The upper bound of Lipschitz function with cell problem

Let us consider the Hamiltonian $H = H(y,p): \mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$,which is continuous and satisfies below.(H1) For each $p\in\mathbb{R}^n, y\mapsto H(y,p)$ is $\mathbb{Z}^n-periodic.$(H2) For y $\in\mathbb{T}^n$, $\lim_{|p|\to \infty} H(y,p) = +\infty$.(H3)For each y $\in\mathbb{T}^n$, $p\mapsto H(y,p)$ is convex.

It was known that there exists a unique constant $\overline{H}(p)\in\mathbb{R}$ such that the following cell problem has a Lipschitz continuous viscosity solution$H(y, p + Dv) = \overline{H}(p)$ in $\mathbb{T}^n$.(1.1)

In this paper, they assume further that H grows quadratically, that is,$|p|^2/2$ - C $\le$ H(y,p) $\le$$|p|^2/2$+ C for all (y,p)$\in$$\mathbb{T}^n\times\mathbb{R}^n$,for some C >1. Then, we also have that$|p|^2/2$ - C $\le$$\overline{H}(p)$$\le$$|p|^2/2$+ C for all p$\in$$\mathbb{R}^n$.

They use two inequalities above to get that, for each Lipschitz function $v_{p}$ solving(1.1), $||Dv_{p}||_{L^{\infty}(\mathbb{T}^n)}$$\leq$ 2(|p| + C).But I cannot get this inequality. Can you help me?