4
$\begingroup$

The classical Rademacher theorem says that any Lipschitz function on a doman in $\mathbb{R}^n$ has the first derivative almost everywhere.

I am wondering if this result can be generalized as follows. Let $X$ be an Alexandrov space with curvature bounded below of finite dimension $n$. It is well known that at almost every point $x$ (with respect to the $n$-Hausdorff measure) the tangent cone at $x$ is isometric to the Euclidean space $\mathbb{R}^{n}$. Let us call such points regular.

Let $f\colon X\to \mathbb{R}$ be a Lipschitz function. We define its differential (if it exists) in the standard way as follows. Let $X_N$ (resp. $\mathbb{R}_N$) denote the space $X$ (resp. $\mathbb{R}$) with the metric multiplied by $N$. Let $$f_N\colon X_N\to \mathbb{R}_N$$ denote the map $f$ between these rescaled spaces. $f_N$ has the same Lipschitz constant as $f$. Fix a point $x\in X$. Then $(X_N,x)$ converges in the Gromov-Hausdorff sense to the tangent cone $T_xX$ at $x$, and $(\mathbb{R}_N,f(x))$ converges to $(\mathbb{R},0)$. Then if there is a limit map $df_x\colon T_xX\to \mathbb{R}$ we call it the differential of $f$ (automatically it will have the same Lipschitz constant as $f$).

Question. Is it true that for almost every regular point $x\in X$ the differential at $x$ of a Lipschitz function $f$ exists and is a linear functional on $\mathbb{R}^n$?

$\endgroup$
4
$\begingroup$

I assume you are interested in the finite-dimensional case.

You can write this function a distance chart (these charts cover almost all points). Apply the standard Rademacher theorem and notice that the distance chart is differentiable almost everywhere. Hence the result follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.