First order weak derivatives of $f(x)=|x|^r$

Question

Let $f(x)=|x|^r$ for a given real number $r$. Show that $f$ has first order weak derivatives on the unit ball $B_1(0)\subset \mathbb{R}^n$ provided that $r > 1-n$.

Does anyone have an idea on how to prove this?

Answer 1

The function is absolutely continuous on almost every line segment (specifically, on every line segment that does not pass through the origin). I.e., it has the ACL property. The partial derivatives are continuous outside of the origin and are homogeneous of degree $r-1 >-n$; therefore, they are integrable. It is a standard fact that ACL+integrable partials => weak derivatives exist and are represented by the partials (i.e., the function is in the Sobolev class $W^{1,1}$).

There are other approaches. For example, you can take a partition of unity $\phi_n$ subordinate to concentric annuli, so that each $f\phi_n$ is smooth. Then observe that the $W^{1,1}$ norm of $\phi_n f$ decays exponentially as $n\to\infty$ (via rescaling), hence the series converges in $W^{1,1}$.