Hahn-Banach extension of positive functional is positive Consider the following lemma from Takesaki's book "Theory of operator algebra I":

Why is the sentence "Any Hahn-Banach extension of a positive linear functional on a $C^*$-subalgebra of $A$ is positive" true?
Attempt: Let $B\subseteq A$ a $C^*$-algebra and $\omega: B \to \mathbb{C}$ positive. If I can show that $\|\omega\|= \omega(b)$ for some positive $b \in B$, then a Hahn-Banach extension $\widetilde{\omega}: A \to \mathbb{C}$ satisfies $\|\widetilde{\omega}\| = \widetilde{\omega}(b)$ as well and thus the lemma implies the result. But is it true that the operator norm of a positive functional is attained at a positive element of the unit ball?
Any help/comments are highly appreciated!
 A: In the unital case, this is always true as the norm of a positive functional is equal to the value of the unit.
If I am not mistaken, this is not true in the non-unital case: consider $A=C_0(0,1]=\{f\in C[0,1]:f(0)=0\}$ and let $\mu$ be the Lebesgue measure on $(0,1]$. The functional $\phi(f)=\int_0^1fd\mu$ is positive and $\|\phi\|_{C_0(0,1]}=\mu((0,1])=1$.
Now let $f\in C_0(0,1]$ be positive and $0\le f(t)\le1$. Since $f(0)=0$, there exists some $\eta\in(0,1)$ such that $0\le f(t)\le 1/2$ on $[0,\eta]$. But then $\phi(f)=\int_0^1fd\mu\le\frac{1}{2}\int_{(0,\eta]}d\mu+\int_{(\eta,1]}d\mu=\frac{1}{2}\mu((0,\eta])+\mu((\eta,1])<\mu((0,\eta])+\mu((\eta,1])=\mu((0,1])=1$, so the norm of $\phi$ is not attained.
I would recommend sticking to the approach of Murphy (prop. 3.3.3 - prop. 3.3.8).
A: The result holds in any C$^*$-algebra. When the subalgebra has the same unit as the algebra, things are easy because
$$
\tilde\omega(1)=\omega(1)=\|\omega\|=\|\tilde\omega\|
$$
which makes $\tilde\omega$ positive.
In the non-unital case, one can prove that if $\omega$ is positive and $\{e_j\}$ is an approximate unit, then $\|\omega\|=\lim_j\omega(e_j)$ (see for instance Lemma I.9.5 in Davidson's C$^*$-Algebras by Example). Now consider a Hahn-Banach extension $\tilde\omega$ to $B\supset A$. By extending once more if $B$ is not unital, we may assume that $B$ is unital. We now use Takesaki's argument. Fix $\varepsilon>0$; there exists $j_0$ such that $\omega(e_j)>\|\omega\|-\varepsilon$ for all $j\geq j_0$. Since $0\leq e_j\leq 1$, $\|e_j+\lambda(1-e_j)\|\leq1$ for all $\lambda$ such that $|\lambda|=1$ (unless I missed something this is not that obvious, so I included a short argument at the end). Choosing $\lambda$ such that $\lambda\tilde\omega(1-e_j)\geq0$,
$$
\|\tilde\omega\|\leq\tilde\omega(e_j)+\lambda\tilde\omega(1-e_j)+\varepsilon=\tilde\omega(e_j+\lambda(1-e_j))+\varepsilon\leq\|\tilde\omega\|+\varepsilon.
$$
Thus
$$
0\leq\lambda\tilde\omega(1-e_j)<\|\tilde\omega\|-\tilde\omega(e_j)+\varepsilon<2\varepsilon. 
$$
This shows that $\lim_j\tilde\omega(1-e_j)=0$. That is,
$$
\tilde\omega(1)=\lim_j\tilde\omega(e_j)=\lim_j\omega(e_j)=\|\omega\|=\|\tilde\omega\|. 
$$
So $\tilde\omega$ is a linear functional that achieves its norm at the identity, which  makes it positive.

Proof that $\|a+\lambda(1-a)\|\leq1$ if $0\leq a\leq1$ and $|\lambda|=1$.
Since $a+\lambda(1-a)\in C^*(1,a)$, it is normal. Then its norm is the maximum of the absolute values of the elements in its spectrum. Thus it is enough to check that $|\alpha+\lambda(1-\alpha)|\leq1$ when $\alpha\in[0,1]$, $|\lambda|=1$.
Noting that
\begin{align}
|\alpha+\lambda(1-\alpha)|^2&=|\alpha^2+(1-\alpha)^2+(2\operatorname{Re}\lambda)\,\alpha(1-\alpha)|
\\[0.3cm]
&\leq
\alpha^2+(1-\alpha)^2+2\alpha(1-\alpha)\\[0.3cm]&=(\alpha+1-\alpha)^2=1,
\end{align}
we conclude that $|\alpha+\lambda(1-\alpha)|\leq1$  and thus $\|a+\lambda(1-a)\|\leq1$.
