TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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

Statement

Let $U \subseteq R^{n}$ be nonempty, open, and piecewise- $C^{1}$ path-connected. If the continuous field $F : U \to R^{n}$ is path-independent, then it is conservative. More precisely, for any basepoint $a \in U$ ,

$ϕ (x) : = \int_{a}^{x} F \cdot d r$

is well-defined, is $C^{1}$ , satisfies $ϕ (a) = 0$ , and has $\nabla ϕ = F$ .

Facts & Assumptions

Given: The domain, field, path independence, and basepoint in the Statement.

[L1]

Piecewise- $C^{1}$ path-connectedness supplies a path in $U$ from $a$ to each $x$ , and path independence makes the integral depend only on its endpoints (Piecewise-C1 path-connected domains, potential functions, conservative fields, and path independence).

[L2]

Vector line integrals add under concatenation, and a constant path has integral zero (Line integrals under reversal and concatenation, Scalar line integrals with respect to arc length and vector-field line integrals).

[L3]

If all partial derivatives exist near a point and are continuous there, then the function is totally differentiable there, with derivative matrix equal to its Jacobian (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L4]

An orientation-preserving oriented piecewise- $C^{1}$ reparametrization leaves a vector line integral unchanged (Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses).

Proof

technique · direct

1.1

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

givenL1L2

1.2

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 $∣ q ∣ < r$ . 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 $σ_{q} (t) = x + t q e_{j}$ , $0 \leq t \leq 1$ , now have domain $[0, 1]$ , so the concatenation in [L2] is defined; append $σ_{q}$ .

givenL1L2L4

2.1

Path independence and [L2] give, for $0 < ∣ q ∣ < r$ , $ϕ (x + q e_{j}) - ϕ (x) = \int_{σ_{q}} F \cdot d r = q \int_{0}^{1} F_{j} (x + t q e_{j}) d t .$

step 1.2L2algebra

3.1

Divide step 2.1 by $q$ . For every $ε > 0$ , continuity of $F_{j}$ at $x$ makes $∣ F_{j} (x + t q e_{j}) - F_{j} (x) ∣ < ε$ uniformly for $0 \leq t \leq 1$ when $q$ is sufficiently small. Therefore the quotient tends to $F_{j} (x)$ , so $\partial_{j} ϕ (x) = F_{j} (x)$ .

givenstep 2.1algebra

4.1

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

step 3.1L3

5.1

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

step 1.1step 4.1L1∎

Depends on

Used by

Conservative, path-independent, and zero-closed-loop conditions are equivalent Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 94 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, Theorem 9.3.3 (standard reference, not scraped)