If you go through the classical proof of the fact that if $\Delta u= 0$ on $B_R$, then for every $R^\prime<R$ there holds $$ \int_{B_{R^\prime}}|\nabla u|^2 \leq \frac{C}{(R-R^\prime)^2}\int_{B_{R}\setminus B_{R^\prime}}u^2 $$ I just don't see why it is not commonly written with $C=1$, as it is very easy to show.
Am I wrong thinking $C=1$ is true (and possibly optimal)?
I remind the reader that proof is simply to integrate $\Delta u$ against $u\chi^2$, and write
$$ \int_{B_{R^\prime}}|\nabla u|^2 \leq \int_{B_{R}}|\nabla (u\chi)|^2 = \int_{B_{R}}|\nabla \chi|^2 u^2 \leq \int_{B_{R}\setminus B_{R^\prime}}|\nabla \chi|^2 u^2 $$ for a radial cut-off such that $\chi(t)=1$ when $t<R^\prime$ and $\chi(R)=0$. Any $C^2$ approximation of the function $$\max\left(0,\min\left(1,1-\frac{t-R^\prime}{R-R^\prime}\right)\right)$$ works, and then taking a limit gives the result. The proofs I see googling online typically use $2$ instead of $1$, or in Guiaquinta's 1985 PUP monograph '$c$', for reasons that elude me.
It could just be of course that no one wants to know if it is $1$ or $\sqrt{17}$, but $1$ takes as long to write as $2$.
