Complementary 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
conjugate function
Last edited: February 2, 2026The conjugate of a function \(f\) is \(f^{*}\qty(y) = \text{sup}_{x \in \text{dom }f} \qty(y^{T}x - f\qty(x))\). \(f^{*}\) is convex, even if \(f\) is not.
(fyi \(\text{sup}_{x} = \max_{x}\))
convex function
Last edited: February 2, 2026a function for which, given any two points, the function between those points sits at (lines are convex!) or below the plane given those points
constituents
For \(f: \mathbb{R}^{n} \to \mathbb{R}\)
requirements
\begin{equation} f\qty(\theta x + \qty(1-\theta) y) \leq \theta f\qty(x) + \qty(1-\theta) f\qty(y) \end{equation}
strictly convex is the strict inequality
additional information
log conditions
- \(f\) is log-linear IFF log \(f\) is affine
- \(f\) is log-concave iff log \(f\) is concave
- \(f\) is log-convex IFF log \(f\) is convex
check if something is a convex function
- check definition
- restrict it to a line: convexity preserve line restriction
- 1st order condition
- 2nd order condition
- show that \(f\) is constructed by operations that preserve fuction convexity
some convex functions
- affine: \(ax + b\)
- exponential: \(e^{ax}\)
- powers on \(R_{++}\): \(x^{\alpha}\) for \(\alpha \geq 1\) or \(\alpha \leq 0\)
- \(|x|^{p}\) for \(p \geq 1\)
- relu
- any norm
- sum of squares: \(\qty {x}^{2}_{2} = x_1^{2} + … + x_{n}^{2}\)
- max function: \(\max \qty(x) = \max \qty {x_1 \dots x_{n}}\)
- softmax: \(\log \qty(\exp x_1 + \dots + \exp x_{n})\)
- general affine function: \(f\qty(X) = tr\qty(A^{T} X) + b\) (“an inner product”)
- spectral norm: \(f\qty(X) = \norm{X}_{2} = \sigma_{\max}\qty(X)\) (the maximum singular value of \(X\)
- logsumexp: \(f\qty(x) = \log \sum_{k=1}^{n} \exp x_{k}\)
- quadratic over linear: \(f\qty(x,y) = \frac{x^{2}}{y}, y >0\)
- quadratic: \(f\qty(x) = \frac{1}{2} x^{T} P x + q^{T} x + r\), with \(P \succeq 0\) is convex
- least squares: \(f\qty(x) = \norm{Ax - b}^{2}_{2}\) is convex for any \(A\)
- inverse product: \(f\qty(x) = \frac{1}{\prod_{i=1}^{n} x_{i}}\)
- inv_pos on \(R_{++}\): \(f\qty(x) = \frac{1}{x}\) is convex if \(x\) is concave and positive
some concave fuctions
- affine
- square root
- fractional powers
- min
- logs: \(\log x\)
- entropy: \(- x \log x\)
- negative part (opposite relu)
- log determinant: \(f\qty(X) = \log \text{det} X\)
- geometric mean: \(f\qty(x) = \qty(\prod_{k=1}^{n} x_{k})^{\frac{1}{n}}\) an \(\mathbb{R}_{++}^{n}\)
sublevel set
\begin{equation} C_{\alpha} = \qty {x \in \text{dom f} \mid f\qty(x) \leq \alpha } \end{equation}
convex set
Last edited: February 2, 2026constituents
- set \(C\)
- \(\theta \in \mathbb{R}\)
requirements
A set \(C\) is a convex set if:
\begin{equation} x_1, x_2 \in C, 0 \leq \theta \leq 1 \implies \theta x_{1} + \qty(1-\theta) x_{2} \in C \end{equation}
definitions
additional information
operations that preserve convexity
Convex sets is a calculus! Methods to showing complexity:
Anything in Euclidian Geometry Crash Course
- apply definition: show \(x_1, x_2 \in C, 0 \leq \theta \leq 1 \implies \theta x_1 + \qty(1-\theta) x_2 \in C\)
- use convex functions
- show that \(C\) is obtained from convex sets via the following operations
intersection
Intersections of any number of convex sets, include infinite, are convex.
dual transformation
Last edited: February 2, 2026introducing new variables
Consider the original problem:
\begin{align} \min_{x}\quad & \norm{Ax - b} \end{align}
Now, we can reformulate and write \(y = Ax - b\). Now, remember the conjugate of the norm is:
\begin{equation} \norm{z} = \begin{cases} 0, \norm{z} \leq 1\\ \infty \end{cases} \end{equation}
And remember the dual is some massaging of the conjugate. And thus we can formulate a “norm conjugate”:
\begin{align} \max_{v}\quad & b^{T}v \\ \textrm{s.t.} \quad & A^{T}v = 0 \\ & \norm{v} \leq 1 \end{align}
