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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?

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

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    $\begingroup$ For disjoint compact sets $A$ and $B$ in $G$ you can find an open $U$ containing $A$ and an identity nbd $V$ s.t $VU^{-1}$ is disjoint from $A$. You can find continuous functions $\phi,\psi$ supported on $U,V$ and equal 1 on $A,e$ correspondingly. The corresponding matrix coef will separate $A$ and $B$. I don't think this is a research level question. $\endgroup$
    – Uri Bader
    Commented Apr 23, 2016 at 17:11
  • $\begingroup$ @user89334 i.e., $\lambda_{\phi,\psi}$ separate $A$ and $B$? $\endgroup$
    – Mambo
    Commented Apr 23, 2016 at 17:27
  • $\begingroup$ Yes (possibly I was careless with the choice of $U$ and $V$ but I think I spelled it correctly, otherwise a little modification will do it, and this should be an easy exercise). $\endgroup$
    – Uri Bader
    Commented Apr 23, 2016 at 17:33
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    $\begingroup$ @user89334 No problem. I posted this at MSE. I didn't get any response. $\endgroup$
    – Mambo
    Commented Apr 24, 2016 at 18:35
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    $\begingroup$ For next time, indicate this in the question's body. You'd get a more simpathic atitude. I will upvote and also go foreward and post my comment as an answer then. $\endgroup$
    – Uri Bader
    Commented Apr 24, 2016 at 18:46

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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.

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  • $\begingroup$ Good luck in your search of $U,V$ when $B = A^{-1}$. $\endgroup$
    – Guntram
    Commented Apr 25, 2016 at 17:15
  • $\begingroup$ Thanks, @Guntram, for pointing the typo so kindly. Fixed. $\endgroup$
    – Uri Bader
    Commented Apr 25, 2016 at 17:43

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