Let $A$ be a commutative $C^\*$-algebra with a discrete group $G$ acting on it. The reduced crossed product is the completion of the algebraic crossed product $C\_c(G,A)$ in the reduced norm $\Vert \cdot\Vert\_r$. Is there a smaller $C^\*$-norm $\Vert \cdot\Vert\_s$ on the algebraic crossed product, meaning that 
$$\Vert f\Vert\_s\le \Vert f\Vert\_r$$
for every $f\in C\_c(G,A)$ and the two norms are in fact different?



Smallest norms on crossed product $C^*$-algebras
The answer is yes. Take $A=\mathbb{C}$ and $G=\mathbb{Z}$. Then $C^\*(\mathbb{Z})\cong C(\mathbb{T})$ canonically, where $\mathbb{T}$ is the unit circle. The reduced norm corresponds to the canonical sup-norm of functions. Under this identification, the $\*$-algebra $C\_c(\mathbb{Z},\mathbb{C})$ corrsponds to the polynomial functions in $C(\mathbb{T})$. Let $\mathbb{T}\_+$ be the set of those $z\in\mathbb{T}$ with nonnegative real part, i.e., the closed upper half-circle. One may consider
$$
\|\cdot\|\_0: C^\*(\mathbb{Z})\cong C(\mathbb{T}) \to [0,\infty),\quad \|f\|\_0=\|f|\_{\mathbb{T}\_+}\|\_\infty.
$$
On $C^\*(\mathbb{Z})$, this (semi-)norm is definitely smaller than the usual norm. But since any polynomial function is determined by its restriction on some arbitrary open set, it follows that this semi-norm defines a norm on $C\_c(\mathbb{Z},\mathbb{C})$ that is smaller than and not equal to the reduced C\*-cross norm. 


