Does there exist any C\*-algebra $(A,\|\cdot\|)$ enjoying the following property?


$\bullet$ There exists a norm $|\cdot|$ on $A$ with $\|\cdot\|\leq|\cdot|$ such that $(A,|\cdot|)$ is a pre C\*-algebra (necessarily non-complete).



A C*-algebra enjoying some different C*-norms
No that's not possible (except the trivial case). Any $\*$-homomorphism between $C^\*$-algebras is automatically contractive, and if it is injective then it is isometric. You can apply this to the identity map of $A$ seens as a map from $(A,\Vert \cdot \Vert)$ to the completion of $(A,\vert \cdot \vert)$ and conclude that two norm are equal.


More concretely it follows from the fact that (for self adjoint elements) the norm in a $C^\*$-algebra coincide with the spectral radius, which force relations between the two norm. Here it forces $\vert \cdot \vert \leqslant \Vert \cdot \Vert$ hence the equality of the two norms.


