TITLE: Recovering the solution of optimization problem from the dual problem
QUESTION [5 upvotes]: In the context of (most of the times convex) optimization problems -
I understand that I can build a Lagrange dual problem and assuming I know there is strong duality (no gap) I can find the optimum of the primal problem from the one of the dual. Now I want to find the primal optimum point (i.e. the point in which the optimum is attained). Somehow, it is assumed that both the optimum of the primal and dual point are achieved at the same point (which is then a saddle point). Why/when is that true? Does strong duality can happen only at a saddle point? Does that require the primal problem to be convex?
Now, from this, it is assumed it's enough for to find the minimizer of the Lagrangian at the value of dual parameters (point of the solution of the dual problem) in order to get the primal optimal point. Why is that true?
Thank you,
Dany

REPLY [5 votes]: I know this question was asked some time ago, but I just stumbled upon it myself wondering the same thing, so I will post what I have found out in case it is helpful to someone else.
I think the answer you seek is in Boyd and Vandenberghe's Convex Optimization (freely available on Boyd's website: http://stanford.edu/~boyd/cvxbook/) in sections 5.1-5.5 (especially, see subsection 5.5.5). Below are my key takeaways from that text.
Let 
\begin{align} 
\min \;\;\;\; &f_0(x)\\
\text{s.t.} \;\;\;\; &f_i(x) \leq 0 \;\;\;\; i = 1,\ldots,m \tag{1} \label{eq:1}\\
 &h_i(x) = 0 \;\;\;\; i = 1,\ldots,p,
\end{align}
where $x \in \mathbb{R}^n$ be the optimization problem (not necessarily convex) we're considering. The Lagrangian of this problem is
$$
L(x, \lambda, \nu) = f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \sum_{i=1}^p \nu_i h_i(x).
$$
The Lagrangian dual is
$$
g(\lambda,\nu) = \inf_{x \in \mathcal{D}} \left( f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \sum_{i=1}^p \nu_i h_i(x) \right),
$$
where $\mathcal{D} := \left( \bigcap_{i=0}^m \operatorname{dom} f_i \right) \cap \left( \bigcap_{i=1}^p \operatorname{dom} h_i \right)$. The Lagrangian dual problem, in turn, is
\begin{align}
\max \;\;\;\; &g(\lambda,\nu)\\
\text{s.t.} \;\;\;\; &\lambda \succcurlyeq 0.
\end{align}
Note that 
\begin{align}
\sup_{\lambda \succcurlyeq 0, \; \nu} L(x,\lambda,\nu) 
&= \sup_{\lambda \succcurlyeq 0, \; \nu} \left( f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \sum_{i=1}^p \nu_i h_i(x) \right) \\
&= 
\begin{cases}
f_0(x) & \text{if } f_i(x) \leq 0, \; i = 1,\ldots,m, \text{ and } h_i(x) = 0, \; i=1,\ldots,p \\
\infty & \text{otherwise}.
\end{cases} 
\end{align}
This means that 
$$
p^\star := \inf_x \sup_{\lambda \succcurlyeq 0, \; \nu} L(x, \lambda, \nu)
$$
is the optimal value of the primal problem \eqref{eq:1}. Note that there is no restriction on the $x$ values over which we take the infimum, since the conditions of \eqref{eq:1} are already included in the $\sup L$. Moreover, by definition,
$$
d^\star := \sup_{\lambda \succcurlyeq, \; \nu} \inf_x L(x,\lambda,\nu)
$$
is the optimal value of the dual problem.
Let us now assume that strong duality holds. Then $d^\star = p^\star$. Furthermore, also assume that both the primal and dual problem optimal values are attained, i.e., that there exist feasible $x^\star$ and $(\lambda^\star,\nu^\star)$ such that $f(x^\star) = d^\star$ and $g(\lambda^\star,\nu^\star) = d^\star$. Then
\begin{align}
f_0(x^\star) 
&= g(\lambda^\star,\nu^\star) \\
&= \inf_x \left( f_0(x) + \sum_{i=1}^m \lambda_i^\star f_i(x) + \sum_{i=1}^p \nu_i^\star h_i(x) \right) \\
&\leq f_0(x^\star) + \sum_{i=1}^m \lambda_i^\star f_i(x^\star) + \sum_{i=1}^p \nu_i^\star h_i(x^\star) \\
&\leq f_0(x^\star)
\end{align}
(for motivations of each equality/inequality, see page 242 in Boyd and Vandenberghe's book). This implies that the inequalities are in fact equalities. We from this, we can conclude that $x^\star$ solves the problem $\inf_x L(x,\lambda^\star,\nu^\star)$. Finally, assume that $\inf_x L(x,\lambda^\star,\nu^\star)$ has a unique solution (which happens e.g. if $L(x,\lambda^\star,\nu^\star)$ is a strinctly convex function of $x$). Then $x^\star$ is that solution, and therefore
$$
x^\star = \arg\min_x L(x,\lambda^\star,\nu^\star).
$$