Lemma 4.6 (00RJ) · ScalingStacks

00RJ

Lemma 4.6. Let $\Phi$ be a subharmonic function on $B_{2}^{n}\times T^{k}=B_{2}\times\mathbb{R}^{k}/\epsilon\mathbb{Z}^{k}$ equipped with the Euclidean metric $g=\sum_{1}^{n}dx_{i}^{2}+\sum_{1}^{k}dy_{j}^{2}$ , where $0<\epsilon\ll 1$ . Let $v$ be the averaging function of $\Phi$ over the $T^{k}$ fibres. Assume $\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.98003pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-6.26338pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.6363pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.45924pt}}\!\int|\Phi|\lesssim 1$ and a Lipschitz bound $\text{Lip}(v)\lesssim 1$ , then on $B_{1}\times T^{k}$ we have $\Phi\leq v+C\epsilon^{1/2}$ .

00RK

Proof. (courtesy of W. Feldman) By passing to the universal cover $B_{2}\times\mathbb{R}^{k}$ , the standard mean value inequality implies

\sup_{B_{3/2}\times T^{k}}\Phi\lesssim\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-6.11674pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.48965pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.31259pt}}\!\int|\Phi|\lesssim 1.

Let $p\in B_{1}\times T^{k}$ , which lifts to a point $p$ in $B_{1}\times\mathbb{R}^{k}$ . Consider the Euclidean ball $B_{g}(p,\epsilon R)\subset B_{3/2}\times\mathbb{R}^{k}$ , where $R\gg 1$ is a parameter to be chosen. Then by the mean value inequality,

\Phi(p)\leq\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-6.11674pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.48965pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.31259pt}}\!\int_{B_{g}(p,\epsilon R)}\Phi.

Define the subset $E\subset B_{g}(\epsilon R)$ as the union of all interior lattice cubes, then

B_{g}(p,\epsilon R)\setminus E\subset B_{g}(p,\epsilon R)\setminus B_{g}(p,\epsilon(R-C)),

and by the lattice periodicity of $\Phi$ we have $\int_{E}\Phi=\int_{E}v$ . By partitioning the integral $\int_{B_{g}(p,\epsilon)}\Phi$ into the contributions from $E$ and $B_{g}(p,\epsilon R)\setminus E$ ,

\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-6.11674pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.48965pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.31259pt}}\!\int_{B_{g}(p,\epsilon R)}\Phi\leq\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-6.11674pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.48965pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.31259pt}}\!\int_{B_{g}(p,\epsilon R)}v+CR^{-1}\sup_{B_{g}(p,\epsilon R)}(\Phi-v)\leq\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-6.11674pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.48965pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-5.31259pt}}\!\int_{B_{g}(p,\epsilon R)}v+CR^{-1}.

By the Lipschitz bound of $v$ , the RHS is bounded above by

v(p)+\text{Lip}(v)\epsilon R+CR^{-1}\leq v(p)+C(\epsilon R+R^{-1}).

Choosing $R=\epsilon^{-1/2}$ gives $\Phi(p)\leq v(p)+C\epsilon^{1/2}$ . ∎