Let consider the following exercise:
Let $V$ and $W$ be Banach spaces and $T\colon V \longrightarrow W$ be a closed linear operator. Then, the following assumptions are equivalent:
(i) R$(T)=W$;
(ii) $T^{\ast}$ is lower bounded, i.e., there exists $r>0$ such that $|\!| T^\ast (f) |\!| \geq r |\!| f|\!|$, for all $f \in W'$;
(iii) $T$ is an open map.
$T^\ast$ denotes the adjoint operator of $T$.
$W^\prime$ denotes the set of bounded linear functionals in $W$.
I have some questions:
(1) The best way to solve this exercise is (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(i)?
(2) My solution to (i)$\Rightarrow$(ii) is ok?
Suppose that $R(T)=W$, i.e., $T$ is surjective. For any $f\in W^\prime$, we have that $$\|f\|=\sup_{w\in W;\|w\|=1}|\langle f,w\rangle_{W^\prime,W}|.$$ Since $T$ is surjective, for any $w\in W$, there exists $v\in V$ such that $w=T(v)$, hence $$\|f\|=\sup_{v\in V;\|T(v)\|=1}|\langle f,T(v)\rangle_{W^\prime,W}|=\sup_{v\in V;\|T(v)\|=1}|\langle T^\ast(f),v\rangle_{V^\prime,V}|\leqslant\sup_{v\in V;\|T(v)\|=1}\|T^\ast(f)\|\|v\|$$ $$\Longrightarrow\quad \|f\|\leqslant\|T^\ast(f)\|\left(\sup_{v\in V;\|T(v)\|=1}\|v\|\right).$$ Then, we conclude that, there exists $$r:=\frac{1}{\sup_{v\in V;\|T(v)\|=1}\|v\|}$$ such that $\|T^\ast(f)\|\geqslant r\|f\|$, for all $f\in W^\prime$.
(3) I'm working for a lot of hours on (ii)$\Rightarrow$(iii) and I can't see how conclude.