The state space is convex, closed under countable mixtures, and complete in the trace norm

[(a) finite mixtures are states] Let $\rho ,\sigma \in D(H)$ and $\lambda \in [0,1]$ real. The combination $\lambda \rho +(1-\lambda )\sigma$ is trace class by [F1] and self-adjoint because $(\lambda \rho +(1-\lambda )\sigma {)}^{\ast }=\lambda {\rho }^{\ast }+(1-\lambda ){\sigma }^{\ast }=\lambda \rho +(1-\lambda )\sigma$ by [F4] with real scalars; it is positive because $⟨(\lambda \rho +(1-\lambda )\sigma )x,x⟩=\lambda ⟨\rho x,x⟩+(1-\lambda )⟨\sigma x,x⟩\ge 0$ by [F2] and [F3]; and its trace is $\lambda \mathrm{tr}\rho +(1-\lambda )\mathrm{tr}\sigma =\lambda +(1-\lambda )=1$ by the linearity of the trace in [F3]. Hence it is a density operator, so $D(H)$ is closed under finite real convex combinations in the reading of [F8]; the finite-combination statement follows by induction on the number of components.

[(b) the mixture series converges in the trace norm] For every $n$ the operator $w_n{\rho }_n$ is self-adjoint by [F4] and positive, and it is trace class with $\mathrm{\parallel }{w}_n{\rho }_n{\mathrm{\parallel }}_1=w_n\mathrm{\parallel }{\rho }_n{\mathrm{\parallel }}_1=w_n\mathrm{tr}{\rho }_n=w_n$ by [F1] and [F3]; hence ${\sum }_n\mathrm{\parallel }{w}_n{\rho }_n{\mathrm{\parallel }}_1={\sum }_n{w}_n=1<\mathrm{\infty }$, so the series is absolutely convergent in the Banach space ${\mathcal{S}}_1(H)$ by [F6] and has a trace-class sum $T$; moreover $\mathrm{\parallel }T-{\sum }_{n<K}{w}_n{\rho }_n{\mathrm{\parallel }}_1\to 0$ because the tail is a convergent series of norms of the omitted terms, by [F6]. The self-adjointness argument below places this sum in $X$.

[(a) $X$ is a real Banach space] If $T_m\in X$ and $\mathrm{\parallel }{T}_m-T{\mathrm{\parallel }}_1\to 0$, then by [F1] and [F4], $\mathrm{\parallel }{T}^{\ast }-T\mathrm{\parallel }\le \mathrm{\parallel }{T}^{\ast }-T_m^{\ast }\mathrm{\parallel }+\mathrm{\parallel }{T}_m-T\mathrm{\parallel }=\mathrm{\parallel }(T_m-T{)}^{\ast }\mathrm{\parallel }+\mathrm{\parallel }{T}_m-T\mathrm{\parallel }=2\mathrm{\parallel }{T}_m-T\mathrm{\parallel }\le 2\mathrm{\parallel }{T}_m-T{\mathrm{\parallel }}_1\to 0$. Thus $T^{\ast }=T$ and $X$ is closed in ${\mathcal{S}}_1(H)$. It is a real-linear subspace because sums and real scalar multiples of self-adjoint operators are self-adjoint by [F4], and it is complete in the restricted trace-norm metric by [F7] applied to the complete space ${\mathcal{S}}_1(H)$ of [F1].

[(c) block members are states] Fix $k$ with $u_k>0$. The finite combination ${\sigma }_k=u_k^{-1}{\sum }_{N_k\le n<N_{k+1}}{w}_n{\rho }_n$ is a convex combination of the operators ${\rho }_n$, $N_k\le n<N_{k+1}$, with nonnegative real coefficients $w_n\mathrm{/}{u}_k$ summing to $1$ (the finite sum of the weights is $u_k$), so ${\sigma }_k\in D(H)$ by step 1.1 and induction.

[(b) the mixture sum is a state] Let $S_K:={\sum }_{n<K}{w}_n{\rho }_n$, the self-adjoint partial sums of step 1.2. Trace-norm convergence and [F1] give $\mathrm{\parallel }T-S_K\mathrm{\parallel }\to 0$, and operator-norm continuity of the adjoint [F4] gives $\mathrm{\parallel }{T}^{\ast }-T\mathrm{\parallel }\le \mathrm{\parallel }{T}^{\ast }-S_K^{\ast }\mathrm{\parallel }+\mathrm{\parallel }{S}_K-T\mathrm{\parallel }=2\mathrm{\parallel }T-S_K\mathrm{\parallel }\le 2\mathrm{\parallel }T-S_K{\mathrm{\parallel }}_1\to 0$. Hence $T$ is self-adjoint. It is positive because for every $x$ the numbers $⟨S_{K}x,x⟩={\sum }_{n<K}{w}_n⟨{\rho }_{n}x,x⟩$ are nonnegative by [F2], while $\mathrm{\mid }⟨(T-S_K)x,x⟩\mathrm{\mid }\le \mathrm{\parallel }T-S_K{\mathrm{\parallel }}_1\mathrm{\parallel }x{\mathrm{\parallel }}^2\to 0$ by [F5], so $⟨Tx,x⟩\ge 0$; and $\mathrm{tr}T={\lim }_K{\sum }_{n<K}{w}_n=1$ because $\mathrm{\mid }\mathrm{tr}T-\mathrm{tr}{S}_K\mathrm{\mid }\le \mathrm{\parallel }T-S_K{\mathrm{\parallel }}_1\to 0$ by [F3] and the trace is linear. Thus $T\in D(H)\subseteq X$, and since $S_K\in X$ this trace-norm convergence is also convergence in $X$.

[(c) grouping of consecutive blocks] Let $T_m:={\sum }_{n<m}{w}_n{\rho }_n$ be the partial sums of the $n$-series; by regrouping finite sums in the real vector space $X$ one has, for every $K$, ${\sum }_{k<K}{u}_k{\sigma }_k={\sum }_{k<K}{\sum }_{N_k\le n<N_{k+1}}{w}_n{\rho }_n={\sum }_{n<N_K}{w}_n{\rho }_n=T_{N_K}\mathrm{.}$ Since $T_m\to T$ by step 1.2 and $N_K\to \mathrm{\infty }$, the sequence $(T_{N_K}{)}_K$ is a subsequence of a convergent sequence in the metric space $X$ and therefore converges to $T$ as well by [F6]; the left series converges absolutely because ${\sum }_k\mathrm{\parallel }{u}_k{\sigma }_k{\mathrm{\parallel }}_1={\sum }_k{u}_k\mathrm{tr}{\sigma }_k={\sum }_k{u}_k={\sum }_n{w}_n=1$ by [F3] and step 2.1. Hence both series have the same sum $T$.

[(d) closedness and completeness] Let ${\rho }_m\in D(H)$ converge to $\rho \in X$ in the trace norm. Then $\rho$ is self-adjoint by definition of $X$; it is positive because $\mathrm{\mid }⟨(\rho -{\rho }_m)x,x⟩\mathrm{\mid }\le \mathrm{\parallel }\rho -{\rho }_m{\mathrm{\parallel }}_1\mathrm{\parallel }x{\mathrm{\parallel }}^2\to 0$ by [F5] while $⟨{\rho }_{m}x,x⟩\ge 0$; and $\mathrm{tr}\rho ={\lim }_m\mathrm{tr}{\rho }_m=1$ by [F3] and trace linearity. Hence $\rho \in D(H)$, so $D(H)$ is closed in $X$; since $X$ is complete by step 1.3, $D(H)$ is complete for the restricted metric by [F7]. This proves (a)-(d).