duality
Last edited: February 2, 2026Lagrange Dual Problem
Given the Lagrange Dual Function, we can formulate the Lagrange Dual Problem:
\begin{align} \max_{\lambda, v}\quad & g\qty(\lambda,v) \\ \textrm{s.t.} \quad & \lambda \succeq 0 \end{align}
- finds the best lower bound on \(p^{*}\), obtained from Lagrange dual function
- a convex optimization problem, even if original primal fiction is not
- dual optimal value denoted \(d^{*}\)
\(\lambda, v\) as “dual feasible” if \(\lambda \succeq 0, \qty(\lambda, v) \in \text{dom } g\).
Example Dual Problems
LPs
\begin{align} \min_{x}\quad & c^{T}x \\ \textrm{s.t.} \quad & Ax = b \\ & x \succeq x \end{align}
Duality for Feasible Problems
Last edited: February 2, 2026Consider feasible problems.
\begin{align} \min_{x}\quad & 0 \\ \textrm{s.t.} \quad & \dots \end{align}
so the optimal is either \(p^{*} = 0\) or \(p^{*} = +\infty\). And thus the Lagrange Dual Problem boils down to checking if \(d^{*} \geq 0\), if so, then the original problem is’nt feasible (since dual gives a lower bound.)
Farkas' Lemma
Last edited: February 2, 2026\begin{equation} Ax \leq 0, c^{T}x < 0 \end{equation}
and
\begin{equation} A^{T}y + c = 0, y \geq 0 \end{equation}
are strong alteratives. This is through duality.
investment arbitrage
- invest \(x_{j}\) in each of \(n\) assets \(1 … n\) with prices \(p_1 … p_{n}\)
- suppose \(V_{ij}\) is the payoff of asset \(j\) and outcome \(i\)
- risk-free (cash): \(p_{1} = 1\), \(V_{i1} = 1\) for all \(i\) as our first investment
Arbitrage means there exists some \(x\) with \(p^{T}x < 0, V x \succeq 0\) (first thing is you borrow money?, and then second thing is you get more money back).
function convexity conditions
Last edited: February 2, 20261st order condition
differentiable \(f\) with convex domain is convex IFF:
\begin{equation} f\qty(y) \geq f\qty(x) + \nabla f\qty(x)^{T} \qty(y-x), \forall x,y \in \text{dom } f \end{equation}
“the function is everywhere above the Taylor approximation” => “first-order Taylor approximation of \(f\) is a global underestimator of \(f\).”
2nd order condition
for twice differentiable \(f\) with convex domain, we have:
- \(f\) is convex IFF \(\nabla^{2} f\qty(x) \succeq 0, \forall x \in \text{dom } f\) (i.e. that the Hessian is PSD)
- if \(\nabla^{2} f\qty(x) \succ 0 \forall x \in \text{dom } f\), then we call \(f\) strictly convex (i.e. that the Hessian is PD)
you may enjoy using Cauchy-Schwartz Inequality to show these.
Geometric Interperattion of the Dual
Last edited: February 2, 2026Let \(\mathcal{G} = \qty {f_{1}\qty(x), f_{0}\qty(x) \mid x \in \mathcal{D}}\) be the set of achievable constraint/objective pairs. The Lagrange Dual Function \(g\qty(\lambda) = \text{inf}_{t,u \in G} \qty(t + \lambda u)\).
We will push this line all the way down, and then push up its slope as much as possible.
