Consider the primal objective to be
$\min_{x\in\mathbb{R}^n}x^\top Px$ subject to $Ax\leq b$
where $P$ is symmetric positive definite. To convert to dual objective, need to first state the Lagrange formula
$L(x,\lambda)=x^\top Px-\lambda^\top (Ax-b)$
then
$g(\lambda)=\inf_{x\in\mathbb{R}^n}L(x,\lambda)$
My questions are
(1) Why we can directly treat $\inf_{x\in\mathbb{R}^n}L(x,\lambda)$ as taking gradient w.r.t. $x$ and set it to $0$ so we get the $x_{min}$, and then plug it back to $L(x,\lambda)$ to get $\inf_{x\in\mathbb{R}^n}L(x,\lambda)$. Does it mean that $\inf_{x\in\mathbb{R}^n}L(x,\lambda)=L(x_{min},\lambda)$ ?
(2) If we did (1), aren't we just finish the primal problem? Then why we need to do it in dual?
(3) After getting the
$g(\lambda)=\inf_{x\in\mathbb{R}^n}L(x,\lambda)=-\frac{1}{4}\lambda^\top AP^{-1}A^\top\lambda-b^\top\lambda$
I know that our dual objective is $\max g(\lambda)$ but I have problem to determine the constraints for the dual objective. Does the constraint $\lambda\geq0$ naturally exist? Why we don't have other constraints and we don't have any $x$ in our dual objective. Then how do we know which $x$ is feasible for our primal?
Thank you very much
(1) Yes, since $L(x,\lambda)$ is a smooth strictly convex function.
(2) Minimizing the original problem (constrained) is not equivalent to minimizing Lagrangian (unconstrained)! they are related but not equivalent!
(3) $\lambda\ge 0$ must be there! There are no other constraints in this case but I would suggest thinking about the case where $P\not \succ 0$!