The $C^*$-enveloping algebra of a $*$-Banach algebra

Question

I read somewhere that if $A$ is a $*$-Banach algebra with an approximate identity, then the norm $$\| a \|_{C^*} := \sup_{\pi} \| \pi(a) \|$$ where $\pi$ ranges over all continuous $*$-representations of $A$ on some Hilbert space, is well-defined. Consequently, the completion of $A$ with respect to this norm becomes a $C^*$-algebra, the enveloping algebra of $A$. Is this true (and if so, easy to prove), is the assumption that $A$ is complete needed, and is there a counter-example when $A$ does not admit an approximate identity?

Answer 1

This requirement is not needed if you demand that the involution in your *-algebra is norm-continuous. Indeed, then there is no problem as the completion of a normed *-algebra in this case is a Banach *-algebra. Actually, most of the authors require the involution to be an isometry, so there is no problem at all.

Answer 2

The proof that $\| \cdot \|_{C^*}$ defines a seminorm is routine; all of the properties lift from the representations. However, you will hit a problem proving that $\| \cdot \|_{C^*}$ separates the points of $A$. It's possible that $A$ has no nontrivial representations at all, even if $A$ is unital, commutative, and finite-dimensional. Let $A = \mathbb{C}^2$ with the $\infty$-norm and pointwise multiplication, and define $(a, b)^* = (\overline{b}, \overline{a})$. You can show that any $*$-representation of $A$ must vanish on the basis elements $(1, 0)$ and $(0, 1)$, and thus all of $A$.

You can work around this problem of degeneracy by requiring the existence of enough $*$-representations to separate the points of $A$, or by defining the $*$-radical as the ideal of $A$ consisting of elements that vanish in every $*$-representation. The the representation norm can be defined on the quotient of $A$ by the $*$-radical. In most common situations (e.g. involving locally compact groups) the existence of enough representations is guaranteed. In others, e.g. some semigroup algebras, the $*$-representations may not separate points.