Assume $A$ is a nonunital C*-algebra, $B$ is a unital C*-algebra and $\phi:A \rightarrow B$ is a contractive completely map. Then $\phi$ can extend to a unital completely positive map $\bar{\phi}: \bar{A} \rightarrow B$ by the formula $$\bar{\phi} (a+\lambda1_{\bar{A}})=\phi(a)+\lambda1_{B}$$ where $\bar{A}$ denotes the unitization of $A$.

There are two quotation below:

  1. The norm of $\bar{\phi}$ may be larger than that of $\phi$ if and only if $||\phi||<1$. I do not know how to verify, and I think oneside is easy.

  2. If we require that $1_{\bar{A}}\longmapsto||\phi||1_{B}$, and this produces a map with the same norm as that of $\phi$? I do not know how to get $||\phi||=||\bar{\phi}||$ here.

  1. Because $\bar\phi$ will be unital and completely contractive, so $\|\bar\phi\|=1$.
  2. Because a completely positive map achieves its norm at the identity. So $$ \|\bar\phi\|=\|\bar\phi(1_{\bar A})\|=\|\phi\|.$$
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