Let $H$ be a Hilbert space. For given operators $a$ and $b$ on $H$, how can we find all solutions of the equation $xb=a$? 



For given operators $a$ and $b$, solve the equation $xb=a$
I will assume that you want $x$ to be a bounded operator on $H$. Then, a necessary condition for such an $x$ to exist, is that there must be some constant $C>0$ satisfying $\Vert a(\eta)\Vert \leq C\Vert b(\eta)\Vert$ for all $\eta\in H$.


Conversely, suppose that $a,b\in B(H)$ and $C>0$ satisfy these conditions. Let $V\_0={\rm ran}(b)=bH$ (which need not be closed), and note that the restriction of $b$ to $\ker(b)^\perp$ is a linear bijection $b\_1: \ker(B)^\perp \to V\_0$.


Let $x\_0 = ab\_1^{-1}: V\_0 \to H$ (this is well-defined and linear by the previous remark), and note that $x\_0b =a$ as functions $H\to H$.


Given $\xi\in V\_0$ there exists $\eta\in\ker(B)^\perp$ such that $\xi=b\_1(\eta) = b(\eta)$, and so $x\_0(\xi)=a(\eta)$. Hence, by the 2nd condition, we have
$$\Vert x\_0(\xi)\Vert \leq C\Vert b(\eta) \Vert \leq C\Vert\xi\Vert $$
and so $x\_0$ is bounded linear. Therefore, if we let $V=\overline{V\_0}$, basic functional analysis tells us that $x\_0$ has a unique continuous extension to a bounded operator $x\_1: V \to H$.


Finally, we can define $x:H\to H$ to be $x\_1y$ for any idempotent $y\in B(H)$ satisfying $yH= V$. (For example, we could take the orthogonal projection from $H$ onto $V$, but there are other possibilities.) Then $xb=x\_1b=a$, as required; and all solutions arise this way.


