Question
Let $\Omega \subset \mathbb{R}^3$ be a bounded domain, and let $\gamma \colon [0,1] \to \Omega$ be a smooth and closed path in $\Omega$. Can we find a constant $C >0$ such that for every vector field $v \in C^2(\Omega) \cap C^0(\overline{\Omega})$ with $\text{curl}\, v = 0$ in $\Omega$ there holds $$ \bigg\lvert \int_0^1 v(\gamma(t)) \cdot \dot\gamma(t) \, dt \bigg\rvert \leq C \int_{\Omega} \lvert v(x) \rvert^2 \, dx\, ? $$
Attempt
To be honest, I do not know how to tackle this problem. I do, however, believe that such a constant exist: the vector field $v$ is curl-free and so the line integral does not change if we deform $\gamma$ in such a way that we stay in the same homotopy class. So maybe we can take curves $\gamma^{\prime}$ that are parallel and homotopic to $\gamma$. But at the moment I have no idea how to come from a line integral to an integral over $\Omega$. Maybe, someone of you has an idea.
Thanks in advance!
This is not true. Define $$ v_n(x) = x\cdot \|x\|_2^n $$ with $\Omega =$ closed unit ball, $\gamma$ a path in the boundary of the ball. Then for $n\to\infty$ the $L^2$-norm on the rhs in your inequality vanishes, while the curve integral stays constant.
If the curve is required to stay in the interior of the domain, then set $\Omega=B_2(0)$ and take a curve in the boundary of the unit ball again, modify $v_n$ to $$ v_n(x) = x\cdot \phi(\|x\|_2^n), $$ where $\phi$ is a smooth function such that $\phi(t)=1$ for all $t\ge1$ and $\phi(t)\ge t$ for $t\in [0,1)$.