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How to recognize, by "analytic" methods, if a $C^0$ vector field $v:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is the gradient of a function $h:\mathbb{R}^n \rightarrow \mathbb{R}$, given that the verification of the path independence of the line integrals may be unpractical and that one cannot verify the symmetry $\partial_{i}v^j =\partial_{j} v^i$ given that $v=(v^1,...,v^n)$ is assumed to be merely continuous?

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  • $\begingroup$ what do you mean by "merely continuous" ? $\endgroup$ Dec 20, 2018 at 19:19
  • $\begingroup$ In what sense impracticable? Do you mean that $v$ is given by explicit formulas, and you want to compute by hand? Practicality depends a lot on how the $v$ is "given". $\endgroup$
    – Ben McKay
    Dec 20, 2018 at 19:35
  • $\begingroup$ @Carlo Beenakker: v is only assumed to be continuous, This is the general case for the gradient of a C1 function. $\endgroup$
    – Nautilus
    Dec 20, 2018 at 19:53
  • $\begingroup$ @Ben McKay: looking for an alternative criterium to the verification of the path independence of the line integrals, $\endgroup$
    – Nautilus
    Dec 20, 2018 at 19:55

1 Answer 1

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Let $(\phi_k)$ be any sequence of mollifiers such that $v*\phi_k\to v$ uniformly for each $C^0$ vector field $v$. Then we have

Theorem. A $C^0$ field $v$ is a gradient field iff for all $k,i,j$ we have $\partial_{i}v_k^j =\partial_{j} v_k^i$, where $v_k:=v*\phi_k$.

Proof. Suppose that $v$ is a gradient field, so that $v=h'$ for some $h$. Then for all $k$ we see that $v_k=v*\phi_k=h'*\phi_k=(h*\phi_k)'$ is a smooth gradient field and hence $\partial_{i}v_k^j =\partial_{j} v_k^i$ for all $k,i,j$.

Vice versa, suppose that for all $k,i,j$ we have $\partial_{i}v_k^j =\partial_{j} v_k^i$, so that $v*\phi_k=v_k=(h_k)'$ for some $h_k$. Without loss of generality, $h_k(x)=\int_0^x (v*\phi_k)(y)\cdot\,dy$ for all $k$, all $x$, and all paths from $0$ to $x$. Since $v*\phi_k\to v$ uniformly, we have $h(x):=\lim_k h_k(x)=\int_0^x v(y)\cdot\,dy$ for all $x$ and all paths from $0$ to $x$. So, $v$ is a gradient field, with $h'=v$. $\Box$

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  • $\begingroup$ Nice concise proof of an interesting criterium. Many thanks! $\endgroup$
    – Nautilus
    Dec 20, 2018 at 21:17
  • $\begingroup$ We remark that your functions, being continuous, are distributions. Hence you can use the classical formulation where, now, the derivatives are used in the distributional sense . $\endgroup$ Dec 21, 2018 at 10:13
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    $\begingroup$ @Parschallen : One may of course say that, by using mollifiers, here we are actually using distributions. To an extent, this would be true, but not quite: (i) We may use here a rather arbitrary, but specific, sequence of mollifiers, not tied to any of the standard versions of distributions -- which may sometimes be convenient. (ii) Here we need not engage the entire heavy machinery of the distribution theory; the proof is elementary and self-contained. (iii) I am not aware of the existence of the theorem in my answer in the distribution theory. $\endgroup$ Dec 21, 2018 at 15:09
  • $\begingroup$ ad(iii). I was unaware of the fact that the criterion for stating a result on MO was not its correctedness but whether Prof. Pinelis was aware of it. He is clearly not aware of de Rham and his theory of currents. Of course for the simple case of the Poincaré lemma relevant here, the whole weight of this machinery is not necessary—- a very short and elementary proof can be given. $\endgroup$
    – user131781
    Jan 1, 2019 at 13:50
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    $\begingroup$ Previous comment continued: I specialize in probability theory, in which I am unaware perhaps of 95% of all developments, and then I may be unaware of at least 99.9% of all developments in all mathematics. Still, I try to give answers on MO when I think they may be of use, and I certainly try to learn more here. In this particular case, it would certainly be good and instructive if you could indeed give here an alternative "very short and elementary proof", where "the whole weight of this machinery is not necessary". $\endgroup$ Jan 1, 2019 at 18:25

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