structural:reformatting
Proof of the converse

Suppose U is an open, path-connected subset of R, and F : U → R is a continuous and path-independent vector field. Fix some element a of U, and define f : U → R by f (x):= ∫ γ F (u) ⋅ d u {\displaystyle f(\mathbf {x}):=\int _{\gamma }\mathbf {F} (\mathbf {u})\cdot \mathrm {d} \mathbf {u} } Here γ is any (differentiable) curve in U originating at a and terminating at x. We know that f is well-defined because F is path-independent.

Let v be any nonzero vector in R. By the definition of the directional derivative, ∂ f (x) ∂ v = lim t → 0 f (x + t v) − f (x) t = lim t → 0 ∫ γ F (u) ⋅ d u − ∫ γ F (u) ⋅ d u t = lim t → 0 1 t ∫ γ F (u) ⋅ d u {\displaystyle {\begin{aligned}{\frac {\partial f(\mathbf {x})}{\partial \mathbf {v} }}&=\lim _{t\to 0}{\frac {f(\mathbf {x} +t\mathbf {v})-f(\mathbf {x})}{t}}\\&=\lim _{t\to 0}{\frac {\int _{\gamma }\mathbf {F} (\mathbf {u})\cdot \mathrm {d} \mathbf {u} -\int _{\gamma }\mathbf {F} (\mathbf {u})\cdot d\mathbf {u} }{t}}\\&=\lim _{t\to 0}{\frac {1}{t}}\int _{\gamma }\mathbf {F} (\mathbf {u})\cdot \mathrm {d} \mathbf {u} \end{aligned}}} To calculate the integral within the final limit, we must parametrize γ. Since F is path-independent, U is open, and t is approaching zero, we may assume that this path is a straight line, and parametrize it as u (s) = x + s v for 0 < s < t. Now, since u' (s) = v, the limit becomes lim t → 0 1 t ∫ 0 t F (u (s)) ⋅ u ′ (s) d s = d d t ∫ 0 t F (x + s v) ⋅ v d s | t = 0 = F (x) ⋅ v {\displaystyle \lim _{t\to 0}{\frac {1}{t}}\int _{0}^{t}\mathbf {F} (\mathbf {u} (s))\cdot \mathbf {u} '(s)\,\mathrm {d} s={\frac {\mathrm {d} }{\mathrm {d} t}}\int _{0}^{t}\mathbf {F} (\mathbf {x} +s\mathbf {v})\cdot \mathbf {v} \,\mathrm {d} s{\bigg |}_{t=0}=\mathbf {F} (\mathbf {x})\cdot \mathbf {v} } where the first equality is from the definition of the derivative with a fact that the integral is equal to 0 at t = 0, and the second equality is from the first fundamental theorem of calculus. Thus we have a formula for ∂ f, (one of ways to represent the directional derivative) where v is arbitrary; for f (x):= ∫ γ F (u) ⋅ d u {\displaystyle f(\mathbf {x}):=\int _{\gamma }\mathbf {F} (\mathbf {u})\cdot \mathrm {d} \mathbf {u} } (see its full definition above), its directional derivative with respect to v is ∂ f (x) ∂ v = ∂ v f (x) = D v f (x) = F (x) ⋅ v {\displaystyle {\frac {\partial f(\mathbf {x})}{\partial \mathbf {v} }}=\partial _{\mathbf {v} }f(\mathbf {x})=D_{\mathbf {v} }f(\mathbf {x})=\mathbf {F} (\mathbf {x})\cdot \mathbf {v} } where the first two equalities just show different representations of the directional derivative. According to the definition of the gradient of a scalar function f, ∇ f (x) = F (x) {\displaystyle \nabla f(\mathbf {x})=\mathbf {F} (\mathbf {x})}, thus we have found a scalar-valued function f whose gradient is the path-independent vector field F (i.e., F is a conservative vector field.), as desired.