Pre-annihilator of the annihilator of a closed subspace $Y$ is $Y$

Question

Let $Z$ be a linear subspace in $X^*$, the dual space of $X$, where $X$ is a normed linear space. We define $$Z_ \bot = \{x \in X : \phi(x) = 0, \forall \phi \in Z \}. \quad (I)$$

Now, I have to prove that for any closed subspace $Y$ in $X$, $(Y^ \bot )_ \bot = Y.$ Where, $ Y ^ \bot$ is the annihilator or orthogonal complement of $Y$. That is, $ Y ^\bot = \{ \phi \in X^*: \phi(y) =0, \forall y \in Y\}$.

Well, my idea is to show that the two sides are subsets of each other: $(Y^ \bot )_ \bot \subseteq Y$ and $Y \subseteq (Y^ \bot )_ \bot$.

We know that $Y^ \bot$ is indeed a subspace of $X^*$. Thus, we can plug in $Y^ \bot$ as $Z$ in the definition (I), and form $$(Y^ \bot )_ \bot = \{x \in X, \phi(x) = 0, \forall \phi \in Y^\bot \} \quad (II).$$ Now, let's pick an arbitrary $y \in Y$ and an arbitrary $\psi \in Y ^\bot$. Obviously, $y$ satisfies the conditions in (II). Thus, we have $ y \in (Y^ \bot )_ \bot. $ Thus, $Y \subseteq (Y^ \bot )_ \bot$. Am I right so far? This whole conclusion, though obvious, seems a bit cyclic to me.

Also, I have difficulty in the proof of the inverse direction. That is, to show that if $ y \in (Y^ \bot )_ \bot $, then $y \in Y.$ Plus, and perhaps more importantly, where should I use the closedness of $Y$? Thanks for any help.

Answer 1

Because $Y$ is closed, if $Y$ is properly contained in $(Y^\perp)_\perp$, it's a closed subspace. Assume we have proper containment. By Hahn-Banach, we can define a nonzero functional on $(Y^\perp)_\perp$ which is zero on $Y$ by choosing any nonzero linear functional on $(Y^\perp)_\perp/Y$, and extending it to a functional on all of $X$. Call such a functional $\phi$. Then $\phi\in Y^\perp$, so $\phi|_{(Y^\perp)_\perp}=0$, a contradiction.