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

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  • $\begingroup$ What type of answer are you looking for? $\endgroup$ – Luca Ghidelli Sep 11 at 20:56
  • $\begingroup$ Any type. Of course, all types will be perfect. $\endgroup$ – Ali Bagheri Sep 11 at 21:07
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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.

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