I have an open, bounded, connected, and simply connected domain $\Omega\subset\mathbb{R}^2$ with smooth boundary. I'd like to prove that there is a constant $M$ so that any two points in the domain can be connected by a smooth curve of length at most $M$.
I was imagining a counterexample of some kind of interior spiral where the path can become arbitrarily long as it spirals inward, but it seems that the smooth boundary would prevent this kind of thing.
One thing I considered was taking any path connecting two points of $\Omega$, covering the path with open balls, taking a finite subcover, connecting the centers to get a chain of straight segment paths, and smoothing the corners with some kind of mollification, but this seems like it may be overkill, and doesn't necessarily give a uniform bound on the total path length.
I also considered that you could move from each point to the nearest boundary point, and connect the paths along the boundary, which seems that a bound like $M=\text{(perimeter of }\Omega)+2(\text{diameter of }\Omega$) could work, but I'm sure this is not precise enough.
I would appreciate any advice or references!
The function $d:\Omega \times \Omega \to [0,\infty),\,(p,q)\mapsto d(p,q) = \inf\{L(\gamma):\gamma\in C^1([0,1],\Omega),\gamma(0) = p, \gamma(1) = q\}$ is the distance between two points in $\Omega$, where $L(\gamma) := \int_0^1 |\gamma'(t)| dt$. The function $d$ is continuous and we want to show that it is bounded. The clue is, that we can continuously extend $d$ to $\overline \Omega \times \overline \Omega$. Then, $d$ is bounded, as it is continuous on a compact set.
For the statement about extendability, let $p\in \Omega$ and take a point $q\in \partial \Omega$ on the boundary. Choose a sequence of points $q_k \in \Omega$ converging to $q$ and apply the continuity of $d$ inside $\Omega\times \Omega$ to get a limit $d(p,q) := \lim_{k\to\infty} d(p,q_k)$ which doesn't depend on the sequence $(q_k)_k$ (this is the difficult part where we one needs to apply the definition of smooth boundary). This yields a continuous extension on $\Omega \times \overline \Omega$. Repeat the argument for the first component to get an extension on $\overline \Omega \times \overline \Omega$.