$\DeclareMathOperator{\tr}{tr}$Let $H$ be a separable complex Hilbert space and $L(H)$ be all bounded linear operators on $H$.
- Define a state to be a bounded linear functional $\phi: L(H)\rightarrow \mathbb{C}$ such that $\phi(I)=1$ and $\phi (A) \ge 0$ for all $A \ge 0$.
- Define a density operator $\rho$ to be a bounded positive ($\ge 0$) linear operator on $H$ such that $\tr(\rho)=1$.
Then how would you prove that the mapping $\rho \mapsto \tr(\cdot \rho)$ is a bijective map from density operators to states on $H$? I can prove everything else except for surjectivity.
If $H$ is infinite-dimensional, there are states on $L(H)$ that are not induced by density operators. Let $K(H)$ denote the set of compact operators on $H$ and consider the functional $\tilde \phi\colon K(H)+\mathbb{C}\mathrm{id}\to \mathbb{C},\,T+\lambda\mathrm{id}\mapsto \lambda$.
By Hahn-Banach, it has an extension $\phi$ to $L(H)$ with equal norm (i.e. norm $1$). Since additionally $\phi(\mathrm{id})=1$, this means that $\phi$ is a state. But if $\phi$ were of the form $\phi=\mathrm{tr}(\,\cdot\,\rho)$ for some density matrix $\rho$, we would have $0=\phi(T)=\mathrm{tr}(T\rho)$ for every compact $T$, hence $\rho=0$, a contradiction.
States of the form $\mathrm{tr}(\,\cdot\,\rho)$ for some density matrix $\rho$ are exactly the normal states, i.e. states that are continuous with respect to the $\sigma$-weak operator topology (there is a long list of equivalent definitions).
One standard reference for these questions is Blackadar's book on operator algebras, in this case Theorem III.2.1.4.