I believe i have a major misunderstanding surrounding quasi-convex constraints in maximisation, when using monotone functions. Can you help me spot my errors please?
The definition of a quasi-convex function that I will be using is:
A function $f$ from $\mathbb{R^n}$ to $\mathbb{R}$ is quasi-convex if, $\forall \alpha \in \mathbb{R^n}$, the set
$$\{x\in D: f(x)<\alpha\},$$
where $D$ is the domain of $f$, is a convex set.
Monotonic Functions are Quasi-convex and Quasi-concave - Wikipedia
- If i take $f: \mathbb{R^{2+}} \to \mathbb{R^+}$ where $f=(x_1x_2)^{1/2}$ why are my lower contour sets not convex? (See attached image). In my mind, the function is monotonic, therefore it is quasi-convex and quasi-concave, but i cannot show quasi convexity using the contours, as per the definition.
Quasi-concave constraint in minimisation problem
- A lecture video times-tamped I was watching made the point that when defining our equality constraint in a minimisation problem, we have to write $f(x_1, x_2) - q = 0$, rather than $q - f(x_1, x_2) = 0$ because we need the constraint to be quasi-concave. But again in the video i believe the constraint is strictly increasing, so surely the constraint is already quasi-convex and quasi-concave? So it doesn't matter?
Where I am possibly going wrong
- I'm misunderstanding the relationship between a monotonic function and quasi-convexity/concavity? Perhaps its different when our domain is in $\mathbb{R^n}$ vs $\mathbb{R}$
- I'm misunderstanding the terms monotonic, strictly increasing, and non decreasing, and they apply differently to quasi-convexity/concavity than I imagine?
- Everyone else is wrong and I'm about to reinvent optimisation theory?