Invariant $C^\ast$ subalgebras of $C(\mathbb R)$ On the bounded continuous functions on $\mathbb R$, $C(\mathbb R)$,  define the $\mathbb R$-group action of translations $\alpha: \mathbb R \times C(\mathbb R) \to C(\mathbb R)$ as $\alpha_x(f)(y) := \alpha(x, f)(y) = f(y - x)$. Further, let us consider $\beta: C(\mathbb R) \to C(\mathbb R), ~\beta(f)(y) = f(-y)$.
Of course, there are plenty of $\alpha$- and/or $\beta$-invariant $C^\ast$ subalgebras of $C(\mathbb R)$. It is not difficult to come up with subalgebras which are $\alpha$- but not $\beta$-invariant, such as $\{ f \in C(\mathbb R): ~f(x) \to 0, ~x \to \infty\}$. Somehow, all examples of $\alpha$- but not $\beta$-invariant $C^\ast$ subalgebras of $C(\mathbb R)$ I can imagine are of the following kind:

*

*They are ideals in some $\alpha$- and $\beta$-invariant subalgebra $B\subseteq C(\mathbb R)$;

*They are of the form $A = A_0 \oplus \mathbb{C} 1$, where $A_0$ is an ideal in some $\alpha$- and $\beta$-invariant subalgebra $B \subseteq C(\mathbb R)$.

To possibly clarify this: In the above example we have

*

*$\{ f \in C(\mathbb R): f(x) \to 0, x \to \infty\}$ is $\alpha$- but not $\beta$-invariant and it is of the first kind; it is an ideal of $C(\mathbb R)$,

*The algebra $\{ f \in C(\mathbb R): f(x) \to c, x \to \infty$ for some $c \in \mathbb C\}$ is of the form $A_0 \oplus \mathbb C1$ with $A_0$ the ideal of $C(\mathbb R)$ from 1.

So my question is:
Are there examples of $\alpha$- but not $\beta$-invariant $C^\ast$ subalgebras of $C(\mathbb R)$ which do not arise in one of the two ways? Or conversely:
If $A \subseteq C(\mathbb R)$ is an $\alpha$-invariant $C^\ast$ subalgebra (say, without unit), is there always some $\alpha$- and $\beta$-invariant subalgebra $B \subseteq C(\mathbb R)$ such that $A$ is an ideal of $B$?
 A: Here is an example.
Consider the closure of graph $\Gamma$ of
$$\sin\left(\frac{1}{|x|-1}\right), \qquad x\in(-1,1)$$
This is a compact space which looks sort of like this:

Basically there are two intervals and that are connected via some line. Chose some homoemorphism of the line in the middle with $\Bbb R$ to embed the algebra $C(\overline\Gamma)$ into $C_b(\Bbb R)$. Now the points on the intervals are explicit examples of "points at infinity" (basically $\overline\Gamma$ is a special compactification of $\Bbb R$ that we have chosen). The $\alpha$-action leaves the points at infinity unchanged and shifts the points on the connecting line forwards and backwards in a way depending on our homeomorphism (but we don't care about the details).
We can choose the identification of the line with $\Bbb R$ in such a way that $\beta$ becomes the map
$$\beta(f)= [(x,y)\mapsto f(-x,y)].$$
Now look at a sub-algebra $A$ of $C(\overline\Gamma)$ defined by:
$$A=\{f\in C(\overline\Gamma) \mid f(1,1) = f(1,-1),  \quad f(-1,-1) = 0\}$$
In other words we want the point at the top right to evaluate to the same thing as the point of the bottom right and the point at the bottom left should evaluate to $0$. This condition is $\alpha$-invariant but not $\beta$-invariant.
Let $g$ be some function on $\overline\Gamma$ with $g(1,1)=1=g(1,-1)$, and $g(-1,1)=1$, $g(-1,-1)=0$. Then
$$(\beta(g)\cdot g) (1,1) = 1\cdot 1, \quad (\beta(g)\cdot g)(1,-1) = 0\cdot 1$$
and clearly $\beta(g)\cdot g\notin A$. But $g\in A$, hence $A$ is not an ideal of any $\beta$-invariant sub-algebra of $C_b(\Bbb R)$.
