[proofplan] We define the null projection $q$ as the supremum of all projections annihilated by $\varphi$, and set $s=1-q$. The main point is to prove that $q$ is itself annihilated by $\varphi$: finite joins of null projections remain null by the [Cauchy-Schwarz inequality](/theorems/432) for positive functionals, and normality then passes from finite joins to the supremum. Once $q$ is known to be null, the minimality of $s$ follows directly from the order definition. The final kernel identity is obtained by proving that $\varphi$ only sees the support corner $sMs$ and that the restriction of $\varphi$ to this corner is faithful. [/proofplan]

[step:Construct the largest projection annihilated by $\varphi$] Let \begin{align*} \mathcal N=\{e\in\mathcal P(M):\varphi(e)=0\} \end{align*} be the set of projections annihilated by $\varphi$. By the complete projection lattice property for von Neumann algebras, [citetheorem:9267], the projection \begin{align*} q=\bigvee_{e\in\mathcal N}e \end{align*} exists in $M$. By definition of the support projection, \begin{align*} s(\varphi)=1-q. \end{align*} For the rest of the proof, write \begin{align*} s=s(\varphi). \end{align*} We first record the Cauchy-Schwarz consequence used repeatedly. If $e\in\mathcal N$ and $a\in M$, then the [Cauchy-Schwarz inequality for positive linear functionals](/theorems/8558), applied to the pair $e,a\in M$, gives \begin{align*} |\varphi(ea)|^2\le \varphi(e^*e)\varphi(a^*a)=\varphi(e)\varphi(a^*a)=0. \end{align*} Thus $\varphi(ea)=0$. Applying this to $a^*$ and using positivity, hence $\varphi(y^*)=\overline{\varphi(y)}$ for all $y\in M$, also gives \begin{align*} \varphi(ae)=\overline{\varphi(ea^*)}=0. \end{align*} [/step]

[step:Show finite joins of null projections are null]Let $e,f\in\mathcal N$, and define the positive operator \begin{align*} a=e+f. \end{align*} Let $q_{e,f}=e\vee f\in\mathcal P(M)$. Since $M\subseteq\mathcal L(H)$, the kernel of $a$ is $eH^\perp\cap fH^\perp$, so the support projection of $a$ is exactly $q_{e,f}$. For each integer $n\ge 1$, let $h_n\in\mathcal P(M)$ be the spectral projection of $a$ for the Borel set $[1/n,\infty)$. Then $h_n\le n a$ in the operator order, and positivity of $\varphi$ gives \begin{align*} 0\le\varphi(h_n)\le n\varphi(a)=n\varphi(e)+n\varphi(f)=0. \end{align*} Thus $\varphi(h_n)=0$ for every $n\ge 1$. The projections $(h_n)_{n\ge 1}$ increase strongly to the support projection $q_{e,f}$. Since $\varphi$ is positive and normal, normal positive functionals preserve increasing bounded suprema of projections, as in [citetheorem:9263]. Hence \begin{align*} \varphi(e\vee f)=\varphi(q_{e,f})=\sup_{n\ge 1}\varphi(h_n)=0. \end{align*} By induction, every finite join of projections in $\mathcal N$ again belongs to $\mathcal N$.[/step]

[guided]The point is to replace the false algebraic decomposition of $e\vee f$ by a spectral approximation from below. Let $e,f\in\mathcal N$, and define \begin{align*} a=e+f. \end{align*} This is a positive operator in $M$, because $e$ and $f$ are positive projections. Let \begin{align*} q_{e,f}=e\vee f \end{align*} be their join in the projection lattice of $M$. In the representation $M\subseteq\mathcal L(H)$, a vector $\xi\in H$ satisfies $a\xi=0$ if and only if \begin{align*} (e\xi,\xi)_H+(f\xi,\xi)_H=0. \end{align*} Both summands are nonnegative, so this is equivalent to $e\xi=0$ and $f\xi=0$. Thus \begin{align*} \ker a=eH^\perp\cap fH^\perp. \end{align*} Therefore the orthogonal complement of $\ker a$ is the closed linear span of $eH+fH$, and the support projection of $a$ is precisely $q_{e,f}=e\vee f$. For each integer $n\ge 1$, let $h_n\in\mathcal P(M)$ be the spectral projection of $a$ for the Borel set $[1/n,\infty)$. Functional calculus gives the operator inequality \begin{align*} h_n\le n a, \end{align*} because the scalar inequality $\mathbb 1_{[1/n,\infty)}(t)\le nt$ holds for every $t\ge 0$. Applying the positive functional $\varphi$ preserves order, so \begin{align*} 0\le\varphi(h_n)\le n\varphi(a)=n\varphi(e)+n\varphi(f)=0. \end{align*} Hence $\varphi(h_n)=0$ for every $n\ge 1$. The projections $h_n$ increase strongly to the support projection of $a$, which we identified as $q_{e,f}$. Since $\varphi$ is positive and normal, [citetheorem:9263] applies to this increasing bounded net of projections and yields \begin{align*} \varphi(q_{e,f})=\sup_{n\ge 1}\varphi(h_n)=0. \end{align*} Thus \begin{align*} \varphi(e\vee f)=0. \end{align*} Repeating the two-projection argument finitely many times proves that every finite join of elements of $\mathcal N$ belongs to $\mathcal N$.[/guided]

[step:Use normality to show the supremum projection is null] Let $\mathcal F$ be the directed set of finite subsets of $\mathcal N$, ordered by inclusion. For each $F\in\mathcal F$, define \begin{align*} q_F=\bigvee_{e\in F}e, \end{align*} with $q_\varnothing=0$. By the previous step, $q_F\in\mathcal N$ for every $F\in\mathcal F$, so \begin{align*} \varphi(q_F)=0. \end{align*} The net $(q_F)_{F\in\mathcal F}$ is increasing and has supremum $q$. Since $\varphi$ is positive and normal, normal positive functionals preserve increasing bounded suprema of projections, as in [citetheorem:9263]. Hence \begin{align*} \varphi(q)=\sup_{F\in\mathcal F}\varphi(q_F)=0. \end{align*} Since $1-s=q$, this proves \begin{align*} \varphi(1-s)=0. \end{align*} [/step]

[step:Deduce the minimality property of the support projection] Let $p\in\mathcal P(M)$ satisfy $\varphi(1-p)=0$. Then $1-p\in\mathcal N$, so by the definition of $q$ as the supremum of all projections in $\mathcal N$, \begin{align*} 1-p\le q. \end{align*} Taking complements in the projection order gives \begin{align*} s=1-q\le p. \end{align*} This proves that $s(\varphi)$ is the least projection $p\in\mathcal P(M)$ such that $\varphi(1-p)=0$. [/step]

[step:Show that $\varphi$ only depends on the support corner] We prove that \begin{align*} \varphi(a)=\varphi(sas) \end{align*} for every $a\in M$. Since $1-s=q$ and $\varphi(q)=0$, the Cauchy-Schwarz consequence from the first step gives \begin{align*} \varphi((1-s)a)=0 \end{align*} and \begin{align*} \varphi(a(1-s))=0 \end{align*} for every $a\in M$. Decompose \begin{align*} a=sas+s a(1-s)+(1-s)as+(1-s)a(1-s). \end{align*} Applying $\varphi$ and using the two vanishing identities gives \begin{align*} \varphi(a)=\varphi(sas). \end{align*} [/step]

[step:Prove faithfulness on the support corner] We claim that the restricted positive functional \begin{align*} \varphi_s:sMs\to\mathbb C \end{align*} defined by $\varphi_s(b)=\varphi(b)$ for $b\in sMs$ is faithful on the von Neumann algebra $sMs$. Let $b\in sMs$ be positive and suppose $\varphi_s(b)=0$. For each integer $n\ge 1$, let $e_n\in\mathcal P(sMs)$ denote the spectral projection of $b$ for the Borel set $[1/n,\infty)$. Then \begin{align*} \frac{1}{n}e_n\le b. \end{align*} By positivity of $\varphi$, \begin{align*} 0\le \frac{1}{n}\varphi(e_n)\le \varphi(b)=0. \end{align*} Thus $\varphi(e_n)=0$, so $e_n\in\mathcal N$. Since $e_n\le s$ and $e_n\le q$ by the definition of $q$, we have \begin{align*} e_n\le s\wedge q=0. \end{align*} Therefore $e_n=0$ for every $n\ge 1$. The spectral projections of $b$ on $[1/n,\infty)$ all vanish, hence the spectrum of the positive operator $b$ is contained in $\{0\}$, and so $b=0$. This proves faithfulness of $\varphi_s$. [/step]

[step:Identify the quadratic kernel with the right support kernel] Let $x\in M$. First suppose \begin{align*} \varphi(x^*x)=0. \end{align*} Using the support-corner identity with $a=x^*x$, we obtain \begin{align*} 0=\varphi(x^*x)=\varphi(sx^*xs)=\varphi((xs)^*(xs)). \end{align*} The element $(xs)^*(xs)$ belongs to the positive cone of $sMs$. By faithfulness of $\varphi_s$ on $sMs$, \begin{align*} (xs)^*(xs)=0. \end{align*} In a $C^*$-algebra, $y^*y=0$ implies $y=0$; applying this to $y=xs$ gives \begin{align*} xs=0. \end{align*} Conversely, suppose $xs=0$. Since $1=s+(1-s)$, we have \begin{align*} x=x(1-s). \end{align*} Therefore \begin{align*} x^*x=(1-s)x^*x(1-s). \end{align*} Applying the support-corner identity to $a=x^*x$ gives \begin{align*} \varphi(x^*x)=\varphi(sx^*xs). \end{align*} But $xs=0$, so \begin{align*} sx^*xs=(xs)^*(xs)=0. \end{align*} Hence \begin{align*} \varphi(x^*x)=0. \end{align*} This proves \begin{align*} \varphi(x^*x)=0 \quad \Longleftrightarrow \quad xs(\varphi)=0, \end{align*} and completes the proof. [/step]