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#1: Initial revision by user avatar Pavel Kocourek‭ · 2023-01-23T14:58:24Z (about 1 year ago)
Is there a "regular" quasi-convex function $f:\Bbb R^2 \to \Bbb R$ that is not a monotone transformation of any convex function?
### Question

> Can you find an example of a  differentiable quasi-convex function $f:\Bbb R^2 \to \Bbb R$ that is *non-degenerate*, but there does not exist any strictly increasing $\phi:\Bbb R \to \Bbb R$ such that $\phi \circ f$ is convex?
>> **Definition.** We say that $f$ is ***non-degenerate*** iff $f'(x_0)=0$ implies $x_0\in \arg\min f(x)$. 

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### Context

Why the assumption that $f$ is non-degenerate is essential is shown in: https://math.stackexchange.com/q/4624119/1134951

I'm convinced that such an example does not exist in one dimension: https://math.stackexchange.com/q/4624226/1134951

I first considered functions in the form $f(x,y)=u(x)+v(y)$ and it seems that those can always be represented as a monotone transformation of a convex function if the construction I have in mind for the one-dimensional case works.

I believe that there must be a way to design a two-variable quasi-convex function that can not be represented as a monotone transformation of a convex function, but I don't readily see how.