Timeline for 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?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2015 at 14:11 | history | edited | David E Speyer | CC BY-SA 3.0 |
edited body
|
| Dec 4, 2010 at 15:30 | comment | added | Theo Buehler | 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... | |
| Dec 4, 2010 at 15:22 | vote | accept | Dick Palais | ||
| Dec 4, 2010 at 15:22 | comment | added | Dick Palais | Thanks again, Theo, now you have done both of the things I asked for---given very nice proof AND an optimal counterexample. :-) | |
| Dec 4, 2010 at 11:45 | history | edited | Theo Buehler | CC BY-SA 2.5 |
Polished the argument
|
| Dec 4, 2010 at 9:17 | history | edited | Theo Buehler | CC BY-SA 2.5 |
Corrected a slip
|
| Dec 4, 2010 at 9:09 | history | edited | Theo Buehler | CC BY-SA 2.5 |
Added answer to the modified question
|
| Dec 4, 2010 at 8:38 | comment | added | Dick Palais | 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. | |
| Dec 4, 2010 at 7:02 | history | answered | Theo Buehler | CC BY-SA 2.5 |