I only know that:
for an LP standard primal $\left\{\begin{matrix} \underset{x}{\operatorname{max}}[c^Tx]\\ Ax \leq b\\ x \geq 0 \end{matrix}\right.$ corresponds the LP dual $\left\{\begin{matrix} \underset{y}{\operatorname{min}}[b^Ty]\\ A^Ty \geq c\\ y \geq 0 \end{matrix}\right.$;
strong LP duality states that, if the primal admits optimal solution $x$, dual admits optimal solution $y$ and results $c^Tx=b^Ty$. And similarly, weak LP duality states $c^Tx \leq b^Ty$.
Unfortunately, as explained here at page 4, to pass from (3) to (6) we need to use respectively conic and Fenchel's duality. I don't know anything about that, with many thanks to my professor, so I can't figure out this proof.
Could you help me? Every link is also welcome.
Thanks in advance.
you can write
$(a + P\zeta )^T x \le d, \quad \forall \zeta \in Z$
as
$a^T x + (P\zeta )^T x \le d, \quad \forall \zeta \in Z$
Since this must hold for any $\zeta \in Z$, it also holds for the max of $\zeta$ in $Z$. Hence (4).
Then take the dual of the max problem (you know how to do this, use your first message). Hence (5).
They then say: we want $min_{w \in W} f(w) \le p$, that is there must exist $\bar{w} \in W$ such that $f(\bar{w}) \le p$.
So nothing fancy :)