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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 $\mathbb{R}\times \mathbb{S}^{n-1}$.)

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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.

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    $\begingroup$ I don't see in the original question any assumption that implies $L>0$. $\endgroup$
    – Deane
    Apr 22 at 18:30
  • $\begingroup$ @Deane : Presumably $\mathbb{R}^+$ in the question means the set of strictly positive real numbers; if it meant the set of non-negative real numbers, then there'd be no point in mentioning $L$ (since one could always use $L=0$). But it would be good to have the question edited to clarify this. $\endgroup$ Apr 22 at 23:36
  • $\begingroup$ Actually, the comment on the other answer confirms this, so I'll submit an edit to the question myself. $\endgroup$ Apr 22 at 23:37
  • $\begingroup$ By ‘the other answer’, I meant the one by Michał Miśkiewicz (I should be clear on this in case more answers are added later). $\endgroup$ Apr 23 at 0:17
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[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.

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    $\begingroup$ I was perhaps too vague in just saying $f$ was bounded, I intended this to mean that $f$ is bounded above and below by positive numbers, rather than allowing it to be bounded below only by $0$. $\endgroup$
    – J.V.Gaiter
    Apr 22 at 23:04

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