Consistent local densities extend to an algebraic state

For $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$, define ${\omega }_0(A):={\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}A)$. This is independent of the chosen localization region: if $A$ is also viewed in ${\mathcal{A}}_{\mathrm{\Delta }}$, embed both representatives into ${\mathcal{A}}_{\mathrm{\Lambda }\cup \mathrm{\Delta }}$; the two partial-trace consistency equations make both pairings equal to the one computed using ${\rho }_{\mathrm{\Lambda }\cup \mathrm{\Delta }}$. Thus ${\omega }_0$ is a well-defined functional on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$.

Linearity follows by placing any two local arguments and their sum in one common finite region. The identity has value ${\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}I)=1$, and for local $A$, [F2] gives ${\omega }_0(A^{\ast }A)={\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}{A}^{\ast }A)\ge 0$. Therefore ${\omega }_0$ is normalized and positive on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$.

For local $A$, its restriction is the normalized finite-matrix state from [F2], whose functional norm is one. Hence $\mathrm{\mid }{\omega }_0(A)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }$.

The bound of step 3.1 makes ${\omega }_0$ continuous and uniquely extendible from the dense subalgebra ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ to a bounded linear functional $\omega$ on $\mathcal{A}$, with $\mathrm{\parallel }\omega \mathrm{\parallel }\le 1$. It remains normalized because $I\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and ${\omega }_0(I)=1$; since $\mathrm{\parallel }I\mathrm{\parallel }=1$, this also gives $\mathrm{\parallel }\omega \mathrm{\parallel }\ge 1$, hence $\mathrm{\parallel }\omega \mathrm{\parallel }=1$. For any $X\in \mathcal{A}$, choose local $X_n\to X$; then $X_n^{\ast }{X}_n\to X^{\ast }X$ by continuity of multiplication and adjoint, and $\omega (X^{\ast }X)={\lim }_n{\omega }_0(X_n^{\ast }{X}_n)\ge 0$. Thus $\omega$ is an infinite algebraic state.

Any state with the prescribed local restrictions agrees with ${\omega }_0$ on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and hence agrees with $\omega$ on its norm closure, since both are continuous and ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is dense. The extension is unique.