Im trying to work through some prelim prep problems and I cant seem to get this one, can anyone post a solution that I can walk through?
Suppose $X$ is a Banach space, $Y$ a normed vector space, $A$ and $B$ are bounded linear operators from $X$ to $Y$, and $A$ is one-to-one.
Prove: There exists a bounded linear operator $C: X\to X$ such that $B=AC\iff B(X)\subset A(X).$
The forward implication is trivial so let me only comment on the backwards one. Since $A$ is injective, there exists a linear map $A^{-1}:A(X) \to X$ that is inverse to $A$. By our assumption the operator $C = A^{-1}B:X \to X$ is well-defined. However, we do not know that $A^{-1}$ is bounded so we cannot immediately conclude $C$ is bounded.
Since $C$ is a map between Banach spaces, by the closed graph theorem, to see that $C$ is bounded it will suffice to prove that if $x_n \to x$ in $X$ and $Cx_n \to y$ in $X$ then $Cx= y$. This isn't too difficult.
Indeed, since $A$ is bounded and $Cx_n \to y$, $Bx_n = ACx_n \to Ay$ as $n \to \infty$. Also, since $B$ is bounded and $x_n \to x$, we know that $Bx_n \to Bx$ as $n \to \infty$. Hence $Ay = Bx$ which implies that $y = A^{-1}Bx = Cx$ as desired.