If $\mathfrak{Y}$ is a closed subspace of a reflexive Banach space $\mathfrak{X}$, then is $\mathfrak{Y}$ reflexive too? Exercise 2.4.8 in Analysis Now by Pedersen:

If $\mathfrak{Y}$ is a closed subspace of a reflexive Banach space $\mathfrak{X}$, show that $\mathfrak{Y}$ and $\mathfrak{X}/\mathfrak{Y}$ are reflexive using the following theorem:
2.4.13. Proposition. Consider a closed subspace $\mathfrak{Y}$ of a normed space $\mathfrak{X}$. Let $I:\mathfrak{Y}\to \mathfrak{X}$ denote the inclusion map and $Q:\mathfrak{X}\to\mathfrak{X}/\mathfrak{Y}$ denote the quotient map. Then we may identify $Q^\ast$ with the inclusion map of $\mathfrak{Y}^\perp$ into $\mathfrak{X}^\ast$ and $I^\ast$ with the quotient map of $\mathfrak{X}^\ast$ onto $\mathfrak{X}^\ast/\mathfrak{Y}^\perp$.

 A: The essence in both parts is this: 

The Hahn-Banach theorem lets us identify each $\psi\in \mathfrak{Y}^{\ast\ast}$ as some distinct $\psi_{\mathfrak{X}}\in \mathfrak{X}^{\ast\ast}$. We've assumed $\mathfrak{X}^{\ast\ast}=\mathfrak{X}$, so each $\psi_{\mathfrak{X}}$ can be identified with a distinct $x\in\mathfrak{X}$. Applying the annihilator of $\mathfrak{Y}$ to the $\psi_{\mathfrak{X}}$'s reveals that their associated elements $x\in\mathfrak{X}$ must be in $\mathfrak{Y}$.

First notice that for $\mathfrak{X}$ normed, linear, the mapping $\Phi:\mathfrak{X}\to\mathfrak{X}^{\ast\ast}$, $\langle \phi,\Phi(x)\rangle\triangleq \phi(x)$ is always isometric so if $\mathfrak{X}$ is Banach we just have to show that $\Phi$ is surjective (hits everything in $\mathfrak{X}^{\ast\ast}$).


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*$\Phi$'s restriction to $\mathfrak{Y}$ is surjective to $\mathfrak{Y}^{\ast\ast}$:
First note that any functional on $\mathfrak{Y}$ can be extended to one on $\mathfrak{X}$ by the Hahn-Banach theorem, so $\mathfrak{Y}^\ast$ can be realized as a subset of $\mathfrak{X}^\ast.$
Take $\psi\in\mathfrak{Y}^{\ast\ast}$. Define a new function $\psi_\mathfrak{X}:\mathfrak{X}^\ast\to\mathbb{C}$:  for any $\nu\in \mathfrak{X}^\ast,$ denote its restriction to $\mathfrak{Y}$ as $\nu|_\mathfrak{Y}\in\mathfrak{Y}^\ast$ and take $\psi_{\mathfrak{X}}(\nu)\triangleq \psi(\nu|_\mathfrak{Y}).$ By construction $\psi_{\mathfrak{X}}$ agrees with $\psi$ on $\mathfrak{Y}^\ast.$
$\psi_{\mathfrak{X}}$ is easily verified to be a bounded linear functional on $\mathfrak{X}^\ast$, i.e. $\psi_\mathfrak{X}\in\mathfrak{X}^{\ast\ast},$ so since $\mathfrak{X}$ is reflexive we can identify $\psi_{\mathfrak{X}}=x_\psi\in\mathfrak{X}$ using $\Phi^{-1}$
Finally, notice that for any $\nu\in\mathfrak{Y}^\perp\subset \mathfrak{X}^\ast$, then by definition $\nu(x_\psi)=\psi_{\mathfrak{X}}(\nu)=0$, so since $\mathfrak{Y}$ is closed we know $x_\psi\in\mathfrak{Y}$.
We've just seen that $\Phi^{-1}$ maps $\mathfrak{Y}^{\ast\ast}$ into to $\mathfrak{Y}$, so what we want holds.

*$\Phi':\mathfrak{X}/\mathfrak{Y}\to(\mathfrak{X}/\mathfrak{Y})^{\ast\ast}$, $\langle \phi,\Phi'(z)\rangle\triangleq \phi(z)$ is surjective:
By Prop. 2.4.13, $(\mathfrak{X}/\mathfrak{Y})^\ast$ is isomorphic to $\mathfrak{Y}^\perp$ through some $\iota:(\mathfrak{X}/\mathfrak{Y})^\ast \to \mathfrak{Y}^\perp$.
Take $\psi\in (\mathfrak{X}/\mathfrak{Y})^{\ast\ast}$. Define a new function $\psi_\mathfrak{X}:\mathfrak{X}^\ast\to\mathbb{C}$: if $\nu\in\mathfrak{X}^\ast$ then decompose $\nu=\nu_\mathfrak{Y}+\nu_\perp$ with $\nu_\mathfrak{Y}\in \mathfrak{Y}^\ast$ and $\nu_\perp\in \mathfrak{Y}^\perp$, and let $\psi_{\mathfrak{X}}(\nu)\triangleq \psi(\iota^{-1}(\nu_\perp))$. By construction $\psi_{\mathfrak{X}} \circ \iota$ agrees with $\psi$ on $(\mathfrak{X}/\mathfrak{Y})^\ast.$
(Here we notice that $\mathfrak{Y}^\ast$ is the orthogonal complement of $\mathfrak{Y}^\perp$ so that $\mathfrak{Y}^\ast +\mathfrak{Y}^\perp=\mathfrak{X}^\ast$).
$\psi_{\mathfrak{X}}$ is easily verified to be a bounded linear functional on $\mathfrak{X}^\ast$, i.e. $\psi_\mathfrak{X}\in\mathfrak{X}^{\ast\ast},$ so since $\mathfrak{X}$ is reflexive we can identify $\psi_{\mathfrak{X}}=x_\psi\in\mathfrak{X}$ using $\Phi^{-1}$.
Notice that for any $\nu\in\mathfrak{Y}^*\subset \mathfrak{X}^\ast$, then by definition $\nu(x_\psi)=\psi_{\mathfrak{X}}(\nu)=0$, so since $\mathfrak{Y}$ is closed we know $x_\psi$ is in the orthogonal complement of $\mathfrak{Y}$. Thus $x_\psi$ is the natural representative of some class in $\mathfrak{X}/\mathfrak{Y}$, i.e. it can be isometrically & isomorphically be identified with some $x_{\psi;\mathfrak{X}/\mathfrak{Y}}\in \mathfrak{X}/\mathfrak{Y}$.
This defines a bounded, linear, injective mapping $\psi\mapsto x_{\psi;\mathfrak{X}/\mathfrak{Y}}$ from $(\mathfrak{X}/\mathfrak{Y})^{\ast\ast}$ into $\mathfrak{X}/\mathfrak{Y}$. Compute: $$\langle\phi,\Phi'(x_{\psi;\mathfrak{X}/\mathfrak{Y}})\rangle =\phi(x_{\psi;\mathfrak{X}/\mathfrak{Y}})=\langle x_\psi, \iota(\phi)\rangle = \psi(\phi).$$
So the mapping $\psi\mapsto x_{\psi;\mathfrak{X}/\mathfrak{Y}}$ is actually $\Phi'^{-1}$. So we are finished.
