[proofplan] We construct the expectation by orthogonally projecting $L^2(M,\tau)$ onto the closed subspace $L^2(N,\tau|_N)$. The main point is that the projected vector $P_Nx$ is represented by an element of $N$, which follows from a left-boundedness estimate against vectors from $N$. The resulting map is characterized by the trace identity $\tau(y^*E_N(x))=\tau(y^*x)$ for all $y\in N$; this identity gives uniqueness, trace preservation, positivity, normality, and bimodularity. [/proofplan] [step:Represent $M$ and $N$ on their trace Hilbert spaces] Let $L^2(M,\tau)$ denote the [Hilbert space](/page/Hilbert%20Space) completion of $M$ for the [inner product](/page/Inner%20Product) \begin{align*} (a,b)_{L^2(M,\tau)}=\tau(b^*a),\qquad a,b\in M. \end{align*} Let \begin{align*} \Lambda_M:M\to L^2(M,\tau),\qquad a\mapsto \Lambda_M(a) \end{align*} be the canonical dense embedding. Since $\tau$ is faithful, $\|\Lambda_M(a)\|_{L^2(M,\tau)}=0$ implies $a=0$. The restriction $\tau|_N:N\to\mathbb C$ is again a faithful normal tracial state. We identify $L^2(N,\tau|_N)$ with the closed subspace of $L^2(M,\tau)$ generated by $\Lambda_M(N)$. Let \begin{align*} P_N:L^2(M,\tau)\to L^2(N,\tau|_N) \end{align*} be the Hilbert space [orthogonal projection](/theorems/437). For $a\in M$, left multiplication by $a$ defines a bounded operator $L_a:L^2(M,\tau)\to L^2(M,\tau)$ satisfying \begin{align*} L_a\Lambda_M(b)=\Lambda_M(ab),\qquad b\in M, \end{align*} and $\|L_a\|_{\mathrm{op}}\le \|a\|_{\mathrm{op}}$. Similarly, right multiplication by $a$ defines a bounded operator $R_a:L^2(M,\tau)\to L^2(M,\tau)$ satisfying \begin{align*} R_a\Lambda_M(b)=\Lambda_M(ba),\qquad b\in M. \end{align*} These are the standard $L^2$-representation facts for finite von Neumann algebras. [/step] [step:Show that the projection of $x$ is left bounded over $N$] Fix $x\in M$, and define \begin{align*} \eta_x=P_N\Lambda_M(x)\in L^2(N,\tau|_N). \end{align*} We prove that $\eta_x$ is left bounded over $N$, meaning that the map initially defined on $\Lambda_M(N)$ by \begin{align*} \Lambda_M(y)\mapsto \eta_x y \end{align*} extends to a bounded operator on $L^2(N,\tau|_N)$. Let $y,z\in N$. Since $R_y$ preserves $L^2(N,\tau|_N)$ and its orthogonal complement, $R_y$ commutes with $P_N$. Therefore \begin{align*} (\eta_x y,\Lambda_M(z))_{L^2(M,\tau)}=(P_N\Lambda_M(xy),\Lambda_M(z))_{L^2(M,\tau)}. \end{align*} Since $\Lambda_M(z)\in L^2(N,\tau|_N)$ and $P_N$ is the orthogonal projection onto this subspace, \begin{align*} (P_N\Lambda_M(xy),\Lambda_M(z))_{L^2(M,\tau)}=(\Lambda_M(xy),\Lambda_M(z))_{L^2(M,\tau)}. \end{align*} Thus \begin{align*} |(\eta_x y,\Lambda_M(z))_{L^2(M,\tau)}|=|\tau(z^*xy)|. \end{align*} By Cauchy-Schwarz in $L^2(M,\tau)$ and the operator norm bound for $L_x$, \begin{align*} |\tau(z^*xy)|\le \|\Lambda_M(xy)\|_{L^2(M,\tau)}\|\Lambda_M(z)\|_{L^2(M,\tau)}\le \|x\|_{\mathrm{op}}\|\Lambda_M(y)\|_{L^2(M,\tau)}\|\Lambda_M(z)\|_{L^2(M,\tau)}. \end{align*} Taking the supremum over all $z\in N$ with $\|\Lambda_M(z)\|_{L^2(M,\tau)}\le 1$ gives \begin{align*} \|\eta_x y\|_{L^2(M,\tau)}\le \|x\|_{\mathrm{op}}\|\Lambda_M(y)\|_{L^2(M,\tau)}. \end{align*} Hence $\eta_x$ is left bounded over $N$. [guided] The goal is to prove that the Hilbert space vector $\eta_x=P_N\Lambda_M(x)$ is not merely an abstract $L^2$-vector, but comes from multiplication by a bounded element of $N$. The way to test this is to estimate the matrix coefficients of the map $\Lambda_M(y)\mapsto \eta_x y$ against arbitrary vectors $\Lambda_M(z)$ from $N$. Fix $y,z\in N$. Right multiplication by $y$ maps $\Lambda_M(N)$ into itself. It also preserves the orthogonal complement of $L^2(N,\tau|_N)$: if $\xi\perp L^2(N,\tau|_N)$ and $n\in N$, then using traciality of $\tau$, \begin{align*} (R_y\xi,\Lambda_M(n))_{L^2(M,\tau)}=(\xi,\Lambda_M(ny^*))_{L^2(M,\tau)}=0, \end{align*} because $ny^*\in N$. Therefore $R_y$ commutes with the orthogonal projection $P_N$. Using this commutation, \begin{align*} \eta_x y=R_yP_N\Lambda_M(x)=P_NR_y\Lambda_M(x)=P_N\Lambda_M(xy). \end{align*} Now pair with $\Lambda_M(z)\in L^2(N,\tau|_N)$. Since $P_N$ is the orthogonal projection onto $L^2(N,\tau|_N)$, projection does not change inner products against vectors in that subspace: \begin{align*} (\eta_x y,\Lambda_M(z))_{L^2(M,\tau)}=(\Lambda_M(xy),\Lambda_M(z))_{L^2(M,\tau)}. \end{align*} By the definition of the $L^2$ inner product, this is \begin{align*} (\Lambda_M(xy),\Lambda_M(z))_{L^2(M,\tau)}=\tau(z^*xy). \end{align*} We now estimate the last expression. Cauchy-Schwarz in the Hilbert space $L^2(M,\tau)$ gives \begin{align*} |\tau(z^*xy)|\le \|\Lambda_M(xy)\|_{L^2(M,\tau)}\|\Lambda_M(z)\|_{L^2(M,\tau)}. \end{align*} Since left multiplication by $x$ is bounded with operator norm at most $\|x\|_{\mathrm{op}}$, we have \begin{align*} \|\Lambda_M(xy)\|_{L^2(M,\tau)}\le \|x\|_{\mathrm{op}}\|\Lambda_M(y)\|_{L^2(M,\tau)}. \end{align*} Combining the two inequalities gives \begin{align*} |(\eta_x y,\Lambda_M(z))_{L^2(M,\tau)}|\le \|x\|_{\mathrm{op}}\|\Lambda_M(y)\|_{L^2(M,\tau)}\|\Lambda_M(z)\|_{L^2(M,\tau)}. \end{align*} Taking the supremum over all $z\in N$ with $\|\Lambda_M(z)\|_{L^2(M,\tau)}\le 1$ proves \begin{align*} \|\eta_x y\|_{L^2(M,\tau)}\le \|x\|_{\mathrm{op}}\|\Lambda_M(y)\|_{L^2(M,\tau)}. \end{align*} Thus the rule $\Lambda_M(y)\mapsto \eta_x y$ is bounded on the dense subspace $\Lambda_M(N)$, so $\eta_x$ is left bounded over $N$. [/guided] [/step] [step:Define $E_N(x)$ as the bounded element represented by $P_Nx$] We use the standard bounded-vector identification for a finite von Neumann algebra in its trace representation: if $\eta\in L^2(N,\tau|_N)$ and the right-module multiplication rule $\Lambda_M(y)\mapsto \eta y$ extends to a bounded operator on $L^2(N,\tau|_N)$, then there exists a unique $a\in N$ such that \begin{align*} \eta=\Lambda_M(a). \end{align*} The hypotheses of this identification are satisfied because $N$ is a finite von Neumann algebra with faithful normal tracial state $\tau|_N$, and the preceding step proved the required boundedness for $\eta_x=P_N\Lambda_M(x)$. Hence there is a unique element $E_N(x)\in N$ such that \begin{align*} \Lambda_M(E_N(x))=P_N\Lambda_M(x). \end{align*} This defines a map \begin{align*} E_N:M\to N,\qquad x\mapsto E_N(x). \end{align*} Linearity of $E_N$ follows from linearity of $P_N$ and injectivity of $\Lambda_M$. [/step] [step:Derive the trace pairing identity and uniqueness] For every $x\in M$ and $y\in N$, since $\Lambda_M(y)\in L^2(N,\tau|_N)$, \begin{align*} (\Lambda_M(E_N(x)),\Lambda_M(y))_{L^2(M,\tau)}=(P_N\Lambda_M(x),\Lambda_M(y))_{L^2(M,\tau)}. \end{align*} By the defining property of orthogonal projection, the right-hand side equals \begin{align*} (\Lambda_M(x),\Lambda_M(y))_{L^2(M,\tau)}. \end{align*} Therefore \begin{align*} \tau(y^*E_N(x))=\tau(y^*x) \end{align*} for all $x\in M$ and $y\in N$. Taking $y=1_N=1_M$ gives \begin{align*} \tau(E_N(x))=\tau(x). \end{align*} Let $F:M\to N$ be another trace-preserving [conditional expectation](/page/Conditional%20Expectation). Since $F$ is an $N$-bimodule map and $y^*\in N$, for every $x\in M$ and $y\in N$ we have \begin{align*} \tau(y^*F(x))=\tau(F(y^*x))=\tau(y^*x). \end{align*} Thus $F$ satisfies the same trace-pairing identity as $E_N$. Therefore, for every $x\in M$ and $y\in N$, \begin{align*} \tau(y^*(F(x)-E_N(x)))=0. \end{align*} Taking $y=F(x)-E_N(x)\in N$ gives \begin{align*} \tau((F(x)-E_N(x))^*(F(x)-E_N(x)))=0. \end{align*} Faithfulness of $\tau|_N$ implies $F(x)=E_N(x)$. Hence the trace-preserving conditional expectation is unique. [/step] [step:Verify projection, unitality, and bimodularity] If $x\in N$, then $\Lambda_M(x)\in L^2(N,\tau|_N)$, so \begin{align*} P_N\Lambda_M(x)=\Lambda_M(x). \end{align*} Thus $E_N(x)=x$ for every $x\in N$. In particular $E_N(1_M)=1_N$, and $E_N$ is idempotent with range $N$. Let $a,b\in N$ and $x\in M$. Left multiplication by $a$ and right multiplication by $b$ preserve $L^2(N,\tau|_N)$ and its orthogonal complement, so both commute with $P_N$. Therefore \begin{align*} \Lambda_M(E_N(axb))=P_N\Lambda_M(axb)=L_aR_bP_N\Lambda_M(x). \end{align*} Using $\Lambda_M(E_N(x))=P_N\Lambda_M(x)$, this becomes \begin{align*} \Lambda_M(E_N(axb))=\Lambda_M(aE_N(x)b). \end{align*} Injectivity of $\Lambda_M$ gives \begin{align*} E_N(axb)=aE_N(x)b. \end{align*} Hence $E_N$ is an $N$-bimodule projection onto $N$. [/step] [step:Prove positivity] Let $x\in M$ be positive. For every $y\in N$, the trace-pairing identity applied to $y^*y\in N$ gives \begin{align*} \tau(y^*yE_N(x))=\tau(y^*yx). \end{align*} Using traciality on both sides, this is \begin{align*} \tau(y^*E_N(x)y)=\tau(y^*xy). \end{align*} Since $x\ge 0$, the element $y^*xy\in M$ is positive, so $\tau(y^*xy)\ge 0$. Hence \begin{align*} \tau(y^*E_N(x)y)\ge 0 \end{align*} for every $y\in N$. First $E_N(x)$ is self-adjoint. Indeed, for every $y\in N$ the trace-pairing identity and $x=x^*$ give \begin{align*} \tau(y^*E_N(x))=\tau(y^*x)=\tau((xy^*)^*)=\overline{\tau(xy^*)}=\overline{\tau(y^*x)}. \end{align*} Applying the same identity with $y$ replaced by $E_N(x)^*y$ and using faithfulness of $\tau|_N$ gives $E_N(x)=E_N(x)^*$. We use the positive-cone test in a finite von Neumann algebra: for a self-adjoint element $a\in N$, if $\tau(y^*ay)\ge 0$ for every $y\in N$, then $a\ge 0$. Applying this test to $a=E_N(x)$ gives \begin{align*} E_N(x)\ge 0. \end{align*} Thus $E_N$ is positive. [/step] [step:Prove normality from monotone convergence] Let $(x_i)_{i\in I}$ be an increasing bounded net of positive elements of $M$, and let $x=\sup_{i\in I}x_i$ in $M$. For every $y\in N$, the net $(y^*x_i y)_{i\in I}$ is increasing and bounded above by $y^*xy$. Since $\tau$ is normal, it preserves bounded increasing suprema of positive elements, so \begin{align*} \sup_{i\in I}\tau(y^*x_i y)=\tau(y^*xy). \end{align*} Using the trace-pairing identity and bimodularity, \begin{align*} \tau(y^*E_N(x_i)y)=\tau(y^*x_i y) \end{align*} for every $i\in I$, and \begin{align*} \tau(y^*E_N(x)y)=\tau(y^*xy). \end{align*} Therefore \begin{align*} \sup_{i\in I}\tau(y^*E_N(x_i)y)=\tau(y^*E_N(x)y) \end{align*} for every $y\in N$. The net $(E_N(x_i))_{i\in I}$ is increasing because $E_N$ is positive, and it is bounded above by $E_N(x)$. Let $e=\sup_{i\in I}E_N(x_i)$ in $N$. Normality of $\tau|_N$ gives \begin{align*} \tau(y^*ey)=\sup_{i\in I}\tau(y^*E_N(x_i)y)=\tau(y^*E_N(x)y) \end{align*} for every $y\in N$. Taking $y=e-E_N(x)$ after rewriting the equality for the self-adjoint difference gives $e=E_N(x)$ by faithfulness. Hence \begin{align*} E_N(x_i)\uparrow E_N(x). \end{align*} This proves that $E_N$ preserves suprema of bounded increasing nets of positive elements. A positive [linear map](/page/Linear%20Map) between von Neumann algebras is normal exactly when it preserves such bounded increasing suprema, so $E_N$ is normal. [/step] [step:Conclude that $E_N$ is the required conditional expectation] We have constructed a linear map $E_N:M\to N$ that fixes $N$ pointwise, is unital, positive, idempotent, and satisfies the $N$-bimodule identity \begin{align*} E_N(axb)=aE_N(x)b \end{align*} for all $a,b\in N$ and $x\in M$. Thus $E_N$ is a conditional expectation from $M$ onto $N$. The preceding step proves that $E_N$ is normal, and the trace-pairing identity with $y=1_N$ proves that it is trace-preserving: \begin{align*} \tau(E_N(x))=\tau(x) \end{align*} for all $x\in M$. The uniqueness argument shows that no other normal conditional expectation with this trace-preserving property can exist. [/step]