Closed Subspaces of Reflexive Spaces are Reflexive

Closed Subspaces of Reflexive Spaces are Reflexive

Theorem 1: Let $X$ be a normed linear space and let $Y \subseteq X$ be a subspace. If $X$ is reflexive and $Y$ is closed then $Y$ is reflexive.
  • Proof: Let $J_X : X \to X^{**}$ and $J_Y : Y \to Y^{**}$ denote the canonical embeddings. Let $\eta \in Y^{**}$ and define $\xi : X^* \to \mathbb{C}$ for all $\varphi \in X^*$ by:
(1)
\begin{align} \quad \xi (\varphi) = \eta (\varphi |_Y) \end{align}
  • We aim to show that $\xi$ is linear and continuous. For linearity, let $\varphi, \psi \in X^*$ and let $\lambda \in \mathbb{C}$. Then:
(2)
\begin{align} \quad \xi (\varphi + \psi) = \eta ([\varphi + \psi] |_Y) = \eta (\varphi |_Y + \psi |_Y) = \eta(\varphi |_Y) + \eta(\psi |_Y) = \xi (\varphi) + \xi (\psi) \end{align}
(3)
\begin{align} \quad \xi (\lambda \varphi) = \eta (\lambda \varphi |_Y) = \lambda \eta ( \varphi |_Y) = \lambda \xi (\varphi) \end{align}
  • And for continuity, we have that:
(4)
\begin{align} \quad \| \xi (\varphi) \| = \| \eta ( \varphi |_Y) \| \leq \| \eta \| \| \varphi |_Y \|= \| \eta \| \| \varphi \| \end{align}
  • (Here $M = \| \eta \|$). Therefore $\xi \in X^{**}$. Now since $X$ is reflexive we have that:
(5)
\begin{align} \quad J(X) = X^{**} \end{align}
  • So $\xi \in J(X)$. So there exists an $x \in X$ such that:
(6)
\begin{align} \quad J_X(x) = \xi \end{align}
  • We want to show that $x \in Y$. Let $\psi \in X^*$ be such that $\psi |_Y = 0$. Then we have that:
(7)
\begin{align} \quad \psi (x) = (J_X(x))(\psi) = \xi(\psi) = \eta (\psi |_Y) = \eta (0) = 0 \end{align}
(8)
\begin{align} \quad \Phi |_Y = \varphi \end{align}
  • Therefore we have that:
(9)
\begin{align} \quad \xi (\Phi) = \eta (\Phi_Y) = \eta(\varphi) = (J_Y(x))(\varphi) = \varphi(x) = \Phi(x) = (J_X(x))(\Phi) \end{align}
  • Therefore $(J_Y(x)) = \eta$ and so $J_Y : Y \to Y^{**}$ is surjective. Hence $J_Y$ is reflexive. $\blacksquare$
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