# Extending $C^*$-norms from $*$-subalgebras

Let $$A$$ be a unital $$*$$-algebra, and $$B$$ a unital $$*$$-subalgebra of $$A$$. In addition, assume that there exists a $$B$$-$$B$$-sub-bimodule $$C \subset A$$, such that $$A \simeq B \oplus C,$$ where $$\simeq$$ means an isomorphism of $$B$$-$$B$$-bimodules. If $$B$$ is endowed with a pre-$$C^*$$-norm $$\|*\|$$, then is it always possible to $$\|*\|$$ to a $$C^*$$-norm for $$A$$?

P.S. If it helps, one can assume that $$A$$ admits a $$C^*$$-norm which does not restrict to $$\|*\|$$ on $$B$$.

Edit: I am not assuming that $$B$$ is complete with respect to $$\|*\\$$.

Ignoring the P.S. the answer is an easy no: $$A$$ could be any unital $$*$$-algebra with a codimension $$1$$ $$*$$-ideal $$C$$. Then let $$B = \mathbb{C}\cdot e$$ where $$e$$ is the unit of $$A$$, so a $$B$$-$$B$$ bimodule is just a vector space. Of course $$B$$ is endowed with a C*-norm but $$A$$ is arbitrary so it need not have a C*-norm. The P.S. doesn't add anything because the C*-norm on $$B$$, if it exists, is unique, so if $$A$$ has a C*-norm then its restriction to $$B$$ must be $$\|\cdot\|$$.
Edit: if the C*-norm on $$B$$ is only a pre-C*-norm, as per the edit, then it need not be unique, but the answer is still no. Let $$A_0$$ be the polynomials in $$x$$ with norm inherited from $$C[0,1]$$ and let $$A = A_0 \oplus \mathbb{C}$$ ($$l^\infty$$ direct sum). Let $$f = (f_0, 2)$$ where $$f_0$$ is the polynomial $$x \mapsto x$$, let $$B$$ be the set of unital polynomials in $$f$$, and let $$C$$ be the $$\mathbb{C}$$ summand. Then $$B$$ is isomorphic to $$A_0$$ as $$*$$-algebras, but the norm on $$A_0$$ transferred to $$B$$ has no extension to $$A$$. That's because $$f -2$$ is not invertible in $$A$$ so the norm of $$f$$ in $$A$$ must be at least $$2$$, whereas its norm in $$B$$ is $$1$$.
• Is the $C^*$-norm unique even in the case of a pre-$C^*$-algebra. To be clear I am not assuming that $A$ or $B$ is complete. Aug 14, 2019 at 21:23