If $X$ is a normed vector space and $U\subseteq X$, how do we show $(U^\perp)_\perp=\overline U$? Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $X$ be a normed $\mathbb K$-vector space. It is easy to see that $$U^\perp:=\{\varphi\in X':\langle x,\varphi\rangle=0\text{ for all }x\in U\}$$ is a closed subspace of $X'$ for all $U\subseteq X$ and $$V_\perp:=\{x\in X:\langle x,\varphi\rangle=0\text{ for all }\varphi\in V\}$$ is a closed subspace of $X$ for all $V\subseteq X'$.

I would like to show


*

*If $U\subseteq X$ and $V:=U^\perp$, then $\overline U=V_\perp$.

*If $V\subseteq X'$ and $U:=V_\perp$, then $\overline V=U^\perp$.

In 1. (resp. 2.) it is easy to show that $U\subseteq V_\perp$ (resp. $V\subseteq U^\perp$) and hence $\overline U\subseteq V_\perp$ (resp. $\overline V\subseteq U^\perp$).
Now I know the proof of the analogue result in the Hilbert space case, but this proof relies on the projection theorem and hence doesn't easily generalize to the present situation.

So, how do we show $V_\perp\subseteq\overline U$ (resp. $U^\perp\subseteq\overline V$)?

 A: Suppose that there exist  $ x \in  (U^\bot)_\bot:=  \{ x \in X :  f(x) =0, \text { for all } f \in U^\bot  \}   $ such that    $ x \notin \overline U $. Evidently,  $ x  $ is not zero. From a known Corollary of Hahn-Banach thm, we can find  $ f \in X^* $ such that      $ f(x)=d(x,\overline U)>0  $ and  $ f|_{\overline U}=0 $.
Since   $ f (U) = \{0\} $ we have that   $ f \in U^\bot $. Moreover, since   $ x \in  (U^\bot)_\bot  $, we get that     $ f(x) =0 $ which is impossible.
We now show that $(V_\bot)^\bot := \{ f  \in X^* : f ( V_\bot ) = \{0\}\} = \overline{V}^{w^*} $ (weak-star closure). It is easy to see that  $ \overline {V}^{w^*} \subset (V_\bot)^\bot$. Suppose that there exist  $ x_0^*    \in  (V_\bot)^\bot  \setminus \overline{V}^{w^*} $. From the strong separation thm, we can find  $ x^{**}: (X^*,w^*) \to \mathbb R \ $  linear and continuous such that
$$ \sup_{ v^*  \in \overline{V}^{w^*}} x^{**} (v^*)<x^{**}(x_0^*) . $$
Since  $ x^{**} $ is  $ w^*  $-continuous, we can find $  x\in X$ such that $ x^{**}(x^*)=x^*(x), $ for all  $ x^* \in X^* $. So,
$$ \sup_{v^*  \in \overline{V}^{w^*}} v^*(x)< x^*_0(x) . $$
This leads to a contradiction:
If   $ x\in  V_\bot, $ since  $ \overline {V}^{w^*} \subset  ( V_\bot )^\bot  $, it must be true that     $ 0 = \sup \{ v^*(x) \colon v^*  \in \overline{V}^{w^*} \}< x^*_0(x)=0 $  which is impossible.
If  $ x \notin V_\bot $, then  $ v_0^*(x)\neq 0  \text{ for some  } v_0^*\in V $.  Hence, (since  $\frac{x^*_0(x)}{v_0^*(x)} v_0^*(x) \in \overline {V}^{w^*} $)
$$ \frac{x^*_0(x)}{v_0^*(x)} v_0^*(x)  \leq \sup \{ v^*(x) \colon v^*  \in \overline{V}^{w^*} \}< x^*_0(x)   $$
which is impossible
A: The proof is greatly facilitated by noting that in general, $M_\perp:=\{x\in X:\langle x,\varphi\rangle=0\text{ for all }\varphi\in M\}$ is norm closed and that $N^\perp:=\{\varphi\in X':\langle x,\varphi\rangle=0\text{ for all }x\in N\}$ is weak$^*$ closed.
For example, Rudin proves the first part like this: by the remark, $\overline V\subseteq (V^{\perp})_{\perp}$. Now, if $x\notin \overline V$ then there is an $x\in V^{\perp}$ such that $\langle x,x^*\rangle \neq 0$ which says that $x\notin (V^{\perp})_{\perp}.$
The second part is similar.
