Let $G$ be a locally compact group. Consider the left regular representation $\lambda$ over $L^2(G)$. Then according to Eymard, Fourier algebra of $G$, $A(G)$ is the set of all coefficients of $\lambda$, i.e. $A(G) = \{\lambda\_{x,y} : x,y \in L^2(G)\}$ where, $\lambda\_{x,y}(g) = \langle \lambda(g)x,y\rangle$. From the definition of $\lambda$, we have $A(G) \subset C(G,\Bbb{C})$. Since $G$ is completely regular, we can separate disjoint compact sets with continuous functions. Does $A(G)$ separate disjoint compact sets too?




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EDIT by YC: this question has also been posed at MSE <https://math.stackexchange.com/questions/1745237/does-fourier-algebra-of-locally-compact-group-separate-compact-sets-of-the-group>



Does Fourier Algebra of locally compact group separate compact sets of the group?
For disjoint compact sets $A$ and $B$ in $G$ you can find an open set $U$ containing $A$ and an identity nbd $V$ such that both have compact closures and $UV^{−1}$ is disjoint from $B$. You can find $[0,1]$-valued continuous functions $\phi$ and $\psi$ supported on $U$ and $V$ correspondingly such that $\phi|\_A=1$ and $\psi(e)=1$. Then it is easy to check that the matrix coefficient $\lambda\_{\psi,\phi}$ is zero on $B$ but poisitive on $A$.
I will not spell out the computation, only note that you use the regularity of the Haar measure and continuity when you prove positivity.


