Suppose $\mathcal{R}$ is a von Neumann algebra and $\mathbb{P}_\mathcal{R}$ is the set of its projection operators. Because the linear span of $\mathbb{P}_\mathcal{R}$ is norm-dense in $\mathcal{R}$ (I'm assured this is the case) it seems reasonable that if a function "behaves like" the restriction to $\mathbb{P}_\mathcal{R}$ of a state on $\mathcal{R}$, then it must be one. Is this true?
To make this precise, call a real function $\phi : \mathbb{P}_\mathcal{R} \rightarrow \mathbb{R}$ "state-like" if:
(1) $\phi$ only takes values in $[0,1]$, and has value $1$ on the identity projection;
(2) $\phi$ is linear in the sense that if $P_1, \ldots, P_n$ are pairwise orthogonal, then $\phi(P_1 + \ldots + P_n) = \phi(P_1) + \ldots + \phi(P_n)$;
(3) $\phi$ is continuous with respect to the operator norm topology.
If $\phi$ is a state-like function on $\mathbb{P}_\mathcal{R}$, must it extend to a state on $\mathcal{R}$?
If the answer is "yes," the first step in proving it is presumably to show that $\phi$ extends to a unique positive linear functional $\phi^+$ on the linear span of $\mathbb{P}_\mathcal{R}$. You would define
$\phi^+(c_0 P_0 + \ldots + c_n P_n) \equiv c_0 \phi(P_0) + \ldots c_n \phi(P_n),$
and then show that this extension of $\phi$ is well-defined. That is, you'd show that if $T \in \mathcal{R}$ can be decomposed in two distinct ways as linear combinations of projections, then the $\phi^+$ values calculated by the above definition are the same for both. But I haven't been able to show this -- it's frustrating because I haven't been able to pin down a "normal form" or "canonical form" for a linear combination of projections. Any thoughts greatly appreciated.