Let $X$ be an inner product space and let $x\in X$. $M=\{z\in X:\langle z,x\rangle=0\}$. I want to find $M^{\perp}$ and $M^{\perp \perp}$.
Clearly, $\{x\}\subset M^\perp$. Thus $M^{\perp \perp}\subset\{x\}^{\perp}=M$. But how to find $M^{\perp}$? Please help.
If $x=0$, then clearly $M=X$ and $M^\bot = \{0\}$, so suppose $x \neq 0$.
The key point here is that any point $y \in X$ can be written as $y=\alpha x + w$, where $w \bot x$.
If you draw a picture, you can guess that $M^\bot = \operatorname{sp} \{x \}$.
To prove this, suppose $y \in \operatorname{sp} \{x \}$, that is $y = \alpha x$. Then $y \bot z$ for all $ z \in M$ and so $y \in M^\bot$.
Now suppose $y \in M^\bot$, and so we can write $y=\alpha x + w$, where $w \bot x$. Since $w \bot x$, we have $w \in M$, and since $y \in M^\bot$, we have $\langle y, w \rangle = \langle \alpha x + w, w \rangle = \|w\|^2 = 0$, and so $y = \alpha x \in \operatorname{sp} \{x \}$.
Since $M^\bot = \operatorname{sp} \{x \}$, we see that $M^{\bot \bot} = \{ z | \langle z, x \rangle = 0 \} = M$.