Suppose $X$ and $Y$ are Banach spaces, and $T \in \mathcal{B}(X,Y)$ Then $$ \mathcal{N}(T^*) = \mathcal{R}(T)^{\perp} \;\;\text{and}\;\; \mathcal{N}(T) = ^{\perp}\mathcal{R}(T^*) $$
The corollary in the title
b) $\mathcal{R}(T)$ is dense in $Y$ iff $T^*$ is one-to-one
The proof is really short but there's a specific bit I don't get (fully at least)
$\mathcal{R}(T)$ is dense in $Y$ iff $\mathcal{R}(T)^{\perp} = \left\{ 0\right\}$
I've tried to prove this bit by exercise, and I think the most promosing was to show that
$$ ^{\perp}(\mathcal{R}(T)^{\perp}) = Y \iff \mathcal{R}(T) = \left\{0\right\} $$
The reason of the equality on the left of the double implication is because of theorem 4.7, which states that the orthogonal of the orthogonal of a set is the norm closure of a set.
However I'm not entirely sure how to exploit this, can you help?
Here is another approach:
Suppose $R(T)$ is dense. Any linear functional that vanishes on $R(T)$ must vanish on the entire space by continuity considerations. That is, $R(T)^\perp$ consists only of the zero functional.
Conversely suppose $R(T)$ is not dense. Then $\overline{R(T)}$ is a proper closed subspace of $Y$ and there exists (per e.g. Hahn-Banach) a nontrivial linear function $\phi$ that vanishes on $\overline{R(T)}$, meaning $R(T)^\perp \not= \{0\}$.