Extending a positive map to the unitisation Let $A$ be a (non-unital) $C^*$-algebra with unitisation $\widetilde{A}$ and $B$ be a unital $C^*$-algebra. Assume that $f: A \to B$ is a positive map. Is it true that $f$ extends to a positive map
$\widetilde{f}: \widetilde{A}\to B?$
When $B= \mathbb{C}$, the result is well-known. As an extension, I would guess that
$$\widetilde{f}: \widetilde{A}\to B: a + \lambda 1 \mapsto f(a) + \|f\|\lambda 1_B$$
works, but I'm having trouble showing that this is a positive map.
 A: The proposition is true and your guess on the extending map is correct.

Lemma: Let $A$ be a unital $C^*$-algebra, $h\in A_{sa}$ s.t. $h\leq1_A$. Then $h_+\leq 1_A$.

Proof of lemma: Let $t\in\sigma(h)$. Since $1-h\ge0$ and $\sigma(1-h)=\{1-s:s\in\sigma(h)\}$ we have that $t\leq1$. Thus $t_+=t\leq1$ if $t\ge0$ and $t_+=0$ if $t\le0$. Since $h_+$ is by definition $f(h)$, where $f:\sigma(h)\to\mathbb{C}$ is the function $f(t)=t_+$, we have $\sigma(f(h))=f(\sigma(h))\subset[0,1]$, i.e. $h_+\leq1$.

Proposition: Let $A$ be a non-unital $C^*$-algebra, $B$ a unital $C^*$-algebra and $\phi:A\to B$ a positive map. Then $\tilde\phi:\tilde{A}\to B$ given by $\tilde\phi(a+\lambda1):=\phi(a)+\lambda\|\phi\|1_B$ is positive.

Proof: Let $a+\lambda1\in\tilde A$ be a positive element. We shall show that $\phi(a)+\lambda\|\phi\|1_B\ge0$. Note that $\lambda\ge0$, since the map $\tilde A\to\mathbb{C}$, $a+\lambda1\mapsto\lambda$ is a $*$-homomorphism, thus maps positive things to positive things. We distinguish two cases: first, if $\lambda=0$ there is nothing to prove, since $\phi$ is positive.
Assume that $\lambda>0$. Since $a+\lambda1\ge0$, we have $1\ge-\frac{1}{\lambda}a$  (note that the element $-\frac{1}{\lambda}a$ is obviously self adjoint in $A$). We write $-\frac{1}{\lambda}a=\big(-\frac{1}{\lambda}a\big)_+-\big(-\frac{1}{\lambda}a\big)_-$ and thus $0\leq\big(-\frac{1}{\lambda}a\big)_+\leq1$ (by our lemma), so $\big{\|}(-\frac{1}{\lambda}a)_+\big{\|}\leq1$. Now
$$\phi(-\frac{1}{\lambda}a)=\phi\big((-\frac{1}{\lambda}a)_+\big)-\phi\big((-\frac{1}{\lambda}a)_-\big)\leq\phi\big((-\frac{1}{\lambda}a)_+\big)\leq\bigg{\|}\phi\big((-\frac{1}{\lambda}a)_+\big)\bigg{\|}\cdot1_B\leq$$ $$\leq\|\phi\|\cdot\|(-\frac{1}{\lambda}a)_+\|\cdot1_B\leq\|\phi\|1_B$$
(where we used the inequality $h\leq \|h\|1$ which is true for all selfadjoint elements)
so $-\phi(a)\leq\lambda\|\phi\|1_B$, which is equivalent to $\tilde\phi(a+\lambda1)\ge0$, which is what we wanted to prove.
