Existence of bounded linear operator $C$ I'm trying to find the solution of the following problem.
Let $X$ be a Banach space and $A,B \in \mathbb{B}(X)$ and $A$ is one to one. Show that there is $C \in \mathbb{B}(X)$ such that $B=AC$ if and only if $Range\, B\subset Range\, A$. 
I'm struggling with proving reverse inclusion.
 A: Clearly if $B = AC$ for some bounded operator $C$, then $B(X) \subseteq A(X)$.
Now, conversely, suppose $B(X) \subseteq A(X).$ Then since $A$ is injective, as a map of sets, it has a left inverse $A^{-1}: A(X) \rightarrow X$. The function $C = A^{-1}B:X \rightarrow X$ will be defined on all of $X$ since $B(X) \subseteq A(x)$. It will also be linear, since if $x,y \in X$, and $x',y' \in X$ so that $Ax' = Bx$ and $Ay' = By$, then 
$$
C(x+y) = A^{-1}(Bx + By) = x' + y' = Cx + Cy.
$$
A similar argument shows that $C$ will also respect scalar multiplication.
To show that $C$ is bounded, we use the closed graph theorem. Suppose $\{x_n\}$ is a sequence in $X$ so that $\{ (x_n, Cx_n )\}$ converges in $X \times X$. Then $\{x_n \}$ converges to some $x$. Also $\{Cx_n\}$ converges to some $x'$. Then $ACx_n = Bx_n.$ Since $B$ is bounded, and so continuous $\{Bx_n\}$ converges to $Bx$. So $\{ACx_n\}$ converges to $Bx$, but also converges to $Ax'$. So $Ax' = Bx$, and $Cx = x'$. Therefore $(x_n, Cx_n ) \rightarrow (x,Cx)$, so the graph of $C$ is closed in $X \times X$, and by the closed graph theorem it must be bounded and continuous.
