SU-EE364A FEB102026
Last edited: February 2, 2026Key Sequence
Notation
New Concepts
Important Results / Claims
Questions
Interesting Factoids
- fun fact: any (even non-convex) problem in two quadratics usually has zero duality gap and thus its dual can be the solution to the primal problem
worst-case robust least-squares
Last edited: February 2, 2026Consider some kind of ellipsoid where your data is constrained:
\begin{align} \mathcal{A} = \qty {\bar{A} + u_1 A_1 + \dots + u_{p} A_{p} \mid \norm{u}_{2} \leq 1} \end{align}
You can form the “worst-case robust least squares”:
\begin{align} \min \text{sup}_{A \in \mathcal{A}} \norm{A x - b}_{2}^{2} = \min \text{sup}_{\norm{u}_{2} \leq } \norm{P\qty(x) u + q\qty(x)}_{2}^{2} \end{align}
This is usually a minimax problem, but taking the dual of the inner maximize thing turns out has zero duality gap.
Approximation and Fitting
Last edited: February 2, 2026china ece
Last edited: February 2, 2026Complementary Slackness
Last edited: February 2, 2026Consider:
\begin{align} f_{0}\qty(x^{*}) &= \text{inf}_{x} \qty(f_{0}\qty(x) + \sum_{i=1}^{m} \lambda_{i}f_{i}\qty(x) + \sum_{i=1}^{p} v_{i}h_{i}\qty(x)) \\ &\leq f_{0}\qty(x) + \dots \\ &\leq f_{0}\qty(x) \end{align}
So the inequality holds strictly
