I need to prove the following statement:
If $\vec{v}$ is a conservative vector field $\vec{v}: \mathbb{R}^n \rightarrow \mathbb{R}^n$, then it is a gradient field, i.e. $\exists$ a function $f$ such that $\vec{v} =\nabla f(\vec{x})$
I've been given the hint to consider the function $$f(\vec{x}) = \int_{C_{px}} \vec{v} \cdot d\vec{r}$$
where $C_{px}$ represents a curve going from $\vec{p}\in \mathbb{R}^n$ to $\vec{x}\in \mathbb{R}^n$.
Now I have noted that since $v$ is conservative, the integral is independent of path and $f$ is thus well-defined. However, I don't know where to progress from here.
Any tips/hints will be appreciated.
Claim 1: $\frac{\partial f}{\partial x}$ equals the $x$-component of $v$.
Claim 2: Similarly, $\frac{\partial f}{\partial y}$ is the $y$-component of $v$ and $\frac{\partial f}{\partial z}$ is the $z$-component. Together this makes $\nabla f=\vec v$.
Hopefully, claim 2 is easy enough to follow. I will therefore focus on claim 1.
Proof of claim 1: Fix a point $(x_0,y_0,z_0)\in \Bbb R^3$. We have $$ \frac{\partial f}{\partial x}=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x,y_0,z_0)-f(x_0,y_0,z_0)}{\Delta x} $$ Because $\vec v$ is conservative, we have $$ f(x_0+\Delta x,y_0,z_0)-f(x_0,y_0,z_0)=\int_{C} \vec v(x,y,z)\cdot d\vec r $$ Where $C$ is a curve going from $(x_0,y_0,z_0)$ to $(x_0+\Delta x,y_0,z_0)$. Also, I gave arguments to $\vec v$ to help remind us that it is a vector-valued function.
At this point I choose to parameterize the curve $C$ as a straight line at unit speed. In that case $\vec v\cdot d\vec r$ becomes $v_xdt$, where $v_x$ is the $x$-component of $\vec v$. I would also like to define the function $$g(a)=\int_{x_0}^{x_0+a}v_x(x_0+t,y_0,z_0)dt$$ and note that since $g(0)=0$, our algebraic manipulations have taken us to $$ \frac{\partial f}{\partial x}=\lim_{\Delta x\to 0}\frac{g(\Delta x)}{\Delta x}=g'(0) $$ All that's left is to differentiate $g$. The fundamental theorem of calculus tells us that $g'(a)=v_x(x_0+a,y_0,z_0)$, and we are finished.