Is it true that a coercive Riemannian metric implies geodesic completeness?

Question

Let $U \subset \mathbb{R}^n$ be open and let $Q:U \to S_{++}^n$ be a smooth positive definite matrix valued function on $U$. Suppose also that as $x \to \partial U$, we have $\|Q\|_2 \to \infty$. That is, $Q(.)$ is coercive on $U$. Define the Riemannian metric on $U$ at $x \in U$ as $g_x(v,w):= v^TQ(x)w$. My question is, is $(U,g)$ geodesically complete? My intuition says yes, since the metric explodes near the boundary, so geodesics will never “pierce” it. They will keep traversing towards the boundary, but doing so at slower and slower “speeds” in the Euclidean, not Riemannian, sense.

Answer 1

Take $f(x)=x^{-1/2} + (1-x)^{-1/2}$. A direct computation shows that $$ \int_0^1 f(x)dx= 2+ 2=4<\infty. $$ Now, consider the Riemannian metric $ds^2=f^2(x)dx^2$ on $(0,1)$. This metric has the property that $$ \lim_{x\to 0+} f(x)=\infty= \lim_{x\to 1-} f(x), $$ i.e. the metric is coercive. But the total length of $(0,1)$ with respect to the Riemannian metric $f^2(x)dx^2$ is $$ \int_0^1 f(x)dx<\infty. $$ Hence, the metric is incomplete. Thus, coercivity does not imply completeness. I think (but I did not check) that to get completeness you need a metric such that $$ ||Q^{-1}(x)||=O \left(d(x, \partial U)\right), x\in U. $$