TITLE: A closed subspace of a reflexive Banach space is reflexive
QUESTION [18 upvotes]: 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 $j_{y}=y^{**}$.

REPLY [15 votes]: 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, $j_Yy = y^{**}$.