# A C*-algebra enjoying some different C*-norms

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).

## 1 Answer

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.

• 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 $\infty$. Feb 19, 2022 at 15:20
• @ Tim Campion, The reference is so helpful. Thanks a lot.
– ABB
Feb 19, 2022 at 15:47