I've recently been learning some basic facts about $C^*$-algebras and their connections to representations of locally compact groups $G$, and I'm currently trying to understand the definition of the $C^*$-algebra of a group, which is obtained from the algebra $L^1(G)$ by completion.
Question: Why do we need to complete $L^1(G)$ to form the $C^*$-algebra of $G$?
It's easy to see that the only axiom of a $C^*$-algebra that could possibly fail is the crucial identity $\|T^*T\| = \|T\|^2$, so in one sense, my question is asking why this axiom fails. By playing around with simple examples it is easy to see that the axiom fails for $G=\mathbb Z$ (Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra?).
More generally, Dixmier has an argument that appears to work for any abelian $G$, though I don't understand it. He considers the left regular action $\lambda$ of $G$ on $L^2(G)$. It is easy to see that $\|\lambda(f)\|=\|\hat f\|_\infty$, where $\hat f$ is the Fourier transform of $f$. Since $\|f\|_\infty\neq\|f\|_1$ in general, this means that $\lambda$ is not isometric. All this I understand. But then Dixmier concludes that $L^1(G)$ is not a $C^*$-algebra, and this is where he loses me.
Can anyone clarify Dixmier's argument? (Is it true that morphisms of $C^*$-algebras are isometries, even though this is not required as part of the definition of a morphism?) Or more generally, does anyone have a good "moral" explanation as to why $L^1(G)$ is not a $C^*$-algebra and we have to pass to its $C^*$-enveloping algebra?
