The set of unital homomorphisms

Question

Let $X$ be a locally compact Hausdorff space, which is not compact. $C_b(X)$ is a Banach algebra of all bounded continuous functions with the sup norm. Let $C_\infty(X)$ be the Banach algebra of continuous functions such that for every $f \in C_\infty(X)$, for every $\epsilon >0$ there exists a compact subset $K$ such that if $x \notin K, |f(x)|_\infty < \epsilon$. Let $p \notin X$ and define $\tilde{X} = X \cup p$. Topology on $\tilde{X}$ is topology on $X$ plus all complements of compact subsets of $X$. Notice that $\tilde{X}$ is compact Hausdorff. Set $\mathcal{B} = C(\tilde{X})$. $\mathcal{B}$ can be identified with the closed subalgebra of $C_b(X)$ generated by all $f \in C_\infty(X)$ and the constant function 1.

I need to determine the set of all unital homomorphisms from $\mathcal{B}$ into $\mathcal{R}$, and I have no clue where to start. Any hints are greatly appreciated.

Edit (next step): thank you all for the hints, I got the first part. Now comes the second part: Let $\mathcal{A} = C_\infty(X)$. It is Banach algebra without unit. I need to determine $\hat{\mathcal{A}}$, a set of homomorphisms into R that are not identically zero. Hint given is to use the set of unital homomorphisms from $\mathcal{B}$ into R. I still have no idea. Help, please?

Answer 1

For a compact Hausdorff space $K$ the nonzero homomorphisms onto $\mathbb R$, usually called characters, are precisely the point-evaluations. That is, $\gamma:C(K)\to\mathbb R$ is a character if and only if there exists $x\in K$ such that $\gamma(f)=f(x)$ for all $f\in C(K)$.

As far as I can tell (would be very happy to be proven wrong) there is no entirely straightforward argument to prove this. For a fixed character $\gamma$, here is one possible path:

• Show that for each $f\in C(K)$ there exists $x_f\in K$ such that $\gamma(f)=f(x_f)$.

• Then show that $\gamma$ is bounded and positive.

• Using that $\gamma$ is positive, show that if $T_f=f^{-1}(\gamma(f))$ then $\{T_f\}_{f\in C(K)}$ has the finite-intersection property and hence $\bigcap_{f\in C(K)}T_f\ne\varnothing$. The intersection necessarily consists of a single point.

None of the above steps is very hard, but they require a bit of trickery.

Another possible path is as follows:

• Show that $\gamma$ is bounded.

• The above implies that $\ker\gamma$ is a closed ideal of $C(K)$

• Prove that closed ideals of $C(K)$ are of the form $\{f:\ f|_L=0\}$ for a fixed closed $L\subset K$.

• Use that $\gamma$ is scalar-valued to show that $\ker\gamma$ is maximal.

• Conclude that the corresponding $L$ is of the form $K\setminus \{x_\gamma\}$.