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.
It seems to me that the end of the proof of the almost splitting could be made easier to follow by explicitly using the lemma I am statin below, however I had no luck trying to come up with a proof so far.
The lemma that I would like to use (in a simplified version) is the following :
Lemma :
For any $\varepsilon>0$, there is an $\eta>0$ such that if $u:B(0,2)\subset\mathbb{R}^n\to\mathbb{R}$ 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 of $(x,y)\in B(0,1)\times B(0,1)$ 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 $B(0,2)$.
Note : The real lemma I need is stated in terms of functions defined on balls in a Riemannian manifold with $\mathrm{Ric}\geq -\eta g$. I chose to state it for function in $\mathbb{R}^n$ because the heart of the difficulty remains in this setting, and it may look more appealing to PDE specialists not too fond of differential geometry.:
- the heart of the difficulty remains in this setting.
- it may look more appealing to PDE specialists not too fond of differential geometry.