Mollifying while conserving symmetries Suppose $B = B(z,v)$ is a function in $L^1_\rm{loc}(\mathbb{R}^d \times S^{d-1})$ depending only on the values of $|z|$ and $|z \cdot v|$ (you don't make any assumptions on how $B$ depends on $|z|$ and $|z \cdot v|$). Is it possible to mollify in the $z$-variable without destroying these symmetries? More precisely: Can you approximate $B$ in $L^1_\rm{loc}(\mathbb{R}^d \times S^{d-1})$ by functions $B_n \in L^1 \cap L^\infty(\mathbb{R}^d \times S^{d-1})$, such that, for every $n$, the map $z \mapsto B_n(z,v)$ is smooth? (Background: This problem originates from the DiPerna-Lions proof of global existence to the Cauchy problem for the Boltzmann equation. But that shouldn't be of any relevance here.)

Okay, the answer is more trivial than I thought:

If $B$ has the above symmetries, then it's in particular symmetric under orthogonal coordinate changes _simultaneous_ in both variables. On the other hand, whenever $|z_1| = |z_2|$ and $|z_1 \cdot v_1| = |z_2 \cdot v_2|$, there exists $Q \in O(d)$ such that $Qv_1 = v_2$ and $Qz_1 = z_2$ (first perform a rotation $R$ that sends $v_1$ to $v_2$, then rotate about the $v_2$-axis such that $Rz_1$ is sent to $z_2$). Hence we just have to choose a rotational symmetric standard mollifier and do a convolution in the $z$ variable.