Does positive definite Hessian imply the Jacobian is injective?

Suppose $f(x):\mathbb{R}^n \mapsto \mathbb{R}$ is an infinitely differentiable function. If $\nabla^2 f(x)$, the hessian of $f$ is positive definite everywhere, does this imply that the gradient(first order derivative) $\nabla f:x\mapsto \nabla f(x)$ from $\mathbb{R}^n$ to $\mathbb{R}^n$ is injective?

It is the case for $n=1$ but not sure if it is the case for general $n$. I have tried to prove thinking along the inverse function theorem. But that one is local while I need a global property. Maybe there is a simple counter example I have not seen.

Write $g = \nabla f$ and $H = \nabla^2 f$.

Let $x,h\in\mathbb R^n$ with $h\ne 0$. Apply Taylor expansion on $g(x+h) - g(x)$. Then, by continuity and positive definiteness of $H$ it follows $$ h^T (g(x+h) - g(x)) = \int_0^1 \underbrace{h^T H(x+th)h}_{>0} \; \mathrm d t > 0. $$ Thus, $g(x)\ne g(x+h)$.