STATEMENT: Let $C_b(\mathbb{R})$ be the $C^*$-algebra of bounded continuous functions on $\mathbb{R}$. Let $A$ be the $C^*$-subalgebra of $C_b(\mathbb{R})$ generated by $C_∞(\mathbb{R})$ together with the function $f(t) = e^{it}$. Determine $\hat{A}$, which is the maximal ideal space of $A$
QUESTION So in this case I used the fact that $C_\infty$ is an ideal, so it is contained in some maximal ideal. I then let $\phi$ be the associated linear functional on $A$. Since our sub algebra is generated by $C_\infty$ and $f(t)$ we see that $\phi$ is uniquely determined by what $\phi(f(t))$ gets mapped to. Since $f(t)$ is invertible it has to be mapped to a nonzero value. Furthermore, the inequality, $$|\phi(f(t))|\leq |\phi||f(t)|=1\;\;\;\;\; \text{and}\;\;\;\;\; |\phi(f(t)^{-1})|\leq |\phi||f(t)^{-1}|=1$$
So that $|\phi(f(t))|=1$, so $f(t)$ has to be mapped to some element on the unit circle in $\mathbb{C}$. Therefore, we can identify the maximal ideal space with the circle in $\mathbb{R}^2$
My question is that when they say determine the maximal ideal space do they just mean up to homeomorphism? If its not too much of a bother it would be helpful if someone outlines how they would solve such a problem.
Let me try to outline the solution.
I. First, we describe the set of characters (continuous nonzero linear multiplicative functionals) $\varphi$ on $A,$ which is the same as $\hat A=\{\ker\varphi\mid\varphi:A\to\mathbb C\ \mbox{is a character}\}.$
As you have mentioned, there is a set of characters $\varphi_z\in\hat A$ which vanish on $C_\infty$ and $\varphi_z(f)=z\in\mathbb T\subseteq\mathbb C.$
If $\varphi$ does not vanish on $C_\infty,$ then $\varphi$ is a character on $C_\infty.$ By the description of the dual of $C_\infty=C_\infty(\mathbb R),$ there is a point $t\in\mathbb R$ such that $\varphi(g)=g(t),\ \forall g\in C_\infty.$ There is a unique extension of $\varphi$ onto $A,$ namely $\varphi(f)=f(t)=e^{it}.$ Indeed, take $g\in C_\infty,\ g(t)\neq 0.$ Then $fg\in C_\infty$ and by multiplicativity of $\varphi$ we have $\varphi(f)=\frac{\varphi(fg)}{\varphi(g)}=\frac{fg(t)}{g(t)}=f(t).$ Since $f$ and $C_\infty$ generate $A,$ we have $\varphi(g)=g(t)$ for all $g\in A.$ Let $\varphi_{t}$ denote the character $\varphi.$
It follows that $\hat A$ is the disjoint union $\{\varphi_z\mid z\in\mathbb T\}\cup\{\varphi_t\mid t\in\mathbb R\}.$
II. The (Gelfand-)topology $\tau$ on $\hat A$ is the weakest topology such that all functions $g\in C_\infty$ and $f$ are continuous as functions on $\hat A$. (Then all sums, products and sup-norm-limits are continuous on $\hat A$).
Let $g\in C_\infty.$ Then $g(\varphi_z)=0$ and $g(\varphi_t)=g(t).$ Function $g$ is continuous on $\hat A\Longleftrightarrow\tau$ contains all pre-images $g^{-1}((a,b))\Longleftrightarrow$ $\tau$ contains all sets $$ \{\varphi_t\mid t\in(c,d)\}\ \mbox{and all sets}\ \{\varphi_t\mid t\in(c,d)\}\cup\{\varphi_z\mid z\in\mathbb T\}\ \ \ \ \ \ \ (1). $$
(The latter are $g$-pre-images of open sets containing $0$).
Function is $f$ continuous on $\hat A\Longleftrightarrow\tau$ contains $f$-pre-images of all open subsets of $\mathbb T\Longleftrightarrow\tau$ contains the unions $$ \{\varphi_z\mid z=e^{it}, t\in(a,b)\}\cup\bigcup_{k\in\mathbb Z}\{\varphi_t\mid t\in(a+2\pi k,b+2\pi k)\}\ \ \ \ \ \ (2). $$
The sets of type (1) and (2) generate the topology $\tau.$
The topology $\tau$ has the following properties: $\tau$ restricted onto $\mathbb T\subseteq\hat A$ is the usual (euclidean) topology, $\tau$ restricted onto $\mathbb R\subseteq\hat A$ is the usual topology, $\mathbb T$ is closed in $\hat A$, $\mathbb R$ is open in $\hat A,$ the closure of $\mathbb R$ is the whole $\hat A.$
Looking at the sets of type (2) you can imagine how the neighborhoods of $\varphi_z$ look like. The neighborhoods of $\varphi_t$ are given by the "usual intervals in $\mathbb R.$"
You can also try to imagine $(\hat A,\tau)$ as Tracing mentioned in comments "...$\mathbb R$ being a spiral which wraps around a circle $\mathbb T$..."