The Boyd Convex Optimization book defines the Lagrange dual function as $$ g(\lambda, \nu) = \inf_{x \in D} L(x, \lambda, \nu) = \inf_{x \in D} (f_0(x) + \lambda^T f(x) + \nu^T h(x) $$ where $f$, $h$ are the inequality and equality constraint functions, and the problem domain is defined as $D=\operatorname{dom} f_0 \cap \operatorname{dom} f \cap \operatorname{dom}g$. Also define the feasible set as $X = \{x \in D | f(x) \leq 0, h(x) =0 \}$, and primal solution value $p^*=\inf_{x\in X}f_0(x)$. I have the following questions:

  1. If I define $g(\lambda, \nu) = \inf_{x \in S} L(x, \lambda, \nu)$, where $S$ is any convex set such that $X \subset S$, it seems that weak/strong duality theorems would still hold. In particular, we still have for any feasible $\tilde x \in X$ and any $\lambda \geq 0, \nu$, $$g(\lambda, \nu) = \inf_{x \in S} L(x, \lambda, \nu) \leq L(\tilde x, \lambda, \nu) \leq f_0(x)$$ hence $g(\lambda, \nu) \leq p^*$ (by ranging over all $\tilde x \in X$). Provided the set $S$ is easy to optimize over, wouldn't this be a useful way to get a potentially tighter dual lower bound on $p^*$ than is typically obtained by minimizing $L$ over $D$, since $\inf_{x \in D} L(x, \lambda, \nu) \leq \inf_{x \in S} L(x, \lambda, \nu)$?

  2. In most of the textbook examples, the domain $D$ is simple (e.g., $x \geq 0$), and $g$ can be found analytically. This may not always be the case in practice. What if $D$ is a more complicated set? Would it make sense to also use Lagrange multipliers and KKT conditions in finding $g$? (Also see this question, basically asking about general approaches for handling (potentially analytically intractable) dual functions)

  1. Yes. Let me give an example. Suppose you'd like to derive the dual of $\min_{x \in S} \{ x^2 : x \geq 1\}$ where $S=\mathbb{R}$. You could consider $D=\mathbb{R}_+$ and have the same problem. The duals are different: $\sup_y \{ y - 0.25 y^2 \}$ vs $\sup_y \{y - 0.25\max\{0,y\}^2\}$. For negative $y$, the second dual provides a tighter dual bound. The practical value is debatable, because I have yet to see someone plug in a random value into the dual to obtain a useful bound.

  2. If you can find $g$ with the use of KKT conditions, I would say $D$ is still simple. In the most extreme case, $D=X$, it makes sense to use KKT conditions.


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