Dear Downvoters: if you leave a comment, you can influence the way this post gets modified, if you don't this post might never satisfy you - even though I keep editing
Let $\Omega \subseteq \mathbb{R}^n$ denote a bounded domain (open and connected) and consider a vector field $g = (g_1,\dots,g_n) \in C^\infty(\overline\Omega)^n$.
Here $C^\infty(\overline\Omega)$ denotes the set of all restrictions of vector fields $u|_{\overline\Omega}$ of functions $u \in C^\infty(\mathbb{R}^n)$ such that $$\sup_{|\alpha| < \infty, x \in \mathbb{R}^n} |D^\alpha u | < \infty.$$ On this space we define the norm $$ \| u\| := \sup_{|\alpha| \leq k, x \in \overline \Omega} |D^\alpha u(x)|. $$ Considering the product space $C^\infty(\overline\Omega)^n$, on this space we define the norm for a vector field $g$ as $$ \| g\| := \sup_{j = 1,\dots,n} \| g_j \| $$
These definitions are taken from Sohr - The Navier-Stokes Equations: An Elementary Functional Analytic Approach.
One version of the FTC for closed curves I am familiar with, works with closed curves and domains, not the closure of domains. However there seems to be a modification of this theorem that I am not able to prove:
If for each closed piecewise $C^1$ curve $\gamma \colon [0,1] \to \overline\Omega$ the curve integral $$ \int_0^1 g(\gamma(t)) \cdot \gamma'(t) dt = 0, $$ then $g$ has the form $g = \nabla U$ with $U \in C^\infty(\overline\Omega)$.
- Do you maybe know how to prove this from the original theorem?
- Can you recommend a reference where I could look the proof of this particular theorem up?
This "theorem" is again referenced in Sohr - The Navier-Stokes Equations: An Elementary Functional Analytic Approach, p.74
EDIT to avoid further confusion:
I know that if for each closed curve the integral vanishes we can define a candidate for a potential function $$ f(x) := \int_p^x g \cdot ds $$ independent of the curve chosen to connect an arbitrary but fixed point $p$ and the point $x$.
I know a proof that does it just like this, but only for domains. When it comes to prove that $f$ is indeed differentiable at $x$ the fact that a domain is open is used.
What I need is some extension property that guarantees me the differentiability in an open neighborhood of boundary points $x \in \partial \Omega$.
Suppose the circulation of the vector field on any closed path is zero. Then the integral on any path depends only on the starting and ending point, but not on the path itself.
Fix a point in the domain and consider the integral on a path, going from the point, as a function of the other end; it's well-defined function.
To check that this function is indeed the potential of the vector field, one might consider a path consisting solely from linear pieces, going along the coordinate axes.
P.S. Of course, it means that each connected part of the domain has it's own potential, and the potential itself is, precisely speaking, not just a function, but a class of them (element of a factor space) given by this particular function.
P.P.S. And of course on the boundary points you have to use the derivative by a vector: $\lim_{h\to 0} h^{-1}(f(\mathbf x +h\mathbf v)-f(\mathbf x))$, where $h: \mathbb R_{>0}$ and $\mathbf v$ gives the direction.