Let $x$ be an operator in $B(H)$. We say a pair $(c,y)$ forms a polar decomposition for $x$ if $y$ is a positive operator, $c$ in $B(H)$ with $x=cy$ such that the restriction of $c$ on $\overline{yH}$ is one to-one. We also say $y$ is a generalized absolute value for $x$.
Does there exists a non-trivial generalized absolute value for $x$? (I mean a positive operator except $|x|$).
Remark. In finite dimensional, the answer is yes. Indeed, any positive operator $y$ with $yH=x^*H$ is a generalized absolute value! So it is natural to ask in infinite dimensional case.
Added points: It is well-known that for a given operator $x\in B(H)$, there exists unique partial isometry $u$ given by $x=u|x|$. Martin proved the pair $(u,|x|)$ is unique in this sense: assume $v$ is a partial isometry and $y$ is a positive operator with $x=vy$ such that $Ker(x)=Ker(y)=Ker(v)$ then, $v=u$ and $y=|x|$ (enter link description here). This problem asks a generalization. Indeed we are looking for a pair $(c,y)$ where $y$ is a positive operator with $x=cy$ where the restriction of $c$ on $\overline{yH}$is still one to one (which implies that $Ker(x)=Ker(y)=Ker(v)$). As you observe, $c$ is one to one but may no longer isometry. With this weaker condition, do we have a non-trivial triple $(x,c,y)$?