Background:
Let $G$ be a locally compact group, with left regular representation $\lambda:G\to B(L^{2}(G))$.
The Fourier algebra $A(G)$ is typically defined to be the space of functions of the form:
$f(s) = \langle \lambda(s)\zeta,\eta\rangle$ for some $\zeta,\eta \in L^{2}(G)$
and the Fourier-Stieltjes algebra $B(G)$ is defined to be the space of all functions of the form
$f(s) = \langle \pi(s)\zeta,\eta\rangle$ for some $\zeta,\eta \in H_{\pi}$, for some continuous unitary representation $(\pi,H_{\pi})$ of $G$.
Body:
I am reading some lecture notes on these algebras which instead do the following:
For any continuous unitary representation $(\pi,H_{\pi})$ of $G$, define $\pi_{\zeta,\eta}:G\to\mathbb{C}$ by $\pi_{\zeta,\eta}(s) = \langle\pi(s)\zeta,\eta\rangle$ and set
$$F_{\pi} = span\{\pi_{\zeta,\eta}:\zeta,\eta\in H_{\pi}\}$$
Then define the Fourier-Stieltjes algebra to be $B(G) = \bigcup_{\pi} F_{\pi}$ where the union is taken over all continuous unitary representations of $G$.
Then the space $A_{\pi}$ is defined to be the closure of $F_{\pi}$ in $B(G)$; and the Fourier Algebra $A(G)$ is defined be the space $A_{\lambda}$.
My questions:
1) Clearly the two definitions of $B(G)$ are identical. But with the latter definition of $A(G)$ we've added both a linear span and closure in the norm topology. I'm going to hazard a guess that it is a very non-trivial result that these definitions are equivalent, and the advantage to using the latter definition is to be able to avoid the issue entirely.
2) Assuming that my guess is correct, is this also true for other $\pi\neq\lambda$? In other words, is $A_{\pi}$ simply the space of all functions of the form
$f(s) = \langle \lambda(s)\zeta,\eta\rangle$ for some $\zeta,\eta \in H_{\pi}$
or is the span and closure necessary for these other spaces $A_{\pi}$?.
Many thanks in advance for any clarification I can get!
The answer is yes to part (1) and no to (2), but I do not have a proof.
If anyone can do better, I'd happily award the bounty.
Edit: The bounty has expired, and the question is no longer urgent, but if someone could give an example that establishes that (2) fails I will eagerly start another bounty and award it to them (as long as that does not violate the rules of this site).