## A Vector Field Giving Rise to a Function Defining the Field

Theorem
Let
$\mathbf{F}$
defined in a polygon ally connected open set
$B \subset \mathbb{R}^n$
.
If the line integral
$\int_C \mathbf{F} \cdot d \mathbf{r}$
is independent of the path in
$B$
from
$x_0$
to
$x$
then the function
$f(x) = \int_{\mathbf{x_0}}^{\mathbf{x}} \mathbf{F} \cdot d \mathbf{x}$
is continuously differentiable and throughout
$B$
the relationship
$\mathbf{F}= grad \: f$
is satisfied.
Proof
Since
$x \in B \subset \mathbb{R}^n$
and
$B$
us open there exists an open ball
$B(x, \delta ) \subset B$
$\delta$
with centre
$x$

For all real numbers
$| \alpha | < \delta$
and unit vector
$\mathbf{u}$
,
$x+ \alpha \mathbf{u} \in B$

The line integral
$\int_C \mathbf{F} \cdot d \mathbf{x}$
is independent of the path so we can choose any piecewise smooth curve in
$B$
from
$x_0$
to
$x$
and extend it to
$\mathbf{x} + \alpha \mathbf{u}$
by a linear segment.
\begin{aligned} f(\mathbf{x} + \alpha \mathbf{u} )- f(\mathbf{x}) &= \int^{\mathbf{x} + \alpha \mathbf{u}}_{\mathbf{x_0}} \mathbf{F} \cdot d \mathbf{x} - \int^{\mathbf{x}}_{\mathbf{x_0}} \mathbf{F} \cdot d \mathbf{x} \\ &= \int^{\mathbf{x} + \alpha \mathbf{u}}_{\mathbf{x}} \mathbf{F} \cdot d \mathbf{x} \\ &= \int^{\alpha}_{0} \mathbf{F}( \mathbf{x}+ \beta \mathbf{u} ) \cdot \mathbf{u} d \beta \end{aligned}

If
$\mathbf{u}= \mathbf{e_k}$
is a base vector in
$\mathbb{R}^n$
then
\begin{aligned} \frac{\partial f}{\partial x_k} (\mathbf{x}) &= lim_{\alpha \rightarrow 0} \frac{f(\mathbf{x} + \alpha \mathbf{u}) - f (\mathbf{x})}{\alpha} \\ &= lim_{\alpha \rightarrow 0} \frac{1}{\alpha} \int^{\alpha}_0 \mathbf{F} ( \mathbf{x} + \beta \mathbf{e_k} ) \cdot \mathbf{e_k} d \beta \\ &= \frac{d}{d \alpha} \int^{\alpha}_0 \mathbf{F} ( \mathbf{x} + \beta \mathbf{e_k} ) \cdot \mathbf{e_k} d \beta |_{\alpha =0} \\ &= \mathbf{F}( \mathbf{x}) \cdot \mathbf{e_k }) \\ &= F_k ( \mathbf{x}) \end{aligned}

Since
$\mathbf{F}$
is differentiable, the partial derivatives are continuous, Hence
$\mathbf{F} = grad \: f$