# Existence of operator with certain properties on a Banach space

I ran across this question, and was a little puzzled by it. I neither know how to solve it, nor its meaning:

Let $X$ be a Banach space, and let $A,B$ be bounded linear operators on $X$ such that $A$ is injective. Prove there is a bounded linear operator $C$ on $X$ with $B=AC$ if and only if $\rm{Range}(B) \subset \rm{Range} (A)$.

At the point in this book where the problem is presented, one has learned the open mapping theorem, closed graph theorem, bounded inverse theorem, and the Hanh-Banach theorem. I tried a few things with the OMT and got nowhere.

But what does this actually mean? Why would one care about this?

And how do you solve it?

If there exists $C$ with $B=AC$, then $\operatorname{Range}(B)=\operatorname{Range}(AC)=A(C(X))\subseteq A(X)=\operatorname{Range}(A)$.
For the converse, we use the closed graph theorem. Suppose $\operatorname{Range}(B)\subseteq\operatorname{Range}(A)$. Note that $C=A^{-1}B$ is well-defined as a linear operator, and it satisfies $B=AC$ (in fact, it is the only such option), o we simply need to prove that $A^{-1}B:X\to X$ is continuous, which from the closed graph theorem reduces to the following: Suppose $x_n\to 0$ and $A^{-1}Bx_n\to p$, for some $p\in X$, and we need to show that $p=0$. Indeed, $A^{-1}Bx_n\to p$ implies $Bx_n\to Ap$, since $A$ is continuous, but $x_n\to 0$ implies $Bx_n\to B0=0$, since $B$ is continuous, so $Ap=0$ and hence $p=0$ because $A$ is injective.
As for "why would one care": This is a generalization of the bounded inverse theorem (which should have its importance clear). Let's deduce it from this exercise: Suppose $A:X\to X$ is bijective and continuous. Take $B=\operatorname{Id}_X$. Then $\operatorname{Range}(B)\subseteq\operatorname{Range}(A)$, so there exists $C$ bounded (and linear) with $\operatorname{Id}_X=B=AC$. The only possibility is $C=A^{-1}$, which is therefore bounded.