[proofplan] We show that every element of $Y^{**}$ lies in the image of the canonical embedding $J_Y: Y \to Y^{**}$. Given $\eta \in Y^{**}$, we use the [Norm-Preserving Extension](/theorems/880) to extend each functional on $Y$ to one on $X$, thereby constructing an element $\xi \in X^{**}$. [Reflexivity](/page/Reflexive%20Space) of $X$ then provides $x \in X$ with $J_X(x) = \xi$. Finally, we verify that $x$ must lie in the closed subspace $Y$ and that $J_Y(x) = \eta$. [/proofplan] [step:Extend an arbitrary bidual element of $Y$ to an element of $X^{**}$] Let $\eta \in Y^{**}$. Define the map \begin{align*} \xi: X^* &\to \mathbb{R} \\ f &\mapsto \eta(f|_Y), \end{align*} where $f|_Y \in Y^*$ denotes the restriction of $f$ to $Y$. This is well-defined since $Y \subset X$ is a subspace and $f|_Y$ is a bounded linear functional on $Y$ with $\|f|_Y\|_{Y^*} \le \|f\|_{X^*}$. The map $\xi$ is linear: for $f, g \in X^*$ and $\alpha, \beta \in \mathbb{R}$, $(\alpha f + \beta g)|_Y = \alpha f|_Y + \beta g|_Y$, so $\xi(\alpha f + \beta g) = \eta(\alpha f|_Y + \beta g|_Y) = \alpha \eta(f|_Y) + \beta \eta(g|_Y) = \alpha \xi(f) + \beta \xi(g)$. The map $\xi$ is bounded: $|\xi(f)| = |\eta(f|_Y)| \le \|\eta\|_{Y^{**}} \|f|_Y\|_{Y^*} \le \|\eta\|_{Y^{**}} \|f\|_{X^*}$. Therefore $\xi \in X^{**}$ with $\|\xi\|_{X^{**}} \le \|\eta\|_{Y^{**}}$. [guided] We want to show that $J_Y: Y \to Y^{**}$ is surjective. The idea is to "lift" any $\eta \in Y^{**}$ into $X^{**}$, use the surjectivity of $J_X$ (reflexivity of $X$), and then bring the result back down to $Y$. How do we lift $\eta$ to $X^{**}$? An element of $X^{**}$ is a bounded linear functional on $X^*$. Given $f \in X^*$, its restriction $f|_Y$ belongs to $Y^*$, and we can evaluate $\eta$ on this restriction. This defines a map $\xi: X^* \to \mathbb{R}$ by $\xi(f) = \eta(f|_Y)$. We verify $\xi \in X^{**}$: - **Linearity:** The restriction map $f \mapsto f|_Y$ from $X^*$ to $Y^*$ is itself linear (restrictions respect addition and scalar multiplication). Since $\eta$ is linear, the composition $\xi = \eta \circ (\cdot|_Y)$ is linear. - **Boundedness:** For any $f \in X^*$, $\|f|_Y\|_{Y^*} = \sup\{|f(y)| : y \in Y, \|y\|_X \le 1\} \le \sup\{|f(x)| : x \in X, \|x\|_X \le 1\} = \|f\|_{X^*}$, since $Y \subset X$. Therefore $|\xi(f)| = |\eta(f|_Y)| \le \|\eta\|_{Y^{**}} \|f|_Y\|_{Y^*} \le \|\eta\|_{Y^{**}} \|f\|_{X^*}$. Hence $\xi \in X^{**}$ and $\|\xi\|_{X^{**}} \le \|\eta\|_{Y^{**}}$. [/guided] [/step] [step:Use reflexivity of $X$ to represent $\xi$ by an element $x \in X$] Since $X$ is reflexive, the canonical embedding $J_X: X \to X^{**}$, defined by $J_X(x)(f) = f(x)$ for all $f \in X^*$, is surjective. Therefore there exists $x \in X$ with $J_X(x) = \xi$. That is, \begin{align*} f(x) = \xi(f) = \eta(f|_Y) \end{align*} for every $f \in X^*$. [/step] [step:Show that $x$ lies in the closed subspace $Y$] Suppose for contradiction that $x \notin Y$. Since $Y$ is a closed subspace of $X$ and $x \notin Y$, we have $\operatorname{dist}(x, Y) = \inf_{y \in Y} \|x - y\|_X > 0$. By the [Norm-Preserving Extension of Linear Functionals](/theorems/880) applied to the subspace $\operatorname{span}(Y \cup \{x\})$ (or directly by a standard corollary of the [Hahn-Banach Theorem](/theorems/879)), there exists $f_0 \in X^*$ such that \begin{align*} f_0|_Y = 0 \quad \text{and} \quad f_0(x) \neq 0. \end{align*} Applying the identity $f(x) = \eta(f|_Y)$ with $f = f_0$: \begin{align*} f_0(x) = \eta(f_0|_Y) = \eta(0) = 0, \end{align*} which contradicts $f_0(x) \neq 0$. Therefore $x \in Y$. [guided] This is the step where the hypothesis that $Y$ is closed is consumed. If $Y$ were not closed, then $x$ could be a limit point of $Y$ lying outside $Y$, and the Hahn-Banach separation would not be available. Why must $x$ lie in $Y$? The identity $f(x) = \eta(f|_Y)$ says that $x$ behaves, with respect to every functional $f \in X^*$, exactly as an element of $Y$ would. More precisely, if $f$ vanishes on $Y$, then $f|_Y = 0$, so $f(x) = \eta(0) = 0$, meaning $f$ also vanishes at $x$. Thus $x$ belongs to the annihilator of $Y^\perp$: \begin{align*} x \in (Y^\perp)_\perp := \{z \in X : f(z) = 0 \text{ for all } f \in Y^\perp\}. \end{align*} The Hahn-Banach theorem guarantees $(Y^\perp)_\perp = \overline{Y}$: if $z \notin \overline{Y}$, then $\operatorname{dist}(z, \overline{Y}) > 0$ and the [Hahn-Banach Theorem](/theorems/879) produces $f \in X^*$ vanishing on $Y$ with $f(z) \neq 0$, so $z \notin (Y^\perp)_\perp$. Since $Y$ is closed, $\overline{Y} = Y$, and therefore $x \in Y$. [/guided] [/step] [step:Verify $J_Y(x) = \eta$ and conclude reflexivity] Since $x \in Y$, we may evaluate the canonical embedding $J_Y: Y \to Y^{**}$, defined by $J_Y(y)(g) = g(y)$ for all $g \in Y^*$. We must show $J_Y(x)(g) = \eta(g)$ for every $g \in Y^*$. By the [Norm-Preserving Extension of Linear Functionals](/theorems/880), every $g \in Y^*$ admits an extension $f \in X^*$ with $f|_Y = g$ and $\|f\|_{X^*} = \|g\|_{Y^*}$. The hypotheses of the norm-preserving extension are satisfied: $X$ is a [normed space](/page/Normed%20Vector%20Space) and $Y \subset X$ is a subspace. Then \begin{align*} J_Y(x)(g) = g(x) = f(x) = \eta(f|_Y) = \eta(g). \end{align*} Since $g \in Y^*$ was arbitrary, $J_Y(x) = \eta$. Since $\eta \in Y^{**}$ was arbitrary, $J_Y$ is surjective, so $Y$ is reflexive. [guided] The final step connects the element $x \in Y$ back to the original bidual element $\eta$. We need to verify that $J_Y(x)$ and $\eta$ agree on every $g \in Y^*$. The key point is that not every $g \in Y^*$ arises as the restriction of a fixed $f \in X^*$ — different $g$'s may require different extensions. But we only need to know that *some* extension exists. The [Norm-Preserving Extension of Linear Functionals](/theorems/880) provides such an extension for each $g \in Y^*$: there exists $f \in X^*$ with $f|_Y = g$. Once we have this extension, the chain of equalities $J_Y(x)(g) = g(x) = f(x) = \eta(f|_Y) = \eta(g)$ follows from the identity $f(x) = \eta(f|_Y)$ established in the second step and the fact that $f|_Y = g$. This completes the proof that $Y$ is reflexive. The argument uses only the reflexivity of $X$, the closedness of $Y$, and the Hahn-Banach theorem — no additional structure on $Y$ is required. [/guided] [/step]