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Let $M$ be a noncompact $C^\infty$ manifold, let $X$ be a complete $C^\infty$ vector field on $M$, and take $f\in C^\infty\big(M;(0,\infty)\big)$ a strictly positive function.

Question: Does anyone know sufficient conditions on the function $f$ implying the completeness of the vector field $fX$ ?

(When $M$ is compact, the vector field $fX$ is complete and has the same integral curves as the vector field $X$, cf. Chapter 2, Section 2 of the book of Ergodic Theory of Cornfeld, Fomin and Sinai.)

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up vote 2 down vote accepted

If $f$ is bounded, then the rate at which one travels along the integral curves of $X$ is only increased by a bounded factor, so it still takes infinite time to get all the way along each integral curve,so $fX$ is complete.

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It seems indeed reasonable; don't you have any textbook reference for this fact? It is also interesting to note that one does not need to impose that f is bounded below by a strictly positive constant (the rate of travel along the integral curves could be as slow as desired, as soon as it is not zero!?) –  Peter F Feb 21 '12 at 3:06
The first step in a proof is to notice that the vector field $X$ restricts to a vector field along each of its integral curves, so we only need to work on one integral curve at a time. So without loss of generality, our manifold is one dimensional. It is also easy to pull back to the universal covering space, because completeness of vector fields is preserved and reflected by covering maps. You can then assume that your manifold is the real number line. You can use the flow of $X$ to parameterize the real number line, so then $X=d/dt$. After that the proof is elementary. –  Ben McKay Feb 21 '12 at 9:14
Indeed, thank you very much for the details. –  Peter F Feb 21 '12 at 11:13
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