Geodesic completeness of manifolds with "warped" metrics Let $(M,g)$ be a geodesically complete Riemannian manifold, and let $f: M\to \mathbb{R}^+$ be a smooth, bounded strictly positive function on $M$; i.e., there exist $L,A\in \mathbb{R}^+$ such that $0<L<f(m)<A<\infty$ for all $m\in M$. Since $f$ is positive, $f^2g$ defines a metric on $M$. Since $(M,g)$ is geodesically complete, is $(M, f^2g)$ also geodesically complete?
(Note: This question was inspired as a generalization of another question posed recently about the geodesic completeness of $\mathbb{R}\times \mathbb{S}^{n-1}$, given a warped metric $ds^2=dr^2+\psi^2d\theta^2$ where $dr$ and $d\theta$ are the metrics from $\mathbb{R}$ and $\mathbb{S}^{n-1}$ respectively and $\psi$ is a positive function on $\mathbb{R}\times \mathbb{S}^{n-1}$.)
 A: Let $L=\inf_{m\in M}(f(m))$ and $A=\sup_{m\in M}(f(m))$. By assumption $L, A$ exist and are positive. We define two metric structures, $d,d': M\times M\to \mathbb{R}^*$, $$d(x,y)=\inf \bigg\{\int_{0}^1( g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt: \gamma \text{ is a piecewise differentiable curve from } a \text{ to } b \bigg\}$$
And $$d'(x,y)=\inf \bigg\{\int_{0}^1( f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt: \gamma \text{ is a piecewise differentiable curve from } a \text{ to } b \bigg\}$$
Since $(M,g)$ is geodesically complete, $d$ makes $M$ into a complete metric space by the Hopf-Rinow theorem. It suffices to show that the metrics $d$ and $d'$ are strongly equivalent, i.e. for all $x,y$ in $M$, there exist $\alpha,\beta\in \mathbb{R}$ such that $\alpha d(x,y)\leq d'(x,y)\leq \beta d(x,y)$.
We can see that for any piecewise differentiable curve $\gamma:[0,1]\to M$ such that $\gamma(0)=x$ and $\gamma(1)=y$ we have $L(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}\leq(f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}\leq A(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}$ implying that $$L\int_{0}^1(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt\leq \int_{0}^1 (f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt\leq A\int_{0}^1(g_{ij}\dot{\gamma} ^i\dot{\gamma} ^j)^{1/2}dt$$ and hence $$Ld(x,y)\leq d'(x,y)\leq Ad(x,y)$$ So $d$ and $d'$ are strongly equivalent. Because $(M,d)$ is a complete metric space and $(M,d')$ is strongly equivalent to $(M,d)$, then it is also a complete metric space and $(M,f^2g)$ is geodesically complete.
A: [This is an answer to the original question]
$\newcommand{\R}{\mathbb{R}}$
Let $M := \R$ with the standard metric $g$, and $f(x) := e^{-x^2}$. Then the whole line $M$ has finite length if given the metric $f^2 g$, which shows it cannot be geodesically complete.
