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Let $X$ be a reflexive Banach Space. Let $Y$ be a closed subspace of it.I need to show that $Y$ is reflexive as well. So as usual I consider the inclusion map $$J: Y \to Y'', J(y)=j_{y}, j_{y}(y')=y'(y)$$, where $Y''$ denotes the bidual space of $Y$. I need to show that the map is onto because everything else is guaranteed (one-one and isometry)

I pick an element say $y^{**} \in Y''$, then define $\tilde{y} \in X''$ such that $\tilde{y}(x')=y^{**}(x'|_{Y})$. Then it is well-defined, linear, continuous. Since $X$ is reflexive there is $x \in X$ such that $j_{x}=\tilde{y}$.

Suppose that $x \not \in Y$. Then since $Y$ is closed, by Hahn-Banach theorem, there is $f:X \to K$ such that $f(x)=1$ and $f|_{Y}=0$. Then $\tilde{y}(f)=f(x)=1=y^{**}(0)=0$ which can't happen. Thus $x=y \in Y$

I need to show that $j_{y}=y^{**}$.

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To show that $j_Yy = y^{**}$, let $y^* \in Y'$ and $x^*\in X'$ a Hahn-Banach extension of $y^*$. We have \begin{align*} (j_Yy)(y^*) &= y^*(y)\\ &= x^*(y)\\ &= j_Xx(x^*)\\ &= \tilde y(x^*)\\ &= y^{**}(x^*|_Y)\\ &= y^{**}(y^*) \end{align*} Hence, $j_Yy = y^{**}$.

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    $\begingroup$ An (obvious) remark: the fact that $Y$ is closed is also important, because it guarantees that $Y$ is again a Banach space. $\endgroup$ Nov 13 '15 at 12:48
  • $\begingroup$ Why is the condition that $X$ be Banach space necessary? You are just using the fact that $Y$ is closed. $\endgroup$ May 6 '21 at 23:11
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    $\begingroup$ It's not necessary per se, but $X$ being reflexive implies that $X$ is Banach, since $X\cong X^{**}$ and the dual space of a normed space is always Banach. $\endgroup$
    – Darsen
    May 11 '21 at 20:17

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