Length of boundary of level set Say we have $\phi\in C^1(\overline{\Omega})$ of a bounded domain $\Omega$ in $\mathbb{R}^2$ and $D=\{x:\phi(x)>0\}$ and $\nabla \phi(x) \neq 0$ for $x$ on the curve satisfying $\phi(x)=0$.
Question: What can we say about an upper bound of $L(\partial D)$? Do we have:
$$\|\phi\|_{C^1}\leq M_1 \quad \Rightarrow \quad L(\partial D) \leq M_2,$$
where $M_1$ is some constant and $M_2$ is only dependent on $M_1$?
Progress: My intuition (so far) is that $L(\partial D)$ is finite and we need a lower bound $|\nabla \phi|>c$ on $\phi^{-1}(0)$ and then $M_2=M_2(c)$.
Indeed for each $x\in \phi^{-1}(0)$ we can find a neighborhood of $x$ and a local $C^1$ parametrization $\psi$ (inverse function theorem) of that local piece of the surface $\phi^{-1}(0)$. The length of this piece is finite and relates to $|\psi'|$ which somehow relates to $(\nabla \phi)^{-1}$.
Then by compactness we add finitely many pieces to get the total surface area. How to more precisely relate $|\psi'|$ to $|\nabla\phi|$?
 A: The answer is no.
Fix $M_1 >0$ and $M_2>0$.
Let $\Gamma$ be any simple $\mathcal{C}^1$ loop in $\Omega$ bounding a topological disk $D$ with length $L(\partial D)>M_2$.
This is possible: for instance, take any small euclidean disk and deform its boundary by waving it smoothly in order to increase its length.
Then $D$ admits such a function $\phi$: this is called a defining function.
Let $\lambda >0$ be small enough such that $\lambda\| \phi\|_{\mathcal{C}^1}\leqslant M_1$.
Then $\lambda \phi$ is still a defining function, but
$$
\|\lambda \phi \|_{\mathcal{C}^1} \leqslant M_1,
$$
while
$$
L(\partial D) > M_2.
$$
In other words: the length of the boundary of $D$ is independent from the choice of a defining function for $D$.
Note that this generalizes to any dimension and as well as to domains inside Riemannian manifolds.

Example of a construction of a defining function:
let $\Gamma$ be any simple loop of class $\mathcal{C}^1$.
Let $\nu$ be the inward unit normal on $\gamma$, which is also of class $\mathcal{C}^1$.
By an application of the inverse function Theorem, there exists $\varepsilon >0$ such that the map
$(x,t) \in \Gamma\times (-\varepsilon,\varepsilon) \mapsto x+t\nu(x) \in \Omega$ is a $\mathcal{C}^1$-diffeomorphism onto its image $U$.
On $U$, consider $\tilde{\phi}(y) = t$ where $y = x+t\nu(x)$ as earlier.
Then $\tilde{\phi}$ is of class $\mathcal{C}^1$, and has gradient of length $1$ everywhere.
It is positive on the part of $U$ inside of $\Gamma$ and negative on the part of $U$ outside of $\Gamma$.
To complete the construction, one needs to extend $\Gamma$ on all of $\bar \Omega$.
I leave the following as an instructive exercise: let $V\subset U$ be a relatively compact neighbourhood of $\Gamma$ in $U$.
Then one can extend $\tilde{\phi}$ as $\phi\colon \bar \Omega\to \Bbb R$ such that

*

*$\phi = \tilde{\phi}$ on $V$,

*$\phi$ is a positive constant on the part of $\bar \Omega \setminus U$ which lies inside $\Gamma$,

*$\phi$ is constant and negative on the part of $\bar \Omega \setminus U$ which lies in the exterior of $\Gamma$,

*And of course, $\phi^{-1}(0) = \Gamma$.

Hint: use cut-off functions.
Remark: here, interior and exterior have to be understood as in the Jordan's curve theorem: the interior part of $\Gamma$ is the bounded connected component of $\Bbb R^2 \setminus \Gamma$, the exterior is the unbounded connected component.
