Consider the Riemannian manifold $\mathbb{R}^n$ and a smooth Riemannian metric $G:\mathbb{R}^n\rightarrow\mathbb{R}^{n\times{n}}$. What is the minimum assumption on $G$ such that the manifold $\mathbb{R}^n$ is complete as a metric space with respect to the Riemannian distance determined by $G$. The Riemannian distance $d_G(x,y)$ between points $x,y\in\mathbb{R}^n$ is defined as follows: $d_G(x,y)=\inf_{\chi\in\Omega(x,y)}\int_0^1{\sqrt{\left(\frac{d\chi(s)}{ds}\right)^TG(\chi(s))\frac{d\chi(s)}{ds}}ds}$, where $\Omega(x,y)$ is the set of all the piecewise smooth paths connecting $x$ to $y$. It can be easily checked that if $G$ satisfies: $\omega_1\Vert{y}\Vert_2^2\leq y^TG(x)y\leq\omega_2\Vert{y}\Vert_2^2$ for any $x,y\in\mathbb{R}^n$ and for some positive constants $\omega_1$ and $\omega_2$, then $\mathbb{R}^n$ is a complete metric space with respect to $d_G$. Moreover, if we can find a change of coordinate such that the metric in the new coordinate satisfies the preceding assumption, again the result is true. I am just wondering if there exist weaker assumptions on $G$ to show the result or there exists counter examples to show that this assumptions are necessary.
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You are assuming that the identity map between the Riemannian metric and the Euclidean metric is biLipschitz, but this is certainly far too strong. Here are two counterexamples:


