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?



How to recognize if a continuous vector field in the Euclidean space is a gradient
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 



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> **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 $v\_k:=v\*\phi\_k$. 
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*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 $h'=v$. $\Box$


