Length of boundary of level set

Question

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 $|\nabla\phi|$?

Answer 1

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.

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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

1. $\phi = \tilde{\phi}$ on $V$,
2. $\phi$ is a positive constant on the part of $\bar \Omega \setminus U$ which lies inside $\Gamma$,
3. $\phi$ is constant and negative on the part of $\bar \Omega \setminus U$ which lies in the exterior of $\Gamma$,
4. 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.