# Map bounded if composition is bounded

Let $X,Y,Z$ Banach spaces and $A:X\rightarrow Y$ and $B:Y\rightarrow Z$ linear maps with $B$ bounded and injective and $BA$ bounded. Prove that $A$ is bounded as well.

If I knew that $B(Y)$ is closed I’d have a bounded linear map $B^{-1}:B(Y)\rightarrow Y$ by the bounded inverse theorem. Therefore $A=B^{-1}BA$ is bounded. How to prove the claim if $B(Y)$ is not closed.

#### Solutions Collecting From Web of "Map bounded if composition is bounded"

This is extended version of Nate Eldredge’s hint:

• Take $\{x_n\}$ such that $\lim_{n\to\infty}x_n= x$,$\lim_{n\to\infty}A(x_n)= y$.
• Show that $\lim_{n\to\infty}B(A(x_n))= B(A(x))$.
• Recall that $B$ is injective.
• Apply closed graph theorem.