MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
I guess I have this all wrong. What's an example of a convex function on a nonconvex domain with more than one minimum? – Deane Yang Jan 31 '11 at 1:55
Igor, I delete my answer, the ref I gave does not prove the statement. – Anton Petrunin Jan 31 '11 at 6:43
up vote 6 down vote accepted

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.


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

share|cite|improve this answer
Quick question: What do you mean by "conjugate"? – Deane Yang Jan 31 '11 at 1:48
@Deane Yang. Sorry. I'm thinking of $f$ transported by $\phi$, so the formula is $f(\phi^{-1}(x))$; I suppose it's a misnomer to call this "conjugation", although it would be nice to have a single term applying in general to the concept of transforming a structure by a homeomorphism. – Bill Thurston Jan 31 '11 at 2:13
@Bill: my apologies, I misspoke (see Another Edit in the statement) -- it has been a while since I thought about this. I think that makes your peanut example not work; I have not quite absorbed your second example... – Igor Rivin Jan 31 '11 at 4:15
@Igor: I actially wondered whether you meant $\Omega \rightarrow \Omega_1$, and I probably should have worded it conditionally on the intended meaning. I suspect you also intended (or should have intended) to make $f$ a proper function --- if so, that would get rid of the second example, and change the game. – Bill Thurston Jan 31 '11 at 4:43
Bill, thanks for the clarifications! – Deane Yang Jan 31 '11 at 14:13

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.

share|cite|improve this answer
This sounds like what Anton has been trying to explain to me. Is it? – Deane Yang Feb 1 '11 at 0:21
I'm not sure I understand your discussion with Anton but it seems that your are talking about different things: you say something about a proper Morse function and Anton says that what you say does not hold for a non-proper one. You are both right. – Sergei Ivanov Feb 1 '11 at 0:39
A nice argument! – Igor Rivin Feb 1 '11 at 1:03
What if $\Omega$ is contractible? – Anton Petrunin Feb 1 '11 at 14:20

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?

share|cite|improve this answer
#1 is wrong: Think of two discs connected by a narrow winding path. – Anton Petrunin Jan 31 '11 at 4:29
Anton, could you say more? I seem to be rather dense about this. Why wouldn't there be another critical point along the path? – Deane Yang Jan 31 '11 at 4:33
Its winding, you go up and down, the level sets are open, and no critical point... – Anton Petrunin Jan 31 '11 at 6:36
Anton, thanks! Unfortunately, I'm still having trouble understanding the example, but I'm sure this is just my limitation, not yours. Are you saying that it's possible to have two disks, each with a local minimum joined with a narrow winding path, where there are no other critical points? I just can't see this. – Deane Yang Jan 31 '11 at 14:11
It appears that what I had in mind is the first situation described by Bill Thurston but I hadn't stated my assumptions properly. – Deane Yang Jan 31 '11 at 14:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.