Question about perpendicular complements in Banach spaces Let $V,W$ be Banach spaces with $T : V \to W$ a bounded linear transformation. Let $T^* : W^* \to V^*$ be the standard adjoint map on the dual spaces. That is, for $g \in W^*$, we have $[T^*g](v) = g(T(v))$. 
Let $R, R^*, N, N^*$ denote the ranges and kernels of $T$ and $T^*$, respectively. Define:
$$N^{* \perp} = \{w \in W | g(w)=0 \text{ for all } g\in N^*\},$$
$$N^{ \perp} = \{f \in V^* | f(v)=0 \text{ for all } v\in N\},$$ 

Assume that $R$ is a closed subspace of $W$ Prove that:
  $$R = N^{* \perp},$$
  $$R^* = N^{\perp}.$$

I have been able to show $\subseteq$ for both cases. For example, if $W \ni y = T(x)$ for some $x \in V$, and we take $g \in N^*$, then of course we have $g(y) = g(T(x)) = [T^*g](x) = 0$ (since $g$ is in the kernel of $T^*$).
The trickier part of the proof is showing $R \supseteq N^{* \perp}$ and $R^* \supseteq N^{ \perp}$, and I know this is where I will need to bring in the fact that $R$ is closed (since I didn't need to use that fact for the $\subseteq$ containments). 
If I start with $w \in N^{*\perp}$ for instance, my hope would be to show that we can produce a sequence $\{w_n\}_{n=1}^\infty \subseteq R$ converging to $w$. By the closedness of $R$, I could then conclude that $w \in R$. But I still don't see how to do this yet.
Hints or suggestions are greatly appreciated. 
 A: The inclusion $R\supseteq {N^*}^\perp$  follows from a more general fact: 

If $T:V\to W$ is a bounded linear operator (without assuming closed range), then $\overline{R}={N^*}^\perp$.



*

*The inclusion $R\subseteq {N^*}^\perp$ is easy, and since ${N^*}^\perp$ is closed, you get  $\overline{R}={N^*}^\perp$. 

*for the converse, take $w\notin \overline{R}$. Use a separation form of the Hahn-Banach theorem to obtain a linear functional $g\in W^*$ that vanishes on $\overline{R}$ and is nonzero on $w$. Since $g\circ T=0$, we have $g\in N^*$. But $g(w)\ne 0$, so $w\notin {N^*}^\perp$.


The part $R^*\supseteq N^\perp$ is harder (at least to me): I will use the open mapping theorem. Namely, since $T:V\to R$ is a surjection between Banach spaces, there is $r>0$ such that every element $w\in R$ with $\|w\|\le r$ can be written as $Tv$ with $\|v\|\le 1$.
Our goal: given $f\in N^\perp $, we want to define $g\in W^*$ such that $g\circ T=f$. First define $g$ on $R$ by $g(Tv)=f(v)$; this is well-defined because $f$ vanishes on the kernel of $T$. Note that when $w\in R$ has norm at most $r$, we have $g(w) \le \|f\|$ by the above consequence of the open mapping theorem. Therefore, $|g(w)|\le r^{-1}\|f\|\|w\|$ for all $w\in R$. By the Hahn-Banach theorem, $g$ extends to a bounded linear functional on $W$, and this extension is the $g$  we wanted. 
