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).



Hidden convexity

> 
> 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](http://dl.dropbox.com/u/5390048/NonconvexNonsingular.jpg)


