---
tag: 242AXRRT28
kind: proof
name: Proof that a curl-free vector field on a convex set is conservative
msc: ["26B12"]
proves: 0B22P010D6
uses: [0GDNHS9J2X, 05R9NF2J49, 0K06X9QXDX, 0AR06W1E1B, 0A8TJ0VHC7, 0KXSK8TEEB, 0NS2QYVC4Y, 0V9A43H7WV, 0NHGY0N0JB, 0TCBZMEYGN, 05AXG3QRKR]
sources:
  - url: https://www.maths.tcd.ie/~zaitsev/topics-ca/wedhorn-Complex-Analysis.pdf
    locator: Lemma 3.11
---

## Proof

We differentiate the explicit formula for $\phi$ under the integral sign.

1. Write $F = (F_1, F_2, F_3)$ and $\partial_i = \partial / \partial x_i$. By the definition of the [[curl|0GDNHS9J2X]], $\nabla \times F = 0$ means $\partial_2 F_3 = \partial_3 F_2$, $\partial_3 F_1 = \partial_1 F_3$ and $\partial_1 F_2 = \partial_2 F_1$. So $\partial_j F_i = \partial_i F_j$ for all $i, j$.
2. Fix $p \in U$. For $x \in U$ and $t \in [0, 1]$, the point $p + t (x - p)$ lies in $U$, since $U$ is [[convex|0K06X9QXDX]]. So $\phi(x) = \int_0^1 F(p + t (x - p)) \cdot (x - p) \, dt$ is defined, with a continuous integrand.
3. Each $F_i$ is $C^1$, so it is differentiable by [[continuous partial derivatives imply differentiability|0KXSK8TEEB]]. Its derivative in a direction $v$ is $\nabla F_i \cdot v$, by the definition of the [[gradient|0NS2QYVC4Y]].
4. Fix $x \in U$ and $j \in \set{1, 2, 3}$. Choose $r > 0$ such that $x + h e_j \in U$ for $\abs{h} < 2 r$, where $e_j$ is the $j$-th basis vector.
5. For $t \in [0, 1]$ and $\abs{h} < 2 r$, let $y_h = x + h e_j$ and $q(t, h) = p + t (y_h - p)$. By step 2, $q(t, h) \in U$.
6. Let $g(t, h) = F(q(t, h)) \cdot (y_h - p) = \sum_i F_i(q(t, h)) \, (y_{h, i} - p_i)$. Then $\phi(x + h e_j) = \int_0^1 g(t, h) \, dt$, and $g$ is continuous on $[0, 1] \times [-r, r]$.
7. For fixed $t$, the curve $h \mapsto q(t, h)$ has derivative $t e_j$. By the [[chain rule along a curve|0A8TJ0VHC7]] and step 3, $\frac{\partial}{\partial h} F_i(q(t, h)) = t \, \partial_j F_i(q(t, h))$.
8. Also $\frac{\partial}{\partial h} (y_{h, i} - p_i)$ is $1$ for $i = j$ and $0$ otherwise. By the [[product rule|0V9A43H7WV]], $\frac{\partial g}{\partial h}(t, h) = t \sum_i \partial_j F_i(q(t, h)) \, (y_{h, i} - p_i) + F_j(q(t, h))$.
9. This [[partial derivative|05AXG3QRKR]] is continuous on $[0, 1] \times [-r, r]$, since the $\partial_j F_i$ are continuous. By [[differentiation under the integral sign|0AR06W1E1B]] at $h = 0$, $\partial_j \phi(x) = \int_0^1 \frac{\partial g}{\partial h}(t, 0) \, dt$.
10. Let $c(t) = p + t (x - p)$, so $c(t) = q(t, 0)$. By step 1, $\frac{\partial g}{\partial h}(t, 0) = t \sum_i \partial_i F_j(c(t)) \, (x_i - p_i) + F_j(c(t))$.
11. The curve $c$ has derivative $x - p$. By the chain rule along a curve and step 3, $\frac{d}{dt} F_j(c(t)) = \sum_i \partial_i F_j(c(t)) \, (x_i - p_i)$ for $t \in (0, 1)$.
12. By the product rule and steps 10 and 11, $\frac{\partial g}{\partial h}(t, 0) = \frac{d}{dt} \big( t \, F_j(c(t)) \big)$ for $t \in (0, 1)$. The function $t \mapsto t F_j(c(t))$ is continuous on $[0, 1]$.
13. By the [[fundamental theorem of calculus|0NHGY0N0JB]], $\partial_j \phi(x) = 1 \cdot F_j(c(1)) - 0 \cdot F_j(c(0)) = F_j(x)$.
14. So the partial derivatives of $\phi$ are $F_1, F_2, F_3$, which are continuous. Hence $\phi$ is of [[class $C^1$|0TCBZMEYGN]], and $\nabla \phi = F$ on $U$ by the coordinate form of the gradient.
15. Therefore $F$ is [[conservative|05R9NF2J49]] on $U$, with the potential $\phi$ of the statement.

## Idea

The candidate potential is the work of $F$ along the straight segment from $p$ to $x$. Differentiating in $x_j$ under the integral produces terms $\partial_j F_i$. The curl-free condition turns them into $\partial_i F_j$, and then the integrand becomes an exact $t$-derivative of $t F_j$, which integrates to $F_j(x)$.