Hahn Banach for functionals on $C^*$-algebras that are $*$-morphisms. Does Hahn-Banach theorem also hold for $*$-morphisms between a $C^*$-algebra and $\mathbb{C}$?
I.e. Let $\mathfrak{U}$ be a $C^*$-algebra and $\mathfrak{P} \subset \mathfrak{U}$ be a $C^*$-subalgebra. Let $u: \mathfrak{P} \rightarrow \mathbb{C}$ be a continuous $*$-homomorphism. Then there is a continuous $*$-homomorphism $\chi: \mathfrak{U} \rightarrow \mathbb{C}$ with $\left. x \right|_{\mathfrak{P}} = u$ and $\|x\| = \|u\|$.
Especially I need an extension for characters of a $C^*$-algebra, to prove the last corollary of the first section of https://projecteuclid.org/download/pdf_1/euclid.tmj/1178245105 .
 A: This is impossible in general.  This is because there exist C$^*$-algebras which have no characters on them (i.e. any simple $C^*$-algebra), but have commutative C$^*$-subalgebras (which always have characters).
That being said, the result in the paper you linked is concerned with commutative $C^*$-algebras, where it is possible, for the inclusion $\mathfrak P\hookrightarrow \mathfrak U$ is an injective $*$-homomorphism of commutative C$^*$-algebras,  hence it yields a surjection $\Omega(\mathfrak U)\to\Omega(\mathfrak P)$ (where $\Omega(\mathfrak U)$ denotes the character space of $\mathfrak U$).
To see this, we may assume $\mathfrak P$ and $\mathfrak U$ are unital.  Then the inclusion $i$ induces a continuous map $\tilde i:\Omega(\mathfrak U)\to\Omega(\mathfrak P)$.  Since $\Omega(\mathfrak U)$ and $\Omega(\mathfrak P)$ are compact, $\tilde i(\Omega(\mathfrak U))$ is closed in $\Omega(\mathfrak P)$.  If $\tilde i(\Omega(\mathfrak U))\neq\Omega(\mathfrak P)$, then by Urysohn's lemma there is a nonzero continuous function $f:\Omega(A)\to\mathbb C$ that vanishes on $\tilde i(\Omega(\mathfrak U))$.  Then $f$ is the Gelfand transform of some nonzero $x\in\mathfrak P$.  But since $\tau(a)=0$ for all $\tau\in\Omega(\mathfrak U)$, $a=0$, a contradiction.  
