[proofplan] We first record the standard structural facts about normal functionals on a von Neumann algebra: they form a norm-closed subspace of $M^*$, normal positive functionals admit GNS vector realizations, and vector functionals in a universal normal representation span the normal functionals. The canonical evaluation map is contractive by definition and is isometric because normal states detect the $C^*$-norm through $x^*x$. Surjectivity is proved by representing an arbitrary functional on $M_*$ as a bounded sesquilinear form on the universal normal representation, converting that form into an operator, and using the bicommutant theorem to identify the operator with an element of $M$. [/proofplan]

[step:Show that the normal functionals form a Banach space] Let $M^*$ be the Banach dual of $M$, with norm \begin{align*} \|\omega\|_{M^*}=\sup\{|\omega(x)|:x\in M,\ \|x\|_{\mathrm{op}}\le 1\}. \end{align*} By definition, \begin{align*} M_*=\{\omega\in M^*:\omega \text{ is normal}\}. \end{align*} Normality is stable under complex linear combinations, so $M_*$ is a linear subspace of $M^*$. We use the standard norm-closedness theorem for normal functionals on a von Neumann algebra: if $(\omega_n)_{n\in\mathbb N}$ is a sequence in $M_*$ and $\omega_n\to\omega$ in the norm of $M^*$, then $\omega$ is normal. For positive functionals this is an immediate consequence of the monotone projection criterion for normality in [citetheorem:9272]; the standard complex form follows by applying the positive case to the Jordan decompositions of the real and imaginary parts. Hence $M_*$ is a closed linear subspace of the [Banach space](/page/Banach%20Space) $M^*$, and therefore $M_*$ is a Banach space. [/step]

[step:Define the canonical evaluation map and prove it is contractive] Define \begin{align*} \kappa:M\to (M_*)^* \end{align*} by \begin{align*} \kappa(x)(\omega)=\omega(x) \end{align*} for $x\in M$ and $\omega\in M_*$. For fixed $x\in M$, the map $\kappa(x):M_*\to\mathbb C$ is linear. If $\omega\in M_*$, then the definition of the dual norm gives \begin{align*} |\kappa(x)(\omega)|=|\omega(x)|\le \|\omega\|_{M^*}\|x\|_{\mathrm{op}}. \end{align*} Thus $\kappa(x)\in (M_*)^*$ and \begin{align*} \|\kappa(x)\|_{(M_*)^*}\le \|x\|_{\mathrm{op}}. \end{align*} Linearity of $\kappa$ follows directly from linearity of each $\omega\in M_*$. [/step]

[step:Use normal states to prove that evaluation preserves the operator norm]We prove the reverse inequality for $\kappa$. Let $x\in M$ with $x\ne 0$. Let $M_*^+$ denote the cone of positive normal linear functionals on $M$. We use the standard separation theorem for normal states on a von Neumann algebra: for every positive element $a\in M$, \begin{align*} \|a\|_{\mathrm{op}}=\sup\{\varphi(a):\varphi\in M_*^+,\ \varphi(1)=1\}. \end{align*} Apply this to the positive element $a=x^*x$. For every $\varepsilon>0$, choose a normal state $\varphi\in M_*^+$ such that \begin{align*} \varphi(x^*x)>\|x^*x\|_{\mathrm{op}}-\varepsilon. \end{align*} Define a linear functional $\omega_\varphi:M\to\mathbb C$ by \begin{align*} \omega_\varphi(y)=\varphi(x^*y) \end{align*} for $y\in M$. Since left multiplication by $x^*$ is ultraweakly continuous and $\varphi$ is normal, $\omega_\varphi$ is normal. Also, \begin{align*} |\omega_\varphi(y)|=|\varphi(x^*y)|\le \|x^*y\|_{\mathrm{op}}\le \|x\|_{\mathrm{op}}\|y\|_{\mathrm{op}} \end{align*} for every $y\in M$, so $\|\omega_\varphi\|_{M^*}\le \|x\|_{\mathrm{op}}$. Therefore \begin{align*} \|\kappa(x)\|_{(M_*)^*}\ge \frac{|\kappa(x)(\omega_\varphi)|}{\|\omega_\varphi\|_{M^*}}\ge \frac{\varphi(x^*x)}{\|x\|_{\mathrm{op}}}. \end{align*} Using $\|x^*x\|_{\mathrm{op}}=\|x\|_{\mathrm{op}}^2$, we obtain \begin{align*} \|\kappa(x)\|_{(M_*)^*}\ge \|x\|_{\mathrm{op}}-\frac{\varepsilon}{\|x\|_{\mathrm{op}}}. \end{align*} Letting $\varepsilon\downarrow 0$ gives \begin{align*} \|\kappa(x)\|_{(M_*)^*}\ge \|x\|_{\mathrm{op}}. \end{align*} Together with contractivity, this proves \begin{align*} \|\kappa(x)\|_{(M_*)^*}= \|x\|_{\mathrm{op}}. \end{align*} Thus $\kappa$ is an isometric embedding.[/step]

[guided]The only non-formal part of the isometry proof is showing that the normal functionals are numerous enough to detect the whole operator norm. For a $C^*$-algebra, the norm of $x$ is encoded by the positive element $x^*x$, because \begin{align*} \|x^*x\|_{\mathrm{op}}=\|x\|_{\mathrm{op}}^2. \end{align*} So we test $x^*x$ against normal states. Let $M_*^+$ denote the cone of positive normal linear functionals on $M$. We use the standard fact that normal states detect the norm of positive elements in a von Neumann algebra: if $a\in M$ is positive, then \begin{align*} \|a\|_{\mathrm{op}}=\sup\{\varphi(a):\varphi\in M_*^+,\ \varphi(1)=1\}. \end{align*} Apply this with $a=x^*x$. Given $\varepsilon>0$, choose a normal state $\varphi$ satisfying \begin{align*} \varphi(x^*x)>\|x^*x\|_{\mathrm{op}}-\varepsilon. \end{align*} The functional that tests $x$ itself is not $\varphi$, but the normal functional obtained by multiplying on the left by $x^*$. Define \begin{align*} \omega_\varphi:M\to\mathbb C \end{align*} by \begin{align*} \omega_\varphi(y)=\varphi(x^*y) \end{align*} for $y\in M$. Left multiplication by $x^*$ is ultraweakly continuous, and $\varphi$ is normal, so their composition $\omega_\varphi$ is normal. Its norm is controlled by the operator norm of $x$: for every $y\in M$, \begin{align*} |\omega_\varphi(y)|=|\varphi(x^*y)|\le \|x^*y\|_{\mathrm{op}}\le \|x\|_{\mathrm{op}}\|y\|_{\mathrm{op}}. \end{align*} Hence $\|\omega_\varphi\|_{M^*}\le \|x\|_{\mathrm{op}}$. Now evaluate $\kappa(x)$ on this particular normal functional: \begin{align*} \kappa(x)(\omega_\varphi)=\omega_\varphi(x)=\varphi(x^*x). \end{align*} Therefore \begin{align*} \|\kappa(x)\|_{(M_*)^*}\ge \frac{|\kappa(x)(\omega_\varphi)|}{\|\omega_\varphi\|_{M^*}}\ge \frac{\varphi(x^*x)}{\|x\|_{\mathrm{op}}}. \end{align*} Using the choice of $\varphi$ and the $C^*$-identity, \begin{align*} \|\kappa(x)\|_{(M_*)^*}\ge \frac{\|x\|_{\mathrm{op}}^2-\varepsilon}{\|x\|_{\mathrm{op}}}=\|x\|_{\mathrm{op}}-\frac{\varepsilon}{\|x\|_{\mathrm{op}}}. \end{align*} Since $\varepsilon>0$ was arbitrary, this gives \begin{align*} \|\kappa(x)\|_{(M_*)^*}\ge \|x\|_{\mathrm{op}}. \end{align*} The opposite inequality was already proved from the definition of the dual norm, so $\kappa$ is isometric.[/guided]

[step:Build the universal normal representation] For each normal state $\varphi\in M_*^+$ with $\varphi(1)=1$, let \begin{align*} \pi_\varphi:M\to \mathcal{L}(H_\varphi) \end{align*} be the normal cyclic representation and let $\xi_\varphi\in H_\varphi$ be its cyclic vector, so that \begin{align*} \varphi(x)=(\pi_\varphi(x)\xi_\varphi,\xi_\varphi)_{H_\varphi} \end{align*} for every $x\in M$. This is the GNS representation for a normal state, as in [citetheorem:9308]. Let $S$ be the set of normal states on $M$. Define the [Hilbert space](/page/Hilbert%20Space) \begin{align*} H_u=\bigoplus_{\varphi\in S}H_\varphi. \end{align*} Define the universal normal representation \begin{align*} \pi_u:M\to\mathcal{L}(H_u) \end{align*} by \begin{align*} \pi_u(x)=\bigoplus_{\varphi\in S}\pi_\varphi(x) \end{align*} for $x\in M$. Since each $\pi_\varphi$ is a normal unital $*$-representation, the direct-sum representation $\pi_u$ is unital and preserves products and adjoints componentwise. It is normal because an increasing bounded net of projections in $M$ is carried by each $\pi_\varphi$ to a net with the correct supremum, by [citetheorem:9271], and suprema of projections are computed componentwise in the Hilbert direct sum. Since normal [states separate positive elements](/theorems/8565), the isometry argument above shows that $\pi_u$ is faithful. For $\eta,\zeta\in H_u$, define the vector functional \begin{align*} \omega_{\eta,\zeta}:M\to\mathbb C \end{align*} by \begin{align*} \omega_{\eta,\zeta}(x)=(\pi_u(x)\eta,\zeta)_{H_u} \end{align*} for $x\in M$. Since $\pi_u$ is normal, each $\omega_{\eta,\zeta}$ belongs to $M_*$. We use the standard vector-functional spanning theorem for the universal normal representation: every element of $M_*$ lies in the norm-closed linear span of the vector functionals $\omega_{\eta,\zeta}$ arising from $\pi_u$. [/step]

[step:Represent a functional on $M_*$ by an operator commuting with the commutant]Let $F\in (M_*)^*$. Throughout this step, Hilbert-space inner products are taken to be linear in the first variable and conjugate-linear in the second variable. Define a sesquilinear form \begin{align*} b_F:H_u\times H_u\to\mathbb C \end{align*} by \begin{align*} b_F(\eta,\zeta)=F(\omega_{\eta,\zeta}) \end{align*} for $\eta,\zeta\in H_u$. The vector functional estimate gives \begin{align*} \|\omega_{\eta,\zeta}\|_{M^*}\le \|\eta\|_{H_u}\|\zeta\|_{H_u}. \end{align*} Therefore \begin{align*} |b_F(\eta,\zeta)|\le \|F\|_{(M_*)^*}\|\eta\|_{H_u}\|\zeta\|_{H_u}. \end{align*} Thus $b_F$ is bounded. By the [Riesz representation theorem](/theorems/218) for bounded sesquilinear forms on a Hilbert space, there is a unique operator $T_F\in\mathcal{L}(H_u)$ such that \begin{align*} (T_F\eta,\zeta)_{H_u}=b_F(\eta,\zeta) \end{align*} for all $\eta,\zeta\in H_u$. Let $\pi_u(M)'$ denote the commutant of $\pi_u(M)$ in $\mathcal{L}(H_u)$, and let $\pi_u(M)''$ denote the commutant of $\pi_u(M)'$ in $\mathcal{L}(H_u)$. Let $R\in\pi_u(M)'$. For all $\eta,\zeta\in H_u$ and $x\in M$, \begin{align*} \omega_{R\eta,\zeta}(x)=(\pi_u(x)R\eta,\zeta)_{H_u}=(R\pi_u(x)\eta,\zeta)_{H_u}=(\pi_u(x)\eta,R^*\zeta)_{H_u}=\omega_{\eta,R^*\zeta}(x). \end{align*} Hence $\omega_{R\eta,\zeta}=\omega_{\eta,R^*\zeta}$. Applying $F$ gives \begin{align*} (T_FR\eta,\zeta)_{H_u}=F(\omega_{R\eta,\zeta})=F(\omega_{\eta,R^*\zeta})=(T_F\eta,R^*\zeta)_{H_u}=(RT_F\eta,\zeta)_{H_u}. \end{align*} Since this holds for all $\eta,\zeta\in H_u$, we have $T_FR=RT_F$. Therefore \begin{align*} T_F\in \pi_u(M)''. \end{align*} Because $\pi_u$ is a faithful normal representation of the von Neumann algebra $M$, its range $\pi_u(M)$ is strongly closed in $\mathcal{L}(H_u)$. The bicommutant theorem for von Neumann algebras therefore applies to the unital self-adjoint algebra $\pi_u(M)$ and gives \begin{align*} \pi_u(M)''=\pi_u(M). \end{align*} Thus there exists a unique $x_F\in M$ such that \begin{align*} T_F=\pi_u(x_F). \end{align*}[/step]

[guided]The goal is to turn $F$, which is an abstract bounded linear functional on $M_*$, into evaluation at an element of $M$. The vector functionals $\omega_{\eta,\zeta}$ provide the bridge, because they are normal functionals and therefore lie in the domain of $F$. We use the convention that Hilbert-space inner products are linear in the first variable and conjugate-linear in the second variable. Define \begin{align*} b_F:H_u\times H_u\to\mathbb C \end{align*} by \begin{align*} b_F(\eta,\zeta)=F(\omega_{\eta,\zeta}). \end{align*} This is well-defined because $\omega_{\eta,\zeta}\in M_*$ for every $\eta,\zeta\in H_u$. The usual vector functional estimate says \begin{align*} \|\omega_{\eta,\zeta}\|_{M^*}\le \|\eta\|_{H_u}\|\zeta\|_{H_u}. \end{align*} Indeed, for every $y\in M$ with $\|y\|_{\mathrm{op}}\le 1$, \begin{align*} |\omega_{\eta,\zeta}(y)|=|(\pi_u(y)\eta,\zeta)_{H_u}|\le \|\pi_u(y)\eta\|_{H_u}\|\zeta\|_{H_u}\le \|\eta\|_{H_u}\|\zeta\|_{H_u}. \end{align*} Taking the supremum over such $y$ proves the estimate. Since $F$ is bounded, \begin{align*} |b_F(\eta,\zeta)|\le \|F\|_{(M_*)^*}\|\omega_{\eta,\zeta}\|_{M^*}\le \|F\|_{(M_*)^*}\|\eta\|_{H_u}\|\zeta\|_{H_u}. \end{align*} Thus $b_F$ is a bounded sesquilinear form. The [Riesz representation theorem](/theorems/221) for bounded sesquilinear forms gives a unique bounded operator $T_F\in\mathcal{L}(H_u)$ satisfying \begin{align*} (T_F\eta,\zeta)_{H_u}=b_F(\eta,\zeta) \end{align*} for all $\eta,\zeta\in H_u$. We now prove that $T_F$ comes from the von Neumann algebra represented by $\pi_u$. To do this, it is enough to show that $T_F$ commutes with every operator in the commutant $\pi_u(M)'$. Let $R\in\pi_u(M)'$. Since $R$ commutes with every $\pi_u(x)$, for every $x\in M$ we compute \begin{align*} \omega_{R\eta,\zeta}(x)=(\pi_u(x)R\eta,\zeta)_{H_u}=(R\pi_u(x)\eta,\zeta)_{H_u}. \end{align*} Using the definition of the adjoint of $R$, \begin{align*} (R\pi_u(x)\eta,\zeta)_{H_u}=(\pi_u(x)\eta,R^*\zeta)_{H_u}. \end{align*} Therefore \begin{align*} \omega_{R\eta,\zeta}=\omega_{\eta,R^*\zeta}. \end{align*} Apply $F$ to this identity: \begin{align*} (T_FR\eta,\zeta)_{H_u}=F(\omega_{R\eta,\zeta})=F(\omega_{\eta,R^*\zeta}). \end{align*} By the definition of $T_F$ again, \begin{align*} F(\omega_{\eta,R^*\zeta})=(T_F\eta,R^*\zeta)_{H_u}=(RT_F\eta,\zeta)_{H_u}. \end{align*} Thus \begin{align*} (T_FR\eta,\zeta)_{H_u}=(RT_F\eta,\zeta)_{H_u} \end{align*} for all $\eta,\zeta\in H_u$, so $T_FR=RT_F$. Since $R\in\pi_u(M)'$ was arbitrary, \begin{align*} T_F\in \pi_u(M)''. \end{align*} Since $\pi_u$ is a faithful normal representation of the von Neumann algebra $M$, the range $\pi_u(M)$ is strongly closed in $\mathcal{L}(H_u)$. Thus $\pi_u(M)$ is a von Neumann algebra acting on $H_u$, and the bicommutant theorem identifies $\pi_u(M)''$ with $\pi_u(M)$. Hence $T_F=\pi_u(x_F)$ for a unique $x_F\in M$, where uniqueness follows from faithfulness of $\pi_u$.[/guided]

[step:Extend equality from vector functionals to all normal functionals] Let $F\in(M_*)^*$ and let $x_F\in M$ be the element obtained in the previous step. For all $\eta,\zeta\in H_u$, \begin{align*} F(\omega_{\eta,\zeta})=b_F(\eta,\zeta)=(T_F\eta,\zeta)_{H_u}=(\pi_u(x_F)\eta,\zeta)_{H_u}=\omega_{\eta,\zeta}(x_F). \end{align*} By the norm-density of the linear span of the vector functionals $\omega_{\eta,\zeta}$ in $M_*$, and since both $F$ and $\kappa(x_F)$ are continuous linear functionals on $M_*$, the equality extends to every $\omega\in M_*$. Hence \begin{align*} F(\omega)=\omega(x_F)=\kappa(x_F)(\omega) \end{align*} for every $\omega\in M_*$. Therefore $F=\kappa(x_F)$, so $\kappa$ is surjective. Since $\kappa$ is both isometric and surjective, it is an isometric isomorphism \begin{align*} M\cong (M_*)^*. \end{align*} [/step]