[proofplan] We use only the defining properties of a normal [conditional expectation](/page/Conditional%20Expectation). First, the fact that $E$ fixes $N$ and has range contained in $N$ gives both surjectivity onto $N$ and idempotence. Next, positivity and unitality imply that $E$ is contractive, while $E(1)=1$ gives the reverse inequality for its norm. Finally, normality is converted into ultraweak continuity by the normality criterion for positive maps between von Neumann algebras. [/proofplan] [step:Use the conditional expectation identities to prove that $E$ is a projection onto $N$] By definition of a conditional expectation onto $N$, the map $E:M\to N$ is linear, positive, unital, and satisfies $E(n)=n$ for every $n\in N$. Since the codomain of $E$ is $N$, one has $\operatorname{Range}(E)\subseteq N$. Conversely, if $n\in N$, then $n=E(n)$, so $n\in\operatorname{Range}(E)$. Therefore $\operatorname{Range}(E)=N$. It remains to check idempotence. Let $x\in M$. Since $E(x)\in N$, the identity property on $N$ gives \begin{align*} E(E(x))=E(x). \end{align*} Thus $E^2=E$, so $E$ is a linear projection from $M$ onto $N$. [guided] The word "projection" here means an idempotent [linear map](/page/Linear%20Map) whose range is the target subspace. We therefore have to verify two separate assertions: the range is exactly $N$, and applying $E$ twice is the same as applying it once. By definition of a conditional expectation onto $N$, the map $E:M\to N$ is linear and fixes the subalgebra $N$ pointwise: \begin{align*} E(n)=n \end{align*} for every $n\in N$. Because the codomain is $N$, every value of $E$ lies in $N$, and hence $\operatorname{Range}(E)\subseteq N$. For the reverse inclusion, take an arbitrary element $n\in N$. Since $E$ fixes $N$, we have $n=E(n)$, so $n$ is in the range of $E$. Therefore $N\subseteq \operatorname{Range}(E)$. Combining the two inclusions gives $\operatorname{Range}(E)=N$. Now let $x\in M$. The element $E(x)$ belongs to $N$ because $E$ maps into $N$. Applying the pointwise identity property on $N$ to the particular element $E(x)$ gives \begin{align*} E(E(x))=E(x). \end{align*} Since this holds for every $x\in M$, we have $E^2=E$. Thus $E$ is a linear projection from $M$ onto $N$. [/guided] [/step] [step:Use the norm formula for positive maps to prove that the projection has norm one] We use the positive-map norm formula for unital $C^*$-algebras: if $A$ and $B$ are unital $C^*$-algebras and $\Phi:A\to B$ is a positive linear map, then $\Phi$ is bounded and \begin{align*} \|\Phi\|=\|\Phi(1_A)\|_{\mathrm{op}}. \end{align*} The hypotheses of this result apply with $A=M$, $B=N$, and $\Phi=E$, because every von Neumann algebra is a unital $C^*$-algebra, $N$ has the same identity element as $M$, and $E$ is positive and linear by the definition of conditional expectation. Since $E$ is unital, \begin{align*} E(1_M)=1_N=1_M. \end{align*} Therefore \begin{align*} \|E\|=\|E(1_M)\|_{\mathrm{op}}=\|1_N\|_{\mathrm{op}}=1. \end{align*} Thus the projection $E$ has norm one. [guided] The delicate point is that states do not detect the $C^*$-norm of an arbitrary non-normal element by the formula $\sup_\omega |\omega(a)|$; that supremum is the numerical radius in general. So instead of estimating $E(u)$ for unitary $u$ through states, we invoke the precise standard result that is designed for this situation: the positive-map norm formula for unital $C^*$-algebras. The result says that if $A$ and $B$ are unital $C^*$-algebras and $\Phi:A\to B$ is a positive linear map, then $\Phi$ is automatically bounded and \begin{align*} \|\Phi\|=\|\Phi(1_A)\|_{\mathrm{op}}. \end{align*} We verify its hypotheses in the present setting. The algebras $M$ and $N$ are von Neumann algebras, hence unital $C^*$-algebras. The statement assumes that $N\subseteq M$ has the same identity element as $M$, so the unit of the codomain is the same element as the unit of the domain. Finally, by definition of a conditional expectation, the map \begin{align*} E:M\to N \end{align*} is positive and linear. Therefore the positive-map norm formula applies to $E$ and gives \begin{align*} \|E\|=\|E(1_M)\|_{\mathrm{op}}. \end{align*} Because $E$ is unital, $E(1_M)=1_N$. Since $N$ is a unital $C^*$-algebra, its identity has operator norm one, and hence $\|E\|=\|1_N\|_{\mathrm{op}}=1$. This proves that the projection obtained in the previous step is a norm-one projection. [/guided] [/step] [step:Apply the ultraweak continuity criterion for normal positive maps] The map $E:M\to N$ is positive by the definition of conditional expectation, and it is normal by hypothesis. Therefore the ultraweak continuity criterion for positive maps between von Neumann algebras, [citetheorem:9275], applies to $E$. It follows that $E$ is ultraweakly continuous on bounded subsets of $M$. To pass from bounded-subset continuity to ordinary ultraweak continuity, let $(x_i)_{i\in I}$ be a net in $M$ converging ultraweakly to $x\in M$. Let $M_*$ denote the predual of the von Neumann algebra $M$. Ultraweak convergence is weak-star convergence for the dual [Banach space](/page/Banach%20Space) $M=(M_*)^*$, and every weak-star convergent net is norm bounded by the [uniform boundedness principle](/theorems/549). Hence the set $\{x_i:i\in I\}\cup\{x\}$ is norm bounded. Applying ultraweak continuity of $E$ on this [bounded set](/page/Bounded%20Set) gives that $E(x_i)$ converges ultraweakly to $E(x)$ in $N$. Thus $E$ is ultraweakly continuous. Combining this with the previous steps, $E$ is an ultraweakly continuous norm-one projection from $M$ onto $N$. [guided] The criterion [citetheorem:9275] is exactly the bridge between the order-theoretic word "normal" and the topological phrase "ultraweakly continuous," but it states continuity on bounded subsets. We first verify that the criterion applies. The map \begin{align*} E:M\to N \end{align*} is positive by the definition of conditional expectation, and it is normal by the hypothesis of the theorem. Since $M$ and $N$ are von Neumann algebras, [citetheorem:9275] gives that $E$ is ultraweakly continuous on every norm-bounded subset of $M$. Now suppose $(x_i)_{i\in I}$ is a net in $M$ such that $x_i$ converges ultraweakly to $x\in M$. Let $M_*$ denote the predual of the von Neumann algebra $M$. Ultraweak convergence is weak-star convergence under the identification $M=(M_*)^*$. A weak-star convergent net is pointwise bounded on $M_*$, so the uniform boundedness principle implies that it is bounded in the operator norm of $M$. Therefore \begin{align*} \{x_i:i\in I\}\cup\{x\} \end{align*} is a norm-bounded subset of $M$. We may now use the bounded-subset conclusion from [citetheorem:9275] on this particular bounded set. It gives that $E(x_i)$ converges ultraweakly to $E(x)$ in $N$. Since this holds for every ultraweakly convergent net in $M$, the map $E$ is ultraweakly continuous. Together with the projection property and the norm-one property already proved, this shows that $E$ is an ultraweakly continuous norm-one projection from $M$ onto $N$. [/guided] [/step]