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Let $f: \mathbb R^n \to \mathbb R$ be a smooth function. Then the first derivative $f^{(1)}$ makes sense as a function $\mathbb R^n \to \mathbb R^n$, and the second derivative makes sense as a function $f^{(2)}: \mathbb R^n \to \{\text{symmetric }n\times n\text{ matrices}\}$. I would like either a proof or a counterexample to the following claim:

Claim: Suppose that for every $x\in \mathbb R^n$, $f^{(2)}(x)$ is invertible as an $n\times n$ matrix. Then $f^{(1)}: \mathbb R^n \to \mathbb R^n$ is one-to-one.

Some comments:

  1. If you replace $\mathbb R$ by $\mathbb C$ and "smooth" by "algebraic", then the only such functions are (inhomogeneous) quadratic in $x$, since over $\mathbb C$ all non-constant functions have zeros. Then $f^{(1)}$ is (inhomogeneous) linear, and so one-to-one.
  2. If you replace "invertible" by "positive (or negative) definite", then $f$ is convex, and so the claim follows. In particular, the claim is true when $n=1$.
  3. This is a special case of the following more general question. If $g: \mathbb R^n \to \mathbb R^n$ is smooth, then its derivative makes sense as a function $g^{(1)}: \mathbb R^n \to \{n\times n\text{ matrices}\}$. If $g^{(1)}(x)$ is invertible for all $x\in \mathbb R^n$, is $g$ necessarily one-to-one? Of course, then $g$ is locally a diffeomorphism, but I don't know if it is globally. I don't think it is.

Oh, and I have no idea what the best tags are.

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3  
For 3, you can take any local but not global diffeomorphism, for instance one that maps the plane onto a horseshoe shape where the two ends overlap. (Or, more explicitly, the complex exponential function.) –  Dylan Thurston Apr 2 '10 at 6:12
8  
Dylan, your example promotes to a counterexample to the original question by composing with complex conjugation: $f(x,y) = e^x \cos(y)$ has gradient $f^{(1)}(x,y) = e^{x-iy}$. –  BCnrd Apr 2 '10 at 6:48

2 Answers 2

up vote 1 down vote accepted

The counter example given in the comments by Brian Conrad (and Dylan Thurston) is very nice. However, it oscillates wildly at $\infty$, and I believe that if you assume some nice properties at $\infty$ you will get a possitive answer.

Translating $f$ by a linear function does not change the assumptions on $f$ and so the claim is equivalent to the fact that all such $f$ has no or a unique critical points.

If we add assumptions such that e.g. the Conley index (or homotopy index) is well-defined and a sphere then this modified claim would follow. Indeed, if two or more critical points existed they would by assumptions be non-degenerate with same Morse index and a small pertubation would yield a Conley index which is a vedge of two or more spheres - a contradiction.

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Regarding comment 3: If you restrict to polynomial functions $g$, then the question becomes a famous open problem (the Jacobian conjecture).

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