Can a quasiconvex function be made convex by composition with a diffeomorphism? Assume we are given a continuous quasiconvex function $f: \mathbb{R}^n \to \mathbb{R}$. Intuitively I feel that quasiconvexity means that there should exist a diffeomorphism $h: \mathbb{R}^n \to \mathbb{R}^n$ such that $f(h)$ is convex. I suppose, this would also mean that quasiconvexity is somehow equivalent to geodesic convexity. Does anybody know of any results in this direction?  
 A: No, there are continuous quasiconvex functions  that cannot be written as a convex function composed with a diffeomorphis. I give three reasons.


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*Every convex function $g:\mathbb{R}^n\to\mathbb{R}$ is locally Lipschitz. So are diffeomorphisms. Therefore, the composition of a diffeomorphism with a convex function is locally Lipschitz. But, for example, $f(x)=\sqrt{\|x\|}$ is a quasiconvex function on $\mathbb{R}^n$ that fails the locally Lipschitz property.


After seeing the above, one may want to either (a) assume $f$ locally Lipschitz, or (b) allow homeomorphisms instead of diffeomorphisms. However, this won't help against the following two obstructions.


*A nonconstant convex function must be unbounded from above (consider its restriction to a line: most of the graph lies above a secant line). For a quasiconvex function this isn't the case: $f(x)=\min(\|x\|,1)$ is quasiconvex. There is no way to write this as a composition of a convex function with a homeomorphism. 

*For a convex function $g$, every level set $g^{-1}(t)$ except at most one has empty interior. This can be proved by drawing a line intersecting two level sets with nonempty interior: the restriction of $g$ to this line will not be convex. On the other hand, it's easy to imagine a monotone (hence quasiconvex) function $f:\mathbb{R}\to\mathbb{R}$ that has two or more level sets with nonempty interior. 
