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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?

Ilya
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No, there are continuous quasiconvex functions that cannot be written as a convex function composed with a diffeomorphis. I give three reasons.

  1. 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.

  1. 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.

  2. 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.

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  • Dear soup, thank you so much for your reply. Could you please help me with your 3 point? Why does $g^{-1}(t)$ have an empty interior? Do you mean relative interior? For example if $g=x^2$, then $g^{-1}(t)$ are intervals and have non-empty interior in their relative topogoly. Thanks! – Ilya Jun 14 '16 at 06:55
  • Regarding your correct 2 point, if we also avoid such "flat" function and still deal with quasiconvex functions, does this help? I would like to concentrate on your first example with the norm. Indeed, there is a problem with being Lipschitz and allowing $h$ to be a homeomorphism solves the problem. Is this the general case? – Ilya Jun 14 '16 at 07:00
  • If $g(x)=x^2$, then $g^{-1}(t)={-\sqrt{t},\sqrt{t}}$ which has empty interior. I'm talking about level sets, not sub-level sets. –  Jun 14 '16 at 07:08
  • Yes, sure, it was my bad) But what if we do not allow flat regions? – Ilya Jun 14 '16 at 07:17