21
$\begingroup$

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?

$\endgroup$
2
  • 7
    $\begingroup$ Reading the question, It reminded me a technical property called the ... assumption of Palais-Smale. Then I looked at the author of the question ... $\endgroup$ Dec 4, 2010 at 14:57
  • 4
    $\begingroup$ this page is so beautiful that I will print it and put in a frame... $\endgroup$ Dec 4, 2010 at 23:10

1 Answer 1

28
$\begingroup$

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.


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.

$\endgroup$
3
  • $\begingroup$ Thanks Theo ! That's a very neat and simple counter-example. It is also exactly the kind of example that I was looking for; it indicates why in the Legendre transform setting one usually assumes that the hessian is positive definite and not just non-singular, which is just the point that I was trying to understand. So now I will edit the question and ask if strengthening the hypothesis in that way is enough to imply that $f$ is proper. $\endgroup$ Dec 4, 2010 at 8:38
  • $\begingroup$ Thanks again, Theo, now you have done both of the things I asked for---given very nice proof AND an optimal counterexample. :-) $\endgroup$ Dec 4, 2010 at 15:22
  • 2
    $\begingroup$ It was a pleasure. It took me a moment to see the second part, that's why the first answer was so terse and not so helpful. By the way, I couldn't resist chuckling at Denis Serre's comment - it occurred to me as well, but I didn't dare... $\endgroup$ Dec 4, 2010 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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