all bounded complex valued functions $f$ that satisfy the following property: $\forall \epsilon >0$ and $\forall r >0$, there are only finitely many pairs of points $x,y \in \Gamma$ such that $d(x,y)\leq r$ and $|f(x)-f(y)|>\epsilon$.

I have proved that this $h(\Gamma)$ is a commutative $C^*$ sub-algebra of $l^{\infty}$. And here are my two questions:

1. Knowing that $h(\Gamma)$ is a commutative $C^{*}$ algebra, the Gelfand transform tells us that it is isometrically isomorphic to $C(\widehat{h(\Gamma)})$, where $\widehat{h(\Gamma)}$ here is the space of all the characters, or equivalently, the space of all the maximal ideals in $h(\Gamma)$. In an enlightened-more discriptive way, how should I think about this space $\widehat{h(\Gamma)}$?
2. Any experienced $C^{*}$ algebraist can tell me anywhere in the literature that talks about this strange $C^{*}$ algebra $h(\Gamma)$?

Answer 1

If $\Gamma$ is such that every ball is finite, then $h(\mathbb N)=\ell^\infty(\Gamma)$.

If $\Gamma$ is such that it has an infinite ball, then $h(\Gamma)$ is the set of bounded functions that are constant off a countable subset of $\Gamma$.

Indeed, choose $r_1>0$ and $x_1\in\Gamma$ such that $B(x_1,r_1)$ is infinite. Then there is a finite set $F_{11}\subset B(x,r_1)$ such that $|f(x_1)-f(x)|\leq1$ for all $x\in B(x_1,r_1)\setminus F_{11}$; and finite set $F_{12}\subset B(x_1,2r_1)$ such that $|f(x_1)-f(x)|\leq1/2$ for all $x\in B(x_1,r_1)\setminus F_{12}$; and, continuing inductively, a finite set $F_{1k}\subset B(x_1,k\,r_1)$ such that $|f(x_1)-f(x)|\leq1/k$ for all $x\in B(x_1,k\,r_1)\setminus F_{1k}$.

Let $F_1=\bigcup_kF_{1k}$. Then, for any $x\in \Gamma\setminus F_1$ we have $|f(x_1)-f(x)|<1/k$ for all $k$ big enough, so $f(x)=f(x_1)$.

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Regarding the spectrum of $h(\Gamma)$, I cannot think of a specific characterization. That's not surprising, though: very few C$^*$-algebras have a spectrum that can be characterized explicitly.

As for the literature, I wouldn't count on this appearing in the literature. Non-separable abelian C$^*$-algebras are objects that don't generate a lot of interest.