Fourier algebra

Remarks

By a theorem of Eymard, every $u \in A(G)$ has the form $u(x) = \ip{\lambda(x)\xi}{\eta}$ with a single pair $\xi, \eta \in L^2(G)$, and $\norm{u}_A = \inf \norm{\xi}_2 \norm{\eta}_2$ over such pairs. With this norm $A(G)$ is a commutative Banach algebra contained in the space $C_0(G)$ of continuous functions vanishing at infinity, and $\norm{u}_\infty \le \norm{u}_A$. It is the predual of the group von Neumann algebra of $G$. Eymard also showed that the Gelfand spectrum of $A(G)$ is $G$, with $x \in G$ giving the character $u \mapsto u(x)$.

Example: $G = \Z$. The Fourier series isomorphism $\ell^2(\Z) \to L^2(\mathbb{T})$ with $\hat\xi(z) = \sum_k \xi(k) z^k$ turns $\lambda(n)$ into multiplication by $z^n$. So $\ip{\lambda(n)\xi}{\eta} = \int_{\mathbb{T}} z^{n} \, \hat\xi(z) \overline{\hat\eta(z)} \, dz$ for the normalized Haar measure $dz$, which is the Fourier coefficient of index $-n$ of $\hat\xi \overline{\hat\eta} \in L^1(\mathbb{T})$. Every function in $L^1(\mathbb{T})$ is such a product, so $A(\Z)$ is the algebra of Fourier coefficient sequences of integrable functions on the circle. More generally, for abelian $G$, the Fourier transform maps $L^1(\widehat{G})$ isometrically onto $A(G)$.

Non-example: the constant function $1$ on $\R$ is not in $A(\R)$, since it does not vanish at infinity. For a compact group $G$ with normalized Haar measure, $1 = \ip{\lambda(\cdot)1}{1}$ does lie in $A(G)$.