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  
\[B(x, \delta ) \subset B\]
  of radius  
\[\delta\]
  with centre  
\[x\]

For all real numbers  
\[ | \alpha |   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.
\[\begin{equation} \begin{aligned} f(\mathbf{x} + \alpha \mathbf{u} )- f(\mathbf{x}) &= \int^{\mathbf{x} + \alpha \mathbf{u}}_{\mathbf{x_0}} \mathbf{F} \cdot d \mathbf{x} - \int^{\mathbf{x}}_{\mathbf{x_0}} \mathbf{F} \cdot d \mathbf{x} \\ &= \int^{\mathbf{x} + \alpha \mathbf{u}}_{\mathbf{x}} \mathbf{F} \cdot d \mathbf{x} \\ &= \int^{\alpha}_{0} \mathbf{F}( \mathbf{x}+ \beta \mathbf{u} ) \cdot \mathbf{u} d \beta \end{aligned} \end{equation}\]

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

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