This is my first overflow question, so let me apologize in advance if this belongs on http://math.stackexchange.com.

Let $G$ be a discrete group.

Let $\lambda:G\to B(\ell^{2}(G))$ be the left regular representation of $G$ and $A(G)$ the Fourier algebra of $G$.

Given $f,g\in A(G)$, and writing $$f(s) = \langle \lambda(s)x,y\rangle, g(s) = \langle \lambda(s)w,z\rangle$$ for some choice $x,y,w,z\in \ell^{2}(G)$, is there a way to choose (in a way which directly comes from $x,y,w,z$) $u,v\in \ell^{2}(G)$ such that

$$f(s) - g(s) = \langle \lambda(s)u,v\rangle$$ for all $s\in G$?

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  • $\begingroup$ I'm a bit too busy to sit and think in detail about this, but one could try to brute-force the problem by examining the proof that VN(G) acting on $\ell^2(G)$ is already in so-called standard form -- this is the fact that allows you to say every normal functional on VN(G) is realised by a pair of vectors in $\ell^2(G)$ $\endgroup$ – Yemon Choi Jul 26 '16 at 11:58
  • $\begingroup$ I appreciate the suggestion. Thank you! I will take a look at this. $\endgroup$ – roo Jul 26 '16 at 18:14

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