[proofplan] The implication from normality to preservation of increasing projection suprema is immediate because projections are positive elements. The passage from increasing projection suprema to complete orthogonal additivity is obtained by applying the projection condition to the directed net of finite orthogonal sums. Conversely, complete orthogonal additivity is reduced to the state case by normalizing the positive functional when $\varphi(1)>0$; the normal-state projection criterion then gives normality, and the zero functional is handled separately. [/proofplan] [step:Restrict normality to increasing nets of projections] Assume that $\varphi$ is normal. Let $(p_i)_{i\in I}$ be an increasing net of projections in $M$, and let $p=\bigvee_{i\in I}p_i$ be its supremum in $\mathcal P(M)$. Since every projection is positive and $p_i\le p\le 1$, the net $(p_i)_{i\in I}$ is a bounded increasing net in $M_+$. By normality of $\varphi$, \begin{align*} \varphi(p)=\sup_{i\in I}\varphi(p_i). \end{align*} Thus condition 1 implies condition 2. [/step] [step:Apply the projection condition to finite orthogonal sums] Assume condition 2. Let $(p_j)_{j\in J}$ be a pairwise orthogonal family of projections in $M$, and define $p=\bigvee_{j\in J}p_j$. Let $\mathcal F(J)$ denote the directed set of finite subsets of $J$, ordered by inclusion. For each $F\in\mathcal F(J)$, define the projection \begin{align*} s_F=\sum_{j\in F}p_j. \end{align*} The sum is a projection because the projections in the family are pairwise orthogonal. If $F\subset G$, then $s_F\le s_G$, so $(s_F)_{F\in\mathcal F(J)}$ is an increasing net of projections. Its supremum is $p$, because $p$ is the least projection dominating every $p_j$, and a projection dominates every $p_j$ if and only if it dominates every finite sum $s_F$. By condition 2, \begin{align*} \varphi(p)=\sup_{F\in\mathcal F(J)}\varphi(s_F). \end{align*} Since $\varphi$ is linear and $s_F=\sum_{j\in F}p_j$, \begin{align*} \varphi(s_F)=\sum_{j\in F}\varphi(p_j). \end{align*} Therefore \begin{align*} \varphi(p)=\sup_{F\in\mathcal F(J)}\sum_{j\in F}\varphi(p_j)=\sum_{j\in J}\varphi(p_j). \end{align*} Hence condition 2 implies condition 3. [guided] The idea is to turn an arbitrary orthogonal family into a single increasing net. The correct directed set is not $J$ itself, since $J$ need not be ordered. Instead, let $\mathcal F(J)$ be the set of all finite subsets of $J$, ordered by inclusion. This is directed because, for $F,G\in\mathcal F(J)$, the finite set $F\cup G$ contains both $F$ and $G$. For each finite subset $F\subset J$, define \begin{align*} s_F=\sum_{j\in F}p_j. \end{align*} Because the projections $p_j$ are pairwise orthogonal, this finite sum is again a projection. If $F\subset G$, then \begin{align*} s_G-s_F=\sum_{j\in G\setminus F}p_j \end{align*} is a positive projection orthogonal to $s_F$, so $s_F\le s_G$. Thus $(s_F)_{F\in\mathcal F(J)}$ is an increasing net of projections. We now identify its supremum. The projection $p=\bigvee_{j\in J}p_j$ dominates every $p_j$, hence it dominates every finite sum $s_F$. Conversely, if $r\in\mathcal P(M)$ is a projection with $s_F\le r$ for every finite $F\subset J$, then taking $F=\{j\}$ gives $p_j\le r$ for every $j\in J$. Therefore $p\le r$ by the definition of $p$ as the least upper bound of the family $(p_j)_{j\in J}$. Hence \begin{align*} p=\bigvee_{F\in\mathcal F(J)}s_F. \end{align*} Condition 2 applies to this increasing projection net, so \begin{align*} \varphi(p)=\sup_{F\in\mathcal F(J)}\varphi(s_F). \end{align*} Linearity of $\varphi$ gives \begin{align*} \varphi(s_F)=\sum_{j\in F}\varphi(p_j) \end{align*} for every finite $F\subset J$. Therefore \begin{align*} \varphi(p)=\sup_{F\in\mathcal F(J)}\sum_{j\in F}\varphi(p_j), \end{align*} which is exactly the meaning of the unordered sum $\sum_{j\in J}\varphi(p_j)$ in the statement. [/guided] [/step] [step:Normalize complete additivity to invoke the normal state criterion] Assume condition 3. If $\varphi(1)=0$, then positivity gives $\varphi(x)=0$ for every $x\in M$ by the [Cauchy-Schwarz inequality](/theorems/432) for positive functionals, so $\varphi$ is normal. Assume now that $\varphi(1)>0$. Define the positive linear functional $\psi:M\to\mathbb C$ by \begin{align*} \psi: M \to \mathbb C, \qquad x \mapsto \frac{\varphi(x)}{\varphi(1)}. \end{align*} Then $\psi(1)=1$, so $\psi$ is a state on $M$. For every pairwise orthogonal family $(p_j)_{j\in J}$ of projections in $M$ with $p=\bigvee_{j\in J}p_j$, condition 3 gives \begin{align*} \psi(p)=\frac{1}{\varphi(1)}\varphi(p)=\frac{1}{\varphi(1)}\sum_{j\in J}\varphi(p_j)=\sum_{j\in J}\psi(p_j), \end{align*} where both unordered sums are suprema of finite partial sums. Thus $\psi$ satisfies the complete-additivity condition in [citetheorem:9306]: it is a state on the von Neumann algebra $M$, and it is completely additive on arbitrary pairwise orthogonal families of projections, with sums interpreted as suprema of finite partial sums. By that theorem, $\psi$ is a normal state. Since $\varphi=\varphi(1)\psi$ and scalar multiples of normal linear functionals are normal, $\varphi$ is normal. Hence condition 3 implies condition 1. [guided] The goal is to use the normal-state criterion, so we first reduce a positive functional to a state. If $\varphi(1)=0$, positivity implies the Cauchy-Schwarz inequality \begin{align*} |\varphi(y^*x)|^2\le \varphi(x^*x)\varphi(y^*y) \end{align*} for all $x,y\in M$. Taking $y=1$ and using $x^*x\le \|x\|_{\mathrm{op}}^2 1$ gives \begin{align*} |\varphi(x)|^2\le \varphi(x^*x)\varphi(1)=0, \end{align*} so $\varphi$ is the zero functional, and the zero functional is normal. Now suppose $\varphi(1)>0$. Define $\psi:M\to\mathbb C$ by \begin{align*} \psi(x)=\frac{\varphi(x)}{\varphi(1)} \end{align*} for $x\in M$. This map is positive and linear because it is a positive scalar multiple of $\varphi$, and it satisfies $\psi(1)=1$. Hence $\psi$ is a state. We check the projection-additivity hypothesis required by [citetheorem:9306]. Let $(p_j)_{j\in J}$ be a pairwise orthogonal family of projections in $M$, and put $p=\bigvee_{j\in J}p_j$. By condition 3, \begin{align*} \varphi(p)=\sum_{j\in J}\varphi(p_j), \end{align*} where the sum means the supremum over finite subsets of $J$. Dividing every finite partial sum by the positive scalar $\varphi(1)$ preserves suprema, so \begin{align*} \psi(p)=\frac{1}{\varphi(1)}\varphi(p)=\sum_{j\in J}\psi(p_j). \end{align*} Thus $\psi$ satisfies exactly the complete orthogonal additivity condition in [citetheorem:9306]. The theorem gives that $\psi$ is normal. Multiplying the identity \begin{align*} \psi(a)=\sup_{i\in I}\psi(a_i) \end{align*} for bounded increasing positive nets by $\varphi(1)$ gives the corresponding identity for $\varphi$, so $\varphi$ is normal. [/guided] [/step] [step:Conclude the equivalence] We have shown that condition 1 implies condition 2, condition 2 implies condition 3, and condition 3 implies condition 1. Therefore all three conditions are equivalent. [/step]