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

  1. 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.

enter image description here

Quasi-concave constraint in minimisation problem

  1. 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

  1. 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 $\mathbb{R}$
  2. I'm misunderstanding the terms monotonic, strictly increasing, and non decreasing, and they apply differently to quasi-convexity/concavity than I imagine?
  3. Everyone else is wrong and I'm about to reinvent optimisation theory?

1 Answer 1


Real-valued Monotonic functions defined on real line or subset of real line are both quasi-concave and quasi-convex, but that is not necessarily the case if the function is defined on $\mathbb{R}^n$ or its subset, where $n\geq 2$. For example, all these are monotonic functions:

  • $f$ defined on $\mathbb{R}^2_+$ and as $f(x_1,x_2)=x_1^\frac{1}{2}x_2^\frac{1}{2}$ is quasi-concave, but not quasi-convex.
  • $f$ defined on $\mathbb{R}^2_+$ and as $f(x_1,x_2)=x_1^2+x_2^2$ is quasi-convex, but not quasi-concave.
  • $f$ defined on $\mathbb{R}^2_+$ and as $f(x_1,x_2)=x_1 + x_2$ is both quasi-convex and quasi-concave.
  • $f$ defined on $\mathbb{R}^2_+$ and as $f(x_1,x_2)=\sqrt{x_1} + x_2^2$ is neither quasi-convex nor quasi-concave.
  • $\begingroup$ Superb! So it actually was option 1 regarding the possible places I thought I might be going wrong! Would you care to distinguish, monotonic, and strictly increasing for me? $\endgroup$
    – CormJack
    Feb 14, 2023 at 17:13

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