[proofplan]
We use the spectral resolution of the positive operator $a$ to identify the support projection as the spectral projection away from $0$. This immediately gives $s(a)a=a=as(a)$ and the order estimate $a\leq \|a\|_{\mathcal{L}(H)}s(a)$ by functional calculus. Minimality for the multiplicative characterization follows because any projection $p$ satisfying $pa=a$ contains the closure of $aH$. Minimality for the order characterization follows by compressing the inequality to $(1-p)H$ and using positivity to force $(1-p)H\subseteq \ker a$.
[/proofplan]
[step:Represent the support projection using the spectral measure of $a$]
Let $c:=\|a\|_{\mathcal{L}(H)}$. Since $a\in M$ is positive, the spectral theorem for positive operators in a von Neumann algebra gives a projection-valued spectral measure $E_a$ with spectral projections in $M$ such that
\begin{align*}
a=\int_{[0,c]}\lambda\,dE_a(\lambda).
\end{align*}
By the definition of the support projection of a positive operator,
\begin{align*}
s(a)=E_a((0,c])=1-E_a(\{0\}).
\end{align*}
The projection $E_a(\{0\})$ is the [orthogonal projection](/theorems/437) onto $\ker a$, so $s(a)$ is the orthogonal projection onto $(\ker a)^\perp$. Since $a$ is self-adjoint, $(\ker a)^\perp=\overline{\operatorname{Range}(a)}$. Hence $s(a)$ is also the orthogonal projection onto $\overline{aH}$.
[/step]
[step:Show that $s(a)$ leaves $a$ unchanged on both sides]
Using spectral calculus and the identity $s(a)=E_a((0,c])$, we have
\begin{align*}
s(a)a=\int_{[0,c]}\mathbb{1}_{(0,c]}(\lambda)\lambda\,dE_a(\lambda).
\end{align*}
Since $\mathbb{1}_{(0,c]}(\lambda)\lambda=\lambda$ for every $\lambda\in[0,c]$, this gives
\begin{align*}
s(a)a=a.
\end{align*}
The same spectral-calculus computation gives
\begin{align*}
as(a)=a.
\end{align*}
Thus $s(a)$ is a projection in $M$ satisfying $s(a)a=a=as(a)$.
[guided]
The point of introducing the spectral measure is that multiplication by $s(a)$ is multiplication by the indicator function of the nonzero part of the spectrum. Let $c:=\|a\|_{\mathcal{L}(H)}$. The spectral representation is
\begin{align*}
a=\int_{[0,c]}\lambda\,dE_a(\lambda),
\end{align*}
and the support projection is
\begin{align*}
s(a)=E_a((0,c]).
\end{align*}
Therefore spectral calculus gives
\begin{align*}
s(a)a=\int_{[0,c]}\mathbb{1}_{(0,c]}(\lambda)\lambda\,dE_a(\lambda).
\end{align*}
The scalar function $\lambda\mapsto \mathbb{1}_{(0,c]}(\lambda)\lambda$ agrees with $\lambda\mapsto \lambda$ on all of $[0,c]$, including at $\lambda=0$. Hence
\begin{align*}
s(a)a=a.
\end{align*}
Because $s(a)$ is a spectral projection of $a$, it commutes with $a$, and the same calculation gives
\begin{align*}
as(a)=a.
\end{align*}
So the support projection does not remove any part of $a$; it removes only the kernel part, where $a$ is already zero.
[/guided]
[/step]
[step:Prove minimality among projections that multiply $a$ as an identity]
Let $p\in M$ be a projection satisfying $pa=a$. Since $p=p^*$ and $a=a^*$, taking adjoints gives
\begin{align*}
ap=a.
\end{align*}
From $pa=a$ we get
\begin{align*}
(1-p)a=0.
\end{align*}
Thus $aH\subseteq pH$. Since $pH$ is closed, this implies
\begin{align*}
\overline{aH}\subseteq pH.
\end{align*}
The projection $s(a)$ is the orthogonal projection onto $\overline{aH}$, while $p$ is the orthogonal projection onto $pH$. Therefore $s(a)\leq p$. Hence $s(a)$ is the smallest projection $p\in M$ such that $pa=a=ap$.
[/step]
[step:Derive the order estimate for the support projection]
If $c=0$, then $a=0$ and $s(a)=0$, so
\begin{align*}
a\leq c\,s(a).
\end{align*}
Assume now that $c>0$. By spectral calculus,
\begin{align*}
c\,s(a)-a=\int_{[0,c]}\bigl(c\mathbb{1}_{(0,c]}(\lambda)-\lambda\bigr)\,dE_a(\lambda).
\end{align*}
The scalar function $\lambda\mapsto c\mathbb{1}_{(0,c]}(\lambda)-\lambda$ is nonnegative on $[0,c]$: it is $0$ at $\lambda=0$, and it is $c-\lambda\geq 0$ for $\lambda\in(0,c]$. Hence $c\,s(a)-a$ is positive, which is exactly
\begin{align*}
a\leq \|a\|_{\mathcal{L}(H)}s(a).
\end{align*}
[/step]
[step:Prove minimality among projections satisfying the order bound]
Let $p\in M$ be a projection satisfying
\begin{align*}
a\leq c\,p.
\end{align*}
If $c=0$, then $a=0$, hence $s(a)=0\leq p$. Suppose $c>0$, and define $q:=1-p$. Then $q\in M$ is a projection, and compressing the inequality by $q$ gives
\begin{align*}
qaq\leq c\,q p q=0.
\end{align*}
Since $a$ is positive, $qaq$ is positive. Thus $qaq=0$.
For every $\xi\in qH$, we have $q\xi=\xi$, so
\begin{align*}
(a\xi,\xi)_H=(qaq\xi,\xi)_H=0.
\end{align*}
Let $a^{1/2}\in M$ denote the positive square root of $a$, obtained by functional calculus. Then
\begin{align*}
0=(a\xi,\xi)_H=(a^{1/2}\xi,a^{1/2}\xi)_H=\|a^{1/2}\xi\|_H^2.
\end{align*}
Hence $a^{1/2}\xi=0$, and therefore $a\xi=a^{1/2}a^{1/2}\xi=0$. Thus $qH\subseteq \ker a$.
Since $1-s(a)$ is the orthogonal projection onto $\ker a$, the inclusion $qH\subseteq \ker a$ gives $q\leq 1-s(a)$. Equivalently, $s(a)\leq p$. Therefore $s(a)$ is the smallest projection $p\in M$ satisfying
\begin{align*}
a\leq \|a\|_{\mathcal{L}(H)}p.
\end{align*}
This completes both characterizations of the support projection.
[/step]
