Behavior of projection operators in Hilbert space

Question

I had a question about the behavior of projection operators in Hilbert space. Let $H$ be an infinite dimensional Hilbert space and $W$ a closed subspace, such that $P : H \rightarrow W$ denotes the projection operator onto $W$. Let $x, y \in H$. Then under what conditions on $H, W$ and $x,y$ do we have that $|\langle{Px, {Py \rangle}}| \leq C |\langle{x, {y \rangle}}|$, where $C$ is independent of $x$ and $y$. I understand this fails in general, but are there some nice conditions under which this will work for $x, y$ in some subspace? Thanks for the help!

Answer 1

Such $C$ cannot exist for any proper closed subspace $W$.

Let $u\in W$ with $\|u\|=1$ and $w\in W^\perp$ with $\|w\|=1$. Put $$ x=u+w,\qquad\qquad y=u-w. $$ Then $Px=Py=u$, so $\langle Px,Py\rangle=\|u\|^2=1$. And $$ \langle x,y\rangle=\langle u+w,u-w\rangle=\|u\|^2-\|w\|^2-2\operatorname{Re}\langle u,w\rangle=0 $$

Answer 2

An operator $A$ such that $$|\langle Ax,Ay\rangle |\le C |\langle x,y\rangle |$$ for any $x,y,$ is of the form $A=\lambda U,$ where $U$ is a unitary operator. In particular $A$ cannot be a nontrivial projection. Indeed, the assumption implies $$|\langle A^*Ax,y\rangle |\le C |\langle x,y\rangle |$$ Fix $x.$ For any $y$ if $y\perp x$ then $y\perp A^*Ax.$ Hence $A^*Ax=a_xx,$ for a nonnegative coefficient $a_x,$ i.e. every vector $x$ is an eigenvector of $A^*A.$ It is possible only if $A^*A=aI$ for some $a\ge 0.$ If $a=0,$ then $A=0.$ If $a>0$ Then $U:=a^{-1/2}A$ is a unitary operator.