While reading about the almost-splitting theorem of Cheeger and Colding, I thought about the following question, which asks about wether the level sets of a function which is close to being a linear form in an average sense behave like the level sets of a linear form.

I would like to know if the following statement is true :

For any $\varepsilon>0$, there is an $\eta>0$ such that if $u:B(0,2)\to\mathbb{R}$ is a smooth function with :

• $\int_{B(0,2)}\left||\nabla u|^2-1\right|\leq\eta$,
• $\int_{B(0,2)}\left|\mathrm{Hess}\, u\right|^2\leq\eta$,

then the set for $(x,y)\in B(0,1)\times B(0,1)$ for which

$\Big | |u(y)-u(x)|-d\left(y,\{u=u(x)\}\right)\Big|>\varepsilon$

has measure less than $\varepsilon$. Here $d\left(y,\{u=u(x)\}\right)= \inf\Big\{d(z,y)\,\Big |\, z\in B(0,R),\,u(z)=u(x)\Big\}$.

I know how to prove it if the $L^2$ bounds are replaced with $L^\infty$ bounds. I also know how to prove that the set of $(x,y)\in B(0,1)\times B(0,1)$ such that

$\left |u(y)-u(x)-\langle\nabla u(x),y-x\rangle\right|\geq \varepsilon\|y-x\|$

has measure less than $\varepsilon$ provided $\eta$ is small enough.

If needed, feel free to assume the integral bounds on balls bigger than $B(0,2)$.