[proofplan] The proof first constructs the candidate limit from scalar matrix coefficients. Monotonicity and the [uniform norm](/page/Uniform%20Norm) bound imply that each quadratic coefficient $(T_i\xi,\xi)_H$ is an increasing bounded scalar net, so it has a limit. The polarization identity turns these limiting quadratic forms into a bounded sesquilinear form, and the [Riesz representation theorem](/theorems/218) gives a bounded self-adjoint operator $T$. Finally, positivity of $T-T_i$ and the Cauchy-Schwarz estimate for positive operators upgrade convergence of quadratic forms to strong convergence, and the von Neumann algebra clause follows from strong operator closedness. [/proofplan] [step:Construct limiting quadratic forms from the monotone scalar nets] For each $\xi\in H$, define the scalar net $q_\xi:I\to\mathbb R$ by \begin{align*}q_\xi(i)=(T_i\xi,\xi)_H.\end{align*} Since $T_i$ is self-adjoint, $q_\xi(i)$ is real for every $i\in I$. If $i\le j$, then $T_i\le T_j$, so $T_j-T_i$ is positive and hence \begin{align*}q_\xi(j)-q_\xi(i)=((T_j-T_i)\xi,\xi)_H\ge 0.\end{align*} Thus $q_\xi$ is increasing. The norm bound gives \begin{align*}|q_\xi(i)|\le \|T_i\|_{\mathcal{L}(H)}\|\xi\|_H^2\le C\|\xi\|_H^2.\end{align*} Therefore $q_\xi$ is an increasing bounded net in $\mathbb R$, so it converges. Define $q:H\to\mathbb R$ by \begin{align*} q(\xi)=\lim_i (T_i\xi,\xi)_H. \end{align*} The same estimate passes to the limit: \begin{align*} |q(\xi)|\le C\|\xi\|_H^2. \end{align*} [guided] The first objective is to find enough scalar limits to reconstruct an operator. For a fixed vector $\xi\in H$, define \begin{align*}q_\xi:I\to\mathbb R,\qquad i\mapsto (T_i\xi,\xi)_H.\end{align*} This is real-valued because every $T_i$ is self-adjoint. The order hypothesis says that if $i\le j$, then $T_j-T_i$ is a positive operator. Applying the definition of positivity to the vector $\xi$ gives \begin{align*}((T_j-T_i)\xi,\xi)_H\ge 0.\end{align*} Equivalently, \begin{align*} (T_i\xi,\xi)_H\le (T_j\xi,\xi)_H. \end{align*} So $q_\xi$ is an increasing scalar net. We also need an upper bound. Since $\|T_i\|_{\mathcal{L}(H)}\le C$, the operator norm estimate gives \begin{align*}|(T_i\xi,\xi)_H|\le \|T_i\xi\|_H\|\xi\|_H\le \|T_i\|_{\mathcal{L}(H)}\|\xi\|_H^2\le C\|\xi\|_H^2.\end{align*} Thus $q_\xi$ is increasing and bounded above by $C\|\xi\|_H^2$. Completeness of $\mathbb R$ for monotone bounded nets gives the limit \begin{align*} q(\xi)=\lim_i (T_i\xi,\xi)_H. \end{align*} Taking limits in the same norm estimate yields \begin{align*} |q(\xi)|\le C\|\xi\|_H^2. \end{align*} This produces a limiting quadratic form, but not yet a linear operator. The next step uses polarization to recover mixed coefficients from these diagonal coefficients. [/guided] [/step] [step:Polarize the limiting quadratic form to obtain a bounded sesquilinear form] Define $b:H\times H\to\mathbb C$ by the complex polarization formula for an [inner product](/page/Inner%20Product) linear in the first variable: \begin{align*}b(\xi,\eta)=\frac{1}{4}\bigl(q(\xi+\eta)-q(\xi-\eta)+i q(\xi+i\eta)-i q(\xi-i\eta)\bigr).\end{align*} For every $i\in I$, define $b_i:H\times H\to\mathbb C$ by \begin{align*}b_i(\xi,\eta)=(T_i\xi,\eta)_H.\end{align*} Applying the same polarization identity to the quadratic form $\zeta\mapsto (T_i\zeta,\zeta)_H$ gives, for all $\xi,\eta\in H$, \begin{align*}b_i(\xi,\eta)=\frac{1}{4}\bigl(q_{\xi+\eta}(i)-q_{\xi-\eta}(i)+i q_{\xi+i\eta}(i)-i q_{\xi-i\eta}(i)\bigr).\end{align*} Passing to the limit in this finite linear combination gives \begin{align*}b(\xi,\eta)=\lim_i (T_i\xi,\eta)_H.\end{align*} Since each $b_i$ is linear in the first variable and conjugate-linear in the second variable, the pointwise limit $b$ is sesquilinear. For the norm bound, use the polarization expression together with the uniform estimate for $q$: \begin{align*}|b(\xi,\eta)|\le \limsup_i |(T_i\xi,\eta)_H|\le \limsup_i \|T_i\|_{\mathcal{L}(H)}\|\xi\|_H\|\eta\|_H\le C\|\xi\|_H\|\eta\|_H.\end{align*} Hence $b$ is a bounded sesquilinear form on $H$. [/step] [step:Represent the bounded form by a self-adjoint operator] The form $b$ is sesquilinear on $H\times H$ and satisfies $|b(\xi,\eta)|\le C\|\xi\|_H\|\eta\|_H$ for all $\xi,\eta\in H$. By the [Riesz representation theorem](/theorems/221) for bounded sesquilinear forms on a [Hilbert space](/page/Hilbert%20Space), there exists a unique operator $T\in\mathcal{L}(H)$ such that \begin{align*} b(\xi,\eta)=(T\xi,\eta)_H \end{align*} for all $\xi,\eta\in H$. Moreover, \begin{align*} \|T\|_{\mathcal{L}(H)}\le C. \end{align*} For every $\xi\in H$, \begin{align*}(T\xi,\xi)_H=b(\xi,\xi)=q(\xi)\in\mathbb R.\end{align*} Since $T\in\mathcal{L}(H)$ and its quadratic form is real on every vector in $H$, the standard self-adjointness criterion for bounded operators gives that $T$ is self-adjoint, equivalently $T=T^*$ where $T^*$ is the Hilbert-space adjoint of $T$. [/step] [step:Show that each difference $T-T_i$ is positive] Fix $i\in I$ and $\xi\in H$. For every $j\in I$ with $j\ge i$, monotonicity gives $T_i\le T_j$, so \begin{align*} (T_i\xi,\xi)_H\le (T_j\xi,\xi)_H. \end{align*} Taking the limit over $j$ in the directed tail $\{j\in I:j\ge i\}$ gives \begin{align*}(T_i\xi,\xi)_H\le q(\xi)=(T\xi,\xi)_H.\end{align*} Thus \begin{align*}((T-T_i)\xi,\xi)_H\ge 0\end{align*} for every $\xi\in H$, so $T-T_i$ is positive. [/step] [step:Upgrade quadratic-form convergence to strong convergence] Fix $\xi\in H$. Since \begin{align*} ((T-T_i)\xi,\xi)_H=(T\xi,\xi)_H-(T_i\xi,\xi)_H=q(\xi)-q_\xi(i), \end{align*} and $q_\xi(i)\to q(\xi)$, we have \begin{align*}((T-T_i)\xi,\xi)_H\to 0.\end{align*} For each $i\in I$, set $A_i=T-T_i$. The previous step shows that $A_i$ is positive. The positive-operator Cauchy-Schwarz estimate gives \begin{align*} \|A_i\xi\|_H^2\le \|A_i\|_{\mathcal{L}(H)}(A_i\xi,\xi)_H. \end{align*} The operators $A_i$ are uniformly bounded, since \begin{align*}\|A_i\|_{\mathcal{L}(H)}\le \|T\|_{\mathcal{L}(H)}+\|T_i\|_{\mathcal{L}(H)}\le 2C.\end{align*} Therefore \begin{align*}\|(T-T_i)\xi\|_H^2\le 2C\,((T-T_i)\xi,\xi)_H\to 0.\end{align*} Hence $T_i\xi\to T\xi$ in $H$ for every $\xi\in H$, which is exactly strong operator convergence. [guided] At this point we know that $T-T_i$ is positive and that its quadratic form tends to zero on each fixed vector. We must turn this scalar convergence into norm convergence of vectors. Fix $\xi\in H$ and define, for each $i\in I$, \begin{align*}A_i=T-T_i.\end{align*} The previous step proves that $A_i$ is positive. Also, \begin{align*}(A_i\xi,\xi)_H=((T-T_i)\xi,\xi)_H=(T\xi,\xi)_H-(T_i\xi,\xi)_H=q(\xi)-q_\xi(i).\end{align*} Since $q_\xi(i)\to q(\xi)$, it follows that \begin{align*}(A_i\xi,\xi)_H\to 0.\end{align*} The missing point is why this forces $\|A_i\xi\|_H\to 0$. Positivity gives the Cauchy-Schwarz estimate for the positive sesquilinear form $(u,v)\mapsto (A_i u,v)_H$: \begin{align*}|(A_i u,v)_H|^2\le (A_i u,u)_H(A_i v,v)_H\end{align*} for all $u,v\in H$. Applying this with $u=\xi$ and $v=A_i\xi$ gives \begin{align*} \|A_i\xi\|_H^4=|(A_i\xi,A_i\xi)_H|^2\le (A_i\xi,\xi)_H(A_i^2\xi,A_i\xi)_H. \end{align*} Using the operator norm bound on $A_i$ in the last factor gives the standard estimate \begin{align*} \|A_i\xi\|_H^2\le \|A_i\|_{\mathcal{L}(H)}(A_i\xi,\xi)_H. \end{align*} It remains to check that the factor $\|A_i\|_{\mathcal{L}(H)}$ is uniformly bounded. Since $\|T\|_{\mathcal{L}(H)}\le C$ and $\|T_i\|_{\mathcal{L}(H)}\le C$, the triangle inequality gives \begin{align*}\|A_i\|_{\mathcal{L}(H)}=\|T-T_i\|_{\mathcal{L}(H)}\le 2C.\end{align*} Therefore \begin{align*}\|(T-T_i)\xi\|_H^2=\|A_i\xi\|_H^2\le 2C(A_i\xi,\xi)_H\to 0.\end{align*} Thus $T_i\xi\to T\xi$ in $H$ for the fixed vector $\xi$. Since $\xi$ was arbitrary, the convergence is strong operator convergence. [/guided] [/step] [step:Use strong closedness for the von Neumann algebra conclusion] Assume now that $M\subseteq\mathcal{L}(H)$ is a von Neumann algebra and that $T_i\in M$ for every $i\in I$. By definition, a von Neumann algebra is closed in the strong operator topology. Since $T_i\to T$ strongly and every $T_i$ belongs to $M$, the strong operator limit $T$ also belongs to $M$. [/step]