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SU-EE364A FEB102026

Last edited: February 2, 2026

Key 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, 2026

Consider 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, 2026

china ece

Last edited: February 2, 2026

Complementary Slackness

Last edited: February 2, 2026

Consider:

\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