Is a linear functional which is positive on a linearly generating complemented lattice of projections necessarily positive? Let $\mathcal{H}$ be a Hilbert space, and $A \subseteq {\rm B}(\mathcal{H})$ a unital C$^*$-algebra. 
Suppose there is a complemented lattice of projections $L \subset \mathcal{P}(A)$ whose linear span is dense in $A$. If $\varphi \in A^*$ has $\varphi(p) \ge 0$ for all $p \in L$, does it follow that $\varphi \ge 0$?
Here complemented lattice means that for any $p, q \in L$, the projections $1-p$ and $p\wedge q$ also belong to $L$ (and hence so too does $p \vee q = 1-(1-p) \wedge (1-q)$). 
Note that this does hold if every element of $A$ can be approximated in norm by a linear combination of mutually orthogonal projections in $L$ (given any $x^*x \in A_+$, such an approximation for $x$ will lead to an approximation of $x^*x$ by a linear combination with positive coefficients), but is there any reason to believe it in general?
 A: Let $A=M_2(\mathbb C)$ and 
$$
L=\left\{
\begin{bmatrix}1&0\\0&0 \end{bmatrix},\ 
\begin{bmatrix}0&0\\0&1 \end{bmatrix},\ 
\begin{bmatrix}1/2&1/2\\1/2&1/2 \end{bmatrix},\ 
\begin{bmatrix}1/2&-1/2\\-1/2&1/2 \end{bmatrix},\ 
\begin{bmatrix}1/2&i/2\\-i/2&1/2 \end{bmatrix} ,\ 
\begin{bmatrix}1/2&-i/2\\i/2&1/2 \end{bmatrix} ,\ 
\begin{bmatrix}1&0\\0&1 \end{bmatrix},\ 
\begin{bmatrix}0&0\\0&0 \end{bmatrix}
\right\}.
$$
The first six matrices are distinct rank-one projections, so their mutual intersections are $0$. The intersections with $I$ and $0$ are trivially in $L$, and all the complements are in $L$. So $L$ is a complemented lattice of projections. 
We have 
$$\tag1
M_2(\mathbb C)=\operatorname{span}\left\{\begin{bmatrix}1&0\\0&0 \end{bmatrix},\ 
\begin{bmatrix}0&0\\0&1 \end{bmatrix},\ 
\begin{bmatrix}1/2&1/2\\1/2&1/2 \end{bmatrix},\ 
\begin{bmatrix}1/2&-i/2\\i/2&1/2 \end{bmatrix}   \right\},
$$
so $L$ spans $A$. 
Call these four matrices in $(1)$, $P,Q,R,S$. Define a linear functional $\varphi$ by 
$$
\varphi(P)=\tfrac45,\ \ \varphi(Q)=\tfrac15,\ \ \varphi(R)=\tfrac{29}{30},\ \ \varphi(S)=\tfrac1{30}.
$$
and extend by linearity. Then $\varphi\geq0$ on $L$. But if you consider the positive matrix
$$
T=\begin{bmatrix} 1/6&-1/2\\ -1/2&3/2\end{bmatrix}=\tfrac23\,P+2Q-R,
$$
you have 
$$
\varphi(T)=\tfrac23\,\tfrac45+2\,\tfrac15-\tfrac{29}{30}=-\tfrac1{30}<0.
$$
