[proofplan] We show that the canonical embedding $J_X: X \to X^{**}$ is surjective, given that both $Y$ and $X/Y$ are reflexive. For an arbitrary $\xi \in X^{**}$, we use the double [adjoint](/page/The%20Adjoint%20of%20an%20Operator) $\pi^{**}$ to project $\xi$ into $(X/Y)^{**}$, apply [reflexivity](/page/Reflexive%20Space) of $X/Y$ to find a coset representative, and then correct by an element of $J_X(Y)$ using reflexivity of $Y$. The correction step relies on showing that the "error" functional $\xi - J_X(x)$ vanishes on $Y^\perp$ and therefore represents an element of $Y^{**}$. [/proofplan] [step:Project $\xi \in X^{**}$ into $(X/Y)^{**}$ and lift to a coset representative] Let $\xi \in X^{**}$. Let $\pi: X \to X/Y$ denote the quotient map. As in the proof of [Reflexivity of Quotients](/theorems/997), the double adjoint $\pi^{**}: X^{**} \to (X/Y)^{**}$ is defined by $\pi^{**}(\xi) = \xi \circ \pi^*$, where $\pi^*(h) = h \circ \pi$ for $h \in (X/Y)^*$, and the commutative identity \begin{align*} \pi^{**} \circ J_X = J_{X/Y} \circ \pi \end{align*} holds. Consider $\pi^{**}(\xi) \in (X/Y)^{**}$. Since $X/Y$ is reflexive, the canonical embedding $J_{X/Y}: X/Y \to (X/Y)^{**}$ is surjective. Therefore there exists $x_0 \in X$ such that \begin{align*} J_{X/Y}(\pi(x_0)) = \pi^{**}(\xi). \end{align*} Equivalently, for every $h \in (X/Y)^*$: \begin{align*} h(\pi(x_0)) = \pi^{**}(\xi)(h) = \xi(\pi^*(h)) = \xi(h \circ \pi). \end{align*} [guided] The strategy is to decompose $\xi$ into two parts: one coming from the quotient $X/Y$ and one coming from the subspace $Y$. The double adjoint $\pi^{**}$ projects $\xi$ down to $(X/Y)^{**}$. Since $X/Y$ is reflexive, the image $\pi^{**}(\xi)$ is represented by some coset $\pi(x_0) = x_0 + Y \in X/Y$. That is, $\xi$ and $J_X(x_0)$ agree when evaluated on functionals in $Y^\perp = \operatorname{Im}(\pi^*)$. Note that $x_0$ is not unique — any other representative of the same coset (i.e., $x_0 + y$ for $y \in Y$) would work equally well. The "coset freedom" in $x_0$ will not matter because the error $\xi - J_X(x_0)$ will be corrected in the next step. [/guided] [/step] [step:Show that the error $\xi - J_X(x_0)$ vanishes on $Y^\perp$] Define the error functional \begin{align*} \eta := \xi - J_X(x_0) \in X^{**}. \end{align*} For every $f \in Y^\perp$, we have $f = h \circ \pi$ for some $h \in (X/Y)^*$ (since the [Dual of a Quotient Space](/theorems/978) isomorphism identifies $(X/Y)^*$ with $Y^\perp$ via $h \mapsto h \circ \pi$). Then \begin{align*} \eta(f) = \xi(f) - J_X(x_0)(f) = \xi(h \circ \pi) - f(x_0) = h(\pi(x_0)) - h(\pi(x_0)) = 0, \end{align*} where the third equality uses the identity from the previous step ($\xi(h \circ \pi) = h(\pi(x_0))$) and the fact that $f(x_0) = (h \circ \pi)(x_0) = h(\pi(x_0))$. Therefore $\eta|_{Y^\perp} = 0$. [guided] Why does $\eta$ vanish on $Y^\perp$? The element $x_0$ was chosen precisely so that $\xi$ and $J_X(x_0)$ agree on all functionals of the form $h \circ \pi$ — and these are exactly the elements of $Y^\perp$. So the error $\eta = \xi - J_X(x_0)$ is "invisible" to any functional that vanishes on $Y$. This means $\eta$ only carries information about how $\xi$ acts on functionals that detect behavior within $Y$. The next step will show that this residual information corresponds to an element of $Y^{**}$. [/guided] [/step] [step:Interpret $\eta$ as an element of $Y^{**}$ via the restriction map] Since $\eta|_{Y^\perp} = 0$, the functional $\eta: X^* \to \mathbb{R}$ factors through the quotient $X^*/Y^\perp$. By the [Dual of a Subspace](/theorems/979) isomorphism, the map \begin{align*} \Theta: X^*/Y^\perp &\to Y^* \\ f + Y^\perp &\mapsto f|_Y, \end{align*} is an isometric isomorphism. Since $\eta$ vanishes on $Y^\perp$, we can define \begin{align*} \tilde{\eta}: Y^* &\to \mathbb{R} \\ g &\mapsto \eta(f), \quad \text{where } f \in X^* \text{ is any extension of } g \text{ (i.e., } f|_Y = g\text{)}. \end{align*} This is well-defined: if $f_1, f_2 \in X^*$ both satisfy $f_1|_Y = f_2|_Y = g$, then $f_1 - f_2 \in Y^\perp$, so $\eta(f_1) - \eta(f_2) = \eta(f_1 - f_2) = 0$. The map $\tilde{\eta}$ is linear (since $\eta$ is linear) and bounded: for any $g \in Y^*$, by the [Norm-Preserving Extension of Linear Functionals](/theorems/880), there exists $f \in X^*$ with $f|_Y = g$ and $\|f\|_{X^*} = \|g\|_{Y^*}$, giving $|\tilde{\eta}(g)| = |\eta(f)| \le \|\eta\|_{X^{**}} \|f\|_{X^*} = \|\eta\|_{X^{**}} \|g\|_{Y^*}$. Therefore $\tilde{\eta} \in Y^{**}$. [guided] The vanishing condition $\eta|_{Y^\perp} = 0$ is essential here. Without it, $\eta$ would not factor through the quotient $X^*/Y^\perp$, and we could not view $\eta$ as acting on $Y^*$. The isomorphism $\Theta: X^*/Y^\perp \xrightarrow{\sim} Y^*$ (from the [Dual of a Subspace](/theorems/979)) has a dual interpretation: an element of $Y^{**}$ corresponds to an element of $(X^*/Y^\perp)^*$, which in turn corresponds to an element of $X^{**}$ that vanishes on $Y^\perp$. This is precisely what $\eta$ is. The well-definedness of $\tilde{\eta}$ is the key subtlety: two extensions $f_1, f_2$ of the same $g \in Y^*$ differ by an element of $Y^\perp$, and $\eta$ vanishes on $Y^\perp$, so $\eta(f_1) = \eta(f_2)$. [/guided] [/step] [step:Use reflexivity of $Y$ to represent $\tilde{\eta}$ and conclude] Since $Y$ is reflexive, the canonical embedding $J_Y: Y \to Y^{**}$ is surjective. Therefore there exists $y_0 \in Y$ with $J_Y(y_0) = \tilde{\eta}$. That is, for every $g \in Y^*$: \begin{align*} g(y_0) = \tilde{\eta}(g). \end{align*} Set $x := x_0 + y_0 \in X$ (note $x_0 \in X$ and $y_0 \in Y \subset X$). We claim $J_X(x) = \xi$. For any $f \in X^*$: \begin{align*} J_X(x)(f) &= f(x) = f(x_0) + f(y_0) = J_X(x_0)(f) + f(y_0). \end{align*} Since $f|_Y \in Y^*$ and $y_0 \in Y$, we have $f(y_0) = (f|_Y)(y_0) = \tilde{\eta}(f|_Y)$. By definition of $\tilde{\eta}$, this equals $\eta(f)$ (taking the extension of $f|_Y$ to be $f$ itself). Therefore \begin{align*} J_X(x)(f) = J_X(x_0)(f) + \eta(f) = J_X(x_0)(f) + \xi(f) - J_X(x_0)(f) = \xi(f). \end{align*} Since $f \in X^*$ was arbitrary, $J_X(x) = \xi$. Since $\xi \in X^{**}$ was arbitrary, $J_X$ is surjective, so $X$ is reflexive. [guided] Let us verify the final chain of equalities carefully. We need $J_X(x_0 + y_0)(f) = \xi(f)$ for all $f \in X^*$. Expanding: \begin{align*} J_X(x_0 + y_0)(f) = f(x_0 + y_0) = f(x_0) + f(y_0). \end{align*} For the second term, $f(y_0) = (f|_Y)(y_0) = J_Y(y_0)(f|_Y) = \tilde{\eta}(f|_Y)$. By definition of $\tilde{\eta}$, we have $\tilde{\eta}(f|_Y) = \eta(f')$ for any $f' \in X^*$ with $f'|_Y = f|_Y$. Taking $f' = f$ itself gives $\tilde{\eta}(f|_Y) = \eta(f) = \xi(f) - J_X(x_0)(f) = \xi(f) - f(x_0)$. Therefore: \begin{align*} f(x_0) + f(y_0) = f(x_0) + \xi(f) - f(x_0) = \xi(f). \end{align*} This decomposition $x = x_0 + y_0$ reflects the short exact sequence $0 \to Y \to X \to X/Y \to 0$: the element $x_0$ captures the "quotient part" of $\xi$ (its projection onto $(X/Y)^{**}$), while $y_0$ captures the "subspace part" (the residual functional on $Y^*$). The reflexivity of both $Y$ and $X/Y$ ensures that both parts can be represented by actual elements, and together they represent $\xi$. [/guided] [/step]