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

  2. Is the set of all states non-empty?

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1 Answer 1

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 $f(x)=\lambda$.

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