[proofplan] We identify trace-class operators with normal functionals by first checking the formula on rank-one operators, where the trace pairing gives vector functionals. Trace-norm approximation by finite-rank operators then shows that every trace-class operator gives a normal functional, and trace duality gives the norm estimate. Conversely, a positive normal functional is reconstructed from its values on rank-one operators; normality over finite-rank projection nets proves that the resulting positive operator is trace-class and represents the functional. Finally, we decompose a general normal functional into positive normal pieces and characterize states by positivity and value $1$ at the identity. [/proofplan]

[step:Fix the rank-one convention and compute the trace pairing]Let $I_H:H\to H$ denote the identity operator on $H$. For $\xi,\eta\in H$, define the rank-one operator \begin{align*} \theta_{\xi,\eta}:H&\to H \end{align*} \begin{align*} \zeta&\mapsto (\zeta,\eta)_H\xi. \end{align*} Thus $\theta_{\xi,\xi}$ is positive and $\theta_{\xi,\xi}\le \|\xi\|_H^2 I_H$. Let $a\in\mathcal{L}(H)$, and let $F\subset H$ be a finite-dimensional subspace containing $\xi$. Choose an [orthonormal basis](/page/Orthonormal%20Basis) $(e_j)_{j\in J}$ of $F$. Since $\theta_{\xi,\eta}a$ has range contained in $\mathbb C\xi\subset F$, its trace is computed inside $F$: \begin{align*} \operatorname{Tr}(\theta_{\xi,\eta}a) = \sum_{j\in J}(\theta_{\xi,\eta}ae_j,e_j)_H. \end{align*} By the definition of $\theta_{\xi,\eta}$, \begin{align*} (\theta_{\xi,\eta}ae_j,e_j)_H = (ae_j,\eta)_H(\xi,e_j)_H. \end{align*} Since $\xi=\sum_{j\in J}(\xi,e_j)_H e_j$, linearity of the [inner product](/page/Inner%20Product) in the first variable gives \begin{align*} \operatorname{Tr}(\theta_{\xi,\eta}a) = (a\xi,\eta)_H. \end{align*} Thus $\omega_{\theta_{\xi,\eta}}$ is the vector functional $a\mapsto (a\xi,\eta)_H$.[/step]

[guided]The first issue is convention. We use the rank-one operator \begin{align*} \theta_{\xi,\eta}:H&\to H \end{align*} \begin{align*} \zeta&\mapsto (\zeta,\eta)_H\xi, \end{align*} with the Hilbert-space inner product linear in the first variable. This convention is chosen so that the trace pairing with $a\in\mathcal{L}(H)$ gives the vector functional $a\mapsto (a\xi,\eta)_H$. To verify this, choose a finite-dimensional subspace $F\subset H$ containing $\xi$, and choose an orthonormal basis $(e_j)_{j\in J}$ of $F$. The operator $\theta_{\xi,\eta}a$ has range contained in $\mathbb C\xi$, hence contained in $F$, so its trace is computed by summing over this finite basis: \begin{align*} \operatorname{Tr}(\theta_{\xi,\eta}a) = \sum_{j\in J}(\theta_{\xi,\eta}ae_j,e_j)_H. \end{align*} Using the definition of $\theta_{\xi,\eta}$ on the vector $ae_j\in H$, \begin{align*} \theta_{\xi,\eta}ae_j=(ae_j,\eta)_H\xi. \end{align*} Therefore \begin{align*} (\theta_{\xi,\eta}ae_j,e_j)_H = (ae_j,\eta)_H(\xi,e_j)_H. \end{align*} Because $\xi\in F$, its orthonormal expansion in $F$ is \begin{align*} \xi=\sum_{j\in J}(\xi,e_j)_H e_j. \end{align*} Applying $a$ and then pairing with $\eta$ gives \begin{align*} (a\xi,\eta)_H = \sum_{j\in J}(\xi,e_j)_H(ae_j,\eta)_H. \end{align*} This is exactly the trace sum above. Hence \begin{align*} \operatorname{Tr}(\theta_{\xi,\eta}a)=(a\xi,\eta)_H. \end{align*} So rank-one trace-class operators produce precisely the vector functionals.[/guided]

[step:Extend normality from finite-rank operators to trace-class operators] Every vector functional $a\mapsto (a\xi,\eta)_H$ is normal on $\mathcal{L}(H)$: if $(a_i)_{i\in I}$ is a bounded net converging strongly to $a$, then $a_i\xi\to a\xi$ in $H$, hence \begin{align*} (a_i\xi,\eta)_H\to (a\xi,\eta)_H. \end{align*} By the previous step, every rank-one operator gives a normal functional, and finite-rank operators give finite sums of such functionals. Let $T\in\mathcal{T}(H)$. By finite-rank density in trace norm, choose a net $(T_i)_{i\in I}$ of finite-rank operators in $\mathcal{T}(H)$ such that \begin{align*} \|T_i-T\|_{\mathcal{T}(H)}\to 0. \end{align*} For every $a\in\mathcal{L}(H)$, the trace-class estimate gives \begin{align*} |\operatorname{Tr}((T_i-T)a)| \le \|T_i-T\|_{\mathcal{T}(H)}\|a\|_{\mathcal{L}(H)}. \end{align*} Hence $\omega_{T_i}\to\omega_T$ in the operator norm on $\mathcal{L}(H)^*$. Since $\mathcal{L}(H)$ is a von Neumann algebra, [citetheorem:9273] applied to $M=\mathcal{L}(H)$ says that the normal functionals form the [Banach space](/page/Banach%20Space) $\mathcal{L}(H)_*$. With the inherited dual norm from $\mathcal{L}(H)^*$, this is a Banach subspace of $\mathcal{L}(H)^*$, hence it is norm-closed in $\mathcal{L}(H)^*$. Therefore $\omega_T$ is normal. The same estimate also proves \begin{align*} \|\omega_T\|_{\mathcal{L}(H)^*}\le \|T\|_{\mathcal{T}(H)}. \end{align*} [/step]

[step:Reconstruct a positive trace-class operator from a positive normal functional]Let $\varphi:\mathcal{L}(H)\to\mathbb C$ be a positive normal functional. Define a sesquilinear form \begin{align*} b_\varphi:H\times H&\to\mathbb C \end{align*} \begin{align*} (\xi,\eta)&\mapsto \varphi(\theta_{\xi,\eta}). \end{align*} Since $\theta_{\xi,\xi}\ge 0$ and $\theta_{\xi,\xi}\le \|\xi\|_H^2 I_H$, positivity of $\varphi$ gives \begin{align*} 0\le b_\varphi(\xi,\xi)\le \varphi(I_H)\|\xi\|_H^2. \end{align*} The [Cauchy-Schwarz inequality](/theorems/432) for positive sesquilinear forms gives \begin{align*} |b_\varphi(\xi,\eta)|^2 \le b_\varphi(\xi,\xi)b_\varphi(\eta,\eta) \le \varphi(I_H)^2\|\xi\|_H^2\|\eta\|_H^2. \end{align*} Thus $b_\varphi$ is bounded. By the Hilbert-space representation theorem for bounded sesquilinear forms, there is a unique positive operator $T_\varphi\in\mathcal{L}(H)$ such that \begin{align*} (T_\varphi\xi,\eta)_H=b_\varphi(\xi,\eta)=\varphi(\theta_{\xi,\eta}) \end{align*} for all $\xi,\eta\in H$. Let $(e_j)_{j\in J}$ be an orthonormal basis of $H$. For each finite subset $F\subset J$, define the finite-rank projection \begin{align*} P_F:H&\to H \end{align*} \begin{align*} \xi&\mapsto \sum_{j\in F}(\xi,e_j)_H e_j. \end{align*} The projections $(P_F)$, ordered by inclusion of finite subsets $F\subset J$, increase strongly to $I$. Since $\varphi$ is normal and positive, \begin{align*} \varphi(P_F)\uparrow \varphi(I_H). \end{align*} But \begin{align*} \varphi(P_F) = \sum_{j\in F}\varphi(\theta_{e_j,e_j}) = \sum_{j\in F}(T_\varphi e_j,e_j)_H. \end{align*} Therefore \begin{align*} \sup_{F\subset J,\ F\text{ finite}}\sum_{j\in F}(T_\varphi e_j,e_j)_H = \varphi(I)<\infty. \end{align*} For a positive operator $S\in\mathcal{L}(H)$, the trace-class criterion says that $S\in\mathcal{T}(H)$ if and only if, for one equivalently every orthonormal basis $(f_j)_{j\in J}$ of $H$, \begin{align*} \sup_{F\subset J,\ F\text{ finite}}\sum_{j\in F}(Sf_j,f_j)_H<\infty, \end{align*} and in that case this supremum equals $\operatorname{Tr}(S)$. Applying this criterion to $S=T_\varphi$ and the basis $(e_j)_{j\in J}$ proves $T_\varphi\in\mathcal{T}(H)$ and \begin{align*} \operatorname{Tr}(T_\varphi)=\varphi(I_H). \end{align*}[/step]

[guided]We first turn the functional into a bounded positive sesquilinear form. Define \begin{align*} b_\varphi:H\times H&\to\mathbb C \end{align*} \begin{align*} (\xi,\eta)&\mapsto \varphi(\theta_{\xi,\eta}). \end{align*} For each $\xi\in H$, the rank-one operator $\theta_{\xi,\xi}$ is positive and satisfies $\theta_{\xi,\xi}\le \|\xi\|_H^2 I_H$. Since $\varphi$ is positive, \begin{align*} 0\le b_\varphi(\xi,\xi)\le \varphi(I_H)\|\xi\|_H^2. \end{align*} The Cauchy-Schwarz inequality for positive sesquilinear forms gives \begin{align*} |b_\varphi(\xi,\eta)|^2 \le b_\varphi(\xi,\xi)b_\varphi(\eta,\eta) \le \varphi(I_H)^2\|\xi\|_H^2\|\eta\|_H^2. \end{align*} Thus $b_\varphi$ is bounded. By the Hilbert-space representation theorem for bounded sesquilinear forms, there is a unique positive operator $T_\varphi\in\mathcal{L}(H)$ such that \begin{align*} (T_\varphi\xi,\eta)_H=\varphi(\theta_{\xi,\eta}) \end{align*} for all $\xi,\eta\in H$. It remains to prove that this bounded positive operator is trace-class. Let $(e_j)_{j\in J}$ be an orthonormal basis of $H$. For each finite subset $F\subset J$, define \begin{align*} P_F:H&\to H \end{align*} \begin{align*} \xi&\mapsto \sum_{j\in F}(\xi,e_j)_H e_j. \end{align*} The projections $P_F$ increase strongly to $I_H$. Because $\varphi$ is positive and normal, [citetheorem:9263] applied in the von Neumann algebra $\mathcal{L}(H)$ gives \begin{align*} \varphi(P_F)\uparrow \varphi(I_H). \end{align*} For each finite set $F\subset J$, \begin{align*} \varphi(P_F)=\sum_{j\in F}\varphi(\theta_{e_j,e_j})=\sum_{j\in F}(T_\varphi e_j,e_j)_H. \end{align*} Hence \begin{align*} \sup_{F\subset J,\ F\text{ finite}}\sum_{j\in F}(T_\varphi e_j,e_j)_H=\varphi(I_H)<\infty. \end{align*} The positive trace-class criterion says that a positive bounded operator $S\in\mathcal{L}(H)$ is trace-class exactly when this supremum of finite diagonal sums is finite for an orthonormal basis, and then the supremum equals $\operatorname{Tr}(S)$. Applying this criterion to $S=T_\varphi$ proves \begin{align*} T_\varphi\in\mathcal{T}(H),\qquad \operatorname{Tr}(T_\varphi)=\varphi(I_H). \end{align*}[/guided]

[step:Verify that the reconstructed operator represents the functional] Let $T_\varphi\in\mathcal{T}(H)$ be the positive trace-class operator constructed above. For every $\xi,\eta\in H$, \begin{align*} \omega_{T_\varphi}(\theta_{\xi,\eta}) = \operatorname{Tr}(T_\varphi\theta_{\xi,\eta}) = (T_\varphi\xi,\eta)_H = \varphi(\theta_{\xi,\eta}). \end{align*} Thus $\omega_{T_\varphi}$ and $\varphi$ agree on all rank-one operators and hence on all finite-rank operators by linearity. Let $a\in\mathcal{L}(H)$. With $(e_j)_{j\in J}$ and $P_F$ as above, the net $(P_FaP_F)$ consists of finite-rank operators and converges strongly to $a$. Indeed, for each $\xi\in H$, \begin{align*} P_FaP_F\xi-a\xi = P_Fa(P_F\xi-\xi)+(P_Fa\xi-a\xi), \end{align*} and both terms converge to $0$ in $H$ because $P_F\xi\to\xi$ and $P_Fa\xi\to a\xi$. Also, \begin{align*} \|P_FaP_F\|_{\mathcal{L}(H)}\le \|a\|_{\mathcal{L}(H)}. \end{align*} We next justify convergence of both functionals on this particular net. Since $P_F\uparrow I_H$ and $\varphi$ is positive and normal, [citetheorem:9263] gives $\varphi(I_H-P_F)\to 0$. The Cauchy-Schwarz inequality for positive functionals gives \begin{align*} |\varphi((I_H-P_F)a)| \le \varphi(I_H-P_F)^{1/2}\varphi(a^*a)^{1/2} \end{align*} and \begin{align*} |\varphi(P_Fa(I_H-P_F))| \le \varphi(P_F)^{1/2}\varphi((I_H-P_F)a^*a(I_H-P_F))^{1/2}. \end{align*} Because $0\le (I_H-P_F)a^*a(I_H-P_F)\le \|a\|_{\mathcal{L}(H)}^2(I_H-P_F)$, the second bound also tends to $0$. Since \begin{align*} a-P_FaP_F=(I_H-P_F)a+P_Fa(I_H-P_F), \end{align*} we obtain $\varphi(P_FaP_F)\to\varphi(a)$. For $\omega_{T_\varphi}$, choose finite-rank operators $R_i\in\mathcal{T}(H)$ with $\|R_i-T_\varphi\|_{\mathcal{T}(H)}\to 0$. For each fixed finite-rank $R_i$, strong convergence of $P_FaP_F$ to $a$ implies $\operatorname{Tr}(R_iP_FaP_F)\to\operatorname{Tr}(R_ia)$ by computing the trace over a finite-dimensional range containing the range of $R_i$. The estimate \begin{align*} |\operatorname{Tr}((T_\varphi-R_i)b)|\le \|T_\varphi-R_i\|_{\mathcal{T}(H)}\|b\|_{\mathcal{L}(H)} \end{align*} for $b\in\{a,P_FaP_F\}$, together with $\|P_FaP_F\|_{\mathcal{L}(H)}\le\|a\|_{\mathcal{L}(H)}$, gives $\omega_{T_\varphi}(P_FaP_F)\to\omega_{T_\varphi}(a)$. Therefore \begin{align*} \varphi(a) = \lim_F \varphi(P_FaP_F) = \lim_F \omega_{T_\varphi}(P_FaP_F) = \omega_{T_\varphi}(a). \end{align*} Hence every positive normal functional has the required trace-class representation. [/step]

[step:Pass from positive normal functionals to all normal functionals] Let $\psi:\mathcal{L}(H)\to\mathbb C$ be a normal linear functional. By the Jordan decomposition for normal functionals on a von Neumann algebra, applied to the real and imaginary parts of $\psi$, there exist positive normal functionals $\psi_1,\psi_2,\psi_3,\psi_4$ on $\mathcal{L}(H)$ such that \begin{align*} \psi=(\psi_1-\psi_2)+i(\psi_3-\psi_4). \end{align*} This structural result applies because $\mathcal{L}(H)$ is a von Neumann algebra and $\psi\in\mathcal{L}(H)_*$ by hypothesis. By the previous step, for each $k\in\{1,2,3,4\}$ there is a positive trace-class operator $T_k\in\mathcal{T}(H)$ such that \begin{align*} \psi_k(a)=\operatorname{Tr}(T_ka) \end{align*} for every $a\in\mathcal{L}(H)$. Define \begin{align*} T:=T_1-T_2+iT_3-iT_4\in\mathcal{T}(H). \end{align*} [Linearity of the trace](/theorems/7814) gives \begin{align*} \psi(a)=\operatorname{Tr}(Ta) \end{align*} for every $a\in\mathcal{L}(H)$. Thus $T\mapsto\omega_T$ maps $\mathcal{T}(H)$ onto the normal predual $\mathcal{L}(H)_*$. Uniqueness follows because if $\omega_T=0$, then for every $\xi,\eta\in H$, \begin{align*} 0=\omega_T(\theta_{\xi,\eta})=(T\xi,\eta)_H. \end{align*} Since this holds for all $\eta\in H$, $T\xi=0$ for all $\xi\in H$, and hence $T=0$. [/step]

[step:Prove the isometry and identify normal states] We already have \begin{align*} \|\omega_T\|_{\mathcal{L}(H)^*}\le \|T\|_{\mathcal{T}(H)}. \end{align*} For the reverse inequality, let $T\in\mathcal{T}(H)$ and use the [polar decomposition](/theorems/3074) in $\mathcal{L}(H)$: \begin{align*} T=u|T|, \end{align*} where $u\in\mathcal{L}(H)$ is a partial isometry and $\|u^*\|_{\mathcal{L}(H)}\le 1$. Then \begin{align*} \|T\|_{\mathcal{T}(H)} = \operatorname{Tr}(|T|) = \operatorname{Tr}(u^*T) = \omega_T(u^*) \le \|\omega_T\|_{\mathcal{L}(H)^*}. \end{align*} Therefore \begin{align*} \|\omega_T\|_{\mathcal{L}(H)^*}=\|T\|_{\mathcal{T}(H)}. \end{align*} It remains to characterize states. If $T\in\mathcal{T}(H)$ is positive, then $T^{1/2}\in\mathcal{T}(H)^{1/2}$ in the sense that $T^{1/2}$ is Hilbert-Schmidt, and $aT^{1/2}$ is Hilbert-Schmidt for every $a\in\mathcal{L}(H)$. Using cyclicity of the trace for a trace-class operator multiplied by a bounded operator, \begin{align*} \omega_T(a^*a)=\operatorname{Tr}(Ta^*a)=\operatorname{Tr}(T^{1/2}a^*aT^{1/2})=\operatorname{Tr}((aT^{1/2})^*(aT^{1/2}))\ge 0. \end{align*} Thus $\omega_T$ is positive. Conversely, if $\omega_T$ is positive, then for every $\xi\in H$, \begin{align*} (T\xi,\xi)_H = \omega_T(\theta_{\xi,\xi}) \ge 0, \end{align*} so $T\ge 0$. Finally, \begin{align*} \omega_T(I_H)=\operatorname{Tr}(T). \end{align*} Thus $\omega_T$ is a normal state exactly when $T\ge 0$ and $\operatorname{Tr}(T)=1$. This completes the identification of the normal predual of $\mathcal{L}(H)$ with $\mathcal{T}(H)$. [/step]