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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 $\|*\\$.

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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 $1$.

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  • $\begingroup$ 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. $\endgroup$ Aug 14 '19 at 21:23

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