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Question

If we let $\vec x = (x_1, \ldots, x_n)$ be a vector in $\mathbb R^n$, we have a definition that a vector field $\textbf F \ : \ \mathbb R^n \to \mathbb R^n$ is called "central" if it can be written on the form $\textbf F(\vec x) = f(|\vec x|)\vec x$ where $f$ is a function from $[0, \infty) \to \mathbb R$.

Can it be shown that central vector fields are conservative in $\mathbb R^n$ if $f$ is continuously differentiable and $\lim\limits_{r\to0}f'(r) = 0$?

My textbook seems to take this for granted, but I can't find a proof anywhere. Is it a definition, or are there steps to prove this?

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    Could you clarify what is meant by conservative vector field here? Because $x' = -x, \; y' = -y$ is a valid example of such vector field, but it is not conservative in the sense to which I am used to. – 2017-02-22
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    As I understand it, it simply a vector field that is the gradient of a function. Why they interchange these terms, I don't know. – 2017-02-22
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    Ah, so it is conservative in a physics sense. This vector field is analogous to the vector field of conservative force, for which the work done by moving particle from point $A$ to point $B$ is independent of the path. I think this pretty answers your question :) – 2017-02-22

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To show that $\mathbf{F}$ is conservative, we'd like to find a potential function $V\colon\mathbb{R}^n\to\mathbb{R}$. That is, we want $V$ to have the property that $\mathbf{F}=\nabla V$. Since our vector field $\mathbf{F}$ is always flowing either directly towards or directly away from the origin, it seems reasonable that $V(x)$ would depend only on the distance of $x$ from the origin. That is, we expect $V$ to be of the form \begin{equation} V(x) = g(|x|) \end{equation} for some sufficiently nice function $g\colon[0,\infty)\to\mathbb{R}$. In this case, we can compute the gradient of $V$. The $i$-th component will be given by \begin{equation} \frac{\partial V}{\partial x_i} = g'(|x|)\frac{1}{2}\left(\sum_{j=1}^nx_j^2\right)^{-1/2}2x_i = x_i\frac{g'(|x|)}{|x|}, \end{equation} so we have \begin{equation} \nabla V(x) = \frac{g'(|x|)}{|x|}x. \end{equation} If $f(|x|)=g'(|x|)/|x|$, then this is what we wanted. So we'll be able to conclude that $\mathbf{F}$ is conservative if we can find a differentiable function $g\colon[0,\infty)\to\mathbb{R}$ with the properties that $g'(r)=rf(r)$ and that $g'(r)/r$ approaches $0$ as $r$ approaches $0$. (The second property makes sure that $\nabla V$ is nice near the origin.) We can find a function $g$ which is differentiable on $(0,\infty)$ and satisfies $g'(r)=rf(r)$, since $f$ is continuous. This function will then be differentiable on $[0,\infty)$ and satisfy $g'(r)/r\to 0$ as $r\to 0$ from the facts that $f$ is differentiable and that $f'(r)\to 0$ as $r\to 0$.