I was reading a paper and encountered the following claim:
Let $f$ be a real-valued square integrable function on $\mathbb R^d$ and let $\mathbb B_d(a) := \{x\in\mathbb R^d: \|x\|_2\leq a\}$ be the $\ell_2$ ball of radius $a>0$. Consider also $\mathbb S_d(a) = \partial\mathbb B_d(a) = \{x\in\mathbb R^d: \|x\|_2=a$ be the $\ell_2$ surface which is the boundary of $\mathbb B_d(a)$. Then for any $x\in\mathbb R^d$, $$ \nabla \left[\int_{\mathbb B_d(a)} f(x+v)dv\right] = \int_{\mathbb S_d(a)}f(x+u)\frac{u}{\|u\|_2}du. $$
I have the following questions:
- The original paper claims that this identity is a conseuqnce of the Stokes theorem. However, the form of Stokes theorem I know concerns $\nabla\times F$, which does not seem to be connected to either side of the identity.
- If I replace the $\ell_2$ ball with an $\ell_p$ ball for $1\leq p\leq \infty$, would similar results sitll hold? In particular, I'm interested in the $\ell_{\infty}$ case where $\mathbb B_d(a) = \{x: |x_i|\leq a\}$ and $\mathbb S_d(a) = \partial\mathbb B_d(a) = \{x: \max_i|x_i| = a\}$. Is it true that $$\nabla \left[\int_{\mathbb B_d(a)} f(x+v)dv\right] = \int_{\mathbb S_d(a)}f(x+u)\frac{u}{a}du. $$
Thank you very much.
To prove part 1 you can use this formula $$ \int_U \nabla fdx=\int_{\partial U}f\nu dS,$$ where $U$ is an open set of $\mathbb{R}^d$ with sufficently regular boundary, let's say $\partial U$ is $C^1$, and also $f \in C^1(\overline U)$. $\nu$ is the unit outer normal vector of $\partial U$. The formula above is a consequence of the Divergence Theorem, that is a particular case of Stokes Theorem. In your case $U=\mathbb{B}_d(a)$, hence $\nu=u/\|u\|_2$. The only thing you need to justify is the passage of the gradient inside the integral and for that you need to see which regularity assumptions you have on $f$.