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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 $2$.

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  • $\begingroup$ This practice is rather common. Usually such an inequality is but one in a long chain of inequalities, where the exactly formula for the constant becomes more complicated but its exact formula is of no consequence to the proof. So a constant $C$ is used, whether it is really $1$ or something more elaborate. Also, the use of the constant $C$ signals more clearly that its exact value doesn't matter. That said, sometimes using the exact value of a constant does lead to useful stronger results that are sometimes missed by people who aren't keeping careful track of the constants. $\endgroup$
    – Deane Yang
    Jul 7, 2014 at 21:08
  • $\begingroup$ @DeaneYang Thank you. Caccioppoli is often used iteratively, so any number larger than one blows up. Is the optimal constant possibly smaller than one? $\endgroup$
    – username
    Jul 7, 2014 at 21:17
  • $\begingroup$ I'm not so familiar with Caccioppoli inequalities, but for an example where getting a constant less than 1 for a similar estimate really matters, see Schoen, Simon, and Yau's paper, "Curvature estimates for minimal hyper surfaces". $\endgroup$
    – Deane Yang
    Jul 7, 2014 at 21:33
  • $\begingroup$ Why is $|\nabla(u\chi)|\le |u||\nabla\chi|$? $\endgroup$ Jul 8, 2014 at 1:32
  • $\begingroup$ @AnthonyQuas: This doesn't hold pointwise; we need to integrate by parts and use the assumption that $\Delta u\ge 0$. $\endgroup$ Jul 8, 2014 at 2:43

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