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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?

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Reading the question, It reminded me a technical property called the ... assumption of Palais-Smale. Then I looked at the author of the question ... –  Denis Serre Dec 4 '10 at 14:57
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this page is so beautiful that I will print it and put in a frame... –  Pietro Majer Dec 4 '10 at 23:10

1 Answer 1

up vote 24 down vote accepted

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.

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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. –  Dick Palais Dec 4 '10 at 8:38
    
Thanks again, Theo, now you have done both of the things I asked for---given very nice proof AND an optimal counterexample. :-) –  Dick Palais Dec 4 '10 at 15:22
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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... –  Theo Buehler Dec 4 '10 at 15:30

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