When is the Fourier algebra $A(G)$ enough close to the Fourier-Stieltjes algebra $B(G)$? I am reading P.Eymard's paper on the Fourier algebras of locally compact groups, and I have several questions about his constructions. I asked one of them in math.stackexchange, so far without success, so I hope there will be somebody here, in MO, who will help.
Let me first remind two definitions (see details here, or here). For a given locally compact group $G$ the Fourier-Stieltjes algebra $B(G)$ is defined as the algebra of matrix coefficients of unitary representations $\pi:G\to B(H)$:
$$
f(t)=\langle\pi(t)x,y\rangle,\qquad x,y\in H.
$$
Similarly, the Fourier algebra $A(G)$ is defined as the algebra of matrix coefficients of the left regular representation $\lambda:G\to B(L_2(G))$. It is known that $A(G)$ and $B(G)$ are subalgebras in the algebra $C(G)$ of continuous functions on $G$:
$$
A(G)\subseteq B(G)\subseteq C(G).
$$
My questions: 

1) P.Eymard in his paper ((3.6), $1^\circ$) says that when $G$ is compact, these algebras coincide:
  $$
A(G)=B(G).
$$ 
  Is this the only case when they coincide?

And

2) $B(G)$ can be interpreted as the dual space for the $C^*$-group algebra $C^*(G)$ for the group $G$ (see 3). So $A(G)$ (being a subset in $B(G)$) can be considered as a set of linear continuous functionals on $C^*(G)$. In which situations the common kernel of these functionals is zero:
  $$
\bigcap_{f\in A(G)} \text{Ker} f=0?
$$

 A: This is just an expanded version of my comments.
There are at least two ways of defining $A(G)$ that one commonly sees: one can define it to be the set of coefficient functions of the left regular representation $\lambda$; or one can define it to be the closure of $B(G)\cap C_c(G)$ within the Banach algebra $B(G)$. If I recall correctly, Eymard starts with the second description and proves it is equivalent to the first one. Anyway, it follows easily from both definitions that $A(G)\subseteq C_0(G)$.
Now $B(G)$ always contains constant functions, since these are the coefficient functions of the trivial one-dimensional representation $G\to \{1\}$; hence if $A(G)=B(G)$ we must have $1\in C_0(G)$, so $G$ is compact. This answers your Question 1.
$\newcommand{\Cst}{{\rm C}^*}$
The pairing betwen $B(G)$ and $\Cst(G)$ is defined as follows: given a continuous unitary representation $\pi:G \to U(H_\pi)$ let $\widetilde{\pi}:\Cst(G) \to B(H_\pi)$ be the canonical extension of $\pi$; then given vectors $\xi,\eta\in H_\pi$, the corresponding functional on $\Cst(G)$ is
$$ a\mapsto \langle \widetilde{\pi}(a)\xi, \eta \rangle $$
In particular, since $A(G)$ may be identified with the set of coefficient functions of the left regular representation $\lambda$,
$$ \bigcap_{f\in A(G)} \ker(f) = \{ a \in \Cst(G) \colon \widetilde{\lambda}(a) = 0 \} = \ker\left( \widetilde{\lambda} : \Cst(G) \to \Cst_r(G) \right) $$
Therefore your Question (2) is equivalent to asking:

Question 2': for which groups $G$ is the canonical homomorphism $\Cst(G)\to \Cst_r(G)$ injective?

A result of Hulanicki says: this occurs if and only if $G$ is amenable.

Hulanicki, A. Groups whose regular representation weakly contains all unitary representations. Studia Math. 24 1964 37--59. MR0191998 (33 #225) Link

Update: a proof of this result can also be found as Theorem 4.21 in Paterson's book Amenability.
