My question is the following:
Starting with the mean value property for harmonic $u \in C^2(\Bbb R^3)$, ie $$u(x) = {1 \over {4\pi R^2}} \int_{\partial B_R(x)}u(y) dy,$$ deduce that if $\phi \in C_0^\infty (\Bbb R^3)$ has total integral $\int_{\Bbb R^3} \phi(x)dx = 1$ and is radial $\phi(x) = \psi(|x|),$ $\psi \in C_0^\infty (\Bbb R^3),$ then $u = \phi_\epsilon * u,$ where $*$ is convolution and $\phi_\epsilon(x) = \epsilon^{-3}\phi(x/\epsilon)$.
I am not far away from this (I showed that, for not necessarily harmonic $u$ then the equality is true in the limit (for not necessarily radial $\phi$), but I can't work out how to do this. Some hints would be most appreciated!
(Please don't just say the answer if you know, as this does not help me; please give a hint!)
Thank you! :)
I suppose enough time passed since the comment exchange that the answer can be posted without spoiling the fun.
The scaling factor $\epsilon$ is mostly a distraction here. You can work with $\phi$ itself, and include $\epsilon$ later (since $\phi_\epsilon$ has all the same properties as $\phi$). Observe that $$u(x)\psi(R) = {1 \over {4\pi R^2}} \int_{\partial B_R(x)}\psi(R) u(y)\, dy = {1 \over {4\pi R^2}} \int_{\partial B_R(x)} \phi(x-y) u(y) dy $$ because when $y\in \partial B_R(x)$, we have $|x-y|=R$, and therefore $\phi(x-y)=\psi(R)$.
Then multiply both sides by $4\pi R^2$ and integrate over $R$. On the right you get the convolution. On the left you get $u(x)$ multiplied by the integral of $\psi$, which is $1$.