If $f:R^n \to R$ is a smooth real-valued function such that $\nabla f : R^n \to R^n$ is a diffeomorphism, what can one conclude about the behavior of $f(x)$ at infinity?

Question

This question may seem peculiar, so let me preface it by saying that it arose while I was trying to understand Legendre transformations better, and in that context it is fairly natural. Anyway, suppose that $f$ is a smooth real-valued function on $R^n$ such that the gradient map, $\nabla f: p \mapsto {\partial f \over \partial x_i}(p)$, is a diffeomorphism of $R^n$ with itself. Of course two necessary conditions for this are that: (1) the hessian matrix ${\partial^2 f \over \partial x_i \partial x_j}(p)$ is everywhere non-singular, and (2) $\nabla f$ is a proper map, i.e., if $M > 0$ the the set where $ ||\nabla f|| \le M $ is compact. Moreover, it is not hard to show that these two conditions are together sufficient for the gradient map to be a diffeomorphism. Now my question is this: if $f$ is such a function does it follow that $f$ is also proper, i.e., that $\lim_{||x||\to \infty} |f(x)| = \infty$ ? That's clearly so if $n = 1$, but that is a very special case. For general $n$ I hope someone can show me a simple proof (but I also wouldn't be too surprised if I were shown a simple counter-example).

Added in response to Theo's simple and very nice counter-example: Suppose that the hessian is not only everywhere non-singular, but even everywhere positive definite. Can one then deduce that $f$ is proper?

Answer 1

The map $f(x,y) = xy$ has $\nabla f(x,y) = \left(\begin{array}{c} y \\\ x\end{array}\right)$ but $f(x,0) \equiv 0$, so $f$ is not proper.

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Edit in response to the modified question:

Positive definiteness of the Hessian implies strict convexity of $f$ and this indeed implies properness of $f$ as follows:

Since you assume that $\nabla f: {\mathbb R}^{n} \to {\mathbb R}^{n}$ is a diffeomorphism, $f$ has a unique minimum, namely the point $x_{0}$ where $\nabla f(x_0) = 0$. If $f$ were not proper, there would be a constant $M$ such that the closed convex set $C = \{x\, :\, f(x) \leq M\}$ is not compact. But then you would find a direction $y \in \mathbb{R}^{n} \smallsetminus \{0\}$ such that $x_{0} + t y \in C$ for all $t \geq 0$. The function $t \mapsto f(x_{0} + t y)$ is bounded and convex on $\mathbb R_{+}$ and it assumes its minimum. Thus it must be constant, in contradiction to the fact that $x_0$ is the unique minimum.