I have a question: if $\phi$ is a state of $\tilde{A}$ (unitization of a $C*$-algebra $A$), is it true that the restriction of $\phi$ to $A$ is also a state?
I'm asking this because I was trying to prove that (notation: $S(B)$ is the set of all states of a $C*$-algebra $B$):
\begin{equation} S(Ã)=\{\tilde{\phi}; \phi \in S(A)\} \cup \{\phi_\infty\}, \end{equation}
where $\phi_\infty(a+\lambda 1)=\lambda$.
What I did was to consider a state $\tau$ of $\tilde{A}$. If the restriction of $\tau$ to $A$ is zero, then is easy to show that $\tau$ equals to $\phi_\infty$.
Now, if the restriction of $\tau$ to $A$ is not zero, i was not able to prove that it is in fact a state of $A$.
Maybe there is another way of showing the equality, but I have no idea.
No, it's not true that the restriction of a state from the unitization is a state.
For instance let $A=C_0(\mathbb R)$ and $\phi(f+\lambda 1)=\frac{f(0)}2+\lambda$. It is trivial to see that $\phi$ is a state, since $$ \phi=\tfrac12\,\big[\phi_0+\phi_\infty\big], $$ where $\phi_0(g)=g(0)$. But $\phi|_A(f)=\frac{f(0)}2$, so $\phi|_A$ is not a state.