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.
