How can it be proven that the set of all states is a convex set for the W*- algebra? 
*

*How can it be proven that the set of all states is a convex set for the W*- algebra?

*Is the set of all states non-empty? 
 A: The key result is this: given a linear functional $f$, the following statements are equivalent:


*

*$f$ is positive;

*$\|f\|=f(1)$.


Let $\phi,\psi$ be states, $\alpha\in[0,1]$. Then $\alpha\phi+(1-\alpha)\psi$ is clearly positive (because convex combinations of non-negative numbers are non-negative) and linear. To show that it is a state, we use the result above: $$\alpha\phi(1)+(1-\alpha)\psi(1)=\alpha+1-\alpha=1,$$
and so the convex combination is also a state. 
If the von Neumann algebra is represented on a Hilbert space, then one can easily obtain lots of states like this: take a vector $\xi$ with $\|\xi\|=1$, and define the map 
$$
x\mapsto\langle x\xi,\xi\rangle.
$$
This map is a state for all unit $\xi$ in the Hilbert space. 
In an abstract C$^*$-algebra we can use the same result above  to prove   not only that states exist, but that they separate points (in the sense that if $f(x)=0$ for all states $f$, then $x=0$). Indeed, given any  $x$ and $\lambda\in\sigma(x)$, we define a linear functional on the two-dimensional space $\mathbb{C}x+\mathbb{C}1$ by $f_x(a x+b1)=\lambda a+b$. As $\lambda a + b\in\sigma(ax+b1)$, we get  $|f_0(ax+b1)|\leq\|ax+b1\|$. So $f_0$ is a linear functional of norm one, with $f_0(1)=1$. By the Hanh-Banach theorem, there exists a functional $f$ on the whole algebra that extends $f_0$ and with $\|f\|=\|f_0\|=1$. As $f(1)=f_0(1)=1$, we get that $f$ is positive by the first result, so $f$ is a state with $f(x)=\lambda$. 
