TITLE: Hidden convexity
QUESTION [7 upvotes]: 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).

REPLY [5 votes]: 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.