Fourier–Stieltjes algebra

Remarks

The set $B(G)$ is closed under sums, by using direct sums of representations, and under products, by using tensor products. With the norm $\norm{\cdot}_B$ it is a commutative Banach algebra with unit the constant function $1$, coming from the trivial representation. Pairing $u$ with $f \in L^1(G)$ by $\int_G f u$ identifies $B(G)$ isometrically with the dual of the full group C*-algebra $C^*(G)$. It is the linear span of the continuous positive-definite functions on $G$. The Fourier algebra $A(G)$ is a closed ideal of $B(G)$, and $A(G) = B(G)$ exactly when $G$ is compact.

Example: $G = \Z$. A unitary representation of $\Z$ is given by one unitary operator $U = \pi(1)$. By the spectral theorem, $\ip{U^n \xi}{\eta} = \int_{\mathbb{T}} z^n \, d\mu(z)$ for a finite complex measure $\mu$ on the circle, and every finite measure arises in this way. So $B(\Z)$ is the algebra of Fourier–Stieltjes coefficient sequences of finite measures on $\mathbb{T}$.

Non-example: the function $u(n) = 2^n$ on $\Z$ is not in $B(\Z)$. Every coefficient function is bounded, since $\abs{\ip{\pi(x)\xi}{\eta}} \le \norm{\xi}\norm{\eta}$, and $u$ is unbounded.