Suppose we have a projection $p$ on a Hilbert space $\mathfrak{H}$. Is the following true: There exists an set $V\subset\mathfrak{H}$ such that $p(x)=x$ if $x\in V$ and zero else? I asked because I want to prove the following: if $p$ and $q$ are two projections and $p\mathfrak{H}\subseteq q\mathfrak{H}$, then $p\leq q$. One observes that we must say something about $||qx||^2-||px||^2$ and we want this to be positive. If one has what I say above, then we are done. Does one has another idea how to prove this?#
Thanks a lot.
I guess that with projection you mean an orthogonal projection: $p^2=p=p^*$. For such a projection the image $im\, p$ and the kernel $ker \, p$ are closed subspaces of $H$. One has $H=im\, p\oplus ker\, p$, an othogonal sum, which means that each vector $x\in H$ can be presented uniquely as $x=y+z$, with $y\in im\, p$ and $z\in ker\, p$, i.e. $y\perp z$. Note that $py=y$ and $pz=0$. Thus $px=y$, which is not necessary $0$.
Assume that $im\, p \subseteq im\, q$. We would like to prove that $p\leq q$, i.e., $\langle (q-p)x,x\rangle \geq 0$ for any $x\in H$. Note first the following: (i) since $p$ is an orthogonal projection one has $|\langle px,x\rangle| \leq \| p\|\| x\|^2\leq \| x\|^2$ $(\forall x\in H)$, and (ii) since $im\, p\subseteq im\, q$ one has $ker\, p\supseteq ker\, q$.
Let $x\in H$ be arbitary and $x=y+z$ the decompostion with respect to $H=im\, q \oplus ker\, q$. Then $$ \langle (q-p)x,x\rangle=\langle qy,y\rangle+\langle qz,y\rangle+\langle qy,z\rangle+\langle qz,z\rangle -(\langle py,y\rangle+\langle pz,y\rangle+\langle py,z\rangle+\langle pz,z\rangle)= $$ $$=\langle qy,y\rangle-\langle py,y\rangle\geq \| y\|^2-|\langle py,y\rangle|\geq \| y\|^2-\| y\|^2=0.$$