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:

- 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.
- 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 $n=1$.
- 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.