4
$\begingroup$

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 $s\in G$?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.