au D(x)+\sum_{y\ne x}\tau D(y)\,{{\langle \varphi_y
 \varphi_x \rangle}}_\Lambda\leq2D(x)+\sum_{y\ne x}2D(y)\,G(x-y),\end{aligned}$$ where, and from now on without stating explicitly, we use the translation invariance of $G(x)$ and the fact that $G(x-y)$ is an increasing limit of ${{\langle \varphi_y\varphi_x \rangle}}_\Lambda$ as $\Lambda\uparrow{{\mathbb Z}^d}$. By [(\[eq:J-def\])]{} and the assumption in Proposition \[prp:GimpliesPix\] that $\theta_0L^{d-q}$, with $q<d$, is bounded away from zero, we obtain $$\begin{aligned}
{\label{eq:Dbd}}
D(x)\leq O(L^{-d}){\mathbbm{1}{\raisebox{-2pt}{$\scriptstyle \{0<\|x\|_\infty\leq L\}$}}}\leq\frac{O(L^{-d+q})}
 {{\vbx{|\!|\!|}}^q}\leq\frac{O(\theta_0)}{{\vbx{|\!|\!|}}^q}.\end{aligned}$$ For the last term in [(\[eq:tildeG-1stbd\])]{}, we consider the cases for $|x|\leq2\sqrt{d}L$ and $|x|\ge2\sqrt{d}L$ separately.

When $|x|\leq2\sqrt{d}L$, we use [(\[eq:Dbd\])]{}, [(\[eq:IR-xbd\])]{} and [(\[eq:conv\])]{} with $\frac12d<q<d$ to obtain $$\begin{aligned}
\sum_{y\ne x}D(y)\,G(x-y)\leq\sum_y\frac{O(L^{-d+q})}{{\vby{|\!|\!|}}^q}\,
 \frac{\theta_0}{{\vbx-y{|\!|\!|}}^q}\leq\frac{O(\theta_0L^{-d+q})}
 {{\vbx{|\!|\!|}}^{2q-d}}\leq\frac{O(\theta_0)}{{\vbx{|\!|\!|}}^q}.\end{aligned}$$

When $|x|\ge2\sqrt{d}L$, we use the triangle inequality $|x-y|\ge|x|-|y|$ and the fact that $D(y)$ is nonzero only when $0<\|y\|_\infty\leq L$ (so that $|y|\leq\sqrt{d}\|y\|_\infty\leq\sqrt{d}L\leq\frac12|x|$). Then, we obtain $$\begin{aligned}
\sum_{y\ne x}D(y)\,G(x-y)\leq\sum_yD(y)\,\frac{2^q\theta_0}{{\vbx{|\!|\!|}}^q}
 =\frac{2^q\theta_0}{{\vbx{|\!|\!|}}^q}.\end{aligned}$$

This completes the proof of the first inequality in [(\[eq:tildeG-bd\])]{}. The second inequality can be proved similarly.

By repeated use of [(\[eq:tildeG-bd\])]{} and Proposition \[prp:conv-star\](i) with $a=b=2q$ (or Proposition \[prp:conv-star\](ii) with $x=x'$ and $y=y'$), we obtain $$\begin{aligned}
{\label{eq:psi-bd}}
\psi_\Lambda(v',v)\leq\delta_{v',v}+\frac{O(\theta_0^2)}{{\vbv-v'{|\!|\!|}}^{2q}}.\end{aligned}$$