# 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.

• @DanielFischer Yes, I just noticed that. I do not see, however, how to get $C$ bounded without $A$ being bounded below, though. – BigMathTimes Oct 15 '15 at 23:44
• If such a $C$ exists, then, since $A$ is injective, we have $C = A^{-1}AC = A^{-1}B$, where $A^{-1}$ is a well-defined linear (not necessarily continuous) operator on $\operatorname{Range} B$. So you need to prove that $A^{-1}B$ is in fact continuous. What famous theorem may you want to throw at it? – Daniel Fischer Oct 15 '15 at 23:47
• @DanielFischer Ah yes of course. Been a spell since I've done much functional analysis. I had forgotten to use such heavy hammers rather than first principles. – BigMathTimes Oct 16 '15 at 1:40

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.