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Here is the theorem,

Let $G$ and $L$ be two closed subspaces in a Banach space $E$. The following properties are equivalent:

(a) $G + L$ is closed in $E$

(b) $G^{\perp} + L^{\perp}$ is closed in $E^*$

(c) $G + L = (G^{\perp} \cap L^{\perp})^{\perp}$

(d) $G^{\perp} + L^{\perp} = (G \cap L)^{\perp}$

In the proof, he writes

(d) $\Rightarrow$ (b) is obvious.

Why is this obvious?

Assuming (d), let $f$ be a limit point in $E^*$. Then there exists a sequence of continuous linear operators $\{g_n + l_n\}$, $g_n \in G^{\perp}$, $l_n \in L^{\perp}$ such that $||g_n + l_n - f||_{E^*} \to 0$. Since $G^{\perp} + L^{\perp} = (G \cap L)^{\perp}$, we must have that $\{g_n + l_n\} \subset (G \cap L)^{\perp}$, so $g_n(x) + l_n(x) = 0$ for all $x \in G \cap L$. I'm not really sure if I'm doing this correctly, or where to go from here. Is this how one normally shows closedness in the dual space (by showing that $\|g_n + l_n - f \|_{E^*} \to 0$ implies that $f \in G^{\perp} + L^{\perp}$)?

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    The perpendicular space $G^\perp$ is closed for any closed subspace $G$. If (d) is true, then (b) follows since it is the perpendicular space of a closed subspace. – 2017-01-06
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    could you provide a reference for $G$ closed $\Rightarrow G^{\perp}$ closed? – 2017-01-06
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    It is not that hard to see, in fact $G$ does not have to be closed. Suppose $\phi_n\to \phi$ in $E^*$ where $\phi_n\in G^\perp$. Then we have that for any $\epsilon > 0$, there exists an $n$ large enough such that for any $x\in G$, $$||(\phi,x)|| \leq ||(\phi-\phi_n,x)|| \leq \epsilon ||x||. $$ Since $\epsilon$ was arbitrary we have $(\phi,x)=0$. – 2017-01-06

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