I'm having some trouble finishing this exercise:

Suppose $A : X \rightarrow X$ is a bounded linear operator on a Banach space $X$. Show that, if $A^*(X^*) = X^*$, then $A$ has a bounded inverse. (Bachman, Functional Analysis, Exercise 17.7)

(Here $X^*$ denotes de dual space of $X$, and $A^*$ the adjoint operator associated to $A$)

Now, given the fact that $X$ is Banach, we can use the bounded inverse theorem to show that if $A$ admits an inverse then it must be bounded. So the problem reduces to showing that $A$ is injective and surjective.

I have already shown that $A$ must be injective. But I am having problems showing that it must be surjective. If I can show that $A(X)$ is closed and $A^*$ is injective then I can prove the surjectivity of $A$, but I'm doubting that all of this has to be a consequence of just $A^*$ being surjective.

Does anyone have any hint?

Thanks in advance :)


It seems just false to me. Consider $A$ equal to right shift in $\ell^2$, that is, $(a_1,a_2,\dots)\mapsto(0,a_1,a_2,\dots)$. Its adjoint is the left shift in $\ell^2\cong(\ell^2)^*$ (identifying by the usual scalar multiplication) which is surjective.


Under the given restrictions, you can show that $A$ is injective and has a closed range, see https://en.wikipedia.org/wiki/Closed_range_theorem.

You cannot show surjectivity of $A$ since this would imply that $A^*$ is injective, and this may not be true.

In fact, any $A$ which is injective and has a closed range has a surjective adjoint.


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