Laplacian of an integral function So in my PDE course I'm asked to prove the following result
Let $U = \{x \in \mathbb{R}^N ,\; x\neq 0\}$ and $f \in C^2(U)$ define the new function:
$$f_{\#}(x) := \int_{S_1(0)}f(|x|\omega)d\sigma(\omega).$$
Show that $\Delta(f_{\#})=(\Delta f)_{\#}$.
I used the divergence theorem to try and find a nice expression for the right hand side of the equality and thought about using something related to the change of variables to polar coordinates. But now I'm stuck and have no idea on how to proceed.
 A: A key thing to note is that $f_{\#}$ is constant on any sphere centered at the origin. So, if we compute $\Delta(f_\#)$ in polar coordinates the spherical Laplacian term vanishes. Hence we have,
$$\Delta(f_\#)=\frac{1}{r^{N-1}}\frac{\partial}{\partial r}\left( r^{N-1}\frac{\partial}{\partial r}\left(\int_{S_1(0)}f(r\omega)d\sigma(\omega)\right) \right)$$
Some work will show that $\frac{\partial}{\partial r}\left(\int_{S_1(0)}f(r\omega)d\sigma(\omega)\right)=\int_{S_1(0)}(\nabla f.n)(r\omega)d\sigma(\omega).$ Thus we have,
$$\Delta(f_\#)=\frac{1}{r^{N-1}}\frac{\partial}{\partial r}\left( r^{N-1}\int_{S_1(0)}(\nabla f.n)(r\omega)d\sigma(\omega).$ \right)$$
Changing coordinates in the integral gives us,
$$\Delta(f_\#)=\frac{1}{r^{N-1}}\frac{\partial}{\partial r}\left( \int_{S_r(0)}(\nabla f.n)(\omega)d\sigma(\omega). \right)$$
For $t<r$ as $\int_{S_t(0)}(\nabla f.n)(\omega)d\sigma(\omega) $ is constant with respect to $r$ we have,
$$\Delta (f_{\#})= \frac{1}{r^{N-1}}\frac{\partial}{\partial r}\left(\int_{S_r(0)}(\nabla f.n)(\omega)d\sigma(\omega)-\int_{S_t(0)}(\nabla f.n)(\omega)d\sigma(\omega)\right) $$
Letting $A_{t,r}=\{x\in\mathbb{R}^N|t\leq |x|\leq r \} $ and applying the Divergence Theorem, we get,
$$\Delta (f_{\#})= \frac{1}{r^{N-1}}\frac{\partial}{\partial r}\left(\int_{A_{t,r}}(\Delta f)(\omega)d\sigma(\omega)\right). $$
Differentiating we get,
$$\Delta (f_{\#})= \frac{1}{r^{N-1}}\int_{S_r(0)}(\Delta f)(\omega)d\sigma(\omega). $$
By a change of coordinates, we have
$$(\Delta f)_{\#}=  \int_{S_1(0)}(\Delta f)(r\omega)d\sigma(\omega)=\Delta(f_{\#}). $$
