Characterization of Canonical Preordering of Projections on Hilbert Space

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\map B \HH$ be the space of bounded linear operators on $\HH$ understood as a $\text C^\ast$-algebra.
Let $\le_{\map B \HH}$ be the canonical preordering of $\map B \HH$.
Let $P, Q$ be Hilbert space projections on $\HH$.
:$(1) \quad$ $P \le_{\map B \HH} Q$
:$(2) \quad$ $P Q = P$
:$(3) \quad$ $Q P = P$
:$(4) \quad$ $\Img P \subseteq \Img Q$
:$(5) \quad$ for all $x \in \HH$ we have $\norm {P x} \le \norm {Q x}$
:$(6) \quad$ $Q - P$ is a Hilbert space projection
== Proof ==
We first note that from Characterization of Projections, $P$ is Hermitian, and that $P, Q$ is are orthogonal projections onto $\Img P$ and $\Img Q$ respectively.
=== $(2)$ is equivalent to $(3)$ ===
From Adjoint is Involutive:
:$P Q = P$
is equivalent to:
:$\paren {P Q}^\ast = P^\ast = P$ 
From Adjoint of Composition of Linear Transformations is Composition of Adjoints, we have:
:$\paren {P Q}^\ast = Q P$
So:
:$P Q = P$ $Q P = P$ 
=== $(3)$ implies $(4)$ ===
Suppose that:
:$Q P = P$
Then for $x \in \Img P$, we have $P x = x$ from Fixed Points of Orthogonal Projection.
Then $\map {Q P} x = Q x = x$. 
So $x \in \Img Q$. 
Hence we get $\Img P \subseteq \Img Q$. 
=== $(4)$ implies $(3)$ ===
Suppose that:
:$\Img P \subseteq \Img Q$
Then for $x \in \HH$, we have $P x \in \Img Q$.
From Fixed Points of Orthogonal Projection, we then have $Q P x = P x$. 
So $Q P = P$. 
=== $(2)$ implies $(6)$ ===
Suppose that: 
:$P Q = P$ 
Since $(2)$ implies $(3)$, we have $Q P = P$
Then we have:
:$\paren {Q - P}^2 = \paren {Q - P} \paren {Q - P} = Q^2 - PQ - QP + P^2$
Since $Q$ and $P$ are idempotent, we have $Q^2 = Q$ and $P^2 = P$.
Since $P Q = P$ and $Q P = P$, we have $-PQ - QP + P = P$. 
So we have:
:$\paren {Q - P}^2 = Q - P$
Further:
:$\paren {Q - P}^\ast = Q - P$ 
from Adjoining is Linear. 
So from Characterization of Projections, we have that $Q - P$ is a Hilbert space projection.
=== $(6)$ implies $(1)$ ===
Suppose that $Q - P$ is a Hilbert space projection.
From Hilbert Space Projection is *-Algebra Projection, $Q - P$ is a $\ast$-algebra projection.
From Spectrum of Projection in *-Algebra, $\map \sigma {Q - P} \subseteq \set {0, 1}$. 
So $Q - P$ is positive.
We hence obtain $P \le_{\map B \HH} Q$.
=== $(1)$ implies $(5)$ ===
Suppose that $P \le_{\map B \HH} Q$. 
Then $Q - P$ is positive .
From Existence and Uniqueness of Positive Nth Root of Positive Element of C*-Algebra, there exists a positive linear transformation $S : \HH \to \HH$ such that:
:$Q - P = S^2$ 
We then have for $x \in \HH$:
So we have:
:$\norm {P x}^2 \le \norm {Q x}^2$
giving:
:$\norm {P x} \le \norm {Q x}$ 
=== $(5)$ implies $(2)$ ===
Suppose that:
:$\norm {P x} \le \norm {Q x}$ for each $x \in \HH$. 
We then have, for each $x \in \HH$:
:$\norm {P \paren {1 - Q} x} \le \norm {Q \paren {1 - Q} x} = \norm {\paren {Q^2 - Q} x} = 0$
since $Q^2 = Q$.
So we have from :
:$P \paren {1 - Q} x = {\mathbf 0}_\HH$
so:
:$P x = P Q x$ for each $x \in \HH$.
So we can conclude $P = P Q$.
== Sources ==
* : $2.1$: $C^\ast$-Algebras
Category:Canonical Preorderings of C*-Algebras