Surjectivity of $\pi^{**}$ Given a Banach space $X$ and a closed subspace $Y$ I'm trying to prove that $\pi^{**}:X^{**} \to (X/Y)^{**}$ is surjective. This is in fact to prove that $X/Y$ is reflexive if $X$ does.
 A: It follows for example with the closed range theorem.
$\pi \colon X \to X/Y$ has closed range (it is surjective, by definition of $X/Y$), so $\pi^\ast \colon (X/Y)^\ast \to X^\ast$ has closed range, and thus $\pi^{\ast\ast}\colon X^{\ast\ast} \to (X/Y)^{\ast\ast}$ has closed range. Since $\pi^\ast$ is injective (that follows from the surjectivity of $\pi$, since generally the kernel of $T^\ast$ is the annihilator of the range $\mathcal{R}(T)$), $\pi^{\ast\ast}$ has dense range (since the kernel of $T$ is the annihilator of the range $\mathcal{R}(T^\ast)$). Dense and closed means the entire space.
We can also see it directly: $\pi^\ast$ is an isometric embedding of $(X/Y)^\ast$ with range $Y^\perp$ (the annihilator of $Y$). Let $\xi \in (X/Y)^{\ast\ast}$ arbitrary. Then $\varphi_0 = \xi \circ (\pi^\ast)^{-1}$ is a continuous linear functional on $\mathcal{R}(\pi^\ast) = Y^\perp$, and by Hahn-banach, we can extend it to a continuous linear functional $\varphi$ on all of $X^\ast$, i.e. $\varphi \in X^{\ast\ast}$, and that satisfies
$$\pi^{\ast\ast}(\varphi) = \varphi\circ \pi^\ast = \varphi_0 \circ \pi^\ast = (\xi \circ (\pi^\ast)^{-1})\circ \pi^\ast = \xi,$$
since the range of $\pi^\ast$ is the domain of $\varphi_0$.
