In a C* algebra, bounded linear functional + norm condition => positive I am reading John B. Conway's book "A Course in Operator Theory." He proves that if $\mathcal{A}$ is a unital $C^*$ algebra, and $\phi: \mathcal{A} \to \mathbb{C}$ is a positive linear functional, then $\phi$ is bounded with $\lVert \phi \rVert = \phi(1).$ He then proves an interesting converse, which is that if $\mathcal{A}$ is a unital $C^*$ algebra and $\phi : \mathcal{A} \to \mathbb{C}$ is a bounded linear functional with $\lVert \phi \rVert = \phi(1),$ then it must be a positive functional. The converse proof is a straightforward application of the Riesz-Markov theorem for complex measures.
The "positive implies bounded" direction of the theorem I have just explained has an analogue for non-unital $C^*$ algebras. If $\mathcal{A}$ is a non-unital $C^*$ algebra with approximate identity $\{e_{\alpha}\},$ and $\phi : \mathcal{A} \to \mathbb{C}$ is a positive linear functional, then one can show that $\phi$ is bounded with $\lVert \phi \rVert = \lim_{\alpha} \phi(e_{\alpha}).$ What I would like to know is if there is a converse statement in the non-unital case: can one prove the following theorem?

Let $\mathcal{A}$ be a non-unital $C^*$ algebra, $\{e_{\alpha}\}$ an approximate identity, and $\phi : \mathcal{A} \to \mathbb{C}$ a bounded linear functional satisfying $\lVert \phi \rVert = \lim_{\alpha} \phi(e_{\alpha}).$ Then $\phi$ is a positive functional.

A natural way to prove this would be to try to extend $\phi$ to the unitization $\mathcal{A}_1$ by $\phi(a + \lambda) = \phi(a) + \lVert \phi \rVert_{\mathcal{A}} \lambda.$ If one could prove that this extension satisfies $\lVert \phi \rVert_{\mathcal{A}} = \lVert \phi \rVert_{\mathcal{A}_1}$ then this would prove the theorem, but I have been unable to prove or disprove this lemma.
 A: This is Lemma I.9.9 in Davidson's C$^*$-Algebras by Example.
The reduction to the unital case is as follows. Assume without loss of generality that $\|\phi\|=1$. Let $\tilde\phi$ be any Hahn-Banach extension of $\phi$ to $\mathcal A_1$. Let $c=\tilde\phi(1)$. We have
$$
|c|=|\tilde\phi(1)|\leq 1. 
$$
Since $\|2e_\alpha-1\|\leq 1$,
$$
|2-c|=\lim_\alpha|2\phi(e_\alpha)-\tilde\phi(1)|=\lim_\alpha|\tilde\phi(2e_\alpha-1)|\leq1.$$ Thus $|c|\leq1$ and $|2-c|\leq 1$; this implies that $c=1$. That is $\tilde\phi(1)=1$. Now the unital version of the theorem gives $\tilde\phi\geq0$, and hence $\phi\geq0$.
A: These are Pedersen, Proposition 3.1.4 and Lemma 3.1.5. The proof of your lemma goes as follows.
For each $x\in\mathcal A$ and $\lambda\in\Bbb C,$
$$\begin{align}\limsup\|\lambda e_\alpha+x\|^2&=\limsup\|(\lambda e_\alpha+x)^*(\lambda e_\alpha+x)\|\\&\le\limsup\||\lambda|^21+\bar\lambda e_\alpha x+\lambda x^*e_\alpha+x^*x\|\\&=\|\lambda1+x\|^2,\end{align}$$
the inequality using 1.3.5, which says that if $0\le x\le y$ then $\|x\|\le\|y\|.$
Therefore, denoting by $\tilde\phi$ your extension of $\phi,$
$$|\tilde\phi(\lambda1+x)|=\lim|\phi(\lambda e_\alpha+x)|\le\|\lambda1+x\|\|\phi\|,$$
so that $\|\tilde\phi\|=\|\phi\|.$
