How is the dual problem for conic programs derived via Lagrangians?

Question

I'm trying to get a better grip of conic programming and the relations between primal and dual problems.

Given a convex problem in standard form, e.g. $\min_x f(x)$ subject to $f_i(x)\le0$, one standard approach (discussed e.g. in these pdf notes) is to introduce the Lagrangian function $$L(x,\lambda) \equiv f(x) + \sum_i \lambda_i f_i(x), $$ and then argue that if $L(x,\lambda)\ge\alpha$ for all $x$, then the solution to the primal problem is at least $\alpha$.

Consider now a standard conic programming problem, in the form (using the notation from these other pdf notes): $$\max_x \{ \langle a,x\rangle, \,\, \phi(x)=b, \,\, x\in K \},$$ where $\phi$ is a linear function, and $K$ a closed convex cone. To argue that the dual problem leads a solution larger than the primal one, the argument they use is a bit different: they show that $$\langle a,x\rangle \le \langle y,b\rangle$$ whenever $\phi^*(y)-b\in K^*$, where $\phi^*$ is the adjoint of $\phi$, and $K^*$ the dual cone of $K$.

I've read that duality in conic programming should also be derivable via the Lagrangian approach (it is mentioned, but not further elaborated, on top of page 7 in these other pdf notes). How would we actually do this, explicitly? I'm not quite seeing how defining a Lagrangian like $L(x,\lambda)=\langle a,x\rangle+\lambda^T (\phi(x)-b)$ will eventually lead me to the dual problem. Mostly because I'm not sure how to encode the $x\in K$ constraint in it.

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A related question is How to derive the dual of a conic programming problem, $\min_{x\in L}\{c^T x: \,\, Ax-b\in K\}$?

Answer 1

The main idea is that we can use the relation $x\in K$ implicitly; we could penalize it directly in the Lagrangian, but we can also use the fact that $x\in K$ to derive a relation on other quantities.

Consider \begin{align} \begin{array} &\nu (P) := & \max & \langle c, x\rangle \\ &\text{s.t.} & Ax = b\\ && x\in K \end{array} \end{align} for linear function $A$ (and adjoint $A^*$). Then the Lagrangian can be formed by \begin{align} L(x;\lambda) &= \langle c,x \rangle + \langle \lambda, b-Ax\rangle = \langle c,x \rangle + \langle \lambda, b\rangle -\langle \lambda,Ax\rangle\\ &=\langle c - A^*\lambda ,x \rangle + \langle \lambda, b\rangle\\ \end{align} and since $x\in K$, we can ensure that $L(x;\lambda) \geq \langle c, x\rangle$ as long as we require $c-A^*\lambda\in K^*$ (since for any $x\in K$ and $y\in K^*$, $\langle x,y \rangle\geq0$).

The dual can be derived as minimizing the upper bound $(b^\top \lambda)$ subject to the constraints that make the upper bound valid ($c-A^*\lambda\in K^*$): \begin{align} \nu(P) \leq \min_\lambda \{\langle b,\lambda\rangle : c - A^*\lambda \in K^*\}. \end{align}