Let $X$ be a Banach space and let $A: X \rightarrow X $ be a bounded linear operator such that $A'(\tilde{X})=\tilde{X}$, show that $A$ has a bounded inverse (on its range).
If someone could proof that the range of $A$ is closed it end the problem, because after that we can just apply the open map theorem.
PS. $A'$ is the adjoint of $A$ and $\tilde{X}$ is the topological dual of $X$. This problem is from Bachman book's Functional Analysis.
Let $Y$ be the closure of the range of $A$. We define $B : X \to Y$ by $B x = A x$ for all $x \in X$.
Let us show that $B' : \tilde Y \to \tilde X$ is invertible. Since the range of $B$ is dense, $B'$ is injective. It remains to show that $B$ is surjective. For any $\tilde x \in \tilde X$, there is $\tilde r \in \tilde X$, such that $A'\tilde r = \tilde x$. Now, let $\tilde y \in Y'$ be the restriction of $\tilde r$ to $Y$. Then, $$ \langle \tilde x, x\rangle = \langle A' \tilde r, x\rangle = \langle \tilde r, Ax\rangle = \langle \tilde y, Bx\rangle = \langle B'\tilde y, x\rangle $$ for all $x \in X$. Hence, $B'\tilde y =\tilde x$ and $B'$ is surjective.
This shows that $B'$ is boundedly invertible by the open mapping theorem. Moreover, the bounded invertibility of $B$ follows.