Let $L$ be a pseudocomplemented distributive lattice. Let $x^*$ denote the pseudocomplement of $x.$ Show that $S ( L )=\{x \ : \ x=x^{**}\} $ is the set of complemented elements of $L$.
It's easy enough to show that any complemented element of $L$ is in $S(L),$ it follows from the fact that if there is a complement it is the same as the pseudocomplement.
I'm having a very hard time showing the reciprocal, that is, any element in $S(L)$ is complemented. Any tips?
To start with, in a bounded distributive lattice, the complement is unique.
So, since $y^*$ is a complement of $y$ and of $y^{**}$, it follows that $y=y^{**}$.
Thus, $y^*=y^{***}$, yielding $x = x^{**}$, whenever $x=y^*$.
EDIT
With pseudo-complementation in place of complementation, that is, if $$x^* = \max \{ z \in L : x \wedge z = 0 \},$$ (this is what you mean, right?), then I'm not sure that is correct.
Check this example:
Here, $a^*=b^*=e, c^*=d, d^*=c$ and $e^*=a$.
It follows that $S(L) = \{ a,c,d,e \}$, but $c$ and $d$ are not complemented ($c \vee d \neq 1$). Did I miss something in your definition?