I have two $C^1$ functions $ f, g : U \subseteq \mathbb{R}^n \rightarrow \mathbb{R} $ and a curve $c : [0, 1] \rightarrow U $ such that $ f(x) = g(x)$ and $\nabla f(x) = \nabla g(x)$ for all points $x$ on the curve $c$.
Are you aware of any conditions that would imply that $ f(x) = g(x) $ for all $x$ in a small neighbourhood of $c$ ?
In some sense, this would be similar to the identity theorem, but ideally without requiring that the functions are analytic and instead knowing some information about the gradient.
Any references are appreciated, thank you!
The condition on its own is untrue. Consider $f(x,y) = x^2 + y^2$ and $g(x,y) = -x^2 - y^2$. The counterexample is the curve $c(t) = (0,0), \forall t \in [0,1]$. Both these are smooth functions so adding more continuity won't solve it.
If you could show the gradient is the same in some small neighborhood of $c$, then you may have a chance using Taylor Series? For example, given $x$ on the curve and $\epsilon$ small, there exists $\theta \in [0,1]$ such that
$$ f(x + \epsilon) - f(x) = \epsilon^T \nabla f(x + \theta\epsilon) = \epsilon^T \nabla g(x + \theta \epsilon) = g(x + \epsilon) - g(x - \epsilon) \\ \implies f(x+\epsilon) = g(x + \epsilon)$$
However, I imagine that proving that gradients are equal in a neighborhood is more work than proving the functions themselves are equal in the neighborhood.