A continuous path-independent field has a potential constructed by line integrals

By [L1], the displayed formula defines one real number $\varphi (x)$ for every $x\in U$. Choosing the constant path at $a$ and using [L2] gives $\varphi (a)=0$.

Fix $x\in U$ and a coordinate $j$. Since $U$ is open, there is $r>0$ such that $x+q{e}_j\in U$ whenever $\mathrm{\mid }q\mathrm{\mid }<r$. Take any path from $a$ to $x$ and reparametrize it by the increasing affine bijection of $[0,1]$ onto its domain; this is an orientation-preserving oriented reparametrization, so [L4] leaves its integral unchanged. Both it and the coordinate segment ${\sigma }_q(t)=x+tq{e}_j$, $0\le t\le 1$, now have domain $[0,1]$, so the concatenation in [L2] is defined; append ${\sigma }_q$.

Path independence and [L2] give, for $0<\mathrm{\mid }q\mathrm{\mid }<r$, $\varphi (x+q{e}_j)-\varphi (x)={\int }_{{\sigma }_q}F\cdot d\mathbf{r}=q{\int }_0^1{F}_j(x+tq{e}_j)dt\mathrm{.}$

Divide step 2.1 by $q$. For every $\epsilon >0$, continuity of $F_j$ at $x$ makes $\mathrm{\mid }{F}_j(x+tq{e}_j)-F_j(x)\mathrm{\mid }<\epsilon$ uniformly for $0\le t\le 1$ when $q$ is sufficiently small. Therefore the quotient tends to $F_j(x)$, so ${\mathrm{\partial }}_j\varphi (x)=F_j(x)$.

Since $x$ and $j$ were arbitrary, all partial derivatives of $\varphi$ are the continuous components of $F$. By [L3], $\varphi$ is totally differentiable everywhere with $\mathrm{\nabla }\varphi =F$, and these derivatives vary continuously; hence $\varphi$ is $C^1$.

Thus $\varphi$ is the normalized potential asserted in the Statement, and $F$ is conservative.