Given a topological group ($G, *, \tau$), a bounded continuous real value function $f$ on $G$ and a fixed $x \in G$, define $L_x(f)$ to be another function such that $L_x(f)(g) = f(x^{-1}g) \forall g \in G$ and call $L_x(f)$ the left translate of $f$ by $x$. It is not hard to see that for any $g \in G$, $L_g$ will map elements from $C_B(G)$ back to itself and the set of all $L_g$ will be a group under composition.
We denote the set of all bounded continuous real value function on $G$ by $C_B(G)$ and one can check that $d^*(f, g) = sup_{x \in G}|f(x) - g(x)|$ will be a metric on $C_B(G)$. Let $\tau_{d^*}$ be the metric topology generated by $d^*$, $\alpha(G)$ = {$L_x|x \in G$}, $e_f(O)$ = {$L_g \in \alpha(G)|L_g(f) \in O$} where $O$ is a open set in ($C_B(G), d*$) and $\mathscr{A}$ = {$e_f(O)|f \in C_B(G), O$ is open in $(C_B(G), d^*)$}.
One can check that $\mathscr{A}$ is a subbase of a topology. We call $\tau_{\mathscr{A}}$ the pointwise convergence topology. For any $f \in C_B(G)$, define $A_f =$ {$L_x(f) | x \in G$}.
(Latest Edit Part 1):
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Define $X_f$ to be the closure of $A_f$ with respect to $\tau_{d^*}$ and call it the left orbit of $f$. We call $f$ left almost periodic iff $X_f$ compact with respect to $\tau_{d^*}$. Now let $A$ be the set of all continuous and left almost periodic functions defined on $G$. Then for each $x \in G$, $L_x (A) \subseteq A$. Further we assign the space $A^A$ the pointwise topology (the same way how we define $\tau_{\mathscr{A}})$. Let $\overline{\alpha(G)}$ be the closure of $\alpha(G)$ with respect to the pointwise topology.
We can see for any $f \in A$, $L_x(f) \in X_f$. In the rest of the post, we only focus on $L_x|_{A}$. In the rest of the post, please assume all $L_x$ is only defined on $A$ instead of the whole $C_B(G)$.
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(End of Latest Edit Part 1)
(Latest Edit Part 2)
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Used fact:
a. If ($X, d$) is compact metric space and $K$ the group of isometries defined on $X$ (under composition). Here we use $d^*$ as well but $d^*(f, h)$ = $sup_{g \in G} d(f(g), h(g))$. Then ($K, \tau_{d^*}$) is a topological group and $\tau_{d^*}$ is the same as the topology generated by pointwise convergence. Further ($K, \tau_{d^*}$) is compact.
b. If we equip the space $\times_{f \in A} X_f$ with the pointwise convergence topology, then $\overline{\alpha(G)}$ will be a compact subset.
c. We observe $L_x$ is always a isometry from $\tau_{d^*}$ to $\tau_{d^*}$ and hence every element in ($\overline{\alpha(G)}$, $\tau_{\mathscr{A}}$) will be isometry. Let $K_f$ be the set of all isometries that maps from $X_f$ to $X_f$. Inside the book, it says the following (please check (c)):
Here I guess it wants me to prove $\alpha(G)$ is isomorphic $K_f$ so that the "natural" mapping is $L_x \rightarrow (L_x)_{f \in A}$ and I did that. Please feel free to provide your comments to this one. Hence $\alpha(G)$ is a topological group.
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(End of Latest Edit Part 2)
For any topological group, being $T_0$ can imply being Hausdorff (check "General Topology" by John Kelly Chapter 3 Exercise S Questions f) so as a corollary if ($G, *, \tau$) is compact, then then mapping $g \rightarrow L_g$ will be a homeomorphism.
Edit: When I have real value function $f$ on $G$, $L_{h_1}(f)$ and $L_{h_2}(f)$, I want another real value function $g$ such that $g$ is uniformly close to $L_{h_1}(f)$ but not $L_{h_2}(f)$. In a compact topological group, I do not know if there are any other tools I can use to "separate" they two by some preimage of continuous function. If I can, then probably I can adjust that function so that it will be uniformly close to $L_{h_1}(f)$.
Edit again: Thanks for Alex's comment, I will add some important details that I missed in the question.
(Latest Edit Part 3)
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Now I will address Alex's questions
$\mathscr{A}$ is a family of subsets of $\alpha(G)$ but each element of $\mathscr{A}$ does not contain mapping from $C_B(G)$ but mapping that maps $C_B(G)$ to $C_B(G)$. For example, you will see in the latest edit that $L_x(f) \in X_f \forall f \in X_f$. My previous post might confuse you because I missed some important details but this time it should be more clear.
In the new definition, I only focus on those $f$ such that ($X_f, d^*$) is compact. Originally, $L_x$ is defined on the whole $C_B(G)$ space and now I only want $L_x|_{A}$. Now given $X_f$ compact, I am trying to show that the mapping $g \rightarrow L_g$ is continuous ... Will post the proof once I am done.
Let $x,y\in G$ such that $L_x$ and $L_y$ are distinct elements of $\alpha(G)$. Therefore there exists a function $f\in C_B(G)$ such that $L_x(f)\ne L_y(f)$, that is $L_x(f)(g)\ne L_y(f)(g)$ for some $g\in G$. Pick $0<\varepsilon<|L_x(f)(g)-L_y(f)(g)|/2$ and put $O_x=\{h\in C_B(G): d^*(h,L_x(f))<\varepsilon\}$ and $O_y=\{h\in C_B(G): d^*(h,L_y(f))<\varepsilon\}$. Then $e_f(O_x)$ and $e_f(O_y)$ are disjoint, $L_x\in e_f(O_x)$, $L_y\in e_f(O_y)$, that is $\alpha(G)$ is Hausdorff.