Let $f: \Bbb R^n\to\Bbb R$, the Hessian of $f$ is postive definite everywhere. Show that $\mathrm{grad} \, f: \Bbb R^n\to\Bbb R^n$ is bijective.

Question

Let $f: \Bbb R^n\to\Bbb R$, the Hessian of $f$ is postive definite everywhere. Show that $\mathrm{gradient} \, f: \Bbb R^n\to\Bbb R^n$ is bijective.

Argue by contradtion, if $\mathrm{grad} \, f(x_1)=\mathrm{grad} \, f(x_2)$ for some $x_1\neq x_2$, then by Rolle's theorem, it is easy to derive a contradition. But how to prove the surjectivity of $\mathrm{grad} \, f=(f_1',\cdots,f_n')$?

Answer 1

It is not true, consider $f(x)=e^{x_1}+\dots + e^{x_n}$, then the Hessian of $f$ at $x$ is $diag(e^{x_1}, \dots, e^{x_n})$ and thus positive definite. However, the gradient $\nabla f(x)=(e^{x_1}, \dots, e^{x_n})$ is not surjective (e.g. the origin is not in the image).