Integration of the derivative of radial function?

Question

If $f(x) = q(|x|)= q(r)$ is a radial function for $x\in \mathbb{R}^n$ then assuming that $q(r)$ vanishes at the boundary, is it a true that, $$\int_{\mathbb{R}^n} f_{x_i}(x) f_{x_j}(x) dx= 0,\quad \forall i\neq j$$ where $f_{x_i} = \frac{\partial f}{\partial x_i}?$ I tried to prove this idenity using integration by parts, but I am not sure how to use the radial assumption. Any hints will be much appreciated.

Answer 1

Explicitly we have that

$$f_{x_i}(x) = \frac{x_i}{|x|}q'(|x|)$$

which means the integral is given by

$$\int_{\Bbb{R^n}}\frac{x_ix_j}{|x|^2}\Bigr(q'(|x|)\Bigr)^2dx$$

Assuming $q'(|x|)\in L^2(\Bbb{R^n})$, the inside is dominated by $\frac{1}{2}\Bigr(q'(|x|)\Bigr)^2$ by AM-GM inequality.

Notice that the integrand is an odd function of both $x_i$ and $x_j$, thus by symmetry:

$$\int_{\Bbb{R^n}}\frac{x_ix_j}{|x|^2}\Bigr(q'(|x|)\Bigr)^2dx = \int_{\Bbb{R^n}}f_{x_i}(x)f_{x_j}(x)dx = 0$$