Can convolution of two radially symmetric function be radially symmetric?

Question

For example, take $x\in R^3$ and let $f(x)$ and $g(x)$ be radially symmetric. Can we prove that $f\ast g$ is also symmetric?

Answer 1

A function $f$ is radially symmetric if $f(Ax) = f(x)$ for all orthogonal $A$. Thus, if $f$ and $g$ are radially symmetric, and the convolution exists, we have for any $A\in O(n)$

$$\begin{align} (f\ast g)(Ax) &= \int_{\mathbb{R}^n} f(y) g(Ax-y)\,dy\\ &= \int_{\mathbb{R}^n} f(y) g(A(x-A^T y))\,dy\\ &= \int_{\mathbb{R}^n} f(y) g(x-A^T y)\,dy\tag{$g$ symmetric}\\ &= \int_{\mathbb{R}^n} f(A^T y) g(x - A^T y)\,dy\tag{$f$ symmetric}\\ &= \int_{\mathbb{R}^n} f(z) g(x-z) \lvert \det A\rvert\,dz\tag{substitute $z = A^Ty$}\\ &= \int_{\mathbb{R}^n} f(z)g(x-z)\,dz\tag{$\lvert\det A\rvert = 1$}\\ &= (f\ast g)(x), \end{align}$$

so $f\ast g$ is also radially symmetric.