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?