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).
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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.
answered Feb 19, 2022 at 14:59
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$\begingroup$ For another description of the norm derived only from the $\ast$-algebra structure, see here. In fact, the "pre-norm" coming from the $\ast$-algebra structure in this sense is quite robust -- it fails to be a submultiplicative seminorm on a general $\ast$-algebra only in that it can take the value $\infty$. $\endgroup$ Commented Feb 19, 2022 at 15:20
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$\begingroup$ @ Tim Campion, The reference is so helpful. Thanks a lot. $\endgroup$– ABBCommented Feb 19, 2022 at 15:47