I was trying to prove the following statement
"A closed subspace $Y$ of a reflexive Banach space $X$ is also reflexive."
I followed the steps similar to this answer on math.stackexchange
- Since the canonical map from $J_X: X \to X^{**}$ (sending $x \in X$ to $g_x \in X^{**}$ defined by $g_x(f)=f(x)$) is an isometry and an injection, it suffices to show that $J_Y: Y \to Y^{**}$ is surjective
- Let $\eta \in Y^{**}$, we define a linear functional $\xi \in X^{**}$, such that $\xi(f)=\eta(f|_Y), \forall f \in X^*$
- It follows from the reflexivity of $X$ that $\xi(f) = g_x(f) = f(x)$ for some $x \in X$
- We show that $x\in Y$ by considering the annihilator of $Y$ as suggested in the post linked above
- Now we see $\eta(f|_Y)=g_y(f|_Y)$ for every $f \in X^*$
My questions are:
How is it possible to show that every linear functional on $Y$ is a restriction of some functional on $X$? The post linked above seems to suggest using the Hahn-Banach theorem (so do some other sources on the internet) but I fail to see how this can be applied, as the Hahn-Banach theorems require the functional to be bounded by some semi-norm.
Why is $X$ being a Banach space required here?
Maybe I am missing something trivial here and this should really be a comment under that post linked above instead of a separate question but I do not have enough reputation to comment.