Hidden convexity

Question

Suppose you are given a domain $\Omega \subset \mathbb{R}^n,$ and a (Morse) function $f: \Omega \rightarrow \mathbb{R},$ all of whose critical points are positive-definite.

The question is: is there a diffeomorphism $\phi: \Omega \rightarrow \Omega$ such that $f \circ \phi$ is a convex function? Another edit The above is a typo (or braino): I actually meant $\phi: \Omega \rightarrow \Omega_1,$ for some $\Omega_1 \subset \mathbb{R}^n.$

EDIT A bit of motivation: this comes from (many years ago) constructing a counterexample to the statement that a convex function has a unique minimum (which is true on a convex set, but not otherwise). If a function has a unique critical point, the question is fairly easy (just by looking at level sets, and sending them to round circles).

Answer 1

Edited in response to edits of the question. Original answer follows new answer.

If you don't require that $f$ is proper, then there are many counterexamples already in $\mathbb R^2$ (see below, in the answer to question as previously worded).

If you require that $f$ is a proper Morse function (that is, $f^{-1}$ of a compact set is compact, then:

1. Every level set is compact (by properness).
2. There is at least one local minimum: minimize in the region bounded by a level surface.
3. There is at most one local minimum: Otherwise find a minimax arc connecting two local minima.
4. Therefore all level surfaces are spheres: start from a critical point and go up.
5. Therefore $f$ is topologically equivalent to $|x|^2$ on $\mathbb R^n$.

So, with these assumptions, the answer is yes.

Original answer: Counterexamples if $f$ is required to be equivlaent to a convex function on a fixed domain $\Omega$, or if $f$ is not required to be proper.

No.

For any convex function, the sets $\left \{ x: f(x) < a \right \}$ are all convex.

Therefore, if the limit of $f$ on the boundary is constant, the boundary of the set must be convex.

Make a diffeomorphism $\phi$ of the plane that sends the unit disk to a peanut-shaped region $\Omega$. If you conjugate $x^2+y^2$ by $\phi$, it can't be made convex by a diffeomorphism of $\Omega$.

You can modify the example if you like to make $f$ a proper Morse function, if you conjugate $\tan ( 2/\pi (x^2+y^2)$ by $\phi$. Without the restriction that $f$ is proper, it's possible to make Morse functions on any domain $\Omega$ on any noncompact manifold of dimension at least 2 to have any chosen discrete set of Morse singularities. This is done by making local models for the function near the chosen singularities, extending it smoothly any way at all as a Morse function, and then pushing all the unwanted singularities to infinity by isotopies along disjoint rays.

For instance, in the plane, there are uncountably many isomorophism classes of foliations that come from level sets of a real-valued function. Each leaf is proper, and the space of leaves is a non-Hausdorff (branched) 1-manifold. A good example is what you get from the function $x*y$ in $\mathbb R^2 \setminus 0$ by passing to the universal cover, that is, in complex notation, the function Re(exp(z))Im(exp(z)). This is not topologically conjugate to a convex function. Here is a contour plot of $\arctan (Re(\exp (z)) Im(\exp(z)))$ (the arctangent is to give better spacing of the contours).

alt text http://dl.dropbox.com/u/5390048/NonconvexNonsingular.jpg

Answer 2

It seems that the following interpretation is not covered yet: the domain is allowed to change but $f$ is not required to be proper. In this case the answer is no. Take any connected domain on the plane and any Morse function with at least two local maxima. Remove all critical points (except possibly local minima) from the domain, now all critical points are local minima.

Near the two local maxima, there are two closed level curves $\gamma_1$ and $\gamma_2$ such that $\gamma_2$ lies in the "lower" component of $\mathbb R^2\setminus\gamma_1$ and vice versa, where by "lower" component of $\mathbb R^2\setminus\gamma_i$ I mean the one where the values of $f$ are locally (near $\gamma_i$) smaller than on $\gamma_i$. This property of a pair of curves is preserved by diffeomorphisms. But it cannot hold for a convex function because, for a convex $f$, the "lower" component must be the bounded (convex) region enclosed by the level curve, but two curves cannot be both inside each other's bounded regions.

Answer 3

This is not an answer, and I also doubt I'm going to say anything Igor hasn't already thought of. But here are some random thoughts:

1. You want to assume $\Omega$ is connected, right? Assuming that, I think it must be straightforward to show that there is only one critical point. Say, because $f$ has to be a perfect Morse function, maybe?

2. Once you know that, then you know that the level sets of $f$ outside but sufficiently close to the critical point must be spheres.

3. It seems easier if we assume that $f$ is constant along the boundary of $\Omega$. Then it seems likely that you can construct a diffeomorphism that turns $f$ not just into a convex function but into the canonical one $x \mapsto |x|^2$.

So, Igor, where does all of this go wrong?