(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 $\Phi(p)\leq v(p)+C\epsilon^{1/2}$. ∎