Operator $A\ \text{bounded from below}\iff A^*\ \text{surjective}$ Let $H$ be a Hilbert space, $A\in H'$ a bounded operator and $A^*$ its adjoint. $A$ is called bounded from below if there exists a $b>0$, such that $b ||h||\le ||Ah||, \forall h\in H$. Then how can one establish that
$$A\ \text{bounded from below}\iff A^*\ \text{surjective}$$
The hint given is to use closed range theorem that $Ran(A)$ is closed iff $Ran(A^*)$ is closed. But I am not sure how to connect closedness and surjectivity. Any help will be appreciated.
 A: I think you can do the following:
Since $A$ is bounded below and $H$ is complete, you have that $Ran(A)$ is closed. Which means that $Ran(A)$ is a Hilbert space.
Again by being bounded below, it follows that $A_0\colon H\to Ran(A)$, the restriction of $A$ into it's range, is a linear isomorphism (bicontinious as well).
Therefore you have that $A_0^* \colon Ran(A)' \to H'$  is also a linear isomorphism of $Ran(A)' = Ran(A)$ into $H'=H$.
And I think you can prove that $A_0^*$ it is only $A^*$ restricted to $Ran(A)$. Therefore, $A^*$, may no longer be injective, but it is surely surjective.
For the other way around, I think you can do something similar.
If $A^*$ is surjective then define $H_0\doteq \ker (A^*)^\perp $, which is a Hilbert space, and $A^*_0\colon H_0\to H$, which is a linear isomorphism since $A^*$ is surjective.
Then $(A_0^*)^*\colon H'\to H_0'$ is a linear isomorphism (bicontinous as always), therefore it is bounded below.
Once again, since $H$ and $H_0$ are Hilbert spaces,it follows that $H'=H$ and $H_0' = H_0\subset H$.
And I think you can prove that $(A^*_0)^*$ coincides with $A^{**} = A$. And you're done.
