Numerical range of selfadjoint elements in non-unital C*-algebras If $a$ is an element of a C*-algebra $A$ then $V(a)=\{\varphi(a): \varphi\text{ is a state of }A\}$ is the numerical range of $a$. If $a$ is selfadjoint and $A$ is unital then it is known that $V(a)=[\min\sigma(a),\max\sigma(a)]$. 
I am interested in a similar formula when $A$ is non-unital. More, precisely I would like to know the relationship between the numerical range and the spectrum of a selfadjoint (in particular, positive) element of a non-unital C*-algebra. 
I see that the same formula cannot be true since $0=\min\sigma(a)\not\in V(a)$ if a is a strictly positive element of $A$.
 A: It is well-known that, since $A$ is non-unital,
$$\tag1
\sigma(a)=\{\tau(a):\ \tau\ \text{ is a character of }C^*(a)\}\cup\{0\}.
$$
Note that characters are states. This shows that $\sigma(a)\setminus\{0\}\subset V(a)$. Take any state $\varphi$ on $A$ and extend it (uniquely!) to $\tilde A$. In $\tilde A$, we have $$(\min\sigma(a)\,I\leq a\leq (\max\sigma(a))\,I.$$ Applying the extension of $\varphi$ to $(1)$, we get 
$$
\min\sigma(a)\leq \varphi(a)\leq\max\sigma(a).
$$
Thus
$$\tag2
\sigma(a)\setminus\{0\}\subset V(a)\subset [\min\sigma(a),\max\sigma(a)].
$$
As $V(a$ is convex, this inclusion is fairly tight: if $0\not\in\{\min\sigma(a),\max\sigma(a)\}$, the convexity implies $V(a)=[\min\sigma(a),\max\sigma(a)]$. 
So the only pathological case is that where either $\min\sigma(a)=0$ or $\max\sigma(a)=0$. Both cases are similar, since they are switched by considering $-a$. Say $\min\sigma(a)=0$. If there exists a state $\varphi$ with $\varphi(a)=0$, then $V(a)=[0,\max\sigma(a)]$. Otherwise, $0\not\in V(a)$; in that case $0$ cannot be an isolated point in $\sigma(a)$ (if $0$ is isolated, there exists a nonzero projection $p$ with $pa=0$; then any state $\psi$ on $pAp$ induces a state $\tilde\psi$ on $A$ by $\tilde\psi(x)=\psi(pxp)$, and $\tilde\psi(a)=0$). So we have that $[t,\max\sigma(a)]\subset V(a)$ for all $t>0$. Thus 
$$
V(a)=(0,\max\sigma(a)].
$$
