# the reason behind dual the inner problem in robust optimization

I have reading some material about Robust optimization. It seems that dual optimization theory is the main techniques to reformulate robust optimization.

for example: $$\min_x\qquad c^Tx$$ $$s.t\qquad a^Tx\leq b$$

where$$\qquad a\in \{a|Da\leq d\}$$

and the problem is equivalent $$\min_x\qquad c^Tx$$ $$s.t\qquad \max_{Da\leq d}a^Tx\leq b$$

and they use dual theory reformulate the max problem to min problem

which is $$\min_x\qquad c^Tx$$ $$s.t\qquad \min_{D^Tp=x,p\geq0}p^Td\leq b$$

and finally it equivalent to $$\min_{x,p}\qquad c^Tx$$ $$s.t$$ $$p^Td\leq b$$ $$D^Tp=x$$ $$p\geq0$$

My question is

1.why we use dual theory to reformulate the inner problem?

2.why we can omit the second minimize in constraints?

2. If you can find one $$p \geq 0$$ for which $$D^Tp = x$$ and $$p^T d \leq b$$, that is already sufficient to conclude that the minimum over $$p\geq 0$$ for which $$D^Tp=x$$ is less than or equal to $$b$$.
1. The variables in the objective function and the subproblem are different. Hence the minimization problem in the constraints does not influence the feasible region of $$x$$ directly.
2. Since the subproblem in the constraints is an optimization problem in the format: $$min \ f(x)\leq b$$ (it also works for $$max \ f(x)\geq b$$), the constraint can be meet as long as we can find a feasible solution meets $$f(x)\leq b$$ (or $$f(x)\geq b$$). Hence, we can just convert the optimization problem into inequality constraints.