Proof. Let \(g: \mathbb{R}^{\Lambda} \rightarrow \mathbb{R}\) be an increasing function, and note that \[F(\varphi) \coloneq \exp(\beta H_{\Lambda, J}^{\xi}(\varphi) - \beta H_{\Lambda, J}^{\xi'}(\varphi)) = \exp\Bigg( \sum_{\substack{xy \in \overline{E}(\Lambda, J) \\ x \in \Lambda, \, y\in V \setminus \Lambda}} \beta J_{x,y} \varphi_x (\xi'_y -\xi_y) \Bigg)\] is an increasing function. Using the FKG inequality [7], we obtain \[\langle g \rangle_{\Lambda, \beta, \rho, J }^{\xi'} = \frac{\langle F g \rangle_{\Lambda, \beta, \rho, J }^{\xi}}{\langle F \rangle_{\Lambda, \beta, \rho, J }^{\xi}} \geq \langle g \rangle_{\Lambda, \beta, \rho, J }^{\xi}.\] ◻

Proof. The proofs of the first two and last two statements are very similar, so we only prove (i) and (iv) here. For (i), let \(z \in V\) and let \(y_0, y_1, \ldots, y_j\) be a walk from \(x_k\) to \(z\) in \(\overline{(R, J)}\). Considering the walk \(x_0, \ldots, x_k, y_1, \ldots, y_j\), then by definition of \(\tilde{A}(x_{0}, R)\), we have \[|\xi_{z}| \leq C \tilde{A}(x_0, R)^{(n-1)^{k+j}} \left(\prod_{i = 1}^{k} f(J_{x_{i-1}, x_i})^{(n-1)^{k+j-i}} \right) \left(\prod_{i = 1}^{j} f(J_{y_{i-1}, y_i})^{(n-1)^{j-i}} \right).\] Hence \(\tilde{A}(x_0, R)^{(n-1)^{k+j}} \prod_{i = 1}^{k} f(J_{x_{i-1}, x_i})^{(n-1)^{k+j-i}}\) satisfies the requirements for \(\tilde{A}(x_k, R)^{(n-1)^{j}}\), so \[\begin{align} \tilde{A}(x_k, R)^{(n-1)^{j}} &\leq \tilde{A}(x_0, R)^{(n-1)^{k+j}} \left(\prod_{i = 1}^{k} f(J_{x_{i-1}, x_i})^{(n-1)^{k+j-i}} \right)\\ &= \left( \tilde{A}(x_0, R)^{(n-1)^{k}} \prod_{i = 1}^{k} f(J_{x_{i-1}, x_i})^{(n-1)^{k-i}} \right)^{(n-1)^j}. \end{align}\] Taking both sides to the power \(1/(n-1)^j\) yields (i).

To prove (iv), let \(y \in R\) and let \(y_0, y_1, \ldots, y_j\) be a walk from \(x_k\) to \(y\) in \(\overline{(R, J)}\). Considering the walk \(x_0, \ldots, x_k, y_1, \ldots, y_j\), then by definition of \(A(x_{0}, R)\), we have \[|h_{y, R}| \leq \left(\sum_{z \in V \setminus R} |J_{y, z}| f(J_{y, z}) \right) C A(x_0,R) \lambda^{k+j+1} \prod_{i=1}^{k} f(J_{x_{i-1}, x_i})\prod_{i=1}^j f(J_{y_{i-1}, y_i}).\] This means that \(A(x_0,R)\lambda^{k} \prod_{i=1}^{k} f(J_{x_{i-1}, x_i})\) satisfies the requirements for \(A(x_k,R)\), which yields (iv). ◻

Proof. By Young’s inequality and the fact that \(|t_x|,|t_y|\leq |\varphi_x|\), \[\label{eq:Young32inequality} \beta (|\varphi_x| |\varphi_y| + |t_x||t_y|) \leq \frac{\beta}{2} (|\varphi_x|^{2} + |\varphi_y|^{2}+|t_x|^2+|t_y|^2)\leq \frac{\beta}{2}(3|\varphi_x|^{2}+|\varphi_y|^{2}).\tag{4}\] If \(|\varphi_y|<C\), then 4 implies that \[\beta (|\varphi_x| |\varphi_y| + |t_x||t_y|) \leq 2\beta |\varphi_x|^2\leq \frac{2\beta}{C^{n-2}}|\varphi_x|^n.\] If \(|\varphi_y|\geq C\), then 4 gives \[\beta (|\varphi_x| |\varphi_y| + |t_x||t_y|) \leq \frac{\beta}{2C^{n-2}}(3|\varphi_x|^{n}+|\varphi_y|^{n})\leq \frac{3\beta}{2C^{n-2}}(|\varphi_x|^{n}+|\varphi_y|^{n}).\] This completes the proof. ◻

Proof. Note that since \(|\varphi_y| \leq \frac{f(J_{x,y})}{C^{n-2}}|\varphi_x|^{n-1}\), we have \[\begin{align} \beta |\varphi_y| (|\varphi_x| + |t|) -\frac{a f(J_{x, y})}{2M_{f}} |\varphi_x|^{n} &\leq \beta f(J_{x, y}) \frac{|\varphi_x|^{n-1}}{C^{n-2}} (|\varphi_x| + |t|)-\frac{a f(J_{x, y})}{2 M_{f}} |\varphi_x|^{n}\\ &\leq f(J_{x, y}) |\varphi_x|^{n} \left( \frac{\beta}{C^{n-2}} \left(1 + \frac{\alpha_0}{|\varphi_x|}\right) - \frac{a}{2M_f} \right). \end{align}\] For every \(C\geq \alpha_0 + (4a^{-1}M_f\beta)^\frac{1}{n-2}\), the above expression is negative when \(|\varphi_x| \geq C\). ◻

Proof of Theorem 1. Fix a finite subset \(\Lambda \subset V\) and let \(\Lambda' \subset \Lambda\). We write \(E\) for \(\{xy \in \overline{E}(\Lambda, J): x,y \in \Lambda\}\) and \(A_x\) for \(A(x, \Lambda, \xi, C, J, f, n)\), where \(C \geq 1\) is a constant to be determined. Define \(\mathcal{C}\) to be the set of vertices \(x \in \Lambda\) such that for some \(m \in \{ 0, 1, \ldots\}\) there exists a walk \(x_0, x_1, \ldots, x_m\) from \(\Lambda'\) to \(x\) in \(\overline{(\Lambda, J)}\) that satisfies \[\label{cluster32membership32criteria} \forall k \in S_m \quad |\varphi_{x_k}| \geq C A_{x_0}^{(n-1)^{k}} \prod_{i = 1}^{k} f(J_{x_{i-1}, x_i})^{(n-1)^{k-i}},\tag{5}\] where \(S_0 = \{0\}\) and \(S_m = \{1, \ldots, m\}\) for \(m \geq 1\). For each \(i \in \{ 0, 1, \ldots\}\), let \(\mathcal{C}_i\subset \mathcal{C}\) denote the set of vertices for which \(i\) is the smallest value of \(m\) such that there exists a walk satisfying 5 . Note that if \(x \in \mathcal{C}\), then \(|\varphi_x| \geq C\), and if \(y \in \Lambda'\), then Lemma 1 implies that \(y\in \mathcal{C}\) if and only if \(y \in \mathcal{C}_0\), that is, if and only if \(|\varphi_y| \geq CA_y\).

Our aim is to prove that there exists \(K \geq 0\) such that for any \(V_1, V_2, \ldots\) pairwise disjoint subsets of \(\Lambda \setminus \Lambda'\) we have \[\begin{align} \label{eq:branching32process32comparison} &\mathrm{d}\nu_{\Lambda, \beta, \rho, J}^{\xi} \left[ \varphi|_{\Lambda'}, \, (\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1} \right] \leq \\ \nonumber &\exp \left(K|\Lambda'\cup V'| \right) \left( \prod_{x \in \Lambda'} e^{\alpha_1 M_f A_{x}^{n}} \mathrm{d} \rho_{\frac{a}{2}}(\varphi_x) \right) \prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}, \end{align}\tag{6}\] where \(\alpha_1=2\beta C^2\), \(V_0 = \Lambda'\), \(V' = \bigcup_{i = 1}^{\infty} V_i\), and \[p_{x, y} = \begin{cases} \int_{|u| \geq C f(J_{x,y})} \mathrm{d} \rho_{\frac{a}{2}}(u) & \text{if } J_{x,y} \neq 0, \\ 0 & \text{if } J_{x,y} = 0. \end{cases}\] Here \((p_{x, y})_{x, y \in V}\) can be interpreted as the offspring distribution of a branching process in the sense that \(y\) is a child of \(x\) with probability \(p_{x, y}\). The distribution of the number of children of vertex \(x\) in this process depends on \(x\), but we can get a uniform control of the distribution by tuning the value of \(C\).

Assume that 6 holds for now. Summing 6 over all possibilities \((V_j)\) for \((\mathcal{C}_j)\) and noting that \(\Lambda' \cap V' = \emptyset\) on the event \(\{(\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\}\), we obtain \[\begin{align} \mathrm{d}\nu_{\Lambda, \beta, \rho, J}^{\xi} \left[ \varphi|_{\Lambda'}\right] \leq \nonumber \left( \prod_{x \in \Lambda'} e^{\alpha_1 M_f A_{x}^{n}} \mathrm{d} \rho_{\frac{a}{2}}(\varphi_x) \right) \sum_{(V_j)} e^{K(|\Lambda'|+|V'|)}\prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}. \end{align}\] We can thus conclude by applying Lemma 5, which bounds the right-hand side.

Let us now prove 6 . Let \(\alpha_0 \geq 1\) be such that \(\rho_{a}([-\alpha_0, \alpha_0]) > 0\). Given \((\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\) with \(\bigcup_{i = 1}^{\infty} V_i = V'\), let \(t \in [-\alpha_0, \alpha_0]^{\Lambda' \cup V'}\) and define \(\tilde{\varphi} \in \mathbb{R}^{\Lambda}\) to be the configuration with \(\tilde{\varphi}_x = t_x\) for \(x \in \Lambda' \cup V'\) and \(\tilde{\varphi_x} = \varphi_x\) for \(x \in \Lambda \setminus (\Lambda' \cup V')\). Comparing the value of the integrand at \(\varphi\) with its value at \(\tilde{\varphi}\) and setting \(\alpha_1 = 2 \beta C^{2}\), we will see that we can choose \(C\) large enough that \[\begin{align} \label{compactified32inequality} &\beta J_{x,y} \varphi_x \varphi_y \leq \beta J_{x,y} \tilde{\varphi}_x \tilde{\varphi}_y\\ \nonumber & + |J_{x,y}| f(J_{x, y}) \left(\frac{a|\varphi_x|^{n}}{2M_f}\mathbb{1}_{x\in \mathcal{C}}+\alpha_1 A_x^n\mathbb{1}_{x\in \Lambda' \setminus \mathcal{C}}+\frac{a|\varphi_y|^{n}}{2M_f}\mathbb{1}_{y\in \mathcal{C}}+\alpha_1 A_y^n \mathbb{1}_{y\in \Lambda' \setminus \mathcal{C}} \right). \end{align}\tag{7}\] We now verify 7 by considering an edge \(xy \in E\) and splitting into cases based on whether \(x\) and \(y\) are in \(\mathcal{C}\), \(\Lambda' \setminus \mathcal{C}\), or \(\Lambda \setminus (\Lambda' \cup \mathcal{C})\). In each case we will use the bound \(\beta J_{x,y}(\varphi_x \varphi_y - \tilde{\varphi}_x \tilde{\varphi}_y) \leq \beta |J_{x,y}| (|\varphi_x| |\varphi_y| + |\tilde{\varphi}_x| |\tilde{\varphi}_y|)\).

• If \(x, y \in \mathcal{C}\), then \(|\varphi_x|, |\varphi_y| \geq C\), so by Lemma 2, if \(C\geq \alpha_0+\left(\frac{4M_f\beta}{a}\right)^{\frac{1}{n-2}}\), then \[\beta (|\varphi_x| |\varphi_y| + |t_x||t_y|) \leq \frac{a f(J_{x, y})}{2 M_f} ( |\varphi_x|^{n} + |\varphi_y|^{n}).\]

• If \(x \in \mathcal{C}\) and \(y \in \Lambda \setminus (\Lambda' \cup \mathcal{C})\), then there exists a walk \(x_0, x_1, \ldots, x_m\) from \(\Lambda'\) to \(x\) in \(\overline{(\Lambda, J)}\) satisfying 5 , but there is no such walk from \(\Lambda'\) to \(y\). Hence the walk \(x_0, \ldots, x_m, y\) does not satisfy 5 , so \[\begin{align} |\varphi_y| &\leq C A_{x_0}^{(n-1)^{m+1}} f(J_{x, y}) \prod_{i = 1}^{m} f(J_{x_{i-1}, x_i})^{(n-1)^{m+1-i}}\\ &= \frac{f(J_{x, y})}{C^{n-2}} \left(C A_{x_0}^{(n-1)^{m}} \prod_{i = 1}^{m} f(J_{x_{i-1}, x_i})^{(n-1)^{m-i}} \right)^{n-1} \leq \frac{f(J_{x, y})}{C^{n-2}} |\varphi_x|^{n-1}. \end{align}\] Combining the latter with Lemma 3 we get \[\beta (|\varphi_x| |\varphi_y| + |t_x||\varphi_y|) \leq \frac{a f(J_{x, y})}{2 M_f} |\varphi_x|^{n}.\]

• If \(x \in \mathcal{C}\) and \(y \in \Lambda' \setminus \mathcal{C}\), then applying Lemma 2 with \(C \geq \alpha_0+\left(\frac{4M_f\beta}{a}\right)^{\frac{1}{n-2}}\) and using that \(|\varphi_y|\leq CA_y\), we get \[\begin{align} \beta (|\varphi_x| |\varphi_y| + |t_x| |t_y|) &\leq \frac{2\beta}{C^{n-2}}(|\varphi_x|^{n}+(CA_y)^n) \leq \frac{a f(J_{x, y})}{2 M_f}|\varphi_x|^{n} + 2\beta C^{2} A_{y}^{n}. \end{align}\]

• If \(x, y \in \Lambda' \setminus \mathcal{C}\), then \(|\varphi_x| \leq CA_x\) and \(|\varphi_y| \leq CA_y \leq CA_{x}^{n-1}f(J_{x, y})\) by Lemma 1, so \[\beta (|\varphi_x| |\varphi_y| + |t_x| |t_y|) \leq 2 \beta C^{2} A_x^n f(J_{x,y}).\]

• If \(x \in \Lambda' \setminus \mathcal{C}\) and \(y \in \Lambda \setminus (\Lambda' \cup \mathcal{C})\), then \(|\varphi_x| \leq CA_x\) and \(|\varphi_y| \leq CA_{x}^{n-1}f(J_{x, y})\), so \[\begin{align} \beta (|\varphi_x| |\varphi_y| + |t_x||\varphi_y|) \leq 2 \beta C^{2} A_x^n f(J_{x,y}). \end{align}\]

The terms \(e^{\beta h_{x,\Lambda} \varphi_x}\) coming from interaction with the spins outside \(\Lambda\) can be bounded similarly, using that if \(x \in \mathcal{C}\), then by definition of \(\mathcal{C}\) and \(A_x\) we have \[|h_{x, \Lambda}| \leq \left(\sum_{y \in V \setminus \Lambda} |J_{x, y}| f(J_{x,y}) \right) \frac{|\varphi_x|^{n-1}}{C^{n-2}},\] so Lemma 3 applies. If \(x \in \Lambda' \setminus \mathcal{C}\) then we bound \(|h_{x, \Lambda}|\) using the definition of \(A_x\). Overall, we obtain \[\label{compactified32inequality322} \beta h_{x,\Lambda} \varphi_x \leq \beta h_{x, \Lambda} \tilde{\varphi}_x + \left( \frac{a|\varphi_x|^{n}}{2M_f}\mathbb{1}_{x\in \mathcal{C}}+\alpha_1 A_x^n\mathbb{1}_{x\in \Lambda' \setminus \mathcal{C}} \right) \sum_{y \in V \setminus \Lambda} |J_{x, y}| f(J_{x,y}).\tag{8}\]

Now 6 follows from Lemma 4 below, which we also state in the case \(n=2\). It only remains to show that we can choose \(C\) and \(\tilde{C}\) of the required form. Note that \(C\) depends on \(\beta\) only through the condition \(C\geq \alpha_0+\left(\frac{4M_f\beta}{a}\right)^{\frac{1}{n-2}}\) when applying Lemmas 2 and 3. We can choose \(\tilde{C} = \log(\alpha_2 \int_{\mathbb{R}} \mathrm{d} \rho_{\frac{a}{2}}(u)) + \alpha_1 M_f\), where \(\alpha_2\) is given by Lemma 5 and does not depend on \(\beta\). Hence the only dependence of \(\tilde{C}\) on \(\beta\) comes from \(\alpha_1 = 2 \beta C^{2}\). ◻

Proof. For ease of notation, write \(P_{\Lambda}(\varphi) = \prod_{x \in \Lambda} e^{-a|\varphi_x|^{n}}\) and \[\pi_{E}(\varphi) = \left(\prod_{xy \in E} e^{\beta J_{x, y} \varphi_x \varphi_y} \right) \left( \prod_{x \in \Lambda} e^{\beta h_{x, \Lambda} \varphi_x}\right).\] Note that ?? gives that \[\begin{align} &\mathrm{d} \nu_{\Lambda, \beta, \rho, J}^{\xi} \left[ \varphi|_{\Lambda'}, \, (\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1} \right] = \frac{1}{Z_{\Lambda, \beta, \rho, J}^{\xi}} \int_{\mathbb{R}^{\Lambda \setminus \Lambda'}} \mathbb{1}_{\{(\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\}} \pi_{E}(\varphi) P_{\Lambda}(\varphi) \prod_{x \in \Lambda} \mathrm{d} \rho_{a}(\varphi_x), \end{align}\] where \(\rho_a\) satisfies \(0 < \rho_{a}(\mathbb{R})<\infty\). We estimate \(\pi_{E}(\varphi)\) by applying 7 to each element of the first product and 8 to each element of the second product. This yields \[\pi_{E}(\varphi) \leq \prod_{x \in \Lambda' \setminus \mathcal{C}} \exp \left( \alpha_1 A_{x}^{n} \sum_{y \in V} |J_{x, y}| f(J_{x, y}) \right) \prod_{x \in \mathcal{C}} \exp \left( \frac{a|\varphi_x|^{n}}{2 M_f} \sum_{y \in V} |J_{x, y}| f(J_{x, y})\right) \pi_{E}(\tilde{\varphi}).\] Using [C2] the above inequality simplifies to \[\begin{align} \pi_{E}(\varphi) &\leq \left( \prod_{x \in \Lambda'} e^{\alpha_1 M_f A_{x}^{n}} \right) \left(\prod_{x \in \mathcal{C}} e^{\frac{a}{2}|\varphi_x|^{n}}\right) \pi_{E}(\tilde{\varphi}). \end{align}\] Combining this with the product over vertices of the terms coming from the single-site measure, and using that \(|t_x| \leq \alpha_0\) for any \(x \in \Lambda' \cup V'\), we get \[\begin{align} \label{eq:pi32P32inequality} \pi_{E}(\varphi) P_{\Lambda}(\varphi) &\leq \left( \prod_{x \in \Lambda'} e^{\alpha_1 M_f A_{x}^{n}} \right) \left( \prod_{x \in \Lambda'\cup V'} e^{-\frac{a}{2}|\varphi_x|^{n}} \right) \pi_{E}(\tilde{\varphi}) P_{\Lambda \setminus (\Lambda' \cup V')}(\tilde{\varphi})\\ &\leq \exp \left(a \alpha_{0}^{n} |\Lambda'\cup V'| \right) \left( \prod_{x \in \Lambda'} e^{\alpha_1 M_f A_{x}^{n}} \right) \left( \prod_{x \in \Lambda'\cup V'} e^{-\frac{a}{2}|\varphi_x|^{n}} \right) \pi_{E}(\tilde{\varphi}) P_{\Lambda}(\tilde{\varphi}) \notag. \end{align}\tag{9}\] We now integrate with respect to \(\varphi_x\) for each \(x \in \Lambda \setminus \Lambda'\) over the event \(\{(\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\}\). Observe that, on this event, if \(y \in V_{i + 1}\) for some \(i \in \{0, 1, \ldots\}\), then \(|\varphi_y| \geq C f(J_{x, y})\) for some \(x \in V_i\) with \(J_{x, y} \neq 0\). Hence, ignoring the requirement for spins outside \(\mathcal{C}\) to be small, we have \[\label{eq:32integral32p32bound} \int_{\mathbb{R}^{V'}} \mathbb{1}_{\{(\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\}} \prod_{x \in V'} e^{-\frac{a}{2}|\varphi_x|^{n}} \mathrm{d} \rho_{a}(\varphi_x) \leq \prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}.\tag{10}\] Note that for any function \(F: \mathbb{R}^{\Lambda' \cup V'} \to \mathbb{R}_{\geq 0}\), \[\begin{align} \nonumber \rho_{a}([-\alpha_0, \alpha_0])^{|\Lambda' \cup V'|}\min_{t \in [-\alpha_0, \alpha_0]^{\Lambda' \cup V'}} F(t) &\leq \int_{[-\alpha_0, \alpha_0]^{\Lambda' \cup V'}} F(t) \prod_{x \in \Lambda' \cup V'}\mathrm{d} \rho_{a}(t_x)\\ \label{eq:min} &\leq \int_{\mathbb{R}^{\Lambda' \cup V'}}F(t) \prod_{x \in \Lambda' \cup V'}\mathrm{d} \rho_{a}(t_x). \end{align}\tag{11}\] Thus, since \(t \in [-\alpha_0, \alpha_0]^{\Lambda' \cup V'}\) is arbitrary in the definition of \(\tilde{\varphi}\), integrating 9 and dividing by \(Z_{\Lambda, \beta, \rho, J}^{\xi}\), applying 11 for \(F\) being \(\pi_{E}(\tilde{\varphi}) P_{\Lambda}(\tilde{\varphi})\), and using 10 yields 6 , with \(K = \max \{0, a \alpha_{0}^{n} - \log(\rho_{a}([-\alpha_0, \alpha_0]))\}\). ◻

Proof. We aim to compare \(\prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}\) with the probability that \(W_i = V_i\) for all \(i \geq 1\), where \((W_{i})_{i \geq 0}\) is the exploration process defined as follows. First set \(W_0 = \Lambda'\) and ensure that \(C\) is chosen large enough that \(\int_{|u| \geq C} \mathrm{d} \rho_{\frac{a}{2}}(u) \leq 1\). Assuming that \(W_i\) has already been constructed, for each vertex \(x \in W_i\) and \(y \in \Lambda \setminus (W_0 \cup \ldots \cup W_i)\), say that the edge \(xy\) is open with probability \(p_{x, y}\), independently of all other edges. Otherwise, say \(xy\) is closed. Then set \(W_{i+1}\) to be the set of vertices \(y \in \Lambda \setminus (W_0 \cup \ldots \cup W_i)\) such that \(xy\) is open for some \(x \in W_i\). Once \(W_i\) has been constructed for all \(i \in \{0, 1, \ldots \}\), set \(W = \bigcup_{i=1}^{\infty} W_i\). With this definition, we have \[\begin{align} &\mathbb{P}[W_{i+1} = V_{i+1} | W_1 = V_1, \ldots, W_i = V_i] =\\ &\prod_{y \in \Lambda \setminus (V_0 \cup \ldots \cup V_{i+1})} \mathbb{P} \left[ \bigcap_{x \in V_i} \{ xy \text{ closed} \}\right] \prod_{y \in V_{i+1}} \mathbb{P} \left[ \bigcup_{x \in V_i} \{ xy \text{ open} \}\right] \geq b^{|V_i|} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}, \end{align}\] where \[b = \inf_{x \in V} \prod_{y \in V \setminus \{x\}} (1-p_{x, y}).\] Combining over all generations of the exploration process, we obtain \[\begin{align} \nonumber \mathbb{P}[(W_{i})_{i \geq 1} = (V_i)_{i \geq 1}] &= \prod_{i = 0}^{\infty} \mathbb{P}[W_{i+1} = V_{i+1} | W_1 = V_1, \ldots, W_i = V_i]\\ \label{finished32exploration} & \geq \prod_{i = 0}^{\infty} b^{|V_i|} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y} = b^{|\Lambda'| + |V'|} \prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}. \end{align}\tag{12}\]

We now claim that \(\sup_{x \in V} \sum_{y \in V \setminus \{x\}} p_{x, y}\) tends to \(0\) as \(C\) tends to infinity, which in turn implies that \(b\) tends to \(1\). For any \(x, y \in V\) with \(J_{x, y} \neq 0\), we have \[\begin{align} p_{x, y} = \int_{|u| \geq C f(J_{x, y})} e^{-\frac{a}{2}|u|^{n}} \, \mathrm{d} \rho_{a}(u) \leq \exp \left( -\frac{a}{2} (Cf(J_{x,y}))^{n} \right) \rho_{a}(\mathbb{R}). \end{align}\] Given \(x,y \in V\) with \(|J_{x, y}| < \delta_f\), we have by [C2] that \(p_{x, y} \leq \rho_{a}(\mathbb{R}) |J_{x, y}|^{\frac{a}{2}C^{n}}\), and we can choose \(C>0\) to be large enough so that \(\rho_{a}(\mathbb{R}) |J_{x, y}|^{\frac{a}{2}C^{n}}\leq \tfrac{|J_{x, y}|}{CM_f}\). Since \(\sum_{y \in V} |J_{x, y}| \leq M_f\), there are at most \(M_f/\delta_f\) vertices \(y\in V\) such that \(|J_{x,y}|\geq \delta_f\) and for these vertices, we can use the bound \(p_{x, y} \leq e^{-a C^{n}/2} \rho_{a}(\mathbb{R})\). The claim follows.

Let \(\alpha_3 = b^{-1} e^{K} \in [1, \infty)\). It follows from summing 12 over \(V_1, V_2, \ldots\) that \[\label{final95bound} \sum_{(V_j)} e^{K(|\Lambda'|+|V'|)}\prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}\leq \alpha_{3}^{|\Lambda'|} \sum_{k = 0}^{\infty} \alpha_{3}^k \mathbb{P}[|W| = k].\tag{13}\] We want to stochastically dominate \((W_i)_{i \geq 0}\) by a branching process \((Z_i)_{i \geq 0}\). By choosing the value of \(C\) to be large enough, we can ensure that \(b \geq 1/2\). We can then define a random variable \(X\) by \[\mathbb{P}[X = k] = \begin{cases} b &\text{if } k = 0,\\ 1 - b - \frac{(1-b)^2}{b} &\text{if } k = 1, \\ (1-b)^k &\text{if } k \in \{2, 3, \ldots\}, \end{cases}\] and let \((Z_i)_{i \geq 0}\) be a branching process with initial population \(|\Lambda'|\) and offspring distribution \(X\). If \(x \in W_i\) for some \(i \in \{0,1,\ldots\}\), let \(X_x\) be the number of vertices \(y \in W_{i+1}\) such that the edge \(xy\) is open. For all \(k \geq 0\), we have \(\mathbb{P}[X_x \geq k] \leq \mathbb{P}[X_x \ge 1]^k \leq (1-b)^k \leq \mathbb{P}[X \geq k]\), which proves the desired stochastic domination. In particular, since \(\alpha_3 \geq 1\) we have that \[\begin{align} \nonumber \sum_{k = 0}^{\infty} \alpha_{3}^k \mathbb{P}[|W| = k] \leq \sum_{k = 0}^{\infty} \alpha_{3}^k \mathbb{P}[T = k + |\Lambda'|], \end{align}\] where \(T\) is the total progeny of \((Z_i)_{i \geq 0}\). Applying Theorem 4 gives that \[\begin{align} \nonumber \sum_{k = 0}^{\infty} \alpha_{3}^k \mathbb{P}[|W| = k] &\leq \alpha_3^{|\Lambda'|} + \sum_{k = |\Lambda'|}^{\infty} \alpha_{3}^k \frac{|\Lambda'|}{k + |\Lambda'|} \mathbb{P}[X_1 + \ldots + X_{k + |\Lambda'|} = k]\\ \label{sum32of32X321} & \leq \alpha_3^{|\Lambda'|} + \frac{1}{2}\sum_{k = |\Lambda'|}^{\infty} \alpha_3^{k} \mathbb{P}[X_1 + \ldots + X_{2k} \geq k], \end{align}\tag{14}\] where \(X_1, X_2,\ldots\) are independent with the same distribution as \(X\). We now bound the latter probability as follows. Setting \(\theta = 2 \log(2\alpha_3)\), we have that \(\mathbb{E}[e^{\theta X}] \rightarrow 1\) as \(b \rightarrow 1\). By increasing the value of \(C\), we can make \(b\) as close to 1 as desired, so by choosing \(C\) large enough, we can ensure that \(\mathbb{E}[e^{\theta X}] < e^{\theta/4}\). We then have by the exponential Markov inequality and independence \[\label{sum32of32X322} \mathbb{P}[X_1 + \ldots + X_{2k} \geq k] \leq e^{-\theta k} \mathbb{E}[e^{\theta X}]^{2k} \leq e^{-\theta k/2} = (2\alpha_3)^{-k}.\tag{15}\] Combining 14 and 15 yields \[\sum_{k=0}^{\infty} \alpha_{3}^{k} \mathbb{P}[|W| = k] \leq \alpha_3^{|\Lambda'|} + 2^{-|\Lambda'|},\] and substituting this in 13 completes the proof. ◻

Proof of Theorem 6. Fix a finite subset \(\Lambda \subset V\) and let \(\Lambda' \subset \Lambda\). We write \(E\) for \(\{xy \in \overline{E}(\Lambda, J): x,y \in \Lambda\}\) and \(A_x\) for \(A(x, \Lambda, \lambda, \xi, C, J, f)\), where \(C \geq 1\) is a constant to be determined. Define \(\mathcal{C}\) to be the set of vertices \(x \in \Lambda\) such that for some \(m \in \{ 0, 1, \ldots\}\) there exists a walk \(x_0, x_1, \ldots, x_m\) from \(\Lambda'\) to \(x\) in \(\overline{(\Lambda, J)}\) that satisfies \[\label{cluster32membership32criteria32n612} \forall k \in S_m \quad |\varphi_{x_k}| \geq C A_{x_0} \lambda^{k} \prod_{i = 1}^{k} f(J_{x_{i-1}, x_i}),\tag{16}\] where \(S_0 = \{0\}\) and \(S_m = \{1, \ldots, m\}\) for \(m \geq 1\). For each \(i \in \{ 0, 1, \ldots\}\), let \(\mathcal{C}_i\subset \mathcal{C}\) denote the set of vertices for which \(i\) is the smallest value of \(m\) such that there exists a walk satisfying 16 . Let \(\alpha_0 \geq 1\) be such that \(\rho_{a}([-\alpha_0, \alpha_0]) > 0\). Given \((\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\) with \(\bigcup_{i = 1}^{\infty} V_i = V'\), let \(t \in [-\alpha_0, \alpha_0]^{\Lambda' \cup V'}\) and define the configuration \(\tilde{\varphi} \in \mathbb{R}^{\Lambda}\) as in the proof of Theorem 1. Setting \(\alpha_1 = \frac{a C^{2}}{2 M_f} \geq 2 \beta C^{2} \lambda\), one can show that when \(C \geq \alpha_0\) \[\begin{align} \label{compactified32inequality32n612} &\beta J_{x,y} \varphi_x \varphi_y \leq\\ \nonumber &\beta J_{x,y} \tilde{\varphi}_x \tilde{\varphi}_y +|J_{x,y}|f(J_{x, y}) \left(\frac{a|\varphi_x|^{2}}{2M_f}\mathbb{1}_{x\in \mathcal{C}}+\alpha_1 A_x^2 \mathbb{1}_{x\in \Lambda' \setminus \mathcal{C}}+\frac{a|\varphi_y|^{2}}{2M_f}\mathbb{1}_{y\in \mathcal{C}}+\alpha_1 A_x^2 \mathbb{1}_{y\in \Lambda' \setminus \mathcal{C}} \right), \end{align}\tag{17}\] and \[\label{compactified32inequality3223240n61241} \beta h_{x,\Lambda} \varphi_x \leq \beta h_{x, \Lambda} \tilde{\varphi}_x + \left( \frac{a|\varphi_x|^2}{2M_f}\mathbb{1}_{x\in \mathcal{C}}+\alpha_1 A_x^2 \mathbb{1}_{x\in \Lambda' \setminus \mathcal{C}} \right) \sum_{y \in V \setminus \Lambda} |J_{x, y}| f(J_{x,y}).\tag{18}\] The first inequality 17 can be verified in a similar way to 7 . Indeed, if \(|\varphi_x|, |\varphi_y| \geq \alpha_0\), then applying Lemma 2 and using that \(a \geq 4 \beta M_f\) gives that \[\begin{align} \beta (|\varphi_x| |\varphi_y| + |t_x||t_y|) \leq \frac{a f(J_{x,y})}{2M_f} (|\varphi_x|^{2} + |\varphi_y|^{2}). \end{align}\] We also use that if \(x\in \mathcal{C}\) and \(y\in \Lambda\setminus \mathcal{C}\), then \(|\varphi_x| \geq \alpha_0\) and \(|\varphi_y| \leq \lambda f(J_{x,y}) |\varphi_x|\), which implies \[\begin{align} \label{Edge32removal32alt32n6123240i41} \beta (|\varphi_x| |\varphi_y| + |t_x||\varphi_y|) \leq 2 \beta \lambda f(J_{x,y}) |\varphi_x|^{2} \leq \frac{a f(J_{x,y})}{2M_f} |\varphi_x|^{2}. \end{align}\tag{19}\] To prove 18 , if \(x \in \mathcal{C}\) then \[\begin{align} |h_{x, \Lambda}|\leq \left(\sum_{y \in V \setminus \Lambda} |J_{x, y}| f(J_{x,y}) \right) \lambda |\varphi_x|, \end{align}\] so 19 applies. If \(x \in \Lambda' \setminus \mathcal{C}\) then we bound \(|h_{x, \Lambda}|\) using the definition of \(A_{x}\) to obtain \[|h_{x, \Lambda}| \leq C \lambda A_x \left(\sum_{y \in V \setminus \Lambda} |J_{x, y}| f(J_{x,y}) \right).\] Having obtained 17 and 18 , we use Lemmas 4 and 5 as in the proof of Theorem 1 to conclude. ◻

Proof of Theorem 8. Let \(f(1) = 1\) and write \(A_x\) for \(A(x, \Lambda, \lambda, \xi, C_2, J, f)\), where \(C_2 \geq 1\) is a constant to be determined. Define \(\mathcal{C}\) to be the set of vertices \(x \in \Lambda\) such that for some \(m \in \{ 0, 1, \ldots\}\) there exists a walk \(x_0, x_1, \ldots, x_m\) from \(\Lambda'\) to \(x\) in \(\overline{(\Lambda, J)}\) that satisfies \[\label{changed32cluster32membership32criteria} \forall k \in S_m \quad |\varphi_{x_k}| \geq C_2 A_{x_0} \lambda^{k},\tag{20}\] where \(S_0 = \{0\}\) and \(S_m = \{1, \ldots, m\}\) for \(m \geq 1\). For each \(i \in \{ 0, 1, \ldots\}\), let \(\mathcal{C}_i\subset \mathcal{C}\) denote the set of vertices for which \(i\) is the smallest value of \(m\) such that there exists a walk satisfying 20 . Given \((\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1}\) with \(\bigcup_{i = 1}^{\infty} V_i = V'\), let \(t \in T^{\Lambda' \cup V'}\) and define the configurations \(\varphi', \tilde{\varphi} \in \mathbb{R}^{V}\) by \[\varphi'_x = \begin{cases} \varphi_x & \text{if } x \in \Lambda,\\ \xi_x & \text{otherwise}, \end{cases} \qquad \tilde{\varphi}_x = \begin{cases} t_x & \text{if } x \in \Lambda' \cup V',\\ \varphi'_x & \text{otherwise}. \end{cases}\] Assumption (i) in the statement of the theorem is analogous to 19 , and (ii) allows us to bound \(U_{xy}(\varphi'_x, \varphi'_y)\) in terms of \(U_{xy}(\tilde{\varphi}_x, \tilde{\varphi}_y)\) when \(x\) or \(y\) is in \(\Lambda' \setminus \mathcal{C}\). Together with the definitions of \(\mathcal{C}\) and \(A(o, \Lambda)\), they imply that for any \(xy \in E\) \[\begin{align} &U_{xy}(\varphi'_x, \varphi'_y) \leq\\ &U_{xy}(\tilde{\varphi}_x, \tilde{\varphi}_y) \exp\left(\frac{a_{x, \Lambda}}{2D}|\varphi_x|^{2} \mathbb{1}_{x \in \mathcal{C}} + F(A_{x}) \mathbb{1}_{x \in \Lambda' \setminus \mathcal{C}} + \frac{a_{y, \Lambda}}{2D}|\varphi_y|^{2} \mathbb{1}_{y \in \mathcal{C}} + F(A_{y}) \mathbb{1}_{y \in \Lambda' \setminus \mathcal{C}}\right). \end{align}\] Combining over all edges and using that each vertex has degree at most \(D\), we get \[\begin{align} \prod_{xy \in E} U_{xy}(\varphi'_x, \varphi'_y) \leq \left(\prod_{xy \in E} U_{xy}(\tilde{\varphi}_x , \tilde{\varphi}_y) \right) \prod_{x \in \mathcal{C}} e^{\frac{a_{x, \Lambda}}{2}|\varphi_x|^{2}} \prod_{x \in \Lambda'} e^{D F(A_{x})}. \end{align}\] By combining the above with the terms coming from the single-site measure and integrating, we can show as in the proof of Lemma 4 that for some \(K\geq0\), \[\begin{align} &\mathrm{d}\nu_{\Lambda, U, \rho}^{\xi} \left[ \varphi|_{\Lambda'}, \, (\mathcal{C}_i)_{i \geq 1} = (V_i)_{i \geq 1} \right] \leq \\ \nonumber &\exp \left(K|\Lambda'\cup V'| \right) \left( \prod_{x \in \Lambda'} e^{DF(A_x)} e^{\frac{-a_{x, \Lambda}}{2}|\varphi_x|^2} \mathrm{d}\mu_{x, \Lambda}(\varphi_x) \right) \prod_{i = 0}^{\infty} \prod_{y \in V_{i+1}} \max_{x \in V_i} p_{x, y}, \end{align}\] with \(p_{x,y}\) defined as in ?? . Applying Lemma 5 concludes the proof. ◻

Proof. Monotonicity in \(\xi\) has already been established in Proposition 3, so we just need to prove that \(\nu_{\Lambda, \beta, \rho, J}^{\xi}[\varphi_x \geq u] \leq \nu_{\Lambda, \beta, \rho, J'}^{\xi}[\varphi_x \geq u]\). Writing \(\sigma_x\) for the sign of \(\varphi_x\) and using that \(\mathbb{1}_{\{\sigma_x = 1\}} = \frac{1}{2}(1 + \sigma_x)\), we have \[\begin{align} \nu_{\Lambda, \beta, \rho, J}^{\xi}[\varphi_x \geq u] &= \nu_{\Lambda, \beta, \rho, J}^{\xi}[|\varphi_x| \geq u] \nu_{\Lambda, \beta, \rho, J}^{\xi}[\sigma_x = 1 \mid |\varphi_x| \geq u]\\ &= \frac{1}{2} \nu_{\Lambda, \beta, \rho, J}^{\xi}[|\varphi_x| \geq u] (1 +\langle \sigma_x \mid |\varphi_x| \geq u \rangle_{\Lambda, \beta, \rho, J}^{\xi}). \end{align}\] Conditional on the absolute value field, \(\sigma\) is distributed according to an Ising model with coupling constants determined by the absolute value field. As the boundary conditions are positive, monotonicity of the Ising model in \(J\) follows by differentiating and using Griffiths’ inequality [6], and monotonicity of the absolute value field was proved in [13]. ◻

Proof of Proposition 10. Using the characterisation 21 of \(\Xi\), \(\xi \in (\mathbb{R}^{+})^{V} \setminus \Xi\) implies that there exists a sequence of vertices \((z_i)_{i \geq 1}\) such that \(\xi_{z_i}^{(n-1)^{-m_i}} \rightarrow \infty\) as \(i \rightarrow \infty\), where \(m_i = d_G(o, z_i)\). Since there are only finitely many vertices at any fixed distance from \(o\), by passing to a subsequence, we may assume that \(1 \leq m_i < m_j\) for any \(i < j\). Given \(i \geq 1\), let \(\Lambda_i = \{ x \in V : d_G(o, x) < m_i\}\) and let \(y_{i,0}, y_{i,1}, \ldots, y_{i,m_i}\) be a walk from \(o\) to \(z_i\) in \(\overline{(\Lambda_i, J)}\). Also let \(\alpha = \frac{\beta}{\tilde{a}n2^{n-1}}\), and for \(j \in \{0, \ldots, m_i\}\), define \[D_{i, j} = \alpha^{\frac{1-(n-1)^{j-m_i}}{n-2}} \xi_{z_i}^{(n-1)^{j - m_i}}.\] Then \(D_{i, m_i} = \xi_{z_i}\) and \(D_{i, j+1} = \alpha^{-1} D_{i, j}^{n-1}\). We will show that there exists \(\varepsilon > 0\) such that for all \(i\) sufficiently large, \(\nu_{\Lambda_i, \beta, \rho, J_G}^{\xi} [\varphi_o \geq D_{i, 0}] \geq \varepsilon\). Since \(D_{i, 0} \geq \min \{1, \alpha^{\frac{1}{n-2}}\} \xi_{z_i}^{(n-1)^{-m_i}} \rightarrow \infty\) as \(i \rightarrow \infty\), this implies that the sequence is not tight. The strategy for the proof is to condition in turn on the events \(\{ \varphi_{y_{i,j}} \geq D_{i, j} \}\). We can then use Lemma 6 to set \(J_{x, y} = 0\) everywhere except for the edge between \(y_{i,j}\) and \(y_{i,j+1}\), meaning that we only have to calculate a one-dimensional integral in each step. We have \[\begin{align} \nu_{\Lambda_i, \beta, \rho, J_G}^{\xi}[\varphi_{o} \geq D_{i,0}] &\geq \nu_{\Lambda_i, \beta, \rho, J_G}^{\xi} [\varphi_{y_{i,j}} \geq D_{i,j} \; \forall j \in \{0, \ldots, m_i - 1\}]\\ &= \prod_{j = 0}^{m_i - 1} \nu_{\Lambda_i, \beta, \rho, J_G}^{\xi} [\varphi_{y_{i,j}} \geq D_{i, j} \mid \varphi_{y_{i,k}} \geq D_{i,k} \; \forall k \in \{j+1, \ldots, m_i - 1\}]. \end{align}\] Define \(J^{(i,j)}\) by \((J^{(i,j)})_{x, y} = 1\) if \(\{x, y\} = \{y_{i,j}, y_{i, j+1}\}\) and \((J^{(i,j)})_{x, y} = 0\) otherwise. Also let \(\xi^{(i, j)}\) be defined by \(\xi^{(i,j)}_{y_{i, j+1}} = D_{i, j+1}\) and \(\xi^{(i,j)}_{x} = 0\) for all \(x \in V \setminus \{y_{i, j+1}\}\). Using the domain Markov property and Lemma 6, we have \[\nu_{\Lambda_i, \beta, \rho, J_G}^{\xi} [\varphi_{y_{i,j}} \geq D_{i, j} \mid \varphi_{y_{i,k}} \geq D_{i,k} \; \forall k \in \{j+1, \ldots, m_i - 1\}] \geq \nu_{\{y_{i, j}\}, \beta, \rho, J^{(i,j)}}^{\xi^{(i,j)}} [\varphi_{y_{i,j}} \geq D_{i, j}].\] We now estimate the probability on the right hand side. Let \(r = \beta D_{i, j+1} \varphi_{y_{i,j}} - \tilde{a}|\varphi_{y_{i,j}}|^{n}\). Then \[\frac{\mathrm{d} r}{\mathrm{d} \varphi_{y_{i,j}}} \bigg|_{\varphi_{y_{i,j}} = D_{i, j}} = \beta D_{i, j+1} - \tilde{a}n D_{i, j}^{n-1} = \beta D_{i, j+1} \left(1 - \frac{1}{2^{n-1}} \right),\] where the second equality is from our choice of \(\alpha\). This is greater than 1 for all \(j\) if \(i\) is large enough. Then because \(r\) is a concave function of \(\varphi_{y_{i,j}}\), \(\frac{\mathrm{d} r}{\mathrm{d} \varphi_{y_{i,j}}} \geq 1\) whenever \(\varphi_{y_{i,j}} \leq D_{i, j}\). Hence, \[\begin{align} \label{denominator32integral} \int_{- \infty}^{D_{i, j}} e^{\beta D_{i, j+1} \varphi_{y_{i,j}} - \tilde{a}|\varphi_{y_{i,j}}|^{n}} \mathrm{d} \varphi_{y_{i,j}} \leq \int_{- \infty}^{\beta D_{i, j} D_{i, j+1} - \tilde{a} (D_{i, j})^{n}} e^{r} \mathrm{d} r = \exp(\beta D_{i, j} D_{i, j+1} - \tilde{a} (D_{i, j})^{n}). \end{align}\tag{24}\] Note that the maximum value of \(r\) occurs when \(\varphi_{y_{i,j}} = (\frac{\beta}{\tilde{a}n} D_{i, j+1})^{1/(n-1)} = 2 D_{i, j}\), and \(r\) is increasing when \(\varphi_{y_{i, j}} < 2 D_{i, j}\). Consequently, when \(\varphi_{y_{i,j}} \in [D_{i, j}, 2 D_{i,j}]\), the value of \(r\) is at least \(\beta D_{i, j} D_{i, j+1} - \tilde{a} (D_{i, j})^{n}\), which implies \[\begin{align} \label{numerator32integral} \int_{D_{i, j}}^{\infty} e^{\beta D_{i, j+1} \varphi_{y_{i,j}} -\tilde{a}|\varphi_{y_{i,j}}|^{n}} \mathrm{d} \varphi_{y_{i,j}} \geq D_{i, j} \exp(\beta D_{i, j} D_{i, j+1} - \tilde{a} (D_{i, j})^{n}). \end{align}\tag{25}\] Combining 24 and 25 we obtain that \[\nu_{\{y_{i,j}\}, \beta, \rho, J^{(i,j)}}^{\xi^{(i,j)}} [\varphi_{y_{i,j}} \geq D_{i, j}] \geq \frac{D_{i, j}}{D_{i, j} + 1}.\] To conclude, we need to take the product over \(j\) and verify that this is bounded below for all \(i\) sufficiently large by a positive constant that does not depend on \(i\). We have \[\begin{align} \nu_{\Lambda_i, \beta, \rho, J_G}^{\xi}[\varphi_{o} \geq D_{i,0}] \geq \prod_{j = 0}^{m_i - 1} \frac{D_{i, j}}{D_{i, j} + 1} = \exp \left( \sum_{j = 0}^{m_i - 1} \log(D_{i, j}) - \log(D_{i, j} + 1) \right). \end{align}\] Taylor expanding \(\log(D_{i, j} + 1)\) around \(D_{i, j}\), we get \(\log(D_{i, j} + 1) \leq \log(D_{i, j}) + \frac{1}{D_{i, j}}\), so \[\begin{align} \nu_{\Lambda_i, \beta, \rho, J_G}^{\xi}[\varphi_{o} \geq D_{i,0}] \geq \exp \left(- \sum_{j=0}^{m_i -1} \frac{1}{D_{i,j}} \right) \geq \exp \left( -\sum_{j=0}^{\infty} \frac{1}{\min\{ 1, \alpha^{\frac{1}{n-2}}\} (\xi_{z_i}^{(n-1)^{-m_i}})^{(n-1)^{j}}} \right). \end{align}\] Since \(\xi_{z_i}^{(n-1)^{-m_i}} \rightarrow \infty\) as \(i \rightarrow \infty\), The last sum above converges for all \(i\) sufficiently large and decreases to \(0\) as \(i \rightarrow \infty\), from which the desired result follows. ◻

Proof of Proposition 11. If \(\xi \notin \Xi\), then using 21 there exists a sequence of vertices \((z_i)_{i \geq 1}\) such that \(|\xi_{z_i}|^{(n-1)^{-|z_i|}} \rightarrow \infty\) as \(i \rightarrow \infty\). We will proceed with the proof in the case where for infinitely many \(i\), \(z_i\) and \(\xi_{z_i}\) are positive (the other cases are similar). By taking an appropriate subsequence, we can assume that \(\xi_{z_i} \geq 0\) and \(1 \leq z_i < z_j\) for all \(1 \leq i < j\).

Define \(\Lambda_i = \{x \in \mathbb{Z} : |x| < z_i \}.\) We first consider the case when there exists a subsequence \((z_{i_{k}})_{k \geq 1}\) such that \(\xi_{-z_{i_k}} < -\xi_{z_{i_k}}\) for all \(k \geq 1\), which implies that \(\nu_{\Lambda_{i_k}, \beta, \rho, J_G}^{\xi}[\varphi_0 \leq 0] \geq \frac{1}{2}\). To see why this is true, note that if \(\xi_{-z_{i_k}} = -\xi_{z_{i_k}}\), then \(\varphi_0\) and \(-\varphi_0\) have the same distribution, so the probability that \(\varphi_0\) is negative is \(1/2\). Now reducing \(\xi_{-z_{i_k}}\) increases the probability of the event \(\{\varphi_0\leq 0\}\) by Proposition 3.

Conditionally on \(\{\varphi_0 \leq 0\}\), the subgraph \(\{-1, -2, \ldots, -(z_{i_k} -1)\}\) has non-positive boundary conditions. Hence \(-\varphi\) is distributed according to a measure with non-negative boundary conditions, and we can apply Proposition 10 to deduce that there exists \(\varepsilon > 0\) such that \(\nu_{\Lambda_{i_k}, \beta, \rho, J_G}^{\xi}[\varphi_{-1} \leq -D_k|\varphi_0 \leq 0] \geq \varepsilon\) for all \(k\) sufficiently large, where \(D_k \rightarrow \infty\) as \(k \rightarrow \infty\). It follows that \(\nu_{\Lambda_{i_k}, \beta, \rho, J_G}^{\xi}[|\varphi_{-1}| \geq D_k] \geq \frac{\varepsilon}{2}\), so the sequence of measures \((\nu_{\Lambda_{i_k}, \beta, \rho, J_G}^{\xi})_{k \geq 1}\) is not tight.

If no such subsequence exists, then for all \(i\) large enough we have that \(\xi_{-z_i} \geq - \xi_{z_i}\), so \(\nu_{\Lambda_i, \beta, \rho, J_G}^{\xi}[\varphi_0 \geq 0] \geq \frac{1}{2}.\) We can now conclude the proof similarly using that if \(\varphi_0 \geq 0\) then we have non-negative boundary conditions on the subgraph \(\{1, 2, \ldots, z_{i_k} -1\}\). ◻

Proof of Proposition 12. Since \(V\) is \(J\)-connected, we just need to show that \(\tilde{A}(o, V, \lambda)\) is finite, as Lemma 1 then implies \(\tilde{A}(x, V, \lambda) < \infty\) for all \(x \in V\). First observe that if \(\lambda \geq \frac{1}{c},\) then the assumptions on \(f\) imply that for any walk in \(\overline{(V, J)}\) consisting of distinct vertices \(x_0, x_1, \ldots, x_m\) with \(m \geq 2\), \[\begin{align} \label{eq:walk32shortening} \lambda f(d_G(x_{m-2}, x_{m-1})^{-r}) f(d_G(x_{m-1}, x_{m})^{-r}) &\geq \lambda f\left(\left(\frac{1}{2} d_{G}(x_{m-2}, x_{m})\right)^{-r}\right)\\ \nonumber &\geq f(d_G(x_{m-2}, x_{m})^{-r}). \end{align}\tag{26}\] Repeatedly applying 26 and using that \(f(J_{x_{i-1}, x_i}) \geq f(d_G(x_{i-1}, x_i)^{-r})\) by Definition 5, we deduce that \[\begin{align} \label{one-step32walk} \lambda^m \prod_{i=1}^{m} f(J_{x_{i-1}, x_i}) \geq \lambda f(d_G(x_{0}, x_{m})^{-r}). \end{align}\tag{27}\] The above inequality is in fact valid for any walk \(x_0, x_1, \ldots, x_m\) with \(m \geq 1\) because repeating a vertex in the walk makes the left hand side of 27 larger. Now consider \(z \in V \setminus \{o\}\) with a walk \(x_0, x_1, \ldots, x_m\) from \(o\) to \(z\) in \(\overline{(V, J)}\). Applying 27 to reduce \(x_0, x_1, \ldots , x_{m}\) to a one-step walk \(o, z\) yields \[\begin{align} |\xi_z| \leq M_{\xi} f(d_G(o, z)^{-r}) \leq M_\xi \lambda^{m} \prod_{i=1}^{m} f(J_{x_{i-1}, x_i}), \end{align}\] so if \(\mathcal{A} \geq \frac{M_{\xi}}{C}\), then \(\mathcal{A}\) satisfies the requirements for \(\tilde{A}(o, V, \lambda)\) for any \(z \in V \setminus \{o\}\). The case \(z = o\) can also be included by increasing the value of \(\mathcal{A}\) further if necessary. ◻

Proof. Fix \(\Lambda' \Subset V\) and \(a \geq 4 \beta M_f \lambda\). By Theorem 6, we have for any \(i\) large enough that \(\Lambda' \subset \Lambda_i\), \[\mathrm{d} \nu_{\Lambda_i, \beta, \rho}^{\xi}[\varphi|_{\Lambda'} = \psi] \leq \left(\prod_{x\in \Lambda'}\exp(\tilde{C} A(x, \Lambda_i, \lambda, \xi)^2)\right) \mathrm{d} \nu_{\Lambda', 0, \rho_{\frac{a}{2}}}^{0} [\psi].\] Taking \(i \rightarrow \infty\) and using ?? , we have that \(\nu\) is \(\frac{a}{2}-\)regular with \(B = \tilde{C} A^2_{\max}\). The fact that \(\nu\) is a Gibbs measure follows from the domain Markov property for the measures \(\nu_{\Lambda_i, \beta, \rho}^{\xi}\). ◻

Proof. We will drop \(\beta, \rho\) from the notation and just write \(\nu_{\Lambda}^{\xi}\) for the finite-volume measure on \(\Lambda\) with boundary conditions \(\xi\). Let \(\lambda = \frac{1}{c}\), where \(c\) is as in Definition 5. Applying Theorem 6 with \(a = 4 \beta M_f \lambda\), together with a union bound, yields \[\nu_{\Lambda}^{\xi^{+}}[F_{\Lambda, \Lambda'}^{\mathsf{c}}] \leq \sum_{x \in \Lambda \setminus \Lambda'} \nu_{\Lambda}^{\xi^{+}}[|\varphi_x| > \xi^{+}_x] \leq \frac{1}{\rho_{\frac{a}{2}}(\mathbb{R})} \sum_{x \in \Lambda \setminus \Lambda'} \exp(\tilde{C} A(x, \Lambda, \lambda)^2 ) \rho_{\frac{a}{2}}[|\varphi_x| > \xi^{+}_x].\] Let \(a' > 0\) be a constant to be determined. Applying Markov’s inequality to the random variable \(e^{a' \varphi_x^{2}}\), we have \[\label{eq:Markov} \nu_{\Lambda}^{\xi^{+}}[F_{\Lambda, \Lambda'}^{\mathsf{c}}] \leq \frac{\rho_{\frac{a}{2}}[e^{a' \varphi_x^2}]}{\rho_{\frac{a}{2}}(\mathbb{R})} \sum_{x \in \Lambda \setminus \Lambda'} \exp(\tilde{C} A(x, \Lambda, \lambda)^2 ) |B_{d_{G}(o, x)}(o)|^{-a'}.\tag{29}\] We will show that there exists a constant \(C' \geq 1\) such that for any finite \(\Lambda \subset V\) containing \(o\) and any \(x \in \Lambda \setminus \{o\}\), \[\label{eq:bound32on32A} A(x, \Lambda, \lambda) \leq \sqrt{C' \log(|B_{d_G(o, x)}(o)|)}.\tag{30}\] Combining 29 and 30 gives for any \(\Lambda'\) containing \(o\) \[\begin{align} \nu_{\Lambda}^{\xi^{+}}[F_{\Lambda, \Lambda'}^{\mathsf{c}}] \leq \frac{\rho_{\frac{a}{2}}[e^{a' \varphi_x^2}]}{\rho_{\frac{a}{2}}(\mathbb{R})} \sum_{x \in \Lambda \setminus \Lambda'} |B_{d_{G}(o, x)}(o)|^{\tilde{C} C' -a'}\leq \frac{\rho_{\frac{a}{2}}[e^{a' \varphi_x^2}]}{\rho_{\frac{a}{2}}(\mathbb{R})} \sum_{i = g(\Lambda')}^{\infty} |B_{i}(o)|^{1+ \tilde{C} C' - a'}, \end{align}\] where \(g(\Lambda') = \min_{x \in V \setminus \Lambda'} d_G(o, x)\). By choosing \(a'\) appropriately, the last sum converges and decreases to \(0\) as \(\Lambda' \nearrow V\).

It remains to prove 30 . Let \(C' \geq 1\) be a constant to be determined and set \[\mathcal{A}_x = \sqrt{C' \log(|B_{d_G(o, x)}(o)|)}.\] We aim to show that \(C'\) can be chosen so that for any \(\Lambda \Subset V\), \(x \in \Lambda \setminus \{o\}\) and any walk \(x_0, \ldots, x_m\) from \(x\) to a vertex \(z \in V \setminus \Lambda\), \[\begin{align} \label{eq:walk32requirement32n612} C\mathcal{A}_{x} \lambda^{m} \prod_{i = 1}^{m} f(J_{x_{i-1}, x_i}) \geq |\xi^{+}_z|, \end{align}\tag{31}\] which implies that \(A(x, \Lambda, \lambda) \leq \mathcal{A}_x\). As in the proof of Proposition 12, we can reduce the walk \(x_0, \ldots, x_m\) to a one-step walk \(x_0, x_m\) by using that \(f(J_{x_{i-1}, x_{i}}) \geq f(d_G(x_{i-1}, x_{i})^{-r})\) and applying the inequality 27 . This gives \[\begin{align} \label{eq:walk32reduction} C\mathcal{A}_{x} \lambda^{m} \prod_{i = 1}^{m} f(J_{x_{i-1}, x_i}) \geq \frac{C\sqrt{C'}}{c} \sqrt{\log(|B_{d_G(o, x)}(o)|)} f(d_G(x, z)^{-r}), \end{align}\tag{32}\] and ?? implies that \[\begin{align} \label{eq:f32ball32assumption} \frac{1}{c_0} \sqrt{\log(|B_{d_G(o, x)}(o)|)} f(d_G(x, z)^{-r}) \geq \sqrt{\log(|B_{d_{G}(o, x) + d_G(x, z)}(o)|)} \geq |\xi^{+}_z|. \end{align}\tag{33}\] Combining 32 and 33 , we see that it is possible to choose \(C'\) so that 31 is satisfied, completing the proof of 30 and of the lemma. ◻

Proof of Proposition 13. We only prove the statements for \(\nu^{+}_{\beta, \rho}\). As we have fixed \(\beta, \rho\) we will drop them from the notation and just write \(\nu_{\Lambda}^{\xi}\) for the finite-volume measure on \(\Lambda\) with boundary conditions \(\xi\). Let \(H\) be an increasing event that depends only on spins inside a finite subset \(\Lambda' \subset V\) and let \((\Lambda_i)_{i \geq 1}\) be a sequence of finite subsets of \(V\) with \(\Lambda_i \nearrow V\) as \(i \rightarrow \infty\). For \(k \geq i\), write \(F_{k,i}\) for the event \(F_{\Lambda_{k}, \Lambda_i}\) defined in Lemma 7. Using the domain Markov property and the fact that if \(\varphi \in F_{k,i}\), then Proposition 3 implies that \(\nu_{\Lambda_i}^{\varphi}[H] \leq \nu_{\Lambda_i}^{\xi^{+}}[H]\), we have for any \(k \geq i\) such that \(\Lambda' \subset\Lambda_i\), \(\nu_{\Lambda_{k}}^{\xi^{+}}[H] \leq \nu_{\Lambda_i}^{\xi^{+}}[H] + \nu_{\Lambda_{k}}^{\xi^{+}}[F_{k,i}^{\mathsf{c}}]\). Sending first \(k\) to infinity, and then \(i\) to infinity, and using Lemma 7, we get that \(\limsup_{k\to\infty}\nu_{\Lambda_{k}}^{\xi^{+}}[H]\leq \liminf_{i\to\infty}\nu_{\Lambda_{i}}^{\xi^{+}}[H]\). Hence \(\lim_{\Lambda \nearrow V} \nu_{\Lambda}^{\xi+}[H]\) exists for any increasing event \(H\) depending only on finitely many spins. As these events generate the \(\sigma-\)algebra, we obtain convergence of \(\nu_{\Lambda}^{\xi+}\) to a measure \(\nu^{+}\) as \(\Lambda \nearrow V\).

The assumption ?? implies that the boundary conditions \(\xi^{+}\) satisfy the assumptions of Proposition 12. Hence, ?? is satisfied with \(\lambda = \frac{1}{c}\), and we can apply Corollary 2 to deduce that \(\nu^{+}\) is an \(a\)-regular Gibbs measure for any \(a \geq 2 \beta M_f \lambda\).

Now consider any \(a>0\) and suppose \(\nu\) is an \(a-\)regular Gibbs measure. Then for any finite \(\Lambda \subset V\), the DLR equation gives that \[\begin{align} \label{eq:32xi324332DLR} \nu[H] = \nu[F_{\Lambda}] \int_{\varphi \in \mathbb{R}^{V}} \nu_{\Lambda}^{\varphi}[H] \mathrm{d} \nu(\varphi | F_{\Lambda}) + \nu[F_{\Lambda}^{\mathsf{c}}] \int_{\varphi \in \mathbb{R}^{V}} \nu_{\Lambda}^{\varphi}[H] \mathrm{d} \nu(\varphi | F_{\Lambda}^{\mathsf{c}}), \end{align}\tag{34}\] where \(F_{\Lambda} = \{\varphi:\varphi_x \leq \xi^{+}_x \, \forall x \in V\setminus \Lambda\}\). We can show that \(\nu[F_{\Lambda}^{\mathsf{c}}] \rightarrow 0\) as \(\Lambda \nearrow V\) in the same way as in the proof of Lemma 7: first using a union bound, regularity, and Markov’s inequality, we have for any \(a' > 0\) \[\nu[F_{\Lambda}^{\mathsf{c}}] \leq \sum_{x \in V \setminus \Lambda} \nu[|\varphi_x| \geq \xi^{+}_x] \leq \frac{B \rho_{a}[e^{a' \varphi_x^2}]}{\rho_{a}(\mathbb{R})} \sum_{x \in V \setminus \Lambda} |B_{d_{G}(o, x)}(o)|^{-a'}.\] It then follows that \(\nu[F_{\Lambda}^{\mathsf{c}}] \rightarrow 0\) as \(\Lambda \nearrow V\), provided that \(a'\) is chosen large enough. Hence taking \(\Lambda \nearrow V\) in 34 and using Proposition 3, we have \(\nu[H] \leq \nu^{+}[H]\). ◻

Proof. We aim to show that for any \(u \in \mathbb{R}^{\Lambda'}\), \[\nu_{\Lambda, \beta, \rho}^{\xi} [\varphi_x \geq u_x, \forall x \in \Lambda'] \leq \nu_{\Lambda', 0, \rho_a}^{0} [\varphi_x + B_x \geq u_x, \forall x \in \Lambda' \mid \varphi_x \geq 0, \forall x \in \Lambda'].\] The probability on the right hand side above is equal to \(\nu_{\Lambda, 0, \zeta}^{0} [\varphi_x \geq u_x, \forall x \in \Lambda']\), so this implies the desired stochastic domination. Applying Theorem 6 with \(\Lambda_{u} := \{x \in \Lambda' : u_x \geq B_x\}\) in place of \(\Lambda'\), we obtain \[\begin{align} \nu_{\Lambda, \beta, \rho}^{\xi} [\varphi_x \geq u_x, \forall x \in \Lambda'] &\leq \nu_{\Lambda, \beta, \rho}^{\xi} [\varphi_x \geq u_x, \forall x \in \Lambda_{u}]\\ &\leq \left(\prod_{x \in \Lambda_u}\exp(\tilde{C} A(x, \Lambda)^2) \right) \nu_{\Lambda_u, 0, \rho_{\frac{a}{2}}}^{0}[\varphi_x \geq u_x, \forall x \in \Lambda_u]\\ &\leq \prod_{x \in \Lambda_u} \frac{1}{\rho_{\frac{a}{2}}(\mathbb{R})} \int_{u_x}^{\infty} \exp \left( \tilde{C} A(x, \Lambda)^2 -\frac{a}{2}|\varphi_x|^{2} \right) \mathrm{d} \rho_{a} (\varphi_x). \end{align}\] The choice of \(B_x\) gives that \(\exp\left(\tilde{C} A(x, \Lambda)^2 -\frac{a}{2}|\varphi_x|^{2} \right) \leq \frac{\rho_{\frac{a}{2}}(\mathbb{R})}{\rho_{a}(\mathbb{R})}\) whenever \(\varphi_x \geq B_x\), so since \(u_x \geq B_x\) for \(x \in \Lambda_u\), \[\begin{align} \nu_{\Lambda, \beta, \rho}^{\xi} [\varphi_x \geq u_x, \forall x \in \Lambda'] &\leq \prod_{x \in \Lambda_{u}} \frac{1}{\rho_{a}(\mathbb{R})} \int_{u_x}^{\infty} \mathrm{d} \rho_{a} (\varphi_x)\\ &\leq \nu_{\Lambda', 0, \rho_a}^{0}[\varphi_x + B_x \geq u_x, \forall x \in \Lambda' \mid \varphi_x \geq 0, \forall x \in \Lambda']. \end{align}\] ◻

Proof. Let \(A\) be an increasing event depending only on vertices in \(\partial \Lambda\). Without loss of generality, assume that \(\nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0}[A] > 0\). Differentiating with respect to \(\beta\) we obtain \[\frac{\mathrm{d} \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0}[A]}{\mathrm{d} \beta} = \sum_{xy \in E} \langle \mathbb{1}_{A} \varphi_x \varphi_y \rangle_{\Lambda, \beta, \tilde{\rho}}^{0} - \langle \mathbb{1}_{A} \rangle_{\Lambda, \beta, \tilde{\rho}}^{0} \langle \varphi_x \varphi_y \rangle_{\Lambda, \beta, \tilde{\rho}}^{0},\] so it suffices to show that \(\langle \varphi_x \varphi_y \mid A \rangle_{\Lambda, \beta, \tilde{\rho}}^{0} - \langle \varphi_x \varphi_y \rangle_{\Lambda, \beta, \tilde{\rho}}^{0} \geq 0\) for all \(xy \in E\).

If \(x, y \in \partial \Lambda\) then \(\varphi_x, \varphi_y \geq 0\), so \(\varphi_x\varphi_y\) is an increasing function and the desired inequality follows from the FKG inequality. Now suppose that \(x, y \in \Lambda \setminus \partial \Lambda\). We use the domain Markov property and the fact that \(\tilde{\rho} = \rho\) on \(\Lambda \setminus \partial \Lambda\) to get \[\langle \varphi_x \varphi_y \mid A \rangle_{\Lambda, \beta, \tilde{\rho}}^{0} = \int_{\eta \in (\mathbb{R}^{+})^{\partial \Lambda}} \langle \varphi_x \varphi_y \rangle_{\Lambda \setminus \partial \Lambda, \beta, \rho}^{\eta} \mathrm{d} \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0} [\eta \mid A].\] Since \(\nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0} [\;\cdot \mid A] \succeq \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0} [\;\cdot \;]\) by the FKG inequality, and \(\langle \varphi_x \varphi_y \rangle_{\Lambda \setminus \partial \Lambda, \beta, \rho}^{\eta}\) is an increasing function of \(\eta\) (which follows by differentiating and using Griffiths’ inequality), the right hand side above is at least \(\langle \varphi_x \varphi_y \rangle_{\Lambda, \beta, \tilde{\rho}}^{0}\).

It remains to consider the case when the edge \(xy\) has one endpoint in \(\partial \Lambda\) and the other endpoint in \(\Lambda \setminus \partial \Lambda\). If \(x \in \Lambda \setminus \partial \Lambda\) and \(y \in \partial \Lambda\), then \[\langle \varphi_x \varphi_y \mid A \rangle_{\Lambda, \beta, \tilde{\rho}}^{0} = \int_{\eta \in (\mathbb{R}^{+})^{\partial \Lambda}} \eta_y \langle \varphi_x \rangle_{\Lambda \setminus \partial \Lambda, \beta, \rho}^{\eta} \mathrm{d} \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0} [\eta \mid A].\] Proposition 3 together with the fact that the boundary conditions \(\eta\) are positive and \(\rho\) is an even measure implies that \(\eta_y \langle \varphi_x \rangle_{\Lambda \setminus \partial \Lambda, \beta, \rho}^{\eta}\) is an increasing function of \(\eta\), so the right hand side is again at least \(\langle \varphi_x \varphi_y \rangle_{\Lambda, \beta, \tilde{\rho}}^{0}\). ◻

Proof. By Lemma 8, we have \(\tilde{\nu}_{(\Lambda | \partial \Lambda), 0, \rho}^{0} = \nu_{(\Lambda | \partial \Lambda), 0, \tilde{\rho}}^{0} \preceq \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0}\), and stochastic domination in the whole of \(\Lambda\) follows from the domain Markov property and monotonicity in boundary conditions. ◻

Proof. Let \(\xi \sim \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0}\), write \(R=\Lambda\setminus\partial\Lambda\) and consider a set \(S\subset \partial \Lambda'\). By Theorem 6, \[\mathrm{d} \nu_{R, \beta, \rho}^{\xi}[\varphi|_{S} = \psi] \leq \prod_{x \in S} e^{\tilde{C} A(x, R, \xi, C)^2} \mathrm{d} \nu_{S, 0, \rho_{\frac{a}{2}}}^{0} [\psi].\] It suffices to prove that \[\label{eq:expected32regularity} \nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0}\left[\prod_{x \in S} e^{\tilde{C} A(x, R, \xi, C)^2}\right]\leq e^{2\tilde{C} |S|}.\tag{35}\] Indeed, since \(S\) is an arbitrary subset of \(\partial \Lambda'\), 35 implies that \(\nu^0_{(\Lambda | \partial \Lambda'),\beta,\tilde{\rho}}\) is a (finite-volume) \(a/2\)-regular measure with constant \(B=2\tilde{C}\), and we can apply the argument of Corollary 3 to deduce that \(\nu_{\partial \Lambda', 0, \zeta}^{0} =\tilde{\nu}^0_{(\Lambda' | \partial \Lambda'),\beta,\rho}\) stochastically dominates \(\nu^0_{(\Lambda | \partial \Lambda'),\beta,\tilde{\rho}}\). Then the desired stochastic domination in the whole of \(\Lambda'\) follows from monotonicity in boundary conditions. By corollary 4, we also get \(\tilde{\nu}^0_{\Lambda',\beta,\rho} \succeq \tilde{\nu}^0_{(\Lambda | \Lambda'),\beta,\rho}\) and \(\nu^0_{\Lambda',\beta,\tilde{\rho}} \succeq \nu^0_{(\Lambda | \Lambda'),\beta,\tilde{\rho}}\).

Let us now prove 35 . Note that \[A(x, R, \xi, C)^2\leq\max \left\{ 1, \max_{z \in \partial \Lambda} \left(\frac{|\xi_z|^2}{C^2\lambda^{2d_{\Lambda}(x, z)}}\right) \right\}\leq 1+\sum_{z \in \partial \Lambda} \frac{|\xi_z|^2}{C^2\lambda^{2d_{\Lambda}(x, z)}}.\] Hence \[\label{eq:32sum32of32A32bound} \sum_{x\in S} A(x, R, \xi, C)^2\leq |S|+\sum_{z\in \partial \Lambda} \frac{|\xi_z|^2}{C^2}\sum_{k\geq d_{\Lambda}(z,S)}\frac{|B_k(z)|}{\lambda^{2k}} \leq |S|+\sum_{z\in \partial \Lambda} \frac{|\xi_z|^2}{\lambda^{d_{\Lambda}(z,S)}},\tag{36}\] provided \(\lambda\geq 2D\), where \(D\) is the maximum degree, and \(d_G(\partial \Lambda, \Lambda')\) is large enough. Now let \(B' > 0\) and let \(\zeta'\) be the single-site measure given by \(\mathrm{d} \zeta'(u) = \mathbb{1}_{\{u \geq B'\}} e^{a |u - B'|^2} \mathrm{d} \zeta(u - B')\). By Corollary 3, we have \(\nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0} \preceq \nu^0_{\partial \Lambda, 0, \zeta'}\) for some choice of \(B'\). Therefore, using that \(\prod_{x \in S} e^{\tilde{C} A(x, R, \xi, C)^2}\) is an increasing function of \(\xi\) for \(\xi \geq 0\), we obtain \[\nu_{(\Lambda | \partial \Lambda), \beta, \tilde{\rho}}^{0}\left[\prod_{x \in S} e^{\tilde{C} A(x, R, \xi, C)^2}\right]\leq \nu_{\partial \Lambda, 0, \zeta'}^{0}\left[\prod_{x \in S} e^{\tilde{C} A(x, R, \xi, C)^2}\right]\leq e^{\tilde{C} |S|}\exp\left(C'\sum_{z\in \partial \Lambda}\lambda^{-d_{\Lambda}(z,S)}\right)\] for some constant \(C'>0\), where in the second inequality we used 36 , independence, and the inequality \(\mathbb{E}(e^{tX})\leq 1+t\mathbb{E}(e^X)\leq e^{t\mathbb{E}(e^X)}\) for any \(t\in (0,1)\) and any random variable \(X\geq 0\). By decomposing according to the points in \(S\) that attain \(d_{\Lambda}(z,S)\) we get that \[\sum_{z\in \partial \Lambda}\lambda^{-d_{\Lambda}(z,S)}\leq \sum_{x\in S} \sum_{k\geq d_G(\partial \Lambda,\Lambda')} |B_k(x)|\lambda^{-k}\leq \tilde{C}|S|/C',\] provided that \(d_{G}(\partial \Lambda,\Lambda')\) is large enough. This implies 35 and concludes the proof. ◻

Proof of Proposition 15. First note that the single-site measures \(\tilde{\rho}_{x, \Lambda}\) satisfy the assumptions of Remark 7 with \(a_{\min} = \frac{a}{2}\), and that the result of Theorem 6 still applies in this case. Therefore, there exists \(\tilde{C}_2 > 0\) such that for all \(\Lambda \Subset V\), \(\Lambda' \subset \Lambda \setminus \partial \Lambda\) and \(\psi \in \mathbb{R}^{\Lambda'}\), \[\label{eq:regularity32for32rho32tilde32n612} \mathrm{d} \nu_{\Lambda, \beta, \tilde{\rho}}^{0}[\varphi|_{\Lambda'} = \psi] \leq e^{\tilde{C}_2 |\Lambda'|} \mathrm{d} \nu_{\Lambda', 0, \rho_{\frac{a}{2}}}^{0} [\psi].\tag{37}\] By the monotonicity in volume of Proposition 16, \(\nu_{\Lambda, \beta, \tilde{\rho}}^{0}\) converges to some measure \(\tilde{\nu}\) as \(\Lambda \nearrow V\). The measure \(\tilde{\nu}\) is \(\frac{a}{2}\)-regular by 37 and is a Gibbs measure by the domain Markov property; hence \(\tilde{\nu} \preceq \nu^{+}_{\beta, \rho}\) by Proposition 13. It remains to prove that \(\nu^{+}_{\beta, \rho} \preceq \tilde{\nu}\), which implies that \(\tilde{\nu}=\nu^{+}_{\beta, \rho}\).

Let \((\Lambda_i)_{i \geq 1}\) be a sequence of finite subsets of \(V\) with \(\Lambda_i \nearrow V\) as \(i \rightarrow \infty\). Setting \(\lambda = 2\) and using 22 , we see that for each \(i \geq 1\) we can find \(k(i) \geq i\) such that \(A(x, \Lambda_{k(i)}, \xi^{+}, C) = 1\) for any \(x \in \Lambda_i\). Now using Corollary 3 and Lemma 8, \[\nu_{(\Lambda_{k(i)}| \partial \Lambda_i), \beta, \rho}^{\xi^{+}} \preceq \nu_{\partial \Lambda_i, 0, \tilde{\rho}}^{0} \preceq \nu_{(\Lambda_i|\partial \Lambda_i), \beta, \tilde{\rho}}^{0}.\] Let \(\Lambda' \Subset V\) and consider an increasing event \(H\) that depends only on spins inside \(\Lambda'\). The domain Markov property and Proposition 3 together with the above stochastic domination imply that whenever \(i\) is large enough that \(\Lambda' \subset \Lambda_i \setminus \partial \Lambda_i\), \[\begin{align} \nu_{\Lambda_{k(i)}, \beta, \rho}^{\xi^{+}} [H] &= \int_{\mathbb{R}^{\partial \Lambda_i}} \nu_{\Lambda_i \setminus \partial \Lambda_i, \beta, \rho}^{\eta}[H] \mathrm{d} \nu_{(\Lambda_{k(i)}| \partial \Lambda_i), \beta, \rho}^{\xi^{+}}[\eta]\\ &\leq \int_{\mathbb{R}^{\partial \Lambda_i}} \nu_{\Lambda_i \setminus \partial \Lambda_i, \beta, \rho}^{\eta}[H] \mathrm{d} \nu_{(\Lambda_{i}|\partial \Lambda_i), \beta, \tilde{\rho}}^{0}[\eta] = \nu_{\Lambda_i, \beta, \tilde{\rho}}^{0}[H]. \end{align}\] Taking \(i \rightarrow \infty\), we have \(\nu^{+}_{\beta, \rho}[H] \leq \tilde{\nu}[H]\).

The desired stochastic domination follows from Corollary 3. ◻

Proof of Proposition 14. First observe that the monotonicity in volume of Proposition 16 implies convergence of the measures \(\tilde{\nu}^0_{\Lambda,\beta,\rho}\) to an infinite-volume limit \(\tilde{\nu}\) as \(\Lambda \nearrow V\). By Corollary 4, we have that \(\tilde{\nu}^0_{\Lambda,\beta,\rho} \preceq \nu^0_{\Lambda,\beta,\tilde{\rho}}\) for any \(\Lambda \Subset V\), so \(\tilde{\nu} \preceq \nu^{+}_{\beta, \rho}\) because \(\nu^0_{\Lambda,\beta,\tilde{\rho}} \rightarrow \nu^{+}_{\beta, \rho}\) by Proposition 15. The proof that \(\nu^{+}_{\beta, \rho} \preceq \tilde{\nu}\) is similar to that of Proposition 15, and in fact, even simpler, as one does not need to use Lemma 8. ◻